Numeric Comparisons to the 2- Dimensional Wave Equation Rob Morien Advanced Dynamics Fall 2005
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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 ,21
21
==
•yxUmmT r&
where ),( yxU•
is velocity and ),( yxU is displacement above or below the undisturbed membrane plane.
( ) ( )∫∫∫∫
=⇒=D
D
dydxyxmdydx
myx ,, ρρ
So, ( )∫∫
∂∂
=D
dydxt
UyxT
2
,21
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 dydxyU
xU
dA22
1
∂∂
+
∂∂
+=
( ) ( )
∫∫
∫∫
∂∂
∂∂ +
+≈
∂∂
+
∂∂
+=
D
yU
xU
D
dydx
dydxyU
xU
dW
21
1
22
22
τ
τ
( ) ( ) ( ) ( )
++−
++= ∫∫∫∫∫ ∂
∂∂
∂∂
∂∂
∂
D
yU
xU
D
yU
xUU
U
dydxdydxWdU2
12
12222
00111
0
τ
Taking 00 =U (the undisturbed membrane plane) as the zero reference potential:
( ) ( )
( ) ( )
−
++=
−
++=
∫∫∫∫∫∫
∫∫∫∫∫
∂∂
∂∂
∂∂
∂∂
DD
yU
xU
D
DD
yU
xUU
U
dydxdydxdydx
dydxdydxWdU
2
12
1
22
22
11
111
0
τ
τ
( ) ( )∫∫ ∂∂
∂∂ +=
D yU
xU dydxV
2211
2τ
This is the potential energy of the membrane. Therefore, VT −=L becomes
( ) ( ) ( ) ( )∫∫∫∫ ∂∂
∂∂
∂∂ +−=
D yU
xU
D tU dydxdydxyx
222 11
2,
21 τ
ρL
or ( ) ( ) ( ) ( )( )∫∫∫∫ ∂∂
∂∂
∂∂ +−
D yU
xU
D tU dydxdydxyx
222 11,21
τρ
Hamilton’s principle states that the actual motion connecting two known states of a system is the one that minimizes the integral:
( )
( )[ ] minimum21
minimum
2
1
2
1
2
1
222 =+−=
==−=
∫ ∫∫
∫∫t
t Dyxt
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
tU
yU
xU
∂∂
=∂∂
+∂∂
τρ
Looking at the coefficient in front of the second derivative of the temporal coordinate
τρ
, and inspecting its units:
( ) [ ][ ]
[ ][ ][ ] [ ]
[ ][ ]2
2
LTLm
Lm
TL,
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 τρ
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 ρτ
=2c .
Then 2
2
22
2
2
2 1tU
cyU
xU
∂∂
=∂∂
+∂∂
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 UtU 22
2
∇−∂∂
0=+++++ FEDcBA yxyyxyxx σσσσσ
with ACBCBBA
−<−=
−
=
2110
01detdet , 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 ( )yxgtU
t
,0
=∂∂
=
(given initial velocity ( )yxg , )
6
The wave equation will have a solution of the form ( ) ( ) ( ) ( )tyxtyxU φψχ=,,
χψφ
χφψ
ψφχ
xxxxU
∂∂
+∂∂
+∂∂
=∂∂
ψφχ2
2
2
2
xxU
∂∂
=∂∂
χψφ
χφψ
ψφχ
yyyyU
∂∂
+∂∂
+∂∂
=∂∂
χφψ
2
2
2
2
yyU
∂∂
=∂∂
χψφ
χφψ
ψφχ
ttttU
∂∂
+∂
∂+
∂∂
=∂
∂
χψφ2
2
2
2
ttU
∂∂
=∂∂
Substituting these values into the wave equation:
χψφ
χφψ
ψφχ
2
2
22
2
2
2 1tcyx ∂
∂=
∂∂
+∂∂
and dividing through the φψχ ,, terms:
φφ
ψψ
χχ 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.
22
2
2
2
22
2
2
11
11
kyx
ktc
−=∂∂
+∂∂
−=∂∂
ψψ
χχ
φφ
The separation process can be continued further to show that both spatial deriva tives are also both equal to constants.
