2.3. Quantitative Properties of Finite Difference Schemes. Reading: Tannehill et al. Sections and

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1 3 Quantitative Properties of Finite Difference Schemes 31 Consistency, Convergence and Stability of FD schemes Reading: Tannehill et al Sections 333 and 334 Three important properties of FD schemes: Consistency An FD representation of a PDE is consistent if the difference between PDE and FDE, ie, the truncation error, vanishes as the grid interval and time step size approach zero, Comment: ie, whenlim(pde FDE) = 0 0 Consistency deals with how well the FDE approximates the PDE Stability For a stable numerical scheme, the errors from any source will not grow unboundedly with time Comments: A concept that is applicable only to marching (time-integration) problems Generally we are much more concerned with stability than consistency Some hard wor is often needed to establish analytically the stability of a scheme Convergence It means that the solution to a FDE approaches the true solution to the PDE as both grid interval and time step size are reduced 1

2 Lax's Equivalence Theorem For a well-posed, linear initial value problem, the necessary and sufficient condition for convergence is that the FDE is stable and consistent The theorem has been proved for initial value problems governed by linear PDE's (Richtmyer and Morton 1967) We will discuss the three concepts one by one 3 Consistency Consistency means PDE FDE 0 when x 0 and t 0 Clearly consistency is the necessary condition for convergence Example: Consider a 1-D diffusion equation: u t u = K ( K > 0 and constant) x We use the forward-in-time and centered-in-space (FTCS) scheme:

3 u u u u + u t x n+ 1 n n n n i i i 1 i i+ 1 = K To show consistency, we need to determine the truncation error τ Using Taylor series expansion method, 3 3 n+ 1 n u ( t) u ( t) u ui = ui + t t! t 3! t 3 3 n n u ( x) u ( x) u ui± 1 = ui ± x + ± + x 3! x 3! x Substituting into the FDE, we have u t u u + + O( t ) = K + O( x ) + t t x therefore t u τ = + + t O( t x ) τ 0 when x 0 and t 0 => the scheme is consistent A counter example: The Dufort-Franel method for the same diffusion equations: 3

4 n+ 1 n 1 n n n+ 1 n 1 ui ui ( ui+ 1+ ui 1) ( ui + ui ) = K t x It's a centered-in-time scheme that is nd-order accurate in both space and time We can find again the truncation error (do it yourself!) 4 3 K( x) u t u ( t) u τ = K HRT x x t 6 t We can see that lim τ 0 except for the second term x, t 0 If t lim = 0 x x, t 0 then the scheme is consistent therefore t must approaches zero faster then x If they approaches zero at the same rate, then lim x, t 0 t = β, then x lim x, t 0 τ = K u β, t our equation becomes t t x u u u + K β = K 4

5 Thus, we are solving the wrong equation In fact this equation is hyperbolic instead of parabolic t Note that if ~1 x you might see spurious waves in your solution, due to the hyperbolic nature of the "new" PDE 33 Convergence General Discussion Definition is given earlier Symbolically, it is n lim u = u( x, t) x, t 0 i Convergence is generally hard to prove, especially for nonlinear problems The Lax's Theorem we presented earlier is very helpful in understanding the convergence for linear systems, and is often extended to nonlinear systems We will also discuss numerical convergence and methods for measuring solution accuracy later We will first show a convergence proof for a diffusion problem Certain concept introduced will be useful later Convergence proof for a 1-D diffusion problem Consider 5

6 u t u = K ( K > 0 and constant) x for 0 x L which has an initial condition: π x uxt (, = 0) = a sin( ) = f( x) L = 1 The BC is u(0, t) = u( L, t) = 0 This is a well-posed, linear initial value problem (notice the IC satisfies the BC as well) First, let's find the analytical solution to the PDE Because the problem in linear, we need only to examine the solution for a single wavenumber, and can assume a solution of the form: π x u( x, t) = A( t)sin( ) L Here A is the amplitude and sin gives the spatial structure and the final solution should be the sum of all wave components Note that this solution satisfies the boundary conditions Substituting the solution into the PDE, we obtain an ODE for the amplitude A : 6

