University of Wollongong
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Faculty of Engineering and Information Sciences
2007
A nonlinear 1-D case backward heat problem:
Regularization and error
Dang Duc Trong
HoChiMinh City National University
Pham Hoang Quan
HoChiMinh City National University
Tran Vu Khanh
University of Wollongong, [email protected]
Nguyen Huy Tuan
HoChiMinh City National University
Publication Details
Trong, D., Quan, P., Khanh, T. & Tuan, N. (2007). A nonlinear 1-D case backward heat problem: Regularization and error.
ZEITSCHRIFT FÜR ANALYSIS UND IHRE ANWENDUNGEN, 26 (2), 231-245.
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A nonlinear 1-D case backward heat problem: Regularization and error
Keywords
1, case, backward, nonlinear, heat, problem, regularization, error
Disciplines
Engineering | Science and Technology Studies
Publication Details
Trong, D., Quan, P., Khanh, T. & Tuan, N. (2007). A nonlinear 1-D case backward heat problem:
Regularization and error. ZEITSCHRIFT FÜR ANALYSIS UND IHRE ANWENDUNGEN, 26 (2),
231-245.
This journal article is available at Research Online: http://ro.uow.edu.au/eispapers/4526
Zeitschrift für Analysis und ihre Anwendungen
Journal for Analysis and its Applications
Volume 26 (2007), 231–245
c European Mathematical Society
°
A Nonlinear Case of the
1-D Backward Heat Problem:
Regularization and Error Estimate
Dang Duc Trong, Pham Hoang Quan, Tran Vu Khanh
and Nguyen Huy Tuan
Abstract. We consider the problem of finding, from the final data u(x, T ) = ϕ(x),
the temperature function u(x, t), x ∈ (0, π), t ∈ [0, T ] satisfies the following nonlinear
system
ut − uxx = f (x, t, u(x, t)), (x, t) ∈ (0, π) × (0, T )
u(0, t) = u(π, t) = 0,
t ∈ (0, T ).
The nonlinear problem is severely ill-posed. We shall improve the quasi-boundary
value method to regularize the problem and to get some error estimates. The approximation solution is calculated by the contraction principle. A numerical experiment
is given.
Keywords. Backward heat problem, nonlinearly Ill-posed problem, quasi-boundary
value methods, quasi-reversibility methods, contraction principle
Mathematics Subject Classification (2000). Primary 35K05, secondary 35K99,
47J06, 47H10
1. Introduction
Let T be a positive number, we consider the problem of finding the temperature
u(x, t), (x, t) ∈ (0, π) × [0, T ] such that
ut − uxx = f (x, t, u(x, t)), (x, t) ∈ (0, π) × (0, T )
u(0, t) = u(π, t) = 0,
t ∈ (0, T )
u(x, T ) = ϕ(x),
x ∈ (0, π),
(1)
(2)
(3)
where ϕ(x), f (x, t, z) are given. The problem is called the backward heat problem, the backward Cauchy problem or the final value problem.
HoChiMinh City National University, Department of Mathematics and Informatics,
227 Nguyen Van Cu, Q. 5, HoChiMinh City, VietNam; [email protected]
232
D. D. Trong et al.
As is known, the nonlinear problem is severely ill-posed, i.e., solutions do not
always exist, and in the case of existence, these do not depend continuously on
the given data. In fact, from small noise contaminated physical measurements,
the corresponding solutions have large errors. It makes difficult to numerical
calculations. Hence, a regularization is in order. The linear case was studied
extensively in the last four decades by many methods. The literature related to
the problem is impressive (see, e.g., [3, 4, 7] and the references therein). In the
pioneering work [7] in 1967, the authors presented, in a heuristic way, the quasireversibility method. They approximated the problem by adding a ”corrector”
into the main equation. In fact, they considered the problem
ut + Au − ²A∗ Au = 0, t ∈ [0, T ]
u(T ) = ϕ.
The stability magnitude of the method is of order ec² . In [1, 12], the problem
is approximated with
−1
ut + Au + ²Aut = 0, t ∈ [0, T ]
u(T ) = ϕ.
The method is useful if we cannot construct clearly the operator A∗ . However,
the stability order in the case is quite as large as that in the original quasireversibility methods. In [10], using the method, so-called, of stabilized quasi
reversibility, the author approximated the problem with
ut + f (A)u = 0, t ∈ [0, T ]
u(T ) = ϕ.
