Week 11: Maximum principles and the eigenvalue problem

We now turn our attention to classical solutions to second-order elliptic PDE and employ more classical techniques to glean information on their behaviour. Throughout this lecture, we shall suppose $\Omega\subset\mathbb{R}^{n}$ is open and bounded.

Write $L$ for the non-divergence form operator

where $a_{ij}$, $b^{i}$ and $c$ are continuous functions on $\overline\Omega$, $(a_{ij}(x))_{i,j=1}^{n}$ is a symmetric matrix for each $x\in\Omega$ and for some constant $\lambda_{0}>0$ the bound

holds for all $x\in\Omega$ and $v\in\mathbb{R}^{n}$.

If a function $u\in C^{2}(\Omega)$ satisfies the inequality $Lu\leq 0$ on $\Omega$, we call $u$ a subsolution to the equation $Lu=0$, and if $Lu\geq 0$, we say that $u$ is a supersolution.

The weak maximum principle

It turns out that in the absence of the zeroth order term of $L$, the pointwise behaviour of subsolutions and supersolutions are in a sense controlled in terms of their boundary data. We shall require some auxiliary results from calculus and linear algebra to establish this.

Lemma 11.1. If $u\in C^{2}(\Omega)$ attains its maximum in $\Omega$, i.e. if there is an $x_{0}\in\Omega$ such that $\sup_{\Omega}u = u(x_{0})$, then $\D u(x_{0})=0$ and the Hessian $\D^{2}u(x_{0})$ is nonpositive definite.

Proof. Fix $v\in\mathbb{R}^{n}$ and consider $t\mapsto u(x_{0}+tv)$ for $t$ sufficiently close to $0$. By classical calculus results, we must have that $\sum_{i=1}^{n}v^{i}\partial_{i}u(x_{0})=\left.\frac{d}{d t}\right|_{t=0}u(x_{0}+tv)=0$ and $\sum_{i,j=1}^{n}v^{i}\partial_{i}\partial_{j}u(x_{0})v^{j}=\left.\frac{d^{2}}{dt^{2}}\right|_{t=0}u(x_{0}+tv)=0$. Since $v$ is arbitrary, this establishes the claim.

Lemma 11.2. Suppose $A$ and $B$ are $n\times n$ non-negative definite matrices and $A$ is symmetric. Then

Proof. By classical linear algebra, there exists an orthonormal basis $\{v_{i}\}_{i=1}^{n}$ for $\mathbb{R}^{n}$ and non-negative numbers $\{\lambda_{i}\}_{i=1}^{n}$ (not all distinct) such that $Av_{i}=\lambda_{i}v_{i}$. By definition of the trace,

$\textup{tr}(A^{T}B) = \sum_{i=1}^{n}v_{i}^{T}A^{T}Bv_{i} = \sum_{i=1}^{n}\lambda_{i}v_{i}^{T}Bv_{i}\geq 0.$

Theorem 11.3 (weak maximum principle). Suppose $u\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$. If $u$ is a subsolution to the equation $Lu=0$ with $c\equiv 0$, then it must attain its maximum on $\partial\Omega$, i.e.

If instead $u$ is a supersolution to the equation $Lu=0$ with $c\equiv 0$, then it must attention its minimum on $\partial\Omega$, i.e.

Proof. First suppose that $u$ is a strict subsolution in the sense that $Lu<0$ on $\Omega$ and assume there exists an $x_{0}\in\Omega$ such that $\max_{\Omega}u=u(x_{0})$. By the preceding lemmas, we compute that

$Lu(x_{0})=\sum_{i,j=1}^{n}a_{ij}(x_{0})(-\partial_{i}\partial_{j}u) + \sum_{i=1}^{n}b^{i}(x_{0})\partial_{i}u(x_{0})\geq 0,$

which contradicts the assumption that $Lu<0$ on $\Omega$ so that we must have $x_{0}\in\partial\Omega$. More generally, suppose $Lu\leq 0$ and define $v:\Omega\rightarrow\mathbb{R}$ by $v(x)=u(x) + \eps e^{kx^{1}}$ for fixed $\eps>0$ and $k\in\mathbb{R}$ to be determined. We compute that

$Lv(x) = Lu(x) + \eps e^{kx^{1}}\left(- k^{2}a_{11}(x) + k b^{1}(x) \right)\leq \eps e^{kx^{1}}\left(-k^{2}\lambda_{0} + k\sup b^{1} \right)<0$

provided $k>\max\{0,\frac{b^{1}}{\lambda_{0}}\}$. Therefore, we see that for such $k$,

