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Cylindrical Laplace Equation Solutions, Slides of Engineering Analysis

Solutions to the cylindrical laplace equation, a partial differential equation used in engineering and physics to describe various phenomena such as heat transfer, potential flow, and electrostatics. The solutions are presented for both homogeneous and non-homogeneous boundary conditions, with eigenfunctions and eigenvalues determined for radial and vertical directions.

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Cylindrical Laplace Solutions February 16, 2009
ME 501B – Engineering Analysis 1
Laplace Equation Solutions in
Laplace Equation Solutions in
Cylindrical Geometry
Cylindrical Geometry
Larry Caretto
Mechanical Engineering 501B
Seminar in Engineering
Seminar in Engineering
Analysis
Analysis
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 = uE(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 u1(x,y)
Solution
here is
u2(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) =
ub(x) to get solution for u(x,0) = ub(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 u2(x,y)
from u1
pf3
pf4
pf5
pf8
pf9

Partial preview of the text

Download Cylindrical Laplace Equation Solutions and more Slides Engineering Analysis in PDF only on Docsity!

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
    • u(r,L) = uN (r)
  • 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