# Integral equation and differential equation

We have $$y'=By+G(y)$$, where $$B=\left[ \begin{array}{cc} P & 0\\ 0 & Q \end{array} \right]$$ and Let $$U(t)=\left[ \begin{array}{cc} e^{Pt} & 0\\ 0 & 0 \end{array} \right]$$ and $$V(t)=\left[ \begin{array}{cc} 0 & 0\\ 0 & e^{Qt} \end{array} \right]$$. Consider the integral equation $$u(t,a)=U(t)a+\int_{0}^{t}U(t-s)G(u(s,a))ds-\int_{t}^{\infty}V(t-s)G(u(s,a))ds$$

And i don't understand why if $$u(t,a)$$ is a continuous solution of this integral equation, then it is a soution of the differential equation?

You have (at least due to my calculations) $$\frac{d}{dt}\int_0^tf(t-s)g(s)\,ds = f(0)g(t) + \int_0^tf'(t-s)g(s)\,ds$$ and therefore also $$\frac{d}{dt}\int_t^\infty f(t-s)g(s)\,ds = -f(0)g(t) + \int_t^\infty f'(t-s)g(s)\,ds$$ Using these rules when differentiating $$u$$ with respect to $$t$$ you get \begin{align*} u_t = U'(t)a + U(0)G(u) + &\int_0^t U'(t-s)G(u(s,a))\,ds + V(0)G(u)\\ &- \int_t^\infty V'(t-s)G(u(s,a))\,ds. \end{align*} Now, $$U'(t) = BU(t)$$, $$V'(t) = BV(t)$$, and $$U(0) + V(0) = I$$. Hence, indeed, \begin{align*} u_t &= BU(t)a + G(u) + B\int_0^t U(t-s)G(u(s,a))\,ds - B\int_t^\infty V(t-s)G(u(s,a))\,ds\\ &= Bu + G(u). \end{align*}