Laplace equation solutions February 11, 2009ME 501B – Engineering Analysis 1Laplace Equation SolutionsLaplace Equation SolutionsLarry CarettoMechanical Engineering 501BSeminar in Engineering Seminar in Engineering AnalysisAnalysisFebruary 11, 20092Overview• Review last class– Laplace equation solutions for homogenous boundary conditions on three boundaries• Solutions of Laplace’s equation for more than one nonzero boundaries– Superposition solutions– Superposition for gradient and other boundary conditions• Cylindrical coordinates3Review Laplace’s Equation• Used to express equilibrium fields of engineering variable like temperature, species concentration, electrostatic potential and ideal fluid flow• Written in general coordinates as 02=∇ uφφφθφθ∂∂+∂∂+∂∂+⎟⎠⎞⎜⎝⎛∂∂∂∂=∇∂∂+∂∂+⎟⎠⎞⎜⎝⎛∂∂∂∂=∇∂∂+∂∂+∂∂=∇urururrurrruSpherezuurrurrrulCylindricazuyuxuuCartesian222222222222222222222222cot1sin11114Similar to Diffusion EquationLaplaceDiffusiontinboundaryopenrutrutLututLxxutu)(),(0),(0),0(001022===≥≤≤∂∂=∂∂α0)0,()(),(0),(0),0(0002222====≤≤≤≤∂∂=∂∂−xuruHxuyLuyuHyLxxuyu5Separation of VariablesLaplaceDiffusion222)()(1)()(1xxXxXttTtT∂∂=∂∂=−αλ22222)()(1)()(1xxXxXyyYyY∂∂=∂∂−=−λtAetTαλ2)(−=)cosh()sinh()(yDyAyYλλ+=6x = 0,L Boundary ConditionsLaplaceDiffusion0),(0),0(0====tLuforLntuforCπλ)cos()sin()(xCxBxXλλ+=⎟⎠⎞⎜⎝⎛=LxnBxXnnπsin)()cos()sin()(xCxBxXλλ+=0),(0),0(0====yLuforLnyuforCπλ⎟⎠⎞⎜⎝⎛=LxnBxXnnπsin)(Laplace equation solutions February 11, 2009ME 501B – Engineering Analysis 27y = 0 Boundary ConditionLaplaceDiffusion00)0cosh()0sinh()0(==+=DhaveMustDAYλλ⎟⎠⎞⎜⎝⎛=LynAyYnnπsinh)(No equivalent condition because of open boundary in time8General Solution; Fitted ConditionLaplaceDiffusionLnxeCtxunnntnnπλλαλ==∑∞=−1)sin(),(2∑∞===10)sin()()0,(nnnxCxuxuλLnxyCyxunnnnnπλλλ==∑∞=1)sin()sinh(),(∑∞===1)sinh()sin()(),(nnnnNHrCxuHxuλλ9Eigenvalue Expansion for CnLaplaceDiffusion∫∫∫⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛LLLdxLxnxuLdxLxndxLxnxu000200sin)(2sinsin)(πππ⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛∫∫∫LHndxLxnxuLdxLxnLHndxLxnxuLNLLNπππππsinhsin)(2sinsinhsin)(002010Review Constant Boundary• If uN(x) = U, the solution for u(x,y) is()∑∞=⎟⎠⎞⎜⎝⎛++⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛+=1)12(sinh12)12(sinh)12(sin4),(nLHnnLynLxnUyxuππππ• u/U = depends on x/L, y/L and H/L• Plot on next page for H/L = 1Cm11 12Review Gradients as Fluxes• Laplace and diffusion equation are based on conservation of fluxes which are (negative) gradients of potential• Laplace equation gives equilibrium where net flux from region should be zero• Provided notes showing that net outflow is zero for first Laplace solution0000000=∂∂−∂∂+∂∂−∂∂=∫∫∫∫====LLHyyHLxHxdxyudxyudyxudyxuoutflowNetLaplace equation solutions February 11, 2009ME 501B – Engineering Analysis 313Review