Mathematical Physics Vol 1

7.8 Examples

413

from where it follows

X ( 0 )= c 1 sin0 + c 2 cos0 = 0 ⇒ c 2 = 0 , X ′ ( 2 )= c 1 k cos2 k + > 0

c 2 k sin2 x = 0 ⇒ cos2 k = 0

and we further obtain

π 4

3 π 4

5 π 4

( 2 n − 1 ) π 4

k =

,

,

,...,

.

Substituting the value k , into (7.337), we obtain

2

( 2 n − 1 ) 16

2 t

π

T n ( t )= c 3 e −

,

that is, the solution

∞ ∑ n = 1

( 2 n − 1 ) 4

2

( 2 n − 1 ) 16

2 t sin

π

π x .

e −

u ( x , t )=

b n |{z} c 1 · c 3

The constants b n are determined from the initial conditions

π x = ϕ ( x )= (

∞ ∑ n = 1

u ( x , 0 )= for 0 < x < 1 , 2 − x , for 1 < x < 2 . . Using the expansion of function ϕ ( x ) into a Fourier series, for constants b n we obtain: b n = 2 Z 1 0 x sin ( 2 n − 1 ) 4 π x d x + Z 2 1 ( 2 − x ) sin ( 2 n − 1 ) 4 π x d x = = 32 ( 2 n − 1 ) 2 π 2 sin ( 2 n − 1 ) 4 π x − 8 x ( 2 n − 1 ) π cos 2 n − 1 4 π x 1 0 + + − 32 ( 2 n − 1 ) 2 π 2 sin ( 2 n − 1 ) 4 π x + 8 ( x − 2 ) ( 2 n − 1 ) π cos 2 n − 1 4 π x 2 1 = = 32 ( 2 n − 1 ) 2 π 2 sin ( 2 n − 1 ) π 4 + cos ( n π ) , and the solution of the initial PDE is b n sin ( 2 n − 1 ) 4 x ,

sin ( 2 n − 1 ) π 4

+ cos ( n π )

∞ ∑ n = 1

32 π 2

2 n − 1 4

2

( 2 n − 1 ) 16

2 t sin

π

π x

e −

u ( x , t )=

(7.338)

( 2 n − 1 ) 2

Problem 271 Find the solution for the PDE

u t = u xx , 0 < x < 3 , t > 0

(7.339)

that satisfies the following conditions - initial

u ( x , 0 )= 4 x − x 2 ,

(7.340)