Abstract ODE; PDE; uniqueness of solution

Question

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.

Answer 1

No, this is not true. There is no backward uniqueness in general.

What you need is the theory of operator semigroups, and here is a simple example.

Consider the operator $Af=f'$ in the space $X=L^2[0,1]$ with domain $$D(A)=\{f\in H^1[0,1]\,,\, f(1)=0\}.$$

Then the solution to the initial value problem is given by $x(t)=T(t)x_0$ with $$T(t)f(s):=\begin{cases} f(t+s), t+s\leq 1,\\ 0, \quad t+s>1.\end{cases}$$

Clearly, $T(t)f=0$ for all $t>1$, but the initial value problem is well-posed in the classical sense ($A$ closed, classical solutions from a dense set of initial values, continuous dependence on initial values).

ADDED: After clarifications on the question, let me add the following: If $A$ is sectorial and hence generates an analytic semigroup, then finite time extinction of soultions cannot happen. Other conditions are difficult to formulate. If you say more on your operator, it can help...

Answer 2

It is true if $A$ is assumed bounded (as can be found in Section 2 of the reference given by Andras Batkai). An "elementary proof" consists in proving the fundamental theorem of calculus, that is for any $C^1$ map $x:I\to X$ from an open interval containing $0$ you have $$\left(\forall t\in I\right)\,\,\,\,x(t)-x(0)=\int_0^t\dot x(s)\mathrm ds.$$ Once you have this property you are done by the usual construction of the exponential $R(t):=\exp(tA)$ of $A$. It even works in the "non-constant" case using the resolvent, which can be built verbatim for bounded operators as in the finite-dimensional case by setting $$ R(t) := \sum_{n=0}^{+\infty} U_n(t)$$ where $$ U_0 := \mathrm{Id}$$ and $$U_{n+1}(t):=\int_0^t A(s)U_n(s)\mathrm ds.$$ For small values of $t$ the operator $R(t)$ is invertible and $R(-t)x(t)$ has a vanishing derivative.

In order to prove the FTC you assume that $\dot x=0$ and prove by a connectedness argument that for all $\varepsilon>0$ the property $$\left(\forall 0\leq s\leq t\right)\,\,\,\,\,\,\|x\left(s\right)-x\left(0\right)\|\leq\varepsilon |s| $$ is true for all $t\in I$, then take the limit $\varepsilon\to0$, so that $x$ must be constant.