# Why can we apply the strong maximum principle?

I am considering the following section of Peter Grindrods „Pattern and Waves“.

We begin by considering a scalar equation $$u_t=\Delta u+f(u,x,t),\quad x\in\Omega\subseteq\mathbb{R}^n, t>0.$$ Here $f$ is smooth (say, continously differentiable).

Suppose $\overline{u}\colon\Omega\times [0,T]\to B$, some bounded subset of $\mathbb{R}$, and $$\overline{u}_t\geq\Delta\overline{u}+f(\overline{u},x,t);$$ then we say that $\overline{u}$ is a super-solution. If $\underline{u}\colon\Omega\times [0,T]\to B$, and $$\underline{u}_t\leq\Delta\underline{u}+f(\underline{u},x,t),$$ then we say that $\underline{u}$ is a sub-solution.

Now suppose that there exist constants $\alpha,\beta (\alpha^2+\beta^2\neq 0)$ such that $$\alpha\overline{u}-\beta\nabla\overline{u}.n\geq \alpha\underline{u}-\beta\nabla\underline{u}.n,\quad x\in\partial\Omega, t>0,$$ (n is the outer normal to $\partial\Omega$); and that $$\overline{u}(x,0)\geq\underline{u}(x,0), x\in\Omega.$$ Then we claim that $$\overline{u}(x,t)\geq \underline{u}(x,t),\quad x\in\Omega.$$

The author also gives a short "proof".

To see this, set $$w=\overline{u}-\underline{u}.$$ Then the mean value theorem implies $$w_t-\Delta w\geq f_u(\underline{u}+\theta (\overline{u}-\underline{u}))w$$ for some mapping $\theta\colon\Omega\times [0,T]\to [0,1]$. The result follows from the strong maximum principle for linear parabolic equations [15] (The point is that $w$ is initially nonnegative and the boundary conditions ensure that if it becomes negative it must do so at an interior point of $\Omega$. We then can construct a contradiction.)

Unfortunately, I have no opportunity to have a look at the mentioned maximum principle cited from book [15] (which is a book by P.C. Fife).

So we set $w=\overline{u}-\underline{u}$ and note that $$w_t-\Delta w\geq f(\overline{u},x,t)-f(\underline{u},x,t)=f_u(\underline{u}+\theta(\overline{u}-\underline{u})w,x,t)w$$ where the last identity comes from applying the mean value theorem.

Up to here, I can follow the proof.

Now, i write this inequality by using a parabolic linear operator in order to bring it in the usual textbook form for maximum principles:

$$Lw = \sum_{i=1}^n\frac{\partial^2 w}{\partial^2 x_i}-f_u(\underline{u}+\theta(\overline{u}-\underline{u},x,t)w-(-w_t)\geq 0.$$ Now, if $$f_u(\underline{u}+\theta(\overline{u}-\underline{u},x,t)\leq 0,$$ I could use the strong maximum principle for the function $-w$, i.e. making some statement about non-negative maxima of the function $-w$ (resp. about negative minima of the function $w$).

However, I can neither find the assumption $$f_u(\underline{u}+\theta(\overline{u}-\underline{u},x,t)\leq 0$$ nor deduce it from the text.

But let's suppose for a moment that $$f_u(\underline{u}+\theta(\overline{u}-\underline{u},x,t)\leq 0.$$

The proof by contradiction which the author mentions maybe works like this:

Assume we have at least one point $(x_0,t_0), t_0>0$ such that $$\overline{u}(x,t)<\underline{u}(x,t).$$ Then $-w(x_0,t_0)>0$.

By initial- and boundary conditions, the point $(x_0,t_0)$ had to be an inner point of $\Omega\times [0,T]$ since $w<0$ can only happen at inner points.

From this, can we deduce that in the interior of $\Omega\times [0,T]$ we have a negative minimum $M$ (say at some point $(x',t')$ which equals $(x_0,t_0)$ in case this is the only point such that $\overline{u}<\underline{u}$) of $w$ and that - by the strong maximum principle - $$w(x,t)=M$$ for all $(x,t)$ which are on the horizontal line on which $(x',t')$ lies? (in particular for boundary points on this line, which gives a contradiction since there $w$ cannot be negative)?

Thats how I did understand the proof.

• Is $f_u\geq 0$? – Jeff Mar 2 '18 at 22:25
• @Jeff I do not know. At least the text does not say that. – Salamo Mar 2 '18 at 22:37
• Is the PDE $u_t = \Delta u + f(u,x,t)$? (there are some typos in your post). Generally the maximum principle only holds when $f_u \leq 0$ (sorry I wrote wrong way in previous comment). Are there any other assumptions on $f$ in your book (there must be something)? – Jeff Mar 2 '18 at 22:45
• @Jeff Sorry for the typos. Yes, the Pde is $u_t=\Delta u+f(u,x,t)$. - Considering $w_t-\Delta w\geq f_u(...)w$, why does the maximal principle only hold for $f_u\leq 0$? I am a bit confused since the maximum principles are always formulated in the form $Lu\leq g$ or $Lu\geq g$ and I am not sure what are $L$ and $g$ here and whether we need $\leq$ or $\geq 0$. - - I will have a closer look at the book. (Unfortunately, I only have the first 100 pages.) – Salamo Mar 2 '18 at 22:50
• For a linear equation $w_t-\Delta w + cw = 0$, maximum principle requires $c\geq 0$. Here $c=-f_u$. A good reference for the maximum principle for parabolic equations is Evans PDE book Chapter 6. – Jeff Mar 2 '18 at 22:58

There is a trick you can use to adjust the sign of the zeroth order term so that the maximum principle applies. You have a solution $w$ of

$$w_t - Lw \geq cw$$

in a parabolic cylinder, with $w\geq 0$ on sides and base, and you want to show $w\geq 0$ in the interior. Define $v = e^{-\lambda t}w$. Then $v$ satisfies

$$v_t + (\lambda-c)v - Lv \geq 0.$$

Provided $c$ is bounded, you can choose $\lambda$ large enough so that $\lambda -c\geq 0$. Now you can apply the standard strong (or weak) maximum principle to conclude that $v \geq 0$ everywhere, and so $w\geq 0$.

• Am I right that your operator L is $Lw=\Delta w$ and your c is $c=f_u(...)$? And the parabolic cylinder is $\Omega\times [0,T]$? – Salamo Mar 3 '18 at 19:47
• How can we conclude that $v\geq 0$ everywhere? – Salamo Mar 3 '18 at 20:02
• Yes. You use the minimum principle. $v$ attains its minimum on the sides and base, where $v\geq 0$ (since $w\geq 0$ on sides and base). There are similar versions of the maximum principle for your boundary conditions. The whole point is the trick to change variables and introduce a zeroth order term $(\lambda-c)v$ with the right sign. – Jeff Mar 3 '18 at 20:04
• Sorry I did not see yet why Minimum principle. You mean if we assume that w is negative, then w (and v) has a negative minimum M in the interior and hence v also has value M on the sides and the base which is a contradiction since there v is nonnegative? – Salamo Mar 3 '18 at 20:10
• But for minimum principle I need $...\leq 0$ and not $\geq 0$, right? – Salamo Mar 3 '18 at 20:13