He shows that, with appropriate conditions on the ”corrector” f (A), the stability magnitude of the method is of order c²−1 .
Sixteen years after the work by Lattes-Lions, in 1983, Showalter presented
the quasi-boundary value method. He considered the problem
ut − Au(t) = Bu(t), t ∈ [0, T ]
u(0) = ϕ,
and approximated the problem with
ut − Au(t) = Bu(t), t ∈ [0, T ]
u(0) + ²u(T ) = ϕ.
According to him, this method gives a better stability estimate than the other
discussed methods. Clark and Oppenheimer, in their paper [4], used the quasiboundary value method to regularize the backward problem with
ut + Au(t) = 0, t ∈ [0, T ]
u(T ) + ²u(0) = ϕ.
1-D Backward Heat Problem
233
The authors show that the stability estimate of the method is of order ²−1 . Very
recently, in [6], the quasi-boundary method was used to solve a backward heat
equation with an integral boundary condition.
Although we have many works on the linear case of the backward problem,
the literature of the nonlinear case is quite scarce. Very recently, in [11], the
authors tranform the problem into the one of minimizing an appropriate functional. However, a sharp error estimate and an effective method of calculation
are not given in [11].
Informally, problem (1)−(3) can be transformed to an integral equation
having the form
u(x, t) =
∞ ·
X
e
(T −t)n2
n=1
ϕn −
Z
T
e
(s−t)n2
¸
fn (u)(s) ds sin nx
t
P
P∞
where ϕ(x) = ∞
n=1 ϕn sin nx, f (u)(x, t) =
n=1 fn (u)(t) sin nx are the expan2
2
sion of ϕ and f (u), respectively. The terms e(T −t)n , e(s−t)n (n large) are the
unstability cause. Hence, to regularize the problem, we have to replace the
terms by better terms. Naturally, we shall replace two terms by
2
2
e−tn
,
βn (ε, t, s) + e−sn2
e−tn
,
αn (ε, t) + e−T n2
where αn , βn are positive functions satisfying
lim αn (ε, t) = lim βn (ε, t, s) = 0.
ε↓0
ε↓0
Many versions of αn , βn are suggested from the quasi-type methods discussed
above.
In the present paper, we shall use an association of the quasi-reversibility
method and the quasi-boundary value method to regularize our problem. In
fact, we approximate problem (1)−(3) by the following problem:
u²t
−
u²xx
=
∞
X
n=1
e−tn
t
T
² +
2
e−tn2
u² (0, t) = u² (π, t) = 0,
²u² (x, 0) + u² (x, T )
∞ µZ
X
= ϕ(x) −
n=1
fn (u² )(t) sin nx,
(x, t) ∈ (0, π) × (0, T )
t ∈ [0, T ]
T
0
¶
²
²
fn (u )(s) ds sin nx,
s
² T + e−sn2
(4)
(5)
x ∈ [0, π],
(6)
where 0 < ² < 1, fn (u)(t) = π2 hf (x, t, u(x, t)), sin nxi and h·, ·i is the inner
product in L2 (0, π). We shall prove that, the (unique) solution u² of (4)−(6)
234
D. D. Trong et al.
satisfies the following equality:
Ã
!
Z T
∞
−tn2
−tn2
X
e
e
ϕ −
f (u² )(s) ds sin nx ,
u² (x, t) =
s
2 n
−T n2 n
−sn
T
²
+
e
² +e
t
n=1
(7)
where ϕn = π2 hϕ(x), sin nxi.
The remainder of the paper is divided into three sections. In Section 2,
we shall show that (4)−(6) is well posed and that the solution u² (x, t) satisfies (7). Then, in Section 3, we estimate the error between an exact solution u 0
of problem (1)−(3) and the approximation solution u² . In fact, we shall prove
that
t
ku² (·, t) − u0 (·, t)k ≤ C² T
(8)
and that there is a t² > 0 such that
²
ku (·, t² ) − u0 (·, 0)k ≤
√
4
µ µ ¶¶− 41
1
8 C T ln
,
²
√
4
(9)
where k · k is the norm in L2 (0, π) and C depends on u0 and f . Finally, a
numerical experiment will be given in Section 4.
