13.5 Laplace’s Equation

INTRODUCTION

Suppose we wish to find the steady-state temperature u(x, y) in a rectangular plate whose vertical edges x = 0 and x = a are insulated, and whose upper and lower edges y = b and y = 0 are maintained at temperatures f(x) and 0, respectively. See FIGURE 13.5.1. When no heat escapes from the lateral faces of the plate, we solve the following boundary-value problem:

A rectangular plate placed in an x y coordinate plane. The bottom edge of the plate coincides with the x axis and the left edge of the plate coincides with y axis. The top right vertex of the rectangular plate is labeled (a, b). A thin strip to the left and right of the vertical edges of the plate is marked out in barred lines and labeled insulated. The top edge of the plate is indicated by an arrow and labeled u = f(x). The bottom edge is labeled u = 0.

FIGURE 13.5.1 Find the temperature u in a rectangular plate

(1)

(2)

u(x, 0) = 0, u(x, b) = f(x), 0 < x < a.(3)

Solution of the BVP

With u(x, y) = X(x)Y(y), separation of variables in (1) leads to

(4)

(5)

The three homogeneous boundary conditions in (2) and (3) translate into X′(0) = 0, X′(a) = 0, and Y(0) = 0. The Sturm–Liouville problem associated with the equation in (4) is then

(6)

Examination of the cases corresponding to λ = 0, λ = –α2 < 0, and λ = α2 > 0, where α > 0, has already been carried out in Example 1 in Section 12.5. For convenience a shortened version of that analysis follows.

For λ = 0, (6) becomes

The solution of the ODE is X = c1 + c2x. The boundary condition X′(0) = 0 then implies c2 = 0, and so X = c1. Note that for any c1, this constant solution satisfies the second boundary condition X′(a) = 0. By imposing c1 ≠ 0, X = c1 is a nontrivial solution of the BVP (6). For λ = –α2 < 0, (6) possesses no nontrivial solution. For λ = α2 > 0, (6) becomes

Applying the boundary condition X′(0) = 0 the solution X = c1 cos αx + c2 sin αx implies c2 = 0 and so X = c1 cos αx. The second boundary condition X′(a) = 0 applied to this last expression then gives –c1α sin αa = 0. Because α > 0, the last equation is satisfied when αa = nπ or α = nπ/a, n = 1, 2, … . The eigenvalues of (6) are then λ0 and λn = = n2π2/a2, n = 1, 2, …. By corresponding λ0 = 0 with n = 0, the eigenfunctions of (6) are

We must now solve equation (5) subject to the single homogeneous boundary condition Y(0) = 0. First, for λ0 = 0 the DE in (5) is simply Y″ = 0, and thus its solution is Y = c3 + c4y. But Y(0) = 0 implies c3 = 0 so Y = c4y. Second, for λn = n2π2/a2, the DE in (5) is Y″ – Y = 0. Because 0 < y < b is a finite interval, we write the general solution in terms of hyperbolic functions:

Why hyperbolic functions? See pages 127 and 701.

Y(y) = c3 cosh(nπy/a) + c4 sinh(nπy/a).

From this solution we see Y(0) = 0 again implies c3 = 0 so Y = c4 sinh(nπy/a).

Thus product solutions un = X(x)Y(y) that satisfy Laplace’s equation (1) and the three homogeneous boundary conditions in (2) and (3) are

where we have rewritten c1c4 as A0 for n = 0 and as An for n = 1, 2, ….

The superposition principle yields another solution

(7)

Finally, by substituting y = b in (7) we see

is a half-range expansion of f in a Fourier cosine series. If we make the identifications A0b = a0/2 and An sinh(nπb/a) = an, n = 1, 2, … , it follows from (2) and (3) of Section 12.3 that

(8)

and

(9)

The solution of the boundary-value problem (1) – (3) consists of the series in (7), with coefficients A0 and An defined in (8) and (9), respectively.

