## PDEs: The Wave Equation (Long Post..)

So far we’ve derived and found solutions for a simple, first order, linear PDE describing the transport of a chemical through a fluid. Today we’re going to derive our first second-order equation, namely the wave equation. This equation is actually rather famous, and applies to a crap-ton of different physical phenomena. You’ve no doubt heard of Schroedinger’s wave equation, which is a fundamental result of quantum mechanics. Well, that’s really just applying quantum mechanical principles to the classical wave equation (see here, for instance). The wave equation can also describe electromagnetic waves, sonic tremors, tidal waves, guitar strings and almost anything “wave-like”, given the proper treatment.

We’re going to start with the simplest of those, guitar strings. First, some assumptions. Suppose that the string is infinitely flexible, meaning that at no point along its length will the string resist being bent a certain way. This implies that the tension of the string at any one point will be tangential to the string at that point. Secondly, assume that the string doesn’t vibrate too much. Thirdly, assume that the string has no defects, that it’s perfectly homogeneous throughout. Lastly, we’re going to assume that each point on the string moves only either up or down as the string vibrates. Let’s summarize these assumptions.

Suppose, as always, that $u$ is the solution of our yet-to-be-determined PDE, where $u$ here represents the height of the string. Here are our assumptions:

Assumption 1: The tension force $T(x,t)$ is tangent to the solution $u(x,t)$.
Assumption 2: $\left| \frac{{\partial u}}{{\partial x}}\right| \ll 1$ for all $x$, i.e. vibrations are much smaller in magnitude than 1.
Assumption 3: The string density $\rho$ is constant (and, of course, positive).
Assumption 4: There is no horizontal component to the string’s movement.

So what physical restrictions do these assumptions imply? We’ve mentioned one force acting on a vibrating string, namely the tension. Clearly if the string has a density, then it must have a mass (remember $\rho = \frac{mass}{area}$). If it has mass, then there must be a gravitational force acting upon it; that’s the second force. Suppose, for convenience, that this string is on Earth; that way the weight of the string points downward. As always, a picture is handy:

Edit: There’s a error in this graph. The right $T(x,t)$ should read $T(x_1,t)$.

Let’s put all of our eggs in one basket and find the total force acting on the string. Keep in mind that the total force should be equal to $m\textbf{a}$ at the end. Let’s start with forces in the vertical direction.

The weight of the string is simply $\textbf{w} = m\textbf{g}$, where $m$ is the mass of the string and $\textbf{g}$ is the acceleration due to gravity. It’s convenient to change this equation so that it involves our $\rho$, since we’ve already defined that. As before, $\rho=\frac{mass}{area}$. In this case, the “area” is just the length of the string. Therefore, given that our imagined interval is $\left[ x_0,x_1\right]$, we have that $m \approx \rho (x_1-x_0)$. Let’s rewrite that as $m \approx \rho \Delta x$, for convenience. (You’ve probably already guessed why it’s only approximately equal; we’ll get to that in a minute)

The tension forces are a bit more complicated. We need to find the projections of the vectors $T$ onto our vertical axis. Just apply simple trigonometry to get $T(x_1,t)\sin (\theta_1 ) - T(x_0,t)\sin (\theta_0 )$.

Adding all of the forces together, we get:

$\textbf{F}_{vertical} = T(x_1,t)\sin (\theta_1 ) - T(x_0,t)\sin (\theta_0 ) - \rho \Delta x g$

Here, $\theta_1$ and $\theta_0$ are the respective angles between the vector and the vertical axis. The minuses appear the way they do because of a choice of coordinate system. You could choose another one and come up with something similar. These $\theta$‘s, however, are troublesome. Let’s see what we can do about them.

Look at the triangle in which $T$ is the hypotenuse:

We can see that $\tan \theta = \frac{\Delta u}{\Delta x}$. Taking $x$ to zero, the right-hand side becomes just the derivative. Therefore $\tan \theta = u_x$. Reformulating the triangle, and replacing $\tan \theta = u_x$, we get:

With this triangle, we can get the following:

$\sin \theta = \frac{u_x}{\sqrt{1+u_x^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+u_x^2}}$.

