Laplace Soln

8/17/2019 Laplace Soln

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College of Engineering and Computer ScienceMechanical Engineering Department

Notes on Engineering Analysis

February 6, 2009 Larry Caretto

Solution of Laplace’s Equation

Laplace’s equation describes the equilibrium distribution of energy, species andelectromagnetic fields under certain assumptions. t is a second!order, partial!differential equation "ith closed boundaries. #he simplest case to consider is a

rectangular region "ith $ and y coordinates such that 0 ≤ $ ≤ L and 0 ≤ y ≤ %.#he dependent &ariable is a potential, u'$,y(, "hich may be temperature,electromagnetic potential, &elocity potential or a range of other physical&ariables. )e assume that the potential is *no"n o&er the complete boundary.n subsequent problems "e "ill consider the case "here the potential gradient,rather than the potential, is *no"n o&er a portion of the boundary. n t"o

dimensions, Laplace’s equation "ith a simple set of boundary conditions may be"ritten as follo"s

)(),(0)0,(),(),0(

0,002

2

2

2

xu H xu xu y Lu yu

H y L x y

u

x

u

N ====

≤≤≤≤=∂

∂+∂

∂+-

%ere, the boundary conditions gi&e u 0 at $ 0, $ L, and y 0. /t y %, theboundary condition is a *no"n function of $. ote that there may be adiscontinuity at edges of the upper boundary '$ 0, y %( and '$ L, y %(.

#he solution of Laplace’s equation proceeds by a method *no"n as theseparation of &ariables. n this method "e postulate a solution that is the productof t"o functions, 1'$( a function of $ only and 'y( a function of the y only. )iththis assumption, our solution becomes.

u'$,t( 1'$('y( +2-

)e do not *no", in ad&ance, if this solution "ill "or*. %o"e&er, "e assume thatit "ill and "e substitute it for u in equation +-. 3ince 1'$( is a function of $ onlyand 'y( is a function of y only, "e obtain the follo"ing result "hen "e substituteequation +2- into equation +-.

[ ] [ ]0

)()(

)()(

)()()()(2

2

2

2

2

2

2

2

2

2

2

2

=∂∂+

∂∂=

∂∂+

∂∂=

∂∂+

∂∂

y

yY x X

x

x X yY

y

yY x X

x

yY x X

y

u

x

u+4-

f "e di&ide the final equation through by the product 1'$('y(, and mo&e the yderi&ati&e to the other side of the equal sign, "e obtain the follo"ing result.


2/27

2

2

2

2 )(

)(

1)(

)(

1

y

yY

yY x

x X

x X ∂∂

−=∂

∂+5-

#he left hand side of equation +6- is a function of $ only the right hand side is afunction of y only. #he only "ay that this can be correct is if both sides equal a

constant. #his also sho"s that the separation of &ariables solution "or*s. norder to simply the solution, "e choose the constant to be equal to −λ2. #hisgi&es us t"o ordinary differential equations to sol&e.

2

2

22

2

2 )(

)(

1)(

)(

1λ λ −=−−=

dy

yY d

yY dx

x X d

x X +7-

8quation +7- sho"s that "e ha&e t"o separate differential equations, each of"hich has a *no"n general solution. #hese equations and their generalsolutions are sho"n belo". 2

)cos()sin()(0)()( 22

2

x B x A x X x X dx

x X d λ λ λ +=⇒=+ +6-

)cosh()sinh()(0)()( 2

2

2

y D yC yY yY dy

yY d λ λ λ +=⇒=− +-

From the solutions in equations +6- and +-, "e can "rite the general solution foru'$,t( 1'$(#'t( as follo"s.

[ ][ ])cosh()sinh()cos()sin(),( y D yC x B x A y xu λ+λλ+λ= +:-

)e no" apply the boundary conditions sho"n "ith the original equation +- toe&aluate the constants /, ;, C, and ust plain λ comes from e$perience. Choosingthe constant to ha&e this form no" gi&es a more con&enient result later. f "e chose the constant

to be simply λ, "e "ould obtain the same result, but the e$pression of the constant "ould bea"*"ard. n solutions to Laplace’s equation in rectangular geometries "e "ill select the constant

as =λ2 to gi&e us an ordinary differential equation "hose solutions results in sines and cosines inthe direction for "hich "e ha&e a 3turm!Liou&ille problem '"ith homogenous boundaryconditions(.2 /s usual, you can confirm that this solution satisfies the differential equation by substituting thesolution into the differential equation.


3/27

3ince sinh'0( 0 and cosh'0( , this boundary condition "ill be satisfied only if< 0. #he third boundary condition that u 0 occurs at $ L. /t this point "eha&e the follo"ing result, using solution in +:- "ith ; 0 and < 0 as foundpre&iously.

[ ][ ])sinh()sin(0),( yC L A y Lu λλ== +-

8quation +- can only be satisfied if the sine term is ?ero. #his "ill be true only if

λL is an integral times π. f n denotes an integer, "e must ha&e

L

nor n L π

=λπ=λ +2-

3ince any integral &alue of n gi&es a solution to the original differential equations,"ith the boundary conditions that u 0 at both boundaries, the most generalsolution is one that is a sum of all possible solutions, each multiplied by adifferent constant. n the general solution for one &alue of n, "hich "e can no"

"rite as /sin'λn$(Csinh'λny(, "ith λn nπ$@L, "e can "rite the product of t"oconstants, /C, as the single constant, Cn., "hich may be different for each &alueof n. #he general solution "hich is a sum of all solutions "ith different &alues ofn is "ritten as follo"s

L

nwith y xC y xu n

n

nnn

π=λλλ=∑

∞

=1

)sinh()sin(),( +4-

#he final boundary condition, the one at y %, states that u'$,%( is a gi&enfunction of $, u'$(. 3etting u'$,%( u'$( in equation +4- gi&es the follo"ingresult.

∑∞

=

π π==

1

sinhsin)(),(n

n N L

H n

L

xnC xu H xu +5-

n re&ie"ing the solution for u'$,y(, "e see that the differential equation for 1'$( isa 3turm!Liou&ille problem. #his is a result of both the form of the differentialequation for 1'$( and the boundary conditions for 1'$(, implied by the boundaryconditions that u'0,y( 0 and u'L,y( 0. #hese boundary conditions can besatisfied only if 1'0( 0 and 1'L( 0, gi&ing the homogenous boundaryconditions required for a 3turm!Liou&ille problem.

