Laplace’s Equation
c 2014 by Philip D. Loewen
UBC M257/316 Lecture Notes A. The Laplacian
For a single-variable function u = u(x), u′ (x) measures slope and u′′ (x) measures
concavity or curvature. When u = u(x, y) depends on two variables, the gradient (a
vector) and the Laplacian (a scalar) record the corresponding quantities:
∇u(x, y) = (ux (x, y), uy (x, y)) ,
∆u(x, y) = uxx (x, y) + uyy (x, y).
(the gradient)
(the Laplacian).
The Laplacian operator ∆ is important enough to deserve an intuitive understanding
on its own. However, we’ll still think of it as being somehow related to concavity/curvature. It certainly does the same job in problems of wave motion or heat
flow in 2D domains, and similar extensions hold in 3 or more dimensions.
Wave Equation. For a 2D membrane stretched across a wire frame around a region
Ω in the (x, y)-plane (like a drum head), the lateral displacement at point (x, y) and
time t is a function u = u(x, y, t) that obeys
utt = c2 (uxx + uyy ) = c2 ∆u,
(x, y) ∈ Ω, t > 0.
(“Curvature drives acceleration.”) [Sketch something.]
Heat Equation. For a 2D metal plate occupying the plane region Ω, sandwiched
between insulation slabs on its flat sides so heat can only flow in the (x, y)-plane, the
temperature at point (x, y) and time t is a function u = u(x, y, t) that obeys
ut = α2 (uxx + uyy ) = α2 ∆u,
(x, y) ∈ Ω, t > 0
(“Curvature drives flow rate.”)
Laplace’s Equation. Given an open set Ω in the (x, y)-plane, Laplace’s Equation
for u = u(x, y) is
def
0 = ∆u = uxx + uyy ,
(x, y) ∈ Ω.
(∗)
Applications:
(1) Steady-state temperature in a 2D region with fixed boundary temperatures.
(The case of a 1D region is easy: u′′ (x) = 0 implies u(x) = mx + c. For a higherdimensional region, however, a much greater variety of solutions is possible.)
(2) Potential fields (electrostatic, gravitational, etc.) in regions free from potential
sources (charges, masses, etc.) obey Laplace’s equation.
(3) Minimal surfaces (e.g., soap films in equilibrium) obey the nonlinear PDE
(1 + u2y )uxx + 2ux uy uxy + (1 + u2x )uyy = 0,
(x, y) ∈ Ω.
When ∇u is so small that terms of second and higher order in its components
are negligible, this is approximated by Laplace’s equation.
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PHILIP D. LOEWEN
Boundary Conditions. Write Γ for the boundary curve of Ω.
(i) Dirichlet (prescribed function values): a function g is given, and we seek a
function u obeying (∗) and
u(x, y) = g(x, y)
for (x, y) ∈ Γ.
(ii) Neumann (prescribed directional derivatives normal to boundary): a function h
is given and we seek u satisfying (∗) and
b
∇u(x, y) • N(x,
y) = h(x, y) for (x, y) ∈ Γ.
b
Here N(x,
y) is the outward unit normal to the curve Γ at point (x, y).
(iii) Mixed (Dirichlet on some segments of Γ, Neumann on the rest).
B. Dirichlet Problem in a Rectangular Box
Consider Ω = {(x, y) : 0 < x < a, 0 < y < b}. Here the boundary curve Γ consists
of the four segments on the sides of the rectangle, and BC’s of Dirichlet type can be
drawn right onto the picture:
y
u(x,b) = f1(x)
u(0,y) = g0(y)
∆u=0
u(a,y)=g1(y)
u(x,0)=f0(x)
x
In detail, functions f0 , f1 , g0 , and g1 are given, and we seek a function u obeying
both ∆u = 0 in Ω and
u(x, 0) = f0 (x),
u(x, b) = f1 (x),
0<x<a
0<x<a
(bottom),
(top),
u(0, y) = g0 (y),
u(a, y) = g1 (y),
0 < y < b (left),
0 < y < b (right).
Trivial Case. If f0 ≡ f1 ≡ g0 ≡ g1 ≡ 0, what’s u? Physical intuition and mathematical outcome agree: u ≡ 0.
