Proc. Indian Acad. Sci. (Math. Sci.) Vol. 121, No. 3, August 2011, pp. 349–367.
c Indian Academy of Sciences
Value functions for certain class of Hamilton Jacobi equations
ANUP BISWAS1, RAJIB DUTTA1 and PROSENJIT ROY2
1 TIFR Centre for Applicable Mathematics, Sharadanagar, Chikkabommasandra,
Bangalore 560 065, India
2 Institut für Mathematik, Universität Zürich, Winterthurerstrasse 190, 8057 Zürich,
Switzerland
E-mail: [email protected]; [email protected];
[email protected]
MS received 1 September 2010; revised 11 October 2010
Abstract.
We consider a class of Hamilton Jacobi equations (in short, HJE) of type
ut +
eu
1
|u x1 |2 + · · · + |u xn−1 |2 + |u xn |m = 0,
2
m
in Rn × R+ and m > 1, with bounded, Lipschitz continuous initial data. We give a
Hopf-Lax type representation for the value function and also characterize the set of
minimizing paths. It is shown that the minimizing paths in the representation of value
function need not be straight lines. Then we consider HJE with Hamiltonian decreasing
in u of type
u t + H1 u x1 , . . . , u xi + e−u H2 u xi+1 , . . . , u xn = 0
where H1 , H2 are convex, homogeneous of degree n, m > 1 respectively and the
initial data is bounded, Lipschitz continuous. We prove that there exists a unique viscosity solution for this HJE in Lipschitz continuous class. We also give a representation
formula for the value function.
Keywords.
function.
Hamilton Jacobi equation; viscosity solution; minimizing paths; value
1. Introduction
Let H : R × Rn → R be a continuous function and g : Rn → R be a bounded, Lipschitz
continuous function. Consider the Hamilton Jacobi equation (HJE)
u t + H (u, Du) = 0 for (x, t) ∈ Rn × R+ ,
u(x, 0) = g(x),
(1.1)
∂u
where Du = ∂ x1 , . . . , ∂∂u
xn . This problem has been studied extensively in [2,4,5] and the
references therein. Some of the open issues related to problem (1.1) are the following:
(i) Does there exist a viscosity solution for HJE (1.1) and is the solution unique?
(ii) Is it possible to give a Hopf-Lax type representation for the value function (solution)?
Are the minimizing paths always straight lines?
349
350
Anup Biswas, Rajib Dutta and Prosenjit Roy
The question (ii) was also asked in [3]. It has been shown in [5] that apart from the usual
hypothesis in H if the H is increasing in u then (1.1) has a unique viscosity solution.
But when the Hamiltonian is decreasing in u then a few results are known. Even in the
general set up when H (u, p) depends on u, getting a Hopf-Lax type representation for u
is quite difficult. For n = 1, an explicit Hopf-Lax type representation is given in [1]. But
the dimensionality was crucially used there and its implementation for higher dimension
is quite difficult. Here we consider a certain class of HJE
ut +
eu
1
|u x1 |2 + · · · + |u xn−1 |2 + |u xn |m = 0
2
m
and give a Hopf-Lax type representation for u(x, t). We also characterize the set of minimizers for the representation and show that the minimizing paths could be of exponential
types. Apart from minimizer characterizations, it is also possible to give the Hopf-Lax
type formula for solution u(x, t) for a more general Hamiltonian (see Remark 2.2). Even
though it is not known, in general, whether the viscosity solution exists or not when
the Hamiltonian is decreasing in u, we give a class of Hamiltonian decreasing in u and
show that there exists a unique viscosity solution for the corresponding HJE. For the
convenience of the reader we recall the definition of viscosity solution,
DEFINITION 1.1
Let U be a bounded, continuous function in Rn × R+ .
1. U is said to be a subsolution of (1.1) if for any (x0 , t0 ) ∈ Rn × R+ , ν ∈ C 1 (Rn × R+ )
such that (x0 , t0 ) is a local maximum for U − ν, with U (x0 , t0 ) = ν(x0 , t0 ), then at
(x0 , t0 ), νt + H (ν, Dν) ≤ 0 and limt→0 U (x, t) ≤ g(x).
2. U is said to be a supersolution of (1.1) if for any (x0 , t0 ) ∈ Rn ×R+ , ν ∈ C 1 (Rn ×R+ )
such that (x0 , t0 ) is a local minimum for U − ν, with U (x0 , t0 ) = ν(x0 , t0 ), then at
(x0 , t0 ), νt + H (ν, Dν) ≥ 0 and limt→0 U (x, t) ≥ g(x).
U is said to be a viscosity solution if it is both sub and supersolution.
The paper is organized as follows: In §2 we consider HJE with Hamiltonian increasing
in u. We prove that there exists a unique viscosity solution for the HJE. Then we characterize the minimizing paths in the representation of u(x, t) in §3. In §4 we consider
HJE with Hamiltonian decreasing in u and prove that there exists a unique viscosity solution for the corresponding HJE. In §5 we provide an example to strengthen the arguments
of §3.
2. Value function for HJE increasing in u
Let g : Rn → R be a bounded, Lipschitz continuous function. Consider the HJE
u t + H (u, Du) = 0 for (x, t) ∈ Rn × R+ ,
u(x, 0) = g(x),
(2.1)
Value functions for HJE
∂u
∂u
∂ x1 , . . . , ∂ xn
where Du =
H (u, p) =
and
n−1
1 2
351
| pi |2 +
i=1
eu
| pn |m for (u, p) ∈ R × Rn .
m
We assume that m > 1 so that H (u, ·) becomes a convex function in Rn . For x ∈ Rn and
t ≥ 0, we define
tx = {γ : [0, t] → Rn : γ is absolutely continuous and γ (t) = x}.
Define
L(x1 , . . . , xn−1 ) =
1
|x1 |2 + · · · + |xn−1 |2 for (x1 , . . . , xn−1 ) ∈ Rn−1 .
