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This is Lemma 5 in the chapter on maximum principles in Friedman's book Partial Differential Equations of Parabolic Type. I am having trouble understanding one of the steps in the proof.

Let $$ \Omega_{0}\equiv\mathbb{R}^{n}\times\left(0,T\right]\text{ and }\Omega\equiv\mathbb{R}^{n}\times\left[0,T\right]. $$ Further let $$ L\equiv\sum_{i,j=1}^{n}a_{ij}\left(x,t\right)\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}+\sum_{i=1}^{n}b_{i}\left(x,t\right)\frac{\partial}{\partial x_{i}}+c\left(x,t\right)-\frac{\partial}{\partial t} $$ be a parabolic operator with continuous coefficients in $\Omega_{0}$. If $c\leq0$ and $Lu\leq0$ in $\Omega_{0}$, $u\left(x,0\right)\geq0$ and $$ \liminf_{\left|x\right|\rightarrow\infty}u\left(x,t\right)\geq0 $$ uniformly w.r.t. $t$ ($0\leq t\leq T$) then $u\left(x,t\right)\geq0$ in $\Omega$.

Proof: For any $\epsilon>0$, we have $u\left(x,0\right)\geq0$ on $t=0$ and hence $u\left(x,0\right)+\epsilon>0$. Similarly, $u\left(x,t\right)+\epsilon>0$ for $\left|x\right|=R$, $0\leq t\leq T$ provided $R$ is sufficiently large (from the $\liminf$ assumption). Since $$ L\left(u+\epsilon\right)=\left[\sum_{i,j=1}^{n}a_{ij}\left(x,t\right)\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}+\sum_{i=1}^{n}b_{i}\left(x,t\right)\frac{\partial}{\partial x_{i}}+c\left(x,t\right)-\frac{\partial}{\partial t}\right]\left(u+\epsilon\right)\leq c\epsilon\leq0, $$ $u\left(x,t\right)+\epsilon>0$ for $\left|x\right|\leq R$, $0\leq t\leq T$ (this is the part I don't understand). Taking $\left(x,t\right)$ fixed and letting $\epsilon\rightarrow0$, it follows that $u\left(x,t\right)\geq0$.

parsiad
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1 Answers1

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The author used the maximum principle for parabolic pde (see Evans Chapter Theorem 9 for example).

Denote $\Omega_T = \{|x|<R\}\times (0,T]$, and $\omega_T = \{|x|\leq R\}\times \{T\}$, then if $L (u+\epsilon) \leq 0$ (Theorem 9 case 2) in $\Omega_T$, we have: $$ \inf_{\Omega_T} (u+\epsilon) = \min_{\bar{\Omega}_T} (u+\epsilon) \geq -\max_{\partial \Omega_T\backslash \omega_T} (u+\epsilon)^- =0. $$ For $(u+\epsilon)^- = 0$ on $\partial \Omega_T\backslash\omega_T$ because $u+\epsilon> 0$ for fixed $0<t<T$. Therefore by above, minimum is attained on the boundary, and we have $$ \inf_{\Omega_T} (u+\epsilon) \geq 0\implies u+\epsilon >0 \text{ in }\Omega_T. $$ For $u+\epsilon \in C^{2,1}(\Omega_T)\cap C (\overline{\Omega}_T)$, you can resolve the case when $t=T$, however I think that the inequality should be weakened to $\geq 0$ at the $t= T$.

Shuhao Cao
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