PDE notes: Wave Equation

2.1. Introduction

The one-dimensional wave equation for small displacement of a perfectly elastic string of length ${l}$ with no frictional forces and no restoring forces is

$\displaystyle \rho_0 u_{tt} = T_0 u_{xx} \ \ \ \ \ (88)$

$\displaystyle u_{tt} = c^2 u_{xx} \ \ \ \ \ (89)$

for ${t\geq0}$ , ${0\leq x \leq l}$ , where ${c^2=\frac{T_0}{\rho_0}}$ . We assume both tension ${T_0}$ and linear density ${\rho_0}$ to be constant. Tension ${T_0}$ has units of ${\frac{mass}{time^2}}$ and linear density has units of ${\frac{mass}{length^2}}$ , so ${c=\sqrt{\frac{T_0}{\rho_0}}}$ has the units of ${\frac{length}{time}}$ , which is also the unit of speed. Thus, ${c}$ turns out to be the velocity of wave propagation along the string. The one-dimensional wave equation models sound waves, water waves, vibrations in solids, and longitudinal or torsional vibrations in a rod, among other things.

Since the PDE in (89) contains the second time derivative, two initial conditions are required. The initial conditions usually take the follwing form:

$\displaystyle \mathrm{Initial \hspace{0.2cm} displacement:} \hspace{0.5cm} u(x,0)= f(x) \ \ \ \ \ (90)$

$\displaystyle \mathrm{Initial \hspace{0.2cm} velocity:} \hspace{0.5cm} v(x,0)=u_t(x,0)= g(x) \ \ \ \ \ (91)$

Typical boundary conditions are of the same form as thus given in the discussion of one-dimensional heat equation. For homogeneous Dirichlet conditions,

$\displaystyle u(0,t)=0=u(l,t) \ \ \ \ \ (92)$

for ${t \geq 0}$ . The end ends of the vibrating strings are fixed.

For homogeneous Neumann conditions,

$\displaystyle u_x(0,t)=0=u_x(l,t) \ \ \ \ \ (93)$

for ${t \geq 0}$ . These conditions are be achieved, for example, by attaching the ends of the string to a frictionless sleeve that moves vertically.

2.2. Solution by separation of variables

We now solve the one-dimensional wave equation with homogeneous Dirichlet boundary conditions. The following problem is defined for ${0 \leq x \leq L}$ and ${t \geq 0}$ :

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_{tt}=c^2u_{xx} \ \ \ \ \ (94)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u(0,t)=0 \ \ \ \ \ (95)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u(L,t)=0 \ \ \ \ \ (96)$

$\displaystyle \mathrm{IC1:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (97)$

$\displaystyle \mathrm{IC2:}\hspace{0.5cm} u_x(x,0) = g(x) \ \ \ \ \ (98)$

Solution Since both the PDE and the boundary conditions are linear and homogeneous, the method of separation of variables is attempted. We look for special product solutions of the form:

$\displaystyle u(x,t)=\phi(x)h(t) \ \ \ \ \ (99)$

Substitute (99) into (94) yields

$\displaystyle \phi(x)h''(t)=c^2h(t)\phi''(x) \ \ \ \ \ (100)$

Divide both sides by ${\phi(x)h(t)c^2}$ separates the variables:

$\displaystyle \frac{1}{h(t)c^2}h''(t)= \frac{1}{\phi(x)}\phi''(x)=-\lambda \ \ \ \ \ (101)$

where ${\lambda}$ is the separation constant. This implies that

$\displaystyle h''=-\lambda hc^2 \ \ \ \ \ (102)$

$\displaystyle \phi''=-\lambda \phi \ \ \ \ \ (103)$

We consider (103) first since it has a complete set of boundary conditions. Letting ${\phi=e^{rx}}$ , then ${\phi''=r^2e^{rx}}$ . So ${r=\pm\sqrt{-\lambda}}$ . Using results from the heat equation, we know that ${\lambda < 0}$ and ${\lambda=0}$ yields trivial solutions. For ${\lambda > 0}$ , the general solution is

$\displaystyle \phi(x) = c_1\mathrm{cos}(\sqrt{\lambda}x)+c_2\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (104)$

Apply ${\phi(0)=0}$ and we arrive at ${c_1=0}$ . Consider ${\phi(L)=0}$ , we arrive at ${c_2\mathrm{sin}(\sqrt{\lambda}L)=0}$ . If ${c_2=0}$ , we get a trivial solution. Therefore, we get a nontrivial solution if and only if

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (105)$

This means that

$\displaystyle \sqrt{\lambda}L= n\pi \ \ \ \ \ (106)$

The eigenvalues are

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=1,2,3... \ \ \ \ \ (107)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_2 \mathrm{sin}(\frac{n\pi x}{L}), n=1,2,3... \ \ \ \ \ (108)$

The time-dependent part of the solution is

$\displaystyle h(t) = c_3\mathrm{cos}(\frac{cn\pi t}{L})+c_4\mathrm{sin}(\frac{cn\pi t}{L}) \ \ \ \ \ (109)$

for ${n=1,2,3...}$ . Thus, for ${n\geq 0}$ , each of the functions

$\displaystyle u_n(x,t)=\phi_n(x)h_n(t) = \mathrm{sin}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (110)$

is a solution to the PDE and satisfies the boundary condition. By superposition principle, we can solve the initial value problem by considering a linear combinations of all product solutions:

$\displaystyle u(x,t) = \sum^{\infty}_{0}\mathrm{sin}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (111)$

