Upload
trinhngoc
View
239
Download
4
Embed Size (px)
Citation preview
CHAPTER 8
Laplace’s Equation and Poisson’s Equation
In this chapter, we consider Laplace’s equation and its inhomogeneous counterpart, Pois-
son’s equation, which are prototypical elliptic equations.1 They may be thought of as time-
independent versions of the heat equation, with and without source terms:
�u(x) = 0 (Laplace’s equation)
��u(x) = f(x) (Poisson’s equation).
We consider these equations in a domain U ⇢ Rn, n � 2, but also on all of Rn. Applications of
these equations include the classical field of potential theory, of importance in electrostatics
and steady incompressible fluid flow. In electrostatics, f(x) in Poisson’s equation represents
a charge density distribution, inducing the electric potential u(x). In two-dimensional steady
fluid flow, u is the velocity potential or stream function, both of which satisfy Laplace’s
equation.
Several properties of solutions of Laplace’s equation parallel those of the heat equation: maxi-
mum principles, solutions obtained from separation of variables, and the fundamental solution
to solve Poisson’s equation in Rn.
1Pierre-Simon Laplace, 1749-1827, made many contributions to mathematics, physics and astronomy.
Simon Denis Poisson, 1781-1840, was a mathematician and physicist known for his contributions to the theory
of electricity and magnetism.
183
184 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
8.1. The Fundamental Solution
To solve Poisson’s equation, we begin by deriving the fundamental solution �(x) for the
Laplacian. This fundamental solution is rather di↵erent from the fundamental solution for
the heat equation, which is designed to solve initial value problems, and consequently has
a singularity at the initial time t = 0. The fundamental solution for the Laplacian, being
time-independent, is used to represent solutions in space alone. To do this, �(x) is chosen to
have a singularity at a point x0
in the domain; since the Laplacian is translation invariant,
we can take x0
= 0. Moreover, the Laplacian is invariant under rotations, so we can seek a
rotationally invariant fundamental solution.
Motivated by the above discussion, we seek rotationally symmetric solutions u(x) = v(r), r =
|x|, of Laplace’s equation. Then
(1.1) �u(x) = v00(r) +n� 1
rv0(r) = 0.
Therefore,
v00
v0= �
n� 1
r.
Integrating, we obtain log v0 = �(n� 1) log r + C. That is, v0 = Arn�1 , where A = logC is the
constant of integration. Integrating again, we get a two-parameter family of solutions
v(r) =
8>>><
>>>:
a
rn�2
+ b, if n � 3
a log r + b, if n = 2.
8.2. SOLVING POISSON’S EQUATION IN Rn 185
The fundamental solution �(x) is defined by setting b = 0, and choosing the constant a to
normalize �(x) depending on the volume ↵(n) of the unit ball:
(1.2) �(x) =
8>>><
>>>:
1
n(n� 2)↵(n)
1
|x|n�2
, if n � 3
�
1
2⇡log |x|, if n = 2.
The purpose of the normalization is to make the formula for the solution of Poisson’s equation
on Rn as simple as possible. (See (2.3) below.) Note that although � has an integrable
singularity at the origin (� is integrable on bounded sets, even though it is not defined at
x = 0), we will see that the singularity of ��� is not integrable, and is in fact the singular
distribution �, which we define carefully in §9.2.
We say a function u is harmonic in an open set U ⇢ Rn if u 2 C2(U) and �u(x) = 0 for each
x 2 U. By construction, �(x) is harmonic in every open set not containing the origin.
8.2. Solving Poisson’s equation in Rn
In this section, we establish a formula for the solution of Poisson’s equation in all space Rn
using the fundamental solution, much as we used the heat kernel to solve the Cauchy problem
for the heat equation.
Let f 2 C2(Rn) have compact support and define
(2.3) u(x) = (� ⇤ f)(x) =
Z
Rn
�(x� y)f(y) dy,
the convolution product of � with f .
186 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
Remark: Note that ��(x) has a non-integrable singularity at the origin. Therefore, we
cannot di↵erentiate under the integral sign in this formula. If we could, we would have
�u(x) = 0, since � is harmonic away from the origin. However, as we now show, the non-
integrable singularity makes the convolution product work to solve Poisson’s equation.
