Math 412-501
Theory of Partial Differential Equations
Lecture 2-11: Review for Exam 2.
Heat conduction in a rectangle
Initial-boundary value problem:
∂ 2u ∂ 2u ∂u
(0 < x < L, 0 < y < H),
+
=k
∂t
∂x 2 ∂y 2
u(x, y , 0) = f (x, y ) (0 < x < L, 0 < y < H),
∂u
∂u
(0, y , t) =
u(L, y , t) = 0 (0 < y < H),
∂x
∂x
∂u
∂u
(x, 0, t) =
u(x, H, t) = 0 (0 < x < L).
∂y
∂y
We search for the solution u(x, y , t) as a
superposition of solutions with separated variables
that satisfy the boundary conditions.
Separation of variables: u(x, y , t) = φ(x)h(y )G (t).
Substitute this into the heat equation:
2
dG
d 2h
d φ
φ(x)h(y )
=k
h(y )G (t) + φ(x) 2 G (t) .
dt
dx 2
dy
Divide both sides by k · φ(x)h(y )G (t) = k · u(x, y , t):
1 dG
1 d 2φ 1 d 2h
·
= · 2 + · 2.
kG dt
φ dx
h dy
1 d 2φ
1 d 2h
It follows that
= −λ,
= −µ,
·
·
φ dx 2
h dy 2
1 dG
·
= −λ − µ, where λ and µ are separation
kG dt
constants.
The variables have been separated:
dG
dt
d 2φ
dx 2
= −(λ + µ)kG ,
= −λφ,
d 2h
dy 2
= −µh.
Proposition Suppose G , φ, and h are solutions of
the above ODEs for the same values of λ and µ.
Then u(x, y , t) = φ(x)h(y )G (t) is a solution of the
heat equation.
Boundary conditions ∂u
∂x (0, y , t) =
hold if φ′ (0) = φ′ (L) = 0.
∂u
∂x (L, y , t)
∂u
Boundary conditions ∂y
(x, 0, t) =
hold if h′ (0) = h′ (H) = 0.
∂u
∂y (x, H, t)
=0
=0
1st eigenvalue problem: φ′′ = −λφ, φ′ (0) = φ′ (L) = 0.
2
Eigenvalues: λn = ( nπ
L ) , n = 0, 1, 2, . . .
Eigenfunctions: φ0 = 1, φn (x) = cos nπx
L , n ≥ 1.
2nd eigenvalue problem: h′′ = −µh, h′ (0) = h′ (H) = 0.
2
Eigenvalues: µm = ( mπ
H ) , m = 0, 1, 2, . . .
Eigenfunctions: h0 = 1, hm (y ) = cos mπy
H , m ≥ 1.
Dependence on t:
G ′ (t) = −(λ + µ)kG (t) =⇒ G (t) = C0 e −(λ+µ)kt
Solution of the boundary value problem:
u(x, y , t) = e −(λn +µm )kt φn (x)hm (y )
mπy
mπ 2
2
= exp −(( nπ
)
+
(
)
)kt
· cos nπx
L
H
L · cos H .
We are looking for the solution of the
initial-boundary value problem as a superposition of
solutions with separated variables.
X∞ X ∞
Cn,m e −(λn +µm )kt φn (x)hm (y )
u(x, y , t) =
n=0
=
∞ X
∞
X
n=0 m=0
m=0
mπy
mπ 2
2
Cn,m exp −(( nπ
)
+
(
)
)kt
· cos nπx
L
H
L · cos H
How do we find coefficients Cn,m ?
From the initial condition u(x, y , 0) = f (x, y ).
X ∞ X∞
mπy
Cn,m cos nπx
f (x, y ) =
L cos H
n=0
m=0
(double Fourier cosine series)
How do we solve the heat conduction problem?
∂ 2u ∂ 2u ∂u
(0 < x < L, 0 < y < H),
=k
+
∂t
∂x 2 ∂y 2
u(x, y , 0) = f (x, y ) (0 < x < L, 0 < y < H),
∂u
∂u
(0 < y < H),
∂x (0, y , t) = ∂x (L, y , t) = 0
∂u
∂u
(0 < x < L).
∂y (x, 0, t) = ∂y (x, H, t) = 0
• Expand f into the double Fourier cosine series:
X ∞ X∞
mπy
f (x, y ) =
Cn,m cos nπx
L cos H .
n=0
m=0
• Write the solution: u(x, y , t) =
∞ X
∞
X
mπy
mπ 2
2
=
)
+
(
)
)kt
cos nπx
Cn,m exp −(( nπ
L
H
L cos H .
n=0 m=0
How do we expand f into the double Fourier cosine
series?
X ∞ X∞
mπy
Cn,m cos nπx
f (x, y ) =
L cos H .
n=0
m=0
If f (x, y ) is smooth then the expansion exists.
How do we determine coefficients?
Multiply both sides by φN (x)hM (y ) and integrate
over the rectangle.
