Mathematical Physics Vol 1

7.8 Examples

407

if the boundary conditions are

u ( 0 , t )= 0 , u x ( π , t )= 0 ,

and the initial conditions are

2

u ( x , 0 )= 3sin

5 x

= f ( x ) .

Solution In this case we will use the method of separation of variables, i.e. we shall assume a solution in the form u ( x , t )= X ( x ) T ( t ) , so that the initial PDE comes down to two ordinary differential equations ˙ T a 2 T = λ ∧ X ′′ X = λ , λ = const . and the respective conditions u ( 0 , t )= X ( 0 ) · T ( t )= 0 ⇒ X ( 0 )= 0 , u x ( π , t )= X ′ ( π ) · T ( t )= 0 ⇒ X ′ ( π )= 0 , u ( x , t )= X ( x ) · T ( 0 )= 3sin 5 x 2 ≡ ϕ ( x ) . Similarly, as in the previous case, from X ′′ − λ X = 0 , for − λ = µ 2 , i . e . X ′′ + µ 2 X = 0 we obtain X ( x )= C 1 sin ( µ x )+ C 2 cos ( µ x ) . Now, from boundary conditions, we obtain: X ( 0 )= 0 ⇒ C 2 = 0 , X ′ ( π )= 0 ⇒ µ C 1 cos ( πµ )= 0 ⇒ πµ = 2 n − 1 2 π , µ = 2 n − 1 2 = n − 1 2 , n = 1 , 2 ,...,

λ n = − n − X n = sin n −

1 2

2

,

1 2

x .

For the function T ( t ) , we obtain d T T

= a 2 λ Z ⇒ ln T = a 2 λ t ⇒ T = exp " − a 2 n − 1 2 2 t # ,