c Anthony Peirce.
Introductory lecture notes on Partial Differential Equations - °
Not to be copied, used, or revised without explicit written permission from the copyright owner.
1
Lecture 30: Sturm-Liouville Problems involving the
Cauchy-Euler Equation - Applications
(Compiled 3 March 2014)
In this lecture we look at eigenvalue problems involving equidimensional or Cauchy-Euler differential operators.
Key Concepts: Eigenvalue Problems, Sturm-Liouville Boundary Value Problems; Cauchy-Euler Equations; Equidimensional equations.
Reference Section: Boyce and Di Prima Section 11.1 and 11.2
30 Variable coefficient BVP - eigenfunctions involving solutions to the Euler Equation:
Example 30.1 Eigenfunctions involving solutions to an Euler Equation:
0
(x2 φ0 ) + λφ = 0
1<x<2
φ(1) = 0, φ(2) = 0
x2 φ00 + 2xφ0 + λφ = 0 A Cauchy-Euler Eq.
(30.1)
Let
φ(x) = xr
r(r − 1) + 2r + λ = r2 + r + λ = 0.
Therefore
−1 ±
r=
√
1 − 4λ
2
= r1 , r2 .
(30.2)
(30.3)
Case I: λ = 14 :
1
1
φ(x) = c1 x− 2 + c2 x− 2 log x
− 12
φ(1) = c1 = 0 φ(2) = c2 2
(30.4)
log 2 = 0 ⇒ c2 = 0
(30.5)
so there is no nontrivial eigenfunction for λ = 1/4.
Case II: λ 6= 14 :
φ(x) = c1 xr1 + c2 xr2
(30.6)
φ(1) = c1 + c2 = 0 c2 = −c1
¢
¡
φ(2) = c1 2r1 − 2r2 = 0
(30.7)
2r1 −r2 = 1
e(r1 −r2 ) ln 2 = 1 = e2πin
(r1 − r2 ) ln√2 = 2πin
r1 − r2 = 1 − 4λ = 2πni/ ln(2)
(30.8)
(30.9)
2
Thus to obtain nontrivial solutions we require 1 − 4λ < 0 which implies λ >
√
1
4
. Thus for λ >
1
4
√
1 − 4λ = i 4λ − 1 = 2πni/ ln(2).
(30.10)
The Eigenvalues are:
λn =
π 2 n2
1
4π 2 n2
+
,
4λ
−
1
=
= (2βn )2
n
4 [ln(2)]2
[ln(2)]2
(30.11)
βn = (nπ/ ln 2).
The corresponding roots r1 and r2 are as follows
1
1
(r1 )n = − + iβn and (r2 )n = − − iβn
2
2
¢
1 ¡
φn (x) = cn x− 2 xiβn − x−iβn
¤
1 £
= cn x− 2 eiβn ln x − e−iβn ln x
= dn x
− 12
= dn x
− 12
(30.12)
(30.13)
(30.14)
sin (βn ln x)
·
¸
ln x
sin nπ
ln(2)
(30.15)
(30.16)
Now let us consider expanding a function f (x) in terms of a ‘Fourier Series’ of these new eigenfunctions in the
following form
∞
X
f (x) =
cn φn (x) =
n=1
∞
X
µ
−1/2
cn x
sin
n=1
nπ ln x
ln 2
¶
(30.17)
In order to determine the coefficients cn we project the function f (x) onto the basis functions φn (x) as follows:
µ
Z2
−1/2
f (x)x
1
sin
mπ ln x
ln 2
¶
dx =
µ
Z2
∞
X
x−1/2 sin
cn
n=1
1
mπ ln x
ln 2
¶
µ
x−1/2 sin
nπ ln x
ln 2
¶
dx
The integrals under the summation can be evaluated by making the simple substitution z = ln x so that dz =
in which case:
µ
Z2
sin
1
mπ ln x
ln 2
¶
µ
sin
nπ ln x
ln 2
¶
dx
=
x
Zln 2 ³
³ nπz ´
mπz ´
ln 2
sin
sin
dz = δmn
ln 2
ln 2
2
0
Substituting this result into (30.17) we obtain the following expression for the Fourier coefficients cn :
2
cn =
ln 2
µ
Z2
f (x)x−1/2 sin
1
nπ ln x
ln 2
¶
dx
dx
x ,
Sturm-Liouville Two-Point Boundary Value problems
3
Example 30.2 A variable coefficient Heat Conduction Problem with Cauchy-Euler Eigenfunctions:
ut = D(x2 ux )x − u
u(1, t) = 0 = u(2; t)
1<x<2 t>0
u(x, 0) = f (x).
