lecture 3 1 short

7/28/2019 Lecture 3 1 Short

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The 2D wave equation Separation of variables Superposition Examples

The two dimensional wave equation

Ryan C. Daileda

Trinity University

Partial Differential EquationsMarch 1, 2012

Daileda The 2D wave equation
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Physical motivation

Consider a thin elastic membrane stretched tightly over a

rectangular frame. Suppose the dimensions of the frame are a band that we keep the edges of the membrane fixed to the frame.

Perturbing the membrane from equilibrium results in somesort of vibration of the surface.

Our goal is to mathematically model the vibrations of the

membrane surface.Daileda The 2D wave equation
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We let

u(x, y, t) = deflection of membrane from equilibrium atposition (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, uniformtension, no resistance to motion, small deflection, etc.) one canshow that u satisfies the two dimensional wave equation

utt = c22u = c2(uxx + uyy) (1)

for 0 < x< a, 0 < y < b.

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Remarks:

For the derivation of the wave equation from Newtons secondlaw, see exercise 3.2.8.

As in the one dimensional situation, the constant c has theunits of velocity. It is given by

c2

=

,

where is the tension per unit length, and is mass density.

The operator

2 =2

x2 +2

y2

is called the Laplacian. It will appear in many of oursubsequent investigations.


Th 2D i S i f i bl S i i E l
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The fact that we are keeping the edges of the membrane fixed is

expressed by the boundary 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 andset 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, (3)where R = [0, a] [0, b].


Th 2D ti S ti f i bl S iti E l
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Solving the 2D wave equation

Goal: Write down a solution to the wave equation (1) subject tothe boundary conditions (2) and initial conditions (3).

We will follow the (hopefully!) familiar process ofusing separation of variables to produce simple solutions to(1) and (2),

and then the principle of superposition to build up a

solution that satisfies (3) as well.

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Fortunately, we have already solved the two boundary valueproblems for X and Y. The nontrivial solutions are

Xm(x) = sinmx, m =m

a

, m = 1, 2, 3, . . .

Yn(y) = sin ny, n =n

b, n = 1, 2, 3, . . .

with separation constants B = 2m and C = 2n .

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q p p p p

Assembling our results, we find that for any pair m, n

1 we have

the normal mode

umn(x, y, t) = Xm(x)Yn(y)Tmn(t)

= sinmxsin ny(Bmn cosmnt + B

mn sinmnt)

where

m = ma

, n = nb

, mn = c2m + 2n .

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Superposition

According to the principle of superposition, because the normalmodes umn satisfy the homogeneous conditions (1) and (2), wemay add them to obtain the general solution

u(x, y, t) =n=1

m=1

sinmxsin ny(Bmn cosmnt + B

mn sin mnt).

We must use a double series since the indices m and n varyindependently throughout the set

N = {1, 2, 3, . . .}.

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Orthogonality (again!)

To compute the coefficients in a double Fourier series we canappeal to the following result.

Theorem

The functions

Zmn(x, y) = sin ma

xsin nb

y , m, n N

are pairwise orthogonal relative to the inner product

f, g = a

0

b0

f(x, y)g(x, y) dy dx.

This is easily verified using the orthogonality of the functionssin(nx/a) on the interval [0, a].


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Using the usual argument, it follows that assuming we can write

f(x, y) =

n=1

m=1

Bmn sin ma

xsin nb

y,

then

Bmn =f,ZmnZmn,Zmn =

a0

b0

f(x, y)sin ma

xsin nb

y dy dx

a0

b0

sin2m

axsin2

n

by dy dx

=4

ab

a0

b0

f(x, y)sinm

axsin

n

by dy dx (4)


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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 C2 function on the rectangle [0, a] [0, b], then

f(x, y) =n=1

m=1

Bmn sin ma

xsin nb

y,

where Bmn is given by (4).

To say that f(x, y) is a C2 function means that f as well asits 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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Theorem (continued)

and the coefficients Bmn and B

mn are given by

Bmn =4

ab

a0

b0

f(x, y)sinm

axsin

n

by dy dx

and

Bmn =4

abmn

a0

b0

g(x, y)sinm

axsin

n

by dy dx.

We have not actually verified that this solution is unique, i.e.

that this is the only solution to the wave equation with thegiven boundary and initial conditions.

Uniqueness can be proven using an argument involvingconservation of energy in the vibrating membrane.


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Example 1

Example

A 2 3 rectangular membrane has c = 6. If we deform it to haveshape given by

f(x, y) = xy(2 x)(3 y),keep its edges fixed, and release it at t = 0, find an expression thatgives the shape of the membrane for t > 0.


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We must compute the coefficients Bmn and B

mn. Sinceg(x, y) = 0 we immediately have

Bmn = 0.

We also have

Bmn =

4

2 3 2

0

30 xy(2 x)(3 y)sin

m

2 xsin

n

3 y dy dx

=2

3

20

x(2 x)sin m2

x dx

30

y(3 y)sin n3

y dy

=

2

316(1 + (

1)m+1)

3m354(1 + (

1)n+1)

3n3

=576

6(1 + (1)m+1)(1 + (1)n+1)

m3n3.

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Example 2

Example

Suppose in the previous example we also impose an initial velocity

given by

g(x, y) = sin2x.

Find an expression that gives the shape of the membrane for t > 0.

Because Bmn depends only on the initial shape, which hasntchanged, we dont need to recompute them.

We only need to find Bmn and add the appropriate terms tothe previous solution.


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Using the values ofmn computed above, we have

Bmn =2

3

9m2 + 4n2

2

0

3

0

sin2x sin m2

x sin n3

y dy dx

=2

3

9m2 + 4n2

20

sin2xsinm

2x dx

30

sinn

3y dy.

The first integral is zero unless m = 4, in which case it evaluatesto 1. Evaluating the second integral, we have

B

4n =

1

336 + n23(1 + (

1)n+1)

n =

1 + (

1)n+1

2n36 + n2

and Bmn = 0 for m = 4.


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Let u1(x, y, t) denote the solution obtained in the previousexample. If we let

u2(x, y, t) =

m=1

n=1

Bmn sinm

2xsin

n

3ysinmnt

=sin2x

2

n=1

1 + (1)n+1n

36 + n2sin

n

3ysin2

36 + n2t

then the solution to the present problem is

u(x, y, t) = u1(x, y, t) + u2(x, y, t).

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lecture 3 1 short

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