Dear friends,

I have a problem with the solution representation formula for first-order (delay) differential equations.

Let $\displaystyle t_{0}\in\mathbb{R}$, $\displaystyle A,f\in C([t_{0},\infty),\mathbb{R})$ and $\displaystyle \alpha\in C([t_{0},\infty),\mathbb{R})$ be a function such that $\displaystyle \alpha(t)\leq t$ for all $\displaystyle t\geq t_{0}$ and $\displaystyle \lim\nolimits_{t\to\infty}\alpha(t)=\infty$.

Let $\displaystyle t_{-1}:=\min\{\alpha(t):t\in[t_{0},\infty)\}$, and $\displaystyle \varphi\in C([t_{-1},t_{0}],\mathbb{R})$.

Consider the following differential equation:

$\displaystyle \begin{cases}

x^{\prime}(t)+A(t)x(\alpha(t))=f(t)\quad\text{for} \ t\in[t_{0},\infty)\\

x(t)=\varphi(t)\quad\text{for}\ t\in[t_{-1},t_{0}].

\end{cases}\rule{3.85cm}{0cm}(1)$

Then, $\displaystyle x$ can be represented in the following form uniquely:

$\displaystyle x(t)=\varphi(t_{0})\mathcal{X}(t,t_{0})+\int_{t_{0 }}^{t}\mathcal{X}(t,\eta)\big[f(\eta)-A(\eta)\varphi(\alpha(\eta))\big]\mathrm{d}\eta\rule{2cm}{0cm}(2)$

for $\displaystyle t\geq t_{0}$, where $\displaystyle f,A,\varphi$ are assumed to be identically zero out of their domains, and thefundamental solution$\displaystyle \mathcal{X}=\mathcal{X}(t,s)$ is the solution of the following differential equation:

$\displaystyle \begin{cases}

x^{\prime}(t)+A(t)x(\alpha(t))=0\quad\text{for}\ t\in[s,\infty)\\

x(s)=1\quad\text{and}\quad x(t)\equiv0\quad\text{for}\ t\in(-\infty,s).

\end{cases}\rule{3.65cm}{0cm}(3)$

In (3), $\displaystyle A,\alpha$ are same as in (1).

I would be glad if some one can help me with the proof of the representation formula (2).

Thanks.