2
Let γ̄ (l) = (γ1 (l), . . . , γn−1 (l)). For (x, t) ∈ Rn × R+ we define
1
t
˙
u(x, t) = infx (m − 1) ln e m−1 {g(γ (0))+ 0 L(γ̄ (θ))dθ}
t
t
m
1
1 t
˙
+
|γ̇n | m−1 e m−1 θ L(γ̄ (l))dl dθ .
m 0
(2.2)
Our main result is the following.
Theorem 2.1. u(x, t) defined in (2.2) is a Lipschitz continuous function and is a viscosity
solution of (2.1).
Proof. We prove the theorem by subsequent lemmas.
Lemma 2.1. Let M1 ≤ g ≤ M2 for all x ∈ Rn . Then M1 ≤ u(x, t) ≤ M2 for all
(x, t) ∈ Rn × R+ . Also there exists a path φ ∈ tx such that
t
˙
1
u(x, t) = (m − 1) ln e m−1 {g(φ(0))+ 0 L(φ̄(θ))dθ }
t
˙
m
1
1 t
+
|φ̇n | m−1 e m−1 θ L(φ̄(l))dl dθ .
m 0
Proof. Taking γ ≡ x in (2.2) we have u(x, t) ≤ g(x) ≤ M2 . Again
u(x, t) ≥ infx g(γ (0)) ≥ M1 .
t
This proves the first part. Let {γk } be the minimizing sequence for (2.2). Since
t
0
and
0
n−1
1 xi − (γk )i (0) 2
˙
L(γ̄k (θ ))dθ ≥
t
2
t
i=1
t
x − (γ ) (0) m
m
k n
n
m−1
|(γ̇k )n (θ )| m−1 dθ ≥ t ,
t
352
Anup Biswas, Rajib Dutta and Prosenjit Roy
we note that {γk (0)} remains in a compact set. Therefore we can get a subsequence,
denoted by the same index {γk (0)}, and a path φ ∈ tx such that
γk → φ uniformly in [0, t],
γ̄˙k → φ̄˙ weakly in L 2 ([0, t], Rn−1 ),
m
(γ̇k )n → φ̇n weakly in L m−1 ([0, t], R).
Again we note that
t
m
1
1 t
˙
|γ̇n | m−1 e m−1 θ L(γ̄ (l))dl dθ
m 0
is convex in (·, . . . , ·, γn ) and (γ1 , . . . , γn−1 , ·). Since lim inf{a(n) + b(n)}
lim inf a(n) + lim inf b(n), it is easy to see that φ is a minimizing path.
Lemma 2.2 (Dynamic Programming Principle). For 0 ≤ s < t, we have
t−s
1
L(γ̄˙ (θ))dθ}
m−1 {u(γ (0),s)+ 0
(m
−
1)
ln
e
u(x, t) = inf
x
t−s
t−s
m
1
1 t−s
L(γ̄˙ (l))dl
θ
m−1
m−1
+
|γ̇n |
e
dθ .
m 0
≥
(2.3)
Proof. Let v(x, t) denote the right-hand side of (2.3). From Lemma 2.1, we can find
φ ∈ tx such that
t
˙
1
u(x, t) = (m − 1) ln e m−1 {g(φ(0))+ 0 L(φ̄(θ))dθ}
t
˙
m
1
1 t
L(φ̄(l))dl
θ
m−1
m−1
+
|φ̇n |
e
dθ .
m 0
Therefore
1
t
˙
1
s
˙
m
m−1
s
1
m−1 θ
˙
L(φ̄(l))dl
t
˙
L(φ̄(l))dl
0 L(φ̄(θ))dθ}
eu(x,t)/(m−1) = e m−1 {g(φ(0))+
t
˙
m
1
1 t
+
|φ̇n | m−1 e m−1 θ L(φ̄(l))dl dθ
m 0
t
˙
0 L(φ̄(θ))dθ+ s L(φ̄(θ))dθ}
= e m−1 {g(φ(0))+
s
t
˙
m
1
1
+
|φ̇n | m−1 e m−1 θ L(φ̄(l))dl dθ
m 0
t
˙
m
1
1 t
|φ̇n | m−1 e m−1 θ L(φ̄(l))dl dθ
+
m
s
s
˙
1
− e m−1 {g(φ(0))+ 0 L(φ̄(θ))dθ}
1
+
m
1
+
m
s
|φ̇n |
e
0
t
s
m
1
|φ̇n | m−1 e m−1
θ
1
dθ × e m−1
dθ.
t
s
˙
L(φ̄(θ))dθ
Value functions for HJE
353
Define ψ(θ ) = φ(θ + s) for θ ∈ [0, t − s]. Then
t−s
1
˙
L(ψ̄(θ))dθ}
0
eu(x,t)/(m−1) ≥ e m−1 {u(ψ(0),s)+
t−s
t−s
˙
m
1
1
+
|ψ˙n (θ )| m−1 e m−1 θ L(ψ̄(l))dl dθ
m 0
≥ ev(x,t)/(m−1) .
x such that
Therefore u(x, t) ≥ v(x, t). Consider > 0. Then there exists φ ∈ t−s
e
v(x,t)+
m−1
t−s
1
˙
≥ e m−1 {u(φ(0),s)+ 0 L(φ̄(θ))dθ}
t−s
˙
m
1
1 t−s
+
|φ˙n (θ )| m−1 e m−1 θ L(φ̄(l))dl dθ.
m 0
(2.4)
φ(0)
Again using Lemma 2.1 we have an absolutely continuous path ψ ∈ s
e
u(φ(0),s)
m−1
= e
such that
s
˙
1
m−1 {g(ψ(0))+ 0 L(ψ̄(θ))dθ}
+
1
m
s
m
1
|ψ˙n (θ )| m−1 e m−1
s
θ
˙
L(ψ̄(l))dl
dθ.
(2.5)
0
Now define a path χ ∈ tx by
ψ(θ ),
for θ ∈ [0, s],
χ (θ ) =
φ(θ − s), for θ ∈ (s, t].
