IOP PUBLISHING
INVERSE PROBLEMS
Inverse Problems 25 (2009) 123013 (75pp)
doi:10.1088/0266-5611/25/12/123013
TOPICAL REVIEW
Carleman estimates for parabolic equations and
applications
Masahiro Yamamoto
Department of Mathematical Sciences, The University of Tokyo, Komaba Meguro,
Tokyo 153-8914, Japan
E-mail: [email protected]
Received 18 June 2009, in final form 14 October 2009
Published 3 December 2009
Online at stacks.iop.org/IP/25/123013
Abstract
In this review, concerning parabolic equations, we give self-contained
descriptions on
(1) derivations of Carleman estimates;
(2) methods for applications of the Carleman estimates to estimates of
solutions and to inverse problems.
Moreover limited to parabolic equations, we survey the previous and recent
results in view of the applicability of the Carleman estimate. We do not intend
to pursue any general treatments of the Carleman estimate itself but by showing
it in a direct manner, we mainly aim to demonstrate the applicability of the
Carleman estimate to the estimation of solutions and inverse problems.
Contents
1.
2.
3.
4.
5.
Introduction
What is a Carleman estimate?
A direct derivation of a Carleman estimate for a parabolic equation
Global Carleman estimate
Applications to the unique continuation and the observability inequality
5.1. Conditional stability for the Cauchy problem
5.2. Observability inequality
6. Applications to inverse problems for parabolic equations
6.1. Local Hölder stability for an inverse coefficient problem
6.2. Global Lipschitz stability for an inverse source problem
7. Overview for Carleman estimates for parabolic equations
7.1. Pointwise Carleman estimate
7.2. General treatment
7.3. Global Carleman estimate
7.4. Carleman estimate with second large parameter
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7.5. H −1 -Carleman estimates
7.6. Carleman estimates for less regular principal terms
7.7. Carleman estimate for degenerate parabolic equations
7.8. Carleman estimates for parabolic systems
8. Overview of results by parabolic Carleman estimates
8.1. Estimation of solutions to parabolic equations
8.2. Inverse problems for parabolic equations
9. Carleman estimates with x-independent weight functions
9.1. Carleman estimate and an application to a parabolic equation backward in time
9.2. Carleman estimate for a forward problem in time and an application to an
inverse source problem
Acknowledgments
References
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1. Introduction
For an inverse problem, in spite of the ill-posedness, one can prove conditional stability
estimates which assure that one can restore the stability if one can restrict a class of solutions
within an a priori bounded set. In practise, such an a priori bounded set can be interpreted
as a physically acceptable constraint set. The conditional stability is not only theoretically
interesting but is also important for stable numerics. There are several methods for proving
the conditional stability and a method by Carleman estimates is one of them.
Recently in the fields of the inverse problem and the control theory, Carleman estimates
are applied in various ways to produce remarkable results. The purposes of this review are as
follows:
(1) self-contained descriptions for deriving Carleman estimates;
(2) discussions of typical methodologies for the application of a Carleman estimate in
establishing the uniqueness and the stability for estimation problems of solutions and
inverse problems of determining coefficients and source terms;
(3) an overview of classical and recent results for Carleman estimates and the applications to
the estimation problem and the inverse problem.
There have already been rich amounts of work for the theory of Carleman estimates, and
a general theory is completed, but here we will expose a direct method for proving a Carleman
estimate. Such a direct derivation may give hints for Carleman estimates for other types of
partial differential equations.
Moreover, there are many works in progress as well as established works in various fields,
even for equations of parabolic type, and so we do not intend to provide a perfect list and we
will make an overview, mainly in terms of the inverse problem.
It is a common and important feature to discuss the theory of Carleman estimates and the
application to the inverse problem uniformly for a wide class of partial differential equations,
including equations of hyperbolic type. However in the current review, we will restrict
ourselves to equations of parabolic type.
Notations. ⊂ Rn : a bounded spatial domain with smooth boundary ∂, Q = × (0, T ).
We understand x = (x1 , . . . , xn ) ∈ Rn and t 0, respectively, as the spatial and time
variables. x = (x2 , . . . , xn ) ∈ Rn−1 , ξ = (ξ2 , . . . , ξn ) ∈ Rn−1 ,
2
Inverse Problems 25 (2009) 123013
∂t =
∂
,
∂t
Topical Review
∂j =
∂
,
∂xj
∇ = (∂1 , · · · , ∂n ),
∇x,t = (∇, ∂t ),
= ∂12 + · · · + ∂n2 .
ω: an arbitrarily fixed sub-domain of . Let α = (α1 , α2 , . . . , αn ) be a multi-index
with αj ∈ N ∪ {0}. We set ∂xα = ∂1α1 ∂2α2 · · · ∂nαn , |α| = α1 + α2 + · · · + αn , and
∂
= ν · ∇.
ν = ν(x) = (ν1 (x), . . . , νn (x)) is the outwards unit normal vector to ∂ at x. Let ∂ν
Let D ⊂ × (0, T ) be a domain with smooth boundary ∂D and let ν be the outwards unit
=
ν · ∇x,t u,
normal vector to ∂D. On ∂D, we set ∂u
∂
ν
12
∂u 2 ∂u 2
|∇x,t u| = + ∂
ν
∂
τ
2
+
where ∂u
is the orthogonal component of ∇x,t to ∂u
. For example, |∇x,t u| = |∂t u|2 + ∂u
∂
τ
∂
ν
∂ν
∂u 2 1
2 on ∂ × (0, T ) ⊂ ∂D and |∇x,t u| = (|∂t u|2 + |∇u|2 ) 21 on × {t} ⊂ ∂D, where ∂u is
∂τ
∂τ
. We use usual function spaces C 1 (Q), H s (), H0s (),
the orthogonal component of ∇u to ∂u
∂ν
s,p
W s,p (), W0 () with s 0 and 1 p ∞ (e.g., Adams [1]),
H 1,0 (Q) = {u ∈ L2 (Q); ∇u ∈ L2 (Q)}
and for m ∈ N,
H 2m,m (Q) = u ∈ L2 (Q); ∂xα ∂tαn+1 u ∈ L2 (Q), |α| + 2αn+1 2m
and ·H 2m,m (Q) is the corresponding norm. For a ∈ Rn , by aT we denote the transpose vector,
and by [a]k we denote the kth component of a. By C(λ) we denote generic constants which
depend on other parameter λ.
If we do not specially state, then we always assume that
aij ∈ C 1 (Q),
aij = aj i ,
1 i, j n,
(1.1)
and that the coefficients {aij } ≡ {aij }1i,j n satisfy the uniform ellipticity: there exists a
constant σ1 > 0 such that
n
aij (x, t)ξi ξj σ1 |ξ |2 ,
ξ = (ξ1 , . . . , ξn ) ∈ Rn ,
(x, t) ∈ Q.
(1.2)
i,j =1
Let bk , c ∈ L∞ (Q), 1 k n. We set
n
∂u
=
aij (∂i u)νj
∂νA
i,j =1
on ∂. A typical parabolic operator which we will discuss is
(Lu)(x, t) ≡ ∂t u(x, t) −
n
aij (x, t)∂i ∂j u(x, t)
i,j =1
−
n
bk (x, t)∂k u − c(x, t)u,
(x, t) ∈ Q.
(1.3)
k=1
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2. What is a Carleman estimate?
A Carleman estimate is an L2-weighted estimate with large parameter for a solution to a partial
differential equation. Here in place of a statement of a general theorem for the Carleman
estimate, we start with a very direct derivation for a simplest heat equation:
∂t u(x, t) = u(x, t) + f (x, t),
x ∈ , t > 0.
(2.1)
2
Our aim is to find an L -weighted estimate (2.2) with large parameter s in some domain
D ⊂ Q, which is called a Carleman estimate. More precisely, we choose a suitable function
ϕ(x, t) satisfying: there exist constants C > 0 and s0 > 0 such that
s(|∇u(x, t)|2 + |u(x, t)|2 ) e2sϕ(x,t) dx dt C
|f (x, t)|2 e2sϕ(x,t) dx dt
(2.2)
D
D
for all s > s0 and all u ∈ C0∞ (D). We note that in Carleman estimate (2.2), the estimate is
valid uniformly for all large s > 0, i.e., s s0 : a fixed constant. In other words, the constant
C > 0 should be independent of s > s0 and u ∈ C0∞ (D).
For applications, the parameter s plays an essential role and it is also important how to
choose a weight function ϕ(x, t) in order to adjust given geometric configurations (see, e.g.,
proposition 5.1 and theorem 5.1 in section 5).
The Carleman estimate was first established by Carleman [30] for proving the unique
continuation for a two-dimensional elliptic equation. Since then there have been great concerns
for the Carleman estimate and applications, and there are remarkable general treatments by
Egorov [44], Hörmander [69, 70], Isakov [85, 86, 90], Tataru [134], Taylor [135] and Trèves
[138]. Here, however, for easier understanding for non-specialists, we will give a heuristic
derivation, which may be useful for the insight into the characters of a Carleman estimate.
We first consider a simple heat equation (2.1). Let us assume that we already find a weight
function ϕ(x, t). For treating the weighted L2-norms, we introduce
w(x, t) = esϕ(x,t) u(x, t),
P w(x, t) = esϕ (∂t − )(e−sϕ w).
Then we rewrite the right-hand side of (2.2) by
2 2sϕ
f e dx dt =
|P w(x, t)|2 dx dt.
D
D
Therefore our task is a lower estimate of P w2L2 (D) . Direct calculations yield
P w = ∂t w − w + 2s∇ϕ · ∇w + (−s∂t ϕ − s 2 |∇ϕ|2 + sϕ)w.
Let us consider formally. That is, assuming that u ∈ C0∞ (D), we will take integration by
parts as we like. One traditional way for obtaining a lower estimate for P w2L2 (D) , is the
decomposition of the operator P into the symmetric part P+ and the antisymmetric part P− :
P w = P+ w + P− w. See e.g., Bukhgeim [19] for a one-dimensional Schrödinger equation.
We consider the formal adjoint operator P ∗ to P:
(P w, v)L2 (D) = (w, P ∗ v)L2 (D) ,
v, w ∈ C0∞ (D).
For example, we have
(∂t w, v)L2 (D) = −(w, ∂t v)L2 (D)
and
(−w, v)L2 (D) = (w, −v)L2 (D)
by integration by parts, the Green theorem and v, w ∈ C0∞ (D). Hence we see that
P ∗ w = −∂t w − w − 2s∇ϕ · ∇w − (sϕ + s 2 |∇ϕ|2 + s(∂t ϕ))w.
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We define the symmetric part P+ and the antisymmetric part P− of P by
P+ = 12 (P + P ∗ ),
P− = 12 (P − P ∗ ).
Then we have P w = P+ w + P− w, and
P+ w = −w − (s 2 |∇ϕ|2 + s∂t ϕ)w
and
P− w = ∂t w + 2s∇ϕ · ∇w + s(ϕ)w.
Hence
f 2 e2sϕ dx dt = P+ w + P− w2L2 (D)
D
= P+ w2L2 (D) + P− w2L2 (D) + 2(P+ w, P− w)L2 (D) 2(P+ w, P− w)L2 (D) . (2.3)
That is, we will estimate the right-hand side of (2.2) from below by means of
2(P+ w, P− w)L2 (D) . We note here that we discarded other terms P+ w2L2 (D) and P− w2L2 (D)
although there may be better possibilities for the decomposing of P w (see section 3 for a
general parabolic equation).
We have
∂t w + 2s∇ϕ · ∇w + s(ϕ)w)L2 (D)
2(P+ w, P− w)L2 (D) = 2(−w − (s 2 |∇ϕ|2 + s∂t ϕ)w,
= 2(−w, ∂t w)L2 (D) + 2(−w, 2s∇ϕ · ∇w)L2 (D) + 2(−w, s(ϕ)w)L2 (D)
− 2((s 2 |∇ϕ|2 + s∂t ϕ)w, ∂t w)L2 (D) − 2((s 2 |∇ϕ|2 + s∂t ϕ)w, 2s∇ϕ · ∇w)L2 (D)
− 2((s 2 |∇ϕ|2 + s∂t ϕ)w, s(ϕ)w)L2 (D) .
By the integration by parts and w ∈ C0∞ (D), we will reduce the orders of derivatives of w.
Henceforth, C > 0 denotes generic constants which are independent of s and may change line
by line. For example, the calculations are as follows:
n −2((s 2 |∇ϕ|2 + s∂t ϕ)w, 2s∇ϕ · ∇w)L2 (D) = −4s
{(s 2 |∇ϕ|2 + s∂t ϕ)w}(∂i ϕ)(∂i w) dx dt
= −2s
n i=1
= 2s
D
(s 2 |∇ϕ|2 + s∂t ϕ)(∂i ϕ)∂i (w 2 ) dx dt
D
i=1
= 2s
i=1
n ∂i ((s 2 |∇ϕ|2 + s∂t ϕ)∂i ϕ)w 2 dx dt
D
{∇(s 2 |∇ϕ|2 + s∂t ϕ) · ∇ϕ + (s 2 |∇ϕ|2 + s∂t ϕ)ϕ}w 2 dx dt.
D
Next the Green formula yields
∇w · ∇((ϕ)w) dx dt
2(−w, s(ϕ)w)L2 (D) = 2s
D
= 2s (ϕ)|∇w|2 dx dt + 2s
∇(ϕ) · w∇w dx dt,
D
and
D
s
Cs
∇(ϕ)
·
w∇w
dx
dt
|w||∇w| dx dt.
D
Hence
D
2(−w, s(ϕ)w)L2 (D) 2s
(ϕ)|∇w|2 dx dt − Cs
D
|w||∇w| dx dt.
D
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Next, noting 2(∂k w)(∂k ∂j w) = ∂j (|∂k w|2 ) and integration by parts, we have
2(−w, 2s∇ϕ · ∇w)L2 (D) = 2
n
2
−∂k w, 2s(∂j w)∂j ϕ L2 (D)
j,k=1
n
=2
(∂k w, 2s(∂k ∂j w)(∂j ϕ))L2 (D) + (∂k w, 2s(∂j w)(∂k ∂j ϕ))L2 (D)
j,k=1
= −2s
n n 2 ∂j ϕ |∂k w|2 dx dt + 4s
(∂j w)(∂k w)(∂j ∂k ϕ) dx dt.
j,k=1 D
j,k=1 D
Therefore, noting that s3 and s are the maximum orders respectively among all the terms w2
and |∇w|2 and estimating any lower order terms as non-principal terms, we have
1
3
{∇(|∇ϕ|2 ) · ∇ϕ}w 2 dx dt
(P+ w, P− w)L2 (D) s
2
D
n (∂j w)(∂k w)(∂j ∂k ϕ) dx dt
+ 2s
j,k=1 D
−C
|w||∇w| dx dt
D
s
s 2 w 2 dx dt − Cs
{∇(|∇ϕ| ) · ∇ϕ}w dx dt + 2s
3
2
2
D
(∂j w)(∂k w)(∂j ∂k ϕ) dx dt
j,k=1 D
−C
n (|∇w|2 + s 2 w 2 ) dx dt.
D
At the last inequality we used also
s|∇w||w| 12 s 2 |w|2 + 12 |∇w|2 .
Hence, since we can consider only sufficiently large s > 0, noting the maximum powers in s
for the terms of |w|2 and |∇w|2 , we can absorb terms of lower powers, so that if ϕ satisfies
{∂i ∂j ϕ}1i,j n is positive definite
(2.4)
and
there exists a constant r1 > 0 such that
∇(|∇ϕ|2 ) · ∇ϕ r1
on D,
(2.5)
then there exist constants C > 0 and s0 > 0 such that
2
3
2
(s|∇w| + s |w| ) dx dt C
f 2 e2sϕ dx dt
D
D
for all s s0 and all w ∈ C0∞ (D). Noting w = esϕ u, we rewrite in terms of u, and
2
3
2 2sϕ
(s|∇u| + s |u| ) e dx dt C
f 2 e2sϕ dx dt
D
(2.6)
D
for all s s0 and all u ∈ C0∞ (D).
The next important step is the choice of the weight function ϕ. The weight function has to
satisfy not only (2.4) and (2.5) but also some geometric condition for meaningful applications
to the unique continuation and the inverse problem. More precisely, in applying a Carleman
estimate, as D we usually consider a level set defined by {(x, t); ϕ(x, t) > δ} with some
constant δ > 0, and such a level set should be a bounded domain at least (see sections 5
and 6).
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For Carleman estimate (2.6), there are not many choices of ϕ. For example, only a typical
choice is
ϕ(x, t) = |x − x0 |2 − β(t − t0 )2 .
(2.7)
Here we assume that |x − x0 | = 0 for any (x, t) ∈ D, and β > 0 and t0 ∈ (0, T ) are arbitrarily
fixed. Then we have ∇(|∇ϕ|2 ) · ∇ϕ = 16|x − x0 |2 > 0 on D and {∂i ∂j ϕ}1i,j n = 2En ,
where
⎛1 0 · · · 0⎞
⎜0
En ≡ ⎝..
.
0
1
.
.
.
0
···
.
.
.
···
⎟
⎠.
0
1
That is, conditions (2.4) and (2.5) are satisfied.
As is seen from sections 5 and 6, the Carleman estimate is applied for the unique
continuation and the inverse problem of determining coefficients and/or right-hand sides.
If we apply the Carleman estimate (2.6) which is derived naively in the case treated in
theorem 5.1 in section 5, then due to the choice (2.7), we must assume some extra geometric
condition. For example, near , the domain has to be convex. However, by the parabolicity
of the equation, such convexity should be deleted, and the Carleman estimate (2.6) does not
produce sharp results, and so in section 3 we will derive more applicable Carleman estimates.
3. A direct derivation of a Carleman estimate for a parabolic equation
Let D ⊂ Q be a bounded domain whose boundary ∂D is composed of a finite number
of smooth surfaces. As is discussed in detail in section 7, there are many papers deriving
Carleman estimates even though we are restricted to parabolic equations, and as for derivations,
I refer to
(1) general way: [46, 84, 86, 88, 89, 134];
(2) direct way: [32, 60, 72, 112, 144].
As for the direct derivations for hyperbolic equations, see also [105, 112], but we will not
discuss the hyperbolic case.
In this survey, we will mainly explain a direct method, because of its flexibility. The
key tool of the direct method is the integration by parts and suitable grouping of the terms
according to the orders of the parameter s in a Carleman estimate. Here we will make
motivating explanations for the grouping in order that the explanations may be more readerfriendly and they may be able to apply the direct method to other types of equations to obtain
possible Carleman estimates.
Before starting the derivations, we note that it is sufficient to prove a Carleman estimate
for one of two types of parabolic equations:
ρ(x, t)∂t u(x, t) −
n
∂i (
aij (x, t)∂j u(x, t))
i,j =1
−
n
c(x, t)u(x, t) = f(x, t)
bk (x, t)∂k u(x, t) −
k=1
and
∂t u(x, t) −
n
i,j =1
aij (x, t)∂i ∂j u(x, t) −
n
bk (x, t)∂k u(x, t) − c(x, t)u(x, t) = f (x, t).
k=1
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Here ρ ∈ C 1 (D) with ρ > 0 on D and bk , bk , c,
c ∈ L∞ (D), we assume that
aij ∈ C 1 (D),
aij = aj i , 1 i, j n,
n
n
aij (x, t)ξi ξj σ1 i=1 ξi2 ,
(x, t) ∈ D, ξ1 , . . . , ξn ∈ R.
i,j =1 This is seen because
n
∂i (
aij (x, t)∂j u(x, t))
ρ(x, t)∂t u(x, t) −
i,j =1
−
n
c(x, t)u(x, t) = f(x, t)
bk (x, t)∂k u(x, t) −
k=1
if and only if
n
n
n
aij
f
1 c
∂i ∂j u −
∂i
aik ∂k u − u = .
∂t u(x, t) −
bk +
ρ
ρ
ρ
ρ
i,j =1
k=1
i=1
Let us set
n
Lu(x, t) = ∂t u −
aij (x, t)∂i ∂j u(x, t) −
i,j =1
n
bk (x, t)∂k u(x, t) − c(x, t)u(x, t)
in Q
k=1
and
L0 u = ∂t u −
n
aij (x, t)∂i ∂j u(x, t)
in Q.
i,j =1
Here we assume that aij, 1 i, j n satisfy (1.1) and (1.2), and bk , c ∈ L∞ (Q), 1 k n.
We consider a parabolic equation Lu = f . Our purpose is to establish a Carleman
estimate
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 u2 e2sϕ dx dt
|∂t u|2 +
⎭
D ⎩ sϕ
i,j =1
|Lu|2 e2sϕ dx dt
C
D
for all large s > 0 and λ > 0 and all u ∈ H 2,1 (Q) satisfying
supp u ∈ D. For it, it suffices to
prove the estimate for L0. Because |L0 u|2 2|Lu|2 + 2| nk=1 bk ∂k u + cu|2 in Q, that is,
|L0 u|2 e2sϕ dx dt
D
n
2 2sϕ
2
2
2
2
bk L∞ (D) |∇u| + cL∞ (D) |u| e2sϕ dx dt.
2 |Lu| e dx dt + 4
D
D
k=1
Hence the Carleman estimate for L0 yields
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 u2 e2sϕ dx dt
|∂t u|2 +
⎭
D ⎩ sϕ
i,j =1
|Lu|2 e2sϕ dx dt
C
D
+C
n
k=1
8
bk 2L∞ (D)
D
|∇u|2 e2sϕ dx dt + Cc2L∞ (D)
|u|2 e2sϕ dx dt.
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Inverse Problems 25 (2009) 123013
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Therefore, we choose s > 0 sufficiently large and we can absorb the second and third terms
on the right-hand side into the left-hand side. Therefore, for the Carleman estimate, only the
terms with derivatives of highest orders in x and t are important if the coefficients of lower
order terms are in L∞ (Q). Thus we consider
L0 u = f
in Q.
(3.1)
The simple method in section 2 is based on the decomposition into the symmetric and the
antisymmetric parts and may be transparent, but for the general parabolic equation with
variable coefficients, it does not work. In this section, we explain a direct method for deriving
a Carleman estimate for a general parabolic equation. Here, for a weight function ϕ, the
important factor is the second large parameter λ > 0 and we search for the weight function ϕ
in the form of eλψ . This form has been recognized as useful and see e.g., [section 8.6, 69], and
is essentially used in several references: [46, 72, 84, 91, 92]. Also see section 7.4. Thanks to
the form eλψ , it is easier to guarantee the positivity of the coefficients of |w|2 and |∇w|2 in
estimating P w2L2 (D) . This kind of form is very useful also in section 9.
Let d ∈ C 2 (D) and |∇d| = 0 on D and let us set
ψ(x, t) = d(x) − β(t − t0 )2 + c0
with t0 ∈ (0, T ), c0 , β > 0 such that inf (x,t)∈Q ψ(x, t) > 0 and
ϕ(x, t) = eλψ(x,t) .
It is not essential that ψ > 0 in Q, but the positivity is convenient in the succeeding arguments.
Remark. For Lu = f , we consider a Carleman estimate, and our argument holds also for the
parabolic inequality
n
C(|∇u(x, t)| + |u(x, t)| + f (x, t)|)
∂t u(x, t) −
a
(x,
t)∂
∂
u(x,
t)
ij
i
j
i,j =1
in Q.
First we assume
u ∈ C0∞ (D).
We further set
σ (x, t) =
n
aij (x, t)(∂i d)(x)(∂j d)(x),
(x, t) ∈ Q
i,j =1
and
w(x, t) = esϕ(x,t) u(x, t)
and
P w(x, t) = esϕ L0 (e−sϕ w) = esϕ L0 u.
(3.2)
The derivation argument consists of
(1) the decomposition of P into the parts P1 and P2, where P1 is composed of second-order
and zeroth-order terms in x, and P2 is composed of first-order terms in t and first-order
terms in x. Here the terms in P w are classified by the highest order of s, λ and ϕ, and
not by the symmetric and antisymmetric parts. Compare P1 , P2 (see (3.4) and (3.5)) with
P+ , P− in section
2.
(2) Estimation of D (|P2 w|2 + 2(P1 w)(P2 w)) dx dt from below.
9
Inverse Problems 25 (2009) 123013
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(3) Another estimate for
P w × (the term u with second highest order of s, λ, ϕ among P w).
D
By our decomposition,
we have to estimate the L2-term of ∂t u, and so we need to estimate
2
also the term D |P2 w| dx dt in step (2). Moreover, the estimate in the second step produces
the estimate of u with desirable order of s, λ, ϕ but not the term of ∇u. This is a natural
consequence with different orders of the derivatives of terms under consideration. Therefore
another estimate in the third step is necessary. This kind of double estimate is also used
in section 9 and in proving the observability inequality by the multiplier method. As for
the multiplier method, see e.g., [107, pp 36–9] where the wave equation is considered but a
principle is similar: the two estimates are obtained from
2
∂t u − u − c(x)u (h(x) · ∇u) dx dt
D
and
D
2
∂t u − u − c(x)u u dx dt
with a suitable vector-valued function h(x), and we added them to obtain an L2-estimate of u.
The second estimate for the wave equation by the multiplier method has a similar purpose to
step (3) in our case.
