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Consider the following evolution equation

$$u_t=\Delta u+u$$ in a bounded and regular open subset $\Omega$ of $\mathbb{R}^N$, with smooth initial conditions $u_0\geq 0$ and homogeneous Dirichlet boundary conditions.

It is known that this equation has a smooth global solution $u$. My goal is to prove that the solution remains nonnegative. So I consider $w=\min(0,u)$ and its energy $E(t):=\int_\Omega w^2 dx$. We know that \begin{align} E(0) &= \int_\Omega w(0,x)^2 dx \\ &=\int_\Omega \min(0,u(0,x))^2 dx \\ &=\int_\Omega \min(0,u_0(x))^2 dx \\ &=0 \end{align}

By differentiating $E(t)$ and using integration parts we get

\begin{align} E'(t) &= 2\int_\Omega ww_t \\ &= 2\int_\Omega wu_t \\ &= 2\int_\Omega w\Delta u+2\int_\Omega wu \\ &= -2\int_\Omega \nabla w \cdot \nabla u+2\int_\Omega w^2 \\ &= -2\int_\Omega |\nabla w|^2+2E(t) \\ &\leq-\frac{2}{c^2}\int_\Omega w^2 dx+2E(t)\\ &\leq\left(2-\frac{2}{c^2}\right)E(t),\quad \text{for almost every} \ t \end{align} where $c$ is the Poincaré constant. Thus $E(t)\leq e^{\left(2-\frac{2}{c^2}\right)t}E(0)=0$ for almost every $t$ which implies that for a.e $t\geq 0$ $w(t,x)=0$ for a.e $x\in \Omega$. But since $w=\min(0,u)$ is continuous then $w(t,x)=0$ for all $t\geq 0$ and for all $x\in \Omega.$ Therefore $u(t,x)\geq 0$ for all $t\geq 0$ and for all $x\in \Omega.$

My concerns are:

1) How can I justify the derivation under integral sign $E'(t)=2\int_\Omega ww_t$ because unlike $u$ which is smooth, $w=\min(0,u)$ has only weak time derivative $w_t=u_t \mathbb{1}_{\{u\leq0\}}.$

2) In the end I proved that for a.e $t\geq 0$, $\int w(t,x)^2dx=0$, thus for a.e $t\geq 0$: $w(t,x)=0$ for a.e $x\in \Omega$. I then concluded by continuity of $w$ that this holds for all $t\geq 0$ and for all $x\in \Omega.$ I am not use but the space negligable sets of $\Omega$ might depend on time $t$. Does this make my argument still valid?

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    $\begingroup$ Can't we apply a comparison principle to show non negativity of the solution? $\endgroup$ – Jonas Lenz Jul 31 '19 at 19:29
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I think it is possible to write your procedure rigurously in the following way. First of all notice that we can re-write $w(t,x)$ as $$ w(t,x)=\dfrac{u-\vert u\vert}{2}. $$ Thus, differentiating we have $$ w_t=\dfrac{1}{2}\big(u_t-\hbox{sgn}(u)u_t\big)=\dfrac{1}{2}(1-\hbox{sgn}(u))u_t. $$ Now, replacing this in $E'(t)$ we obtain $$ E'(t)=2\int w w_t=\int (\Delta u+u)(1-\hbox{sgn}(u))w $$ Finally, integrating by parts we obtain $$ E'(t)=-\int (1-\hbox{sgn}(u))\nabla u\cdot\nabla w+\int (1-\hbox{sgn}(u))uw, $$ where the Dirac's delta dissapears because it is multiply by $w$, which is zero on the set $\{u=0\}$. Finally, noticing that $$ (1-\hbox{sgn}(u))u=2w \quad \hbox{and} \quad (1-\hbox{sgn}(u))\nabla u=2\nabla w, $$ you obtain the result proceeding exactly as you did on your question.

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  • $\begingroup$ Hi, you still used the same derivation under integral sign $E'(t)=2\int w w_t$ using weak derivatives. How can we justify that? $\endgroup$ – David Lingard Aug 1 '19 at 13:49
  • $\begingroup$ There is no problem there, as the proof shows. The important part is that you are inside the integral and that $w$ has a weak-derivative. $\endgroup$ – Sharik Aug 1 '19 at 15:33
  • $\begingroup$ But it's a time weak derivative inside a space integral... $\endgroup$ – David Lingard Aug 1 '19 at 20:08

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