Z LZ
0
=
=
∞ X
∞
X
n=0 m=0
∞ X
∞
X
H
f (x, y )φN (x)hM (y ) dx dy
0
Cn,m
Z LZ
0
Cn,m
n=0 m=0
Z
H
φN (x)hM (y )φn (x)hm (y ) dx dy
0
L
φN (x)φn (x) dx
0
Z
H
hM (y )hm (y ) dy .
0
We have the orthogonality relations:
Z L
φN (x)φn (x) dx = 0,
n 6= N,
0
Z
0
H
hM (y )hm (y ) dy = 0,
m 6= M.
Hence
Z LZ
0
H
f (x, y )φN (x)hM (y ) dx dy
0
= CN,M
Z
0
L
φ2N (x) dx
It remains to recall that
Z
Z L
φ20 (x) dx = L,
0
0
Z
0
L
H
h02 (x) dx
= H,
Z
0
Z
0
H
2
hM
(y ) dy .
L
φ2N (x) dx = ,
2
H
2
hM
(x) dx =
H
,
2
N ≥ 1,
M ≥ 1.
How do we expand f (x, y ) into the double series?
X ∞ X∞
mπy
Cn,m cos nπx
f (x, y ) =
L cos H ,
n=0
m=0
where
Z LZ H
1
C0,0 =
f (x, y ) dx dy ,
LH 0 0
Z LZ H
nπx
2
dx dy , n ≥ 1,
f (x, y ) cos
Cn,0 =
LH 0 0
L
Z LZ H
2
mπy
C0,m =
dx dy , m ≥ 1,
f (x, y ) cos
LH 0 0
H
Z LZ H
mπy
4
nπx
cos
dx dy .
Cn,m =
f (x, y ) cos
LH 0 0
L
H
Laplace’s equation in a semicircle
Solve Laplace’s equation inside a semicircle of
radius a (0 < r < a, 0 < θ < π) subject to the
boundary conditions: u = 0 on the diameter and
u(a, θ) = g (θ).
Laplace’s equation in polar coordinates (r , θ):
1 ∂ 2u
∂ 2 u 1 ∂u
+
+ 2 2 = 0.
∂r 2 r ∂r
r ∂θ
u = 0 on the diameter =⇒
u(r , 0) = u(r , π) = 0
u(0, θ) = 0
(0 < r < a),
(0 < θ < π).
Separation of variables: u(r , θ) = h(r )φ(θ).
Substitute this into Laplace’s equation:
d 2h
1 dh
d 2φ
1
φ(θ)
+
h(r
)
= 0.
φ(θ)
+
dr 2
r dr
r2
dθ2
Divide both sides by r −2 h(r )φ(θ) = r −2 u(r , θ):
dh 1 d 2φ
1 2 d 2h
· r
+r
= − · 2.
h
dr 2
dr
φ dθ
It follows that
1 2 d 2h
dh 1 d 2φ
+r
= − · 2 = λ = const.
· r
h
dr 2
dr
φ dθ
The variables have been separated:
2
dh
d 2φ
2 d h
+r
= λh,
= −λφ.
r
dr 2
dr
dθ2
Proposition Suppose h and φ are solutions of the
above ODEs for the same value of λ. Then
u(r , θ) = h(r )φ(θ) is a solution of Laplace’s
equation.
Boundary conditions u(r , 0) = u(r , π) = 0 hold if
φ(0) = φ(π) = 0.
Boundary condition u(0, θ) = 0 holds if
h(0) = 0.
Euler’s (or equidimensional) equation
dh
d 2h
+
r
− λh = 0 (r > 0)
r
dr 2
dr
2
λ > 0 =⇒ h(r ) = C1 r p + C2 r −p (λ = p 2 , p > 0)
λ = 0 =⇒ h(r ) = C1 + C2 log r
Eigenvalue problem: φ′′ = −λφ, φ(0) = φ(π) = 0.
Eigenvalues: λn = n2 , n = 1, 2, . . .
Eigenfunctions: φn (θ) = sin nθ.
Dependence on r :
r 2 h′′ + rh′ = λh,
=⇒ h(r ) = C0 r p
h(0) = 0.
√
(p = λ)
Solution of Laplace’s equation:
u(r , θ) = r n sin nθ,
n = 1, 2, . . .
A superposition of the solutions with separated
variables is a series
X∞
cn r n sin nθ,
u(r , θ) =
n=1
where c1 , c2 , . . . are constants.
Substituting the series into the boundary condition
u(a, θ) = g (θ), we get
X∞
g (θ) =
cn an sin nθ.
n=1
Hence cn = bn a−n , n = 1, 2, . . . , where
X∞
g (θ) =
bn sin nθ.
n=1
Solution.
u(r , θ) =
where
X∞
n=1
X∞
n=1
bn
r n
a
sin nθ,
bn sin nθ
is the Fourier sine series of the function g (θ) on
[0, π], that is,
Z
2 π
g (θ) sin nθ dθ, n = 1, 2, . . .
bn =
π 0