SOLUTION OF Partial Differential Equations. (PDEs)

Transcription

1 SOLUTION OF Partial Differential Equations (PDEs) Mathematics is the Language of Science PDEs are the expression of processes that occur across time & space: (x,t), (x,y), (x,y,z), or (x,y,z,t)

2 Partial Differential Equations (PDE's) A PDE is an equation which includes derivatives of an unknown function with respect to or more independent variables

3 Partial Differential Equations (PDE's) PDE's describe the behavior of many engineering phenomena: Wave propagation Fluid flow (air or liquid) Air around wings, helicopter blade, atmosphere Water in pipes or porous media Material transport and diffusion in air or water Weather: large system of coupled PDE's for momentum, pressure, moisture, heat, Vibration Mechanics of solids: stress-strain in material, machine part, structure Heat flow and distribution Electric fields and potentials Diffusion of chemicals in air or water Electromagnetism and quantum mechanics 3

4 Partial Differential Equations (PDE's) Weather Prediction heat transport & cooling advection & dispersion of moisture radiation & solar heating evaporation air (movement, friction, momentum, coriolis forces) heat transfer at the surface To predict weather one need "only" solve a very large systems of coupled PDE equations for momentum, pressure, moisture, heat, etc. 4

5 Modelización Numérica del Tiempo 360x80x3 x nvar Conservación de energía, masa, momento, vapor de agua, ecuación de estado de gases. resolución /Δt Duplicar la resolución espacial supone incrementar el tiempo de cómputo en un factor 6 v = (u, v, w), T, p, ρ = /α y q Condición inicial H+0 H+4 D+0 D+ D+ sábado 5// D+3 D+4 D+5 5

6 Partial Differential Equations (PDE's) Learning Objectives ) Be able to distinguish between the 3 classes of nd order, linear PDE's. Know the physical problems each class represents and the physical/mathematical characteristics of each. ) Be able to describe the differences between finite-difference and finite-element methods for solving PDEs. 3) Be able to solve Elliptical (Laplace/Poisson) PDEs using finite differences. 4) Be able to solve Parabolic (Heat/Diffusion) PDEs using finite differences. 6

7 Partial Differential Equations (PDE's) Engrd 4 Focus: Linear nd-order PDE's of the general form u u u A + B + C + D= 0 x y x y u(x,y), A(x,y), B(x,y), C(x,y), and D(x,y,u,,) The PDE is nonlinear if A, B or C include u, u/ x or u/ y, or if D is nonlinear in u and/or its first derivatives. Classification B 4AC < 0 > Elliptic (e.g. Laplace Eq.) B 4AC = 0 > Parabolic (e.g. Heat Eq.) B 4AC > 0 > Hyperbolic (e.g. Wave Eq.) Each category describes different phenomena. Mathematical properties correspond to those phenomena. 7

8 Partial Differential Equations (PDE's) Elliptic Equations (B 4AC < 0) [steady-state in time] typically characterize steady-state systems (no time derivative) temperature torsion pressure membrane displacement electrical potential closed domain with boundary conditions expressed in terms of u u u u(x,y), (in terms of and ) η x y Typical examples include 0 u u u + = u u x y D(x,y,u,, ) x y A =, B = 0, C = ==> B 4AC = 4 < 0 Laplace Eq. Poisson Eq. 8

9 Elliptic PDEs Boundary Conditions for Elliptic PDE's: Dirichlet: u provided along all of edge Neumann: u η provided along all of the edge (derivative in normal direction) Mixed: u provided for some of the edge and u η for the remainder of the edge Elliptic PDE's are analogous to Boundary Value ODE's 9

10 Elliptic PDEs u(x,y =) or u/ y given on boundary y u(x=0,y) or u/ x given on boundary Estimate u(x) from Laplace Equations u(x=,y) or u/ x given on boundary u(x,y =0) or u/ y given on boundary x 0

11 Parabolic PDEs Parabolic Equations (B 4AC = 0) [first derivative in time ] variation in both space (x,y) and time, t typically provided are: initial values: u(x,y,t = 0) boundary conditions: u(x = x o,y = y o, t) for all t u(x = x f,y = y f, t) for all t all changes are propagated forward in time, i.e., nothing goes backward in time; changes are propagated across space at decreasing amplitude.

