wave_equation

Numeric Comparisons to the 2-
Dimensional Wave Equation
Rob Morien
Advanced Dynamics
Fall 2005

2
Introduction
In many physical situations it is desired to determine how a disturbance
propagates through a particular medium, such as water, thin membranes and even
extended bodies. It is the purpose of this paper to investigate the evolution of a
disturbance through a thin membrane such as a square drumhead.
Given an initial deflection above the undisturbed membrane plane it is possible to
determine the elevation above or below the undisturbed plane at any time t through use of
the two-dimensional wave equation. Before making these calculations, the wave
equation is derived using the Lagrangian and Hamilton’s principle.
After the wave equation has been derived it is possible to choose various initial
deflections and prescribe various boundary and initial conditions to determine the
different eigenvalues (frequencies) of each particular situation. This is done both
analytically and numerically through the use of the central difference method.
Comparisons of these two methods will be shown for a prescribed time step with
the analytic solutions provided by Maple and central difference calculations provided by
hand.

3
Analytic Method
Given a stretched rectangular membrane, it is desired to determine the elevation of some
point ( )tyxU ,, at any time t above the membrane plane given an initial disturbance.
The Lagrangian is defined as VT −=L , where T is the kinetic energy, and V is the
potential energy. Thus it is required to derive T and V.
( )
2
2
,
2
1
2
1






==
•
yxUmmT r&
where ),( yxU
•
is velocity and ),( yxU is displacement above or below the undisturbed
membrane plane.
( ) ( )∫∫
∫∫
=⇒=
D
D
dydxyxm
dydx
m
yx ,, ρρ
So, ( )∫∫ 





∂
∂
=
D
dydx
t
U
yxT
2
,
2
1
ρ
In this case, ( )yx,ρ is the membrane density per unit area and is assumed constant.
To get the potential energy, V, recall that VW ∆=→21 .
∫∫∫ +=⋅=→ dyFdxFdW yx
2
1
21
r
r
rF
Let τ=F , where τ is the membrane’s tension per unit length and is assumed constant
and uniform throughout the membrane.
Referring to figure 2,
( )0,0 ( )0,a
( )ba,( )b,0
b
a da
db
Figure 1 Figure 2

4
( ) ( ) ( ) ( )areadabddbadabdW ττττ ==+= , where in this case the area refers to the surface
area. It is found from the equation dydx
y
U
x
U
dA
22
1 





∂
∂
+





∂
∂
+=
( ) ( )
∫∫
∫∫
∂
∂
∂
∂
+
+≈






∂
∂
+





∂
∂
+=
D
y
U
x
U
D
dydx
dydx
y
U
x
U
dW
2
1
1
22
22
τ
τ
( ) ( ) ( ) ( )















 +
+−







 +
+= ∫∫∫∫∫
∂
∂
∂
∂
∂
∂
∂
∂
D
y
U
x
U
D
y
U
x
UU
U
dydxdydxWdU
2
1
2
1
2222
00111
0
τ
Taking 00 =U (the undisturbed membrane plane) as the zero reference potential:
( ) ( )
( ) ( )








−
+
+=








−







 +
+=
∫∫∫∫∫∫
∫∫∫∫∫
∂
∂
∂
∂
∂
∂
∂
∂
DD
y
U
x
U
D
DD
y
U
x
UU
U
dydxdydxdydx
dydxdydxWdU
2
1
2
1
22
22
11
111
0
τ
τ
( ) ( )∫∫ ∂
∂
∂
∂
+=
D y
U
x
U
dydxV
22
11
2
τ
This is the potential energy of the membrane.
Therefore, VT −=L becomes
( ) ( ) ( ) ( )∫∫∫∫ ∂
∂
∂
∂
∂
∂
+−=
D y
U
x
U
D t
U
dydxdydxyx
222 11
2
,
2
1 τ
ρL
or ( ) ( ) ( ) ( )( )∫∫∫∫ ∂
∂
∂
∂
∂
∂
+−
D y
U
x
U
D t
U
dydxdydxyx
222 11
,
2
1
τρ
Hamilton’s principle states that the actual motion connecting two known states of a
system is the one that minimizes the integral:
( )
( )[ ] minimum
2
1
minimum
2
1
2
1
2
1
222
=+−=
==−=
∫∫∫
∫∫
t
t D
yxt
t
t
t
t
dtdydxUUU
dtdtVTI
τρ
L
Therefore, the equation of motion becomes:

