# How to deal with an integro-differential equation of this form - fixed points?

I've encountered an integro-differential equation of the following form: $$\frac{dx(t)}{dt} = \int_0^t ds\ f_{1}(s) - \int_0^t ds\ f_{2}(s) x(t - s)$$

The functions $$f_{1}(t)$$ and $$f_{2}(t)$$ are known explicitly, and the goal is to solve for $$x(t)$$ given some initial condition $$x(0)$$.

My question is, what are some approaches to getting information about possible solutions $$x(t)$$? I've had two ideas:

(a) Fixed points for this equation.

Might there exist a constant solution $$x(t) = x^{\star}$$? In this case, we'd set the LHS of the above equation to $$0$$ and solve for a constant $$x^{\star}$$ which gives: $$x^{\star} = \frac{\int_{0}^t ds\ f_{1}(s)}{\int_0^t ds\ f_{2}(s)}$$

But this must be wrong since there is time dependence explicitly in this. I wanted to ask is there a method for finding a fixed point solution $$x^{\star}$$? (maybe the implication of the above is that there aren't any fixed points?)

(b) Getting a second order DE and solving that.

If you differentiate both sides of the integro-differential equation you get: $$\frac{d^2x(t)}{dt^2} = f_{1}(t) - f_{2}(t) x(0)$$ which looks a lot simpler then the initial equation. This looks like a simple second-order DE, with the only unusual feature being that the DE explicitly depends on the initial condition $$x(0)$$. I am confused here also because naively integrating the above equation gives a solution like: $$\frac{dx(t)}{dt} = \int_0^t ds\ f_{1}(s) - x(0) \int_0^t ds\ f_{2}(s) + c$$ Where the constant $$c$$ is constrained by the original integro-differential equation to be $$c=0$$. This is weird to me, because my naive integration has a factor $$x(0)$$ outside the integral, rather than a $$x(t-s)$$ underneath the integral as in the original DE. What am I getting wrong in this naive integration of $$\ddot{x}(t)$$?

Your equation can be written as $$\dot x=F_1(t)+(f_2*x)(t)$$ where the last term is the convolution product. Without knowing more about $$f_2$$ you can say nothing more about the equation. Its derivative is $$\ddot x=f_1(t)+(f_2*\dot x)(t)=f_1(t)+(\dot f_2*x)(t)$$ and so on, so that if there is a linear DE with constant coefficients for $$f_2$$ the last term can be annihilated in some linear product resulting in a linear DE for $$x$$.
Fixed points, that is constant solutions, have to satisfy $$\dot x=0$$, which would require $$0=F_1(t)+\int_0^tf_2(s)x^*\,ds=F_1(t)+x^*F_2(t),$$ that is, $$F_1$$ has to be a constant multiple of $$F_2$$, and thus the same has to hold for their derivatives, $$0=f_1(t)+x^*f_2(t)$$.