Mathematical Physics Vol 1

7.8 Examples

425

The boundary conditions (7.398) become

u ( x , y , t )= X ( x ) Y ( y ) T ( t ) ⇒ u | x = 0 = X ( 0 ) Y ( y ) T ( t )= 0 ⇒ X ( 0 )= 0 ⇒ A = 0 , u | x = a = X ( a ) Y ( y ) T ( t )= 0 ⇒ X ( a )= 0 ⇒ B sin kt = 0 ⇒ ⇒ k m a = m π , u | y = 0 = X ( x ) Y ( 0 ) T ( t )= 0 ⇒ Y ( 0 )= 0 ⇒ C = 0 , u | y = b = X ( x ) Y ( b ) T ( t )= 0 ⇒ Y ( b )= 0 ⇒ D sin pt = 0 ⇒ ⇒ p n b = n π . Without losing in generality we can assume that B = 1 and D = 1, and thus we obtain:

m π x a

n π y b

X ( x )= sin

Y ( y )= sin

(7.410)

∧

.

Now, for function F we obtain

m π x a

n π y b

F mn = sin

sin

(7.411)

.

Given that

p 2 = ν 2 − k 2 , λ = c ν = c p p 2 + k 2 ⇒ m π 2 n π 2 2 2

λ mn c r

= c π r

a

+

b

m a

+

n b

⇒

λ mn = c π r

m a

+

n b

2

2

(7.412)

.

Now the solution of equation (7.401) is

T mn ( t )= B mn cos λ mn t + B ∗ mn sin λ mn t ,

(7.413)

and for u mn ( x , y , t ) we obtain

n π y b

m π x a

u mn ( x , y , t )=( B mn cos λ mn t + B ∗ mn sin λ mn t ) sin

sin

(7.414)

,

that is, for the total solution u

∞ ∑ m = 1

∞ ∑ n = 1

m π x a

n π y b

( B mn cos λ mn t + B ∗ mn sin λ mn t ) sin

u ( x , y , t )=

sin

(7.415)

.

Thus, the solution was obtained in the form of a double Fourier series. The coefficients are determined from the initial conditions (7.398) and (7.399)

∞ ∑ m = 1

∞ ∑ n = 1

n π y b

m π x a

u ( x , y , 0 )=

B mn sin

sin

= f ( x , y ) .

(7.416)

If we introduce the substitution

∞ ∑ n = 1

n π y b

K m ( y )=

B mn sin

(7.417)

,