12 Parabolic PDEs Parabolic Equations (B 4AC = 0) [first derivative in time ] Typical example: Heat Conduction or Diffusion (the Advection-Diffusion Equation) u u u D: = k + D(x,u, ) t x x A = k, B = 0, C = 0 > B 4AC = 0 u u u u u D: = k + D(x,y,u,, ) t + x y x y = k u + D

13 Parabolic PDEs t u(x=0,t) given on boundary for all t Estimate u(x,t) from the Heat Equation u(x=,t) given on boundary for all t u(x,t =0) given on boundary as initial conditions for u/ t x 3

14 Parabolic PDEs c(x,t) = concentration at time, t, and distance, x. x=0 Δx x=l An elongated reactor with a single entry and exit point and a uniform cross-section of area A. A mass balance is developed for a finite segment Δx along the tank's longitudinal axis in order to derive a differential equation for concentration (V = A Δx). 4

15 Parabolic PDEs x=0 Δx x=l Δc c(x) c(x) V = Qc(x) Q c(x) + Δx DA Δt x x Flow in Flow out Dispersion in c(x) c(x) + DA + Δx kvc(x) x x x Decay Dispersion out reaction As Δt and Δx ==> 0 Δc c c Q c ==> = D Δt t x A x kc 5

16 Hyperbolic PDEs Hyperbolic Equations (B 4AC > 0) [nd derivative in time ] variation in both space (x, y) and time, t requires: initial values: u(x,y,t=0), u/ t (x,y,t = 0) "initial velocity" boundary conditions: u(x = x o,y = y o, t) for all t u(x = x f,y = y f, t) for all t all changes are propagated forward in time, i.e., nothing goes backward in time. 6

17 Hyperbolic PDEs Hyperbolic Equations (B 4AC > 0) [nd derivative in time] Typical example: Wave Equation u u u u D : + D(x, y, u,, ) = 0 x c t x t A =, B = 0, C = -/c ==> B 4AC = 4/c > 0 Models vibrating string water waves voltage change in a wire 7

18 Hyperbolic PDEs t u(x=0,t) given on boundary for all t Estimate u(x,t) from the Wave Equation u(x=,t) given on boundary for all t x u(x,t =0) and u(x,t=0)/ t given on boundary as initial conditions for u/ t Now PDE is second order in time 8

19 Numerical Methods for Solving PDEs Numerical methods for solving different types of PDE's reflect the different character of the problems. Laplace - solve all at once for steady state conditions Parabolic (heat) and Hyperbolic (wave) equations. Integrate initial conditions forward through time. Methods Finite Difference (FD) Approaches (C&C Chs. 9 & 30) Based on approximating solution at a finite # of points, usually arranged in a regular grid. Finite Element (FE) Method (C&C Ch. 3) Based on approximating solution on an assemblage of simply shaped (triangular, quadrilateral) finite pieces or "elements" which together make up (perhaps complexly shaped) domain. In this course, we concentrate on FD applied to elliptic and parabolic equations. 9

20 Finite Difference for Solving Elliptic PDE's Solving Elliptic PDE's: Solve all at once Liebmann Method: Based on Boundary Conditions (BCs) and finite difference approximation to formulate system of equations Use Gauss-Seidel to solve the system 0 u + = u u x D(x,y,u,, ) x y u y Laplace Eq. Poisson Eq. 0

21 Finite Difference Methods for Solving Elliptic PDE's. Discretize domain into grid of evenly spaced points. For nodes where u is unknown: u ui,j ui,j+ ui+,j = + Δ x ( Δx) O( x ) w/ Δ x = Δ y = h, substitute into main equation u ui,j ui,j + ui,j + = + Δ y ( Δy) u u u i,j + u i+,j + u i,j + u i,j+ 4u i,j + = + x y h O( y ) 3. Using Boundary Conditions, write, n*m equations for u(x i=:m,y j=:n ) or n*m unknowns. 4. Solve this banded system with an efficient scheme. Using Gauss-Seidel iteratively yields the Liebmann Method. O(h )

22 Elliptical PDEs The Laplace Equation x u u + = y 0 The Laplace molecule j+ j j- i- i i+ If Δx= Δythen T + T + T + T 4T = 0 i+,j i,j i,j+ i,j i,j

23 Elliptical PDEs The Laplace molecule: T + T + T + T 4T = 0 i+,j i,j i,j+ i,j i,j T = 0 o C T = 00 o C,,,3,,,3 3, 3, 3,3 T = 0 o C T = 00 o C T T T3 T T T3 T3 T3 T T T T T T T T 33 The temperature distribution can be estimated by discretizing the Laplace equation at 9 points and solving the system of linear equations. T T = Excel 3

