Wave equation and classification of PDEs February 25, 2009ME 501B Engineering Analysis 1Wave Equation and Introduction Wave Equation and Introduction to Classification of to Classification of PDEsPDEsLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisFebruary 25, 20092Overview• 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– General analysis– Parabolic equations– Elliptic equations– Hyperbolic equations3Course Items• Notes on wave equation on web site• Midterm – Wednesday, March 11– 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 questions4Review Gradients• Gradients of Laplace equation solutions often proportional to flux terms– Heat flux and temperature gradient– Diffusion flux and mass fraction gradient– Velocity and velocity potential in ideal flow– In constant potential plot, lines perpen-dicular to the potential are flux lineszfyfxfffgrad∂∂+∂∂+∂∂=∇= kjizyx ∂∂+∂∂+∂∂=∇ kji5 6Review Interpretation of 02=∇ u• 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 zero012=⋅−=∇∫∫∫∫∫dAkudVSurfaceVolumeEnclosednv• Example of this result shown last week• Result applies to any problem in any geometry with Laplace’s equationWave equation and classification of PDEs February 25, 2009ME 501B Engineering Analysis 27Review Complex Variable• Cauchy-Riemann conditionsyuiyvxvixudzdfthenyuxvandyvxuIf∂∂−∂∂=∂∂+∂∂=∂∂−=∂∂∂∂=∂∂• Equivalent to Laplace equation– Function u(x,y) that satisfies Laplace equa-tion 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)8Review Additional Results• 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 proofs9Review Wave Equation• Wave phenomena: u(x,t) is wave amplitude varying with space, x, and time, t• c is wave speed22222xuctu∂∂=∂∂• Can solve by usual separation of variables technique• Also have D’Alambert solution with arbitrary functions F and G with coordinates ξ = x + ct and η = x – ct10Review Separation of Variables• Usual assumption u(x,t) = X(x)T(t)222222)()(1)()(11λ−=∂∂=∂∂xxXxXttTtTcResult is function of t equal to function of x[][])cos()sin()cos()sin()()(),(xDxCctBctAxXtTtxuλλλλ++==• Use above solution as starting point– Boundary conditions at x = 0 and x = L– Initial conditions on u and ∂u/∂tat x = 011Review Separation of Variables∑∞=⎟⎠⎞⎜⎝⎛⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=1sincossin),(nnnLxnLctnBLctnAtxuπππ∫⎟⎠⎞⎜⎝⎛ππ=LmdxLxmxgmA0sin)(2∫⎟⎠⎞⎜⎝⎛=LmdxLxmxfLB0sin)(2π• 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) = 0speedwaveisctLxxuctu0,022222≥≤≤∂∂=∂∂12General Solution• Substitute Amand Bmfrom equations just found and substitute into previous solution• Examine case where g(x) = 0 so An= 0∑∞=⎟⎠⎞⎜⎝⎛⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=1sincossin),(nnnLxnLctnBLctnAtxuπππ∑∞=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=1sincos),(nnLxnLctnBtxuππWave equation and classification of PDEs February 25, 2009ME 501B Engineering Analysis 313General Solution for g(x) = 0• From trig identities for sin(x ± 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∑∞=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=1sincos),(nnLxnLctnBtxuππ∑∞=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛+=1)(sin)(sin21),(nnLctxnLctxnBtxuππ14Similar Solution for f(x) = 0• From trig identities for cos(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∑∞=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=1sinsin),(nnLxnLctnAtxuππ∑∞=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+−⎟⎠⎞⎜⎝⎛−=1)(cos)(cos21),(nnLctxnLctxnAtxuππ15D’Alambert Solution• Wave phenomena: u(x,t) is wave amplitude varying with space, x, and time, t• c is wave speed22222xuctu∂∂=∂∂)()()()( ctxGctxFGFu −−+=+=ηξ• D’Alambert solution, shown below, uses arbitrary functions F and G with coordinates ξ = x + ct and η = x – ct• Proof of solution based on transforming derivatives16Derive D’Alambert Solutionηξηηξξ∂∂−∂∂=∂∂∂∂+∂∂∂∂=∂∂ccttt• Transform equation from (x,t) to (ξ,η) using ξ = x + ct and η = x – ct()ηξηηξξ∂∂+∂∂=∂∂∂∂+∂∂∂∂=∂∂)1(1xxx• Apply transforms to u, ∂u/∂t, and ∂u/∂x[])(')()(ξηξξξFGFu=+∂∂=∂∂[])(')()(ηηξηηGGFu=+∂∂=∂∂17Derive D’Alambert Solution II• Continue transformations)(')(')]()([)1()]()([)1(ηξηηξξηξηηξξGFGFGFuxuxxu+=∂+∂+∂+∂=∂∂∂∂+∂∂∂∂=∂∂[])(')(')]()([)]()([ηξηηξξηξηηξξGFcGFcGFcututtu−=∂+∂−∂+∂=∂∂∂∂+∂∂∂∂=∂∂18Derive D’Alambert Solution III• Second derivatives satisfy wave equation[][])''''()''()''(222GFcGFccGFcctuctuctuttuttuttu+=−∂∂−−∂∂=∂∂∂∂−∂∂∂∂=∂∂∂∂∂∂+∂∂∂∂∂∂=∂∂∂∂=∂∂ηξηξηηξξ[] [])''''()''()1()''()1(22GFGFGFtuttutxuxxu+=+∂∂++∂∂=∂∂∂∂∂∂+∂∂∂∂∂∂=∂∂∂∂=∂∂ηξηηξξ22222xuctu∂∂=∂∂Wave equation and classification of PDEs February 25, 2009ME 501B Engineering Analysis 419Solution with Initial Conditions• Define u(x,0) = f(x) and ∂u/∂t|0= g(x)• Solution, u(x,t) uses f(x ± ct)[]∫+−+−++=ctxctxdgcctxfctxfxtuνν)(21)()(21),(• Terms f(x + ct) and f(x – ct) satisfy wave