Classification/multidimensional PDEs March 2, 2009ME 501B Engineering Analysis 1Classification of Classification of PDEsPDEsand and Multidimensional Multidimensional PDEsPDEsLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisMarch 2, 20092Overview• Review last class– Wave equation solutions by separation of variables and D’Alembert approach• Characteristics and classification of partial differential equations– General analysis– Parabolic equations– Elliptic equations– Hyperbolic equations• Solving a wave equation problem3Midterm Exam• Wednesday, March 11• Covers material on diffusion and Laplace equations• Includes material up to and including tonight’s February 23 lecture and homework for Monday, March 2• Open book and notes, including homework solutions• Focus on working with existing solutions4Review Wave Equation• 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 = 05Review General Solution∑∞=⎟⎠⎞⎜⎝⎛⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=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≥≤≤∂∂=∂∂6Review Use of Trig Identities• 2 sin z cos y = sin(z + y) + sin(z – y) • 2 sin z sin y = cos(z – y) – cos(z + y)∑∞=⎪⎩⎪⎨⎧⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+π−⎟⎠⎞⎜⎝⎛−π=1)(cos)(cos21),(nnLctxnLctxnAtxu∑∞=⎟⎠⎞⎜⎝⎛⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛=1sincossin),(nnnLxnLctnBLctnAtxuπππ⎪⎭⎪⎬⎫⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−π+⎟⎠⎞⎜⎝⎛+π+LctxnLctxnBn)(sin)(sinsin zcos ysin yClassification/multidimensional PDEs March 2, 2009ME 501B Engineering Analysis 27Review D’Alembert Solution• Wave phenomena: u(x,t) is wave amplitude varying with space, x, and time, t• c is wave speed• Over any x region and t ≥ 022222xuctu∂∂=∂∂• D’Alembert solution using ξ = x + ct and η= x – ct, u(x,0) = f(x) and ∂u/∂x|t=0= g(x)[]∫+−+−++=ctxctxdgcctxfctxftxuνν)(21)()(21),(8Review D’Alambert II• We see that the solution obtained by separation of variables agrees with the D’Alambert solution for one case• Solution shows propagation of wave shapes without damping– Look at meaning of f(x + ct) and f(x – ct)• As time increases f(x + ct) retains the shape of the initial condition and moves to the left• Similarly f(x – ct) retains its shape and moves to the right9Review Wave Propagation-5 -4 -3 -2 -1 0 1 2 3 4 5distance, xtime, tt = 0ct=1ct = 2ct = 3x+ctx+ctx-ctx-ct10Review Boundaries• With g(x) = 0 solution is Fourier sine series which is periodic, odd function∑∞=⎟⎠⎞⎜⎝⎛==1sin)()0,(nnLxnBxfxuπ-1-0.500.51-2 -1.5 -1 -0.5 0 0.5 1 1.5 2xf(x)Actual solutionPeriodic extensions⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛−⎟⎠⎞⎜⎝⎛=53sin52sin2sin22022ππππmmmmBm11Review Time Evolution• Look at evolution when ct = 0.42)()(),(ctxfctxftxu−++=-1-0.500.51-2 -1.5 -1 -0.5 0 0.5 1 1.5 2xf(x)• For larger values of x ± ct, periodic extensions move into 0 ≤ x ≤ L = 1Initial Conditionf(x+ct)/2 = f(x+0.4)/2f(x-ct)/2 = f(x-0.4)/2-0.500.510 0.2 0.4 0.6 0.8 1xf(x+ct)ct = 0.450-0.500.510 0.2 0.4 0.6 0.8 1xf(x-ct)ct = 0.450-0.500.510 0.2 0.4 0.6 0.8 1xu(x,t)ct = 0.450Phase behavior of sine function causes initial wave form to be reflected at bound-ariesClassification/multidimensional PDEs March 2, 2009ME 501B Engineering Analysis 313Review Characteristics0,,,,22222=⎟⎟⎠⎞⎜⎜⎝⎛∂∂∂∂+∂∂+∂∂∂+∂∂yuxuuyxDyuCyxuBxuA• Used to classify second-order PDEsAACBBdxdy242−±−=• Analysis gives slope of “characteristics”• Characteristics slopes gives region of influence and domain of dependence14Review Classification of PDEs• The general second-order PDE in two variables is classified as follows– If B2–4AC < 0the PDE is called ellipticand has no real characteristic directions– If B2–4AC = 0the PDE is parabolic and has one repeated characteristic direction– If B2–4AC > 0the PDE is hyperbolic and has two real characteristic directionsAACBBdxdy242−±−=1502222=−∂∂+∂∂DyuxuLaplace/Poisson/Helmholtz equations (A = C = 1, B = 0, B2– 4AC < 0) are elliptic (no real characteristics)022=+∂∂DxuαDiffusion equation (A = α>0, B = C = 0, B2–4AC = 0) is parabolic (one characteristic)Wave equation (B2–4AC > 0) is hyperbolic (two real characteristics)0,,,,22222=⎟⎟⎠⎞⎜⎜⎝⎛∂∂∂∂+∂∂+∂∂∂+∂∂yuxuuyxDyuCyxuBxuA16Wave Equation Characteristics• Compute characteristic directions022222=∂∂−∂∂xuctu0,,,,22222=⎟⎠⎞⎜⎝⎛∂∂∂∂+∂∂+∂∂∂+∂∂tuxuuxtDxuCxtuBtuAWave equation: A = 1; B = 0; C = –c2; B2–4AC = c2> 0()ccAACBBdtdx±=−−±−=−±−=12))(1(4002422217Wave Equation Characteristics II• Compute characteristic directions with order of variables reversed022222=∂∂−∂∂tuxuc0,,,,22222=⎟⎠⎞⎜⎝⎛∂∂∂∂+∂∂+∂∂∂+∂∂tuxuuxtDtuCtxuBxuAWave equation: A = c2; B = 0; C = –1; B2–4AC = c2> 0()cccAACBBdxdt 12))(1(400242222±=−−−±−=−±−=18Review Wave Propagation-5 -4 -3 -2 -1 0 1 2 3 4 5distance, xtime, tt = 0ct=1ct = 2ct = 3x+ctx+ctx-ctx-ctCharacteristic slope dx/dt = 1/cCharacteristic slope dt/dx = -1/cClassification/multidimensional PDEs March 2, 2009ME 501B Engineering Analysis 419Behavior of Equation Types• Domain of dependence for u(x1,y1)– 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) • Importance for specifying boundary conditions and for numerical solutionsRight-running charac-teristic (dy/dx > 0 )Left-running charac-teristic (dy/dx < 0 )x1, y1Initial Condition CurveDomain of dependence20Hyperbolic Equations• Domain of dependence shown on previous chart• Region of influence is region of characteristics leaving x1, y1•