# Does this ODE system have solution?

Let $A(t)$ and $B(t)$ be matrices with each element in $L^\infty(0,T).$ Let $A(t)$ have an inverse. I know nothing else about this inverse.

Let $c(t)$ be a vector in $L^2(0,T).$

Let $u(t) \in \mathbb{R}^n$ be unknown. Does the system $$A(t)u'(t) + B(t)u + c(t)= 0$$ have a unique solution?

Again, I don't know that $A(t)^{-1}$ is in $L^\infty.$

• no unique solution without initial condition.... May 24, 2013 at 21:21

Let $n=1$. Let $A(t)=1$, $B(t)=0$ and let $c(t)=0$ for all $t\in(0,T)$ for any $T>0$. Then $A(t)$ has an inverse for $t\in (0,T)$. Observe that $u(t)=k$ for all $t\in (0,T)$, for any $k\in \mathbb{R}$ solves the differential equation $$A(t)u'(t) + B(t) + c(t) = 1*0 + 0 + 0 = 0 \ \ \forall t\in (0,T).$$ You may wish to impose boundary conditions on the problem if you seek a uniqueness result. I also recommend the first two chapters of the book by Coddington and Levinson "Theory of Ordinary DIfferential Equations" to address this type of question.
Let us try for $(u(t),t)$, and invert $A(t)$. Then we have an autonomous ODE on $\mathbb R^{n+1}$, or a flow equation for a vector field. We have local existence and uniqueness if the coefficient functions are locally Lipschitz. We have existence alone, by Peano's theorem, if the coefficients are continuous, but we loose uniqueness in general. Your coefficients are measurable. So nothing is true in general.