## 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:

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