Positivity of $g'(t)=g(t)\int_0^t a(t-\tau)g'(\tau)d\tau-g(t)a(t)$

Question

I have

$$f(t)=g(t)\int_0^t a(t-\tau)f(\tau)d\tau+g(t)a(t)$$ $$g'(t)=-f(t)$$

$a(t)$ is positive decreasing, and $g(0)$ is positive.

I must prove the positivity of $g$.

Note that $$g(t)=-\int_0^t f(\tau)d\tau+g(0)$$

$$g'(t)=g(t)\int_0^t a(t-\tau)g'(\tau)d\tau-g(t)a(t)$$

I could not proceed.

Answer 1

Your last equation is useful:

$$\frac{g'(t)}{g(t)} = \int_0^t a(t-\tau)g'(\tau) \ d \tau - a(t)$$

and now integrate both sides with respect to $t$ to obtain:

$$\ln(g(s)) = \int_0^s \left( \int_0^t a(t-\tau) g'(t) \ d \tau - a(t)\right) \ dt+ \ln (g(0))$$

which is justified since you assume that $g(0) > 0$, and therefore the logarithm makes sense. Now exponentiating both sides gives an expression of the form $g(s) = \exp(\text{stuff})$, which is evidently positive.