Page 56 - 6637

Solved Problems

in the velocity as a function of x, the coefficients in the expansion will decay as
2
1/n .

∞ (
nπc nπx v for |x − ξ| < d
X
u (x, 0) = b sin =
t
n
L L 0 for |x − ξ| > d.
n=1
Z L
nπc 2 nπx
b = u (x, 0) sin dx
n
t
L L 0 L
2 Z ξ+d nπx
b = v sin dx
n
nπc ξ−d L

4Lv nπd nπξ
= sin sin
2 2
n π c L L
The solution for u(x, t) is,

∞
4Lv X 1 nπd nπξ nπx nπct
u(x, t) = sin sin sin sin .
2
π c n 2 L L L L
n=1

b) The form of the solution is again,
∞

X nπx nπct
u(x, t) = b sin sin
n
L L
n=1
We determine the coefficients in the expansion from the initial velocity.

∞ π(x−ξ)
(
nπc nπx v cos for |x − ξ| < d
X
u (x, 0) = b sin = 2d
t
n
L L 0 for |x − ξ| > d.
n=1
Z L
nπc 2 nπx
b = u (x, 0) sin dx
t
n
L L 0 L
2 Z ξ+d π(x − ξ) nπx
b = v cos sin dx
n
nπc ξ−d 2d L

2
8dL v nπd nπξ L
 cos sin for d 6= ,
2
2
nπ c(L −4d n ) L L 2n
2 2
b =
n
v 2nπd nπξ L
 2nπd + L sin sin for d =
2 2
n π c L L 2n
The solution for u(x, t) is,
∞
8dL v 1 nπd nπξ nπx nπct L
2 X
u(x, t) = cos sin sin sin for d 6= ,
2
2 2
2
π c n(L − 4d n ) L L L L 2n
n=1
52