7 da () t dt π = L KA dln( A ) π = K dt L A() t = A(0)exp K π / L t ( ) It says that the amplitude of the solutions for all wave numbers decreases with time From the IC, A (0) = a, so we have (, ) exp ( / ) π x u x t = a K π L t sin( ) L and uxt (, ) = u ( xt, ) which is the analytical solution to the original diffusion equation A numerical approximation to the diffusion equation should converge to this solution as x, t 0 Consider the forward-in-time centered-in-space (FTCS) scheme we derived earlier 7

8 Goal: Show that u t i u(x,t) as x, t 0 First, find the numerical solution This time, we use the FDE and substitute a discrete Fourier series into the equation Let the IC be given by π x f x u a for i= 0, 1,,, J J 0 i ( i) = i = sin( ) = 0 L where J+1 = total number of grid points used to represent the initial condition The coefficient a is given by a discrete Fourier transform: J π xi a = f( xi)sin( ) for =0, 1,,, J J L i= 0 Note 1: L = J x As x 0, J, the discrete Fourier series becomes continuous and a a Note : The number of harmonics or Fourier wave components that can be represented is a function of the number of grid points (J+1), which is the number of degrees of freedom Spectral methods represent fields in terms of spectral components, whose amplitudes are solved for The wavenumber for the wave components is π in the above equations L Recall that wavelength 8

9 π π L λ = = = wn ( π / L) where is the number (index) of wave components Longest wavelength =, corresponding to wavenumber zero (=0) Next longest wave = L ( =1 ) Shortest wave = L/J = J x/j = x Comments: A x wave is the shortest wave that can be resolved on any grid and it taes at least 3 points to represent a wave 9

10 x waves often have some special properties They are also represented most poorly by numerical methods recall that smooth fields are more accurately represented by a finite number of grid points As in the continuous case, we examine only one wavenumber (the solution is the sum of all waves), so for our discrete problem, assume a solution of the form u n i π xi = A( n)sin L (It satisfies BC) and n is the time level We also have from IC A (0) = a Substituting this into the FDE and letting u u u u + u t x n+ 1 n n n n i i i 1 i i+ 1 = K, 10

11 S i π xi = sin L, we have A( n+ 1) Si A( n) Si KA( n) = [ S i+ 1 S i + S i 1 ] t ( x) Since x i = i x, x i+1 = (i+1) x therefore S i+ 1 π ( x+ x) = sin L Using standard trigonometric identities, we can write the above in the form of a recursion relation: π x A( n+ 1) = A( n) 1 4µ sin L K t where µ = ( x) π x If we let M() 1 4µ sin L, then we have A ( n+ 1) = M( ) A ( n) 11

12 Write it out for n =0, 1,,, n: A (1) = M( ) A (0) = M( ) a () = ( ) (1) = [ ( )] A M A M a A ( n) = M( ) A ( n 1) = [ M( )] n a we therefore have the solution of u for wave mode : n n π xi ( ui ) = A( n) Si = a [ M( )] sin L Definition: M() is nown as the amplification factor, and if M() 1, the solution will not grow in time as n This had better to be the case because the amplitude of the analytical solution is supposed to always decrease with time For our problem we can see that M() 1 means π x L 1 4µ sin 1 If we tae the maximum possible value of sin ( ) = 1, 1 1 4µ 1 1

13 µ 1/ This condition needs to be met for all to prevent solution growth Based on the definition of µ, the condition becomes ( x) t K This imposes an upper bound on the t that can be used for a given value of x, and such a condition is unnown as the Stability Constraint Now we have our solution, let's chec convergence for single mode Be definition of convergence, we tae n n 4 π lim ( ui ) lim a 1 t sin sin x, t 0 x, t 0 K x πxi = ( x) L L 4K π x or if we let f( x) sin ( x) L, n π xi lim ( u ) = lim a [ 1 t f( x) ] sin L n x, t 0 i x, t 0 13

