# Ngô Quốc Anh

## November 17, 2013

### Existence result for quasilinear second order elliptic PDEs via fixed point theorems

Filed under: PDEs — Ngô Quốc Anh @ 0:33

Today, we show how fixed point theorems can be used to obtain some existence results for elliptic PDEs. This topic is adapted from Chapter 9 in Evan’s book. The fixed point theorem that we are going to use is the so-called Schaefer Fixed Point Theorem.

Theorem (Schaefer Fixed Point Theorem). Let $X$ be a real Banach space. Suppose $A : X\to X$ is a continuous and compact mapping. Assume further that the set $\{u \in X:u=\lambda A(u) \quad \text{ for some } 0 \leqslant \lambda \leqslant 1\}$

is bounded (in $X$). Then $A$ has a fixed point.

Following is the PDE that we are going to demonstrate. $\left\{ \begin{array}{rcl} - \Delta u + \mu u &=& f(\nabla u) \quad \text{ in } \Omega , \hfill \\ u &=& 0 \quad \text{ on }\partial \Omega , \hfill \\ \end{array} \right.$

where $\Omega \subset \mathbb R^n$ is bounded with smooth boundary $\partial \Omega$ and $f : \mathbb R^n \to \mathbb R$ is a smooth Liptschitz continuous function satisfying the following growth condition $|f(\vec p)| \leqslant C(|\vec p|+1)$

for some constant $C>0$ and all $\vec p \in \mathbb R^n$. We claim that

Theorem. If $\mu>0$ is large enough, there exists a function $u\in H_0^1(\Omega) \cap H^2(\Omega)$ solving the above PDE.

To prove this result, we do as follows.

Step 1.  Construction of the mapping $A$. First, given $u \in H_0^1(\Omega)$, from the growth condition for $f$, clearly $f \in L^2(\Omega)$. Then the following PDE $\left\{ \begin{array}{rcl} - \Delta w + \mu w &=& f \quad \text{ in } \Omega , \hfill \\ w &=& 0 \quad \text{ on }\partial \Omega , \hfill \\ \end{array} \right.$

has a unique weak solution $w \in H_0^1(\Omega)$. Moreover, the following Schauder estimate holds $\displaystyle \|w\|_{H^2} \leqslant C \|f\|_{L^2}$

for some constant $C>0$. Then we can define a mapping $A : H_0^1(\Omega) \to H_0^1(\Omega) , \quad u \mapsto w.$

This mapping obeys $\displaystyle \|A(u)\|_{H^2} \leqslant C \|f\|_{L^2} \leqslant C\big( \|u\|_{H_0^1}+1\big).$

Step 2. Proving the continuity and compactness of $A$.

First, we take a sequence $\{u_k\}_k \subset H_0^1(\Omega)$ in such a way that $u_k \to u$ in $H_0^1(\Omega)$. The continuity follows once we can show that $A(u_k) \to A(u)$ in $H_0^1(\Omega)$. To do so, thanks to the a prior estimate in Step 1, we know that $\displaystyle \sup_k \|A(u_k)\|_{H^2} \leqslant \infty.$

Then, up to a subsequence, $\{A(u_k)\}_k$ converges weakly to some $w$ in $H^2(\Omega)$. The compact embedding (known as extension of the Rellich-Kondrachov theorem) $H^2(\Omega) \hookrightarrow H^1(\Omega)$ implies that $A(u_k) \to w$ strongly in $H_0^1(\Omega)$.

The from the weak form $\displaystyle\int_\Omega {\nabla (A({u_k}))\nabla vdx} + \mu \int_\Omega {A({u_k})vdx} = \int_\Omega {f(\nabla {u_k})vdx}$

for any $v \in H_0^1(\Omega)$, we can pass to the limit to get $\displaystyle\int_\Omega {\nabla w\nabla vdx} + \mu \int_\Omega {wvdx} = \int_\Omega {f(\nabla u)vdx}$

for each $v \in H_0^1(\Omega)$. (The strong convergence $A(u_k) \to w$ in $H_0^1(\Omega)$ will help us to pass to the limit the first two integrals while the last integral makes use of the growth condition for $f$ plus the strong convergence $u_k \to u$ in $H_0^1(\Omega)$.) Thus, we have just proved that $A(u)=w$, completing the proof of the continuity of $A$.

Second, to prove the compactness of $A$, we observe that if the sequence $\{u_k\}_k \subset H_0^1(\Omega)$ is bounded in $H_0^1(\Omega)$, then the a priori estimate tells us that $\{A(u_k)\}_k$ is bounded in $H^2(\Omega)$. Again, the compact embedding $H^2(\Omega) \hookrightarrow H^1(\Omega)$ helps us to conclude that up to a subsequence $A(u_k)$ possesses a strongly convergence subsequence in $H_0^1(\Omega)$.

Step 3. Proving the boundedness of the subset $K$. We now show that if $\mu$ is large, then the set $K=\{u \in H_0^1(\Omega):u=\lambda A(u) \quad \text{ for some } 0 \leqslant \lambda \leqslant 1\}$

is bounded in $H_0^1(\Omega)$. Indeed, we first pick $u \in K$. Since $u/\lambda = A(u)$ for some $\lambda \in [0,1]$, we know that $u$ solves $-\Delta u + \lambda u =\lambda f(\nabla u)$

a.e. in $\Omega$. Using $u$ as a test function, thanks to $\lambda \in [0,1]$, we obtain $\begin{array}{lcl} \displaystyle\int_\Omega {|\nabla u{|^2}dx} + \mu \int_\Omega {{u^2}dx} &=& \displaystyle\lambda \int_\Omega {f(\nabla u)udx} \hfill \\ &\leqslant & \displaystyle C\int_\Omega {(|\nabla u| + 1)udx} \hfill \\ &\leqslant &\displaystyle\frac{1}{2}\int_\Omega {|\nabla u{|^2}dx} + C\left( {\int_\Omega {{u^2}dx} + \int_\Omega {1dx} } \right). \hfill \\ \end{array}$

where the tricky part is to make use of $\displaystyle C(x + 1)y = Cxy + Cy \leqslant \frac{1}{2}{x^2} + 2{C^2}{y^2} + C{y^2} + C \leqslant \frac{1}{2}{x^2} + (2{C^2} + C)({y^2} + 1).$

Thus, $\displaystyle\frac{1}{2}\int_\Omega {|\nabla u{|^2}dx} + (\mu - C)\int_\Omega {{u^2}dx} \leqslant C|\Omega |$

from now we know why the set $K$ is bounded if $\mu$ is large.

Step 4. Applying the Schaefer theorem. This is a direct use of the theorem, then we conclude the result.

## 1 Comment »

1. thanks for this interesting post

Comment by Josef — November 21, 2013 @ 18:09

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