INTRODUCTION

In this section we are going to consider boundary-value problems involving forms of the heat and wave equation in polar coordinates and a form of Laplace’s equation in cylindrical coordinates. There is a commonality throughout the examples and most of the exercises—the boundary-value problem possesses radial symmetry.

Radial Symmetry

The two-dimensional heat and wave equations

expressed in polar coordinates are, in turn,

(1)

where u = u(r θ, t). To solve a boundary-value problem involving either of these equations by separation of variables we must define u = R(r)Θ(θ)T(t). As in Section 13.8, this assumption leads to multiple infinite series. See Problem 18 in Exercises 14.2. In the discussion that follows we shall consider the simpler, but still important, problems that possess radial symmetry—that is, problems in which the unknown function u is independent of the angular coordinate θ. In this case the heat and wave equations in (1) take, in turn, the forms

(2)

where u = u(r, t). Vibrations described by the second equation in (2) are said to be radial vibrations.

The first example deals with the free undamped radial vibrations of a thin circular membrane. We assume that the displacements are small and that the motion is such that each point on the membrane moves in a direction perpendicular to the xy-plane (transverse vibrations)—that is, the u-axis is perpendicular to the xy-plane. A physical model to keep in mind while studying this example is a vibrating drumhead.

EXAMPLE 1 Radial Vibrations of a Circular Membrane

Find the displacement u(r, t) of a circular membrane of radius c clamped along its circumference if its initial displacement is f(r) and its initial velocity is g(r). See FIGURE 14.2.1.

SOLUTION

The boundary-value problem to be solved is

Substituting u = R(r)T(t) into the partial differential equation and separating variables gives

(3)

Note in (3) we have returned to our usual separation constant −λ. The two equations obtained from (3) are

(4)

and (5)

Because of the vibrational nature of the problem, equation (5) suggests that we use only λ = α² > 0, α > 0. Now (4) is not a Cauchy–Euler equation but is the parametric Bessel differential equation of order ν = 0; that is, rR″ + R′ + α²rR = 0. From (13) of Section 5.3 the general solution of the last equation is

R(r) = c₁J₀(αr) + c₂Y₀(αr). (6)

The general solution of the familiar equation (5) is

T(t) = c₃ cos aαt + c₄ sin aαt.

Now recall, the Bessel function of the second kind of order zero has the property that Y₀(αr) → −∞ as r → 0⁺, and so the implicit assumption that the displacement u(r, t) should be bounded at r = 0 forces us to define c₂ = 0 in (6). Thus R(r) = c₁J₀(αr).

Since the boundary condition u(c, t) = 0 is equivalent to R(c) = 0, we must have c₁J₀(αc) = 0. We rule out c₁ = 0 (this would lead to a trivial solution of the PDE), so consequently

J₀(αc) = 0. (7)

If x_n = α_nc are the positive roots of (7), then α_n = x_n/c and so the eigenvalues of the problem are λ_n = = /c² and the eigenfunctions are c₁J₀(α_nr). Product solutions that satisfy the partial differential equation and the boundary condition are

u_n = R(r)T(t) = (A_n cos aα_nt + B_n sin aα_nt) J₀(α_nr), (8)

where we have done the usual relabeling of the constants. The superposition principle then gives

(9)

The given initial conditions determine the coefficients A_n and B_n.

Setting t = 0 in (9) and using u(r, 0) = f(r) gives

(10)

This last result is recognized as the Fourier–Bessel expansion of the function f on the interval (0, c). Hence by a direct comparison of (7) and (10) with (8) and (15) of Section 12.6, we can identify the coefficients A_n with those given in (16) of Section 12.6:

(11)

Next, we differentiate (9) with respect to t, set t = 0, and use u_t(r, 0) = g(r):

This is now a Fourier–Bessel expansion of the function g. By identifying the total coefficient aα_nB_n with (16) of Section 12.6 we can write

(12)

Finally, the solution of the given boundary-value problem is the series (9) with coefficients A_n and B_n defined in (11) and (12), respectively. ≡

Standing Waves

Analogous to (11) of Section 13.4, the product solutions (8) are called standing waves. For n = 1, 2, 3, …, the standing waves are basically the graph of J₀(α_nr) with the time-varying amplitude

A_n cos aα_nt + B_n sin aα_nt.

