INTEGRAL TRANSFORM SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS

9. INTEGRAL TRANSFORM SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS

Laplace Transform Solutions We have seen (Chapter 8, Section 9) that tak- ing the Laplace transform of an ordinary differential equation converts it into an algebraic equation. Taking the Laplace transform of a partial differential equation reduces the number of independent variables by one, and so converts a two-variable partial differential equation into an ordinary differential equation. To illustrate this, we solve the following problem.

Example 1.

A semi-infinite bar (extending from x = 0 to x = ∞), with insulated sides, is ◦

initially at the uniform temperature u = 0 . At t = 0, the end at x = 0 is brought to u = 100 ◦ and held there. Find the temperature distribution in the bar as a

function of x and t. The differential equation satisfied by u is

We are going to take the t Laplace transform of (9.1); the variable x will just be a parameter in this process. Let U be the Laplace transform of u, that is,

(9.2) − U (x, p) = u(x, t)e pt dt.

By Chapter 8, equation (9.1) we have

= pU − u t=0 = pU

∂t

since u = 0 when t = 0. Also

2 ∂x = ∂x 2 L(u) = ∂x 2

660 Partial Differential Equations Chapter 13

(remember that x is just a parameter here; we are taking a t Laplace transform). The transform of (9.1) is then

Now if we think of p as a constant and x as the variable, this is an ordinary differ- ential equation for U as a function of x. Its solutions are

e (√p/α)x ,

U=

e −( √p/α)x .

To find the correct combination of these solutions to fit our problem, we need the Laplace transforms of the boundary conditions on u since these give the conditions on U . Using L1 (see Laplace Transform Table, page 469) to find the transforms, we have

u = 100 at x = 0, U = L(100) = 100 at x = 0; (9.5)

as x → ∞. Since U → 0 as x → ∞, we see that we must use the solution e −( √p/α)x from (9.4)

u→0

as x → ∞, U → L(0) = 0

and discard the positive exponential solution. We determine the constant multiple of this solution which fits our problem from the condition that U = 100/p at x = 0. Thus we find that the U solution satisfying the given boundary conditions is

e −( √p/α)x .

We find u by looking up the inverse transform of (9.6); it is, by L22

and this is the solution of the problem. Fourier Transform Solutions In the examples in Sections 2, 3 and 4, we ex-

panded a given function in a Fourier series. This was possible because the function was to be represented by a series over a finite interval. We could then take that interval as the period for the Fourier series. If we are dealing with a function which is given over an infinite interval (and not periodic), then instead of representing it by a Fourier series we represent it by a Fourier integral (Chapter 7, Section 12). Let us do this for a specific problem.

Example 2. An infinite metal plate (Figure 9.1) covering the first quadrant has the edge along the y axis held at 0 ◦ , and the edge along the x axis held at

100 ◦ , 0 < x < 1,

u(x, 0) =

x > 1.

Find the steady-state temperature distribution as a function of x and y.

Section 9 Integral Transform Solutions of Partial Differential Equations 661

The differential equation and its solutions are the same as in the semi-infinite plate problem dis- cussed in Section 2, equations (2.1), (2.6), and (2.7). As in that problem, we assume u → 0 as y → ∞, and use only the e −ky terms. Since u = 0 when x = 0, we use only the sine solutions. The basis functions we want are then u = e −ky sin kx. We do not have any requirement here which de- termines k as we did in Section 2. We must then

Figure 9.1 allow all k’s and try to find a solution in the form

of an integral over k. Instead of coefficients b n in a series, we have a coefficient function B(k) to determine. Remember that k > 0 since e −ky must tend to zero as y → ∞. Thus we try to find a solution of the form

u(x, y) =

B(k)e −ky sin kx dk.

When y = 0, we have

u(x, 0) =

B(k) sin kx dk.

