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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?

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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*}

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