Theorem 8.1. If f 2 C2(Rn) has compact support and u(x) is given by (2.3), then
��u(x) = f(x), x 2 Rn.
Proof. Changing variables in (2.3), we have
u(x) =
Z
Rn
�(y)f(x� y) dy.
Therefore,
�u(x) =
Z
Rn
�(y)�yf(x� y) dy =
Z
Rn
�(y)�yf(x� y) dy.
In this integral, and in subsequent calculations, we use subscripts to indicate the variable of
di↵erentiation or integration. Thus �y indicates that the Laplacian is taken with respect to
the y variables. We would like to integrate by parts to put � back on �(y). However, �(y)
has a singularity at y = 0, so we have to treat the integral as an improper integral. For ✏ > 0,
let U✏ = Rn� B(0, ✏), where B(x, r) denotes the open ball of radius r centered at x 2 Rn.
Then
�u(x) =
Z
B(0,✏)
�(y)�xf(x� y) dy +
Z
U✏
�(y)�yf(x� y) dy.
The first integral approaches zero as ✏! 0, since � is integrable at the origin. We use Green’s
second identity on the second integral, observing that (since f has compact support) the
8.3. PROPERTIES OF HARMONIC FUNCTIONS 187
integrand is identically zero outside a large enough ball centered at x. Thus,
Z
U✏
�(y)�yf(x� y) dy =
Z
@B(0,✏)
✓�(y)
@f
@⌫y(x� y)� f(x� y)
@�
@⌫y(y)
◆dSy +
Z
U✏
��(y)f(x� y) dy.
Incidentally, there is no contribution from the boundary of the support of f, since the integrand
is zero there. The final integral is zero since � is harmonic in U✏. The integral on the boundary
has two terms. The first term converges to zero as ✏ ! 0, since � is integrable at the origin.
We need to prove that the second integral converges to f(x).
Since the unit normal ⌫ is outward with respect to U✏, on the sphere @B(0, ✏) we have ⌫ = �y/✏.
Therefore, since �(y) is a function of r = |y|,
@�
@⌫(y) = �
@�
@r(y)
�����|y|=✏
=1
n↵(n)✏n�1
.
Note that this formula holds for n � 2 even though the formula for � is di↵erent for n = 2.
Thus,
�
Z
@B(0,✏)
f(x� y)@�
@⌫y(y) dSy = �
Z
@B(0,✏)
f(x� y) dSy1
|@B(0, ✏)|,
where |@B(0, ✏)| = n↵(n)✏n�1 denotes the measure of the sphere in Rn. This integral converges
to f(x) because f is continuous.
8.3. Properties of Harmonic Functions
188 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
In this section, we state and prove the mean value property of harmonic functions, and use it
to prove the maximum principle, leading to a uniqueness result for boundary value problems
for Poisson’s equation. We state the mean value property in terms of integral averages.
Theorem 8.2. (Mean Value Property) Suppose u 2 C2(U). Then u is harmonic in U if and
only if it has the mean value property:
(3.4) u(x) = �
Z
@B(x,r)
u(y)dS = �
Z
B(x,r)
u(y)dy,
for every ball B(x, r) ⇢ U.
Proof: Suppose u is harmonic. Then, for B(x, r) ⇢ U,
0 =
Z
B(x,r)
�u(y)dy =
Z
@B(x,r)
@u
@⌫dS =
Z
@B(x,r)
ru(y) ·y � x
rdSy.
Now let
(3.5) �(r) = �
Z
@B(x,r)
u(y)dSy.
Since limr!0
�(r) = u(x), we complete the proof by using (3.5) to show that �(r) is constant.
To do so, we calculate directly that �0(r) = 0. Let y = x + rz, z 2 B(0, 1), to facilitate
di↵erentiating the integral. Then
�0(r) =d
dr�
Z
@B(0,1)
u(x+ rz) dSz
= �
Z
@B(0,1)
ru(x+ rz) · z dSz
= �
Z
@B(x,r)
ru(y) ·y � x
rdSy = 0, by (3.5)
Thus, �(r) is constant, so that �(r) = lims!0
�(s) = u(x).
8.3. PROPERTIES OF HARMONIC FUNCTIONS 189
xr U
U
u = 0∆
Figure 8.1. A bounded region U.
