Solving system ODEs with one unknown variable coefficient.

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Martina 2015-12-1

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评论： Torsten 2015-12-2

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 2015-12-1

编辑：Martina 2015-12-1

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 2015-12-2

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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