The two dimensional wave equation

1. The 2D wave equation Separation of variables Superposition Examples
The two dimensional wave equation
Ryan C. Daileda
Trinity University
Partial Differential Equations
March 1, 2012
Daileda The 2D wave equation

2. The 2D wave equation Separation of variables Superposition Examples
Physical motivation
Consider a thin elastic membrane stretched tightly over a
rectangular frame. Suppose the dimensions of the frame area×b
and that we keep the edges of the membrane fixed to the frame.
Perturbing the membrane from equilibrium results in some
sort of vibration of the surface.
Our goal is to mathematically model the vibrations of the
membrane surface.
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3. The 2D wave equation Separation of variables Superposition Examples
We let
u(x,y,t) =
deflection of membrane from equilibrium at
position (x,y) and timet.
For a fixedt, the surfacez=u(x,y,t) gives the shape of the
membrane at timet.
Under ideal assumptions (e.g. uniform membrane density, uniform
tension, no resistance to motion, small deflection, etc.) one can
show thatusatisfies thetwo dimensional wave equation
utt=c
2
∇
2
u=c
2
(uxx+uyy) (1)
for 0<x<a, 0<y<b.
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4. The 2D wave equation Separation of variables Superposition Examples
Remarks:
For the derivation of the wave equation from Newton’s second
law, see exercise 3.2.8.
As in the one dimensional situation, the constantchas the
units of velocity. It is given by
c
2
=
τ
ρ
,
whereτis the tension per unit length, andρis mass density.
The operator
∇
2
=
∂
2
∂x
2
+
∂
2
∂y
2
is called theLaplacian. It will appear in many of our
subsequent investigations.
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5. The 2D wave equation Separation of variables Superposition Examples
The fact that we are keeping the edges of the membrane fixed is
expressed by theboundary conditions
u(0,y,t) =u(a,y,t) = 0, 0≤y≤b,t≥0,
u(x,0,t) =u(x,b,t) = 0, 0≤x≤a,t≥0.(2)
We must also specify how the membrane is initially deformed and
set into motion. This is done via theinitial conditions
u(x,y,0) =f(x,y), (x,y)∈R,
ut(x,y,0) =g(x,y), (x,y)∈R, (3)
whereR= [0,a]×[0,b].
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6. The 2D wave equation Separation of variables Superposition Examples
Solving the 2D wave equation
Goal:Write down a solution to the wave equation (1) subject to
the boundary conditions (2) and initial conditions (3).
We will follow the (hopefully!) familiar process of
usingseparation of variablesto produce simple solutions to
(1) and (2),
and then theprinciple of superpositionto build up a
solution that satisfies (3) as well.
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7. The 2D wave equation Separation of variables Superposition Examples
Separation of variables
We seek nontrivial solutions of the form
u(x,y,t) =X(x)Y(y)T(t).
Plugging this into the wave equation (1) we get
XYT
′′
=c
2

X
′′
YT+XY
′′
T

.
If we divide both sides byc
2
XYTthis becomes
T
′′
c
2
T
=
X
′′
X
+
Y
′′
Y
.
Because the two sides are functions of different independent
variables, they must be constant:
T
′′
c
2
T
=A=
X
′′
X
+
Y
′′
Y
.
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8. The 2D wave equation Separation of variables Superposition Examples
The first equality becomes
T
′′
−c
2
AT= 0.
The second can be rewritten as
X
′′
X
=−
Y
′′
Y
+A.
Once again, the two sides involve unrelated variables, so both are
constant:
X
′′
X
=B=−
Y
′′
Y
+A.
If we letC=A−Bthese equations can be rewritten as
X
′′
−BX= 0,
Y
′′
−CY= 0.
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9. The 2D wave equation Separation of variables Superposition Examples
The first boundary condition is
0 =u(0,y,t) =X(0)Y(y)T(t), 0≤y≤b,t≥0.
Since we want nontrivial solutions only, we can cancelYandT,
yielding
X(0) = 0.
When we perform similar computations with the other three
boundary conditions we also get
X(a) = 0,
Y(0) =Y(b) = 0.
There are no boundary conditions onT.
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10. The 2D wave equation Separation of variables Superposition Examples
Fortunately, we have already solved the two boundary value
problems forXandY. The nontrivial solutions are
Xm(x) = sinμmx, μm=
mπ
a
, m= 1,2,3, . . .
Yn(y) = sinνny, νn=
nπ
b
, n= 1,2,3, . . .
with separation constantsB=−μ
2
mandC=−ν
2
n.
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11. The 2D wave equation Separation of variables Superposition Examples
Recall thatTmust satisfy
T
′′
−c
2
AT= 0
withA=B+C=−
−
μ
2
m+ν
2
n