7
kxBkxAkx
cossin02
2
+⇒=+∂∂ 2χ
χ
pyDpyCpy
cossin022
2
+⇒=+∂∂
ψψ
Recalling the boundary conditions, ( ) ( ) ( ) ,00,0,00 === ψχχ a and ( ) 0=aψ ,
000cos0sin
000cos0sinm
m
=⇒=+
=⇒=+
DDC
BBA
( ) ⇒== 0sin kaAaχ this is true only if a
mk
π= and m any integer
( ) ⇒== 0sin pbCaψ this is true only if b
np
π= and n any integer
Then ( )tyxU ,, becomes ( )tb
yna
xmφ
ππsinsin .
The solution to ( )tφ leads to what is known as the “t ime function.”
( ) 02222
2
=++∂∂
φφ
pkct
. Let νλ c= and 22 pk +=ν
⇒=+∂∂
022
2
φλφ
t( )tBtB mnmnmnmn λλ cossin* +
22
2
2
2
2
22
bn
am
cb
na
mcmn
22
+=+= πππ
λ
( ) ( )tBtBb
xna
xmtyxU mnmnmnmn λλ
ππcossinsinsin,, * +=
From the superposition theorem, the sum of finitely many solutions ( )tyxU ,, is also a solution. ( )tyxU ,, then becomes a double sum.
( )∑∑∞
=
∞
=
+1 1
* sinsincossinm n
mnmnmnmn byn
axm
tBtBππ
λλ
From the initial condition ( ) ( )yxftyxU ,,, = , ( )0=t
( ) ( )yxftb
yna
xmByxU
m nmnmn ,cossinsin0,,
1 1
== ∑∑∞
=
∞
=
λππ
, which is known as a double
Fourier series. Recall from a Fourier series, the constant nB can be determined by
8
( )∫L
dxL
xnxf
L 0
sin2 π
. It follows that mnB can also by determined by
( )∫ ∫b a
dydxb
yna
xmyxf
ab 0 0
sinsin,4 ππ
.
( ) ( )∑∑ ∫ ∫∞
=
∞
=
=
1 1 0 0
cossinsinsin,4
0,,m n
b a
mntbyn
axm
dydxb
ynins
axm
yxfab
yxU λππππ
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 nmnmnmnmnmnmn
tm nmnmnmnmn
byn
axm
BB
byn
axm
tBtBt
ππλλλλ
ππλλ
( )yxgb
yna
xmB
m nmnmn ,sinsin
1 1
* == ∑∑∞
=
∞
=
ππλ where ( )yxg , is some initial velocity. *
mnB can
be determined through the use of “Fourier’s trick”:
( )∫ ∫=b a
mnmn dydx
byn
axm
yxgab
B0 0
* sinsin,4 ππλ
Then given an initial displacement, ( )yxf , , and some initial velocity, ( )yxg , , the solution to the two-dimensional wave equation will be of the form:
( ) ( ) ( ) tb
yna
xmdydx
byn
axm
yxfyxgab
tyxU mnm n
b a
mn
λππππ
λcossinsinsinsin,,
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 tb
yna
xmBtyxU λ
ππ
2
2
2
2
bn
am
cmn += πλ
9
( )∫ ∫=b a
mn dydxb
yna
xmyxf
abB
0 0
sinsin,4 ππ
( )( )
∫∫∫∫
∫ ∫∫ ∫
∫ ∫
−=
−=
−
ππππ
πππ π
π π
ππ
ππ
ππππ
0
2
02
0
2
02
2
0 02
2
0 02
2
0 0
sinsin2
sin2sin2
sin2sin2
sinsin2
2sinsin2sinsin
22
sinsinsin2sin2
sinsin2sinsinsin2sin24
