Solutions to Laplace’s Equation in CylindricalCoordinates and Numerical solutions

Lecture 8

1 Introduction

Solutions to Laplace’s equation can be obtained using separation of variables in Cartesianand spherical coordinate systems. In this lecture separation in cylindrical coordinates isstudied, although Laplaces’s equation is also separable in up to 22 other coordinate systemsas previously tabulated. Thus one chooses the system in which the appropriate boundaryconditions can be applied. The application of the boundary conditions (Dirichlet of Neu-mann on a closed surface) results in the unique solution to a specific problem.

In cylindrical coordinates the divergence of the gradient of the potential yields;

∇2V (ρ, φ, z) = ρ ∂2V∂ρ2 + ∂V

∂ρ+ (1/ρ) ∂2V

∂φ2 + ∂2V∂z2 = 0

Look for solutions by separation of variables;

V = R(ρ)Ψ(φ)Z(z)

As previously, one finds 2 separation constants, k and ν, which result in the three ode equa-tions with 2 of these providing 2 eigenfunctions.

d2Zdz2 − k2Z = 0

d2Ψdφ2 + ν2Ψ = 0

d2Rdρ2 + (1/ρ) dR

dρ+ (k2 − (ν/ρ)2)R = 0

Solution of the later 2 equations result in eigenfunctions. Solutions are;

Z ∝ e±kz

Ψ ∝ e±iνφ

R ∝ Jν(kρ), Nν(kρ)

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Figure 1: An example of the Cylindrical Bessel function Jν(x) as a function of x showingthe oscillatory behavior

2 Bessel Functions

In the last section, Jν(kρ), Nν(kρ) are the 2 linearly independent solutions to the separatedode radial equation. This is Bessel’s equation with Bessel functions as solutions. Besselfunctions oscillate, but are not periodic like harmonic functions, see Figure 1. Thus theharmonic function solution for Ψ and the Bessel function solution for R result in eigenfunc-tions when the boundary conditions are imposed. The Bessel functions, Jν(x), are regularat x = 0, while the Bessel (Neumann) functions, Nν(x), are singular at x = 0.

From the requirement that the solution is single valued as φ → 2π, ie the solution must notchange when φ is replaced by φ + 2π, the values of ν are integral and this produces eigen-funtions in Ψ. The series solution for the Bessel function Jν can be found by the method ofFrobenius. However, the second linearly independent equation is not easily obtained whenn is an integer. The series representation of the Bessel function takes the form;

Jν(x) =∞∑

s=0

(−1)s

s!(ν + s)!(x/2)ν+2s = xν

2nn!− xν+2

2ν+2(ν + 1)!+ · · ·

For integral values of ν one can show J−ν = (−1)νJν . Bessel functions also satisfy therecurrence relations;

Jν−1(x) + Jν+1(x) = νx Jν(x)

Jν−1(x) = xν Jν(x) +

dJν(x)dx

ddx

[xνJν(x)] = xνJν−1(x)

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At times an integral representation is useful.

Jν(x) = (1/π)π∫

0

dθ eix cos(θ) cos(νθ)

Bessel functions are orthogonal, but they are not normalized.

∞∫

0

kdk Jν(kρ) Jν(kρ′) = δ(ρ − ρ′)/ρ

a∫

0

ρdρ Jν(ανkρ/a) Jν(ανk′ρ/a) = (a2/2)[Jν+1(ανk)]2 δkk′

In the above ανk are the zeros of the Bessel function of order ν where k numbers these zeros.As the Bessel functions form a complete set, any function can be expanded in a Bessel seriesfor a finite interval in ρ or a Bessel integral if the space is infinite.

