Multiple independent variables March 4, 2009ME 501B – Engineering Analysis 1Multidimensional Partial Multidimensional Partial Differential EquationsDifferential EquationsLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisMarch 4, 20092Overview• Review last class– Characteristics and classification of partial differential equations• Problems in more than two independent variables– Solution by separation of variables– Problems with multiple nonhomogenousboundary conditions– Solutions for rectangular geometry– Homework problem for cylindrical geometry3Review General and Hyperbolic• 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) • Areas for hyperbolic equations shown belowRight-running charac-teristic (dy/dx > 0 )Left-running charac-teristic (dy/dx < 0 )x1, y1Initial Condition CurveDomain of dependence4Review Elliptic and Parabolic• Imaginary characteristics for elliptic equations like Laplace and Poisson’s– 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, t1is entire domain 0 ≤ x ≤ L and 0 ≤ t < t1– Region of influence at x1, t1is entire region 0 ≤ x ≤ L and t > t15Review Multidimensional• 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– E. g., time is a parabolic coordinate6Review Multidimensional IISu −=∇2φφφθφθ∂∂+∂∂+∂∂+⎟⎠⎞⎜⎝⎛∂∂∂∂=∇∂∂+∂∂+⎟⎠⎞⎜⎝⎛∂∂∂∂=∇∂∂+∂∂+∂∂=∇urururrurrruSpherezuurrurrrulCylindricazuyuxuuCartesian222222222222222222222222cot1sin1111Sutu+∇α=∂∂2Suctu+∇=∂∂2222Laplace Diffusion WaveMultiple independent variables March 4, 2009ME 501B – Engineering Analysis 27Review 2D Diffusion• Two-dimensional diffusion equation for u(x,y,t)0),,(),0,(),,(),,0(),()0,,(000=====≤≤≤≤≥tHxutxutyLutyuyxfyxuHyLxt()()2222)()(11)()(11yyYyYxxXxXttTtT ∂∂+∂∂=∂∂α• Use separation of variable approach with all variables u(x,y,t) = X(x)Y(y)T(t)22221yuxutu∂∂+∂∂=∂∂α8Review 2D Diffusion II• Boundary conditions give general solution as sum of all eigenfunctions⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∑∑∞=∞=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−HymLxneCtyxunmtHmLnnmππαππsinsin),,(1122⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛==∑∑∞=∞=HymLxnCyxfyxunmnmππsinsin),()0,,(11• Eigenfunction expansion for t = 0∫∫⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π=HLnmdxdyHymLxnyxfHLC00sinsin),(49Look at f(x,y) = U, a Constant∫∫⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π=HLnmdxdyHymLxnyxfHLC00sinsin),(4• When we substitute f(x,y) – U, we can separate the x and y integrationsGeneral Result⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛π⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛π=⎟⎠⎞⎜⎝⎛π⎟⎠⎞⎜⎝⎛π=∫∫∫∫LHHLnmdxLxndyHymHLUdxdyHymLxnUHLC0000sinsin4sinsin410Look at f(x,y) = constant U II• Both integrals are effectively the same– for z = x, W = L and for z = y, W = H()[]⎪⎩⎪⎨⎧π=−ππ−=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛ππ−=⎟⎠⎞⎜⎝⎛π∫nevennoddnWnnWWzppWdxWzpWW021coscossin0020016224sinsin4π=ππ=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛π⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛π=∫∫mnnLmHHLUdxLxndyHymHLUCLHnmCnm= 0 zero for even m or n11Result for f(x,y) = U, a Constant• Replace n by 2n + 1 and m by 2m + 1 to get odd indices only and define new parameters βnand γmas followsmnoldoldnmmnUmnUmnUCmnγβ=π++=π=π+=γπ+=β16)12)(12(1616)12()12(22⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∑∑∞=∞=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−HymLxneCtyxunmtHmLnnmππαππsinsin),,(1122⎟⎠⎞⎜⎝⎛γ⎟⎠⎞⎜⎝⎛βγβ=∑∑∞=∞=α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−HyLxeUtyxumnnmmntHLmnsinsin16),,(00222212Dimensionless Parameters222222222LtHLtHLmnmneeα⎟⎟⎠⎞⎜⎜⎝⎛γ+β−α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−=• Modify exponential argument as follows• Where• Substitute into u(x,y,t) and divide by U⎟⎠⎞⎜⎝⎛γ⎟⎠⎞⎜⎝⎛βγβ=∑∑∞=∞=α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−HyLxeUtyxumnnmmnLtHLmnsinsin16),,(0022222π+=γπ+=β )12()12( mnmnMultiple independent variables March 4, 2009ME 501B – Engineering Analysis 313Important Parameters⎟⎠⎞⎜⎝⎛γ⎟⎠⎞⎜⎝⎛βγβ=∑∑∞=∞=α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−HyLxeUtyxumnnmmnLtHLmnsinsin16),,(0022222• Result shows that u/U is a function of x/L, y/H, L/H, and αt/L2• Can simplify double summation in this case by splitting exponential term222222222222222HtLtLtHLLtLtHLmnmnmneeeeeαγ−αβ−αγ−αβ−α⎟⎟⎠⎞⎜⎜⎝⎛γ+β−==14Product Solution• Can now separate n and m sums⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛γγ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛ββ=⎟⎠⎞⎜⎝⎛γγ⎟⎠⎞⎜⎝⎛ββ=⎟⎠⎞⎜⎝⎛γγ⎟⎠⎞⎜⎝⎛ββ=∑∑∑∑∑∑∞=αγ−∞=αβ−∞=∞=αγ−αβ−∞=∞=αγ−αβ−000000sin4sin4sinsin16sinsin16),,(222222222222mmmHtnnnLtnmmmHtnnLtnmmmHtnnLtHyeLxeHyeLxeHyeLxeUtyxumnmnmnProduct of one-dimen-sionalsolutions15Nonzero 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 boundaries16Nonzero Boundaries II• Solve Laplace equation for w (with superposition if required)• Solution for v is same as previous solution for u with zero boundaries• Initial condition for u found from u(x,y,t) = v(x,y,t) + w(x,y) by setting t = 0⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=∑∑∞=∞=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−HymLxneCtyxvnmtHmLnnmππαππsinsin),,(112217Nonzero