Remember, however, that we assumed $\left| u_x\right| \ll 1$. Therefore we can do a bit of fudge work and just assume $u_x^2 \approx 0$. Rewriting, we get that $\sin \theta = u_x$ and $\cos \theta = 1$. That’s convenient, because now we can get rid of the $\theta$‘s in our force equation:

$\textbf{F}_{vertical}=T(x_1,t)u_x (x_1,t) - T(x_0,t)u_x (x_0,t) - \rho \Delta x g$

Or, more concisely:

$\textbf{F}_{vertical}=\left. {T(x,t)} u_x(x,t) \right|_{x_0 }^{x_1 } - \rho \Delta x g = m\textbf{a} = \rho \Delta x \textbf{a}$

Now we have to be careful. We said above that the length of the string is approximately $x_1-x_0 = \Delta x$. This would be correct if the string were pulled completely taut. But it isn’t. We have a vibrating string that looks like a wave. Thus for every small interval $dx$, the length of the string is a bit different. This tells us that we cannot simply define $m = \rho \Delta x$. Instead, we have to look at infinitesimal changes in $x$. Therefore, instead of $m = \rho \Delta x$ we have $m = \int\limits_{x_0 }^{x_1 } {\rho \cdot dx}$. Well, while we’re replacing our $\Delta x$‘s with $dx$‘s, we might as well do the same for the $T u_x$ term, which we can just write as $T(x,t)u_x(x,t)$. Putting it all together, our force equation reduces to:

$\textbf{F}_{vertical}=T(x,t)u_x(x,t) - \int\limits_{x_0 }^{x_1 } {\rho g \cdot dx} = \int\limits_{x_0 }^{x_1 } {\rho\textbf{a}\cdot dx}$

Note that the $\int\limits_{x_0 }^{x_1 } {\rho g \cdot dx}$ term simply reduces to $\rho g \Delta x$ anyway. If we had just assumed that, however, we could’ve made a serious error.

There’s another problem. We haven’t specified what $\textbf{a}$ is. Well, that’s easy enough. Any freshman physics student could tell you that the acceleration $\textbf{a}$ is simply the second derivative of position with respect to time. In our case, that means $\textbf{a}=u_{tt}$. For convenience, let’s also let $x = x_1-x_0$. Thus the equation becomes:

$\textbf{F}_{vertical}=T(x,t)u_x(x,t) - \rho g x = \int\limits_{x_0 }^{x_1 } {\rho u_{tt}(x,t)\cdot dx}$

Let’s look at the horizontal force component now, taking our approximation of $\cos \theta$ into account. This time we only have two tension forces, and we get:

$\textbf{F}_{horizontal} = T(x_1,t)\cos \theta_1 - T(x_0,t)\cos \theta_0 = T(x_1,t) - T(x_0,t)$

However, by our last assumption, the string has no horizontal component. Therefore the above quantity is zero! Which means that the tension on the string at $x_0$ and $x_1$ and, indeed, anywhere on the string, is the same. Thus we can let our $T(x,t)$ in the above equation just be some constant $T$. Since the horizontal component is zero, the equation for the total force on the string is:

$\textbf{F}=Tu_x(x,t) - \rho g x = \int\limits_{x_0 }^{x_1 } {\rho u_{tt}(x,t)\cdot dx}$

Like in this post, we need to differentiate to get rid of the integral. Differentiating with respect to $x$, we get:

$Tu_{xx}(x,t) - \rho g = \rho u_{tt}(x_1,t) - \rho u_{tt}(x_0,t)$

Let $\rho u_{tt}(x_1,t) - \rho u_{tt}(x_0,t) = \rho u_{tt}(x,t)$. Normalizing the equation, and removing the $(x,t)$‘s for clarity’s sake (they’re implied, though; only $T$ and $\rho$ are constant), we get:

$u_{tt} - \frac{T}{\rho}u_{xx}=-g$

And there you have it, the famous wave equation.