F (ρ) =∫

kdk A(k) Jν(kρ)

3 Integral Representation

Let;

fn = (1/π)π∫

0

dω cos(x sin(ω) − nω)

Then look at the relations for fn+1 and fn−1. Subtracting these we obtain;

fn+1 − fn−1 = (1/π)π∫

0

dω [cos(x sin(ω) − nω) cos(ω) + sin(x sin(ω) − nω) sin(ω) −

cos(x sin(ω) − nω) cos(ω) + sin(x sin(ω) − nω) sin(ω)

fn+1 − fn−1 = (2/π)π∫

0

dω sin[x(ω) − nω)] sin(ω)

The term on the right is twice f ′

n. Thus the recursion relation for the Bessel function isreproduced, so identify fn → Jn. The integral representation is;

Jn(x) = (1/π)π∫

0

dω cos(x sin(ω) − nω) = (1/2π)π∫

−π

dω ei(x sin(ω)−nω)

This implies that the Bessel function, Jn, is the nth Fourier coefficient of the expansion;

ei2 sin(ω) =∞∑

n=−∞

Jneinω. The following expressions for the generating function are obtained.

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cos(z sin(ω)) =∞∑

n=−∞

Jn cos(nω)

sin(z sin(ω)) =∞∑

n=−∞

Jn sin(nω)

Observe that J−n = (−1)n Jn

4 Generating function

Begin with the series in the last section;

cos(z sin(ω)) =∞∑

n=−∞

Jn cos(nω)

sin(z sin(ω)) =∞∑

n=−∞

Jn sin(nω)

Let ω = π/2 to obtain;

cos(z) = J0 − 2J2 + 2J4 + · · ·

sin(z) = 2J1 − 2J3 + · · ·

Also let eiω = t so that eiω − e−iω = 2i sin(ω). Then;

eiz sin(ω) = ez(t−1/t)/2 =∞∑

n=−∞

Jn tn

This is the generating function for Jn.

G(z, t) = ez(t−1/t)/2 =∞∑

n=−∞

Jn(z) tn

One then obtains the value of Jn(x) by determining the coefficient of the nth power in theseries. Thus;

dn

dtn[G(z, t)]|t=0 = n! Jn(z)

The generating function can be used to establish the Bessel power series, and the recursionrelations.

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L

a

V = Vo

V = 0

V = 0x

z

y

Figure 2: The geometry of a cylinder with one endcap held at potential V = V0(ρ) and theother sides grounded

5 Limiting values

Limiting Values for the Bessel functions are needed to apply boundary conditions.

Jn(x) limx→0 →(x/2)ν

Γ(ν + 1)

N0(x) limx→0 → (2/π)ln(x)

Nν(x) limx→0 → (1/π)Γ(ν)(x/2)−ν

Jν(x) limx→∞ →√

2/πx cos(x − νπ/2 − π/4)

Nν(x) limx→∞ →√

2/πx sin(x − νπ/2 − π/4)

H1ν (x) = Jν + iNν limx→∞ →

√

2/πx ei(x−νπ/2−π/4)

H2ν (x) = Jν − iNν limx→∞ →

√

2/πz e−i(x−νπ/2−π/4)

6 Example

Find the solution for the interior of a cylindrical shell with the top end cap held at a potentialV = V0(ρ) and all the other surfaces grounded, Figure 2. The solution has the form;

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Figure 3: A graphical representation of the above solution

V =∑

νn

AνknJν(knρ)sinh(knz)eiνφ

In this case the solution is independent of the angle φ so ν = 0. Note that Nν is excludedin the solution because it must be finite as ρ = 0. Also choose sinh(knz) to satisfy theboundary condition at z = 0. The reduced solution is;

V =∑

n

An J0(knρ)sinh(knz)

Now V = 0 for ρ = a. This means that;

J0(kna) = 0

The values of kna are the zeros of the Bessel function J0(kna). The first few are, α0n =2.4048, 5.5201, 8.6537, · · · . Then when z = L, An is obtained using the orthogonality of theBessel functions. The graphic form of the solution is shown in Figure 3.

An = 2a2 [J1(kna)]2 sinh(knL)

a∫

0

ρdρ J0(knρ) V0(ρ)

7 Example

As another example, find the potential inside a cylinder when the potential is specified onthe end caps and the cylindrical wall is at zero potential, Figure 4. Note that the figureshows the coordinate origin at the center of the cylinder. Thus the solution is symmetric for±z in order to match the boundary conditions. The boundary conditions are;

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a

V = 0

z

y

x

L

−LV = −Vo sin( )

V = Vo sin( )

φ

φ

Figure 4: The geometry of a problem with endcaps held at potential V = ±V0 sin(φ)