Gradient Boundary• Zero gradient at x = 0)(),(0)0,(),(0,00),0(2222xuHxuxuyLuxuHyLxyuxuNy====∂∂≤≤≤≤=∂∂+∂∂x = 0 x = Ly = H y = 0ux(0,y) = 0 u(L,y) = 0u(x,H) = uN(x) u(x,0) = 014Review Zero Gradient Solution• General solution • Solution for uN(x) = U, a constantLnyxCyxunnnnn2)12()sinh()cos(),(0πλλλ+==∑∞=()()LndxxxuLHCnLnNnn2)12(cos)(sinh20πλλλ+==∫LnHyxLUyxunnnnnnn2)12()sinh()sinh()cos()1(2),(0πλλλλλ+=−=∑∞=15Rectangular Laplace Summary• Problem: homogenous boundary conditions except at y = H• Separation of variables solution gives u(x,y) = X(x)Y(y) = [Asin(λx) + Bcos(λx)] •[Csinh(λy) + Dcosh(λy)] (Start here!)• Boundary conditions at x = 0 and x = L give A or B and eigenvalue, λ• Eigenfunction expansion at y = H gives coefficients in infinite series solution of all eigenfunctions16Exercise• Solve problem from last class with boundary condition at y = H changed to a gradient• Use previous solution of similar problem as starting point – in this case have most of the work donex = 0 x = Ly = H y = 0u = 0 u = 0∂u/∂y= gN(x) u = 002222=∂∂+∂∂yuxuLnyxCyxunnnnnπλλλ==∑∞=1)sinh()sin(),(17Exercise Solution• Start with potential solution• Take the gradient• Apply y = H boundary conditionLnyxCyxunnnnnπλλλ==∑∞=1)sinh()sin(),(LnyxCyyxunnnnnnπλλλλ==∂∂∑∞=1)cosh()sin(),(LnHxCyHxuxgnnnnnnNπλλλλ==∂∂=∑∞=1)cosh()sin(),()(18Exercise Solution II• Get eigenfunction expansion()()LndxxxgLHCnLnNnnnπλλλλ==∫0sin)(cosh2• For constant gradient, gN(x) = GN()()()()()[]()LHGLHxGLHdxxGCnnnnNnnLnnNnnLnNnλλλλλλλλλλcosh112coshcos2coshsin200−−=−==∫()LnLHyxGyxunnnnnnNπλλλλλ==∑∞= ,...5,3,12cosh)sinh()sin(4),(Laplace equation solutions February 11, 2009ME 501B – Engineering Analysis 419More than One Nonzero Boundary• Laplace’s equation for 0 ≤ x ≤ L and 0 ≤y ≤ H with boundary conditions shown• Do not have homogenous boundary conditions in any coordinate direction• Solution is sum of two simpler solutionsx = 0 x = Ly = H y = 0u = 0 u = uE(y)u = uN(x) u = 020Superposition Solution• Solution is u(x,y) = u1(x,y) + u2(x,y)x = 0 x = Ly = H y = 0u = 0 u = 0u = uN(x) u = 0x = 0 x = Ly = H y = 0u = 0 u = uE(y)u = 0u = 0•u1(x,y) has u(L,y) = 0•u2(x,y) has u(H,x) = 021Superposition Solution II• u(x,y) = u1(x,y) + u2(x,y)•u1 and u2satisfy Laplace’s equation and their sum satisfies boundary conditions–u(x,0) = u1(x,0) + u2(x,0) = 0 + 0 = 0–u(x,H) = u1(x,H) + u2(x,H) = uN+ 0 = uN– u(0,y) = u1(0,y) + u2(0,y) = 0 + 0 = 0– u(L,y) = u1(L,y) + u2(L,y) = 0 + uE= uE• Solution for u2is same as solution for u1with x and y (and L and H) interchanged22Superposition Solution III• Sum two solutions shown belowx = 0 x = Ly = H y = 0u = 0 u = 0u = uN(x) u = 0x = 0 x = Ly = H y = 0u = 0 u = uE(y)u = 0 u = 0•u1(x,y) found previously• Swap x and y to get u2(x,y) from u123Superposition Solution IV• u(x,y) = u1(x,y) +