;ecause "e ha&e a 3turm!Liou&ille problem for 1'$(, "e *no" that the solutionssin'nπ$@L( form a complete orthogonal set o&er the region 0 A $ A L. )e can usethis fact in equation +5- to get a solution for Cm. f "e multiply both sides by

sin'mπ$@L(, "here m is another integer, and integrate from a lo"er limit of ?ero toan upper limit of L, "e get the follo"ing result.


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∫ ∑∫

∫ ∑∫

π

π=

π

π

π=

π

π

π=

π

∞

=

∞

=

L

m

n

L

n

L

n

n

L

N

dx L

xm

L

H mC dx

L

H n

L

xn

L

xmC

dx L

H n

L

xn

L

xmC dx

L

xm xu

0

2

1 0

0 10

sinsinhsinhsinsin

sinhsinsinsin)(

+7-

n the second ro" of equation +7- "e can re&erse the order of summation andintegration because these operations commute. )e then recogni?e that theintegrals in the summation all &anish unless m n, lea&ing only the sine!squaredintegral to e&aluate. 3ol&ing for Cm and e&aluating the last integral

4 in equation+7- gi&es the follo"ing result.

π

π

=

π

π

π

=∫

∫

∫

L

H m L

dx L

xm xu

dx

L

xm

L

H m

dx L

xm xu

C

L

N

L

L

N

m

sinh

sin)(2

sinsinh

sin)(0

0

2

0+6-

For any initial condition, then, "e can perform the integral on the right hand sideof equation +6- to compute the &alues of Cm and substitute the result intoequation +4- to compute u'$,y(. For e$ample, consider the simplest case "hereu'$( B, a constant. n this case "e find Cm from the usual equation for theintegral of sin'a$(.

( )[ ]π−π=

ππ+

ππ−=

ππ−=

π=

π=

π ∫ ∫

mm

U

L

m

m

L

L

Lm

m

L

L

U

L

xm

m

L

L

U dx

L

xm

L

U dx

L

xm xu

LC

L

H m

N N

L

N

L

N

L

N m

cos120

coscos2

cos2

sin2

sin)(2

sinh

000

+-

3ince cos'mπ( is "hen m is e&en and = "hen m is odd, the factor of =cos'mπ( is ?ero "hen m is e&en and 2 "hen m is odd. #his gi&es the final resultfor Cm sho"n belo".

π

=evenm

odd mm

U

C

N

m

0

4

+:-

3 Bsing a standard integral table, and the fact that the sine of ?ero and the sine of an integralmultiple of π is ?ero, "e find the follo"ing result

2

2sin

42

2sin

42sin

00

2 L

L

Lm

m

L L

L

xm

m

L xdx

L

xm L L

=

−=

−=

∫

π

π

π

π

π


5/27

3ubstituting this result into equation +4- gi&es the follo"ing solution to thediffusion equation "hen u'$,%( u'$( B, a constant, and all other boundary&alues of u are ?ero.

L

n

H n

y xU t xu n

n n

nn N π=λ

λ

λλ

π

= ∑∞

=

5,3,1 )sinh(

)sinh()sin(4),( +9-

f "e substitute the equation for λn into the summation terms "e get the follo"ingresult.

∑∞

=

π

π π

π=

5,3,1 sinh

sinhsin4

),(n

N

L

H nn

L

yn

L

xn

U t xu +20-

f "e compare this equation to equation +9- in the notes on the solution of thediffusion equation, "e see that the sine terms are the same. #he e$ponential

time term, 2max

22

x

t n

e

απ− , in the diffusion equation, has been replaced by the

hyperbolic sine terms in equation +20-. #his similarity arises from the fact that "eha&e the same ordinary differential equation, "ith the same ?ero boundaryconditions, for the &ariable 1'$( in each problem. #his is an important feature ofseparation of &ariable solutions. #he contribution to the solution from a term li*e

2

2

xu

∂∂ "ill be the same in different partial differential equations "here the

separation of &ariables solution "or*s and that the boundary conditions for the

term are the same.

8quation +20- sho"s that the ratio u'$,y(@B depends on the dimensionlessparameters $@L, y@L, and %@L. #hus "e can compute uni&ersal plots of thesolution for different &alues of %@L. / contour plot of the solution u'$,y( "here B and %@L is sho"n in Figure belo".


6/27

Figure 1. MATLAB Plot of Laplace Equation Solution

#he e$pansion of the boundary condition that u'y,%( u'$( by eigenfunctions ofa 3turm!Liou&ille solution required homogenous boundary conditions. #hisrequirement for of ?ero boundary conditions at all boundaries e$cept the one at y

0 appears to be a limitation on the approach used here. %o"e&er, "e cangenerali?e the solution done here to more general problems by using theprinciple of superposition. n this approach "e loo* to the solution of fourproblems li*e the one done abo&e and use the results to obtain the solution tothe follo"ing general problem.

)(),()()0,()(),()(),0(

0,002

2

2

2

xu H xu xu xu yu y Lu yu yu

H y L x y

u

x

u

N S W W ====

≤≤≤≤=∂∂

+∂∂

+2-

'#he subscripts , 3, 8, and ) refer to the north, "est, east and "est sides ofthe rectangular region.( #o sol&e this general problem, "e sol&e four separateproblems li*e "e >ust did for equation +- for four separate potentials, u'$,y(, u2,'$,y(, u4'$,y(, and u5'$,y(. #he desired result is the sum of these potentials.

u'$,y( u'$,y( D u2'$,y( D u4'$,y( D u5'$,y( +22-

8ach of the subscripted functions, ui'$,y( represents a solution of the differentialequation in equation +2-, but "ith different boundary conditions. #he different


7/27

boundary conditions for the different dependent &ariables, u i'$,y(, are sho"nbelo".

