Mathematical Physics Vol 1

7.6 The variable separation method

369

Thus, for u n we obtain

π n l

¯ C

a 2 λ

( C n = ¯ C n

n t sin

u n = C n e −

x ,

(7.210)

n ) .

Finally, according to the superposition principle, for u ( x , t ) we obtain

∞ ∑ n = 1

∞ ∑ n = 1

π n l

a 2 λ

n t sin

C n e −

u ( x , t )=

u n =

x .

(7.211)

Initial conditions (7.202) are used for determining the constants C n , which yields

∞ ∑ n = 1

π n l

u ( x , 0 )= ϕ ( x )=

x .

(7.212)

C n sin

Once again, we have reduced the problem to expansion of known functions into Fourier series C n = 2 l l Z 0 ϕ ( ξ ) sin π n l ξ d ξ . (7.213) We have thus solved the given task.

Solving equations of elliptic type by Fourier method

Example 243 Observe the Laplace equation in spherical coordinates, where u = u ( r , ϕ , θ )

∂ ∂ r

∂ u ∂ r

∂ ∂θ

∂ u ∂θ

∂ 2 u ∂ϕ 2

1 r 2

1 r 2 sin θ

1 r 2 sin θ

r 2

∆ u =

sin θ

= 0 .

(7.214)

+

+

In the case of spherical symmetry, i.e. if u = u ( r ) , we have ∂ u ∂θ = 0 , ∂ u ∂ϕ = 0 i ∂ u ∂ r = d u d r and the Laplace equation becomes ∂ ∂ r r 2 ∂ u ∂ r = d d r r 2 d u d r = 0 ⇒

(7.215)

,

(7.216)

C 1 r

u = −

+ C 2 ,

(7.217)

where C 1 and C 2 are arbitrary constants. Assume, for example, that C 1 = − 1, and C 2 = 0, which yields u o = 1 r (7.218) the basic solution of the Laplace equation in the space .

R Note that the function u o satisfies the Laplace equation everywhere, except at point r = 0.