We use this result to obtain the integral average over the ball B(x, r) :
Z
B(x,r)
u(y)dy =
Z r
0
Z
@B(x,⇢)
u(y) dS d⇢
= u(x)
Z r
0
n↵n⇢n�1 d⇢ = ↵nr
nu(x).
Conversely, suppose u has the mean value property. Then, as above, we get
�0(r) =r
n�
Z
B(x,r)
�u(y)dy = 0, since �(r) = u(x) is constant.
Thus, �RB(x,r)
�u(y)dy ⌘ 0, and letting r ! 0, we obtain �u(x) = 0, since �u is continuous in
U. This completes the proof of the theorem.
The mean value property of harmonic functions is peculiar to solutions of Laplace’s equation,
and has no counterpart for more general elliptic equations. However, it simplifies the proofs
of key results that do generalize, in particular the Maximum Principle, which we now state in
both weak and strong form:
190 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
Theorem 8.3. Let U ⇢ Rn be open and bounded, and suppose u 2 C2(U)\C(U) is harmonic
in U . Then
1. Weak form: maxx2U
u(x) = maxx2@U
u(x).
2. Strong form: If U is connected, then either u = constant in U, or
u(x) < maxy2@U
u(y) for all x 2 U.
Proof: We prove the strong form first. The weak form then follows easily. Suppose U is
connected and there is a point x0
2 U such that
u(x0
) = maxx2U
u(x) = M.
Choose r so that B(x, r) ⇢ U. Then by the mean value property,
u(x0
) = M = �
Z
B(x0,r)
u(y) dy.
But u(y) M everywhere, so equality implies that u(y) = M throughout B(x0
, r). Thus,
the set S = {x 2 U : u(x) = M} is non-empty and open. However, S is also relatively
closed in U: Let xn 2 S converge to x 2 U as n ! 1. Then, by continuity of u, we have
u(x) = limn!1u(xn) = M, so x 2 S. But the only non-empty open and closed set in U is U
itself, so we have S = U, implying that u is constant in U. The weak form (1) follows from
(2).
Remarks: 1. The corresponding Minimum Principle follows by applying the maximum
principle to �u.
8.3. PROPERTIES OF HARMONIC FUNCTIONS 191
2. If U is not connected, i.e., U = U1
[U2
with U1
, U2
disjoint open sets in Rn, then the weak
maximum principle holds but the strong maximum principle breaks down. To see this, define
u(x) = k, for x 2 Uk, k = 1, 2. Then u(x) is harmonic, but fails to satisfy either conclusion of
part (2).
We can apply the maximum principle to theDirichlet Problem, which is the following boundary
value problem on a bounded open set U ⇢ Rn :
(3.6)�u(x) = 0, x 2 U
u(x) = g(x), x 2 @U.
Theorem 8.4. Suppose g is continuous and u 2 C2(U) \ C(U) is a solution of the Dirichlet
problem. If U is connected and g satisfies: g(x) � 0 for all x 2 @U, and g(x) > 0, for some
x 2 @U, then
u(x) > 0 for all x 2 U.
Proof: From the weak minimum principle, we have min@U u = min@U g. But the strong
version gives either u(x) > min@U g, for all x 2 U, or u(x) = constant. In either case, the
conclusion follows.
8.3.1. Uniqueness of Solutions of Boundary Value Problems. As with the heat
equation, we can prove uniqueness of solutions of boundary value problems from the maximum
principle, or from energy considerations. Let U ⇢ Rn be open and bounded. Consider the
192 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
boundary value problem with Dirichlet boundary conditions:
(3.7)
��u = f in U
u = g on @U,
where f 2 C(U), g 2 C(@U).
Theorem 8.5. There is at most one solution u 2 C2(U)\C(U) of the boundary value problem
(3.7).
Proof: Let u1
, u2
be solutions. Applying the weak maximum principle to u1
� u2
and to
u2
� u1
, both of which satisfy (3.7) with f = 0; g = 0, we deduce that u1
= u2
.
Alternatively, the energy approach sets u = u1
� u2
and applies a version of Green’s identity:
0 =
Z
U
u�u dx =
Z
@U
u@u
@⌫dS �
Z
U
|ru|2 dx.