The standing waves at different values of time are represented by the dashed graphs in FIGURE 14.2.2. The zeros of each standing wave in the interval (0, c) are the roots of J₀(α_nr) = 0 and correspond to the set of points on a standing wave where there is no motion. This set of points is called a nodal line. If (as in Example 1) the positive roots of J₀(α_nc) = 0 are denoted by x_n, then x_n = α_nc implies α_n = x_n/c and consequently the zeros of the standing waves are determined from

Now from Table 5.3.1, the first three positive zeros of J₀ are (approximately) x₁ = 2.4, x₂ = 5.5, and x₃ = 8.7. Thus for n = 1, the first positive root of

Since we are seeking zeros of the standing waves in the open interval (0, c), the last result means that the first standing wave has no nodal line. For n = 2, the first two positive roots of

Thus the second standing wave has one nodal line defined by r = x₁c/x₂ = 2.4c/5.5. Note that r ≈ 0.44c < c. For n = 3, a similar analysis shows that there are two nodal lines defined by r = x₁c/x₃ = 2.4c/8.7 and r = x₂c/x₃ = 5.5c/8.7. In general, the nth standing wave has n − 1 nodal lines r = x₁c/x_n, r = x₂c/x_n, ..., r = x_n−1c/x_n. Since r = constant is an equation of a circle in polar coordinates, we see in Figure 14.2.2 that the nodal lines of a standing wave are concentric circles.

Use of Computers

It is possible to see the effect of a single drumbeat for the model solved in Example 1 by means of the animation capabilities of a computer algebra system. In Problem 21 in Exercises 14.2 you are asked to find the solution given in (9) when

Some frames of a “movie” of the vibrating drumhead are given in FIGURE 14.2.3.

Laplacian in Cylindrical Coordinates

From FIGURE 14.2.4 we can see that the relationship between the cylindrical coordinates of a point in space and its rectangular coordinates is given by

We saw in Section 14.1 that the Laplacian of a function u in polar coordinates is

It follows with no extra work that the Laplacian of a function u in cylindrical coordinates is then

Laplace’s equation is then

(13)

If u does not depend on the polar angle , then the radially symmetric form of Laplace’s equation is

(14)

EXAMPLE 2 Steady Temperatures in a Circular Cylinder

Find the steady-state temperature in the circular cylinder shown in FIGURE 14.2.5.

SOLUTION

The boundary conditions suggest that the temperature u has radial symmetry. Accordingly, u(r, z) is determined from

Using u = R(r)Z(z) and separating variables gives

(15)

and rR″ + R′ + λrR = 0 (16)

Z″ − λZ = 0.(17)

For the choice λ = α² > 0, α > 0, the general solution of (16) is

R(r) = c₁J₀(αr) + c₂Y₀(αr),

and since a solution of (17) is defined on the finite interval [0, 2], we write its general solution as

Z(z) = c₃ cosh αz + c₄ sinh αz.

As in Example 1, the assumption that the temperature u is bounded at r = 0 demands that c₂ = 0. The condition u(2, z) = 0 implies R(2) = 0. This equation,

J₀(2α) = 0, (18)

defines the positive eigenvalues λ_n = of the problem. Last, Z(0) = 0 implies c₃ = 0. Hence we have R = c₁J₀(α_nr), Z = c₄ sinh α_nz,

u_n = R(r) Z(z) = A_n sinh α_nz J₀(α_nr)

and u(r, z) = sinh α_nz J₀(α_nr).

The remaining boundary condition at z = 4 then yields the Fourier–Bessel series

u₀ = sinh 4α_nJ₀(α_nr),

so that in view of (18) the coefficients are defined by (16) of Section 12.6,

To evaluate the last integral we first use the substitution t = α_nr, followed by = tJ₀(t).

From

we obtain

Finally, the temperature in the cylinder is

≡

Do not conclude from two examples that every boundary-value problem in cylindrical coordinates gives rise to a Fourier–Bessel series.