<0. It follows that for any choice
ofmandnthe general solution forTis
Tmn(t) =Bmncosλmnt+B
∗
mn
sinλmnt,
where
λmn=c
q
μ
2
m
+ν
2
n
=cπ
r
m
2
a
2
+
n
2
b
2
.
These are thecharacteristic frequenciesof the membrane.
Daileda The 2D wave equation

12. The 2D wave equation Separation of variables Superposition Examples
Assembling our results, we find that for anypair m,n≥1 we have
thenormal mode
umn(x,y,t) =Xm(x)Yn(y)Tmn(t)
= sinμmxsinνny(Bmncosλmnt+B
∗
mn
sinλmnt)
where
μm=
mπ
a
,νn=
nπ
b
,λmn=c
q
μ
2
m
+ν
2
n
.
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13. The 2D wave equation Separation of variables Superposition Examples
Remarks:
Note that the normal modes:
oscillate spatially with frequencyμmin thex-direction,
oscillate spatially with frequencyνnin they-direction,
oscillate in time with frequencyλmn.
Whileμmandνnare simply multiples ofπ/aandπ/b,
respectively,
λmnis not a multiple of any basic frequency.
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14. The 2D wave equation Separation of variables Superposition Examples
Superposition
According to the principle of superposition, because the normal
modesumnsatisfy the homogeneous conditions (1) and (2), we
may add them to obtain the general solution
u(x,y,t) =
∞
X
n=1
∞
X
m=1
sinμmxsinνny(Bmncosλmnt+B
∗
mnsinλmnt).
We must use a double series since the indicesmandnvary
independently throughout the set
N={1,2,3, . . .}.
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15. The 2D wave equation Separation of variables Superposition Examples
Initial conditions
Finally, we must determine the values of the coefficientsBmnand
B
∗
mnthat are required so that our solution also satisfies the initial
conditions (3). The first of these is
f(x,y) =u(x,y,0) =
∞
X
n=1
∞
X
m=1
Bmnsin
mπ
a
xsin
nπ
b
y
and the second is
g(x,y) =ut(x,y,0) =
∞
X
n=1
∞
X
m=1
λmnB
∗
mnsin
mπ
a
xsin
nπ
b
y.
These are examples ofdouble Fourier series.
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16. The 2D wave equation Separation of variables Superposition Examples
Orthogonality (again!)
To compute the coefficients in a double Fourier series we can
appeal to the following result.
Theorem
The functions
Zmn(x,y) =sin
mπ
a
xsin
nπ
b
y, m,n∈N
are pairwise orthogonal relative to the inner product
hf,gi=
Z
a
0
Z
b
0
f(x,y)g(x,y)dy dx.
This is easily verified using the orthogonality of the functions
sin(nπx/a) on the interval [0,a].
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17. The 2D wave equation Separation of variables Superposition Examples
Using the usual argument, it follows thatassumingwe can write
f(x,y) =
∞
X
n=1
∞
X
m=1
Bmnsin
mπ
a
xsin
nπ
b
y,
then
Bmn=
hf,Zmni
hZmn,Zmni
=
Z
a
0
Z
b
0
f(x,y) sin
mπ
a
xsin
nπ
b
y dy dx
Z
a
0
Z
b
0
sin
2
mπ
a
xsin
2
nπ
b
y dy dx
=
4
ab
Z
a
0
Z
b
0
f(x,y) sin
mπ
a
xsin
nπ
b
y dy dx (4)
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18. The 2D wave equation Separation of variables Superposition Examples
Representability
The question of whether or not a given function is equal to a
double Fourier series is partially answered by the following result.
Theorem
If f(x,y)is a C
2
function on the rectangle[0,a]×[0,b], then
f(x,y) =
∞
X
n=1
∞
X
m=1
Bmnsin
mπ
a
xsin
nπ
b
y,
where Bmnis given by(4).
To say thatf(x,y) is aC
2
functionmeans thatfas well as
its first and second order partial derivatives are all continuous.
While not as general as the Fourier representation theorem,
this result is sufficient for our applications.
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19. The 2D wave equation Separation of variables Superposition Examples
Conclusion
Theorem
Suppose that f(x,y)and g(x,y)are C
2
functions on the rectangle
[0,a]×[0,b]. The solution to the wave equation(1)with
boundary conditions(2)and initial conditions(3)is given by
u(x,y,t) =
∞
X
n=1
∞
X
m=1
sinμmxsinνny(Bmncosλmnt+B
∗
mn
sinλmnt)
where
μm=
mπ
a
,νn=
nπ
b
,λmn=c
q
μ
2
m+ν
2
n,
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20. The 2D wave equation Separation of variables Superposition Examples
Theorem (continued)
and the coefficients Bmnand B
∗
mnare given by
Bmn=
4
ab
Z
a
0
Z
b
0
f(x,y) sin
mπ
a
xsin
nπ
b
y dy dx
and
B
∗
mn=
4
abλmn
Z
a
0
Z
b
0
g(x,y) sin
mπ
a
xsin
nπ
b
y dy dx.
We have not actually verified that this solution isunique, i.e.
that this is theonlysolution to the wave equation with the
given boundary and initial conditions.
Uniqueness can be proven using an argument involving
conservation of energyin the vibrating membrane.
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21. The 2D wave equation Separation of variables Superposition Examples
Example 1
Example
A2×3rectangular membrane has c= 6. If we deform it to have
shape given by
f(x,y) =xy(2−x)(3−y),
keep its edges fixed, and release it at t= 0, find an expression that
gives the shape of the membrane for t>0.
Daileda The 2D wave equation

