Solving the wave equation

The wave equation is a ubiquitous partial differential equation found in many areas of physics. We will attempt to solve it within this post.

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, it is actually solvable.

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:

$$ \frac{\partial^2 u}{\partial x^2} = \frac{d^2 f}{dx^2} g(t) = f'' g $$

$$ \frac{\partial^2 u}{\partial t^2} = \frac{d^2 g}{dt^2} f(x) = g'' f $$

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:

$$ f'' = C_1 f $$

$$ g'' = c^2 C_1 g $$

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.

$$ f'' = -C_2 f $$

$$ g'' = -c^2 C_2 g $$

The solution to the first differential equation is:

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

And:

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

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 due to the principle of linearity, we have:

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

Which is the solution of the differential equation.