Semiconcavity
• It may seem reasonable to define a “weak” solution of
in Rn × (0, +∞)
ut + H(Du) = 0,
(1)
u=g
on Rn × {t = 0}
to be a Lipschitz function which agrees with g on Rn × {t = 0}, and solves the
PDE a.e. on Rn × (0, ∞).
– However, this turns out to be an inadequate definition, as such weak solutions
would not in general be unique.
Example. Consider the one-dimensional Hamilton-Jacobi equation
ut + u2x = 0.
Three global solutions of this equation are
u1 (x, t) =0.
u2 (x, t) =x − t.
u3 (x, t) = − x − t.
Since the minimum and maximum of any two solutions are again solutions, we have
further solutions
u4 (x, t) = max{u2 , u3 } = |x| − t,
|x| − t
|x| < t
u5 (x, t) = min{u1 , u4 } =
0
|x| ≥ t.
Thus u1 and u5 are two different solutions which have the same Cauchy data
(2)
u1 (x, 0) ≡ u5 (x, 0) ≡ 0.
Note that u5 is Lipschitz and solves the PDE a.e. (everywhere except on the lines
x = ±t). There are in fact infinitely many Lipschitz functions satisfying the initial
condition (2) and solves the PDE a.e.
• This example shows we must presumably require more of a weak solution than
merely that it satisfy the PDE a.e.
• We will look to Hopf’s formula for a further clue as to what is needed to ensure
uniqueness.
• The following Lemma shows that u inherits a kind of “one-sided” second derivative estimates from the initial function g.
Proposition 1. Suppose that there exists a constant C such that
(3)
g(x + z) − 2g(x) + g(x − z) ≤ C|z|2
∀x, z ∈ Rn .
Define u by Hopf’s formula. Then
u(x + z, t) − 2u(x, t) + u(x − z, t) ≤ C|z|2
∀x, z ∈ Rn , t > 0.
Typeset by AMS-TEX
1
2
Proof. Choose y ∈ Rn so that
x−y
u(x, t) = tL
+ g(y).
t
Then, putting y + z and y − z in the formulae for u(x + z, t) and u(x − z, t)
u(x + z, t) − 2u(x, t) + u(x − z, t)
x−y
x−y
+ g(y + z) − 2 tL
+ g(y)
≤ tL
t
t
x−y
+ tL
+ g(y − z)
t
=g(y + z) − 2g(y) + g(y − z) ≤ C|z|2
by (3).
Definition. A function g : Rn → R is said to be semiconcave if there exists a
constant C such that (3) holds.
• (3) holds if g ∈ C 2 and supRn |D2 g(x)| < ∞.
• g is semiconcave iff the function x 7→ g(x) − C2 |x|2 is concave for some constant
C.
• As a semiconcavity condition for u will turn out to be important, we pause to
identify some other circumstances under which it is valid.
• We will no longer assume g to be semiconcave, but will suppose the Hamiltonian
H to be uniformly convex.
Definition. A C 2 convex function H : Rn → R is called uniformly convex with
constant θ > 0 if D2 H ≥ θI.
• We now prove that even if g is not semiconcave, the uniform convexity of H
forces u to become semiconcave for times t > 0;
this is a kind of mild regularizing effect for Hopf’s solution of the initial value
problem (1).
Proposition 2. Suppose that H : Rn → R is uniformly convex with constant
θ > 0 and u(x, t) is defined by Hopf’s formula. Then
u(x + z, t) − 2u(x, t) + u(x − z, t) ≤
1 2
|z|
θt
∀x, z ∈ Rn , t > 0.
To prove Proposition 2, we first observe the following.
Lemma 3. Suppose that L = H ∗ : Rn → R is semiconcave with constant θ∗ > 0.
Then
u(x + z, t) − 2u(x, t) + u(x − z, t) ≤
θ∗ 2
|z|
t
∀x, z ∈ Rn , t > 0.
3
Proof of Lemma 3. Choose y ∈ Rn so that u(x, t) = tL x−y
+ g(y). Then, using
t
the same value of y in the formulae for u(x + z, t) and u(x − z, t)
u(x + z, t) − 2u(x, t) + u(x − z, t)
x−y
x+z−y
+ g(y) − 2 tL
+ g(y)
≤ tL
t
t
x−z−y
+ g(y)
+ tL
t
x+z−y
x−y
x−z−y
=t L
− 2L
+L
.
t
t
t
By assumption on the semiconcavity of L, we obtain
z 2
θ∗
(4) u(x + z, t) − 2u(x, t) + u(x − z, t) ≤ tθ∗ ≤ |z|2 ∀x, z ∈ Rn , t > 0. t
t
We therefore proceed to prove the following.
Lemma 4. Suppose that H : Rn → R is uniformly convex with constant θ > 0.
