Applications 7.5. of Conformal Mapping.
Let us now apply what we have learned about analytic/conformal maps. We begin with boundary value problems for the Laplace equation, and then present some applications in fluid mechanics. We conclude by discussing how to use conformal maps to construct Green's functions for the two-dimensional Poisson equation. Applications to Harmonic Functions and Laplace's Equation We are interested in solving a boundary value problem for the Laplace equation on a domain R 2 . Our strategy is to map it to a corresponding boundary value problem on the unit disk D that we know how to solve. To this end, suppose we know a conformal map = g(z) that takes z to D. As we know, the real and imaginary parts of an analytic function F () defined on D are harmonic. Moreover, according to Proposition 7.31, the composition f (z) = F (g(z)) defines an analytic function whose real and imaginary parts are harmonic functions on . Thus, the conformal mapping can be regarded as a change of variables that preserves the property of harmonicity. In fact, this property does not even require the harmonic function to be the real part of an analytic function, i.e., we need not assume the existence of a harmonic conjugate. Proposition 7.37. If U (, ) is a harmonic function of , , and = + i = (x, y) + i (x, y) = g(z) is any analytic function, then the composition u(x, y) = U ((x, y), (x, y)) is a harmonic function of x, y. Proof : This is a straightforward application of the chain rule: U U u = + , x x x 2U 2u = x2 2 2u 2U = y 2 2 x y
2
(7.81)
(7.82)
u U U = + , y y y x y
2
2 U 2U +2 + x x 2 +2 2 U 2U + y y 2
U 2 U 2 + + , x2 x2 + U 2 U 2 + . y 2 y 2
2
2
Using the CauchyRiemann equations =- , x y = , y x
for the analytic function = + i , we find, after some algebra, u = 1/19/12 2u 2u + 2 = x2 y x
2
+
x 255
2
2U 2U + 2 2
= | g (z) |2 U,
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(7.83)
Peter J. Olver
where | g (z) |2 = ( /x ) + ( /x ) . We conclude that whenever U (, ) is any harmonic function, and so solves the Laplace equation U = 0 (in the , variables), then u(x, y) is a solution to the Laplace equation u = 0 in the x, y variables, and is thus also harmonic. Q.E.D. This observation has immediate consequences for boundary value problems arising in physical applications. Suppose we wish to solve the Dirichlet problem u = 0 in , u=h on , (7.84)
2
2
on a simply connected domain C. Let = g(z) = p(x, y) + i q(x, y) be a oneto-one conformal mapping from the domain to the unit disk D, whose existence is guaranteed by the Riemann Mapping Theorem 7.33. (Although its explicit construction may be problematic.) Then the change of variables formula (7.82) will map the harmonic function u(x, y) on to a harmonic function U (, ) on D. Moreover, the boundary values of U = H on the unit circle D correspond to those of u = h on by the same change of variables formula: h(x, y) = H(p(x, y), q(x, y)), for (x, y) . on D. (7.85)
We conclude that U (, ) solves the Dirichlet problem U = 0 in D, U =H (7.86)
But we already know how to solve the Dirichlet problem (7.86) the on unit disk by the Poisson integral formula (4.116)! We conclude that the solution to the original boundary value problem is given by the composition formula u(x, y) = U p(x, y), q(x, y) . In summary, the solution to the Dirichlet problem on a unit disk can be used to solve the Dirichlet problem on more complicated planar domains -- provided we are in possession of an appropriate conformal map. Example 7.38. According to Example 7.24, the analytic function + i = = z-1 x2 + y 2 - 1 2y = +i 2 + y2 z+1 (x + 1) (x + 1)2 + y 2 (7.87)
maps the right half plane R = { x = Re z > 0 } to the unit disk D = { | | < 1 }. Proposition 7.37 implies that if U (, ) is a harmonic function in the unit disk, then u(x, y) = U 2y x2 + y 2 - 1 , 2 + y2 (x + 1) (x + 1)2 + y 2 (7.88)
is a harmonic function on the right half plane. (This can, of course, be checked directly by a rather unpleasant chain rule computation.) To solve the Dirichlet boundary value problem u = 0, x > 0, u(0, y) = h(y), (7.89)
on the right half plane, we adopt the change of variables (7.87) and use the Poisson integral formula to construct the solution to the transformed Dirichlet problem U = 0, 1/19/12 2 + 2 < 1, 256 U (cos , sin ) = H(),
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(7.90)
Peter J. Olver
on the unit disk. The relevant boundary conditions are found as follows. Using the explicit form x+ iy = z = (1 + )(1 - ) 1 + - - | |2 1 - 2 - 2 + 2 i 1+ = = = 1- | 1 - |2 | 1 - |2 ( - 1)2 + 2
for the inverse map, we see that the boundary point = + i = e i on the unit circle D will correspond to the boundary point iy = 2 2 i sin = i cot = 2 2 + 2 ( - 1) 2 (cos - 1)2 + sin (7.91)
on the imaginary axis R = { Re z = 0 }. Thus, the boundary data h(y) on R corresponds to the boundary data H() = h cot 1 2 on the unit circle. The Poisson integral formula (4.116) can then be applied to solve (7.90), from which we are able to reconstruct the solution (7.88) to the boundary value problem (7.88) on the half plane. Let's look at an explicit example. If the boundary data on the imaginary axis is provided by the step function u(0, y) = h(y) 1, 0, y > 0, y < 0,
then the corresponding boundary data on the unit disk is a (periodic) step function H() = 1, 0, 0 < < , - < < 0.
After some tedious algebra, we find that the corresponding solution in the right half plane is simply 1 1 1 y 1 u(x, y) = + ph z = + tan-1 , 2 2 x an answer that, in hindsight, we should have been able to guess. Remark : The solution to the preceding Dirichlet boundary value problem is not, in fact, unique, owing to the unboundedness of the domain. The solution that we pick out by using the conformal map to the unit disk is the one that remains bounded at . The unbounded solutions would correspond to solutions on the unit disk that have a singularity in their boundary data at the point -1; see Exercise . 1/19/12 257
c 2012 Peter J. Olver
According to (4.119), the corresponding solution in the unit disk is 2 2 1 - 1 tan-1 1 - - , 2 + 2 < 1, > 0, 2 1 U (, ) = , 2 + 2 < 1, = 0, 2 2 2 - 1 tan-1 1 - - , 2 + 2 < 1, < 0. 2