Partial Differential Equations I
Fall 2007
Prof. D. Slepčev∗
Chris Almost†
Disclaimer: These notes are not the official course notes for this class. These notes
have been transcribed under classroom conditions and as a result accuracy (of the
transcription) and correctness (of the mathematics) cannot be guaranteed.
Contents
Contents
1
0 Examples of PDE
2
1 Linear PDE
1.1 Transport equation
1.2 Laplace’s equation
1.3 Heat equation . . .
1.4 Wave equation . . .
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3
3
4
18
27
2 Nonlinear first order PDE
2.1 Method of characteristics . .
2.2 Hamilton-Jacobi equations .
2.3 Conservation Laws . . . . . .
2.4 Rankine-Hugoniot condition
2.5 Existence of solutions . . . .
2.6 Viscosity solutions . . . . . .
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Index
∗
†
53
[email protected]
[email protected]
1
2
PDE I
0
Examples of PDE
An example of a simple one dimensional PDE is u t + u x = 0, where u is a function
of x (which we think of as space) and t (which we think of as time).
1. The transport equation is u t + V · Du = 0, where V is a vector field. It is
called the transport equation because the value of u is constant under the
action of the vector field V . This is a first order linear PDE.
2. The continuity equation is u t + div(V · u) = 0. Recall that div(u) = tr(Du) for
u : Rn+1 → Rn .
3. A conservation law (in one dimension) is a PDE of the form u t + F (u) x = 0,
where F is some function (not necessarily linear). We say that u is the conserved quantity in this case. It can be seen that the transport and continuity
equations are conservation laws.
4. A Hamilton-Jacobi equation is a PDE of the form u t + H(Du) = 0, where
H is a given scalar field. Solutions to these equations are the solutions to
Hamiltonian dynamics. These PDE are not necessarily linear.
5. Laplace’s equation is −u x x = 0 in one dimension, and −∆u = 0 in higher
dimensions, where ∆ = tr(D2 u) is the Laplacian. This equation describes
equilibrium solutions in many systems. It is an elliptic PDE.
6. The heat equation is u t − ∆u = 0. The Laplace equation gives the steadystate solutions to the heat equation. The heat equation is an example of a
parabolic PDE, it smooths its initial and boundry conditions, and information propagates infinitely fast.
7. The wave equation is u t t − ∆u = 0 and describes (small) waves moving in a
system. Information does not propagate infinitely fast, and the wave equation does not smooth its initial conditions (but regularity is maintained).
8. Schrödinger’s equation is iu t + ∆u = 0. It is really a system of two real PDE.
9. The minimal surface equation (describing, for example, soap film across a
bent closed loop) is


Du
 = 0.
− div  p
1 + |Du|2
10. u t − ∆u + f (u) = 0 is the reaction-diffusion equation. There is also nonlinear
diffusion given by u t − ∆um = 0.
11. There are also important systems of equations. If u is a vector-valued function then the system of PDE
u t + Du · u = −Dp + ν∆u,
div u = 0
are the Naiver-Stokes equations that describe fluids.
Linear PDE
3
There are numerous questions that arise in the study of PDE. Intrinsic mathematical questions include the existence of solutions, uniqueness of solutions, and
continuity with respect to data (i.e. stability). If a problem has a unique, regular
solution then it is said to be well-posed. What are the properties of the solutions?
1
Linear PDE
Any PDE can be written as F (x, u, Du, D2 u, . . . ) = 0. The PDE is said to be linear if
F is a linear function of u, Du, D2 u, . . . , and homogeneous if no term is a function
of x alone, otherwise it is inhomogeneous. For homogeneous linear PDE, linear
combinations of solutions are also solutions.
1.1
Transport equation
The transport equation is a first order linear equation. To solve it we will need a
few facts about ODE. Let V be a smooth bounded vector field on Rn . Consider the
IVP
dx
= V (x, t), x(0) = x 0 .
dt
It has a unique smooth solution defined everywhere, so we may define a mapping
Φ(x 0 , t) := x(t). The mapping Φ(·, T ) : Rn → Rn is injective. (Indeed, consider a
change of variable τ = T − t. Then
dx
dτ
=
dx
dt
·
dt
dτ
= −V (x, T − τ).
This change of variable “runs time backwards.” Structurally, the backwards equation is the same as the original, and if Φ(·, T ) were not injective then the backwards equation would not have a unique solution, a contradiction.) Similarly,
Φ(·, T ) is surjective (by the same argument using existence instead of uniqueness). Since V is smooth, Φ is smooth, and we can solve the equation backwards
in time, so Φ(·, t) is a smooth bijection for all t.
The transport equation is
u t + V · D x u = 0,
u(x, 0) = g(x)
(T)
where g is smooth and V = V (x) is as above. Another way of writing (T) is
D x,t u·(V, 1) = 0, so (T) is the assertion that the directional derivative of u vanishes
in the direction (V, 1). Fix a point (x 0 , t 0 ) ∈ Rn+1 at which we would like to
determine the value of u. Consider the system of ODE
dx
ds
= V (x),
dt
x(0) = x 0 ,
ds
= 1,
t(0) = t 0 .
Then t(s) = s + t 0 and ddsx = V (x) has a unique smooth solution by the theory of
ODE. The curve (x(s), s + t 0 ) is a characteristic curve for the PDE. Notice that
d
ds
u(x(s), s + t 0 ) = D x u ·
dx
dt
+ u t = Du · V + u t ,
4
PDE I
which is identically zero if u is a solution to (T). Therefore u(x(s), s + t 0 ) is
constant as a function of s. Whence the value when s = 0 is the same as the
value when s = −t 0 , so u(x 0 , t 0 ) = g(x(−t 0 )). In the notation of the aside on
ODE, u(t 0 , x 0 ) = g(Φ(·, t 0 )−1 (x 0 )).
1.1.1 Theorem. Let V be a smooth bounded vector field on Rn and g be a smooth
function Rn → R. Then (T) has a unique smooth solution.
Consider now the inhomogeneous form
u t + V · D x u = f (x, t),
u(x, 0) = g(x).
Then the directional derivative of u in the direction (V, 1) is now f (x, t). On the
= f (x(s), s + t 0 ). If
solution curves to the same ODE, z(s) := u(x(s), s + t 0 ) has dz
ds
f is smooth then we can solve this equation (in principle) when x is known. We
could go further and solve
u t + V · D x u = h(x, t, u),
u(x, 0) = g(x)
by the similar methods.
1.1.2 Example (u t + 2u x = u, u(x , 0) = g (x )).
Write ddsx = 2, x(0) = x 0 , which has solution x(s) = x 0 +2s, so Φ(x 0 , t 0 ) = x 0 +2t 0 .
The characteristic curves are straight lines. The inverse mapping is Φ(·, t)−1 : x 7→
x − 2t. Taking z(s) = u(x(s), s), we have
dz
ds
= Du · V + u t = u(x(s), s) = z.
Whence z(t) = z(0)e t , where z(0) = u(x 0 , 0) = g(x 0 ), and u(x 0 + 2s, s) = z(s) =
g(x 0 )es , so by renaming variables we see that u(x, t) = g(x − 2t)e t is the solution
to the PDE.
Warning: The symbol x is used in two different ways, as a variable independent
of t and as a function of s. When x appears as a function of s then it will always
be written x(s). It is hoped that this is not too confusing.
1.2
Laplace’s equation
The Laplace equation is −∆u = 0 in Rn . In n = 1 the solutions are exactly the
linear functions. In n = 2 the second derivatives “balance each other out,” so we
get solutions like u(x, y) = x 2 − y 2 that are convex in one variable and concave
in the other. The Laplace equation gives the equilibrium solutions to the heat
equation, which will be discussed in the next section. A related equation is the
Poisson equation, −∆u = f for a function f . It describes the equilibrium solution
to the heat equation u t − ∆u = f , where f describes the source of heat.
1.2.1 Definition. A C 2 function u for which −∆u = 0 is called a harmonic function.
Laplace’s equation
5
Solution to the Poisson equation
Notice that the Laplace equation is translation invariant, i.e. if a ∈ Rn and u is
a solution then u(· − a) is also a solution. Moreover, it is invariant under all
orthogonal transformations, i.e. if R is an orthogonal matrix and u is a solution
then u(R·) is also a solution. Indeed, let w(x) = u(Rx) and notice that
∂w
=
∂ yi
n
X
∂u
k=1
∂ xk
R ki ,
so
∂ 2w
∂ yi2
=
n X
n
X
∂ 2u
k=1 `=1
∂ x k ∂ x`
R ki R`i
and it follows that
−∆w = −
n X
n
X
∂ 2u
i=1 k,`=1
∂ x k ∂ x`
R ki R`i = −
n
X
∂ 2u
k,`=1
∂ x k ∂ x`
δk,` = −∆u = 0.
To solve the Laplace equation we p
first seek a radial solution, a solution of the
form u(x) = v(r), where r = |x| = x 12 + . . . .x n2 on x ∈ Rn \ {0} (we wish to
avoid trivial constant solutions). Then
∂u
∂ xi
= v0
∂r
∂ xi
= v0
xi
r
,
Whence
00
−∆u = −v − v
∂ 2u
so
0
∂ xi
n
r
+v
0
2
1
r
= v 00
 x ‹2
i
r
00
=− v +
+ v0
n−1
r
1
r
v
0
− v0
xi xi
r2 r
.
,
and for u to be a solution we need to solve v 00 + n−1
v 0 = 0. We can do this
r
(exercise), and the solutions are
¨
c1 log r + c2 n = 2
v(r) = c1
+ c2
n≥3
r n−2
We are looking only for one particular special solution, so we may set c2 = 0. We
require that
Z
−Du(x) · ν ds = 1,
∂ B(0,r)
where ν is the outward pointing normal vector field, for various reasons having
to do with the Gauss-Green Theorem that I didn’t really understand. In n = 2 this
gives
Z
Z 2π
1
1 x x
1 = −c1
· ds = −c1
dθ = −c1 · 2π,
|r| r r
r
∂ B(0,r)
0
1
so we take c1 = − 2π
. The fundamental solution to Laplace’s equation is
(
Φ(x) =
1
log |x|
− 2π
1
(n−2)S(n)|x|n−2
n=2
n≥3
6
PDE I
where S(n) is the surface area of B(0, 1) in Rn . Recall that
n
S(n) = nVol(B(0, 1)) = n
Notice that
(
DΦ(x) =
π2
Γ( 2n + 1)
1 1 x
− 2π
|x| |x|
.
n=2
1
1
x
− S(n)
|x|n−1 |x|
n≥3
so for n = 2,
Z
DΦ(x) ·
∂ B(0,")
x
|x|
dS =
Z
−
∂ B(0,")
=
Z
−
∂ B(0,")
=−
1 1
2π "
1 1 x
·
x
2π |x| |x| |x|
1 1
2π "
dS
dS
2π" = −1
and for n ≥ 3 the computation is basically the same and the result is also −1.
Therefore
Z
∂ B(0,")
∂ν Φ dS = −1,
a fact which we will need later.
1.2.2 Convolution. Let k be a smooth function on Rn such that
R
for all |α| ≥ 0. Let f ∈ L 1 (Rn ) (so Rn | f | < ∞). Then
k ∗ f (x) :=
Z
k(x − y) f ( y)d y =
Rn
and
Z
∂ αk
∂ xα
is bounded
k(z) f (x − z)dz = f ∗ k(x)
Rn
∂ αk ∗ f
(x) =
∂ αk
∗ f (x).
∂ xα
∂ xα
It follows that k ∗ f is also smooth, and ∆(k ∗ f ) = 0 if ∆k = 0.
1.2.3 Dominated Convergence Theorem. Suppose that f k ∈ L 1 (Rn ) and f k → f
a.e. If thereR is a nonnegative
g ∈ L 1 (Rn ) such that | f k | ≤ g for all k then f ∈
R
1
n
L (R ) and Rn f k → Rn f .
1.2.4 Theorem. Let f ∈ Cc2 (Rn ). Then
Z
u(x) :=
Rn
is a solution to −∆u = f .
Φ(x − y) f ( y)d y = Φ ∗ f
Laplace’s equation
7
This theorem gives a solution to Poisson’s equation when f is sufficiently nice.
It may be extended to larger classes of f with hard work.
Notation (®). We write A( f , g, . . . ) ® B( f , g, . . . ) if there a (positive?) constant C
independent of the entries f , g, . . . such that A ≤ C B.
PROOF: Consider that
Z
|Φ(x)| d x =
2D
R
Z
¨
(log r)r d r ®
C
0
B(0,R)
R2 (| log R| + 1)
R2
n=2
n≥3
so Φ ∈ L 1 (B(0, R)) for all R > 0. Therefore u is well-defined for all x ∈ Rn , using
the fact that f is bounded.
Let {e1 , . . . , en } denote the standard orthonormal basis for Rn . Now
u(x + hei ) − u(x)
h
=
Z
f (x + hei − y) − f (x − y)
Φ( y)
h
Rn
Z
Φ( y)
→
Rn
∂f
∂ xi
(x − y)d y =
∂u
∂ xi
dy
(x)
by the Dominated Convergence Theorem, since
Φ( y)
f (x + hei − y) − f (x − y)
≤ Φ( y)kD f k∞
h
and since we can reduce the integral over Rn to the integral over some compact
set. Repeating this argument we get that
∂ 2u
∂ xi∂ x j
=
Z
Φ( y)
Rn
∂2f
∂ xi∂ x j
(x − y)d y,
so
∆u =
Z
Φ( y)∆ f (x − y)d y.
Rn
Let " > 0. Since Φ not defined at 0, we will break up the integral into an
integral over B(0, ") and an integral over the rest. Let A" = Rn \ B(0, ").
Z
Φ( y)∆ f (x − y)d y ≤
B(0,")
Z
|Φ( y)| · |∆ f (x − y)|d y
B(0,")
2
Z
® kD f k∞
|Φ( y)|d y
B(0,")
¨
2
® kD f k∞
" 2 | log " + 1|
"2
n=2
n≥3
8
PDE I
which goes to 0 as " → 0. By integration by parts,
Z
Φ( y)∆ f (x − y)d y
A"
=−
Z
∇Φ( y) · ∇ f (x − y)d y −
Z
Φ( y)∇ f (x − y) · ν dS y
∂ B(0,")
A"
But the magnitude of far righthand term is
Z
® kD f k∞
¨
|Φ( y)|dS y ® kD f k∞
∂ B(0,")
"| log "|
1
" n−1 = "
" n−2
n=2
n≥3
which goes to zero as " → 0. Integrating the other term by parts give
Z
Z
∆Φ( y) f (x − y)d y +
f (x − y)∇Φ( y) · ν dS y
∂ B(0,")
A"
=0+
Z
( f (x) − y · D f (x − θ y))∇Φ( y) · ν dS y
∂ B(0,")
= f (x)
Z
∇Φ( y) · ν dS y −
∂ B(0,")
= − f (x) −
Z
y · D f (x − θ y))∇Φ( y) · ν dS y
∂ B(0,")
Z
y · D f (x − θ y))∇Φ( y) · ν dS y
∂ B(0,")
using the Taylor expansion of f around x. Finally, the magitude of the last term is
(
1
"
n=2
® "kD f k∞ " 1 n−1
= "kD f k∞
"
n
≥3
" n−1
which goes to zero as " → 0. Putting it all together, ∆u = − f , so u satisfies the
Poisson equation.
ƒ
Mean-value property of harmonic functions
1.2.5 Theorem. Let u be a harmonic function on an open set Ω and assume that
B(x, R) ⊆ Ω. Then
Z
Z
u(x) = −
u( y)dS y
and
u(x) = −
∂ B(x,R)
y( y)d y.
B(x,R)
PROOF: For the first equality, consider that
Z
ψ(r) = −
∂ B(x,r)
R
u( y)dS y =
u( y)dS y
∂ B(x,r)
r n−1 S(n)
,
Laplace’s equation
9
for 0 < r ≤ R, is a continuous function of r. We have lim r→0+ ψ(r) = u(x), which
can be proved along the lines of the previous theorem, noting that
Z
ψ(r) = −
u(x) + ( y − x)Du(θ r x + (1 − θ r ) y)dS y
∂ B(x,r)
for some θ r ∈ (0, 1). Therefore ψ is well-defined and continuous on [0, R]. We
will show that ψ is constant. To take the derivative we must first introduce a
y−x
change of variables z ← R . Then
R
Z
r n−1 ∂ B(0,1) u(x + rz)dSz
ψ(r) =
=−
u(x + rz)dSz .
r n−1 S(n)
∂ B(0,1)
Thus
dψ
dr
Z
d
=−
∂ B(0,1)
Z
=−
dr
u(x + rz)dSz
∇u(x + rz) · z dSz
Chain rule
∂ B(0,1)
Z
=−
∇u( y) ·
y−x
∂ B(x,r)
R
B(x,r)
=
=
r
dS y
∆u( y)d y
Divergence Theorem
S(n)r n−1
r
R
B(x,r)
∆u( y)d y
S(n) = nVol(B(0, 1))
n Vol(B(0, 1))r n
Z
r
= −
∆u( y)d y = 0
n B(x,r)
u is harmonic
and it is seen that ψ is constant. In particular, ψ(R) = ψ(0) = u(x). The second
equality is proved below.
ƒ
1.2.6 Coarea Formula. Let η : Rn → R be Lipschitz and for all λ ∈ R {x ∈ Rn |
η(x) = λ} is, for a.e. λ, either a smooth (n − 1)-dimensional
hypersurface in Rn
R
n
or empty. Let f : R → R be continuous and such that Rn f exists. Then
Z
Z∞Z
f · |∇η| =
f dSdλ.
−∞
{η=λ}
1.2.7 Example. Take η = |x|. Then |∇η| = 1 (except at x = 0 where the gradient
does not exist), and so
Z
Z ∞Z
f =
Rn
f ( y)dS y d r.
0
{|x|=r}
10
PDE I
To finish the proof the theorem, we take η( y) = | y − x| and note that
Z
Z
u( y)d y =
B(x,R)
=
=
=
u( y)1B(x,R) d y
Rn
Z ∞Z
u( y)1B(x,r) dS y d r
0
| y−x|=r
Z RZ
u( y)dS y
∂ B(x,r)
0
ZR
u(x)S(n)r n−1 d r
0
= u(x)nVol(B(0, 1))
rn
n
= u(x)Vol(B(x, r))
R
1.2.8 Theorem. If u ∈ C 2 (Ω) satisfies u(x) = −∂ B(x,R) u( y)d y for every B(x, R) ⊆
Ω (this is the mean-value property) then u is harmonic in Ω.
PROOF: Assume that u has the mean-value property but is not harmonic. Then
there is some point z ∈ Ω such that ∆u(z) 6= 0. Then since ∆u is continuous there
R > 0 such that, without loss of generality, ∆u > 0 on B(z, R). Take ψ as in the
proof of the last theorem. The averaging property says simply that ψ is constant
(indeed, ψ ≡ u(z)). But for 0 < r < R,
Z
r
0
ψ (r) = −
∆u( y)d y > 0
n B(z,r)
since ∆u is positive on B(z, R).
ƒ
Regularity property of harmonic functions
1.2.9 Mollifiers. Mollifiers are functions that are used for “smoothing.” For this
class a mollifier is a function η : Rn → R such that
1. η ∈ C ∞ (Rn )
2. η ≥ 0
R
3. Rn η = 1
4. η is radially symmetric and η(|x|) is non-increasing.
5. η has compact support.
For example, we may take
(
η(x) =
cn exp( |x|21−1 ) |x| < 1
0
|x| ≥ 1
Laplace’s equation
11
were the constant cn is chosen so that
R
Rn
η" (x) :=
and notice that
R
B(0,")
η = 1. For " > 0 take
1
x‹
"
"
η
n
η" ( y)d y = 1.
1.2.10 Regularization. Let f be locally integrable on Ω. A regularization of f is
f " (x) = η" ∗ f (x) defined on Ω" = {x ∈ Ω | dist(x, ∂ Ω) > "} (notice that
Z
Z
f " (x) =
f (x − y)η" ( y)d y =
B(0,")
f ( y)η" (x − y)d y,
B(x,")
so we require that B(x, ") ⊆ Ω). Some facts are true.
1. f " ∈ C ∞ (Ω" )
2. f " → f a.e. on Ω as " → 0
P
P
3. If | f | p is locally integrable (i.e. f ∈ Lloc
(Ω)) then f " → f in Lloc
(Ω).
4. If f ∈ C(Ω) then f " → f uniformly on compact subsets of Ω.
5. If f ∈ C k (Ω) then f " → f in C k (K) for all K ⊂⊂ Ω.
1.2.11 Theorem. Let u ∈ C(Ω) satisfy the mean-value property. Then u is infinitely differentiable (i.e. u ∈ C ∞ (Ω)).
PROOF: We have for x ∈ Ω"
u" (x) = η" ∗ u(x)
Z
=
=
=
=
η" (x − y)u( y)d y
B(x,")
Z "Z
0
Z"
0
Z"
η" (x − y)u( y)dS y d r
∂ B(x,r)
η" (r)
Z
u( y)dS y d r
η" (r)u(x)
0
= u(x)
u is radially symmetric
∂ B(x,r)
Z
Z
1dS y d r
by MVP
∂ B(x,r)
η" (x − y)d y
B(x,")
= u(x)
Therefore u = u" ∈ C ∞ (Ω" ). Letting " → 0 gives the result.
ƒ
12
PDE I
Maximum principle
1.2.12 Theorem (Maximum Principle). Let Ω be a bounded open set and let u ∈
C 2 (Ω) ∩ C(Ω) be harmonic.
1. (Weak maximum principle) maxΩ u = max∂ Ω u.
2. (Strong maximum principle) If Ω is connected and there is x 0 ∈ Ω such that
u(x 0 ) = maxΩ u then u is constant on Ω.
PROOF: Clearly the strong maximum principle implies the weak maximum principle. If x 0 ∈ Ω is a point such that u(x 0 ) = maxΩ u then u is constant on the connected component containing x 0 , so it attains this maximum value on the boundry
of this component (and hence on the boundry of Ω).
Suppose that Ω is connected and there is x 0 ∈ Ω such that u(x 0 ) = maxΩ u. Let
r
be
such that B(x 0 , r) ⊆ Ω. Since u is harmonic, by the MVP we have u(x 0 ) =
R
−
u(x)d x. But u(x 0 ) ≥ u(x) for all x ∈ B(x 0 , r), so u ≡ u(x 0 ) on B(x 0 , r).
B(x 0 ,r)
(Indeed, assume that there is y0 ∈ B(x 0 , r) such that u( y0 ) < u(x 0 ). Let " :=
u(x 0 )−u( y0 ) and notice that by continuity there is δ > 0 such that u( y) < u(x 0 )−
"
for all y ∈ B( y0 , δ). But then
2
Z
Z
Z
u(x)d x =
B(x 0 ,r)
u(x)d x +
B( y0 ,δ)
u(x)d x
B(x 0 ,r)\B( y0 ,δ)

