 # Numerical Analysis Problems

## Transcribed Text

1. Find the explicit form for the iteration matrix I − Q−1A in the Gauss-Seidel method when A = 2 −1 −1 2 −1 . . . . . . . . . −1 2 −1 −1 2 2. Let A be positive denite, and let b a xed vector. For any x, the residual vector is r = b − Ax, and the error vector is e = A−1 b − x. Show that the inner product of the error vector with the residual vector is positive unless Ax = b. 3. An integral equation is an equation involving an unknown function with an argumentation. For example, here is a typical integral equation x(t) = Z t 0 cos(s + x(s))ds + e 0 By dierentiating this integral equation, obtain an equivalent initial-value problem for the unknown function. 4. Prove that when the fourth-order Runge-Kutta method is applied to the problem x 0 = λx, the formula for advancing this solution will be x(t + h) = [1 + hλ + 1 2 h 2λ 2 + 1 6 h 3λ 3 + 1 24 h 4λ 4 ]x(t). 5. Let θ ∈ [0, 1] be a constant, and denote tn+θ = (1 − θ)tn + θtn+1. Consider the generalized midpoint method xn+1 = xn + hf(tn+θ,(1 − θ)xn + θxn+1) Show that the method is absolutely stable when θ ∈ [1/2, 1]. 6. The partial dierential equation ∂ 2u ∂x2 + ∂ 2u ∂y2 + a(x, y)u = f(x, y) is called a Helmholtz equation. When the coecient function a(x, y) = 0 it is reduced to the Poisson equation. Consider the special case of a nonpositive constant coecient a, and the corresponding boundary value problem ∂ 2u ∂x2 + ∂ 2u ∂y2 + au = f(x, y), 0 < x, y < 1 u(x, y) = g(x, y), x = 0, or1, or y = 0 or 1 Derive a dierence scheme to solve this boundary value problem. Implement the scheme in Matlab and use it solve the problem with a = −2, g(x, y) = 0, and f(x, y) = xy[(x 2 − 7)(1 − y 2 ) + (1 − x 2 )(y 2 − 7)] 1 7. Write a Matlab function that implements the Crank-Nicolson method for the problem ut = uxx = f(x) 0 < x < L, t > 0 u(x, 0) = g(x), 0 < x < L u(0, t) = u(L, t) = 0, t > 0 Test your function on the problem with the following data: L = 1, f(x) = 0, g(x) = sin(πx). Compare the numerical solution to the exact solution u(x, t) = e −π 2 t sin(πx).

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Problem 6
function [u,telapsed ] = P6(n,to_plot)
% INPUTS
% n is the number of points to use, usually a power of 2
% to_plot --indicates if you want to see plots (1) or none (0)

% OUTPUTS
% solution u
% telapsed -- elapsed time.

rv=[];
iter=0;
if nargin < 1
n=16;
end
if nargin < 2
to_plot = 0;
end

G = numgrid('S',n+1);
A = delsq(G);

%make f vector and exact solutions
x=linspace(0,1,n+1);
y=linspace(0,1,n+1);
h=x(2)-x(1);

%modify A
a=-2;
A=A-a*h^2*speye(size(A));

[X,Y]=meshgrid(x,y);
f_large=X.*Y.*((X.^2-7).*(1-Y.^2) + (1-X.^2).*(Y.^2-7));
f=f_large(2:end-1,2:end-1);
f_vec=-h^2*reshape(f,(n-1)*(n-1),1);...
\$45.00 for this solution

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