Dear Mathoverflower's.

I have a somewhat vague question regarding an abstract ODE in a Banach space.
Suppose $A:D(A) \subset X \rightarrow X$ is some linear operator (lets assume its closed) and maybe add some other conditions.

Suppose $x_0 \in X$ is non-zero and suppose one has the relevant theory to show the existence of a solution of $x'(t)=Ax(t)$ for $t>0$ with $x(0)=x_0$.

So the question that came up is can $x(t)$ converge to zero in finite time? I assumed the answer was no and i did the ``usual ODE proof'' where one assumes it is and then runs the ODE backwards with initial condition $x_0=0$ and then uses uniqueness of solution to get a contradiction. So I guess my question is, is x(t)=0 the unique solution of $x'(t)=-Ax(t)$, $x(0)=0$ ? At first glance I thought this was completely obvious but I can't seem to prove it.

thanks Craig

I have a somewhat vague question regarding an abstract ODE in a Banach space.
Suppose $A:D(A) \subset X \rightarrow X$ is some linear operator (let's assume it's closed) and maybe add some other conditions.

Suppose $x_0 \in X$ is non-zero and suppose one has the relevant theory to show the existence of a solution of $x'(t)=Ax(t)$ for $t>0$ with $x(0)=x_0$.

So the question that I came up is: can $x(t)$ converge to zero in finite time?

I assumed the answer was no and I did the ``usual ODE proof'' where one assumes it is and then runs the ODE backwards with initial condition $x_0=0$ and then uses uniqueness of solution to get a contradiction. So I guess my question is, is x(t)=0 the unique solution of $x'(t)=-Ax(t)$, $x(0)=0$ ? At first glance I thought this was completely obvious but I can't seem to prove it.