"‹
≤ u(x 0 ) −
|B( y0 , δ)| + u(x 0 )(|B(x 0 , r)| − |B( y0 , δ)|)
2
"
= u(x 0 )|B(x 0 , r)| − |B( y0 , δ)|
2
a contradiction of the MVP.) Let A := {x ∈ Ω | u(x) = u(x 0 )}. Then A is open by
the argument above, and A is closed (relative to Ω) since it is u−1 ({u(x 0 )}) and u
is continuous. Therefore A = Ω since Ω is connected, and u is constant on Ω, again
by continuity.
ƒ
1.2.13 Theorem (Comparison). Let u and v be solutions of the Poisson equation
−∆u = f on a bounded open set Ω. Assume that u, v ∈ C 2 (Ω) ∩ C(Ω) and v ≥ u
on ∂ Ω. Then v ≥ u on Ω.
PROOF: Consider w := v − u. Then −∆w = 0, w ∈ C 2 (Ω) ∩ C(Ω), and w ≥ 0 on
∂ Ω. By the maximum principle (minimum principle), w ≥ 0 on Ω.
ƒ
1.2.14 Theorem (Uniqueness). Let g ∈ C(∂ Ω) and f ∈ C(Ω). Then there is at
most one solution u ∈ C 2 (Ω) ∩ C(Ω) of
¨
−∆u = f on Ω
u= g
on ∂ Ω
PROOF: Assume that u1 and u2 are solutions. Then w := u1 − u2 is harmonic and
w ≡ 0 on ∂ Ω. By the maximum and minimum principles it follows that w ≡ 0 on
Ω.
ƒ
Laplace’s equation
13
Local estimates
1.2.15 Theorem. Let u be harmonic on B(x, r). Let α = (α1 , . . . , αn ) be a multiindex of order k = |α|. Then
|Dα u(x)| ≤
(2n+1 nk)k
1
ωn
r n+k
kuk L 1 (B(x,r)) ,
where n is the dimension and ωn is the volume of the unit ball.
PROOF: We prove the case where k = 1. Then Dα u = u x i for some i. Notice that
all derivatives of harmonic functions are harmonic functions.
Z
|∂ x i u(x)| = −
B(x, 2r )
=
=
2n
∂ x i u( y)d y
Z
ωn r n
2n
ωn r n
=
B(x, 2r )
u x i ( y)d y
Z
u · νi dS y
∂ B(x, 2r )
2n nωn r n−1
kuk L ∞ (B(x, r ))
2
ωn r n 1 2n−1
n
2n 2
≤
kuk L 1 (B(x,r))
r ωn r n
The proof for all other k is by induction.
ƒ
1.2.16 Theorem (Liouville). If u : Rn → R is harmonic and bounded then u is
constant.
PROOF: Let x ∈ Rn .
|∂ x i u(x)| ≤
c
r r+1
kuk L 1 (B(x,r)) ≤
c
r r+1
M ωn r n =
c̃
r
for all r > 0, since u is bounded. Therefore the left hand side is zero, and u must
be constant since all partials are zero.
ƒ
1.2.17 Theorem (Harnack’s Inequality). Let u ≥ 0 be harmonic on an open set
Ω. Let V be a connected open set such that V ⊆ Ω and V is compact. There exists
C > 0 depending only on V and Ω such that supV u ≤ C infV u.
PROOF: Let r = 14 dist(V , ∂ Ω), and x, y ∈ V . If |x − y| < r then
Z
Z
1
u(x) = −
u(z)dz =
u(z)dz
ωn 2n r n B(x,2r)
B(x,2r)
Z
Z
1
1
ωn r n
1
≥
u(z)dz
=
u(z)dz = n u( y).
−
ωn 2n r n B( y,r)
ωn 2n r n 1
2
B( y,r)
14
PDE I
Cover V with the balls B(z, 2r ) and choose a finite sub-cover. Let K be the number
of balls required. Now if x and y are farther apart than r we can show that
ƒ
u(x) ≥ 21nK u( y), from which the inequality follows.
Green’s function
R
Recall that a solution to −∆u = f (for sufficiently nice f ) is u = Φ ∗ f = Rn Φ(x −
y) f ( y)d y. Let Ω be a bounded open set with boundary C 1 , and consider the
boundary value problem
¨
−∆u = f on Ω
u=0
on ∂ Ω
R
Is there is function G such that u = Ω G(x − y) f ( y)d y is a solution to this problem? If u is a solution to Poisson’s equation on Rn then
Z
Z
u(x) =
Φ(x − y) f ( y)d y = −
Φ( y − x)∆u( y)d y.
Rn
Rn
There is no longer equality when Rn is replaced with Ω. We will look for a correction term.
1.2.18 Green’s Identity. Let u, v ∈ C 2 (Ω) ∩ C 1 (Ω), where ∂ Ω is C 1 . Then
Z
Z
∂v
∂u
u∆v − v∆u =
u
−v
dS.
∂
ν
∂
ν
Ω
∂Ω
Let Ω" := Ω x," := Ω \ B(x, ").
Z
Φ( y − x)∆u( y)d y
Ω"
=−
Z
∇Φ( y − x) · ∇u( y)d y +
Z
∂ Ω"
Ω"
=
Z
∆Φ( y − x)u( y)d y −
Z
Ω"
=
u( y) ·
∂ Ω"
Z
−u( y) ·
∂Ω
+
∂Φ
∂ν
( y − x) + Φ( y − x) ·
Z
u( y) ·
∂ B(0,")
=
Φ( y − x) ·
Z
−u( y) ·
∂Ω
∂Φ
∂ν
∂Φ
∂ν
∂Φ
∂ν
∂u
∂ν
( y − x)dS y +
∂u
∂ν
∂u
∂ν
Z
Φ( y − x) ·
∂ Ω"
∂u
∂ν
( y)dS y
( y)dS y
( y − x) − Φ( y − x) ·
( y − x) + Φ( y − x) ·
( y)dS y
∂u
∂ν
( y)dS y
( y)dS y − u(x) + O(")
so
u(x) = −
Z
Ω
Φ( y − x)∆u( y)d y −
Z
u( y) ·
∂Ω
∂Φ
∂ν
( y − x) + Φ( y − x) ·
∂u
∂ν
( y)dS y .
Laplace’s equation
15
But we don’t know how to compute ∂∂ νu . For each point x we introduce a corrector
function ϕ x ( y) defined by
¨
−∆ϕ x = 0
in Ω
ϕ x ( y) = Φ( y − x) on ∂ Ω
Then
Z
u∆ϕ − ϕ ∆u =
x
Z
x
Ω
u
∂ ϕx
∂Ω
− Φ( y − x)
∂ν
∂u
∂ν
( y)dS y ,
which contains the term for which we are looking. Plugging into above, defining
Green’s function G(x, y) = Φ( y − x) − ϕ x ( y),
Z
Z
Z
∂Φ
∂ ϕx
x
u(x) =
−Φ( y − x)∆ud y + ϕ ∆ud y +
−u( y)
( y − x) + u( y)
( y)dS y
∂ν
∂ν
Ω
Ω
∂Ω
Z
Z
∂G
= − G(x, y)∆ud y −
u( y)
(x, y)dS y .
∂ νy
Ω
∂Ω
If a solution exists to the problem (a big “if” at this point)
¨
−∆u = f on Ω
u= g
on ∂ Ω
then
u(x) =
Z
G(x, y) f ( y)d y −
Ω
Z
g( y)
∂Ω
∂G
∂ νy
(x, y)dS y .
If instead we are given data about ∂∂ νu (Von Neumann boundary conditions)
then the problem term is u on the boundary. In this case we define a different type
of corrector via
(
−∆ϕ x = 0
in Ω
∂ ϕx
∂y
( y) =
∂Φ
(y
∂y
− x) on ∂ Ω
1.2.19 Proposition. Let Ω be a bounded, connected open set with C 1 boundary.
Then G(x, y) = G( y, x) for all x, y ∈ Ω.
PROOF: Let v(z) = G(x, z) and w(z) = G( y, z). We need to show that v( y) = w(x).
Let Ω" = Ω \ (B(x, ") ∪ B( y, ")).
Z
0=
v∆w − w∆v dz
Ω"
=
Z
∂ Ω"
=
v∂ν w − w∂ν v dSz
Z
∂ B(x,")∪∂ B( y,")
−v∂ν w + w∂ν v dSz
16
PDE I
Now
Z
∂ B(x,")
−(Φ(z − x) − ϕ x (z))∂ν w + w∂ν (Φ(z − x) − ϕ x (z)) dSz
R
converges to ∂ B(x,") ∂ν Φ(z − x) · w dSz as " → 0, as before, and this last term
goes to w(x), as before. An analogous argument for the other ball shows that the
whole thing goes to w(x) − v( y).
ƒ
This proposition allows to extend the definition of G to the boundary of Ω in
both variables (but not simultaneously).
n
Green’s function on R+
We need to solve
¨
n
−∆ϕ x = 0
in R+
n
ϕ x = Φ( y − x) on ∂ R+
Take ϕ x ( y) = Φ( ỹ − x) where ỹ = ( y1 , . . . , yn−1 , − yn ) is the reflection of y
through {x n = 0}. The idea here is that Φ is a function with the desired boundary
values, but fails to be harmonic at x. Instead we take Φ reflected through {x n = 0}
then it becomes harmonic on the upper half space and has the correct boundary
conditions. We take G(x, y) = Φ( y − x) − Φ( ỹ − x). Then for y on the boundary
∂G
∂ νy
(x, y) = −
2x n
1
S(n) |x − y|n
for n ≥ 3.
1.2.20 Theorem. Let g ∈ C b (Rn−1 × {0}). Then for
Z
u(x) := −
Rn−1 ×{0}
∂G
∂ νy
(x, y) · g( y)d y
we have the following
n
n
1. u ∈ C ∞ (R+
) ∩ L ∞ (R+
)
n
2. −∆u = 0 on R+
3. lim x→x 0 ,x∈R+n u(x) = g(x 0 ) for x 0 ∈ Rn−1 × {0}.
PROOF: Note that
∂G
∂ νy
n
(x, y) is smooth on R+
× (Rn−1 × {0}), and in fact it is har-
monic in x and in y for y on the boundary. An explicit computation shows that
R
∂G
n
(x, y)d y = 1 for all x ∈ R+
.
ƒ
Rn−1 ×{0} ∂ ν
y
Laplace’s equation
17
Energy Methods
Consider the boundary value problem
¨
−∆u = 0
u= g
E(u) =
R
Ω
in Ω
on ∂ Ω
|∇u|2 d x is the associated energy functional for the problem.
1.2.21 Theorem (Uniqueness). The Poisson problem for a bounded domain Ω
has at most one solution in C 2 (Ω) ∩ C(Ω).
PROOF: Assume that u and v are both solutions and consider w := u − v. Then
−∆w = 0 in Ω and w ≡ 0 on ∂ Ω. We have
Z
Z
Z
∂w
2
E(w) =
|∇w| d x = − ∆w · w d x +
w
dS = 0,
∂ν
Ω
Ω
∂Ω
so ∇w ≡ 0 a.e. in Ω. Since w is C 2 , it is zero everywhere on Ω.
ƒ
Now consider the boundary value problem
¨
−∆u = f in Ω
u= g
on ∂ Ω
R
In this case E(u) = Ω 12 |∇u|2 − u f d x is the associated energy functional for the
problem. The admissible set of functions for the problem is
A = {u ∈ C 2 (Ω) ∩ C(Ω) | u = g on ∂ Ω}.
1.2.22 Theorem. If u is a solution of the above BVP then u minimizes the energy
E over the admissible set A . Conversely, if there is a minimizer u of E over A
then u solves the BVP.
PROOF: Assume that u minimizes E over A . Consider i(τ) := E(u + τv) where v
is such that u + τv ∈ A for all τ ∈ R, e.g. v ∈ C 2 (Ω) ∩ C(Ω) and v ≡ 0 on ∂ Ω.
Then τ = 0 must be a critical point of i. We have
i(τ) = E(u + τv)
Z
1
1
=
|∇u|2 + τ∇u · ∇v + τ2 |∇v|2 − u f − τv f d x
2
2
ZΩ
0 = i 0 (0) =
=
∇u · ∇v − f v d x
ZΩ
(−∆u)v − f v d x +
Ω
=
Z
Ω
Z
∂Ω
v(−∆u − f )d x
(∇u)v · ν dS
18
PDE I
Since this is true for every v, we must have −∆u − f ≡ 0. Therefore u solves the
BVP. Conversely, consider w ∈ A and E(w) − E(u). Let v = w − u, so w = v + u.
Then
Z
Z
1
1
2
E(w) − E(u) =
|∇(u + v)| − (u + v) f d x −
|∇u|2 − u f d x
2
2
Ω
ZΩ
1
=
∇u · ∇v + |∇v|2 − v f d x
2
ZΩ
Z
1
2
= (−∆u)v + |∇v| − v f d x +
v(∇u · ν)dS
2
∂Ω
ZΩ
1
=
|∇v|2 d x ≥ 0
2
Ω
so u minimizes E. Notice that the difference is strictly positive if w 6= u.
1.3
ƒ
Heat equation
The (inhomogeneous) heat equation IBVP is