$\max_{x\in\overline\Omega}u(x) \leq \max_{x\in\overline\Omega}\left(u(x) + \eps e^{kx^{1}}\right)\leq \max_{x\in\partial\Omega}\left(u(x) + \eps e^{kx^{1}} \right)\leq \max_{\partial\Omega} u + \eps \max_{x\in\partial\Omega}e^{kx^{1}}\xrightarrow{\eps\searrow 0} \max_{\partial\Omega}u.$

On the other hand, we clearly have $\max_{\overline{\Omega}}u \geq \max_{\partial\Omega} u$. This establishes the claim for subsolutions. If $u$ is a supersolution, then clearly $-u$ is a subsolution so that

$\min_{\overline{\Omega}}u = -\max_{\overline{\Omega}}(-u) = -\max_{\partial\Omega}(-u) = \min_{\partial\Omega}u.$

Remark 11.4. We deduce that if $Lu=0$ on $\Omega$, then $u$ attains its minimum and maximum on $\partial\Omega$.

In the case where $c\geq 0$, we can apply the weak maximum principle to extract the following maximum principle.

Corollary 11.5. Suppose $u\in C^{2}(\Omega)\cap C^{0}(\overline{\Omega})$. If $u$ is a subsolution to the equation $Lu=0$ with $c\geq 0$, then we have the inequality

where $u^{+}(x):=\max\{u(x),0)\}$. If instead $u$ is a supersolution to the equation $Lu=0$ with $c\geq 0$, then it satisfies the inequality

where $u^{-}(x):=-\min\{u(x),0\}$.

Proof. First note that we have the inequality $u\leq u^{+}$ on $\overline{\Omega}$ so that

$\max_{\overline\Omega}u\leq \max_{\overline\Omega}u^{+}.$

Let $U:=\{x\in\Omega: u^{+}>0\}$. If $U$ is empty, then $u<0$ and we are done. Otherwise, $\max_{\overline\Omega}u^{+}= \max_{\overline{U}}u^{+}$, and since on $U$,

$-\sum_{i,j=1}^{n}a_{ij}\cdot \partial_{i}\partial_{j}u + \sum_{i=1}^{n} b^{i}\cdot \partial_{i}u \leq -cu\leq 0,$

the weak maximum principle implies that

$\max_{\overline\Omega}u \leq \max_{\overline{U}}u^{+} = \max_{\partial U} u^{+}.$

Since $\partial U = (\overline{U}\cap\partial\Omega)\cap \{x\in\Omega:u(x)=0\}$, we see that $\max_{\partial U}u^{+}=\max_{\partial\Omega}u^{+}$, establishing the claim for subsolutions. As for supersolutions, we again replace $u$ with $-u$ and apply the above argument again to obtain

$-\min_{\overline\Omega}u = \max_{\overline\Omega}(-u) \leq \max_{\partial\Omega} (-u)^{+} = \min_{\partial\Omega} u^{-}\Leftrightarrow \min_{\overline\Omega}u\geq -\min_{\partial\Omega}u^{-}.$

The strong maximum principle

In the case where $c\equiv 0$, the weak maximum principle tells us that a (sub)solution to $Lu=0$ on $\Omega$ attains its maximum on $\partial\Omega$. We now show that if the maximum is actually attained in the interior of $\Omega$, then $u$ must actually be constant. We first prove the following technical lemma.

Lemma 11.6 (Hopf). Suppose $u\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega})$ is a subsolution to the equation $Lu=0$ on $\Omega$ and there exists an $x_{0}\in\partial\Omega$ such that $u(x_{0})>u(x)$ for all $x\in \Omega$. If $\Omega$ satisfies the interior ball condition at $x_{0}$, i.e. there exists an open ball $B\subset\Omega$ with $x_{0}\in\partial B$, and if either $c\equiv 0$ or $c\geq 0$ and $u(x_{0})\geq 0$, then $\frac{\partial u}{\partial \nu}(x_{0})>0$, where $\nu$ is the outer unit normal to $B$ at $x_{0}$.

Proof. We may assume without loss of generality that we are dealing with the second case (otherwise replace $u$ with $u-u(x_{0})$). Moreover, we may suppose that $B=B(0,r)$ for some $r>0$. Now, define $v:\overline{B(0,r)}\rightarrow\mathbb{R}$ by

$v(x)=e^{-k|x|^{2}} - e^{-k r^{2}}.$

Note that $\left.v\right|_{\partial B(0,r)}\equiv 0$. Moreover, we compute that $\partial_{i}v(x)=-2kx^{i} e^{-k|x|^{2}}$ and $\partial_{i}\partial_{j}v(x)=-2k\delta_{ij}e^{-k|x|^{2}}+4k^{2}x^{i}x^{j}e^{-k|x|^{2}}$ so that

