the laplace equation chris olm and johnathan wensman december 3, 2008
TRANSCRIPT
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The Laplace EquationThe Laplace Equation
Chris Olm and Johnathan WensmanChris Olm and Johnathan Wensman
December 3, 2008December 3, 2008
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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
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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
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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
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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
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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
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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
.
↓ ↓
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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
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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
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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
↘
↘
↘
↓□
□
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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
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Applying the general Applying the general solutionsolution
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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
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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
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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 ''
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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'
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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
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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
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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
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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
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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
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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' ''
'
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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
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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