2. The well-posedness of problem (4)−(6)
In the section, we shall study the existence, the uniqueness and the stability of
a (weak) solution of problem (4)−(6). In fact, one has
Theorem 2.1. Let ϕ ∈ L2 (0, π) and let f ∈ L∞ ([0, π] × [0, T ] × R) satisfy
|f (x, y, w) − f (x, y, v)| ≤ k|w − v|
for a k > 0 independent of x, y, w, v. Then problem (4)−(6) has uniquely a
weak solution u² ∈ C([0, T ]; L2 (0, π)) ∩ L2 (0, T ; H01 (0, π)) ∩ C 1 (0, T ; H01 (0, π))
satisfying (7). The solution depends continuously on ϕ in C([0, T ]; L 2 (0, π)).
Proof. The proof is divided into three steps. In Step 1, we shall prove that
problem (4)−(6) is equivalence to problem (7). In Step 2, we prove the existence
and the uniqueness of a solution of (7). Finally in Step 3, the stability of the
solution is given.
Step 1. Prove that (4)-(6) is equivalence (7). We divide this step into two
parts.
Part A. If u² ∈ C([0, T ]; L2 (0, π)) satisfies (7), then u² is solution of (4)−(6).
1-D Backward Heat Problem
235
¶
2
e−tn
²
fn (u )(s) ds sin nx ,
s
² T + e−sn2
(10)
For 0 ≤ t ≤ T , we have
²
u (x, t) =
∞ µ
X
n=1
2
e−tn
ϕn −
² + e−T n2
Z
T
t
where u² ∈ C([0, T ]; L2 (0, π) ∩ C 1 ((0, T ); H01 (0, π)) ∩ L2 (0, T ; H01 (0, π))) can be
verified directly. In fact, u² ∈ C ∞ ((0, T ]; H01 (0, π))). Moreover, one has
u²t (x, t)
=
2
∞ µ
X
−n2 e−tn
n=1
∞
X
+
=−
² + e−T n2
n=1 ²
∞
X
2
π
e
t
T
n=1
∞ µ
X
ϕn −
−tn2
+
Z
e−tn2
T
t
¶
2
−n2 e−tn
²
fn (u )(s)ds sin nx
s
² T + e−sn2
fn (u² )(t) sin nx
­
®
n2 u² (x, t), sin nx sin nx
e−tn
2
²
¶
fn (u )(t) sin nx
t
² T + e−tn2
¶
2
∞ µ
X
e−tn
²
²
fn (u )(t) sin nx
= uxx (x, t) +
t
² T + e−tn2
n=1
+
n=1
(11)
and
²
²
²u (x, 0) + u (x, T ) = ϕ −
∞ µZ
X
n=1
T
¶
²
²
fn (u )(s)ds sin nx.
s
² T + e−sn2
0
(12)
So u² is the solution of (4)−(6).
Part B. If u² satisfies (4)−(6), then u² is a solution of (7).
In fact, taking the inner product of the equation (4) with respect to sin nx
we get in view of (4)
2
d ²
e−tn
un (t) + n2 u²n (t) = t
fn (u² )(t),
dt
² T + e−tn2
(13)
where we recall that
u²n (t) =
It follows that
®
®
2­
2­ ²
u (x, t), sin nx , fn (u² )(t) =
f (x, t, u² (x, t)), sin nx .
π
π
u²n (t)
=
2
e−tn u²n (0)
+
Z
t
2
e−(t−s)n fn (u² )(s) ds.
0
(14)
236
D. D. Trong et al.
Hence, we have the Fourier expansion
∞ µ
X
¶
2
e−sn
²
u (x, t) =
f (u )(s) ds sin nx
+
e
s
2 n
T + e−sn
²
0
n=1
¶
Z t
2
∞ µ
X
e−tn
²
−tn2 ²
f (u )(s) ds sin nx.
(15)
=
e
un (0) +
s
−sn2 n
0 ²T + e
n=1
²
2
e−tn u²n (0)
Z
t
−(t−s)n2
Hence
²
u (x, T ) =
∞ µ
X
2
e−T n u²n (0)
n=1
+
Z
¶
2
e−T n
²
fn (u )(s) ds sin nx.
s
² T + e−sn2
T
0
(16)
Substituting (16) into (6) gives
∞ ³
X
¡
²+e
−T n2
n=1
∞
X
¢ ² ´
un (0) sin nx = ϕ −
n=1
µZ
T
0
¶
2
² + e−T n
²
fn (u )(s) ds sin nx.
s
² T + e−sn2
We obtain
u²n (0)
1
=
ϕn −
² + e−T n2
Z
T
0
1
f (u² )(s) ds.
