The 2D wave equation Separation of variables Superposition Examples
The two dimensional wave equation
Atife Caglar
University of Houston
Partial Differential Equations
Lecture 11
Caglar The 2-D wave equation
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Solving the 2D Wave Equation: Separation of Variables, Superposition, and Applications, Lecture notes of Mathematics
A comprehensive explanation of solving the two-dimensional wave equation, a fundamental concept in physics and engineering. It delves into the method of separation of variables, superposition principle, and their application to model the motion of vibrating membranes. Detailed examples and explanations, making it an excellent resource for students studying partial differential equations and related fields.
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The two dimensional wave equation
Atife Caglar
University of Houston
Partial Differential Equations Lecture 11
Vibrating membranes
Goal: Model the motion of an ideal elastic membrane. Set up: Assume the membrane at rest is a region of the xy -plane and let
u(x, y , t) =
vertical deflection of membrane from equilib- rium at position (x, y ) and time t.
For a fixed t, the surface z = u(x, y , t) gives the shape of the membrane at time t. Under ideal assumptions (e.g. uniform membrane density, uniform tension, no resistance to motion, small deflection, etc.) one can show that u satisfies the two dimensional wave equation
utt = c^2 ∆u = c^2 (uxx + uyy ).
We must also specify how the membrane is initially deformed and set into motion. This is done via the initial conditions
u(x, y , 0) = f (x, y ), (x, y ) ∈ R, ut (x, y , 0) = g (x, y ), (x, y ) ∈ R.
New goal: solve the 2-D wave equation subject to the boundary and initial conditions just given.
As usual, we will:
Use separation of variables to find separated solutions satisfying the homogeneous boundary conditions; and
Use the principle of superposition to build up a series solution that satisfies the initial conditions as well.
Separation of variables
We seek nontrivial solutions of the form
u(x, y , t) = X (x)Y (y )T (t).
Plugging this into utt = c^2 (uxx + uyy ) we get
XYT ′′^ = c^2
X ′′YT + XY ′′T
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
T ′′^ − c^2 AT = 0,
X ′′ X
Y ′′
Y
+ A.
We have already solved the two boundary value problems for X and Y. The nontrivial solutions are
X = Xm(x) = sin(μmx), μm =
mπ a , m ∈ N,
Y = Yn(y ) = sin(νny ), νn = nπ b
, n ∈ N,
with separation constants B = −μ^2 m and C = −ν n^2.
Since T ′′^ − c^2 AT = 0, and A = B + C = −
μ^2 m + ν n^2
T = Tmn(t) = Bmn cos(λmnt) + B mn∗ sin(λmnt),
where
λmn = c
μ^2 m + ν n^2 = cπ
m^2 a^2
n^2 b^2
These are the characteristic frequencies of the membrane.
Normal modes
Assembling our results, we find that for any pair m, n ∈ N we have the normal mode umn(x, y , t) = Xm(x)Yn(y )Tmn(t) = sin(μmx) sin(νny ) (Bmn cos(λmnt) + B mn∗ sin(λmnt)) = Amn sin(μmx) sin(νny ) cos(λmnt − φmn)
Remarks: Note that the normal modes: oscillate spatially with frequency μm = m/ 2 a in the x-direction, oscillate spatially with frequency νn = n/ 2 b in the y -direction, oscillate temporally with frequency λmn/ 2 π. While μm and νn are simply multiples of π/a and π/b, respectively, λmn is not a multiple of any basic frequency.
Representability
Which functions are given by double Fourier series?
The following result partially answers this first question. Theorem If f (x, y ) is a C 2 function on the rectangle [0, a] × [0, b], then
f (x, y ) =
∑^ ∞
n=
∑^ ∞
m=
Bmn sin
( (^) mπ a
x
sin
( (^) nπ b
y
for appropriate Bmn.
To say that f (x, y ) is a C 2 function means that f as 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.
Orthogonality (again!)
How can we compute the coefficients in a double Fourier series?
The following result helps us answer this second question.
Theorem The functions
Zmn(x, y ) = sin
( (^) mπ a x
sin
( (^) nπ b y
, m, n ∈ N
are pairwise orthogonal relative to the inner product
〈f , g 〉 =
∫ (^) a
0
∫ (^) b
0
f (x, y )g (x, y ) dy dx.
This is easily verified using the orthogonality of the functions sin(nπx/p) on the interval [0, p].
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 vibrating membrane problem is given by u(x, y , t) =
∑^ ∞
n=
∑^ ∞
m=
sin(μmx) sin(νny ) (Bmn cos(λmnt) + B mn∗ sin(λmnt))
where μm = m aπ , νn = n bπ , λmn = c
μ^2 m + ν^2 n , and
Bmn =
ab
∫ (^) a
0
∫ (^) b
0
f (x, y ) sin(μmx) sin(νny ) dy dx,
B mn∗ =
abλmn
∫ (^) a
0
∫ (^) b
0
g (x, y ) sin(μmx) sin(νny ) dy dx.
Example
A 2 × 3 rectangular 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.
The coefficients λmn are given by
λmn = c
μ^2 n + ν n^2 = 6π
m^2 4
n^2 9
= π
9 m^2 + 4n^2.
Assembling all of these pieces yields
u(x, y , t) =
π^6
∑^ ∞
n=
∑^ ∞
m=
(1 + (−1)m+1)(1 + (−1)n+1) m^3 n^3
sin
( (^) mπ 2
x
× sin
( (^) nπ 3 y
cos
π
9 m^2 + 4n^2 t
Example
Suppose in the previous example we also impose an initial velocity given by g (x, y ) = 8 sin 2πx. Find an expression that gives the shape of the membrane for t > 0.
Since we have the same initial shape, Bmn don’t change. We only need to find B mn∗ and add the appropriate terms to the previous solution.
Using λmn computed above, we have
B mn∗ =
2 · 3 π
9 m^2 + 4n^2
0
0
8 sin(2πx) sin
( (^) mπ 2
x
sin
( (^) nπ 3
y
dy dx
3 π
9 m^2 + 4n^2
0
sin(2πx) sin
( (^) mπ 2
x
dx
0
sin
( (^) nπ 3
y
dy.
The first integral is zero unless m = 4, i.e. B mn∗ = 0 for m 6 = 4.