Solution to Laplace’s Equation in Cylindrical Coordinates ...

1 Solution to laplace s Equation in CylindricalCoordinatesLecture 81 IntroductionWe have obtained general solutions for laplace s Equation by separtaion of variables in Carte-sian and spherical coordinate systems. The last system we study is Cylindrical Coordinates ,but remember Laplaces s Equation is also separable in a few (up to 22) other coordinatesystems. As you know, choose the system in which you can applythe appropriate boundryconditions. It is only through application of the boundry conditions (Dirichlet of Neumannon a closed surface) that one finds a unique Solution to the problem studied.

2 In cylindricalcoordinates apply the divergence of the gradient on the potential to get laplace s Equation . 2V( , , z) = 2V 2+ V + (1/ ) 2V 2+ 2V z2= 0We look for a Solution by separation of variables;V=R( ) ( )Z(z)As previously, this yields 2 separation constants,kand , which will lead to 2 eigen-function equations. The three separated ode equations are;d2 Zdz2 k2Z= 0d2 d 2+ = 0d2Rd 2+ (1/ )Rd + (k2 ( / )2)R= 0 The later 2 equations can be set up as eigenfunction equations. The solutions are;Z e kz e i R J (k ),or/andN (k )1 Figure 1: An example of the Cylindrical Bessel functionJ(x) as a function ofxshowing theoscillaltory behavior2 Bessel FunctionsIn the last section,J (k ), N (k ) are the 2 linearly independent solutions to Bessel s equa-tion.

3 Bessel functions oscillate but not harmonically, seeFigure 1. Thus we expect thatthe harmonic function solutions for and the Bessel function solutions forRwill be theeigenfunctions when the boundry conditions are imposed. The Bessel functions,J (x), areregular atx= 0, while the Bessel functions,N (x), are singular atx= limiting values of the Bessel functions are;limx 0J (x) (x2) limx 0N (x) [(2/ )ln(x) = 0(2/x) ( ) otherwise]limx J (x) 2 xcos(x /2 /4)limx N (x) 2 xsin(x /2 /4)From the requirement that the Solution be single valued as 2 ,iethe Solution ismust not change when is replaced by + 2 , the values of are integral and this produceseigenfuntions of.

4 The series Solution for the Bessel functionJ can be found by the methodof Frobenius. However, the second linearly independent Equation is not easily obtained whennis an integer, and another technique is required. The Besselfunction takes the form;2J (x) = s=0( 1)ss!( +s)!(x/2) +2s=x 2nn! x +22 +2( + 1)!+ For integral values of one can showJ = ( 1) J . The Bessel functions also satisfy therecurrence relations;J 1(x) +J +1(x) = xJ (x)J 1(x) =x J (x) +dJ dxddx[x J (x)] =x J 1At times an integral representation is (x) = (1/ ) 0d eix cos( )cos( )The Bessel functions are orthogonal.

5 0kdk J (k )J (k ) = (r r )/ra 0 d J ( k /a)J ( k /a) = (a2/2)[J +1( k)]2 kk In the above kare the zeros of the Bessel function of order wherekorders these the Bessel functions form a complete set, any function maybe expanded in a Bessel seriesor integral for an infinite ( ) = kdk A(k)J (k )3 ExamplesWe find the Solution for the interior of a Cylindrical shell with the top end cap held at apotentialV=V0( ) and all the other surfaces grounded, Figure 2. The solutionwe seekhas the form;V= nA knJ (kn )sinh(knz)ei In this case the Solution is independent of the angle so we take = 0.

6 Note that we havenot includedN in the Solution because we want it to be finite as = 0. Also we have chosensinh(knz) to satisfy the boundary condition atz= 0. The reduced Solution is;3 LaV = VoV = 0V = 0xzyFigure 2: The geometry of a cylinder with one eddcap held at potentialV=V0( ) and theother sides groundedV= nAnJ0(kn )sinh(knz)NowV= 0 for =a. This means that;J0(kna) = 0 The values ofknaare the zeros of the bessel functionJ0(kna). The first few are, 0n= , , , . Then atZ=Lwe findAnusing the orthogonality of the [J1(kna)]2sinh(knL)a 0 d J0(kn )V0( )The graphic form of the Solution is shown in figure another example we find the potential inside a cylinder when the potential is spec-ified on the end caps and the Cylindrical wall is at zero potential, figure 4.

7 The boundryconditions are that;V=V0sin( )z = LV= V0sin( )z = -LV= 0 =aThe Solution must have the form;4 Figure 3: A graphical representation of the above solutionaV = 0zyxL LV = Vo sin( )V = Vo sin( ) Figure 4: The geometry of the problem with endcaps held at potentialV=V0 sin( )5 V = f( ) zacbyxV = 0 Figure 5: The geometry for the problem of two concentric cylindersV= nA nJ (kn )sin( )sinh(knz)Here we have discarded solutions inN n(k ) which are infinite at the origin. To match theboundry atz= Lwe need to have a termsin( ) which requires = 1. Then we requirethat the Bessel function,J1n(kna) = 0 which determines the zeros of the Bessel function oforder 1.

8 We write these as 1nso thatkn= 1n/a. The Solution then has the form;V= nAnJ1( 1n /a)sinh( z/a)sin( )Finally we match the boundry condition atz= LwhereV=V0sin( ). Use orthorgonal-ity to obtain;(L/2)[J1( 1n)]2An=1sinh( L/a)a 0 d J1( 1n /a)V0As another example we look at a Solution for concentric cylinders with the boundryconditions;r=a, candz= 0V = 0z=b V=f( )This geometry is shown in Figure 5. We choose a Solution to have the form;V= nAnsinh(knz)G0(kn )Here we have written a superposition of the Bessel and Neumann functions;6G0= [J0(kn )J0(knc) N0(kn )N0(knc)]So that at =c, the Cylindrical surface of the inner cylinder,G0vanishes.

9 Note we havechosen = 0 because the potential is independent of ,iethe problem is aximuthally sym-metric. Now we must choose the values ofknto makeG0= 0 when =a. This will selecta set of zeros, n, of the function,G0, and in fact make the functions,G0, a completeorthogonal set. This points out that we separated the solutions of the radial ode into aform which was regular at = 0 and one which was not. But we could have separated thesolutions so that,G0, was one of the two linearly independent solutions, and thusit wouldhave similar oscillating properties as the functionJ . Of course the location of the zeroswould be different.

10 Use orthogonality to obtain the coefficients in the above An= [1/sinh(knb)]a c d V( )G( n /a)Here;H=a c d G2( n /a)Finally consider the problem with the Cylindrical wall heldat potentialV=f(z) andthe endcaps grounded. This geometry is shown in figure 6. The boundry conditions are;z= 0, b V= 0 =a V=f(z)In this case we cannot use the hyperbolic function inzto match the boundry if we letk ikthen 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 thezcoordinate.