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Lloyd N. Trefethen

Ruth J. Williams
July 1984
Technical Report #127

Conformal mapping solation of Laplace's equation on a polygon
with oblique derivative boundary conditions

Lloyd N. Trefethen*

Department of Mathematics
Massachusetts Institute of Technology

Ruth J. Williams**

Department of Mathematics
University of California at San Diego


Wc consider Laplace's equation in a polygonal domain together with the boundary
conditions that along each side, the derivative in the direction at a specified oblique angle
from the normal should be zero. First we prove that solutions to this problem can always
be constructed by taking the real part of an analjrtic function that maps the domain onto
another region with straight sides oriented according to the angles given in the boundary
conditions. Then we show that this procedure can be carried out successfully in practice
by the numerical calculation of Schwarz-Christoffel transformations. The method is illus-
trated by application to a Hall effect problem in electronics, and to a reflected Brownian
motion problem motivated by qucucing theory.

Key phrases: Laplace equation, conformal mapping, Schwarz-Christoffel map, oblique derivative, liall effea,
Brownian motion, queueing theory

AMS (MOS) Subjea Qassification: 3000, 35J25, 60105, 65E05, 65N99

•Research supported b>- an NSF Nbthematical Sciences Postdoctoral Fellcw-shjp, and bj- tlie U. S. Dept. of
Energy under contract DE-Aa)2-76-ER03077-V,

"Research supponed in p-an by NSF Grant DMS 8319562. This work was performed whJle both authors
were at the Courant Institute of Muthematical Sciences, New York University.

1. The oblique derivative problem and conformal mapping

Let n be a polygonal domain in the complex plane C, by which we mean a possibly
unbounded simply connected open subset of C whose boundary dCi consists of a finite
number of straight lines, rays, and Une segments. Let these sides be denoted by
Tj, . . . ,r„, n>l, where T^ is the complex open interval (24,2;+,), for some sequence of
vertices z^ € CU{a=}, and let (I he everywhere to the left of dil as dil is traversed in the
order Tj.r^, • • • . For convenience we set zq^z„ and z„+^^z^. We permit geometries in
which one or more slits are cut into fi by viewing the two sides of such a sUt as distinct
boundary segments, e.g. T^ and r^+i. The closure O is then a manifold that cannot be
embedded in the plane (although it can be reduced to an embeddable region by a confor-
mal map), and a function in C(fi) will have distinct values on T^ and r^+i corresponding
to Umits on the slit taken from different sides. One could also go further and let fl be a
polygonal Riemann surface, but we will not do this.

Let a set of real numbers Sj e„ be given. At any point z e r^, let u„{z) = — (z)


denote the inward normal derivative of a function u, and u^ = — its tangential deriva-
tive along r^ in the positive (counterclockwise) sense with respect to ft. This paper is con-
cerned with the following oblique derivative problem:

Problem O. Find functions u d C^(ft)nC^(ftU Ui^i) satisfying Laplace's equation

A«(z) = 0, z€ft (1.1)

together with the homogeneous oblique derivative boundary conditions

cos9iM„(z) + sin9iU„(z) = 0, r e T^. (1.2)

The situation is illustrated in Figure la. If 64 = (modir), we have a Neumann condition
on side r^, while 6^ ^ ^(mod'iT) gives a tangential condition, which we do not rule out.

Obviously any constant function is a solution to Problem O. K ft were a smooth
domain with a continuous single-valued obliquity function 6(ct) for a e aft, the constants
would be the only solutions, as can be shown by consideration of Ricmann-Hilbcrt prob-
lems [12,13,20]. More generally, if ft were smooth and e(a) changed continuously by ATtt
as a traversed aft, there would be a solution space of finite dimension raax{l,A'}. But
Problem O is quite different from the analogous problem on a smooth domain, for no





u = Ref


Figure 1. Problem O and its solution by conformal mapping. The derivative of u at
the boundary of il in the direction given by the arrow must be zero.

boundary conditions have been specified at the vertices, nor have we required regularity
or boundcdness there. As a result the space of solutions has infinite dimension, so that in
any particular application, additional conditions will be needed to ensure uniqueness. We
have intentionally omitted such conditions from the statement of Problem O for the rea-
son that their most natural form varies from problem to problem. The statements of
Problems O^ and Og in Sections 2 and 3 below incorporate appropriate additional condi-
tions for the two applications considered there.

The purpose of this paper is to construct nonconstant solutions to Problem O by the
following process of conformal mapping, illustrated in Figure 1. Suppose an analytic
function / in HUUil^i is found that maps CL onto another polygonal domain f(Cl). (/
may fail to be globally or locally one-to-one, in which case we view f(fl) not as a subset of
C but as a Ricmann surface.) Suppose further that each image /(P^) consists of one or
more sides of /(H) oriented at an angle 9^ (modir) counterclockwise from the real axis.
For each point z e T^, either /'(z) = 0, or / is conformal at z. Assuming the latter, we sec
that / maps any curve that meets T^ at z at an angle 6^ clockwise from the inward normal
(Figure la) onto a curve that meets /(PJ at /(z) with a vertical slope (Figure lb). In
other words, the derivative of / at z with respect to arc length in the direction given by 6^
is pure imaginary. Now set

u(,) = Re/(z). (1.3)

Then the derivative of u in the direction given by 6^ is zero. Of course the same is true if
/'(z) = 0. Thus in cither case (1.3) satisfies (1.1) and (1.2), and wc have found a solution
to Problem O.


Conversely, every solution to Problem O is the real part of an analytic function that
maps onto a domain with straight sides. The following theorem is proved in the

THEOREM 1. A function u{z) defined m HU (Ji^^i 'f be the function (IJ) defined by an arbitrary choice ofC^. T^, p. and {mj subject to the
restrictions listed above. Then u = Ref with /=«{»o

1 3

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