Solutions to Laplace s Equation in Cylindrical Coordinates and Numerical solutions Lecture 8 1 Introduction Solutions to Laplace s equation can be obtained using separation of variables in Cartesian and spherical coordinate systems. In this lecture separation in cylindrical coordinates is studied, although Laplaces s equation is also separable in up to 22 other coordinate systems as previously tabulated. Thus one chooses the system in which the appropriate boundary conditions can be applied. The application of the boundary conditions (Dirichlet of Neumann 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; 2 V (ρ, φ, z) = ρ 2 V ρ 2 + V ρ + (1/ρ) 2 V φ 2 + 2 V z 2 = 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 equations with 2 of these providing 2 eigenfunctions. d 2 Z dz 2 k 2 Z = d 2 Ψ dφ 2 + ν 2 Ψ = d 2 R dρ 2 + (1/ρ) dr dρ + (k2 (ν/ρ) 2 )R = Solution of the later 2 equations result in eigenfunctions. Solutions are; Z e ±kz Ψ e ±iνφ R J ν (kρ), N ν (kρ) 1
Figure 1: An example of the Cylindrical Bessel function J ν (x) as a function of x showing the oscillatory behavior 2 Bessel Functions In the last section, J ν (kρ), N ν (kρ) are the 2 linearly independent solutions to the separated ode radial equation. This is Bessel s equation with Bessel functions as solutions. Bessel functions oscillate, but are not periodic like harmonic functions, see Figure 1. Thus the harmonic function solution for Ψ and the Bessel function solution for R result in eigenfunctions when the boundary conditions are imposed. The Bessel functions, J ν (x), are regular at x =, while the Bessel (Neumann) functions, N ν (x), are singular at x =. From the requirement that the solution is single valued as φ 2π, ie the solution must not change when φ is replaced by φ + 2π, the values of ν are integral and this produces eigenfuntions in Ψ. The series solution for the Bessel function J ν can be found by the method of Frobenius. However, the second linearly independent equation is not easily obtained when n is an integer. The series representation of the Bessel function takes the form; J ν (x) = s= ( 1) s s!(ν + s)! (x/2)ν+2s = xν 2 n n! x ν+2 2 ν+2 (ν + 1)! + For integral values of ν one can show J ν recurrence relations; J ν 1 (x) + J ν+1 (x) = ν x J ν(x) J ν 1 (x) = x ν J ν(x) + dj ν(x) dx d dx [xν J ν (x)] = x ν J ν 1 (x) = ( 1) ν J ν. Bessel functions also satisfy the 2
At times an integral representation is useful. J ν (x) = (1/π) π dθ e ix cos(θ) cos(νθ) Bessel functions are orthogonal, but they are not normalized. a kdk J ν (kρ) J ν (kρ ) = δ(ρ ρ )/ρ ρdρ J ν (α νk ρ/a) J ν (α νk ρ/a) = (a 2 /2)[J ν+1 (α νk )] 2 δ kk In the above α νk 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 series for a finite interval in ρ or a Bessel integral if the space is infinite. F(ρ) = kdk A(k) J ν (kρ) 3 Integral Representation Let; f n = (1/π) π dω cos(xsin(ω) nω) Then look at the relations for f n+1 and f n 1. Subtracting these we obtain; f n+1 f n 1 = (1/π) cos(xsin(ω) nω) cos(ω) + sin(xsin(ω) nω) sin(ω) f n+1 f n 1 = (2/π) π π dω [cos(xsin(ω) nω) cos(ω) + sin(xsin(ω) nω) sin(ω) dω sin[x(ω) nω)] sin(ω) The term on the right is twice f n. Thus the recursion relation for the Bessel function is reproduced, so identify f n J n. The integral representation is; J n (x) = (1/π) π dω cos(xsin(ω) nω) = (1/2π) π π i(x sin(ω) nω) dω e This implies that the Bessel function, J n, is the n th Fourier coefficient of the expansion; e i2 sin(ω) = J n e inω. The following expressions for the generating function are obtained. 3
cos(z sin(ω)) = sin(z sin(ω)) = Observe that J n = ( 1) n J n J n cos(nω) J n sin(nω) 4 Generating function Begin with the series in the last section; cos(z sin(ω)) = sin(z sin(ω)) = Let ω = π/2 to obtain; J n cos(nω) J n sin(nω) cos(z) = J 2J 2 + 2J 4 + sin(z) = 2J 1 2J 3 + Also let e iω = t so that e iω e iω = 2i sin(ω). Then; e iz sin(ω) = e z(t 1/t)/2 = This is the generating function for J n. G(z, t) = e z(t 1/t)/2 = J n t n J n (z) t n One then obtains the value of J n (x) by determining the coefficient of the n th power in the series. Thus; d n dt n [G(z, t)] t= = n! J n (z) The generating function can be used to establish the Bessel power series, and the recursion relations. 4
