Sobolev Spaces

The motivation behind weak solutions to partial differential equations (PDEs) in divergence form, based on the formula for integration by parts, is well known. In this way, one can develop the notion of weak derivative for suitable classes of Lebesgue integrable functions. This section discusses some fine properties of functions found within one such class, the so-called Sobolev space $W^{k,p}(\Omega)$ with given $k\in\mathbb{Z}_{\geq 0}$ , $1\leq p\leq \infty$ , and domain $\Omega \subset \mathbb{R}^n$ .

Weak Maximum Principles

In studying the existence of weak solutions to quasilinear PDE, one can consider the method of sub and super solutions. It is a constructive approach in which a weak maximum principle plays a key role. An example of this principle reads as follows.

Theorem 1. (Weak Maximum Principle) Let $\Omega\subset \mathbb{R}^n$ be a bounded domain with Lipschitz boundary, and let $u\in H^{1}(\Omega)$ be such that $u\geq 0$ on $\partial \Omega$ in the sense that $\text{min}(u,0)\in H_0^1(\Omega)$ . Furthermore, suppose that $-\Delta u+\lambda u\geq 0$ in $\Omega$ weakly, for some $\lambda \geq 0$ , which is to say that

$\displaystyle \int_{\Omega}\nabla u\cdot \nabla \phi +\lambda u\phi\mathrm{d}x\geq 0$

for all $\phi\in H_0^{1}(\Omega)$ with $\phi \geq 0$ a.e in $\Omega$ . Then $u\geq 0$ a.e in $\Omega$ .

I met this result when I attended The University of Oxford’s course on Fixed Point Methods for Nonlinear PDEs (HT 2020). The conclusion follows by taking $\phi=-\text{min}(u,0)$ in the weak definition of $-\Delta u+\lambda u\geq 0$ as given in the statement of the theorem, to then conclude $u=\text{max}(u,0)$ , which is obviously non-negative a.e. This idea can be taken further in the setting of slightly more elaborate PDEs formulated over bounded open sets where the constant $\lambda$ may be negative, which the following result of mine shows.

Theorem 2. (A Generalisation) Let $\Omega\subset \mathbb{R}^n$ be a bounded open set. Consider a pair $(c,\lambda)$ of real constants such that $0\leq c<C_{Poin}^{-1}$ and $\lambda> -C_{Poin}^{-2}(1-cC_{Poin})$ . Further, let $\textbf{b}\in L^{\infty}(\Omega,\mathbb{R}^n)$ be a vector field for which $\|\textbf{b}\|_{L^{\infty}(\Omega,\mathbb{R}^n)}\leq c.$ Suppose $u\in H^1(\Omega)$ is such that $u\geq 0$ on $\partial \Omega$ in the sense that $\text{min}(u,0)\in H_0^1(\Omega)$ , and $-\Delta u-\textbf{b}\cdot\nabla u+\lambda u\geq 0$ holds in $\Omega$ weakly, which is to say

$\displaystyle \int_{\Omega}\nabla u\cdot \nabla \phi-\phi\text{ }\textbf{b}\cdot \nabla u +\lambda u\phi\text{ }\mathrm{d}x\geq 0$

for all $\phi\in H_0^{1}(\Omega)$ with $\phi \geq 0$ a.e in $\Omega$ . Then $u\geq 0$ a.e in $\Omega$ .

Note: Here we define $\|\textbf{b}\|_{L^{\infty}(\Omega,\mathbb{R}^n)}:=\text{ess sup}_{\Omega}|\textbf{b}|$ , while $C_{Poin}>0$ denotes the Poincare constant associated with Poincare’s inequality on $H_0^1(\Omega)$ .

Proof. Write $u=u^{+}+u^{-}$ where we let $u^{+}:=\text{max}(u,0)$ and $u^{-}:=\text{min}(u,0)$ . Then, by assumption we know that $\phi:=-\text{min}(u,0)=-u^{-}\in H_0^{1}(\Omega)$ and is non-negative a.e in $\Omega$ . Consequently, the weak interpretation of $-\Delta u-\textbf{b}\cdot\nabla u+\lambda u\geq 0$ implies

$\displaystyle \int_{\Omega}\nabla (u^{+}+u^{-})\cdot \nabla (-u^{-})+u^{-}\text{ }\textbf{b}\cdot \nabla (u^{+}+u^{-}) +\lambda (u^{+}+u^{-})(-u^{-})\text{ }\mathrm{d}x\geq 0$

which simplifies to

$\displaystyle \lambda \|u^{-}\|_{L^2(\Omega)}^2+\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq \int_{\Omega}u^{-}\textbf{b}\cdot\nabla u^{-}\mathrm{d}x.$