We have
n
n
aij (x, t)∂i ∂j w + 2sλϕ
aij (x, t)(∂i d)∂j w
P w = ∂t w −
i,j =1
i,j =1
− s 2 λ2 ϕ 2 σ w + sλ2 ϕσ w + sλϕw
n
aij ∂i ∂j d − sλϕw(∂t ψ) in D.
(3.3)
i,j =1
Here we note that we have specified all the dependence of coefficients on s, λ and ϕ. We set
A1 = sλ2 ϕσ + sλϕ
n
aij ∂i ∂j d − sλϕ(∂t ψ) ≡ sλ2 ϕa1 (x, t; s, λ).
i,j =1
Then
P w = ∂t w −
n
aij (x, t)∂i ∂j w + 2sλϕ
i,j =1
n
aij (x, t)(∂i d)∂j w − s 2 λ2 ϕ 2 σ w + A1 w
in D.
i,j =1
We note that a1 depends on s and λ but
|a1 (x, t; s, λ)| C
for (x, t) ∈ D and all sufficiently large λ > 0
and
s > 0.
Here and henceforth by C, C1, etc, we denote generic constants which are independent of s, λ
and ϕ but may change line by line.
Then taking into consideration the orders of (s, λ, ϕ), we divide P w as follows:
P1 w = −
n
aij (x, t)∂i ∂j w − s 2 λ2 ϕ 2 wσ (x, t) + A1 w
(3.4)
i,j =1
and
P2 w = ∂t w + 2sλϕ
n
i,j =1
10
aij (x, t)(∂i d)∂j w.
(3.5)
Inverse Problems 25 (2009) 123013
Topical Review
By f esϕ 2L2 (D) = P1 w + P2 w2L2 (D) , we have
2
2 (P1 w)(P2 w) dx dt + P2 wL2 (D) f 2 e2sϕ dx dt.
D
(3.6)
D
We estimate
n (P1 w)(P2 w) dx dt = −
aij (∂i ∂j w)(∂t w) dx dt
D
i,j =1 D
−
n aij (∂i ∂j w)2sλϕ
i,j =1 D
s 2 λ2 ϕ 2 σ w(∂t w) dx dt −
D
(A1 w)(∂t w) dx dt +
D
6
n
2s 3 λ3 ϕ 3 σ w
D
+
≡
ak (∂k d)(∂ w) dx dt
k,=1
−
n
(A1 w)2sλϕ
D
aij (∂i d)(∂j w) dx dt
i,j =1
n
aij (∂i d)(∂j w) dx dt
i,j =1
Jk .
(3.7)
k=1
Now, applying the integration by parts, aij = aj i and u ∈ C0∞ (D) and assuming that λ > 1
and s > 1 are sufficiently large, we reduce all the derivatives of w to w, ∂i w, ∂t w. We continue
the estimation of Jk, k = 1, . . . , 6.
n |J1 | = −
aij (∂i ∂j w)(∂t w) dx dt i,j =1 D
n n
=
(∂i aij )(∂j w)(∂t w) dx dt +
aij (∂j w)(∂i ∂t w) dx dt i,j =1 D
i,j =1 D
n =
(∂i aij )(∂j w)(∂t w) dx dt +
aij ((∂j w)(∂i ∂t w) + (∂i w)(∂j ∂t w)) dx dt
i>j D
i,j =1 D
n
aii (∂i w)(∂i ∂t w) dx dt +
D i=1
n
1
C
|∇w||∂t w| dx dt + (∂t aij )(∂i w)(∂j w) dx dt 2 D i,j =1
D
C
|∇w||∂t w| dx dt + C
|∇w|2 dx dt.
(3.8)
D
D
Here we used
n
aij ((∂j w)(∂i ∂t w) + (∂i w)(∂j ∂t w)) dx dt +
aii (∂i w)(∂i ∂t w) dx dt
i>j
D
=
n
1
2 i,j =1
D i=1
aij ∂t ((∂j w)(∂i w)) dx dt.
D
11
Inverse Problems 25 (2009) 123013
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Next
J2 = −
n n 2sλϕaij ak (∂k d)(∂ w)(∂i ∂j w) dx dt
i,j =1 k,=1 D
= 2sλ
n n
λ(∂i d)ϕaij ak (∂k d)(∂ w)(∂j w) dx dt
D i,j =1 k,=1
+ 2sλ
n n
ϕ∂i (aij ak ∂k d)(∂ w)(∂i w) dx dt
D i,j =1 k,=1
+ 2sλ
n n
ϕaij ak (∂k d)(∂i ∂ w)(∂j w) dx dt.
D i,j =1 k,=1
We have
(first term) = 2sλ2
2
n
ϕ
aij (∂i d)(∂j w) dx dt 0,
D i,j =1
and similarly to J1, we can estimate
(third term) = sλ
n n ϕaij ak (∂k d)∂ ((∂i w)(∂j w))
i,j =1 k,=1 D
= −sλ2
ϕσ
D
ϕ
D
aij (∂i w)(∂j w) dx dt
i,j =1
− sλ
n
n n
∂ (aij ak ∂k d)(∂i w)(∂j w) dx dt.
i,j =1 k,=1
Hence
J2 −
sλ2 ϕσ
D
n
aij (∂i w)(∂j w) dx dt
i,j =1
2
n
2
2
−C
sλϕ|∇w| dx dt + 2sλ
ϕ
aij (∂i d)(∂j w) dx dt
D
D i,j =1
n
− sλ2 ϕσ
aij (∂i w)(∂j w) dx dt − C
sλϕ|∇w|2 dx dt.
D
i,j =1
1 2 2 2
2
s λ ϕ σ ∂t (w ) dx dt |J3 | = −
2
D
1
2 2
2
2 2 2
2
= s λ ϕ{λ(∂t ψ)ϕ}σ w dx dt +
s λ ϕ (∂t σ )w dx dt 2 D
D
C
s 2 λ3 ϕ 2 w 2 dx dt.
D
12
(3.9)
D
(3.10)
Inverse Problems 25 (2009) 123013
J4 = −
3 3 3
2s λ ϕ σ w
D
s 3 λ3 ϕ 3
D
s 3 λ3
D
σ aij (∂i d)∂j (w 2 ) dx dt
n
3ϕ 2 {λ(∂j d)ϕ}σ aij (∂i d)w 2 dx dt
i,j =1
s 3 λ3 ϕ 3
+
n
i,j =1
aij (∂i d)(∂j w) dx dt
i,j =1
=−
=
n
Topical Review
D
n
∂j (σ aij ∂i d)w 2 dx dt
i,j =1
3s 3 λ4 ϕ 3 σ 2 w 2 dx dt − C
D
s 3 λ3 ϕ 3 w 2 dx dt.
(3.11)
D
|J5 | = (A1 w)(∂t w) dx dt = sλ2 ϕa1 w(∂t w) dx dt D
D
1 = sλ2 ϕa1 ∂t (w 2 ) dx dt 2 D
1
= sλ2 ϕ(∂t a1 )w 2 dx dt +
sλ3 ϕ(∂t ψ)a1 w 2 dx dt 2 D
D
C
sλ3 ϕw 2 dx dt.
(3.12)
D
n
2
|J6 | = sλ ϕa1 × 2sλϕw
aij (∂i d)(∂j w) dx dt D
i,j =1
n
2 3 2
= 2a1 s λ ϕ
aij (∂i d)w(∂j w) dx dt D
i,j =1
n
2 3 2
2
= a1 s λ ϕ
aij (∂i d)∂j (w ) dx dt D
i,j =1
= − ∂j (a1 s 2 λ3 ϕ 2 aij (∂i d))w 2 dx dt D
C
s 2 λ4 ϕ 2 w 2 dx dt.
(3.13)
D
We remark that in estimating |J6 |, we need integrationby parts. If we will apply a simpler
way by the Cauchy–Schwarz inequality to estimate like D s 2 λ3 ϕ 2 |∇w||w| dx dt, then we lose
orders of s, λ and we cannot continue the estimation.
Hence, by (3.7)–(3.13), we obtain
n
(P1 w)(P2 w) dx dt 3 s 3 λ4 ϕ 3 σ 2 w 2 dx dt −
sλ2 ϕσ
aij (∂i w)(∂j w) dx dt
D
−C
D
sλϕ|∇w|2 dx dt − C
D
D
i,j =1
(s 3 λ3 ϕ 3 + s 2 λ4 ϕ 2 )w 2 dx dt − C
D
|∇w||∂t w| dx dt.
D
13
Inverse Problems 25 (2009) 123013
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Consequently
s λ ϕ σ w dx dt −
3 4 3
3
2
D
2
sλ ϕσ
D
2
n
aij (∂i w)(∂j w) dx dt
i,j =1
(P1 w)(P2 w) dx dt + C
sλϕ|∇w|2 dx dt
D
+ C (s 3 λ3 ϕ 3 + s 2 λ4 ϕ 2 )w 2 dx dt + C
|∇w||∂t w| dx dt.
D
D
(3.14)
D
Moreover for all large s > 0, by definition (3.5) of P2 and an inequality: |α+β|2 12 |α|2 −|β|2 ,
we obtain
2
n
1
1
2
2
∂t w + 2sλϕ
|P2 w| dx dt =
|P2 w| dx dt aij (∂i d)(∂j w) dx dt
D
D sϕ
D sϕ i,j =1
2
n
1
1
sλ2 ϕ aij (∂i d)(∂j w) dx dt,
|∂t w|2 dx dt − C
2 D sϕ
D
i,j =1
that is,
ε
D
1
|∂t w|2 dx dt C
sϕ
|P2 w|2 dx dt + Cε
D
sλ2 ϕ|∇w|2 dx dt
D
for any ε > 0. Hence by (3.14) and (3.6), we have
n
3 4 3 2 2
2
sλ ϕσ
aij (∂i w)(∂j w) dx dt + ε
3 s λ ϕ σ w dx dt −
D
D
C
i,j =1
D
1
|∂t w|2 dx dt
sϕ
f 2 e2sϕ dx dt + C
sλϕ|∇w|2 dx dt + Cε
sλ2 ϕ|∇w|2 dx dt
D
+ C (s 3 λ3 ϕ 3 + s 2 λ4 ϕ 2 )w 2 dx dt + C
|∇w||∂t w| dx dt.
D
D
D
D
Now we note that the factor with the maximal order in s, λ, ϕ of w2 is s 3 λ4 ϕ 3 σ 2 , the maximal
1
. For example, since we can
factor of |∇w|2 is sλ2 ϕσ and the maximal order of |∂t w|2 is sϕ
3 3 3
2 4 2
2
choose s, λ large, the term (s λ ϕ + s λ ϕ )w is of lower order.
Here, since the Cauchy–Schwarz inequality implies that
|∂t w||∇w| = s − 2 ϕ − 2 λ− 2 |∂t w|s 2 ϕ 2 λ 2 |∇w| 1
1
1
1
1
1
1 1
1
|∂t w|2 + sλϕ|∇w|2 ,
2 sλϕ
2
we have
n
3 s 3 λ4 ϕ 3 σ 2 w 2 dx dt −
sλ2 ϕσ
aij (∂i w)(∂j w) dx dt
D
D
i,j =1
C
1
+ ε−
|∂t w|2 dx dt
λ
sϕ
D
C
f 2 e2sϕ dx dt + C
sλϕ|∇w|2 dx dt + Cε
sλ2 ϕ|∇w|2 dx dt
D
D
D
3 3 3
2 4 2
2
+ C (s λ ϕ + s λ ϕ )w dx dt.
D
14
(3.15)
Inverse Problems 25 (2009) 123013
Topical Review
The first and second terms on the left-hand side have different signs and so we need another
estimate. Thus we will execute another estimation for
n
sλ2 ϕσ
aij (∂i w)(∂j w) dx dt
D
by means of
i,j =1
(P1 w + P2 w) × (sλ2 ϕσ w) dx dt.
D
Here we have chosen the factor sλ2 ϕσ w for obtaining the term of |∇w|2 with desirable
(s, λ, ϕ)-factor sλ2 ϕ. That is, multiplying
n
n
aij (∂i d)(∂j w) −
aij ∂i ∂j w − s 2 λ2 ϕ 2 σ w + A1 w = f esϕ
∂t w + 2sλϕ
i,j =1
i,j =1
with sλ2 ϕσ w, we have
n
(∂t w)(sλ2 ϕσ w) dx dt +
2sλϕ
aij (∂i d)(∂j w)sλ2 ϕσ w dx dt
D
⎛
⎝
−
D
D
n
i,j =1
⎞
aij ∂i ∂j w ⎠ sλ2 ϕσ w dx dt −
s 3 λ4 ϕ 3 σ 2 w 2 dx dt
D
i,j =1
(A1 w)(sλ2 ϕσ w) dx dt ≡
+
5
D
Ik =
f esϕ sλ2 ϕσ w dx dt.
(3.16)
D
k=1
Now, in terms of the integration by parts and w ∈ C02 (D), noting that |∂t ϕ| = |λ(∂t ψ)ϕ| Cλϕ and ∂i ϕ = λ(∂i d)ϕ, etc, we estimate the terms.
1 2
2
|I1 | = sλ ϕσ ∂t (w ) dx dt C
sλ3 ϕw 2 dx dt.
(3.17)
D 2
D
n
|I2 | = s 2 λ3 ϕ 2 σ
aij (∂i d)∂j (w 2 ) dx dt D
i,j =1
n
= −
s 2 λ3 {2λ(∂j d)ϕ 2 }σ aij (∂i d)w 2 dx dt
D
i,j =1
n
s 2 λ3 ϕ 2 ∂j (σ aij (∂i d))w 2 dx dt −
i,j =1
C
s 2 λ4 ϕ 2 w 2 dx dt.
(3.18)
D
I3 = −
D
=
sλ2
D
n
sλ2
n
sλ2
ϕσ aij (∂i w)(∂j w) dx dt +
i,j =1
sλ2 ϕσ
D
ϕσ aij w(∂i ∂j w) dx dt
i,j =1
n
i,j =1
D
n
∂i (ϕσ aij )w(∂j w) dx dt
i,j =1
aij (∂i w)(∂j w) dx dt − C
sλ3 ϕ|∇w||w| dx dt.
(3.19)
D
15
Inverse Problems 25 (2009) 123013
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I4 = −
s 3 λ4 ϕ 3 σ 2 w 2 dx dt.
(3.20)
D
2
2
2
|I5 | C sλ ϕ × sλ ϕσ w dx dt C
s 2 λ4 ϕ 2 w 2 dx dt.
D
(3.21)
D
Hence, by (3.16)–(3.21), we obtain
n
2
sλ ϕσ
aij (∂i w)(∂j w) dx dt −
s 3 λ4 ϕ 3 σ 2 w 2 dx dt
D
i,j =1
D
C
|f e sλ ϕσ w| dx dt + C
s λ ϕ w dx dt + C
sλ3 ϕ|∇w||w| dx dt
D
D
C
f 2 e2sϕ dx dt + C
s 2 λ4 ϕ 2 w 2 dx dt + C
λ2 |∇w|2 dx dt.
(3.22)
sϕ
2
2 4 2
2
D
D
D
D
At the last inequality, we argue as follows: by
sλ3 ϕ|∇w||w| = (sλ2 ϕ|w|)(λ|∇w|) 12 s 2 λ4 ϕ 2 w 2 + 12 λ2 |∇w|2 ,
we have
1
sλ ϕ|∇w||w| dx dt 2
D
(s 2 λ4 ϕ 2 w 2 + λ2 |∇w|2 ) dx dt.
3
D
Furthermore
|f esϕ sλ2 ϕσ w| 12 f 2 e2sϕ + 12 s 2 λ4 ϕ 2 σ 2 w 2 12 f 2 e2sϕ + Cs 2 λ4 ϕ 2 w 2 .
Finally we consider 2 × (3.22) + (3.15). Using (1.2) and σ0 ≡ inf (x,t)∈Q σ (x, t) > 0, we
obtain
3 4 3 2 2
s λ ϕ σ0 w dx dt + (σ0 σ1 − Cε)
sλ2 ϕ|∇w|2 dx dt
D
D
1
C
2
|∂t w| dx dt C
f 2 e2sϕ dx dt
+ ε−
λ
D sϕ
D
+ C (sλϕ + λ2 )|∇w|2 dx dt + C (s 3 λ3 ϕ 3 + s 2 λ4 ϕ 2 )w 2 dx dt.
(3.23)
D
D
Therefore, first choosing ε > 0 sufficiently small such that σ0 σ1 − Cε > 0 and then taking
λ > 0 sufficiently large such that ε − Cλ > 0, we can absorb the second and third terms on the
right-hand side of (3.23) into the left-hand side and we obtain
1
|∂t w|2 dx dt C
s 3 λ4 ϕ 3 w 2 dx dt +
sλ2 ϕ|∇w|2 dx dt +
f 2 e2sϕ dx dt.
sϕ
D
D
D
D
(3.24)
Noting w = uesϕ , we have
1
|∂t u|2 + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 u2 e2sϕ dx dt C
f 2 e2sϕ dx dt.
sϕ
D
D
(3.25)
Moreover we assume that for t ∈ [0, T ], the boundary of the domain D∩{t} ⊂ R is composed
of a finite number of smooth surfaces. Then we can include the terms of ∂i ∂j u in the Carleman
estimate by means of the a priori estimate for an elliptic equation as follows. By (3.2) and
(3.3) and |A1 (x, t)| Csλ2 ϕ, we have
2
n
2
2 2 2
2
4 4 4 2
2 2sϕ
a
∂
∂
w
) in Q.
ij i j C(|∂t w| + s λ ϕ |∇w| + s λ ϕ w + |f | e
i,j =1
n
16
Inverse Problems 25 (2009) 123013
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Hence, by (3.24),
2
n
1
1 2
2
2
3 4 3 2
|∂
dx dt
a
∂
∂
w
dx
dt
C
w|
+
sλ
ϕ|∇w|
+
s
λ
ϕ
w
ij i j t
D sϕ i,j =1
D sϕ
2 2sϕ
+C
f e dx dt C
f 2 e2sϕ dx dt
D
for all large s > 0 and λ > 0.
Moreover we have
∂i ∂j w ∂ i ∂j ϕ
w
= √ −
∂i ∂j √
3 w
ϕ
ϕ
2ϕ 2
3
1
(∂i ϕ)(∂j ϕ)w,
−
3 {(∂j w)(∂i ϕ) + (∂i w)(∂j ϕ)} +
5
2
2ϕ
4ϕ 2
and
n
i,j =1
aij ∂i ∂j
w
√
ϕ
+
3
g
=√ −
ϕ
w
5
4ϕ 2
(3.26)
D
n
n
i,j =1
aij ∂i ∂j ϕ
3
2ϕ 2
aij (∂i ϕ)(∂j ϕ) −
i,j =1
1 i, j n,
(3.27)
w
n
1 3
ϕ2
aij (∂i w)(∂j ϕ)
i,j =1
where we set g = ni,j =1 aij ∂i ∂j w. Since w(·, t) ∈ H01 (D ∩ {t}) for all t ∈ [0, T ], we apply
a usual a priori estimate for the Dirichlet problem for the elliptic equation (e.g., Gilbarg and
Trudinger [62]), so that
2
n ∂i ∂j √w (x, t) dx
ϕ D∩{t} i,j =1
2
n
2
a
∂
∂
ϕ
ij
i
j
i,j =1
g(x, t)
dx + C
C
|w(x, t)|2 dx
ϕ
ϕ3
D∩{t}
D∩{t}
2
2
n
n
w(x, t)2 1
+C
aij (∂i ϕ)(∂j ϕ) dx + C
aij (∂i w)∂j ϕ dx.
5
3
ϕ
D∩{t}
D∩{t} ϕ i,j =1
i,j =1
(3.28)
On the other hand, (3.27) yields
2
2
1
2
∂i ∂j √w + |∂i ∂j ϕ| w 2
|∂i ∂j w(x, t)| dx C
3
ϕ
ϕ
D∩{t} ϕ
D∩{t}
1
1
2
2
2
2
2
2 2
+ 3 (|∂j w| |∂i ϕ| + |∂i w| |∂j ϕ| ) + 5 |∂i ϕ| |∂j ϕ| w (x, t) dx.
ϕ
ϕ
(3.29)
Since ∂i ϕ = λ(∂i d)ϕ and ∂i ∂j ϕ = λ(∂i ∂j d)ϕ + λ2 (∂i d)(∂j d)ϕ, we see by λ > 1 that
|∂i ϕ(x, t)| Cλϕ(x, t),
|∂i ∂j ϕ(x, t)| Cλ2 ϕ(x, t),
1 i, j n, (x, t) ∈ D.
(3.30)
17
Inverse Problems 25 (2009) 123013
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Hence, by ϕ 1, estimates (3.28) and (3.29) yield
n 1
g 2 (x, t)
|∂i ∂j w(x, t)|2 dx C
dx
ϕ(x, t)
D∩{t} ϕ(x, t)
i,j =1 D∩{t}
+C
(λ4 w 2 + λ2 |∇w|2 )(x, t) dx.
D∩{t}
Integrating in t, we have
n 1
|∂i ∂j w(x, t)|2 dx dt
sϕ
D
i,j =1
2
n
1 C
a
∂
∂
w
dx
dt
+
C
(λ4 w 2 + λ2 |∇w|2 ) dx dt.
ij i j sϕ
D
D
i,j =1
With (3.24) and (3.26), we obtain
n
1 |∂i ∂j w|2 dx dt C
f 2 e2sϕ dx dt
sϕ
D
D
i,j =1
for all large s > 0 and λ > 0. Thus, we can complete the derivation of a Carleman estimate
for u ∈ C0∞ (D). Noting that
esϕ ∂i ∂j u = ∂i ∂j w − sλϕ((∂i d)(∂j w) + (∂j d)(∂i w))
+ {s 2 λ2 ϕ 2 (∂i d)(∂j d) − sλ2 ϕ(∂i d)(∂j d) − sλϕ(∂i ∂j d)}w,
and writing the estimate in terms of u again by (3.24), we state the Carleman estimate here as
a theorem.
Theorem 3.1. Let d ∈ C 2 () satisfy |∇d| = 0 on and let
ψ(x, t) = d(x) − β(t − t0 )2
with β > 0 and 0 < t0 < T . We assume that ∂D is smooth and for t ∈ [0, T ], the boundary
of the domain D ∩ {t} ⊂ Rn is composed of a finite number of smooth surfaces. There exists a
constant λ0 > 0 such that for arbitrary λ λ0 , we can choose a constant s0 (λ) > 0 satisfying:
there exists a constant C = C(s0 , λ0 ) > 0 such that
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 u2 e2sϕ dx dt
|∂t u|2 +
⎭
D ⎩ sϕ
i,j =1
|Lu|2 e2sϕ dx dt
(3.31)
C
D
for all s > s0 and all u satisfying
u ∈ H 2,1 (Q),
supp u ∈ D.
(3.32)
The constant C > 0 in (3.31) depends continuously on max1i,j n aij C 1 (Q) , bi L∞ (Q) ,
cL∞ (Q) . This dependence holds also in theorems 3.2 and 3.3.
Here we note:
(1) By a usual density argument (i.e., the approximation of any u satisfying (3.32) by a
sequence un ∈ C0∞ (D)), we can transfer (3.31) for un ∈ C0∞ (D) to the Carleman
estimate for all u satisfying (3.32).
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Inverse Problems 25 (2009) 123013
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(x, t) > 0, the proof
(x, t) = d(x) − β(t − t0 )2 + c0 where inf (x,t)∈Q ψ
(2) In the case of ψ
2sϕ
2seλψ
is already completed. In the case of c0 = 0, we have e = e
= exp(2(se−λc0 )eλψ ),
λc0
so that by replacing s0 (λ) by s0 (λ)e , we can reduce the case of c0 = 0 to the previous
case.
Thus theorem 3.1 follows.
Even if supp u ⊂ D does not hold, we can follow the previous argument without omitting
boundary integral terms which are produced by each integration by parts, and we can prove
the following Carleman estimate.
Theorem 3.2. Let d ∈ C 2 () satisfy |∇d| = 0 on and let ψ(x, t) = d(x) − β(t − t0 )2 with
β > 0 and 0 < t0 < T . We assume that ∂D is smooth and for t ∈ [0, T ], the boundary of
the domain D ∩ {t} ⊂ Rn is composed of a finite number of smooth surfaces. There exists a
constant λ0 > 0 such that for arbitrary λ λ0 , we can choose a constant s0 (λ) > 0 satisfying:
there exists a constant C = C(s0 , λ0 ) > 0 such that
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 u2 e2sϕ dx dt
|∂t u|2 +
⎭
D ⎩ sϕ
i,j =1
|Lu|2 e2sϕ dx dt + C eC(λ)s
(|∇x,t u|2 + |u|2 ) dS dt
(3.33)
C
D
∂D
for all s > s0 and all u ∈ H
2,1
(D).
See notations in section 1 as for the definition of |∇x,t u|2 .