5
( ) ( )0, =+− yyxxtt UUUyx τρ or
( )⇒+= yyxxtt UUU τρ 2
2
2
2
2
2
t
U
y
U
x
U
∂
∂
=
∂
∂
+
∂
∂
τ
ρ
Looking at the coefficient in front of the second derivative of the temporal coordinate






τ
ρ
, and inspecting its units:
( ) [ ]
[ ]
[ ][ ]
[ ] [ ]
[ ]
[ ]2
2
LT
Lm
L
m
T
L,
2
2
==
τ
ρ yx
. These units are consistent with velocity units squared. This
is analogous to the equation for velocity, where velocity = distance/time.
It can be asserted that the constant
τ
ρ
is none other than the phase velocity of the wave
squared. Therefore, this is the speed of the transverse waves traveling across the
membrane.
For convenience, let
ρ
τ
=2
c .
Then 2
2
22
2
2
2
1
t
U
cy
U
x
U
∂
∂
=
∂
∂
+
∂
∂
This is the 2-dimensional wave equation to be solved to determine the evolution of the
wave.
Since this equation is a second order partial differential equation of the form U
t
U 2
2
2
∇−
∂
∂
0=+++++ FEDcBA yxyyxyxx σσσσσ
with ACB
CB
BA
−<−=





−
=




 2
1
10
01
detdet , the wave equation is a hyperbolic partial
differential equation.
It can be solved through the use of separation of variables, but before attacking this
problem, it is best to first prescribe the initial and boundary conditions.
For the boundary condition:
Let 0=U on the boundary of the membrane for all 0≥t .
Since this is a second order PDE, 2 initial conditions are required:
Let ( ) ( )yxfyxU ,0,, = (given initial displacement ( )yxf , )
and ( )yxg
t
U
t
,
0
=
∂
∂
=
(given initial velocity ( )yxg , )

6
The wave equation will have a solution of the form ( ) ( ) ( ) ( )tyxtyxU φψχ=,,
χψ
φ
χφ
ψ
ψφ
χ
xxxx
U
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
ψφ
χ
2
2
2
2
xx
U
∂
∂
=
∂
∂
χψ
φ
χφ
ψ
ψφ
χ
yyyy
U
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
χφ
ψ
2
2
2
2
yy
U
∂
∂
=
∂
∂
χψ
φ
χφ
ψ
ψφ
χ
tttt
U
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
χψ
φ
2
2
2
2
tt
U
∂
∂
=
∂
∂
Substituting these values into the wave equation:
χψ
φ
χφ
ψ
ψφ
χ
2
2
22
2
2
2
1
tcyx ∂
∂
=
∂
∂
+
∂
∂
and dividing through the φψχ ,, terms:
φ
φ
ψ
ψ
χ
χ 1111
2
2
22
2
2
2
tcyx ∂
∂
=
∂
∂
+
∂
∂
Now, it is necessary that both sides of this equation be equal to the same constant or else
one side would be allowed to vary while the other side remain unchanged.
2
2
2
2
2
2
2
2
2
11
11
k
yx
k
tc
−=
∂
∂
+
∂
∂
−=
∂
∂
ψ
ψ
χ
χ
φ
φ
The separation process can be continued further to show that both spatial derivatives are
also both equal to constants.