Then using (2.4), (2.5) and the same steps as above we have
e
v(x,t)+
m−1
u(x,t)
≥ e m−1 ,
> 0 being arbitrary we have v(x, t) ≥ u(x, t). This completes the proof.
Lemma 2.3. The function u defined in (2.2) is Lipschitz in Rn × R+ and limt→0 u(x, t) =
g(x).
Proof. Let x1 , x2 ∈ Rn , and from Lemma 2.1 there exists φi ∈ txi (i = 1, 2) such that
e
u(xi ,t)
m−1
t
1
˙
= e m−1 {g(φi (0))+ 0 L(φ̄i (θ))dθ}
t
˙
m
1
1 t
+
|φ̇i n (θ )| m−1 e m−1 θ L(φ̄i (l))dl dθ.
m 0
1
t
˙
Since u(x, t) ≤ M2 we note that e m−1 0 L(φ̄i (θ))dθ ≤ e
x
constant of e m−1 in [M1 , M2 ]. Define ψi ∈ txi by
t
φ̇2 dθ for θ ∈ [0, t]
ψ1 (θ ) = x1 −
θ
and
ψ2 (θ ) = x2 −
θ
t
φ̇1 dθ for θ ∈ [0, t].
M2 +|M1 |
m−1
. Let K 1 be the Lipschitz
354
Anup Biswas, Rajib Dutta and Prosenjit Roy
Therefore ψ1 (0) − φ2 (0) = x1 − x2 and ψ2 (0) − φ1 (0) = x2 − x1 . Hence
t
1
eu(x1 ,t)/(m−1) − eu(x2 ,t)/(m−1) ≤ e m−1 {g(ψ2 (0))+
1
L(ψ̄˙ 2 (θ))dθ}
0
t
− e m−1 {g(φ1 (0))+
1
= e m−1
≤ e
t
0
L(φ̄˙ 1 (θ))dθ
M2 +|M1 |
m−1
0
L(φ̄˙ 1 (θ))dθ}
g(ψ2 (0))
g(φ1 (0)) e m−1 − e m−1
K 1 K 2 |x1 − x2 |,
where K 2 is the Lipschitz constant of g(·). Similarly we can show that
eu(x2 ,t)/(m−1) − eu(x1 ,t)/(m−1) ≤ e M2 /(m−1) K 1 K 2 |x1 − x2 |.
Therefore there exists a constant K such that
|eu(x1 ,t)/(m−1) − eu(x2 ,t)/(m−1) | ≤ K |x1 − x2 |.
(2.6)
From (2.3) we have u(x, t) ≤ u(x, s) for 0 ≤ s < t. From Lemma 2.1 there exists
x such that
φ ∈ t−s
u(x,t)
t−s
1
˙
e m−1 = e m−1 {u(φ(0),s)+ 0 L(φ̄(θ))dθ}
t−s
˙
m
1
1 t−s
+
|φ̇n (θ )| m−1 e m−1 θ L(φ̄(l))dl dθ.
m 0
Therefore from Lemma 2.2 we have
eu(x,t)/(m−1) − eu(x,s)/(m−1)
1
t−s
˙
u(x,s)
= e m−1 {u(φ(0),s)+ 0 L(φ̄(θ))dθ} − e m−1
t−s
˙
m
1
1 t−s
+
|φ̇n (θ )| m−1 e m−1 θ L(φ̄(l))dl dθ
m 0
t−s
u(φ(0),s)
u(x,s) ˙
1
≥ e m−1 0 L(φ̄(θ))dθ e m−1 − e m−1
t−s
u(x,s) ˙
1
+ e m−1 e m−1 0 L(φ̄(θ))dθ − 1
m
1 t−s
+
|φ̇n (θ )| m−1 dθ
m 0
t−s
K4
˙ ))dθ
≥ −K 3 |x − φ(0)| +
L(φ̄(θ
m−1 0
m
1 t−s
+
|φ̇n (θ )| m−1 dθ (for some constant K 3 , K 4 )
m 0
φ̄(t − s) − φ̄(0)
K 4 (t − s)
≥ −K 3 |x − φ(0)| +
L
m−1
t −s
m
t − s φn (t − s) − φn (0) m−1
+
m t −s
m
K4
1
L( p̄) − | pn | m−1
≥ −(t − s) sup sup p, η −
m−1
m
η≤K 3 p
= −K 5 |t − s| (for some constant K 5 ).
Value functions for HJE
355
This proves that for any s, t ∈ R+ and x ∈ Rn ,
|eu(x,t)/(m−1) − eu(x,s)/(m−1) | ≤ K 5 |t − s|.
(2.7)
Now let (x1 , s), (x2 , t) ∈ Rn × R+ . Then
|eu(x1 ,s)/(m−1) − eu(x2 ,t)/(m−1) | ≤ |eu(x1 ,s)/(m−1) − eu(x2 ,s)/(m−1) |
+ |eu(x2 ,s)/(m−1) − eu(x2 ,t)/(m−1) |
≤ max{K , K 5 }(|x1 − x2 | + |t − s|).
u
Since the range of u(·, ·) is in [M1 , M2 ] and e m−1 is not 0, we can have a constant M
such that
|u(x1 , s) − u(x2 , t)| ≤ M (|x1 − x2 | + |t − s|) .
(2.8)
Lemma 2.4. u is a viscosity solution of HJE (2.1).
Proof. Let (x0 , t0 ) ∈ Rn × R+ and ν be a C 1 -function such that u(x0 , t0 ) = ν(x0 , t0 )
and u − ν has a local maximum at (x0 , t0 ). Let B be a ball centered at (x0 , t0 ) such that
u(x, t) ≤ ν(x, t) for (x, t) ∈ B. Let 0 ≤ t < t0 , λi > 0, h = t0 − t and x = x0 − h · λ
with λ = (λ1 , . . . , λn ). Define
def
¯ tx = {γ ∈ tx : γ is a straight line}.