22. The 2D wave equation Separation of variables Superposition Examples
We must compute the coefficientsBmnandB
∗
mn
. Since
g(x,y) = 0 we immediately have
B
∗
mn
= 0.
We also have
Bmn=
4
23
Z
2
0
Z
3
0
xy(2−x)(3−y) sin
mπ
2
xsin
nπ
3
y dy dx
=
2
3
Z
2
0
x(2−x) sin
mπ
2
x dx
Z
3
0
y(3−y) sin
nπ
3
y dy
=
2
3

16(1 + (−1)
m+1
)
π
3
m
3

54(1 + (−1)
n+1
)
π
3
n
3

=
576
π
6
(1 + (−1)
m+1
)(1 + (−1)
n+1
)
m
3
n
3
.
Daileda The 2D wave equation

23. The 2D wave equation Separation of variables Superposition Examples
The coefficientsλmnare given by
λmn= 6π
r
m
2
4
+
n
2
9
=π
p
9m
2
+ 4n
2
.
Assembling all of these pieces yields
u(x,y,t) =
576
π
6
∞
X
n=1
∞
X
m=1

(1 + (−1)
m+1
)(1 + (−1)
n+1
)
m
3
n
3
sin
mπ
2
x
×sin
nπ
3
ycosπ
p
9m
2
+ 4n
2
t

.
Daileda The 2D wave equation

24. The 2D wave equation Separation of variables Superposition Examples
Example 2
Example
Suppose in the previous example we also impose an initial velocity
given by
g(x,y) = sin 2πx.
Find an expression that gives the shape of the membrane for t>0.
BecauseBmndepends only on the initial shape, which hasn’t
changed, we don’t need to recompute them.
We only need to findB
∗
mnand add the appropriate terms to
the previous solution.
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25. The 2D wave equation Separation of variables Superposition Examples
Using the values ofλmncomputed above, we have
B
∗
mn=
2
3π
√
9m
2
+ 4n
2
Z
2
0
Z
3
0
sin 2πxsin
mπ
2
xsin
nπ
3
y dy dx
=
2
3π
√
9m
2
+ 4n
2
Z
2
0
sin2πxsin
mπ
2
x dx
Z
3
0
sin
nπ
3
y dy.
The first integral is zero unlessm= 4, in which case it evaluates
to 1. Evaluating the second integral, we have
B
∗
4n=
1
3π
√
36 +n
2
3(1 + (−1)
n+1
)
nπ
=
1 + (−1)
n+1
π
2
n
√
36 +n
2
andB
∗
mn= 0 form6= 4.
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26. The 2D wave equation Separation of variables Superposition Examples
Letu1(x,y,t) denote the solution obtained in the previous
example. If we let
u2(x,y,t) =
∞
X
m=1
∞
X
n=1
B
∗
mn
sin
mπ
2
xsin
nπ
3
ysinλmnt
=
sin 2πx
π
2
∞
X
n=1
1 + (−1)
n+1
n
√
36 +n
2
sin
nπ
3
ysin 2π
p
36 +n
2
t
then the solution to the present problem is
u(x,y,t) =u1(x,y,t) +u2(x,y,t).
Daileda The 2D wave equation