7
kxBkxAk
x
cossin02
2
+⇒=+
∂
∂ 2
χ
χ
pyDpyCp
y
cossin02
2
2
+⇒=+
∂
∂
ψ
ψ
Recalling the boundary conditions, ( ) ( ) ( ) ,00,0,00 === ψχχ a and ( ) 0=aψ ,
000cos0sin
000cos0sin
m
m
=⇒=+
=⇒=+
DDC
BBA
( ) ⇒== 0sin kaAaχ this is true only if
a
m
k
π
= and m any integer
( ) ⇒== 0sin pbCaψ this is true only if
b
n
p
π
= and n any integer
Then ( )tyxU ,, becomes ( )t
b
yn
a
xm
φ
ππ
sinsin .
The solution to ( )tφ leads to what is known as the “time function.”
( ) 0222
2
2
=++
∂
∂
φ
φ
pkc
t
. Let νλ c= and 22
pk +=ν
⇒=+
∂
∂
02
2
2
φλ
φ
t
( )tBtB mnmnmnmn λλ cossin*
+
22
2
2
2
2
22
b
n
a
m
c
b
n
a
m
cmn
22
+=+= π
ππ
λ
( ) ( )tBtB
b
xn
a
xm
tyxU mnmnmnmn λλ
ππ
cossinsinsin,, *
+=
From the superposition theorem, the sum of finitely many solutions ( )tyxU ,, is also a
solution. ( )tyxU ,, then becomes a double sum.
( )∑∑
∞
=
∞
=
+
1 1
*
sinsincossin
m n
mnmnmnmn
b
yn
a
xm
tBtB
ππ
λλ
From the initial condition ( ) ( )yxftyxU ,,, = , ( )0=t
( ) ( )yxft
b
yn
a
xm
ByxU
m n
mnmn ,cossinsin0,,
1 1
== ∑∑
∞
=
∞
=
λ
ππ
, which is known as a double
Fourier series. Recall from a Fourier series, the constant nB can be determined by

8
( )∫
L
dx
L
xn
xf
L 0
sin
2 π
. It follows that mnB can also by determined by
( )∫∫
b a
dydx
b
yn
a
xm
yxf
ab 0 0
sinsin,
4 ππ
.
( ) ( )∑∑ ∫∫
∞
=
∞
=







=
1 1 0 0
cossinsinsin,
4
0,,
m n
b a
mnt
b
yn
a
xm
dydx
b
yn
ins
a
xm
yxf
ab
yxU λ
ππππ
which is a solution to the wave equation whenever the initial velocity of some initial
deflection is zero. It is also common to determine the position of any point given an
initial velocity not equal to zero. Therefore, *
mnB must be determined. This is done by
taking the time derivative of ( )tyxU ,, at t = 0.
( )
( )∑∑
∑∑
∞
=
∞
=
=
∞
=
∞
=
−=






+
∂
∂
1 1
*
01 1
*
sinsin0sin0cos
sinsincossin
m n
mnmnmnmnmnmn
tm n
mnmnmnmn
b
yn
a
xm
BB
b
yn
a
xm
tBtB
t
ππ
λλλλ
ππ
λλ
( )yxg
b
yn
a
xm
B
m n
mnmn ,sinsin
1 1
*
== ∑∑
∞
=
∞
=
ππ
λ where ( )yxg , is some initial velocity. *
mnB can
be determined through the use of “Fourier’s trick”:
( )∫∫=
b a
mn
mn dydx
b
yn
a
xm
yxg
ab
B
0 0
*
sinsin,
4 ππ
λ
Then given an initial displacement, ( )yxf , , and some initial velocity, ( )yxg , , the
solution to the two-dimensional wave equation will be of the form:
( ) ( ) ( ) t
b
yn
a
xm
dydx
b
yn
a
xm
yxfyxg
ab
tyxU mn
m n
b a
mn
λ
ππππ
λ
cossinsinsinsin,,
14
,,
1 1 0 0
∑∑ ∫∫
∞
=
∞
=












+=
For the examples in this paper, all functions will be assumed to start with initial velocity,
( ) 0, =yxg .
As an example, choose c = 1, ( ) yxyxyxfba 2sinsinsin2sin,,2, −=== ππ
( ) ∑∑
∞
=
∞
=
=
1 1
cossinsin,,
m n
mnmn t
b
yn
a
xm
BtyxU λ
ππ
2
2
2
2
b
n
a
m
cmn += πλ