Simplest Nontrivial Case. All but one boundary function are zero. Suppose
f1 (x) 6≡ 0, whereas f0 ≡ g0 ≡ g1 ≡ 0. Follow our usual 6-step process.
1: Splitting. Not required here.
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Laplace’s Equation
3
2: Eigenfunctions. Look for simple nontrivial solutions in product for u(x, y) =
X(x)Y (y). Substitute into PDE/BC, remembering that separation of variables is
worse than futile (it’s misleading) on nonhomogeneous conditions. So only three
BC’s give useful info, namely,
Y (0) = 0 (from the bottom),
X(0) = 0 (from the left),
X(a) = 0 (from the right).
In PDE, substitution leads to
0 = X ′′ (x)Y ′′ (y) + X(x)Y ′′ (y) ⇐⇒
X ′′ (x)
Y ′′ (y)
=−
= −λ
X(x)
Y (y)
for some separation constant λ. Since we have a pair of BC’s for X and only one for
Y , it is the X-component that provides a well-formed eigenvalue problem:
X ′′ (x) + λX(x) = 0,
0 < x < a;
X(0) = 0 = X(a).
The corresponding eigenfunctions are well known (FSS):
nπx
, n = 1, 2, . . . .
Xn (x) = sin
a
3: Superposition. Separation has done all it can for us. The solution we expect
will not have simple separated form, but rather combine all possible product-form
solutions found above like this:
∞
X
nπx
u(x, y) =
Yn (y) sin
.
(∗∗)
a
n=1
Finding those coefficient functions Yn (y) will complete the solution.
4: Auxiliary Information (formerly “Initialization”). Any series of form (∗∗)
will satisfy the BC’s on the left and right sides. On the bottom, we need
∞
X
nπx
0 = u(x, 0) =
Yn (0) sin
.
a
n=1
Here is a FSS expansion for the zero function, giving
Yn (0) = 0,
n = 1, 2, . . . .
(†)
On the top, we need
nπx
.
Yn (b) sin
f1 (x) = u(x, b) =
a
n=1
∞
X
This is a FSS expansion for the function f1 . Standard coefficient formulas give
Z
2 a
nπx
Yn (b) =
f1 (x) sin
dx.
a 0
a
File “2014notes”, version of 5 August 2014, page 3.
(‡)
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PHILIP D. LOEWEN
5: Propagation. Plug series into PDE to get
" # ∞
2
X
nπ
nπx
′′
0 = uxx + uyy =
−
.
Yn (y) + Yn (y) sin
a
a
n=1
This is a FSS expansion on 0 < x < a, with coefficients independent of x, for the
zero function. Again it requires that all coefficients must vanish, i.e.,
2
nπ
′′
Yn (y) = 0,
n = 1, 2, . . . .
Yn (y) −
a
Guess Yn = esy and plug in to get s = ±
nπ
, and hence the general solution
a
Yn (y) = An enπy/a + Bn e−nπy/a .
Now use (†) to get 0 = Yn (0) = An + Bn , so Bn = −An ,
h
i
Yn (y) = An enπy/a − e−nπy/a .
Plug in y = b and appeal to (‡) to get
2
Yn (b)
= nπb/a
An = nπb/a
−nπb/a
e
−e
a e
− e−nπb/a
Z
0
a
nπx
dx.
f1 (x) sin
a
6: Conclusion. The series solution we want is
∞
i nπx h
X
nπy/a
−nπy/a
,
sin
u(x, y) =
An e
−e
a
n=1
with constants An given in terms of f1 by the integral formula above. This appearance
is typical: a sum of products, with one factor being an eigenfunction and the other
some kind of exponential.
////
Eigenfunction Setup. Notice that the key to Step 2 above was the presence of
homogeneous BC’s on the opposite faces (left and right) of the domain Ω. These
gave us the FSS eigenproblem encoded in the series solution; the homogeneous BC
on the bottom edge helped us only in Step 4, and a nonhomogeneous one could have
been handled in Step 4 with very little extra work.
RTFT. In the textbook by Trench, please read Section 12.3 and try problems 5, 9,
19, 31, 34.
Symmetry Argument #1. Suppose u = u(x, y) obeys Laplace’s Equation in the
given rectangle. Define v = u(x, b − y). Notice that
vy (x, y) = −uy (x, b − y),
vyy (x, y) = (−1)2 uyy (x, b − y) = uyy x, b − y,
while vxx (x, y) = uxx (x, b − y). Now when 0 < y < b, of course 0 < b − y < b also, so
we get
vxx (x, y) + vyy (x, y) = uxx (x, b − y) + uyy (x, b − y) = 0.
