Modeling an Elastic String

A Solution to the Wave Equation

With Finite Boundary Conditions

Colin Mitchell

Now that we have formed our model,

with , we can solve it using the method of separation of variables. So we want to seek a solution of the form

.

So inserting

into the governing equation gives us

Since the left hand side is dependent only on x, and the right hand side is dependent only on t, then both sides must be equal to a constant. So we can form ordinary differential equations in both functions,

Let , so that the equations reduce,

.

Since can be any number , we notice that different cases will give different solutions to each ODE. Basic intuition allows us to understand that our model will have a sinusoidal shape, and as such we should seek a solution of sines and cosines. For , we will get solutions of the form

which will satisfy the governing PDE, but will not give a proper model. For , we will get solutions of the form

The boundary conditions will give a trivial solution, and this form will also not work. So we must assume that . First, let us solve for X. The solution for is of the form

Using the left boundary condition , we get

Using the right boundary condition , we get

Since we do not want the trivial solution , we must have . This then tells us that , so . So we reach

.

This is a sequence of solutions, each of which satisfies the PDE and the boundary conditions. We notice that we will have constants in both X and T, and so we will omit the constant of amplification in X, and T will define the amplitude of the vibrations. So we can write summation of each as the overall form of X,

Now we will focus on T. Similarly, the solution to will take the form . Since we know that , we will replace it in this equation to get

It is also apparent here that T is also a sequence of solutions, and so we can combine it with X to reach the solution

.

Our final step is to fit the solution to the initial conditions and . Using the first condition, we get

Now, we want to find the coefficients . If we take , multiply by , and integrate from 0 to L, we get

.

Using the integral

,

we can reduce this to

.

Now we must consider the initial condition . Inserting the solution into this, we get

Again, we can multiply by and integrate to get

,

and reducing with the integral identity, we get