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Lecture 22

Higher-dimensional PDEs

Relevant section of text: Chapter 7

We now examine some PDEs in higher dimensions, i.e., R2 and R3. In general, the heat and wave

equations in higher dimensions are given by

u

t= k2u (heat equation), (1)

2u

t2= c22u (wave equation). (2)

The higher-dimensional form of the heat equation comes from the application of the conservation of

thermal energy to regions in Rn, along with Fouriers law of heat conduction and the Divergence

Theorem. A derivation of the wave equation in higher dimensions would require a detailed discussion

of stress and strain in solids, which we forego.

In what follows, we examine the two-dimensional wave equation, since it leads to some interesting

and quite visualizable solutions.

Vibrating rectangular, homogeneous membrane

The vibration of a thin membrane is, for small displacements, modelled well by the two-dimensional

wave equation 2u

t2= c2

2u

x2+

2u

y2

. (3)

Here u(x,y,t) represents the displacement of the membrane from its equilibrium position, which is

assumed to be the surface u(x, y) = 0.

The simplest problem is that of a rectangular membrane, represented by the region

0 x L, 0 y H, (4)

which we shall simply denote as D R2. Sometimes, we shall also write the above region as [0, L] [0, H].

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Well assume that the membrane is clamped, leading to the following boundary conditions on

u(x,y,t):

u(x, 0, t) = 0,

u(L,y,t) = 0,

u(x,H,t) = 0,

u(0, y , t) = 0. (5)

Well also need initial conditions:

u(x,y, 0) = (x, y) (initial displacement)

u

t(x,y, 0) = (x, y) (initial velocity). (6)

The above boundary conditions are homogeneous if two solutions u1(x,y,t) and u2(x,y,t) satisfy

the BCs, then a linear combination also satisfies the BCs. This suggest that we can try the separation

of variables method to produce an infinite set of solutions.

Well first try a separation of spatial and time-dependent solutions:

u(x,y,t) = (x, y)h(t). (7)

Substitution into Eq. (3) yields

h = c2xxh + c2yyh. (8)

We may separate solutions as follows:

h

c2h=

xx + yy

= , > 0, (9)

where we have introduced the negative separation constant , based upon past results. The time-dependent equation for h(t),

h + c2h = 0, (10)

has oscillatory solutions, which is desirable. (We have skipped a longer analysis of the problem which

would have involved an examination of all cases, i.e., (i) = 0, (ii) < 0 and (iii) > 0.)

The spatial function satisfies the equation

2

x2+

2

y2= , (11)

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or, simply,

2 + = 0, (12)

which is a two-dimensional regular Sturm-Liouville eigenvalue equation, with boundary conditions,

(0, y) = 0, (x, 0) = 0, (L, y) = 0, (x, H) = 0. (13)

These are homogeneous BCs, suggesting that we can attempt a separation-of-variables-type solu-

tion for (x, y), i.e.,

(x, y) = f(x)g(y). (14)

(Note that this implies that our original function u(x,y ,t) has been separated into three functions,

u(x,y,t) = f(x)g(y)h(t). (15)

We could have started with this form, but it probably would not have been clear why it should work.)

Substitution of (14) into (11) yields

fg + f g = fg. (16)

In an attempt to separate variables, we now divide by the term f g to give

f

f+

g

g= . (17)

We have separated variables: the first term involves x and the second term involves y, but this doesnt

allow us to solve for either f or g. What we have to do is to perform another separation a separationof x and y terms as follows,

f

f= g

g. (18)

Now we have all x-dependent parts on the left and y-dependent parts on the right. We now introduce

a second separation constant:f

f= g

g= , > 0, (19)

which yields the following equations for f and g:

d2fdx2

+ f = 0, f(0) = f(L) = 0,

d2g

dy2+ ( )g = 0, g(0) = g(H) = 0. (20)

This is a system of two Sturm-Liouville eigenvalue problems:

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1. An equation for f with eigenvalue ,

2. An equation for g with eigenvalue .

In fact, the system is coupled because of the appearance of in both eigenvalues. It might seem

somewhat confusing that we have to deal with another separation constant. However, all will be fine

if we proceed systematically.

