INTRODUCTION

In the preceding sections we solved one-dimensional forms of the heat and wave equations. In this section we extend the method of separation of variables to certain problems involving the two-dimensional heat and wave equations.

Heat and Wave Equations in Two Dimensions

Suppose the rectangular region in FIGURE 13.8.1(a) is a thin plate in which the temperature u is a function of time t and position (x, y). Then, under suitable conditions, u(x, y, t) can be shown to satisfy the two-dimensional heat equation

.(1)

On the other hand, suppose Figure 13.8.1(b) represents a rectangular frame over which a thin flexible membrane has been stretched (a rectangular drum). If the membrane is set in motion, then its displacement u, measured from the xy-plane (transverse vibrations), is also a function of time t and position (x, y). When the displacements are small, free, and undamped, u(x, y, t) satisfies the two-dimensional wave equation

.(2)

As the next example will show, solutions of boundary-value problems involving (1) and (2) lead to the concept of a Fourier series in two variables. Because the analysis of problems involving (1) and (2) are quite similar, we illustrate a solution only of the heat equation (1).

EXAMPLE 1 Temperatures in a Plate

Find the temperature u(x, y, t) in the plate shown in Figure 13.8.1(a) if the initial temperature is f(x, y) throughout and if the boundaries are held at temperature zero for time t > 0.

SOLUTION

We must solve

subject to

To separate variables for the PDE in three independent variables x, y, and t we try to find a product solution u(x, y, t) = X(x)Y(y)T(t). Substituting, we get

(3)

Since the left side of the last equation in (3) depends only on x and the right side depends only on y and t, we must have both sides equal to a constant –λ:

and so X″ + λX = 0(4)

(5)

By the same reasoning, if we introduce another separation constant –µ in (5), then

Y″ + µY = 0 and T′ + k(λ + µ)T = 0.(6)

Now the homogeneous boundary conditions

Thus we have two Sturm–Liouville problems, one in the variable x,

,(7)

and the other in the variable y,

(8)

The usual consideration of cases (λ = 0, λ = –α² < 0, λ = α² > 0, µ = 0, λ = –β² < 0, and so on) leads to two independent sets of eigenvalues defined by sin λb = 0 and sin µc = 0. These equations in turn imply

(9)

The corresponding eigenfunctions are

(10)

After substituting the values in (9) into the first-order DE in (6), its general solution is T(t) = . A product solution of the two-dimensional heat equation that satisfies the four homogeneous boundary conditions is then

where A_mn is an arbitrary constant. Because we have two sets of eigenvalues, we are prompted to try the superposition principle in the form of a double sum

.(11)

At t = 0 we want the temperature f(x, y) to be represented by

(12)

Finding the coefficients A_mn in (12) really does not pose a problem; we simply multiply the double sum (12) by the product sin (mπx/b) sin (nπy/c) and integrate over the rectangle defined by 0 ≤ x ≤ b, 0 ≤ y ≤ c. It follows that

.(13)

Thus, the solution of the boundary-value problem consists of (11) with the A_mn defined by (13).≡

The series (11) with coefficients (13) is called a sine series in two variables, or a double sine series. The cosine series in two variables of a function f(x, y) is a little more complicated. If the function f is defined over a rectangular region defined by 0 ≤ x ≤ b, 0 ≤ y ≤ c, then the double cosine series is given by

where

See Problem 2 in Exercises 13.8 for a boundary-value problem leading to a double cosine series.

13.8 Exercises Answers to selected odd-numbered problems begin on page ANS-36.

In Problems 1 and 2, solve the heat equation (1) subject to the given conditions.

1. u(0, y, t) = 0, u(π, y, t) = 0
   u(x, 0, t) = 0, u(x, π, t) = 0
   u(x, y, 0) = u₀
2. 

   u(x, y, 0) = xy

In Problems 3 and 4, solve the wave equation (2) subject to the given conditions.

3. u(0, y, t) = 0, u(π, y, t) = 0
   u(x, 0, t) = 0, u(x, π, t) = 0
   u(x, y, 0) = xy(x – π)(y – π)
   
4. u(0, y, t) = 0, u(b, y, t) = 0
   u(x, 0, t) = 0, u(x, c, t) = 0
   u(x, y, 0) = f(x, y)
   

In Problems 5–7, solve Laplace’s equation

(14)

for the steady-state temperature u(x, y, z) in the rectangular parallelepiped shown in FIGURE 13.8.2.

5. The top (z = c) of the parallelepiped is kept at temperature f(x, y) and the remaining sides are kept at temperature zero.
6. The bottom (z = 0) of the parallelepiped is kept at temperature f(x, y) and the remaining sides are kept at temperature zero.
7. The parallelepiped is a unit cube (a = b = c = 1) with the top (z = 1) and bottom (z = 0) kept at constant temperatures u₀ and –u₀, respectively, and the remaining sides kept at temperature zero.