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?

  • $\begingroup$ Have you answerd this question? $\endgroup$
    – S. Maths
    Commented May 15, 2020 at 1:57
  • $\begingroup$ No, not yet. I found a way how to solve this problem, but it's not so nice and elegant. Do you know how to solve it? $\endgroup$
    – Oleg
    Commented May 16, 2020 at 19:51
  • $\begingroup$ I have an idea but not checked it yet. What way you meant? $\endgroup$
    – S. Maths
    Commented May 17, 2020 at 4:15
  • $\begingroup$ There are so-called Carleman inequalities but they are usually formulated for unbounded domains. One can try to obtain a Carleman inequality for a bounded domain, which would then yield the desired result. But the details are not completely clear to me, so if there is an alternative easier method, I would be very happy to hear about it. $\endgroup$
    – Oleg
    Commented May 17, 2020 at 13:25

1 Answer 1


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).

  • $\begingroup$ thank you very much for your solution! let me have a look at it and let me have a look at the book $\endgroup$
    – Oleg
    Commented May 19, 2020 at 18:09

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