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Laplace’s Equation
5
Meanwhile,
v(x, 0) = u(x, b) = f1 (x),
v(x, b) = u(x, 0) = f0 (x) = 0,
v(0, y) = u(0, b − y) = g0 (b − y) = 0,
v(a, y) = u(a, b − y) = g1 (b − y) = 0.
Now the function u is known, so v is too, and v solves a problem very similar to
the u-problem except that it has a nontrivial temperature on the bottom edge of
Ω. [Sketch a pictorial representation for the v-problem.] Putting this back into the
notation of the original u-problem lets us cover the case where g0 ≡ f1 ≡ g1 ≡ 0 but
f0 6≡ 0:
h
i nπx nπ(b−y)/a
−nπ(b−y)/a
u(x, y) =
An e
−e
sin
,
a
n=1
Z a
nπx
2
f0 (x) sin
dx.
where
An = nπb/a
a
a e
− e−nπb/a 0
∞
X
Symmetry Argument #2. Switching letters x ↔ y, a ↔ b, f ↔ g changes the
appearance but not the validity of the solution. (Physically it makes sense too: we’re
just flipping our metal plate with its steady temperature distribution along the axis
y = x.) This swap leaves the PDE unchanged, but gives a BC where it’s g1 (y) that
is nontrivial. So for the case f0 ≡ g0 ≡ f1 ≡ 0, the solution is
nπy
,
sin
u(x, y) =
Bn e
−e
b
n=1
Z a
2
nπy
Bn = nπa/b
g1 (y) sin
dy.
b
b e
− e−nπa/b 0
∞
X
h
nπx/b
−nπx/b
i
(5)
Approaching this problem directly, we would separate variables as shown above in
the PDE, but the separated BC’s would lead to the homogeneous conditions Y (0) =
0 = Y (b). So in this situation, the eigenvalue problem of interest would involve the
unknowns Y :
Y ′′ (y) − λY (y) = 0, 0 < y < b;
Y (0) = 0 = Y (b).
nπy
The natural series form to postulate would then be u(x, y) =
Xn (x) sin
,
b
n=1
and this is precisely what we see in line (5) above.
∞
X
Practice. Without lengthy calculation, find a series solution formula for the case
where f1 ≡ g1 ≡ f0 ≡ 0 but g0 6≡ 0.
Splitting and Superposition. To handle arbitrary f0 , f1 , g0 , and g1 , consider four
subproblems of the form above, each with one nonzero boundary function. Write a
series solution for each subproblem, then add them up.
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PHILIP D. LOEWEN
C. Laplace’s Equation in Polar Coordinates (Pizza Problems)
In standard polar coordinates, where
x = r cos θ,
y = r sin θ,
the Laplacian of a given function u = u(r, θ) is
1
1
∆u = urr + ur + 2 uθθ .
r
r
p
[Proof idea: Define U (x, y) = u( x2 + y 2 , tan−1 (y/x)), compute ∆U = Uxx + Uyy
with chain rule, express result in terms of r and θ.] Hence Laplace’s equation is
equivalent to
0 = r 2 ∆u = r 2 urr + rur + uθθ .
Separation of this PDE with u(r, θ) = R(r)Θ(θ) leads to
r 2 R′′ (r) + rR′ (r)
Θ′′ (θ)
=−
=σ
R(r)
Θ(θ)
for some separation constant σ, and this produces two linked ODE problems:
(1)
r 2 R′′ (r) + rR′ (r) − σR(r) = 0,
(2)
Θ′′ (θ) + σΘ(θ) = 0.
Depending on what sort of BC’s are present, either of these could be the ODE of
choice in an eigenvalue problem.
