# Ngô Quốc Anh

## March 12, 2013

### The generalized maximum principle

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 9:09

Let us start our notes with a very fundamental maximum principle for any strongly second order elliptic operator. We have

Theorem (Maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle L[u] = \sum\limits_{i,j = 1}^n {{a_{ij}}(x)\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}}} + \sum\limits_{k = 1}^n {{b_k}(x)\frac{{\partial u}}{{\partial {x_k}}}} \geqslant 0$

in a domain $D$ where $L$ is uniformly elliptic. Suppose the coefficients $a_{ij}$ and $b_k$ are uniformly bounded. If $u$ attains a maximum $M$ at a point of $D$, then $u = M$ in $D$.

In order to memorize the above result, let us think about the parabola $y=x^2$ with $x\in[-1,1]$ and $L=\Delta$. In this one-dimentional case, $L[y]=2 \geqslant 0$ which confirms that $y$ only achieves its maximum at $x=\pm 1$.

As you may know the operator $L$ is only assumed to be strongly elliptic which only effects the coefficients $a_{ij}$. Regarding to the coefficients $b_k$, we only assume these are uniformly bounded. However, the uniform ellipticity of the operator L and the boundedness of the coefficients are not really essential as you can check in the proof. Besides, the domain $D$ need not be bounded in this version.

Now for operators of the form $(L + h)$, we still have a result analogous to the above.

Theorem (Maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle (L + h)[u] >0$

with $h <0$, with $L$ uniformly elliptic in $D$, and with the coefficients of $L$ and $h$ bounded. If $u$ attains a non-negative maximum $M$ at an interior point of $D$, then $u = M$.

Clearly, the assumption $h<0$ and $M \geqslant 0$ are crucial as we may face some difficulty as raised in this note. Counterexamples are easily obtained if $h > O$. For example, the function $u = \exp(-r^2)$ has an absolute maximum at $r= 0$.

Now we turn to the so-called strong maximum principle, or the Hopf maximum principle. We know from the maximum principle that the maximum point only occurs on the boundary. The following tells us further the behavior of the function at that maximum point.

Theorem (Hopf maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle L[u] = \sum\limits_{i,j = 1}^n {{a_{ij}}(x)\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}}} + \sum\limits_{k = 1}^n {{b_k}(x)\frac{{\partial u}}{{\partial {x_k}}}} \geqslant 0$

in a domain $D$ in which $L$ is uniformly elliptic. Suppose that $u < M$ in $D$ and that $u= M$ at a boundary point $P$. Assume that $P$ lies on the boundary of a ball $K_1$ in $D$. If $u$ is continuous in $D \cup P$ and an outward directional derivative $\frac{\partial u}{\partial \nu}$ exists at $P$, then

$\displaystyle\frac{\partial u}{\partial \nu}>0$

at $P$ unless $u \equiv M$.

The above result also works for the operator $L + h$ with $h \leqslant 0$. We have

Theorem (Hopf maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle (L+h)[u] \geqslant 0$

in a domain $D$ in which $L$ is uniformly elliptic and $h(x) \leqslant 0$. Suppose that $u < M$ in $D$ and that $u= M$ at a boundary point $P$, and that $M \geqslant 0$. Assume that $P$ lies on the boundary of a ball $K_1$ in $D$. If $u$ is continuous in $D \cup P$ and an outward directional derivative $\frac{\partial u}{\partial \nu}$ exists at $P$, then

$\displaystyle\frac{\partial u}{\partial \nu}>0$

at $P$ unless $u \equiv M$.

We are now in a position to talk about the generalized maximum principle which is the main point of this notes.

Keep in mind that from now on we do not assume that $h$ is nonpositive. Let $w(x)$ be a given positive function on $D \cup \partial D$ and define

$\displaystyle v(x)=\frac{u(x)}{w(x)}.$

A computation shows that

$\begin{array} {lcl}\dfrac{1}{w}(L + h)[u] &=& \displaystyle\frac{1}{w}\left[ {\sum\limits_{i,j = 1}^n {{a_{ij}}(x)\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}}} + \sum\limits_{k = 1}^n {{b_k}(x)\frac{{\partial u}}{{\partial {x_k}}}} + hu} \right] \hfill \\ &=& \displaystyle\sum\limits_{i,j = 1}^n {{a_{ij}}(x)\frac{{{\partial ^2}v}}{{\partial {x_i}\partial {x_j}}}} + \sum\limits_{k = 1}^n {\left( {\frac{2}{w}\sum\limits_{l = 1}^n {{a_{kl}}} (x)\frac{{\partial w}}{{\partial {x_l}}} + {b_k}(x)} \right)\frac{{\partial v}}{{\partial {x_k}}}} + \frac{1}{w}(L + h)[w]v \hfill \\ &\geqslant& 0 \end{array}$

in $D$. Then, if $w$ satisfies the inequality

$\displaystyle (L + h)[w] <0$

in $D$, we may apply the maximum principle above to the function $v$ to obtain the following result

Theorem (Generalized maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle (L+h)[u] \geqslant 0$

in a domain $D$ in which $L$ is uniformly elliptic. If there exists a function $w(x)$ such that

$\begin{array} {rcl} w(x) &>& 0 \quad \text{ on } D \cup \partial D, \hfill \\ (L + h)[w] &\leqslant& 0 \quad \text{ in }D, \end{array}$

then $\frac{u(x)}{w(x)}$ cannot attain a non-negative maximum in $D$ unless it is a constant. If $\frac{u(x)}{w(x)}$ attains its non-negative maximum at a point $P$ on $\partial D$ which lies on the boundary of a ball in $D$ and if $\frac{u}{w}$ is not constant, then

$\displaystyle\frac{\partial}{\partial \nu}\left(\frac{u}{w}\right)>0$

at $P$ where $\frac{\partial}{\partial\nu}$ is any outward directional derivative.

As an application of the generalized maximum principle, we can prove the following non-existence for some Yamabe-type equations.

Theorem. Suppose that $M$ is a complete, non-compact Riemannian manifold of dimension $n$. Let $a, b \in C^0(M)$ satisfy

$b \geqslant 0 \quad \text{ on } M$

and

$\lambda_1(\Delta+a)>0 \quad \text{ on } {\rm supp}(a^+).$

Then the differential equation

$\Delta u +au=bu^\sigma, \quad \sigma>1$

has no nontrivial non-negative ground states, i.e., $u$ is entire and $\lim_{|x| \to +\infty} u(x) =0$

See also:

Source: Murray H. Protter, Hans F. Weinberger, Maximum Principles in Differential Equations, Springer, 1984

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