Homework 6 MA/CS 375, Fall 2005 Due December 15

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1 Homework 6 MA/CS 375, Fall 25 Due December 5 This homework will count as part of your grade so you must work independently. It is permissible to discuss it with your instructor, the TA, fellow students, and friends. However, the programs/scripts and report must be done only by the student doing the project. Please follow the guidelines in the syllabus when preparing your solutions.. (3 pts.) In this problem we eamine the dynamics of repeated matri multiplication and use spectral analysis to understand it. i Let: W = Generate a random 5-vector. Compute W n, n = :. Plot you observe? For n = 2 : 2 : print W n W n. What do you observe? W n W n versus n. What do ii Use the Matlab command eig to compute the eigenvalues and eigenvectors of W. Use the results to eplain your observations from part (i). iii Repeat (i) for the matri: W = iv Repeat (ii) for the results of part (iii) Solutions: For the first (symmetric) matri the starting vector is generated using =rand(5,). Then we can compute powers of W applied to in the following way.. >>W=zeros(5,); W(:,)=; nw=zeros(,); nw()=norm(); >>for n=2: >>W(:,n)=W*W(:,n-); nw(n)=norm(w(:,n)); >>end >>plot(nw(2:)./nw(:), LineWidth,2); grid on;

2 For a plot, one obtains W n / W n n The normalized vectors as described in the problem can be obtained by >> A=W./repmat(nW,5,); >> A(:,2:2:) which yields the vector for varying n Using Matlab to perform the eigenvalue decomposition, e e e e >>[S,L]=eig(W); The matrices of eigenvectors and eigenvalues are S =

3 L = By repeatedly multiplying a vector by the matri, the ratio of successive norms converges to the modulus of the largest eigenvalue, whereas the normalized resulting vector converges (to within a sign) of the corresponding eigenvector. This is called the power method of determining the largest eigenvalue/vector pair. The second matri is not symmetric and as such may not necessarily have real eigenvalues, however, since its entries are real, comple eigenvalues must occur in conjugate pairs. Using the same method as before, one obtains the plot.2. W n / W n n The normalized vector is

4 The matrices of eigenvectors and eigenvalues are S = L =.5 +.i.5.i i. +.5i i..5i i..i While again the ratio of successive norms does converge to the modulus of the largest eigenvalue, the function oscillates initially. The normalized vector produced does not appear in the matri of eigenvectors. The fact that there are two eigenvalues with modulus suggests that the W n vector may be a combination of the two corresponding eigenvectors. We can find the epansion coefficients of the W n vector in terms of the eigenvectors by multiplying it by S. >>c=s\a(:,); This produces the vector c = i i. +.i..i Notice how the last two entries are effectively zero. This means that the solution vector has no component in the span of the eigenvectors corresponding to the eigenvalues ± 2. Clearly, the solution 3 vector is a linear combination of the two eigenvectors corresponding to the eigenvalues ±i. 2. (3 pts.) The method of successive overrelaation or SOR is a way to accelerate the convergence of the Gauss-Seidel method for solving A = b. Let ω be a parameter and (k) be the kth iterate. Then (k+) is computed via the formula: (k+) i = (b i i j= a ij (k+) j n j=i+ a ij (k) j ), a ii (k+) i = ( ω) (k) i + ω (k+) i. Note that if ω = then SOR is identical to Gauss-Seidel. Modify the book s program itermeth.m to accept a new parameter ω as an input and replace the Gauss-Seidel option by SOR with parameter 4

5 ω. Use your method to solve A = b where A is the tridiagonal matri whose diagonal entries are 2. and whose offdiagonal entries are. Considering ω =.2 :.2 :.8 record the number of iterations required to converge to a tolerance of 4. What is the best choice? How much of a difference does it make? Change the diagonal element to 2. and repeat the eperiment. How do your results change? Solution: The modifications to itermeth.m which change the Gauss-Seidel method to the SOR method are in the function declaration function [, iter]= itermeth(a,b,,nma,tol,p,w) and on line 8 U = eye(n); beta = ; alpha = w; Alternately, one could write a simple SOR script which eploits the tridiagonal structure for the given problem function [,iter]=sor3band(a,b,w,,tol,mait) Solve the symmetric positive definite tridiagonal system A=b using the Successive Over-Relaation method n=length(a); L=spdiags(diag(A,-),-,n,n); U=L ; D=spdiags(diag(A),,n,n); =; =+rand(n,); iter=; Mw=D+w*L; Nw=(-w)*D-w*L ; wb=w*b; while (norm(-)>tol & iter<mait) Do iteration =; =Mw\(Nw*+wb); iter=iter+; end 5