Thus, ru = 0 in U, so that u is constant. Since u = 0 on @U, we conclude that u = 0 in U,
so that u1
= u2
.
8.4. Separation of Variables for Laplace’s Equation.
If the domain U ⇢ Rn has special geometry, then separation of variables can work on Laplace’s
equation. Examples include rectangular domains, spheres and cylinders. Here, we treat two
examples to illustrate di↵erences from the heat and wave equations, and then make some
remarks about other domains.
8.4. SEPARATION OF VARIABLES FOR LAPLACE’S EQUATION. 193
Of course, if the boundary conditions are homogeneous, then u = 0 is a solution, generally the
only solution. So, it is more natural to consider linear boundary conditions ↵u+ �@u
@⌫= g on
@U that are non-homogeneous over at least part of the boundary. If ↵ 6= 0, then the energy
method above can be used to prove that this problem has at most one solution.
8.4.1. Laplace’s equation in a rectangle. We consider Laplace’s equation �u = 0 in a
rectangular domain U = (0, a)⇥ (0, b) ⇢ R2 with a mixture of Dirichlet and Neumann bound-
ary conditions, representing parts of the boundary where we specify either the temperature u
or the heat flux, which is proportional to the normal derivative.
y
x
∆u = 0
b
0 a
νu f=
u = fu = f
u = f
1
2
3
4
Figure 8.2. Example of a boundary value problem for Laplace’s equation.
In this problem, we can formulate an eigenvalue problem if boundary conditions on opposite
sides of the rectangular boundary are homogeneous. We use this observation to implement
a solution strategy. We split the boundary value problem into four problems, setting the
boundary condition to zero on three sides in each problem. To illustrate the process, consider
194 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
the example of Figure 8.2. For j = 1, 2, 3, 4, let (Pj) be the problem with fk = 0, k 6= j, and
let uj be the solution of (Pj) Then by linearity, the solution of the full problem is
u = u1
+ u2
+ u3
+ u4
.
We solve (P4
) in detail to illustrate this approach.
Let u = u4
= v(x)w(y). From the boundary conditions, we guess v(x) = sin n⇡xa, so that
u4
(x, y) = sin n⇡xa
w(y). Then �u(x, y) = 0 leads to an ODE for w(y) :
w00(y)�⇣n⇡
a
⌘2
w(y) = 0,
with general solution
wn(y) = An coshn⇡y
a+Bn sinh
n⇡y
a.
Now we need to satisfy a homogeneous boundary condition at y = b,
@u
@⌫(x, b) = 0 : w0
n(0) =n⇡
a
✓An sinh
n⇡b
a+Bn cosh
n⇡b
a
◆= 0.
Thus,
Bn = � tanhn⇡b
aAn.
Now we can form a series to satisfy the nonzero boundary condition at y = 0 :
(4.8) u4
(x, y) =1X
n=1
An sinn⇡x
a
✓cosh
n⇡y
a� tanh
n⇡b
asinh
n⇡y
a
◆.
On y = 0,
u(x, 0) = f4
(x) =1X
n=1
An sinn⇡x
a,
8.4. SEPARATION OF VARIABLES FOR LAPLACE’S EQUATION. 195
from which we get formulae for the coe�cients An :
An = 2
a
R a
0
f4
(x) sin n⇡xa
dx.
The solution u4
is then given by the series (4.8). Similarly, we can obtain u1
, u2
, u3
, and finally
put these series together to get the solution u of the original problem. Note that the series
for u2
is a sine series like (4.8), but the series for u1
and u3
have the form
u(x) =P1
n=0
vn(x) sin(n+ 1
2
)⇡yb
because of the combination of homogeneous boundary conditions at y = 0, b.
8.4.2. Laplace’s equation on spherical and cylindrical domains. In spherical and
cylindrical domains, it is natural to use curvilinear coordinates, i.e., polar coordinates and
cylindrical coordinates, respectively. Since the Laplacian in these coordinates has non-constant
coe�cients, the ODE’s that result will also have non-constant coe�cients. Moreover, the
dimension of the space makes a di↵erence to the type of equation that results. This leads to the
study of special functions, specifically Legendre functions (solutions of Legendre’s equation)
and Bessel functions (solutions of Bessel’s equation). These special functions are typically
expressed as series solutions of ODEs, using the method of Frobenius. Some details may be
found in the PDE book of Strauss [45], in Engineering Mathematics books, such as Je↵ery
[26], Kreyszig [30], and in texts typically named PDEs and Boundary Value Problems.
