Supose we have
$v'(t) = A(t)v(t)+b(t),$ with $v(0)=0$
where $v(t),b(t) \in \mathbb{R}^n $, $A$ is a $n\times n$ real matrix and $t \in [0,1]$
Now, if A and b are discontinuous but integrable functions on the interval, then is there a unique solution for this equation?
I am having problems passing this diferential equation to an integral equation, as I guess this should be the first step. If someone would care to help me! Thanks! (This is an exercise)
Overkill...If $x,y$ are solutions, we get $$ \vert x(t)-y(t)\vert\le\vert x(0)-y(0)\vert+ \int_{0}^{t}\Vert A(s)\Vert\vert x(s)-y(s)\vert ds=R(t) $$ and $ \dot R(t)=\Vert A(t)\Vert\vert x(t)-y(t)\vert \le \Vert A(t)\Vert R(t) $ so that Gronwall's provides $$\vert x(t)-y(t)\vert\le R(t)\le e^{\int_{0}^{t}\Vert A(s)\Vert ds}R(0)=e^{\int_{0}^{t}\Vert A(s)\Vert ds}\vert x(0)-y(0)\vert, $$ Lipschitz continuity with respect to the initial data and uniqueness.
A simple local existence theorem in a Banach space setting. Let $E$ be a Banach space, $I$ be an interval of $\mathbb R$, $\Omega$ be an open subset of $E$, $F:I\times \Omega\longrightarrow E$ be a continuous mapping such that for each $(t_0,x_0)\in I\times \Omega$ there exists a neighborhood $J$ of $t_0$, a neighborhood $V$ of $x_0$, a function $\alpha\in L^1_{loc}(J)$ such that, for $t\in J,x_1,x_2\in V$, $$ \vert F(t,x_1)-F(t,x_2)\vert\le \alpha (t) \omega(\vert x_1-x_2\vert) $$ where $ \omega:{\mathbb R_+}^*\longrightarrow {\mathbb R_+}^*, \ \text{non-decreasing},\ \omega(0_+)=0,\ \int_0^1\frac{dr}{\omega(r)}=+\infty. $ Then the IVP for the ODE $$ \dot x(t)=F(t,x(t)),\quad x(t_0)=x_0, $$ has a unique local solution.