V = V0 sin(φ) z = L

V = −V0 sin(φ) z = -L

V = 0 ρ = a

V = 0 on the (x, y) plane when z = 0

The solution must have the form;

V =∑

νn

Aνn Jν(knρ) sin(νφ) sinh(knz)

Here solutions cannot contain Nνn(kρ) which are infinite at the origin. To match the bound-ary at z = ±L a a term sin(νφ) which requires ν = 1 is needed. Then require that theBessel function, J1n(kna) = 0. This determines the zeros of the Bessel function of order 1.Write these as α1n so that kn = α1n/a. The solution has the form;

V =∑

n

An J1(α1nρ/a) sinh(αz/a) sin(φ)

Finally match the boundary condition at z = ±L where V = V0 sin(φ). Use orthogonalityto obtain;

(L/2)[J2(α1n)]2An = 1sinh(αL/a)

a∫

0

ρdρ J1(α1nρ/a) V0

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ρV = f( )

z

ac

b

y

x

V = 0

Figure 5: The geometry for the problem of two concentric cylinders

8 Example

As another example, look for a solution of concentric cylinders with the boundary conditions;

r = a, c and z = 0 V = 0

z = b V = f(ρ)

This geometry is shown in Figure 5. Choose the solution to have the form;

V =∞∑

n

An sinh(knz) G0(knρ)

Here G0 is a superposition of the Bessel and Neumann functions.

G0 = [J0(knρ)J0(knc)

−N0(knρ)N0(knc)

]

This choice results in G0 = 0 when ρ = c, ie on the cylindrical surface of the inner cylinder.Choose ν = 0 because the potential is independent of φ, ie the problem is azimuthally sym-metric. Now choose the values of kn to make G0 = 0 when ρ = a. This selects a set of zeros,ανn, of the function, G0. Because G0 is a solution to the cylindrical Sturm-Liouville prob-lem, these functions form a complete orthogonal set, as do all solutions to Bessel’s equationwhich satisfy the boundary conditions. This points out that the solutions of the radial odeare placed in forms one of which is regular at ρ = 0 and one which is not. However, otherforms are possible such as G0 above. As all Bessel functions, G0 has oscillating properties,but of course the location of the zeros differs for the various representations. Finally, useorthogonality to obtain the coefficients in the above equation.

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H An = [1/sinh(knb)]a∫

c

ρdρ V (ρ) G(αnρ/a)

Here;

H =a∫

c

ρdρ G2(αnρ/a)

9 Example

Consider the problem with the cylindrical wall held at potential V = f(z) and the endcapsgrounded. This geometry is shown in Figure 6. The boundary conditions are;

z = 0, b V = 0

ρ = a V = f(z)

In this case the hyperbolic function in z cannot be used to match the boundary conditions.However let k → ik so that this function becomes harmonic at the expense of making theargument of the Bessel function imaginary. Note here that the problem is 2-D since the so-lution is independent of angle. Thus there is only one eigenfunction, and this now occurs forthe z coordinate. An eigenvalue of k = nπ/b is determined for this harmonic equation. Theradial ode with a Bessel function solution of imaginary argument result in an eigenfunction.

sin(nπz/b) with n integral

The solution has the form;

V =∞∑

n=1

Ansin(nπz/b) J0(inπρ/b)

In this case use the orthogonality of the harmonic functions rather than the Bessel functionsto find the expansion coefficients. This results in;

An = 2b J0(inπa/b)

b∫

0

dz f(z) sin(nπz/b)

Bessel functions of imaginary argument are called Modified Bessel functions. They do notoscillate as a function of the argument and cannot be eigenfunctions.

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zV = f( )

z

b

y

x

aV = 0

V = 0

Figure 6: The geometry for the problem where a potential is placed on the cylindrical surfaceand the end caps grounded

10 Dual Boundary Conditions

The equation for the potential is ∇2 V = 0. In cylindrical coordinates, Figure 7;

(1/ρ) ∂∂ρ

[ρ ∂V∂ρ

] + (1/ρ2) ∂2V∂θ2 + ∂2V

∂z2 = 0

Assume a solution of the form;

V =∑

n=0

Vn(z, ρ)

[

sin(nθ)cos(nθ)

]

In the above, one uses the harmonic functions sine, cosine or a combination of both tosatisfy the angular boundary conditions. Use sin(θ) as an example.

