The Wave Equation

u_{tt} = c^2 u_{xx}

Originally solved by Jean le Rond d'Alembert in 1747, the function \(u\) describes the "height" of a string (or wave) at a point in space and time \((x,t)\). Note, we are considering a 1-dimensional string. Here (when considering \(c = 1\)), we seek a function whose second temporal partial derivative is equal to its second spatial partial derivative. Immediately, that should set off light bulbs in your head of some sort of \(\sin\) and \(\cos\) functions.

Boundary Conditions on the 1-D String

There are several boundary conditions to consider but we will mostly be concerned with three kinds:

endpoints held at 0 (homogenous)
endpoints that vary with time
force specified on the end points

To interpret these, let's consider the first boundary condition: imagine a guitar string that is clamped down at the ends (as usual). If you pull the string up and release, it will send "standing" waves through the string. For the second boundary condition, imagine two people holding opposite ends of a rope, where they are raising their end up and down. Intuitively, this will send "waves" through the string. As for the last boundary condition, we consider the endpoints of a string connected to a frictionless sleeve. As the sleeve moves up and down, it will drag the string with it. In that way, we have the force specified at the boundary, instead of the explicit position of the string at the endpoint. In the rest of this page, I will only consider the first boundary condition.

The Vibrating Wave (Guitar String)

Consider the following IBVP:

\text{PDE: } u_{tt} = \alpha^2u_{xx}, \quad 0 < x < L, \quad 0 < t < \infty

\text{Boundary conditions: } u(0, t) = 0 = u(L, t)

\text{Initial conditions: } u(x,0) = \phi(x), \quad u_t(x, 0) = \psi(x)

Note that we have two initial conditions now, this is due to the fact that our temporal derivative \(u_{tt}\) is of 2nd order, and therefore the solution to an ODE in \(t\) will produce two coefficients. It makes sense if we think about the guitar string example again: we don't only need the initial position of the string, but we also need to know how "fast" the string is moving from its initial position since that will affect the evolution of the string.

We solve this problem through separation of variables, so we assume

\[u(x,t) = T(t)X(x)\]

Plugging this into \(u_{tt} = \alpha^2u_{xx}\) we get

T''(t)X(x) = \alpha^2T(t)X''(x)

Separating variables gives

\frac{T''(t)}{\alpha^2T(t)} = \frac{X''(x)}{X(x)} = k

We now can write this as two separate equations:

T''(t) - k\alpha^2T(t) = 0

\[X''(x) - kX(x) = 0\]

Unlike the heat equation, here we cannot immediately conclude that \(k < 0\). Both ODEs are second-order homogeneous linear ODEs, so each has three solution types:

\(k = 0: T(t) = At + B, \quad X(x) = Cx + D\)
\(k < 0: T(t) = A\sin(\alpha \beta t) + B\cos(\alpha \beta t), \quad X(x) = C\sin(\beta x) + D\cos(\beta x)\)
\(k > 0: T(t) = Ae^{\alpha \beta t} + Be^{-\alpha \beta t}, \quad X(x) = Ce^{\beta x} + De^{-\beta x}\)

After checking each case, only \(k < 0\) produces a non-trivial bounded solution. Thus,

u(x,t) = [C\sin(\beta x) + D\cos(\beta x)] [A\sin(\alpha \beta t) + B\cos(\alpha \beta t)]

Applying Boundary Conditions

With \(u(0,t) = 0\), we get \(D = 0\). So,

u(x,t) = C\sin(\beta x)[A\sin(\alpha \beta t) + B\cos(\alpha \beta t)]

Applying \(u(L,t) = 0\) gives \(\sin(\beta L) = 0 \implies \beta_n = \frac{n\pi}{L}\).

Therefore, the general solution is

u(x,t) = \sum_{n=1}^{\infty} \sin(\tfrac{n\pi x}{L}) [a_n \sin(\tfrac{n\pi \alpha t}{L}) + b_n \cos(\tfrac{n\pi \alpha t}{L})]

Using Initial Conditions

From \(u(x,0) = \phi(x)\), we find

\phi(x) = \sum_{n=1}^{\infty} b_n \sin(\tfrac{n\pi x}{L})

Using orthogonality of sine functions, this gives

b_n = \frac{2}{L} \int_0^L \phi(x) \sin(\tfrac{n\pi x}{L}) dx

From \(u_t(x,0) = \psi(x)\), we compute

a_n = \frac{2}{n\pi \alpha} \int_0^L \psi(x) \sin(\tfrac{n\pi x}{L}) dx

Final Solution

The final solution to the finite vibrating string with homogenous Dirichlet boundary conditions and arbitrary initial conditions is:

u(x,t) = \sum_{n=1}^{\infty} \sin(\tfrac{n\pi x}{L}) [a_n \sin(\tfrac{n\pi \alpha t}{L}) + b_n \cos(\tfrac{n\pi \alpha t}{L})]

with

a_n = \frac{2}{n\pi \alpha} \int_0^L \psi(x) \sin(\tfrac{n\pi x}{L}) dx, \quad b_n = \frac{2}{L} \int_0^L \phi(x) \sin(\tfrac{n\pi x}{L}) dx