L z V = Vo V = x V = a y Figure 2: The geometry of a cylinder with one endcap held at potential V = V (ρ) and the other sides grounded 5 Limiting values Limiting Values for the Bessel functions are needed to apply boundary conditions. J n (x) lim x (x/2)ν Γ(ν + 1) N (x) lim x (2/π)ln(x) N ν (x) lim x (1/π)Γ(ν)(x/2) ν J ν (x) lim x 2/πxcos(x νπ/2 π/4) N ν (x) lim x 2/πxsin(x νπ/2 π/4) H 1 ν (x) = J ν + in ν lim x 2/πxe i(x νπ/2 π/4) H 2 ν(x) = J ν in ν lim x 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 potential V = V (ρ) and all the other surfaces grounded, Figure 2. The solution has the form; 5
Figure 3: A graphical representation of the above solution V = νn A νkn J ν (k n ρ)sinh(k n z)e iνφ In this case the solution is independent of the angle φ so ν =. Note that N ν is excluded in the solution because it must be finite as ρ =. Also choose sinh(k n z) to satisfy the boundary condition at z =. The reduced solution is; V = n A n J (k n ρ)sinh(k n z) Now V = for ρ = a. This means that; J (k n a) = The values of k n a are the zeros of the Bessel function J (k n a). The first few are, α n = 2.448, 5.521, 8.6537,. Then when z = L, A n is obtained using the orthogonality of the Bessel functions. The graphic form of the solution is shown in Figure 3. A n = 2 a 2 [J 1 (k n a)] 2 sinh(k n L) a ρdρ J (k n ρ) V (ρ) 7 Example As another example, find the potential inside a cylinder when the potential is specified on the end caps and the cylindrical wall is at zero potential, Figure 4. Note that the figure shows 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; 6
V = Vo sin( φ ) z a L y x V = L V = Vo sin( φ ) Figure 4: The geometry of a problem with endcaps held at potential V = ±V sin(φ) V = V sin(φ) z = L V = V sin(φ) z = -L V = ρ = a V = on the (x, y) plane when z = The solution must have the form; V = νn A νn J ν (k n ρ) sin(νφ) sinh(k n z) Here solutions cannot contain N νn (kρ) which are infinite at the origin. To match the boundary at z = ±L a a term sin(νφ) which requires ν = 1 is needed. Then require that the Bessel function, J 1n (k n a) =. This determines the zeros of the Bessel function of order 1. Write these as α 1n so that k n = α 1n /a. The solution has the form; V = n A n J 1 (α 1n ρ/a) sinh(αz/a) sin(φ) Finally match the boundary condition at z = ±L where V = V sin(φ). Use orthogonality to obtain; (L/2)[J 2 (α 1n )] 2 A n = 1 sinh(αl/a) a ρdρ J 1 (α 1n ρ/a) V 7
z b c a V = f( ρ ) V = y x 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 = V = z = b V = f(ρ) This geometry is shown in Figure 5. Choose the solution to have the form; V = A n sinh(k n z) G (k n ρ) n Here G is a superposition of the Bessel and Neumann functions. G = [ J (k n ρ) J (k n c) N (k n ρ) N (k n c) ] This choice results in G = when ρ = c, ie on the cylindrical surface of the inner cylinder. Choose ν = because the potential is independent of φ, ie the problem is azimuthally symmetric. Now choose the values of k n to make G = when ρ = a. This selects a set of zeros, α νn, of the function, G. Because G is a solution to the cylindrical Sturm-Liouville problem, these functions form a complete orthogonal set, as do all solutions to Bessel s equation which satisfy the boundary conditions. This points out that the solutions of the radial ode are placed in forms one of which is regular at ρ = and one which is not. However, other forms are possible such as G above. As all Bessel functions, G has oscillating properties, but of course the location of the zeros differs for the various representations. Finally, use orthogonality to obtain the coefficients in the above equation. 8
Here; a H A n = [1/sinh(k n b)] c H = a c ρdρ G 2 (α n ρ/a) ρdρ V (ρ) G(α n ρ/a) 9 Example Consider the problem with the cylindrical wall held at potential V = f(z) and the endcaps grounded. This geometry is shown in Figure 6. The boundary conditions are; z =, b V = 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 the argument of the Bessel function imaginary. Note here that the problem is 2-D since the solution is independent of angle. Thus there is only one eigenfunction, and this now occurs for the z coordinate. An eigenvalue of k = nπ/b is determined for this harmonic equation. The radial 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 A n sin(nπz/b) J (inπρ/b) In this case use the orthogonality of the harmonic functions rather than the Bessel functions to find the expansion coefficients. This results in; A n = 2 bj (inπa/b) b dz f(z) sin(nπz/b) Bessel functions of imaginary argument are called Modified Bessel functions. They do not oscillate as a function of the argument and cannot be eigenfunctions. 9
