4 Green’s Functions

4 Green's Functions

In this section, we are interested in solving the following problem. Let be an open, bounded

subset of Rn. Consider

-u = f u=g

x Rn x .

(4.1)

4.1 Motivation for Green's Functions

Suppose we can solve the problem,

-yG(x, y) = x G(x, y) = 0

y y

(4.2)

for each x . Then, formally, we can say that for u a solution of (4.1),

u(x) = xu(y) dy

= - yG(x, y)u(y) dy

=

yG(x, y) ? yu(y) dy -

G

(x,

y)u(y)

dS(y)

=-

G(x, y)yu(y) dy +

G(x,

y)

u

(y)

dS

(y)

-

G

(x,

y)u(y)

dS(y)

=

G(x, y)f (y) dy -

G

(y)g(y)

dS(y).

Now, we do know that the fundamental solution of Laplace's equation (y) satisfies

-y(y) = 0

and, moreover,

-y(x - y) = x.

Of course, (x - y) does not satisfy our boundary conditions, but we will use that as a starting ground to try and construct a solution of (4.2), and, ultimately (4.1). Below, we will also make the formal argument given above more precise.

Recalling the definition of distributional derivative, we will start by looking at

(x - y)yu(y) dy.

We would like to integrate this term by parts. However, we know that (x - y) has a singularity at y = x. Therefore, in order to integrate by parts, we must proceed as follows.

Fix x and > 0 such that dist(x, ) < and therefore, B(x, ) . Let V - B(x, ).

1

x

V

Let be the fundamental solution of Laplace's equation. That is,

(x) =

-

1 2

ln

1

|x|

1

n(n-2)(n) |x|n-2

n=2 n 3.

Suppose u C2(). By the Divergence Theorem, we have

(y - x)u(y) dy = -

V

y(y - x) ? yu(y) dy +

V

V

(y

-

x)

u

dS(y)

=

y(y - x)u(y) dy -

V

V

(y

-

x)u(y)

dS(y)

+

V

(y

-

x)

u

dS(y).

where

u

denotes

the

derivative

of

u

in

the

outer

normal

direction

to

V.

Now on V ,

y(y - x) = 0. Therefore,

(y - x)u(y) dy = -

V

V

(y

-

x)u(y)

dS(y)

+

V

(y

-

x)

u

dS(y).

Now, we note that

lim (y - x)u(y) dy = (y - x)u(y) dy.

0+ V

We make the following claims about the limits of the other two terms as 0+.

Claim 1.

lim

0+

-

V

(y

-

x)u(y)

dS(y)

=-

(y

-

x)u(y)

dS(y)

-

u(x).

Claim 2.

lim

0+

V

(y

-

x)

u

(y)

dS(y)

=

(y

-

x)

u

dS(y).

Assuming these claims for a moment, we conclude that for any u C2(),

u(x) =

(y

-

x)

u

(y)

-

(y

-

x)u(y)

dS(y) -

(y - x)u(y) dy.

(4.3)

2

We would now like to use the representation formula (4.3) to solve (4.1). If we knew u

on and u on and

u

on , then we could solve for u.

But, we don't know all this

information. We know u on and u on .

We proceed as follows. For each x , we introduce a corrector function hx(y) which

satisfies the following boundary-value problem,

yhx(y) = 0 hx(y) = (y - x)

y y .

(4.4)

Now suppose we can find such a (smooth) function hx which satisfies (4.4). Then using the same analysis as above, we have

hx(y)u(y) dy = -

yhx(y) ? yu(y) dy +

hx(y)

u

(y)

dS(y)

=

yhx(y)u(y) dy -

hx

(y)u(y)

dS(y)

+

hx(y)

u

(y)

dS(y).

Now using the fact that hx is a solution of (4.4), we conclude that

0=

(y

-

x)

u

(y)

-

hx

(y)u(y)

dS(y) -

hx(y)u(y) dy.

(4.5)

Now subtracting (4.5) from (4.3), we conclude that

u(x) = -

(y

-

x)

-

hx

(y)

u(y) dS(y) -

[(y - x) - hx(y)]u(y) dy.

Let G(x, y) = (y - x) - hx(y).

Then, u can be written as

u(x) = -

G

(x,

y)u(y)

dS(y)

-

G(x, y)u(y) dy.

(4.6)

We define this function G as the Green's function for . That is, the Green's function for a domain Rn is the function defined as

G(x, y) = (y - x) - hx(y) x, y , x = y,

where is the fundamental solution of Laplace's equation and for each x , hx is a solution of (4.5). We leave it as an exercise to verify that G(x, y) satisfies (4.2) in the sense of distributions.

Conclusion: then

If u is a (smooth) solution of (4.1) and G(x, y) is the Green's function for ,

u(x) = -

G

(x,

y)g(y)

dS(y)

+

G(x, y)f (y) dy.

(4.7)

3

We will show below that conversely a function of the form (4.7) will give us a solution of (4.1). First, however, we prove the two claims given above.

Proof of Claim 1.

-

V

(y

-

x)u(y)

dS(y)

=

-

(y

-

x)u(y)

dS(y)

+

B(x,

)

(y

-

x)u(y)

dS(y).

Now

y(y)

=

-

1 n(n)

y |y|n

and the outward normal on B(x, ) is

=

y |y

- -

x x|

.

Therefore,

(y

-

x)

=

y (y

-

x)

?

=

-

1 n(n)

y-x |y - x|n

?

y |y

- -

x x|

=

-

1 n(n)

?

|y

1 - x|n-1

.

Therefore,

B(x,

)

(y

-

x)u(y)

dS(y)

=

-

1 n(n)

B(x,

)

|y

1 - x|n-1

u(y)

dS(y)

=

-

1 n(n)

n-1

u(y) dS(y)

B(x, )

= - - u(y) dS(y).

B(x, )

As 0+,

Therefore, we have and the claim follows.

- u(y) dS(y) u(x).

B(x, )

lim

0+

B(x,

)

(y

-

x)u(y)

dS(y)

=

-u(x),

Proof of Claim 2. Now we know

V

(y

-

x)

u

(y)

dS(y)

=

(y

-

x)

u

(y)

dS(y)

-

B(x,

)

(y

-

x)

u

(y)

dS(y).

4

We just need to show that

B(x,

)

(y

-

x)

u

(y)

dS(y)

0

as 0+.

Substituting in the explicit formula for for n 3 (the case n = 2 can be handled similarly), we see that

B(x,

)

(y

-

x)

u

(y)

dS(y)

1 n(n)(n

-

2)

1 B(x, ) |y - x|n-2

u

(y)

dS(y)

u

1

L(B(x, )) n(n) n-2

dS(y)

B(x, )

C -

dS(y)

B(x, )

=C .

Therefore, as 0+,

B(x,

)

(y

-

x)

u

(y)

dS(y)

0

as claimed. Therefore, the claim follows. Above we have proven the following theorem.

Theorem 3. If u C2() is a solution of

-u = f u=g

x Rn x ,

where f and g are continuous, then

u(x) = -

g(y)

G

(x,

y)

dS

(y)

+

f (y)G(x, y) dy

for x , where G(x, y) is the Green's function for .

Corollary 4. If u is harmonic in and u = g on , then

u(x) = -

g

(y)

G

(x,

y)

dS

(y).

(4.8)

4.2 Finding Green's Functions

Finding a Green's function is difficult. However, for certain domains with special geometries, it is possible to find Green's functions. We show some examples below.

Example 5. Let R2+ be the upper half-plane in R2. That is, let

R2+ {(x1, x2) R2 : x2 > 0}.

5

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