Cylindrical Laplace Solutions February 16 2009 Overview Laplace Equation Solutions in Cylindrical Geometry Review last class Superposition solutions Introduction to radial coordinates Additional solutions of Laplace s equation in radial coordinates Larry Caretto Mechanical Engineering 501B Homogenous boundaries in z direction Gradient boundary conditions in the cylinder Hollow cylinder Combinations of boundary conditions Superposition Seminar in Engineering Analysis February 16 2009 2 Review Superposition Review Superposition Solution Laplace s equation for 0 x L and 0 y H with boundary conditions shown Do not have homogenous boundary conditions in any coordinate direction Superposition sum of simpler solutions y H u uN x u 0 u uE y u 0 y 0 x 0 x L 3 Sum two solutions with one nonzero boundary y H u uN x u 0 u 0 Solution to this problem is u1 x y u 0 x 0 x L y 0 Review General Superposition Review x and y Swap u uE y u 0 x 0 u 0 x L y 4 0 Sum two solutions shown below y H u u x Can obtain solution for several nonzero boundaries by creating a separate solution for each nonzero boundary N u 0 Three dimensional problems can have up to have six separate solutions for u x y z Swap x and y and swap H and L in solution for u 0 y u L y 0 to get solution for y x 0 u x H 0 Set y H y in solution for u x H ub x to get solution for u x 0 ub x x 0 u 0 u 0 Swap x and y to get u2 x y from u1 5 ME 501B Engineering Analysis y H u 0 Solution here is u2 x y u1 x y found previously x L y 0 u 0 u uE y u 0 x 0 y H u 0 x L y 6 0 1 Cylindrical Laplace Solutions February 16 2009 Review x y Swap Solution n n L u1 x y Cn sin n x sinh n y n 1 n 1 L u N 2 sinh n H L Bn H x sin n x dx u E Swap location of one zero boundary y H u uN x u1 x y found u 0 u 0 previously with uN x u 0 x 0 x L y 0 n n H u 2 x y Bn sin n y sinh n x Cn Review Coordinate Transform 2 sinh n L H y sin n y dy 0 0 u x y Cn sin n x sinh n y Bn sin n y sinh n x n 1 u2 x y u1 x H y u 0 Use uS x in place of uN x x 0 7 Review y H y Solution u1 x y Cn sin n x sinh n y n 1 Cn L 2 sinh n H L 0 n 1 Bn 2 n n L u r L uN r L u2 x y Bn sin n x sinh n H y L sinh n H L 0 u 0 u uS x x L y 8 0 Cylindrical Laplace u N x sin n x dx y H u 0 R n n L z 0 Solve Laplace s equation in a cylinder for u r z Zero boundaries at the sides and bottom u R z 0 u r 0 0 Specified top boundary 1 u 2u u r L uN r r 0 r r r z 2 Finite solution at r 0 u S x sin n x dx 9 10 What do We Expect What do We Expect II Note similarity to radial diffusion equation Also similar to Laplace equation for u x y 1 u 2u r 0 u r 0 0 u r L u N r r r r z 2 2 u 2u 0 u x 0 0 u x H u N x x 2 y 2 u is finite at r 0 for both problems 1 u 2u 0 r r r r z 2 u R z u R z 1 u 1 u r 0 r r r t u R z u R z Separation of variables result for P r in Laplace s equation should be similar to result for diffusion equation Separation of variables result for Z z in radial equation should be similar to result for Y y in rectangular coordinates Bessel function eigenfunctions in r direction 11 ME 501B Engineering Analysis Hyperbolic sine cosine solution for Z z 12 2 Cylindrical Laplace Solutions February 16 2009 Separation of Variables Boundary Conditions Proposed solution u r z P r Z z Separation of variables and ODE solutions give starting point for solution 1 d dP r 1 d 2 Z z r 2 rP r dr dr Z z dz2 d 2 Z z 2 Z z 0 dz2 Z z Asinh z B cosh z d dP r r 2 rP r 0 dr dr P r CJ 0 r DY0 r u r z P r Z z u N r u r L Cm sinh m L J 0 m r Cm 0 R rJ m r u N r dr 0 R sinh m L r J 0 m r dr 2 0 0 m R m 0 14 Example uN r U a Constant Region is 0 r R and p r r is weight function rJ u r z Cm sinh m z J 0 m r 13 Radial equation for P r is a SturmLiouville problem so we use eigenfunction expansion for y L boundary m 1 u r 0 0 for all r requires Z 0 0 Z 0 0 A sinh 0 B cosh 0 B 0 Finite solution at r 0 requires D 0 u R z 0 for all z requires P R 0 P R 0 if J0 R 0 so mR m0 the zeros of J0 Solution is sum of all eigenfunctions m 1 Boundary Condition at z L R m r u N r dr 0 R2 2 sinh m L J1 m R 2 15 r R R Cm rJ 0 m r Udr 0 R2 J1 m R 2 2 RJ1 m R sinh m L 2U m rJ1 m r 2U m r 0 2 R sinh m L J1 m R 2 mR m0 2U 2 sinh m L R 2 J1 m R sinh m L m RJ1 m R Substitute Cm equation into general solution u r z Cm sinh m z J 0 m r m 1 16 Example uN r U a Constant Cm 2U sinh m L m RJ1 m R Cm sinh m0 2U m0 m0 L RJ1 m 0 R sinh R R R 2U L m0 J1 m0 R z r u r z Cm sinh m 0 J 0 m 0 R R m 1 Cm 2U m 1 m 0 J 1 m 0 u r z sinh m 0 sinh m 0 ME 501B Engineering Analysis …