Forward Euler solution improvement
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I would like to know if this code can be improved or is well-written as it is? it does not give errors. I'll appreciate any comments.Thanks
%part a
clear; clc;
% Parameters
alpha = 10;
beta_values = [2, 4];
h = 0.01;
t = 0:h:20;
N = length(t);
% Initial conditions
y1_0 = 0;
y2_0 = 2;
for b = 1:length(beta_values)
beta = beta_values(b);
y1 = zeros(1, N);
y2 = zeros(1, N);
% Initial values
y1(1) = y1_0;
y2(1) = y2_0;
% Forward Euler method
for n = 1:N-1
f1 = alpha ...
- y1(n) ...
- (4*y1(n)*y2(n)) / (1 + y1(n)^2);
f2 = beta * y1(n) * ...
(1 - y2(n) / (1 + y1(n)^2));
y1(n+1) = y1(n) + h * f1;
y2(n+1) = y2(n) + h * f2;
end
figure;
plot(t, y1, 'LineWidth', 1.5);
xlabel('t');
ylabel('y_1(t)');
title(['Forward Euler: y_1 vs t, \beta = ', num2str(beta)]);
grid on;
figure;
plot(y1, y2, 'LineWidth', 1.5);
xlabel('y_1');
ylabel('y_2');
title(['Phase Portrait: y_2 vs y_1, \beta = ', num2str(beta)]);
grid on;
end
%part b
% Parameters
alpha = 10;
beta_values = [3.4, 3.5, 3.6];
h = 0.01;
t = 0:h:100;
N = length(t);
% Initial conditions
y1_0 = 0;
y2_0 = 2;
for b = 1:length(beta_values)
beta = beta_values(b);
y1 = zeros(1, N);
y2 = zeros(1, N);
% Initial values
y1(1) = y1_0;
y2(1) = y2_0;
% Forward Euler method
for n = 1:N-1
f1 = alpha ...
- y1(n) ...
- (4*y1(n)*y2(n)) / (1 + y1(n)^2);
f2 = beta * y1(n) * ...
(1 - y2(n) / (1 + y1(n)^2));
y1(n+1) = y1(n) + h * f1;
y2(n+1) = y2(n) + h * f2;
end
figure;
plot(t, y1, 'LineWidth', 1.5);
xlabel('t');
ylabel('y_1(t)');
title(['y_1 vs t, \beta = ', num2str(beta)]);
grid on;
figure;
plot(y1, y2, 'LineWidth', 1.5);
xlabel('y_1');
ylabel('y_2');
title(['Phase Portrait y_2 vs y_1, \beta = ', num2str(beta)]);
grid on;
end
0 Comments
Accepted Answer
Alan Stevens
about 15 hours ago
It's clear and runs quickly. However, it's probably neater to pre-define your gradient functions like this:
% Gradient functions
f1 = @(alpha,y1,y2) alpha - y1 - (4*y1*y2) / (1 + y1^2);
f2 = @(beta,y1,y2) beta * y1 * (1 - y2 / (1 + y1^2));
and then replace the forward Euler equations simply by:
% Forward Euler method
for n = 1:N-1
y1(n+1) = y1(n) + h * f1(alpha,y1(n),y2(n));
y2(n+1) = y2(n) + h * f2(beta,y1(n),y2(n));
end
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