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I am studying the principle of comparison for heat equations and I am having some problems understanding the proof.

The theorem is:

Let $\Omega\subset\mathbb{R}^n$ be open and bounded, $0<T<\infty$ and let $u,v\in{C}(\overline\Omega\times[0,T))$ solve $u_t-\Delta{u}=0$ and $v_t-\Delta{v}=0$ respectively in $\Omega_T:=\Omega\times(0,T)$. Then if $u\leq{v}$ on $\partial\Omega\times(0,T)\cup\Omega\times\{t=0\}=:\partial_p\Omega_T$, $u\leq{v}$ on $\overline\Omega\times[0,T]$.

The Proof first defines $w=u-v$, $w^+=(u-v)^+=max\{0, u-v\}$ and considers the integral of the product of $w^+$ and $w_t-\Delta{w}=0$ on $\Omega_S$, for $0<S<T$. It is stated that $w_tw^+=\frac{\partial}{\partial{t}}\cdot\frac{1}{2}(w^+)^2$ and that $\nabla{w}\nabla{w^+}=|\nabla(w^+)|^2$, nearly everywhere. Here I don't understand how we get this results.

Then the proof moves on to the integral 0=$\int_{0}^{S}\int_{\Omega}(w_tw^+-\Delta{w}w^+)=\int_{0}^{S}\int_{\Omega}\frac{\partial}{\partial{t}}\frac{(w^+)^2}{2}+|\nabla(w^+)^2|dxdt-\int_{0}^{S}\int_{\partial\Omega}\frac{\partial{w}}{\partial{n}}w^+dSdt$, since I didn't understand the step before, I also don't understand this one now. I am also not quite sure how to get $\int_{0}^{S}\int_{\partial\Omega}\frac{\partial{w}}{\partial{n}}w^+dSdt$.

Lastly, $\int_{0}^{S}\int_{\Omega}\frac{\partial}{\partial{t}}\frac{(w^+)^2}{2}+|\nabla(w^+)^2|dxdt-\int_{0}^{S}\int_{\partial\Omega}\frac{\partial{w}}{\partial{n}}w^+dSdt\geq\frac{1}{2}\int_{\Omega}(w^+(S))^2dx-\frac{1}{2}\int_{\Omega}(w^+(0=\frac{1}{2}\int_{\Omega}(w^+(S))^2dx))^2dx$, this follows because $w^+=0$ on $\partial\Omega\times(0,T)$ and also on $\Omega\times\{t=0\}$. This means that $w^+=0$ on $\Omega\times\{t=S\}$, for all $0<S<T$.Therefore $u\leq{v}$ on $\Omega_T$.

This last step I understand, my problem is the first two. Any help would be greatly appreciated.

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1 Answer 1

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I agree that this "proof " looks fishy. A simpler alternative classical proof is outlined below.

Definition. A strict sub-solution of the heat equation on the closed "tin-can" $R=\overline{\Omega}\times[0,T]$ is a smooth function $u$ that satisfies $u_t-\triangle u<0$ on the interior of $R$.

That is, $u$ satisfies the inhomogeneous equation $u_t= \triangle u -f$ where $f>0$. In physical terms, $f$ represents the internal rate in $R$ at which heat is being withdrawn from the system. Intuitively we expect this energy loss $f$ forces the system to "wind down" as time progresses. Here is a precise statement that is in accord with that intuition.

Lemma (Weak Maximum Principle). A strict sub-solution has no local maximum in the interior of $R$.

Proof of lemma. At any hypothetical interior max, the critical point condition would force $u_t=0$, which would force $\triangle u>0$, which violates the Second Derivative Test for a local max (stating that at a local max no second-order directional derivative can be positive.)

The lemma can easily be strengthened to state that the maximum cannot occur on the interior of the top face $\Omega \times \{T\}$. (Essentially the same proof works, since in this case at the endpoint time, the occurrence of a maximum requires $u_t\geq 0$.)

Proof of Comparison Principle.

To prove the Comparison Principle, you want to show that the function $w= u-v$ that is negative on the bottom and sides of the can stays negative in the interior of the can and on the top lid. If we knew that $w$ were actually a strict sub-solution of the heat equation, the lemma would give us the desired conclusion immediately.

The final technical detail in the proof is showing that you can perturb our given $w$ arbitrarily slightly to make it a strict sub-solution that is still negative on the bounding bottom and sides. That can be done by constructing practically any convenient explicit perturbing sub-solution that is small on these bounding edges, and adding it to $w$.

Example: Perturb $w$ by adding a very small multiple of $ h=||x||^2 e^{-Mt}$. Note that $h$ has the property that $h_t -\triangle h= -Mh - Ce^{-Mt}$ is negative, and any sufficiently small multiple of it will not change the boundary values of $w$ significantly.

P.S.The ideas involved in the proof of the Maximum Principle are useful because they generalize easily to apply to many nonlinear equations too, such as $u_t - \sqrt{1+u_x^2 + u_y^2} (\triangle u)=0$.

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  • $\begingroup$ Thank you, this prove is easier to understand. The one I posted is from my lecturers notes. $\endgroup$
    – Jaime02
    Apr 11, 2023 at 16:03

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