Mathematical Physics Vol 1

Chapter 7. Partial differential equations

410

For λ n determined in this way, from (7.323), for function T ( t ) we obtain T n = b n e − n 2 4 t and thus u n ( x , t )= X n ( x ) T n ( t )= c n cos nx 2 a n exp − n 2 t 4 . Note that there is a solution also for n = 0, which we will denote by u 0 = b 0 = a 0 / 2. The final solution (applying the principle of superposition) is of the form

exp −

4

cos

2

∞ ∑ n = 1

n 2 t

a 0 2

nx

u ( x , t )=

b n · c n |{z} a n

(7.324)

+

.

From the initial condition u ( x , 0 )= ϕ ( x )= x we obtain the equation for determining the constants a n ϕ ( x )= a 0 2 + ∞ ∑ n = 1 a n cos nx 2 = x , where Euler coefficients of the Fourier series are

2 L Z

1 π Z

ϕ ( x ) cos

2

x cos

2

2 π

L

nx

nx

a n =

d x =

d x .

0

0

By partial integration ( R uv = uv − R v d u , u = x , d v cos nx

2 d x )we obtain

1 π  

d x 

0 2 π

sin

2 n

2 n Z

2 π

2 x

nx 2

*

 =

a n =

sin (

)

n

0

0

= " −

2 π 0 #

2 n −

2 n

cos

2

nx

4 π n 2

( cos ( n π ) − 1 ) .

=

Given that cos ( n π )=( − 1 ) n , for a

n we obtain

4 π n 2

a n = [( − 1 ) n − 1 ] . Note that even coefficients are ( n = 2 m ) a 2 m = 0, because ( − 1 )

n − 1 = 1 − 1 = 0, , because ( − 1 ) 2 m − 1 − 1 = − 2.

8 π ( 2 m − 1 ) 2

and odd coefficients are ( n = 2 m − 1) a 2 m Separating even and odd members

− 1 = −

exp −

4

a n cos

2

∞ ∑ n = 1 ∞ ∑ n = 1

n 2 t

a 0 2 a 0 2

nx

u ( x , t )=

+

=

cos

x exp −

4

∞ ∑ n = 1

( 2 m − 1 ) 2 t

( 2 m − 1 ) 2

2 t )+

a 2 m cos ( mx ) exp ( − m

a 2 m

=

+

,

− 1

bearing in mind that the even members a 2 m = 0, finally we obtain

cos

x exp −

4

∞ ∑ m = 1

( 2 m − 1 ) 2 t

8 π

1 ( 2 m − 1 ) 2

( 2 m − 1 ) 2

u ( x , t )= π −

.

Note, finally, that

= π .

lim t → ∞ u ( x , t )

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