8.9 Solving the Wave Equation

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The ideas here are essentially the same.

Example 8.9.1. A wire is originally horizontal with tension H going from

−_ to _, and is held into the triangular shape shown in figure 8.10 Find the

shape of the wire at times following its release.

Solution: By symmetry we need only worry about terms which are cosine

terms in space. In fact the problem is basically silly, but let’s plug through it

so we can see what is happening and deal with more complicated questions.

194 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

Figure 8.10: a plucked string

First, as before we assume that a function f(x, t) decribing the wire shape is

separable into f(x, t) = p(x)q(t) Then the wave equation can be written

q¨p =

_

H

p¨q

which can be written:

_

H

¨q

q

=

¨p

p

= k

for some constant k. The usual arguments show k must be negative, I write

it therefore as −K2. This gives

¨q = −K2 H

_

q

and

¨p = −K2 p

The former gives

q(t) = an cos(K

s

H

_

t) + bn sin(K

s

H

_

t)

as the general solution and the latter gives

p(x) = An cos(Kx) + Bn sin(Kx)

Again we observe that any linear combination of these for different choices

of a will give a solution-in-waiting.

Now we have to match the boundary condition which is that at t = 0 we

have the triangular function

f(x, 0) = p(x) = h − |

hx

_

|

8.9. SOLVING THE WAVE EQUATION 195

and this requires that we express p by its Fourier expansion. Since the

function is symmetric we can forget about the sine terms and obtain the

expansion in cosine terms only.

This will give us a set of integral values for K and corresponding An.

The constraints on q derive from the fact that we started with the string at

rest. Thus

@f

@t

____

0

= 0

This gives us

p(x)q˙(0) = 0

which tells us that q contains cosine terms only.

We have the Fourier series for |x| is

_

2

−

4

_

(cos(x) +

cos(3x)

9

+

cos(5x)

25

+ · · · )

So the Fourier series for

h

_

|x|

is

h

2

−

4h

_2 (cos(x) +

cos(3x)

9

+

cos(5x)

25

+ · · · )

and that for

h −

h

_

|x|

is

h

2

+

4h

_2 (cos(x) +

cos(3x)

9

+

cos(5x)

25

+ · · · )

This gives an expression for p(x) as a sum of cosine terms and values of K

which are the odd integers.

We have then that the solution is of the form

a0 +

X1

n=1

fn(x, t)

where

fn(x, t) = an cos(nx) cos(n

s

H

_

t)

196 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

with a0 = h/2 and an = 0 when n is even, and

an =

4h

_2n2

when n is odd.

Remark 8.9.1. The above calculation really is rather silly. If we assume

that the function f(x, t) is separable,

f(x, t) = p(x)q(t)

then the answer has to be that the wire preserves its shape indefinitely except

that is is scaled by some time varying function. And the time varying function

has to be something which starts off at a maximum of 1 and oscillates. The

only point of interest is to decide on the form of the time variation, which

comes from the expansion for p. Note that this gives the amplitude of the

various harmonics.

Remark 8.9.2. Note that if we have the wave equation

fxx =

_

H

ftt

a solution cos(x) cos(

q

H

_ t) can be written:

1

2

(cos(x + vt) + cos(x − vt))

which is an average of two waves going in opposite directions with velocity

v =

s

H

_

This is telling us that the propagation of a transverse wave along the wire

will be at a speed which is proportional to the square root of the tension

and inversely as the square root of the density. This should not come as too

much of a surprise.

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