24 Solution of Elliptic PDE's: Additional Factors Primary (solve for first): u(x,y) = T(x,y) = temperature distribution Secondary (solve for second): heat flux: T T qx = k and qy = k x y obtain by employing: T Ti+,j Ti,j T Ti,j+ Ti,j x Δx y Δy then obtain resultant flux and direction: qn = qx + q qy y θ = tan qx qx > 0 qy θ = tan +π qx qx < 0 (with θ in radians) 4

25 Elliptical PDEs: additional factors T = 0 o C T = 00 o C T = 0 o C T = 00 o C T T qx = k k x T T qy = k k y k' = 0.49 cal/s cm C At point, (middle left): q x ~ (50-0)/( 0cm) = -.5 cal/(cm s) q y ~ (50-4.3)/( 0cm) = cal/(cm s) n x y T i+,j i,j Δx T q = q + q = =.85cal /(cm s) i,j+ i,j Δy qy θ= tan = tan = = 5.5 qx.5 o o o 5

26 Solution of Elliptic PDE's: Additional Factors Neumann Boundary Conditions (derivatives at edges) employ phantom points outside of domain use FD to obtain information at phantom point, T,j + T -,j + T 0,j+ + T 0,j- 4T 0,j = 0 [*] If given T x to obtain then use T T T = x Δx, j i, j T T, j = T, j Δx x T Substituting [*]: T,j Δ x + T0,j+ + T0,j 4T0,j = 0 x Irregular boundaries use unevenly spaced molecules close to edge use finer mesh 6

27 Elliptical PDEs: Derivative Boundary Conditions The Laplace molecule: T + T + T + T 4T = 0 i+,j i,j i,j+ i,j i,j T = 75 o C T = 00 o C,,,3,,,3 3, 3, 3,3 Insulated ==> T/ y = 0 T = 50 o C Derivative (Neumann) BC at (4,): T = y T3, - T5, Δy T T5, = T3, Δy y T + T + T + T 4T = 0 Substitute into: 4, 4,0 3, 5, 4, To obtain: T T4, T4,0 + T3, Δy 4T4, = 0 y 0 7

28 Parabolic PDE's: Finite Difference Solution Solution of Parabolic PDE's by FD Method use B.C.'s and finite difference approximations to formulate the model integrate I.C.'s forward through time for parabolic systems we will investigate: explicit schemes & stability criteria implicit schemes - Simple Implicit - Crank-Nicolson (CN) - Alternating Direction (A.D.I), D-space 8

29 Parabolic PDE's: Heat Equation Prototype problem, Heat Equation (C&C 30.): T T D = k Find T(x,t) t x T T T D = k + Find T(x,y,t) t x y Given the initial temperature distribution as well as boundary temperatures with k' k = = Coefficient of thermal diffusivity Cρ k' = coefficient of thermal conductivity where: C = heat capacity ρ = density 9

30 Parabolic PDE's: Finite Difference Solution Solution of Parabolic PDE's by FD Method. Discretize the domain into a grid of evenly spaces points (nodes). Express the derivatives in terms of Finite Difference Approximations of O(h ) and O(Δt) [or order O(Δt )] x T y T T t Finite Differences 3. Choose h = Δx= Δy, and Δt and use the I.C.'s and B.C.'s to solve the problem by systematically moving ahead in time. 30

31 Parabolic PDE's: Finite Difference Solution Time derivative: Explicit Schemes (C&C 30.) Express all future (t + Δt) values, T(x, t + Δt), in terms of current (t) and previous (t - Δt) information, which is known. Implicit Schemes (C&C ) Express all future (t + Δt) values, T(x, t + Δt), in terms of other future (t + Δt), current (t), and sometimes previous (t - Δt) information. 3

32 Parabolic PDE's: Notation Notation: Use subscript(s) to indicate spatial points. Use superscript to indicate time level: T m+ i = T(x i, t m+ ) Express a future state, T m+ i, only in terms of the present state, T m i T T -D Heat Equation: = k t x m m m T Ti Ti + Ti+ = + O( Δx) Centered FDD x ( Δx) m+ i m i T T T = + O( Δt) Forward FDD t Δt Solving for T i m+ results in: T i m+ = T im + λ(t i-m T im + T i+m ) with λ = k Δt/(Δx) T i m+ = (-λ) T im + λ (T i- m + T i+m ) 3