in Ω × (0, T )
u t − ∆u = f
u= g
on ∂ Ω × (0, T )

u(x, 0) = u0 (x) for all x ∈ Ω
where g is assumed to be constant in time. Conditions on Ω × {0} are called
initial conditions and conditions on ∂ Ω × (0, T ) are lateral boundary conditions.
The energy functional for this problem is
E(u) =
Z
Ω
1
2
|∇u|2 − u f d x.
Then
dE
dt
=
Z
∇u · ∇u t − f u t d x
Ω
=
Z
(−∆u − f )u t d x +
Z
∂
=−
u t (∇u · ν)dS
ΩΩ
Z
(∆u + f )2 d x ≤ 0
Ω
Something about dissipating energy, therefore solutions will converge as t → ∞
to a solution to the Poisson problem.
Heat equation
19
Physical motivation
We consider a simple physical model where u is interpreted
as the temperature of
R
a domain Ω. The energy of this system at a time t is Ω ρcu(x, t)d x, where ρ is
the density and c is the specific heat capacity, and we assume these quantities do
not depend on the time or temperature. The change in energy in Ω is the flow of
energy through the boundary,
Z
Z
dE
=−
q · ν dS = − div q d x,
dt
∂Ω
Ω
R
where q is the heat flux density. But we have ddEt = ρc Ω u t , so combining these,
R
ρcu t + div q d x = 0 and we conclude ρcu t + div q ≡ 0. Fourier’s law of heat
Ω
conduction says that q = −k∇u for some constant k depending on the material.
k
Therefore ρcu t − k∆u = 0, or u t − ρc
∆u = 0. By scaling (in x) it suffices to study
the equation u t − ∆u = 0, the heat equation.
Now suppose that energy is not completely conserved in this system. Then
Z
Z
dE
=−
q · ν dS + ρc f (x)d x,
dt
∂Ω
Ω
k
and through a similar analysis we can show that u satisfies u t − ρc
∆u = f .
Consider
¨
vt − ∆v = 0 on Ω × [0, ∞)
∂ν v = 0
on ∂ Ω × [0, ∞)
R
and assume that v is a smooth solution. Then Ω v is constant in time, i.e.
d
dt
Z
Ω
v dx =
Z
∆v d x =
Ω
Z
∂Ω
∂ν v dS = 0.
Scaling properties
If u is a solution to the heat equation then u(λx, λ2 t) is also a solution for any
λ > 0. This is a property of any elliptic equation. We look for a solution that
preserves this scaling, i.e. u such that λk u(λx, λ2 t) is constant in λ (for some k).
Total heat is conserved, so we need the following integral to be independent of λ.
Z
Z
Z
1
k
2
k
2
k−n
λ u(λx, λ t)d x = λ
u( y, λ t) n d y = λ
u( y, λ2 t)d y,
λ
where y = λx is a change of variable. Total heat is independent of the time
variable, so we must take k = n.
We want λn u(λx, λ2 t) to be constant in λ. Taking λ = p1t we get
u(x, t) =
1
n
t2
u
x
,
1
.
p
t
20
PDE I
Let ϕ(z) = u(z, 1), the similarity profile. Plugging this into the heat equation, we
get
n − n −1
x
1 − n −1
x
x
x
− 2n −1
2
2
0 = u t − ∆u = − t
ϕ p
− t
∇ϕ p
·p −t
∆ϕ p
2
2
t
t
t
t
so 0 = − 2n ϕ(z) − 21 ∇ϕ(z) · z − ∆ϕ(z). We now further require that ϕ be radially
symmetric. This reduces the last equation to an ODE which we can solve. The
ϕ r (where r = |z|
Laplacian of a radial function is given by ∆z ϕ(z) = ϕ r r + n−1
r
and we abuse notation ϕ(r) = ϕ(z)), so the ODE is
ϕr r +
n−1
r
1
n
ϕ r + ϕ r · r + ϕ = 0,
2
2
where we require that ϕ and the derivatives go to zero sufficiently fast as r → ∞.
To solve it we divide by r n−1 , giving (r n−1 ϕ r ) r + 21 (r n ϕ) r = 0. The decay we
desire of ϕ and ϕ r as r → ∞ gives that r n−1 ϕ r + 12 r n ϕ = 0. Solving we get that
R
1 2
n
ϕ(r) = ce− 4 r . We choose c so that Rn ϕ(z)dz = 1, i.e. c −1 = (4π) 2 since
Z
e
−
x2
1
4
···e
−
2
xn
4
d x1 · · · d x n =
Z
e
Rn
n
2
− x4
dx
n
= (4π) 2 .
R
The last integral is computed by noting that
Z
e−
x2+ y2
4
dx d y =
R2
Z
2π Z ∞
0
r2
e− 4 r d r dθ = 4π
∞
e−s ds = 4π.
0
0
Φ(x, t) =
Z
1
|x|
(4πt)
n
2
e− 4t
is the fundamental solution to the heat equation.
Cauchy problem
Let g ∈ L ∞ (Rn ) ∩ C(Rn ) and consider the Cauchy problem for the heat equation,
the initial value problem
¨
u t − ∆u = 0 on Rn × (0, ∞)
u(x, 0) = g
for all x ∈ Rn
1.3.1 Theorem. Every function of the form
Z
u(x, t) = Φ(·, t) ∗ g =
Rn
is a solution to the IVP above. In particular,
1. u ∈ C ∞ (Rn × (0, ∞))
Φ(x − y, t)g( y)d y
Heat equation
21
2. lim x→x 0 ,t→0+ u(x, t) = g(x 0 ) for all x 0 ∈ Rn .
PROOF: By its definition u is smooth on Rn × (0, ∞) (use the same arguments as
before for space, and time is even simpler).
Z
Z
u t − ∆u =
Φ t (x − y, t)g( y)d y −
Rn
∆Φ(x − y, t)g( y)d y = 0,
Rn
since Φ is itself a solution.
Z
|u(x, t) − g(x 0 )|| = Rn
Z
Φ(x − y, t)(g( y) − g(x 0 ))d y Φ(x − y, t)|g( y) − g(x 0 )|d y
≤
Rn
=
Z
Φ(x − y, t)|g( y) − g(x 0 )|d y
B(x 0 ,δ)
+
Z
Φ(x − y, t)|g( y) − g(x 0 )|d y
Rn \B(x 0 ,δ)
The left is at most kgk L ∞ (B(x 0 ,δ)) . The right term is at most
Z
Φ(x − y, t)d y.
2kgk L ∞ (Rn )
Rn \B(x 0 ,δ)
Choose δ > 0 so that |g(x) − g(x 0 )| < 21 " when |x − x 0 | < δ (by continuity of g).
Choose δ0 so that if t < δ0 then
Z
"
.
Φ(x − y, t)d y <
4kgk
L ∞ (Rn )
Rn \B(x ,δ)
0
Then when |x − x 0 | < δ and t < δ0 we have |u(x, t) − g(x 0 )| < ".
ƒ
Non-homogeneous Cauchy problem
The non-homogeneous problem Cauchy is
¨
u t − ∆u = f on Rn × (0, ∞)
u(x, 0) = g
for all x ∈ Rn
By linearity it suffices that we solve the problem for when g ≡ 0, since we already
know how to solve the problem when f ≡ 0.
1.3.2 Theorem. Let f ∈ C 2,1 (Rn × [0, ∞)). The function
Z tZ
Φ(x − y, t − s) f ( y − s)d y ds
v(x, t) =
0
Rn
solves the non-homogeneous problem (with g ≡ 0). This is a special case of
Duhamel’s principle.
22
PDE I
1.3.3 Duhamel’s principle. Let k ≥ 1 and consider the following non-homogeneous
problem, where L is a linear partial differentiable operator in the space variables
only.
¨
∂ tk u(x, t) + Lu(x, t) = f (x, t)
∂ ti u(·, 0) = 0
i = 0, . . . , k − 1
Consider the related homogeneous problem

