Mathematical Physics Vol 1

7.8 Examples

419

which in the given special case for Q Q ( x )= ( x ,

if 0 < x < 1 , 2 − x , if 1 < x < 2 .

(7.364)

yields for q n

q n = Z

n π x 2

2

Q ( x ) sin

d x =

0

= Z =

d x + Z

n π x 2

n π x 2

1

2

x sin

( 2 − x ) sin

d x =

0

1

2 2

1

n π x

n π x

4 n 2 π 2

2 xn π cos

sin

(7.365)

2 −

+

0

+ −

2

n π x 2

n π x

4 n 2 π 2

2 x n π

sin

cos

+

=

1

n π x 2

8 n 2 π 2

sin

=

.

Substituting the respective expressions into (7.359), we obtain ∞ ∑ n = 1 T ′ n ( t ) sin n π x 2 = − ∞ ∑ n = 1 n π x 2 2 T n ( t ) sin n π x 2 + ∞ ∑ n = 1 q n sin n π x 2 , By simplifying and grouping the respective members next to sin npx 2 , we obtain T ′ n ( t )+ n 2 π 2 4 T n ( t )= q n , (7.366) which is a linear differential equation for T n . Its solution is

4 n 2 π 2

π 2 )

( n

2 t

q n + b n e −

T n ( t )=

The final solution is

∞ ∑ n = 1

2 t sin

n π x 2

4 n 2 π 2

π 2 )

( n

q n + b n e −

u ( x , t )=

(7.367)

.

The initial condition yields

∞ ∑ n = 1

q n + b n sin

n π x 2

4 n 2 π 2

2 x − x 2 =

.

If we introduce the substitution

4 n 2 π 2

c n =

q n + b n ,

(7.368)

we obtain

∞ ∑ n = 1

n π x 2

2 x − x 2 =

c n sin

.

This expression represents the Fourier sine series, and thus c n c n = Z 2 0 ( 2 x − x 2 ) sin n π x 2 d x = = 16 n 3 π 3 1 − cos n π 2 ,

(7.369)