# Abstract ODE; PDE; uniqueness of solution

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.

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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...

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thanks you very much for the counterexample (and sorry for taking so long to get back to you). I got caught up in grading finals and then forgot about my question. So my $A$ is nice and the question i am interested in is the decay to zero in finite time. – Craig May 11 '14 at 7:45
@Craig Well, in my opinion this is a nice operator (simply the first derivative), and you reach zero in finite time. – András Bátkai May 11 '14 at 9:11
Good point. Let me rephrase the question. What conditions on $A$ would not allow this to happen? – Craig May 11 '14 at 14:01
Thanks for the update Andras. So this problem was arising in the context of someone talking to me about Lumer–Phillips (or maybe it was Hille-Yosida) theory. I had in my mind various PDE's which I knew wouldn't have the finite time extinction and I figured that would also be the case for general $A$ (which you showed it wasn't). So I guess I am saying lets assume anything that one generally assumes when applying the above mentioned theories. Is this enough to rule out the finite extinction? (I am being vary vague since I don't really know the area at all). – Craig May 12 '14 at 2:32
@Craig: This is a very nice dissipative operator fitting all the theories and corresponding to an extremly simple hyperbolic PDE. However, if you have a parabolic PDE (like the heat equation) or a reversible PDE (like the undamped wave equation or the linear Schrödinger equation), then this cannot happen. – András Bátkai May 12 '14 at 8:06

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.

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En anglais, on dit "the resolvent operator" ou "the resolvent" pour la résolvante. Dans la théorie du contrôle, on utilise aussi "state transition matrix" pour la même chose. – Willie Wong Apr 24 '14 at 8:45
@WillieWong Merci, j'ai corrigé. Je n'ai pas osé utiliser la traduction "évidente", ça ne marche pas tout le temps ;) – Loïc Teyssier Apr 24 '14 at 8:50
thank you very much for the results and sorry about the delay. – Craig May 11 '14 at 7:45