Hey Cheng
Kindly find the code below to use the Galerkin method with hat functions to solve the given boundary value problem (BVP) in MATLAB.
Also have compared the results with the exact solution ( y = t^3 )
function galerkin_bvp()
% Define the intervals
intervals = [1/20, 1/40];
% Exact solution
exact_solution = @(t) t.^3;
% Loop over each interval
for h = intervals
% Discretize the interval
t = 0:h:1;
n = length(t);
% Initialize the matrix A and vector b
A = zeros(n, n);
b = zeros(n, 1);
% Fill the matrix A and vector b using the Galerkin method
for i = 2:n-1
A(i, i-1) = 1/h^2;
A(i, i) = -2/h^2;
A(i, i+1) = 1/h^2;
b(i) = 6 * t(i);
end
% Boundary conditions
A(1, 1) = 1;
b(1) = 0;
A(n, n) = 1;
b(n) = 1;
% Solve the linear system
y_approx = A \ b;
% Compute the exact solution
y_exact = exact_solution(t)';
% Compute the absolute errors
abs_errors = abs(y_exact - y_approx);
% Display the results
fprintf('Results for h = %f:\n', h);
fprintf('Max Absolute Error: %e\n', max(abs_errors));
fprintf('Mean Absolute Error: %e\n\n', mean(abs_errors));
% Plot the results
figure;
plot(t, y_exact, 'b', 'LineWidth', 2); hold on;
plot(t, y_approx, 'ro--');
legend('Exact Solution', 'Galerkin Approximation');
title(['Galerkin Method with h = ', num2str(h)]);
xlabel('t');
ylabel('y');
grid on;
end
end
Results for h = 0.050000:
Max Absolute Error: 2.722128e-15
Mean Absolute Error: 1.475812e-15
Results for h = 0.025000:
Max Absolute Error: 8.523030e-15
Mean Absolute Error: 4.261312e-15