0),(),()0,()(),0(0),0(),()0,()(),(

0),(),(),0()()0,(

0)0,(),(),0()(),(

4444

3333

2222

1111

========

========

y Lu H xu xuand xu yu yu H xu xuand xu y Lu

H xu y Lu yuand xu xu

xu y Lu yuand xu H xu

E

W

S

N

+24-

#he solution posed in equation +22- "ith the boundary conditions in equation +24-is a complete solution to the differential equation and boundary conditions inequation +20-. 3ince each function ui'$,y( satisfies the homogenous differentialequation, the sum of these four functions also satisfies the differential equation.Furthermore, equation +25- sho"s that the boundary conditions on the indi&idualui'$,y( solutions, gi&e the follo"ing boundary conditions on u'$,y(.

)()(000),0(),0(),0(),0(),0(

)(0)(00),(),(),(),(),(

)(00)(0)0,()0,()0,()0,()0,(

)(000)(),(),(),(),(),(

4321

4321

4321

4321

xu xu yu yu yu yu yu

xu xu y Lu y Lu y Lu y Lu y Lu

xu xu xu xu xu xu xu

xu xu H xu H xu H xu H xu H xu

E E

W W

S S

N N

=+++=+++==+++=+++==+++=+++=

=+++=+++=

+25-

#hus, the solution proposed in equation +22-, "ith the boundary conditions inequation +24- satisfies the differential equation and the boundary conditions of theoriginal problem in equation +2-.

#he solution for u'$,y( is the one found abo&e and is gi&en by equation +4-.

∑∞

= π

π

= 1)1(

1 sinhsin),(n

n L

yn

L

xnC y xu +27-

#he Cn coefficient in this equation is gi&en by equation +6-, "hich is copiedbelo".

π

π

=∫

L

H n L

dx L

xn xu

C

L

N

n

sinh

sin)(20)1( +26-

#he solution for u4'$,y( is similar to the one for u'$,y( e$cept that the roles of the$ and y coordinates, and the corresponding ma$imum lengths, % and L, arere&ersed. 8$changing $ and y and re&ersing % and L in equations +27- and +26-gi&e us the solution for u4'$,y(.

∑∞

=

ππ=

1

)3(3 )sinh()sin(),(

n

n H

xn

H

ynC y xu +2-


8/27

#he Cn coefficient in this equation is gi&en by equation +6-, "ith an interchangeof $ and y, an interchange of L and %, and use of u)'y( as the boundarycondition. #he resulting equation is sho"n belo".

π

π

= ∫ H

Ln H

dy

H

yn yu

C

H

W

n

sinh

sin)(20)3( +2:-

#he $ dependence of the solution for u2'$,y( "ill be the same as in the solutionfor u'$,y(. #his holds because the solution obtained by separation of &ariables isa product of t"o separate solutions 1'$( and 'y( and the $ boundary conditionsfor u2'$,y( are the same as those for u'$,y(. f "e ma*e a coordinatetransformation to a ne" coordinate, y’ % = y, the boundary conditions in the y’coordinate system for u2 "ill ha&e the same form as those for u in the ycoordinate system. Furthermore, the differential equation has only the second

deri&ati&e in y. )e can find the second deri&ati&e for the ne" y’ coordinate asfollo"s.

( )22

2

2

''')1(

'

'

')1(

'

'

y

u

y

u

y y

u

y y

y

y

uand

y

u

y

u

y

y

y

u

∂∂=

∂∂−

∂∂−=

∂∂

∂∂

∂∂=

∂∂

∂∂−=

∂∂

∂∂=

∂∂

+29-

#hus Laplace’s equation has the same form, 2

2

2

2

y

u

x

u

∂∂

+∂∂

, for both the y and the y’

coordinate system. #his means that the solution for u2 in the y’ coordinate

system "ill be the same as u in the y coordinate system. #hus "e can ta*e thesolution for u in equations +27- and +26- and replace y in that solution by y’ % =y to get the correct solution for u2. ')e also replace the old boundary conditionfor u by the one for u3.(

∑∞

=

−π π=

1

)2(2

)(sinhsin),(

nn

L

y H n

L

xnC y xu +40-

π

π

=∫

L

H n L

dx L

xn xu

C

L

S

n

sinh

sin)(20)2( +4-

8$cept for the substitution of the Esouth boundary condition, u3'$( that applies aty 0, the equation for C'2( is the same as that for C'(. )e can apply the samecoordinate transformation in the $ direction and use the same logic to obtain thesolution for u5 from the solution for u4, by replacing $ by L = $ in that solution.


9/27

∑∞

=

−π π=

1

)4(4

)(sinhsin),(

n

n H

x Ln

H

ynC y xu +42-

π

π

=

∫

H

Ln H

dy H

yn yu

C

H

E

n

sinh

sin)(2

0)4(

+44-

#hese results sho" ho" "e can obtain a solution "ith general boundaryconditions, "hile retaining the ability to de&elop eigenfunction e$pansions "hichrely upon the e$istence of ?ero boundary conditions.

Solution of Laplace! equation "aed on comple# $aria"le

Background on complex variables = / comple$ &ariable is a quantity that has areal and an imaginary part. 3uch numbers are usually "ritten as ? $ D iy "here

i 1− . )e can also "rite comple$ &ariables as reiθ, "here r 22 y x + and θ tan!'y@$(. #he relationship bet"een these t"o representations of a comple$number is based on 8uler’s formula

eiθ cos'θ( D i sin'θ( +50-

#his gi&es the relationships that $ r cos'θ( and y r sin'θ(.

#he rules for operations "ith comple$ numbers are sho"n belo". n these rules

?, ?2, and ?4 are comple$ numbers defined such that ?* $* D i y*, r *e!θ* and c is

a real number.

?4 ? ± ?2 G $4 $ ± $2 and y4 y ± y2 +5-

?2 c? G $2 c$ and y2 cy +52-

2121321213213 x y y x yand y y x x x z z z +=−=⇒= +54-

213213213 θ+θ=θ=⇒= and r r r z z z +55-

2

2

2

2

212132

2

2

2

21213

2

13

y x

y x x y yand

y x

y y x x x

z

z z

+

−=

+

+=⇒= +57-

213

2

1

3

2

1

3 θ−θ=θ=⇒= and

r

r r

z

z z +56-

#he comple$ con>ugate of a comple$ number ?, "hich is denoted as z or ?H, isfound by setting i to =i. #hus if ? $ D iy reiθ, z $ ! iy re!iθ. From thedefinitions of r and the comple$ con>ugate, "e see that r z z .


