The Laplace EquationThe Laplace Equation

Chris Olm and Johnathan WensmanChris Olm and Johnathan Wensman

December 3, 2008December 3, 2008

Introduction (Part I)Introduction (Part I)

We are going to be solving the Laplace We are going to be solving the Laplace equation in the context of electrodynamicsequation in the context of electrodynamics

Using spherical coordinates assuming Using spherical coordinates assuming azimuthal symmetryazimuthal symmetry– Could also be solving in Cartesian or cylindrical Could also be solving in Cartesian or cylindrical

coordinatescoordinates– These would be applicable to systems with These would be applicable to systems with

corresponding symmetrycorresponding symmetry Begin by using separation of variables Begin by using separation of variables

– Changes the system of partial differential Changes the system of partial differential equations to ordinary differential equations equations to ordinary differential equations

Use of Legendre polynomials to find the Use of Legendre polynomials to find the general solutiongeneral solution

Introduction (Part II)Introduction (Part II)

We will then demonstrate how to We will then demonstrate how to apply boundary conditions to the apply boundary conditions to the general solution to attain general solution to attain particular solutionsparticular solutions– Explain and demonstrate using Explain and demonstrate using

“Fourier’s Trick”“Fourier’s Trick”– Analyzing equations to give us a Analyzing equations to give us a

workable solution workable solution

The Laplace EquationThe Laplace Equation

Cartesian coordinatesCartesian coordinates

– V is potentialV is potential– Harmonic!Harmonic!

Spherical coordinatesSpherical coordinates

– r is the radiusr is the radius is the angle between the z-axis and the vector we’re is the angle between the z-axis and the vector we’re

consideringconsidering is the angle between the x-axis and our vector is the angle between the x-axis and our vector

02

2

2

2

2

2

z

V

y

V

x

V

0sin

1sin

sin

112

2

222

2

2

V

r

V

rr

Vr

rr

Azimuthal SymmetryAzimuthal Symmetry

Assuming azimuthal symmetry Assuming azimuthal symmetry simplifies the systemsimplifies the system

In this case, decoupling V from ΦIn this case, decoupling V from Φ

becomesbecomes

0sinsin

12

V

r

Vr

r

0sin

1sin

sin

112

2

222

2

2

V

r

V

rr

Vr

rr

Potential FunctionPotential Function

Want our function in terms of r Want our function in terms of r and θand θ

SoSo

– Where R is dependent on rWhere R is dependent on r– Θ is dependent on θΘ is dependent on θ

)()(),( rRrV

ODEs!ODEs!

Plugging in we get:Plugging in we get:

– These two ODEs that must be equal These two ODEs that must be equal and opposite:and opposite:

0sinsin

11 2

d

d

d

d

dr

dRr

dr

d

R

kdr

dRr

dr

d

R

21

kd

d

d

d

sin

sin

1

kRdr

dRr

dr

d

2

sinsin

kd

d

d

d

.

↓ ↓

General Solution for General Solution for R(r)R(r) Assume Assume

– Plugging in we getPlugging in we get

SoSo

– We can deduce that the equation is We can deduce that the equation is solved when solved when kk==ll or or kk=-(=-(ll+1)+1)

– So our general solution for R(r) is So our general solution for R(r) is

lrR

llll krrlllrdr

dlrr

dr

dRr

dr

d )1()*()'*( 1122

1

1)(

ll

rBArrR

k1 )(ll

General Solution for General Solution for Θ(θ)Θ(θ) Legendre polynomialsLegendre polynomials

– The solutions to the Legendre The solutions to the Legendre differential equation, where differential equation, where ll is an is an integerinteger

– OrthogonalOrthogonal– Most simply derived using Most simply derived using

Rodriques’s formula: Rodriques’s formula:

In our case x=cosIn our case x=cosθ so θ so

lx

dx

d

lxP l

l

ll 0],)1[(!2

1)( 2

)(cos)( lP

Θ(θ) Part 2Θ(θ) Part 2

Let Let ll=0:=0:

Let Let ll=3:=3:

0)0(sin)1(sin)(cossin 0

d

d

d

d

d

dP

d

d

d

d

0)sin()(cos)10(0 0 P

00

32

3

33 )1(cos)cos(!32

1sin)(cossin

d

d

d

d

d

dP

d

d

d

d

sin

2

3sincos

2

15sincos

2

3cos

2

5sin 23

d

d

d

d

d

d

cossin3)sincossin(cos15sin2

3sincos

2

15 33222

d

d

sin

2

3sincos

2

1512)(cos)13(3 2

3P

cossin3)sincossin(cos15sinsincos518 332

↘

↘

↘

↓□

□

The General Solution The General Solution for Vfor V Putting together R(r), Θ(θ) and Putting together R(r), Θ(θ) and

summing over all summing over all ll

becomes becomes

)(cos),(0

1 l

ll

lll P

r

BrArV

)()(),( rRrV

Applying the general Applying the general solutionsolution

Example 1Example 1

The potential is specified on a hollow The potential is specified on a hollow sphere of radiussphere of radius RR

What is the potential on the What is the potential on the inside of the sphere?inside of the sphere?

