20.5 Poisson Integral Formulas

INTRODUCTION

The success of the conformal mapping method depends on the recognition of the solution to the new Dirichlet problem in the image region R′. It would therefore be helpful if a general solution could be found for Dirichlet problems in either the upper half-plane y ≥ 0 or the unit disk z ≤ 1. The Poisson integral formula for the upper half-plane provides such a solution by expressing the value of a harmonic function u(x, y) at a point in the interior of the upper half-plane in terms of its values on the boundary y = 0.

Formulas for the Upper Half-Plane

To develop the formula, we first assume that the boundary function is given by u(x, 0) = f(x), where f(x) is the step function indicated in FIGURE 20.5.1. The solution of the corresponding Dirichlet problem in the upper half-plane is

(1)

An image of a horizontal line labeled x. The graph also shows a triangle labeled a b z with three acute angles. Angle a is labeled Arg (z minus a), angle z is labeled theta (z), and the exterior angle b is labeled Arg (z minus b). The side a b is labeled u = u subscript i. The side a b extended beyond a and b is labeled u = 0.

FIGURE 20.5.1 Boundary conditions on y = 0

Since Arg(z − b) is an exterior angle in the triangle formed by z, a, and b, Arg(z − b) = θ(z) + Arg(z − a), where 0 < θ(z) < π, and we can write

(2)

The superposition principle can be used to solve the more general Dirichlet problem in FIGURE 20.5.2. If u(x, 0) = ui for xi−1 ≤ x ≤ xi and u(x, 0) = 0 outside the interval [a, b], then from (1),

(3)

An image of a horizontal line labeled x. The graph also shows a triangle with three vertices labeled a = x subscript 0, z, and x subscript n = b with three acute angles. The triangle is divided by three line segments intersecting from the vertex z to the opposite side at x subscript 1, x subscript 2, and x subscript n minus 1, respectively. Angle a = x subscript 0 z x subscript 1 is labeled theta subscript 1, angle x subscript 1 z x subscript 2 is labeled theta subscript 2, and angle x subscript n minus 1 z x subscript n = b is labeled theta subscript n. a = x subscript 0 x subscript 1 is labeled u = u subscript 1, x subscript 1 x subscript 2 is labeled u = u subscript 2, and x subscript n minus 1 x subscript n = b is labeled u = u subscript n. The side a = x subscript 0 x subscript n = b extended beyond a = x subscript 0 and x subscript n = b is labeled u = 0.

FIGURE 20.5.2 General boundary conditions on y = 0

Note that Arg(z − t) = tan−1(y/(x − t)), where tan−1 is selected between 0 and π, and therefore d/dt Arg(z − t) = y/((x − t)2 + y2). From (3),

Since u(x, 0) = 0 outside of the interval [a, b], we have

(4)

A bounded piecewise-continuous function can be approximated by step functions, and therefore our discussion suggests that (4) is the solution to the Dirichlet problem in the upper half-plane. This is the content of Theorem 20.5.1.

THEOREM 20.5.1 Poisson Integral Formula for the Upper Half-Plane

Let u(x, 0) be a piecewise-continuous function on every finite interval and bounded on −∞ < x < ∞. Then the function defined by

is the solution of the corresponding Dirichlet problem on the upper half-plane y ≥ 0.

There are a few functions for which it is possible to evaluate the integral in (4), but in general, numerical methods are required to evaluate the integral.

EXAMPLE 1 Solving a Dirichlet Problem

Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary condition u(x, 0) = x when x < 1, and u(x, 0) = 0 otherwise.

SOLUTION

By the Poisson integral formula,

Using the substitution s = x − t, we can show that

which can be simplified to

≡

In most of the examples and exercises u(x, 0) is a step function, and we will use the integrated solution (3) rather than (4). If the first interval is (−∞, x1), then the term Arg(z − x1) − Arg(z − a) in the sum should be replaced by Arg(z − x1). Likewise, if the last interval is (xn−1, ∞), then Arg(z − b) − Arg(z − xn−1) should be replaced by π − Arg(z − xn−1).

EXAMPLE 2 Solving a Dirichlet Problem

The conformal mapping f(z) = z + 1/z maps the region in the upper half-plane and outside the circle z = 1 onto the upper half-plane v ≥ 0. Use this mapping and the Poisson integral formula to solve the Dirichlet problem shown in FIGURE 20.5.3(a).