Here, we give the detailed solution of Laplace’s equation in a disk, leading to Poison’s formula,
a representation of the solution as an integral, which we eventually interpret in terms of
196 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
Green’s functions. The disk has the advantage that we do not need special functions to solve
the eigenvalue problem.
Consider the Dirichlet problem in a disk of radius a > 0 :
(4.9)�u = 0 in U = B(0, a) ⇢ R2
u = f on @U
In plane polar coordinates,
x = r cos ✓; y = r sin ✓,
we have the problem for u = u(r, ✓) :
(4.10)
urr +1
rur +
1
r2u✓✓ = 0 0 ✓ 2⇡, 0 < r < a
u(a, ✓) = f(✓), 0 ✓ 2⇡.
The boundary @U is the circle r = a, whereas the boundaries for the variables r, ✓ also include
r = 0, ✓ = 0, 2⇡. At the disk center, r = 0, we exclude non-physical solutions by insisting
that solutions remain bounded as r ! 0+. The boundaries ✓ = 0, 2⇡ represent the same line
within the disk, across which the solution should be as smooth as elsewhere in the domain.
These boundaries are accommodated by making the solution 2⇡ periodic in ✓. Similarly, the
boundary function f is treated as a 2⇡ periodic function of ✓.
Let
u(r, ✓) = R(r)H(✓).
8.4. SEPARATION OF VARIABLES FOR LAPLACE’S EQUATION. 197
Substituting into the PDE, we obtain
r2(R00(r) + 1
rR0(r))
R(r)+
H 00(✓)
H(✓)= 0,
whence, separating r from ✓,
H 00(✓)
H(✓)= ��;
r2(R00(r) + 1
rR0(r))
R(r)= �.
We then have an eigenvalue problem for H, in which the boundary condition is that H(✓) is
2⇡ periodic:
(4.11) H 00(✓) + �H(✓) = 0, H(0) = H(2⇡).
The corresponding equation for R(r) is
(4.12) r2R00(r) + rR0(r) = �R(r).
We can solve the eigenvalue problem (4.11), with the result H = H0
= A0
/2 = constant, for
� = 0, and
(4.13) H = Hn(✓) = An cosn✓ +Bn sinn✓, n = 1, 2, ..., with � = �n = n2.
Note that each eigenvalue �n, n � 1, has two independent eigenfunctions.
Setting � = n2 in (4.12), we get
(4.14) r2R00(r) + rR0(r)) = n2R(r), 0 < r < a.
For n = 0, the general solution of (4.14) is
R(r) = C0
+D0
log r.
198 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
However, we seek solutions that are bounded as r ! 0, so we set D0
= 0, and consider only
the solution
R = R0
(r) = 1.
The arbitrary multiple C0
will be incorporated into A0
.
For n � 1, we seek solutions R(r) = r↵. Substituting into (4.14), we find ↵ = ±n. However,
r�n is unbounded at the origin, so we retain only
Rn = rn, n � 1.
Again, the arbitrary coe�cient multiplying this solution will be incorporated into Hn(✓).
So far, we have solutions
u0
(r, ✓) =A
0
2; un(r, ✓) = rn(An cosn✓ +Bn sinn✓), n � 1.
These functions are harmonic in the ball B(0, a), and they reduce to functions of ✓ alone for
r = a.
We form a series u =P1
n=1
un:
(4.15) u(r, ✓) =A
0
2+
1X
n=1
rn(An cosn✓ +Bn sinn✓),
and set r = a to satisfy the boundary condition u(a, ✓) = f(✓). Thus,
f(✓) =A
0
2+
1X
n=1
an(An cosn✓ +Bn sinn✓).
8.4. SEPARATION OF VARIABLES FOR LAPLACE’S EQUATION. 199
Consequently, anAn, anBn are Fourier coe�cients of the 2⇡ periodic function f :
(4.16)
0
BB@An
Bn
1
CCA =1
an⇡
Z2⇡
0
f(✓)
0
BBB@
cosn✓
sinn✓
1
CCCAd✓, n � 0.