)(),( 0VRV

Applying boundary Applying boundary conditions and conditions and intuitionintuition We know it must take the formWe know it must take the form

And on the surface of the sphere must And on the surface of the sphere must be Vbe V00, also all B, also all Bll must be 0, so we get must be 0, so we get

The question becomes are there any AThe question becomes are there any All which satisfy this equation?which satisfy this equation?

)(cos),(

l

0l1lll

l Pr

BrArV

0l

ll

l cosθPAθ),V( rr

Yes! Yes! But doing so is trickyBut doing so is tricky First we note that the Legendre First we note that the Legendre

polynomials are a complete set of polynomials are a complete set of orthogonal functionsorthogonal functions– This has a couple consequences we This has a couple consequences we

can exploitcan exploit

ll'l

ll'

= if 12

2

if 0dPPdxxPxP

1

1 0

llll sincoscos ''

Applying this propertyApplying this property

We can multiply our general solution We can multiply our general solution by Pby Pll’’’’(cos θ) sin θ and integrate (cos θ) sin θ and integrate (Fourier’s Trick) (Fourier’s Trick)

dPVr2

12A

dPV12

2rA

dPdPP

0

o

0

o

00

l

l

lll

ll

l'

llll

l

sincos

sincos'

sincosθVsincoscosrA

''

o'

Solving a particular Solving a particular casecase

We could plug this into our equation giving We could plug this into our equation giving

In scientific terms this is unnecessarily In scientific terms this is unnecessarily cumbersome (in layman's terms this is a cumbersome (in layman's terms this is a hard integral we don’t want or need to do)hard integral we don’t want or need to do)

2V 20 /)( sin

dP2R2

12A

0

2 l

lll sincos/sin

The better wayThe better way

Instead let’s use the half angle formula to Instead let’s use the half angle formula to rewrite our potential asrewrite our potential as

Plugging THIS into our equation givesPlugging THIS into our equation gives

Now we can practically read off the values of ANow we can practically read off the values of A ll

cos121V0 coscos100 PP21V

dPPP2

1

r2

12A

0

10 coscosl

lll sincos

cos12

1V0

coscos 100 PP2

1V

Getting the final Getting the final answeranswer

Plugging these into our general Plugging these into our general solution we getsolution we get

R2

1A

2

1A 10

cosθ

cosθcosθ

cosθPrAθ),V(0l

ll

l

R

r1

2

1

rPR2

1Pr

2

1

r

100

Example 2Example 2

Very similar to the first exampleVery similar to the first example The potential is specified on a hollow The potential is specified on a hollow

sphere of radius Rsphere of radius R

What is the potential on the outside What is the potential on the outside of the sphere?of the sphere?

)(),( 0VRV

Proceeding as beforeProceeding as before

Must be of the formMust be of the form

All AAll All must be 0 this time, and again at must be 0 this time, and again at the surface must be Vthe surface must be V00, so, so

andand

)(cos),(

l

0l1lll

l Pr

BrArV

cosl

lll, P

r

BrV

01

)(),( oVRV

Fourier’s Trick againFourier’s Trick again

By applying Fourier’s Trick again we By applying Fourier’s Trick again we can solve for Bcan solve for Bll

As far as we can solve without a specific potentialAs far as we can solve without a specific potential

dPV2

12rB

dPV12

2

r

B

0

o1

0

o1

l

l

ll

l

lll

sincos

sincos' ''

'

ConclusionConclusion

By solving for the general solution we By solving for the general solution we can easily solve for the potential of can easily solve for the potential of any system easily described in any system easily described in spherical coordinatesspherical coordinates

This is useful as the electric field is the This is useful as the electric field is the gradient of the potentialgradient of the potential– The electric field is an important part of The electric field is an important part of

electrostaticselectrostatics

ReferencesReferences

1)Griffiths, David. Introduction to Electrodynamics. 3rd 1)Griffiths, David. Introduction to Electrodynamics. 3rd ed. Upper Saddle River: Prentice Hall, 1999.ed. Upper Saddle River: Prentice Hall, 1999.

2)Blanchard, Paul, Robert Devaney, and Glen Hall. 2)Blanchard, Paul, Robert Devaney, and Glen Hall. Differential equations. 3rd ed. Belmont: Thomson Differential equations. 3rd ed. Belmont: Thomson Higher Education, 2006.Higher Education, 2006.

3) White, J. L., “Mathematical Methods Special 3) White, J. L., “Mathematical Methods Special Functions Legendre’s Equation and Legendre Functions Legendre’s Equation and Legendre Polynomials,” Polynomials,”

http://www.tmt.ugal.ro/crios/Support/ANPT/Curs/http://www.tmt.ugal.ro/crios/Support/ANPT/Curs/math/s8/s8legd/s8legd.html, accessed 12/2/2008.math/s8/s8legd/s8legd.html, accessed 12/2/2008.

4) 4) WeissteinWeisstein, Eric W., Eric W. "Laplace's Equation--Spherical "Laplace's Equation--Spherical Coordinates." From Coordinates." From MathWorldMathWorld--A 5) Wolfram Web --A 5) Wolfram Web Resource. Resource. http://mathworld.wolfram.com/LaplacesEquationSpherhttp://mathworld.wolfram.com/LaplacesEquationSphericalCoordinates.htmlicalCoordinates.html, accessed 12/2/2008, accessed 12/2/2008