We can solve the eigenvalue equation for f very easily weve done it many times! The solutions

are

fn(x) = sinnx

L

, n =

nL

2. (21)

For each value of n, n = 1, 2, , the equation for g becomes the eigenvalue problem

d2g

dy2+ ( n)g = 0, g(0) = g(H) = 0. (22)

For simplicity, well define

n = n, (23)

so that the above equation becomes

d2g

dy2+ ng = 0, g(0) = g(H) = 0. (24)

For each n, n = 1, 2, , the solutions are known:

gnm(y) = sinmy

H

, (25)

with eigenvalues

nm =m

H

2, m = 1, 2, . (26)

Note that we had to introduce another index, m, so that the eigenvalue is now doubly indexed.

We now resubstitute for nm in terms of :

n =m

H

2, (27)

so that the eigenvalues of the 2D Sturm-Liouville equation (11) become

nm =n

L

2+m

H

2. (28)

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The associated eigenfunctions will be given by

nm(x, y) = fn(x)gnm(y)

= sinnx

L

sin

myH

. (29)

We now go back to the time-dependent equation for h(t):

h + c2h = 0. (30)

The functions hnm(t) associated with the spatial functions nm(x, y) will satisfy the equation

hnm + nmc2hnm(t) = 0. (31)

The general solution is given by

hnm(t) = anm cos(nmct) + bnm sin(nmct). (32)The product solutions, the normal modes of vibration of the membrane, are given by

unm(x,y,t) = nm(x, y)hnm(t)

= sinnx

L

sin

myH

anm cos(

nmct) + bnm sin(

nmct)

, n, m = 1, 2, .(33)

Each mode is composed of a spatial profile nm(x) that is modulated in time with frequency

nm = cmn = c n

L2

+ m

H2

1/2

. (34)

We now examine the first few modes nice pictures of these modes are to be found in the textbook.

1. n = m = 1. This is the mode with the lowest frequency,

11 = c

L

2+

H

21/2. (35)

The spatial profile,

11(x, y) = sinxL siny

H , (36)

has no zeros in the interior of the rectangle. As a result, this mode has no nodes the entire

membrane executes a uniform up-and-down motion.

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2. n = 2, m = 1. The spatial profile

21(x, y) = sin

2x

L

sin

yH

, (37)

has a zero at x = L/2, which implies that the vibrational mode has a node along the line

x = L/2, 0 y H. As a result, the membrane is divided into two regions, the displacements

of which will lie on opposite sides of the xy-plane. The frequency of this mode is

21 = c

2

L

2+

H

21/2. (38)

3. n = 1, m = 2. The spatial profile

12(x, y) = sinx

L

sin

2y

H

, (39)

has a zero at y = H/2, which implies that the vibrational mode has a node along the line

y = H/2, 0 x L. As a result, the membrane is divided into two regions, the displacementsof which will lie on opposite sides of the xy-plane. The frequency of this mode is

12 = c

L

2+

2

H

21/2. (40)

Note that if L < H, then 21 > 12 and vice versa. If L = H, then 21 = 12 and the two

modes have the same frequency of vibration. In this case, the membrane is a square, and the

two modes are identical, up to a rotation of /2 around the origin.

4. n = 2, m = 2. As in the previous case, the spatial profile

22(x, y) = sin

2x

L

sin

2y

H

, (41)

has a zero at y = H/2, which implies that the vibrational mode has a node along the line

y = H/2, 0 x L. But it also has a zero at x = L/2, which implies that the mode has anode along this line. As a result, the membrane is divided into four regions, the displacements

of which will lie on opposite sides of the xy-plane. The frequency of this mode is

12 = c

2

L

2+

2

H

21/2. (42)

Note that the frequency of this mode, 22, is greater than the frequencies 12 and 21.