EXAMPLE 3 Steady Temperatures in a Circular Cylinder

Find the steady-state temperatures u(r, z) in the circular cylinder defined by if the boundary conditions are

SOLUTION

Because of the nonhomogeneous condition specified at r = 1 we do not expect the eigenvalues of the problem to be defined in terms of zeros of a Bessel function of the first kind. As we did in Section 14.1 it is convenient in this problem to use λ as the separation constant. Thus from (15) of Example 2 we see that separation of variables now gives the two ordinary differential equations

You should verify that the two cases and lead only to the trivial solution u = 0. In the case the DEs are

The first equation is the parametric form of Bessel’s modified DE of order ν = 0. The solution of this equation is We immediately define because the modified Bessel function of the second kind is unbounded at r = 0. Therefore,

Review pages 292–293 of Section 5.3. See also Figures 5.3.3 and 5.3.4.

Now the eigenvalues and eigenfunctions of the Sturm–Liouville problem

are and Z(z) = c₃ sin nπz. Thus product solutions that satisfy the PDE and the homogeneous boundary conditions are

Next we form

The remaining condition at r = 1 yields the Fourier sine series

From (5) of Section 12.3 we can write

and so

The steady-state temperature is then

. ≡

REMARKS

Because Bessel functions appear so frequently in the solutions of boundary-value problems expressed in cylindrical coordinates, they are also referred to as cylinder functions.

14.2 Exercises Answers to selected odd-numbered problems begin on page ANS-37.

1. Find the displacement u(r, t) in Example 1 if f(r) = 0 and the circular membrane is given an initial unit velocity in the upward direction.
2. A circular membrane of radius 1 is clamped along its circumference. Find the displacement u(r, t) if the membrane starts from rest from the initial displacement f(r) = 1 − r², 0 < r < 1. [Hint: See Problem 10 in Exercises 12.6.]
3. Find the steady-state temperature u(r, z) in the cylinder in Example 2 if the boundary conditions are u(2, z) = 0, 0 < z < 4, u(r, 0) = u₀, u(r, 4) = 0, 0 < r < 2.
4. If the lateral side of the cylinder in Example 2 is insulated, then

   

   1. Find the steady-state temperature u(r, z) when u(r, 4) = f(r), 0 < r < 2.
   2. Show that the steady-state temperature in part (a) reduces to u(r, z) = u₀z/4 when f(r) = u₀. [Hint: Use (12) of Section 12.6.]

In Problems 5–8, find the steady-state temperature u(r, z) in a finite cylinder defined by if the boundary conditions are as given and is a constant.

5. 
6. 
7. 
8. 
9. The temperature in a circular plate of radius c is determined from the boundary-value problem

   

   Solve for u(r, t).

10. Solve Problem 9 if the edge r = c of the plate is insulated.
11. When there is heat transfer from the lateral side of an infinite circular cylinder of radius 1 (see FIGURE 14.2.6) into a surrounding medium at temperature zero, the temperature inside the cylinder is determined from

    

    Solve for u(r, t).

    

12. When there is heat transfer from the lateral side of a semi-infinite circular cylinder of unit radius into a surrounding medium at temperature zero, the steady-state temperature inside the cylinder is determined from the boundary-value problem

    

where h and u₀ are constants. Solve for u(r, z).

In Problems 13 and 14, use the substitution u(r, t) = v(r, t) + ψ(r) to solve the given boundary-value problem. [Hint: Review Section 13.6.]

13. A circular plate is a composite of two different materials in the form of concentric circles. See FIGURE 14.2.7. The temperature u(r, t) in the plate is determined from the boundary-value problem

    

    

14. a constant
    
15. The horizontal displacement u(x, t) of a heavy chain of length L oscillating in a vertical plane satisfies the partial differential equation

    

    See FIGURE 14.2.8.

    1. Using −λ as a separation constant, show that the ordinary differential equation in the spatial variable x is xX″ + X′ + λX = 0. Solve this equation by means of the substitution x = τ²/4.
    2. Use the result of part (a) to solve the given partial differential equation subject to

       

       [Hint: Assume the oscillations at the free end x = 0 are finite.]

       

16. Consider the boundary-value problem

    

    1. Use the substitution u(r, t) = v(r, t) + Bt in the preceding problem to show that v(r, t) satisfies

       

       Here B is a constant to be determined.

