Chapter Seven
Harmonic Functions
7.1. The Laplace equation. The Fourier law of heat conduction says that the rate at which
heat passes across a surface S is proportional to the flux, or surface integral, of the
temperature gradient on the surface:
k 4ðT 6ð dA.
XðXð
S
Here k is the constant of proportionality, generally called the thermal conductivity of the
substance (We assume uniform stuff. ). We further assume no heat sources or sinks, and we
assume steady-state conditions the temperature does not depend on time. Now if we take
S to be an arbitrary closed surface, then this rate of flow must be 0:
k 4ðT 6ð dA =ð 0.
XðXð
S
Otherwise there would be more heat entering the region B bounded by S than is coming
out, or vice-versa. Now, apply the celebrated Divergence Theorem to conclude that
Å»ð4ð 6ð 4ðTÞðdV =ð 0,
XðXðXð
B
where B is the region bounded by the closed surface S. But since the region B is completely
arbitrary, this means that
/ð2T /ð2T /ð2T
4ð 6ð 4ðT =ð +ð +ð =ð 0.
/ðx2 /ðy2 /ðz2
This is the world-famous Laplace Equation.
Now consider a slab of heat conducting material,
7.1
in which we assume there is no heat flow in the z-direction. Equivalently, we could assume
we are looking at the cross-section of a long rod in which there is no longitudinal heat
flow. In other words, we are looking at a two-dimensional problem the temperature
depends only on x and y, and satisfies the two-dimensional version of the Laplace equation:
/ð2T /ð2T
+ð =ð 0.
/ðx2 /ðy2
Suppose now, for instance, the temperature is specified on the boundary of our region D,
and we wish to find the temperature TÅ»ðx, yÞð in region. We are simply looking for a solution
of the Laplace equation that satisfies the specified boundary condition.
Let s look at another physical problem which leads to Laplace s equation. Gauss s Law of
electrostatics tells us that the integral over a closed surface S of the electric field E is
proportional to the charge included in the region B enclosed by S. Thus in the absence of
any charge, we have
E 6ðdA =ð 0.
XðXð
S
But in this case, we know the field E is conservative; let dð be the potential function that
is,
E =ð 4ðjð.
Thus,
E 6ðdA =ð 4ðjð 6ðdA.
XðXð XðXð
S S
Again, we call on the Divergence Theorem to conclude that jð must satisfy the Laplace
equation. Mathematically, we cannot tell the problem of finding the electric potential in a
7.2
region D, given the potential on the boundary of D, from the previous problem of finding
the temperature in the region, given the temperature on the boundary. These are but two of
the many physical problems that lead to the Laplace equation You probably already know
of some others. Let D be a domain and let að be a given function continuous on the
boundary of D. The problem of finding a function jð harmonic on the interior of D and
which agrees with að on the boundary of D is called the Dirichlet problem.
7.2. Harmonic functions. If D is a region in the plane, a real-valued function uÅ»ðx, yÞð
having continuous second partial derivatives is said to be harmonic on D if it satisfies
Laplace s equation on D :
/ð2u /ð2u
+ð =ð 0.
/ðx2 /ðy2
There is an intimate relationship between harmonic functions and analytic functions.
Suppose f is analytic on D, and let fÅ»ðzÞð =ð uÅ»ðx, yÞð +ð ivÅ»ðx, yÞð. Now, from the Cauchy-Riemann
equations, we know
/ðu /ðv
=ð , and
/ðx /ðy
/ðu /ðv
=ð ?ð .
/ðy /ðx
If we differentiate the first of these with respect to x and the second with respect to y, and
then add the two results, we have
/ð2u /ð2u /ð2v /ð2v
+ð =ð ?ð =ð 0.
/ðx/ðy /ðy/ðx
/ðx2 /ðy2
Thus the real part of any analytic function is harmonic! Next, if we differentiate the first of
the Cauchy-Riemann equations with respect to y and the second with respect to x, and then
subtract the second from the first, we have
/ð2v /ð2v
+ð =ð 0,
/ðx2 /ðy2
and we see that the imaginary part of an analytic function is also harmonic.