2 n
−sn
² +e
s
T
(17)
Replacing (17) in (15), we receive (7). This completes the proof of Step 1.
Step 2. The existence and the uniqueness of solution of (7).
Put
G(w)(x, t) = ϕ(x, t) −
∞ Z
X
n=1
T
t
2
e−tn
fn (w)(s) ds sin nx
s
² T + e−sn2
2
P
e−tn
for w ∈ C([0, T ]; L2 (0, π)), where ϕ(x, t) = ∞
n=1 ²+e−T n2 ϕn sin nx. We claim
2
that, for every w, v ∈ C([0, T ]; L (0, π)), m ≥ 1, we have
µ ¶2m
k
(T − t)m C m
kG (w)(·, t) − G (v)(·, t)k ≤
|kw − vk|2 ,
²
m!
m
m
2
(18)
where C = max{T, 1} and |k · k| is the supremum norm in C([0, T ]; L2 (0, π)).
We shall prove the latter inequality by induction.
1-D Backward Heat Problem
237
For m = 1, we have
kG(w)(·, t) − G(v)(·, t)k2
"Z
#2
2
∞
T
πX
e−tn
=
(fn (w)(s) − fn (v)(s)) ds
s
2 n=1 t ² T + e−sn2
!2 Z
Ã
∞ Z T
T ¡
−tn2
X
¢2
π
e
f
(w)(s)
−
f
(v)(s)
ds
≤
ds
s
n
n
2
2 n=1 t
² T + e−sn
t
Z T
∞
¡
¢2
πX 1
fn (w)(s) − fn (v)(s) ds
(T − t)
≤
2
2 n=1 ²
t
Z TZ π
¡
¢2
1
f (x, s, w(x, s)) − f (x, s, v(x, s)) dx ds
= 2 (T − t)
²
0
t
Z TZ π
2
k
≤ 2 (T − t)
|w(x, s) − v(x, s)|2 dx ds
²
0
t
k2
= C 2 (T − t)|kw − vk|2 .
²
Thus (18) holds.
Suppose that (18) holds for m = j. We prove that (18) holds for m = j + 1.
We have
kGj+1 (w)(·, t) − Gj+1 (v)(·, t)k2
¸2
∞ ·Z T
¯
¯
π1 X
j
j
¯
¯
fn (G (w))(s) − fn (G (v))(s) ds
≤
2 ²2 n=1 t
Z TX
∞
¯
¯
π1
¯fn (Gj (w))(s) − fn (Gj (v))(s)¯2 ds
≤
(T
−
t)
2 ²2
t n=1
Z T
°
°
1
°f (·, s, Gj (w)(·, s)) − f (·, s, Gj (v)(·, s))°2 ds
≤ 2 (T − t)
²
t
Z T
° j
°
1
2
°G (w)(·, s) − Gj (v)(·, s)°2 ds
≤ 2 (T − t)k
²
t
µ ¶2j Z T
k
(T − s)j
1
dsC j |kw − vk|2
≤ 2 (T − t)k 2
²
²
j!
t
µ ¶2(j+1)
j+1
k
(T − t)
≤
C j+1 |kw − vk|2 .
²
(j + 1)!
Therefore, by the induction principle, we have
µ ¶m m
k
T2 √ m
m
m
√
|kG (w) − G (v)k| ≤
C |kw − vk|
²
m!
for all w, v ∈ C([0, T ]; L2 (0, π)).
238
D. D. Trong et al.
We consider G : C([0, T ]; L2 (0, π)) → C([0, T ]; L2 (0, π)). There exists√a pos¡ ¢m T m2 C m
√
itive integer m0 such that Gm0 is a contraction since limm→∞ k²
=
m!
m0
0. It follows that the equation G (w) = w has a unique solution u² ∈
C([0, T ]; L2 (0, π)). In fact, one has G(Gm0 (u² )) = G(u² ). Hence Gm0 (G(u² )) =
G(u² ). By the uniqueness of the fixed point of Gm0 , one has G(u² ) = u² , i.e.,
the equation G(w) = w has a unique solution u² ∈ C([0, T ]; L2 (0, π)). From
Part A, Step 1, we complete the proof of Step 2.
