Consider linear system of ODE given by \begin{eqnarray} u'_i(t)&=&\sum\limits_{i=1}^{n}a_{ij}(t)u_j+b_i(t) &\quad i=1,2,\ldots,n\\ u_i(0)&=&u_i^0 &\quad i=1,2,\ldots,n \end{eqnarray}
Suppose $a_{ij},b_i \in L^{\infty}(\mathbb{R}^+)$ and $u_i^0\in \mathbb{R}$ for $i=1,2,\ldots,n,$ then do we have the existence of solution? ?If so how to prove it?
In other words, can we relax continuity assumption on $a_{ij},b_i$ in the existence proof?
As the other answers demonstrate, there are obstacles to solving $u'=Au+b$ as differential equation with differentiable solutions. However, the associated integral equation $$ u(t)=u(0)+\int_0^t(A(s)u(s)+b(s))\,ds $$ is well-defined as fixed-point equation over continuous functions. As $A$ is essentially bounded, $L={\rm ess}\sup_t\|A(t)\|<\infty$ and $$ \|u\|_L=\sup_t e^{-2L|t|}\|u(t)\| $$ is a norm on a closed subspace of continuous functions that makes the fixed-point iteration to the above equation contracting with factor $\frac12$. Thus by the fixed-point theorem there is a unique solution.
Following the thoughts on my comment:
Let's say $u_1 = u$ and $a_{11} = 2\chi_{[0,+\infty)}$. For $t < 0$, one has $u' = 0$, and so $u(t) = A$ for some $A \in \mathbb{R}$.
On the other hand, if $t \geq 0$, then $u' = 2u$, and so $u(t) = Be^{2t}$ for some $B \in \mathbb{R}$.
Let's say one asks for $u(0) = 1$. Then $B = 1$. How do we find $A$? As we're treating with ODES, $u$ has to be differentiable. For that, is has to be continuous, so $A=u(0^-) = u(0^+) = 1$. Then $A = 1$.
With these results, it is easy to check that $u$ is not differentiable on $x = 0$. So...there is no solution (in the classical sense) in this case.
Of course one can extend the notion of differentiabilitty, but that's out of the scope of this question, I guess.
Unfortunately this is not correct. You will at least need $a_{ij}$ and $b_i$ to be continuous. Let me give a counterexample for $n=1$: Let $b=0$, $a(t)=0$ for $t\leq 0$, $a(t)=1$ for $t>0$. Then let us assume we have a differentiable function $u:(-\varepsilon, \varepsilon)\rightarrow \mathbb{R}$ (not necessarily with a continuous derivative) satisfying $$u'(t)=a(t) u(t),\ \ u(0)=1.$$ Since $u$ itself has to be continuous, we find a $\delta>0$ small, such that for all $t\in[-\delta,\delta]$ we have $$u(t)\in\left[\frac{1}{2},\frac{3}{2}\right].$$ Then by the differential equation $$u'(0)=a(0)\cdot u(0)=0\cdot 1=0.$$ for all $0<t\leq\delta$ we have $$u'(t)=a(t)\cdot u(t) = u(t)\geq \frac{1}{2}.$$ Hence the derivative has to jump. But this is not possible, see e.g. Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
