Powers D.L. Boundary Value Problems and Partial Differential Equations

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318 Chapter 5 Higher Dimensions and Other Coordinates

The function φ(r) = J

(λr) is a solution of Eq. (6), and we wish to choose λ

so that Eq. (7) is satisﬁed. Then we must have

(λa) = 0

, n =1, 2,...,

where α

are the zeros of the function J

. Thus the eigenfunctions and eigen-

values of Eqs. (6), (7), and (8) are

(r) = J

(λ

r), λ





. (9)

TheseareshownontheCD.

Returning to Eq. (5), we determine that the time factors T

are

(t) = exp



−λ



We may now assemble the general solution of the partial differential equa-

tion (1), under the boundary condition (2) and boundedness condition (8),

as a general linear combination of our product solutions:

v(r, t) =

∞



n=1

(λ

r) exp



−λ



. (10)

It remains to determine the coefﬁcients a

so as to satisfy the initial condi-

tion (3), which now takes the form

v(r, 0) =

∞



n=1

(λ

r) = f (r), 0 < r < a. (11)

While this problem is not a routine exercise in Fourier series or even a reg-

ular Sturm–Liouville problem (see Section 2.7, especially Exercise 6 there), it

is nevertheless true that the eigenfunctions of Eqs. (6) and (7) are orthogonal,

as expressed by the relation



(r)φ

(r)rdr=0 (n = m)



(λ

r)J

(λ

r)rdr=0 (n = m).

More importantly, the following theorem gives us justiﬁcation for Eq. (11).

5.6 Temperature in a Cylinder 319

Theorem.

If f (r) is sectionally smooth on the interval 0 < r < a, then at every

point r on that interval,

∞



n=1

(λ

r) =

f (r+) +f (r−)

, 0 < r < a,

where the λ

are solutions of J

(λa) =0 and



f (r)J

(λ

r)rdr



(λ

r)rdr

. (12)



The CD shows an animation of a Bessel series converging.

Now we may proceed with the problem at hand. If the function f (r) in the

initial condition (3) is sectionally smooth, the use of Eq. (12) to chose the

coefﬁcients a

guarantees that Eq. (11) is satisﬁed (as nearly as possible), and

hence the function

v(r, t) =

∞



n=1

(λ

r) exp



−λ



(13)

satisﬁes the problem expressed by Eqs. (1), (2), (3), and (8).

By way of example, let us suppose that the function f (r) = T

,0< r < a.It

is necessary to determine the coefﬁcients a

by formula (12). The numerator

is the integral



(λ

r)rdr.

This integral is evaluated by means of the relation (see Exercise 6 of Sec-

tion 5.5)



(x)



=xJ

(x). (14)

Hence, we ﬁnd



(λ

r)rdr=

(λ



(λ

a) =

(α

). (15)

The denominator of Eq. (12) is known to have the value (Exercise 5)



(λ

r)rdr=

(λ

(α

). (16)

320 Chapter 5 Higher Dimensions and Other Coordinates

n α

(α

)

(α

)

1 2.405 +0.5191 +1.6020

2 5.520 −0.3403 −1.0647

3 8.654 +0.2715 +0.8512

4 11.792 −0.2325 −0.7295

Tab le 2 Values for Eq. (18)

Figure 8 Graphs of the solution of the example problem. The function v(r, t) as

giveninEq.(18)isshownvsr for times chosen so that kt/a

takes the values 0,

0.01, 0.1, and 0.5. T

=100 and a = 1.

Putting together the numerator and denominator from Eqs. (15) and (16),

we ﬁnd that the coefﬁcients we need are

(α

)

. (17)

Thus, the solution to the heat conduction problem is

v(r, t) = T

∞



n=1

(α

)

(λ

r) exp



−λ



. (18)

InTable2arelistedtheﬁrstfewvaluesoftheratio2/[α

(α

)]. Figure 8

shows graphs of v(r, t) as a function of r for several times. Also, see Exercise 1.

An animation is shown on the CD.

5.7 Vibrations of a Circular Membrane 321

EXERCISES

1. Use Eq. (18) to ﬁnd an expression for the function v(0, t)/T

.Evaluatethe

function for

=0.1, 0.2, 0.3.

(The ﬁrst two terms of the series are sufﬁcient.)

2. Write out the ﬁrst three terms of the series in Eq. (18).

3. Solve the heat problem consisting of Eqs. (1)–(3) if f (r) is

f (r) =







, 0 < r <

< r < a.

4. Let φ(r) = J

(λr) so that φ(r) satisﬁes Bessel’s equation of order 0. Multiply

through the differential equation by rφ



and conclude that



(rφ



)



+λ





=0.

5. Assuming that λ is chosen so that φ(a) = 0, integrate the equation in Exer-

cise 4 over the interval 0 < r < a to ﬁnd



(r)rdr=

2λ



aφ



(a)



6. Use Exercise 5 to validate Eq. (16).

5.7 Vibrations of a Circular Membrane

We shall now attempt to solve the problem of describing the displacement of a

circular membrane that is ﬁxed at its edge.