Step 3. The solution of the problem (4)−(6) depends continuously on ϕ in
2
L (0, π).
Let u and v be two solutions of (4)−(6) corresponding to the final values ϕ
and ω. From (7) one has in view of the inequality (a + b)2 ≤ 2(a2 + b2 )
¶2
2
∞ µ
X
e−tn
2
|ϕn − ωn |
ku(·, t) − v(·, t)k ≤ π
² + e−T n2
n=1
¶2 (19)
2
∞ µZ T
X
¯
¯
e−tn
¯fn (u)(s) − fn (v)(s)¯ds .
+π
s
2
T + e−sn
²
t
n=1
2
One has, for s > t and α > 0,
e−tn
2
α+e−sn
=
1
t
t
2
2
(αesn +1) s (α+e−sn )1− s
α = ², s = T , we get
t
≤ α s −1 . Letting
2
t
e−tn
−1
T
.
2 ≤ ²
−T
n
²+e
(20)
s
Letting α = ² T , we get
2
s
t
e−tn
≤ ²T −T .
s
2
−sn
²T + e
(21)
Hence, from (19) it follows that
t
ku(·, t) − v(·, t)k2 ≤ 2²2( T −1) kϕ − ωk2
Z
2
2 Tt
+ 2k (T − t)²
T
s
²−2 T ku(·, s) − v(·, s)k2 ds.
t
So, we have
t
²−2( T ) ku(·, t) − v(·, t)k2 ≤ 2²−2 kϕ − ωk2
Z
2
+ 2k (T − t)
T
t
s
²−2 T ku(·, s) − v(·, s)k2 ds.
Using Gronwall’s inequality we have
¡
¢
t
ku(·, t) − v(·, t)k ≤ 2² T −1 exp k 2 (T − t)2 kϕ − ωk.
This completes the proof of Step 3 and the proof of our theorem.
1-D Backward Heat Problem
239
3. Regularization of problem (1)−(3)
We first have a uniqueness result
Theorem 3.1. Let ϕ, f be as in Theorem 2.1. Then problem (1)−(3) has at
most one (weak) solution u ∈ W , where
W = C([0, T ]; L2 (0, π)) ∩ L2 (0, T ; H01 (0, π)) ∩ C 1 ((0, T ); L2 (0, π)).
(x, t, z)| ≤ M for all (x, t, z) ∈ (0, π) ×
Proof. Let M > 0 be such that | ∂f
∂z
(0, T ) × R. Let u1 (x, t) and u2 (x, t) be two solutions of problem (1)−(3) such
that u1 , u2 ∈ W.
Put w(x, t) = u1 (x, t) − u2 (x, t). Then w satisfies the equation
wt (x, t) − wxx (x, t) = f (x, t, u1 (x, t)) − f (x, t, u2 (x, t)).
Since f is Lipschizian, we have (wt − wxx )2 ≤ M 2 w2 . Now w(0, t) = w(π, t) = 0
and w(x, T ) = 0. Hence by the Lees–Protter theorem ([8, p. 373]), w = 0 which
gives u1 (x, t) = u2 (x, t) for all t ∈ [0, T ]. The proof is completed.
Despite the uniqueness, problem (1)−(3) is still ill-posed. Hence, a regularization has to resort. We have the following result.
Theorem 3.2. Let ϕ, f, uε be as in Theorem 2.1.
a) If we can find a u and a subsequence (uεj ) in (C[0, T ]; L2 (0, π)) such that
uεj → u in C([0, T ]; L2 (0, π)),
then u is the unique solution of Problem (1)−(3).
b) If problem (1)−(3) has a weak solution
u ∈ W (defined in Theorem 3.1)
R T P∞ 2sn2 2
which satisfies 0
fn (u)(s)ds < ∞. Then
n=1 e
µ 2
¶
√
3k T (T − t) t
²
²T
ku(·, t) − u (·, t)k ≤ M exp
2
R T P∞ 2sn2 2
for every t ∈ [0, T ], where M = 3ku(0)k2 + 6π 0
fn (u)(s)ds
n=1 e
²
and u is the unique solution of problem (4)−(6).
Proof. a) We present an outline of the proof.
The function uεj satisfiesP(4), (5) (with ε replaced by P
εj ) subject to the
∞
∞
j
initial condition uεj (x, 0) =
ϕ
sin
nx
and
u(x,
0)
=
n=1 n
n=1 un (0) sin nx.