This is the first of equations (12.14) in Chapter 7, if we identify k with α, u(x, 0) with f s (x), and B(k) with

s (α). Thus the given temperature on the x axis is a Fourier sine transform of the desired coefficient function, so B(k) can be found as the inverse transform. Using the second of equations (12.14) in Chapter 7, we get

B(k) =

u(x, 0) sin kx dx. π

g s (k) =

f s (x) sin kx dx =

For the given u(x, 0) in (9.8), we find

2 1 1 200 cos kx

(1 − cos k). Finding B(k) corresponds to evaluating the coefficients in a Fourier series. Substi-

B(k) =

100 sin kx dx = −

0 πk

tuting (9.12) into (9.9), we get the solution to our problem in the form of an integral instead of a series:

200 ∞ 1 − cos k

(9.13) u(x, y) =

e −ky sin kx dk.

An integral can, of course, be evaluated numerically just as a convergent series can be approximated by calculating a few terms. However, (9.13) can be integrated;

a convenient way to do it is to recognize that it is a Laplace transform of f (k) = [(1 − cos k) sin kx]/k, where x is just a parameter and y corresponds to p and k to t. From L19 and L20

(9.14) u(x, y) =

arc tan

2 y This can also be written in polar coordinates as (Problem 1)

100 r 2 − cos 2θ

u=

− arc tan

2 sin 2θ

662 Partial Differential Equations Chapter 13

PROBLEMS, SECTION 9

1. Verify that (9.15) follows from (9.14). Hint: Use the formulas for tan (α ± β), tan 2α, etc., to condense (9.14) and then change to polar coordinates. You may find

Show that if you use principal values of the arc tangent, this formula does not give the correct boundary conditions on the x axis, whereas (9.15) does.

2. A metal plate covering the first quadrant has the edge which is along the y axis insulated and the edge which is along the x axis held at

( 100(2 − x), for 0 < x < 2,

u(x, 0) =

for x > 2.

Find the steady-state temperature distribution as a function of x and y. Hint: Follow the procedure of Example 2, but use a cosine transform (because ∂u/∂x = 0 for x = 0). Leave your answer as an integral like (9.13).

3. Consider the heat flow problem of Section 3. Solve this by Laplace transforms (with respect to t) by starting as in Example 1. You should get

Solve this differential equation to get

100 sinh (p 1/2 /α)x

p sinh (p 1/2

/α)l

pl

Assume the following expansion, and find u by looking up the inverse Laplace trans- forms of the individual terms of U:

sinh (p – /α)x x 2 sin (πx/l) sin (2πx/l) sin (3πx/l) p sinh (p 1/2

3[p + (9π 2 α 2 /l 2 )] ··· Your answer should be (3.15).

4. ◦ A semi-infinite bar is initially at temperature 100 for 0 < x < 1, and 0 for x > 1. Starting at t = 0, the end x = 0 is maintained at 0 ◦ and the sides are insulated.

Find the temperature in the bar at time t, as follows. Separate variables in the

sin kx and e − α 2 k 2 t cos kx. Discard the cosines since u = 0 at x = 0. Look for a solution

heat flow equation and get elementary solutions e − α 2 k 2 t

u(x, t) = − B(k)e α 2 k 2 t sin kx dk.

and proceed as in Example 2. Leave your answer as an integral. 5. A long wire occupying the x axis is initially at rest. The end x = 0 is oscillated up

and down so that

y(0, t) = 2 sin 3t,

t > 0.

Find the displacement y(x, t). The initial and boundary conditions are y(0, t) = 2 sin 3t, y(x, 0) = 0, ∂y/∂t| t=0 = 0. Take Laplace transforms of these conditions and of the wave equation with respect to t as in Example 1. Solve the resulting differential equation to get

6e − (p/v)x

Y (x, p) =

p 2 +9

Section 10 Miscellaneous Problems 663

Use L3 and L28 to find

( 2 sin 3

`t − x v ´, x < vt, 0,

y(x, t) =

x > vt. 6. Continue the problem of Example 2 in the following way: Instead of using the

explicit form of B(k) from (9.12), leave it as an integral and write (9.13) in the form

u(x, y) = 200

e −ky sin kx dk

sin kt dt.

Change the order of integration and evaluate the integral with respect to k first. (Hint: Write the product of sines as a difference of cosines.) Now do the t integration and get (9.14).

7. Continue with Problem 4 as in Problem 6.