Therefore for small h we have
e
ν(x0 ,t0 )
m−1
= e
u(x0 ,t0 )
m−1
h
1
e m−1 {u(γ (0),t0 −h)+
≤ inf
x
0
L(γ̄˙ (θ))dθ}
¯ h 0
1
+
m
⎡
h
|γ̇n |
m
m−1
e
h
1
m−1 θ
L(γ̄˙ (l))dl
dθ
0
1
= inf ⎣e m−1 {u(y,t0 −h)+h L(
x0 −y
h
)}
y∈Rn
⎛
⎞⎤
x0 −y
m
1
m − 1 (x0 )n − yn m−1 ⎝ e m−1 h L( h ) − 1 ⎠⎦
+
m h
L( x0 −y )
h
≤
e
1
m−1 {u(x,t0 −h)+h L(λ̄)}
m
m−1
+
|λn | m−1
m
e
1
m−1 {ν(x,t0 −h)+h L(λ̄)}
m
m−1
|λn | m−1
+
m
≤
1
1
e m−1 h L(λ̄) − 1
L(λ̄)
e m−1 h L(λ̄) − 1
L(λ̄)
.
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Anup Biswas, Rajib Dutta and Prosenjit Roy
Therefore if ν is C 2 using Taylor’s expansion we have from above
e
ν(x0 ,t0 )
m−1
≤ e
ν(x0 ,t0 )
m−1
n
ν(x0 ,t0 )
1
+
−λi hνxi (x0 , t0 ) − hνt (x0 , t0 ) + h L(λ̄) e m−1
m−1
i=1
1
m
e m−1 h L(λ̄) − 1
m−1
2
|λn | m−1
+ O(h ) +
.
m
L(λ̄)
Hence dividing both sides by h and letting h ↓ 0 we have
n
ν(x0 ,t0 )
1
−λi νxi (x0 , t0 ) − νt (x0 , t0 ) + L(λ̄) e m−1
0 ≤
m−1
i=1
m
1 m−1
|λn | m−1 .
+
m−1 m
Similarly we could choose λi ≤ 0 to get the above estimate. Hence λ being arbitrary
we have
ν(x0 ,t0 ) m − 1
m
− m−1
m−1
≤ 0,
|λn |
νt (x0 , t0 ) + sup λ, Dν(x0 , t0 ) − L(λ̄) − e
m
λ
i.e
1
eν(x0 ,t0 )
|νxi |2 +
|νxn |m ≤ 0.
2
m
n−1
νt (x0 , t0 ) +
i=1
Since the result can be deduced for ν in C 1 , this proves that u is a subsolution.
Now we prove that u is a supersolution. Let (x0 , t0 ) ∈ Rn × R+ be a point and ν be a
function on Rn × R+ such that ν(x0 , to ) = u(x0 , t0 ) and u − ν has a local minimum
at (x0 , t0 ). We have to show that
C1
1
eν(x0 ,t0 )
|νxn |m ≥ 0.
|νxi |2 +
2
m
n−1
νt (x0 , t0 ) +
(2.9)
i=1
If this is not true then there exists θ > 0 such that
1
eν(x,t)
|νxn |m ≤ −θ ∀ (x, t) ∈ B1
|νxi |2 +
2
m
n−1
νt (x, t) +
(2.10)
i=1
and
ν(x, t) ≤ u(x, t) ∀ (x, t) ∈ B1 ,
(2.11)
where B1 is a small closed ball around (x0 , t0 ). From (2.3) we note that for small h > 0
x
there exists a path γ ∈ h 0 such that
γ (l) ∈ B1 ∀ l ∈ [0, h]
Value functions for HJE
357
and
e
u(x0 ,t0 )
m−1
=
h
1
e m−1 {u(γ (0),t0 −h)+
+
1
m
h
0
m
L(γ̄˙ (θ))dθ}
1
|γ̇n | m−1 e m−1
h
θ
L(γ̄˙ (l))dl
dθ .
(2.12)
0
Define φ(s) = γ (s − t0 + h) for s ∈ [t0 − h, t0 ]. Then
e
ν(φ(t0 ),t0 )
m−1
=
t0
t0 −h
=
1
− e m−1
t0
t0 −h
t
˙
{ν(φ(t0 −h),t0 −h)+ t 0−h L(φ̄(θ))dθ}
0
˙
d 1 {ν(φ(s),s)+ t0 L(φ̄(θ))dθ}
s
e m−1
ds
t
˙
1
1
0
˙
e m−1 {ν(φ(s),s)+ s L(φ̄(θ))dθ} [νt + Dν, Dφ − L(φ̄)]ds.
m−1
From (2.10) we have
˙
νt (φ(s), s) + Dν(φ(s), s), Dφ(s) − L(φ̄(s))
−
m
m − 1 − ν(φ(s),s)
e m−1 |φ̇n (s)| m−1 ≤ −θ,
m
i.e.
ν(φ(s),s)
1
˙
e m−1 [νt (φ(s), s) + Dν(φ(s), s), Dφ(s) − L(φ̄(s))]
m−1
≤−
ν(φ(s),s)
m
θ
1
e m−1 + |φ̇n (s)| m−1 ,
m−1
m
i.e.
e
ν(φ(t0 ),t0 )
m−1
1
− e m−1
≤−
t0
t0 −h
+
1
m
t
˙
{ν(φ(t0 −h),t0 −h)+ t 0−h L(φ̄(θ))dθ}
0
t
˙
1
θ
0
e m−1 {ν(φ(s),s)+ s L(φ̄(θ))dθ}
m−1
t0
t0 −h
m
1
|φ̇n (s)| m−1 e m−1
t
s
0
˙
L(φ̄(l))dl
ds.
Therefore using (2.11) and (2.12) we have
t0
t
˙
1
θ
0
0≤−
e m−1 {ν(φ(s),s)+ s L(φ̄(θ))dθ} .
m
−
1
t0 −h
This is a contradiction. Therefore (2.9) is true. Hence u is a supersolution.
