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).


\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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