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Here's one way to get the hypergeometric function for the "simpler" equation: Consider the operator $x^3 (1 + t\partial_t)(\partial^3_{xxx} + \frac{6}x \partial^2_{xx} + \frac{6}{x^2} \partial_x)$, we …

Firstly, your definition of pointwise $C^{1,\alpha}$ is strictly speaking "incorrect" in that it doesn't give the desired conclusion. My interpretation of what you wrote is, with all the quantifiers i …

The purpose of this answer is to extend Christian Remling's answer to dimension $d = 3$. There are two steps. (N.B. below the cut I show how to replace part 1 by a different argument that works in all …

NB: Evidently I used a different convention on divergence. In terms of matrix components my convention is $(\mathrm{div} M)_j = \sum_{i = 1}^2 \partial_{x^i} M_{ij}$. If your convention is $\sum_{i = …

First, if $fg$ were holomorphic and nontrivial, then $\{g = 0\}$ has to be discrete. This implies that $g$ has to be either $\geq 0$ or $\leq 0$. So we can assume WLOG $g \geq 0$. Restricting away f …

Assume without loss of generality that $0\in \Omega$. Let $f_1, f_2$ be radial, with support contained in $\Omega$. By the mean value property of harmonic functions, you have $$ \int f_2 \varphi …

In many ways the natural generalization of ODEs are hyperbolic PDEs (in that they admit wellposed initial value problems). What you have here is one of those ways. Roughly speaking for a linear hyperb …

Forget about $x,t$. Consider a $C^1$ mapping $\phi:(\zeta,\eta)\mapsto (u,v)$. Locally if $|d\phi|\neq 0$ we can invert it. Let the inverse be $\psi: (u,v)\mapsto (\zeta,\eta)$, so $\psi\circ\phi(\zet …