There is even more excitement. Suppose we are given a function jð harmonic in a simply
connected region D. Then there is a function f analytic on D which is such that Re f =ð jð.
Let s see why this is so. First, define g by
7.3
/ðjð /ðjð
gÅ»ðzÞð =ð ?ð i .
/ðx /ðy
We ll show that g is analytic by verifying that the real and imaginary parts satisfy the
Cauchy-Riemann equations:
/ðjð /ð2jð /ð2jð /ðjð
/ð /ð
=ð =ð ?ð =ð ?ð ,
/ðx /ðx /ðy /ðy
/ðx2 /ðy2
since jð is harmonic. Next,
/ðjð /ð2jð /ð2jð /ðjð
/ð /ð
=ð =ð =ð ?ð ?ð .
/ðy /ðx /ðy/ðx /ðx/ðy /ðx /ðy
Since g is analytic on the simply connected region D, we know that the integral of g around
any closed curve is zero, and so it has an antiderivative GÅ»ðzÞð =ð u +ð iv. This antiderivative
is, of course, analytic on D, and we know that
/ðjð /ðjð
/ðu /ðu
GvðÅ»ðzÞð =ð ?ð i =ð ?ð i .
/ðx /ðy /ðx /ðy
Thus, uÅ»ðx, yÞð =ð jðÅ»ðx, yÞð +ð hÅ»ðyÞð. From this,
/ðjð
/ðu
=ð +ð hvðÅ»ðyÞð,
/ðy /ðy
and so hvðÅ»ðyÞð =ð 0, or h =ð constant, from which it follows that uÅ»ðx, yÞð =ð jðÅ»ðx, yÞð +ð c. In other
words, Re G =ð u, as we promised to show.
Example
The function jðÅ»ðx, yÞð =ð x3 ?ð 3xy2 is harmonic everywhere. We shall find an analytic
function G so that Re G =ð jð. We know that GÅ»ðzÞð =ð Å»ðx3 ?ð 3xy2Þð +ð iv, and so from the
Cauchy-Riemann equations:
/ðv /ðu
=ð ?ð =ð 6xy
/ðx /ðy
7.4
Hence,
vÅ»ðx, yÞð =ð 3x2y +ð kÅ»ðyÞð.
To find kÅ»ðyÞð differentiate with respect to y :
/ðv /ðu
=ð 3x2 +ð kvðÅ»ðyÞð =ð =ð 3x2 ?ð 3y2,
/ðy /ðx
and so,
kvðÅ»ðyÞð =ð ?ð3y2, or
kÅ»ðyÞð =ð ?ðy3 +ð any constant.
If we choose the constant to be zero, this gives us
v =ð 3x2y +ð kÅ»ðyÞð =ð 3x2y ?ð y3,
and finally,
GÅ»ðzÞð =ð u +ð iv =ð Å»ðx3 ?ð 3xy2Þð +ð iÅ»ð3x2y ?ð y3Þð.
Exercises
1. Suppose jð is harmonic on a simply connected region D. Prove that if jð assumes its
maximum or its minimum value at some point in D, then jð is constant in D.
2. Suppose jð and að are harmonic in a simply connected region D bounded by the curve C.
Suppose moreover that jðÅ»ðx, yÞð =ð aðÅ»ðx, yÞð for all Å»ðx, yÞð 5ð C. Explain how you know that
jð =ð að everywhere in D.
3. Find an entire function f such that Re f =ð x2 ?ð 3x ?ð y2, or explain why there is no such
function f.
4. Find an entire function f such that Re f =ð x2 +ð 3x ?ð y2, or explain why there is no such
function f.
