Mathematical Physics Vol 1

7.8 Examples

401

and initial conditions

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

Solution The solution of the equation is of the form

∞ ∑ n = 1 h

A n cos

at + B n sin

at i sin

∞ ∑ n = 1

π n l

π n l

π n l

u =

u n ( x , t )=

x .

where

n π Z 0

n π Z 0

l Z 0

l Z 0

n π x

n π x l

2 l

2 l

l n π ·

l n π

2 l n 2 π 2

2 l

ϕ ( x ) · sin

A n =

d x =

x · sin

d x =

t sin t ·

d t =

t sin t d t =

l ·

n π

2 l n 2 π 2

2 l n 2 π 2

( sin n π − n π · cos n π − sin0 + 0 · cos0 )=

( sin t − t · cos t ) n π · ( − 1 ) n + 1 =

=

=

0

2 l n 2 π 2 ·

2 l n π ·

= ( − 1 ) n + 1 . The required coefficients are of the form:

l Z 0

l Z 0

n π x l

n π x l

2 l n π ·

2 π an

2 π an

( − 1 ) n + 1 , B

ψ ( x ) · sin

A n =

d x =

0 · sin

d x = 0 .

n =

The solution of the given partial differential equation is of the form

∞ ∑ n = 1

( − 1 ) n + 1 n

an π t

n π x l

2 l π

u ( x , t )=

cos

sin

l ·

.

Problem 262

Find the solution of equation

2 u

u tt = a

xx ,

that satisfies boundary

u ( 0 , t )= 0 , u ( l , t )= 0 , 0 ≤ x ≤ l

and initial conditions

u ( x , 0 )= π − x , u t ( x , 0 )= 0 .