In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 2 on the positive horizontal axis and a point labeled minus 2 on the negative horizontal axis. The graph also shows a semicircle centered at the origin. The top of the semicircle is labeled u = 0 and the left and right of the semicircle are labeled u = 1. The exterior of the semicircle is shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled 2 on the positive horizontal axis and a point labeled minus 2 on the negative horizontal axis. The center is labeled U = 0 and the first and the second quadrant are labeled U = 1. The first and second quadrants are shaded.

FIGURE 20.5.3 Image of Dirichlet problem in Example 2

SOLUTION

Using the results of Example 3 in Section 20.2, we can transfer the boundary conditions to the w-plane. See Figure 20.5.3(b). Since U(u, 0) is a step function, we will use the integrated solution (3) rather than the Poisson integral. The solution to the new Dirichlet problem is

and therefore

which can be simplified to . ≡

Formula for the Unit Disk

A Poisson integral formula can also be developed to solve the general Dirichlet problem for the unit disk.

THEOREM 20.5.2 Poisson Integral Formula for the Unit Disk

Let u(eiθ) be bounded and piecewise continuous for −π ≤ θ ≤ π. Then the solution to the corresponding Dirichlet problem on the open unit disk z < 1 is given by

(5)

Geometric Interpretation

FIGURE 20.5.4 shows a thin membrane (such as a soap film) that has been stretched across a frame defined by u = u(eiθ). The displacement u in the direction perpendicular to the z-plane satisfies the two-dimensional wave equation

and so at equilibrium, the displacement function u = u(x, y) is harmonic. Formula (5) provides an explicit solution for the displacement u and has the advantage that the integral is over the finite interval [−π, π]. When the integral cannot be evaluated, standard numerical integration procedures can be used to estimate u(x, y) at a fixed point z = x + iy with z < 1.

An image of two ellipses. The boundary of the upper ellipse is labeled u = u (x y) above the ellipse and frame below the ellipse. The lower ellipse shows a horizontal line intersecting at the center labeled mod(z) is less than 1 and a slanting line intersecting both the ellipse and the horizontal line at the center from top right to bottom left. The two ellipses are joined with two vertical dashed lines at the two ends.

FIGURE 20.5.4 Thin membrane on a frame

EXAMPLE 3 Displacement of a Membrane

A frame for a membrane is defined by u(eiθ) = θ for −π ≤ θ ≤ π. Estimate the equilibrium displacement of the membrane at (−0.5, 0), (0, 0), and (0.5, 0).

SOLUTION

From (5), we get u(x, y) = dt. When (x, y) = (0, 0), we get

For the other two values of (x, y), the integral is not elementary and must be estimated using a numerical integration procedure. Using Simpson’s rule, we obtain (to four decimal places) u(−0.5, 0) = 2.2269 and u(0.5, 0) = 0.9147. ≡

Fourier Series Form

The Poisson integral formula for the unit disk is actually a compact way of writing the Fourier series solution to Laplace’s equation that we developed in Chapter 14. To see this, first note that un(r, θ) = rn cos nθ and vn(r, θ) = rn sin nθ are each harmonic, since these functions are the real and imaginary parts of zn. If a0, an, and bn are chosen to be the Fourier coefficients of u(eiθ) for −π < θ < π, then, by the superposition principle,

(6)

is harmonic and u(1, θ) = (a0/2) + (an cos nθ + bn sin nθ) = u(eiθ). Since the solution of the Dirichlet problem is also given by (5), we have

EXAMPLE 4 Solving a Dirichlet Problem

Find the solution of the Dirichlet problem in the unit disk satisfying the boundary condition u(eiθ) = sin 4θ. Sketch the level curve u = 0.

SOLUTION

Rather than working with the Poisson integral (5), we will use the Fourier series solution (6), which reduces to u(r, θ) = r4 sin 4θ. Therefore, u = 0 if and only if sin 4θ = 0. This implies u = 0 on the lines x = 0, y = 0, and y = ±x.

If we switch to rectangular coordinates, u(x, y) = 4xy(x2 − y2). The surface u(x, y) = 4xy(x2 − y2), the frame u(eiθ) = sin 4θ, and the system of level curves were sketched using graphics software and are shown in FIGURE 20.5.5. ≡

An image of a three-dimensional plot that looks like a Viking helmet with a square piece of paper pasted around it. The thickest horn, which is at the back, of the helmet, is labeled frame u (e^i theta) = sin 4 theta. The piece of paper is labeled z plane. The helmet appears to be standing on its two legs, which are on a sketch of a circle divided into 8 parts. Each part has a set of contour lines toward the circumference of the circle. The contour lines are labeled level curves.

