Midterm Review March 9, 2009ME 501B – Engineering Analysis 1Midterm ReviewMidterm ReviewLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisMarch 9, 20092Overview• Review last class– General approach– Diffusion equation in two space dimensions– Three-dimensional Laplace equation• Review for midterm– Sturm-Liouville solution eigenvalues– General approach for PDEs– Transformations and superposition– Diffusion equation– Laplace equation– Rectangular and cylindrical coordinates3Review 2D Diffusion• Two-dimensional diffusion equation for u(x,y,t)0),,(),0,(),,(),,0(),()0,,(000=====≤≤≤≤≥tHxutxutyLutyuyxfyxuHyLxt22221yuxutu∂∂+∂∂=∂∂α⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∑∑∞=∞=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−HymLxneCtyxunmtHmLnnmππαππsinsin),,(1122∫∫⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=HLnmdxdyHymLxnyxfHLC00sinsin),(4ππ4Review f(x,y) = U, a Constant∫∫⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π=HLnmdxdyHymLxnyxfHLC00sinsin),(4⎪⎩⎪⎨⎧=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∫∫otherwisenandmoddmndxdyHymLxnUHLCHLnm016sinsin4200πππ⎟⎠⎞⎜⎝⎛γ⎟⎠⎞⎜⎝⎛βγβ=∑∑∞=∞=α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−HyLxeUtyxumnnmmntHLmnsinsin16),,(002222πγπβ)12()12(+=+=mnmn5Review Nonzero Boundaries• Sturm-Liouville eigenfunction expan-sions require zero boundary conditions• For nonzero boundaries, split solution as in 1D case u(x,y,t) = v(x,y,t) + w(x,y)– 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 boundaries6Review 3D Laplace• Use combination of separation of variables and superposition• Start with basic solution that has homogenous boundary conditions at five surfaces and u(x,y,W) = uW(x,y)0)0,,(),,(),0,(),,(),,0(0000222222=====≤≤≤≤≤≤=∂∂+∂∂+∂∂yxuzHxuzxuzyLuzyuWzHyLxzuyuxuMidterm Review March 9, 2009ME 501B – Engineering Analysis 27Review 3D Laplace II• Solution of Laplace equation in x, y, z similar to diffusion solution in x, y, t• Have hyperbolic cosine in z direction instead of exponential time decay⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π+π=∑∑∞=∞=HymLxnHzmLznCzyxunmnmsinsinsinh),,(11∫∫⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π+π=HLWnmdxdyHymLxnyxuHWmLWnHLC00sinsin),(sinh48Midterm Exam• Wednesday, March 11• Covers material on diffusion and Laplace equations (2 independent variables only)• Includes material up to and including homework for Monday, March 2• Open book and notes, including homework solutions and integral tables– Typical possible integrals: sin2ax, xsin(ax) 9Sturm-Liouville[]0)()()(=++⎟⎠⎞⎜⎝⎛yxpxqdxdyxrdxdλ0)(0)(2121=+=+==bxaxdxdybydxdykaykll• Can expand any f(x) in terms of complete set of eigenfunctions, ym()∫∫==bammbammmmmdxxyxyxpdxxfxyxpyyfya)()()()()()(,),(∑∞==0)()(mmmxyaxf10Partial Differential Equations• Solving PDEs– Basic idea is to get solution to PDE as sum of eigenfunctions than can be used to represent an initial or boundary condition• Ability to get such a set of eigenfunctions assured if we have a Sturm Liouville problem• Key element of such a problem is homogenous differential equation and boundary conditions• Will use various transforms to get this problem• For diffusion equation use u(x,t) = v(x,t) + w(x)• For Laplace’s equation use superposition11PDE Boundaries• Boundary conditions– Fixed value (first kind or Dirichlet) (e.g., u = 0 at x = 0)– Fixed gradient (second kind or Newmann) (e.g., ∂u/∂x = 0 at x = L)– Mixed (third kind) a (∂u/∂y)y=H+ b uy=H= 0– Use transforms like u = v + w (diffusion) or superposition (Laplace) if boundary conditions do not equal zero– Problems in cylinder and sphere require u to be finite at r = 012Solving PDE Start• Perform necessary operations if boundaries not homogenous– Diffusion: define u(x,t) = v(x,t) + w(x)• v satisfies diffusion equation with zero boundary conditions; w satisfies boundary• Can also uses this for “source term”– Laplace equation use superposition• Solution is sum of two or more solutions each of which has only one nonzero boundary• Each solution has consistent boundary condition kind for zero and nonzero partsMidterm Review March 9, 2009ME 501B – Engineering Analysis 313Separation of Variables• Not needed if existing separation of variables solution is available– Set variable in PDE, u, as product of two functions of one variable: u = F(x1)G(x2)– Substiute product into PDE and differentiate– Divide by original product solution to get two terms, one with F only and one with G– Set one term to a –λ2to get eigenfunctions– Solve resulting pair of ODEs14Separation of Variables Result• Starting solutions– Diffusion equation, rectangular geometry[])cos()sin(),(212xCxCetxutλ+λ=αλ−– Diffusion equation, cylindrical geometry[])()(),(02012rYCrJCetrutλ+λ=αλ−– Diffusion equation, spherical geometry⎟⎠⎞⎜⎝⎛+=−rrCrrCetrutλλαλcossin),(212.15Separation of Variables Result II• Rectangular Laplace starting solutions– Rectangular, nonzero BC at y = 0 or y = H– Cylindrical, nonzero BC at z = 0 or z = H[][ ])cosh()sinh()cos()sin(),( yDyCxBxAyxuλ+λλ+λ=[][])cos()sin()cosh()sinh(),( yDyCxBxAyxuλ+λλ+λ=[][])()(coshsinh),(00rDYrCJzBzAzruλ+λλ+λ=[][ ]0)()(cossin),(00=λ+λλ+λ= RDKRCIzBzAzru– Rectangular, nonzero BC at x = 0 or x = L– Cylindrical, nonzero BC at r = Roor R = Ri16Fitting Boundary Conditions• Start with homogenous boundary con-ditions (variable, gradient or sum = 0)• Eliminate some constants in starting solutions– Constants will be zero or be related to each other, e.g. C = –DI0(λRo)/K0(λRo)– One boundary condition will lead to eigen-values and eigenfunctions– Final result should be product solution with one eigenfunction and one constant17Eigenfunction Expansions• Most general solution is sum of infinite series of eigenfunctions, each with it’s own constant: ΣnCnFn(λξ)Gn(η)• Use nonzero boundary condition (initial condition in diffusion equation) to fit constants via eigenfunction expansion()() (