10/27

)e can ha&e some comple$ &ariables, "hich are functions of other comple$&ariables. 3uch a functional relationship is usually "ritten in terms of the realpart, u'$,y( and the imaginary part &'$,y( as follo"s

f'?( u'$,y( D i&'$,y( +5-

For e$ample, if f'?( ?2 '$ D iy(2 $2 D2i$y D 'iy(2 $2 = y2 D 2i$y, "e "ouldha&e u'$,y( $2 = y2 and &'$,y( 2$y. )hen "e consider the deri&ati&e of acomple$ function, using the usual definition of a deri&ati&e from the first course incalculus, "e ha&e the follo"ing result.

z

z f z z f

z dz

z df

∆−∆+

→∆= )()(

0

lim)(+5:-

3ince ? represents a t"o!dimensional space 'containing both the $ and ydirections( "e may ha&e different definitions of this deri&ati&e depending on thedirections in "hich "e approach the limit. /n important result of comple$

analysis, *no"n as the Cauchy!Iiemann conditions states that f'?( isdifferentiable at ? if the follo"ing conditions hold.

y

ui

y

v

x

vi

x

u

dz

df then

y

u

x

vand

y

v

x

u If

∂∂

−∂∂

=∂∂

+∂∂

=∂∂

−=∂∂

∂∂

=∂∂

+59-

n the pre&ious e$ample of f'?( ?2 "here u'$,y( $2 = y2 and &'$,y( 2$y. )eha&e the follo" results "hen "e apply the Cauchy!Iiemann conditions

y y

u y

x

vand x

y

v x

x

u2222 =

∂∂

−==∂∂

=∂∂

==∂∂

+70-

#hus the Cauchy!Iiemann conditions are satisfied and "e can "rite thederi&ati&e df@d? as follo"s.

)2(222 yi x y

ui

y

v

dz

df or yi x

x

vi

x

u

dz

df −−=

∂∂

−∂∂

=+=∂∂

+∂∂

= +7-

ote that this result is simply the result "e "ould ha&e obtained by ta*ing thederi&ati&e of f'?( ?2 directly as df@d? 2? 2'$ Diy(. Functions that satisfy theCauchy!Iiemann conditions and are continuous are called analytic functions.

#he Cauchy!Iiemann conditions pro&ide the lin* bet"een the comple$ analysis

and the solutions to Laplace’s equation. f the ta*e x∂∂ of the first condition and

y∂∂ of the second condition, "e ha&e the follo"ing results.

2

222

2

2

y

u

x y

v

y

u

x

v

yand

y x

v

x

u

y

v

x

u

x ∂∂−=

∂∂∂⇒

∂∂−=

∂∂∂∂

∂∂∂=

∂∂⇒

∂∂=∂∂∂∂

+72-


11/27

3ince x y

v

y x

v

∂∂

∂=

∂∂

∂ 22

, subtracting the t"o results in equation +72- sho"s that u

satisfies Laplace’s equation.

02

2

2

2

2

222

2

2

=∂

∂+∂

∂⇒

∂

∂−=∂∂

∂−

∂∂

∂=∂

∂

y

u

x

u

y

u

x y

v

y x

v

x

u+74-

n a similar "ay, "e could ta*e y∂∂ of the first Cauchy!Iiemann condition and

x∂∂ of the second condition, and use the same manipulations of equations +72-and +74- to sho" that the imaginary part of the analytical function also satisfiesLaplace’s equation.

02

2

2

2

=∂∂

+∂∂ y

v

x

v+75-

/lthough "e ha&e approached this result by considering the requirements for acomple$ &ariable to ha&e a deri&ati&e, "e ha&e obtained an important result forsolutions of Laplace’s equations. f "e ha&e one solution, u'$,y( to Laplace’sequation, "e "ill also ha&e a companion solution, &'$,y( related to the originalsolution by the Cauchy!Iiemann conditions of equation +59-, repeated belo".

y

y xu

x

y xvand

y

y xv

x

y xu

∂∂−=

∂∂

∂∂=

∂∂ ),(),(),(),(

+59-

)e can sho" that these solutions are mutually perpendicular as follo"s. #he&ector gradients of the t"o components, u and & can be e$pressed in terms ofthe unit &ectors in the $ and y directions 'i and %( as follo"s.

ji ji y

v

x

vvand

y

u

x

uu

∂∂

+∂∂

=∇∂∂

+∂∂

=∇ +77-

)e find the angle bet"een these t"o gradients by ta*ing their dot product.Iecalling that i&i %&% and %&i iJ % 0, "e get the follo"ing result for the dotproduct of these t"o &ectors.

y

u

y

v

x

v

x

u

y

u

x

v

y

v

x

u

y

u

y

v

x

v

x

uvu

∂∂

∂∂+

∂∂

∂∂=⋅

∂∂

∂∂+

∂∂

∂∂+⋅

∂∂

∂∂+⋅

∂∂

∂∂=∇⋅∇ ji j jii +76-

)e can use the Cauchy!Iiemann conditions to sho" that this &ector dot productis ?ero. 3ubstituting those conditions from equation +59- gi&es

0=

∂∂

−∂∂

+∂∂

∂∂

=∂∂

∂∂

+∂∂

∂∂

=∇⋅∇ x

v

y

v

x

v

y

v

y

u

y

v

x

v

x

uvu +76-


12/27

3olutions of Laplace’s equations that ha&e continuous, second!order partialderi&ati&es are *no"n as harmonic functions. #he physical relationships thatlead to Laplace’s equation commonly use the concept of a potential. Fore$ample, in a t"o!dimensional, in&iscid flo", "e can define a &elocity potential,

Φ, such that the t"o &elocity components, &$ and &y are gi&en by the equations

yvand

xv y x ∂

Φ∂−≡

∂Φ∂−≡ +77-

For an incompressible fluid flo", the continuity 'conser&ation of mass( equationrequires the follo"ing relationship hold among the &elocity components in a t"o!dimensional flo".

0=∂

∂+

∂∂ y

v

x

v y x +76-

3ubstituting equations +77- into equation +76- sho"s that the &elocity potentialsatisfies Laplace’s equation.

