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Laplace Equation Solutions inLaplace Equation Solutions in
Cylindrical GeometryCylindrical Geometry
Larry Caretto Mechanical Engineering 501B Seminar in EngineeringSeminar in Engineering AnalysisAnalysis February 16, 2009
2
Overview
- Review last class
- Superposition solutions
- Introduction to radial coordinates
- Additional solutions of Laplace’s equation in radial coordinates - Homogenous boundaries in z direction - Gradient boundary conditions in the cylinder - Hollow cylinder - Combinations of boundary conditions - Superposition
3
Review Superposition
- Laplace’s equation for 0 ≤ x ≤ L and 0 ≤ y ≤ H with boundary conditions shown
- Do not have homogenous boundary conditions in any coordinate direction
- Superposition sum of simpler solutions
x = 0 x = L
y = H
y = 0
u = 0 u = u (^) E(y)
u = uN (x)
u = 0 4
Review Superposition Solution
- Sum two solutions with one nonzero boundary
x = 0 x = L
y = H
y = 0
u = 0 u = 0
u = uN (x)
u = 0
x = 0 x = L
y = H
y = 0
u = 0 u = uE(y)
u = 0
u = 0
- Solution to this problem is u 1 (x,y)
- Solution here is u 2 (x,y)
5
Review General Superposition
- Can obtain solution for several nonzero boundaries by creating a separate solution for each nonzero boundary - Three-dimensional problems can have up to have six separate solutions for u(x,y,z)
- Swap x and y and swap H and L in solution for u(0,y) = u(L,y) = 0 to get solution for y(x,0) = u(x,H) = 0
- Set y = H – y in solution for u(x,H) = u (^) b (x) to get solution for u(x,0) = u (^) b (x)
6
Review x and y Swap
- Sum two solutions shown below
x = 0 x = L
y = H
y = 0
u = 0 u = 0
u = uN (x)
u = 0
x = 0 x = L
y = H
y = 0
u = 0 u = uE(y)
u = 0
u = 0
- u 1 (x,y) found previously
- Swap x and y to get u 2 (x,y) from u (^1)
7
Review x-y Swap Solution
H u xy B y x n n n
= (^) ∑ n κ (^) n κn κ =^ π
∞ = 1
2 (, ) sin( )sinh( )
L u xy C x y n n n n n n
= (^) ∑ λ λ λ =^ π
∞ (^1) = 1 (, ) sin( )sinh( )
∫ (^ )
= • L N n
n
n
u x xdx
HL
C
0
()sin
sinh
2
λ
λ (^ )
∫ (^ )
= • H E n
n
n
u y ydy
LH
B
0
( )sin
sinh
2
κ
κ
∑[^ ]
∞
= + 1
( , ) sin( )sinh( ) sin( )sinh( ) n
uxy Cn λnx λny Bn κny κnx
8
Review Coordinate Transform
- Swap location of one zero boundary
x = 0 x = L
y = H
y = 0
u = 0 u = 0
u = uN (x)
u = 0
x = 0 x = L
y = H
y = 0
u = 0 u = 0
u = 0
u = u (^) S(x)
- u 1 (x,y) found previously with uN (x)
- u 2 (x,y) = u 1 (x,H–y)
- Use u (^) S(x) in place of u (^) N (x)
9
Review y ← H – y Solution
u xy B x ( H y) n n L
n
= ∑n λn λn − λ=^ π
∞ = 1
2 (, ) sin( )sinh ( )
L u xy C x y n n n
= (^) ∑ n λn λn λ =^ π
∞ = 1
1 (, ) sin( )sinh( )
= ∫ ( )
L N n n
Cn (^) HL u x xdx 0
()sin sinh
(^2) λ λ
= ∫ ( )
L S n n
Bn (^) HL u x xdx 0
()sin sinh
(^2) λ λ 10
Cylindrical Laplace
- Solve Laplace’s equation in a cylinder for u(r,z)
- Zero boundaries at the sides and bottom - u(R,z) = 0 - u(r,0) = 0
- Specified top boundary
- Finite solution at r = 0
L
z = 0
R
u(r,L) = uN (r)
2
2
∂
z
u r
u r r r
11
What do We Expect?
