Finite Difference Methods for

Transcription

1 CE 601: Numerical Methods Lecture 33 Finite Difference Methods for PDEs Course Coordinator: Course Coordinator: Dr. Suresh A. Kartha, Associate Professor, Department of Civil Engineering, IIT Guwahati.

2 We have discussed the following finitedifference methods to solve Parabolic PDEs. o Forward time Centered Space (FTCS) o Backward time Centered Space (BTCS) 2 o Crank Nicholson method (where O ( t ) and O 2 ( x ) for derivatives are used).

3 The next important method is: Alternating Direction Implicit Method (ADI) for Multi spatial dimension Problems If you have multi spatial dimension problems say f f f = + α t x y If you use FTCS, then the FDE will be: n n n ( n) ( n) ( n) f ( 1) ( ) 1, 2., 1, 1, 2. n n i j fi j + fi+ j fi j fi, j + f + i+ 1, j fi, j = fi, j + α t. + (1) x y

4 Similarly, you can frame fully implicit FDE using BTCS, i.e. α t ( n+ 1) α t ( n+ 1) 2α t 2α t ( n+ 1) f 2 i 1, j + f 2 i, j f i, j x y x y α t + α t + + f f f = y x ( n 1) ( n 1) ( n) i, j 1 i 1, j i, j (2) On applying eq.(2) at each grid node (i,j), we will get a system of linear equation. This system will be here banded dand pentadiagonal. You can use appropriate solvers to solve each banded matrix system.

5 However the implicit technique is computationally tedious. We can go for Alternate Direction Implicit Method (ADI). This method involves two steps: Recall earlier we said 1 n+ ( n+ 1) ( n) 2 1 f f f + t 2 t t i, j i, j i, j That is this quantity consist of half of implicit quantity and half of explicit quantity.

6 Since For = α + t x y f f f f 1 f f = + t 2 t t n + 12 ( n + 1) ( n ) i, j i, j i, j f = α + x i j ( n + 1) ( n ) f y, i, j

7 i.e. Step 1 f f f 2. f + f f 2. f + f = α. + t x y ( n+ 1) ( n) ( n+ 1) ( n+ 1) ( n+ 1) ( n) ( n) ( n) i, j i, j i 1, j i, j i+ 1, j i 1, j i, j i+ 1, j Step 2 ( n ) ( n ) ( n ) ( n+ 1) ( n+ 1) f 1 f f f f = + = α + t i, j 2 t i, j t i, j x y i, j i, j f f f 2. f + f f 2. f + f i.e.. t x y ( n+ 2) ( n+ 1) ( n+ 1) ( n+ 1) ( n+ 1) ( n+ 2) ( n+ 2) ( n+ 2) i, j i, j i 1, j i, j i+ 1, j i 1, j i, j i+ 1, j = α + So using these two steps ADI is implemented in the numerical scheme.

8 Hyperbolic PDE We have seen before that hyperbolic PDEs are also Propagation problems We have to use marching methods to solve hyperbolic PDEs. It has definite (finite) propagation speed for the physical information propagation. You have distinct t domain of dependence d and range of influence. Some examples of hyperbolic PDE: f f + u = 0 (The Convection Equation) t x f 2 f = c (The Wave Equation) t x

9 The convection equation is usually used in o Fluid mechanics o Heat transfer Wave equation is used for o Vibrating systems (strings, acousticfields) As usual we can have o Dirichlet B.C. o Neumann B.C. o Mixed B.C. for various hyperbolic PDEs.

10 Recall as hyperbolic PDE has finite range of influence as well as domain of dependence. The finite physical propagation speed is approximated as x cn = t dx Actual is. dt f f dx For + u = 0; we have = u t x dt f 2 f dx For = c, we have =± c t x dt

11 Convection Equation df dt The derivative in the convection equation can be approximated using > Forward Time or > Backward Time or > Centered Time while using FDM. f f In + u = 0; t x f xx suggest something on physical convection. The physical information propagation speed is dx u dt =

12 The solution at a point depends only on the information in the domain of dependence d specified by upstream characteristic paths. f The first derivative (spatial) should be x approximated dby one sided d approximations in the direction from which physical information is propagated. Theyareupwind approximations.

13 x f fi fi 1 i x We can also use centered space approximations with acceptable results. i.e. f fi f x i 2 x Upwind Methods For the convection equation f t f + u = x 0 dx dt = u Information propagates with If u > 0, information propagates from left to right. If u < 0, information propagates from right to left. + 1 i 1

14 First order upwind scheme: For u > 0, i.e. f t ( n ) ( n ) i f + u. = 0; x i ( n+ 1) ( n) ( n) ( n) fi fi fi fi 1 i.e. + u = 0 t x ( ) 1 or, o, f = f c. f f ( n+ 1) ( n) ( n) ( n) i i i i u t where c = Convection Number x From the stability point of view, you require c 1.0.

15 This is because the actual speed Numerical speed C n x =. tt u = dx. dt The convection number suggest that you require atleast a certain computational speed c n that is not less than the actual convection speed. Or, if the actual speed exceeds numerical speed, results fluctuates. This again boils down to the criteria that you cannot have large values of Δt. If Δt is large, then computational convection speed reduces.