FIGURE 20.5.5 Level curves in Example 4

20.5 Exercises Answers to selected odd-numbered problems begin on page ANS-49.

In Problems 1–4, use the integrated solution (3) to the Poisson integral formula to solve the given Dirichlet problem in the upper half-plane.

  1. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis and a point labeled minus 1 on the negative horizontal axis. u = 0 is labeled at the right of 1 and left of minus 1 and u = 1 is labeled at the left of 1 and right of minus 1 on the horizontal axis. The first and second quadrants are shaded.

    FIGURE 20.5.6 Dirichlet problem in Problem 1

  2. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis and a point labeled minus 2 on the negative horizontal axis. u =0 is labeled at the right of 1 and left of minus 2, u = 1 is labeled at the left of 1, and u = 5 is labeled at the right of minus 2 on the horizontal axis. The first and second quadrants are shaded.

    FIGURE 20.5.7 Dirichlet problem in Problem 2

  3. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis, a point labeled minus 2 at the left of the negative horizontal axis, and a point labeled minus 1 at the right of the negative horizontal axis. u =0 is labeled at the left of 1 and the left of minus 2, u = 1 is labeled at the right of minus 1, u = 5 is labeled at the right of 1, and u = minus 1 is labeled between minus 1 and minus 2 on the horizontal axis. The first and second quadrants are shaded.

    FIGURE 20.5.8 Dirichlet problem in Problem 3

  4. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the positive horizontal axis, a point labeled minus 2 at the left of the negative horizontal axis, and a point labeled minus 1 at the right of the negative horizontal axis. u = 0 is labeled at the right of 1, u = 1 is labeled at the left of 1, u = 1 is labeled at the right of minus 1 and left of minus 2, and u = minus 1 is labeled between minus 1 and minus 2 on the horizontal axis. The first and second quadrants are shaded.

    FIGURE 20.5.9 Dirichlet problem in Problem 4

  5. Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary condition u(x, 0) = x2 when 0 < x < 1, and u(x, 0) = 0 otherwise.
  6. Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary condition u(x, 0) = cos x. [Hint: Let s = t − x and use the Section 19.6 formulas

    for a > 0.]

In Problems 7–10, solve the given Dirichlet problem by finding a conformal mapping from the given region R onto the upper half-plane v ≥ 0.

  1. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 on the horizontal axis and a point labeled i on the vertical axis. u = 0 is labeled at the right of 1 on the horizontal axis and bottom of i on the vertical axis, u = 1 is labeled at the left of 1 on the horizontal axis, and u = 5 is labeled at the top of i on the vertical axis. The first quadrant is shaded and labeled R.

    FIGURE 20.5.10 Dirichlet problem in Problem 7

  2. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a horizontal line labeled u = 0 passes through the horizontal axis from the origin to 3 and a vertical line labeled u = 1 passes through the vertical axis from the origin to top. A vertical line labeled u = 1 intersecting the horizontal axis at 3. The region bound by the lines is shaded and labeled R.

    FIGURE 20.5.11 Dirichlet problem in Problem 8

  3. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a circle centered at the origin. The sections of the circle in first and fourth quadrants are labeled u = 1 and the sections of the circle in the second and third quadrants are labeled u = 0. The region bound by the circle is shaded and labeled R.

    FIGURE 20.5.12 Dirichlet problem in Problem 9

  4. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph showing a semicircle centered at the origin passing through the positive horizontal axis at 1. The section of the semicircle in the first quadrant is labeled u = 0 and the section of the semicircle in the second quadrant is labeled u = 1. The positive horizontal axis below the semicircle is labeled u = 1 and the negative horizontal axis below the semicircle is labeled u = 0. The region bound by the semicircle is shaded and labeled R.

    FIGURE 20.5.13 Dirichlet problem in Problem 10

  5. A frame for a membrane is defined by u(eiθ) = θ2/θ2 for −π ≤ θ ≤ π. Use the Poisson integral formula for the unit disk to estimate the equilibrium displacement of the membrane at (−0.5, 0), (0, 0), and (0.5, 0).
  6. A frame for a membrane is defined by u(eiθ) = e−|θ| for −π ≤ θ ≤ π. Use the Poisson integral formula for the unit disk to estimate the equilibrium displacement of the membrane at (−0.5, 0), (0, 0), and (0.5, 0).
  7. Use the Poisson integral formula for the unit disk to show that u(0, 0) is the average value of the function u = u(eiθ) on the boundary z = 1.

In Problems 14 and 15, solve the given Dirichlet problem for the unit disk using the Fourier series form of the Poisson integral formula, and sketch the system of level curves.

  1. u(eiθ) = cos 2θ
  2. u(eiθ) = sin θ + cos θ