# Strong maximum principle for heat equation. Positivity of solution

I have a non-negative solution $u \in L^2(0,T;H^1) \cap H^1(0,T;(H^1)')$ of the heat equation $$u_t-\Delta u =0$$ on bounded $C^1$ domain $\Omega$, with the boundary condition $$\frac{\partial u(t,x)}{\partial \nu(x)} = a(t,x) - b(t,x)u(t,x)$$ on $\partial\Omega$, with initial data $u(0) = u_0$.

I have that $u_0 >0$ a.e. and also bounded from abve. Furthermore, $a$ and $b$ are bounded from above a.e. and $a, b \geq 0$ a.e.

Can I deduce that $u$ is actually strictly positive? I.e. $u > 0$ a.e.?? Another problem i have is I am not sure that solutions are in in $C^2$ (which seems to be necessary to apply the maximum principles).

I tried searching for strong maximum principles but I couldn't see one that I could apply for this problem.

• Are you willing to assume that $a$ is differentiable wrt time? – Delio Mugnolo Apr 23 '16 at 23:20
• @DelioMugnolo Sure if it helps. – ChristopherSail Apr 24 '16 at 7:37
• One way to get this (possibly overkill) is from the parabolic Harnack inequality. – Nate Eldredge Apr 25 '16 at 23:32
• @NateEldredge The problem is that I need $u \in C^{1,2}((0,T)\times \Omega)$ to apply max principle/Harnack which I don't see how to obtain. – ChristopherSail Apr 26 '16 at 16:10

## 1 Answer

Your conjecture is correct, but the answer is going to be somewhat technical.

First of all, we are going to make the Ansatz that the solution is actually smoother wrt time (this will be justified a posteriori), so that the boundary condition can be differentiated wrt to time to yield the dynamic b.c. $$\frac{\partial Lu}{\partial t}(t,x)=a(t,x)$$ where the boundary operator $L$ is defined by $$Lu(t,x):=\frac{\partial u}{\partial \nu}(t,x)+b(t,x)u(t,x).$$ In this way, you can turn your problem into an abstract Cauchy problem $$\frac{d\mathcal u}{d t}(t)={\mathcal A}{\mathcal u}(t)+\Phi(t)$$ where $\mathcal A$ is the operator matrix $$\mathcal A:=\begin{pmatrix}\Delta & 0\\ 0 & 0\end{pmatrix}$$ with domain $$D(\mathcal A):=\left\{\begin{pmatrix}u\\ Lu\end{pmatrix}:u\in H^\frac{3}{2}(\Omega):\Delta u\in L^2(\Omega) \right\}$$ and the inhomogeneous term is $$\Phi(t):=\begin{pmatrix}0\\ a(t,\cdot)\end{pmatrix}$$ Now, you are in the setting of the Klaus-Jochen Engel's theory of one-sided coupled operator matrices and you can apply a result - essentially due to Engel, but you can find it in a ready-to-use-version in an old paper by Kramar and myself (Proposition 8.10 in "Theory and applications of one-sided coupled operator matrices", 2003) that claims that positivity (and in fact irreducibility) of the semigroup $e^{t\mathcal A}$ generated by $\mathcal A$ follows from positivity (and in fact irreducibility) of the semigroup generated by $\Delta$ with homogeneous Robin boundary conditions plus the strong maximum principle that holds for the Helmholtz equation with homogeneous Robin boundary conditions. (If you are not very familiar with the theory of operator semigroups: positivity and irreducibility of the semigroup generated by $\mathcal A$ correspond to an affirmative answer to your question in the case $a\equiv 0$.) (Everything so far is precisely true in the one-dimensional case: in the $n$-dimensional case you need a slight generalization of Engel's theory presented in a more recent paper by myself, "Asymptotics of semigroups generated by operator matrices", 2014.)

Now, validity of your conjecture simply follows from the Duhamel formula that tells you that $${\mathcal u}(t)=e^{t\mathcal A}{\mathcal u}_0+\int_0^t e^{(t-s)\mathcal A}\Phi(s)\ ds\ .$$

• Thank you. I will need time to take a look at the papers you cited but I think you need to change $a$ to $a'$ in your displayed equations. – ChristopherSail Apr 24 '16 at 20:44
• Well, $e^{t\mathcal A}$ is analytic, therefore so is $u$ wrt time; but when it comes to its space regularity, the solution is a priori only in the domain of $\Delta^k$ ($\Delta$ being the distributional Laplacian without b.c.), so one cannot apply any boundary regularity result. I am pretty sure $u$ is $C^\infty$ wrt space on compact subsets of $\Omega$, but I cannot go further. – Delio Mugnolo Apr 25 '16 at 22:57
• @S.Cho Yes, this is known. As long as $p$ is $L^\infty$, an easy way to see this is to invoke Theorem 3.6 in ejde.math.txstate.edu/Volumes/2006/118/mugnolo.pdf (formally speaking, you have to work with a slightly different sesquilinear form, where you have added the term $\int_\Omega p u\bar{v} dx$, but it's very easy to see that the proof of Theorem 3.6 goes through.) – Delio Mugnolo Dec 5 '18 at 7:05
• @S.Cho I haven't checked the details, but I'm almost sure a perturbation argument should do. Basically, you consider the operator matrix featuring the terms $p$ and $q$: this is a bounded, positive operator, so by the Dyson-Phillips series representation of the perturbed semigroup you get positivity again. – Delio Mugnolo Dec 5 '18 at 20:31
• Are $p>0$ and $q>0$? Ok, nevermind, then just apply Trotter-Kato. The semigroups generate by the operator matrix featuring $p$, $q$ is positive. But again, this is unnecessarily complicated. It's much easier to work with forms. – Delio Mugnolo Dec 6 '18 at 9:02