358
Anup Biswas, Rajib Dutta and Prosenjit Roy
Remark 2.1. We note that the Hamiltonian in (2.1) is increasing in u and so the viscosity
solution of (2.1) is unique. Hence u is the unique solution of (2.1) and given by (2.2).
Remark 2.2. Let 1 < i < n and H1 : Ri → R+ , H2 : Rn−i → R+ be two convex
homogeneous functions of order n, m > 1 respectively that vanish only at 0 and have
homogeneous Legendre transformations. It is easy to note that we can give a Hopf-Lax
type representation for the viscosity solution of HJE of type
u t + H1 (u x1 , . . . , u xi ) + eu H2 (u xi+1 , . . . , u xn ) = 0,
u(x, 0) = g(x).
Let L 1 , L 2 be the Legendre transformations of H1 , H2 respectively. Then
t
1
u(x, t) = infx (m − 1) ln e m−1 {g(γ (0))+ 0 L 1 (χ̇1 (θ))dθ}
t
+
1
m−1
t
1
L 2 (χ̇2 (θ ))e m−1
t
θ
L 1 (χ̇1 (l))dl
dθ ,
0
where χ1 (·) denotes the path corresponding to the first i coordinates of γ (·) and χ2 (·)
denotes the path corresponding to the rest of the coordinates of γ (·).
3. Characterization for minimizing paths
For γ ∈ tx define
def
J (γ ) =
t
1
e m−1 {g(γ (0))+
1
+
m
t
|γ̇n |
0
L(γ̄˙ (θ))dθ}
m
m−1
e
t
1
m−1 θ
L(γ̄˙ (l))dl
dθ .
(3.1)
0
The minimizer of J defines u and from Lemma 2.1 we note that there exists a minimizer
Z (·) of J in tx . In this section we deduce some characterizations for the minimizing
paths that would reduce the set of controls in (2.2). Let α(·) be a smooth curve from
[0, t] to R with the property that α(0) = 0 = α(t). Let τ > 0 and we define Yτ (l) =
(Z 1 (l), . . . , Z n−1 (l), Z n (l) + τ α(l)). Then
t
˙
1
J (Yτ ) = e m−1 {g(Yτ (0))+ 0 L(Ȳτ (θ))dθ}
1
+
m
t
| Ż n + τ α̇|
m
m−1
e
t
1
m−1 θ
L(Ȳ˙τ (l))dl
dθ
0
First we note that if Z i (0) = xi for some i, 1 ≤ i ≤ n, then Z i (l) = xi for all l ∈ [0, t]
since it is a minimizer of J . So it is better to assume that Z i (0) = xi just by removing all
those i for which this does not hold. Z being a minimizer, we have
dJ (Yτ ) = 0.
dτ τ =0
Value functions for HJE
359
Therefore from the above we have
t
t
˙
1
1
| Ż n | m−1 sign( Ż n (θ ))α̇(θ )e m−1 θ L( Z̄ (l))dl dθ = 0,
0
α being arbitrary and we have
1
1
| Ż n | m−1 sign( Ż n (θ ))e m−1
t
θ
L( Z̄˙ (l))dl
= constant = A
(3.2)
almost everywhere in [0, t]. We can assume Ż n to be continuous in [0, t] and hence (3.2)
holds everywhere in [0, t]. Again sign of Ż n depends on sign of A. If sign of A is negative
Ż n < 0 is everywhere which readily implies that Z n is decreasing in [0, t]. Since Z n (0) =
xn , we can assume that Ż n (l) = 0 for all [0, t]. Also we note that Ż n is monotone in [0, t].
Now taking logarithm on both sides of (3.2) and differentiating we get
d2 Z n (θ )
= Ż n (θ )L( Z̄˙ (θ )),
dθ 2
a.e. in [0, t].
(3.3)
Fix i, 1 ≤ i < n, such that Z i (0) = xi . Then as before perturbing in the i-th direction
we have
(mG + Z n (l) − Z n (0)) Ż i (l) = m Bi ,
where G =
e
g(Z (0))
m−1
Ż n (0)
a.e. in [0, t],
(3.4)
and Bi is some constant. This implies that Ż i is continuous in [0, t].
Now we note that if Bi = 0, Z n being a monotone function we have Ż i = 0 a.e. in [0, t]
what contradicts the assumption that Z i (0) = xi . This implies that Bi = 0 and hence
(mG + Z n (l)− Z n (0)) = 0 and Ż i (l) = 0 for all l ∈ [0, t]. Since (mG + Z n (l)− Z n (0)) is
a monotone function, Ż i is also monotone and hence differentiable. After differentiating
(3.4), we get
Ż n (l) = −
m Bi d2 Z i (l)
, a.e. in [0, t].
( Ż i )2 dl 2
(3.5)
Therefore from (3.3) and (3.5) we get
n−1
d2 Z n (l)
m d2 Z i (l)
=
−
Bi
2
dl 2
dl 2
a.e. in [0, t].
i=1
After integration we have
Ż n (l) − Ż n (0) = −
n−1
m
Bi ( Ż i (l) − Ż i (0)),
2
i=1
i.e.
n−1
i=1
Bi Ż i (l) =
n−1
i=1
Bi Ż i (0) −
2
[ Ż n (l) − Ż n (0)].
m
(3.6)
360
Anup Biswas, Rajib Dutta and Prosenjit Roy
Now using (3.6) we have from (3.4),
n−1
n−1
2
Bi Ż i (0) − ( Ż n (l) − Ż n (0)) (mG + Z n (l) − Z n (0)) = m
Bi2 ,
m
i=1
i.e.
i=1
n−1
n−1
m
m2 2
Bi Ż i (0) + Ż n (0) − Ż n (l) (mG + Z n (l) − Z n (0)) =
Bi .
