Wave Equation

The wave equation is

$$\nabla^2\psi = {1\over v^2} {\partial^2\psi\over \partial t^2},$$ (1)

where $\nabla^2$ is the Laplacian.

The 1-D wave equation is

$${\partial^2\psi\over \partial x^2} = {1\over v^2} {\partial^2\psi\over \partial t^2}.$$ (2)

In order to specify a wave, the equation is subject to boundary conditions

$$\psi(0,t) = 0$$ (3)

$$\psi(L,t) = 0,$$ (4)

and initial conditions

$$\psi(x,0) = f(x)$$ (5)

$${\partial \psi\over \partial t}(x,0) = g(x).$$ (6)

The wave equation can be solved using the so-called d'Alembert's solution, a Fourier Transform method, or Separation of Variables.

d'Alembert devised his solution in 1746, and Euler subsequently expanded the method in 1748. Let

$$\xi \equiv x-at$$ (7)

$$\eta \equiv x+at.$$ (8)

By the Chain Rule,

$${\partial^2\psi\over \partial x^2} = {\partial^2\psi\over \p... ...tial \xi\partial \eta} + {\partial 2\psi\over \partial \eta^2}$$ (9)

$${1\over v^2} {\partial^2\psi\over \partial t^2} = {\partial... ...al \xi\partial \eta} + {\partial^2\psi\over \partial \eta^2}.$$ (10)

The wave equation then becomes

$${\partial^2\psi\over \partial \xi\partial \eta} = 0.$$ (11)

Any solution of this equation is of the form

$$\psi(\xi,\eta) = f(\eta)+g(\xi) = f(x+vt)+g(x-vt),$$ (12)

where $f$ and $g$ are any functions. They represent two waveforms traveling in opposite directions, $f$ in the Negative $x$ direction and $g$ in the Positive $x$ direction.

The 1-D wave equation can also be solved by applying a Fourier Transform to each side,

$$\int_{-\infty}^\infty {\partial^2\psi(x,t)\over\partial x^2}... ...fty {\partial^2 \psi(x,t)\over\partial t^2} e^{-2\pi ikx}\,dx,$$ (13)

which is given, with the help of the Fourier Transform Derivative identity, by

$$(2\pi ik)^2 \Psi(k,t) = {1\over v^2} {\partial^2\Psi(k,t)\over \partial t^2},$$ (14)

where

$$\Psi(k,t) \equiv {\mathcal F}[\psi(x,t)] = \int_{-\infty}^\infty \psi(x,t)e^{-2\pi ikx}\,dx.$$ (15)

This has solution

$$\Psi(k,t)=A(k)e^{2\pi i kvt}+B(k)e^{-2\pi ikvt}.$$ (16)

Taking the inverse Fourier Transform gives

$$\psi(x,t) \equiv \int_{-\infty}^\infty \Psi(k,t)e^{2\pi ikx}\,dx$$

$$= \int_{-\infty}^\infty [A(k)e^{2\pi ikvt}+B(k)e^{-2\pi ikvt}]e^{-2\pi ikx}\,dk$$

$$= \int_{-\infty}^\infty A(k)e^{-2\pi ik(x-vt)}\,dk+\int_{-\infty}^\infty B(k)e^{-2\pi ik(x+vt)}\,dk$$

$$= f_1(x-vt)+f_2(x+vt),$$ (17)

where

$$f_1(u)\equiv {\mathcal F}[A(k)] = \int_{-\infty}^\infty A(k)e^{-2\pi iku}\,dk$$ (18)

$$f_2(u)\equiv {\mathcal F}[B(k)] = \int_{-\infty}^\infty B(k)e^{-2\pi iku}\,dk.$$ (19)

This solution is still subject to all other initial and boundary conditions.

The 1-D wave equation can be solved by Separation of Variables using a trial solution

$$\psi(x,t)=X(x)T(t).$$ (20)

This gives

$$T{d^2X\over dx^2}={1\over v^2} X {d^2T\over dt^2}$$ (21)

$${1\over X}{d^2X\over dx^2} = {1\over v^2} {1\over T}{d^2T\over dt^2} = -k^2.$$ (22)

So the solution for $X$ is

$$X(x)=C\cos(kx)+D\sin(kx).$$ (23)

