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Solved Problems

Since the initial position and velocity of the string is zero, we have

0
u (0) = u (0) = 0.
n
n
First we solve the differential equation on the range 0 < t < δ. The homogeneous
solutions are

nπct nπct
cos , sin .
L L
Since the right side of the ordinary differential equation is a constant times sin(πt/δ),

which is an eigenfunction of the differential operator, we can guess the form of a
particular solution, p (t).
n

πt
p (t) = d sin
n
δ
We substitute this into the ordinary differential equation to determine the multi-
plicative constant d.

3
2
8dδ L v nπd nπξ πt
p (t) = − cos sin sin
n
2
3
2 2 2
2
2 2
π (L − c δ n )(L − 4d n ) L L δ
The general solution for u (t) is
n

3
2

8dδ L v cos nπd sin nπξ sin πt
nπct nπct L L δ
u (t) = a cos + b sin − .
n
2 2 2
2
2
2 2
3
L L π (L − c δ n )(L − 4d n )
We use the initial conditions to determine the constants a and b. The solution for
0 < t < δ is

2
3
8dδ L v cos nπd sin nπξ
L L L nπct πt
u (t) = sin − sin .
n
2
2 2
2
3
2 2 2
π (L − c δ n )(L − 4d n ) δcn L δ
The solution for t > δ, the solution is a linear combination of the homogeneous
solutions. This linear combination is determined by the position and velocity at
t = δ. We use the above solution to determine these quantities.
4
2
8dδ L v nπd nπξ nπcδ
u (δ) = cos sin sin
n
2
2 2
3
2
2 2 2
π δcn(L − c δ n )(L − 4d n ) L L L
3
2
8dδ L v nπd nπξ nπcδ
0
u (δ) = cos sin 1 + cos
n 2 2 2 2 2 2 2 2
π δ(L − c δ n )(L − 4d n ) L L L
The fundamental set of solutions at t = δ is

nπc(t − δ) L nπc(t − δ)
cos , sin
L nπc L
From the initial conditions at t = δ, we see that the solution for t > δ is
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