9
( )∫∫=
b a
mn dydx
b
yn
a
xm
yxf
ab
B
0 0
sinsin,
4 ππ
( )
( )
∫∫∫∫
∫∫∫∫
∫∫
−=
−=
−
ππππ
πππ π
π π
ππ
ππ
ππ
ππ
0
2
0
2
0
2
0
2
2
0 0
2
2
0 0
2
2
0 0
sinsin
2
sin2sin
2
sin2sin
2
sinsin
2
2
sinsin2sinsin
2
2
sinsinsin2sin
2
sinsin2sinsinsin2sin
2
4
dxmxxdy
ny
ydxmxxdy
ny
y
dydx
ny
mxyxdydx
ny
mxyx
dydx
b
yn
a
xm
yxyx
nmB , = 2
2
π
( )I I I I1 2 3 4−
4
0
2
0
1,
2
1,0
sinsin
2,
2
2,0
sin2sin
I
m
m
dxmxx
I
m
m
dxmxx
=




=
≠
=
=




=
≠
=
∫
∫
π
π
π
π
1
2
2
2
0
2
0
2
0
2,
2,0
2,
2,0
2
sin
2
2
sin
22
2
sin
2
2
sin
2
sinsin
I
n
n
n
n
dy
nyy
dy
nyy
dy
ny
y
=



=
≠
=



=
≠
==
/⋅
/
=
∫
∫∫
ππ
π
π
π
π
π
π
π
π
π
ππ
3
2
2
2
0
2
0
2
0
4,
4,0
4,
4,0
2
sin
2
4
sin
22
2
sin
2
22
sin
2
sin2sin
I
n
n
n
n
dy
nyy
dy
nyy
dy
ny
y
=



=
≠
=



=
≠
==
/⋅
/
=
∫
∫∫
ππ
π
π
π
π
π
π
π
π
π
ππ
nmB , = 2
2
π
( )I I I I1 2 3 4− =





==
==
≠≠
2m&4n1-
2m&21
4,2or2,10
n
nm
4,2or2,1,0
1,1 4,22,2
≠≠=
−==
nmB
BB
mn

10
( )
5
5
14
4
44
2
22
2,2
2222
2
2
2
2,2
=
=
+
=+=+=
λ
π
π
π
π
ππ
π
ππ
πλ c
( )
8
44
4
164
2
42
4,2
222
2
2
2
4,2
=
+=+=+=
λ
π
π
ππ
π
ππ
πλ c
( ) t
yx
Bt
yx
BtyxU 4,24,22,22,2 cos
2
4
sin
2
sincos
2
2
sin
2
sin,, λ
π
π
π
π
λ
π
π
π
π
+=
( ) tyxtyxtyxU 8cos2sin2sin5cossinsin,, −=
Substituting any value for x, y, and t (anywhere within the membrane), this equation will
tell you the height of the membrane at any time t.

11
Numerical Method
Solutions to the wave equation can be approximated using numerical methods. The
central difference approximation scheme provides nearly accurate solutions with
minimum residual error.






∂
∂
+
∂
∂
=
∂
∂
2
2
2
2
2
2
2
y
U
x
U
c
t
U
Substituting these approximations back into the wave equation:
( ) ( ) ( ) 







∆
+−
+
∆
+−
=
∆
+− −+−+
−+
2
1,,1,
2
,1,,12
2
1
,,
1
, 222
y
UUU
x
UUU
c
t
UUU
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
( )
( ) ( ) ( ) ( )
( ) ( ) 







∆∆
+−∆++−∆
=
∆
+− −+−+
−+
22
1,,1,
2
,1,,1
2
2
2
1
,,
1
, 222
yx
UUUxUUUy
c
t
UUU k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
Let yxh ∆=∆=
iˆ
jˆ
h
h
1=j
1=i 1ii =
1jj =
1, +ji
ji ,1+ji,
1, −ji
ji ,1−
( )
( )
( )2
1,,1,
2
2
2
,1,,1
2
2
2
1
,,
1
,
2
2
2
2
2
y
UUU
y
U
x
UUU
x
U
t
UUU
t
U
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
∆
+−
=
∂
∂
∆
+−
=
∂
∂
∆
+−
=
∂
∂
−+
−+
−+
These values are the central difference approximation
for each derivative. The superscripts represent the
time iteration number.