- Note similarity to radial diffusion equation
- u is finite at r = 0 for both problems
2
2
u Rz u z
z
u r
u r r r = R
u Rz u z
t
u r
u r r r = R
α
- Separation of variables result for P(r) in Laplace’s equation should be similar to result for diffusion equation - Bessel function eigenfunctions in r direction 12
What do We Expect? II
- Also similar to Laplace equation for u(x,y)
0 (, 0 ) 0 ( , ) ( )
2
2 ur urL u r z
u r
u r r r N
- Separation of variables result for Z(z) in radial equation should be similar to result for Y(y) in rectangular coordinates - Hyperbolic sine/cosine solution for Z(z)
2 0 ( ,^0 )^0 ( , ) (^ )
2 2
2 ux uxH u x y
u x
u = = = N ∂
19 20
Important Observation
- In solving PDEs by separation of variables, individual terms have the same behavior in any equation (diffusion and Laplace so far) - Terms like ∂^2 u/∂x^2 , with homogenous boundary conditions give eigenfunctions that are sines and/or cosines - Terms like (1/r) ∂ [r∂u/∂r] / ∂r, with homogenous boundary conditions give Bessel functions as eigenfunctions
21
Two Boundary Changes
- Laplace’s Equation in two-dimensional cylindrical region 0 ≤ z ≤ L and 0 ≤ r ≤ R - Boundary conditions u(R,z) = u (^) R (z) Other boundaries: u(r,0) = 0, ∂u/∂z|z=L = 0 - u(0,z) is finite (zero gradient)
- Here we have homogenous boundary conditions in z direction - Want Sturm-Liouville solution in this direction to get eigenfunction expansions for nonzero boundary at r = R 22
Separation of Variables
- Use u(r,z) = P(r)Z(z) to solve equation
2 2
− 1 = =− λ dz
dZz dr Zz
rdPr dr
d rPr
( ) (^2) rP(r) 0 P(r) CI 0 (r) DK 0 (r ) dr
rdPr dr
d −λ = = λ + λ
( 2 ) (^2) () 0 () sin( ) cos( ) 2 Zz Zz A z B z dz
dZz+λ = = λ + λ
2
2
∂
z
u r
u r r r
- Pick separation of variables constant to give sine and cosine solution in z
( ) (^2) rP(r) 0 P(r) CI 0 (r) DK 0 (r ) dr
rdPr dr
d −λ = = λ + λ
( 2 ) (^2) () 0 () sin( ) cos( ) 2 Zz Zz A z B z dz
dZz+λ = = λ + λ
23
Modified Bessel Functions
- Iν(x) = i-νJν(ix), Kν(x) = i-νYν(ix), i^2 = -
- Satisfy modified differential equation
2 (^22 )^0
2 (^2) + −x + y= dx
dy x dx
d y
x ν
