Multiple independent variables March 4 2009 Overview Multidimensional Partial Differential Equations Review last class Characteristics and classification of partial differential equations Problems in more than two independent variables Larry Caretto Mechanical Engineering 501B Solution by separation of variables Problems with multiple nonhomogenous boundary conditions Solutions for rectangular geometry Homework problem for cylindrical geometry Seminar in Engineering Analysis March 4 2009 2 Review General and Hyperbolic Domain of dependence for u x1 y1 Imaginary characteristics for elliptic equations like Laplace and Poisson s The area in x y space whose u values affect the value of u x1 y1 Region of influence of u x1 y1 The area in x y space whose u values are affected by the value of u x1 y1 Areas for hyperbolic equations shown below Left running characteristic dy dx 0 x1 y1 Domain of dependence Right running characteristic dy dx 0 Initial Condition Curve Entire solution region is both domain of dependence and region of influence Parabolic equations typically involve time and space as coordinates Domain of dependence at x1 t1 is entire domain 0 x L and 0 t t1 Region of influence at x1 t1 is entire region 0 x L and t t1 3 Review Multidimensional 4 Review Multidimensional II Can have equations in three space dimensions and time Classification as elliptic parabolic or hyperbolic does not apply to equations with more than two dimensions Coordinates can have elliptic like parabolic like and hyperbolic like behavior in multidimensional equations Laplace Diffusion 2u S u 2u S t Cartesian Cylindrical Sphere 2u E g time is a parabolic coordinate 5 ME 501B Engineering Analysis Review Elliptic and Parabolic 2u Wave 2u t 2 c 2 2 u S 2u 2u 2u x 2 y 2 z 2 2u 1 u 1 2u 2u r r r r r 2 2 z 2 1 2 u 1 2u 1 2u cot u 2 r 2 2 2 r r r r sin 2 r 2 2 r 6 1 Multiple independent variables March 4 2009 Review 2D Diffusion Review 2D Diffusion II Two dimensional diffusion 1 u 2u 2u equation for u x y t t x 2 y 2 Boundary conditions give general solution as sum of all eigenfunctions t 0 0 x L 0 y H u x y 0 f x y u 0 y t u L y t u x 0 t u x H t 0 u x y t Cnm e n x m y u x y 0 f x y Cnm sin sin L H n 1 m 1 C nm 4 HL Look at f x y U a Constant for z x W L and for z y W H W p z p z sin cos dx p W 0 W 0 4 HL HL n x m y U sin sin dxdy L H 00 Cnm 9 Result for f x y U a Constant u x y t Cnm e n 2 m 2 t L H n 1 m 1 n x m y sin sin L H Cnm u x y t 16U 16U nold mold 2 2n 1 2m 1 2 n m n 0 m 0 2 2 2n m2 t L H e n m 16U y x sin n sin m H L ME 501B Engineering Analysis Dimensionless Parameters Modify exponential argument as follows Replace n by 2n 1 and m by 2m 1 to get odd indices only and define new parameters n and m as follows 16U 2W W odd n cos n 1 n n even n 0 H L 4U 4U 2 H 2 L 16 n x m y sin dx dy sin HL H L HL m n mn 2 0 0 10 Cnm 0 zero for even m or n H L 4U m y n x sin dy sin dx HL H 0 L 0 m 2m 1 W W Cnm n 2n 1 8 Both integrals are effectively the same n x m y sin dxdy General L H Result When we substitute f x y U we can separate the x and y integrations n x m y dxdy sin L H f x y sin Look at f x y constant U II f x y sin 00 HL 00 7 HL n x m y sin sin L H Eigenfunction expansion for t 0 1 1 T t 1 2 X x 1 2Y y 2 T t t X x x Y y y 2 4 HL n 2 m 2 t L H n 1 m 1 Use separation of variable approach with all variables u x y t X x Y y T t Cnm 11 2 2 n2 m2 t L H e 2 2 L t n2 m 2 2 H L e Substitute into u x y t and divide by U u x y t 16 U n 0 m 0 Where 2 L2 t n2 m 2 2 H L e n m x y sin n sin m L H n 2n 1 m 2m 1 12 2 Multiple independent variables March 4 2009 Important Parameters u x y t 16 U 2 L2 n2 m 2 e H t L2 n m n 0 m 0 Product Solution Can now separate n and m sums x y sin n sin m L H Result shows that u U is a function of x L y H L H and t L2 Can simplify double summation in this case by splitting exponential term 2 L2 t 2n m 2 2 H L e e 2n t 2 L e 2 2 m L t H 2 L2 e n2 t L2 e 16 n 0 2m t H2 n2 t 2 m t u x y t e L y x e H 16 sin n sin m U L H n m n 0 m 0 e n2 t L2 n 2 2 m t 2 Product y x e H sin n sin m of one L H m 0 dimenm 2 sional t 2 t n2 m solutions 2 e L2 e H y x 4 sin n 4 sin m H L m 0 m n 0 n 13 14 Nonzero Boundaries Nonzero Boundaries II Sturm Liouville eigenfunction expansions require zero boundary conditions For nonzero boundaries split solution as in 1D case u x y t v x y t w x y Solve Laplace equation for w with superposition if required Solution for v is same as previous solution for u with zero boundaries v satisfies diffusion equation with zero boundary conditions w satisfies Laplace s and diffusion equation with nonzero boundary conditions u satisfies diffusion equation with u x y t w x y at boundaries v x y t Cnm e n 2 m 2 t L H n 1 m 1 n x m y sin sin L H Initial condition for u found from u x y t v x y t w x y by setting t 0 15 16 Nonzero Boundaries III …