Vn(z, ρ) = 1/π2π∫

0

dθ V (z, ρ, θ) sin(nθ)

This results in the pde for Vn.

(1/ρ) ∂∂ρ

[ρ ∂Vn∂ρ

] − (n2/ρ2) Vn + ∂2Vn

∂z2 = 0

Apply a Hankel transform. That is, expand the function, V (ρ, θ, z) using the completenessintegral given in the section on properties of Bessel functions.

∞∫

0

ρ dρ Jn(λρ) [(1/ρ) ∂∂ρ

[ρ ∂Vn∂ρ

] − (n2/ρ2) Vn + ∂2Vn

∂z2 ] = 0

V n(λ, z) =∞∫

0

ρ dρ Vn Jn(λρ)

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x

z

ya

F

Figure 7: The geometry of a flat, conducting disk of radius a held at a potential F

The solution is symmetric for z → −z, so choose z > 0. Use eigenfunctions for Jn(λρ)and cos(nθ). This means the functional form for z is e−λz. Choose n = 0 to simplify theexposition of the Hankel transform given above.

Vn=0(ρ, z) =∞∫

0

λ dλ A(λ) e−λz J0(λρ)

The boundary conditions are applied in two steps.

Step 1 - z = 0 and ρ < a

V0(ρ, 0) = F (ρ) =∞∫

0

λ dλ A(λ) J0(λρ)

Step 2 - z = 0 and ρ > a

∂V0∂ρ

|z=0 =∞∫

0

λ2 dλ A(λ) J ′

0(λρ) = −Eρ = 0

In the second equation above, Eρ is the field in the ρ direction. The above equations thenform a pair of integral equations which must be simultaneously solved.

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11 Numerical Solutions

Separation of variables provides an analytic solution when the boundaries of the potentialsurfaces are the same as those obtained by choosing each variable of the separation geometryas a constant in order to project out a surface in the coordinate system. Of course Laplace’sequation also must separate into ode equations each involving only one of these variables.While analytic solutions provide insight into more realistic problems, they are not alwaysuseful in obtaining detailed information. Thus techniques to obtain accurate numerical so-lution of Laplace’s (and Poisson’s) equation are necessary.

First consider a result of Gauss’ theorem. Integrate Laplace’s equation over the volumewhich bounds the potential surface.

∫

dτ ∇2V =∫

~∇V · d~σ = 0

In the above, ~σ is the surface which encloses the volume τ . In the case of a spherical surface,d~σ = R2dΩ r. Substitute this in the above to write;

R2 ddR

∫

dΩ ~∇V = 0

This equation means that∫

dΩ V is constant and . In Cartesian coordinates divide spaceinto a grid with cells of the dimensions (δx, δy, δz). From the above analysis, the potentialat the center of a cell will approximately equal the average of the potential over the enclosingsurface. In Cartesian coordinates the potential at a point is approximately the average ofthe sum of the potentials over its nearest neighbors.

Vl,m,n = 1/6[Vl−1,m,n + Vl+1,m,n + Vl,m−1,n + Vl,m+1,n + Vl,m−1,n + Vl,m,n−1 + Vl,m,n+1]

Begin a numerical solution by taking the exact potential values on the surface and assigninitial values to the interior potential at all grid points. These initial values can be any guess.The average values at each point are then obtained, but one keeps the correct potential val-ues on the surface. The process is then iterated to convergence. This procedure technique iscalled the “relaxation method”. It is stable with respect to iteration and converges rapidlyto the correct potential inside a volume. The technique (finite element analysis) is generallyapplied to any process which is described by Laplace’s equation. This includes a number ofphysical problems in addition to electrostatics.

If charge is present, a solution to Poisson’s equation is required. For a sphere of radius, r,the potential at the center relative to the surface is;

∆V = ρr2/(6ǫ0)

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Figure 8: An 3-D numerical example showing contour lines of constant potential of a geom-etry having grounded metal boundaries and wires held at potential

This is included in the equation above when computing the average. As an example, Figure8 shows a numerical valuation of the potential of a set of grounded metal boundaries andwires which are held at constant potential.

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