- Equation above transforms to
2
⎟⎟^ =
z y
dz z
dy
z
dz
d
λ
ν
- Solution is z = AIν(λx) + BKν(λ x)
Since J ν ~ x ν , I ν and K ν are real
24
Modified Bessel Function Plots
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 x
Kn(x)
n = 0 n = 1n = 4
0
2
4
6
8
10
12
14
16
18
20
0 1 2 3 4 5 6 7 x
In(x)
n = 0n = 1 n = 4 All Kn become infinite as x approaches 0
I 0 (0) = 1
25
Boundary Conditions
- u(r,0) = PZ = 0 for all r requires Z(0) = 0
- Z(0) = 0 = A sin (0) + B cos (0) = B = 0
- Finite solution at r = 0 requires D = 0
- ∂u/∂z|z=L = 0 for all r requires dZ/dz|z=L = λA cos (λL) = 0; 2λL = nπ (odd integer n)
- P(r) is finite at r = 0 only if B = 0
- Solution is sum of all eigenfunctions
L
u rz C zI r m m m
(, ) m sin( m ) ( m) ( 2 1 ) 2 0
0
π = (^) ∑ λ λ λ = +
∞
= (^26)
Boundary Condition at r = R
- Equation for Z(z) is a Sturm-Liouville problem; use eigenfunction expansion in sin(λmz) for r = R boundary
∑
∞
=
0
( ) ( , ) sin( ) 0 ( ) m
uR z uRz Cm λmzI λmR
∫[^ ]
∫
= L
m m
L m R m
I R z dz
zu zdz
C
0
2 0
0
( ) sin( )
sin( ) ()
L m m 2
λ =( 2 + 1 )^ π
27
Boundary Condition at r = R II
- Denominator integral in Cm equation
π π
π
λ
λ λ λ λ
( 2 1 )
2 2
( 2 1 )
cos(^221 ) 1
sin( ) cos( ) cos( )^1 (^00)
⎜⎝⎛^ + ⎟⎠⎞− =−
=− =− −
n
LU L
n
n U
zUdz U z U L m
m
L m
L m m
[ sin( )] 2 sin( 2 )cos( ) 2 sin( 2 )cos( ) 2 0 0
z^2 dz x z z L L L L m
m m
L m
L m m m ⎥ = − = ⎦
⎤ ⎢⎣ =⎡^ −
λ λ λ λ λ λ
- Numerator integral in Cm equation for u (^) R = U
0
28
Boundary Condition at r = R III
- Result for uR (z) = U, a constant
∞ = (^) ⎟ ⎠
⎜ ⎞ ⎝
⎟ ⎠
⎜ ⎞ ⎝
⎟ ⎛^ + ⎠
⎜ ⎞ ⎝
⎛ +
(^00)
0
2 ( 2 1 ) (^21 )
2
( 2 1 ) 2
sin(^21 ) ( ,)^4 n L n I n R
L
I n r L
n z urz U π
π π
π
[ ] (^21 ) ( )
4 ( ) 2
( 2 1 )
2
( ) sin( )
sin( )
0 0 0
2 0
0 n I R
U I RL
n
LU
I R z dz
zUdz C m L m m m
L m m π λ λ
π λ λ
λ
= = + =
L m n 2
λ =( 2 + 1 )^ π
29 30
Exercise
- Solve previous problem changing the boundary condition at z = L from a zero gradient to a zero potential
- Pick starting point from previous solution
0
(, 0 ) (, ) 0 ( , ) () (^1 ) 2 =
= = =
∂
+∂ ∂
∂ ∂
∂ uisfiniteat r
ur urL uRz u z z
u r
r u r r
R