12
( )
( ) ( )







 +−++−
=
∆
+− −+−+
−+
4
1,,1,
2
,1,,1
2
2
2
1
,,
1
, 222
h
UUUhUUUh
c
t
UUU
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
( ) 2
,1,1,,1,12
2
1
,,
1
, 42
h
UUUUU
c
t
UUU k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji −+++
=
∆
+− −+−+
−+
Solving for the displacement at time 1+kt :
( ) ( ) ( )
( ) ( ) k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
k
ji
U
h
tc
UU
h
tc
UUU
h
tc
UUUU
h
tc
U
,2
22
1
,,2
22
1
,,,2
22
1,1,,1,12
22
1
,
42
010
101
010
24







 ∆
−+−










∆
=
−+
∆
−+++
∆
=
−
−
−+−+
+
Letting
( )
2
1
2
22
=
∆
h
tc
, the last term will drop out.
2c
h
t =∆
U U U U U Ui j
k
i j
k
i j
k
i j
k
i j
k
i j
k
, , , , , ,( )+
− + + −
−
= + + + −1
1 1 1 1
11
2
Or,
1
,,
1
,
010
101
010
2
1 −+
−










= k
ji
k
ji
k
ji UUU
To get the first time step, take
0=∂
∂
tt
U
.
Then
1
,
0
,
1
,
1
010
101
010
4
1
ji
jiji
U
UU −










= . ( )jiji yxVtU ,1
, ∆= , so
( )jijiji yxVtUU ,
010
101
010
4
1 0
,
1
, ∆+










=
Assume the membrane is raised at the center with an initial displacement
( ) ( )( )( )( )2222,0
−+−+= yyxxyxU and bound at the edges.

13
This would look like:
Assume the initial velocity is zero, that a = 2, b = 2, and c = 1. It is desired to find how
the displacement varies with time. Allow
2
1
=∆=∆= yxh to obtain 9 interior nodes.
The time increments are found from 354.0
2
2/1
2
===∆
c
h
t
( )jijiji yxgtUU ,
010
101
010
4
1 0
,
1
, ∆+










= since ( )yxg , is zero everywhere.
The values in the charts below show the elevation at each node. The values in the second
chart are analytical values obtained by Maple from the double infinite series (from 1 to
10):

14
( ) ∑∑
∞
=
∞
=
=
1 1
cossinsin,,
m n
mnmn t
b
yn
a
xm
BtyxU λ
ππ
( )
( )( )
( )( )( )( )∫∫
∫∫
−+−+=
=
2
0
2
0
0 0
sinsin2222
22
4
sinsin,
4
dydx
b
yn
a
xm
yyxx
dydx
b
yn
a
xm
yxf
ab
B
b a
mn
ππ
ππ
( )( ) ( )( )
∫ ∫
∫ ∫






−





−=
−+−+=
2
0
2
0
22
2
0
2
0
2
sin4
2
sin
2
sin4
2
sin
sin22sin22
dy
ynyn
ydx
xmxm
x
dy
b
xn
yydx
a
xm
xx
ππππ
ππ
= − −( )( )I I I I1 2 3 4
I2 = ∫
2
0
2
sin4 dx
xmπ
=
8 1(cos( ) )m
m
π
π
−
=
πm
16
( )m odd=
I x
m x
dx1
2
0
2
2
= ∫ sin( )
π
=
− − − −8 2 2 12 2
3 3
(( )cos( ) ( sin( ) ))m m m m
m
π π π π
π
=
−
=
− =






8 4
8
2 2
3 3
( )
( )
( )
m
m
m odd
m
m even
π
π
π
I4 = ∫
2
0
2
sin4 dy
ynπ
=
8 1(cos( ) )n
n
π
π
−
=
πn
16
( )n odd=

