8.4 The Wave Equation

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Remark 8.4.1. One of the earliest applications of PDEs is to the study of

vibrating systems. As with bees and heat, I look at the simplest case first.

This is done out of kindness for your soft mathematical gums and not because

it is the most interesting.

Suppose we have an elastic string suspended as both ends. If anyone deforms

the string and lets it go at time t = 0, the string will oscillate. At any later

time t there is some function specifying the shape of the string, constrained

to take fixed values at the end points. So the displacement vertically of a

horizontal string can be written f(x, t), for x 2 [a, b], t 2 R+.

What can we say about the local dynamics of the string? Naturally we shall

put in some simplifying assumptions such as zero air resistance and internal

damping, so the string will be assumed to be perfectly elastic. It is fair to

hope that over short times the flagrant falsity of the assumptions will not

172 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

affect the general behaviour significantly. I also assume each point of the

string moves only vertically.

Let _ denote the density of the string, supposed constant, and let T(x, t)

denote the tension in the string. Let H(x, t) and V (x, t) denote the horizontal

and vertical components of this tension. The assumption that the string

moves only vertically ensures that H is constant. The acceleration of a small

bit of string of length _x at location x at time t is, by definition,

@2f

@t2 , ftt

This by Newton’s Law is the force divided by the mass,

8 t 2 R+ 8x 2 [a, b], __xftt = V (x + _x, t) − V (x, t)

or more perspicuously:

8 t 2 R+ 8x 2 [a, b],

V (x + _x, t) − V (x, t)

_x

= _

@2f

@t2

Taking limits we obtain:

8 t 2 R+ 8x 2 [a, b],

@V

@x

= _

@2f

@t2

Now V (x, t) = H(x, t) tan(_), where _ is the angle made by the string at

(x, t) That is

8 t 2 R+ 8x 2 [a, b], V (x, t) = H

@f

@x

where H is the (constant) horizontal component of T. Substituting for V

above we obtain:

8 t 2 R+ 8x 2 [a, b],

@2f

@x2 =

_

H

@2f

@t2 (8.4)

Equation 8.4 is the wave equation.

Remark 8.4.2. I shall defer actually solving it generally until later, but

curiosity might make us want to see if we could solve the problem:

Example 8.4.1. Take a steel string _ units long and fixed at the end points

at height zero. Suppose its density per unit length is _, given, and its tension

8.4. THE WAVE EQUATION 173

is set at some value, say by hanging a kilogram weight off one end prior to

fixing it. Twang it in the middle. What pitch would you get?

Solution: If you reflect briefly on the fact that you might hope to solve this

problem getting an answer in cycles per second having measured the length

in metres, the density in kilograms per meter and the tension in kilograms,

you will see why Mathematics was known in some quarters as Greek Magic.

To try to solve the above problem, imagine the simplest possible solution

of the wave equation. I incline to think that if we could freeze the wire at

its maximum amplitude it would look, if we took the origin in the middle,

rather like cos(x). This would allow it to be about the least complicated

shape that had the ends at ±_/2 fixed at value zero. As time changed,

the wave would flatten down to zero then turn upside down. We can get

this effect by multipying by sin(2_!t) where ! is the frequency in cycles per

second. This suggests a trial solution to be:

f(x, t) = cos(x) sin(2_!t)

Differentiating partially twice for x we get

fxx = −f(x, t)

and differentiating twice partially with respect to t we get

ftt = −f(x, t)(4_2!2)

So the wave equation is satisfied with

1

4_2!2 =

_

H

which gives

! =

1

2_

s

H

_

If the length of the wire were different, we should simply scale it by changing

the units.

Exercise 8.4.1. Find the fundamental frequency of vibration of a wire half

a metre long with a density of 0.033 kilograms per metre with a tension of

ten kilograms.

174 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

Figure 8.4: Another solution to the vibrating string

Remark 8.4.3. You ought to take time off to think about the fact that

you can actually compute an answer to this. On the one hand a friend does

the experiment. He or she gets some wire and measures the weight of a cut

length. They then hang a known weight off one end, then drive a peg in to

stop one end moving, the other end already being fixed. Now they twang it.

He or she connects up a microphone to a CRT oscilloscope and measures the

frequency, or listens to it and compares it with tuning forks. The output is

a number to some accuracy.

You, meanwhile, take the numbers they have given you and perform certain

acts of writing down squiggly marks on paper. Then you do some arithmetic,

and bingo, you too come out with a number and can tell your friend the

readings of the oscilloscope. How come this amazing relationship between

the squiggled marks on the paper and the physical set-up? That it works is

well known. That it is amazing that it works is something you might not

have thought about, but surely it is incredible.

Remark 8.4.4. It is clear that there are other solutions to how the string can

move: if we placed a finger about a third of the way along we might plausibly

force a vibrating string to give a solution like figure 8.4 where the vertical

amplitude has been shown (much exaggerated) at four different times. Some

reflection will show a discrete but infinite set of plausible solutions, looking

rather like Fourier terms. The one sketched is

cos(3t) sin(2_!t)

More generally we can get solutions

an cos((2n + 1)t) sin(2_!t), n 2 Z

8.5. SCHR ¨ODINGER’S EQUATION 175

It is also the case that the sum of any solutions is also a solution. In other

words there is a whole infinite dimensional vector space of solutions, and we

have looked only at some basis elements of the space. I leave you to brood

on this.

Exercise 8.4.2. Prove the claim that the set of solutions is a linear space.

Remark 8.4.5. We can imagine the problem of drumming: I take a thin

membrane and attach it to a rigid circle or maybe a square. It is a 2-

dimensional version of the string. Now I give it a good smack in the middle.

What pitch is the resulting sound? In order to work the answer out, I should

need to have a two dimensional version of equation 8.4. Can we make a stab

at setting up a 2-d wave equation?

Remark 8.4.6. I might make a guess at:

@2f

@x2 +

@2f

@y2 =

1

u2

@2f

@t2 (8.5)

and

@2f

@x2 +

@2f

@y2 +

@2f

@z2 =

1

u2

@2f

@t2 (8.6)

in three dimensions, for some constant u.

It is beyond the scope of the course to deal with these, but you might like

to experiment to see if you can persuade yourself that these are plausible

equations for describing waves in two and three dimensions.

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