One gets (see [5])
¸
Z t
2
∞ ·
X
e−tn
−tn2 j
εj
εj
e
ϕn +
fn (u )ds sin nx.
u (x, t) =
s/T
−sn2
ε
+
e
0
n=1
j
240
D. D. Trong et al.
Letting ε ↓ 0, we shall get
¶
Z t
∞ µ
X
−(t−s)n2
−tn2
u(x, t) =
fn (u) ds sin nx.
un (0) +
e
e
0
n=1
On the other hand, letting ε ↓ 0 in (6), we get u(x, T ) = ϕ(x). Hence u is the
solution of problem (1)–(3) as desired.
b) The exact solution u satisfies
¶
Z T
∞ µ
X
−(t−s)n2
−(t−T )n2
u(x, t) =
fn (u)(s) ds sin nx
(22)
ϕn − e
e
u(x, T ) =
n=1
∞ µ
X
t
e
−T n2
un (0) +
Z
T
e
¶
∞
X
fn (u)(s)ds sin nx =
ϕn sin nx,
−(T −s)n2
0
n=1
n=1
where we recall un (0) = π2 hu(x, 0), sin nxi (see [5]). Hence
Z T
2
−T n2
e−(T −s)n fn (u)(s) ds = ϕn .
e
un (0) +
(23)
0
From (7), (22) and (23), we get
¯
Z T
s
−tn2
−tn2
¯
T e
²
²e
²
ϕ
−
fn (u)(s) ds
|un (t) − un (t)| = ¯¯ −T n2
s
n
(² + e−T n2 )
e
e−sn2 (² T + e−sn2 )
t
¯
Z T
2
¡
¢ ¯
e−tn
²
fn (u)(s) − fn (u )(s) ds¯¯
−
s
2
T + e−sn
²
t
¯
Z T
2
¯ ²e−tn2
²e−tn
¯
≤¯
un (0) +
f (u)(s) ds
−sn2 (² + e−T n2 ) n
² + e−T n2
0 e
¯
Z T
s
2
¯
² T e−tn
¯
(24)
−
f
(u(s)ds
s
n
2
2
¯
e−sn (² T + e−sn )
t
Z T
2
¯
¯
e−tn
¯fn (u)(s) − fn (u² )(s)¯ds.
+
s
2
−sn
²T + e
t
From (20), (21) and (24), we have
|un (t) − u²n (t)|
¯
¯
¯
¯
Z T
¯ fn (u)(s) ¯
s
t
s ¯ fn (u)(s) ¯
−
¯
¯
≤²·²
|un (0)| +
²·²
² T .² T T ¯¯ −sn2 ¯¯ ds
¯ e−sn2 ¯ ds +
e
0
t
Z T
¯
¯
s
t
+
² T − T ¯fn (u)(s) − fn (u² )(s)¯ds
t
¯
Z T¯
Z T
¯ fn (u)(s) ¯
¯
s ¯
t
t
t
¯
¯
T
²− T ¯fn (u)(s) − fn (u² )(s)¯ds.
≤ ² T |un (0)| + 2² T
¯ e−sn2 ¯ ds + ²
0
t
t
T
−1
Z
T
t
T
−1
1-D Backward Heat Problem
241
We have in view of the inequality (a + b + c)2 ≤ 3(a2 + b2 + c2 )
ku(·, t) − u² (·, t)k2 =
∞
πX
|un (t) − u²n (t)|2
2 n=1
µZ T ¯
∞
∞
¯ ¶2
X
t
3π X 2 t
1
¯
¯
2
2
≤
² T
² T |un (0)| + 6π
¯ −sn2 fn (u)(s)¯ds
2 n=1
e
0
n=1
¶2
µ
Z T
∞
¯
s ¯
3π X 2 t
²
−
² T ¯fn (u)(s) − fn (u )(s)¯ds
² T
+
2 n=1
t
Z TX
∞
2
2 Tt
2
2 Tt
≤ 3² ku(0)k + 6πT ²
e2sn fn2 (u)(s) ds
0
t
Z
T
n=1
s
²−2 T kf (·, s, u(·, s)) − f (·, s, u² (·, s))k2 ds
+ 3(T − t)²2 T
t
µ
Z TX
∞
2
2 Tt
2
≤²
3ku(0)k + 6πT
e2sn fn (u(s))2 ds
0
2
+ 3k T
Z
T
²
−2 Ts
t
n=1
²
2
¶
ku(·, s) − u (·, s)k ds .