15
I y
n y
dy3
2
0
2
2
= ∫ sin( )
π
=
− − − −8 2 2 12 2
3 3
(( ) cos( ) ( sin( ) ))n n n n
n
π π π π
π
=
−
=
− =






8 4
8
2 2
3 3
( )
( )
( )
n
n
n odd
n
n even
π
π
π
B
m m m m
m
m
m
n n n n
n
n
n
mn =
− − − −

 −
− 

 ×
− − − −

 −
− 


8 2 2 1 8 1
8 2 2 1 8 1
2 2
3 3
2 2
3 3
(( ) cos( ) ( sin( ) )) (cos( ) )
(( )cos( ) ( sin( ) )) (cos( ) )
π π π π
π
π
π
π π π π
π
π
π
Maple Gives the displacements easily:
B[m,n]:=Int(Int((x+2)*(x-2)*(y+2)*(y-
2)*sin(m*Pi*x/2)*sin(n*Pi*y/2),x=0..2),y=0..2);
:=B ,m n
d
⌠
⌡



0
2
d
⌠
⌡



0
2
( )+x 2 ( )−x 2 ( )+y 2 ( )−y 2 




sin
m π x
2





sin
n π y
2
x y
u(x,y,t):=Sum(Sum(B[m,n]*sin(m*Pi*x/2)*sin(n*Pi*y/2)*cos(t*
Pi*sqrt(m^2/4+n^2/4)),m=1..infinity),n=1..infinity);
( )u , ,x y t ∑
=n 1
∞
∑
=m 1
∞
d
⌠
⌡



0
2
d
⌠
⌡



0
2
( )+x 2 ( )−x 2 ( )+y 2 ( )−y 2 




sin
m π x
2





sin
n π y
2
x y







:=





sin
m π x
2





sin
n π y
2







cos
t π +m2
n2
2







At Time t = 0, u(x,y,0);
u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*
Pi*(1/2)/2),m=1..20),n=1..20));u(1.0,0.5,0):=evalf(sum(sum(
B[m,n]*sin(m*Pi*(1)/2)*sin(n*Pi*.5/2),m=1..10),n=1..10));u(
1.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*.

16
5/2),m=1..10),n=1..10));u(0.5,1.0,0):=evalf(sum(sum(B[m,n]*
sin(m*Pi*.5/2)*sin(n*Pi*1/2),m=1..10),n=1..10));u(1.0,1.0,0
):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1/2),m=1..10
),n=1..10));u(1.5,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5
/2)*sin(n*Pi*1/2),m=1..10),n=1..10));u(0.5,1.5,0):=evalf(su
m(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*Pi*1.5/2),m=1..10),n=1..1
0));u(1.0,1.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*
Pi*1.5/2),m=1..10),n=1..10));u(1.5,1.5,0):=evalf(sum(sum(B[
m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 15.22971493
:=( )u , ,1.0 0.5 0 11.99631509
:=( )u , ,1.5 0.5 0 6.253747587
:=( )u , ,0.5 1.0 0 11.99631508
:=( )u , ,1.0 1.0 0 9.775603853
:=( )u , ,1.5 1.0 0 5.096078137
:=( )u , ,0.5 1.5 0 6.253747591
:=( )u , ,1.0 1.5 0 5.096078137
:=( )u , ,1.5 1.5 0 2.656614646
At Time t = .354, u(x,y,.354);
Pi*(1/2)/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..20),n=1..20
));u(1.0,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1)/2)*sin(n
*Pi*.5/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10))
;u(1.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*P
i*.5/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u
(0.5,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*Pi*1
/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0
,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1/2)*c
os(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.0,
0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1/2)*cos(
Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1.5,0):
=evalf(sum(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*Pi*1.5/2)*cos(Pi
*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.5,0):=e
valf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1.5/2)*cos(Pi*.3
54*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,0):=eval
f(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*cos(Pi*.35
4*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 13.73168329