The initial conditions in (97) and (98) are satisfied if,

$\displaystyle f(x)=\sum^{\infty}_{0}A_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (112)$

$\displaystyle g(x)=\sum^{\infty}_{0}B_n\frac{cn\pi}{L}\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (113)$

We can consider the fact that ${\mathrm{sin}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{sin}(\frac{n \pi x}{L})\mathrm{sin}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m \\ \end{array} \right. \ \ \ \ \ (114)$

Multiply (112) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = A_{n}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (115)$

Solving for ${A_n}$ yields,

$\displaystyle A_n=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (116)$

Multiply (113) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = B_n\frac{cn\pi}{L}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (117)$

Solving for ${B_n}$ yields,

$\displaystyle B_n\frac{cn\pi}{L}=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (118)$

Therefore, the PDE with homogeneous Dirichlet boundary conditions has a simple explicit solution.

${\square}$

The product solutions are also called the normal modes of vibration. The coefficients of ${t}$ inside the product solutions, namely ${\frac{n\pi c}{L}, n=1,2,3...}$ , are called the frequencies. The fundamental mode of the string has a frequency of ${\frac{\pi c}{L}}$ . The ${n}$ th overtone is just the ${n}$ th integral multiple of the fundamental.

We now consider the motion of a vibrating string governed by the homogeneous Neumann boundary conditions for the wave equation:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_{tt}=c^2u_{xx} \ \ \ \ \ (119)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u_x(0,t)=0 \ \ \ \ \ (120)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u_x(L,t)=0 \ \ \ \ \ (121)$

$\displaystyle \mathrm{IC1:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (122)$

$\displaystyle \mathrm{IC2:}\hspace{0.5cm} u_x(x,0) = g(x) \ \ \ \ \ (123)$

Solution Again, we use the method of separation of variables. We look for production solutions of the form:

$\displaystyle u(x,t)=\phi(x)h(t) \ \ \ \ \ (124)$

with

$\displaystyle h''=-\lambda hc^2 \ \ \ \ \ (125)$

$\displaystyle \phi''=-\lambda \phi \ \ \ \ \ (126)$

The general solution for (126) is

$\displaystyle \phi(x) = c_1\mathrm{cos}(\sqrt{\lambda}x)+c_2\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (127)$

We also need

$\displaystyle \phi'(x) = \sqrt{\lambda}(c_2\mathrm{cos}(\sqrt{\lambda}x)-c_1\mathrm{sin}(\sqrt{\lambda}x)) \ \ \ \ \ (128)$

Apply ${\phi'(0)=0}$ and we arrive at ${c_2=0}$ . Apply ${\phi'(L)=0}$ and we arrive at ${\sqrt{\lambda} c_1\mathrm{sin}(\sqrt{\lambda}x)=0}$ . Since ${c_1=0}$ yields a trivial solution, we get a nontrivial solution if and only if

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (129)$

This means that

$\displaystyle \sqrt{\lambda}L= n\pi \ \ \ \ \ (130)$

The eigenvalues are

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=1,2,3... \ \ \ \ \ (131)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_1 \mathrm{cos}(\frac{n\pi x}{L}), n=1,2,3... \ \ \ \ \ (132)$

The time-dependent part of the solution is

$\displaystyle h(t) = c_3\mathrm{cos}(\frac{cn\pi t}{L})+c_4\mathrm{sin}(\frac{cn\pi t}{L}) \ \ \ \ \ (133)$

for ${n=0,1,2,3...}$ . Thus, for ${n\geq 0}$ , each of the functions

$\displaystyle u_n(x,t)=\phi_n(x)h_n(t) = \mathrm{cos}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (134)$

is a solution to the PDE and satisfies the boundary condition. By superposition principle, we can solve the initial value problem by considering a linear combinations of all product solutions:

$\displaystyle u(x,t) = A_0+\sum^{\infty}_{1}\mathrm{cos}(\frac{n\pi x}{L})(A_n\mathrm{cos}(\frac{cn\pi t}{L})+B_n\mathrm{sin}(\frac{cn\pi t}{L})) \ \ \ \ \ (135)$

The initial conditions in (97) and (98) are satisfied if,

$\displaystyle f(x)=A_0+\sum^{\infty}_{1}A_n\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (136)$

$\displaystyle g(x)=A_0+\sum^{\infty}_{1}B_n\frac{cn\pi}{L}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (137)$

for ${0<x<L}$ . We use the fact that ${\mathrm{cos}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{cos}(\frac{n \pi x}{L})\mathrm{cos}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m\neq 0 \\ L & n=m=0 \end{array} \right. \ \ \ \ \ (138)$

Multiply (136) by ${\mathrm{cos}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{cos}(\frac{m \pi x}{L})\mathrm{dx} = A_{n}\int_{0}^{L}\mathrm{cos^2}(\frac{m\pi x}{L})\mathrm{dx} \ \ \ \ \ (139)$

Solving for ${A_n}$ yields,

$\displaystyle A_0= \frac{1}{L} \int_{0}^{L}f(x) \mathrm{dx} \ \ \ \ \ (140)$

$\displaystyle A_n=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{cos}(\frac{n \pi x}{L})\mathrm{dx}, n\geq 1 \ \ \ \ \ (141)$

Multiply (137) by ${\mathrm{cos}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle B_n\frac{cn\pi}{L}=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{cos}(\frac{n \pi x}{L})\mathrm{dx}, n\geq 1 \ \ \ \ \ (142)$

${\square}$