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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