33 Parabolic PDE's: Explicit method t = t = 5 t = 4 t = 3 t = t = t = Initial temperature 33

34 Parabolic PDE's: Example - explicit method Example: The -D Heat Equation T t k = 0.8 cal/s cm oc, 0-cm long rod, Δt = secs, Δx =.5 cm (# segs. = 4) = k t x T t I.C.'s: T(0 < x < 0, t = 0) = 0 B.C.'s: T(x = 0, all t) = 00 T(x = 0, all t) = C Heat Eqn. Example 50 C with λ = k Δt/ (Δx) = 0.6 X = 0 0 C X = 0 cm ( ) + m+ m m m m i = i +λ i i + i T T T T T 34

35 Parabolic PDE's: Example - explicit method Example: The -D Heat Equation ( ) + m+ m m m m i = i +λ i i + i T T T T T Starting at t = 0 secs. (m = 0), find results at t = secs. (m = ): T = T 0 + λ(t 0 0 +T 0 +T 0 ) = [00 (0)+0] = 6. T = T 0 + λ(t 0 +T 0 +T 0 3 ) = [0 (0)+0] = 0 T 3 = T λ(t 0 +T 0 3 +T 0 4 ) = [0 (0)+50] = 3. From t = secs. (m = ), find results at t = 4 secs. (m = ): T = T + λ(t 0 +T +T ) = [00 (6.)+0] = 38.7 T = T + λ(t +T +T 3 ) = 0+0.6[6. (0)+3.] = 0.3 T 3 = T 3 + λ(t +T 3 +T 4 ) = [0 (3.)+50] = 9.3 T T = k t x 35

36 Parabolic PDE's: Explicit method t = 6 t = 4 t = Left Bndry 00 C Right Bndry 50 C t = 0 0 C Initial temperature T = T 0 + λ(t 00 T 0 +T 0 ) = [00 (0)+0] = 6. T = T 0 + λ(t 0 T 0 +T 30 ) = [0 (0)+0] = 0 T 3 = T λ(t 0 T 3 0 +T 40 ) = [0 (0)+50] = 3. 36

37 Parabolic PDE's: Explicit method t = 6 t = 4 t = Left Bndry 00 C Right Bndry 50 C t = 0 0 C Initial temperature T = T + λ(t 0 T +T ) = [00 (6.)+0] = 38.7 T = T + λ(t T +T 3 ) = [6. (0)+3.] = 0.3 T 3 = T 3 + λ(t T 3 +T 4 ) = [0 (3.)+50] =

38 Parabolic PDE's: Stability We will cover stability in more detail later, but we will show that: The Explicit Method is Conditionally Stable : For the -D spatial problem, the following is the stability condition: k Δ t ( Δ λ= or Δt x) ( Δx) k λ / can still yield oscillation (D) λ /4 ensures no oscillation (D) λ = /6 tends to optimize truncation error We will also see that the Implicit Methods are unconditionally stable. Excel: Explicit 38

39 Parabolic PDE's: Explicit Schemes Summary: Solution of Parabolic PDE's by Explicit Schemes Advantages: very easy calculations, simply step ahead Disadvantage: low accuracy, O (Δt) accurate with respect to time subject to instability; must use "small" Δt's requires many steps!!! 39

40 Parabolic PDE's: Implicit Schemes Implicit Schemes for Parabolic PDEs Express T i m+ terms of T j m+, T im, and possibly also T j m (in which j = i and i+ ) Represents spatial and time domain. For each new time, write m (# of interior nodes) equations and simultaneously solve for m unknown values (banded system). 40

41 The -D Heat Equation: T t = k x T Simple Implicit Method. Substituting: T m+ Ti m+ Ti m+ Ti+ + = + O( Δx) Centered FDD x ( Δx) m+ m T Ti Ti = + O( Δt) Backward FDD t Δt Δt results in: λ T + (+ λ)t λ T = T with λ = k ( Δ x) m+ m+ m+ m i i i+ i. Requires I.C.'s for case where m = 0: i.e., T i0 is given for all i.. Requires B.C.'s to write st and last interior nodes (i=0 and n+) for all m. 4