Dirichlet Problem

A boundary-value problem in which we seek a solution to an elliptic partial differential equation such as Laplace’s equation ∇2u = 0 within a region R (in the plane or 3-space) such that u takes on prescribed values on the entire boundary of the region is called a Dirichlet problem. This problem was named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805–1859). In addition to being remembered for his contributions to the theory of Fourier series, Dirichlet is credited with being one of the first mathematicians to give the formal definition of a function as we know it today. In Problem 1 in Exercises 13.5 you are asked to show that the solution of the Dirichlet problem for a rectangular region

is

(10)

In the special case when f(x) = 100, a = 1, b = 1, the coefficients An are given by An = 200 . With the help of a CAS the plot of the surface defined by u(x, y) over the region R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 is given in FIGURE 13.5.2(a). You can see in the figure that boundary conditions are satisfied; especially note that along y = 1, u = 100 for 0 ≤ x ≤ 1. The isotherms, or curves, in the rectangular region along which the temperature u(x, y) is constant can be obtained using the contour plotting capabilities of a CAS and are illustrated in Figure 13.5.2(b). The isotherms can also be visualized as the curves of intersection (projected into the xy-plane) of horizontal planes u = 80, u = 60, and so on, with the surface in Figure 13.5.2(a). Notice that throughout the region the maximum temperature is u = 100 and occurs on the portion of the boundary corresponding to y = 1. This is no coincidence. There is a maximum principle that states a solution u of Laplace’s equation within a bounded region R with boundary B (such as a rectangle, circle, sphere, and so on) takes on its maximum and minimum values on B. In addition, it can be proved that u can have no relative extrema (maxima or minima) in the interior of R. This last statement is clearly borne out by the surface shown in Figure 13.5.2(a).

Two graphs. The first graph is a surface graph plotted on an x y u(x y) coordinate system. The x and y axes are marked from 0 to 1 with increments of 0.2. The u(x y) axis is marked from 0 to 100 with increments of 20. The surface is created with the mesh of the curves that go up from points on y axis, go up steeply to the right, touch the line of u(x y) = 100 and comes down to the right gradually and end on the x axis. The second graph has isotherms. The x and y axes are marked similarly as the first graph. The curve of 10 goes down to the right from x = 1, up to y = 0.2 and goes up to the right to x and y = 1. The rest of the curves for 20, 40, 60 and 80 also have similar trends with higher values of y.

FIGURE 13.5.2 Surface is graph of partial sums when f(x) = 100 and a = b = 1 in (10)

Superposition Principle

A Dirichlet problem for a rectangle can be readily solved by separation of variables when homogeneous boundary conditions are specified on two parallel boundaries. However, the method of separation of variables is not applicable to a Dirichlet problem when the boundary conditions on all four sides of the rectangle are nonhomogeneous. To get around this difficulty we break the boundary-value problem

(11)

into two problems, each of which has homogeneous boundary conditions on parallel boundaries, as shown.

Suppose u1 and u2 are the solutions of Problems 1 and 2, respectively. If we define u(x, y) = u1(x, y) + u2(x, y), it is seen that u satisfies all boundary conditions in the original problem (11). For example,

u(0, y) = u1(0, y) + u2(0, y) = 0 + F(y) = F(y)

u(x, b) = u1(x, b) + u2(x, b) = g(x) + 0 = g(x)

and so on. Furthermore, u is a solution of Laplace’s equation by Theorem 13.1.1. In other words, by solving Problems 1 and 2 and adding their solutions we have solved the original problem. This additive property of solutions is known as the superposition principle. See FIGURE 13.5.3.

Three shaded rectangles are graphed on x y coordinate planes. The top right of the rectangle is (a, b). The bottom of the rectangle is f(x), the top is marked g(x). The first rectangle is labeled delta^2 u = 0. The second rectangle is labeled delta^2 u subscript 1 = 0. The third rectangle is labeled delta^2 u subscript 2 = 0.

FIGURE 13.5.3 Solution u = Solution u1 of Problem 1 + Solution u2 of Problem 2

We leave as exercises (see Problems 13 and 14 in Exercises 13.5) to show that a solution of Problem 1 is

,

where

and that a solution of Problem 2 is

,

where

13.5 Exercises Answers to selected odd-numbered problems begin on page ANS-34.

In Problems 1–10, solve Laplace’s equation (1) for a rectangular plate subject to the given boundary conditions.

