ECE257 Numerical Methods andECE257 Numerical Methods andScientific ComputingScientific ComputingPartial Differential EquationsPartial Differential EquationsECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutTodayToday’’s class:s class:••Finite DifferenceFinite Difference––Parabolic EquationsParabolic EquationsECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutPartialPartial Differential Equation (PDE Differential Equation (PDE))€ A∂2U∂x2+ B∂2U∂x∂y+ C∂2U∂y2+ D = 0• A PDE involves partial derivatives of an unknown function oftwo or more independent variables• In many engineering applications, PDEs tend to involve twoindependent variables (x,y) and have the following generalform:• Where A, B, and C are functions of x and y and D is a functionof x,y,U, and€ ∂U∂x€ ∂U∂yECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutThreeThree Types of Types of PDEsPDEsB2-4AC Category Examples< 0 Elliptical= 0 Parabolic> 0 Hyperbolic € Laplace Equation∂2φ∂x2+∂2φ∂y2= 0Heat transfer Equationk∂2T∂x2=∂T∂tWave Equation∂2z∂x2=1C2∂2z∂t2ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutParabolic PDEsParabolic PDEsFrom Numerical Methods for Engineers, Chapra and Canale, Copyright © The McGraw-Hill Companies, Inc.ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of Connecticut € q x( )ΔyΔzΔt − q x + Δx( )ΔyΔzΔt = ΔxΔyΔzρCΔT q is the heat flux function, ρ is the material density, C is the heat capacity of the material Heat Conduction ExampleHeat Conduction Example• A thin insulated rod: the temperatures at its ends are known• Heat in - Heat out = Heat storage••Take the limitTake the limit€ q x( )Δt − q x + Δx( )Δt = ΔxρCΔT€ q x( )− q x + Δx( )Δx=ρCΔTΔt€ −∂q∂x=ρC∂T∂tECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutHeat Conduction ExampleHeat Conduction Example••Substituting FourierSubstituting Fourier’’s laws law••Elliptical PDEs are bounded in all dimensionsElliptical PDEs are bounded in all dimensions••Parabolic PDEs are bounded in only spatial dimensionsParabolic PDEs are bounded in only spatial dimensions––Time is open-endedTime is open-ended€ qx= −kρC∂T∂x€ k∂2T∂x2=∂T∂tECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExplicit MethodExplicit Method€ ∂2T∂x2=T x + Δx,t( )− 2T x,t( )+ T x − Δx,t( )Δx2+ O Δx2( )••Central difference approximation of partial secondCentral difference approximation of partial secondderivativederivative••Forward finite divided difference approximation ofForward finite divided difference approximation ofpartial derivativepartial derivative••Substitute into Heat Conduction EquationSubstitute into Heat Conduction Equation€ kT x + Δx,t( )− 2T x,t( )+ T x − Δx,t( )Δx2=T x,t + Δt( )− T x,t( )Δt€ ∂T∂t=T x,t + Δt( )− T x,t( )Δt+ O Δt( )ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExplicit MethodExplicit Method••Solve for Solve for T(x,t+T(x,t+∆∆t)t)••A three-point formula that allows us to solve at aA three-point formula that allows us to solve at anode for a future time based on present values at thenode for a future time based on present values at thenode and its neighborsnode and its neighbors€ kT x + Δx,t( )− 2T x,t( )+ T x − Δx,t( )Δx2=T x,t + Δt( )− T x,t( )Δt€ T x,t + Δt( )= T x,t( )+ kΔtΔx2T x + Δx,t( )− 2T x,t( )+ T x − Δx,t( )( )€ Til +1= Til+λTi+1l− 2Til+ Ti−1l( )ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExplicit MethodExplicit MethodFrom Numerical Methods for Engineers, Chapra and Canale, Copyright © The McGraw-Hill Companies, Inc.ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExplicit Method (Continued)Explicit Method (Continued))2( 3 )2( 2 )2( 1 ,0020304031301020302120001020111TTTTTiTTTTTiTTTTTil+−+==+−+==+−+===λλλt=0, l=0l=1l=2x0=0x4=Lx1x2x3The variables at the R.H.S of equ.s areknown.ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExplicit Method (ContinuedExplicit Method (Continued))••EExplicit method: to compute values at each node ofa future time step (l+1) based on the present valuesof the node and its neighboring nodes• The equations can be made at each interior node.• At l =0, the value of T is given by the initialcondition.• At the two ends (i = 0 & i = n), i.e, the edge nodes,the value of T is given by the boundary conditions.ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExampleExample € 0.835∂2T∂x2=∂T∂t, T x,0( )= 0,T 0,t( )= 100,T 10,t( )= 50 € Δx = 2 Δt = 0.1€ T11= T10+ 0.020875 T20− 2T10+ T00( )= 0 + 0.020875 0 − 2 0( )+ 100( )= 2.0875€ ⇒λ= kΔtΔx2= 0.8350.122= 0.020875€ T21= T20+ 0.020875 T30− 2T20+ T20( )= 0 + 0.020875 0 − 2 0( )+ 0( )= 0ECE 257 Numerical Methods and Scientific ComputingFall 2004Lecture 21John A. ChandyDept. of Electrical and Computer EngineeringUniversity of ConnecticutExampleExample€ T31= T30+ 0.020875 T40− 2T30+ T20( )= 0 + 0.020875 0 − 2 0( )+ 0( )= 0€ T41= T40+ 0.020875 T50− 2T40+ T30( )= 0 + 0.020875 50 − 2 0( )+ 0( )= 1.04375€ T12= T11+ 0.020875 T21− 2T11+ T01( )= 2.0875 + 0.020875 0 − 2 2.0875( )+ 100( )= 4.0878ECE 257 Numerical