Mathematical Physics Vol 1

Chapter 7. Partial differential equations

426

we obtain

∞ ∑ m = 1

m π x a

f ( x , y )=

K m ( y ) sin

(7.418)

,

and thus for fixed y we obtain

a Z 0

m π x a

2 a

K m ( y )=

f ( x , y ) sin

d x .

(7.419)

Thus, we have

b Z 0

n π y b

2 b

B mn =

K m ( y ) sin

d y ,

(7.420)

which yields the generalized Fourier formula

b Z 0

a Z 0

m π x a

n π y b

4 ab

B mn =

f ( x , y ) sin

sin

d x d y .

(7.421)

From the second condition we obtain ∂ u ∂ t t = 0 = ∞ ∑ m = 1 ∞ ∑ n = 1 B ∗ mn λ mn sin

m π x a

n π y b

sin

= g ( x , y ) ⇒

(7.422)

b Z 0

a Z 0

m π x a

n π y b

4 ab

B ∗ mn =

g ( x , y ) sin

sin

d x d y .

(7.423)

Problem 277 Solve the wave equation

∂ 2 ψ ∂ t 2

1 c 2

∆ ψ − (7.424) where the unknown function ψ is a function of a set of variables ( r , t ) , a r = r ( r , ϕ , θ ) . r , ϕ and θ are spherical coordinates, t is time, and c a constant. = 0 ,

Solution We will look for the solution applying the method of separation of variables, assuming that it has the form ψ ( r , t )= P ( r ) · T ( t ) . (7.425) Starting from (7.425), the equation (7.424) becomes T · ∆ P − 1 c 2 P · ¨ T = 0 , that is ∆ P P = 1 c 2 ¨ T T .