# Nonnegativity of solution of $u_t=\Delta u+u$

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?

• Can't we apply a comparison principle to show non negativity of the solution? – Jonas Lenz Jul 31 '19 at 19:29

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.
• 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? – David Lingard Aug 1 '19 at 13:49
• 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. – Sharik Aug 1 '19 at 15:33