Rewriting (22) gives

$${1\over T}{d^2T\over dt^2} = -v^2k^2 \equiv-\omega^2,$$ (24)

so the solution for $T$ is

$$T(t)=E\cos(\omega t)+F\sin(\omega t),$$ (25)

where $v\equiv\omega/k$. Applying the boundary conditions $\psi(0,t)=\psi(L,t)=0$ to (23) gives

$$C=0 \qquad kL=m\pi,$$ (26)

where $m$ is an Integer. Plugging (23), (25) and (26) back in for $\psi$ in (21) gives, for a particular value of $m$,

$$\psi_m(x,t) = [E_m\sin(\omega_m t)+F_m\cos(\omega_m t)]D_m\sin\left({m\pi x \over L}\right)$$

$$\equiv [A_m\cos(\omega_m t)+B_m\sin(\omega_m t)]\sin\left({m\pi x\over L}\right).$$ (27)

The initial condition $\dot\psi(x,0)=0$ then gives $B_m=0$, so (27) becomes

$$\psi_m(x,t)=A_m\cos(\omega_m t) \sin\left({m\pi x\over L}\right).$$ (28)

The general solution is a sum over all possible values of $m$, so

$$\psi(x,t)=\sum_{m=1}^\infty A_m\cos(\omega_m t) \sin\left({m\pi x\over L}\right).$$ (29)

Using Orthogonality of sines again,

$$\int_0^L \sin\left({l\pi x\over L}\right)\sin\left({m\pi x\over L}\right)\, dx = {\textstyle{1\over 2}}L\delta_{lm},$$ (30)

where $\delta_{lm}$ is the Kronecker Delta defined by

$$\delta_{mn}\equiv\cases{1 & $m=n$\cr 0 & $m\not= n$\cr},$$ (31)

gives

$$\int_0^L \psi(x,0)\sin\left({m\pi x\over L}\right)\,dx\&= \sum_{l=1}^\infty A_l\sin\left({l\pi x\over L}\right)\sin\left({m\pi x\over L}\right)\,dx$$

$$= \sum_{l=1}^\infty A_l {\textstyle{1\over 2}}L \delta_{lm}= {\textstyle{1\over 2}}L A_m,$$ (32)

so we have

$$A_m = {2\over L}\int_0^L \psi(x,0)\sin\left({m\pi x\over L}\right)\, dx.$$ (33)

The computation of $A_m$s for specific initial distortions is derived in the Fourier Sine Series section. We already have found that $B_m=0$, so the equation of motion for the string (29), with

$$\omega_m\equiv vk_m={vm\pi\over L},$$ (34)

is

$$\psi(x,t)=\sum_{m=1}^\infty A_m\cos\left({vm\pi t\over L}\right)\sin \left({m\pi x\over L}\right),$$ (35)

where the $A_m$ Coefficients are given by (33).

A damped 1-D wave

$${\partial^2\psi\over\partial x^2} = {1\over v^2}{\partial^2\psi\over\partial t^2} + b{\partial\psi\over\partial t},$$ (36)

given boundary conditions

$$\psi(0,t) = 0$$ (37)

$$\psi(L,t) = 0,$$ (38)

initial conditions

$$\psi(x,0) = f(x)$$ (39)

$${\partial \psi\over \partial t}(x,0) = g(x),$$ (40)

and the additional constraint

$$0 < b < {2\pi \over Lv},$$ (41)

can also be solved as a Fourier Series.

$$\psi(x,t) = \sum_{n=1}^\infty \sin\left({n\pi x\over L}\right)e^{-v^2bt/2} [a_n\sin(\mu_nt)+b_n\cos(\mu_nt)],$$ (42)

where

$$\mu_n \equiv {\sqrt{4v^2n^2\pi^2-b^2L^2v^4}\over 2L} = {v\sqrt{4n^2\pi^2-b^2L^2v^2}\over 2L}$$ (43)

$$b_n = {2\over L}\int_0^L \sin\left({n\pi x\over L}\right)f(x)\,dx$$ (44)

$$a_n = {2\over L\mu_n}\left\{{\int_0^L \sin\left({n\pi x\over L}\right)\left[{g(x)+ {v^2b\over 2} f(x)}\right]\,dx}\right\}.$$ (45)

To find the motion of a rectangular membrane with sides of length $L_x$ and $L_y$ (in the absence of gravity ), use the 2-D wave equation

$${\partial^2z\over\partial x^2}+{\partial^2z\over\partial y^2}= {1\over v^2}{\partial^2z\over\partial t^2},$$ (46)

where $z(x,y,t)$ is the vertical displacement of a point on the membrane at position ($x,y$) and time $t$. Use Separation of Variables to look for solutions of the form

$$z(x,y,t)=X(x)Y(y)T(t).$$ (47)

