Solving the wave equation

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The wave equation is a ubiquitous partial differential equation found in many areas of physics. In this guide, we will sketch out the basic approach to solving it.

First, let's state the problem. The wave equation is given by:

$$ \frac{\partial^2 u}{\partial t^2} - c^2 \frac{\partial^2 u}{\partial x^2} = 0 $$

Unlike many partial differential equations, the wave equation is actually solvable, and as such it not only carries significant physical importance, but also serves as a useful demonstration of the methods of solving PDEs.

First, notice that the equation is linear. That means all the terms of $u(x, t)$ and its derivatives are at most, scaled by a constant, and added (or subtracted) together. This is a powerful fact, because that means any two solutions can be added (or subtracted) together and scaled by constants to form a new solution that satisfies the equation:

$$ u_1(x, t) \pm u_2(x, t) = u_3(x, t) $$

We are going to prove that a solution to the wave equation can be put in the general form:

$$ u(x, t) = A \sin(x - vt + \phi_0) \pm B \sin(x + vt + \phi_0) $$

To do this, we will first say that our solution $u(x, t)$ can be written as a product of two functions, one of time, and one of space:

$$ u(x, t) = f(x) g(t) $$

Now, we can take the derivatives of $u(x, t)$ with respect to each variable:

$$ \begin{align*} \frac{\partial^2 u}{\partial x^2} = \frac{d^2 f}{dx^2} g(t) = f'' g \\ \frac{\partial^2 u}{\partial t^2} = \frac{d^2 g}{dt^2} f(x) = g'' f \end{align*} $$

We can plug these derivatives into the equation, to have:

$$ g''f - c^2 f''g = 0 $$

We can move the second term to the right side of the equation:

$$ g''f = c^2 f''g $$

And we can rearrange the terms to make each side in terms of only one variable - this is called separation of variables:

$$ \frac{g''}{g} \frac{1}{c^2} = \frac{f''}{f} $$

Now, recognize that the left hand side depends on $g$, and the right hand side depends on $f$, but the values of the derivatives are the same, up to a scaling constant. The only way two derivatives of otherwise unrelated variables would be the same is if both yield a constant, which we'll call $C_1$:

$$ \frac{f''}{f} = \frac{g''}{g}\frac{1}{c^2} = C_1 $$

This yields a set of two ordinary differential equations:

$$ \begin{align*} f'' &= C_1 f \\ g'' &= c^2 C_1 g \end{align*} $$

If we set $C_2 = -C_1$, and rewrite both equations in terms of $C_2$, the equations immediately start to look familiar: they are the differential equations of both the sine and cosine functions.

$$ \begin{align*} f'' &= -C_2 f \\ g'' &= -c^2 C_2 g \end{align*} $$

The solution to the first differential equation is:

$$ f(x) = C_2 \sin (x) $$

And that of the second is:

$$ g(t) = C_2 \sin(ct) $$

So if we plug that back into $u(x, t)$, we have:

$$ u(x, t) = f(x) g(t) = C_2 \sin (x)C_2 \sin(ct) $$

If we use a trig identity, where $C_3 = \frac{(C_2)^2}{2}$, this becomes:

$$ u(x, t) = C_3 \cos(x - ct) - C_3\cos (x + ct)) $$

Since we can write any cosine as a sine with a phase shift, we have:

$$ u(x, t) = C_3 \sin(x - ct + \phi_0) - C_3 \sin(x + ct + \phi_0) $$

Replacing $C_3$ with arbitrary constants $A$ and $B$, and changing the subtraction to $\pm$, which we can do by simply defining $B = -C_3$, we have:

$$ u(x, t) = A \sin (x - vt + \phi_0) \pm B \sin(x + vt + \phi_0) $$

Which is the solution of the wave equation. Note that it is not the only solution, as the specific solution to the wave equations depends on the boundary conditions, which are additional constraints on the form of $u(x, t)$ - we did not provide boundary conditions to the problem for simplicity. However, it is a solution, and indeed, an important solution, because it describes standing waves in physics, like the vibrations of a string or a drumhead, as well as radio waves in a metal box. To learn more, please feel free to take a look at the full guide to PDEs.