dxmxxdyny
ydxmxxdyny
y
dydxny
mxyxdydxny
mxyx
dydxb
yna
xmyxyx
nmB , = 2
2π
( )I I I I1 2 3 4−
40
20
1,2
1,0sinsin
2,2
2,0sin2sin
Im
mdxmxx
Im
mdxmxx
=
=
≠=
=
=
≠=
∫
∫
π
π
π
π
12
2
2
0
2
0
2
0
2,2,0
2,2,0
2sin
22
sin
222
sin22
sin2
sinsin
Inn
nn
dynyy
dynyy
dyny
y
=
=≠
=
=≠
==
/⋅/
=
∫
∫∫
πππ
ππ
ππ
ππ
π
π
ππ
32
2
2
0
2
0
2
0
4,4,0
4,4,0
2sin
24
sin
222
sin2
22sin
2sin2sin
Inn
nn
dynyy
dynyy
dyny
y
=
=≠
=
=≠
==
/⋅/
=
∫
∫∫
πππ
ππ
ππ
ππ
π
π
ππ
nmB , = 2
2π
( )I I I I1 2 3 4− =
====
≠≠
2 m & 4n1-2m & 21
4,2or2,10
n
nm
4,2or 2,1,0
1,1 4,22,2
≠≠=
−==
nmB
BB
mn
10
( )5
514
444
222
2,2
2222
2
2
2
2,2
=
=+
=+=+=
λ
ππ
ππ
πππ
πππλ c
( )8
444164
242
4,2
222
2
2
2
4,2
=
+=+=+=
λ
ππ
πππ
πππλ c
( ) tyx
Btyx
BtyxU 4,24,22,22,2 cos24
sin2
sincos22
sin2
sin,, λππ
ππ
λππ
ππ
+=
( ) 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
22
2
2
yU
xU
ctU
Substituting these approximations back into the wave equation:
( ) ( ) ( )
∆
+−+
∆
+−=
∆
+− −+−+−+
21,,1,
2,1,,12
2
1,,
1, 222
y
UUU
x
UUUc
t
UUU kji
kji
kji
kji
kji
kji
kji
kji
kji
( )( ) ( ) ( ) ( )
( ) ( )
∆∆
+−∆++−∆=
∆
+− −+−+−+
221,,1,
2,1,,1
22
2
1,,
1, 222
yx
UUUxUUUyc
t
UUU kji
kji
kji
kji
kji
kji
kji
kji
kji
Let yxh ∆=∆=
i
j
h
h
1=j1=i 1ii =
1jj =
1, +ji
ji ,1+ji,
1, −ji
ji ,1−
( )
( )
( )21,,1,
2
2
2,1,,1
2
2
2
1,,
1,
2
2
2
2
2
y
UUU
yU
x
UUU
xU
t
UUU
tU
kji
kji
kji
kji
kji
kji
kji
kji
kji
∆
+−=
∂∂
∆
+−=
∂∂
∆
+−=
∂∂
−+
−+
−+
These values are the central difference approximation for each derivative. The superscripts represent the time iteration number.
12
( )( ) ( )
+−++−=
∆
+− −+−+−+
41,,1,
2,1,,1
22
2
1,,
1, 222
hUUUhUUUh
ct
UUU kji
kji
kji
kji
kji
kji
kji
kji
kji
( ) 2,1,1,,1,12
2
1,,
1, 42
h
UUUUUc
t
UUU kji
kji
kji
kji
kji
kji
kji
kji −+++
=∆
+− −+−+−+
Solving for the displacement at time 1+kt :
( ) ( ) ( )
( ) ( ) kji
kji
kji
kji
kji
kji
kji
kji
kji
kji
kji
Uh
tcUU
htc
UUUh
tcUUUU
htc
U
,2
221
,,2
22
1,,,2
22
1,1,,1,12
221
,
42010101010
24
∆−+−
∆
=
−+∆
−+++∆
=
−
−−+−+
+
Letting ( )
21
2
22
=∆
htc
, the last term will drop out. 2c
ht =∆
U U U U U Ui jk
i jk
i jk
i jk
i jk
i jk
, , , , , ,( )+− + + −
−= + + + −11 1 1 1
112
Or, 1,,
1,
010101
010
21 −+ −
= k
jik
jik
ji UUU
To get the first time step, take 0=∂
∂
ttU
.
