Can two solutions of a PDE with different initial conditions coincide after finite time? Consider an ODE
$$
\dot x=f(x),\quad x(0)=x_0,
$$
where $f$ is a smooth function. It is well-known that if $y$ is another solution to this ODE with different initial condition $y_0\neq x_0$, then the trajectories of $x(t)$ and $y(t)$ do not intersect - in particular, for any finite $t$ we have $x(t)\neq y(t)$.

I wonder if the same is true for PDEs. I'm interested in the following simple example:
$$
\partial_t u=\partial_{xx} u+f(t,(u(t,x)),
$$
with periodic boundary conditions. Here $x$ lives on a circle $[0,1]$, $t\ge0$ and $f$ is a nice function. Is it true that if $u,v$ are two solutions of this PDE with $u(0)\neq v(0)$, then $u(t)\neq v(t)$ for any $t$? (equivalent statement: if $u(1)=v(1)$ is it true that $u(0)=v(0)$?) How can we prove this?

UPD: the case $f\equiv0$ can be found inn Evans' book and it involves taking second derivative of the energy. Is it possible to extend this technique to the case $f\neq0$?
Tychonoff gives an infinite family of solutions of the heat equation on $(x,t) \in \Bbb{R} \times \Bbb{R}$, all of which are $0$ everywhere at time $t \leq 0$ and different subsequently.  These solutions are also discussed on MSE.

Such solutions are backwards from what you request -- they are nonunique at a prior time, and then subsequently become unique. However, by the choice $f(t, u(t,x)) = 2\partial_t u(t,x)$, time reversal is implemented.  Tychonoff's solutions are then distinct on $t &lt;0$ and identically zero on $t \geq 0$.  It would be no great feat to adapt these to $(x,t) \in S^1 \times \Bbb{R}$.
Tychonoff gives an infinite family of solutions of the heat equation on $(x,t) \in \Bbb{R} \times \Bbb{R}$, all of which are $0$ everywhere at time $t \leq 0$ and different subsequently.  These solutions are also discussed on MSE.

Such solutions are backwards from what you request -- they are nonunique at a prior time, and then subsequently become unique. However, by the choice $f(t, u(t,x)) = 2\partial_t u(t,x)$, time reversal is implemented.  Tychonoff's solutions are then distinct on $t &lt;0$ and identically zero on $t \geq 0$.  It would be no great feat to adapt these to $(x,t) \in S^1 \times \Bbb{R}$.