Thanks to the large parameters λ > 0 and s > 0, we can derive a Carleman estimate for a
weakly coupled parabolic system whose principal parts are not coupled: let u = (u1 , . . . , uN )T
and aijk ∈ C 1 (Q), 1 i, j n, 1 k N , satisfy (1.1) and (1.2), and bik , ck ∈ L∞ (Q),
1 i n, 1 k, N. Here and henceforth (u1 , . . . , uN )T denotes the transpose of a
vector (u1 , . . . , uN ). We set
[Au] =
n
aij ∂i ∂j u
i,j =1
−
N n
k=1 i=1
bik ∂i uk
−
N
ck uk ,
1 N.
(3.34)
k=1
Then
Theorem 3.3. Let d ∈ C 2 () satisfy |∇d| = 0 on and let ψ(x, t) = d(x) − β(t − t0 )2 with
β > 0 and 0 < t0 < T . We assume that ∂D is smooth and for t ∈ [0, T ], the boundary of
the domain D ∩ {t} ⊂ Rn is composed of a finite number of smooth surfaces. There exists a
constant λ0 > 0 such that for arbitrary λ λ0 , we can choose a constant s0 (λ) > 0 satisfying:
there exists a constant C = C(s0 , λ0 ) > 0 such that
⎫
⎧ ⎛
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 |u2 | e2sϕ dx dt
|∂t u|2 +
⎭
D ⎩ sϕ
i,j =1
|∂t u − Au|2 e2sϕ dx dt
(3.35)
C
D
for all s > s0 and all u satisfying
u ∈ H 2,1 (Q)N ,
supp u ∈ D.
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4. Global Carleman estimate
In sections 2 and 3, we proved Carleman estimates in a domain D which is not necessarily
the same as × (0, T ). Moreover for applications to inverse problems (see sections 5.1 and
6.1), D is given by ϕ. In other words, when we apply the Carleman estimate in section 3,
first we have to choose ϕ and then D is determined where the results concerning the inverse
problems are valid, and not vice versa. Imanuvilov [72] proved a global Carleman estimate
which holds over × (0, T ) for functions without compact supports. See also [32, 60, 71].
The global Carleman estimate is very useful for proving an observability inequality which
yields the null exact controllability and a stability estimate in determining coefficients over .
In this section, we present the global Carleman estimate by Imanuvilov.
Henceforth let ω ⊂ be an arbitrary sub-domain.
We recall that
H 2,1 (Q) = {u ∈ L2 (Q); ∂t u, ∂i u, ∂i ∂j u ∈ L2 (Q), 1 i, j n}.
We consider a boundary value problem for a parabolic operator:
Lu(x, t) ≡ ∂t u −
n
aij (x, t)∂i ∂j u −
i,j =1
l1 (x)
n
bi (x, t)∂i u − c(x, t)u = f
in Q
(4.1)
i=1
∂u
+ l2 (x)u = 0
∂νA
on ∂ × (0, T ).
(4.2)
Here we define the conormal derivative with respect to aij by
assume (1.2) and
aij ∈ C 1 (Q),
∂u
∂νA
=
n
i,j =1
bi , c ∈ L∞ (Q),
aij = aj i ,
aij (∂i u)νj . We
1 i, j n.
Suppose that l1 , l2 ∈ C 2 (∂) and
either l1 > 0
or
l1 = 0
and
l2 = 1
on ∂.
(4.3)
For the global Carleman estimate, we need a special weight function. The existence of such a
function is proved in [60, 72, 77].
Lemma 4.1. Let ω0 be an arbitrarily fixed sub-domain of such that ω0 ⊂ ω. Then there
exists a function d ∈ C 2 () such that
d(x) > 0 x ∈ ,
d|∂ = 0,
|∇d(x)| > 0,
x ∈ \ω0 .
(4.4)
Example. Let = {x; |x| < 1} and let 0 ∈ ω0 . Then d(x) = 1 − |x|2 satisfies (4.4).
We set
ϕ(x, t) =
eλd(x)
,
t (T − t)
α(x, t) =
eλd(x) − e2λdC()
,
t (T − t)
(4.5)
where λ > 0. Moreover we set
σ2 =
n
i,j =1
aij C 1 (Q) +
n
bi L∞ (Q) + cL∞ (Q) ,
i=1
The following is the global Carleman estimate.
20
σ3 =
n
i,j =1
aij C 1 (Q) .
(4.6)
Inverse Problems 25 (2009) 123013
Topical Review
Theorem 4.1. There exists a number λ0 > 0 such that for an arbitrary λ λ0 , we can choose
a constant s0 (λ) 0 satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 |u|2 e2sα dx dt
|∂t u|2 +
⎭
Q ⎩ sϕ
i,j =1
|Lu|2 e2sα dx dt + C
s 3 λ4 ϕ 3 u2 e2sα dx dt
C
Q
ω×(0,T )
for all s > s0 and all u ∈ H 2,1 (Q) satisfying (4.2). Here the constant C > 0 depends
continuously on σ2 , λ0 , but is independent of s and the constant λ0 depends continuously on
σ3 .
By (4.5), we note that
lim ϕ(x, t)k e2sα(x,t) = lim ϕ(x, t)k e2sα(x,t) = 0
t↑T
t↓0
2sα(x,t)
behaves with the same order of
for any
k C 0 and near t = 0, T , the weight function e
exp − t (T −t) with C > 0. Thanks to this decay of the weight function, we need not assume
that u vanishes at t = 0, T . Moreover for the Carleman estimate, it is sufficient that u satisfies
one boundary condition (4.2) on the lateral boundary ∂ × (0, T ). Therefore, this Carleman
estimate holds over × (0, T ) for u without compact supports.
For the case where we are given overdetermining data on an arbitrary part ⊂ ∂, we
need another weight function and see [72, 78].
Lemma 4.2. Let = ∅, ⊂ ∂ be an arbitrary relatively open subset. Then there exists a
function d0 ∈ C 2 () such that
x ∈ ,
|∇d0 (x)| > 0,
x ∈ ,
d0 (x) > 0,
n
aij (x, t)(∂i d0 )(x)νj (x) 0,
x ∈ ∂\, 0 < t < T
(4.7)
i,j =1
if aij ∈ C 1 (Q), 1 i, j n satisfy (1.1) and (1, 2).
We note that d0 satisfies (4.7) for all aij satisfying (1.1) and (1.2). In other words, d0 does
not depend on choices of aij.
We set
eλd0 (x)
eλd0 (x) − e2λd0 C()
,
α0 (x, t) =
.
(4.8)
ϕ0 (x, t) =
t (T − t)
t (T − t)
Example. Let us consider a special case where aij = 0 if i = j and aii = 1 and
= {x ∈ ∂; (x − x0 , ν(x)) 0}
= {x ∈ Rn ; |x| < R},
n
with an arbitrarily fixed x0 ∈ R \. Here (·, ·) denotes the scalar product in Rn . Then we
can take d0 (x) = |x − x0 |2 .
By α, we have
Theorem 4.2. There exists a number λ0 > 0 such that for an arbitrary λ λ0 , we can choose
a constant s0 (λ) 0 satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎛
⎞
n
1 ⎝
2
2⎠
2
2
3 4 3
2
|∂t u| +
|∂i ∂j u| + sλ ϕ0 |∇u| + s λ ϕ0 |u| e2sα0 dx dt
sϕ
0
Q
i,j =1
2 2sα0
C(λ)s
|Lu| e
dx dt + C e
(|∂t u|2 + |∇u|2 + |u|2 ) dS dt
C
Q
×(0,T )
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Inverse Problems 25 (2009) 123013
Topical Review
for all s > s0 and all u ∈ H 2,1 (Q) satisfying (4.2). Here the constant C > 0 depends
continuously on σ2 , λ0 , but is independent of s and the constant λ0 depends continuously on
σ3 .
2 ∂u 2
+ where ∂u is the orthogonal
Here on × (0, T ) we note that |∇u|2 = ∂u
∂ν
∂τ
∂τ
∂u
component of ∇u to ∂ν .
The proof is done on the basis of the decomposition of the operator P w = esα L(e−sα w)
into P1 and P2 defined by (3.4) and (3.5). In fact, the proof in section 3 is an imitation of the
original proofs of theorems 4.1 and 4.2.
We conclude this section with the corresponding Carleman estimates to theorems 4.1 and
4.2 for a weakly coupled parabolic system:
∂t u = Au
in Q
with
l1 (x)
n
aij (∂i u )νj + l2 (x)u = 0
on ∂ × (0, T ), 1 N .
(4.9)
i,j =1
Here A is defined by (3.34) and let ϕ0 , α0 , ϕ and α be defined by (4.5) and (4.8). Then
Theorem 4.3. There exists a number λ0 > 0 such that for an arbitrary λ λ0 , we can choose
a constant s0 (λ) 0 satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 |u|2 e2sα dx dt
|∂t u|2 +
⎭
Q ⎩ sϕ
i,j =1
|∂t u − Au|2 e2sα dx dt + C
s 3 λ4 ϕ 3 |u|2 e2sα dx dt
C
Q
ω×(0,T )
for all s > s0 and all u ∈ H (Q) satisfying (4.9). Here the constant C > 0 depends
continuously on σ2 , λ0 , but is independent of s and the constant λ0 depends continuously on
σ3 .
2,1
N
Theorem 4.4. There exists a number λ0 > 0 such that for an arbitrary λ λ0 , we can choose
a constant s0 (λ) 0 satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎛
⎞
n
1 ⎝
2
2⎠
2
2
3 4 3
2
|∂i ∂j u| + sλ ϕ0 |∇u| + s λ ϕ0 |u| e2sα0 dx dt
|∂t u| +
Q sϕ0
i,j =1
|∂t u|2 + |∇u|2 + |u|2 dS dt
|∂t u − Au|2 e2sα0 dx dt + C eC(λ)s
C
Q
×(0,T )
for all s > s0 and all u ∈ H (Q) satisfying (4.9). Here the constant C > 0 depends
continuously on σ2 , λ0 , but is independent of s and the constant λ0 depends continuously
on σ3 .
2,1
N
As for the derivation of a Carleman estimate, see also Choulli [34], section 3.2.
5. Applications to the unique continuation and the observability inequality
Originally the Carleman estimate has been invented for proving the uniqueness in a Cauchy
problem for an elliptic equation by Carleman [30] and as the first application of the Carleman
estimate in this section, we will discuss the methodology for the uniqueness and the conditional
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Inverse Problems 25 (2009) 123013
Topical Review
stability for a parabolic equation by a local version of the Carleman estimate: theorem 3.2.
We emphasize that whenever we apply a local version of the Carleman estimate, it is essential
to introduce a cut-off function (e.g., (5.15), (5.21), (6.9)) and apply the Carleman estimate to
the product of a solution to a partial differential equation by the cut-off function.
5.1. Conditional stability for the Cauchy problem
Let ⊂ ∂ be an arbitrary sub-boundary of ∂. That is, let contain a non-empty set
{x ∈ Rn ; |x − x0 | < ρ} ∩ ∂ with some x0 ∈ Rn and ρ > 0. We consider a Cauchy problem
for a parabolic equation:
n
n
(Lu)(x, t) ≡ ∂t u −
aij (x, t)∂i ∂j u −
bi (x, t)∂i u − c(x, t)u = f, (x, t) ∈ Q,
i,j =1
i=1
(5.1)
u|×(0,T ) = g,
n
∂u
|×(0,T ) ≡
aij (∂i u)νj = h,
∂νA
i,j =1
(5.2)
where aij , 1 i, j n satisfy (1.1) and (1.2), and
bi , c ∈ L∞ (Q),
1 i, j n.
(5.3)
Remark. As is seen in theorem 7.4, we have an H −1 Carleman estimate if aij are Lipschitz
continuous on Q, and so for all the discussions in section 5, we can relax the regularity (1.1)
to the Lipschitz continuous aij on Q.
Cauchy problem. Let u satisfy (5.1). Then determine u in some domain D ⊂ Q by u|×(0,T )
∂u
and ∂ν
|×(0,T ) .
A
As for the uniqueness results, there are a lot of works. Therefore, we will not list them
comprehensively even though we restrict ourselves to parabolic equations. Mizohata [122]
wrote one of the early papers for the parabolic case, and we refer to [93, 110, 130, 131]. See
also the monographs [44, 69, 70, 85, 90, 105, 146] and a survey paper [141] by Vessella,
and references therein. In this section, we give accounts of methods for applying Carleman
estimates to prove stability results in the Cauchy problem.
One introduces a suitable cut-off function and extends Cauchy data in a suitable Sobolev
space to reduce the problem to functions with compact supports and then one can apply a local
Carleman estimate (e.g., theorem 3.1) to obtain a stability estimate of u by data on × (0, T ).
This argument is quite traditional and is valid for other types of partial differential equations,
and see e.g., [[90], sections 3.2 and 3.3]. More precisely, applying theorem 3.1, we can prove
the following proposition by the same argument as in theorem 3.2.2 in [[90], section 3.2].
Proposition 5.1. Let d ∈ C 2 () and |∇d| = 0 on , and ψ(x, t) = d(x) − β(t − t0 )2 with
t0 ∈ (0, T ). We set
Q(ε) = {(x, t) ∈ × (0, T ); ψ(x, t) > ε}
with ε > 0. Moreover, we assume that
Q(0) ⊂ Q ∪ ( × [0, T ])
with sub-boundary ⊂ ∂. Then for any small ε > 0, there exist constants C > 0 and
θ ∈ (0, 1) such that
F θ + CF0
uH 2,1 (Q(ε)) Cu1−θ
H 1,0 (Q) 0
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Inverse Problems 25 (2009) 123013
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where F0 = f L2 (Q) + gL2 (0,T ;H 32 ()) + gH 1 (0,T ;L2 ()) + hL2 (0,T ;H 12 ()) + hH 1 (0,T ;L2 ()) .
Using the local estimate for a parabolic equation (e.g., [[109], section 10, chapter IV]),
we can replace the factor uH 1,0 (Q) in the above stability estimate by a weaker norm uL2 (Q)
(see e.g., [[90], theorem 3.3.10]: version in 2006) and here we omit the details.
Proposition 5.1 gives an estimate of the solution in Q(ε) by data on × (0, T ), and
Q(ε) and are determined by the function d(x). In other words, for an arbitrarily given
sub-boundary , proposition 5.1 does not work for estimating the solution in a given domain.
Moreover, since we use theorem 3.1 for functions with compact supports (also in [90]), we
need some extension argument for gaining compact supports, which breaks the best possibility
of the regularity of Cauchy data. Therefore on the basis of theorem 3.2, which is a Carleman
estimate for functions without compact supports, we will show a sharper estimate. Moreover
our choice of sub-domain for the stability is more flexible.
Theorem 5.1 (conditional stability). Let ⊂ ∂ be an arbitrary non-empty sub-boundary of
∂. For any ε > 0 and an arbitrary bounded domain 0 such that 0 ⊂ ∪ , ∂0 ∩ ∂ is a
non-empty open subset of ∂ and ∂0 ∩ ∂ , there exist constants C > 0 and θ ∈ (0, 1)
such that
uH 2,1 (0 ×(ε,T −ε)) Cu1−θ
(f L2 (Q) + gH 1 (×(0,T )) + hL2 (×(0,T )) )θ
H 1,0 (Q)
+ C(f L2 (Q) + gH 1 (×(0,T )) + hL2 (×(0,T )) ).
(5.4)
In the theorem, we are given Cauchy data u, ∇u on × (0, T ), and we estimate u in
a sub-domain 0 ⊂ over a time interval (ε, T − ε). The theorem does not directly give
an estimate when ε = 0, but we can derive an estimate for ε = 0 by an argument similar
to theorem 5.2 below and we do not discuss details. Estimate (5.4) immediately implies the
∂u
= 0 on × (0, T ), because
uniqueness of the solution in × (0, T ) to Lu = 0 with u = ∂ν
A
ε > 0 is arbitrary.
This is a kind of interior estimate and usually the interior estimate is of Hölder type.
Remark. At best we can expect the Lipschitz stability for the Cauchy problem, but it seems
impossible to prove the Lipschitz stability and difficult to obtain the maximum exponent
θ ∈ (0, 1) in (5.4). Moreover in (5.4), we cannot replace the norm of g by a weaker norm
gL2 (0,T ;H 1 ()) . That is, with such a weaker norm, any stability fails in a reasonably general
infinite-dimensional subspace of L2 ( × (0, T )). For simplicity, we consider Lu = ∂t u − u.
We define an infinite-dimensional Hilbert space H by
H = {h ∈ L2 ( × (0, T )); supp h ⊂ (∂0 ∩ ∂) × (ε, T − ε)}.
Contrarily assume the stability, in particular, that uH 2,1 (0 ×(ε,T −ε)) are bounded for bounded
gL2 (0,T ;H 1 ()) and hL2 (×(0,T )) . Then there exists some function C0 ∈ C([0, ∞)3 ) such
that
, gL2 (0,T ;H 1 ()) , hL2 (×(0,T )) ).
uH 2,1 (0 ×(ε,T −ε)) C0 (u1−θ
H 1,0 (Q)
Let u(h) be a unique solution to
⎧
⎨∂t u(x, t) = u(x, t),
u(x, 0) = 0,
⎩ ∂u
(x, t) = h(x, t),
∂ν
(5.5)
x ∈ , 0 < t < T ,
x ∈ ,
x ∈ ∂, 0 < t < T .
Then, by example 2 (p 81, vol II) in [120], we see that
u(h)L2 (0,T ;H 32 ()) + u(h)H 34 (0,T ;L2 ()) ChL2 (×(0,T )) .
24
(5.6)
Inverse Problems 25 (2009) 123013
3
2
Topical Review
3
4
Here H () and H (0, T ; L2 ()) are defined in [1, 120] for example. By (5.6) and the trace
theorem, we have
u(h)L2 (0,T ;H 1 ()) ChL2 (×(0,T )) .
(5.7)
Let {hn }n∈N ⊂ H be an arbitrary bounded sequence: hn L2 (×(0,T )) M. Applying (5.6)
and (5.7) in (5.5) and noting u(hn )H 1,0 (Q) u(hn )L2 (0,T ;H 32 ()) CM, we have
u(hn )H 2,1 (0 ×(ε,T −ε)) C(M),
n ∈ N.
(5.8)
Since the embedding H 2,1 (0 × (ε, T − ε)) −→ L2 (0 × (ε, T − ε)) is compact (e.g., [136,
p 271]), applying theorem 16.2 (p 101, vol I) and proposition 2.1 (p 7, vol II) in [120], we see
that the embedding
3 3
H 2,1 (0 × (ε, T − ε)) −→ H 2 , 4 (0 × (ε, T − ε))
is compact. Hence by (5.8), we can extract a subsequence hn ∈ H, denoted by
3 3
the same notations, such that u(hn ) −→ u∞ in H 2 , 4 (0 × (ε, T − ε)) with some
3 3
,
u∞ ∈ H 2 4 (0 × (ε, T − ε)). By the trace theorem, we obtain that hn −→ ∂u∂ν∞ in
L2 ((∂0 ∩ ∂) × (ε, T − ε)). This means that an arbitrary bounded sequence in H contains a
convergent subsequence, and so dim H < ∞. This is a contradiction. Thus the stability does
not hold if we replace gH 1 (×(0,T )) by gL2 (0,T ;H 1 ()) .
In theorem 5.1, in order to estimate u, we have to assume the a priori bound uH 1,0 (Q) .
Estimate (5.4) is called a conditional stability estimate (see also remarks after the statement
of theorem 6.1).
Proof. Unlike proposition 5.1, we are given a sub-boundary and a sub-domain 0 , and
requested to estimate the solution in 0 × (ε, T − ε) by Cauchy data on × (0, T ). Thus,
according to the geometry of 0 and , we have to choose a suitable weight function ϕ, that
is, d(x). For this, we first choose a bounded domain 1 with smooth boundary such that
1 ,
= ∂ ∩ 1 ,
∂\ ⊂ ∂1 .
(5.9)
such that ∂ ∩ = We note that 1 is constructed by taking a union of and a domain and that 1 \ contains some non-empty open set. Choosing ω0 ⊂ 1 \, we apply
lemma 4.1 to obtain d ∈ C 2 (1 ) satisfying
d(x) = 0, x ∈ ∂1 ,
d(x) > 0, x ∈ 1 ,
|∇d(x)| > 0, x ∈ 1 ∩ .
(5.10)
Then, since 0 ⊂ 1 , we can choose a sufficiently large N > 1 such that
4
dC(1 ) } ∩ ⊃ 0 .
N
Moreover we choose β > 0 such that
{x ∈ 1 ; d(x) >
(5.11)
(5.12)
2βε2 > dC(1 ) > βε2 .
√
√
We arbitrarily fix t0 ∈ [ 2ε, T − 2ε]. We set ϕ(x, t) = eλψ(x,t) with fixed large parameter
2 λ > 0 and ψ(x, t) = d(x) − β(t − t0 )2 , μk = exp λ Nk dC(1 ) − βε
, k = 1, 2, 3, 4, and
N
D = {(x, t); x ∈ , ϕ(x, t) > μ1 }.
Then we can verify that
√
√
ε
ε
⊂ D ⊂ × (t0 − 2ε, t0 + 2ε).
0 × t0 − √ , t0 + √
(5.13)
N
N
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Inverse Problems 25 (2009) 123013
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√ε , t0 + √ε . Then, by (5.11) we have x ∈ and
N
N
In fact, let (x, t) ∈ 0 × t0 −
d(x) >
4
dC(1 ) ,
N
so that
4
βε2
dC(1 ) −
,
N
N
that is, ϕ(x, t) > μ4 , which implies that (x, t) ∈ D by the definition. Next let (x, t) ∈ D.
2
Then d(x) − β(t − t0 )2 > N1 dC(1 ) − βε
. Therefore
N
d(x) − β(t − t0 )2 >
1
βε2
dC(1 ) +
> β(t − t0 )2 .
N
N
2
> β(t − t0 )2 , that is, 2βε2 > β(t − t0 )2 . The
Applying (5.12), we have 2 1 − N1 βε2 + βε
N
verification of (5.13) is completed.
Next we have
∂D ⊂ 1 ∪ 2 ,
(5.14)
2 = {(x, t); x ∈ , ϕ(x, t) = μ1 }.
1 ⊂ × (0, T ),
dC(1 ) −
In fact, let (x, t) ∈ ∂D. Then x ∈ and ϕ(x, t) μ1 . We separately consider the cases
x ∈ and x ∈ ∂. First let x ∈ . If ϕ(x, t) > μ1 , then (x, t) is an interior point of D.
Therefore if x ∈ , then ϕ(x, t) = μ1 . Next let x ∈ ∂. Let x ∈ ∂\. Then x ∈ ∂1
by the third condition in (5.9), and d(x) = 0 by the second condition in (5.10). On the other
hand, ϕ(x, t) μ1 yields that
1
βε2
,
d(x) − β(t − t0 )2 = −β(t − t0 )2 dC(1 ) −
N
N
that is, 0 β(t − t0 )2 N1 (βε2 − dC(1 ) ), which is impossible by (5.12). Therefore x ∈ .
In terms of (5.13) the verification of (5.14) is completed.
We apply theorem 3.2 in D with suitably fixed λ > 0. Henceforth C > 0 denotes generic
constants depending on λ, but independent of s and choices of g, h, u. For it, we need a
cut-off function because we have no data on ∂D\( × (0, T )). Let χ ∈ C ∞ (Rn+1 ) such that
0 χ 1 and
1,
ϕ(x, t) > μ3 ,
(5.15)
χ (x, t) =
0,
ϕ(x, t) < μ2 .
We set v = χ u, and have
⎛
⎞
n
n
n
Lv = χf + χ u − 2
aij (∂i χ )∂j u − ⎝
aij ∂i ∂j χ ⎠ u −
bi ∂i χ u in D.
i,j =1
i,j =1
i=1
By (5.14) and (5.15), we see that
v = |∇v| = 0
on 2 .
Hence theorem 3.2 yields
⎧ ⎛
⎫
⎞
⎨
n
⎬
1⎝
|∂i ∂j v|2 ⎠ + s|∇v|2 + s 3 |v|2 e2sϕ dx dt
|∂t v|2 +
⎭
D ⎩s
i,j =1
⎛
⎞
n
n
C
f 2 e2sϕ dx dt + C χ u − 2
aij (∂i χ )∂j u − ⎝
aij ∂i ∂j χ ⎠ u
D
D
i,j =1
i,j =1
2
n
bi ∂i χ u e2sϕ dx dt + C eCs
(|∂t v|2 + |∇v|2 + |v|2 ) dS dt
−
1
i=1
26
(5.16)
Inverse Problems 25 (2009) 123013
Topical Review
for all s s0 . By (5.15), the second integral on the right-hand side does not vanish only if
μ2 ϕ(x, t) μ3 and so
⎛
⎞
n
n
n
χ u − 2
aij (∂i χ )∂j u − ⎝
aij ∂i ∂j χ ⎠ u −
bi ∂i χ u e2sϕ dx dt
D
i,j =1
i,j =1
i=1
C e2sμ3 u2H 1,0 (Q) .