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Lecture 23

Vibrating rectangular membrane (conclusion)

In the previous lecture, using separation of variables, we obtained the normal modes of vibration for

this problem,

unm(x,y,t) = nm(x, y)hnm(t)

= sinnx

L

sin

myH

anm cos(

nmct) + bnm sin(

nmct)

, n, m = 1, 2, .(43)

Each mode is composed of a spatial profile nm(x) that is modulated in time with frequency

nm = c

mn = c

nL

2+m

H

21/2. (44)

Because of the homogeneous boundary conditions, the general solution to the vibrating membrane

problem may be written as a superposition of these normal mode solutions,

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

m=1

nm(x, y)hnm(t)

=

n=1

m=1

sinnx

L

sin

myH

anm cos(

nmct) + bnm sin(

nmct)

. (45)

Imposition of the initial condition u(x,y, 0) = (x, y) implies that

(x, y) =

n=1

m=1anm sin

nx

L sin

my

H =

n=1

m=1

amnnm(x, y). (46)

We now make use of the fact that the product basis nm(x, y) forms an orthogonal set over the

region D = [0, L] [0, H], i.e.,

Dn1,m1(x, y)n2,m2(x, y) dxdy =

H0

L0

sinn1x

L

sin

m1yH

sin

n2xL

sin

m2yH

dxdy

= L0

sinn1xL

sinn2xL

dx H0

sinm1yH

sinm2yH

dy=

L2

H2

, (n1, m1) = (n2, m2)

0, (n1, m1) = (n2, m2).(47)

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We also state, without proof, that this product basis sometimes called a tensor product basis is

complete in the function space L2[D], the set of functions f : D R (or C) that are square-integrableon D, i.e.,

D

|f(x, y)|2 dA < . (48)

As such, we may multiply both sides of Eq. (46) by kl(x, y) for a k

1 and l

1, and integrate over

D to give

akl =4

LH

D

(x, y)nm(x, y) dxdy. (49)

Likewise, the initial velocity condition,

u

t(x,y, 0) = (x, y), (50)

yields

bkl =1

c

nm

4

LH D (x, y)nm(x, y) dxdy. (51)This concludes our discussion of the (clamped) rectangular vibrating membrane problem.

The relevance of the vibrating membrane problem to quantum mechanics

The 1D vibrating string and 2D rectangular vibrating membrane problems are quite relevant to quan-

tum mechanics, as we now discuss briefly. In one-dimension, the time-dependent Schrodinger equation

for a particle in a box is given by the eigenvalue/BVP,

2

2md2

dx2 = E, (0) = (L) = 0. (52)

The wavefunction (x) provides a description of a particle that is confined inside a potential We

rewrite Eq. (52) asd2

dx2 +

2mE

2 = 0, (0) = (L) = 0. (53)

The solutions of this BVP are well known to us:

n =2mE

2=

n22

L2, n(x) = sin

nx. (54)

This implies the existence of a discrete set of energy eigenvalues,

En =2

2m

nL

2, n = 1, 2, . (55)

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In two dimensions, we consider a particle confined in the two-dimensional region [0, L] [0, H]which is surrounded by potential walls of infinite height. The time-dependent Schrodinger equation

for the particle is2

2m2 = E, (56)

with the same boundary conditions as the clamped rectangular membrane. It may be rewritten as

2 + 2mE2

= 0. (57)

We have derived the solutions earlier:

nm =2mEnm

2=n

L

2+m

H

2, nm(x, y) = sin

nxL

sin

mH

. (58)

The energy eigenvalues of these wavefunctions are

Enm =2

2mn

L2 + m

H2 . (59)

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Vibrating circular membrane

Relevant section of text: 7.7

We now consider the problem of a clamped, vibrating circular membrane, i.e., a drum. The

vertical displacement satisfies the 2D wave equation,

2ut2

= c22u. (60)

Since the drum is circular, it is convenient to use polar coordinates, i.e., u = u(r, ). We assume that

the drum has radius a > 0. The clamping of the drum along the outer boundary implies the following

boundary condition

u(a,,t) = 0, /2 /2, t 0. (61)

As was the case for Laplaces equation on a circular region, we can impose only one boundary condition

at this time.

