Solving system ODEs with one unknown variable coefficient.

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Martina
Martina 2015년 12월 1일
댓글: Torsten 2015년 12월 2일
Hi all, I am trying to solve this system of ODEs: $$ S'(a)= {\tau}R(a)-\lambda(a)S(a)\\ I'(a)=\lambda(a)S(a)-{\gamma}I(a)+rL(a)\\ R'(a)=(1-q)\gammaI(a)-{\tau}R(a)\\ L'(a)=q\gammaI(a)-rL(a) $$
I know the value of all the parameters apart from $\lambda$ which is more over a-dependent. I have tried to solve it analitically using this solve but it seems not possible. So I want to tried with a numerical approach but I don't know how to procede because of the presence of this $\lambda(a)$. Do you have any hints-suggestion ?
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Martina
Martina 2015년 12월 1일
편집: Martina 2015년 12월 1일
Thank you for your comment. Actually no, my starting point was a system of pdes with age and time variables. My goal is to find the steady states and procede with my analysis from there. By definition, I know that the non trivial one is the one that solves the system that I have written above (dropping the time variable). I tried to solve it analytically, in order to obtain a symbolic solution (with like an integral for $\lambda(a)$) but it seems not possible. Hence I want to try the numerical approach. To be complete, $\lambda$ has a expression.. which is the following: $$ \lambda(a)=f(a)\int_{15}^{65}g(s)I(s)ds $$ But I would like to treat it like a simple paramter which depends on the variable a, but probably it is not possible..
Best,
Martina
Torsten
Torsten 2015년 12월 2일
Even if an analytical solution existed for your linear ODE system with variable coefficients, it would be so complicated that it does not help for interpretation.
A numerical solution is only possible if you supply an explicit function lambda=lambda(a).
Best wishes
Torsten.

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