7.5
7.3. Poisson s integral formula. Let Cð be the disk bounded by the circle
C_ð =ð áðz : z =ð _ðâð. Suppose jð is harmonic on Cð and let f be a function analytic on Cð and
| |
such that Re f =ð jð. Now then, for fixed z with z <ð _ð, the function
| |
fÅ»ðsÞð z
gÅ»ðsÞð =ð
_ð2 ?ð s z
is analyic on Cð. Thus from Cauchy s Theorem
fÅ»ðsÞð z
gÅ»ðsÞðds =ð ds =ð 0.
Xð Xð
_ð2 ?ð s z
C_ð C_ð
We know also that
fÅ»ðsÞð
1
fÅ»ðzÞð =ð ds.
Xð
s ?ð z
2^ði
C_ð
Adding these two equations gives us
1 1 z
fÅ»ðzÞð =ð +ð fÅ»ðsÞðds
Xð
s ?ð z
2^ði
_ð2 ?ð s z
C_ð
_ð2 ?ð z
| |2 fÅ»ðsÞðds.
1
=ð
Xð
2^ði
Å»ðs ?ð zÞðÅ»ð_ð2 ?ð s z Þð
C_ð
Next, let LðÅ»ðtÞð =ð _ðeit, and our integral becomes
2^ð
_ð2 ?ð z
| |2
1
fÅ»ðzÞð =ð fÅ»ð_ðeitÞði_ðeitdt
Xð
2^ði
Å»ð_ðeit ?ð zÞðÅ»ð_ð2 ?ð _ðeit z Þð
0
2^ð
_ð2 ?ð z fÅ»ð_ðeitÞð
| |2
=ð dt
Xð
2^ð
Å»ð_ðeit ?ð zÞðÅ»ð_ðe?ðit ?ð z Þð
0
2^ð
_ð2 ?ð z fÅ»ð_ðeitÞð
| |2
=ð dt
Xð
2^ð
_ðeit ?ð z
| |2
0
Now,
7.6
2^ð
_ð2 ?ð z jðÅ»ð_ðeitÞð
| |2
jðÅ»ðx, yÞð =ð Re f =ð dt.
Xð
2^ð
_ðeit ?ð z
| |2
0
Next, use polar coordinates: z =ð reiSð :
2^ð
_ð2 ?ð r2 jðÅ»ð_ðeitÞð
jðÅ»ðr, SðÞð =ð dt.
Xð
2^ð
_ðeit ?ð reiSð
| |2
0
Now,
_ðeit ?ð reiSð =ð Å»ð_ðeit ?ð reiSðÞðÅ»ð_ðe?ðit ?ð re?ðiSðÞð =ð _ð2 +ð r2 ?ð r_ðÅ»ðeiÅ»ðt?ðSðÞð +ð e?ðiÅ»ðt?ðSðÞðÞð
| |2
=ð _ð2 +ð r2 ?ð 2r_ð cosÅ»ðt ?ð SðÞð.
Substituting this in the integral, we have Poisson s integral formula:
2^ð
_ð2 ?ð r2 jðÅ»ð_ðeitÞð
jðÅ»ðr, SðÞð =ð dt
Xð
2^ð
_ð2 +ð r2 ?ð 2r_ð cosÅ»ðt ?ð SðÞð
0
This famous formula essentially solves the Dirichlet problem for a disk.
Exercises
2^ð
1
5. Evaluate Xð dt. [Hint: This is easy.]
_ð2+ðr2?ð2r_ð cosÅ»ðt?ðSðÞð
0
6. Suppose jð is harmonic in a region D. If Å»ðx0, y0Þð 5ð D and if C Ðð D is a circle centered at
Å»ðx0, y0Þð, the inside of which is also in D, then jðÅ»ðx0, y0Þð is the average value of jð on the
circle C.
7. Suppose jð is harmonic on the disk Cð =ð áðz : z ²ð _ðâð. Prove that
| |
1
jðÅ»ð0, 0Þð =ð jðdA.
XðXð
^ð_ð2
Cð
7.7