Chapter 29
Lecture 33 Variable
coefficient BVP eigenfunctions involving
solutions to the Euler
Equation:
Example 29.1 Eigenfunctions involving solutions to an Euler Equation:
(x2 φ ) + λφ = 0
1<x<2
φ(1) = 0 φ(2) = 0
x2 φ + 2xφ + λφ = 0 An Euler Eq.
(29.1)
Let
φ(x) = xr
r(r − 1) + 2r + λ = r2 + r + λ = 0.
Therefore
r=
29.1
−1 ±
√
1 − 4λ
= r1 , r2 .
2
(29.2)
(29.3)
Cases:
λ = 14 :
1
1
φ(x) = c1 x− 2 + c2 x− 2 log x
− 12
φ(1) = c1 = 0 φ(2) = c2 2
(29.4)
log 2 = 0 ⇒ c2 = 0
(29.5)
167
Lecture 33 Variable coefficient BVP - eigenfunctions involving solutions to
the Euler Equation:
so there is no Eigenfunction for λ = 1/4.
λ = 14 :
φ(x) = c1 xr1 + c2 xr2
(29.6)
φ(1) = c1 + c2 = 0 c2 = −c1
φ(2) = c1 2r1 − 2r2 = 0
(29.7)
(29.8)
2r1 −r2 = 1
ρ(r1 −r2 ) ln 2 = 1 = ρ2πin
(r1 − r2 ) ln√2 = 2πin
r1 − r2 = 1 − 4λ = 2πni/ ln(2)
(29.9)
Thus to obtain nontrivial solutions we require 1 − 4λ < 0 λ >
√
√
1 − 4λ = i 4λ − 1 = 2πni/ ln(2).
1
4
. For λ >
1
4
(29.10)
The Eigenvalues are:
λn =
1
4π 2 n2
π 2 n2
,
4λ
−
1
=
= (2βn )2
+
n
4 [ln(2)]2
[ln(2)]2
βn = (nπ/ ln 2). (29.11)
The corresponding roots r1 and r2 are as follows
1
1
(r1 )n = − + iβn and (r2 )n = − − iβn
2
2
− 12
iβn
−iβn
x −x
φn (x) = cn x
1
= cn x− 2 ρiβn ln x − ρ−iβn ln x
1
= dn x− 2 sin (βn ln x)
ln x
− 12
= dn x sin nπ
ln(2)
choose dn = 1 or normalize so that
2
1
(29.12)
(29.13)
(29.14)
(29.15)
(29.16)
φ2n (x) dx = 1.
Example 29.2 Solving a variable coefficient Heat Conduction Problem:
ut = D(x2 ux )x − u
u(1, t) = 0 = u(2; t)
168
1<x<2 t>0
u(x, 0) = f (x).
(29.17)
29.2. SOLVING THE HEAT EQUATION BY EXPANDING IN
EIGENFUNCTIONS INVOLVING SOLUTIONS TO AN EULER
EQUATION:
Let
u(x, t) = X(x)T (t)
(29.18)
Ṫ (t)
DT (t)
(29.19)
(x2 X )
=
X
(x2 X )
= −λ
X
Ṫ (t)
+1 =
DT (t)
Ṫ + D(1 + λ)T = 0
(x2 X ) + λX = 0
X(1) = 0 = X(2)
−1
(29.20)
T (t) = cρ−D(1+λ)t
ln x (πn)2
−1
λn = 14 + [ln(2)]
2 ; Xn (x) = x 2 sin nπ ln 2 (29.21)
n = 1, 2, . . .
− 12
u(x, t) = x
∞
ln x
sin nπ
ln 2
∞
ln x
cn sin nπ
ln 2
−D(1+λn )t
cn ρ
n=1
1
f (x) = u(x, 0) = x− 2
(29.22)
(29.23)
n=1
29.2
Solving the heat equation by expanding in
eigenfunctions involving solutions to an Euler
Equation:
1
1
Δu = urr + ur + 2 uθθ = 0
r
r
u(r, 0) = 0, u(r, α) = f (r)
1 < r < 2, 0 < θ < α (29.24)
(29.25)
u(1, θ) = 0, u(2, θ) = 0
(29.26)
u(r, θ) = R(r)Θ(θ)
+ rR
Θ
= −
= −λ2
R(r)
Θ
(29.27)
r2 R
(because of Homog. BC)
(29.28)
169
Lecture 33 Variable coefficient BVP - eigenfunctions involving solutions to
the Euler Equation:
Θ − λ2 Θ = 0
Θ = c cosh λθ + D sinh λθ
Θ(0) = 0
Θ(0) = c = 0 ⇒ Θ(θ) = D sin hλθ
2
2
R: r R + rR + λ R = 0() R(1) = 0 = R(2). Although we can easily
see that dividing through by r we can reduce () to S-L form, let us use the
integrating factor
Θ:
μ() =
1 ρ
P
Q
dr
P
=
1 ρ
r2
r
dr
r2
=
1 ln r
1
ρ = .
r2
r
(29.29)
Therefore
1
λ2
R.
− · () ⇒ rR R = −(rR ) =
r
r
(29.30)
Now let us look for Eigenvalues and Eigenfunctions to (). Let
R(r) = rγ ⇒ γ(γ − 1) + γ + λ2 = γ 2 + λ2 = 0 γ = ±iλ. (29.31)
R(r) = c1 riλ + c2 r−iλ
riλ = ρiλ ln r
(29.32)
= A cos(λ ln r) + B sin(λ ln r)
(29.33)
R(1) = A cos [λ(ln 1)] + B sin(λ ln 1) = A = 0
(29.34)
R(2) = B sin [λ ln 2] = 0 ⇒ λn ln 2 = nπ
n = 1, 2, . . .
(29.35)
ln r and the corresponding Eigenfunctions are Rn = sin nπ ln 2 . Therefore
u(r, θ) =
∞
Bn sinh(λn θ) sin(λn ln r).
(29.36)
n=1
Now match BC:
f (r) = u(r, α) =
∞
n=1
ln r
Bn sinh(λn α) sin nπ
ln 2
x
= ln r
.(29.37)
dx0 ? = 1r dr
Now
2
1
1
sin
r
mπ ln r
ln 2
sin
nπ ln r
ln 2
ln 2 nπx mπx 0
sin
dr =
sin
dx =
ln 2
ln 2
ln 2
2
0
Therefore
2
Bn =
ln 2 sinh nπα
ln 2
170
2
1
f (r)
sin
r
m?n?π ln r
r
dr.
(29.39)
m = n
.
(29.38)
m=n