COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

Complimentary software to accompany the textbook:

COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
Tele. (800) 832-0034; FAX: (508) 443-8000, E-mail: mkt@jbpub.com, http://www.jbpub.com/


CHAPTER 10 APPLICATIONS of HARMONIC FUNCTIONS

Section 10.1 Preliminaries

In most applications involving harmonic functions it is required to find a harmonic function that takes on prescribed values along certain contours. We will assume that the reader is familiar with the material in Sections 2.5, 3.3, 5.1, and 5.2.

 

Load Maple's "contourplot" procedure.
Make sure this is done only ONCE during a Maple session.

> with(plots):

 

Warning, the name changecoords has been redefined


Example 10.1, Page 387. A specific example of the general example. Find a function
`u(x,y)` that is harmonic in the vertical strip `1 < x < 2` and takes on the boundary values:
u(1,y) = 6 for all y ,
u(2,y) = 1 for all y .

> a:='a': b:='b': x:='x': y:='y': U:='U': U1:='U1': U2:='U2':
a := 1:
b := 2:
U1 := 6:
U2 := 1:
U := proc(x,y)
global a,b,U1,U2;
U1 + (U2 - U1)/(b-a)*(x-a);
end:
`a ` = a, ` b ` = b;
`U(a,y) ` = U1, ` U(b,y) ` = U2;
`U(x,y) ` = U(x,y);

`a ` = 1, `   b ` = 2

`U(a,y) ` = 6, `   U(b,y) ` = 1

`U(x,y) ` = 11-5*x

 

> contourplot(U(x,y), x=1..2, y=-1..1,
title=`11 - 5x = C`,
axes=boxed, grid=[20,20],
scaling=constrained);

[Maple Plot]

 


Example 10.2, Page 388. A specific example of the general example. Find a function
`u(x,y)` that is harmonic in the sector 0 < Arg(z) < 2*pi/3 and takes on the boundary values:
u(x,0) = 1 for 0 < x ,
u(x,y) = 9 for all points on the ray 0 < r , theta = alpha .

> a:='a': C1:='C1': C2:='C2': x:='x': y:='y': U:='U':
a := 2*Pi/3:
C1 := 1:
C2 := 9:
U := proc(x,y)
global a,C1,C2;
C1 + (C2 - C1)/a * arctan(y,x);
end:
`Angle `, alpha = a;
`U(x,0) ` = C1, ` U(x,y) ` = C2;
`U(x,y) ` = U(x,y);

`Angle   `, alpha = 2/3*Pi

`U(x,0) ` = 1, `   U(x,y) ` = 9

`U(x,y) ` = 1+12/Pi*arctan(y,x)

> contourplot(U(x,y), x=-2..2, y=0..2,
title=`1 + 12 arctan(y,x)/Pi = C`,
axes=boxed, grid=[20,20],
scaling=constrained);

[Maple Plot]

 


Example 10.3, Page 389. A specific example of the general example. Find a function
`u(x,y)` that is harmonic in the annulus `1 < |z| < R` and takes on the boundary values:
u(x,y) = 1 when abs(z) = 1 ,
u(x,y) = 6 when abs(z) = 4 .

> K1:='K1': K2:='K2': R:='R': x:='x': y:='y': U:='U':
R := 4:
K1 := 1:
K2 := 6:
U := proc(x,y)
global K1,K2,R;
K1 + ((K2 - K1)*ln(sqrt(x^2 + y^2)))/ln(R);
end:
`R ` = R;
`For |z| = 1, U(x,y) ` = K1;
`For |z| = R, U(x,y) ` = K2;
`U(x,y) ` = U(x,y);

`R ` = 4

`For |z| = 1,   U(x,y) ` = 1

`For |z| = R,   U(x,y) ` = 6

`U(x,y) ` = 1+5*ln(sqrt(x^2+y^2))/ln(4)

 

> contourplot(U(x,y), x=-4.2..4.2, y=-4.2..4.2,
title=`1 + 5 ln((x^2+y^2)^1/2)/ln(4) = C`,
axes=boxed, grid=[20,20],
scaling=constrained);

[Maple Plot]

 

End of Section 10.1.