PDEs: Some Solutions of the Wave Equation

In this post we derived the wave equation for a one-dimensional object moving in two-dimensional space (namely, a string vibrating up and down). This time we’re going to derive a family of solutions for this one-dimensional wave equation.

So the equation we derived previously looked like this: u_{tt} - \frac{T}{\rho}u_{xx}=0. This equation is often rewritten as: u_{tt} - c^2u_{xx}=0, where c=\sqrt{\frac{T}{\rho}}. We want to find solutions u describing how the string looks at any point in space and time. Well, we know in general what a wave looks like. It looks more or less like a sine or cosine. Suppose our solution looks like this:

u(x,t) = \cos\left(nx\right)T(t), where T(t) is the time component of the solution. In other words, we’re assuming that the space and time components of the wave are independent. Let’s differentiate:

u_{tt} = \cos\left(nx\right)T''(t)
u_{xx} = -n^2\cos\left(nx\right)T(t)

Plugging these in, the PDE becomes: \cos\left(nx\right)\left(T''(t) + c^2n^2T(t)\right)=0

Since we already have our space component, let’s focus on the time component: T''(t) + c^2n^2T(t)=0

This is just a second order ODE with constant coefficients. And it’s in quite a nice form. Which function’s derivative looks like the original function? That would, of course, be the exponential function. Let’s assume for a second that T(t) = e^{rt}. Then T''(t) = r^2e^{rt}, and the time component equation becomes: r^2e^{rt} + c^2n^2e^{rt}=0. Dividing out the exponentials, we get a simple quadratic equation: r^2 + c^2n^2=0. This equation is called the “characteristic equation” (not to be confused with the equation of the characteristic curve! These are different things!). Solving, we get that r = \pm\sqrt{-c^2n^2} = \pm icn.

Thus T(t) = \exp\left(\pm icnt\right). Euler’s formula gives us an equivalent equation: T(t) = \cos\left(\pm cnt\right) + i \sin\left(\pm cnt\right). Thus one solution is T_1(t) = \cos\left(cnt\right) + i \sin\left(cnt\right) and the other is T_2(t) = \cos\left(cnt\right) - i \sin\left(cnt\right). You can obtain T_2 by noting that cosine is an even function, and sine is an odd function.

Now we need to alter the T‘s to get rid of their imaginary parts.

Very quickly, the principle of superposition says that if u_1, u_2, u_3, ..., u_n are n solutions of the linear equation \mathcal{L}[u]=0, then \sum\limits_{i = 1}^j {A_i u_i } (for any index j between 1 and n and for any constants A_i) is also a solution of \mathcal{L}.

Let’s use this to get a family of real solutions. Let’s compute T_1 + T_2 and T_1 - T_2, both of which are solutions:

T_1 + T_2 = 2\cos\left(cnt\right)
T_1 - T_2 = 2 i \sin\left(cnt\right)

This is convenient, because now we can just multiply \frac{A}{2} to the first equation and \frac{B}{2i} to the second, and then add the two. That gives us the solution T(t) = A\cos\left(cnt\right) + B\sin\left(cnt\right), where A and B can be any real number.

Finally, plugging T back into the solution u, we get:

u(x,t) = \cos\left(nx\right)T(t) = \cos\left(nx\right)\left(A\cos\left(cnt\right) + B\sin\left(cnt\right)\right)


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