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### 4.7One dimensional wave equation

Note: 1 lecture, §9.6 in [EP], §10.7 in [BD]

Imagine we have a tensioned guitar string of length . Suppose we only consider vibrations in one direction. That is, let denote the position along the string, let denote time, and let denote the displacement of the string from the rest position. See Figure 4.18.

The equation that governs this setup is the so-called one-dimensional wave equation:

for some constant . Assume that the ends of the string are ﬁxed in place:

Note that we have two conditions along the axis as there are two derivatives in the direction.

There are also two derivatives along the direction and hence we need two further conditions here. We need to know the initial position and the initial velocity of the string. That is,

for some known functions and .

As the equation is again linear, superposition works just as it did for the heat equation. And again we will use separation of variables to ﬁnd enough building-block solutions to get the overall solution. There is one change however. It will be easier to solve two separate problems and add their solutions.

The two problems we will solve are

 (4.10)

and

 (4.11)

The principle of superposition implies that solves the wave equation and furthermore and . Hence, is a solution to

 (4.12)

The reason for all this complexity is that superposition only works for homogeneous conditions such as , , or . Therefore, we will be able to use the idea of separation of variables to ﬁnd many building-block solutions solving all the homogeneous conditions. We can then use them to construct a solution solving the remaining nonhomogeneous condition.

Let us start with (4.10). We try a solution of the form again. We plug into the wave equation to obtain

Rewriting we get

Again, left hand side depends only on and the right hand side depends only on . Therefore, both equal a constant, which we will denote by .

We solve to get two ordinary diﬀerential equations

The conditions implies and implies that . Therefore, the only nontrivial solutions for the ﬁrst equation are when and they are

The general solution for for this particular is

We also have the condition that or . This implies that , which in turn forces . It is convenient to pick (you will see why in a moment) and hence

Our building-block solutions are

We diﬀerentiate in , that is

Hence,

We expand in terms of these sines as

Using superposition we can just write down the solution to (4.10) as a series

Exercise 4.7.1: Check that and .

Similarly we proceed to solve (4.11). We again try . The procedure works exactly the same at ﬁrst. We obtain

and the conditions , . So again and

This time the condition on is . Thus we get that and we take

Our building-block solution will be

We expand in terms of these sines as

And we write down the solution to (4.11) as a series

Exercise 4.7.2: Fill in the details in the derivation of the solution of (4.11). Check that the solution satisﬁes all the side conditions.

Putting these two solutions together, let us state the result as a theorem.

Theorem 4.7.1. Take the equation

 (4.13)

where

and

Then the solution can be written as a sum of the solutions of (4.10) and (4.11). In other words,

Example 4.7.1: Let us try a simple example of a plucked string. Suppose that a string of length 2 is plucked in the middle such that it has the initial shape given in Figure 4.19. That is

The string starts at rest (). Suppose that in the wave equation for simplicity.

We leave it to the reader to compute the sine series of . The series will be

Note that is the sequence for . Therefore,

The solution is given by

A plot for is given in Figure 4.20. Notice that unlike the heat equation, the solution does not become “smoother,” the “sharp edges” remain. We will see the reason for this behavior in the next section where we derive the solution to the wave equation in a diﬀerent way.

Make sure you understand what the plot, such as the one in the ﬁgure, is telling you. For each ﬁxed , you can think of the function as just a function of . This function gives you the shape of the string at time .

#### 4.7.1Exercises

Exercise 4.7.3: Solve

Exercise 4.7.4: Solve

Exercise 4.7.5: Derive the solution for a general plucked string of length , where we raise the string some distance at the midpoint and let go, and for any constant (in the equation ).

Exercise 4.7.6: Imagine that a stringed musical instrument falls on the ﬂoor. Suppose that the length of the string is 1 and . When the musical instrument hits the ground the string was in rest position and hence . However, the string was moving at some velocity at impact (), say . Find the solution for the shape of the string at time .

Exercise 4.7.7 (challenging): Suppose that you have a vibrating string and that there is air resistance proportional to the velocity. That is, you have

Suppose that . Derive a series solution to the problem. Any coeﬃcients in the series should be expressed as integrals of .

Exercise 4.7.101: Solve

Exercise 4.7.102: Solve

Exercise 4.7.103: Solve

Exercise 4.7.104: Let’s see what happens when . Find a solution to , , , .