u( r,z)= [ Asin( λz )+Bcos( λz)][CI 0 ( λr)+DK 0 ( λr)]
- D = 0 to eliminate K 0 (= –∞ at r = 0)
- B = 0 for u(r,0) = 0
- λ = nπ/L (n an integer) for u(r,L) = 0
37
Results for uR(z) = U
⎟⎟⎠
⎞ ⎜⎜⎝
⎛ +
⎟⎟⎠
⎞ ⎜⎜⎝
⎛ +
L
R R K n R
L
R R I n R F i
i n 0 (^00)
0 (^00) ( 2 1 )
( 2 1 )
π
π ⎟ ⎠
⎜ ⎞ ⎝
⎟− ⎛^ + ⎠
⎜ ⎞ ⎝
= ⎛^ + L FK n R L G I n R n 0 0 n 0 0
( 2 1 )π ( 2 1 ) π
∞ = +
⎥⎦
⎤ ⎢⎣
⎡ ⎟⎟⎠
⎞ ⎜⎜⎝ − ⎛^ + ⎟⎟⎠
⎞ ⎜⎜⎝ ⎟ ⎛^ + ⎠ ⎜ ⎞ ⎝
⎛ +
0
0 (^00) 0 (^00) ( 2 1 )
( 2 1 ) ( 2 1 ) 2 sin(^21 ) (,) 4 n n
n n G
L
R R FK n r L
R R I n r L
n z
U
urz
π π π
π
∞ = +
⎥⎦
⎤ ⎢⎣
⎡ (^) ⎟ ⎠
⎜ ⎞ ⎝
⎟− ⎛^ + ⎠
⎜ ⎞ ⎝
⎟ ⎛^ + ⎠
⎜ ⎞ ⎝
⎛ +
0
0 0 ( 2 1 )
sin(^21 ) (^21 ) (^21 ) ( ,)^4 n (^) n
n n G
L
FK n r L
I n r L
n z
urz U
π π π
π
⎟⎟⎠
⎞ ⎜⎜⎝
⎛
0
0 0
, , , R
R L
R L
z R
f r U
u (^) i 38
39
Another Hollow Cylinder
- Laplace’s Equation in two-dimensional region 0 ≤ z ≤ L and Ri ≤ r ≤ R 0 - Boundary conditions u(r,L) = uN (z) with all other boundaries zero: u(r,0) = u(Ri ,z) = u(R 0 ,z) = 0
2
2 2
2 (^2) ∂ =
z
u u r r
u r
r r θ
- Laplace’s equation for cylinder with no angular variations
0
40
Separation of Variables
- Proposed solution u(r,z) = P(r)Z(z)
- Usual separation of variables process and ODE solutions give 2 2
(^2) () ()
() 1 ()
(^1) = − =− λ dz
dZz dr Zz
rdPr dr
d rPr
( ) (^2) rP(r) 0 P(r) CJ 0 (r) DY 0 (r ) dr
rdPr dr
d +λ = = λ + λ
( 2 ) (^2) () 0 () sinh( ) cosh( ) 2 Zz Zz A z B z dz
dZz −λ = = λ + λ
41
Boundary Conditions
- Z(0) = 0 = A sinh (0) + B cosh (0) = B = 0
- u(Ri,z) and u(R 0 ,z) = 0 with u = PZ for all z requires P(Ri) = P(R 0 ) = 0 - CJ 0 (λRi ) + DY 0 (λRi ) = CJ 0 (λR 0 ) + DY 0 (λR 0 ) = 0
0 0 0 0
0 0
D
C
J R Y R
J Ri Y Ri
- Must have zero determinant zero to avoid trivial solution C = D = 0 42
Boundary Conditions
- This is eigenvalue equation for λ
- Substitute αm = λmR 0 into this result to show that eigenvalues depend on radius ratio, Ri/R 0
0 0 0
0 0 0 0 0 0 0
i
i i i J R Y R
J R Y R
J R Y R
J R Y R
Det
0
0 0 0 0
R
R
Y J Y
R
R
J αm i αm αm αm i
43
Finding Eigenvalues = f(Ri /R 0 )
-0.
-0.
-0.
-0.
0.
0.
0.
0 20 40 60 80 100
radius ratio = 0. radius ratio = 0.5radius ratio = 0.