Then
1,
0,
1,
1
010101
010
41
jijiji U
UU −
= . ( )jiji yxVtU ,1
, ∆= , so
( )jijiji yxVtUU ,010101
010
41 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 21
=∆=∆= yxh to obtain 9 interior nodes.
The time increments are found from 354.022/1
2===∆
ch
t
( )jijiji yxgtUU ,010101
010
41 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 tb
yna
xmBtyxU λ
ππ
( )
( )( ) ( )( )( )( )∫ ∫
∫ ∫
−+−+=
=
2
0
2
0
0 0
sinsin222222
4
sinsin,4
dydxb
yna
xmyyxx
dydxb
yna
xmyxf
abB
b a
mn
ππ
ππ
( )( ) ( )( )
∫ ∫
∫ ∫
−
−=
−+−+=
2
0
2
0
22
2
0
2
0
2sin4
2sin
2sin4
2sin
sin22sin22
dyynyn
ydxxmxm
x
dyb
xnyydx
axm
xx
ππππ
ππ
= − −( )( )I I I I1 2 3 4
I 2 = ∫2
0 2sin4 dx
xmπ
=8 1(cos( ) )m
mππ
− =
πm16
( )m odd=
I xm x
dx12
0
2
2= ∫ sin( )
π
=− − − −8 2 2 12 2
3 3
(( )cos( ) ( sin( ) ))m m m mm
π π π ππ
=
−=
− =
8 4
8
2 2
3 3
( )( )
( )
mm
m odd
mm even
ππ
π
I 4 = ∫2
0 2sin4 dy
ynπ
=8 1(cos( ) )n
nππ
− =
πn16
( )n odd=
15
I yn y
dy32
0
2
2= ∫ sin( )
π
=− − − −8 2 2 12 2
3 3
(( ) cos( ) ( sin( ) ))n n n nn
π π π ππ
=
−=
− =
8 4
8
2 2
3 3
( )( )
( )
nn
n odd
nn even
ππ
π
Bm m m m
mmm
n n n nn
nn
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 ππ
ππ
π π π ππ
ππ
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 x2
sin
n π y2
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 x2
sin
n π y2
x y
:=
sin
m π x2
sin
n π y2
cos
t π + m 2 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(sum(sum(B[m,n]*sin(m*Pi*.5/2)*sin(n*Pi*1.5/2),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),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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*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*Pi*.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)*cos(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):=evalf(sum(sum(B[m,n]*sin(m*Pi*1/2)*sin(n*Pi*1.5/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));u(1.5,1.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*cos(Pi*.354*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:= ( )u , ,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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*.708*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*.708*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*.708*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*.708*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*.708*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*.708*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*.708*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)*cos(Pi*.708*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)*cos(Pi*.708*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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*1.062*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*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*Pi*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)*cos(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)*cos(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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*1.416*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(
19
n*Pi*.5/2)*cos(Pi*1.416*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*1.416*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*1.416*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*1.416*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*1.416*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*1.416*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)*cos(Pi*1.416*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)*cos(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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*1.77*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*1.77*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*1.77*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*1.77*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*1.77*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*1.77*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*1.77*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):=evalf(sum(sum(B[m,n]*sin(m*Pi*1.5/2)*sin(n*Pi*1.5/2)*cos(Pi*1.77*sqrt(m^2/4+n^2/4)),m=1..10),n=1..10));
:= ( )u , ,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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*2.124*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*2.124*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*2.124*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*2.124*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*2.124*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*2.124*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*2.124*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)*cos(Pi*2.124*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)*cos(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); u(0.5,0.5,0):=evalf(sum(sum(B[m,n]*sin(m*Pi*(1/2)/2)*sin(n*Pi*(1/2)/2)*cos(Pi*2.478*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*2.478*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*2.478*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*2.478*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*2.478*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*2.478*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*2.478*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)*cos(Pi*2.478*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)*cos(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,
010101
010
21 −+ −
= k
jik
jik
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