02

2

2

2

=∂Φ∂+

∂Φ∂

y x+7-

#hus "e should be able to define a second &ariable, Ψ, "hich also satisfiesLaplace’s equation.

02

2

2

2

=∂Ψ∂+

∂Ψ∂

y x+7:-

Furthermore, lines of constant Ψ "ill be perpendicular to lines of constant Φ. naddition, the solutions for Ψ and Φ "ill satisfy the Cauchy!Iiemann conditions

y

y x

x

y xand

y

y x

x

y x

∂Φ∂−=

∂Ψ∂

∂Ψ∂=

∂Φ∂ ),(),(),(),(

+79-

From these conditions and the definitions of the &elocity potential, Φ, in equation+77-, "e ha&e the follo"ing definitions of the &elocity components in terms

gradients of the Ψ.

x yvand

y xv y x ∂Ψ∂=∂Φ∂−≡∂Ψ∂−=∂Φ∂−≡ +60-

n fluid mechanics, Ψ is called the stream function.


13/27

'mportant reult from comple# anal(i

#he solutions to Laplace’s equation can be used to construct a comple$ &ariable,F'?( u'$,y( D i&'$,y(. #he Cauchy theorem for integration of comple$ &ariablescan be applied to F'?( in a solution region of arbitrary 't"o!dimensional( shape to

obtain important results for Laplace’s equation in t"o dimensions. '3ee section6.6 of Kreys?ig for proofs of these statements.(

. #he solutions to Laplace’s equation ha&e a minimum and a ma$imum onthe boundary.

2. f the boundary has a specified &alue '


14/27

Iegardless of the boundary conditions, the solution proceeds by separation of&ariables. )e postulate a solution that is the product of t"o functions, '?( afunction of the ? coordinate only and M'r( a function of the radial coordinate, r,only. )ith this assumption, our solution becomes.

u'r,t( M'r('?( +62-

)e substitute equation +62- for u in equation +6-. 3ince M'r( is a function of ronly and '?( is a function of ? only, "e obtain the follo"ing result.

[ ]

[ ]r

r r

r r z !

r

z ! r r

r r r

ur

r r

z

z ! r

z

z ! r

z

u

∂∂∂∂=

∂∂∂∂=

∂∂∂∂

=∂∂=

∂∂=

∂∂

)(1)(

)()(11

)()(

)()(2

2

2

2

2

2

+64-

3ubstituting these results into equation +62- and di&iding the result through by the

product M'r('?(, gi&es follo"ing.

2

2

2

2)(

)(

1)(1

)(

10

)(

)(

1)(1

)(

1

z

z !

z ! r

r r r r r z

z !

z ! r

r r r r r ∂

∂−=

∂∂

∂∂

⇒=∂

∂+

∂∂

∂∂

+65-

8quation +65- says that a function of r only equals a function of ? only. #his canonly be true if each term equals a constant. /s before, "e "ant to choose theconstant to gi&e us a 3turm!Liou&ille problem in the direction that hashomogenous boundary conditions. )e set the term "ith homogenous boundary

conditions to be equal to −λ2. #hen the other term "ill be equal to λ2. For a solid

cylinder, no boundary condition is required at r 0. t is sufficient to require thatu remain finite at r 0 to get a 3turm!Liou&ille problem. f "e ha&e a ?eroboundary condition in the radial direction, "e can get a solution to the radialproblem in terms of ;essel functions and the solutions in the ? direction "ill be interms of hyperbolic sines and cosines. f "e ha&e homogenous boundaryconditions in the ? directions, "e can obtain eigenfunction solutions in terms ofsines and cosines in this direction and the radial solution "ill be in terms ofmodified ;essel functions.

/s usual in solutions of Laplace’s equation, "e can use superposition if "e donot ha&e a sufficient number of homogenous boundary conditions.

)e start "ith a simple set of boundary conditions for a solid cylinder u'I,?( u'0,I( 0 and u'r,%( u'%( this gi&es a 3turm Liou&ille problem in the radial

direction so "e "ill "ant to set the radial term equal to λ2 this requires us to setthe ? term to λ2 to satisfy equation +65-.


15/27

z B z A z ! z ! dz

z ! d λ+λ=⇒=λ− coshsinh)(0)()( 22

2

+67-

0)()()(0)()(

002 =λ+λ=⇒=λ+ " DY "C# r r r

dr

r d r

dr

d +66-

%ere N0 and 0, are the ;essel functions of the first and second *ind "ith ?eroorder.5 #hus, our general solution for u'$,t( 1'$('?( becomes

[ ][ ])()(coshsinh)()(),( 00 r DY r C# z B z Ar z ! z r u λ+λλ+λ== +6-

/s r O 0, 0'r( O !P to *eep the solution finite, "e require that C 2 0. #o satisfythe condition that u'I,?( 0, "e must satisfy the follo"ing equation.

)()(0),( 0 " # z ! z "u λ== +6:-

8quation +6:- defines a transcendental equation similar to that in equation +-"hich required sin'Q$ma$( 0. #hat equation has a simple solution since "e *no"that the sin'nR( 0 "hen n is an integer. )e do not ha&e such a simple resultfor the equation that N0'QI( 0. %o"e&er, the points at "hich N0 0 can bedetermined and are a&ailable in tables. )e use the symbol Smn to indicate the m

th

point "here Nn is ?ero. #hat is "e define Smn by the follo"ing equation.

∞==α ,,10)( m # mnn +69-

For e$ample, the first fi&e points at "hich N 0 is ?ero are for S0 2.505:4, S20 7.7200:, S40 :.6744, S50 .974, and S70 5.94092. #here are an infinite

number of such &alues such that N0'Qmr(, "ith QmI Sm0 pro&ides a complete setof orthogonal eigenfunctions that can be used to represent any other function

o&er 0 A r A I. #hese αmn &alues are called the ?eros of Nn.

)e can apply the boundary condition that u'r,0( 0 to equation +6- '"ith < setequal to ?ero to eliminate the 0 term( to obtain.

