Solving System of two second order ODE's

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chen 2015-1-9

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评论： Amos 2015-1-11

Hi , I have tried solving the following system of ODE's (eq1 attached) using Matlab. first i reduced it into a system of four 1st order ODE's (eq2 attached). thani tried to solve it using ode45() in the following manner:

        function xprime = Mirage(t,x);
        k=2;
        xprime=[x(1); k*(1-x(1).^2)./(1+k*x(2)); x(3) ; -(k*x(1)*x(3))./(1+k*x(2))];
        end
        x0=[1 1 1 1];
        tspan=[0,20];
        [t,x]=ode45(@Mirage,tspan,x0)

did i set the vector x well in the function? where did i wrong? i will appreciate your help.

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Answer 1

Amos 2015-1-9

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In order to get a system of first order equations, you substitute y' by t and z' by s, right? So your independent functions are then y,t,z,s. These should be ordered in a vector, e.g. x=[y,t,z,s]. The function xprime obviously returns the derivative of the four entries of x. But then x(1)' should be set to x(2), as you have y'=t and not y'=y. The same holds for x(3)' which should not be set to x(3) but rather to x(4).

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chen 2015-1-10

在 MATLAB Online 中打开

ok i've changed the function code:

function xprime = Mirage(t,x);
k=0.0001;
        xprime=[x(2); k*(1-x(2).^2)./(1+k*x(1)); x(4) ; (k*x(2)*x(4))./(1+k*x(1))];
end

the result of x is a matrix . each row is another iteration. i get values for the first two terms [y,t..] but no values for the two last terms [..,z,s] z stays on its initial condition as well s.

finally i want to plot y(z) like so: plot(x(:,3),x(:,1),'r');

again, i'll appreciate your help

Amos 2015-1-11

So what is actually your question?

When I run the code, I get a four column matrix for x, as you expected. It is true that t=x(2) is stationary, but this is just a consequence of t' being proportional to 1-t. So, if you start with t=1, t' is zero and therefore t doesn't change.

By the way, I think in your corrected version Mirage(t,x), there is a minus sign missing in the last entry of xprime.

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