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Wave Equation--1-Dimensional


The one-dimensional wave equation is given by

 (partial^2psi)/(partialx^2)=1/(v^2)(partial^2psi)/(partialt^2).
(1)

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

psi(0,t)=0
(2)
psi(L,t)=0,
(3)

and initial conditions

psi(x,0)=f(x)
(4)
(partialpsi)/(partialt)(x,0)=g(x).
(5)

The one-dimensional wave equation can be solved exactly by d'Alembert's solution, using a Fourier transform method, or via separation of variables.

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

xi=x-vt
(6)
eta=x+vt.
(7)

By the chain rule,

(partial^2psi)/(partialx^2)=(partial^2psi)/(partialxi^2)+2(partial^2psi)/(partialxipartialeta)+(partial^2psi)/(partialeta^2)
(8)
1/(v^2)(partial^2psi)/(partialt^2)=(partial^2psi)/(partialxi^2)-2(partial^2psi)/(partialxipartialeta)+(partial^2psi)/(partialeta^2).
(9)

The wave equation then becomes

 (partial^2psi)/(partialxipartialeta)=0.
(10)

Any solution of this equation is of the form

 psi(xi,eta)=f(eta)+g(xi)=f(x+vt)+g(x-vt),
(11)

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 one-dimensional wave equation can also be solved by applying a Fourier transform to each side,

 int_(-infty)^infty(partial^2psi(x,t))/(partialx^2)e^(-2piikx)dx=1/(v^2)int_(-infty)^infty(partial^2psi(x,t))/(partialt^2)e^(-2piikx)dk,
(12)

which is given, with the help of the Fourier transform derivative identity, by

 (2piik)^2Psi(k,t)=1/(v^2)(partial^2Psi(k,t))/(partialt^2),
(13)

where

Psi(k,t)=F_x[psi(x,t)](k)
(14)
=int_(-infty)^inftypsi(x,t)e^(-2piikx)dx.
(15)

This has solution

 Psi(k,t)=A(k)e^(2piikvt)+B(k)e^(-2piikvt).
(16)

Taking the inverse Fourier transform gives

psi(x,t)=int_(-infty)^inftyPsi(k,t)e^(2piikx)dk
(17)
=int_(-infty)^infty[A(k)e^(2piikvt)+B(k)e^(-2piikvt)]e^(-2piikx)dk
(18)
=int_(-infty)^inftyA(k)e^(-2piik(x-vt))dk+int_(-infty)^inftyB(k)e^(-2piik(x+vt))dk
(19)
=f_1(x-vt)+f_2(x+vt),
(20)

where

f_1(u)=F_k[A(k)](u)=int_(-infty)^inftyA(k)e^(-2piiku)dk
(21)
f_2(u)=F_k[B(k)](u)=int_(-infty)^inftyB(k)e^(-2piiku)dk.
(22)

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

The one-dimensional wave equation can be solved by separation of variables using a trial solution

 psi(x,t)=X(x)T(t).
(23)

This gives

 T(d^2X)/(dx^2)=1/(v^2)X(d^2T)/(dt^2)
(24)
 1/X(d^2X)/(dx^2)=1/(v^2)1/T(d^2T)/(dt^2)=-k^2.
(25)

So the solution for X is

 X(x)=Ccos(kx)+Dsin(kx).
(26)

Rewriting (25) gives

 1/T(d^2T)/(dt^2)=-v^2k^2=-omega^2,
(27)

so the solution for T is

 T(t)=Ecos(omegat)+Fsin(omegat),
(28)

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

 C=0    kL=mpi,
(29)

where m is an integer. Plugging (◇), (◇) and (29) back in for psi in (◇) gives, for a particular value of m,

psi_m(x,t)=[E_msin(omega_mt)+F_mcos(omega_mt)]D_msin((mpix)/L)
(30)
=[A_mcos(omega_mt)+B_msin(omega_mt)]sin((mpix)/L).
(31)

The initial condition psi^.(x,0)=0 then gives B_m=0, so (31) becomes

 psi_m(x,t)=A_mcos(omega_mt)sin((mpix)/L).
(32)

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

 psi(x,t)=sum_(m=1)^inftyA_mcos(omega_mt)sin((mpix)/L).
(33)

Using orthogonality of sines again,

 int_0^Lsin((lpix)/L)sin((mpix)/L)dx=1/2Ldelta_(lm),
(34)

where delta_(lm) is the Kronecker delta defined by

 delta_(mn)={1   m=n; 0   m!=n,
(35)

gives

int_0^Lpsi(x,0)sin((mpix)/L)dx=sum_(l=1)^(infty)A_lsin((lpix)/L)sin((mpix)/L)dx
(36)
=sum_(l=1)^(infty)A_l1/2Ldelta_(lm)
(37)
=1/2LA_m,
(38)

so we have

 A_m=2/Lint_0^Lpsi(x,0)sin((mpix)/L)dx.
(39)

The computation of A_ms 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 (◇), with

 omega_m=vk_m=(vmpi)/L,
(40)

is

 psi(x,t)=sum_(m=1)^inftyA_mcos((vmpit)/L)sin((mpix)/L),
(41)

where the A_m coefficients are given by (◇).

A damped one-dimensional wave

 (partial^2psi)/(partialx^2)=1/(v^2)(partial^2psi)/(partialt^2)+b(partialpsi)/(partialt),
(42)

given boundary conditions

psi(0,t)=0
(43)
psi(L,t)=0,
(44)

initial conditions

psi(x,0)=f(x)
(45)
(partialpsi)/(partialt)(x,0)=g(x),
(46)

and the additional constraint

 0<b<(2pi)/(Lv),
(47)

can also be solved as a Fourier series.

 psi(x,t)=sum_(n=1)^inftysin((npix)/L)e^(-v^2bt/2)[a_nsin(mu_nt)+b_ncos(mu_nt)],
(48)

where

mu_n=(sqrt(4v^2n^2pi^2-b^2L^2v^4))/(2L)=(vsqrt(4n^2pi^2-b^2L^2v^2))/(2L)
(49)
b_n=2/Lint_0^Lsin((npix)/L)f(x)dx
(50)
a_n=2/(Lmu_n){int_0^Lsin((npix)/L)[g(x)+(v^2b)/2f(x)]dx}.
(51)

See also

d'Alembertian, d'Alembert's Solution, Korteweg-de Vries Equation, Laplacian, Telegraph Equation, Wave Equation, Wave Equation--Disk, Wave Equation--Rectangle, Wave Equation--Triangle

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References

Abramowitz, M. and Stegun, I. 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 and 271, 1953.Zwillinger, D. (Ed.). CRC Standard Mathematical Tables and Formulae. Boca Raton, FL: CRC Press, p. 417, 1995.Zwillinger, D. Handbook of Differential Equations, 3rd ed. Boston, MA: Academic Press, p. 130, 1997.

Cite this as:

Weisstein, Eric W. "Wave Equation--1-Dimensional." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/WaveEquation1-Dimensional.html

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