Existence of solution for linear system of ODE 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?

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.
