Mathematical Physics Vol 1

Chapter 7. Partial differential equations

408

and thus

t !

u n ( x , t )= X n ( x ) T n ( t )= sin n − 1 2

x exp − α 2 n − 1 2 2

and the solution is

t ! .

b n sin n −

1 2

x exp − α 2 n − 1 2 2

∞ ∑ n = 1

u ( x , t )=

The constants b n are determined from the initial condition u ( x , 0 )= 3sin 5 x 2 , i.e. u ( x , 0 )= ∞ ∑ n = 1 b n sin n − 1 2 x = 3sin 5 x 2 . It is necessary to expand the function ϕ ( x ) into a Fourier series, and thus

6 π · (

3sin

2

sin n −

1 2

x d x =

2 π Z

π / 2 , for n = 3 , 0 , otherwise .

π

5 x

b n =

0

Thus, only b 3̸ = 0, while other constants are equal to zero. Finally, the solution is u ( x , t )= 3sin 5 x 2 exp " − 5 α 2 2 t # . Check:

t ! ,

α 2 sin

2

exp −

2

2

5 α

75 4 75 4

5 x

u t ( x , t )= −

t ! ,

sin

2

exp −

2

2

5 α

5 x

u xx ( x , t )= −

and thus

2 u

u t = α

xx ,

i.e. the solution is correct.

Problem 268 Solve the PDE

u t = u xx , 0 ≤ x ≤ 2 π , t > 0 .