Couple ODE System not enough Input arguments, Why?

Hey all,

I have a coupled system of ODE's that I want to solve and followed the instructions. However, there seems to be an error with the input arguments. Could somebody help?

Best

% Define the boundary value problem
solinit = bvpinit(linspace(0,1,10), @guess);

% Solve the boundary value problem
sol = bvp4c(bvpfun, bcfun, solinit);

% Plot the solution
x = linspace(0,1,100);
y = deval(sol, x);
plot(x, y(1,:), 'b-', 'LineWidth', 2);
hold on
plot(x, y(2,:))
xlabel('x');
ylabel('y');
title('Solution of Second Order ODE with BVP');
grid on;

function dydx = bvpfun(x, y)
    dydx = zeros(3, 1); % Initialize dydx as a column vector
    dydx(1) = y(3) + y(1);
    dydx(2) = y(1) - 2/y(3);
    dydx(3) = y(1) - 1;
    dydx = [dydx(1); dydx(2); dydx(3)];
end

function res = bcfun(ya,yb)
    res(1) = ya(1);
    res(2) = ya(2);
    res(3) = ya(3);
end

function y = guess(x)
    y = [0; 0; 0];
end

[Sam Chak]

From the dynamics

the condition given at the boundary , , causes the singularity .

[Thanh Hoang]

Hey,

thanks for pointing this out, there was a tippo from my side it should have been y2' = y1 - y3 / 2

Answer 1

[Aquatris]

You should use @ symbol before the functions in your bvp4c call. Now you have another error, good luck solving it :D hint: what happens when y(3) = 0.

% Define the boundary value problem
    solinit = bvpinit(linspace(0,1,10), @guess);
    % Solve the boundary value problem
    sol = bvp4c(@bvpfun, @bcfun, solinit);

Error using bvp4c (line 196)
Unable to solve the collocation equations -- a singular Jacobian encountered.

    % Plot the solution
    x = linspace(0,1,100);
    y = deval(sol, x);
    plot(x, y(1,:), 'b-', 'LineWidth', 2);
    hold on
    plot(x, y(2,:))
    xlabel('x');
    ylabel('y');
    title('Solution of Second Order ODE with BVP');
    grid on;
function dydx = bvpfun(x, y)
    dydx = zeros(3, 1); % Initialize dydx as a column vector
    
    dydx(1) = y(3) + y(1);
    dydx(2) = y(1) - 2/y(3);
    dydx(3) = y(1) - 1;
    dydx = [dydx(1); dydx(2); dydx(3)];
end
function res = bcfun(ya,yb)
res(1) = ya(1);
res(2) = ya(2);
res(3) = ya(3);
end
function y = guess(x)
y = [0; 0; 0];
end

Answer 2

[Torsten]

If all boundary conditions are given at x = 0 as in your case, you have an initial value problem instead of a boundary value problem. In this case, use "ode45" instead of "bvp4c".

Answer 3

[Sam Chak]

Hi @Thanh Hoang

I have added these four lines to your original code. The modifications made are minimal.

%% ----- added these 4 lines -----
ivpfun  = @bvpfun;
xspan   = [0, 1];
yinit   = [0; 0; 0];
sol     = ode45(ivpfun, xspan, yinit);
%% ----- added these 4 lines -----

% sol = bvp4c(@bvpfun, @bcfun, solinit);

% Plot the solution
x = linspace(0,1,100);
y = deval(sol, x);
plot(x, y(1,:), 'b-', 'LineWidth', 2);
hold on
plot(x, y(2,:))
xlabel('x');
ylabel('y');
title('Solution of Second Order ODE with BVP');
grid on;

function dydx = bvpfun(x, y)
    dydx    = zeros(3, 1);  % Initialize dydx as a column vector
    dydx(1) = y(3) + y(1);
    dydx(2) = y(1) - y(3)/2;
    dydx(3) = y(1) - 1;
    dydx    = [dydx(1); 
               dydx(2); 
               dydx(3)];
end