Plugging (47) into (46) gives

$$YT {d^2X\over dx^2}+XT{d^2Y\over dy^2}={1\over v^2} XY {d^2T\over dt^2},$$ (48)

where the partial derivatives have now become complete derivatives. Multiplying (48) by $v^2/XYT$ gives

$${v^2\over X}{d^2X\over dx^2} +{v^2\over Y}{d^2Y\over dy^2} = {1\over T}{d^2T\over dt^2}.$$ (49)

The left and right sides must both be equal to a constant, so we can separate the equation by writing the right side as

$${1\over T} {d^2T\over dt^2}=-\omega^2.$$ (50)

This has solution

$$T(t)=C_\omega\cos(\omega t)+D_\omega\sin(\omega t).$$ (51)

Plugging (50) back into (49),

$${v^2\over X}{d^2X\over dx^2} +{v^2\over Y}{d^2Y\over dy^2}=-\omega^2,$$ (52)

which we can rewrite as

$${1\over X}{d^2X\over dx^2}=-{1\over Y}{d^2Y\over dy^2}-{\omega^2\over v^2} = -{k_x}^2$$ (53)

since the left and right sides again must both be equal to a constant. We can now separate out the $Y(y)$ equation

$${1\over Y}{d^2Y\over dy^2}={k_x}^2-{\omega^2\over v^2}\equiv -{k_y}^2,$$ (54)

where we have defined a new constant $k_y$ satisfying

$${k_x}^2+{k_y}^2={\omega^2\over v^2}.$$ (55)

Equations (53) and (54) have solutions

$$X(x)=E\cos(k_xx)+F\sin(k_xx)$$ (56)

$$Y(y)=G\cos (k_yy)+H\sin (k_yy).$$ (57)

We now apply the boundary conditions to (56) and (57). The conditions $z(0,y,t)=0$ and $z(x,0,t)=0$ mean that

$$E=0\qquad G=0.$$ (58)

Similarly, the conditions $z(L_x,y,t)=0$ and $z(x,L_y,t)=0$ give $\sin(k_xL_x)=0$ and $\sin(k_yL_y)=0$, so $L_xk_x=p\pi$ and $L_yk_y=q\pi$, where $p$ and $q$ are Integers. Solving for the allowed values of $k_x$ and $k_y$ then gives

$$k_x={p\pi\over L_x} \qquad k_y={q\pi\over L_y}.$$ (59)

Plugging (52), (56), (57), (58), and (59) back into (22) gives the solution for particular values of $p$ and $q$,

$$z_{pq}(x,y,t) = [C_\omega\cos(\omega t)+D_\omega\sin(\omega ... ...)}\right]\left[{H_q\sin\left({q\pi y\over L_y}\right)}\right].$$ (60)

Lumping the constants together by writing $A_{pq}\equiv C_\omega F_pH_q$ (we can do this since $\omega$ is a function of $p$ and $q$, so $C_\omega$ can be written as $C_{pq}$) and $B_{pq}\equiv D_\omega F_p H_q$, we obtain

$$z_{pq}(x,y,t) = [A_{pq}\cos(\omega_{pq} t)+B_{pq}\sin(\omega... ...ft({p\pi x\over L_x}\right)\sin\left({q\pi y\over L_y}\right).$$ (61)

Plots of the spatial part for modes (1, 1), (1, 2), (2, 1), and (2, 2) follow.

The general solution is a sum over all possible values of $p$ and $q$, so the final solution is

$$z(x,y,t) = \sum_{p=1}^\infty\sum_{q=1}^\infty [A_{pq}\cos(\o... ...ft({p\pi x\over L_x}\right)\sin\left({q\pi y\over L_y}\right),$$ (62)

where $\omega$ is defined by combining (55) and (59) to yield

$$\omega_{pq} \equiv \pi v\sqrt{\left({p\over L_x}\right)^2+\left({q\over L_y}\right)^2}.$$ (63)

Given the initial conditions $z(x,y,0)$ and ${\partial z\over \partial t} (x,y,0)$, we can compute the $A_{pq}$s and $B_{pq}$s explicitly. To accomplish this, we make use of the orthogonality of the Sine function in the form

$$I\equiv \int_0^L \sin\left({m\pi x\over L}\right)\sin \left({n\pi x\over L}\right)\,dx = {\textstyle{1\over 2}}L\delta_{mn},$$ (64)

where $\delta_{mn}$ is the Kronecker Delta. This can be demonstrated by direct Integration. Let $u\equiv \pi x/L$ so $du=(\pi /L)\,dx$ in (64), then

$$I={L\over \pi} \int_0^\pi \sin(mu)\sin(nu)\,du.$$ (65)