By (5.11) and the definition of D, we can directly verify that if (x, t) ∈ 0 × t0 − √εN , t0 + √εN ,
then ϕ(x, t) > μ4 . Therefore, noting (5.13) and (5.15), we see that
(the left-hand side of (5.16))
⎧ ⎛
⎫
⎞
t0 + √ε ⎨
n
⎬
N
1⎝
|∂i ∂j v|2 ⎠ + s|∇v|2 + s 3 |v|2 e2sϕ dx dt
|∂t v|2 +
⎭
t0 − √εN
0 ⎩ s
i,j =1
⎧ ⎛
⎫
⎞
t0 + √ε ⎨
n
⎬
N
1
⎝|∂t u|2 +
e2sμ4
|∂i ∂j u|2 ⎠ + s|∇u|2 + s 3 |u|2 dx dt.
⎭
t0 − √ε
0 ⎩ s
i,j =1
N
Hence (5.16) yields
⎧ ⎛
⎫
⎞
t0 + √ε ⎨
n
⎬
N
1
⎝|∂t u|2 +
e2sμ4
|∂i ∂j u|2 ⎠ + s|∇u|2 + s 3 |u|2 dx dt
⎭
t0 − √εN
0 ⎩ s
i,j =1
f 2 e2sϕ dx dt + C e2sμ3 u2H 1,0 (Q)
C
D
Cs
+C e
(|∂t v|2 + |∇v|2 + |v|2 ) dS dt.
×(0,T )
Setting
F = f L2 (Q) + gH 1 (×(0,T )) + hL2 (×(0,T )) ,
and noting
2
|∇v| dS dt 2
(χ 2 |∇u|2 + |∇χ |2 u2 ) dS dt
×(0,T )
×(0,T )
2 2 ∂u ∂u 2
C
|u| + + dS dt,
∂ν
∂τ
×(0,T )
we have
t0 + √ε ⎧
⎨
N
t0 − √εN
0
⎩
|∂t u|2 +
n
|∂i ∂j u|2 + |∇u|2 + |u|2
i,j =1
⎫
⎬
⎭
dx dt
Cs e−2s(μ4 −μ3 ) u2H 1,0 (Q) + C eCs F 2
for all s s0 . By sups>0 se−s(μ4 −μ3 ) < ∞, we estimate se−2s(μ4 −μ3 ) by Ce−s(μ4 −μ3 ) on the
right-hand side. Moreover, replacing C by C eCs0 , we can have
u2H 2,1 (0 ×(t0 − √ε
N
,t0 + √εN ))
C e−s(μ4 −μ3 ) u2H 1,0 (Q) + C eCs F 2
(5.17)
for all s 0. By the argument, we note that the constant C in (5.17) is independent also of t0.
27
Inverse Problems 25 (2009) 123013
√
√
In (5.17), taking t0 = 2ε +
2ε +
(m+1)ε
√
N
Topical Review
jε
√
,
N
j = 0, 2, . . . , m such that
√
2ε +
mε
√
N
T −
√
2ε T and summing up over j , we obtain
(5.18)
u2H 2,1 (0 ×(ε,T −ε)) C e−s(μ4 −μ3 ) u2H 1,0 (Q) + C eCs F 2
√
for all s 0 by noting that N is sufficiently large and replacing 2ε by ε.
First let F = 0. Then letting s → ∞ in (5.18), we see that u = 0 in 0 × (ε, T − ε), so
that the conclusion of theorem 5.1 holds true. Next let F = 0. First let F uH 2,1 (Q) . Then
(5.18) implies uH 2,1 (0 ×(ε,T −ε)) C eCs F for s 0, which already proves the theorem.
Second let F < uH 2,1 (Q) . We choose s > 0 minimizing the right-hand side of (5.18), that
is,
e−s(μ4 −μ3 ) u2H 1,0 (Q) = eCs F 2 .
By F = 0, we can choose
s=
uH 1,0 (Q)
2
> 0.
log
C + μ4 − μ3
F
Then (5.18) gives
C
C+μ −μ
μ4 −μ3
uH 2,1 (0 ×(ε,T −ε)) 2CuH 1,04(Q)3 F C+μ4 −μ3 .
The proof of theorem 5.1 is completed.
For better estimation, we used theorem 3.2 for functions without compact supports. If we
use a Carleman estimate for functions with compact supports, then we have to extend g and
∂H
|×(0,T ) = h. Setting u = u − H and
h to functions H in Q such that H |×(0,T ) = g and ∂ν
A
taking the cut-off function χ defined by (5.15), we can obtain a stability estimate (proposition
5.1) for the Cauchy problem with stronger norms of data: gL2 (0,T ;H 32 ())∩H 1 (0,T ;L2 ()) and
hL2 (0,T ;H 12 ())∩H 1 (0,T ;L2 ()) (see, e.g., [[90], theorems 3.2.2, 3.2.3 and 3.3.10]).
Remark (Conditional stability and numerical algorithm). The conditional stability is
important not only from the viewpoint of the mathematical analysis, but also from the
viewpoint of the numerical analysis. For example in the Cauchy problem, we consider a
∂u
|×(0,T ) . By
numerical algorithm for reconstructing u|D by Cauchy data u|×(0,T ) and ∂ν
A
possible instability of the Cauchy problem, for a stable numerical algorithm, regularization
techniques are used. We do not describe the details here, see [8, 48, 65, 68, 137]. Among
the regularization techniques, the Tikhonov regularization is widely used where the original
problem is changed to a minimization problem to which some regularizing term is attached
with factor parameter γ > 0. One important subject in the regularization theory is how
to choose a regularizing parameter γ > 0 for gaining the convergence rates of ‘regularized
solutions’ and there have been many results (e.g., [8, 48, 65, 68, 137], and references therein).
In [33], it is proved that a conditional stability estimate gives such a choice strategy, and so
this is one significance of the conditional stability.
Next we consider an estimate of u(x0 , t0 ) for x0 ∈ ∂\ and t0 > 0. An argument for
obtaining an estimate on the boundary does not seem popular. Here we consider a simple case
for convenience. Let x = (x2 , . . . , xn ),
= (0, 2) × {x ; |x | < R}
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Inverse Problems 25 (2009) 123013
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and
let usT be given data of u in (1, 2) × {x ; |x | < R} × (0, T ). Then we would like to estimate
u 0, 0, 2 . We set
⎧
⎨DT = {(x , t); |x | < R, 0 < t < T },
(5.19)
⎩F = sup (u(x1 , ·, ·)H 1 (DT ) + ∂1 u(x1 , ·, ·)L2 (DT ) ).
1x1 2
Let m ∈ N satisfy
m > max
n+1
,3 .
2
Then our estimate is
Theorem 5.2.
CM
u 0, 0, T m−3 .
2
log F1 2(m+1)2
provided that F < 1 and
uH m+2 (Q) M.
Since we are interested in the rate of the stability estimate as F → 0, it is sufficient to
consider the case of F < 1.
The estimate at a boundary point is of logarithmic rate and much weaker than the interior
estimate given by theorem 5.1, and we need an a priori bound of solutions in Sobolev space
H m+2 of higher order.
We note that F is not a norm of boundary data but by the method of theorem 5.1, we can
estimate F in terms of boundary data on x1 = 2 after we enlarge the domain Q in x and t and
assume the parabolic equation with Cauchy data for the extended domain. Moreover, we can
repeatedly apply the argument here to obtain an estimate for u(0, x , t) for |x | < R − ε and
ε < t < T − ε with arbitrarily fixed ε > 0. For estimates on a sub-boundary for elliptic
equations, see [47, 132].
Proof. The proof consists of two steps:
(1) Hölder stability estimate in a sub-domain with distance ρ from the sub-boundary x1 = 0
whose Hölder exponent depends on ρ. See (5.29).
(2) Integration of the Hölder estimates over the distance variable ρ.
Step (1) is a direct consequence of the Carleman estimate theorem 3.2 in view of a suitable
cut-off function.
We choose β > 0 and γ > 0 such that
γ R 2 > 1,
We set
βT 2
> 1.
4
(5.20)
T 2
ψ(x, t) = x1 − γ |x | − β t −
2
2
and
Q(ξ ) = {(x, t); x1 < 1 + ξ, ψ(x, t) > ξ }.
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Inverse Problems 25 (2009) 123013
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Then,
by (5.20), we see that Q(ξ ) ⊂ × (0, T ) for ξ ∈ (0, 1). For applying theorem 3.2, for
η ∈ 0, 13 , we need a cut-off function χη ∈ C ∞ (Rn+1 ) satisfying 0 χη 1 and
1,
ψ(x, t) 3η,
χη (x, t) =
(5.21)
0,
ψ(x, t) 2η.
Moreover
C
,
η
C
|∂i ∂j χη (x, t)| 2 ,
η
|∇x,t χη (x, t)| (5.22)
1 i, j n, (x, t) ∈ R
n+1
.
χ ∈ C ∞ (R) satisfy 0 χ 1 and
Let χη can be constructed as follows. Let 0, ξ 0,
ψ(x,t)−2η , condition (5.21) is satisfied.
χ
χ (ξ ) = 1, ξ 1 . By setting χη (x, t) = η
We set
vη = χη u.
Then
on ∂Q(η)\{(1 + η, x , t); (x , t) ∈ DT }.
vη = |∇vη | = 0
Direct calculations yield
n
Lvη = χη u − 2
⎛
aij (∂i χη )∂j u − ⎝
i,j =1
n
⎞
aij ∂i ∂j χη ⎠ u −
i,j =1
n
(5.23)
bi ∂i χη u in Q(η).
i=1
In terms of (5.22), fixing λ > 0 we can apply theorem 3.2 to have
1
|∂t vη |2 + s|∇vη |2 + s 3 |vη |2 e2sϕ dx dt
Q(η) s
⎛
⎞
n
n
C
aij (∂i χη )∂j u − ⎝
aij ∂i ∂j χη ⎠ u
χη u − 2
Q(η) i,j =1
i,j =1
2
n
bi ∂i χη u e2sϕ dx dt + C eCs F 2
−
(5.24)
i=1
for all s s0 . Here F is defined by (5.19). Here and henceforth C > 0, Cj > 0 denote
generic constants which are dependent on λ but independent of s and a respective choice of u.
By (5.21) and (5.22), the first term on the right-hand side of (5.24) is bounded by
C 2sμ1
e u2H 1,0 (Q) ,
η4
where we set μ1 = e2λη and μ2 = e3λη . Since
|u|2 dx dt |v|2 e2sϕ dx dt
e2sμ2
Q(3η)∩Q(η)
Q(η)
C
4 e2sμ1 u2H 1,0 (Q) + C eCs F 2
η
for all s s0 . Hence
C
u2L2 (Q(3η)∩Q(η)) 4 e−2s(μ2 −μ1 ) M 2 + C eCs F 2
η
30
(5.25)
Inverse Problems 25 (2009) 123013
Topical Review
for all s s0 . Since Q(3η) = (Q(3η) ∩ Q(η)) ∪ (Q(3η) ∩ {1 + 3η > x1 1 + η}) and
uL2 (Q(3η)∩{1+3η>x1 >1+η}) F by (5.19) and 0 < η < 13 , we obtain
C −2s(μ2 −μ1 ) 2
e
M + C eCs F 2 + CF 2
η4
C1
4 e−2s(μ2 −μ1 ) M 2 + C1 eC1 s F 2
η
u2L2 (Q(3η)) for all s s0 . Replacing C2 by C1 es0 C1 , we have
C2
u2L2 (Q(3η)) 4 e−2s(μ2 −μ1 ) M 2 + C2 eC2 s F 2
(5.26)
η
for all s 0.
First let F = 0. Then theorem 5.1 implies that u 0, 0, T2 = 0, and so already the
conclusion of the theorem is verified. Next we assume that F = 0. Without loss of generality,
we can assume that M > 1. Since μ2 − μ1 = e2λη (eλη − 1) eλη − 1 λη, in terms of
(5.26), we have
C2
uL2 (Q(3η)) 2 e−sλη M + C2 eC2 s F
(5.27)
η
for all s 0. We choose s > 0 minimizing the right-hand side of (5.27): e−ληs M = eC2 s F ,
that is,
M
1
log
> 0.
s=
C2 + λη
F
Then
C2
λη
C3
uL2 (Q(3η)) 2 M C2 +λη F C2 +λη ,
(5.28)
η
1
where C3 > 0 is independent of η ∈ 0, 3 .
We apply the interpolation inequality and the Sobolev embedding in terms of 2m > n + 1
(e.g., [1]), we see that
m
1
uL∞ (Q(3η)) C4 uH m (Q(3η)) C5 uHm+1m+1 (Q(3η)) uLm+1
2 (Q(3η)) .
1
Here, since Q(3η) is congruent with each other in η ∈ 0, 3 by translation in x1, the constants
C4 and C5 are independent of η ∈ 0, 13 . Consequently noting M > 1, we have
C2
λη 1
λη
m 2
sup |u(x, t)| C6 M m+1 η−2 M C2 +λη F C2 +λη m+1 C7 Mη− m+1 F (m+1)(C2 +λη) .
(x,t)∈Q(3η)
Since F < 1, we see that
λη
F (m+1)(C2 +λη) F C8 η
with some constant C8 > 0 which is independent of η. Hence
sup
|u(x, t)| C9 Mη− m+1 F C8 η .
2
(x,t)∈Q(3η)
Let ξ = (ξ2 , . . . , ξn ) ∈ Rn−1 . In particular,
2
u 3η + γ |ξ |2 , ξ , T C9 Mη− m+1
F C8 η ,
2
1
1
|ξ | < √ , 0 < η < .
γ
3
(5.29)
In terms of the change of independent variables: (x1 , x ) −→ (η, ξ ) defined by
x1 = 3η + γ |ξ |2 ,
x = ξ ,
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Inverse Problems 25 (2009) 123013
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we have
2
u x, T dx
1
2
2
γ |x | <x1 <1,|x |< √γ
2 1 − γ |x |2 3
3
T
2
dη dξ .
u 3η + γ |ξ | , ξ ,
=3
2 √1
|ξ |<
γ
0
Therefore by (5.29) we obtain
γ |x |2 <x1 <1,|x |< √1γ
2
u x, T dx 3R 2 C 2
9
2 1
3
η− m+1 M 2 F 2C8 η dη.
4
(5.30)
0
4
< 1, we see that
Here by noting that F < 1 and m+1
1
1
3
3
4
4
1
η− m+1 F 2C8 η dη =
η− m+1 e−2C8 (log F )η dη
0
0
m−3
m−3
1 − m+1
C10 log
η
e
.
m+1
F
0
Here (·) is the gamma function. Setting Q0 = (x1 , x ); γ |x |2 < x1 < 1, |x | <
have
! !
m−3
!
!
1 − 2(m+1)
!u ·, T !
C11 M log
.
!
2 !L2 (Q0 )
F
∞
4
− m+1
−C10 (log
1
F
)η dη = √1
γ
, we
Applying the interpolation inequality in Q0, we obtain
! !
!
T 2
T !
!
!
u
·,
sup u x,
C
12 !
2 2 !H m (Q0 )
x∈Q0
m
1
! ! ! m+1
! m+1
!
!
T !
T !
!
!
!
!
u ·,
C13 !u ·,
.
2 !H m+1 (Q0 ) !
2 !L2 (Q0 )
Since (0, 0) ∈ Q0 and
! !
!
!
!u ·, T !
C13
uC([0,T ];H m+1 ()) C13 uH 1 (0,T ;H m+1 ())
!
2 !H m+1 ()
by the Sobolev embedding, the proof of the theorem is completed.
5.2. Observability inequality
In section 5.1, we consider Cauchy problems where we are not given boundary values on some
part of the boundary ∂. In this section, assuming that we know the boundary condition on
the whole lateral boundary ∂ × (0, T ), but not an initial value, we discuss the estimation of
the solution by extra boundary data or interior data of the solution.
We consider a boundary value problem for a parabolic operator:
Lu(x, t) ≡ ∂t u −
n
i,j =1
l1 (x)
32
∂u
+ l2 (x)u = 0
∂νA
aij (x, t)∂i ∂j u −
n
bi (x, t)∂i u − c(x, t)u = 0 in Q
(5.31)
i=1
on ∂ × (0, T ).
(5.32)
Inverse Problems 25 (2009) 123013
We assume (1.2) and
aij ∈ C 1 (Q),
aij = aj i ,
Suppose that l1 , l2 ∈ C 2 (∂) and
either l1 > 0 or l1 = 0 and
Topical Review
bi , c ∈ L∞ (Q),
l2 = 1
1 i, j n.
on ∂.
5.2.1. Estimation of solution. Let 0 < t0 < T be given, and ⊂ ∂ a sub-boundary, ω ⊂ be a sub-domain. Let u satisfy (5.31) and (5.32). Estimate u(·, t0 ) in by means of data of u
on × (0, T ) or in ω × (0, T ).
The global Carleman estimate in section 4 provides answers. That is, by (4.5) and (4.8),
we see in theorems 4.1 and 4.2 that for any ε > 0,
−c0
α0 (x, t), α(x, t) ,
ε < t < T − ε,
ε(T − ε)
where c0 = maxx∈ (e2λd0 C() − eλd0 (x) ) or c0 = maxx∈ (e2λdC() − eλd(x) ). Therefore
theorems 4.1 and 4.2 imply
uH 1 (ε,T −ε;L2 ()) CuL2 (ω×(0,T )
(5.33)
and
!
!
! ∂u !
!
!
uH 1 (ε,T −ε;L2 ()) C uH 1 (×(0,T ) + !
(5.34)
∂ν ! 2
A
L (×(0,T ))
respectively. These establish the Lipschitz stability in determining u(·, t0 ), t0 > 0. As for
other types of results on the estimation of solutions, see section 8. We note that a Carleman
estimate yields an energy estimate called an observability inequality for conservative systems
such as hyperbolic equations and see [96, 103, 105, 133].
Finally, we should mention the null exact controllability as very important application of
the global Carleman estimate. More precisely, we consider
Ly(x, t) = q(x, t),
(x, t) ∈ Q
(5.35)
with the same boundary condition (5.32) and
y(x, 0) = y0 (x),
x ∈ .
(5.36)
5.2.2. Null exact controllability. Find q ∈ L2 (Q) such that supp q ⊂ ω × (0, T ) and
y(x, T ) = 0 for x ∈ .
This problem has been considered comprehensively for many years and [54, 128,
129] are early papers. We can also discuss the null exact controllability for semilinear
parabolic equations, and the Navier–Stokes equations, the Burgers equation and the Boussinesq
equation. Imanuvilov [71, 72] applied the global Carleman estimate to prove the null exact
controllability: there exists such a control q and the norm is estimated by y0. See also Lebeau
and Robbiano [113] as an early paper.
Also for hyperbolic equations which we do not treat in this survey, there are many results
on the exact controllability. See [67, 107, 119], and references therein. Since Imanuvilov
[71], there have been substantial amounts of works for the null exact controllability for the
parabolic equations on the basis of the Carleman estimates. As very limited references on the
basis of the Carleman estimate, we refer to the following papers:
(1) Parabolic equations: [37, 55, 56, 63, 64, 127].
(2) Parabolic equations whose principal coefficients are discontinuous or of bounded
variations: [13–15, 42, 115].
(3) The Navier–Stokes equations: [38, 53, 57, 60, 66, 73, 74].
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6. Applications to inverse problems for parabolic equations
6.1. Local Hölder stability for an inverse coefficient problem
We consider
∂t u = div(p(x)∇u(x, t)),
(x, t) ∈ Q.
(6.1)
Let 0 < t0 < T be fixed arbitrarily and let 0 ⊂ be a sub-domain such that
0 ≡ ∂0 ∩ ∂ = ∅ is a sub-boundary (i.e., a relatively open subset of ∂).
Now we discuss the following.
Inverse coefficient problem: Determine p(x), x ∈ from u(·, t0 ) in and u|0 ×(0,T ) and
∇u|0 ×(0,T ) .
Our subjects are the uniqueness and the stability for the inverse coefficient problem.
This is an inverse problem with a finite number of measurements which requires a couple of
single inputs of initial value and boundary value (i.e., u|0 ×(0,T ) ) and the measurements of
the corresponding data of u(·, t0 ) and ∇u|0 ×(0,T ) . On the other hand, we can consider other
formulation with many boundary measurements which is based on the Dirichlet-to-Neumann
map, and such inverse problems have been studied comprehensively. However, the inverse
problems by the Dirichlet-to-Neumann map are outside the scope of this review and see e.g.,
a monograph by Isakov [90], Katchlov et al [95] as reference books.
Inverse problems with a finite number of measurements have been solved firstly by
Bukhgeim and Klibanov [20], whose methodology is composed of a Carleman estimate and
an integral inequality by the maximality of the weight function at t = t0 when we are given
spatial data u(·, t0 ). For a wide class of partial differential equations, their method is valid
to yield the uniqueness in the inverse problem, provided that a suitable Carleman estimate is
prepared. Since [20], there have been many works by their methodology: [78, 81, 82, 84, 85,
87, 97–100, 106] and see more references in section 8. As for the parabolic case, [85] is one
of earlier works on the uniqueness, and Imanuvilov and Yamamoto [78] proved the Lipschitz
stability which holds over the whole domain . For the determination of p(x) in a hyperbolic
equation ∂t2 u = div(p(x)∇u), see [106].
In this section, we present a method on the basis of Carleman estimates for inverse
coefficient problems with a finite number of measurements. Imanuvilov and Yamamoto
[78, 81, 82] modified the method in [20] and proved the Lipschitz stability for an inverse
coefficient problem for a hyperbolic equation. We modify the method in [81] to discuss
an inverse parabolic problem. We will apply two versions of Carleman estimates shown in
sections 3 and 4, and the local Carleman estimate in section 3 yields the local estimate of
Hölder type in determining a coefficient, while the global Carleman estimate in section 4
produces a global Lipschitz estimate over .
Let u satisfy (6.1) and v satisfy
∂t v = div(q(x)∇v(x, t)),
(x, t) ∈ Q.
(6.2)
We define an admissible set of unknown coefficients p, q by
A = {p ∈ C 2 (); p > 0 on , pC 2 () M},
(6.3)
where M > 0 is an arbitrarily fixed constant. This means that unknown coefficients are
uniformly bounded by the C 2 ()-norms. Moreover we assume that both solutions u, v are
sufficiently smooth, so that we can take t-derivatives necessary times.
For the application of the Carleman estimate theorem 3.2, we introduce a weight function
and a sub-domain defined by the weight function. We set x = (x2 , . . . , xn ), x = (x1 , x ) ∈ Rn ,
34
Inverse Problems 25 (2009) 123013
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and
1 2 δ
(6.4)
(δ) = (x, x ); 0 < x1 < − |x | +
γ
γ
with some γ > 0 and δ > 0. We assume that (4δ) ⊂ and 0 ⊂ {x1 = 0}. For a given
, we can apply the argument in section 5.1 with the special choice d for the weight function,
but for simplicity, we consider the domain in the form (δ).
Our weight function is
ϕ(x, t) = eλψ(x,t) ,
ψ(x, t) = −γ x1 − |x |2 − β(t − t0 )2 .
2m
(6.5)
We can make other choices such as ψ(x, t) = −γ x1 − |x | − β(t − t0 ) with m ∈ N to
prove the stability in a wider domain (δ), but we choose a simple function given by (6.5).
We set
2
Q(δ) = {(x, t); x1 > 0, ϕ(x, t) > e−λδ }
= {(x, t); x1 > 0, ψ(x, t) > −δ}
(6.6)
for δ > 0. We note that Q(δ) ∩ {t = t0 } = (δ). We choose β > 0 and δ > 0 such that
"
"
4δ
4δ
< t0 < T −
.
(6.7)
β
β
Then, noting that (4δ) ⊂ , we see that Q(4δ) ⊂ Q.
We set
y = u − v,
f = p − q,
R=v
in Q.
Then
∂t y = div(p∇y) + div(f ∇R)
in Q.
(6.8)
Since y has not a compact support, we need a cut-off function. By Q(4δ) ⊃ Q(3δ) ⊃ Q(2δ) ⊃
Q(δ), there exists χ ∈ C ∞ (Rn+1 ) such that 0 χ 1 and
1,
(x, t) ∈ Q(2δ),
χ (x, t) =
(6.9)
0,
(x, t) ∈ Q(4δ)\Q(3δ).
Thanks to the sufficient smoothness of u and v, we can take
z1 = χ ∂t y,
z2 = χ ∂t2 y.
Direct calculations yield
∂t zk − div(p∇zk ) = χ div f ∇∂tk R + (∂t χ ) ∂tk y − p(χ )∂tk y
−2p(∇χ ) · ∇ ∂tk y − (∇p · ∇χ )∂tk y, k = 1, 2.
(6.10)
Setting
Gk (∇x,t χ , χ )(x, t) = Gk (x, t) = (∂t χ ) ∂tk y − p(χ )∂tk y
− 2p(∇χ ) · ∇ ∂tk y − (∇p · ∇χ )∂tk y, k = 1, 2,
(6.11)
we see that
Gk (x, t) = 0 if ∇x,t χ (x, t) = χ (x, t) = 0,
2
|G1 (x, t)| + |G2 (x, t)| C
|∂tk y(x, t)| + |∇∂tk y(x, t)| .
(6.12)
k=1
Moreover (6.4) and (6.9) imply
zk = |∇zk | = 0 on ∂Q(4δ)\(0 × (0, T )), k = 1, 2.
(6.13)
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Inverse Problems 25 (2009) 123013
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We set
F2 =
3
∂tk (u − v)2L2 (0 ×(0,T ))
k=1
!
