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Wave Equation--Rectangle


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

 (partial^2z)/(partialx^2)+(partial^2z)/(partialy^2)=1/(v^2)(partial^2z)/(partialt^2),
(1)

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).
(2)

Plugging (2) into (1) gives

 YT(d^2X)/(dx^2)+XT(d^2Y)/(dy^2)=1/(v^2)XY(d^2T)/(dt^2),
(3)

where the partial derivatives have now become complete derivatives. Multiplying (3) by v^2/XYT gives

 (v^2)/X(d^2X)/(dx^2)+(v^2)/Y(d^2Y)/(dy^2)=1/T(d^2T)/(dt^2).
(4)

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/T(d^2T)/(dt^2)=-omega^2.
(5)

This has solution

 T(t)=C_omegacos(omegat)+D_omegasin(omegat).
(6)

Plugging (5) back into (◇),

 (v^2)/X(d^2X)/(dx^2)+(v^2)/Y(d^2Y)/(dy^2)=-omega^2,
(7)

which we can rewrite as

 1/X(d^2X)/(dx^2)=-1/Y(d^2Y)/(dy^2)-(omega^2)/(v^2)=-k_x^2
(8)

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

 1/Y(d^2Y)/(dy^2)=k_x^2-(omega^2)/(v^2)=-k_y^2,
(9)

where we have defined a new constant k_y satisfying

 k_x^2+k_y^2=(omega^2)/(v^2).
(10)

Equations (◇) and (◇) have solutions

 X(x)=Ecos(k_xx)+Fsin(k_xx)
(11)
 Y(y)=Gcos(k_yy)+Hsin(k_yy).
(12)

We now apply the boundary conditions to (11) and (12). The conditions z(0,y,t)=0 and z(x,0,t)=0 mean that

 E=0    G=0.
(13)

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=ppi and L_yk_y=qpi, where p and q are integers. Solving for the allowed values of k_x and k_y then gives

 k_x=(ppi)/(L_x)    k_y=(qpi)/(L_y).
(14)

Plugging (◇), (◇), (◇), (◇), and (14) back into (◇) gives the solution for particular values of p and q,

 z_(pq)(x,y,t)=[C_omegacos(omegat)+D_omegasin(omegat)][F_psin((ppix)/(L_x))][H_qsin((qpiy)/(L_y))].
(15)

Lumping the constants together by writing A_(pq)=C_omegaF_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)=D_omegaF_pH_q, we obtain

 z_(pq)(x,y,t)=[A_(pq)cos(omega_(pq)t)+B_(pq)sin(omega_(pq)t)]sin((ppix)/(L_x))sin((qpiy)/(L_y)).
(16)
WaveEquationRectangle

Plots of the spatial part for modes are illustrated above.

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)^inftysum_(q=1)^infty[A_(pq)cos(omega_(pq)t)+B_(pq)sin(omega_(pq)t)]sin((ppix)/(L_x))sin((qpiy)/(L_y)),
(17)

where omega is defined by combining (◇) and (◇) to yield

 omega_(pq)=pivsqrt((p/(L_x))^2+(q/(L_y))^2).
(18)

Given the initial conditions z(x,y,0) and (partialz)/(partialt)(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=int_0^Lsin((mpix)/L)sin((npix)/L)dx=1/2Ldelta_(mn),
(19)

where delta_(mn) is the Kronecker delta. This can be demonstrated by direct integration. Let u=pix/L so du=(pi/L)dx in (◇), then

 I=L/piint_0^pisin(mu)sin(nu)du.
(20)

Now use the trigonometric identity

 sinalphasinbeta=1/2[cos(alpha-beta)-cos(alpha+beta)]
(21)

to write

 I=L/(2pi)int_0^picos[(m-n)u]du+int_0^picos[(m+n)u]du.
(22)

Note that for an integer l!=0, the following integral vanishes

int_0^picos(lu)du=1/l[sin(lu)]_0^pi
(23)
=1/l[sin(lpi)-sin0]
(24)
=1/lsin(lpi)
(25)
=0,
(26)

since sin(lpi)=0 when l is an integer. Therefore, I=0 when l=m-n!=0. However, I does not vanish when l=0, since

 int_0^picos(0·u)du=int_0^pidu=pi.
(27)

We therefore have that I=Ldelta_(mn)/2, so we have derived (◇). 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)[int_0^(L_x)z(x,y,0)sin((ppix)/(L_x))dx]sin((qpiy)/(L_y))dy.
(28)

Now plug in z(x,y,t), set t=0, and prime the indices to distinguish them from the p and q in (28),

 I=sum_(q^'=1)^inftyint_0^(L_y)[sum_(p^'=1)^inftyA_(p^'q^')int_0^(L_x)sin((ppix)/(L_x))sin((p^'pix)/(L_x))dx]×sin((qpiy)/(L_y))sin((q^'piy)/(L_y))dy.
(29)

Making use of (◇) in (29),

 I=sum_(q^'=1)^inftyint_0^(L_y)sum_(p^'=1)^inftyA_(p^'q^')(L_x)/2delta_(p,p^')sin((qpiy)/(L_y))sin((q^'piy)/(L_y))dy,
(30)

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

 I=(L_x)/2sum_(q=1)^inftyA_(pq^')(L_y)/2delta_(q,q^')=(L_xL_y)/4A_(pq).
(31)

Equating (30) and (31) and solving for A_(pq) then gives

 A_(pq)=4/(L_xL_y)int_0^(L_y)[int_0^(L_x)z(x,y,0)sin((ppix)/(L_x))dx]sin((qpiy)/(L_y))dy.
(32)

An analogous derivation gives the B_(pq)s as

 B_(pq)=4/(omega_(pq)L_xL_y)int_0^(L_y)[int_0^(L_x)(partialz)/(partialt)(x,y,0)sin((ppix)/(L_x))dx]sin((qpiy)/(L_y))dy.
(33)

See also

Wave Equation, Wave Equation--1-Dimensional, Wave Equation--Disk, Wave Equation--Triangle

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Cite this as:

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

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