$Lv(x)=e^{-k|x|^{2}}\left(2k\cdot\left(\textup{tr}\ (a_{ij}(x))_{i,j=1}^{n} - \sum_{i=1}^{n}b^{i}(x)x^{i}\right) - 4k^{2}\sum_{i,j=1}^{n}a_{ij}(x)x^{i}x^{j} + c(x) \right) - c(x) e^{-kr^{2}}\leq e^{-k|x|^{2}}\left(-4\lambda_{0}|x|^{2}k^{2} + 2k\cdot \left(\textup{tr}\ (a_{ij}(x))_{i,j=1}^{n} - \sum_{i=1}^{n}b^{i}(x)x^{i} + c(x)\right) \right).$

In particular, if $x\in B(0,r)\backslash \overline{B(0,\frac{r}{2})}$, we have that

$Lv(x)\leq e^{-k|x|^{2}}\left(-\lambda_{0}r^{2}k^{2} + 2k\cdot\left(\sup|\textup{tr}\ (a_{ij})| + \sup|b|\cdot r + \sup |c| \right) \right)\leq 0$

for sufficiently large $k$. Moreover, setting $w(x)=u(x)+ \eps v(x) - u(x_{0})$ for sufficiently small $\eps>0$ such that $u(x_{0})\geq u(x) + \eps v(x)$ on $\partial B(0,\frac{r}{2})$, since

$Lw = Lu + \eps Lv - cu(x_{0}) \leq 0,$

we may apply the weak maximum principle to $w$ to conclude that for all $x\in \overline{B(0,r)}\backslash B(0,\frac{r}{2})$,

$w(x)\leq 0.$

In particular, we see that

$\frac{\partial w}{\partial \nu}(x_{0})=\lim_{h\searrow 0} \frac{w(x_{0}-h\nu)-w(x_{0}) }{-h} = \lim_{h\searrow 0}\frac{-w(x_{0}-h\nu) }{h}\geq 0\Leftrightarrow \frac{\partial u}{\partial \nu}(x_{0})\geq -\eps\frac{\partial v}{\partial \nu}(x_{0})=2\eps k r\cdot e^{-k r^{2}}>0.$

Theorem 11.7 (strong maximum principle). Suppose $u\in C^{2}(\Omega)\cap C^{0}(\overline\Omega)$, $c\equiv 0$ on $\Omega$ and $\Omega$ is connected. If $u$ is a subsolution to the equation $Lu=0$ on $\Omega$ and $\max_{\overline{\Omega}}u=u(x_{0})$ for some $x_{0}\in \Omega$, then $u$ is constant. Likewise, if $u$ is a supersolution to the equation $Lu=0$ on $\Omega$ and $\min_{\overline{\Omega}}u=u(x_{0})$ for some $x_{0}\in\Omega$, then $u$ is constant.

Proof. Suppose $u$ is a subsolution and let $C:=\{x\in\Omega:u(x)=\max_{\overline{\Omega}}u\}$ and $U:=\Omega\backslash C=\{x\in\Omega:u(x)<\max_{\overline{\Omega}}u\}$. Suppose $U$ is nonempty, fix $y\in U$ such that $\dist(y,C)<\dist(y,\partial\Omega)$ and let $B$ be the largest ball centred at $y$ such that $B\subset U$ so that $\partial B\cap C$ is nonempty. Let $y_{0}\in \partial B\cap C$. By construction, $U$ satisfies the interior ball condition at $y_{0}$ so that we may apply the Hopf lemma to conclude that $\frac{\partial u}{\partial \nu}(y_{0})>0$, but $y_{0}$ is a local maximum so that $\D u(y_{0})=0$, which is a contradiction.

Similarly, we have the following result in the case where $c\geq 0$. The proof is analogous to that of Theorem 11.7, using the latter part of the Hopf lemma.

Theorem 11.8 (strong maximum principle). Suppose $u\in C^{2}(\Omega)\cap C^{0}(\overline\Omega)$, $c\geq 0$ on $\Omega$ and $\Omega$ is connected. If $u$ is a subsolution to the equation $Lu=0$ on $\Omega$ and $\max_{\overline{\Omega}}u=u(x_{0})\geq 0$ for some $x_{0}\in \Omega$, then $u$ is constant. Likewise, if $u$ is a supersolution to the equation $Lu=0$ on $\Omega$ and $\min_{\overline{\Omega}}u=u(x_{0})\leq 0$ for some $x_{0}\in\Omega$, then $u$ is constant.

Harnack inequality

In addition to the maximum principles, we have the following inequality for nonnegative solutions to $Lu=0$ on $\Omega$, which tells us that the supremum of such a solution is controlled by its infimum away from $\partial\Omega$.

Theorem 11.9. For any connected $U\Subset \Omega$, there exists a constant $C>0$ depending only on $U$ and $L$ such that whenever $u\in C^{2}(\Omega)$ is nonnegative and solves $Lu=0$ on $\Omega$,