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 be a real Banach space. Suppose is a continuous and compact mapping. Assume further that the set
is bounded (in ). Then has a fixed point.
Following is the PDE that we are going to demonstrate.
where is bounded with smooth boundary and is a smooth Liptschitz continuous function satisfying the following growth condition
for some constant and all . We claim that
Theorem. If is large enough, there exists a function solving the above PDE.
To prove this result, we do as follows.
Step 1. Construction of the mapping . First, given , from the growth condition for , clearly . Then the following PDE
has a unique weak solution . Moreover, the following Schauder estimate holds
for some constant . Then we can define a mapping
This mapping obeys
Step 2. Proving the continuity and compactness of .
First, we take a sequence in such a way that in . The continuity follows once we can show that in . To do so, thanks to the a prior estimate in Step 1, we know that
Then, up to a subsequence, converges weakly to some in . The compact embedding (known as extension of the Rellich-Kondrachov theorem) implies that strongly in .
The from the weak form
for any , we can pass to the limit to get
for each . (The strong convergence in will help us to pass to the limit the first two integrals while the last integral makes use of the growth condition for plus the strong convergence in .) Thus, we have just proved that , completing the proof of the continuity of .
Second, to prove the compactness of , we observe that if the sequence is bounded in , then the a priori estimate tells us that is bounded in . Again, the compact embedding helps us to conclude that up to a subsequence possesses a strongly convergence subsequence in .
Step 3. Proving the boundedness of the subset . We now show that if is large, then the set
is bounded in . Indeed, we first pick . Since for some , we know that solves
a.e. in . Using as a test function, thanks to , we obtain
where the tricky part is to make use of
from now we know why the set is bounded if is large.
Step 4. Applying the Schaefer theorem. This is a direct use of the theorem, then we conclude the result.