I have recently started reading into PDE theory, and came across the following question. Consider the PDE $$ \begin{cases} \partial_t \rho = \Delta (\rho + \rho^2) & \text{ on } (0,T) \times \mathbb{R}^d,\\ \rho(0,\cdot ) = \rho_0, & \end{cases} $$ for some (smooth) $\rho_0\colon \mathbb{R}^d \to [0,\infty)$ and $T > 0$. This equation is quasilinear (linear in its highest derivatives) and uniformly parabolic, so we know there is a unique classical solution (see the references below, for example). The way this is usually proved is with a fixed point argument: We first rewrite $\Delta (\rho^2) = 2 \rho \Delta \rho + 2 (\nabla \rho)^2 $, so $$ \partial_t \rho = (1 + 2 \rho) \Delta \rho + 2 (\nabla \rho)^2, $$ and define an operator $$ \mathcal{F} \colon u \mapsto \text{ the unique solution of } \partial_t \rho = (1 + 2u) \Delta \rho + 2 (\nabla u)^2, $$ which is well-defined on a suitable domain by linear PDE theory. Then a solution to the original problem is exactly a fixed point of $\mathcal{F}$.

Up to here this all feels very natural to me, and at this point I would expect that one could show that $\mathcal{F}$ is a contraction on a suitable Banach space of functions and apply Banach's fixed point theorem. However, in the books I checked (see below) I only found proofs using Schauder's fixed point theorem (i.e. Brower's theorem for Banach spaces). This makes me wonder, is it difficult to show contractivity of this map on a suitable space? Because if it was possible I would think it should be the preferred method as it gives a much more constructive way to obtain a solution (for example, because it can be obtained via fixed point iteration).

Lieberman, Gary M., Second order parabolic differential equations, Singapore: World Scientific. xi, 439 p. (1996). ZBL0884.35001.

Friedman, A., Partial differential equations of parabolic type, Moskau: Verlag ’Mir’. 427 S. (1968). ZBL0173.12701.

  • $\begingroup$ Usually the contraction method requires some smallness of parameters, at least for a strightforward application, and this is not necessary for other fixed point theorems. This could be an explanation. However there situations where a clever use of contraction method allows to remove any smallness. $\endgroup$ Feb 27, 2022 at 11:39


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