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

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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
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CHAPTER 3 ANALYTIC and HARMONIC FUNCTIONS

Section 3.3 Harmonic Functions

Let be a real-valued function of the two real variables and . The partial differential equation

is known as L aplace's equation and is sometimes referred to as the potential equation. If , , , , , and are all continuous and if satisfies Laplace's equation, then is called a harmonic function . Harmonic functions are important in the areas of applied mathematics, engineering, and mathematical physics. They are used to solve problems involving steady state temperatures, two-dimensional electrostatics, and ideal fluid flow. In Chapter 10 we will see how complex analysis techniques can be used to solve some problems involving harmonic functions. We begin with an important theorem relating analytic and harmonic functions.

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

Theorem 3.8 Let be an analytic function in the domain . Assume that all of the second-order partial derivatives of and are continuous. Then both and are harmonic functions in . In other words, the real and imaginary parts of an analytic function are harmonic.

Example for Page 115. Show that the complex function
is NOWHERE analytic.

> U:='U': V:='V': x:='x': y:='y':
U := proc(x,y) x^2 + y^2 end:
V := proc(x,y) 2*x*y end:
`F(x,y)` = U(x,y) + I*V(x,y);
`U(x,y)` = U(x,y);
`V(x,y)` = V(x,y); ` `;
`Look at the Cauchy-Riemann equations.`;
`Ux(x,y)` = diff(U(x,y),x);
`Vy(x,y)` = diff(V(x,y),y);
print(`Ux = Vy `,diff(U(x,y),x) = diff(V(x,y),y),
evalb(diff(U(x,y),x) = diff(V(x,y),y)));
` `;
`Uy(x,y)` = diff(U(x,y),y);
`Vx(x,y)` = diff(V(x,y),x);
print(`Uy = -Vx `,diff(U(x,y),y) = - diff(V(x,y),x),
evalb(diff(U(x,y),y) = - diff(V(x,y),x)));

>

Therefore, is differentiable only when . Since is NOT differentiable in any open neighborhood, is NOWHERE analytic.

If we are given a function that is harmonic in the domain and if we can find another harmonic function , such that the partial derivatives for and satisfy the Cauchy-Riemann equations throughout , then we say that is a harmonic conjugate of . It then follows that the function is analytic in .

Example 3.11, Page 115. Show that both and are harmonic functions, and is the harmonic conjugate of .

> f:='f': U:='U': V:='V': w:='w': x:='x': y:='y': z:='z':
f := z -> z^2:
`f(z) ` = f(z);
`f(z) is an analytic function.`;
`f(x + I y) ` = f(x+I*y);
`f(x + I y) ` = evalc(f(x+I*y)); ` `;
U := proc(x,y) x^2 - y^2 end:
V := proc(x,y) 2*x*y end:
`The real and imaginary parts are harmonic functions.`;
`Re(f(z) = U(x,y)` = U(x,y);
`Im(f(z) = V(x,y)` = V(x,y);

And we can show that both and satisfy Laplace's equation.

> `U(x,y) ` = U(x,y); `V(x,y) ` = V(x,y);
`Verify Laplace's equation.`;
`Uxx(x,y) ` = diff(U(x,y),x$2);
`Uyy(x,y) ` = diff(U(x,y),y$2);
print(`0 = Uxx + Uyy `,
0 = diff(U(x,y),x$2) + diff(U(x,y),y$2),
evalb(0 = diff(U(x,y),x$2) + diff(U(x,y),y$2)));
`Vxx(x,y) ` = diff(V(x,y),x$2);
`Vyy(x,y) ` = diff(V(x,y),y$2);
print(`0 = Vxx + Vyy `,
0 = diff(V(x,y),x$2) + diff(V(x,y),y$2),
evalb(0 = diff(V(x,y),x$2) + diff(V(x,y),y$2)));

Hence, both and are harmonic functions.

Example 3.12, Page 115. Show that both and are harmonic functions, and is the harmonic conjugate of .

