--- /dev/null
+function LorenzAttractor(T)
+%%
+% Nonlinear ODE example from nonlinear dynamics to solve the
+% Lorenz system using a Runge-Kutta 4 time integrator
+%
+% d(y1)/dt = 10*(y2-y1)
+% d(y2)/dt = 28*y1 - y2 - y1*y3
+% d(y3)/dt = y1*y2 - 8/3*By3
+%
+% INPUT:
+%
+% T - final time
+%
+% OUTPUT:
+%
+% NONE - this makes a movie
+%
+%%
+% adhoc tactic to set the time step
+ num_steps = 2500;
+ dt = T/num_steps;
+%%
+% set the initial condition and allocate memory for the solution vector
+ y = zeros(3,num_steps);
+ ini = 5*ones(3,1);
+%%
+% save the initial condition and enter the Runge-Kutta loop to solve the ODE
+ y(:,1) = ini;
+ for i = 2:num_steps
+ k1 = RHS(y(:,i-1));
+ k2 = RHS(y(:,i-1) + 0.5*dt.*k1);
+ k3 = RHS(y(:,i-1) + 0.5*dt.*k2);
+ k4 = RHS(y(:,i-1) + dt.*k3);
+ y(:,i) = y(:,i-1) + (dt/6).*(k1 + 2.*k2 + 2.*k3 + k4);
+ end
+%%
+% visualize the solution progessively to make a movie
+ clf;
+ hold on;
+ xlim([-20 20]);
+ ylim([-30 30]);
+ zlim([0 50]);
+ grid on;
+ for n = 1:(length(y)/7)
+ plot3(y(1,1:7*n),y(2,1:7*n),y(3,1:7*n),'b');
+ drawnow
+ view(-37.5-n, 24)
+ pause(0.02)
+ end
+end
+
+%%%
+% auxiliary function to get the right hand side
+%%%
+ function dy = RHS(y)
+%%
+% evaluate the right hand side of the nonlinear ODE for the Lorenz
+% attractor. The input y is a vector with three components.
+ dy = zeros(3,1);
+ dy(1) =
+ dy(2) =
+ dy(3) =
+ end
\ No newline at end of file
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--- /dev/null
+function [L,U,P] = lufact(A)
+%%
+% Computes the LU factorization of the matrix A. The factorization is done
+% in-place and then the L and U matrices are extracted at the end for
+% output. The factorization is PA = LU
+%
+% INPUT:
+% A - n by n square, non-singular matrix
+%
+% OUTPUT:
+% L - lower triangular matrix with ones along the main diagonal
+% U - upper triangular matrix
+% P - permutation matrix for pivoting
+%%
+% get the size of the system
+ n = length(A);
+%%
+% initialize the pivoting matrix
+ P = eye(n);
+%%
+% Vectorized in-place LU factorization (with row pivoting) that keeps
+% track of the total permutations by scrambleing the matrix P
+ for j = 1:n-1
+ A
+ [v i] = max(abs(A(j:n, j)));
+ A([j i+j-1],:) = A([i+j-1 j],:);
+ P([j i+j-1],:) = P([i+j-1 j],:);
+ A
+%%
+% Find the index of the largest pivot element in the current column
+
+
+%%
+% Swap the rows within the in-place array as well as the permutation matrix P
+
+
+
+%%
+% Perform the in-place elimination and save the new column of L
+ i = j+1:n; % indices for the "active" matrix portion
+ A(i,j) = A(i,j)/A(j,j);
+ A(i,i) = A(i,i) - A(i,j)*A(j,i);
+ %A, return
+ end
+%%
+% Extract L and U from the in-place form
+ U = triu(A);
+ L = eye(n) + tril(A,-1);
+end
--- /dev/null
+%% Section 1
+
+N = 5;
+
+% create three random vectors
+main = N*ones(N,1) + rand(N,1);
+upper = rand(N-1,1);
+lower = rand(N-1,1);
+% use the ’diag’ command to create tridiagonal matrix
