III Stochastic Calculus and Applications - Stochastic differential equations

4Stochastic differential equations

III Stochastic Calculus and Applications

4.3 Representations of solutions to PDEs

Recall that in Advanced Probability, we learnt that we can represent the solution

to Laplace’s equation via Brownian motion, namely if

is a suitably nice domain

and

∂D → R

is a function, then the solution to the Laplace’s equation on

with boundary conditions g is given by

u(x) = E

[g(B

)],

where T is the first hitting time of the boundary ∂D.

A similar statement we can make is that if we want to solve the heat equation

∂u

∂t

= ∇

with initial conditions u(x, 0) = u

(x), then we can write the solution as

u(x, t) = E

(

√

)]

This is just a fancy way to say that the Green’s function for the heat equation is

a Gaussian, but is a good way to think about it nevertheless.

In general, we would like to associate PDEs to certain stochastic processes.

Recall that a stochastic PDE is generally of the form

= b(X

) dt + σ(X

) dB

for some

→ R

and

→ R

d×m

which are measurable and locally

bounded. Here we assume these functions do not have time dependence. We

can then associate to this a differential operator L defined by

L =

i,j

∂

where a = σσ

Example. If b = 0 and σ =

√

2I, then L = ∆ is the standard Laplacian.

The basic computation is the following result, which is a standard application

of the Itˆo formula:

Proposition.

Let

x ∈ R

, and

a solution to

(

σ, b

). Then for every

f : R

× R

→ R that is C

in R

and C

in R

, the process

= f(t, X

) − f(0, X

) −



∂

∂s

+ L



f(s, X

) ds

is a continuous local martingale.

We first apply this to the Dirichlet–Poisson problem, which is essentially to

solve

−Lu

. To be precise, let

U ⊆ R

be non-empty, bounded and open;

f ∈ C

(

) and

g ∈ C

(

∂U

). We then want to find a

u ∈ C

(

) =

(

)

∩C

(

)

such that

−Lu(x) = f(x) for x ∈ U

u(x) = g(x) for x ∈ ∂U.

= 0, this is called the Dirichlet problem; if

= 0, this is called the Poisson

problem.

We will have to impose the following technical condition on a:

Definition

(Uniformly elliptic)

We say

U → R

d×d

is uniformly elliptic if

there is a constant c > 0 such that for all ξ ∈ R

and x ∈

U, we have

a(x)ξ ≥ c|ξ|

is symmetric (which it is in our case), this is the same as asking for the

smallest eigenvalue of a to be bounded away from 0.

It would be very nice if we can write down a solution to the Dirichlet–Poisson

problem using a solution to

(

σ, b

), and then simply check that it works. We

can indeed do that, but it takes a bit more time than we have. Instead, we shall

prove a slightly weaker result that if we happen to have a solution, it must be

given by our formula involving the SDE. So we first note the following theorem

without proof:

Theorem.

Assume

has a smooth boundary (or satisfies the exterior cone

condition),

a, b

are H¨older continuous and

is uniformly elliptic. Then for

every H¨older continuous

U → R

and any continuous

∂U → R

, the

Dirichlet–Poisson process has a solution.

The main theorem is the following:

Theorem.

Let

and

be bounded measurable and

σσ

uniformly elliptic,

U ⊆ R

as above. Let

be a solution to the Dirichlet–Poisson problem and

solution to E

(σ, b) for some x ∈ R

. Define the stopping time

= inf{t ≥ 0 : X

6∈ U}.

Then ET

< ∞ and

u(x) = E

g(X

) +

f(X

) ds

In particular, the solution to the PDE is unique.

Proof.

Our previous proposition applies to functions defined on all of

, while

u is just defined on U. So we set



x ∈ U : dist(x, ∂U ) >



, T

= inf{t ≥ 0 : X

6∈ U

and pick

∈ C

(

) such that

. Recalling our previous notation,

let

= (M

)

= u

t∧T

) − u

) −

t∧T

) ds.

Then this is a continuous local martingale that is bounded by the proposition,

and is bounded, hence a true martingale. Thus for

x ∈ U

and

large enough,

the martingale property implies

u(x) = u

(x) = E

u(X

t∧T

) −

t∧T

Lu(X

) ds

= E

u(X

t∧T

) +

t∧T

f(X

) ds

We would be done if we can take

n → ∞

. To do so, we first show that

[

]

< ∞

Note that this does not depend on

and

. So we can take

= 1 and

= 0,

and let v be a solution. Then we have

E(t ∧ T

) = E

−

t∧T

Lv(X

) ds

= v(x) − E(v(X

t∧T

)).

Since

is bounded, by dominated/monotone convergence, we can take the limit

to get

E(T

) < ∞.

Thus, we know that t ∧ T

→ T

as t → ∞ and n → ∞. Since

|f(X

)| ds

≤ kfk

∞

E[T

] < ∞,

the dominated convergence theorem tells us

t∧T

f(X

) ds

→ E

f(X

) ds

Since u is continuous on

U, we also have

E(u(X

t∧T

)) → E(u(T

)) = E(g(T

)).

We can use SDE’s to solve the Cauchy problem for parabolic equations as

well, just like the heat equation. The problem is as follows: for

f ∈ C

(

), we

want to find u : R

× R

→ R that is C

in R

and C

in R

such that

∂u

∂t

= Lu on R

× R

u(0, ·) = f on R

Again we will need the following theorem:

Theorem.

For every

f ∈ C

(

), there exists a solution to the Cauchy problem.

Theorem.

Let

be a solution to the Cauchy problem. Let

be a solution to

(σ, b) for x ∈ R

and 0 ≤ s ≤ t. Then

(f(X

) | F

) = u(t − s, X

In particular,

u(t, x) = E

(f(X

)).

In particular, this implies X

is a continuous Markov process.

Proof.

The martingale has

∂

∂t

, but the heat equation has

∂

∂t

− L

. So we set

g(s, x) = u(t − s, x). Then



∂

∂s

+ L



g(s, x) = −

∂

∂t

u(t − s, x) + Lu(t − s, x) = 0.

So g(s, X

) − g(0, X

) is a martingale (boundedness is an exercise), and hence

u(t − s, X

) = g(s, X

) = E(g(t, X

) | F

) = E(u(0, X

) | F

) = E(f(X

) | X

There is a generalization to the Feynman–Kac formula.

Theorem

(Feynman–Kac formula)

Let

f ∈ C

(

) and

V ∈ C

(

) and

suppose that u : R

× R

→ R satisfies

∂u

∂t

= Lu + V u on R

× R

u(0, ·) = f on R

where V u = V (x)u(x) is given by multiplication.

Then for all t > 0 and x ∈ R

, and X a solution to E

(σ, b). Then

u(t, x) = E



f(X

) exp



V (X

) ds



is the Laplacian, then this is Schr¨odinger equation, which is why Feynman

was thinking about this.