(30.18)
Let
u(x, t) = X(x)T (t)
2
(30.19)
0 0
(x X )
Ṫ (t)
=
−1
DT (t)
X
(30.20)
0
(x2 X 0 )
Ṫ (t)
+1=
= −λ
DT (t)
X
Ṫ + D(1 + λ)T = 0
¾
0
(x2 X 0 ) + λX = 0
X(1) = 0 = X(2)
(30.21)
T (t) = ce−D(1+λ)t
¡
¢
(πn)2
x
− 12
λn = 41 + [ln(2)]
sin nπ ln
2 ; Xn (x) = x
ln 2
n = 1, 2, . . .
µ
¶
ln x
cn e−D(1+λn )t sin nπ
ln 2
n=1
µ
¶
∞
X
ln x
− 12
f (x) = u(x, 0) = x
cn sin nπ
ln 2
n=1
1
u(x, t) = x− 2
∞
X
(30.22)
(30.23)
(30.24)
where
2
cn =
ln 2
µ
Z2
−1/2
f (x)x
sin
1
nπ ln x
ln 2
¶
dx
31 Solving Laplace’s equation using Cauchy-Euler eigenfunctions
Figure 1. Annular sector subtending an arc of α radians between the radii a and b: 1 = b < a = 2
1
1
∆u = urr + ur + 2 uθθ = 0
r
r
u(r, 0) = 0, u(r, α) = f (r)
1 < r < 2, 0 < θ < α
(31.1)
(31.2)
u(1, θ) = 0, u(2, θ) = 0
(31.3)
u(r, θ) = R(r)Θ(θ)
Θ00
r R + rR0
=−
= −λ2
R(r)
Θ
(31.4)
2
00
(because of Homog. BC)
(31.5)
4
Θ00 − λ2 Θ = 0
Θ = C cosh λθ + D sinh λθ
Θ(0) = 0
Θ(0) = C = 0 ⇒ Θ(θ) = D sinh λθ
R: r2 R00 + rR0 + λ2 R = 0(?), R(1) = 0 = R(2) Although we can easily see that dividing through by r we can
Θ:
reduce (?) to S-L form, let us use the integrating factor
µ(r) =
1 R
e
P
Q
P dr
1 R
e
r2
=
r
r2
dr
=
1 ln r
1
e
= .
r2
r
(31.6)
Therefore − 1r · (?) and some rearrangement implies:
0
−rR00 − R0 = − (rR0 ) =
λ2
R
r
Now let us look for Eigenvalues and Eigenfunctions to (?). Let
R(r) = rγ ⇒ γ(γ − 1) + γ + λ2 = γ 2 + λ2 = 0
R(r) = c1 r
iλ
+ c2 r
−iλ
r
iλ
=e
γ = ±iλ.
(31.7)
iλ ln r
(31.8)
= A cos(λ ln r) + B sin(λ ln r)
(31.9)
R(1) = A cos [λ(ln 1)] + B sin(λ ln 1) = A = 0
(31.10)
R(2) = B sin [λ ln 2] = 0 ⇒ λn ln 2 = nπ n = 1, 2, . . .
¡ ln r ¢
and the corresponding Eigenfunctions are Rn = sin nπ ln
2 . Therefore
u(r, θ) =
∞
X
(31.11)
Bn sinh(λn θ) sin(λn ln r).
(31.12)
n=1
Now match BC:
¶
µ
ln r
f (r) = u(r, α) =
Bn sinh(λn α) sin nπ
ln 2
n=1
∞
X
Now making the substitution x = ln r so that dx =
dr
r
we can reduce the orthogonality integrals for the Fourier
coefficients to the form:
µ
¶
µ
¶
½
Z2
Zln 2 ³
³ nπx ´
1
mπ ln r
nπ ln r
mπx ´
0
sin
sin
sin
dr =
sin
dx =
ln 2
r
ln 2
ln 2
ln 2
ln 2
2
1
(31.13)
m 6= n
.
m=n
(31.14)
0
Therefore
2
¢
¡
Bn =
ln 2 sinh nπα
ln 2
Z2
1
f (r)
sin
r
µ
nπ ln r
r
¶
dr.
(31.15)