!2
2
!∂ k
!
! ∂ (u − v)!
+
! ∂ν t
! 2
L (0 ×(0,T ))
k=1
!
!2
!∂ k
!
!
∂
+!
(u
−
v)
! ∂τ t
! 2
.
L (0 ×(0,T ))
Hence, by fixing λ > 0 large, theorem 3.2 implies
⎛
⎞
n
2
⎝1
|∂i ∂j zk |2 + s|∇zk |2 + s 3 zk2 ⎠ e2sϕ dx dt C
χ 2 div f ∇∂tk R e2sϕ dx dt
s i,j =1
Q(4δ)
Q(4δ)
G2k e2sϕ dx dt + C eCs F 2 ,
k = 1, 2
+C
Q(4δ)
for alllarge s > 0. By ∇zk = (∇χ )∂tk y + χ ∇∂tk y and ∂i ∂j zk = χ ∂i ∂j ∂tk y + (∂i χ ) ∂j ∂tk y +
(∂j χ ) ∂i ∂tk y + (∂i ∂j χ )∂tk y, we have
⎛
⎞
n
1
2
2
2
∂i ∂j ∂ k y + s ∇∂ k y + s 3 ∂ k y ⎠ e2sϕ dx dt
χ2 ⎝
t
t
t
s i,j =1
Q(4δ)
χ 2 (|∇f |2 + f 2 ) e2sϕ dx dt + C eCs F 2
C
Q(4δ)
n
2
2
2
2 1 |∂i χ |2 ∂j ∂tk y + |∂j χ |2 ∂i ∂tk y + |∂i ∂j χ |∂tk y +C
Q(4δ) s i,j =1
2
2 k 2
k = 1, 2
+ G + s|∇χ | ∂ y
e2sϕ dx dt,
k
t
for all large s > 0. By (6.9) and (6.12), we see that
(the left-hand side)
C
Q(4δ)
where
(|∇(χf )|2 + |χf |2 ) e2sϕ dx dt + Cs 2 M12 e2sμ1 + C eCs F 2 ,
⎧
2 1
⎪
!
!
! j k !2
12
⎪
⎨M1 =
!∇ ∂ u! ∞ + !∇ j ∂ k v !2 ∞
+ M,
t
t
L (Q)
L (Q)
(6.14)
k=0 j =0
⎪
⎪
⎩
μ1 = e−2λδ > 0.
Therefore
n
2
2 2 2
2
2
2
2
2
4
2
2
4
2
χ
(∂i ∂j ∂t y + |∂i ∂j ∂t y| ) + s ∇∂t y + s |∇∂t y| + s |∂t y| +s |∂t y| e2sϕ dx dt
Q(4δ)
i,j =1
C
Q(4δ)
s(|∇(χf )|2 + |χf |2 ) e2sϕ dx dt + Cs 2 M12 e2sμ1 + C eCs F 2
(6.15)
for all large s > 0. We set
a(x) = (u − v)(x, t0 ) = y(x, t0 ),
36
b(x) = v(x, t0 ),
x ∈ .
(6.16)
Inverse Problems 25 (2009) 123013
Topical Review
Equation (6.8) yields
div(f ∇b) = ∂t y(·, t0 ) − div(p∇a),
∇ div(f ∇b) = ∇∂t y(·, t0 ) − ∇ div(p∇a)
in .
Hence
div(χf ∇b) = χ ∂t y(·, t0 ) − χ div(p∇a) − ∇χ · f ∇b,
∂i div(χf ∇b) = χ ∂i ∂t y(·, t0 ) + (∂i χ )∂t y(·, t0 ) − ∂i (χ div(p∇a)) − ∂i (∇χ · f ∇b).
(6.17)
Moreover by (6.9) and (6.12), equations (6.17) yield
(|div(χf ∇b)|2 + |∇ div(χf ∇b)|2 ) e2sϕ(x,t0 ) dx
(4δ)
χ 2 (x, t0 )(|∂t y(x, t0 )|2 + |∇∂t y(x, t0 )|2 ) e2sϕ(x,t0 ) dx + CM12 e2sμ1
C
(4δ)
+ C eCs a2H 3 ((4δ)) .
(6.18)
For j = 0, 1, we estimate
χ ∇ j (∂t y)esϕ 2H 1 (Q(4δ))
=
(|∂t (χ ∇ j (∂t y)esϕ )|2 + |∇(χ ∇ j (∂t y)esϕ )|2 + |χ ∇ j (∂t y)esϕ |2 ) dx dt.
Q(4δ)
Hence (6.9) and (6.15) yield
χ ∇ j (∂t y)esϕ 2H 1 (Q(4δ))
2
χ 2 ∇ j ∂t2 y + s 2 |∇ j (∂t y)|2 + |∇ j +1 (∂t y)|2 e2sϕ dx dt
Q(4δ)
+C
(|∂t χ |2 + |∇χ |2 )|∇ j (∂t y)|2 e2sϕ dx dt
Q(4δ)
s(|χf |2 + |∇(χf )|2 ) e2sϕ dx dt + Cs 2 M12 e2sμ1 + C eCs F 2 ,
C
j = 0, 1
Q(4δ)
for all large s > 0. Applying the trace theorem in Q(4δ) and noting
Q(4δ) ∩ {t = t0 } = (4δ),
we have
χ 2 (x, t0 )(|∂t y(x, t0 )|2 + |∇∂t y(x, t0 )|2 ) e2sϕ(x,t0 ) dx
(4δ)
C
1
χ ∇
j
(∂t y)esϕ 2H 1 (Q(4δ))
j =0
+ Cs 2 M12 e2sμ1 + C eCs F 2 ,
C
s(|χf |2 + |∇(χf )|2 ) e2sϕ dx dt
Q(4δ)
j = 0, 1
(6.19)
for all large s > 0. Equations (6.18) and (6.19) imply
(|div(χf ∇b)|2 + |∇ div(χf ∇b)|2 ) e2sϕ(x,t0 ) dx
(4δ)
s(|χf |2 + |∇(χf )|2 ) e2sϕ dx dt
C
Q(4δ)
Cs
+ C e (F 2 + a2H 3 ((4δ)) ) + Cs 2 M12 e2sμ1 .
(6.20)
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Inverse Problems 25 (2009) 123013
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We have to estimate f 2H 1 ((δ)) by the left-hand side. For it, we need an estimate for the
first-order partial differential operator. This can be done by the method of characteristics, but
we use another Carleman estimate because we have to estimate with the weight e2sϕ(x,t0 ) .
Lemma 6.1. We set
(P0 g)(x) = div(g∇b),
Assume that
x ∈ (4δ).
⎧
n
⎪
⎨γ ∂1 b(x) + 2
(∂i b)xi < 0,
i=2
⎪
⎩
∂1 b(0, x ) > 0,
x ∈ (4δ),
(6.21)
(0, x ) ∈ 0 .
Then there exists a constant C > 0 such that
s 2 (|∇g|2 + g 2 ) e2sϕ(x,t0 ) dx
(4δ)
C
(|∇(P0 g)(x)|2 + |(P0 g)(x)|2 ) e2sϕ(x,t0 ) dx
(6.22)
(4δ)
for all large s > 0 and all g ∈ H 2 ((4δ)) such that |g| = |∇g| = 0 on ∂(4δ)\{x1 = 0}.
For the proof, we will show the following lemma for a general case.
Lemma 6.2. Let ⊂ Rn be a bounded domain with smooth boundary ∂ and let
ν = (ν1 , . . . , νn ) be the unit outwards normal vector to ∂ and let us consider a first-order
partial differential operator
(P0 g)(x) =
n
pj (x)∂j g(x) + p0 (x)g(x),
j =1
where p0 ∈ W 1,∞ (), pj ∈ C 1 (), j = 1, . . . , n. We assume that ϕ0 ∈ C 1 () satisfies
n
x ∈ .
pj (x)∂j ϕ0 (x) > 0,
(6.23)
j =1
Then there exist s1 > 0 and C1 = C1 (s0 , ) > 0 such that
s 2 (|∇g|2 + |g|2 ) e2sϕ0 (x) dx C1 (|∇(P0 g)|2 + |P0 g|2 ) e2sϕ0 (x) dx
for all s > s0 and g ∈ H () satisfying
2
|∇g| = |g| = 0
on
⎧
⎨
⎩
x ∈ ∂;
n
j =1
⎫
⎬
pj (x)νj (x) 0 .
⎭
Proof of lemma 6.2. First let p0 (x) = 0, x ∈ .
esϕ0 P0 (e−sϕ0 w). Then
|P0 g|2 e2sϕ0 (x) dx =
|Q0 w|2 dx.
We have
Q0 w = P0 w − sq0 w,
38
(6.24)
We set w = esϕ0 g, Q0 w =
Inverse Problems 25 (2009) 123013
where q0 (x) =
obtain
n
j =1
Topical Review
pj (x)∂j ϕ0 (x). Therefore, by (6.23), (6.24) and integration by parts, we
Q0 w2L2 () = P0 w2L2 () + s 2 q0 w2L2 () − 2s
s2
q0 (x)2 w(x)2 dx − s
n
C2 s 2
n
w(x)2 dx − s
pj (∂j w)q0 w dx
j =1
pj q0 ∂j (w 2 ) dx
j =1
n
pj q0 w 2 νj dS + s
∂ j =1
w 2 dx − s
(C2 s 2 − C3 s)
∂∩{
∂j (pj q0 )w 2 dx
j =1
(C2 s 2 − C3 s)
n
n
j =1
pj νj 0}
⎛
⎞
n
⎝
pj νj ⎠ q0 w 2 dS
j =1
w 2 dx.
By (6.23), we have q0 > 0 on ∂, so that the right-hand side is greater than or equal to
(C2 s 2 − C3 s) w 2 dx. Thus by taking s > 0 sufficiently large, we have
2
2 2sϕ0 (x)
s |g| e
dx C4
|P0 g|2 e2sϕ0 (x) dx.
Next, setting h = P0 g, we observe
n
P0 (∂k g) = ∂k h −
(∂k pj )(x)(∂j g)(x),
k = 1, . . . , n.
j =1
By (6.24), we can apply the previous argument to ∂k g, k = 1, . . . , n, so that we have
2
n
2
2 2sϕ0
2 2sϕ0
s
|∇g| e
dx C4
|∇h| e
dx + C4
(∂k pj )(x)(∂j g)(x) e2sϕ0 dx
j,k=1
|∇(P0 g)|2 e2sϕ0 dx + C5
|∇g|2 e2sϕ0 dx.
C5
Consequently, choosing s > 0 large, we can absorb the second term on the right-hand side
into the left-hand side, so that we can obtain
|∇g|2 e2sϕ0 dx C6
|∇(P0 g)|2 e2sϕ0 dx.
s2
Next let p0 be not zero. Then we already proved that
2
2
2 2sϕ0 (x)
s (|∇g| + |g| ) e
dx C1 (|∇(P0 − p0 )g|2 + |(P0 − p0 )g|2 ) e2sϕ0 (x) dx
2
2C1 (|∇(P0 g)| + |P0 g|2 ) e2sϕ0 (x) dx
+ 2C1 (|∇(p0 g)|2 + |p0 g|2 ) e2sϕ0 (x) dx.
Taking s > 0 large, we can absorb the second term on the right-hand side into the left-hand
side. Thus the proof of lemma 6.2 is completed.
In lemma 6.2, we set = (4δ), pj = ∂j b, j = 1, . . . , n, ϕ0 (x) = exp(λ(−γ x1 −|x |2 )).
Then (6.23) and (6.24) are satisfied by (6.21), so that the proof of lemma 6.1 is completed.
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Inverse Problems 25 (2009) 123013
Topical Review
Applying lemma 6.1 to χf and P0 g = div(g∇b), in terms of (6.20), we have
s 2 (|∇(χf )|2 + |χf |2 ) e2sϕ(x,t0 ) dx C
s(|∇(χf )|2 + |χf |2 ) e2sϕ dx dt
(4δ)
Q(4δ)
+ C eCs (F 2 + a2H 3 ((4δ)) ) + Cs 2 M12 e2sμ1
for all large s > 0. Since
ϕ(x, t) ϕ(x, t0 ),
x ∈ , 0 t T ,
(6.25)
we see that
s2
(|∇(χf )|2 + |χf |2 ) e2sϕ(x,t0 ) dx CsT
(|∇(χf )|2 + |χf |2 ) e2sϕ(x,t0 ) dx dt
(4δ)
(4δ)
+ C eCs F 2 + a2H 3 ((4δ)) + Cs 2 M12 e2sμ1 ,
and taking s > 0 sufficiently large such that s s0 , we can absorb the first term on the
right-hand side into the left-hand side, and
s2
(|∇(χf )|2 + |χf |2 ) e2sϕ(x,t0 ) dx
(4δ)
C eCs (F 2 + a2H 3 ((4δ)) ) + Cs 2 M12 e2sμ1
for all s s0 . We set F12 = F 2 + a2H 3 ((4δ)) . Since (δ) ⊂ (4δ) and e2sϕ(x,t0 ) 1 for
x ∈ , by (6.9) we obtain
−λδ
e2se
(|∇f |2 + |f |2 ) dx (|∇(χf )|2 + |χf |2 ) e2sϕ(x,t0 ) dx C eCs F12 + CM12 e2sμ1
(δ)
(4δ)
−λδ
for all s s0 . We set μ2 = e . Then
(|∇f |2 + |f |2 ) dx C eCs F12 + CM12 e−2s(μ2 −μ1 )
(6.26)
(δ)
for all s s0 . We note that μ2 − μ1 > 0.
Replace C by C eCs0 , we obtain (6.26) for all s 0. We can assume that F1 = 0. We
choose s 0 minimizing the right-hand side of (6.26):
eCs F12 = M12 e−2s(μ2 −μ1 ) ,
that is,
M1
2
log
.
C + 2(μ2 − μ1 )
F1
Then by (6.26), we obtain
2(μ2 −μ1 )
C
√
C+2(μ −μ )
C+2(μ −μ )
f H 1 ((δ)) 2CM1 2 1 F1 2 1 .
s=
To sum up, we state
Theorem 6.1. Let u and v satisfy (6.1) and (6.2) respectively. For 0 < δ < δ , let (δ ) ⊂ and 0 ⊂ {x1 = 0}. We assume
2
! k !
!∂ u!
t
L∞ (Q)
! !
!
!
!
!
+ !∂tk v !L∞ (Q) + !∇∂tk u!L∞ (Q) + !∇∂tk v !L∞ (Q) M
k=0
and let A and (δ) be defined by (6.3) and (6.4) with δ > 0. Moreover we assume that
b = u(·, t0 ) or b = v(·, t0 ) satisfies (6.21). Then, for δ ∈ (0, δ ), there exist constants C > 0
40
Inverse Problems 25 (2009) 123013
and θ ∈ (0, 1) such that
p − qH 1 ((δ)) CM
+
2
k=1
1−θ
Topical Review
(u − v)(·, t0 )H 3 ((δ )) +
!
!
!
!∂ k
! ∂ (u − v)!
!
! ∂ν t
3
∂tk (u − v)L2 (0 ×(0,T ))
k=1
L2 (0 ×(0,T ))
!
!
!∂ k
!
!
+ ! ∂t (u − v)!
!
∂τ
θ
(6.27)
L2 (0 ×(0,T ))
for p, q ∈ A.
In the estimation in (δ), we need data of solution at t = t0 > 0 in a larger domain
(δ ). For the uniqueness, we need not take such a larger domain. That is, the above argument
produces the uniqueness: if u = v and ∇u = ∇v on 0 × (0, T ) and u(·, t0 ) = v(·, t0 ) in
(δ), then p(x) = q(x) for x ∈ (δ). Estimate (6.27) of coefficients holds if unknown
coefficients and solutions are in a priori bounded subsets in suitable norms and (6.27) is called
a conditional stability estimate for the inverse problem. The exponent θ in (6.27) depends on
the a priori bound M, and θ → 0 as M → ∞, which means that (6.27) becomes worse if M
is larger.
Remark. As is remarked already, our argument can yield an estimate of Hölder type in the
case where 0 is an arbitrary sub-boundary. On the other hand, by assuming one boundary
condition on the whole ∂ × (0, T ), Cauchy data on an arbitrary sub-boundary 0 yield the
Lipschitz stability under a suitable positivity condition like (6.21), and the argument is similar
to that in section 6.2.
For applying a Carleman estimate, we cannot choose t0 = 0. Therefore this is not an
inverse problem to a usual initial value/boundary value problem. The uniqueness in the case
of t0 = 0 is a longstanding open problem. If the coefficients are independent of t and the
observation area is sufficiently large, then one can change the inverse parabolic problem to
an inverse problem for a hyperbolic equation by an integral transform in t, and prove the
uniqueness and the stability for the case t0 = 0 (see [[100], theorem 4.7, [105]]). Also see
[[90], section 9.2]. Moreover for the inverse problems by Carleman estimates, we always need
a positivity condition such as (6.21).
We can apply the global Carleman estimates theorems 4.1 and 4.2 to prove the Lipschitz
stability over in determining p(x). As for the Lipschitz stability in determining principal
coefficients in hyperbolic equations, see [106], but for the parabolic case, the argument for
the Lipschitz stability is simpler than those hyperbolic case, thanks to the global Carleman
estimate with a singular weight function.
6.2. Global Lipschitz stability for an inverse source problem
In this section, as one example of applications of the global Carleman estimate in section 4,
we discuss an inverse source problem for a weakly coupled system of parabolic equations. In
particular, if one boundary condition is given on the whole ∂ × (0, T ) (see (6.29)), then the
global Carleman estimate gives the Lipschitz stability rather directly.
We consider
⎞T
⎛
n
n
aij1 (x)∂i ∂j u1 , . . .
aijN ∂i ∂j uN ⎠
∂t u(x, t) = ⎝
i,j =1
+
n
i,j =1
Bi (x, t)∂i u(x, t) + C(x, t)u(x, t) + R(x, t)f(x),
(x, t) ∈ Q,
(6.28)
i=1
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u|∂×(0,T ) = 0,
(6.29)
where u = (u1 , . . . , uN )T , f = (f1 , . . . , fN )T , where ·T denotes the transpose of a vector, and
Bi, 1 i n and C are N × N matrix functions and aijk ∈ C 1 (), 1 i, j n, 1 k N ,
satisfy (1.1) and (1.2), Bi ∈ L∞ (Q)N×N , C ∈ L∞ (Q)N×N , R is a given N × N matrix
function. A weakly coupled parabolic system (6.28) appears for example, as linearized
reaction-diffusion equations.
For describing an application argument by a global Carleman estimate to an inverse
problem, we discuss the following.
6.2.1. Inverse source problem. Let ⊂ ∂ be an arbitrarily fixed sub-boundary and let
∂u
(x, t),
t0 ∈ (0, T ) be any fixed time. Determine f(x), x ∈ by u(x, t0 ), x ∈ and ∂ν
A
x ∈ , 0 < t < T .
The main result is stated as follows.
Theorem 6.2. We assume that u, ∂t u ∈ H 2,1 (Q), R, ∂t R ∈ L∞ (Q) and that
det R(x, t0 ) = 0,
x ∈ .
(6.30)
Then there exists a constant C > 0 depending on , R, aijk , Bi , C such that
!
!
! ∂
!
!
!
fL2 ()N C u(·, t0 )H 2 ()N + !
(∂t u)!
∂νA
L2 (×(0,T ))N
(6.31)
for any f ∈ L2 ()N .
Proof. The proof is a modification of Imanuvilov and Yamamoto [78]. Setting z = ∂t u and
⎞T
⎛
n
n
n
aij1 ∂i ∂j u1 , . . . ,
aijN ∂i ∂j uN ⎠ +
Bi ∂i u + Cu,
Au = ⎝
i,j =1
i,j =1
i=1
we have
∂t z = Az + (∂t R)f
in Q
(6.32)
z = 0 on ∂ × (0, T )
(6.33)
and
R(x, t0 )f(x) = ∂t u(x, t0 ) − (Au)(x, t0 ),
x ∈ .
(6.34)
For the proof, it is sufficient to prove the estimate of f by the Neumann data over a shorter
time interval centred at t0:
!
!
! ∂
!
!
!
fL2 ()N C u(·, t0 )H 2 ()N + !
(∂t u)!
∂νA
L2 (×(t0 −t1 ,t0 +t1 ))N
with 0 < t0 − t1 < t0 + t1 < T . Hence by the change of t-variables, it is sufficient to prove
(6.31) in the case of t0 = T2 .
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Fixing λ > 0 in theorem 4.4 and applying to (6.32) with (6.33),
1 2 2
2
3 3
2
∂t u + sϕ0 |∇∂t u| + s ϕ0 |∂t u| e2sα0 dx dt
Q sϕ0
2
∂
2 2sα0
Cs
|(∂t R)f| e
dx dt + C e
C
∂ν (∂t u) dS dt
A
Q
×(0,T )
(6.35)
for all large s > 0.
On the other hand, in terms of (6.34), for estimating f, we have to estimate ∂t u ·, T2 with
the weight function:
2
∂t u x, T e2sα0 (x, T2 ) dx.
2 By limt→0 e2sα0 (x,t) = 0 for x ∈ , we have
2
T
2 ∂
2 2sα0 (x,t)
∂t u x, T e2sα0 (x, T2 ) dx =
|∂t u(x, t)| e
dx dt
2 0 ∂t
T
2
=
(2(∂t u(x, t) · ∂t2 u(x, t)) + 2s(∂t α0 )|∂t u(x, t)|2 ) e2sα0 dx dt
0
2|∂t u(x, t)|∂t2 u(x, t) + Csϕ02 |∂t u(x, t)|2 e2sα0 dx dt.
Q
Here we used |∂t α0 (x, t)| Cϕ0 (x, t)2 , (x, t) ∈ Q. By the Cauchy–Schwarz inequality, we
have
1
√ |∂t u(x, t)|∂t2 u(x, t) = √ |∂t u(x, t)| × s ϕ0 ∂t2 u(x, t)
s ϕ0
2
1
2 ∂t2 u(x, t) + s 2 ϕ0 |∂t u(x, t)|2 ,
s ϕ0
and the application of (6.35) yields
2
1 2 2
2 2
2
∂t u x, T e2sα0 (x, T2 ) dx C
∂t u + s ϕ0 |∂t u| e2sα0 dx dt
2
2 Q s ϕ0
2
∂
C
|(∂t R)f|2 e2sα0 dx dt + C eCs
(∂
u)
t dS dt.
s Q
×(0,T ) ∂νA
By (6.34), (6.30) and ∂t R ∈ L∞ (Q), we see that
C
T
|f(x)|2 e2sα0 (x, 2 ) dx |f(x)|2 e2sα0 dx dt
s Q
! 2
!2
!
∂
T !
Cs !
Cs
!
+Ce
+ C e !u ·,
∂ν (∂t u) dS dt.
2 !H 2 ()N
A
×(0,T )
By the definition of α0 , we have
T
,
α0 (x, t) α0 x,
2
Therefore
(x, t) ∈ Q.
|f(x)| e
2 2sα0 (x,t)
Q
(6.36)
T
|f(x)|2 e2sα0 (x, 2 ) dx,
dx dt T
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that is,
C
T
1−
|f(x)|2 e2sα0 (x, 2 ) dx
s
! 2
!2
!
∂
T !
Cs !
!
Ce
(∂t u) dS dt
!u ·, 2 ! 2 N +
×(0,T ) ∂νA
H ()
for all large s > 0.
theorem 6.2.
Taking s > 0 sufficiently large, we can complete the proof of
6.2.2. Technical comment I. Although both are based on Carleman estimates, the original
method since [20] is different from ours and does not work in our inverse problems.
We first explain their method for the determination of ρ(x) in
1
∂t u =
u in Q.
ρ(x)
Let u satisfy
1
u in Q.
ρ (x)
We assume that u = u and ∇u = ∇
u on 0 × (0, T ) and u(·, t0 ) = u(·, t0 ), u(·, t0 ) = 0 on
. Setting y = u − u, R = u and f = ρ − ρ, we obtain
∂t
u=
∂t y =
f
1
y +
R
ρ(x)
ρ
ρ
in Q
and
y(·, t0 ) = 0
in ,
y = |∇y| = 0
on 0 × (0, T ).
By u(·, t0 ) = 0 on , we choose t1 > 0 such that R(x, t) = 0 for x ∈ and
0 < t0 − t1 t t0 + t1 < T . The key in [20] is first division by R and second differentiation
in t as follows: setting y = Rz, we obtain
1
∂t R
2∇R
1 R
f
z −
z+
· ∇z +
z+
in × (t0 − t1 , t0 + t1 ).
ρ(x)
R
ρR
ρ R
ρ
ρ
t
Since ρfρ is independent of t and z(x, t) = t0 ∂t z(x, η) dη by z(·, t0 ) = 0, setting w = ∂t z and
differentiating in t, we obtain
t
∂t R
∂t R
2∇R
1 R
1
w −
w+
· ∇w +
w − ∂t
∂t w =
w dη
ρ(x)
R
ρR
ρ R
R
t0
t
t
∇R
R
1
2
·
∇w dη + ∂t
w dη in × (t0 − t1 , t0 + t1 )
(6.37)
+ ∂t
ρ
R
ρ
R
t0
t0
∂t z =
and
w = |∇w| = 0
on 0 × (0, T ).