1
Then L = H ∗ : Rn → R is semiconcave with constant 4θ
> 0; i.e.,
1
q1 + q2
+ |q1 − q2 |2 , ∀q1 , q2 ∈ Rn
(5)
L(q1 ) + L(q2 ) ≤ 2L
2
4θ
Substituting (5) into (4) completes the proof of Proposition 2.
Proof of Lemma 4. Fix q1 , q2 ∈ Rn . We choose p1 , p2 ∈ Rn so that
L(q1 ) = p1 · q1 − H(p1 ),
L(q2 ) = p2 · q2 − H(p2 ).
Then
1
1
1
1
1
L(q1 ) + L(q2 ) = (p1 · q1 + p2 · q2 ) − H(p1 ) − H(p2 ).
2
2
2
2
2
We note first using Taylor’s formula, the uniform convexity of H implies
1
1
θ
p1 + p2
≤ H(p1 ) + H(p2 ) − |p1 − p2 |2 .
H
2
2
2
8
Hence
1
θ
1
1
p1 + p2
L(q1 )+ L(q2 ) ≤ (p1 · q1 + p2 · q2 ) − H
− |p1 − p2 |2
2
2
2
2
8
q1 + q2
p1 + p2
p1 + p2
·
−H
=
2
2
2
1
θ
+ [(2p1 · q1 + 2p2 · q2 ) − (p1 + p2 ) · (q1 + q2 )] − |p1 − p2 |2
8
4
1
θ
q1 + q2
2
+ (p1 − p2 ) · (q1 − q2 ) − |p1 − p2 |
≤L
2
4
8
2
2 |q1 − q2 |
θ
θ
q1 + q2
+
≤L
+ |p1 − p2 |2 − |p1 − p2 |2
2
θ
4
8
8
1
q1 + q2
+ |q1 − q2 |2 ,
=L
2
8θ
which is (6).
(6)
4
Example. Consider the initial value problem
(7)
ut + 12 |Du|2 = 0,
in Rn × (0, +∞)
u = |x|
on Rn × {t = 0}
Here H(p) = 12 |p|2 and so L(q) = 12 |q|2 .
The Hopf-Lax formula for the unique, weak solution of (7) is
(8)
u(x, t) = minn
y∈R
|x − y|2
+ |y| .
2t
(i) The minimum can be found by consideration
Dy
|x − y|2
+ |y|
2t
y−x
y
+
,
t
|y|
=
(y 6= 0);
if y 6= 0, this expression equals zero if
x=y+
y
t,
|y|
y = (|x| − t)
x
6= 0;
|x|
at these points y, we have
(9)
|x − y|2
+ |y| =
2t
|x| − 2t ,
3t
2
if |x| ≥ t
− |x|,
if |x| < t.
(ii) On the other hand, at y = 0, we obtain
(10)
|x|2
|x − y|2
+ |y|
.
=
2t
2t
y=0
– Comparing the values in (9) and (10), we conclude that
x
(i) the minimum in (8) is attained at y = (|x| − t) |x|
, if |x| ≥ t,
(i) the minimum in (8) is attained at y = 0, if x < t.
– Consequently, the Hopf-Lax formula defines
u(x, t) =
(
|x| − 2t ,
if |x| ≥ t
|x|2
2t ,
if |x| < t.
Observe that the solution becomes semiconcave at times t > 0, even though the
initial function g(x) = |x| is not semiconcave.
5
Example. Consider the initial value problem
(11)
ut + 12 |Du|2 = 0,
in Rn × (0, +∞)
u = −|x|
on Rn × {t = 0}
Then the Hopf-Lax formula for the unique, weak solution of (11) is
(12)
u(x, t) = minn
y∈R
Now, if y 6= 0,
Dy
|x − y|2
− |y| .
2t
|x − y|2
+ |y|
2t
=
y−x
y
−
,
t
|y|
and this expression equals zero if
x=y−
y
t,
|y|
y = (|x| + t)
x
6= 0;
|x|
at these points y, we have
(12)
t
|x − y|2
− |y| = −|x| − .
2t
2
On the other hand, at y = 0, we obtain
(13)
|x − y|2
|x|2
+ |y|
.
=
2t
2t
y=0
Comparing the values in (12) and (13), we conclude that the Hopf-Lax formula
defines
t
(x ∈ Rn , t ≥ 0).
u(x, t) = −|x| − ,
2
The initial function g(x) = −|x| is semiconcave, and the solution remains so for
times t > 0.
6
Weak Solutions; Uniqueness
In this section we show that the semiconcavity conditions of the sorts discussed for
the Hopf-Lax solution u in Lemmas 3 and 4 can be utilized as uniqueness criteria.