17
:=( )u , ,1.0 0.5 0 11.48679972
:=( )u , ,1.5 0.5 0 5.850152116
:=( )u , ,0.5 1.0 0 11.48679970
:=( )u , ,1.0 1.0 0 8.957095820
:=( )u , ,1.5 1.0 0 4.516788103
:=( )u , ,0.5 1.5 0 5.850152118
:=( )u , ,1.0 1.5 0 4.516788106
:=( )u , ,1.5 1.5 0 2.207949353
At Time t = .708, u(x,y,.708);
Pi*(1/2)/2)*cos(Pi*.708*sqrt(m^2/4+n^2/4)),m=1..20),n=1..20
*Pi*.5/2)*cos(Pi*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10))
i*.5/2)*cos(Pi*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u
/2)*cos(Pi*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0
os(Pi*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.0,
Pi*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1.5,0):
*.708*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.5,0):=e
valf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1.5/2)*cos(Pi*.7
08*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,0):=eval
f(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*cos(Pi*.70
8*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 -14.79433063
:=( )u , ,1.0 0.5 0 -4.233097635
:=( )u , ,1.5 0.5 0 -3.371121281
:=( )u , ,0.5 1.0 0 -4.233097637
:=( )u , ,1.0 1.0 0 6.555284339
:=( )u , ,1.5 1.0 0 2.912539267
:=( )u , ,0.5 1.5 0 -3.371121279
:=( )u , ,1.0 1.5 0 2.912539273

18
:=( )u , ,1.5 1.5 0 1.127100152
At Time t = 1.062, u(x,y,1.062);
Pi*(1/2)/2)*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..20),n=1..2
0));u(1.0,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1)/2)*sin(
n*Pi*.5/2)*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10
));u(1.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n
*Pi*.5/2)*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10)
);u(0.5,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*P
i*1/2)*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u
(1.0,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1/
2)*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5
,1.0,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1/2)
*cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1
.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*Pi*1.5/2)*
cos(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.
5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1.5/2)*co
s(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,
0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*co
s(Pi*1.062*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 -4.761423251
:=( )u , ,1.0 0.5 0 -16.21698752
:=( )u , ,1.5 0.5 0 -3.595003391
:=( )u , ,0.5 1.0 0 -16.21698752
:=( )u , ,1.0 1.0 0 -15.82088364
:=( )u , ,1.5 1.0 0 -4.459870815
:=( )u , ,0.5 1.5 0 -3.595003394
:=( )u , ,1.0 1.5 0 -4.459870825
:=( )u , ,1.5 1.5 0 0.1312790189
At Time t = 1.416, u(x,y,1.416);

19
2)*cos(Pi*1.416*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5
*cos(Pi*1.416*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1
cos(Pi*1.416*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.
s(Pi*1.416*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,
s(Pi*1.416*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 -2.994597905
:=( )u , ,1.0 0.5 0 -5.290612631
:=( )u , ,1.5 0.5 0 -5.195011744
:=( )u , ,0.5 1.0 0 -5.290612625
:=( )u , ,1.0 1.0 0 -15.77018009
:=( )u , ,1.5 1.0 0 -6.221225508
:=( )u , ,0.5 1.5 0 -5.195011752
:=( )u , ,1.0 1.5 0 -6.221225517
:=( )u , ,1.5 1.5 0 -1.729165708
At Time t = 1.77, u(x,y,1.77);
*Pi*.5/2)*cos(Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10))
i*.5/2)*cos(Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u
/2)*cos(Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0
os(Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.0,
Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1.5,0):
*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.5,0):=e

20
valf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1.5/2)*cos(Pi*1.
77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,0):=eval
f(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*cos(Pi*1.7
7*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 -1.187083434
:=( )u , ,1.0 0.5 0 -1.528967272
:=( )u , ,1.5 0.5 0 -4.670921694
:=( )u , ,0.5 1.0 0 -1.528967268
:=( )u , ,1.0 1.0 0 -3.629418086
:=( )u , ,1.5 1.0 0 -13.59326888
:=( )u , ,0.5 1.5 0 -4.670921687
:=( )u , ,1.0 1.5 0 -13.59326889
:=( )u , ,1.5 1.5 0 -11.68786223
At t = 2.124, u(x,y,2.124);
2)*cos(Pi*2.124*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5
*cos(Pi*2.124*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1
cos(Pi*2.124*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.
s(Pi*2.124*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,
s(Pi*2.124*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 0.9685162845
:=( )u , ,1.0 0.5 0 1.547781108
:=( )u , ,1.5 0.5 0 0.7297354742
:=( )u , ,0.5 1.0 0 1.547781103