4 ;essel functions, li*e sines and cosines, are solutions to differential equations. /lthough ;esselfunctions occur less frequently than sines and cosines, they are functions that can be found intables or in computer function calculations. #he general form of ;essel’s equation used for

obtaining series solutions is 0)(

)(1)(2

22

2

2

=

−

++ x y x x

dx

xdy

xdx

x yd ν

. #his equation can be

transformed into 02

2

=

+−+

y z $ z dz

dy z

dz

d ν "hose solution is /N ν'*?( D ; ν'*?(. )e see that

this second equation has the same form as 0)()( 2 =+

∂∂

∂∂

r r r

r r r

λ , pro&ided that "e set ν

0. #his gi&es the result abo&e that the solutions are N0 and 0. #he factor of r in the Q2rM'r( term

is a "eighting factor that must be included in the definition for the inner product of solutions to theradial portion of the diffusion equation.


16/27

[ ] 0)()(0cosh0sinh)()0()0,( 00 =λ=λλ+λ== r B# r C# B Ar ! r u +0-

#his gi&es ; 0. #he general solution to the problem is the sum of alleigen&alue solutions, each multiplied by a different constant.

01

0 )()sinh(),( mmm

mmm "r # z C z r u α=λλλ= ∑∞

=+-

)e still ha&e to satisfy the boundary condition that u'r,%( u'r(. )e can do thisby using an eigenfunction e$pansion. 3etting ? % in equation +- gi&es thefollo"ing equation for the boundary condition at y %.

∑∞

=λλ==

10 )()sinh()(),(

mmmm N r # H C r u H r u +2-

#he &alues of Cm are found from the general equation for orthogonaleigenfunction e$pansions "hich includes a "eighting function. %ere the"eighting function p'r( is equal to r. For any initial condition, u0'r(, "e find the&alues of Cm from the follo"ing equation. 'n the second equation belo", theintegral in the denominator has been e&aluated using integral tables and the fact

that N0'λmI( 0.(

[ ] [ ]212

0

0

0

2

0

0

0

)(2

)sinh(

)(

)()sinh(

)()(

" # "

H

dr ur r#

dr r # r H

dr r ur r#

C

mm

"

N m

"

mm

"

N m

m

λλ

λ=

λλ

λ=

∫

∫

∫ +4-

f "e consider the simple case "here u'r( is a constant equal to B. "e ha&e thefollo"ing result for Cm.

[ ] [ ]

[ ] )sinh()(

2

)(

)sinh(

2

)(

)(

)sinh(

2

)(

)(

)sinh(

2

)(2

)sinh(

)(

01012

12

1

2

1

0

1

22

1

20

0

H #

U

" "#

H

U

" # "

" "#

H

U

" #

r r#

H "

U

" # "

H

Udr r r#

C

mmmmm

m

m

m

m

m

m

"r

r m

m

mmm

"

m

m

λαα=λλ

λ

=λ

λλ

λ

=

λ

λλ

λ=

λλ

λ=

=

=∫

+5-

)ith this result for Cm our solution for u'r,t( sho"n in equation +- "ith theconstant boundary condition u'r(, is "ritten as sho"n belo", using the result

from equation +- that λm αm0@I.


17/27

∑∞

=

α

α

ααα

=1

000

0010

sinh

sinh)(

2),(

m

mm

mmm "

z

"

r #

"

H #

U z r u

+7-


18/27

"ur # z C z r u mm "

mmmm

0

00 )()sinh(),(

α=λ+λλ= ∑∞

=+-

/pplying the same integration that "as used to obtain equation +5- gi&es thefinal solution for u'r( B, a constant, as

( )∑∞

=

α

α

ααα

−=−

1000

0010

sinh

sinh)(

2),(

mmm

mmm

" "

"

z

"

r #

"

H #

uU ut r u

+:-


19/27

H

mr I z C z r u m

mmmm

π=λλλ= ∑∞

=10 )()sin(),( +:2-

#he usual eigenfunction e$pansion gi&es the boundary condition that u'I,?( uI'?(. 3ol&ing this eigenfunction e$pansion for Cm gi&es.

π

π

=

π

π

π

=∫

∫

∫

H

"m HI

dz H

z m z u

dx H

z m

H

"m I

dz H

z m z u

C

H

"

H

H

"

m

0

0

0

20

0

sin)(2

sin

sin)(

+:4-

#he solutions in equations +- and +:2- 'along "ith the corresponding equationsfor Cm in equations +4- and +:4-( can be used along "ith the principle ofsuperposition to obtain solutions to Laplace’s equation in a solid cylinder for morecomple$ boundary conditions.

+ollo, c(linder ,ith "oundar( condition at outer radiu * f "e replace theproblem in equation +9- that "e >ust sol&ed for a solid cylinder "ith a similarproblem for the hollo" cylinder, "e ha&e to include a boundary condition at theinner radius. #his gi&es the problem sho"n belo".

)(),(0),(),()0,(

0,01

2

2

z u z "u z "u H r ur u

H z "r " z

u

r

ur

r r

ooi

oi

====

≤≤≤≤=∂∂+

∂∂

∂∂

+:5-

#he solution to this equation proceeds in e$actly the same "ay as the solution to

the equation +9- problem. %o"e&er, "e can not eliminate the K0 term in thiscase. )ith the eigenfunction solutions in the ? direction, a typical t"odimensional solution could be "ritten as follo"s.

[ ] ( ) ( ) H

m z r C z r & Dr I C z r u mmmmmmmmmm

π=λλ=λλ+λ= sin)(sin)()(),( 00 +:7-

/pplying the inner!radius boundary condition that u'Ii,?( 0 to equation +:7-gi&es the follo"ing relationship bet"een Cm and


20/27

3ubstituting this result into equation +:7- and summing o&er all &alues of m gi&esthe general solution for u'r,?(.

H

mr z C z r u m

mmmm

π=λλ=∑∞

=1)()sin(),( +::-

)e can fit the boundary condition that u'I0,?( uo'?( by an eigenfunctione$pansion of this boundary condition. n the same "ay that equation +:4- "asderi&ed, "e can "rite

( ) ( )om

H

o

H

om

H

o

m " H

dz H

z m z u

dx H

z m "

dz H

z m z u

C

∫

∫

∫

π

=

π

π

= 0

0

2

0

sin)(2

sin

sin)(

+:9-

Bsing the definition of the radial solution from equation +:- gi&es the solution tothe problem in equation +:5- as follo"s.

