ñòð. 58 |

valued, and

2

2d R dR

âˆ’ m2 R = 0,

r +r (6.138)

dr 2 dr

whose solutions are R = rÂ±m when m = 0, and 1, ln r, when m = 0. The

general solution is therefore a sum of these

(Am r |m| + Bm r âˆ’|m| )eimÎ¸ .

Ï‡ = A0 + B0 ln r + (6.139)

m=0

The singular terms, ln r and râˆ’|m| , are not solutions at the origin, and should

be omitted when that point is part of the region where 2 Ï‡ = 0.

Example: Dirichlet problem in the interior of the unit circle. Solve 2 Ï‡ = 0

in â„¦ = {r âˆˆ R2 : |r| < 1} with Ï‡ = f (Î¸) on âˆ‚â„¦ â‰¡ {|r| = 1}.

Î¸â€™

r,Î¸

Dirichlet problem in the unit circle.

174 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

We expand

âˆž

Am r |m| eimÎ¸ ,

Ï‡(r.Î¸) = (6.140)

m=âˆ’âˆž

and read oï¬€ the coeï¬ƒcients from the boundary data as

1 2Ï€

eâˆ’imÎ¸ f (Î¸ ) dÎ¸ .

Am = (6.141)

2Ï€ 0

Thus,

âˆž

1 2Ï€

r |m| eim(Î¸âˆ’Î¸ ) f (Î¸ ) dÎ¸ .

Ï‡= (6.142)

2Ï€ 0 m=âˆ’âˆž

We can sum the geometric progression

âˆž

reâˆ’i(Î¸âˆ’Î¸ )

1

|m| im(Î¸âˆ’Î¸ )

r e = +

1 âˆ’ rei(Î¸âˆ’Î¸ ) 1 âˆ’ reâˆ’i(Î¸âˆ’Î¸ )

m=âˆ’âˆž

1 âˆ’ r2

= . (6.143)

1 âˆ’ 2r cos(Î¸ âˆ’ Î¸ ) + r 2

Therefore,

1 âˆ’ r2

1 2Ï€

Ï‡(r, Î¸) = f (Î¸ ) dÎ¸ . (6.144)

1 âˆ’ 2r cos(Î¸ âˆ’ Î¸ ) + r 2

2Ï€ 0

This is known as the Poisson kernel formula.

If we set r = 0 in the Poisson formula we ï¬nd

1 2Ï€

Ï‡(0, Î¸) = f (Î¸ ) dÎ¸ . (6.145)

2Ï€ 0

We deduce that if 2 Ï‡ = 0 in some domain then the value of Ï‡ at a point

in the domain is the average of its values on any circle centred on the chosen

point and lying wholly in the domain.

From this is should be clear that Ï‡ can have no local maxima or minima

within â„¦. The same result holds in Rn , and a formal theorem to this eï¬€ect

can be proved:

Theorem (The mean-value theorem for harmonic functions): If Ï‡ is harmonic

( 2 Ï‡ = 0) within the bounded (open, connected) domain â„¦ âˆˆ Rn , and is

continuous on its closure â„¦, and if m â‰¤ Ï‡ â‰¤ M on âˆ‚â„¦, then m < Ï‡ < M in

â„¦ â€” unless, that is, m = M , when Ï‡ is constant.

6.4. LAPLACEâ€™S EQUATION 175

6.4.2 Green Functions

The Green function for the Laplacian in the entire Rn is given by the sum

over eigenfunctions

dn k eikÂ·(râˆ’r )

g(r, r ) = . (6.146)

(2Ï€)n k 2

It obeys

âˆ’ 2 g(r, r ) = Î´ n (r âˆ’ r ). (6.147)

r

We can evaluate the integral for any n by using Schwingerâ€™s trick to turn the

integrand into a Gaussian:

dn k ikÂ·(râˆ’r ) âˆ’sk2

âˆž

g(r, r ) = ds e e

(2Ï€)n

0

Ï€ n 1 âˆ’ 1 |râˆ’r |2

âˆž

= ds e 4s

(2Ï€)n

s

0

1 âˆž n 2

dt t 2 âˆ’2 eâˆ’t|râˆ’r | /4

=

2n Ï€ n/2 0

1âˆ’n/2

|r âˆ’ r |2

1 n

âˆ’1

= n n/2 Î“ . (6.148)

2Ï€ 2 4

Here, Î“(x) is Eulerâ€™s gamma function:

âˆž

dt txâˆ’1 eâˆ’t .

Î“(x) = (6.149)

0

For three dimensions we ï¬nd

1 1

g(r, r ) = , n = 3. (6.150)

4Ï€ |r âˆ’ r |

In two dimensions the Fourier integral is divergent for small k, and one has

to use

1

Î“(x) = Î“(x + 1) (6.151)

x

and

ax = ea ln x = 1 + a ln x + Â· Â· Â· (6.152)

to examine the behaviour of g(r, r ) near n = 2:

1 Î“(n/2)

1 âˆ’ (n/2 âˆ’ 1) ln(Ï€|r âˆ’ r |2 ) + O (n âˆ’ 2)2

g(r, r ) =

4Ï€ (n/2 âˆ’ 1)

1 1

âˆ’ 2 ln |r âˆ’ r | âˆ’ ln Ï€ + Â· Â· Â· .

= (6.153)

4Ï€ n/2 âˆ’ 1

176 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

ñòð. 58 |