k
n
∂ t U(x, t, s) + LU = 0 on R × [0, ∞)
i
∂ u(·, s, s) = 0
i = 0, . . . , k − 2
 t
k−1
∂ t u(·, s, s) = f (·, s).
If U is a solution to the homogeneous problem then then
Z t
u(x, t) =
U(x, t, s)ds
0
is a solution to the inhomogeneous problem.
PROOF: By the FTC and the chain rule,
Z t
u t (x, t) = U(x, t, t) +
U t (x, t, s)ds =
0
Z
t
U t (x, t, s)ds
0
..
.
∂ tk u(x, t) = ∂ tk−1 U(x, t, t) +
Z
t
∂ tk U(x, t, s)ds
0
= f (x, t) +
Z
t
∂ tk U(x, t, s)ds
0
and
Lu =
t
Z
LU(x, t, s)ds =
0
t
Z
−∂ tk U(x, t, s)ds.
0
so the given u solves the non-homogeneous problem.
ƒ
Maximum principle for Dirichlet problem
Now we let Ω be a bounded open set and we consider Ω T := Ω × (0, T ], the
parabolic cylinder over Ω. The parabolic boundary is
PΩ t := Ω × {0} ∪ ∂ Ω × [0, T ].
The Dirichlet problem for this domain is
¨
u t − ∆u = 0
u= g
in Ω T
on PΩ T
Heat equation
23
1.3.4 Definition. A function u ∈ C 2,1 (Ω T )∩C(Ω T ) is a sub-solution to the Dirichlet
problem if
¨
u t − ∆u ≤ 0 in Ω T
u≤ g
on PΩ T
and a super-solution is defined analogously.
1.3.5 Comparison principle. If u is a sub-solution to the problem and v is a
super-solution to the problem then u ≤ v on Ω T .
PROOF: Assume that there is (x̂, t̂) ∈ Ω T such that u(x̂, t̂) > v(x̂, t̂). Consider, for
" > 0, v " (x, t) := v(x, t)+"t, a (strict) super-solution. For small enough " we still
have u(x̂, t̂) > v " (x̂, t̂). Let (x 0 , t 0 ) be a point where u − v " reaches is maximum
on Ω T . Then (x 0 , t 0 ) ∈ Ω T , so
Du(x 0 , t 0 ) = Dv " (x 0 , t 0 ) and
D2 u(x 0 , t 0 ) ≤ D2 v " (x 0 , t 0 )
by the necessary conditions for the maximization of a multi-variable function. We
also know that u t (x 0 , t 0 ) ≥ vt" (x 0 , t 0 ), so
0 ≥ u t − ∆u ≥ vt" − ∆v " = " + vt − ∆v ≥ " > 0,
a contradiction.
ƒ
1.3.6 Comparison principle. If u is a sub-solution of the heat equation and v is a
super-solution of the heat equation and u ≤ v on PΩ T then u ≤ v on all of Ω T .
1.3.7 Theorem (Weak Maximum Principle). Let u ∈ C 2,1 (Ω T ) ∩ C(Ω T ) be a solution to the heat equation u t − ∆u = 0 on Ω T . Then max PΩT u = maxΩT u and
min PΩT u = minΩT u.
PROOF: Define v ≡ max PΩT u. Then v is a constant function, so it is a solution to
the heat equation, and v ≥ u on PΩ T . By the comparison principle, u ≤ v on all of
ΩT .
ƒ
1.3.8 Theorem (Uniqueness). There is at most one solution u ∈ C 2,1 (Ω T )∩C(Ω T )
to the initial/boundary value problem

u t − ∆u = 0
u= g

u = u0
in Ω × (0, T ]
on ∂ Ω × [0, T ]
on Ω × {0}
PROOF: If u and v are both solutions then by the comparison principle u ≤ v and
v ≤ u (so u ≡ v) on all of Ω T .
ƒ
24
PDE I
Maximum principle for Cauchy Problem
For the theorems in the last subsection we required that Ω was a bounded open
set. What about for Ω = Rn ?
1.3.9 Theorem. Let u ∈ C 2,1 (Rn × (0, T ]) ∩ C(Rn × [0, T ]) be a solution to
¨
u t − ∆u = 0
on Rn × (0, ∞)
u(x, 0) = g(x) for all x ∈ Rn
2
where g is continuous and bounded from above. Assume that u(x, t) ≤ Ae a|x| for
some A, a ∈ R for all x ∈ Rn and t ∈ [0, T ]. Then supRn ×[0,T ] u = supRn g.
PROOF: Consider, for t < 0,
ψ(z, t) :=
|z|2
‚
1
n
(−t) 2
exp
Œ
.
−4t
Then ψ t − ∆ψ = 0.
There are two cases. If 4aT < 1 then there exists " > 0 such that 4a(T +") < 1.
Let y ∈ Rn and µ > 0 and consider
v(x, t) := u(x, t) − µψ(x − y, t − (T + ")).
on Rn × [0, T ]. Then vt − ∆V = 0 on Rn × (0, T ] and v ≤ u and v ≤ g on Rn × {0}.
When |x − y| = r,
Œ
‚
µ
r2
v(x, t) = u(x, t) −
n exp
4(T + " − t)
(T + " − t) 2
‚
Œ
µ
r2
a|x|2
≤ Ae
−
n exp
4(T + " − t)
(T + " − t) 2
‚
Œ
µ
r2
a(| y|2 +r 2 )
≤ Ae
−
n exp
4(T + ")
(T + ") 2
But a < 4(T1+") , so this last quantity is bounded by a constant, and we may choose
r such that
max
v = max v ≤ sup g.
P(B( y,r)×[0,T ])
Rn
B( y,r)×{0}
But
max
B( y,r)×[0,T ]
µ
≥ v( y, t) = u(x, t) −
so taking µ → 0 we obtain supRn
n
(T + " − t) 2
g ≥ u( y, t). Conclude by pasting.
ƒ
Let
(
ϕ(z) =
exp(− z12 ) z 6= 0
0
z=0
(P∞
and u(x, t) =
Then u t − ∆u = 0 on Rn × (0, ∞) and u(·, 0) = 0.
n=0 ϕ
0
(n)
2n
x
(t) (2n)!
t >0
t = 0.
Heat equation
25
Regularity of solutions
The non-homogeneous heat equation u t − ∆u = f does not necessarily have
smooth solutions. Indeed, take h ∈ C 2,1 \ C 3,1 . Then h t − ∆h =: f˜, where f˜ ∈ C
and f˜ is not differentiable (h(x, t) = |x|3 should work).
1.3.10 Theorem. Assume u ∈ C 2,1 (Ω T ) solves the heat equation u t − ∆u = 0 in
Ω T . Then u ∈ C ∞ (Ω T ).
PROOF: Let C(x, t, r) be the parabolic cylinder
C(x, t, r) := {( y, s) ∈ Rn+1 | kx − yk < r, t − r 2 < s ≤ t}.
For (x̂, t̂) ∈ Ω T . Choose r > 0 so that C = C(x̂, t̂, r) ⊆ Ω T . Choose smaller
cylinders C 0 and C 00 corresponding to 34 r and 21 r. Choose a smooth cut-off function
ξ such that ξ ≡ 1 on C 0 and ξ ≡ 0 outside of C. Define
¨
v(x, t) :=
u(x, t)ξ(x, t) if (x, t) ∈ C
0
otherwise
Then v is defined on Rn × (−∞, t̂] and
vt − ∆v = ξu t + uξ t − ξ∆u − 2∇ξ∇u − u∆ξ = uξ t − 2∇ξ∇u − u∆ξ =: f˜
so v solves vt − ∆v = f˜ on its domain, and f˜ ∈ C 1,1 . (Without loss of generality
we are assuming that t̂ > 0 and we choose r > 0 so that t̂ − r 2 > 0.) Notice that
v(·, 0) = 0 on Rn , so v is a solution to the non-homogeneous Cauchy problem.
Since the boundary conditions are compactly supported,
v(x, t) =
Z tZ
0
Φ(x − y, t − s) f˜( y, s)d y ds
Rn
by uniqueness of solutions to the heat equation. Note further that f˜ is supported
on C \ C 0 . For (x, t) ∈ C 00 we have
u(x, t) = v(x, t)
ZZ
=
Φ(x − y, t − s)((ξ t − ∆ξ)u − 2∇ξ∇u)d y ds
C\C 0
=
ZZ
=:
Φ(x − y, t − s)(ξ t − ∆ξ)u − 2∇Φ(x − y, t − s)∇ξu d y ds
ZZC
K(x, y, t, s)u( y, s)d y ds
C
Now K is supported on C \ C 0 and is smooth, so u is smooth at (x, t).
ƒ
26
PDE I
1.3.11 Theorem. For all k, ` ≥ 0 there is a constant Ck,` such that for each multiindex α with |α| = k,
max |Dαx D`t u| ≤
C(x,t, 2r )
Ck,`
kuk L 1 (C(x,t,r))
k+2`+n+2
r
whenever u t − ∆u = 0 on C(x, t, r).
Energy methods
1.3.12 Theorem (Uniqueness). There is at most one solution u ∈ C 2 (Ω T )∩C(Ω T )
of
¨
u t − ∆u = f in Ω T
u= g
on PΩ T
where f and g are continuous, Ω is bounded, and ∂ Ω is C 1 .
PROOF: Assume that w1 and w2 are both solutions. Then u := w1 − w2 solves
¨
u t − ∆u = 0 in Ω T
u=0
on PΩ T .
It suffices to show that u ≡ 0 is the only solution to this problem. The energy (or
entropy) functional for the problem is
Z
u2 (x, t)d x.
e(t) = E(u) :=
Ω
Then
de
dt
=2
Z
uu t d x = 2
Ω
Z
u · ∆u d x = −2
Z
Ω
|∇u|2 d x ≤ 0,
Ω
and e(0) = 0, so e(t) = 0 for all t ≥ 0. Therefore
for all times.
R
Ω
u2 d x = 0, so u is zero on Ω
ƒ
1.3.13 Poincare’s Inequality. Given Ω bounded with C 1 boundary there is a
constant λ such that, for all u ∈ C 1 (Ω) ∩ C(Ω) for which u ≡ 0 on ∂ Ω,
Z
Z
u2 (x)d x ≤ λ
Ω
|∇u|2 d x
Ω
1.3.14 Theorem. Let u be a solution of

u t − ∆u = 0 in Ω × (0, ∞)
u=0
on ∂ Ω × (0, ∞)

u= g
on Ω × {0}
Wave equation
27
where g is continuous and is compatible with the other boundary condition. There
is a constant, C > 0, independent of g, such that such that
Z
u2 (x, t)d x ≤ e−C t
Ω
Z
g 2 (x)ds.
Ω
PROOF: RAs in the proof of the last theorem we consider the energy. This time
e(0) = Ω g 2 (x)d x and
de
dt
= −2
Z
2
|∇u| d x ≤ −
Ω
2
Z
2
u2 (x, t)d x = − e(t).
λ Ω
λ
2
By Gronwall’s inequality e(t) ≤ e− λ e(0).
1.4
ƒ
Wave equation
d’Alambert’s formula (n = 1)
1.4.1 Theorem (d’Alambert Formula). Let g ∈ C 2 (R) and h ∈ C 1 (R). The function
Z x+t
1
u(x, t) =
g(x + t) + g(x − t) +
h( y)d y
2
x−t
is C 2 (R × [0, ∞)) and solves

