So I know that you can use Picard Lindelof to show uniqueness of a solution to an ODE, but if given an ordinary delayed differential equation, could you still use it? How would that establish uniqueness for all values of, say, t, if I were trying to show uniqueness for something like
$$y'(t) = ay(t-1)$$ with $y(t) = y_0$ on $[-1,0]$?
I'm reading "An Introduction to Delay Differential Equations with Sciences Applications to the Life Sciences," and the existence and uniqueness part,,, doesn't quite make much sense to me, even though I understand what the method of steps itself is, so… :/
Also, side question. Is it sufficient to show that $y'(t)$ and its derivative are continuous to show existence/uniqueness?
(I'm trying to show uniqueness for the solution of a different DDE, but I want to try to understand it first, which is why I added a much simpler problem instead)
Best Answer
For this example, you do not need much. You can apply what is called forward propagation (or the method of steps).
First consider $t\in[0,1]$. In that case, we have that
$$y'(t)=\alpha y_0(t-1),$$
and this has the simple unique solution $$y(t)=y_0(0)+\alpha\int_0^ty_0(s-1)ds.$$ Let us call it $y_1(t)$.
Now, consider $t\in[1,2]$. In that case, we have that
$$y'(t)=\alpha y_1(t-1),$$
and we have that the unique solution on $[1,2]$ is given by
$$y(t)=y_1(1)+\alpha\int_1^ty_1(s-1)ds,$$
and so on.
All linear systems with delays can be shown to have a unique solution using this approach. In the general single-delay case we have that
$$\dot{x}(t)=A_0x(t)+A_1x(t-h),\ x(s)=\varphi(s),\ s\in[-h,0].$$
Then, we have that
$$x(t)=\exp(A_0t)\varphi(0)+\int_0^t\exp(A(t-s))A_1\varphi(s-h)ds,\ t\in[0,h],$$
which is the unique solution. Propagating further we have that
$$x(t)=\exp(A_0(t-h))x(h)+\int_h^t\exp(A(t-s))A_1x(s-h)ds,\ t\in[h,2h],$$
etc.