+A = diag(main,0) + diag(upper,1) - diag(lower,-1);
+
+known_x = ones(N,1); % make known solution vector of 1s
+manuf_b = A*known_x; % manufacture the right-hand-side
+
+A
+
+%% Section 2
+
+N = 5;
+x = linspace(1,2, N);
+h = x(2) - x(1);
+
+% Build tridiag matrix A
+main = ones(N-2,1);
+for i=2:N-1
+ main(i-1,1) = (2/(x(i)^2)) - (2/(h^2));
+end
+lower = ones(N-3,1);
+for i=2:N-2
+ lower(i-1,1) = (1/(h^2)) - (2/(x(i+1)*h));
+end
+upper = ones(N-3,1);
+for i=2:N-2
+ upper(i-1,1) = (1/(h^2)) + (2/(x(i)*h));
+end
+
+A = diag(main,0) + diag(upper,1) + diag(lower,-1);
+
+% Build constant matrix b
+b = zeros(N-2, 1);
+for i=2:N-1
+ b(i-1) = (2*log(x(i)))/(x(i)^2);
+end
+b(1) = b(1) + -((1/(h^2)) - (2/(x(1)*h)))*0.5;
+b(end) = b(end) + -((1/(h^2)) + (2/(x(end)*h)))*log(2);
+
+answer = thomas(A, b);
+answer = [0.5;answer;log(2)];
+
+plot(x, answer);
+hold on;
+real_values = zeros(N, 1);
+for i=1:N
+ real_values(i) = (4/x(i)) - (2/(x(i)^2)) + log(x(i)) - (3/2);
+end
+plot(x, real_values);
+
+
+norm(answer - real_values, inf)
\ No newline at end of file
--- /dev/null
+function [x] = thomas(A,b)
+ % from wikipedia
+ n = length(b);
+ c_prime = zeros(n-1, 1);
+ d_prime = zeros(n, 1);
+ c_prime(1) = A(1,2) / A(1,1);
+ d_prime(1) = b(1) / A(1,1);
+ for i=2:n-1
+ c_prime(i) = A(i,1 + i) / (A(i,i) - (A(i, i-1)*c_prime(i-1)));
+ d_prime(i) = (b(i) - (A(i, i-1)*d_prime(i-1))) / (A(i, i) - (A(i, i-1)*c_prime(i-1)));
+ end
+ d_prime(n) = (b(n) - (A(n, n-1)*d_prime(n-1))) / (A(n, n) - (A(n, n-1)*c_prime(n-1)));
+
+ % backward sub
+ x = zeros(n, 1);
+ x(n) = d_prime(n);
+ for j=n-1:-1:1
+ x(j) = d_prime(j) - (c_prime(j)*x(j+1));
+ end
+
+end
\ No newline at end of file
--- /dev/null
+function [x] = thomas(A,b)
+ % from wikipedia
+ n = length(b);
+ c_prime = zeros(n-1, 1);
+ d_prime = zeros(n, 1);
+ c_prime(1) = A(1,2) / A(1,1);
+ d_prime(1) = b(1) / A(1,1);
+ for i=2:n-1
+ c_prime(i) = A(i,1 + i) / (A(i,i) - (A(i, i-1)*c_prime(i-1)));
+ d_prime(i) = (b(i) - (A(i, i-1)*d_prime(i-1))) / (A(i, i) - (A(i, i-1)*c_prime(i-1)));
+ end
+ d_prime(n) = (b(n) - (A(n, n-1)*d_prime(n-1))) / (A(n, n) - (A(n, n-1)*c_prime(n-1)));
+
+ % backward sub
+ x = zeros(n, 1);
+ x(n) = d_prime(n);
+ for j=n-1:-1:1
+ x(j) = d_prime(j) - (c_prime(j)*x(j+1));
+ end
+
+end
\ No newline at end of file
--- /dev/null
+function [t,y] = RungeKutta4(f,t0,T,y0,h)
+%%
+% Classical four-stage Runge-Kutta time integration technique
+% to solve the initial value problem
+%
+% y' = f(t,y), y(t0) = y0
+%
+% INPUT:
+%
+% f - right-hand-side function that can depend on
+% the time t and the function y
+% t0 - intial time
+% T - final time
+% y0 - initial solution value
+% h - fixed time step size
+%
+% OUTPUT:
+%
+% t - vector of time values
+% y - solution vector of approximate values
+%
+%%
+% discretize the time variable and set the problem size
+ t = t0:h:T;
+ y = zeros(length(t),1);
+%%
+% save the intial values
