2 8-1 General RemarksThe governing equations in fluid mechanics and heat transfer can be reduced to elliptic form for particular applications. Such examples are the steady-state heat conduction equation, velocity potential equation for incompressible, inviscid flow, and the stream function equation.Typical elliptic equations in a two-dimensional Cartesian system are Laplace’s equations,and Poisson’s equation
3 8-2 Finite Difference Formulations (1) Five-point formula---central differenceLaplace’s equations:F.D.which is accurate to
4 8-2 Finite Difference Formulations (2) General form for five-point formula
5 8-2 Finite Difference Formulations (3) From eqs. (1) and (2), we can get
6 8-2 Finite Difference Formulations (4) In order to explore various solution procedures, first consider a square domain with Dirichlet B.C.s. For instance, a simple 6x6 grid system subject to the following B.C.s. is considered:x=0 u=u2, y=0 u=u1x=L u=u4, y=H u=u3
7 8-2 Finite Difference Formulations (5) The interior grid points produces sixteen equations with sixteen unknowns. The equations are:
8 8-2 Finite Difference Formulations (6) Solution algorithms:(1) The Gauss-Seidel Iteration Method (point-by-point iteration method):(a) The finite difference equation is given by(b) For the computation of the first point,say (2,2), it follows that
9 8-2 Finite Difference Formulations (7) (c) For point (3,2), one has(d) The general formulation is
10 8-2 Finite Difference Formulations (8) (2) SOR Method:orWhat is the optimum value of ω?One such relation, for the solution of elliptic equations in arectangular domain subject to Dirichlet B.C.s with constantstep size, is
11 8-2 Finite Difference Formulations (9) (3) Line-by-line iteration method:(a) In this formulation, it results in three unknowns at points(i-1,j), (i,j), and (i+1,j). It becomes(b) This equation, applied to all i at constant j, results in asystem of linear equations which, in compact form, has atridiagonal matrix coefficient. The solution of each row atconstant j can be solved by TDMA method.
12 8-2 Finite Difference Formulations (10) (c) Grid points employed in the line-by line iteration method.
13 8-2 Finite Difference Formulations (11) (4) Line-by line SOR method:There is no simple way to determine the value of optimum ω.In practice, trial and error is used to compute ωopt for a particular problem.
14 8-2 Finite Difference Formulations (12) (5) The Alternating Direction Implicit (ADI) Method:(a) An iteration cycle is considered complete once the resultingtridiagonal system is solved for all rows and then followedby columns, or vice versa. It follows
15 8-2 Finite Difference Formulations (13) (b) Grid points used in ADI method:
16 8-2 Finite Difference Formulations (14) (c) ADI with SOR Method:
18 9-1 General RemarksEquations of motion in fluid mechanics are frequently reduced to parabolic formulations.Boundary layer equations are examples of such formulations.In addition, the unsteady heat conduction equation is also parabolic.
19 9-2 Finite Difference Formulations (1) A typical parabolic second-order PDE is the unsteady heat conduction equation, which is considered first in one-space dimension. It has the following form(1) FTCS (forward time/central space) method:(i) is expressed by a forward difference approximation whichis of order :
20 9-2 Finite Difference Formulations (2) (ii) Using the second-order central differencing of orderfor the diffusion term , eq. (8-1) can be approximated by(iii) Eq. (8-2) is also called explicit formulation,which is of order It will beshown that the solution is stable for
21 9-2 Finite Difference Formulations (3) (2) BTCS (backward/central space) method:(i)The above equation can be solved by TDMA.
22 9-2 Finite Difference Formulations (4) (ii) Eq. (8-3) is defined as being implicit, since more than oneunknown appears in the finite difference equation. As a result,a set of simultaneous equations needs to be solved, whichrequire more computation time per time step. Implicit methodsgreater advantage on the stability of the finite differenceequations, since most are unditionally stable. Therefore, alarger step size in time is permitted.