Example (A). Pizza slice with first bite gone: polar region
Ω = {(r, θ) : a < r < b, 0 < θ < α} .
u(r
,α)
=0
u(b,θ)=g(θ)
∆u=0
u(a,θ)=f(θ)
u(r,0)=0
Here α ∈ (0, 2π) and a > 0, b > a are some preassigned constants. If the BC’s on
the flat sides are homogeneous, i.e.,
u(r, 0) = 0,
u(a, θ) = f (θ),
u(r, α) = 0,
u(b, θ) = g(θ),
a < r < b,
0 < θ < α,
then separation of variables in the BC’s leads to
Θ(0) = 0 = Θ(α),
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Laplace’s Equation
7
so equation (2) participates in the eigenvalue problem
Θ′′ (θ) + σΘ(θ) = 0, 0 < θ < α,
Θ(0) = 0 = Θ(α).
nπθ
Thus we get FSS eigenfunctions Θn (θ) = sin
, and a series-form solution
α
∞
X
nπθ
u(r, θ) =
Rn (r) sin
.
α
n=1
The BC’s give
Z
nπθ
2 α
nπθ
f (θ) sin
, so Rn (a) =
dθ,
f (θ) = u(a, θ) =
Rn (a) sin
α
α
α
0
n=1
Z
∞
X
nπθ
2 α
nπθ
Rn (b) sin
g(θ) = u(b, θ) =
, so Rn (b) =
g(θ) sin
dθ.
α
α
α
0
n=1
∞
X
Plugging the series into the PDE leads to
"
# 2
∞
X
nπ
nπθ
2
2 ′′
′
0 = r urr + rur + uθθ =
r Rn + rRn −
,
Rn sin
α
α
n=1
whence
0=r
2
Rn′′
+
rRn′
−
nπ
α
2
Rn .
This equation has Euler type: a function Rn (r) = r s gives a solution iff
2
nπ
nπ
s(s − 1) + s −
= 0, i.e., s = ± .
α
α
So the general solution is
Rn (r) = An r nπ/α + Bn r −nπ/α ,
An , Bn ∈ R.
When functions f , g are given in detail, the right-hand sides in the system below are
known constants, and it is possible to solve for An , Bn :
Z
nπθ
2 α
nπ/α
−nπ/α
f (θ) sin
dθ,
An a
+ Bn a
= Rn (a) =
α 0
α
Z
nπθ
2 α
nπ/α
−nπ/α
g(θ) sin
dθ.
An b
+ Bn b
= Rn (b) =
α 0
α
The series solution is then
∞ h
i nπθ X
nπ/α
−nπ/α
sin
u(r, θ) =
An r
+ Bn r
.
α
n=1
Consider same region with other BC’s later (see (D) below).
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PHILIP D. LOEWEN
Example (B). Pizza slice before first bite taken: polar region
Ω = {(r, θ) : 0 < r < b, 0 < θ < α} .
u(r
,α)
=0
u(b,θ)=g(θ)
∆u=0
u(r,0)=0
This is the limiting case a → 0+ of (1); we use the same boundary conditions.
Mathematically, there is no place for the function f (θ) describing u-values on the
curved inner boundary shown in problem (A), so we don’t have enough information
to determine both constants An and Bn above. However, problems of physical interest
typically have bounded solutions. The requirement that u(r, θ) behave well near
the boundary (which includes the origin, where r = 0) forces us to choose all Bn = 0,
so the solution simplifies to
u(r, θ) =
∞
X
n=1
An r
nπ/α
nπθ
sin
α
,
2
An = n
b α
Z
α
0
nπθ
g(θ) sin
dθ.
α
Example (B′ ). Infinite pizza slice missing one bite. Practice: how should the
solution shown in part (A) be modified in the limiting case b → +∞?
Example (C). Annulus. Solve ∆u = 0 in the polar region Ω = {(r, θ) : a < r < b},
where a > 0 and b > a are given constants and
u(a, θ) = f (θ),
u(b, θ) = g(θ),
−π < θ < π.
u(b,θ)=g(θ)
∆u=0
f(θ)
Main Idea: Periodicity implicit in the polar coordinate representation provides boundary conditions not written explicitly in the problem statement, namely,
u(r, −π) = u(r, π),
uθ (r, −π) = uθ (r, π),
File “2014notes”, version of 5 August 2014, page 8.
a < r < b.
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Laplace’s Equation
9
Separating u(r, θ) = R(r)Θ(θ) provides two pieces of boundary information about Θ,
so we focus the ODE for Θ in line (2) above. This produces an eigenvalue problem
for Θ:
Θ′′ (θ) + λΘ(θ) = 0, −π < θ < π;
Θ(−π) = Θ(π), Θ′ (−π) = Θ′ (π).