Equation (6.37) is a parabolic equation with memory term. Thanks to an inequality
t
2
w(x, η) dη e2sϕ dx dt C
|w(x, t)|2 e2sϕ dx dt.
s
t0
Q(δ)
Q(δ)
where ϕ ∈ C 2 (Q) has to satisfy
ϕ(x, t) < ϕ(x, t0 )
44
x ∈ , t = t0 ,
(6.38)
Inverse Problems 25 (2009) 123013
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we can apply the argument by a Carleman estimate in proving the unique continuation also
to (6.37) and we can conclude that w = 0 in (δ) with δ > 0, which implies that ρ = ρ in
(δ). In fact, inequality (6.38) is crucial in the paper [20] by Bukhgeim and Klibanov and
essential for the application of a Carleman estimate to inverse problems. As for the proof of
(6.38), see [100, 105].
Remark. Under a weaker condition ϕ(x, t) ϕ(x, t0 ) for (x, t) ∈ Q(δ), we can prove
t
2
w(x, η) dη e2sϕ dx dt C
|w(x, t)|2 e2sϕ dx dt
Q(δ)
t0
Q(δ)
1
s
[100]. For this inverse problem, the factor on the right-hand side of (6.38) is not necessary.
However for some inverse problems, the factor is important (see, e.g., [31]).
The above is the original argument in Bukhgeim and Klibanov [20], but their method
does not work for inverse problems with a single measurement for (6.1) involving derivatives
of an unknown coefficient and a parabolic system (6.28). It is easy to understand the situation
for (6.1). As for (6.28) we will explain below. By (6.30), we can choose t1 > 0 such that
det R(x, t) = 0, x ∈ , 0 < t0 − t1 t t0 − t1 < T . For simplicity, we assume that
u(x, t0 ) = 0 in . Setting u = Rv and dividing by R, we obtain
∂t v = R −1 ARv − (R −1 ∂t R)v + f
Let R = (rij )1i,j n and R
−1
in × (t0 − t1 , t0 + t1 ).
(6.39)
= (r )1i,j n . Then
N
n
−1
k pk
[R ARv]p =
aij r rk ∂i ∂j v + (lower order terms).
ij
i,j =1
k,=1
If aijk depend on k ∈ {1, . . . , N }, then R −1 ARv is not necessarily weakly coupled, so that we
cannot directly apply theorem 4.4 to the resulting system (6.39). Also see section 7.8.
6.2.3. Technical comment II. For the inverse problem, it is essential that the weight function
ϕ attains the maximum at t = t0 where the spatial data of solutions are given. See (6.25) and
(6.36). The maximality at t = t0 is an alternative essential fact to inequality (6.38) and the
corresponding part in the method by Bukgheim and Klibanov [20].
7. Overview for Carleman estimates for parabolic equations
In sections 5 and 6, we show how to apply the Carleman estimates in sections 3 and 4
to the estimation of solutions to parabolic equations and inverse problems. We know that
relevant Carleman estimates produce stability results for these problems. In this section, also
for possible novel applications, we summarize the existing results on Carleman estimates
themselves for parabolic equations. Carleman estimates are used for various subjects and the
number of related papers is very large, so we do not intend to provide a perfect list of the
papers. Moreover, it is reasonable to treat Carleman estimates not only for parabolic equations
but also for a wider class of partial differential equations, but we concentrate on parabolic
equations.
7.1. Pointwise Carleman estimate
In Lavrent’ev et al [112] (also Klibanov and Timonov [105]), a pointwise Carleman estimate
was proved for a parabolic equation. According to [105, 112], we will present the pointwise
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Inverse Problems 25 (2009) 123013
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Carleman estimate. First we choose a weight function similar to (6.5) whose level set is of a
parabolic shape:
|x |2 (t − t0 )2 1
+
+
c1
c2
4
3
,
D = (x, t); x1 > 0, ψ(x, t) <
4
ψ(x, t) = x1 +
(7.1)
(7.2)
where c1 > 0 and c2 > 0 are suitable chosen constants. The boundary ∂D consists of two
parts:
∂D = 1 ∪ 2 ,
|x |2 (t − t0 )2
1
+
<
1 = (x, t); x1 = 0,
,
c1
c2
2
|x |2 (t − t0 )2
1
.
+
=
2 = (x, t); x1 > 0, x1 +
c1
c2
2
We set
ϕ(x, t) = ψ(x, t)−λ
(7.3)
where λ > 0 is a large second parameter in the Carleman estimate in a different form from the
Carleman estimates in sections 2– 4. We note that ψ < 1 on D.
Theorem 7.1 ([105, 112]). There exist sufficiently large positive constants λ0 and s0
which depend on σ1 , max1i,j n aij C 1 (Q) , ψ, D, such that if λ λ0 , then for all
u ∈ C 2,1 (D) = {u; u, ∂t u, ∂i u, ∂i ∂j u ∈ C(D), 1 i, j n} and all s s0 , we have
the following pointwise Carleman estimate:
⎞
⎛
n
1
−λ
|∂i ∂j u(x, t)|2 ⎠
|Lu(x, t)|2 e2sψ C ⎝|∂t u(x, t)|2 +
s
i,j =1
−λ
+ s|∇u(x, t)|2 + s 3 |u(x, t)|2 e2sψ + div U + ∂t V in D,
(7.4)
where the vector-valued function (U, V ) satisfies
|(U, V )(x, t)| Cs 3 (|∂t u|2 + |∇u|2 + |u|2 ) e2sψ
n
C −λ
+
|∂i ∂j u(x, t)|2 e2sψ ,
s i,j =1
−λ
(7.5)
and the constant C > 0 depends on σ1 , max1i,j n aij C 1 (Q) , ψ, D.
The estimate without the terms ∂t u(x, t) and ∂i ∂j u(x, t) was proved in [112], and in
[105] the estimate with such terms is shown thanks to the a priori estimate for boundary value
problems for the elliptic equation.
If we take the integrations of (7.4) over D, then in terms of the Gauss theorem for the last
two terms on the right-hand side of (7.4), we can obtain a Carleman estimate similar to (2.6)
and theorems 3.1 and 3.2. The pointwise Carleman estimate is very useful because one can
estimate the boundary terms and so we need not assume that functions have compact supports.
For other types of equations, such pointwise Carleman estimates were proved in [112]
and see also [5, 59, 105, 145]. In [5], Carleman estimates are proved for ultrahyperbolic
equations.
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7.2. General treatment
According to Isakov [90], we overview the general treatment. Let m = (m1 , . . . , mn+1 ) ∈ Nn+1
such that
m1 = · · · = mq > mq+1 · · · mn+1 ,
∇q = (∂1 , . . . , ∂q , 0, . . . , 0)
with some q ∈ {1, 2, . . . , n + 1}. Moreover let t = xn+1 , α = (α1 , . . . , αn+1 ) ∈ (N ∪ {0})n+1
αn+1
and |α| = α1 + · · · + αn+1 , ∂ α = ∂1α1 · · · ∂nαn ∂n+1
. We set |α : m| = mα11 + · · · + mαn+1
.
n+1
We consider a partial differential operator:
aα (x, t)∂ α .
(7.6)
A=
|α:m|1
If m1 = · · · = mn+1 , then the operator (7.6) is called isotropic and otherwise call (7.6)
anisotropic. For hyperbolic and elliptic operators, we have m1 = m2 = · · · = mn+1 = 2,
while in the parabolic case, m1 = · · · = mn = 2 and mn+1 = 1. For the isotropic case, a
general theory for Carleman estimates has been completed for functions with compact supports
(Hörmander [69]): a Carleman estimate holds if a quadratic form made by the weight function
and the coefficients of (7.6) satisfies some positivity condition called the strong pseudoconvexity (e.g., [[69], section 8.5]). For the anisotropic case, Isakov [86, 90] established a
general theory for Carleman estimates for functions with compact supports. For the partial
differential operator (7.6), we set
√ |α|
αn+1
aα (x, t) −1 ζ1α1 · · · ζn+1
,
ζ = (ζ1 , . . . , ζn+1 ) ∈ Cn+1 .
(7.7)
Am (x, t, ζ ) =
|α:m|=1
We assume that aα ∈ C 1 (D) if |α : m| = 1 and aα ∈ L∞ (D) if |α : m| < 1. Let ϕ ∈ C 2 (D)
such that ∇q ϕ = 0 on D. Then
Theorem 7.2 ([90]). Suppose that either
Am (x, ξ ) = 0 for all ξ ∈ Rn+1 \{0}
or
the coefficients of Am are real-valued.
We assume
q
i,j =1
∂Am ∂Am
(∂i ∂j ϕ)
∂ζi ∂ζj
for all (x, t) ∈ D, if
ζ =ξ+
√
−1s∇q ϕ,
1
+ Im
s
q
∂Am
(∂k Am )
∂ζk
k=1
ζ = 0,
D
(7.8)
Am (x, ζ ) = 0.
Then there exists a constant C > 0 such that
s
|∂ α u|2 e2sϕ dx dt C
|Au|2 e2sϕ dx dt
|α:m|<1
>0
(7.9)
D
for all large s > 0 and u ∈ C0∞ (D).
See also [88, 89]. It is not always simple to find a weight function satisfying (7.8). For
the realization for a parabolic equation, one can apply the Holmgren transform of the variables
(e.g., [130]). We can easily show that our choice (6.5) satisfies (7.8) if γ > 0 is sufficiently
large. As is seen in section 6, the stability can be proved in a level set bounded by ψ(x, t).
The level set by ψ with large γ is a flat paraboloidal domain whose volume is proportional
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Inverse Problems 25 (2009) 123013
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to γ1 , and for large γ , the volume is small. Therefore the sub-domain where the uniqueness
holds, is smaller if we choose large γ > 0, and in terms of the parabolicity of the equation,
it is not desirable to choose large γ . In order to prove the uniqueness in a larger domain, we
need to divide such a domain into several congruent parabolic sub-domains and repeat the
estimates (see e.g., [90, pp 63–4]). However, as is remarked after the proof of theorem 5.1,
such a patchwork argument may lose the best regularity. Thus, for Carleman estimates, we
have to choose ψ not only for validating a Carleman estimate but also for gaining a desirable
shape of the level set where we can work for inverse problems. As for a general theory for
Carleman estimates for functions without compact supports, see [134].
7.3. Global Carleman estimate
In section 4, we stated a global Carleman estimate by Imanuvilov [72] and here we refer to
the succeeding papers. In the general theory, under requirements for a weight function, we
can deduce a Carleman estimate. In other words, it is another difficult task to verify that a
Carleman estimate holds true for given (x, t)-domain (e.g., × (0, T )). Thus, it is usually
necessary to construct a special weight function. Among the achievements after [72], we
present such a Carleman estimate with a regular weight function, which is proved in Yuan and
Yamamoto [144]. More precisely, a global Carleman estimate with the same type of theorem
4.2 is proved similarly:
Theorem 7.3. Assume that ⊂ ∂ is an arbitrary sub-boundary, aij ∈ C 1 (), aij = aj i ,
1 i, j n satisfy (1.1) and (1.2). Let d0 ∈ C 2 () be a function constructed in lemma 4.2
2
and 0 < t0 < T . Let ϕ(t, x) = eλ(d0 (x)−β|t−t0 | +M1 ) , where M1 > sup0<t<T β(t − t0 )2 . Then
there exist positive constants λ0 , s0 and C = C(λ0 , s0 ) such that
⎧ ⎛
⎫
⎞
⎨
n
⎬
1 ⎝
|∂i ∂j u|2 ⎠ + sλ2 ϕ|∇u|2 + s 3 λ4 ϕ 3 |u|2 e2sϕ dx dt
|∂t u|2 +
⎭
Q ⎩ sϕ
i,j =1
|Lu|2 e2sϕ dx dt + Csλ
C
Q
∂u 2 2sϕ
e dS dt
ϕ ∂νA ×(0,T )
for all s > s0 , λ > λ0 and all u ∈ H 2,1 (Q) satisfying
u(·, 0) = u(·, T ) = 0 in u = 0 on ∂ × (0, T ).
The constants λ0 , s0 and C continuously depend on ni,j =1 aij C 1 () , T, , σ1 .
We recall the definition of σ1 in (1.2).
7.4. Carleman estimate with second large parameter
Carleman estimates are valid for parabolic systems such as a thermoelasticity system and a
elasticity system with residual stress. For this system, by the coupling in terms of the highestorder derivatives, when we use a weight function with only a parameter s > 0, unlike a weakly
coupled system (6.28), we cannot obtain a Carleman estimate. As the weight function ϕ, we
search for ϕ = eλψ with large parameter λ > 0 (see e.g., theorem 3.1), and λ is the second large
parameter. With the Carleman estimate with the second large parameter λ > 0, we can still
manage the highest-order terms to obtain a Carleman estimate. As general treatment proving
a Carleman estimate with the second large parameter, see [91, 92]. By our direct derivation,
Carleman estimates with second large parameter are direct consequences (theorems 3.1–3.3,
4.1–4.4 and 7.3).
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7.5. H −1 -Carleman estimates
Henceforth, identifying the dual of L2 () with itself, by H − () we denote the dual of
H0 (): H − () = (H0 ()) (e.g., [1]). So far, we establish Carleman estimates for Lu = f
by L2 (Q)-norm of f and we call it an L2-Carleman estimate. For the exact null controllability
for a semilinear parabolic equation with nonlinear term F (x, t, u, ∇u) and an inverse source
problem of determining f ∈ H − () with > 0, an L2-Carleman estimate does not work.
Let the right-hand side of Lu = f be given by
f = f0 +
n
∂i fi ,
f1 , . . . , fn ∈ L2 (Q).
i=1
We are interested in a Carleman estimate where the right-hand side is estimated by
f0 esϕ L2 (0,T ;H −1 ()) + ni=1 fi esϕ L2 (Q) , which we call an H −1 -Carleman estimate. More
precisely, we consider
Lu = ∂t u −
n
∂i (aij (x, t)∂j u)
i,j =1
−
n
∂i (bi (x, t)u) − c(x, t)u = f
in Q,
(7.10)
i=1
u|∂×(0,T ) = 0,
u(·, 0) = u0 .
(7.11)
In addition to (1.2), assume
⎧
aij , 1 i, j n are Lipschitz continuous on Q, aij = aj i ,
⎪
⎪
⎨
r > 2n, 1 i n,
bi ∈ L∞ (0, T ; Lr ()),
1
2n
⎪
∞
−μ,r
⎪
1
⎩c ∈ L (0, T ; W
,1 .
()),
0 μ < , r1 > max
2
3 − 2μ
Here and henceforth, for r2 > 1, identifying the dual of Lr1 () with Lr2 (), where
by W −μ1 ,r1 () we denote the dual of W0μ1 ,r2 ():
(7.12)
1
r1
+ r12 = 1,
μ ,r
W −μ1 ,r1 () = W0 1 2 ()
(e.g., [1]). We say that u ∈ L2 (Q) is a weak solution to the problem (7.10) and (7.11) if for
any z ∈ L2 (0, T ; H 2 ()) with L∗ z ∈ L2 (Q), z|∂×(0,T ) = 0 and z(·, T ) = 0, the following
equality holds:
(u, L∗ z)L2 (Q) = (f, z)L2 (Q) + (u0 , z(·, 0))L2 () .
Let d be given in lemma 4.1 and let ϕ, α and σ2 be defined by (4.5) and (4.6). In Imanuvilov
and Yamamoto [80, 83], the following H −1 - Carleman estimate is proved.
Theorem 7.4.
(i) There exists a positive constant λ0 such that for an arbitrary λ λ0 , we can choose
s0 (λ) > 0 satisfying: there exists a constant C > 0 such that
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Inverse Problems 25 (2009) 123013
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Q
{(sϕ)1−2 |∇u|2 + (sϕ)3−2 |u|2 }e2sα dx dt
C f esα 2L2 (0,T ;H − ()) +
(sϕ)3−2 |u|2 e2sα dx dt
(7.13)
ω×(0,T )
for all s > s0 , ∈ [0, 1] and each solution u ∈ L2 (Q) to (7.10) and (7.11). Here the
constants λ0 , s0 and C are continuously dependent on σ2 and independent of s.
(ii) If
f (x, t) = f0 (x, t) +
n
∂i fi (x, t)
i=1
with fi ∈ L2 (Q), 1 i n, then for ∈ R, there exists a positive constant λ0 such
that for an arbitrary λ λ0 , we can choose s0 (λ) > 0 satisfying: there exists a constant
C > 0 such that
{(sϕ)−1 |∇u|2 + (sϕ)+1 |u|2 ) e2sα dx dt
Q
n
C f0 (sϕ) 2 esα 2L2 (0,T ;H −1 ()) +
fi (sϕ) 2 esα 2L2 (Q)
i=1
(sϕ)1+ |u|2 e2sα dx dt
+
(7.14)
ω×(0,T )
for all s > s0 , and each solution u ∈ L2 (Q) to (7.10) and (7.11). Here the constants λ0 ,
s0 and C are continuously dependent on σ2 and independent of s.
Moreover if we further assume that aij ∈ W 1,∞ () and bi = 0, 1 i n and
u|ω×(0,T ) = 0, then
1 $
t (T − t)
s
sα 2
2
2
t (T − t)∂t (ue )L2 (0,T ;H −1 ()) +
|∇u| +
|u| e2sα dx dt
s
s
t (T − t)
Q
Cf esα 2L2 (0,T ;H −1 ())
(7.15)
for all s s0 and each solution u ∈ L2 (Q) to (7.10) and (7.11).
Moreover we refer to [77] which proves an H −1 Carleman estimate for functions with
Dirichletboundary values on ∂ × (0, T ). In the Carleman estimate in [77], the term
non-zero
1
2
|∇u|
+ sϕ|u|2 e2sα is estimated with sharp norm of u|∂×(0,T ) . Furthermore, we refer to
sϕ
[76] as for a transferring argument from an L2 Carleman estimate to an H −1 Carleman estimate
for functions with compact supports. See [75] for H −1 Carleman estimated for hyperbolic
equations.
7.6. Carleman estimates for less regular principal terms
For the direct derivation of a Carleman estimate in section 3, we assume that aij ∈ C 1 (Q). In
H −1 Carleman estimate, we relax the regularity of aij to the Lipschitz continuity on Q (see
(7.12)). In applications, it is also significant that the coefficients of the principal term are
discontinuous. We consider
∂t u = div(a(x)∇u(x, t))
50
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Inverse Problems 25 (2009) 123013
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where a is of piecewise C1. Under some geometrical constraints on the interface of the
break of the continuity of a, Carleman estimates are proved: [13, 15, 16, 42, 116, 117, 126].
Especially [116] is a most updated work which establishes a Carleman estimate with generous
geometrical constraints on the interface. See [14, 115] for a case of coefficients of bounded
variations.
7.7. Carleman estimate for degenerate parabolic equations
So far, we assume that {aij }1i,j n is positive definite, that is, that (1.2) holds with σ1 > 0. In
some applications such as laminar flow on flat plates and population genetics, degenerate
parabolic equations appear and see [21–24], where Carleman estimates are proved for
degenerate parabolic equations and applied to the controllability. Here we state it according
to [25] in a simple case.
Let 1 < α < 2 and let us consider
∂u
∂
xα
+ f (x, t),
0 < x < 1, t > 0,
(7.16)
∂t u(x, t) =
∂x
∂x
∂u
u(1, t) = 0,
xα
(0, t) = 0,
t > 0.
(7.17)
∂x
We set
a(x) = x α ,
Ha1 (0, 1)
0 x 1,
= {u ∈ L2 (0, 1); u is absolutely continuous in (0, 1],
√ ∂u
∈ L2 (0, 1), u(1) = 0},
a
∂x
and
∂u
∈ H 1 (0, 1) ,
D(A) = u ∈ L2 (0, 1); a
∂x
x 2−α − 2
∂u
∂
xα
,
ϕ(x, t) =
,
Lu = ∂t u −
∂x
∂x
(2 − α)2 t 4 (T − t)4
0 x 1, 0 < t < T .
In terms of the classical Hardy inequality, we can prove a Carleman estimate for a degenerate
parabolic equation (7.16).
Theorem 7.5. Let 1 < α < 2. There exist constants s0 > 0 and C > 0 such that
12
4
1
1
x 2−α u2 + s
x α−2 u2
s3
t (T − t)
t (T − t)
Q
4
1
α
2
−2
8
2
+s
x |∂x u| + s (t (T − t)) |∂t u| e2sϕ dx dt
t (T − t)
4
T 1
2 2sϕ
C
|Lu| e dx dt + C1
s
|ux (1, t)|2 e2sϕ(1,t) dt
t (T − t)
Q
0
for all s s0 and u ∈ C([0, T ]; Ha1 (0, 1)) ∩ L2 (0, T ; D(A)) ∩ H 1 (0, T ; L2 (0, 1)) satisfying
(7.16) and (7.17).
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7.8. Carleman estimates for parabolic systems
For weakly coupled parabolic systems (i.e., (6.28)), as seen in sections 3 and 4, Carleman
estimates are easily derived under the assumption that the coefficients of terms of lower orders
are in L∞ (Q). However, the Carleman estimate is very difficult for strongly coupled systems
whose principal terms are different:
∂t u =
N n
aijk (x, t)∂i ∂j uk + f ,
1 N.
(7.18)
k=1 i,j =1
Only a general theory by Calderón is available for proving Carleman estimates for systems
of partial differential equations (e.g., [44, 146]), but the Calderón theorem seems not to be
applied to general strongly coupled parabolic systems. The difficulty for proving a Carleman
estimate for a general parabolic system (7.18) consists in that the estimate in step (2) in section
3 does not work, so that we cannot obtain an estimate similar to (3.15) unlike a single parabolic
equation. However, the estimate in step (3) in section 3 can be executed to obtain an estimate
corresponding to (3.22). Here we will show it. We first assume conditions for aijk :
aijk ∈ C 1 (Q),
1 i, j n, 1 k, N,
(7.19)
and there exists a constant σ4 > 0 such that
N n
aijk (x, t)ξik ξj σ4
k,=1 i,j =1
N n
|ξik |2 ,
(x, t) ∈ Q, ξik ∈ R, 1 i n, 1 k N,
k=1 i=1
(7.20)
aijk = ajki = aijk ,
1 i, j n, 1 k, N.
(7.21)
Condition (7.20) means that the part of second-order derivatives in x is uniformly elliptic and
satisfies the Legendre condition (e.g., [61]). Condition (7.21) is the symmetry of aijk also
in k, , and is satisfied for example by a parabolic system describing a compressible fluid
dynamics.
As in section 3, let D ⊂ Q be a domain with smooth boundary ∂D, d ∈ C 2 (D) with
|∇d| = 0 on D, and let
ψ(x, t) = d(x) − β(t − t0 )2 + c0
with t0 ∈ (0, T ), β > 0, c0 > 0 satisfying inf (x,t)∈Q ψ(x, t) > 0. Then we have
Proposition 7.1. There exist constants λ0 > 0 and s0 > 0 such that we can choose
C = C(s, λ) > 0 such that
2
N N n
2sϕ
k
∂t u −
sλ2 ϕ|∇u|2 e2sϕ dx dt C
a
∂
∂
u
dx dt
ij i j k e
D
D =1 k=1 i,j =1
s 3 λ4 ϕ 3 |u|2 e2sϕ dx dt
+C
D
for all s > s0 , λ > λ0 and u ∈ H 2,1 (Q)N such that supp u ∈ D.
This is not a Carleman estimate because the right-hand side contains the zeroth-order term
3 4 3
s
λ ϕ |u|2 e2sϕ dx dt. This is similar to (3.22) and unlike a single parabolic equation, we
D
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Inverse Problems 25 (2009) 123013
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cannot estimate in step (2) in section 3 (i.e., (3.15)) by means of more coupling with aijk , so that
we cannot absorb the term D s 3 λ4 ϕ 3 |u|2 e2sϕ dx dt into the left-hand side. The proposition
corresponds to the pseudoconvexity in Hörmander [69] (p 203). Similarly to theorem 8.7.1
and corollary 8.7.1 in [69], by our proposition we can prove: let aijk ∈ C ∞ (Q). There exists
a finite-dimensional space M in H 2,1 (D)N such that if u ∈ H 2,1 (D)N satisfies
∂t u =
N n
in D, 1 N
aijk ∂i ∂j uk
k=1 i,j =1
and
|u| = |∇u| = 0 on ∂D,
then u ∈ M. If we have a Carleman estimate, then we can conclude that M = {0}, that is, the
unique continuation from ∂D holds. Moreover if M = {0}, then we can prove a conditional
stability estimate by an argument in section 5.
Proof. We set
w = (w1 , . . . , wN )T = esϕ u,
ϕ = eλψ ,
⎛
⎛
⎞T ⎞
N N n
n
⎜
⎟
aijk1 ∂i ∂j (e−sϕ wk ), . . . ,
aijkN ∂i ∂j (e−sϕ wk )⎠ ⎠ .