42 Parabolic PDE's: Simple Implicit Method t = t = 5 t = 4 t = 3 t = t = t = Initial temperature 4

43 Parabolic PDE's: Simple Implicit Method m+ m Explicit Method i- i i+ ( ) + m+ m m m m i = i +λ i i + i T T T T T Simple Implicit Method m + m i- i i+ ( ) m m m m i = λ i λ i + λ i+ + T T T T Δt with λ= k for both ( Δx) 43

44 Parabolic PDE's: Simple Implicit Method t = 6 t = 4 t = Left Bndry 00 C Right Bndry 50 C t = 0 0 C Initial temperature At the Left boundary: (+λ)t m+ - λt m+ = T m + λ T 0 m+ Away from boundary: -λt i- m+ + (+λ)t i m+ - λt i+ m+ = T i m At the Right boundary: (+λ)t i m+ - λt i- m+ = T i m + λ T i+ m+ 44

45 Parabolic PDE's: Simple Implicit Method m = 3; t = 6 m = ; t = 4 m = ; t = m = 0; t = 0 Let λ = 0.4 Left Bndry 00 C T 0 C Initial temperature Right Bndry 50 C At the Left boundary: (+λ)t - λt = T 0 + λt 0 T.8 T 0.4 T = *00 = 40 Away from boundary: -λt i- + (+λ)t i λt i+ = T i 0 T 3 T T T 0 = T T4 T 7.6 T 6.4 = T T4-0.4 T +.8 T 0.4 T 3 = 0 = T +.8 T T 4 = 0 = 0 At the Right boundary: (+λ)t 3 λt = T λt 4.8 T i m+ 0.4 T i- m+ = *50) = 0 45

46 Parabolic PDE's: Simple Implicit Method m = 3; t = 6 m = ; t = 4 m = ; t = m = 0; t = 0 Let λ = 0.4 Left Bndry 00 C T 0 C Initial temperature Right Bndry 50 C At the Left boundary: (+λ)t - λt = T + λt 0 T.8 T 0.4 T = *00 = 78.5 Away from boundary: -λt i- + (+λ)t i λt i+ = T i T 3 T T T 6.4 = T T4 T 38.5 T 4. = T T4-0.4*T +.8*T 0.4*T 3 = 6.4 = *T +.8*T 3 0.4*T 4 = 4.03 = 4.03 At the Right boundary: (+λ)t 3 λt 4 = T 3 + λt 4.8*T 3 0.4*T 4 = *50 = 3.0 Excel: Implicit 46

47 Parabolic PDE's: Crank-Nicolson Method Implicit Schemes for Parabolic PDEs Crank-Nicolson (CN) Method (Implicit Method) Provides nd-order accuracy in both space and time. Average the nd-derivative in space for t m+ and t m. m m m m+ m+ m+ T Ti Ti + Ti+ Ti Ti + T i+ = O( x) + + Δ x ( Δx) ( Δx) T t m+ i m i T T = + O( Δt ) Δt (central difference in time now) m+ m+ m+ m m m i i i+ i i i+ λ T + ( +λ)t λ T =λ T + ( λ)t λt Requires I.C.'s for case where m = 0: T i0 = given value, f(x) Requires B.C.'s in order to write expression for T 0 m+ & T i+ m+ 47

48 Parabolic PDE's: Crank-Nicolson Method x i- x i x i+ t m+ t m+/ t m Left Bndry 00 C Right Bndry 50 C 0 C Initial temperature m+ m+ m+ i i i+ m m m i i i+ λ T + ( +λ) T λ T =λ T + ( λ)t λt Crank-Nicolson 48

49 Parabolic PDE's: Implicit Schemes Summary: Solution of Parabolic PDE's by Implicit Schemes Advantages: Unconditionally stable. Δt choice governed by overall accuracy. [Error for CN is O(Δt ) ] May be able to take larger Δt fewer steps Disadvantages: More difficult calculations, especially for D and 3D spatially For D spatially, effort same as explicit because system is tridiagonal. 49

50 Stability Analysis of Numerical Solution to Heat Eq. Consider the classical solution of the Heat Equation: T T = k t x To find the form of the solutions, try: at T(x,t) = e sin( ωx) Substituting this into the Heat Equation yields: - a T(x,t) = - k ω T(x,t) OR a = k ω kω t T(x,t) = e sin( ωx) Each sin component of the initial temperature distribution decays as exp{- k ω t) 50

