18 October 2002 Physics 217, Fall 2002 1Today in Physics 217: separation of variables IV Separation in cylindrical coordinates Example of the split cylinder (solution sketched at right) More on orthogonality of trig functions and Fourier’s trickEV0V =0VV=18 October 2002 Physics 217, Fall 2002 2Separation of variables in cylindrical coordinatesIn Griffiths problem 3.23, on homework 6, you did the setup of the separation solution to all Laplace-equation problems in cylindrical geometry. Recall that the Laplace equation in cylindrical coordinates isIf you know a priorithat Vdoesn’t depend on z– infinite cylinder, boundary conditions independent of z–then the last term drops out22222 2110VVVVsss sszφ∂∂ ∂ ∂∇= + + =∂∂∂∂222110VVsss ssφ∂∂ ∂+=∂∂∂18 October 2002 Physics 217, Fall 2002 3Separation of variables in cylindrical coordinates (continued)So you tried a solution of the formand solved the resulting radial and angular ordinary differential equations to obtain particular solutions:and with an additional radial solution for m= 0:() ()(),:Vs Ssφφ=Φ222210sd dS dsmmSds dsdφΦ+==−Φ()2mmddSss mS SsCrDrds ds−=⇒=+()222cos sindmAmBmdφφφφΦ=− Φ ⇒ Φ = +()00 00lnddSss SsCsDds ds=⇒ = +18 October 2002 Physics 217, Fall 2002 4Separation of variables in cylindrical coordinates (continued)From the periodicity of the angular solution you also deduced that m= 0,1,2,3,…The most general solution is a linear combination of all of these solutions, for all values of m, which you wrote asYou applied this solution in a concrete example, Griffiths problem 3.24, in which you were able to avoid the use of Fourier’s trick. Now for one in which you can’t.()()()000,lncos sinmmmm m mmVs C s DCs Ds A m B mφφφ∞−==+++ +∑18 October 2002 Physics 217, Fall 2002 5The split cylinderExample. A long, thin-wall conducting cylindrical tube with radius R– a small section of which is shown at right – is split in half lengthwise. The two halves are insulated from one another; one is held at potential V0and the other is grounded. Find the potential inside the tube. xyzR0VV=0V =18 October 2002 Physics 217, Fall 2002 6The split cylinder (continued)The solution:Boundary conditions:Apply the last one first: the term approaches infinity at s= 0 unless Similarly, 0. at ,0.0 at , 2. finite at 0iV V s Rii V s Riii V sφπφπ π===→===→=()()()001,lncos sinmmmm m mmVs C s DCs Ds A m B mφφφ∞−==+++ +∑mmDs−0.mD=00.C=18 October 2002 Physics 217, Fall 2002 7More on trig function orthogonalityNow apply the first two boundary conditions. This will be conveniently done in concert with the application of Fourier’s trick to extract the rest of the coefficients inside the sum. And this will be a good place to fill in some more details about the orthogonality of the trig functions. Last Friday we showed thatObviously we also have0sin sin2mnmy nydyππδ=∫20sin sin .mnmy nydyππδ=∫18 October 2002 Physics 217, Fall 2002 8More on trig function orthogonality (continued)But what about cosines? It goes just like the derivation for sines, starting with two integrations by parts:ThusFor m = n, Integrate this by parts once:22200022001cos cos cos sin sin sin1sin cos cos cosmmx nxdx mx nx mx nxdxnnmmmx nx mx nxdxnn nπππππ=+=− +∫∫∫()2222001 cos cos 0 , or cos cos 0 .mmx nxdx mx nxdx m nnππ−= =≠∫∫2222001cos cos .mmxdx wdwmππ=∫∫=0=018 October 2002 Physics 217, Fall 2002 9More on trig function orthogonality (continued)Thusand()22222000222200cos cos sin sin11cos sin .22mmmmmwdw w w wdww w dw dw mππππππ=− +=+==∫∫∫∫22020coscos cos .mnmxdxmx nxdxππππδ==∫∫=018 October 2002 Physics 217, Fall 2002 10More on trig function orthogonality (continued)Then there are products of sines and cosines to consider:Recall the trig identitySo, once again, this shows that2220001cos sin sin sin sin cosnmx nxdx nx mx mx nxdxmmπππ=−∫∫=0()sin sin cos cos sin :uv u v u v+= +()()2220002200cos sin sin cos sincos1cossin 0.nnmx nxdx m n xdx mx nxdxmmmnxnnmx nxdxmmmnπππππ=− + ++−==+∫∫∫∫()20cos sin 0 .mx nxdx m nπ=≠∫18 October 2002 Physics 217, Fall 2002 11More on trig function orthogonality (continued)For the case m = n, recall the trig identitySo it’s zero whether m = n or not.Summary: Now back to that cylinder:2cos sin sin2 :uu u=2244000011 1cos sin sin 2 sin cos 0.24 4mmmx mxdx mxdx udu ummππππ== =−=∫∫∫220020cos cos sin sin ,cos sin 0 .mnmx nxdx mx nxdxmx nxdxππππδ===∫∫∫18 October 2002 Physics 217, Fall 2002 12The split cylinder (continued)What’s left of the solution:Let’s apply the first two boundary conditions, while extracting the Am:()()1,cossinmmm mmVs Cs A m B mφφφ∞==+∑()()()2210001000,cos cos sin coscos 0sin 0mmm mmmmmmnmnnn nnVR nd C R A m B m ndVnd CRAVnACR ACnππππφφφ φ φφφφφ πδφπ∞=∞==+=+=⇒=∑∫∫∑∫, as we just found.18 October 2002 Physics 217, Fall 2002 13The split cylinder (continued)Now work on the Bs by integrating everything with ()()()()22100010000,sin cos sin sinsin 0cos110 if is even, 2 if it's odd.mmm mmmmmmnmnnnnnnnVR nd C R A m B m ndVnd CR BVnBCRnVBC Rnnππππφφφ φ φφφφφ πδφππ∞=∞==+=+−=−− ==∑∫∫∑∫sin :nφ, as we just found.18 October 2002 Physics 217, Fall 2002 14The split cylinder (continued)So replace mwith 2m+1 in the sum:()()0101,3,5,...21002,sin21sin21sin 2 1 .21nnnmmmmmmmmVBCnRVs BCs