on R × (0, ∞)
u t t − u x x = 0
u(x, 0) = g(x) for all x ∈ R

u t (x, 0) = h(x) for all x ∈ R
PROOF: Exercise, recalling the solution of the transport equation and noting that
(∂ t + ∂ x )(∂ t − ∂ x )u = 0.
ƒ
Notice that for h ≡ 0 the value of u(x, t) is just the average of the values of
g(x + t) and g(x − t). Even for h non-trivial, the value of u depends only on the
values of h between x + t and x − t. Information propagates at finite speed and
there we do not have the smoothing properties of the heat equation.
1.4.2 Example (Wave equation on the half line).
The problem is, where g(0) = 0 and h(0) = 0,

ut t − ux x = 0


u(x, 0) = g

u (x, 0) = h

 t
u(0, t) = 0
on R+ × (0, ∞)
x ≥0
x ≥0
t ≥0
28
PDE I
Define g̃(x) to be g(x) for x ≥ 0 and −g(−x) for x < 0, and h̃ similarly. Then
there is a solution ũ given by d’Alambert’s formula. We have
ũ(0, t) =
so
(
u(x, t) =
1
2
1
(g(x
2
1
(g(x
2
g̃(t) + g̃(−t) +
t
Z
h̃( y)d y
= 0,
−t
+ t) + g(x − t) +
+ t) − g(t − x) +
R x+t
R x−t
x+t
t−x
h( y)d y)
x ≥t >0
h( y)d y)
t > x >0
is a solution to the above problem. (Notice that h̃ is odd.)
Spherical means
1.4.3 Lemma.
R
R
1. If a is continuous on B(x, r) and Φ(r) = B(x,r) a( y)d y then Φ0 (r) = ∂ B(x,r) a( y)d y.
2. If a ∈ C 1 (U) and B(x, r) ⊂ U and ϕ(r) =
R
−
∇a( y) · ν dS y .
∂ B(x,r)
R
−
∂ B(x,r)
a( y)dS y then ϕ 0 (r) =
R
PROOF:
1. Φ(r) = r n B(0,1) a(x + rz)dz by the change of variable y = x + rz.
When a is continuously differentiable,
Z
Z
Φ r (r) = nr r−1
a(x + rz)dz + r n
∇a(x, +rz) · z dz
B(0,1)
=
=
=
n
Z
r
B(0,1)
a( y)d y +
Z
B(x,r)
n
r
Z
y−x
∇a( y) ·
r
B(x,r)
Z
Z
a( y)d y −
B(x,r)
n
a( y) d y +
r
B(x,r)
dy
Z
a( y)
∂ B(x,r)
y−x
r
· ν dS y
a( y)dS y
∂ B(x,r)
2. With the same change of variables, dSz =
ϕ(r) =
1
Z
nα(n)
Z
=−
and
∇a(x + rz) · zdSz
∂ B(0,1)
∇a( y) ·
∂ B(x,r)
let
1
dS y
r n−1
y−x
r
dS y
ƒ
Let n ≥ 2 and m ≥ 2 and u ∈ C m (Rn × [0, ∞)). For x ∈ Rn and t ≥ 0 and r > 0
Z
U(x, r, t) = −
∂ B(x,r)
u( y, t)dS y
Wave equation
and
29
Z
G(x, r) = −
Z
g( y)dS y
and
H(x, r) = −
∂ B(x,r)
h( y)dS y .
∂ B(x,r)
1.4.4 Lemma (Euler-Poisson-Darboux Equation).
For u and U as above, if u solves

on Rn × [0, ∞)
u t t − ∆u = 0
u(x, 0) = g(x) for all x ∈ Rn

u t (x, 0) = h(x) for all x ∈ Rn
then for fixed x ∈ Rn , U(x, ·, ·) ∈ C m (R+ × [0, ∞)) and solves

n−1
+
U t t − U r r − r U r = 0 on R × [0, ∞)

U(x, r, 0) = G(x, r)
U t (x, r, 0) = H(x, r)
for all r > 0
for all r > 0
PROOF: For r > 0, and computing as for the MVP for the Laplacian, one obtains
Z
Z
r
U r (x, r, t) = −
∇u · ν dS y = U r (x, r, t) = −
∆u( y, t)d y.
n B(x,r)
∂ B(x,r)
Whence, buy the product and quotient rules,
Z
Z
r
1
U r r (x, r, t) = −
∆u( y, t)d y +
−1 −
∆u( y, t)d y.
n ∂ B(x,r)
n
B(x,r)
Note that lim r→0+ U r (x, r, t) = 0 and lim r→0+ U r r (x, r, t) = 1n ∆u(x, t) so we may
extend U to r = 0 as well. Now
Z
Z
r
1
Ur = −
u t t d y,
ut t d y =
n B(x,r)
nα(n)r n−1 B(x,r)
so multiplying both sides by r n−1 and taking the derivative,
Z
r n−1
n−1
(r U r ) r =
u t t dS y
−
nα(n) ∂ B(x,r)
r n−1 U r r + (n − 1)r n−2 U r = r n−1 U t t
which is the EPD equation.
ƒ
Kirchhoff’s formula (n = 3)
Let Ũ = r U, G̃ = r G, and H̃ = r H. Then
Ũ t t = r U t t
and
2
Ũ r r = r( U r + U r r ),
r
30
PDE I
so, if and only if n = 3,
Ũ t t − Ũ r r = r(U t t −
2
r
U r − U r r ) = 0.
By the example solution to the wave equation on the half line, for r < t,
Ũ(x, r, t) =
1
2
(G̃(r + t) − G̃(t − r)) +
1
2
r+t
Z
H̃( y)d y.
t−r
Then U(x, t, r) = 1r Ũ(x, r, t), so
u(x, t) = lim+
r→0
G̃(t + r) − G̃(t − r)
2r
+
1
2r
Z
t+r
H̃( y)d y = G̃ 0 (t) + H̃(t).
t−r
We have therefore derived Kirchhoff’s formula
d
(t G(t)) + t H(t)
dt Z
Z
d
=
t−
g( y)dS y + t−
h( y)dS y
dt
∂ B(x,t)
∂ B(x,t)
Z
Z
Z
y−x
=−
g( y)dS y + t−
∇g( y) ·
dS y + t−
h( y)dS y
t
∂ B(x,t)
∂ B(x,t)
∂ B(x,t)
Z
u(x, t) =
=−
th( y) + g( y) + ∇g( y) · ( y − x)dS y
∂ B(x,t)
It can be checked that this is truly a solution to the wave equation in dimension
n = 3.
Method of descent (n = 2)
In the case n = 2, let ḡ(x̄) := g(x) and h̄(x̄) := h(x), where x̄ := (x 1 , x 2 , x 3 )
and x̄ := (x 1 , x 2 ). Let ū be a solution to the wave equation in dimension 3
with initial data ḡ and h̄ given by Kirchhoff’s formula. Is u defined by u(x, t) =
ū((x 1 , x 2 , 0), t) a solution the wave equation in dimension 2? Indeed it is, since
ū x 3 x 3 ((x 1 , x 2 , 0), t) = 0 and the values of ū, as given by the formula, are invariant
under translation in the third coordinate.
Remark. If the data are constant in the x 3 direction but we don’t have a formula
for the solution to the problem, it is enough that we know that the solution is
unique for us to conclude that the solution is constant in the x 3 direction. Indeed,
otherwise translation in the x 3 direction would give different solutions.
Wave equation
31
See the text for the derivation of the following formula.
Z
Z
g( y)
1 d
h( y)
2
2
u(x, t) =
t −
dy +t −
dy
p
p
2 dt
t 2 − | y − x|2
t 2 − | y − x|2
B(x,t)
B(x,t)
Z
t g( y) + t 2 h( y) + t∇g( y) · ( y − x)
1
= −
dy
p
2 B(x,t)
t 2 − | y − x|2
Non-homogeneous problem
We now consider the non-homogeneous problem

n
u t t − ∆u = f in R × (0, ∞)
u(·, 0) = 0
on Rn

u t (·, 0) = 0
on Rn .
We apply Duhamel’s principle and consider the problem

in Rn × (s, ∞)
U t t − ∆U = 0
U(·, s, s) = 0
on Rn

U t (·, s, s) = f (·, s) on Rn .
We need certain smoothness properties for this to hold (this is the reason that it
is Duhamel’s principle and not theorem), but we are only concerned with finding
some solution and so we are not too concerned.
1.4.5 Theorem. Let either
1. n = 1 and f ∈ C 2 (R × [0, ∞)); or
n
2. n ≥ 2 and f ∈ C b 2 c+1 (Rn × [0, ∞));
(smoothness at 0 is required), then u(x, t) :=
and solves the non-homogeneous problem.
Rt
0
U(x, t, s)ds is in C 2 (Rn × [0, ∞))
Energy methods
As usual, let Ω be a bounded domain and Ω T = Ω×(0, T ] be the parabolic cylinder.
Energy methods can be used to show uniqueness of the solution to the following
problem (but not existence).

u t t − ∆u = f in Ω T


u(·, 0) = g
on Ω

u
(·,
0)
=
h
on Ω

 t
u = g̃
on ∂ Ω × [0, T ].
32
PDE I
1.4.6 Theorem. There is at most one solution u ∈ C 2 (Ω T ) ∩ C 1 (Ω T ) solving the
above problem.
PROOF: As usual, by linearity it suffices to show that the zero function is the only
solution to the homogeneous problem when the initial and boundary data is zero.
Let
Z
1
|∇u|2 + u2t d x
E(u) =
2 Ω
be the energy for the problem. Then
Z
Z
Z
dE
=
∇u · ∇u t + u t u t t d x =
u t (−∆u + u t t )d x +
u t ∇u · ν dS = 0,
dt
Ω
Ω
∂Ω
since u = 0 on ∂ Ω × [0, T ] somehow implies that u t = 0 on that surface. Whence
∇u = 0 and u t = 0 for all t > 0, so u ≡ 0 since u(·, 0) = 0.
ƒ
Next we prove that “information propagates at finite speed” through the wave
equation.
1.4.7 Theorem. Let C x 0 ,t 0 = {(x, t) | 0 ≤ t ≤ t 0 , kx − x 0 k ≤ t 0 − t}. If u t t − ∆u =
0 on Ω T , (x 0 , t 0 ) ∈ Ω T , C x 0 ,t 0 ⊆ Ω T , and u(·, 0) = u t (·, 0) = 0 on B x 0 ,t 0 then
u(x 0 , t 0 ) = 0.
PROOF: Define the energy for this problem to be
Z
1
e(t) =
u2 + |∇u|2 d x.
2 B(x ,t −t) t
0
0
Then
de
dt
=
Z
1
u t u t t + ∇u · ∇u t d x −
2
B(x 0 ,t 0 −t)
=
Z
u t (u t t − ∆u)d x +
B(x 0 ,t 0 −t)
1
2
Z
∂ B(x 0 ,t 0 −t)
Z
∂ B(x 0 ,t 0 −t)
u2t + |∇u|2 dS
2u t ∂ν u − u2t − |∇u|2 dS ≤ 0
since
|2u t ∂ν u| ≤ 2|u t ||∇u| ≤ u2t + |∇u|2
by trivial inequalities. Therefore energy is dissipating. Since e(0) = 0 there was
no energy to begin with, so e ≡ 0 on [0, t 0 ] and u(x 0 , t 0 ) = 0.
ƒ
2
2.1
Nonlinear first order PDE
Method of characteristics
The general first order PDE is F (Du, u, x) = 0, for x ∈ Rn . The PDE is said to be
quasi-linear if F has the form
F (p, z, x) = b(z, x) · p − c(z, x).
Method of characteristics
33
Suppose we are given boundary data g on some n − 1 dimensional sub-manifold
Γ. If we have a solution u to the quasi-linear PDE then consider z(s) = u(x(s)), the
value of the solution along curves passing through the boundary manifold. The
method of characteristics considers the associated system of ODE
dx
ds
= b(z(s), x(s)),
dz
ds
= c(z(s), x(s))
since by the chain rule
dz
ds
=
d
ds
u(x(s)) = b(z(s), x(s)) · Du(x(s)) = c(z(s), x(s)).
By solving this system of ODE we obtain a solution to the problem.
2.1.1 Example (u x + 2u y = u 2 , u(x , 0) = g (x )).
dy
= z 2 . Then x(s) = x 0 + s, y(s) = 2s, and (solving the
Take ddsx = 1, ds = 2, and dz
ds
ODE for z),
g(x 0 )
z0
=
z(s) =
.
1 − sz0
1 − sg(x 0 )
Therefore, for a general point (x, y),
y
g(x − 2 )
y
u(x, y) = z( 2 ) =
y
y
1 − 2 g(x − 2 )
,
y
where we take the z associated with x 0 = x − 2 . Assuming g is positive and
bounded, we require that 0 ≤ y < 2kgk∞ , and we can solve the problem in this
strip. In general the solution blows up at certain points.
In the two-dimensional case, for a graph of the solution (x, y, u(x, y)), tangent
vectors are (1, 0, u x ) and (0, 1, u y ), and a normal vector is (−u x , −u y , 1). Notice
that the coefficients to the problem (as a vector in R3 , e.g. (1, 2, u2 ) in the case of
the example above)) are tangent to the graph of the solution. In general, Γ is said
to be non-characteristic if the vector field (b, c) : R1+n → Rn+1 is not tangent to Γ
at any point.
In the fully non-linear case we must also keep track of how the derivatives
of u change along characteristics. To this end let p(s) := Du(x(s)), so that dz
=
ds
p(s) · ddsx . There is no structure of F to exploit to eliminate p. Differentiating the
original equation, for all i = 1, . . . , n,
0=
∂F
∂ xi
=
n
X
∂F
j=1
∂ pj
(Du, u, x)
Therefore
d pi
ds
=
d
ds
∂u
∂ x j∂ xi
+
∂F
∂z
(Du, u, x)
∂u
∂ xi
(x) +
∂F
∂ xi
X
n
dxj ∂ u
(x(s)) =
(x(s)).
∂ xi
ds ∂ x j ∂ x i
j=1
∂u
(Du, u, x).
34
PDE I
Note that taking the derivative of the original equation always gives a quasilinear equation. We use this observation to close the system of ODE, setting
d xi
= ∂∂ pF (p, z, x). The system becomes a system of 2n + 1 ODE
ds
i
ṗi = − ∂∂ Fz (p, z, x)pi − ∂∂ xF (p, z, x) so ṗ = −(Dz F )p − D x F
i