This is a standard problem: we know that its eigenfunctions are precisely the set
of all basis functions associated with the Full Fourier Series on [−π, π]. Hence we
postulate a solution of the form
u(r, θ) =
1
2 A0 (r)
+
∞
X
[An (r) cos(nθ) + Bn (r) sin(nθ)] ,
n=1
for some coefficient functions An and Bn to be determined.
To see how the coefficients evolve with r, plug that series into Laplace’s Equation:
0 = r 2 urr + rur + uθθ
=
1
2
∞
2 ′′
X
2 ′′
′
r A0 (r) + rA0 (r) +
r An (r) + rA′n (r) − n2 An (r) cos(nθ)
n=1
+
∞
X
n=1
r 2 Bn′′ (r) + rBn′ (r) − n2 Bn (r) sin(nθ)
In this full Fourier series for the zero function, all coefficients must be 0. For case
n ≥ 1, this gives a pair of identical Euler-type equations. Exactly as in problem (A)
above, we find the general solutions
An (r) = an r n + bn r −n ,
Bn (r) = cn r n + dn r −n ,
an , bn , cn , dn ∈ R.
For case n = 0, a shrewd observation gives
!
0 = r 2 A′′0 (r) + rA′0 (r) = r
d
(rA′0 (r)) .
dr
Hence rA′0 (r) = b0 for some b0 ∈ R, and this implies
A′0 (r) =
b0
=⇒ A0 (r) = a0 + b0 ln(r), a0 , b0 ∈ R.
r
Thus we may express our series solution as
u(r, θ) =
1
2
∞
X
[a0 + b0 ln(r)] +
an r n + bn r −n cos(nθ)
+
n=1
∞
X
n=1
n
cn r + dn r −n sin(nθ).
(1)
Now using standard coefficient formulas on the BC’s
f (θ) = u(a, θ),
g(θ) = u(b, θ)
will give 2 × 2 systems of linear equations to solve for (an , bn ) and (cn , dn ). [Try it!]
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PHILIP D. LOEWEN
Example (C′ ). Steady temperature in a disk of radius b > 0, with given boundary
temperature g(θ). This is the limiting case a → 0+ of (C) above. Again we must
apply the boundedness requirement imposed by the physical interpretation (steady
temperature). This takes the place of the prescribed temperature f on the inner
edge; now boundedness requires bn = 0, dn = 0 for all n. (Please think about b0
separately.) The result is
u(r, θ) = 12 a0 +
∞
X
r n (an cos(nθ) + cn sin(nθ)) .
n=1
Now the BC u(b, θ) = g(θ) gives
Z
1 π
n
b an =
g(θ) cos(nθ) dθ,
π −π
Z
1 π
n
b cn =
g(θ) sin(nθ) dθ,
π −π
n = 0,1, 2, 3, . . . ,
n = 1, 2, 3, . . . .
• Averaging Property: Midpoint temperature is average of boundary temperatures. (Proof. Plug in r = 0, recall definition of Fourier Coefficient a0 .)
Important Consequence: For any solution u of Laplace’s equation, in any 2D
domain Ω, there can be no local min and no local max for u in interior of Ω.
def
Reason: Suppose P0 = (x0 , y0 ) is a point where a local min occurs. Choose a
little disk with centre P0 where all the boundary values are higher than u(P0 ).
Then ∆u = 0 in that disk, and the averaging property just proved is violated.
This can’t happen.
Physics: Steady temperature in a 2D region can’t have isolated local extrema.
This makes sense—“hot spots” would be unstable. Likewise for a soap-film
stretched between curved wires—simple bumps get pulled down by surface tension.
• Poisson Kernel Formula (Optional): Plug integral coefficients straight into
the series and interchange sum and integral to get the following:
"
Z
∞ n Z π
X
1 1 π
r
1
u(r, θ) = ·
g(t) dt +
g(t) cos(nt) dt cos(nθ)
2 π −π
b
π
−π
n=1
#
Z π
1
+
g(t) sin(nt) dt sin(nθ)
π −π
"
#
Z
∞ n
1 X r
1 π
g(t)
[ cos(nt) cos(nθ) + sin(nt) sin(nθ)] dt
+
=
π −π
2 n=1 b
"
#
Z
∞ n
1 π
1 X r
=
g(t)
cos(n(t − θ)) dt
+
π −π
2 n=1 b
"
#
Z
∞ n
1 X r
1 π
g(t) − +
cos(n(t − θ)) dt
=
π −π
2 n=0 b
File “2014notes”, version of 5 August 2014, page 10.