Symmetric Vibrations

To begin with, we treat the simple case in which the initial conditions are in-

dependent of θ . Thus the displacement v(r, t) satisﬁes the problem

∂

∂r



∂v

∂r



∂

∂t

, 0 < r < a, 0 < t, (1)

v(a, t) = 0, 0 < t, (2)

322 Chapter 5 Higher Dimensions and Other Coordinates

v(r, 0) = f (r), 0 < r < a, (3)

∂v

∂t

(r, 0) = g(r), 0 < r < a. (4)

We start immediately with separation of variables, assuming v(r, t) =

φ(r)T(t). The differential equation (1) becomes

(rφ



)



T =

φT



and the variables may be separated by dividing by φT.Thenweﬁnd

(rφ



(r))



rφ(r)



(t)

T(t)

The two sides must both be equal to a constant (say, −λ

), yielding two linked,

ordinary differential equations



+λ

T = 0, 0 < t, (5)

(rφ



)



+λ

rφ = 0, 0 < r < a. (6)

The boundary condition Eq. (2) is satisﬁed if

φ(a) = 0. (7)

Of course, because r = 0 is a singular point of the differential equation (6), we

add the requirement



φ(r)



bounded at r =0, (8)

which is equivalent to requiring that |v(r, t)| be bounded at r = 0. We

recognize that Eq. (6) is Bessel’s equation, of which the function φ(r) =

(λr) is the solution bounded at r = 0. In order to satisfy the boundary con-

ditionEq.(7),wemusthave

(λa) =0,

which implies that

, n =1, 2,..., (9)

where α

are the zeros of the function J

. Thus the eigenfunctions and eigen-

values of Eqs. (6)–(8) are

(r) = J

(λ

r), λ





5.7 Vibrations of a Circular Membrane 323

The rest of our problem can now be dispatched easily. Returning to Eq. (5),

we see that

(t) = a

cos(λ

ct) + b

sin(λ

ct),

and then for each n = 1, 2,...wehaveasolutionofEqs.(1),(2),and(8):

(r, t) = φ

(r)T

(t).

The most general linear combination of the v

would be

v(r, t) =

∞



n=1

(λ



cos(λ

ct) + b

sin(λ

ct)



. (10)

The initial conditions Eqs. (3) and (4) are satisﬁed if

v(r, 0) =

∞



n=1

(λ

r) = f (r), 0 < r < a,

∂v

∂t

(r, 0) =

∞



n=1

(λ

r) = g(r), 0 < r < a.

As in the preceding section, the coefﬁcients of these series are to be found

through the integral formulas



f (r)J

(λ

r)rdr, b



g(r)J

(λ

r)rdr,





(λ



rdr.

With the coefﬁcients determined by these formulas, the function given in

Eq. (10) is the solution to the vibrating membrane problem that we started

with.

General Vibrations

Having seen the simplest case of the vibrations of a circular membrane, we

return to the more general case. The full problem was

∂

∂r



∂u

∂r



∂

∂θ

∂

∂t

, 0 < r < a, 0 < t. (11)

u(a,θ,t) = 0, 0 < t, (12)



u(0,θ,t)



bounded, 0 < t, (13)

u(r,θ +2π, t) = u(r,θ,t), 0 < r < a, 0 < t, (14)

324 Chapter 5 Higher Dimensions and Other Coordinates

u(r,θ,0) = f (r,θ), 0 < r < a, (15)

∂u

∂t

(r,θ,0) = g(r,θ), 0 < r < a. (16)

Following the procedure suggested in Section 5.4, we assume that u has the

product form

u =φ(r,θ)T(t)

and we ﬁnd that Eq. (11) separates into two linked equations:



+λ

T =0, 0 < t,(17)

∂

∂r



∂φ

∂r



∂

∂θ

=−λ

φ, 0 < r < a.(18)

If we separate variables of the function φ by assuming φ(r,θ)= R(r)Q(θ ),

Eq. (18) takes the form







Q +



=−λ

RQ.

The variables will separate if we multiply by r

and divide by RQ. Then the

preceding equation may be put in the form

r(rR



)



+λ

=−



=µ

Finally we obtain two problems for R and Q:



+µ

Q =0, −π<θ≤ π, (19a)

Q(θ +2π)=Q(θ ), (19b)







−

R +λ

rR =0, 0 < r < a, (20)



R(0)



bounded,

R(a) =0.

As we observed before, the problem (19) has the solutions

=0, Q

=1,

, Q

=cos(mθ) and sin(mθ), m = 1, 2, 3,....

Also, the differential equation (20) will be recognized as Bessel’s equation, the

general solution of which is (using µ = m)

R(r) = CJ

(λr) +DY

(λr).

5.7 Vibrations of a Circular Membrane 325

In order for the boundedness condition in Eq. (20) to be fulﬁlled, D must be

zero. Then we are left with

R(r) = J

(λr).

(Because any multiple of a solution is another solution, we can drop the con-

stant C.) The boundary condition of Eq. (20) becomes

R(a) =J

(λa) =0,

implying that λa must be a root of the equation

(α) =0.