∑∫ ∞

=

π

π

π

−

π

π

π

π

−

π

π

π

=1

0

0

0

0

0

0

0

0

0 sin

sin)(2

),(m

o

i

i

o

i

i

H

o

H

"m &

H

"m

&

H

"m I

H

"m I

H

r m &

H

"m &

H

"m I

H

r m I

H

z m

H

dz H

z m z u

z r u +90-

/s an e$ercise you can find the solution to a modification of the problem inequation +:5- "here the radial boundary conditions are modified to ha&e u'Io,?( 0 and u'Ii,?( ui'?(.

+ollo, c(linder ,ith non-ero "oundar( condition on top * n the pre&iousproblem of a hollo" cylinder "e had an eigenfunction e$pansion in the ?direction. #he follo"ing problem "ill use an eigenfunction e$pansion in the radialdirection.

)(),(0),(),()0,(

0,01

2

2

r u H r u z "u z "ur u

H z "r " z

u

r

ur

r r

N oi

oi

====

≤≤≤≤=∂∂+∂∂∂∂

+9-

#he solution to this problem proceeds in the same "ay as the solution of theinitial radial geometry problem that lead to equation +6- copied belo".


21/27

)()(coshsinh)()(),( 00 r DY r C# z B z Ar z ! z r u λ+λλ+λ== +6-

/lthough the boundary condition that u'r,0( 0 gi&es ; 0, "e can no longerdrop the 0 term because our present solution domain does not include r 0.#hus the boundary conditions that u'Ii,?( u'Io,?( 0 require that the radial

solution, M'r( &anish at the inner and outer radius. #his gi&es us the follo"ingpair of equations.

0)()(0)()( 0000 =λ+λ=λ+λ ooii " DY "C# " DY "C# +92-

#his is a pair of homogenous algebraic equations that has the matri$ equationsho"n belo".

=

λλλλ

0

0

)()(

)()(

00

00

D

C

"Y " #

"Y " #

oo

ii+94-

#his pair of equations "ill ha&e a tri&ial solution C < 0 unless the determinantof the coefficients &anishes.

0)()()()()()(

)()(0000

00

00 =λλ−λλ=

λλλλ

iooi

oo

ii "Y " # "Y " #

"Y " #

"Y " # Det +95-

8quation +95- is a transcendental equation that can be sol&ed for the eigen&alues

for this problem. f "e define α λIo, "e can re"rite equation +95- as follo"s.

0)()( 0000 =

αα−α

α

o

i

o

i

"

"Y # Y

"

" # +97-

#hus, the eigen&alues in this problem depend on the ratio of the inner radius tothe outer radius. / plot of equation +97- for three different radius ratios is sho"nin Figure 2.

Figure . Plot of Equation /02 a a Function of 3 λ

)o


22/27

;ecause the determinant in the equation for C and < is ?ero, there are an infinitenumber of solutions for these &ariables, but all solutions ha&e the same ratio of Cto


23/27

/s usual, the general solution for u'r,?( is the sum of all the eigenfunctionsolutions as sho"n in equation +99-. n the general solution "e can combine theconstant product /8 into a single constant Cm, "hich "ill be different for eacheigenfunction.

∑∞

= λλ= 10 )()sinh(),(

mmmm z z C z r u +00-

#he boundary condition that u'r,%( u'r( can be found from an eigenfunctione$pansion. #he normali?ation integral in this e$pansion is gi&en belo".

[ ][ ]20

020

202

0)(

)]()([2)(

0

imm

mim

"

"

m " #

" # " # dr r r

iλπλλ−λ

=λ∫ +0-

Bsing this result, the solution for Cm is

[ ]

[ ]

)]()([2)sinh(

)()()(

)()sinh(

)()(

020

20

0

2

0

2

0

0

0

0

0

" # " # H

dr r r ru " #

dr r r H

dr r r ru

C mimm

"

"

m N imm

"

"

mm

"

"

m N

mi

i

i

λ−λλ

λλπλ

=

λλ

λ

=∫

∫

∫ +02-

#he case of a constant boundary potential at the top of the cylinder, u'r,%( B,gi&es the follo"ing integral.8rror Ieference source not found

)(

)]()([2)()()(

02

000

00

00

imm

mim

"

"m

"

"m N " #

" # " # U dr r rU dr r r ru

ii λπλ

λ−λ

=λλ ∫ ∫ +04-

3ubstituting this result into equation +02- gi&es.

)]()()[sinh(

)(

000

0

" # " # L

"U# C

mimm

imm λ+λλ

λπ= +05-

3ubstituting this e$pression for Cm into equation +00- gi&es the follo"ing solutionto the problem in equation +9-, "here u'r,%( u'r( B, a constant the &alues of

λm αm@Io are found from equation +97-.

∑∞

= λ+λλλ

λλ

π=1 000

00

)()(

)()(

)sinh(

)sinh(),(

m mim

mim

m

m

" # " #

r " #

H

z U z r u +07-

Bsing the roots of the eigen&alue equation, αm λmI, allo"s us to re"rite thisequation as follo"s.

7 Carsla" and Naeger, Conduction of Heat in Solids, T$ford Bni&ersity Mress, 97:.


24/27

( )∑∞

= α+

α

αα

α

α

α

π=1

00

0

00

00

0

0

)(

sinh

sinh

),(

mm

im

mimi

m

m

m

# "

" #

"

r "

"

" #

"

H

"

z

U z r u +07b-

Bsing the definition of M0 from equation +9:-, "ith λ replaced by α@I, gi&es the fullsolution as follo"s.

( )∑∞

= α+

α

αα−

αα

α

α

α

π=1

00

00000 )()(

sinh

sinh),(

mm

o

im

omm

omm

o

im

om

om

# "

" #

"

r Y #

"

r # Y

"

" #

"

H

"

z

U

z r u

+07c-

#his equation sho"s the solution as a relation bet"een the dimensionlesspotential ratio, u'r,?(@B, as a function of dimensionless spatial coordinates ?@Io and r@Io. #hat solution also depends on t"o dimensionless geometric propertyratios, %@Io, and Ii@Io.

4radient "oundar( condition

#his section presents an o&er&ie" of handling gradient boundaryconditions to guide you in their use. ou should be able to supply thedetails of the results pro&ided here based on pre&ious material in these

notes and detailed e$amples of problems "ith gradient boundaryconditions in the notes on solutions of the diffusion equation.

