Chapter 22
Lecture 26 - Laplace’s
Equation
Steady State Solutions of the Heat or Wave Equations that do not vary with
∂u
∂2u
time so that
=0= 2.
∂t
∂t
2D:
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
(22.1)
∂2u ∂2u ∂2u
+ 2 + 2 = 0.
∂x2
∂y
∂z
(22.2)
Δu =
3D:
Δu =
• No initial conditions required.
• Only boundary conditions.
In Polar Coordinates: Δu =
1 ∂2u
∂ 2 u 1 ∂u
+
+
= 0.
∂r2
r ∂r
r2 ∂θ2
129
Lecture 26 - Laplace’s Equation
22.1
Summary
In this course we have studied the solution of the second order linear PDE.
∂u
= α2 Δu
Heat:
Parabolic
T = αX 2
∂t
∂2u
∂t2
= c2 Δu
Δu
= 0
Wave:
Hyperbolic
T 2 − c2 X 2 = A(22.3)
X 2 + Y 2 = A.
Laplace’s Eq.: Elliptic
Important:
1. These equations are second order because they have at most 2nd partial derivatives.
2. These equations are all linear so that a linear combination of solutions
is again a solution.
22.2
Laplace’s Equation
(1) 2D Steady-State Heat Conduction and (2) Static Deflection of a Membrane.
ut = α2 (uxx + uyy ) −→ u(x, y, t) inside a domain D.
(22.4)
• Steady-State Solution satisfies:
Δu = uxx + uyy = 0
BC:
(x, y) ∈ D
u prescribed on ∂D.
(22.5)
(22.6)
• We consider domains D that are rectangular, circular, pizza slices.
22.3
Rectangular Domains
u(0, y) = g1 (y) Δu = uxx + uyy = 0 u(a, y) = g2 (y)
130
(22.7)
22.4. IDEA
22.4
Idea
• We want to use separation of variables so we need homogeneous boundary conditions.
• Since the equation is linear we can break the problem into simpler
problems which do have sufficient homogeneous BC and use superposition to obtain the solution to (22.7).
Pictorially:
22.5
Solution to Problem (1A) by Separation of
Variables
(1A)
uxx + uyy = 0
u(0, y) = 0 = u(a, y) = u(x, b);
(22.8)
u(x, 0) = f1 (x). (22.9)
Let
u(x, y) = X(x)Y (y).
(22.10)
X (x)Y (y) + X(x)Y (y) = 0
(22.11)
X (x) = −Y (y) = const = ±λ2
X(x)
Y (y)
BC
(22.12)
÷XY:
−λ2 :
X + λ2 X = 0
X = A cos λx + B sin λx
X(0) = 0 = X(a)
Y − λ2 Y = 0
Y = C cosh λx + D sinh λx Y (0) = . . . Y (b) = 0
131
Lecture 26 - Laplace’s Equation
• Because sin and cos have an ∞ # of real roots the choice −λ2 is
good for BC’s for Problems (A) and (C).
+λ2 :
X − λ2 X = 0
Y + λ2 Y = 0
X = A cosh(λx) + B sinh(λx)
Y = C cos(λy) + D sin(λy)
X(0) = . . . X(a) = . . .
(22.13)
.
Y (0) = 0 = Y (b)
• Again because sin and cos have an ∞ # of real roots the choice
+λ2 is good for BC’s for Problems (B) and (D).
Back to Solving (1A):
X(0) = 0 ⇒ A = 0
(22.14)
X(a) = B sin(λa) = 0 ⇒
λn
= nπ
.
a n=
1, 2, . .(22.15)
.
nπx
Xn (x) = sin a
u(x, b) = X(x)Y (b) = 0 ⇒ Y (b) = 0
(22.16)
Y (b) = C cosh(λb) + D sinh(λb) = 0 ⇒ c = −D tan h(λa) (22.17)
Y (y) = −D tan h(λb) cosh(λy) + D sinh(λy)
sinh(λy) cosh(λb) − cosh(λy) sinh(λb)
= D
cosh(λb)
D
sinh λ(y − b) = D̄ sinh λ(y − b).
=
cosh(λb)
(22.18)
(22.19)
(22.20)
Note: We could save ourselves the time by building the BC y(b) = 0
directly into the solution by letting
Yn (y) = D̄ sinh λn (y − b)
directly.
Now the functions: un (x, y) = sin
nπx sinh
(22.21)
nπ
(y − b) n = 1, 2, . . .
a
a
satisfy all the homogeneous BC of Problem (1A). In order to match
the BC u(x, 0) = f1 (x) we need to superimpose all these solutions.
u(x, y) =
f1 (x) = u(x, 0) =
∞
Bn sin
n=1
∞ nπx a
−Bn sinh
n=1 bn
132
sinh
nπb
a
nπ
a
(y − b) (22.22)
sin
nπx (22.23)
a
22.5. SOLUTION TO PROBLEM (1A) BY SEPARATION OF
VARIABLES
where
−Bn sinh
nπb
a
2
= bn =
a
a
f1 (x) sin
nπx a
0
dx.
(22.24)
Therefore
2
= −
a sinh nπb
a
Bn
∞
u(x, y) =
Bn sin h
a
f1 (x) sin
0
nπx a
dx
(22.25)
nπx (y − b) sin
;
a
a
nπ
n=1
Bn = −
a
f1 (x) sin
0
22.5.1
2
a sinh nπb
a
nπx a
dx.(22.26)
Specific Example
Let fL (x) = 1 =
∞
n=1
bn =
bn sin
nπx .
a
2 1 + (−1)n+1 = −Bn sinh
anπ
nπb
a
.
(22.27)
Therefore
u(x, y) =
∞
nπx nπ
1 2 [1 + (−1)n+1 ]
nπb sin
sinh
(y − b) . (22.28)
−
a
nπ sinh a
a
a
n=1
133