> f:='f': U:='U': V:='V': w:='w': x:='x': y:='y': z:='z':
f := z -> z^3:
`f(z) ` = f(z);
`f(z) is an analytic function.`;
`f(x + I y) ` = f(x+I*y);
`f(x + I y) ` = evalc(f(x+I*y)); ` `;
U := proc(x,y) x^3 - 3*x*y^2 end:
V := proc(x,y) 3*x^2*y - y^3 end:
`The real and imaginary parts are harmonic functions.`;
`Re(f(z)) = U(x,y)` = U(x,y);
`Im(f(z)) = V(x,y)` = V(x,y);

And we can show that both and satisfy Laplace's equation

> `U(x,y) ` = U(x,y); `V(x,y) ` = V(x,y);
`Verify Laplace's equation.`;
`Uxx(x,y) ` = diff(U(x,y),x$2);
`Uyy(x,y) ` = diff(U(x,y),y$2);
print(`0 = Uxx + Uyy `,
0 = diff(U(x,y),x$2) + diff(U(x,y),y$2),
evalb(0 = diff(U(x,y),x$2) + diff(U(x,y),y$2)));
`Vxx(x,y) ` = diff(V(x,y),x$2);
`Vyy(x,y) ` = diff(V(x,y),y$2);
print(`0 = Vxx + Vyy `,
0 = diff(V(x,y),x$2) + diff(V(x,y),y$2),
evalb(0 = diff(V(x,y),x$2) + diff(V(x,y),y$2)));

Hence, both and are harmonic functions.

Theorem 3.9 (Construction of a conjugate)

Let be harmonic in an -neighborhood of the point ( ). Then there exists a conjugate harmonic function defined in this neighborhood such that is an analytic function.

METHOD. Construction of the harmonic conjugate of .
Activate the following procedure before doing Example 3.13, Page 117.

> conj:='conj': U:='U': V:='V': x:='x': y:='y':
conj := proc(U)
local lap,v1,v2,v3,v4;
lap := diff(U,x$2)+diff(U,y$2);
v1 := int(diff(U,x), y);
v2 := - diff(U,y) - diff(v1,x);
v3 := int(v2,x);
v4 := v1 + v3;
if lap=0 then
RETURN(v4)
else
RETURN(`U(x,y) was Not harmonic.`)
fi
end:

Example 3.13, Page 117. Show that is a harmonic function and find the harmonic conjugate .

> U := proc(x,y) x*y^3 - x^3*y end:
`U(x,y) ` = U(x,y);
`Verify Laplace's equation.`;
`Uxx(x,y) ` = diff(U(x,y),x$2);
`Uyy(x,y) ` = diff(U(x,y),y$2);
print(`0 = Uxx + Uyy `,
0 = diff(U(x,y),x$2) + diff(U(x,y),y$2),
evalb(0 = diff(U(x,y),x$2) + diff(U(x,y),y$2)));
` `; `The harmonic conjugate is:`;
V := conj(U(x,y)):
`V(x,y) ` = V;

Example 3.14, Page 119. Show that is the scalar potential function for the fluid flow: .

> f:='f': F:='F': U:='U': V:='V': x:='x':
X:='X': y:='y': Y:='Y': z:='z': Z:='Z':
assume(X,real); assume(Y,real);
F := z -> z^2:
`F(z) ` = F(z);
f := z -> subs(Z=z, diff(F(Z), Z)):
`f(z) = F '(z)`;
`f(z) ` = f(z);
`f(x + I y) ` = f(x+I*y);
v := conjugate(f(X+I*Y)):
V := proc(x,y) subs({X=x,Y=y},v) end:
`V(x,y) ` = V(x,y);
`F(x + I y) ` = F(x+I*y);
`F(x + I y) ` = evalc(F(x+I*y));
U := proc(x,y) x^2-y^2 end:
`U(x,y) ` = U(x,y);

> v:='v': x:='x': y:='y':
v := proc(x,y) 2*x*y end:
`v(x,y) ` = v(x,y);
contourplot(v(x,y), x=0..5, y=0..5,
title=`The streamlines 2xy = C`,
scaling=constrained,
axes=boxed, grid=[30,30]);

>

End of Section 3.3.