( ) ( ) 0 (^0 ) ⎟⎟= ⎠
⎞ ⎜⎜⎝ − ⎛ ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ R Y J Y R R J R i αm i αm αm αm
44
Boundary Conditions
- Second row of matrix gives CJ 0 (λR 0 ) + DY 0 (λR 0 ) = 0 or C = –DY 0 (λR 0 ) / J 0 (λR 0 )
- Radial solution is P(r) = CJ 0 (λr) + DY 0 (λr)
- Combine these to get P(r) = –D [ Y 0 (λR 0 ) / J 0 (λR 0 ) ] J 0 (λr) + DY^0 (λr)
- Multiply by J 0 (λR 0 ) and add m subscript
0 0 0 0
0 0
D
C
J R Y R
J Ri Y Ri
45
Boundary Conditions II
- Define new constant: C (^) m = –D/J 0 (λmR 0 )
∑^ [^ ]
∞
= − 1 0 0 0 0 0 0
( , ) sinh( ) ( ) ( ) ( ) ( ) m m^ m m m m m
urz C λzY λR J λr J λRY λr
[ ( ) ( ) ( ) ( )]
( )
()
( ) ( )
() ( ) ( )
0 0 0 0 0 0 0 0
0 0 0
0 0 0
Y R J r J R Y r J R
Pr D
DY r J R
Pr DY R J r
m m m m m
m m
m m
λ λ λ λ λ
λ λ
λ λ
=− −
=− +
Pm ( r)=CmP 0 (λm r)=Cm[Y 0 (λmR 0 )J 0 (λmr)−J 0 (λmR 0 )Y 0 ( λmr)]
- Solution is sum of all eignefunctions
46
Boundary Condition at z = L
- Equation for P(r) is a Sturm-Liouville problem; use eigenfunction expansion in P (^) m(r) = Y 0 (λmR 0 )J 0 (λmr) – J 0 (λmR 0 ) · Y 0 (λmr) at z = L
[ ]
[ ]
sinh( ) 2 [ ( ) ( )]
( ) () ( )
sinh( ) ( )
() ( )
02 02 0
0 2 0
0 2
0
0
0
0
L J R J R
J R ru rP rdr
L rP r dr
ru rP rdr C m m i m
R R
m m i N m R R
m m
R R
N m m i
i
i λ λ λ
πλ λ λ
λ λ
λ
−
= =
∞
= = 1
( ) (, ) sinh( ) 0 ( ) m
uN r urL Cm λmLP λmr
47
Solution for uN = U, a Constant
[ ]
[ ]
sinh( )[ ( ) ( )]
( )
sinh( ) 2 [ ( ) ( )]
( ) ( )^2 [ ( ) ( )]
sinh( ) 2 [ ( ) ( )]
( ) ( )
0 0 0
0
02 02 0
(^20) 0 2 0 0 0
02 02 0
0 2 0
0
L J R J R
UJ R
L J R J R
J R U J R J R J R
L J R J R
J R rUP rdr C
m mi m
mi
m mi m
m mi m mi mi m
m mi m
R
m m mi Ri m
λ λ λ
π λ
λ λ λ
πλ λ πλ λ λ λ
λ λ λ
πλ λ λ
= +
−
⎥⎦
⎤ ⎢⎣
⎡ −
= −
∑
∞
(^1 )
0 0
sinh( )
sinh( )
m (^) m i m
m i m m
m
J R J R
J R P r
L
z
urz U
Integrals from Carslaw and Jaeger, Conduction of Heat in Solids, Oxford, 1959.
48
Solution for uN = U, a Constant
∞ = (^) ⎟⎟+ ⎠
⎞ ⎜⎜⎝
⎛
⎟⎟⎠
⎞ ⎜⎜⎝
⎛ ⎟⎟⎠
⎞ ⎜⎜⎝
⎛
⎟⎟⎠
⎞ ⎜⎜⎝
⎛
⎟⎟⎠
⎞ ⎜⎜⎝
⎛
(^10) (^00)
(^0000)
0
0 sinh ( )
(,) sinh m m i m
m i m
m
m J RR J
R P r R J R
R
L
L
z R
L
U
urz α α
α α
α
α π
∑
∞
(^1 )
0 0
sinh( )
sinh( )
m (^) m i m
m i m m
m
J R J R
J R P r
L
z
urz U
- Convert to dimensionless form; we previously showed that αm = f(Ri/R 0 )
U(r,z)/U depends on z/L, r/R 0 , Ri/R 0 , and L/R