(See Table 1.) For each ﬁxed integer m, α

,α

,...are the ﬁrst, second,

third, . . . solutions of the preceding equation. The values of λ for which J

(λr)

solves the differential equation and satisﬁes the boundary condition are

, m =0, 1, 2,..., n = 1, 2, 3,....

Now that the functions R and Q are determined, we can construct φ.For

m =1, 2, 3,...and n =1, 2, 3,..., both of the functions

(λ

r) cos(mθ), J

(λ

r) sin(mθ) (21)

are solutions of the problem Eq. (18), both corresponding to the same eigen-

value λ

.Form = 0andn = 1, 2, 3,...,wehavethefunctions

(λ

r), (22)

which correspond to the eigenvalues λ

. (Compare with the simple case.) The

function T(t) that is a solution of Eq. (17) is any combination of cos(λ

ct)

and sin(λ

ct).

Now the solutions of Eqs. (11)–(14) have any of the forms

(λ

r) cos(mθ)cos(λ

ct), J

(λ

r) sin(mθ)cos(λ

ct),

(23)

(λ

r) cos(mθ)sin(λ

ct), J

(λ

r) sin(mθ)sin(λ

ct)

for m = 1, 2, 3,... and n = 1, 2, 3,.... In addition, there is the special case

m =0, for which solutions have the form

(λ

r) cos(λ

ct), J

(λ

r) sin(λ

ct). (24)

The CD shows a few of these “standing waves” animated.

The general solution of the problem Eqs. (11)–(14) will thus have the form

of a linear combination of the solutions in Eqs. (23) and (24). We shall use

326 Chapter 5 Higher Dimensions and Other Coordinates

several series to form the combination:

u(r,θ,t) =



(λ

r) cos(λ

ct)



m,n

(λ

r) cos(mθ)cos(λ

ct)



m,n

(λ

r) sin(mθ)cos(λ

ct)



(λ

r) sin(λ

ct)



m,n

(λ

r) cos(mθ)sin(λ

ct)



m,n

(λ

r) sin(mθ)sin(λ

ct). (25)

When t =0, the last three sums disappear, and the cosines of t in the ﬁrst three

sumsareallequalto1.Thus

u(r,θ,0) =



(λ

r) +



m,n

(λ

r) cos(mθ)



m,n

(λ

r) sin(mθ)

= f (r,θ), 0 < r < a, −π<θ≤ π. (26)

We expect to fulﬁll this equality by choosing the a’s and b’s according to

some orthogonality principle. Since each function present in the series is an

eigenfunction of the problem

∇

φ =−λ

φ, 0 < r < a,

φ(a,θ)= 0,

φ(r,θ +2π) = φ(r,θ), 0 < r < a,

we expect it to be orthogonal to each of the others (see Section 5.4, Exercise 7).

This is indeed true: Any function from one series is orthogonal to all of the

functions in the other series and also to the rest of the functions in its own

series. To illustrate this orthogonality, we have



(λ

r)J

(λ

r) cos(mθ)dA



(λ

r)J

(λ



−π

cos(mθ)dθrdr= 0, m = 0. (27)

5.7 Vibrations of a Circular Membrane 327

There are two other relations like this one involving functions from two dif-

ferent series.

We already know that the functions within the ﬁrst series are orthogonal to

each other:



−π



(λ

r)J

(λ

r)rdrdθ =0, n = q.

Within the second series we must show that, if m = p or n = q,then

0 =



−π



(λ

r) cos(mθ)J

(λ

r) cos(pθ)rdrdθ. (28)

(Recall that rdrdθ = dA in polar coordinates.) Integrating with respect to θ

ﬁrst,weseethattheintegralmustbezeroifm = p, by the orthogonality of

cos(mθ) and cos(pθ).Ifm = p, the preceding integral becomes



(λ

r)J

(λ

r)rdr

after the integration with respect to θ .Finally,ifn = q,thisintegraliszero;the

demonstration follows the same lines as the usual Sturm–Liouville proof. (See

Section 2.7.) Thus the functions within the second series are shown orthogonal

to each other. For the functions of the last series, the proof of orthogonality is

similar.

Equipped now with an orthogonality relation, we can determine formulas

for the a’s and b’s. For instance,



−π



f (r,θ)J

(λ

r)rdrdθ

2π



(λ

r)rdr

. (29)

The A’s and B’s are calculated from the second initial condition.

It should now be clear that, while the computation of the solution to the

original problem is possible in theory, it will be very painful in practice. Worse

yet, the ﬁnal form of the solution Eq. (25) does not give a clear idea of what u

looks like. All is not wasted, however. We can say, from an examination of the

λ’s, that the tone produced is not musical — that is, u is not periodic in t.Also

we can sketch some of the fundamental modes of vibration of the membrane

corresponding to some low eigenvalues (Fig. 9). The curves represent points

for which displacement is zero in that mode (nodal curves).

EXERCISES

1. Verify that each of the functions in the series in Eq. (10) satisﬁes Eqs. (1),

(2), and (8).

2. Derive the formulas for the a’s and b’s of Eq. (10).