# Can the orbits of a semigroup touch without one being included in the other?

Question. Let $(S(t))_{t \ge 0}$ be a continuous semigroup of linear operators on some Banach space $X$. Might there exist $f, g\in X$ and $0<t_0<t_1$ such that $$S(t_0)f=S(t_1)g$$ but $$S(t_0-\varepsilon_0)f\ne S(t_1-\varepsilon_1)g$$ for all $0<\varepsilon_0\le t_0$ and $0<\varepsilon_1\le t_1$ (in particular, $f\ne g$)?

Pictorially, I am wondering if the following configuration is possible:

Of course we know that, when the evolution is given by a group, this is not possible: orbits either coincide or are disjoint. This is the case of autonomous ODE systems or of the Schrödinger equation. But here we have a semi group, such as the heat one, which only goes forward in time, not backwards. So the only obvious thing that we can say is that, as soon as they touch, orbits merge into one. But in principle I don't see why they should coincide in the past.

Added: After some searches, I have found that for the special case of the heat equation, the answer is negative. This is commonly referred to as backward uniqueness property. Here is a simplified version, which takes into account classical solutions on bounded domains:

Theorem (Taken from Evans's book on PDE, 2nd ed., pag.64) Let $U\subset \mathbb{R}^n$ be an open and bounded domain. Suppose $u, \bar{u}$ are classical solutions of $$\begin{cases} u_t=\Delta u & \text{in }U\times(0, T) \\ u=0 &\text{on }\partial U \times [0, T] \end{cases}$$ If at time $T$ we have $$u(x, T)=\bar{u}(x, T),\quad \forall x\in U,$$ then $u\equiv \bar{u}$ on the whole parabolic cylinder $U\times (0, T]$.

The property holds in much more general functional settings, as I read here (look for the keyword backward uniqueness).

All of this leaves the general question open. Is the backward uniqueness property true for all continuous linear semigroups? I guess that the answer should be negative, otherwise this would not be regarded as a special feature of the heat equation. However, I cannot find an explicit example.

-

## 2 Answers

I believe this is a counterexample:

Consider the semigroup on $L_2[0,\infty)$ given by $$S(t) f(x) = f(x+t)$$ Let $f = I_{[0,1)}$, and $g = 2I_{[0,2)}$. Then $S(1)f = S(2)g = 0$, but $S(1-\epsilon_1)f \ne S(2-\epsilon_2)g$.

-

I think that it is a good question. To simplify, let us study the equation $$u_t=-\Lambda u,$$ where $\Lambda=\sqrt{-\Delta}$. This equation is locally well-posed (both, forward and backward in time) for analytic initial data in a complex strip around the real axis. Then, at some time $T>0$, two solutions with a different initial data (at time $t=0$) can not coincide. Otherwise the backward problem posed at time $T$ would be ill-posed.

I think that for the heat equation the situation is similar.

-
This example is a bit too complicated for me. Could you do something more basic? It's fine to discuss the heat equation only. – Giuseppe Negro May 24 '13 at 13:31
The point of this second equaton is that I know how to solve it forward and backward (because is a first order equation both in space and time you can apply a Cauchy-Kowalevsky Theorem). For the heat equation all this is much more difficult. You have to solve backward a heat equation and you know that your initial data is an entire function. if this problem is well-posed the solution can not touch themselves. – guacho May 24 '13 at 13:50
I agree that it is more difficult. We are lucky that someone already did this for us. See the update to the question. – Giuseppe Negro May 27 '13 at 19:17