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Laplace’s Equation
11
For real x, y with |x| < 1, geometric series calculation shows
∞
X
n
x cos(ny) =
n=0
∞
X
n iny
ℜe x e
n=0
= ℜe
= ℜe
∞
X
n=0
iy
1
1 + xe
·
iy
1 − xe
1 + xeiy
xeiy
=
n
1 − x cos(y)
.
1 − 2x cos(y) + x2
Therefore
∞
X
xn cos(ny) −
n=0
1
1 − x cos y
1 1 − 2x cos y + x2
=
−
2
1 − 2x cos y + x2
2 1 − 2x cos y + x2
=
1
1 − x2
2 1 − 2x cos y + x2
Use calc above with x = r/b and y = t − θ to get a famous and useful formula:
u(r, θ) =
Z
π
−π
b2 − r 2
1
g(t) dt,
2π b2 − 2br cos(t − θ) + r 2
0 < r < b, θ ∈ R.
1
b2 − r 2
Defining P (~z, t) =
makes this formula look like matrix2π b2 − 2br cos(t − θ) + r 2
vector multiplication:
Z π
u(~z) =
P (~z , t)g(t) dt.
(∗)
−π
Interpretation: Let Y be the set of all integrable 2π-periodic functions g = g(θ).
Let X be the collection of all well-behaved functions u = u(x, y) that satisfy
Laplace’s Equation on r < b. Define a linear operator A: X → Y like this:
A[u] = g ⇐⇒ g(θ) = lim− u(r, θ).
r→b
E.g., if u(r, θ) = r cos θ then g = A[u] is the function
g(θ) = b cos θ,
−π < θ < π.
It’s easy to find g once u is given, but the really interesting problem is usually
just the opposite, namely, solve for u in A[u] = g. Ideally, we would like to have
an “inverse” for A, so we could just write
u = A−1 [g].
Formula (∗) does exactly this, with P replacing A−1 . (Analogy: For n-dimensional
Z π
n
X
vectors, u = P g iff uk =
Pkt gt ; for functions, u(~z ) =
P (~z, t)g(t) dt.)
t=1
−π
Froese’s notes are a good source for this. (See link on course home page.)
File “2014notes”, version of 5 August 2014, page 11.
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12
PHILIP D. LOEWEN
Example (D). Standard bitten slice, different boundary conditions. Again we solve
∆u = 0 in the polar region
Ω = {(r, θ) : 1 < r < b, 0 < θ < α}
u(b,θ)=0
u(r
,α)
=k
(r)
where a = 1, b > 1 and α ∈ (0, 2π) are given.
∆u=0
u(a,θ)=0
u(r,0)=h(r)
This time, however, nonhomogeneous boundary data are given on the flat sides of
the domain:
u(r, 0) = h(r) u(r, α) = k(r)
u(1, θ) = 0
u(b, θ) = 0
1 < r < b,
0 < θ < α.
[Shortcut: If both given functions h and k are constant, careful choices of the constants A and B together with the substitution u(r, θ) = A + Bθ + w(r, θ) will reduce
this problem to an instance of (A) above. But if one of these functions is nonconstant,
the method shown below seems inevitable.]
A new eigenvalue problem. Separating u(r, θ) = R(r)Θ(θ) in the homogeneous
BC gives
R(1)Θ(θ) = 0 = R(b)Θ(θ),
i.e.,
R(1) = 0 = R(b).
These homogeneous conditions on R force us to build our eigenvalue problem using
the ODE in line (1) of the separation-of-variables result above. We arrive at this
eigenvalue problem for the function R:
r 2 R′′ (r) + rR′ (r) + λR(r) = 0,
1 < r < b;
R(1) = 0 = R(b).
(11)
This is not a FSS problem, because the ODE has form different from the one familiar
so far. To find eigenfunctions will take grinding case-by-case analysis. Since the ODE
has Euler type, guess R(r) = r p and plug in:
r 2 p(p − 1)r p−2 + r pr p−1 + λr p = 0 ⇐⇒ p2 + λ = 0.
