Wave equation and classification of PDEs February 25 2009 Overview Wave Equation and Introduction to Classification of PDEs Review last class Wave equation solutions by separation of variables and D Alembert approach Wave equation solution with boundaries Characteristics and classification of partial differential equations Larry Caretto Mechanical Engineering 501B Seminar in Engineering Analysis February 25 2009 General analysis Parabolic equations Elliptic equations Hyperbolic equations 2 Course Items Review Gradients Gradients of Laplace equation solutions often proportional to flux terms Notes on wave equation on web site Midterm Wednesday March 11 Heat flux and temperature gradient Diffusion flux and mass fraction gradient Velocity and velocity potential in ideal flow In constant potential plot lines perpendicular to the potential are flux lines f f f grad f f i j k x y z Covers material on diffusion and Laplace equations Includes material up to and including lecture and homework for March 2 Open textbook and notes including homework solutions Use existing solutions to answer questions i 3 j k x y z 4 Review Interpretation of 2u 0 When v k grad u is a flux that is the gradient of a scalar Laplace s equation for u says that the net inflow of v is zero 1 2udV v ndA 0 k Surface Enclosed Volume Example of this result shown last week Result applies to any problem in any geometry with Laplace s equation 5 ME 501B Engineering Analysis 6 1 Wave equation and classification of PDEs February 25 2009 Review Complex Variable Review Additional Results Cauchy Riemann conditions If u v x y and v u x y df u v v u i i dz x y x y then Equivalent to Laplace equation Function u x y that satisfies Laplace equation in two dimensions has associated function v x y that satisfies Laplace Lines of u x y and v x y are perpendicular Typically if u is a potential e g temperature v is a corresponding flux Cauchy theorem for complex integration shows Laplace equation solutions Have maximum and minimum on boundary If boundary is a constant at all points then solution is the same constant in region Dirichlet problem has unique solution Neumann problem does not Kreyszig section 18 6 has proofs 7 Review Wave Equation 8 Review Separation of Variables Wave phenomena u x t is wave amplitude varying with 2 2 u space x and time t 2 u c c is wave speed t 2 x 2 Can solve by usual separation of variables technique Also have D Alambert solution with arbitrary functions F and G with coordinates x ct and x ct Usual assumption u x t X x T t Result is 1 1 2T t 1 2 X x 2 function of t X x x 2 c 2 T t t 2 equal to u x t T t X x function of x A sin ct B cos ct C sin x D cos x Use above solution as starting point Boundary conditions at x 0 and x L Initial conditions on u and u t at x 0 9 Review Separation of Variables u 2 Solution for u x t with initial and boundary conditions u x 0 f x u x 0 g x u 0 t u L t 0 c2 u 2t x 2 0 x L t 0 c is wave speed L 2 m x g x sin dx m L 2 m x f x sin dx L0 L L Bm 0 11 ME 501B Engineering Analysis General Solution 2 n ct n x n ct u x t An sin sin Bn cos L L L n 1 Am 10 Substitute Am and Bm from equations just found and substitute into previous solution n ct n x n ct u x t An sin sin Bn cos L L L n 1 Examine case where g x 0 so An 0 n ct n x u x t Bn cos sin L L n 1 12 2 Wave equation and classification of PDEs February 25 2009 General Solution for g x 0 Similar Solution for f x 0 n ct n x u x t An sin sin L L n 1 n ct n x u x t Bn cos sin L L n 1 From trig identities for cos x y From trig identities for sin x y cos x y cos x cos y sin y sin x cos x y cos x cos y sin y sin x cos x y cos x y 2 sin x sin y sin x y sin x cos y sin y cos x sin x y sin x cos y sin y cos x sin x y sin x y 2 sin x cos y n x ct 1 n x ct u x t Bn sin sin L L 2 n 1 u x t n x ct 1 n x ct An cos L cos L 2 n 1 13 D Alambert Solution Derive D Alambert Solution Wave phenomena u x t is wave amplitude varying with 2 2 u space x and time t 2 u c c is wave speed t 2 x 2 D Alambert solution shown below uses arbitrary functions F and G with coordinates x ct and x ct u F G F x ct G x ct Proof of solution based on transforming derivatives 15 Derive D Alambert Solution II Transform equation from x t to using x ct and x ct c c t t t 1 1 x x x Apply transforms to u u t and u x u F G F u F G G 16 Derive D Alambert Solution III Continue transformations u u u F G 1 x x x F G 1 F G u u u F G c t t t F G c c F G Second derivatives satisfy wave equation 2u u u u u u c c t 2 t t t t t t t t c c F G c c F G c 2 F G 2 17 ME 501B Engineering Analysis 14 2u u u u x 2 x x t t t t F G F G 1 F G 1 u 2t 2u c2 2 x 18 3 Wave equation and classification of PDEs February 25 2009 Solution with Initial Conditions Solution with Initial Conditions II Define u x 0 f x and u t 0 g x Solution u x t uses f x ct Integral term satisfies wave equation Details of derivation in wave equation notes t x ct u t x 1 f x ct f x ct 1 g d 2 2c x ct Terms f x ct and f x ct satisfy wave equation since any function of these arguments satisfies the equation Solution gives u 0 x f x as required 2 t 2 2 x 2 x ct x ct g d c g x ct g x ct x ct x ct g d c g x ct g x ct 2 x ct …