14 It can be shown that if f(x) is a complex-valued function of a real argument, say x, such that lim f ( x) = a, then, x 0 lim [1 ± tf( x)] n = e ± at (we will show this later) x, t 0 sin( y) We now thatlim = 1, therefore y 0 y π x sin K( π) L π lim f ( x) = lim = K a L π x = L L x 0 x 0 Also lim a x 0 = a, therefore x, t 0 π π x j n lim ( ui ) = aexp K t sin L L Interestingly, this solution is identical to our analytical solution derived earlier! Therefore the numerical solution converges to the PDE solution when x, t 0 14

15 Finally, we show here (noting that n t = t) nn ( 1) nn ( 1)( n )! 3! 3 a 1 a 1 3 = 1± at + 1 ( n t) ± 1 1 ( n t) +! n 3! n n n lim(1 ± f t) = 1 ± fn t+ ( f t) + 3 ( f t) + x 0 3 ( at) ( at) = 1 ± at + ± +! 3! ± at = e 34 Numerical Convergence Reading: Fletcher (handout), Sections 41, 41, 441 Numerical Convergence Convergence is often hard to demonstrate theoretically True analytical solution is hard or even impossible to find is one of the reasons We can, however, find out the convergence of a given scheme numerically We compute solutions at successively higher resolutions and see how the error changes with the resolution Does τ 0 when x 0? 15

16 And how fast τ decreases? The procedure can be very expensive (remember the cost factor increase as doubles) A typical measure of error is the L norm or RMS error: L [ u u ] = n i x i where u is a true solution or a 'converged' numerical solution when exact solution is not available Example: 1-D diffusion equation using FTCS scheme: 4 t u ( x) u τ = K + 4 t 1 x Maing use of 4 u u u u u u = K = K K K = = 4 t x t t x x t x we have 4 K( x) 1 u τ = s + O x + t 4 6 x 4 ( ) K t where s = ( x) 16

17 Table 41 (from Fletcher) shows the error reduction with x for two values of s The above figure shows plots of log 10 (err) as a function of log 10 ( x) 17

18 Recall that τ = A ( x) n log τ = log A + n log x This is a straight line in log-log diagram with a slope of n and intercept A Thus the slope of the line gives the rate of convergence In the above figure, we see that when s = 1/6, the error line has a steeper slope and the error is smaller for all x This is because for this value of s, the first term in τ drops out and the scheme because 4th-order accurate The scheme is second-order accurate for all other values of s Note that you can choose x and t such that s=1/6 only when K is constant in the entire domain In cases where no exact solutions is available, a so-called 'grid-convergence' or reference solution is often sought and this solution can be used in the place of true solution in the estimating the solution error 18

19 An example from Straa et al (1993) It shows a reference solution obtained at x=5 m, for a density current resulting from a dropping cold bubble 19

20 Figure (from Straa et al 1993) Graph of θ' L norms ( C) from self-convergence tests with the compressible reference model (REFC) The bold solid line labeled with 'self-convergence solutions' represents the L norms for spatial truncation errors of solutions made with t = constant and varying grid spacings The L norms were computed against a 50 m reference solution The bold dashed lines labeled with, for example, '000 m solutions' represent L norms for temporal truncation errors of solutions made with x =constant (eg 000 m) and varying time steps The reference solutions for these computations were made using a time step consistent with t = 15 s times a constant in each of the cases The solid fines labeled O(l) and O() represent first- and second-order convergence, respectively 0

21 Richardson Extrapolation As the grid becomes very fine, the error behaves much lie that predicted from the leading terms in τ Further refinements are expensive, so we use another technique to improve the solution Richardson Extrapolation Consider two numerical solutions obtained at x a and x b With FTCS scheme (assuming s 1/6), τ a 4 K( xa ) 1 u 4 = s + O x + t 6 x 4 ( a ) 4 K( xb ) 1 u 4 τ b = s + O x + t 6 x 4 ( b ) Find a linear combination of the two solutions, u a and u b u c = a u a + b u b where a + b =1 and a and b are chosen so that the leading terms in τ a and τ b cancel and the scheme becomes 4thorder accurate (Of course this assumes that u x is the same in both case, which is reasonably assumption only 4 4 at relatively high resolutions when the solution is well resolved) If x b = x a /, then 4a + b = 0 with a + b = 1 a = -1/3, b = 4/3 and u c = -1/3 u a + 4/3 u b 1