Homework Set 4 - SOLUTIONS
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1 MATH 55 Scientific Computing II Spring 6 Homework Set 4 - SOLUTIONS. For N even define a discrete grid x h for [, π] by x k = k h where h = π/n and k N. Show e i( N +j)x k = e i( N j)x k for j =,,..., N. This shows that on such a grid, wave numbers greater than N are aliased to wave numbers below N. ANS: There are a number of ways to show this result, one of which is by first noting that e inx πk k in = e N = e iπk = cos (kπ) + i sin (kπ) =. Multiplying on the right by this expression gives e i( N j)x k e inx k = e i N x k e ijx k e inx k = e i( N +j)x k. An alternative would be to simply subtract the two expressions, and using a trig identity to arrive at.
2 timing. Given a real vector v = (v,..., v N ) T the Discrete Sine Transform of v is given by ˆv = P v, where P is an (N ) (N ) matrix with P i,j = (/ N) sin (ijπ/n) for i, j =,,..., N. We showed that ˆv could be computed using an FFT, implemented by the M-file dst.m. For a randomly chosen v and values of N given by N = [64, 96, 8, 56, 368, 5, 4, 874, 48, 3477, 496] compute ˆv by direct matrix-vector multiplication and also by dst.m. Using tic and roc in MATLAB, plot the cpu time on a semilogy plot, and discuss the results. To obtain a reasonably accurate timing, execute each method 5 times and then take the average. ANS: Below is the graph. For both multiplication by P and using the DST, except for N small, the DST is clearly faster. In the case of small N, we can see that the DST is actually slower. This is most likely due to the fact that for small N matrix-vector multiplication is extremely fast because of the presence of L cache, a small amount of memory close to the arithmetic units with low latency, i.e. data moves in and out very rapidly. - Multplication by P vs. DST P DST N Let us try to get a clearer picture of the timings. Below is a graph of timings with the time divided by N in the case of multiplication by P case, and divided by N log N in the DST case. Focusing on the larger values of N, on this log scale the multiplication by P is nearly a constant, confirming the expected N scaling. For the DST it is not as clear because the optimality of the DST varies with N, the best case being when N is a power of, which is quite apparent in the results. Here is a copy of the code:
3 scaled timing -6 Scaled timings: (P time)/n vs. (DST time)/(nlog(n)) P DST N N = [ ] ; time_p = zeros(length(n),); time_dst = zeros(length(n),); num_trials = 5; for i = :length(n) n = N(i) v = rand(n-,); [P,x] = createp(n); tic for j = :num_trials v = P*v; time_p(i) = toc/num_trials; tic for j = :num_trials v = dst(v); time_dst(i) = toc/num_trials;
4 3. Consider the -point one-dimensional BVP { u + u = (π sin πx π cos πx)e x u() = u() =. The exact solution is u(x) = e x sin (πx). (a) Write a MATLAB script to solve the problem by the FFT method, using the Discrete Sine Transform as implemented by dst.m applied to the nd order centered FD scheme, assuming σ is a constant, D v i + σv i = f i, where D = D + D. Assume a meshsize h = / p, where p is a positive integer. For p = : 4, plot the exact solution (u(x) vs. x) and the numerical solution (v i vs. x i ), including the boundary points. The 4 plots should appear separately in one figure, with axes labeled and a title for each indicating p. Investigate subplot in MATLAB for how to have multiple plots in a single figure window. (b) For p = : 5 present a table with the following data - column : h; column : u h v h ; column 3: u h v h /h, where h = /n. Discuss the trs in each column. Include a copy of your code. ANS: Here is the graph for part (a): N: N: N: 8 N: The table for part (b) is below. We see clear second order accuracy up to n = ( ) = 496. After that roundoff error begins to be significant and we lose the convergence to a constant in column 3.
5 h inf_error inf_error/h^ e- 93e- 6.37e-.5e e e- e-.3945e e- 6.5e e e- 3.5e e e- 65e-.36e e- 7.85e e e- 3.96e-3.398e e-.953e e e e e e e-4.843e e-.444e e e-.7e-4.436e e- 6.35e e-9.744e+ 3.58e e e+ Here is a copy of the code: n = [:5] ; hvals = zeros(length(n),); err = zeros(length(n),); for i = :length(n) N = ^(n(i)); h = /N; hvals(i) = h; xh = h*(:n) ; lam = *(-cos(xh(:n)*pi))/(h^); % eigenvalues of A_h = exp(xh).*sin(pi*xh); % true soln u on xh grid fh = exp(xh).*(pi^*sin(pi*xh)-*pi*cos(pi*xh)); % rhs fh_int = fh(:n); % rhs is f evaluated at N- interior pts ftil = dst(fh_int); % dst of rhs vtil = ftil./(lam+); % soln in Fourier space by simple division = dst(vtil); % transform back to physical space = [;;]; % set BCs err(i) = max(abs(-)); if (i < 5) subplot(,,i) plot(xh,, -,xh,, * ) leg(,, location, northwest )
6 title([ N:,numstr(N)]); format short e disp( ) disp( h inf_error inf_error/h^ ) disp( ) disp([hvals err err./(hvals.^)])
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