The Wave Equation

Wave Equation in Cartesian Coordinates

The wave equation is

$$ \frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u $$

where the Laplacian, $ \nabla^2 $, in cartesian coordinates, is

$$ \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$

In one spacial dimension, we have

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

In this example we will solve it on the region $ 0 \leq x \leq L $ and $ 0 \leq t \leq T, $ with boundary conditions $ u(0,t) = u(L,t) = 0. $

Separate variables

$$ u(x,t) = f(x)g(t) $$ $$ c^2 f''(x)g(t) - f(x)g''(t) = 0 $$ $$ c^2 \frac{f''(x)}{f(x)} - \frac{g''(t)}{g(t)} = 0 $$

Since the left-hand side separates into a function of $ x $ and a function of $ t $, the only way this equality can hold for all values of $ x $ and $ t $ is if both terms equal the same constant.

Let

$$ f''(x) = \lambda f(x) $$ $$ g''(t) = c^2 \lambda g(t) $$

The boundary conditions give

$$ f(0)g(t) = f(L)g(t) = 0 $$ $$ f(0) = f(L) = 0 $$

Assume a solution of the form $ f(x) = e^{kx} $

$$ k^2 e^{kx} = \lambda e^{kx} \quad \Rightarrow \quad k^2 = \lambda $$

i) $ \lambda > 0 $

$$ k = \pm \sqrt{\lambda} $$ $$ f(x) = c_1 e^{\sqrt{\lambda}x} + c_2 e^{-\sqrt{\lambda}x} $$ $$ f(0) = 0 = c_1 + c_2 \quad \Rightarrow \quad c_1 = -c_2 $$ $$ f(L) = c_1 \left( e^{\sqrt{\lambda}L} - e^{-\sqrt{\lambda}L} \right) = 0 $$

Either $ c_1 = 0 $, which implies $ f(x) = 0 $, and we are not interested in the trivial solution, or

$$ e^{2\sqrt{\lambda}L} = 1 \quad \Rightarrow \quad 2\sqrt{\lambda}L = \ln 1 = 0 $$

Therefore there is no nonzero solution in this case.

ii) $ \lambda = 0 $

$$ k = 0 \quad \Rightarrow \quad f''(x) = 0 $$ $$ \therefore \, f(x) = c_1 x + c_2 $$ $$ f(0) = c_2 = 0 $$ $$ f(L) = c_1 L = 0 \quad \Rightarrow \quad c_1 = 0 $$

Hence $ f(x) = 0 $, and there is again no nontrivial solution.

iii) $ \lambda < 0 $

$$ k = \pm i\sqrt{-\lambda} \quad |\lambda| = - \lambda $$

Let $ \lambda = -b^2 $ where $ b \ge 0 $

$$ f(x) = c_1 e^{i b x} + c_2 e^{-i b x} $$ $$ f(0) = c_1 + c_2 = 0 \quad \Rightarrow \quad c_1 = -c_2 $$ $$ f(L) = c_1 \left( e^{i b L} - e^{-i b L} \right) = 0 $$

This holds if and only if either $ c_1 = 0 $, which implies the trivial solution, or

$$ e^{2 i b L} = 1 $$

Recall that

$$ e^{ix} = 1 \Leftrightarrow x = 2\pi n, \quad n \in \mathbb{Z} $$

since

$$ e^{ix} = \cos x + i\sin x $$ $$ \sin(2\pi n) = 0, \quad \cos(2\pi n) = 1 $$

Therefore,

$$ 2bL = 2\pi n $$

$$ b = \frac{n\pi}{L} $$

where $ n $ is a non-negative integer, since we defined $ b \ge 0 $

$$ \lambda_n = -\left(\frac{n\pi}{L}\right)^2, \quad n = 0, 1, 2, ... $$

Then

$$ f(x) = c_1 \left( e^{ibx} - e^{-ibx} \right) = 2c_1 i \sin(bx) $$ $$ = 2c_1 i \sin\left(\frac{n\pi x}{L}\right) $$

For a real solution, let $ \displaystyle c_1 = \frac{a}{2i}, $ where $ a $ is a real constant. Then

$$ f_n(x) = a_n \sin\left(\frac{n\pi x}{L}\right) $$

$ n = 0 $ gives $ f(x) = 0 $, and we can see more clearly why $ n $ is non-negative; since $ \sin(-x) = - \sin x $, $ n < 0 $ does not produce new solutions, since the sign can be absorbed into the constant coefficients. Hence $ n $ is a positive integer

The boundary conditions therefore quantize the allowed spatial modes:

$$ \lambda_n = -\left(\frac{n\pi}{L}\right)^2, \quad n = 1, 2, 3, ... $$

The time equation is

$$ g''(t) = c^2 \lambda_n g(t) $$ $$ g''(t) = -\left(\frac{n\pi c}{L}\right)^2 g(t) $$

$$ g_n(t) = b_{n,1}\cos\left(\frac{n\pi ct}{L}\right) + b_{n,2}\sin\left(\frac{n\pi ct}{L}\right) $$

Therefore

$$ u_n(x,t) = \left( b_{n,1}\cos\left(\frac{n\pi ct}{L}\right) + b_{n,2}\sin\left(\frac{n\pi ct}{L}\right) \right) \sin\left(\frac{n\pi x}{L}\right) $$

By superposition, the most general solution is

$$ u(x,t) = \sum_{n=1}^{\infty} u_n(x,t) $$ $$ \therefore \, u(x,t) = \sum_{n=1}^{\infty} \left( b_{n,1}\cos\left(\frac{n\pi ct}{L}\right) + b_{n,2}\sin\left(\frac{n\pi ct}{L}\right) \right) \sin\left(\frac{n\pi x}{L}\right) $$

It's easier to do iii) in the trig function approach, I just used complex exponentials to be explicit

$ \lambda < 0 $, let $ \lambda = -b^2 $ where $ b \ge 0 $

$$ \Rightarrow \quad f(x) = c_1 \sin(bx) + c_2 \cos(bx) $$ $$ f(0) = c_2 = 0 $$ $$ f(L) = c_1 \sin(bL) = 0 $$ $$ bL = n\pi, \quad n \in \mathbb{Z}^+ $$ $$ \lambda_n = -\left(\frac{n\pi}{L}\right)^2 $$ $$ f_n(x) = a_n \sin\left(\frac{n\pi x}{L}\right) $$