P w = esϕ ⎝∂t (e−sϕ w) − ⎝
k=1 i,j =1
k=1 i,j =1
Moreover we set
n
σk (x, t) =
aijk (x, t)(∂i d)(x)∂j d(x),
μk (x, t) =
i,j =1
n
aijk (x, t)∂i ∂j d(x)
i,j =1
and
A = sλ2 ϕ
N
σk wk + sλϕ
k=1
N
μk wk − sλϕ(∂t ψ)w ,
1 N.
k=1
Direct calculations yield
[P w] = ∂t w −
N n
aijk ∂i ∂j wk + 2sλϕ
k=1 i,j =1
− s 2 λ2 ϕ 2
N
N n
aijk (∂i d)(∂j wk )
k=1 i,j =1
σk wk + A ,
1 N.
(7.22)
k=1
Here we recall that [P w] denotes the th component of P w. Now as in step (3) in
section 3, for 1 N, we multiply [P w] = f esϕ with sλ2 ϕw , sum up over = 1, . . . , N
and integrate by parts over (x, t) ∈ D. The estimation is done by using (7.19)–(7.21), similarly
to estimates (3.17)–(3.21). Henceforth let λ > 1 and s > 1. For example, we have
N n aijk (∂i ∂j wk )sλ2 ϕw dx dt
−
k,=1 i,j =1 D
=
N n k,=1 i,j =1 D
sλ2 (∂i ϕ)aijk (∂j wk )w dx dt +
σ4
sλ2 ϕ|∇w|2 dx dt − C
D
N n k,=1 i,j =1 D
sλ2 ϕaijk (∂j wk )(∂i w ) dx dt
sλ3 ϕ|∇w||w| dx dt
D
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Inverse Problems 25 (2009) 123013
and
N n
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aijk (∂i d)w (∂j wk )
k,=1 i,j =1
=
N n
aijk (∂i d)(w (∂j wk ) + wk (∂j w )) +
k> i,j =1
=
=
1
2
N
n
aijkk (∂i d)wk (∂j wk )
k=1 i,j =1
aijk (∂i d)∂j (wk w ) +
k= i,j =1
N
N n
1
2
N
n
aijk (∂i d)∂j (wk w )
k= i,j =1
n
1
a k (∂i d)∂j (wk w ),
2 k,=1 i,j =1 ij
so that
n N 2 3 2 k
s
λ
ϕ
a
(∂
d)w
(∂
w
)
dx
dt
i
j k
ij
k,=1 i,j =1 D
N n
1
= s 2 λ3 ϕ 2 aijk (∂i d)∂j (wk w ) dx dt 2 k,=1 i,j =1
N n
1
= s 2 λ3 ∂j ϕ 2 aijk (∂i d) wk w dx dt 2 k,=1 i,j =1
C
s 2 λ4 ϕ 2 |w|2 dx dt.
D
Hence we have
2
2
3
sλ ϕ|∇w| dx dt − C
sλ ϕ|∇w||w| dx dt − C (s 3 λ4 ϕ 3 + s 2 λ4 ϕ 2 )|w|2 dx dt
σ4
D
D
D
2
sϕ
sλ ϕ|(fe · w)| dx dt.
D
Moreover
sλ3 ϕ|∇w||w| = sλ2 ϕ|w|λ|∇w| 12 s 2 λ4 ϕ 2 |w|2 + 12 λ2 |∇w|2
and
|sλ2 ϕ(f esϕ · w)| 12 |f esϕ |2 + 12 s 2 λ4 ϕ 2 |w|,
noting that s > 1 and λ > 1 we can choose only terms with the maximal orders in s, λ to
obtain
C
σ4 −
sλ2 ϕ|∇w|2 dx dt C
|f|2 e2sϕ dx dt + C
s 3 λ4 ϕ 3 |w|2 dx dt.
sϕ
D
D
D
Choosing s > 0 large, we can
sλ2 ϕ|∇w|2 dx dt C
|f|2 e2sϕ dx dt + C
s 3 λ4 ϕ 3 |w|2 dx dt.
D
D
D
Rewriting the obtained estimate in terms of u, we complete the proof of the proposition.
Here we look over other types of parabolic systems which are not strongly coupled but
are important in the mathematical physics.
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7.8.1. The Navier–Stokes equations. We consider the linearized Navier–Stokes equations
describing the velocity field v = (v1 , v2 , v3 )T in the incompressible viscous fluid:
∂t v − γ v + (b(x, t) · ∇)v + ∇p = f(x, t),
div v(x, t) = 0,
v(x, t) = 0,
(x, t) ∈ Q
(7.23)
(x, t) ∈ Q
(7.24)
x ∈ ∂, 0 < t < T .
(7.25)
Here ⊂ R3 is a bounded domain with smooth boundary, γ > 0 is a constant describing the
viscosity, and for simplicity we assume that the density is one, and we set b = (b1 , . . . , bn )T ,
[(b · ∇)v] = nj=1 bj ∂j v .
Setting b = v, equation (7.23) is the Navier–Stokes equations. We assume that
b ∈ L∞ (0, T ; W 1,∞ ())3 ,
bL∞ (0,T ;W 1,∞ ())3 M
with arbitrarily fixed constant M > 0. Let ω ⊂ be a sub-domain such that ∂ω ⊃ ∂. Let
d ∈ C 2 () be constructed in lemma 4.1 and let α be defined by (4.5).
Theorem 7.6. There exists a constant λ0 = λ0 (, ω, T ) > 0 such that for λ λ0 , we can
choose constants C = C(λ, M) > 0 and s0 = s0 (λ0 , M) > 0 such that
s5
s3
s4
|v|2 + 3
|∇v|2 + 4
|rot v|2
5
5
3
t (T − t)
t (T − t)4
Q t (T − t)
s2
+ 2
|∇rot v|2 e2sα dx dt
t (T − t)2
s
C
|rot f|2 e2sα dx dt + C eCs ∂t (rot v)2L2 (ω×(0,T )) + v2L2 (0,T ;H 3 (ω))
Q t (T − t)
for all s s0 and all v ∈ H 2,1 (Q) satisfying (7.23)–(7.25) and rot v ∈ H 2,1 (Q).
In theorem 7.6, we have to assume that ∂ω ⊃ ∂ and it is quite strong. For any subdomain ω ⊂ , we can prove the following Carleman estimate similarly to [77] where b = 0
is assumed. See also [35]. We set n = 2, 3,
H = {v ∈ L2 (Q)n ; div v = 0, (v · ν) = 0 on ∂}
and
V = v ∈ H01 (Q)n ; v ∈ H .
Theorem 7.7. Let ∈ C ∞ [0, T ], (t) > 0 for 0 < t < T , and (t) =
α1 (x, t) =
t,
T − t,
0 t T4 ,
3
4T t T,
and
eλd(x) − e2λdC()
.
8 (t)
We assume that f ∈ L2 (0, T ; H ). There exists a constant λ0 > 0 such that for λ > λ0 , there
exist constants C > 0 and s0 > 0 satisfying
sϕ|rot v|2 e2sα1 dx dt +
s 2 ϕ 2 |v|2 e2sα1 dx dt
Q
Q
C
|f| e
2 2sα1
Q
T
dx dt +
0
(sϕ|rot v|2 + s 2 ϕ 2 |v|2 ) e2sα1 dx dt
ω
for all s s0 and v ∈ L2 (0, T ; V ) ∩ H 2,1 (Q)n with v(·, 0) ∈ V .
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These Carleman estimates can be applied to inverse problems by a method in section 6
and see [35].
The Carleman estimates for the Navier–Stokes equations have been studied related to the
controllability and see [38, 53, 57, 60, 66, 73, 74].
7.8.2. Carleman estimates for parabolic systems by data of one compoment. Let
(cij )1i,j 2 ∈ L∞ (Q) and (Aij )1i,j 2 ∈ L∞ (Q)n . Consider the following reaction-diffusion
system with convection terms:
⎧
⎨∂t u = u + c11 u + c12 v + A11 · ∇u + A12 · ∇v + f in Q,
∂t v = v + c21 u + c22 v + A21 · ∇u + A22 · ∇v + g in Q,
(7.26)
⎩
u=v=0
on ∂ × (0, T ).
Our main interest is to derive a Carleman estimate of solution (u, v) to (7.26) by solely
observing u in ω × (0, T ). We assume
⎧
⎨ω ⊂ with ∂ω ∩ ∂ = γ and |γ | = 0,
(x, t) ∈ γ × (0, T ),
|A12 (x, t) · ν(x)| = 0,
(7.27)
⎩
c12 C 2 (ω×(0,T )),
A11 C 1 (ω×(0,T ))n M,
A12 C 2 (ω×(0,T ))n ,
where M > 0 is an arbitrarily fixed constant. Then in [12], the following Carleman estimate
is proved.
Theorem 7.8. Let ω ⊂ be a sub-domain such that ω ⊂ . Under (7.27), there exist
ψω ∈ C 2 () with ψω < 0 on and two positive constants s0 and C which depend on
T , M, , ω and the L∞ (Q)-norms of cij , Aij , such that there exist positive constants C1 (s)
and C such that the following Carleman estimate holds:
(sρ)−1 e2sαω (|∂t u|2 + |∂t v|2 + |u|2 + |v|2
Q
+ (sρ)2 |∇u|2 + (sρ)2 |∇v|2 + (sρ)4 |u|2 + (sρ)4 |v|2 ) dx dt
2
2
C1 (s) uH 2,1 (ω×(0,T )) + f L2 (ω×(0,T )) + C
e2sαω (|f |2 + |g|2 ) dx dt
Q
for all s s0 and any solution (u, v) to (7.26). Here we set
αω (x, t) =
ψω (x)
,
t (T − t)
ρ(t) =
1
.
t (T − t)
(7.28)
This is a Carleman estimate for a two-component system with extra data in ω × (0, T )
of only one component. In [6, 39], it is assumed that A11 = A12 = 0, and the proof can be
completed by directly substituting v by means of u in ω × (0, T ). By the first-order coupling
in the parabolic system (7.26), we need the first and second conditions in (7.27). In [12], also
the following Carleman estimate for a reaction-diffusion system with three components by
one-component observation is proved. That is, we consider
⎧
∂t u(x, t) = u + c11 (x, t)u + c12 (x, t)v + c13 (x, t)w + f (x, t) in Q,
⎪
⎪
⎨∂ v(x, t) = v + c (x, t)u + c (x, t)v + c (x, t)w + g(x, t) in Q,
t
21
22
23
(7.29)
∂t w(x, t) = w + c31 (x, t)u + c32 (x, t)v + c33 (x, t)w + h(x, t) in Q,
⎪
⎪
⎩
u=v=w=0
on ∂ × (0, T ).
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Inverse Problems 25 (2009) 123013
We assume
Topical Review
⎧
(c )
∈ W 2,∞ (Q),
cij W 2,∞ (Q) M,
⎪
⎪ ij 1i,j 3 2
⎪
⎪
|γ | = 0,
,
∂ω
∩
∂
=
γ
and
ω
is
of
class
C
⎨
c
∇c12 − 12 ∇c13 · ν = 0 on γ × [0, T ],
⎪
⎪
⎪
c13
⎪
⎩
c12 W 3,∞ (ω×(0,T )) , c13 W 3,∞ (ω×(0,T )) M, c13 = 0 on Q.
(7.30)
Then we show a Carleman estimate with extra data of one component.
Theorem 7.9. Under (7.30), there exist ψω ∈ C 2 () with ψω < 0 on and a constant s0 > 0
which depends on T , M, , ω and the L∞ ()-norms of cij, 1 i, j 3 such that we can
choose constants C1 (s) > 0 and C > 0 satisfying:
(sρ)−1 e2sαω (|∂t u|2 + |∂t v|2 + |∂t w|2 + |u|2 + |v|2 + |w|2
Q
+ (sρ)2 |∇u|2 + (sρ)2 |∇v|2 + (sρ)2 |∇w|2 + (sρ)4 u2 + (sρ)4 v 2 + (sρ)4 w 2 ) dx dt
C1 (s) u2H 4,2 (ω×(0,T )) + f 2H 2,1 (ω×(0,T )) + g2L2 (ω×(0,T )) + h2L2 (ω×(0,T ))
+ C (|f 2 | + |g|2 + |h|2 ) e2sαω dx dt
Q
for all s s0 and all (u, v, w) satisfying (7.29). Here αω and ρ are defined by (7.28).
We also refer to [7] as a Carleman estimate with observation data of a limited number of
components.
8. Overview of results by parabolic Carleman estimates
8.1. Estimation of solutions to parabolic equations
Let L be a parabolic operator defined by (1.3) and let us consider a parabolic equation
Lu = f in Q. In addition to classical initial value/boundary value problems, there are several
possibilities of formulations, which are meaningful also from practical viewpoints.
(1) Continuation problem. Let ⊂ ∂ be a sub-boundary and
let D be a given sub-domain
∂u in Q. Determine u|D by extra data such as u|×(0,T ) or ∂ν
.
A ×(0,T )
(2) State estimation. Let boundary values be given over the whole ∂ × (0, T ). Let
0 t0 , t1 T be given. Determine u(x, t0 ), x ∈ by extra data such as u|ω×(t1 ,T ) .
(3) Backward problem. Determine u(x, t0 ), x ∈ by u(x,T), x ∈ .
As other papers on related problems, see [[26], chapters 10 and 11, [93]]. The continuation
problem is important, for example, for the inverse heat conduction problem (e.g., [3]) and the
state estimation is related to the controllability.
For the continuation problem, as is stated in section 5, the Carleman estimate is a main
method. In the case of D ⊂ Q, a Carleman estimate and the cut-off technique yield a Hölder
stability estimate. As other cases, we have
(1) u(x, t0 ), x ∈ ∂, t0 > 0.
(2) u(x, 0), x ∈ .
(3) u(x, 0), x ∈ ∂.
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Case (1) is discussed in theorem 5.2, and we can argue case (2) similarly, and in both cases, the
stability is of single logarithmic rate. Case (3) needs more delicate application of the method
of theorem 5.2 and we can expect the stability of double logarithmic rate.
As for related results with the continuation problem, see [27, 41, 51, 139–141]. In [41],
assuming that u|×(0,T ) = 0, the authors prove a Hölder stability estimate for uL2 (D×(0,T1 ))
where ∂D touches and T1 < T , and apply the estimate for the determination of shapes of
unknown sub-boundary. The situation in [41] is different from theorem 5.2 because u|×(0,T )
is given, and so the stability is much better than theorem 5.2.
For the state estimation, we refer to [102, 118, 142]. In [118], assuming the zero Dirichlet
boundary condition on ∂ × (0, T ), a logarithmic conditional stability estimate is proved in
∂u
|×(t1 ,T ) , where t1 > 0 is given, ω ⊂ is an
determining u(x, 0), x ∈ by u|ω×(t1 ,T ) or ∂ν
A
arbitrary sub-domain and ⊂ ∂ is an arbitrary sub-boundary. In [142], the case of t1 = 0 is
considered. For a parabolic inequality |Lu| |f | in Q, Klibanov [102] proved a logarithmic
∂u
|∂×(0,T ) .
conditional stability estimate in determining u(x, 0), x ∈ by u|∂×(0,T ) and ∂ν
A
See also [104].
8.2. Inverse problems for parabolic equations
Since Bukhgeim and Klibanov [20] applying Carleman estimates to inverse problems of
determining coefficients or source terms, there are many papers not only for parabolic equations
but also for other types of equations such as hyperbolic equations. However, we here restrict
the scope to inverse problems for parabolic equations by Carleman estimates. A typical
formulation for the inverse problem is as follows. Let
∂t u =
n
∂i (aij (x)∂j u) + c(x)u + R(x, t)f (x)
in Q.
i,j =1
Let ω be an arbitrary sub-domain, γ , ⊂ ∂ be sub-boundaries. Let 0 < t0 < T and R be
given. Determine f (x) or c(x) or aij (x) by
{u(·, t0 )| , u|ω×(0,T ) , u|×(0,T ) }
or
∂u
u(·, t0 )| ,
|γ ×(0,T ) , u|×(0,T ) .
∂νA
After [20], we should refer to [19, 85, 87, 90, 98, 100, 105]. In particular, theorem 6.4.1
(p 152) in [85] proves the uniqueness in inverse problems for a parabolic equation by the
method in [20]. Khaı̆darov [97] proved a stability estimate locally in and also see [87]. As
monographs, see [34, section 3], [90, chapter 9] as for accounts for wider classes of inverse
problems for parabolic equations. Imanuvilov and Yamamoto [78] established the Lipschitz
stability in determining f (x) over in the case of = ∂. The key is the global Carleman
estimate (theorems 4.1 and 4.2) and see the proof of theorem 6.2 as a simplified argument in
[78]. Yamamoto and Zou [143] proved the Lipschitz stability in determining c(x), x ∈ and
for it, the assumption u(x, t0 ) > 0 for x ∈ is essential. This condition is very restrictive
and there are two disadvantages in the results by Carleman estimates:
(1) Some positivity at t0 is indispensable.
(2) We cannot choose t0 = 0.
We cannot choose t0 = 0, because as is pointed out in section 6, for the application of a
Carleman estimate, the weight function must gain the maximum at t0 when spatial data u(·, t0 )
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Inverse Problems 25 (2009) 123013
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are given and so if we take t0 = 0, then the solution u must be continued over t = 0 but the
parabolic equation is not reversible in time and the continuation is impossible in general.
The condition that t0 > 0 means that we have to guarantee the positivity condition at
an intermediate time t0 of the heat process. In the case of determination of c(x), by positive
Dirichlet boundary values, the maximum principle guarantees that u(·, t0 ) > 0 on .
In [144], the determination of multiple coefficients aij (x), 1 i, j n, x ∈ is
discussed. Since we have to determine n(n+1)
functions aij, we need data by repeating
2
observations suitably and keep some independence or non-degeneracy of repeated u(·, t0 )
at t0 > 0. The values of u at t0 cannot be chosen directly, so that in [144], we add -times
control functions hi (x, t) supported in ω × (0, T ), i = 1, . . . , :
Lu = hi ,
i = 1, . . . , ,
in order that the corresponding solutions ui satisfy the desired non-degeneracy condition.
The non-degeneracy condition is achieved in view of the approximate controllability. The
2
n
Lipschitz stability requires (n+1)
-times suitable controls and the corresponding observation
2
.
data, and with special choice we can reduce the number of the observations to n(n+3)
2
As for the inverse problems of determining coefficients in parabolic systems with data of
limited numbers of components, we refer to [12, 39, 40]. The key Carleman estimates are
theorems 7.8 and 7.9.
Choulli et al [35], and Imanuvilov and Yamamoto [79] discussed inverse problems
of determining source terms in the linearized Navier–Stokes equations on the basis of
theorem 7.6.
As for works concerning the determination of nonlinearity in a parabolic equation by
Carleman estimates, see [[18, 43, 94, 99, 101] and [105], chapter 4].
As for the inverse problems for parabolic equations with discontinuous coefficients of the
principal term, see [11, 15, 16, 126].
Finally, we mention a method for the numerical reconstruction of a coefficient by changing
the coefficient inverse problem to the unique continuation for partial differential equation (6.37)
with integral terms. We refer to [9, 10].
9. Carleman estimates with x-independent weight functions
In addition to a ‘spacelike’ Carleman estimate considered so far, one can consider a ‘timelike’
Carleman estimate. Such a Carleman estimate is discussed in [[112], section 2, chapter 4]
and [114] with the weight function (α − t)−2s with some α ∈ R and applied to a backward
parabolic problem. In this section, we will prove Carelman estimates with x-independent
weight functions eλψ(t) , where λ > 0 is a large parameter, and apply them to prove the
conditional stability for a backward parabolic problem in time and an inverse problem of
determining a source term.
We consider the parabolic operator whose elliptic part is the divergence form:
n
∂i (aij (x, t)∂j u) = f (x, t).
Lu(x, t) = ∂t u(x, t) −
i,j =1
In this section, in place of (1.1) we assume
aij ∈ C 1 ([0, T ]; L∞ ()),
and we keep the assumption (1.2).
aij = aj i ,
1 i, j n
(9.1)
Remark. By the density argument (e.g., [[83], corollary 2.1]), all the following arguments
work if aij are Lipschitz continuous in t ∈ [0, T ] to L∞ ().
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9.1. Carleman estimate and an application to a parabolic equation backward in time
First we use a simple weight function: we set
ϕ(t) = eλt
where λ > 0 is fixed suitably. We note that ∂t ϕ = λϕ.
Remark. As a method with a similar spirit, we can refer to the weighted energy method.
Concerning the weighted energy method, there are many papers and see monographs [4, 17,
114, 124], and references therein. Except for [123] for equations of hyperbolic types, all the
papers use just t as a weight function, and do not use the second large parameter λ. In [123],
λ
the weight function est is used to prove properties for the asymptotic behaviour.
The serious difference to the existing papers is the introduction of the second large
parameter λ > 0. Such a second large parameter is very flexible and offers a lot of
possibilities for better estimates. Also in this section, we see that the weight function eλt
gives estimates of Carleman type but the simple weight t or another choice without the second
large parameter cannot produce such estimates. As is shown in this section, our Carleman
estimate yields conditional stability estimates by a usual cut-off argument (see e.g., [69]) for
parabolic equations backwards in time whose elliptic part is not symmetric and the coefficients
are also dependent on t.
Set
v = esϕ u,
P v = esϕ L(e−sϕ v) = esϕ f.
Assume that
u|∂×(0,T ) = 0.
−sϕ
sϕ
Then e ∂t (e
v) = ∂t v − sλϕv, e
sϕ
n
−sϕ
i,j =1 ∂i (aij ∂j (ve
⎛
P v = esϕ L(e−sϕ v) = ∂t v − ⎝sλϕv +
n
)) =
(9.2)
n
i,j =1 ∂i (aij ∂j v),
⎞
∂i (aij ∂j v)⎠ = esϕ f.
and
(9.3)
i,j =1
We have
esϕ f 2L2 (Q) =
⎛
(∂t v) ⎝−sλϕv −
|∂t v|2 dx dt + 2
Q
Q
n
⎞
∂i (aij ∂j v)⎠ dx dt
i,j =1
2
n
sλϕv +
+
∂
(a
∂
v)
i ij j dx dt
Q
i,j =1
⎛
⎞
n
|∂t v|2 dx dt + 2
∂t v ⎝−
∂i (aij ∂j v)⎠ dx dt + 2 (∂t v)(−sλϕ)v dx dt
Q
Q
i,j =1
|∂t v|2 dx dt + I1 + I2 .
≡
Q
(9.4)
Q
Thus
f 2 e2sϕ dx dt I1 + I2
(9.5)
Q
and
|∂t v| dx dt f 2 e2sϕ dx dt + |I1 + I2 |.
2
Q
60
Q
(9.6)
Inverse Problems 25 (2009) 123013
Topical Review
Henceforth Cj > 0 denotes generic constants which are independent of s, λ. We assume that
s > 1 and λ > 1. By noting aij = aj i , the assumption (9.2) and integration by parts yield
n
n
|I1 | = −2 (∂t v)
∂i (aij ∂j v) dx dt = 2
(∂i ∂t v)aij ∂j v dx dt Q
Q i,j =1
i,j =1
n = 2
aij ((∂j v)∂i ∂t v + (∂i v)∂j ∂t v) dx dt + 2
aii (∂i v)∂i ∂t v dx dt Q i>j
i=1 Q
n
2
= 2
aij ∂t ((∂i v)(∂j v)) dx dt +
aii ∂t ((∂i v) ) dx dt Q i=1
Q i>j
n
=
aij ∂t ((∂i v)∂j v) dx dt Q i,j =1
n
n
%
&t=T
= −
aij (∂i v)(∂j v) t=0 dx (∂t aij )(∂i v)∂j v dx dt +
Q i,j =1
i,j =1
C1
|∇v|2 dx dt + C1 (|∇v(x, T )|2 + |∇v(x, 0)|2 ) dx.
(9.7)
Q
On the other hand,
I2 = −sλ
2(∂t v)vϕ dx dt = −sλ
∂t (v 2 )ϕ dx dt
Q
Q
% 2 &t=T
2
ϕv t=0 dx
= sλ (∂t ϕ)v dx dt − sλ
Q
ϕv 2 dx dt − sλ (eλT |v(x, T )|2 + |v(x, 0)|2 ) dx.
sλ2
Q
Hence
e
sϕ
f 2L2 (Q)
sλ
ϕv dx dt − C1
2
|∇v| dx dt − sλ
2
Q
(eλT |v(x, T )|2 + |v(x, 0)|2 ) dx
2
Q
− C1
(9.8)
(|∇v(x, T )|2 + |∇v(x, 0)|2 ) dx.
(9.9)
We have to estimate Q |∇v|2 dx dt. For it, we argue similarly to step (3) in section 3, and we
consider Q (P v)v dx dt:
(P v)v dx dt =
Q
(∂t v)v dx dt −
Q
sλϕv 2 dx dt −
Q
n
v∂i (aij ∂j v) dx dt
Q i,j =1
= J1 + J2 + J3 .
We have
1
2
|J1 | = (∂t v)v dx dt = ∂t (v ) dx dt 2
Q
Q
1
%
&
1
t=T
2
=
|v(x, t)| t=0 dx (|v(x, T )|2 + |v(x, 0)|2 ) dx.