    2. Now use the substitution v(r, t) = w(r, t) + ψ(r) to solve the boundary-value problem in part (a). [Hint: You may need to review Section 3.5.]
    3. What is the solution u(r, t) of the first problem?
17. Suppose heat is lost from the flat surfaces of a very thin circular plate into a surrounding medium at temperature zero. If the linear law of heat transfer applies, the heat equation assumes the form

    

    where h is a positive constant. See FIGURE 14.2.9. Find the temperature if the edge is kept at temperature zero and if initially the temperature of the plate is unity throughout.

    

18. In this problem we consider the general case—that is, with θ dependence—of the vibrating circular membrane of radius c:

    

    1. Assume that u = R(r)Θ(θ)T(t) and the separation constants are −λ and −ν. Show that the separated differential equations are

       

    2. Let λ = α² and ν = β² and solve the separated equations in part (a).
    3. Determine the eigenvalues and eigenfunctions of the problem.
    4. Use the superposition principle to determine a multiple series solution. Do not attempt to evaluate the coefficients.

Discussion Problem

19. Discuss how to solve the partial differential equation

    

    subject to the boundary conditions indicated in FIGURE 14.2.10.

    

Computer Lab Assignments

20. 1. Consider Example 1 with a = 1, c = 10, g(r) = 0, and f(r) = 1 − r/10, 0 < r < 10. Use a CAS as an aid in finding the numerical values of the first three eigenvalues λ₁, λ₂, λ₃ of the boundary-value problem and the first three coefficients A₁, A₂, A₃ of the solution u(r, t) given in (9). Write the third partial sum S₃(r, t) of the series solution.
    2. Use a CAS to plot the graph of S₃(r, t) for t = 0, 4, 10, 12, 20.
21. Solve Problem 9 with boundary conditions u(c, t) = 200, u(r, 0) = 0. With these imposed conditions, one would expect intuitively that at any interior point of the plate, u(r, t) → 200 as t → ∞. Assume that c = 10 and that the plate is cast iron so that k = 0.1 (approximately). Use a CAS as an aid in finding the numerical values of the first five eigenvalues λ₁, λ₂, λ₃, λ₄, λ₅ of the boundary-value problem and the five coefficients A₁, A₂, A₃, A₄, A₅ in the solution u(r, t). Let the corresponding approximate solution be denoted by S₅(r, t). Plot S₅(5, t) and S₅(0, t) on a sufficiently large time interval [0, T]. Use the plots of S₅(5, t) and S₅(0, t) to estimate the times (in seconds) for which u(5, t) ≈ 100 and u(0, t) ≈ 100. Repeat for u(5, t) ≈ 200 and u(0, t) ≈ 200.
22. Consider an idealized drum consisting of a thin membrane stretched over a circular frame of radius 1. When such a drum is struck at its center, one hears a sound that is frequently described as a dull thud rather than a melodic tone. We can model a single drumbeat using the boundary-value problem solved in Example 1.
    1. Find the solution u(r, t) given in (9) when c = 1, f(r) = 0, and

       

    2. Show that the frequency of the standing wave u_n(r, t) is f_n = aλ_n/2π, where λ_n is the nth positive zero of J₀(x). Unlike the solution of the one-dimensional wave equation in Section 13.4, the frequencies are not integer multiples of the fundamental frequency f₁. Show that f₂ ≈ 2.295f₁ and f₃ ≈ 3.598f₁. We say that the drumbeat produces anharmonic overtones. As a result the displacement function u(r, t) is not periodic, and so our ideal drum cannot produce a sustained tone.
    3. Let a = 1, b = , and ν₀ = 1 in your solution in part (a). Use a CAS to graph the fifth partial sum S₅(r, t) at the times t = 0, 0.1, 0.2, 0.3, …, 5.9, 6.0 on the interval [−1, 1]. Use the animation capabilities of your CAS to produce a movie of these vibrations.
    4. For a greater challenge, use the 3D plotting capabilities of your CAS to make a movie of the motion of the circular drumhead that is shown in cross section in part (c). [Hint: There are several ways of proceeding. For a fixed time, either graph u as a function of x and y using r = or use the equivalent of Mathematica’s RevolutionPlot3D.]