23 9-2 Finite Difference Formulations (5) (3) CTCS (central time/central space) method: (The Crank-Nicolson method)(i) If the diffusion term of eq. (8-1) is replaced by the averageof the central differences at time levels n and n+1, thediscretized equation would be of the form:Note: The left side of eq. (8-4) is a central difference ofstep , i.e., , which is
25 9-2 Finite Difference Formulations (7) (ii) The method may be thought of as the addition of two stepcomputations as follows:Using the explicit method,while using the implicit method,Adding eqs. (8-5a) and (8-5b), we can get eq. (8-4).(iii) This implicit method is unconditionally stable and is oforder , that is a second-order accurate scheme.Example
26 9-3 Parabolic Equations in Two-Space Dimensions (1) Consider the model equation(1) FTCS (or Explicit) method:which is of orderStability analysis indicates that the method is stable forwhereIf Δx=Δy, i.e., dx=dy=d, then
27 9-3 Parabolic Equations in Two-Space Dimensions (2)
28 9-3 Parabolic Equations in Two-Space Dimensions (3) (2) Implicit (BTCS) method:(i) Consider an implicit formulation for which the FDE is
29 9-3 Parabolic Equations in Two-Space Dimensions (4) (ii) The 2-D FDEs in the ADI formulation areand
30 9-3 Parabolic Equations in Two-Space Dimensions (5) This method is of order and is unconditionallystable. The above equations are written in the tridiagonal form aswhere
31 9-3 Parabolic Equations in Two-Space Dimensions (6)
33 10-1 Stability considerations (1) At a starting point for stability analysis, consider the simple explicit approximation to the heat equationThis may be solved for unj to yield
34 10-1 Stability considerations (2) Let D be the exact solution of this equation(2), N the numerical solution of equation(1) and A the analytical solution of the PDE :Then ,we may writeDiscretization error=A-DRound-off error=N-D
35 10-1 Stability considerations (3) The equation of stability of a numerical method examine the error growth while computations are being performed.The equation of stability is usually answered by using a Fourier analysis. This method is also referred to as a von Neumann analysis.
36 10-2 Fourier or von Neumann analysis (1) Consider eq.(1) and let εbe the round-off error. The numerical solution actually computed may be written N=D+ε----(3)N must satisfy eq.(1). Substituting eq.(3) into eq.(1), yields
37 10-2 Fourier or von Neumann analysis (2) since the exact solution must satisfy the difference eq.(i.e. Eq(1)), therefore
38 10-2 Fourier or von Neumann analysis (3) In this case, the exact solution D and the error εmust both satisfy the same difference equation. This means that the numerical error and exact numerical solution both posses the same growth property in time and either could be used to examine stability.
39 10-2 Fourier or von Neumann analysis (4) Any perturbation of the input values at the nth time level will either be prevented from growing without bound for a stable system or will grow large for an unstable system
40 10-2 Fourier or von Neumann analysis (5) Consider a distribution of error at any time in a mesh. We choose to view this distribution a time t=0 for convenience. This error distribution is shownε (x,0)x
41 10-2 Fourier or von Neumann analysis (6) We assume the error ε (x,t) can be written as a series of the form
42 10-2 Fourier or von Neumann analysis (7) Since the difference equation is linear, superposition may be used, and we may examine the behavior of a single term of the series given in eq.(4).consider the term
43 10-2 Fourier or von Neumann analysis (8) For an assessment of numerical stability, we are interested in the variation of with time. Therefore, we external eq.(2) by assuming the amplitude bm is a function of time. Moreover, it is reasonable to assume an exponential variation with time; error tend to grow or diminish exponentially with time. Therefore, we write
44 10-2 Fourier or von Neumann analysis (8) where km is real, but “a” may be complex.If eq(6) is substituted into eq(1), we obtain……(7)
45 10-2 Fourier or von Neumann analysis (9) If we divide by eateikmx and utilize the relation
46 10-2 Fourier or von Neumann analysis (10) We can get:
47 10-2 Fourier or von Neumann analysis (11) ∴the error will not grow from one time step to the next, if