Since $\|\textbf{b}\|_{L^{\infty}(\Omega,\mathbb{R}^n)}\leq c$ we find

$\displaystyle \int_{\Omega}u^{-}\textbf{b}\cdot\nabla u^{-}\mathrm{d}x\leq c\|u^{-}\|_{L^2(\Omega)}\|\nabla u^{-}\|_{L^2(\Omega)},$

giving

$\displaystyle \lambda \|u^{-}\|_{L^2(\Omega)}^2+\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq c\|u^{-}\|_{L^2(\Omega)}\|\nabla u^{-}\|_{L^2(\Omega)}.$

By Poincare’s inequality on $H_0^1(\Omega)$ , which reads $\|v\|_{L^2(\Omega)}\leq C_{Poin}\|\nabla v\|_{L^2(\Omega)}$ for all $v\in H_0^{1}(\Omega)$ , we arrive at

$\displaystyle \lambda \|u^{-}\|_{L^2(\Omega)}^2+\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq cC_{Poin}\|\nabla u^{-}\|_{L^2(\Omega)}^2$

$\displaystyle \lambda \|u^{-}\|_{L^2(\Omega)}^2+(1-cC_{Poin})\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq 0.$

Now, if $\lambda\geq 0$ , then we see immediately that $\nabla u^{-}=0$ a.e in $\Omega$ and so $u^{-}=0$ a.e in $\Omega$ , by virtue of Poincare’s inequality. And so, $u=u^{+}$ which is non-negative a.e. On the other hand, if $0>\lambda>-C_{Poin}^{-2}(1-cC_{Poin})$ we find by Poincare’s inequality,

$\displaystyle 0\geq \lambda \|u^{-}\|_{L^2(\Omega)}^2+(1-cC_{Poin})\|\nabla u^{-}\|_{L^2(\Omega)}^2\geq \left(\lambda C_{Poin}^2+1-cC_{Poin}\right)\|\nabla u^{-}\|_{L^2(\Omega)}^2.$

Because $\lambda C_{Poin}^{2}>-(1-cC_{Poin})$ , it follows that we must have $\nabla u^{-}=0$ a.e in $\Omega$ once more. This completes the proof of the theorem.

Remark 1. Of course, in the above result we may replace $\textbf{b}$ with $-\textbf{b}$ without changing the conclusion. This result, together with the Lax-Milgram Theorem which one could use to prove a solution operator is well-defined, we can develop examples of quasilinear PDE such as

$\displaystyle -\Delta u+\textbf{b}\cdot\nabla u +\lambda u=f(u)$

formulated over a bounded open set $\Omega\subset\mathbb{R}^n$ , with

homogeneous Dirichlet boundary condition on $\partial \Omega$ ,
essentially bounded vector field $\textbf{b}$ as in Theorem 2,
suitably negative $\lambda$ as given by Theorem 2, and
$f$ a nonlinear function,

for which the method of sub and super solutions applies to prove the existence of a weak solution.

Let $\textbf{A}(x)=(a_{ij}(x))_{1\leq i,j\leq n}$ be a matrix of measurable functions defined almost-everywhere over an open set $\Omega\subset\mathbb{R}^n$ . If there exists a constant $\alpha>0$ such that

$\displaystyle \eta\cdot\textbf{A}(x)\eta\geq \alpha \|\eta\|^2$

for all $\eta\in\mathbb{R}^n$ and almost every $x\in \Omega$ , we’ll say $A$ is $\alpha$ -coercive over $\Omega$ . Assume $c$ is an essentially bounded function over $\Omega$ with $\|c\|_{L^{\infty}(\Omega)}\leq \gamma$ for some $\gamma\geq 0$ . With this assumption, let’s proceed to generalise Theorem 2 to operators in divergence form given by

$\displaystyle Lu:=-\partial_{i}(a_{ij}\partial_ju)+\textbf{b}\cdot\nabla u+cu.$

Theorem 3. Let $\Omega\subset \mathbb{R}^n$ be a bounded open set. Consider a triple $(\alpha,\beta,\gamma)$ of real constants such that the matrix $\textbf{A}=(a_{ij})_{1\leq i,j\leq n}$ is $\alpha$ -coercive over $\Omega$ and $\alpha-\beta C_{Poin}-\gamma C_{Poin}^2>0$ . Further, assume $\textbf{b}\in L^{\infty}(\Omega,\mathbb{R}^n)$ is a vector field for which $\|\textbf{b}\|_{L^{\infty}(\Omega,\mathbb{R}^n)}\leq \beta$ , and let $c$ be an essentially bounded function over $\Omega$ with $\|c\|_{L^{\infty}(\Omega)}\leq \gamma$ . Suppose $u\in H^1(\Omega)$ is such that $u\geq 0$ on $\partial \Omega$ in the sense that $\text{min}(u,0)\in H_0^1(\Omega)$ , and that $-\partial_{i}(a_{ij}\partial_ju)+\textbf{b}\cdot\nabla u+cu\geq 0$ holds in $\Omega$ weakly, which is to say

$\displaystyle \int_{\Omega}\nabla \phi\cdot \textbf{A}\nabla u+\phi\text{ }\textbf{b}\cdot \nabla u + cu\phi\text{ }\mathrm{d}x\geq 0$

for all $\phi\in H_0^{1}(\Omega)$ with $\phi \geq 0$ a.e in $\Omega$ . Then $u\geq 0$ a.e in $\Omega$ .