Definition. A Lipschitz continuous function u : Rn × (0, ∞) → R is a weak
solution of the initial-value problem
ut + H(Du) = 0,
in Rn × (0, +∞)
(14)
u=g
on Rn × {t = 0}
provided
(a) u(x, 0) = g(x), (x ∈ Rn ),
(b) ut (x, t) + H(Du(x, t)) = 0 for a.e. (x, t) ∈ Rn × (0, ∞), and
(c) u(x + z, t) − u(x, t) ≤ D + Ct |z|2
for constants D, C ≥ 0 and all x, z ∈ Rn , t > 0.
Theorem 5 (Uniqueness of Weak Solutions). Assume that H is C 2 , convex,
n
and lim|p|→∞ H(p)
|p| = ∞. Assume g : R → R is Lipschitz continuous. Then there
exists at most one weak solution of the initial-value problem (14).
• In view of Lemma 1 and Lemma 2, we have the following.
Corollary. Suppose H is C 2 , convex, and lim|p|→∞ H(p)
|p| = ∞. If either
(i) g is semiconcave, or
(ii) H is uniformly convex,
then
x−y
u(x, t) = minn tL
+ g(y)
y∈R
t
is the unique weak solution of the initial value problem (14).
• We know that the “characteristic slope”, which is the derivative ẋ of the extremal
x, is bounded independent of its endpoints, namely, ẋ = Hp (ux ), and thus
R ≡ max{|DH(p)| : |p| ≤ Lip (u)}.
is the supremum of the “characteristic slopes”.
– Thinking of data as being carried along the extremal to determine the solution at
its endpoints, it seems likely that the solution at u(x, t) is determined completely
by the data on the base of a cone with sufficiently steep sides:
{(y, 0)||x − y| ≤ M t, t ≥ 0},
M ≥R
Define a cone of determinacy to be any cone of the form
Kx,t = {(z, s)|0 ≤ s ≤ t, |x − z| ≤ M (t − s)}.
It will be seen that the values of a solution u in a cone of determinacy are
completely determined by the values on its base Kx,t ∩ {s = 0}.
– Both the uniqueness of the solution and the continuous dependence on the initial
data follow from the following crucial lemma.
7
Lemma 6. Let H be of class C 2 and convex. Let u and u
e be weak solutions of
the Hamilton-Jacobi equation for H in a common cone of determinacy
Kx,t = {(z, s)|0 ≤ s ≤ t, |x − z| ≤ R(t − s)},
with
(15)
R ≡ max{|DH(p)| : |p| ≤ max(Lip (u), Lip (e
u))}.
Assume that u and u
e have same constant of semiconcavity D +
X ∂ 2H
A = max
.
Kx,t
∂pk ∂pk
C
t
in x. Define
k
Write w = u − u
e. For any smooth function φ : R → [0, ∞), set v = φ(w) and set
Z
E(s) ≡
v(y, s)dy;
BR(t−s) (x)
here BR(t−s) (x) is the horizontal plane section of Kx,t with altitude s. Then, there
holds
Z σ
C
(16)
E(σ) ≤ E(ε) exp
A(D + )ds , ∀σ > ε.
s
ε
In other words,
Z
t−AC e−At
ϕ(w) dx
BR(t−s) (x)
is a nonincreasing function of t in the interval 0 < s ≤ t.
Proof. Step 1. Assume that u and u
e are two weak solutions of (14). Let w = u− u
e.
Oberve that for any point (y, s)
et (y, s)
wt (y, s) =ut (y, s) − u
= − H(Du(y, s)) + H(De
u(y, s))
Z 1
d
=−
H(rDu(y, s) + (1 − r)De
u(y, s)) dr
dr
0
n Z 1
X
=−
Hpk (rDu(y, s) + (1 − r)De
u(y, s)) dr · (uxk (y, s) − u
exk (y, s))
=−
k=1
Z 1
0
DH(rDu(y, s) + (1 − r)De
u(y, s)) dr · (Du(y, s) − De
u(y, s)).
0
Consequently, setting the n-vector
Z 1
b(y, s) = −
DH(rDu(y, s) + (1 − r)De
u(y, s)) dr,
0
we have
(17)
wt + b · Dw = 0,
a.e..
Step 2. Write v =: φ(w) ≥ 0, where φ : R → [0, ∞) is any smooth function.
Multiply (17) by φ0 (w) to discover
vt + b · Dv = 0,
a.e..
hence
(18)
vt + div(vb) = (div b)v.
8
Step 3. Assume first that u, u
e ∈ C 2 . Since u and u
e have the same constant of
DC
semiconcavity + t in x,
D2 ux (x, t), D2 u
ex (x, t) ≤ D +
(19)
C
I,
s
and hence, noting that H is convex implies D2 H ≥ 0,
div b(y, s) =
Z
1
n
X
[Hpk p` (rDu + (1 − r)De
u)](rux` xk + (1 − r)e
ux` xk ) dr
0 k,`=1
C
.