+ y(1) = y0;
+ t(1) = t0;
+%%
+% use Runge-Kutta 4 stage with the given timestep h to solve the problem and
+% save the information for later analysis
+ for j = 2:length(t)
+ k_1 = f(t(j-1), y(j-1));
+ k_2 = f(t(j-1) + (h/2), y(j-1) + ((h/2)*k_1));
+ k_3 = f(t(j-1) + (h/2), y(j-1) + ((h/2)*k_2));
+ k_4 = f(t(j-1) + h, y(j-1) + ((h*k_3)));
+ y(j) = y(j-1) + ((h/6)*(k_1 + (2*k_2) + (2*k_3) + k_4));
+
+
+
+
+ end
+end
--- /dev/null
+function [S,x] = SimpsonsRule(f,a,b,n)
+%%
+% Implementation of the (possibly) composite Simpsons's quadrature
+% rule to approximate the definite integral
+%
+% b
+% /\
+% |
+% | f(x) dx
+% |
+% \/
+% a
+%
+% INPUT:
+%
+% f - anonymous function integrand
+% a - lower integration bound
+% b - upper integration bound
+% n - number of intervals (must be an even integer)
+%
+% OUTPUT:
+%
+% In - approximation to the definite integral on the given interval [a,b]
+% x - set of n+1 quadrature nodes
+%
+%%
+% Simpson's rule needs at least two intervals
+ if n == 1
+ error('Simpsons rule requires n >= 2');
+ end
+%%
+% check if the number of intervals is even
+ if mod(n,2) ~= 0
+ error('The number of intervals n must be even');
+ end
+%%
+% Find the interval spacing h
+ h = (b-a)/n;
+
+%%
+% build the set of uniformly spaced quadrature nodes
+ x = zeros(n+1,1);
+ x(1) = a;
+ for i=2:n+1
+ x(i) = x(i-1) + h;
+
+%%
+% Initialize the integral value to zero
+ S = f(x(1)) + f(x(end));
+ for i=2:n
+ if mod(i, 2) == 0
+ S = S + 4*f(x(i));
+ else
+ S = S + 2*f(x(i));
+ end
+ end
+
+ S = (h/3) * S;
+
+%%
+% Approximate the integral with (composite) Simpson's rule
+
+
+
+
+
+
+
+
+
+
+end
--- /dev/null
+function [In,t] = adaptativeSimpson(f,a,b,tol)
+%%
+% Adaptive Simpson's rule to approximate a definite integral. This uses
+% a posterioi error analysis to recursively apply Simpson's rule where it
+% is needed
+%
+% INPUT:
+%
+% f - integrand function
+% a - lower bound of interval
+% b - upper bound of the interval
+% tol - error tolerance for adaptivity
+%
+% OUTPUT:
+%
+% In - adaptive approximation of the definite integral to
+% within the error tolerance
+% t - vector of the adapted quadrature nodes
+%
+%%
+%%
+% use recursive bisection with error estimation to compute the integral approximation
+ m = 0.5*(b+a); % find the current midpoint
+ [In,t] = do_integral(a,f(a),b,f(b),m,f(m),tol);
+%%
+% this is a recursive function within Matlab so it calls itself
+ function [In,t] = do_integral(a,fa,b,fb,m,fm,tol)
+ %%
+ % need the two midpoints of the sub-intervals and the function
+ % evalutions for the recursion
+ xL = 0.5*(a+m);
+ fxL = f(xL);
+ xR = 0.5*(b+m);
+ fxR = f(xR);
+ %%
+ % save the five nodes at the current level of the recursion
+ t = [a;xL;m;xR;b];
+ %%
+ % get the h value for the whole interval
+ h = 0.5*(b-a);
+ %%
+ % compute the Simpson's rule on the whole interval
+ S_coarse = h/3*(fa + 4*fm + fb);
+ %%
+ % compute the Simpson's rule on the two subintervals
