Mathematical Physics Vol 1

Chapter 7. Partial differential equations

412

From here we obtain: a 0 = 2 Z 1 0

x − x 2 d x = 2 x − x 2 cos ( n π x ) d x = sin ( n π x ) − x 2 − x x 2 2 −

3

1

x 3

1 3

(7.329)

=

,

0

a n = 2 Z

1 0

(7.330)

2 n 3 π 3

1

1 − 2 x n 2 π 2

sin ( n π x )

= 2

(7.331)

n π −

=

0

2 n 3 π 3

( 1 +( − 1 ) n ) .

(7.332)

= −

Thus, the solution of the PDE is

∞ ∑ n = 1

1 3

2 π 3

( − 1 )( n + 1 ) − 1 n 3

e − n 2 π 2 t cos

( n π x ) .

u ( x , t )=

(7.333)

+

Problem 270 Find the solution for the PDE

u t = u xx , 0 < x < 2 , t > 0 ,

(7.334)

that satisfies the following conditions u ( x , 0 )= ( x ,

for 0 < x < 1 , 2 − x , for 1 < x < 2 . ,

u ( 0 , t )= u x ( 2 , t )= 0 .

(7.335)

Solution As in previous problems, we start from the assumption u ( x , t )= X ( x ) T ( t ) ,

(7.336)

which leads to a system of ordinary DE

˙ T = λ T , X ′′ = λ X .

The solution of the first DE is

λ t

, λ = − k 2 ,

T = ce

(7.337)

while the solution of the second differential equation is X ( x )= c 1 sin kx + c 2 cos kx boundary conditions (7.335), for function X , are

u ( 0 , t )= X ( 0 ) · T ( t )= 0 ⇒ X ( 0 )= 0 , u x ( 2 , t )= X ′ ( 2 ) · T ( t )= 0 ⇒ X ′ ( 2 )= 0 ,