Solution to Laplace's Equation in Cylindrical Coordinates 1 ...

LzV = VoV = 0xV = 0ayFigure 2: The geometry of a cylinder with one eddcap held at potential V = V 0 (ρ) and theother sides groundedIn this case the solution is independent of the angle φ so we take ν = 0. Note that wehave not included N ν in the solution because we want it to be finite as ρ = 0. Also we havechosen sinh(k n z) to satisfy the boundary condition at z = 0. The reduced solution is;V = ∑ nA n J 0 (k n ρ)sinh(k n z)Now V = 0 for ρ = a. This means that;J 0 (k n a) = 0The values of k n a are the zeros of the bessel function J 0 (k n a). The first few are, α 0n =2.4048, 5.5201, 8.6537, · · ·. Then at Z = L we find A n using the orthogonality of the Besselfunctions.A n = 1a 2 [J 1 (k n a)] 2 sinh(k n L)∫ a0ρdρ J 0 (k n ρ) V 0 (ρ)The graphic form of the solution is shown in figure 3.As another example we find the potential inside a cylinder when the potential is specifiedon the end caps and the cylindrical wall is at zero potential, figure 4. The boundryconditions are that;V = V 0 sin(φ)z = LV = −V 0 sin(φ) z = -L4

Figure 3: A graphical representation of the above solutionV = Vo sin( φ )zaLyxV = 0−LV = −Vo sin( φ )Figure 4: The geometry of the problem with endcaps held at potential V = V 0 ± sin(φ)5

zbcaV = f( ρ )V = 0yxFigure 5: The geometry for the problem of two concentric cylindersV = 0ρ = aThe solution must have the form;V = ∑ νnA νn J ν (k n ρ) sin(νφ) sinh(k n z)Here we have discarded solutions in N νn (kρ) which are infinite at the origin. To matchthe boundry at z = ±L we need to have a term sin(νφ) which requires ν = 1. Then werequire that the Bessel function, J 1n (k n a) = 0 which determine the zeros of the Bessel functionof order 1. We write these as α 1n so that k n = α 1n /a. The solution then has the form;V = ∑ nA n J 1 (α 1n ρ/a) sinh(αz/a) sin(φ)Finally we match the boundry condition at z = ±L where V = V 0 sin(φ). Use orthorgonalityto obtain;(L/2)[J 1 (α 1n )] 2 A n = 1sinh(αL/a)∫ a0ρdρ J 1 (α 1n ρ/a) V 0As another example we look at a solution for concentric cylinders with the boundryconditions;r = a, c and z = 0 V = 0z = b V = f(ρ)This geometry is shown in figure 56

We choose a solution to have the form;∑V = ∞ A n sinh(k n z) G 0 (k n ρ)nHere we have written a superposition of the Bessel and Neumann functions;G 0 = [ J 0(k n ρ)J 0 (k n c) − N 0(k n ρ)N 0 (k n c) ]So that at ρ = c, the cylindrical surface of the inner cylinder, G 0 vanishes. Note wehave chosen ν = 0 because the potential is independent of φ, ie the problem is aximuthallysymmetric. Now we must choose the values of k n to make G 0 = 0 when ρ = a. This willselect a set of zeros, α νn , of the function, G 0 , and in fact make the functions, G 0 , a completeorthogonal set. This points out that we separated the solutions of the radial ode into aform which was regular at ρ = 0 and one which was not. But we could have separated thesolutions so that, G 0 , was one of the two linearly independent solutions, and thus it wouldhave similar oscillating properties as the function J ν . Of course the location of the zeroswould be different. Use orthogonality to obtain the coefficients in the above equation.Here;H A n = [1/sinh(k n b)]H =∫ acρdρ G 2 (α n ρ/a)∫ acρdρ V (ρ) G(α n ρ/a)Finally consider the problem with the cylindrical wall held at potential V = f(z) andthe endcaps grounded. This geometry is shown in figure 6. The boundry conditions are;z = 0, b V = 0ρ = a V = f(z)In this case we cannot use the hyperbolic function in z to match the boundry conditions.However if we let k → ik then the hyperbolic function becomes harmonic at the expense ofmaking the argument of the Bessel function complex. Note here that the problem is 2-Dso we expect only one eigenfunction and this now occurs for the z coordinate. Then theradial ode with complex Bessel function solutions cannot be eigenfunctions. The eigenvalueof k = nπ/b is determined by the harmonic form;sin(nπz/b)n integralThe solution has the form;7

V = 0zaz V = f( )yxV = 0Figure 6: The geometry for the problem where a potential is placed on the cylindrical surfaceand the end caps groundedV = ∞ ∑n=1A n sin(nπz/b) J 0 (inπρ/b)In this case we use the orthogonality of the harmonic functions rather than the Besselfunctions. The value of the coefficients are;A n =2bJ 0 (inπa/b)∫ b0dz f(z) sin(nπz/b)8