Mathematical Physics Vol 1

Chapter 7. Partial differential equations

406

Thus, it is necessary to expand the function ϕ ( x ) into a Fourier series, and then, comparing with (7.320), obtain the constants b n , i.e. b n = Z 2 0 ϕ ( x ) sin n π x 2 d x = = Z 2 0 2sin π x 2 − sin ( π x )+ 4sin ( 2 π x ) sin n π x 2 d x = = 2 Z 2 0 sin π x 2 sin n π x 2 d x − Z 2 0 sin ( π x ) sin n π x 2 d x + + 4 Z 2 0 sin ( 2 π x ) sin n π x 2 d x . From the condition of orthogonality (see Example 186, on p. 270) it follows Z 2 0 sin π x 2 sin n π x 2 d x = ( 1 , for n = 1 , 0 , otherwise .

d x = ( d x = (

Z 2 0

sin ( π x ) sin sin ( 2 π x ) sin

2 2

n π x

1 , for n = 2 , 0 , otherwise . 1 , for n = 4 , 0 , otherwise .

Z 2 0

n π x

Finally, for constants b n we obtain: b 1 = 2 Z 2 0 sin π x 2 2

sin

2 2

n π x

d x − 0 + 0 = 2 , d x + 0 = − 1 ,

b 2 = 0 − Z b 4 = 0 − 0 + 4 Z 0

sin ( π x ) sin 2

n π x

sin ( 2 π x ) sin

2

n π x

d x = 4 ,

0

b n = 0 if n̸ = 1 , 2 , 4 ,

and the solution is u ( x , t )= 2sin π x 2

exp −

t − sin ( π x ) exp −

t − sin ( 2 π x ) exp − π 2 t .

π 2 16

π 2 4

Problem 267 Find the solution of the following PDE

2 u

u t = α

xx , 0 ≤ x ≤ π , t > 0 ,