Hence
²
−2 Tt
²
2
2
ku(·, t) − u (·, t)k ≤ M + 3k T
where M = 3ku(0)k2 +6πT
ity, we get
R T P∞
0
n=1
Z
T
t
s
²−2 T ku(·, s) − u² (·, s)k2 ds ,
2
e2sn fn2 (u)(s)ds. Using Gronwall’s inequal-
t
²−2 T ku(·, t) − u² (·, t)k2 ≤ M e3k
2 T (T −t)
.
This completes the proof of Theorem 3.2.
Remark 3.3.
1. From Part a), we conclude that if problem (1)−(3) does not have any
exact solution u ∈ W , then one has
n
o
ε
lim inf max ku (·, t) − ψ(·, t)k > 0
ε↓0
0≤t≤T
for every ψ ∈ C([0, T ]; L2 (0, π)).
2. If f (x, t, u) ≡ 0, we have the linear homogeneous problem, the error
estimate is as in [2].
3. From (23), one has
Z T
2
T n2
un (0) = ϕn e
−
esn fn (u)(s)ds .
0
242
D. D. Trong et al.
¡ R T sn2
¢2
P
P
2 2T n2
e fn (u)(s)ds < ∞. Hence the as< ∞, then ∞
If ∞
n=1
n=1 ϕn e
0
sumptions of f in Theorem 3.2 are reasonable.
One has
Theorem 3.4. Let ϕ, f be as in Theorem 2.1 and letR u P
∈ W be a solution of
T
∂u
2sn2 2
2
2
problem (1)−(3) such that ∂t ∈ L ((0, T );L (0, π)) and 0 ∞
fn (u)(s)ds <
n=1 e
∞. Then for all ² > 0 there exists a t² such that
ku(·, 0) − u² (·, t² )k ≤
√
4
µ µ ¶¶− 41
1
8C T ln
,
²
√
4
where
C = max
½
exp
µ
3k 2 T 2
2
¶µ
¶12 ¾
Z TX
∞
2sn2 2
3ku0 (·, 0)k + 6πT
e fn (u)(s)ds , N (25)
2
0
n=1
and
N=
µZ
T
0
Proof. We have u(x, t) − u(x, 0) =
°2 ¶1
°
2
°
° ∂u
° (·, s)° ds .
°
° ∂t
Rt
∂u
(x, s)ds.
0 ∂s
(26)
It follows that
°2
Zt °
° ∂u
°
° ds = N 2 t.
ku(·, 0) − u(·, t)k ≤ t °
(·,
s)
° ∂t
°
2
0
Using Theorem 3.2 and (25)–(26), we have
√
t
ku(·, 0) − u² (·, t)k ≤ ku(·, 0) − u(·, t)k + ku(·, t) − u² (·, t)k ≤ C( t + ² T ).
For every ², there exists t² such that
ln t > − 1t for every t > 0, we get
²
√
t²
t² = ² T , i.e.,
ku(·, 0) − u (·, t² )k ≤
√
4
ln t²
t²
=
2 ln ²
.
T
Using inequality
µ µ ¶¶− 41
1
8C T ln
.
²
√
4
This completes the proof of Theorem 3.4.
Remark 3.5. Using the Galerkin method (see, e.g., [9]), we can show that the
assumption on ut holds if u(·, 0) ∈ H01 (0, π).
In the case of nonexact data, one has
1-D Backward Heat Problem
243
Theorem 3.6. Let ϕ, f be as in Theorem 2.1. Assume that the exact solution
u of (1)−(3) corresponding to ϕ satisfies
u ∈ W,
∂u
∈ L2 ((0, T ); L2 (0, π))
∂t
R T P∞ 2sn2 2
fn (u)(s)ds < ∞. Let ϕ² ∈ L2 (0, π) be a measured data such
and 0
n=1 e
that kϕ² − ϕk ≤ ². Then there exists a function u² satisfying
¶
µ 2
√
3k T (T − t) t
²
ku (·, t) − u(·, t)k ≤ (2 + M ) exp
² T , for every t ∈ (0, T )
2
µ µ ¶¶− 41
√
√
¡
¢
1
4
4
²
ku (·, 0) − u(·, 0)k ≤ 8 T ln
exp(k 2 T 2 ) + C ,
²
R
P
T
∞
2sn2 2
fn (u)(s)ds, and C is defined in
where M = 3ku(·, 0)k2 + 6πT 0
n=1 e
(25)–(26).