• Case λ < 0: Write λ = −s2 for some s > 0 and get p = ±s, so the general
solution is
R(r) = Ar s + Br −s ,
A, B ∈ R.
File “2014notes”, version of 5 August 2014, page 12.
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Laplace’s Equation
13
Now 0 = R(1) = A + B gives B = −A, so R(r) = A(r s − r −s ), and 0 = R(b) =
A(bs − b−s ) forces A = 0 (since bs > 1 > b−s ). Thus only trivial solutions appear
when λ < 0.
• Case λ = 0: Repeated roots, so
R(r) = A + B ln(r),
A, B ∈ R.
Now 0 = R(1) = A gives R(r) = B ln(r) and 0 = R(b) = B ln(b) forces B = 0
(since ln(b) > 0). Only trivial solutions here too.
• Case λ > 0: Write λ = ω 2 for some ω > 0, so p2 = −ω 2 and p = ±iω. Recall
r iω = eiω ln(r) = cos(ω ln(r)) + i sin(ω ln(r)),
and that both real and imaginary parts give a solution. The general solution is
R(r) = A cos(ω ln(r)) + B sin(ω ln(r)),
A, B ∈ R.
Now 0 = R(1) = A gives R(r) = B sin(ω ln(r)), so
0 = R(b) = B sin(ω ln(b)).
This time nonzero values of B can occur, provided ω > 0 satisfies
sin(ω ln(b)) = 0,
i.e., ω ln(b) = nπ, n = 1, 2, 3, . . . .
Thus we have a sequence of eigenvalues
2
nπ
2
λn = ωn =
,
ln(b)
n = 1, 2, 3, . . . ,
and the corresponding eigenfunctions are (multiples of)
nπ
Rn (r) = sin
ln(r) ,
n = 1, 2, 3, . . . .
ln(b)
Expansion Formulas. Dividing the ODE in (11) by r puts it into Sturm-Liouville
form:
1
1
′′
′
′
′
0 = rR + R + λ
R = (rR (r)) + λ
R,
R(1) = 0 = R(b).
r
r
Hence for any reasonable f = f (r) defined for 1 < r < b, we have
nπ
ln(r) ,
f (r) =
bn Rn (r) =
bn sin
ln(b)
n=1
n=1
Z b
nπ
dr
2
f (r) sin
ln(r)
,
⇐⇒ bn =
ln(b) r=1
ln(b)
r
∞
X
∞
X
File “2014notes”, version of 5 August 2014, page 13.
1 < r < b,
(∗)
n = 1, 2, 3, . . . .
Typeset at 16:38 August 5, 2014.
14
PHILIP D. LOEWEN
Back in problem (D), we postulate an eigenfunction-series solution:
nπ
Θn (θ) sin
u(r, θ) =
ln(r)
ln(b)
n=1
∞
X
for some Θn to be determined. Plugging in θ = 0 and θ = α will give two opportunities to use (∗), one with h and the other with k, and these will reveal the values of
Θn (0) and Θn (α). The evolution of Θn will be governed by the PDE, i.e.,
2
0 = r urr + rur + uθθ =
∞
X
n=1
Θn (θ) r 2 Rn′′ (r) + rRn′ (r) + Rn (r)Θ′′n (θ)
Now remember the eigenvalue problem:
r
2
Rn′′ (r)
+
rRn′ (r)
=−
nπ
ln b
2
Rn (r).
Hence we have
0=
∞
X
n=1
"
′′
Θ (θ) −
nπ
ln b
2
#
nπ
Θn (θ) sin
ln(r) .
ln(b)
This is another chance to use (∗), this time with the zero function as the expansion
result. The coefficient formulas give, for each n,
2
nπ
′′
Θn (θ), i.e., Θn (θ) = An enπθ/ ln b + Bn e−nπθ/ ln b .
0 = Θn (θ) −
ln b
Answer:
∞ h
i nπ
X
nπθ/ ln b
−nπθ/ ln b
sin
u(r, θ) =
An e
+ Bn e
ln(r) ,
ln(b)
n=1
with An and Bn determined by using (∗) with h and k.
File “2014notes”, version of 5 August 2014, page 14.
Typeset at 16:38 August 5, 2014.