21
:=( )u , ,1.0 1.0 0 1.719660241
:=( )u , ,1.5 1.0 0 0.279360720
:=( )u , ,0.5 1.5 0 0.7297354758
:=( )u , ,1.0 1.5 0 0.2793607222
:=( )u , ,1.5 1.5 0 -5.769634646
At t = 2.478, u(x,y,2.478);
2)*cos(Pi*2.478*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5
*cos(Pi*2.478*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(0.5,1
cos(Pi*2.478*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.0,1.
s(Pi*2.478*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,
s(Pi*2.478*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:=( )u , ,0.5 0.5 0 2.487143177
:=( )u , ,1.0 0.5 0 3.642616396
:=( )u , ,1.5 0.5 0 8.242865097
:=( )u , ,0.5 1.0 0 3.642616399
:=( )u , ,1.0 1.0 0 4.730398447
:=( )u , ,1.5 1.0 0 9.747117351
:=( )u , ,0.5 1.5 0 8.242865090
:=( )u , ,1.0 1.5 0 9.747117338
:=( )u , ,1.5 1.5 0 10.05139973

22
?t (0.5,0.5) (1.0,0.5) (1.5,0.5) (0.5,1.0) (1.0,1.0) (1.5,1.0) (0.5,1.5) (1.0,1.5) (1.5,1.5)
0 14.0625 11.25 6.5625 11.25 9 5.25 6.5625 5.25 3.0625
0.354 12.375 18.4063 5.875 10.4063 8.25 4.65625 5.875 4.65625 2.625
0.708
1.062
1.416
1.77
2.124
2.478
?t (0.5,0.5) (1.0,0.5) (1.5,0.5) (0.5,1.0) (1.0,1.0) (1.5,1.0) (0.5,1.5) (1.0,1.5) (1.5,1.5)
0 15.230 11.996 6.254 11.996 9.776 5.096 6.254 5.096 2.657
0.354 13.732 11.487 5.850 11.487 8.957 4.517 5.850 4.517 2.208
0.708 -14.794 -4.233 -3.371 -4.233 6.555 2.913 -3.371 2.913 1.127
1.062 -4.761 -16.217 -3.595 -16.217 -15.821 -4.460 -3.595 -4.460 .131
1.416 -2.995 -5.291 -5.195 -5.291 -15.770 -6.221 -5.195 -6.221 -1.729
1.77 -1.187 -1.529 -4.671 -1.529 -3.629 -13.593 -4.671 -13.593 -11.688
2.124 .969 1.548 .730 1.548 1.720 .279 .730 .279 -5.770
2.478 2.487 3.643 8.242 3.643 4.730 9.747 8.242 9.747 10.051
To attain the values for each succeeding time step,
1
,,
1
,
010
101
010
2
1 −+
−










= k
ji
k
ji
k
ji UUU was
used.
Discrepancies in the values of the two charts could be due to using a 7-digit mantissa
instead of a 14-digit mantissa in Maple leading to significant round off errors. Another
possibility of obtaining different values between the two charts might arise from not
using enough terms in the limits of the double sum. Future research will be done to
investigate this phenomena as well as comparisons to the round and triangular drumheads
answering the ultimate question, “Can You Hear the Shape of a Drum?”

23
Works Cited
l Erwin Kreyszig, Advanced Engineering Mathematics, 8th edition
l David J. Griffiths, Quantum Mechanics, 2nd edition
l Gerald-Wheatley, Applied Numerical Analysis, 7th edition
l Michael D. Greenberg, Foundations of Applied Mathematics
l Discussions with Prof. Chris Papadopolous and Prof. Paul Lyman
l http://mathnt.mat.jhu.edu/zelditch/Teaching/F2005110.302/PDF%20Lectures/Dru
mundBessel.pdf
l http://www.falstad.com/mathphysics.html
l http://www.kettering.edu/%7Edrussell/demos.html

wave_equation

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