Proof. Let v ² be the solution of problem (4)−(6) corresponding to ϕ and let w ²
be the solution of problem (4)−(6) corresponding to ϕ² , where ϕ, ϕ² are in right
hand side of (6).
Using Theorem 3.4, there exists a t² such that
√
t²
(27)
t² = ² T
and
²
kv (·, t² ) − v(·, 0)k ≤
Put
u² (·, t) =
(
√
4
√
4
8C T
µ µ ¶¶− 14
1
ln
.
²
w² (·, t), 0 < t < T
w² (·, t² ), t = 0 .
Using Theorem 3.2 and Step 3 in Theorem 2.1, we get
ku² (·, t) − u(·, t)k ≤ kw ² (·, t) − v ² (·, t)k + kv ² (·, t) − u(·, t)k
µ 2
¶
√
t
3k T (T − t)
≤ (2 + M ) exp
²T ,
2
for every t ∈ (0, T ). From (27)–(28) and Step 3 in Theorem 2.1, we have
ku² (·, 0) − u(·, 0)k ≤ kw ² (·, t² ) − v ² (·, t² )k + kv ² (·, t² ) − u(·, 0)k
µ µ ¶¶− 14
√
√
t²
1
4
4
2
2
≤ 2² T exp(k T ) + 8C T ln
²
µ µ ¶¶− 14
√
√
¡
¢
1
4
4
≤ 8 T ln
exp(k 2 T 2 ) + C ,
²
where C is defined in (25)–(26). This completes the proof of Theorem 3.6.
(28)
244
D. D. Trong et al.
4. A numerical experiment
We consider the equation
−uxx + ut = f (u) + g(x, t),
where g(x, t) = 2et sin x − e4t sin4 x, u(x, 1) = ϕ0 (x) ≡ e sin x and
 4
u,



− e30 u + e41 ,
f (u) = e30e−1 e41e−1

u + e−1 ,


 e−1
0,
u ∈ [−e10 , e10 ]
u ∈ (e10 , e11 ]
u ∈ (−e11 , −e10 ]
|u| > e11 .
¡ 99 ¢
t
The exact solution
of
the
equation
is
u(x,
t)
=
e
sin
x.
Especially,
u
x, 100 ≡
¡ 99 ¢
u(x) = exp 100 sin x. Let ϕ² (x) ≡ ϕ(x) = (² + 1)e sin x. We have
kϕ² − ϕk2 =
µZ
π
2 2
2
² e sin x dx
0
¶ 12
= ²e
³ π ´ 12
2
.
¡ 99 ¢
≡ u² (x) having the following
We find the regularized solution u² x, 100
form:
u² (x) = vm (x) = w1,m sin x + w2,m sin 2x + w3,m sin 3x ,
where v1 (x) = (² + 1)e sin x, w1,1 = (² + 1)e, w2,1 = 0, w3,1 = 0 and

w
2
i,m+1
=
e−tm+1 i
2 wi,m
²+e−tm i
 t = 1 − am,
m
a=
−
2
π
1
,
40000
2
R tm
e−tm+1 i
s
2
tm+1 ² tm
+e−si
¡R π
0
¢
4
(vm
(x) + g(x, s)) sin ix dx ds
m = 1, 2, . . . , 4000,
i = 1, 2, 3.
Put a² = ku² −uk the error between the regularization solution u² and the exact
solution u. Letting ² = ²1 = 10−5 , ² = ²2 = 10−7 , ² = ²3 = 10−11 , numerical
results are given as follows.
²
²1 = 10−5
²2 = 10−7
u²
a²
2.430605996 sin x − 0.000171846090 sin 3x 0.32664942510
2.646937077 sin x − 0.002178680692 sin 3x 0.05558566020
²3 = 10−11 2.649052245 sin x − 0.004495263004 sin 3x 0.05316693437
1-D Backward Heat Problem
245
Acknowledgement. The authors wish to thank the referee for their valuable
criticisms and suggestions, leading to the present improved version of our paper.
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Received May 29, 2005; revised February 13, 2006