Mathematical Physics Vol 1

7.8 Examples

409

The boundary conditions are

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

and the initial condition is

u ( x , 0 )= x = f ( x ) .

R Note. In this case, as in the several previous cases, we shall use the method of separation of variables, which leads to the problem of main values. The first part of the solution is the same. The difference is only in boundary conditions, which lead to different solutions.

Solution We start from the assumption that the solution can be represented in the form u ( x , t )= X ( x ) T ( t ) , and by division by T · X , we obtain

˙ T T

X ′′ X

= λ , λ = const .

(7.321)

=

From boundary conditions we obtain:

′ ( 0 ) T ( t )= 0 ⇒ X ′ ( 0 )= 0 (because T ( t ) is not equal to zero for each t ) , ′ ( 2 π ) T ( t )= 0 ⇒ X ′ ( 2 π )= 0 (because T ( t ) is not equal to zero for each t ) . (7.322)

u x ( 0 , t )= X u x ( π , t )= X

From (7.321) it follows d T d t

d T T

= λ d t ⇒ T = Ce λ t .

= λ T ( t ) ⇒

(7.323)

Also, from (7.321) and the boundary conditions (7.322), we obtain X ′′ − λ X = 0 , X = ce α t ⇒ X ′′ = c α 2 e α t ⇒ ce α t ( λ 2 − 1 )= 0 ⇒ α = ± √ λ , X ( x )= c 1 sin ( α x )+ c 2 cos ( α x ) , α = − µ 2 , 0 c 1 cos2 πα − α c 2 sin2 πα = α c 2 sin2 πα = 0 . As we have already emphasized, c 2̸ = 0 (non-trivial solution), and thus sin2 πα = 0 ⇒ 2 πα = n π , ⇒ α = n 2 , X ′ ( 0 )= α c 1 cos0 − α c 2 sin0 = α c 1 = 0 ⇒ c 1 = 0 , X ′ ( 2 π )= α >

n = 1 , 2 ,...

Thus, λ n = − n 2

2 , and it follows that

X n ( x )= a n cos

2

nx

.