2
2
i=1
i=1
Define
n−1
m
Bi Ż i (0) + Ż n (0),
C =
2
def
def
D =
i=1
n−1
2 m
2
(3.7)
Bi2 .
(3.8)
i=1
Therefore integrating the above equality we have
D
D
mG + Z n (l) − Z n (0) −
= mG + xn − Z n (0) −
C
C
C
× e− D {C(t−l)−xn +Z n (l)} .
(3.9)
We can rewrite (3.9) as follows:
D
D
= Z n (l) − mG + xn − Z n (0) −
−mG + Z n (0) +
C
C
C
× e− D {C(t−l)−xn +Z n (l)}
= F(Z n (l)) (say).
We show that F is invertible. First we note from (3.4) that
Bi = G Ż i (0) ∀i.
Therefore
m
C
mG − 1 = 2
D
> 0.
n−1
2
i=1 G Ż i (0) + Ż n (0)
m n−1
2
i=1 G Ż i (0)
2
−1
Therefore for the fix l we have
C
C
C
F (z) = 1 + mG − 1 + (xn − Z n (0)) e− D {C(t−l)−xn +z} > 0.
D
D
This implies that F is invertible in z and the inverse depends on l, Z (0), D Z (0). Let us
denote this inverse by F l,Z (0),D Z (0) and hence from (3.9) we have
D
.
(3.10)
Z n (l) = F l,Z (0),D Z (0) −mG + Z n (0) +
C
Value functions for HJE
361
So we define a class of function as follows:
def
Vtx = {F p,q : R+ × R → R : continuous paths},
(3.11)
where the following conditions are satisfied.
(a) l ∈ R+ , p, q ∈ Rn and (q1 , . . . , qn−1 ) = 0.
(b) qn = 0 and if C, D are defined by (3.7), (3.8) respectively, by replacing Z (0), D Z (0)
g( p)
m−1
with p, q respectively. Let G( p, q) = e qn . Then for l ∈ [0, t] the following are
satisfied:
D
D
p,q
mG( p, q) + F
− pn −
l, −mG( p, q) + pn +
C
C
D
= mG( p, q) + xn − pn −
C
C
D
× e− D {C(t−l)−xn +F (l,−mG( p,q)+ pn + C )} .
(c) F p,q 0, −mG( p, q) + pn + CD = pn and F p,q t, −mG( p, q) + pn +
(d) For all l ∈ [0, t], F p,q l, −mG( p, q) + pn + CD = pn − mG( p, q).
p,q
D
C
= xn .
If (q1 , . . . , qn−1 ) = 0 then Vtx denotes all straight lines γ with γ (t) = xn . For
(q1 , . . . , qn−1 ) = 0 we define
V̄tx = {(x1 , . . . , xn−1 , γ ) : γ ∈ Vtx }.
For qn = 0 we define
V̄tx = {(Z 1 , . . . , Z n−1 , xn )},
where Z i denotes straight lines with property Z i (t) = xi . For q other than the above two
cases we define
V̄tx = {(Z 1 , . . . , Z n ) : Z n ∈ Vtx },
where Z i is defined as follows:
t
mG( p, q)qi
ds,
Z i (l) = xi −
(mG(
p,
q) + Z n (s) − pn )
l
if qi = 0 otherwise Z i (l) = xi . The minimizer of J is contained in V̄tx . We also note that
the only straight lines that V̄tx contains is given by the first two cases. At the end of the
paper we will provide an example to show that there are points (x, t) where J does not
attend minimum on the straight lines.
4. Value function for HJE decreasing in u
In this section we consider a Hamiltonian of the form
e−u
1
| pi |2 +
| p n |m ,
H (u, p) =
2
m
n−1
i=1
for (u, p) ∈ R × Rn ,
(4.1)
362
Anup Biswas, Rajib Dutta and Prosenjit Roy
and the HJE
u t + H (u, Du) = 0 for (x, t) ∈ Rn × R+ ,
u(x, 0) = g(x).
(4.2)
Let tx be as before. For (x, t) ∈ Rn × R+ , we define
t
−1
˙
def
u(x, t) = infx −(m − 1) ln e m−1 {g(γ (0))+ 0 L(γ̄ (θ)dθ)}
t
t
m
1
1 t
− m−1
L(γ̇ (l))dl
θ
m−1
−
|γ̇n (θ )|
e
dθ . (4.3)
m 0
(We have taken the convention − ln(a) = ∞ for a ≤ 0.)
Theorem 4.1. The function u defined by (4.3) is Lipschitz continuous and a viscosity
solution of (4.2).
Proof. Proof of the theorem is same as the proof of Theorem 2.1. For convenience
we give some sketch. Lemmas 2.1 and 2.2 could be shown along the same lines. For
Lemma 2. we show only the Lipschitz continuity in t. First we note that u(x, t) ≤ u(x, s)
for 0 ≤ s ≤ t. Again if φ is a minimizing path for u(x, t), then
t−s
−1
e m−1 {u(φ(0),s)+
˙
L(φ̄(θ)dθ)}
0
≥e
−u(x,t)
m−1
−M2
≥ e m−1 .
Therefore
−1
e m−1
t−s
0
˙
L(φ̄(θ)dθ)
≥e
M1 −M2
m−1
.
In a similar manner we can show that
u(x2 ,t) − u(x1 ,t)
≤ K |x1 − x2 | ∀ x1 , x2 ∈ Rn ,
e m−1 − e− m−1
For some constant K , consider
u(x,s)
u(x,t)
u(x,s)
u(φ(0),s)
e− m−1 − e− m−1
t−s
u(φ(0),s)
˙
1
+ e− m−1 1 − e− m−1 0 L(φ̄)dθ
t−s
˙
m
1
1 t−s
+
|φ̇n | m−1 e− m−1 θ L(φ̄)dl dθ
m 0
≥ −K |x − φ(0)|
1 t−s ˙
t−s
˙
−1
+ e m−1 {u(φ(0),s)+ 0 L(φ̄(θ)dθ)} e m−1 0 L(φ̄)dθ − 1
t−s
˙
m
1
1 t−s
|φ̇n | m−1 e− m−1 θ L(φ̄)dl dθ
+
m 0
t−s
−M2
˙ )dθ
≥ −K |x − φ(0)| + e m−1
L(φ̄(θ
0
t−s
M1 −M2 1
m
m−1
m−1
|φ̇n |
.