Consider the simple modification to the problem in equation +- "here theboundary conditions that u'0,y( u'L,y( 0 is replaced by the conditionthat the gradient of u is ?ero at both boundaries.

)(),(0)0,(

0,00

),(),0(

2

2

2

2

xu H xu xu x

u

x

u

H y L x y

u

x

u

N y L y

===∂∂

=∂∂

≤≤≤≤=∂

∂+

∂

∂

+06-

)e can still apply the separation of &ariables approach that led to equation +:-,ho"e&er the ?ero gradient boundary conditions gi&e the eigenfunctions ascosines so that the solution to equation +06- has the follo"ing form in place ofthe solution in equation +4-.

L

nwith y xC u y xu n

n

nnn

π=λλλ+= ∑

∞

=10 )sinh()cos(),( +0-


25/27

n this solution, unli*e the solution "ith sine eigenfunctions, the n 0 eigen&aluecorresponds to a non?ero solution, cos'0( . )e ha&e separated out the n 0eigenfunction in this solution, because it has a different form. #his different formarises from the basic separation of &ariables solution in equation +7-. )hen "e

ha&e n 0, the separation constant λ 0 and the ordinary differential equations

that result from the separation of &ariables ha&e the follo"ing form in place ofequation +7-.

00)(

)(

10

)(

)(

1 002

02

02

02

0

=π

=λ== L

n

dy

yY d

yY dx

x X d

x X +0:-

#he solutions in this case are

DCy yY B Ax x X +=+= )()( 00 +09-

#he condition that ∂u@∂$ 0 at $ 0 and $ L requires d1 0'$(@d$ 0. #his issatisfied if / 0. #he condition that u 0 at y 0 requires 0'0( 0 "hich gi&es< 0. #hus the solution for the n 0 eigen&alue is

yC BCy yY x X y xu 0000 )()(),( === +0-

)ith this solution for u0, the general solution for u'$,y( in equation +0- becomes

L

nwith y xC yC y xu n

n

nnn

π=λλλ+= ∑

∞

=10 )sinh()cos(),( +-

)e ha&e the usual eigenfunction e$pansion for finding the initial conditions, but

in this case "e ha&e a separate normali?ation integral for the n 0 eigenfunction,cos'0( . #hus our general equation for the constants Cn is

≥

π

=

=

π

∫

∫

1cos)(2

0)(1

sinh

0

0

ndx L

xn xu

L

ndx xu L

C L

H n L

N

L

N

n +2-

#his difference in the prefactors @L in the n 0 case and 2@L for the n U coefficients is the same as the difference for the cosine terms in Fourier series.

f the boundary conditions in problem +06- are changed to ha&e u'0,y( 0instead of ha&ing a gradient boundary at $ 0, "hile retaining the gradientboundary condition at $ L, the solution becomes

L

nwith y xC y xu n

nnnn

2

)12()sinh()sin(),(

0

π+=λλλ=∑

∞

=+4-


26/27

/lternati&ely, if the boundary conditions in problem +06- are changed to ha&eu'L,y( 0 instead of ha&ing a gradient boundary at $ L, "hile retaining thegradient boundary condition at $ 0, the solution becomes

L

nwith y xC y xu n

n

nnn

2

)12()sinh()cos(),(

0

π+=λλλ=∑

∞

=

+5-

Mroblems "ith gradients in the y!direction "ill ha&e similar results to those sho"nabo&e "ith the sine or cosine terms in the y direction and the sinh term in the $direction.

4radient pro"lem in c(lindrical coordinate. Consider the follo"ingproblem, "hich is the same as the problem first sol&ed for equation +6-, e$ceptthat the condition that u'I,?( 0 is replaced by a ?ero gradient condition at r I.

)(),(00)0,(

01

0,02

2

r u H r ur

ur u

z

u

r

ur

r r

H z "r

N

"r

==∂∂=

=

∂

∂+

∂

∂

∂

∂≤≤≤≤

=

+7-

3olution by separation of &ariables gi&es the same result sho"n in equation +6-.#he 0 term in the radial solution is dropped out to allo" the solution to remain

finite as r approaches ?ero. #his lea&es the radial solution as M'r( N0'λr(. #hegradient boundary condition at r I requires the radial solution deri&ati&e,dM'r(@dr 0 at r I. #his radial boundary condition requires:

0)()()(

10 =λλ−=λ

===

" #

dr

r d#

dr

r d

"r "r

+6-

n this problem, the eigen&alues, λm αm@I, "here αm are the ?eros of the;essel function N. 8$cept for the difference in the definition of the eigenfunction,the solution to the problem is the same as the solution gi&en for the problem"here u'I,?( 0 sho"n in equation +-.

"r # H C z r u mm

m

mm1

1

0 )()sinh(),( α

=λλ=∑∞

=+-

#he equation for Cm "ith these eigen&alues is sho"n belo".

8 #he deri&ati&e of N0 is an application of the general equation for ;essel function deri&ati&escombined "ith the result that N !m'$( '!(

mNm'$(

( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( ) x # x # x # x x # xdx

d

dx

xd# x # x x # x

dx

d m

mm

m1110

00

001 −====⇒= −−−


27/27

[ ] )()sinh(

)(2

)()sinh(

)()(

20

0

02

0

2

0

0

0

" # H

dr ur r# "

dr r # r H

dr r ur r#

C mm

"

N m

"

mm

"

N m

m λλ

λ=

λλ

λ=

∫

∫

∫ +:-

#he only difference bet"een this equation and the Cm equation in +4- is theappearance of N0 instead of N in the denominator of the second term. #hiscomes about because the normali?ation integral has the follo"ing result.

[ ] [ ])()(2

)( 2120

2

0

2

0 " # " # "

dr r # r mm

"

m λ+λ=λ∫ +9-

n the problem sol&ed here the eigen&alues "ere gi&en by the equation N 'λmI( 0 in the Cm equation in +4-, the eigen&alues "ere gi&en by the equation N 0'λmI( 0.

Mroblems of more that one non?ero gradient boundary can be handled bysuperposition. n such cases it is necessary to use a combination of solutions"here one solution has a ?ero gradient boundary and the other solution has thedesired &alue for the gradient.

Laplace Soln

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