Pn
ż = j=1 p j ∂∂ pF (p, z, x)
so ż = (D p F ) · p
j

∂F
so ẋ = D p F
ẋ i = ∂ p (p, z, x)

i
2.1.2 Example (u x 1 u x 2 = u). We consider the equation u x 1 u x 2 = u with initial
date u(0, x 2 ) = x 22 on Γ = {x 1 = 0}. We wish to solve on Ω = {x 1 > 0}. Then
F (p, z, x) = p1 p2 − z, so our system becomes

 ṗ1 = p1 , ṗ2 = p2
ż = p1 p1 + p2 p1 = 2p1 p2

ẋ 1 = p2 , ẋ 2 = p1
Then pi (s) = pi (0)es , so x i (s) = x i (0) + p3−i (0)(es − 1) and
z(s) = z(0) + p1 (0)p2 (0)(e2s − 1).
On Γ x 1 (0) = 0 and z(0) = (x 2 (0))2 . Since Γ is a line, we may compute the partial
in that direction, i.e. p2 (0) = 2x 2 (0). Using the original equation, p1 (0) = 21 x 2 (0).
Therefore
x 1 (s) = 2x 2 (0)(es − 1)
p1 (s) =
1
2
x 2 (0)es
x 2 (s) =
1
2
x 2 (0)(es + 1)
z(s) = (x 2 (0))2 e2s
p2 (s) = 2x 2 (0)es
From a general point (x 1 , x 2 ), so (solving) we need es =
u(x 1 , x 2 ) = z(s) =
4x 2 + x 1
4
4x 2 +x 1
.
4x 2 −x 1
Finally,
2
.
Existence of a local solution
At this point we are not even sure that a solution exists. First we consider “flat”
boundary data u = g on Γ = {x n = 0}—we will consider the general case later.
This data for the PDE translates to initial data (at a point x 0 ) x(0) = x 0 , z(0) =
g(x 0 ), pi (0) = g x i (x 0 ) for i = 1, . . . , n − 1, and F (p(0), z(0), x(0)) = 0 (which
determines pn (0)).
2.1.3 Definition. For boundary conditions u = g on Γ, a triple (p, z, x)
1. satisfies the compatibility conditions (at x)
a) x ∈ Γ;
Method of characteristics
35
b) z = g(x); and
c) pi = g x i (x) for i = 1, . . . , n − 1.
2. is admissible if the compatibility conditions hold and F (p, z, x) = 0.
3. is non-characteristic if it is admissible and F pn (p, z, x) 6= 0.
Boundary data are non-characteristic if they are non-characteristic at every
point of Γ. When data are non-characteristic there is no problem of information
trying to propagate along the boundary conditions and possibly conflicting.
2.1.4 Lemma. If (p0 , z0 , x 0 ) is non-characteristic, then there exists a neighbourhood W of x 0 in Γ, and a function q( y) on W such that
F (q( y), g( y), y) = 0
for all y ∈ W and such that (q( y), g( y), y) is admissible for all y ∈ W .
PROOF: Let G : Rn × Rn → Rn be defined by
¨
G (p, y) =
i
pi − g x i ( ŷ)
for i = 1, . . . , n − 1
F (p, g( y), y)
i=n
where ŷ is the projection of y onto Γ. Then G(p0 , x 0 ) = 0, and to apply the
Implicit Function Theorem we need ∂∂ Gp 6= 0. But

1

0

∂G
=
.

∂ p  ..
F p1
0
1
..
.
F p2
...
...
..
.
...

0

0
.. 

. 
F pn
Therefore det ∂∂ Gp (p0 , z0 , x 0 ) = F pn (p0 , z0 , x 0 ) 6= 0 since (p0 , z0 , x 0 ) is non-characteristic,
so by the Implicit Function Theorem there is a neighbourhood W of x 0 in Γ and a
function q such that G(q( y), y) = 0.
ƒ
2.1.5 Lemma. With data as in the previous lemma, there exists
1. an open neighbourhood V ⊆ W of x 0 on Γ;
2. an open interval I containing zero; and
3. an open neighbourhood U of x 0 in Rn
such that for all x ∈ U there is a unique s ∈ I and y ∈ V such that x = x( y, s).
Furthermore, the mapping x 7→ ( y, s) is C 2 .
36
PDE I
PROOF: For x = x 0 we must have y = x 0 and s = 0. We have a mapping x :
( y, s) 7→ x : W × I0 → Rn given by the solution to the characteristic ODE. As
before,


1 0 . . . 0 F p1


0 1 . . . 0 F p2 
. .
∂x
.. 

.. .. . . . ...
(x 0 , 0) = 
. 

∂ ( y, s)


0 0 . . . 1 F pn−1 
0 0 . . . 0 F pn
x
= F pn (p0 , z0 , x 0 ) 6= 0.
and det ∂ (∂y,s)
ƒ
Suppose F (Du, u, x) = 0 near x 0 ∈ Rn with u = g on Γ, an (n − 1)-dimensional
manifold. Assume that en is not tangent to Γ at x 0 . Locally (near x 0 ), there exists
a function h : Rn−1 → R such that Γ = {x n − h(x̂) = 0} (where x̂ = (x 1 , . . . , x n−1 )).
Then ν (normal to Γ at x 0 ) is parallel to (−D x̂ h, 1). Let ŷ := x̂ and yn = x n −h(x̂),
so Γ maps to the set Γ̃ = { yn = 0}. Call y = Φ(x), and notice that


1
0
... 0


1
. . . 0
 0

DΦ =  .
..
.. 
..

..
.
.
.

−h x 1 −h x 2 . . . 1
Let Ψ = Φ−1 , so that x = Ψ( y), and define v( y) := u(Ψ( y)), so that u(x) =
v(Φ(x)). In particular, these problems are equivalent and locally we can solve the
curved problem from a solution to the straight problem. More specifically,
D x u(x) = DΦ(x) · D y (v(Φ(x)).
(Note the order of multiplication. This is done because for these lectures we are
taking D to be a column vector.) Whence
0 = F (Du, u, x) = F (DΦ(x) · D y v(Φ(x)), v(Φ(x)), Ψ(Φ(x)))
= F (DΦ(Ψ( y)) · D y v( y), v( y), Ψ( y))
Take G(p, z, y) = F (DΦ(Ψ( y))·p, z, Ψ( y)), so that new problem is to solve G(Dv, v, y) =
0 with v( y) = g(Φ( y)) on { yn = 0}.
The compatibility conditions for the u problem for a triple (p, z, x 0 ) are that
p · b = ∇ b g(x 0 ) and z = g(x 0 ) for every tangent vector b to Γ at x 0 . Admissibility
is the additional requirement that F (p, z, x 0 ) = 0, and non-characteristic is the
further additional requirement that D p F (p, z, x 0 ) · ν 6= 0 (i.e. that the information
does not try to propagate tangent to the boundary data).
To check that a point remains non-characteristic in the transformed problem,
we must check that
n
X
∂ Φn
G pn =
Fp j
= D p F · (−D x̂ h, 1) 6= 0
∂ xj
j=1
(see sheet of corrections)
Hamilton-Jacobi equations
37
2.1.6 Example (Conservation Laws). Let F : R → Rn be a vector field. The associated conservation law is u t + div(F (u)) = 0 for t > 0, with initial data u(·, 0) =
g(·) at t = 0. Let y = (x, t) ∈ Rn+1 and q = (p, qn+1 ), so that p = D x u = Du and
qn+1 = u t . Write G(q, z, y) = qn+1 + F 0 (z) · p so that the conservation is of the form
G((Du, u t ), u, (x, t)) = 0. The system of characteristic equations is

0
 ẏ = (F (z), 1)
q̇ = (−(F 00 (z) · p)p, 0)

ż = qn+1 + F 0 (z) · p = 0
The equations for x and z form a closed system, so we do not need to worry about
tracking the derivative along the characteristics. (This is also evident because
the conservation law is quasi-linear.) Now z is constant along characteristics,
t = s, and ẋ = F 0 (z0 ), so x(t) = F 0 (g(x 0 ))t + x 0 . Notice that if F 0 (g(x 0 )) 6=
F 0 (g(x̃ 0 )) then the characteristics through those points (which are straight lines)
must intersect at some time.
2.2
Hamilton-Jacobi equations
The Hamilton-Jacobi equation is u t + H(Du, x) = 0 for t > 0, with initial data
u(·, 0) = g(·) at t = 0. H : R2n → R is the Hamiltonian. Let y = (x, t), q =
(p, qn+1 ), and G(q, z, y) = qn+1 + H(p, x). The characteristic equations are

 ẏ = (H p (p, x), 1)
q̇ = (−H x (p, x), 0)

ż = H p · p + qn+1
As in the example above we may take t = s, and notice that
¨
ẋ = H p (p, x)
ṗ = −H x (p, x)
is a closed system of equations. This is the Hamiltonian system of ODE—it describes the characteristics. Notice that from the chain rule,
dH
dt
= H p · ṗ + H x · ẋ = −H p · H x + H x · H p = 0,
so the Hamiltonian is constant along characteristics.
2.2.1 Example. Consider H =
1
|p|2
2m
(
+ V (x). The Hamiltonian system is
ẋ =
1
p
m
ṗ = −∇V
We may interpret x at the position of a particle of mass m and V as a potential
acting on the particle. Then p is the momentum and we derive Newton’s Law
38
PDE I
mẍ = ṗ = −∇V , which is the force described by the potential. For gravity we
c
would take V (x) = − |x|
.
1
1
2.2.2 Example (u t + 2 (u x )2 = 0, g (x ) = − 2 x 2 ).
In this example we have H(p, x) = 21 p2 , so the system is