+e
m 0
e− m−1 − e− m−1 =
Value functions for HJE
363
Rest of the proof is similar to Lemma 2.3. Proof of subsolution and supersolution would
be similar to Lemma 2.4.
We can also characterize the minimizer for (4.3) in the same way as §3. But our goal is
to show that u defined by (4.3) is the unique viscosity solution for (4.2).
4.1 Uniqueness
First we consider the HJE
n−1
1
ut +
(m − 1)m m−1
2(m − 1 − m
1
m−1
u)
|u xi |2 + |u xn |m = 0 for (x, t) ∈ Rn × R+ ,
i=1
u(x, 0) = h(x).
(4.4)
1
m−1
The above HJE is valid whenever m
u = m − 1. Define
s
t − m−1
e
F(t) =
ds ∀ t ∈ R.
1
0 m m−1
(4.5)
Therefore F(·) is an increasing C 1 function. First we show that if u is a viscosity solution
of (4.2) then F(u) is a viscosity solution of (4.4) with initial condition F(g). Let (x0 , t0 ) ∈
Rn × R+ and ν be a C 1 function such that F(u) − ν has a local maximum at (x0 , t0 ) with
F(u(x0 , t0 )) = ν(x0 , t0 ). Then u − F −1 (ν) has a local maximum at (x0 , t0 ). Since u is
subsolution of (4.2), we have
(F −1 (ν(x0 , t0 )))t + H (F −1 (ν(x0 , t0 ), DF −1 (ν(x0 , t0 )) ≥ 0.
A direct calculation shows that
n−1
1
νt +
(m − 1)m m−1
2(m − 1 − m
1
m−1
ν)
|νxi |2 + |νxn |m ≥ 0,
i=1
at the point (x0 , t0 ). We note that F (F −1 (s)) =
1
m−1−m m−1 s
1
(m−1)m m−1
> 0 if s is in the image of
F. This proves that F(u) is a subsolution for (4.4). Similarly we can show that F(u) is a
supersolution for (4.4).
We now prove that viscosity solution of (4.4) is unique whenever the value function is
Lipschitz, bounded with supremum strictly less than m−1
1 . Let u and v be two viscosity
m m−1
solutions (4.4) with the same initial data. Then there exists σ > 0 such that
sup (u − v) ≥ σ > 0.
(4.6)
R n ×R +
Now we consider a function : (Rn )2 × (R̄+ )2 → R defined as follows
x − y2 + |t − s|2
2α 2
− λ(t + s) − (x2 + y2 ),
(x, y, t, s) = u(x, t) − v(y, s) −
(4.7)
364
Anup Biswas, Rajib Dutta and Prosenjit Roy
where α, λ, > 0. Note that, u and v being bounded the supremum of is attended
inside a compact set, say at (x̄, ȳ, t¯, s̄). Again for λ ∈ (0, λ0 ], ∈ (0, 0 ], for λ0 , 0 small,
we can have
σ
(4.8)
(x̄, ȳ, t¯, s̄) ≥ sup (x, x, t, t) ≥ ,
2
R n ×R +
and this implies that u(x̄, t¯) > v( ȳ, s̄). Since (x̄, ȳ, t¯, s̄) ≥ (0, 0, 0, 0), we have
λ(t¯ + s̄) +
x̄ − ȳ2 + |t¯ − s̄|2
+ (x̄2 + ȳ2 ) ≤ C,
2α 2
(4.9)
for some constant C depending on the bounds of u and v. This implies that
1
3
(x̄ + ȳ) = 4 4 (x̄ + ȳ)
1
≤
3
2
2 (x̄ + ȳ)2
+
2
2
1
≤ C1 2 .
(4.10)
Let h(x) be the initial condition for u, v. Therefore from (4.8) we have
σ
≤ u(x̄, t¯) − v( ȳ, s̄)
2
= u(x̄, t¯) − h(x̄) + h(x̄) − v(x̄, t¯) + v(x̄, t¯) − v( ȳ, s̄)
≤ K 1 t¯ + K 2 t¯ + K 2 (x̄ − ȳ + |t − s|),
where K 1 , K 2 are the Lipschitz constants of u, v respectively. Using (4.9) we note that
for small α > 0, t¯ > 0. Similarly we have s̄ > 0. Also note that the choice of α is
independent of , λ.
Now we observe that u − ν has a maximum at (x̄, t¯) for
def
ν(x, t) = v( ȳ, s̄) +
x − ȳ2 + |t − s̄|2
+ λ(t + s̄) + (x2 + ȳ2 ) + K ,
2α 2
where K is a constant that ensures u(x̄, t¯) = ν(x̄, t¯). Since u is a viscosity solution we
have from (4.4),
λ+
t¯ − s̄
x̄ − ȳ
+ H1 (u(x̄, t¯),
+ 2 x̄) ≤ 0,
α2
α2
(4.11)
where
1
H1 (u, p) =
(m − 1)m m−1
1
2(m − 1 − m m−1 u)
n−1
×
| pi |2 + | pn |m ,
for (u, p) ∈
m−1
1
, ∞ × Rn .
m m−1
i=1
We also observe that v − ν has a minimum at ( ȳ, s̄) for
def
ν(y, s) = u(x̄, t¯) −
x̄ − y2 + |t¯ − s|2
− λ(t¯ + s) − (x̄2 + y2 ) + K ,
2α 2
Value functions for HJE
365
where K is a constant that ensures v( ȳ, s̄) = ν( ȳ, s̄). Since v is a viscosity solution of
(4.4), we have
t¯ − s̄
x̄ − ȳ
−λ + 2 + H1 v( ȳ, s̄),
−
2
ȳ
≥ 0.