6 Iterations Matri Matri ω The relaation parameter which minimizes the number of iterations needed from the first matri was found to be ω =.66. The number of iterations needed to solve the second system is minimized when ω is at its upper limit of.8. For all values of ω, the convergence of the SOR method first system is superior due to the greater diagonal dominance. In most practical computations, the optimal relaation parameter is determined from information about the eigenvalues of the matri. 3. (4 pts.) In this problem we compare three different schemes for solving the first order wave equation with periodic boundary conditions: We introduce space and time discretizations u t = c u, a b, u(a, t) = u(b, t), u(, ) = f() () m = a + m h, h = b a, m =,..., M M t n = n k, h = T, n =,..., N N (a) The forward-time, forward-space scheme (FF scheme): u n+ m u n m k = c un m+ u n m h (2) u n+ m = u n m + α(u n m+ u n m), α = ck, m =,..., M, n =,..., N (3) h 6

7 (b) The forward-time, centered-space scheme (CF scheme): u n+ m u n m k = c un m+ u n m 2h (4) u n+ m = u n m + α 2 (un m+ u n m ), m =,..., M, n =,..., N (5) (c) The leap-frog (or centered-time, centered-space scheme (LF scheme): u n+ m um n 2k = c un m+ u n m 2h (6) u n+ m = u n m + α(u n m+ u n m ), m =,..., M, n =,..., N (7) Since the leap-frog scheme is two-levels deep in time, it is called a multi-step scheme. To get it started, we need to specify data for the first two time steps. Here we will do it using the scheme (5). Because of the periodic boundary conditions, the solution can be assumed to etend periodically in outside the computational domain. Thus, we can identify the values UM+k n = un k, for k =,, needed in the above formulas. If we write U n = (u n, u n 2,..., u n M) then the above 3 schemes can be written in matri form: (a) The FF scheme (3) (b) The CF scheme (5) (c) The LF scheme (7) U n+ = A F U n (8) U n+ = A C U n () U n+ = A L U n + U n () You must implement these schemes to solve equation () with a =, b =, T = 2 with initial condition u(, ) = f() = sech(5 sin(π)) 7

8 (a) Use M = 2, 4 and N = 3, and c =.. Compute the numerical solutions at time t = T and plot each against the eact solution u eact (, t) = sech(5 sin(π( + ct))) How do the computations with the two different values of M differ? (b) Now use M = 2, N = 3 and M = 4, N = 6. Compare the performance of the schemes () and (3) Solutions: The Matlab code for these three schemes is function advect(m,n,t,ab,c,f,method) Solve the D advection (first order wave) equation u_t=c*u_, u(,)=f() using difference discretization in time and space Parameters: M - number of grid points N - number of time steps ab - 2 vector containg the left and right endpoints c - wave velocity eact - the eact solution in (,t) method - which method to use FF CF or LF m=(:m) ; h=(ab(2)-ab())/m; =ab()+m*h; k=t/n; e=ones(m,); alpha=c*k/h; u=feval(f,); eact=feval(f,+c*t); if method== FF Use forward time and forward space A=spdiags([e*(-alpha) e*alpha],:,m,m); A(M,)=alpha; for n=:n u=a*u; u=u; end 8

9 plot(,eact,,u, LineWidth,2) else A=spdiags([-e*alpha/2 e e*alpha/2],-:,m,m); A(M,)=alpha/2; A(,M)=-alpha/2; if method== CF Use forward time and centered space for n=:n u=a*u; u=u; end plot(,eact,,u, LineWidth,2) else Use Leap-frog u=a*u; Use CF scheme to obtain u(,t=k) A=spdiags([-e *e e]*alpha,-:,m,m); A(M,)=alpha; A(,M)=-alpha; for n=2:n u2=a*u+u; u=u; u=u2; end plot(,eact,,u2, LineWidth,2) end end label(, FontSize,6); ylabel( u(,t), FontSize,6); title([ method=,method, M=,num2str(M),, N=,num2str(N)], FontSize,6); The results for part (a) are

10 method=ff M=2, N= method=ff M=4, N=3.8 2 u(,t) method=cf M=2, N=3 u(,t) method=cf M=4, N=3 u(,t) u(,t) method=lf M=2, N= method=ff M=4, N=3.8 2 u(,t) u(,t) The CF scheme is unstable in both cases, while the FF and LF schemes are stable in the case where the α < and unstable when α >. The main difference is that the attenuation of the computed wave using the FF scheme is substantial whereas the LF scheme produces a result which is much closer to the eact solution. The LF scheme has the advantage of being second order in spatial and temporal discretization whereas the FF scheme is first order in both. In both cases, there appears to be a slight phase lag between the computed and eact solutions. The results for part (b) are

11 method=ff M=2, N=3 method=ff M=4, N=6.8.8 u(,t).6.4 u(,t) method=lf M=2, N=3 method=lf M=4, N=6.8.8 u(,t).6.4 u(,t) Refinement of the grid and time step size improves both the error in the FF and LF schemes, but the improvement is slight in the FF case due to the lower order of convergence and the LF scheme ehibits far less dissipation.