 ẋ = p
ṗ = 0

ż = p2 + q
Whence p(t) = p0 = g 0 (x 0 ) is constant, and x(t) = x 0 + g 0 (x 0 )t. For (x, t) ∈ R2
x
we take x 0 defined by x = x 0 + g 0 (x 0 )t. For this example x = x 0 − x 0 t so x 0 = 1−t
.
2
We have z(t) = ((p0 ) + q0 )t + z0 , where
1
1
q0 = u t (x 0 , 0) = − (u x (x 0 , 0))2 = − (g 0 (x 0 ))2
2
2
from the PDE, and z0 = u(x 0 , 0) = g(x 0 ). Therefore
u(x, t) = z(t) =
0
1
2
0
2
(g (x 0 )) − (g (x 0 ))
2
t + g(x 0 ) =
1 x2
2 t −1
.
Lagrangian description of classical mechanics
Let L : Rn × Rn → R, L = L(q, x), the Lagrangian. Let the set of all paths x :
[0, t] → Rn , such that x ∈ C 2 [0, t] and x(0) = x 0 and x(t) = x t , be denoted
A x 0 ,x t . Let
Z t
I[x] :=
L(ẋ(s), x(s))ds,
0
the action functional. We would like to find the minimum of I over A . Of course,
we need certain conditions on L in order for a minimum to exist.
2.2.3 Theorem (Euler-Lagrange Equations).
Assume x ∈ A is a minimizer of I. Then for all x ∈ [0, t],
−
d
ds
Dq L(ẋ(s), x(s)) + D x L(ẋ(s), x(s)) = 0.
PROOF: Since x is a minimizer, any small perturbation of the curve will increase
I, i.e. I[x + τv] ≥ I[x] for all v : [0, t] → Rn such that v ∈ C 2 [0, t] with v(0) =
v(t) = 0. Define
i(τ) = I[x + τv] =
Z
0
t
L(ẋ + τv̇), x + τv)ds,
Hamilton-Jacobi equations
39
which has a minimum at τ = 0. Therefore
Z t
0
0 = i (0) =
Dq L(ẋ, x) · v̇ + D x L(ẋ, x) · v ds =
0
Z t
−
0
d
ds
Dq L + D x L
by integration by parts. Since C 2 is dense in L 2 , the E-L equations hold.
v ds,
ƒ
1
2.2.4 Example (L(q , x ) = 2 m|q |2 − V(x )).
In this case Dq L = mq and D x L = −∇V , so the E-L equations are
−
d
ds
mẋ(s) = ∇V (x(s)),
mẍ = −∇V.
or
There is a clear connexion between the Lagrangian and Hamiltonian descriptions
(in this case it is given by p = mq).
Given a Lagrangian L, assume that for every x, p ∈ Rn there exists a unique
vector q = q(p, x) such that p = Dq L(q, x) and that q depends smoothly on p
(the idea is that q(p(s), x(s)) = ẋ(s)). Given a solution x(s) to the associated
E-L equations, define p(s) = Dq L(ẋ(s), x(s)), the generalized momentum. Define
H(p, x) = p · q(p, x) − L(q(p, x), x). Then
∂H
∂ pi
= qi (p, x) +
n
X
k=1
pk
∂ qk
− Dqk L(q, x)
∂ pi
∂ qk
∂ pi
= qi (p, x) = ẋ i (s),
which is the first Hamilton equation, and
∂H
∂ xi
=
n
X
k=1
pk
∂ qk
∂ xi
− Dqk L(q, x)
∂ qk
∂ xi
− D x i L(q, x) = −D x i L(q, x).
By the E-L equations, this is equal to ṗi (s).
Legendre transform
For this section we consider only Hamiltonians and Lagrangians that do not depend on the space variables. The Hamilton-Jacobi equations become
¨
u t + H(Du)
u(·, 0) = g(·)
We further assume that L(q) is convex on Rn , and
lim
|q|→∞
L(q)
|q|
= ∞.
2.2.5 Definition. The Legendre transform of L is defined to be
L ∗ (p) = sup {p · q − L(q)}.
q∈Rn
40
PDE I
Formally, the supremum is attained at q = q(p) = (Dq L)−1 (p) (noting that
Dq L : Rn → Rn and taking the inverse to be the matrix inverse). Therefore L ∗ (p) =
pq(p) − L(q(p)) = H(p) by the definition of the Hamiltonian associated with the
Lagrangian L.
2.2.6 Theorem (Convex duality). Assume that L satisfies the assumptions above.
Define H = L ∗ . Then
1. H(p) is convex on Rn and
lim
H(p)
|p|
|p|→∞
= ∞.
2. L = H ∗ .
PROOF:
1. H(p) = supq∈Rn {p · q − L(q)} is a supremum of linear functions, so it is
p
convex. For the particular value q = λ |p| we have
H(p) = λ|p| − L(λ
p
|p|
) ≥ λ|p| − kLk L ∞ (B(0,λ)) ,
so
lim inf
|p|→∞
H(p)
|p|
≥λ
and the limit is ∞.
2. For all p, q ∈ Rn , H(p) + L(q) ≥ p · q, so subtracting H(p) from both sides
and taking the supremum gives L(q) ≥ H ∗ (q). For the other inequality,
H ∗ (q) = sup {p · q − sup {r · p − L(r)}}
p∈Rn
r∈Rn
= sup { infn {p · (q − r) − L(r)}}
p∈Rn r∈R
≥ infn {Dq L(q) · (q − r) − L(r)}
r∈R
≥ infn {L(r) − Dq L(q) · (r − q)}
r∈R
≥ L(q)
by convexity
If L is not differentiable at every point, instead of Dq L use a supporting
hyper-surface. (For every point there is at least one hyper-surface containing
that point for which the graph of the convex function L lies entirely on one
side.)
ƒ
Hamilton-Jacobi equations
41
Hopf-Lax formula
We will give a formula for the solution to the H-J equation when H is smooth,
depends only on p, is convex, and satisfies
H(p)
lim
|p|→∞
|p|
= ∞.
We further assume that g is Lipschitz. We consider the modified action functional
I[w] =
t
Z
L(ẇ(s))ds + g(w(0))
0
where the set of admissible functions are
A x,t = {w ∈ C 1 ([0, t], Rn ), w(t) = x}.
Our candidate for the solution is u(x, t) = infw∈A x,t I[w].
2.2.7 Theorem (Hopf-Lax formula).
u(x, t) = minn {t L(
y∈R
x− y
)+
t
g( y)}
PROOF: Consider the line joining y to x that takes an amount of time t. It has
equation w(s) = y + st (x − y) and is clearly admissible. Since u is defined as an
infimum,
Z t
u(x, t) ≤
L(
0
x− y
)ds
t
+ g( y) = t L(
x− y
)+
t
g( y)
and so u(x, t) is at most the right hand side of the formula. Conversely, by Jensen’s
inequality,
Z
t
L(ẇ(s))
0
ds
t
Z
≥L
t
ẇ(s)
ds
L is convex
t
x − w(0)
0
=L
by the FTC
t
Therefore, adding g(w(0)) to both sides and noting that taking the infimum over
all w ∈ A x,t amounts to, on the right hand side, taking the infimum over all
starting points y ∈ Rn .
u(x, t) = inf
w∈A x,t
Z
0
t
x− y
L(ẇ(s))ds + g(w(0)) ≥ infn {t L( t ) + g( y)}.
y∈R
Finally, the infimum is actually a minimum since g is Lipschitz, and hence grows
at most linearly, while L grows super-linearly and pwns g.
ƒ
42
PDE I
2.3
Conservation Laws
Push-forward of a measure
Let Φ : Rn → Rn . Given a measure µ on Rn , the push-forward of µ in Φ is Φ# µ. For
ξ ∈ Cc (Rn ),
Z
Z
ξ dΦ# µ =
ξ ξ ◦ Φ dµ.
Consider the system of ODE
¨
ẏ(t) = V ( y(t), t)
y(0) = y0
where V is Lipschitz in y, and let Φ t : y0 → y(t) be the unique solution map.
Let g( y) denote the density of a material at a location y ∈ Rn at time 0, and let
u(x, t) denote the density at x ∈ Rn at time t, where the idea is that the points in
Rn have been transported according to Φ t . For a conservation law we require that
for every domain Ω,
Z
Z
u(x, t)d x =
Φ t (Ω)
g( y)d y.
Ω
For every ψ ∈ Cc∞ (Rn ) we have (noting the push-forward of measures)
Z
Z
ψ(x)u(x, t)d x =
Differentiating with respect to t,
Z
Z
ψ(x)u t (x, t)d x =
Therefore
Z
ψ(Φ t ( y))g( y)d y.
∇ψ(Φ t ( y))·V (Φ t ( y), t)g( y)d y =
Z
∇ψ(x)·(V (x, t)u(x, t))d x.
ψ(x)(u t (x, t), div(V (x, t)u(x, t)))d x = 0
for all smooth functions of compact support, so u t + div(V u) = 0. We restrict our
attention to one space dimension, u t + div F (u) x = 0, where F : R → R. We may
also write u t + f (u)u x = 0, where f = F 0 . The characteristics are ẋ = f (z), ż = 0,
so the characteristics are straight lines x(t) = x 0 + t f (g(x 0 )).
Burger’s equation is the conservation law
‚ 2Œ
u
ut +
= u t + uu x = 0.
2 x
For initial data g( y) = 1(−∞,0] , there are many “almost everywhere solutions”
given by placing the shock line at different angles. Such a notion of solution is not
useful because they are not unique (and in this case there is not even a preferred
solution).
Rankine-Hugoniot condition
43
Weak solutions
If u t + F (u) x = 0 with initial data g on R × {0}, then for any φ ∈ Cc∞ (R × [0, ∞)),
Z
∞Z ∞
0
φ(x, t)(u t + F (u) x )d x d t = 0.
−∞
If u were a classical solution then we could integrate by parts to get
Z ∞Z ∞
Z∞
φ t u + φ x F (u)d x d t +
0
−∞
φ(x, 0)g(x)d x = 0.
−∞
2.3.1 Definition. u ∈ L ∞ (R × [0, ∞)) is an integral solution of the conservation
law if the equation above holds for every φ ∈ Cc∞ (R × [0, ∞)).
This notion gives a unique (preferred) solution to Burger’s equation with initial
data g = 1(−∞,0] . The “shock line” is x = 12 t. (This may not be correct.)
2.4
Rankine-Hugoniot condition
We now give a characterization of piece-wise continuously differentiable integrable solutions u to the scalar convervation law u t + F (u) x = 0 with initial data
g (at time 0). Let γ = γ(t) be the curve of non-differentiability (and possibly discontinuity) of u, Ω L be the domain to the left (smaller t) and ΩR be the domain to
the right (larger t). Then R × [0, ∞) = Ω L ∪ γ ∪ ΩR . For (x, t) ∈ γ let
u L (x, t) =
lim
( y,s)∈Ω L
( y,s)→(x,t)
u( y, s) and uR (x, t) =
lim
( y,s)∈ΩR
( y,s)→(x,t)
u( y, s).
Let ϕ be smooth with compact support and such that ϕ(x, t) 6= 0 and supp(ϕ) ∩
R × {0} = ∅. Then since u is an integral solution, and ϕ is zero at time zero,
Z ∞Z
uϕ t + F (u)ϕ x d x d t = 0,
0
R
so
Z
uϕ t + F (u)ϕ x d x d t +
Ω L ∩supp(ϕ)
Z
uϕ t + F (u)ϕ x d x d t = 0.
ΩR ∩supp(ϕ)
Let V = (F (u)ϕ, uϕ), a vector field. Then
div V = F (u) x ϕ + F (u)ϕ x + u t ϕ + uϕ t ,
and by the divergence theorem,
Z
F (u) x ϕ + F (u)ϕ x + u t ϕ + uϕ t d x d t =
Ω L ∩supp(ϕ)
Z
γ
(F (u L )ϕ, u L ϕ) · ν dS.
44
PDE I
Now a continuously differentiable integral solution is a classical solution, so on
Ω L , u t + F (u) x )ϕ = 0. Applying the same reasoning to ΩR we get
Z
Z
(F (u L )ϕ, u L ϕ) · ν dS −
(F (uR )ϕ, uR ϕ) · ν dS = 0.
γ
γ
Therefore, for every such ϕ, since ν = (1, γ̇),
Z
((F (uR ) − F (uR )) − (u L − uR )γ̇)ϕ dS = 0,
γ
and so F (u L ) − F (uR ) = (u L − uR )γ̇. This is the Rankine-Hugoniot condition. Conversely, if we have a piece-wise continuously differentiable function that is a solution inside its domains of differentiability and such that the R-H condition holds,
then it is an integral solution. Note in particular that a continuous piece-wise
continuously differentiable solution is always an integral solution.
2.4.1 Example (Burger’s equation, revisited).
Do this.
2.4.2 Example (Riemann problem).
Let F be a convex function with F 00 > 0. We consider the PDE u t + F (u) x = 0 with
initial data u L 1(−∞,0] + uR 1(0,∞) , where u L , uR ∈ R. The characteristics are straight
lines that collide, and if u L > uR then the slope of the shock is given by the R-H
conditions, and it is
F (u L ) − F (uR )
γ̇ =
=: σ.
u L − uR
The corresponding solution is
¨
u(x, t) =
x ≤ σt
x > σt.
uL
uR
If u L < uR then the above is still a solution. Notice that F 0 (u L ) < σ < F 0 (uR ) since
F is convex, so the shock line is in the “gap” where there are no characteristics
emanating from the initial data. This type of shock (with characteristics emanating
from it) is a non-physical shock. We look for another solution with characteristics
passing through (0, 0), i.e. a solution for which u(x, t) = ϕ( xt ). Plugging in, we
obtain
x‹ x
  x ‹‹  x ‹ 1
0
ϕ0
−ϕ 0
+
F
ϕ
= 0.
t t2
t
t t
It suffices that F 0 (ϕ( xt )) =
corresponding solution is
u(x, t) =
x
.
t
Take ϕ = (F 0 )−1 on the range [u L , uR ], and the

u L
(F 0 )−1 ( xt )