(4.12)
α
α2
Now subtracting (4.12) from (4.11) we have
x̄ − ȳ
x̄ − ȳ
+ 2 x̄ − H1 v( ȳ, s̄),
− 2 ȳ ≤ 0.
2λ + H1 u(x̄, t¯),
α2
α2
(4.13)
As ↓ 0, we note that x̄, ȳ → 0 from (4.10). Also from (4.9), we note that x̄ − ȳ →
z ∈ Rn , may be for a subsequence, as ↓ 0. u, v being bounded, as ↓ 0 let u(x̄, t¯) → ū
and v( ȳ, s̄) → v̄ where ū ≥ v̄. Therefore letting ↓ 0 in (4.13) we have
z
z
2λ + H1 ū, 2 − H1 v̄, 2 ≤ 0.
α
α
But this is a contradiction as the left-hand side is > 0 due to monotonicity of H1 in u.
Therefore our assumption (4.6) is wrong and hence u ≤ v. Similarly we can prove that
v ≤ u. This also proves that (4.2) has a unique viscosity solution.
Therefore we have the following theorem:
Theorem 4.2. The Hamilton-Jacobi equation (4.2) has a unique bounded, Lipschitz
continuous viscosity solution and this is given by (4.3).
Remark 4.1. Let H1 , H2 be the same as in Remark 2.2. Then
u t + H1 (u x1 , . . . , u xi ) + e−u H2 (u xi+1 , . . . , u xn ) = 0
has a unique viscosity solution. The Hopf-Lax type representation is given by
−1
t
u(x, t) = infx −(m − 1) ln e m−1 {g(γ (0))+ 0 L 1 (χ̇1 (θ))dθ}
t
1
−
m−1
t
−1
L 2 (χ̇2 (θ ))e m−1
t
θ
L 1 (χ̇1 (l))dl
dθ .
0
Remark 4.2. Let H1 , H2 be as above. Then it is easy to check that for α = 0 the unique
viscosity solution of
u t + H1 (u x1 , . . . , u xi ) + eαu H2 (u xi+1 , . . . , u xn ) = 0
has representation
u(x, t) = infx
t
(m − 1) α {g(γ (0))+ t L 1 (χ̇1 (θ))dθ}
0
ln e m−1
α
t
t
α
α
+
L 2 (χ̇2 (θ ))e m−1 θ L 1 (χ̇1 (l))dl dθ .
m−1 0
366
Anup Biswas, Rajib Dutta and Prosenjit Roy
5. Example
Consider the HJE
ut +
|u x1 |2
2
+
eu
2
2 |u x2 |
u(x, 0)
= 0 in R2 × R+
⎫
⎬
= g1 (x1 ) + g2 (x2 ) ⎭
,
(5.1)
where g1 : R → R is defined by
⎧
if |x| ≥ 1,
⎨ 0,
g1 (x) = 1 + x, if − 1 ≤ x ≤ 0,
⎩
1 − x, if 0 ≤ x ≤ 1
and g2 : R → R is defined by
⎧
⎨ 0,
g2 (x) = ln(1 + x),
⎩
ln 2,
Define
and
if x ≤ 0,
if 0 ≤ x ≤ 1,
if 1 ≤ x.
'
(
t x1 − y1 2
u 1 (x, t) = inf g1 (y1 ) + + g2 (x2 )
2
t
y1 ∈R
(5.2)
'
2 (
x
−
y
t
2
2
.
u 2 (x, t) = inf ln eg1 (x1 )+g2 (y2 ) + 2
t
y2 ∈R
(5.3)
If the minimizer of J (defined for HJE (5.1)) is attended on straight lines for all points
(x, t) ∈ R2 × R+ , then
u(x, t) = min{u 1 (x, t), u 2 (x, t)}
would be the viscosity solution of (5.1). We note that u 1 (x, t) = g2 (x2 ) if |x| ≥ 1. Also
for 0 ≤ x ≤ 1, by direct calculation we have for t ≤ 2,
⎧
t
⎪
⎪
⎨ 1 − x1 − 2 + g2 (x2 ), if 0 ≤ x1 ≤ t − 1,
u 1 (x, t) =
2
⎪
⎪
⎩ (1 − x1 ) + g2 (x2 ),
if t − 1 ≤ x1 ≤ 1.
2t
Similarly for x2 ≤ 0, u 2 (x, t) = g1 (x1 ). Consider t > 0 small such that te2 < 1. By
direct calculation we see that for 0 ≤ x2 ≤ 1,
⎧ ⎪
x22
⎪
g
(x
)
⎪
1
1
+
,
if 0 ≤ x2 ≤ eg1 (x1 ) t,
⎨ ln e
2t
u 2 (x, t) =
⎪
t g1 (x1 )
⎪
⎪
, if eg1 (x1 ) t ≤ x2 ≤ 1.
⎩ g1 (x1 ) + ln 1 + x2 − e
2
Now in {(x1 , x2 ) | 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1}, g1 , g2 = 0 and so it is easy to show that
none of u 1 , u 2 is a viscosity solution. So u cannot be a viscosity solution if it is defined
Value functions for HJE
367
as a minimum of u 1 , u 2 . This proves that the minimizer of J may not be attained on
straight lines.
Acknowledgements
The authors would like to thank Prof. Adimurthi and Prof. G D Verrappa Gowda for their
help and fruitful discussions. The authors would also like to thank the referee for his useful
comments. This research is supported in part by the Microsoft Research India Fellowship.
References
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Differential Equations 5 (2000) 97–119
[2] Bardi M and Capuzzo Dolcetta I, Optimal control and viscosity solutions of HamiltonJacobi-Bellman equations, Systems and Control: Foundations and Applications (1997)
(Boston, MA: Birkhäuser Boston Inc.)
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