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I was curious whether anyone knows of some examples in PDE where a standard fixed point argument fails to show the existence of a solution but one can apply one of the more advanced fixed point arguments to obtain a solution.

By standard fixed point i mean any of the fixed point theorems in a beginning PDE/functional analysis book and by the more advanced I mean some of these pseudo contractions versions.

thanks

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    $\begingroup$ Would Schauder's fixed point theorem be standard or advanced? $\endgroup$
    – Dirk
    Nov 24, 2012 at 13:32
  • $\begingroup$ In a now-deleted answer, Craig says: I would consider Schauders fixed point theorem as one of the standard ones. The less standard ones (or the ones I am looking for) are some of these pseudocontraction versions. $\endgroup$
    – S. Carnahan
    Dec 3, 2012 at 15:10

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The graphical minimal surface equation is a great example of a PDE where Leray-Schauder fixed point theory is applied:

$$ \left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}u = 0 $$

This represents the condition that $graph(u)$ is a minimal surface, or in other words is a critical point for the area functional.

For surfaces in $\mathbb{R}^3$, existence for the a slightly more general problem (Plateau's problem) was established by Douglass-Rado in the 1930's, using beautiful conformal methods. However, the higher dimensional problem required the introduction of the Leray-Schauder fixed point theorem. One version of this says

For $\mathscr{B}$ a Banach space, and $T : \mathscr{B} \times [0,1] \to \mathscr{B}$ continuous, compact and so that $T(x,0) = 0$ for all $x \in \mathscr{B}$. Suppose also that there is some $M > 0$ so that if $x = T(x,\sigma)$ for $(x,\sigma) \in \mathscr{B} \times[0,1]$ then $$ \Vert x \Vert_\mathscr{B} \leq M.$$ Then, there is a fixed point for $T(\cdot, 1)$.

One of the reasons that I find this theorem very neat is because in some sense it treats a priori estimates as just that! In other words, you never need to show a solution to some relaxed equation holds or something like that, you just have to show that if there is a solution, then you have some estimates (and of course compactness).

To use this, we define the operator $\hat T : C^{1,\beta}(\overline \Omega) \times [0,1] \to C^{2,\beta'}(\overline\Omega) $ as the solution operator to the linear PDE, solving for some $v$ satisfying $$ \left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}v = 0 \text{ in $\Omega$} $$ $$ v = \sigma \varphi \text{ on $\partial\Omega$} $$ Linear existence theory shows that this map is well defined and if we then compose it with the map $C^{2,\beta'} \hookrightarrow C^{1,\beta}$, we have a map $T: C^{1,\beta} (\overline\Omega) \times [0,1] \to C^{1,\beta} (\overline\Omega)$, which is compact because the inclusion $C^2\to C^{1,\beta}$ is.

Furthermore, if $\sigma =0$, the $0$ solution clearly works.

Thus, to prove that Leray-Schauder applies, one must show that the a priori estimate holds. This is a bit delicate, so I won't go into the details, but only remark that one needs to assume some geometric conditions on the boundary (mean convexity). If you're interested, you can find the details in Gilbarg-Trudinger, starting with 11.3 but sort of jumping around for the various bounds.

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  • $\begingroup$ In a now-deleted answer, Craig says: Thank you very much for the detailed example. Actually I very much enjoy this example. $\endgroup$
    – S. Carnahan
    Dec 3, 2012 at 15:10

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