Malthus Model - Solution Differential Equation

Question

I have this equation of a time dependent Malthus model with a term representing a time dependent immigration: $$N'(t)=r(t)N(t) + m(t)$$ with $r(t)$ and $m(t)$ both continuous and periodic with Period $T$. I have to prove that the function $$N_\infty (t)=\int_{-\infty}^t exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(s)ds$$ is a solution of the ODE above.

I tried to derive $N_\infty (t)$ to obtain our ODE, but unsuccessful. I think that I do somethings wrong on the calculation. I know that the derivative of an integral is so computed: Let $G(t):=\int_a^x f(t)dt$. Then $G'(t)=f(x)$

So in our case I did: $$N'_\infty (t)= exp\biggl(\int_s ^t r(\sigma)d\sigma\biggr)m(t)$$ But I think I miss something. Someone can help me please?

Answer 1

The derivative of $N_\infty(t)$ is incorrect. To calculate its derivative the Leibniz integral rule should be used because the variable $t$ is in the integral sign as well as in the integrand.

$$ N^\prime_\infty (t) = exp\left(\int_t^t r(\sigma)\,d\sigma\right) m(t) + \int_{-\infty}^t r(t)\, exp\left(\int_s^t r(\sigma)d\sigma\right) m(s)\,ds $$

Now the second term is equal to r(t)N(t) $$ r(t)N(t) = \int_{-\infty}^t r(t)\, exp\left(\int_s^t r(\sigma)d\sigma\right) m(s)\,ds $$

Moreover $$ exp\left(\int_t^t r(\sigma)\,d\sigma\right) = exp(0) = 1 $$

Then, replacing in $N^\prime_\infty(t)$ $$ N^\prime_\infty (t) = m(t) + r(t)N(t) $$

which is the ODE. So that solution satisfies it.

Answer 2

With $a$ in the domain of definition let $R(t) = \int_{a}^{t}r(s)ds$ so that $R'(t) = r(t)$. Then if we use the theory of an integrating factor

$$\frac{d}{dt}\left(N(t)\exp(-R(t))\right) = \exp(-R(t))m(t).$$

Integrating this with respect to $t$ gives us $$N(t)\exp(-R(t))-N(a) = \int_{a}^{t}\exp(-R(s))m(s)ds.$$ Therefore the solution is given by

$$N(t) = N(a)\exp(R(t))+\exp(R(t))\int_{a}^{t}\exp(-R(s))m(s) = N(a)\exp\left(\int_{a}^{t}r(s)ds\right)+\exp\left(\int_{a}^{t}r(s)ds\right) \int_{a}^{t}\exp(-R(s))m(s)ds.$$