If we substitute the coe�cients given by (4.16) back into the series (4.15), we get the solution
u in terms of the data, and we can sum the series, just as we summed the series to get the
Dirichlet kernel in proving pointwise convergence. Thus,
u(r, ✓) =1
2⇡
Z2⇡
0
f(�)(1 + 21X
n=1
⇣ra
⌘n
(cosn✓ cosn�+ sinn✓ sinn�)) d�
=1
2⇡
Z2⇡
0
f(�)(1 + 21X
n=1
⇣ra
⌘n
cosn(✓ � �)) d�.
After some manipulation of the sum of the geometric seriesP1
n=�1�ra
�nein(✓��), we obtain
Poisson’s Formula for the solution of the Dirichlet problem in a disk:
(4.17) u(r, ✓) =1
2⇡
Z2⇡
0
a2 � r2
a2 � 2ar cos(✓ � �) + r2f(�) d�.
This integral has the form of a convolution product of the Poisson kernel
P (r, ) =1
2⇡
a2 � r2
a2 � 2ar cos + r2
with the boundary data f( ) = u(a, ). The formula reduces to the mean value property of
harmonic functions when r = 0. In the special case f ⌘ 1, we have the solution u = 1 from
which we conclude thatZ
2⇡
0
P (r, ) d = 1.
Note that you could also guess this by integrating the series term-by-term.
200 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
Just as for fundamental solutions, which are singular integral kernels, the Poisson kernel,
P (r, ✓��) is singular at the very place the function u(r, ✓) is to be evaluated on the boundary:
r = a, ✓ = �. The singularity is needed in order for the convolution to converge to the boundary
data: for f continuous,
lim(r,✓)!(a,�)
u(r, ✓) = f(�).
A more geometric interpretation of Poisson’s formula generalizes to higher dimensions. Con-
sider polar coordinates for
x = (r cos ✓, r sin ✓) 2 B(0, a), and x0 = (a cos�, a sin�) 2 @B(0, a).
ϕ
θ
x
0
x'
a
r
x - x'||
Figure 8.3. Geometric Interpretation of Poisson’s Formula.
Then (see Fig. 8.3) a2 � r2 = |x0|
2
� |x|2, and |x0� x|2 = r2 + a2 � 2ar cos(✓ � �).
Thus,
u(x) = �
Z
|x0|=a
|x0|
2
� |x|2
|x0� x|2
u(x0) ds(x0), |x| < a.
8.4. SEPARATION OF VARIABLES FOR LAPLACE’S EQUATION. 201
The Poisson kernel is an example of a Green’s function, which we study in detail in the next
chapter.
Problems
1. Prove the weak maximum principle using an argument similar to the proof for the heat
equation.
2. Prove the weak maximum principle from the strong form.
3. Consider Poisson’s equation on a bounded open set U 2 Rn with Robin boundary condi-
tions:
�u(x) = f(x), x 2 U,@u
@⌫(x) + ↵u(x) = g(x), x 2 @U.
(a) Prove that if ↵ > 0, then the energy method can be used to show uniqueness of solutions
u 2 C2(U) \ C(U).
(b) For ↵ = 0, show that solutions are unique up to a constant.
(c) Design an example to show that uniqueness can fail if ↵ < 0. (Hint: Choose n = 1.)
4. Derive Poisson’s formula (4.17) by summing the series for u(r, ✓). Provide the details.
5. In Rn let Vr = |B(0, r)| = ↵(n)rn, Sr = |@B(0, r)|. Explain why
Vr =r
nSr.
6. Suppose u 2 C2(U) has the mean value property:
202 8. LAPLACE’S EQUATION AND POISSON’S EQUATION
For all x 2 U, u(x) = �R@B(x,r)
u(y) dSy for all r > 0 such that B(x, r) ⇢ U.
Write a careful proof by contradiction that �u = 0 in U.
7. Suppose U ⇢ Rn is open, bounded and connected, and u 2 C2(U) \ C(U) satisfies
�u = 0 in U, u|@U = g.
Prove: If g 2 C(@U), g(x) � 0 for all x, and g(x) > 0 for some x 2 @U, then
u(x) > 0 for all x 2 U.