uR
x
t
< F 0 (u L )
x
t
> F (uR ).
F 0 (u L ) <
0
x
t
< F (uR )
Existence of solutions
45
This solution is the rarefaction wave.
There are a number of heuristic reasons for preferring the rarefaction wave solution to the non-physical wave solution. One is stability (consider smooth initial
data approximating the discontinuous initial data). Another is that of vanishing
viscosity (consider u t + F (u) x = "u x x and we hope that solutions u" to this equation converge to the solution the conservation law as " → 0). Such conditions are
called entropy conditions.
2.4.3 Definition.
1. The Lax entropy condition at a shock is
F 0 (u L ) >
F (u L ) − F (uR )
u L − uR
> F 0 (uR )
when u L > uR .
2. For general (non-convex) F , the Oleinik entropy condition is
a) if u L > uR then the secant is above the graph of F ; and
b) if uR > u L then the secant is below the graph of F .
3. u satisfies the entropy condition if there is C > 0 for all x ∈ R and h > 0 and
t > 0 then u(x + h, t) − u(x, t) ≤ ct h. The third condition allows only jumps
downwards. It says (for small t) that the gradient decays as 1t .
4. A pair (η, ψ) is an entropy-entropy-flux pair if η > 0 and ψ0 = η0 F . The idea
is that u solves u t + F 0 (u) x = 0 only if η(u) t + ψ(u) x = 0.
2.5
Existence of solutions
Rx
Let h(x) = 0 g( y)d y. Then if there is a solution u to the conservation law then
consider w x = u. The function w satisfies w t + F (w x ) = 0 with initial data h. This
is a Hamilton-Jacobi PDE in w. We have a formula
x−y
+ h( y)
w(x, t) = min t L
y
t
where L = F ∗ . The candidate for u is
u(x, t) =
∂
x−y
min t L
+ h( y) .
∂x y
t
Suppose the minimum is attained at y(x, t). Then y(x, t) is increasing in x. Since
y(x, t) in increasing in x, it is differentiable in x a.e., so
x − y(x, t)
0
u(x, t) = L
(1 − y x ) + h0 ( y(x, t)) y x .
t
Since the function z 7→ t L( 1t (x − y(z, t))) + h( y(z, t)) has a minimum at z = x,
we obtain that
x − y(x, t)
h0 ( y(x, t)) y x = L 0
yx ,
t
46
PDE I
so
u(x, t) = L 0
x − y(x, t)
t
=G
x − y(x, t)
t
,
where G = (F 0 )−1 . This is the Lax-Oleinik formula.
Aside: We show that w(x, t) (defined above) is Lipschitz in x. Consider
w(x 2 , t) − w(x 1 , t)
x2 − y
x 1 − y(x 1 , t)
= min t L
+ h( y) − t L
− h( y(x 1 , t))
y
t
t
≤ h(x 2 − x 1 + y(x 1 , t)) − h( y(x 1 , t))
≤ Lip(h)|x 2 − x 1 |
so (doing the reverse as well) |w(x 2 , t) − w(x 1 , t)| ≤ Lip(h)|x 2 − x 1 |.
2.5.1 Theorem. Under the assumptions of Theorem 1 (F smooth, F (0) = 0, F 00 >
0, F super-linear at infinity, g ∈ L ∞ ) u defined above is an integral solution to the
conservation law.
PROOF: We have w t + F (w x ) = 0 a.e. Let ϕ ∈ Cc∞ (R × [0, ∞)) be a test function.
Then
Z ∞Z
w t ϕ x + F (w x )ϕ x d x d t = 0,
0
R
and integrating by parts,
Z
∞Z
0
wt ϕx d x d t =
R
=
Z
∞Z
0
R
∞Z
Z
0
Therefore
Z
∞Z
0
Z
−wϕ x t d x d t −
R
w x ϕt d x d t −
Z
R
uϕ t + F (u)ϕ x d x d t +
R
wϕ x d x w x ϕd x R
Z
t=0
t=0
gϕd x = 0
R
and u is an integral solution.
ƒ
2.5.2 Lemma. Under the assumptions of the theorem,
1. u given by the formula above satisfies the entropy condition (E3); and
2. there is C such that u defined above satisfies
u(x + h, t) − u(x, t) ≤
for all x ∈ R, t > 0, h > 0.
C
t
h
Viscosity solutions
47
PROOF: We claim that there is a constant C̃ such that for every x
x − y(x, t) ≤ C̃.
t
Assuming this claim, modify G outside the interval [−C̃, C̃] into G̃ so that G̃ is
Lipschitz. Then
x − y(x, t)
u(x, t) = G̃
t
x − y(x + h, t)
≥ G̃
t
x + h − y(x + h, t)
h
≥ G̃
− Lip(G̃)
t
t
h
≥ u(x + h, t) − Lip(G̃).
ƒ
t
2.6
Viscosity solutions
The most general second order equation is F (D2 u, Du, u, x) = 0, where x ∈ Rn
and u : Ω ⊆ Rn → R is the unknown. This is too general a class to investigate.
We are interested in those PDE whose solutions satisfy the comparison principle.
We have seen that Laplace’s equation is of this type, and it can be shown that
the Hamilton-Jacobi equation is of this type. The comparison principle will imply
uniqueness of solutions.
2.6.1 Definition. We say that u is a sub-solution if F (D2 u, Du, u, x) ≤ 0 in Ω, and
u is a super-solution if F (D2 u, Du, u, x) ≥ 0 in Ω. We may also speak of strict suband super-solutions, for which the inequality is strict.
Let u be a sub-solution and v be a super-solution and u ≤ v on ∂ Ω. When is
u ≤ v in Ω? For which equations does the comparison principle hold? Assume
that u − v has a positive maximum at some x 0 ∈ Ω. Then Du = Dv and D2 u ≤ D2 v
at x 0 . (Recall that for a matrix M , we say M ≥ 0 if x T M x ≥ 0 for all x ∈ Rn .)
Assume for the moment that v is a strict super-solution. Then at x 0 ,
F (D2 u, Du, u, x 0 ) ≤ 0 < 0F (D2 v, Dv, v, x 0 ),
and we have comparisons among the entries.
2.6.2 Definition. A function F : Sym(n) × Rn × R × Ω → R is
1. degenerate elliptic if for all X , Y ∈ Sym(n), p ∈ Rn , r ∈ R, and x ∈ Ω, X ≤ Y
implies F (X , p, r, x) ≥ F (Y, p, r, x); and
2. proper elliptic if F is degenerate elliptic and non-decreasing in r.
48
PDE I
2.6.3 Examples.
1. The Poisson equation −∆u = f is given by F (X , x) = − tr X − f (x), and is
proper. Note that ∆u = f is not degenerate elliptic.
2. −∆u + u = 0 is proper while −∆u − u is degenerate elliptic but not proper.
3. H(Du) = 0 is trivially proper.
4. The following PDE are proper.
a) u t + H(Du) = 0
b) u t − ∆u = 0
c) div( p
Du
1+|Du|2
d) u t − div( p
)=0
Du
1+|Du|2
)=0
2.6.4 Definition. Let u : Ω → R and let
u∗ (x) := lim+ sup{u( y) | y ∈ Ω, |x − y| < r}
r→0
and u∗ similarly with an infimum instead. Then u∗ : Ω → {∞} is the upper semicontinuous envelope of u and u∗ : Ω → {−∞} is the lower semi-continuous envelope
of u.
We have u∗ ≤ u ≤ u∗ , and u∗ is the largest lower semi-continuous with this
property, and u∗ is the smallest upper semi-continuous function with this property.
It is not hard to show that an upper semi-continuous function reaches its maximum
on a compact set.
2.6.5 Definition. Let F (D2 u, Du, u, x) be degenerate elliptic and proper. u : Ω →
R is a viscosity sub-solution of F (D2 u, Du, u, x) = 0 if u is upper semi-continuous
and for every ϕ ∈ C 2 (Ω) such that u − ϕ has a local maximum at some x̄ ∈ Ω, we
have F (D2 ϕ(x̄), Dϕ(x̄), u(x̄), x̄) ≤ 0.
The idea is that we must test against all C 2 functions that touch u from above
(since we take derivatives of ϕ, we may add constant in such a way to guarantee
that u − ϕ is non-positive with maximum value 0). Super-solutions are defined
analogously. In this case we test against functions touching u from below.
2.6.6 Definition. u is a viscosity sub-solution to the boundary value problem
¨
F (D2 u, Du, u, x) = 0
on Ω
u = gon ∂ Ω, Ω = Ω̃ × [0, T ]
if it is a sub-solution and u ≤ g on ∂ Ω.
2.6.7 Proposition. If u is a classical sub-solution to the PDE then it is a viscosity
sub-solution.
Viscosity solutions
49
PROOF: If F (D2 u, Du, u, x) ≤ 0 then for any ϕ ∈ Cc2 touching u from above at x,
Du(x) = Dϕ(x) and D2 u(x) ≤ D2 ϕ(x), so
F (D2 ϕ(x), Dϕ(x), u(x), x) ≤ F D2 u, Du, u, x) ≤ 0.
ƒ
2.6.8 Proposition. If u is a viscosity solution and u is C 2 then it is a classical
solution.
PROOF:
2.6.9 Example. Take F (X , p, z, x) = |p|2 − 1, so the corresponding PDE is |Du|2 −
1. In one dimension, consider the domain Ω = (0, 2). There is no smooth function
u such that u(0) = u(2) = 0 and |u0 | = 1. But the function
¨
x
2− x
x ∈ (0, 1)
x ∈ (1, 2)
Then u is a viscosity solution to the PDE. (There are no nontrivial functions to
check for super-solution.)
The function u fails to be a sub-solution to the PDE associated with F̃ = −F
(i.e. −|Du|2 + 1 = 0), but −u is a solution. Crazy.
Viscosity solutions to Hamilton-Jacobi equations are appropriate for problems
arising in optimal (stochastic) control (and hence finance), but they may not be
the “right” solution for problems with other motivation.
The main reference on viscosity solutions is a paper by Crandall, Ishii, and
Lions, “User’s guide to viscosity solutions.”
Comparison
We consider the Hamiltonian H(Du, x). Suppose there is continuous w : [0, ∞) →
[0, ∞) with w(0) = 0 and w > 0 on (0, ∞) such that for all p ∈ Rn and x, y ∈ Ω,
|H(p, x) − H(p, y)| ≤ w((1 + |p|)|x − y|).
2.6.10 Theorem (Comparison). Let u be a viscosity sub-solution and v be a viscosity super-solution of u t + H(Du, x) = 0 on Ω × (0, T ), where Ω is bounded and
open. Assume further that u is upper semi-continuous on Ω×[0, T ] and v is lower
semi-continuous on Ω × [0, T ]. If u ≤ v on the parabolic boundary and H is as
above then u ≤ v in Ω × [0, T ).
2.6.11 Theorem. Let F be proper, F = F (X , p, x), such that the statement above
holds uniformly in X (so w does not depend on X ). Then comparison holds for
u t + F (D2 u, Du, x) = 0.
50
PDE I
Stability and Existence
Suppose −∆un = 0 are a sequence of classical solutions to the Laplace equation.
If un → u uniformly (i.e. in C) then can we conclude −∆u = 0? In this case, yes,
because of the mean-value property.
2.6.12 Definition. For a sequence of functions {un },
∗
lim sup un (x) = lim sup{un ( y) | n ≥ m, y ∈ Ω, |x− y| ≤
m→∞
1
}
m
= sup{lim sup unk (x nk )}
k→∞
The last equality is an exercise. Also show (via diagonalization) that the supremum is attained by some sequence.
2.6.13 Lemma. Let un be a sequence of upper semi-continuous functions on Ω,
and let x̂ ∈ Ω. Assume that u ≥ lim sup∗ un , u is USC, and that u(x̂) = (lim sup∗ un )(x̂).
Assume ϕ ∈ C 2 (Ω) and u − ϕ has a strict local maximum at x̂. Then there exists
a sequence nk → ∞ and x k → x̂ such that unk − ϕ has a local maximum at x k and
unk (x k ) → u(x̂).
This lemma also holds if instead un → u uniformly.
PROOF: There is r > 0 such that
u(x̂) − ϕ(x̂) > u(x) − ϕ(x)
for all x ∈ B(x̂, r). Let x̂ n be the location of the maximum of un − ϕ on B(x̂, r).
(This value may be on the boundary of the ball.) Since u(x̂) = (lim sup∗ un )(x̂),
there is a sequence (x̃ k , nk ) → (x̂, ∞) such that u(x̂) = limk→∞ unk (x̃ k ). Without
loss of generality, assume nk = k (by throwing away some of the un ’s). Let x̄ be
an accumulation point of x̂ n . There exists a subsequence ñk such that x̂ ñk → x̄.
ñk (x̂ ñk ) − ϕ(x̂ ñk )
> uñk (x̃ ñk ) − ϕ(x̃ ñk )
so
u(x̄) − ϕ(x̄) ≥ lim sup uñk (x̂ ñk ) − ϕ(x̂ ñk )
k→∞
≥ lim sup uñk (x̃ ñk ) − ϕ(x̃ ñk )
k→∞
= u(x̂) − ϕ(x̄)
This contradicts the strictness of x̂ unless x̄ = x̂. Therefore the desired sequence
is x̂ ñk .
ƒ
2.6.14 Theorem. Let un be a viscosity sub-solution of Fn (D2 u, Du, u, x) = 0, where
Fn is proper for each n. Assume F proper and such that F ≤ lim inf∗ Fn . Let u =
lim sup∗ un . If u is finite then u is a viscosity sub-solution of F (D2 u, Du, u, x) = 0.
Viscosity solutions
51
PROOF: Let ϕ be a text function such that u − ϕ has a local maximum at some
x̂. Without loss of generality we may assume the maximum is strict (Indeed,
let ϕ̃(x0 = ϕ(x) + 41 |x − x̂|4 , so that u = ϕ̃ has a strict local maximum at x̂.
We will show that F (D2 ϕ(x̂), Dϕ(x̂), u(x̂), x̂) ≤ 0, but it is enough to show that
F (D2 ϕ̃(x̂), Dϕ̃(x̂), u(x̂), x̂) ≤ 0.)
By the previous lemma there exists a subsequence x̂ k → x̂ and nk → ∞ such
that unk − ϕ has a local maximum at x̂ k . Then for each k,
F (D2 ϕ(x̂ k ), Dϕ(x̂ k ), unk (x̂ k ), x̂ k ) ≤ 0
But the entries converge (respectively) to D2 ϕ(x̂), Dϕ(x̂), u(x̂), and x̂. By the
assumption on F , we get the desired inequality.
ƒ
Read about Perron’s method for existence.
Index
action functional, 38
admissible, 35
admissible set, 17
Lax-Oleinik formula, 46
Legendre transform, 39
linear, 2, 3
lower semi-continuous envelope, 48
Burger’s equation, 42
mean-value property, 10
method of characteristics, 33
minimal surface equation, 2
mollifier, 10
Cauchy problem, 20
characteristic curve, 3
compatibility conditions, 34
conservation law, 2, 37, 42
conserved quantity, 2
continuity equation, 2
Naiver-Stokes equations, 2
non-characteristic, 33, 35
non-physical shock, 44
nonlinear diffusion, 2
degenerate elliptic, 47
Dirichlet problem, 22
Duhamel’s principle, 21
Oleinik entropy condition, 45
energy, 17, 18, 26, 32
entropy, 26
entropy condition, 45
entropy-entropy-flux pair, 45
parabolic boundary, 22
parabolic cylinder, 22, 25
Poisson equation, 4
proper elliptic, 47
push-forward, 42
first order, 2
fundamental solution, 5, 20
quasi-linear, 32
generalized momentum, 39
Green’s function, 15
radial solution, 5
Rankine-Hugoniot condition, 44
rarefaction wave, 45
reaction-diffusion equation, 2
regularization, 11
Hamilton-Jacobi equation, 2, 37
Hamiltonian, 37
harmonic function, 4
heat equation, 2
homogeneous, 3
Schrödinger’s equation, 2
similarity profile, 20
sub-solution, 23, 47
super-solution, 23, 47
inhomogeneous, 3
initial conditions, 18
integral solution, 43
translation invariant, 5
transport equation, 2, 3
Kirchhoff’s formula, 30
upper semi-continuous envelope, 48
Lagrangian, 38
Laplace equation, 4
Laplace’s equation, 2
Laplacian, 2
lateral boundary conditions, 18
Lax entropy condition, 45
viscosity sub-solution, 48
wave equation, 2
well-posed, 3
53