Now use the trigonometric identity

$$\sin\alpha\sin\beta ={\textstyle{1\over 2}}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]$$ (66)

to write

$$I = {L\over 2\pi} \int_0^\pi \cos[(m-n)u]\,du+\int_0^\pi \cos [(m+n)u]\,du.$$ (67)

Note that for an Integer $l\not= 0$, the following Integral vanishes

$$\int_0^\pi\cos(lu)\,du = {1\over l}[\sin(lu)]^\pi_0 = {1\over l}[\sin(l\pi)-\sin 0]$$

$$= {1\over l}\sin(l\pi) = 0,$$ (68)

since $\sin (l\pi)=0$ when $l$ is an Integer. Therefore, $I=0$ when $l\equiv m-n\not =0$. However, $I$ does not vanish when $l=0$, since

$$\int_0^\pi \cos(0\cdot u)\,du = \int_0^\pi du = \pi.$$ (69)

We therefore have that $I=L \delta_{mn}/2$, so we have derived (64). Now we multiply $z(x,y,0)$ by two sine terms and integrate between 0 and $L_x$ and between 0 and $L_y$,

$$I=\int_0^{L_y}\left[{\int_0^{L_x} z(x,y,0)\sin\left({p\pi x\... ...L_x}\right)\,dx}\right]\sin\left({q\pi y\over L_y}\right)\,dy.$$ (70)

Now plug in $z(x,y,t)$, set $t=0$, and prime the indices to distinguish them from the $p$ and $q$ in (70),

$$I = \sum_{q'=1}^\infty\int_0^{L_y}\left[{\sum_{p'=1}^\infty A_{p'q'}\... ...in\left({p\pi x\over L_x}\right)\sin\left({p'\pi x\over L_x}\right)\,dx}\right]$$

$$\phantom{=} \mathop{\times}\sin\left({q\pi y\over L_y}\right)\sin\left({q'\pi y\over L_y}\right)\,dy.$$ (71)

Making use of (64) in (71),

$$I=\sum_{q'=1}^\infty\int_0^{L_y}\sum_{p'=1}^\infty A_{p'q'} ... ...\pi y\over L_y}\right)\sin\left({q'\pi y\over L_y}\right)\,dy,$$ (72)

so the sums over $p'$ and $q'$ collapse to a single term

$$I = {L_x\over 2}\sum_{q=1}^\infty A_{pq'} {L_y\over 2}\delta_{q,q'} = {L_xL_y\over 4} A_{pq}.$$ (73)

Equating (72) and (73) and solving for $A_{pq}$ then gives

$$A_{pq} = {4\over L_xL_y} \int_0^{L_y}\left[{\int_0^{L_x} z(x... ...L_x}\right)\,dx}\right]\sin\left({q\pi y\over L_y}\right)\,dy.$$ (74)

An analogous derivation gives the $B_{pq}$s as

$$B_{pq} = {4\over \omega_{pq}L_xL_y} \int_0^{L_y}\left[{\int_... ...L_x}\right)\,dx}\right]\sin\left({q\pi y\over L_y}\right)\,dy.$$ (75)

The equation of motion for a membrane shaped as a Right Isosceles Triangle of length $c$ on a side and with the sides oriented along the Positive $x$ and $y$ axes is given by

$\psi(x,y,t)=[C_{pq}\cos(\omega_{pq} t)+D_{pq}\sin(\omega_{pq} t)]$
$\times \left[{\sin\left({p\pi x\over c}\right)\sin\left({q\pi y\over c}\right)-\sin\left({q\pi x\over c}\right)\sin\left({p\pi y\over c}\right)}\right],$
	(76)

where

$$\omega_{pq} = {\pi v\over c}\sqrt{p^2+q^2}$$ (77)

and $p$, $q$ Integers with $p>q$. This solution can be obtained by subtracting two wave solutions for a square membrane with the indices reversed. Since points on the diagonal which are equidistant from the center must have the same wave equation solution (by symmetry), this procedure gives a wavefunction which will vanish along the diagonal as long as $p$ and $q$ are both Even or Odd. We must further restrict the modes since those with $p< q$ give wavefunctions which are just the Negative of $(q, p)$ and $(p, p)$ give an identically zero wavefunction. The following plots show (3, 1), (4, 2), (5, 1), and (5,3).

References

Abramowitz, M. and Stegun, C. A. (Eds.). ``Wave Equation in Prolate and Oblate Spheroidal Coordinates.'' §21.5 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 752-753, 1972.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 124-125, 1953.