2 2 61
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Next
Topical Review
sλϕv 2 dx dt C2
sλϕv 2 dx dt
|J2 | = −
Q
and
J3 = −
n ∂i (aij ∂j v)v dx dt =
i,j =1 Q
Q
n aij (∂i v)∂j v dx dt σ1
i,j =1 Q
|∇v|2 dx dt
(9.10)
Q
by (1.2). Hence
λ(P v)v dx dt σ1
λ|∇v|2 dx dt − C2
sλ2 ϕv 2 dx dt
Q
Q
Q
1
− λ (|v(x, T )|2 + |v(x, 0)|2 ) dx.
2 (9.11)
On the other hand,
λ|(P v, v)L2 (Q) | P vL2 (Q) (λvL2 (Q) ) 1
λ2
P v2L2 (Q) + v2L2 (Q)
2
2
1
λ2
f esϕ 2L2 (Q) + v2L2 (Q) .
2
2
Hence (9.11) yields
σ1
λ|∇v|2 dx dt C2
sλ2 ϕv 2 dx dt
Q
+
1
2
Q
|f esϕ |2 dx dt +
Q
λ2
2
1
v 2 dx dt + λ
2
Q
(|v(x, T )|2 + |v(x, 0)|2 ) dx.
Estimating the first term on the right-hand side by (9.9), we obtain
σ1
λ|∇v|2 dx dt
Q
sϕ 2
2
|f e | dx dt + C3
|∇v| dx dt + C3
λ2 v 2 dx dt
C3
Q
Q
Q
+ C3 λ v(·, T )2L2 () + v(·, 0)2L2 () + C3 ∇v(·, T )2L2 () + ∇v(·, 0)2L2 ()
+ C3 sλ eλT v(·, T )2L2 () + v(·, 0)2L2 () .
(9.12)
Adding (9.12) and (9.9), we have
sλ2 ϕv 2 dx dt + σ1
λ|∇v|2 dx dt
Q
Q
|f esϕ |2 dx dt + C4
|∇v|2 dx dt + C4
λ2 v 2 dx dt
C4
Q
Q
Q
+ C4 ∇v(·, T )2L2 () + ∇v(·, 0)2L2 () + C4 sλ eλT v(·, T )2L2 () + v(·, 0)2L2 () .
By noting ϕ = eλt 1, we take s > 0 and λ > 0 large to absorb the second and third terms
on the right-hand side into the left-hand side, and we obtain
sλ2 ϕv 2 dx dt +
λ|∇v|2 dx dt C5
|f esϕ |2 dx dt
Q
Q
C(λ)s
+ C5 e
62
u(·, 0)2H 1 ()
Q
+ u(·, T )2H 1 () .
Inverse Problems 25 (2009) 123013
Topical Review
Hence
2
2
2 2sϕ
(sλ ϕu + λ|∇u| ) e dx dt C5
|f esϕ |2 dx dt
Q
Q
+ C5 eC(λ)s u(·, 0)2H 1 () + u(·, T )2H 1 () .
(9.13)
for all large s > 0 and λ > 0.
Next we will estimate |∂t v|2 . Since u = e−sϕ v, we have ∂t u = −sλϕe−sϕ v + e−sϕ ∂t v, we
have
1
2
|∂t u|2 e2sϕ 2sλ2 ϕv 2 +
|∂t v|2 .
sϕ
sϕ
For all large s > 0 and λ > 0, we have
1
2
2 2sϕ
2
2
|∂t u| e dx dt |∂t v|2 dx dt
2sλ ϕv dx dt +
Q sϕ
Q
Q sϕ
2
2
sλ ϕv dx dt + C
|∂t v|2 dx dt
C
Q
Q
2
2
sλ ϕv dx dt + C
f 2 e2sϕ dx dt + C|I1 + I2 |
C
Q
Q
by (9.6). By (9.7), (9.8) and (9.13), we have
1
|∂t u|2 e2sϕ dx dt C
sλ2 ϕv 2 dx dt + C
f 2 e2sϕ dx dt + C
|∇v|2 dx dt
Q sϕ
Q
Q
Q
+ CeC(λ)s u(·, 0)2H 1 () + u(·, T )2H 1 ()
f 2 e2sϕ dx dt + CeC(λ)s u(·, 0)2H 1 () + u(·, T )2H 1 () .
(9.14)
C
Q
Moreover, in place of (9.2), we can consider other boundary condition:
∂u
+ p(x)u = 0,
x ∈ ∂, 0 < t < T
(9.15)
∂νA
where p ∈ C(∂). In the above arguments, only in (9.7) and (9.10), we need to modify: that
is,
n
∂i (aij ∂j v) dx dt
I1 = −2 (∂t v)
Q
=2
n
i,j =1
Q i,j =1
=2
n
n
(∂i ∂t v)aij ∂j v dx dt − 2
aij (∂j v)νi (∂t v) dS dt
∂×(0,T ) i,j =1
aij (∂i ∂t v)∂j v dx dt − 2
Q i,j =1
∂×(0,T )
∂v
(∂t v) dS dt.
∂νA
By (9.15), we obtain
∂v
−2
(∂t v) dS dt = 2
pv(∂t v) dS dt
∂×(0,T ) ∂νA
∂×(0,T )
=
p∂t (v 2 ) dS dt = −
[pv 2 ]t=T
t=0 dS.
∂×(0,T )
∂
Thus also in the case (9.15), we have
n
(∂i ∂t v)aij ∂j v dx dt −
[pv 2 ]t=T
I1 = 2
t=0 dS
Q i,j =1
∂
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Inverse Problems 25 (2009) 123013
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and by the trace theorem: uL2 (∂) CuH 1 () , we can obtain the same estimate
|I1 | C1
|∇v|2 dx dt + C1 u(·, 0)2H 1 () + u(·, T )2H 1 () .
Q
On the other hand, for J3 in (9.10), by (9.15) we have
n aij (∂i v)(∂j v) dx dt −
J3 =
i,j =1 Q
=
∂×(0,T )
n
pv 2 dS dt
aij (∂i v)(∂j v) dx dt +
i,j =1 Q
∂×(0,T )
σ1
Q
∂v
v dS dt
∂νA
|∇v|2 dx dt − C1 v2L2 (0,T ;L2 (∂)) .
We fix sufficiently small ε0 > 0. By the trace theorem and the interpolation inequality (e.g.,
[1]), for ε > 0 there exists C(ε) > 0 such that
v2L2 (0,T ;L2 (∂)) Cv2 2
Hence, setting ε =
1
L (0,T ;H 2 +ε0 ())
2
εvL2 (0,T ;H 1 ()) + C(ε)v2L2 (Q) .
σ1
,
2
we have
σ1
2
J3 |∇v| dx dt − C(σ1 )
|v|2 dx dt.
2 Q
Q
Therefore, in the case of (9.15), we obtain (9.14).
Finally, in view of (9.2) or (9.15), we apply an a priori estimate for the boundary value
problem for
n
∂i (aij ∂j u(·, t)) = f (·, t) − ∂t u(·, t) in −
i,j =1
for t ∈ [0, T ], and we prove the following:
Theorem 9.1. We set
ϕ(t) = eλt .
Then there exists λ0 > 0 such that for any arbitrary λ λ0 we can choose a constant s0 (λ) > 0
satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎞
⎛
n
1 ⎝
2
2⎠
2
2
2
|∂i ∂j u| + λ|∇u| + sλ ϕu e2sϕ dx dt
|∂t u| +
Q sϕ
i,j =1
|Lu|2 e2sϕ dx dt
C
Q
+ CeC(λ)s u(·, 0)2H 1 () + u(·, T )2H 1 ()
for all s > s0 and all u ∈ H 2,1 (Q) satisfying
u=0
on ∂ × (0, T )
(9.16)
or
∂u
+ p(x)u = 0
∂νA
with p ∈ C(∂).
64
on ∂ × (0, T )
(9.17)
Inverse Problems 25 (2009) 123013
Topical Review
Next we apply theorem 9.1 to the backwards parabolic problem and establish a conditional
stability estimate. Let aij ∈ C 1 (Q), 1 i, j n, satisfy (1.1) and (1.2), bi , c ∈ L∞ (Q). We
consider
n
aij (x, t)∂i ∂j u(x, t)
(Lu)(x, t) ≡ ∂t u(x, t) −
i,j =1
−
n
bi (x, t)∂i u(x, t) − c(x, t)u = 0,
(x, t) ∈ Q
(9.18)
i=1
with (9.16) or (9.17).
Parabolic equation backward in time: Let 0 t0 < T . Determine u(x, t0 ), x ∈ from
u(x,T), x ∈ .
This is a determination problem of the history of a heat process, by means of values at
a final time T, and there are definitely many applications in the mathematical physics (e.g.,
[4, 124]). Moreover there are available numerical treatments (e.g., [8, 45, 52, 111]). In
[49, 50], a sharp backward uniqueness result is proved: let |∂t u + u| C(|∇u| + |u|) in
2
{|x| > R} × (0, T ) and let |u(x, t)| C eC|x| . If u(x, 0) = 0, |x| R, then u ≡ 0 in
{|x| > R} × (0, T ).
As is well known, the backward parabolic equation is ill-posed: small errors in data
may cause huge deviations in solutions u(·, t0 ). However, if we assume an a priori bound
for u(·, 0), then we can restore the stability. The conditional stability can be formulated as
follows. Let t0 ∈ [0, T ) and let us set
ζ
UM,ζ = a ∈ H0 (); aH ζ () M
0
with M > 0 and ζ 0. Then by the conditional stability we mean: can we choose a function
ω ∈ C[0, ∞) such that ω 0, ω is strictly monotone increasing and limη↓0 ω(η) = 0
satisfying
u(·, t0 )L2 () ω(u(·, T )L2 () )
provided that u(·, 0) ∈ UM,ζ ?
There are several methods yielding the conditional stability in cases of
(i) t0 > 0 and ζ = 0;
(ii) t0 = 0 and ζ > 0.
In case (i), it is known that
' θ(
ω(η) = O η T
(e.g., [112]). In particular, in [[112], theorem 1, p 99], it is proved that
uL2 (t0 ,T ;L2 ()) Cu(·, T )θL2 ()
with θ ∈ (0, 1), under an a priori boundedness condition on u(·, 0) in UM,1 . On the other
hand, in case (ii),
⎛
θ ⎞
1
⎠
ω(η) = O ⎝
with
θ ∈ (0, 1)
log η1
(e.g., exercise 3.1.2 (p 44) in [90] for ζ = 2 and theorems 3 and 4 in [102] for ζ = 1).
Among the methods for the conditional stability for the backwards parabolic equation,
we shall first refer to the logarithmic convexity and see [2, 4, 28, 29], [[36], chapter 7], [124],
and references therein. See also [[90], pp 44–9] as for an argument for the stability in the
65
Inverse Problems 25 (2009) 123013
Topical Review
backwards problem which was shown in [2], but that method does not work for a parabolic
inequality |Lu| |f | in Q. Our method on the basis of the Carleman estimate is more effective
than the logarithmic convexity: for example, we can treat also semilinear equations directly.
Second, we refer to the method by the t-analyticity by Kreı̆n and Prozorovskaya [108].
Hence the method requires the t-analyticity of solution u(·, t) which implies that we have to
assume that the coefficients aij , bj and c are t-analytic.
Third, we can refer to a weighted energy method or a method by a Carleman estimate and
see [4, 102, 112, 114, 123, 124], for example.
Now we apply theorem 9.1 to establish a conditional stability estimate for u(·, t0 )L2 ()
in the case of ζ = 0 and t0 > 0.
Theorem 9.2. Let u ∈ H 2,1 (Q) satisfy (9.18) with (9.16) or (9.17). For any t0 ∈ (0, T ),
there exist constants θ ∈ (0, 1) and C > 0 depending on t0, max1i,j n aij C 1 (Q) ,
max1in bi L∞ (Q) , cL∞ (Q) , T and such that
u(·, t0 )L2 () Cu1−θ
u(·, T )θH 1 () .
L2 (Q)
(9.19)
Thanks to the large parameters s, λ in theorem 9.1, we can argue similarly also for
a semilinear parabolic equation Lu = F (u, ∇u, x, t) with suitable F and an a priori
boundedness assumption.
Remark. As is seen from the proof, we can clarify the dependence of θ on t0 (see (9.25)), and
after integrating (9.19) with respect to t0 from 0 to T, we can see
− 12
1
.
(9.20)
uL2 (×(0,T )) C(1 + u(·, 0)L2 () ) log
u(·, T )H 1 ()
In particular, (9.20) implies
uL2 (×(0,T )) = O log
1
u(·, T )H 1 ()
− 12
as u(·, T )H 1 () −→ 0, provided that u(·, 0)L2 () M.
By suitable regularity of aij , bi , c, we apply the a priori estimate for the initial
value/boundary value problem (e.g., [58, 109, 125]), we can have
uL2 (Q) Cu(·, 0)L2 () ,
u(·, T )H 2 () Cu(·, 0)L2 () .
Hence thanks to the interpolation inequality:
1
1
u(·, T )H 1 () Cu(·, T )H2 2 () u(·, T )L2 2 () ,
we can improve (9.19) as follows: for any μ ∈ (0, 2), there exists a constant Cμ,t0 > 0 such
that
1− θ (2−μ)
θ (2−μ)
4
u(·, t0 )H μ () Cμ,t0 u(·, 0)L2 ()4 u(·, T )L2 ()
.
(9.21)
Remark. Our result proves the backwards uniqueness if aij : [0, T ] −→ L∞ (),
1 i, j n, are Lipschitz continuous on t ∈ [0, T ]. This regularity in t is the best possible
because in Miller [121], there is a counter-example constructed for the non-uniqueness in a
backwards problem for a parabolic equation with Hölder continuous coefficient in t.
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Inverse Problems 25 (2009) 123013
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Proof. In theorem 9.1, we fix λ and we choose t1 , t2 such that 0 < t2 < t1 < t0 . We set
δk = eλtk , k = 0, 1, 2. Since theorem 9.1 requires u(·, 0) as data on the right-hand side, we
need a cut-off function. Let χ ∈ C ∞ (R) such that 0 χ 1 and
1,
t > t1 ,
(9.22)
χ (t) =
0,
t < t2 .
Setting w = χ u, we have w(·, 0) = 0 and Lw = χ (t)u in Q, and w = 0 or
∂ × (0, T ). Applying theorem 9.1 and we obtain
⎫
⎧⎛
⎞
n
⎬
⎨
1 ⎝
|∂i ∂j w|2 ⎠ + |∇w|2 + s|w|2 e2sϕ dx dt
|∂t w|2 +
⎭
Q s ⎩
i,j =1
|χ (t)|2 |u|2 e2sϕ dx dt + C eCs w(·, T )2H 1 ()
C
∂w
∂νA
+ pw = 0 on
Q
for all large s > 0. By (9.22), we see that
t1
|χ (t)|2 |u|2 e2sϕ dx dt C
|u|2 e2sϕ dx dt C e2sδ1 u2L2 (Q) ,
Q
t2
so that
T 1
|∂t u|2 + s|u|2 dx dt
e2sδ0
t0 s
T 1
2
2
C
|∂t w| + s|w| e2sϕ dx dt
s
0
C e2sδ1 u2L2 (Q) + C eCs u(·, T )2H 1 ()
for all s s0 . Consequently
T 1
2
2
|∂t u| + s|u| dx dt C e2s(δ1 −δ0 ) u2L2 (Q) + C eCs u(·, T )2H 1 ()
s
t0 (9.23)
for all s s0 . Therefore we have
T 2
2
|u(x, t0 )| dx = −
∂t
|u(x, t)| dx dt +
|u(x, T )|2 dx
t0
=−
T
t0
2|u(x, t)||∂t u(x, t)| dx dt + u(·, T )2L2 () .
Hence (9.23) and the Cauchy–Schwarz inequality yield
T 1
u(·, t0 )2L2 () |∂t u|2 + s|u|2 dx dt + u(·, T )2L2 ()
t0 s
C e−2s(δ0 −δ1 ) u2L2 (Q) + C eCs u(·, T )2H 1 ()
(9.24)
for all s s0 . Replacing C by C eCs0 , we have (9.24) for all s 0. Choosing s 0
minimizing the right-hand side of (9.24), we obtain
C
C+2(δ −δ1 )
u(·, t0 )L2 () CuL2 (Q)0
Thus the proof of (9.19) is completed.
2(δ0 −δ1 )
C+2(δ −δ )
0 1
u(·, T )H 1 ()
.
(9.25)
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9.2. Carleman estimate for a forward problem in time and an application to an inverse
source problem
For simplicity, we assume that aij, 1 i, j n are t-independent. In the argument in proving
theorem 9.1, we choose
ϕ(t) = e−λt
with λ > 0. Then we can similarly prove
Theorem 9.3. We set
ϕ(t) = e−λt .
Then there exists λ0 > 0 such that for any arbitrary λ λ0 we can choose a constant s0 (λ) > 0
satisfying: there exists a constant C = C(s0 , λ0 ) > 0 such that
⎞
⎛
n
1 ⎝
2
2⎠
2
2
2
|∂i ∂j u| + λ|∇u| + sλ ϕu e2sϕ dx dt C
|Lu|2 e2sϕ dx dt
|∂t u| +
sϕ
Q
Q
i,j =1
∂u 2sϕ
C(λ)s
2
e dS dt
+C e
u(·, 0)H 1 () + C
sλ(|u| + |∂t u|) ∂νA ∂×(0,T )
for all s > s0 and all u ∈ H 2,1 (Q) satisfying
u(·, T ) = 0
in .
This Carleman estimate can yield a conditional stability estimate on u(·, t) with t > 0
which is a solution to the initial value/boundary value problem, but of course, such an
application of theorem 9.3 is not interesting because other classical methods for the wellposed forward problem produce better and well-known results.
Our interest here is the application of theorem 9.3 to an inverse source problem. For
convenience, we consider a simple case but the treatment for a general L is similar.
Let x = (x1 , x ) ∈ Rn and x = (x2 , ..., xn ) ∈ Rn−1 , = (0, ) × D , D ⊂ Rn−1 be a
bounded domain with smooth boundary ∂D . We consider
∂t u(x, t) = u(x, t) + f (x , t)R(x, t),
u(x, 0) = 0,
∂u
(x, t) = 0,
∂ν
x ∈ , t > 0
x∈
(9.26)
(9.27)
x ∈ ∂, t > 0.
(9.28)
Inverse heat source problem: Let R be given and let t0 > 0. Determine f (x , t), x ∈ D ,
0 < t < t0 by u|∂×(0,t0 ) .
Although we can discuss the conditional stability, we consider only the uniqueness.
Theorem 9.4. We assume that u, ∂1 u ∈ H 2,1 (Q) and
R ∈ C 2 (Q),
|R(x, t)| = 0,
x ∈ , 0 t t0 .
(9.29)
If u|∂×(0,t0 ) = 0, then f (x , t) = 0, x ∈ D , 0 t t0 .
Proof. For arbitrary small ε > 0, we choose t1 , t2 such that 0 < t0 − ε < t1 < t2 < t0 . We set
δk = e−λtk , k = 0, 1, 2. Let χ ∈ C ∞ (R) be a cut-off function such that 0 χ 1 and
1,
t < t1 ,
(9.30)
χ (t) =
0,
t t2 .
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Inverse Problems 25 (2009) 123013
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Setting u = Rw on × [0, t2 ], we have
2∇R
∂t R
w−
· ∇w
∂t w − w +
R
R
R
−
x ∈ , 0 < t < t2
(9.31)
w = f (x , t),
R
and
∂w
|∂×(0,t0 ) = 0.
w|∂×(0,t0 ) =
∂ν
Differentiating both sides of (9.31) with respect to x1 and setting y = ∂1 w, we obtain
⎧
∂t R
2∇R
R
⎪
∂t y − y +
y−
· ∇y −
y
⎪
⎪
⎪
R
R
R
⎨
∂t R
2∇R
R
w − ∂1
· ∇w − ∂1
w = 0,
x ∈ , 0 < t < t0 ,
+∂1
⎪
⎪
R
R
R
⎪
⎪
⎩
y|∂×(0,t0 ) = 0.
Setting v = χy, we have
2∇R
R
∂t R
v−
· ∇v −
v
∂t v − v +
R
R
R
∂t R
2∇R
χ w + ∂1
· χ ∇w
= −χ (t)y − ∂1
R
R
R
χ w,
x ∈ , 0 < t < t0
+ ∂1
R
v|∂×(0,t0 ) = 0.
(9.32)
(9.33)
By y = ∂1 w, and u(0, x , t) = 0, x ∈ D , 0 < t < t0 , by the assumption, we see that
x1
y(η, x , t) dη.
(9.34)
w(x1 , x , t) =
0
Hence (9.32) implies
2∇R
R
∂t R
v−
· ∇v −
v
∂t v − v +
R
R Rx1
x1
∂t R
2∇R
·
= −χ y − ∂1
v(η, x , t) dη + ∂1
∇v(η, x , t) dη
R
R
0
0
x1
R
v(η, x , t) dη,
x ∈ , 0 < t < t0
(9.35)
+ ∂1
R
0
v(x, t0 ) = v(x, 0) = 0,
x ∈ .
(9.36)
Apply theorem 9.3 to (9.35), in terms of (9.33) and (9.36), we obtain
t0 (λ|∇v|2 + sλ2 ϕ|v|2 ) e2sϕ dx dt
0
t0
C
|χ y| e
0
t0
+C
0
0
x1
2 2sϕ
Since
dx dt + C
0
x1
0
0
x1
2
v(η, x , t) dη e2sϕ dx dt
2
∇v(η, x , t) dη e2sϕ dx dt.
2
∇ v(η, x , t) dη j
t0
(9.37)
|∇ j v(η, x , t)|2 dη
0
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Inverse Problems 25 (2009) 123013
Topical Review
for j = 0, 1, we have
2
t0 x1
2sϕ(t)
j
e
∇
v(η,
x
,
t)
dη
dx dt
0
0
dx1
t0 0
2
0
t0
|∇ v(η, x , t)| dη e2sϕ(t) dx dt
j
2 2sϕ(t)
|∇ v(η, x , t)| e
dη dx dt.
0
j
D
2
0
Hence, taking s > 0 and λ > 0 large, we can absorb the second and third terms on the
right-hand side of (9.37) into the left-hand side:
t0 t0 2
2
2 2sϕ
(λ|∇v| + sλ ϕ|v| ) e dx dt C
|χ y|2 e2sϕ dx dt
(9.38)
0
0
for all large s > 0 and λ > 0. We fix such λ > 0. By (9.30), we see that
t0 |χ y|2 e2sϕ dx dt e2sδ1
|y|2 dx dt CM 2 e2sδ1 ,
0
Q
where we set M = ∂1 uL2 (Q) . Hence with (9.38), we obtain
t0 −ε t0 −ε 2sδε
2
2
e
(|∇v| + s|v| ) dx dt (|∇v|2 + s|v|2 ) e2sϕ dx dt
0
0
t0 (|∇v|2 + s|v|2 ) e2sϕ dx dt CM 2 e2sδ1
0
for all large s > 0, where we set δε = e−λ(t0 −ε) . Consequently
vL2 (0,t0 −ε;H 1 ()) CM 2 e−2s(δε −δ1 )
for all large s > 0. Since δε − δ1 = e−λ(t0 −ε) − e−λt1 > 0 by t0 − ε < t1 , letting s → ∞, we
obtain v = 0 in × (0, t0 − ε). Equation (9.34) implies w = 0 in × (0, t0 − ε), with which
(9.31) means f = 0 in D × (0, t0 − ε). Since ε > 0 is arbitrary, the proof of the theorem is
completed.
Remark. For the proof, the original argument in [20] works but the argument in section 6.2
is not applicable.
We conclude this section with strongly coupled parabolic systems: we assume that
aijk ∈ C 1 (Q), 1 i, j n, 1 k, N satisfy (7.19)–(7.21) and that bik , ck ∈ L∞ (Q).
Then we consider a strongly coupled parabolic system whose principal parts are different:
[Lu] (x, t) ≡ ∂t u (x, t) −
N n
∂i aijk (x, t)∂j uk (x, t)
k=1 i,j =1
−
N n
bik (x, t)∂i uk (x, t) −
k=1 i=1
(x, t) ∈ Q, 1 N.
N
ck (x, t)uk (x, t) = f (x, t),
k=1
(9.39)
Since the strong coupling occurs only for x and the weight functions are independent of
x in the Carleman estimates discussed in this section, we can follow the arguments for theorems
9.1 and 9.3, so that we can prove Carleman estimates for the strongly coupled system (9.39).
70
Inverse Problems 25 (2009) 123013
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Acknowledgments
The author is very grateful to the anonymous reviewer for very valuable and constructive
comments. The author thanks Professor J-M Coron (Université de Paris VI), Professor V
Isakov (Wichita State University), Professor M V Klibanov (University of North Carolina at
Charlotte), Professor J Le Rousseau (Université d’Orleans), Professor L Rosier (Université
de Nancy) and Professor X Zhang (Academy of Mathematics and System Sciences) for
information of references and comments. The author thanks Professor V G Romanov (Sobolev
Institute of Mathematics, Novosibirsk), Dr Igor Trooshin (Institute for Problems of Precision
Mechanics and Control, Russia) and Mr Ying Zhang (Fudan University) for very careful
reading and invaluable comments.
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