Consider heat equation with a drift (=reaction-diffusion equation)
$$
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+f(t,u(t,x)), \quad t\ge0,\, x\in [0,1]
$$
with periodic or Dirichlet boundary conditions. Here $f$ is globally bounded and Lipschitz in the second argument. Is it true that if $u$,$v$ are two solutions of this PDE with $u(1)=v(1)$, then $u(0)=v(0)$? How can we prove this?


The case $f=0$ is classical; it can be found, e.g., in Evans book. There, the proof goes by showing that the second derivative of log energy is nonnegative. However, it is not clear to me at all how this technique can be adapted here, since $f$ might have no time derivatives.


An equivalent formulation of the above problem is the following: suppose we are given a bounded function $w\colon [0,1]\times \mathbb{R}\_+\to\mathbb{R}$ with Dirichlet or periodic boundary conditions. Suppose that $w(0)\equiv 0$ and
$$
\Bigl|\frac{\partial w}{\partial t}+\frac{\partial^2 w}{\partial x^2}\Bigr|\le C|w|
$$
everywhere. Is it true that $w(t)\equiv0$ for all $t\ge0$?


UPD: I was told that one can use an appropriate Carleman bound. Do you know whether there are any Carleman estimates suitable for this problem?



Backward uniqueness for a heat equation with a drift
The result follows by an extension of the method of **logarithmic convexity** which is well-known for the heat backward problem.


Let $H$ be a Hilbert space. Consider the following inequality
\begin{equation}
\|\partial\_t u + Au\| \leq \alpha\|u\|, \qquad \text{ on } (0,T), \qquad (1)
\end{equation}
with $\alpha=\mathrm{const}>0$.


We assume that $$A=A\_+ + A\_-, \qquad (2)$$
where $A\_+$ is a linear symmetric operator on $H$ with domain $D(A)$ and $A\_-$ is skew-symmetric such that:
\begin{align}
\|A\_- u\|^2 &\leq c(\|A\_+ u\| \|u\|+ \|u\|^2), \qquad \qquad \qquad \qquad (3)\\
\partial\_t \langle A\_+ u, u\rangle &\leq 2\langle A\_+ u, \partial\_t u\rangle + c(\|A\_+ u\|\|u\|+\|u\|^2) \,\qquad (4).
\end{align}



> 
> **Theorem.** Let $u(t)\in D(A)$, $u\in C^1([0,T];H)$ be a solution of the differential inequality $(1)$, where $A$ satisfies the conditions $(2)$-$(4)$. Then
>  \begin{equation}
> \|u(t)\| \leq C\_1 \|u\_0\|^{1- \lambda(t)} \|u(T)\|^{\lambda(t)}, \qquad 0\leq t\leq T, \qquad (\*)
> \end{equation}
>  for a constant $C\_1=C\_1(\alpha,T)>0$ and $\lambda(t)=\dfrac{1-e^{-Ct}}{1-e^{-CT}}$ with $C$ depending on $\alpha$.
> 
> 
> 


In your case, $A=A\_+=\Delta$,$\; A\_-=0$ and it is easy to check that conditions $(1)$-$(4)$ hold. From $(\*)$ it is clear that if $w(0)=0$ then $w(t) \equiv 0$ for all $t\ge 0$, which answers your question (second formulation).


This is Theorem 3.1.3 pp 51 in



> 
> V. Isakov, Inverse Problems for Partial Differential Equations. Springer, (2017).
> 
> 
> 


It also applies to more general elliptic operators, (see Example 3.1.6 in the same reference).