≤A D +
s
(20)
Step 4. Compute for s > 0:
Z
Z
0
E (s) =
vt dy − R
=
Z
BR(t−s) (x)
v dσ
∂BR(t−s) (x)
[−div(vb) + (div b)v] dy − R
BR(t−s) (x)
=−
≤
Z
Z
v(b · ν + R) dσ +
v dσ,
by (18)
∂BR(t−s) (x)
Z
∂BR(t−s) (x)
Z
(div b)v dy
BR(t−s) (x)
(div b)v dy,
by (19),
BR(t−s) (x)
(21)
C
≤A D +
E(s),
s
by (20).
By (21) and Gronwall’s inequality, we obtain (16).
Step 5. Now we assume that u and u
e are only Lipschitz continuous and approximate u and u
e by C 2 functions. Namely, choose ε > 0 and define uε = ηε ∗ u,
u
eε = ηε ∗ u
e, where ηε is the standard mollifier in the x and t variables. Then
(22)
kDuε kL∞ ≤ Lip(u),
(23)
Duε → Du,
kDe
uε kL∞ ≤ Lip(e
u),
De
uε → De
u,
a.e., as ε → 0,
and
(24)
D2 uεx (x, t), D2 u
eεx (x, t) ≤ D +
C
I, ∀ε > 0, x ∈ Rn , s > 0.
t
9
Step 6. Write
bε (y, s) =
(25)
Z
1
DH(rDuε (y, s) + (1 − r)De
uε (y, s)) dr.
0
Then (17) becomes
wt + bε · Dw = (bε − b) · Dw
a.e.;
hence
vt + div(vbε ) = (div bε )v + (bε − b) · Dv
(26)
a.e.
Now, analogously to (20), we have
Z 1 X
n
[Hpk p` (rDuε + (1 − r)De
uε )](ruεx` xk + (1 − r)e
uεx` xk ) dr
div bε (y, s) =
0 k,`=1
C
.
≤A D +
s
(27)
Step 7. Analogously to (21), compute for a.e. s > 0
Z
Z
0
E (s) =
vt dy − R
v dσ
=
Z
BR(t−s) (x)
∂BR(t−s) (x)
−div(vbε ) + (div bε )v + (bε − b) · Dv dy
BR(t−s) (x)
−R
=−
≤
Z
Z
v dσ,
by (26)
∂BR(t−s) (x)
v(bε · ν + R) dσ +
∂BR(t−s) (x)
Z
Z
(div bε )v + (bε − b) · Dv dy
BR(t−s) (x)
(div bε )v + (bε − b) · Dv dy,
by (15), (22),
BR(t−s) (x)
Z
C
≤A D +
E(s) +
(bε − b) · Dv dy,
s
BR(t−s) (x)
by (27).
The last term on the right hand side tends to zero as ε → 0; indeed,
|Dv(y, s)| ≤ 2
sup
|φ0 (w)|kDwkL∞ ≤ 2
w∈Bλ (0)
sup
(Lip(u) + Lip(e
u)),
w∈BλR (0)
where λ = (Lip(u) + Lip(e
u)) ≥ |u(y, s) − u
e(y, s)|. And
Z 1
|bε (y, s)−b(y, s)| =
[DH(rDu + (1 − r)De
u) − DH(rDuε + (1 − r)De
uε )] dr
0
2
≤ max |D H(p)|
p∈Bλ (0)
≤
Z
1
[r(Du − duε ) + (1 − r)(De
u − De
uε )]dr
0
1
max |D2 H(p)|{|Du(y, s) − Duε (y, s)| + |De
u(y, s) − De
uε (y, s)|};
2 p∈Bλ (0)
10
hence, by the dominated convergence theorem,
Z
|bε (y, s) − b(y, s)|dy → 0,
as ε → 0.
BR(t−s) (x)
Thus
C
E (s) ≤ A D +
E(s),
s
0
This and Gronwall’s inequality yields (16).
for a.e. s.
Proof of Theorem 5. Fix 0 < ε < σ < t and choose the function φ(x) such that
φ(x) = 0,
if |x| ≤ ε(Lip(u) + Lip(e
u)),
φ(x) > 0,
otherwise.
Since u = u
e on Rn × {0},
v = φ(w) = φ(u − u
e) = 0,
at {t = ε}.
Thus E(ε) = 0, and hence by (16), E(σ) = 0. Hence
|u − u
e| ≤ ε(Lip(u) + Lip(e
u))
in BR(t−σ) (x).
This inequality is valid for all ε > 0, and so
u≡u
e in BR(t−σ) (x),
∀t > ε.
Therefore, in particular,
u(x, t) = u
e(x, t).