b V = z a z V = f( ) y x V = Figure 6: The geometry for the problem where a potential is placed on the cylindrical surface and the end caps grounded 1 Dual Boundary Conditions The equation for the potential is 2 V =. In cylindrical coordinates, Figure 7; (1/ρ) ρ [ρ V ρ ] + (1/ρ2 ) 2 V θ 2 + 2 V z 2 = Assume a solution of the form; V = [ ] sin(nθ) V n (z, ρ) n= cos(nθ) In the above, one uses the harmonic functions sine, cosine or a combination of both to satisfy the angular boundary conditions. Use sin(θ) as an example. 2π V n (z, ρ) = 1/π dθ V (z, ρ, θ) sin(nθ) This results in the pde for V n. (1/ρ) ρ [ρ V n ρ ] (n2 /ρ 2 ) V n + 2 V n z 2 = Apply a Hankel transform. That is, expand the function, V (ρ, θ, z) using the completeness integral given in the section on properties of Bessel functions. ρ dρ J n (λρ) [(1/ρ) ρ [ρ V n ρ ] (n2 /ρ 2 ) V n + 2 V n z 2 ] = V n (λ, z) = ρ dρ V n J n (λρ) 1
z x F a y 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 >. Use eigenfunctions for J n (λρ) and cos(nθ). This means the functional form for z is e λz. Choose n = to simplify the exposition of the Hankel transform given above. V n= (ρ, z) = λ dλ A(λ) e λz J (λρ) The boundary conditions are applied in two steps. Step 1 - z = and ρ < a V (ρ, ) = F(ρ) = Step 2 - z = and ρ > a V ρ z= = λ dλ A(λ) J (λρ) λ 2 dλ A(λ) J (λρ) = E ρ = In the second equation above, E ρ is the field in the ˆρ direction. The above equations then form a pair of integral equations which must be simultaneously solved. 11
11 Numerical Solutions Separation of variables provides an analytic solution when the boundaries of the potential surfaces are the same as those obtained by choosing each variable of the separation geometry as a constant in order to project out a surface in the coordinate system. Of course Laplace s equation 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 always useful in obtaining detailed information. Thus techniques to obtain accurate numerical solution of Laplace s (and Poisson s) equation are necessary. First consider a result of Gauss theorem. Integrate Laplace s equation over the volume which bounds the potential surface. dτ 2 V = V d σ = In the above, σ is the surface which encloses the volume τ. In the case of a spherical surface, d σ = R 2 dω ˆr. Substitute this in the above to write; R 2 d dr dω V = This equation means that dωv is constant and. In Cartesian coordinates divide space into a grid with cells of the dimensions (δx, δy, δz). From the above analysis, the potential at the center of a cell will approximately equal the average of the potential over the enclosing surface. In Cartesian coordinates the potential at a point is approximately the average of the sum of the potentials over its nearest neighbors. V l,m,n = 1/6[V l 1,m,n + V l+1,m,n + V l,m 1,n + V l,m+1,n + V l,m 1,n + V l,m,n 1 + V l,m,n+1 ] Begin a numerical solution by taking the exact potential values on the surface and assign initial 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 values on the surface. The process is then iterated to convergence. This procedure technique is called the relaxation method. It is stable with respect to iteration and converges rapidly to the correct potential inside a volume. The technique (finite element analysis) is generally applied to any process which is described by Laplace s equation. This includes a number of physical 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 = ρr 2 /(6ǫ ) 12
Figure 8: An 3-D numerical example showing contour lines of constant potential of a geometry having grounded metal boundaries and wires held at potential This is included in the equation above when computing the average. As an example, Figure 8 shows a numerical valuation of the potential of a set of grounded metal boundaries and wires which are held at constant potential. 13