Proof. Write $u=u^{+}+u^{-}$ where we let $u^{+}:=\text{max}(u,0)$ and $u^{-}:=\text{min}(u,0)$ . Then, by assumption we know that $\phi:=-\text{min}(u,0)=-u^{-}\in H_0^{1}(\Omega)$ and is non-negative a.e in $\Omega$ . Consequently, the weak interpretation of $-\partial_i(a_{ij}\partial_ju)+\textbf{b}\cdot\nabla u+cu\geq 0$ implies

$\displaystyle \int_{\Omega}\nabla (-u^{-})\cdot\textbf{A}\nabla (u^{+}+u^{-})-u^{-}\text{ }\textbf{b}\cdot \nabla (u^{+}+u^{-}) +c (u^{+}+u^{-})(-u^{-})\text{ }\mathrm{d}x\geq 0$

which simplifies to

$\displaystyle \int_{\Omega}c(u^{-})^2\mathrm{d}x+\int_{\Omega}\nabla u^{-}\cdot \textbf{A}\nabla(u^{-})\mathrm{d}x\leq -\int_{\Omega}u^{-}\textbf{b}\cdot\nabla u^{-}\mathrm{d}x.$

Since $\|\textbf{b}\|_{L^{\infty}(\Omega,\mathbb{R}^n)}\leq \beta$ we find

$\displaystyle -\int_{\Omega}u^{-}\textbf{b}\cdot\nabla u^{-}\mathrm{d}x\leq \beta\|u^{-}\|_{L^2(\Omega)}\|\nabla u^{-}\|_{L^2(\Omega)}.$

By coercivity of $\textbf{A}$ we have

$\displaystyle \int_{\Omega}\nabla u^{-}\cdot \textbf{A}\nabla(u^{-})\mathrm{d}x\geq \alpha \|\nabla(u^{-})\|_{L^2(\Omega)}^2.$

Therefore,

$\displaystyle \int_{\Omega}c(u^{-})^2\mathrm{d}x+\alpha\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq \beta\|u^{-}\|_{L^2(\Omega)}\|\nabla u^{-}\|_{L^2(\Omega)}.$

By Poincare’s inequality on $H_0^1(\Omega)$ , we arrive at

$\displaystyle \int_{\Omega}c(u^{-})^2\mathrm{d}x+\alpha\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq \beta C_{Poin}\|\nabla u^{-}\|_{L^2(\Omega)}^2.$

$\displaystyle \int_{\Omega}c(u^{-})^2\mathrm{d}x +(\alpha-\beta C_{Poin})\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq 0.$

Since $c\geq -\|c\|_{L^{\infty}(\Omega)}\geq -\gamma$ almost-everywhere in $\Omega$ , we find

$\displaystyle -\gamma\|u^{-}\|_{L^2(\Omega)}^2+(\alpha-\beta C_{Poin})\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq 0$

and so Poincare’s inequality implies

$\displaystyle (\alpha-\beta C_{Poin}-\gamma C_{Poin}^2)\|\nabla u^{-}\|_{L^2(\Omega)}^2\leq 0.$

But, it is assumed that the triple $(\alpha,\beta,\gamma)$ satisfies $\alpha-\beta C_{Poin}-\gamma C_{Poin}^2> 0$ . Thus, we see immediately $\nabla u^{-}=0$ a.e in $\Omega$ and so $u^{-}=0$ a.e in $\Omega$ , by virtue of Poincare’s inequality. Consequently, $u=u^{+}$ which is non-negative a.e in $\Omega$ , as required.

Remark 2. In both results the conditions on the constants can be interpreted geometrically. In the first instance we have inequality constraints on $(c,\lambda)$ that yield an unbounded region delineated by three straight lines in $\mathbb{R}^2$ : $c=0$ , $c=C_{Poin}^{-1}$ and $\lambda=C_{Poin}^{-1}c-C_{Poin}^{-2}$ . On the other hand, in the second result the inequality constraints on $(\alpha,\beta,\gamma)$ define an unbounded region within the first quadrant of $\mathbb{R}^3$ with part of its boundary found in a plane passing through the origin with normal $\textbf{n}=(1,-C_{Poin},-C_{Poin}^2)$ . Notably, projecting this plane in the horizontal plane where $\gamma=1$ , we observe an unbounded region which contains the one given by the constraints on $(c,\lambda)$ in Theorem 2.

Weak Maximum Principles

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