  1. u(0, y) = 0, u(a, y) = 0
    u(x, 0) = 0, u(x, b) = f(x)
  2. u(0, y) = 0, u(a, y) = 0
    u(x, b) = f(x)
  3. u(0, y) = 0, u(a, y) = 0
    u(x, 0) = f(x), u(x, b) = 0

  4. u(x, 0) = x, u(x, b) = 0
  5. u(0, y) = 0, u(1, y) = 1 – y
  6. u(0, y) = g(y),
  7. u(π, y) = 1
    u(x, 0) = 0, u(x, π) = 0
  8. u(0, y) = 0, u(1, y) = 0
    u(x, 1) = f(x)
  9. u(0, y) = 0, u(1, y) = 0
    u(x, 0) = 100, u(x, 1) = 200
  10. u(0, y) = 10y,
    u(x, 0) = 0, u(x, 1) = 0

In Problems 11 and 12, solve Laplace’s equation (1) for the semi-infinite plate extending in the positive y-direction. In each case assume that u(x, y) is bounded at y → ∞.

  1. A semi-infinite vertical plate of width pi is plotted on an x y coordinate plane such that the bottom edge coincides with the x axis and the left edge coincides with the y axis. The bottom edge is indicated by an arrow and labeled u = f(x). The left edge of the plate is labeled u = 0. The right edge of the plate is labeled u = 0.

    FIGURE 13.5.4 Semi-infinite plate in Problem 11

  2. A semi-infinite vertical plate of width pi is plotted on an x y coordinate plane such that the bottom edge coincides with the x axis and the left edge coincides with the y axis. The bottom edge is indicated by an arrow and labeled u = f(x). A thin strip to the left and the right of the vertical edge of the plate is marked out in barred lines and labeled insulated.

    FIGURE 13.5.5 Semi-infinite plate in Problem 12

In Problems 13 and 14, solve Laplace’s equation (1) for a rectangular plate subject to the given boundary conditions.

  1. u(0, y) = 0, u(a, y) = 0
    u(x, 0) = f(x), u(x, b) = g(x)
  2. u(0, y) = F(y), u(a, y) = G(y)
    u(x, 0) = 0, u(x, b) = 0

In Problems 15 and 16, use the superposition principle to solve Laplace’s equation (1) for a square plate subject to the given boundary conditions.

  1. u(0, y) = 1, u(π, y) = 1
    u(x, 0) = 0, u(x, π) = 1
  2. u(0, y) = 0, u(2, y) = y(2 – y)
    u(x, 0) = 0, u(x, 2) =
  3. In Problem 16, what is the maximum value of the temperature u for 0 ≤ x ≤ 2, 0 ≤ y ≤ 2?

Computer Lab Assignments

    1. In Problem 1 suppose a = b = π and f(x) = 100x(π – x). Without using the solution u(x, y) sketch, by hand, what the surface would look like over the rectangular region defined by 0 ≤ x ≤ π, 0 ≤ y ≤ π.
    2. What is the maximum value of the temperature u for 0 ≤ x ≤ π, 0 ≤ y ≤ π?
    3. Use the information in part (a) to compute the coefficients for your answer in Problem 1. Then use the 3D-plot application of your CAS to graph the partial sum S5(x, y) consisting of the first five nonzero terms of the solution in part (a) for 0 ≤ x ≤ π, 0 ≤ y ≤ π. Use different perspectives and then compare with part (a).
    1. Use the contour-plot application of your CAS to graph the isotherms u = 170, 140, 110, 80, 60, 30 for the solution of Problem 9. Use the partial sum S5(x, y) consisting of the first five nonzero terms of the solution.
    2. Use the 3D-plot application of your CAS to graph the partial sum S5(x, y).
  3. Use the contour-plot application of your CAS to graph the isotherms u = 2, 1, 0.5, 0.2, 0.1, 0.05, 0, –0.05 for the solution of Problem 10. Use the partial sum S5(x, y) consisting of the first five nonzero terms of the solution.

Discussion Problems

  1. Solve the Neumann problem for a rectangle:
    1. Explain why a necessary condition for a solution u to exist is that g satisfy

      This is sometimes called a compatibility condition. Do some extra reading and explain the compatibility condition on physical grounds.

    2. If u is a solution of the BVP, explain why u + c, where c is an arbitrary constant, is also a solution.
  2. Consider the boundary-value problem

    .

    Discuss how the following answer was obtained:

    Carry out your ideas.