51 Stability Analysis Consider FD schemes as advancing one step with a "transition equation": {T m+ } = [A] {T m } with [A] a function of λ = k Δt/ (Δx) { } with T = T,T,,T,,T m m m m m T L i L n with zero boundary conditions First step can be written: {T } = [A] {T 0 } w/ {T 0 } = initial conditions Second step as: {T } = [A] {T } = [A] {T 0 } and m th step as: {T m } = [A] {T m- } = [A] m {T 0 } (Here "m" is an exponent on [A]) 5

52 Stability Analysis {T m } = [A] m {T 0 } For the influence of the initial conditions and any rounding errors in the IC (or rounding or truncation errors introduced in the transition process) to decay with time, it must be the case that A <.0 If A >.0, some eigenvectors of the matrix [A] can grow without bound generating ridiculous results. In such cases the method is said to be unstable. Taking r = A = A = maximum eigenvalue of [A] for symmetric A (the "spectral norm"), the maximum eigenvalue describes the stability of the method. 5

53 Stability Analysis Illustration of Instability of Explicit Method (for a simple case) Consider D spatial case: T i m+ = λt i-m + (- λ)t im + λt i+ m Worst case solution: T im = r m (-) i (high frequency x-oscillations in index i) Substitution of this solution into the difference equation yields: r m+ (-) i = λ r m (-) i- + (-λ) r m (-) i + λ r m (-) i+ r = λ (-) - + (-λ) + λ (-) + or r = 4 λ If initial conditions are to decay and nothing explodes, we need: < r < or 0 < λ < /. For no oscillations we want: 0 < r < or 0 < λ < /4. 53

54 Stability of the Simple Implicit Method Consider D spatial: Worst case solution: m+ m+ m+ m i i i + i λ T + (+ λ)t λ T = T i m m ( ) i T = r Substitution of this solution into difference equation yields: m+ i m+ i m+ i + m i ( ) ( ) ( ) ( ) ( ) λr + λ r λr = r r [ λ ( ) + ( + λ) λ ( )+] = or r = /[ + 4 λ] i.e., 0 < r < for all λ > 0 54

55 Stability of the Crank-Nicolson Implicit Method Consider: m + m + m + m m m i i i + i i i + λ T + ( + λ)t λ T = λ T + ( λ )T + λt Worst case solution: i m m ( ) i T = r Substitution of this solution into difference equation yields: m+ i m+ i m+ i + ( ) ( ) ( ) ( ) λr + + λ r λr = m i m i m i + ( ) ( ) ( ) ( ) λr + λ r +λr r [ λ ( ) + ( + λ) λ ( )+] = λ ( ) + ( λ) + λ ( )+ or r = [ λ ] / [ + λ] i.e., r < for all λ > 0 55

56 Stability Summary, Parabolic Heat Equation Roots for Stability Analysis of Parabolic Heat Eq r(explicit) r(implicit) r(c-n) analytical r λ 56

57 Parabolic PDE's: Stability Implicit Methods are Unconditionally Stable : Magnitude of all eigenvalues of [A] is < for all values of λ. Δx and Δt can be selected solely to control the overall accuracy. Explicit Method is Conditionally Stable : Explicit, -D Spatial: k Δ t ( Δ λ= or Δt x) ( Δx) k λ / can still yield oscillation (D) λ /4 ensures no oscillation (D) λ = /6 tends to optimize truncation error Explicit, -D Spatial: k Δ λ= t or Δt h h 4 4k (h = Δx = Δy) 57

58 Parabolic PDE's in Two Spatial dimension T T T D = k + Find T(x,y,t) t x y Explicit solutions : Stability criterion Δt ( Δ x) + ( Δy) 8k kδt h if h = Δ x = Δ y ==> λ = or Δt h 4 4k Implicit solutions : No longer tridiagonal 58

59 Parabolic PDE's: ADI method Alternating-Direction Implicit (ADI) Method Provides a method for using tridiagonal matrices for solving parabolic equations in spatial dimensions. Each time increment is implemented in two steps: first direction second direction 59

60 Parabolic PDE's: ADI method Provides a method for using tridiagonal matrices for solving parabolic equations in spatial dimensions. Each time increment is implemented in two steps: y i- y i y i+ Explicit Implicit x i- x i x i+ first half-step y i- y i y i+ x i- x i x i+ second half-step ADI example t γ+ t γ+/ t γ 60