+ S_Left = h/6*(fa + 4*fxL + fm);
+ S_Right = h/6*(fm + 4*fxR + fb);
+ S_fine = S_Left + S_Right;
+ %%
+ % error estimate
+ E = S_coarse - S_fine;
+ %%
+ % check against the user set tolerance
+ if abs(E) < 10*tol
+ % exit when tolerence is met
+ In = S_fine;
+ else
+ %%
+ % bisect again if the tolernce is note met
+ [IL,tL] = do_integral(a,fa,m,fm,xL,fxL,tol);
+ [IR,tR] = do_integral(m,fm,b,fb,xR,fxR,tol);
+ In = IL + IR;
+ % append the node edges together (this avoids duplicates)
+ t = [tL;tR(2:end)];
+ end
+ end
+end
--- /dev/null
+function [t,y] = forwardEuler(f,t0,T,y0,h)
+%%
+% Forward Euler time integration technique to solve the
+% initial value problem
+%
+% y' = f(t,y), y(t0) = y0
+%
+% INPUT:
+%
+% f - right-hand-side function that can depend on
+% the time t and the function y
+% t0 - intial time
+% T - final time
+% y0 - initial solution value
+% h - fixed time step size
+%
+% OUTPUT:
+%
+% t - vector of time values
+% y - solution vector of approximate values
+%
+%%
+% discretize the time variable and set the problem size
+ t = t0:h:T;
+ y = zeros(length(t),1);
+%%
+% save the intial values
+ y(1) = y0;
+ t(1) = t0;
+%%
+% use forward Euler with the given timestep h to solve the problem and
+% save the information for later analysis
+ for j = 2:length(t)
+% this is an explicit method so we always use values from the current solution
+ y(j) = y(j-1) + h*f(t(j-1),y(j-1));
+ end
+end
--- /dev/null
+%lab3
+N = 1000;
+x = linspace(0,1,N+1);
+h = x(2) - x(1);
+f_prime = zeros(1, N+1);
+
+f_1 = @(x) (1);
+f_2 = @(x) (x);
+f_3 = @(x) (x.^2);
+f_4 = @(x) (exp(sin(4*x)));
+
+for i=1:N-1
+ f_prime(i+1) = (f_4(x(i+2)) - f_4(x(i))) / (2*h);
+end
+
+f_prime(1) = (f_4(x(2)) - (f_4(x(1)))) / (h);
+f_prime(end) = (f_4(x(end)) - f_4(x(end-1))) / h;
+
+%f_prime(1) = ((-3*f_4(x(1))) + (4*f_4(x(2))) - (f_4(x(3)))) / (2*h);
+%f_prime(end) = ((f_4(x(end-2))) - (4*f_4(x(end-1))) + (3*f_4(x(end)))) / (2*h);
+
+f_prime_exact = zeros(1, N+1);
+for i=1:N+1
+ f_prime_exact(i) = 4*exp(sin(4*x(i))) * cos(4*x(i));
+end
+
+f_prime
+f_prime_exact
+
+max(abs(f_prime_exact - f_prime))
\ No newline at end of file
--- /dev/null
+function plot_composite_quad(f,t)
+%%
+% Script that will plot a function and the interval boundaries for use
+% with a compositie quadrature rule
+%
+% INPUT:
+%
+% f - integrand function
+% t - set of nodes used for the composite quadrature rule
+%
+% OUTPUT:
+%
+% NONE - Produces plot to screen
+%
+%%
+ clf
+ p1 = fplot(f,[t(1) t(end)],'-k','LineWidth',2);
+ hold on
+ p2 = stem(t,f(t),'fill','color',[0.9100 0.4100 0.1700],'MarkerSize',9,'LineWidth',1.5);
+ set(gcf,'Position',[500, 60, 1250, 1250])
+ set(gca,'FontSize',16);
+ legend([p1 p2],{'$(x+1)^2\cos\left(\frac{2x+1}{x-3.3}\right)$','Quadrature nodes'}...
+ ,'interpreter','latex','fontsize',24,'Location','northwest')
+end
\ No newline at end of file
projects / TANA21.git / commitdiff