In order to determine a unique solution, well need initial conditions, i.e., initial displacement and

velocity:

u(r,, 0) = (r, ),u

t(r,, 0) = (r, ). (62)

Since the boundary condition is homogeneous, well try separation of variables, i.e.,

u(r,,t) = (r, )h(t). (63)

Substitution into (60) yields

h = c2(2)h. (64)

Separating variables:h

c2h=2

= . (65)

This yields the following ODEs:

h + c2h = 0, (66)

and

2 + = 0. (67)

Note that these ODEs have the same form as those for the rectangular vibrating membrane. If > 0,

the solutions for h(t) are oscillatory, i.e.,

h(t) = c1 cos(c

t) + c2 sin(c

t). (68)

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We now examine the eigenvalue equation (67) for (r, ). In polar coordinates, the Laplacian operator

has the form

2 = 1r

r

r

r

+

1

r22

2, (69)

so that the eigenvalue equation assumes the form

1

r

r

r

r

+

1

r22

2 + = 0. (70)

We now assume a separation-of-variables solution for , i.e.,

(r, ) = f(r)g(), (a, ) = 0. (71)

Substitution into (70) yieldsg

r

d

dr

r

df

dr

+

f

r2d2g

d2+ f g = 0. (72)

Now divide by f g and multiply by r2:

r

f

d

dr

r

df

dr

+

1

g

d2g

d2+ r2 = 0. (73)

We havent quite separated variables, so we do so now:

1g

d2g

d2=

r

f

d

dr

r

df

dr

+ r2 = , (74)

where we have introduced another separation constant, . This yields the two equations,

d2g

d2

+ g = 0, (75)

and

rd

dr

r

df

dr

+ (r2 )f = 0, f(a) = 0. (76)

The equation for g() will have the following periodicity conditions,

g() = g() dgd

() = dgd

(). (77)

Periodic solutions to (75) exist for 0:

gm() = cos(m), m 0 and sin(m), m 1. (78)

The eigenvalues are given by m = m2.

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Lecture 24

Vibrating circular membrane (contd)

Relevant section of text: 7.7

We now turn to the radial equation (76) for f(r). If we divide by r, the equation is cast into

Sturm-Liouville eigenvalue form:

d

dr

r

df

dr

m

2

rf + rf = 0, 0 < r a. (79)

Here, p(r) = r, q(r) = m2/r and (r) = r. Note that r = 0 is a singular point because q(r) isundefined there. As such this Sturm-Liouville eigenvalue problem is not regular. Nevertheless, we

claim that solutions exist for this problem as is the case for regular S-L problems, that is,

1. A set of eigenvalues nm, n = 1, 2, 3, , m = 0, 1, 2, ,

2. A set of corresponding eigenfunctions fnm: For each fixed m, the functions fnm(r) are orthogonal

with respect to the weight function (r) = r, i.e.,a0

fnm(r)fnm(r) rdr = 0, for n = n. (80)

To produce these solutions, we make the change of variable z =

r in Eq. (79). The result is

(Exercise):

z2d2F

dz2+ z

dF

dz+ (z2 m2)F = 0, m = 0, 1, 2, . (81)

where F(z) = f(z/). (In the lecture, I wrote, erroneously, F(z) = f(r). Sorry about that!) Notethat this removes the parameter from the ODE of course, it is now hidden in the independent

variable z.

You may recognize Eq. (81) it is Bessels equation, which is studied in AMATH 351. Very briefly,

z = 0 is a regular singular point. The standard practice is to assume a Frobenius series solution of

the form

zpn=0

anzn =

n=0

an+pzn+p, a0 = 0, (82)

where p is to b e determined. One substitutes the Frobenius series into Eq. (81), collects like termsin zk, and demands that the series begins with a term that starts with a nonzero term involving a0.

The result is a quadratic equation in p. For Eq. (81), the indicial equation for p is

p2 m2 = 0 p = m. (83)

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For m = 0, the two distinct values of p, i.e., p1 = m and p2 = m, yield two linearly independentseries solutions,

Jm(z) = zm[am,0 + am,1z + ],

Ym(z) = Jm(z) = zm[bm,0 + bm,1z + ]. (84)

The actual values of the coefficients am,i and bm,i are not important for our discussion.

For m = 0, we have p1 = p2 = 0, so only one solution is produced by the series solution method:

J0(z) = 1 + a0,1z + . (85)

A second, linearly independent solution may be produced by the variation of parameters applied to

J0(z). The result is the following function

Y0(z) =2

ln z +

. (86)

It is also important to mention that for all m 0, the functions Jm(z) are oscillatory for z > 0.(Another important result that is shown in AMATH 351 is that as z , the spacing of consecutivezeros of Jm(z) approaches .)

For m = 0, 1, 2, , the general solution to the Bessel equation (81) will have the form,

Fm(z) = c1Jm(z) + c2Ym(z). (87)

We then transform back to the r variable to produce the general solution f(r) to Eq. (76),

fm(r) = c1Jm(

r) + c2Ym(

r). (88)

We now apply these solutions to the vibrating circular membrane problem. Recall that the

membrane being clamped at r = a, i.e., f(a) = 0, gave us one boundary condition. Before considering

that boundary condition, we consider another condition which deals with the singular nature of the

point r = 0: It is the condition

f must be finite at r = 0, i.e., |f(0)| < . (89)

Recall that we imposed such a condition on solutions to Laplaces equation on the circular disk.

The point r = 0 corresponds to the point z = 0. From a look at the behaviour solutions Jm(z)

above, we note that, in all cases,

|Ym(z)| as z 0. (90)

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This implies that these functions must be excluded from the general solution fm(r) in Eq. (88), i.e.,

c2 = 0 so that

f(r) = c1Jm(

r). (91)

We now return to the boundary condition f(a) = 0. It becomes the condition,

Jm(a) = 0. (92)

Given that the functions Jm(z) are oscillatory for z > 0, it follows that the argument

a is a zero of

the function Jm(z). In other words, a = zmn > 0, (93)

where zmn denotes the nth zero of the Bessel function Jm(z). The zeros of the Bessel functions are

presented in standard mathematical tables. (Indeed, the results of this lecture give an idea of why

these zeros are so important.) For example, approximate values of the first three zeros of J0(z) are

z0,1 = 2.40482, z0,2 = 5.52007, z0,3 = 8.65372. (94)

The graph of J0(x) for 0 x 20 is presented below. (Just for interests sake, the numerical valuesused to plot the graph were computed using the series expansion for J0(z).)

-1

-0.75

-0.5

-0.25

0

0.25

0.5

0.75

1

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x

Bessel function J_0(x)

For each m 0, the above conditions produce an infinite sequence of eigenvalues mn given by

mn =zmn

a

2, n = 1, 2, . (95)

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The resulting radial functions,

fmn(r) = Jm(

mnr), (96)

have n positive zeros, with the nth zero located at r = a.

Note: Before going on to construct the full solutions to the clamped vibrating mem-

brane problem, let us stop for a moment and look again at the clamping condition (93)

that produces the eigenvalues mn in (95). This condition is analogous to the boundary

condition

sin(

L) = 0, (97)

encountered in the 1D wave equation (clamped string, zero-endpoint conditions), which

implied the result,

L = n, n = 1, 2, , (98)

yielding the eigenvalue condition,

n =n

L

2, n = 1, 2, . (99)

This was a relatively straightforward result, since the zeros of the sin function are so well

known. Comparing Eqs. (99) with (95),

1. the zeros zmn of Jm(z) correspond to the zeros zn = n of sin(z) and

2. a corresponds to L.

The solutions to the clamped vibrating membrane will then have the form

umn(r,,t) = fmn(r)gm()hmn(t), m = 0, 1, 2, , (100)

where

fmn(r) = Jm(

mnr),

gm() = am cos m + bm sin m,hmn(t) = cmn cos(

mnct) + dmn sin(

mnct). (101)

These solutions correpond to the normal modes of vibration of the circular membrane. We shall return

to them in the next lecture.

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The general solution to the vibration problem will be given by

u(r,,t) =

m=0

n=1

umn(r,,t). (102)

If we assume that the membrane is initially at rest, i.e., u/t = 0, then the sin(

mnct) terms

vanish, so that the general solution can be expressed in the following form,

u(r,,t) =

m=0

n=1

AmnJm(

mnr) cos(m)cos(

mnct)

+

m=0

n=1

BmnJm(

mnr)sin(m)cos(

mnct). (103)

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