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You have a logarithmic extension of the classical Cauchy-Lipschitz theorem: the equation $$\dot x=f(x),\quad x(0)=x_0\tag{ODE}$$ has a unique solution if $f$ has a modulus of continuity $\omega$ ($ \v …

We define $$\Psi(x)= \sum_{k\ge 0} 2^{-k}\psi_{\sigma_k}(x-x_k),\quad \psi_{\sigma}(y)=\exp{-{\vert x\vert}^{-\frac{1}{s-1}}}, $$ where $(x_k)_{k\ge 0}$ is dense in $\mathbb R^d$ and $(\sigma_k)_{k\g …

$$ f(t,x)=e^{t/2}\sum_{n\ge 0} e^{-(n+\frac12)t}\langle{g},{H_{n}}\rangle H_{n}(x) $$ The function $F(t,x)=e^{-t/2}f(t,x)$ is the solution of $$ \frac{\partial F}{\partial t}+\bigl(-\frac{d^2}{dx^2}+ …

Let $f,g$ be smooth functions from $\mathbb R$ to $\mathbb R$. Then $$ \frac{(g\circ f)^{(n)}}{n!}=\sum_{1\le r\le n}\frac{(g^{(r)}\circ f)}{r!} \sum_{\substack{(n_1,\dots, n_r)\in {\mathbb N^*}^r\\n_ …

Let us assume that $n=2$: in that case I claim that there is indeed a closed form. As a preliminary to the proof, I would like to begin with a remark on first-order linear ODE; of course in one dimens …

Let us consider the ODE $\hskip3pt \dot x=F(t,x)\hskip3pt $ in an infinite-dimensional Banach space $E$, where the flux $F$ is defined and continous from the whole $\mathbb R\times E$ into $E$. (1 …

For $g:B_1\rightarrow B_2$, $f:B_2\rightarrow B_3$, $B_j$ Banach spaces , $g,f$ smooth, we have for $n\ge 1$ $$ \frac{(f\circ g)^{(n)}}{n!}=\sum_{n_1+\dots+n_r=n\atop r\ge 1, n_j\ge 1} \frac{f^{(r)}\c …

If the matrix $A(t)=(a_{ij}(t))$ is invertible, e.g. if there exists $B(t)\in L^\infty$, with $B(t) A(t)=Id$, then you get a linear non-characteristic system of type $$ \dot u +C(t) u=g(t),\quad C\in …

Let $T_m$ be a standard Fourier multiplier homogeneous with degree $m$ in $\mathbb R^n$ (this the case of your fractional Laplacean). Then $$ \Vert T_m u\Vert_{L^p(\mathbb R^n)}\le C_{m,n}\Vert u\Ver …

Let me define for $x\in\mathbb R$, $ F(x)=\int_{\mathbb R} e^{-π t^2}\cos(x t^3) dt. $ I claim that $F(x)>0$ for all $x\in\mathbb R$. Well, it is obvious for $x=0$ since $F(0)=1$ and also for $x$ near …

Maybe more a lengthy comment than an answer. Let me work with temperate distributions (dual of the Schwartz space $\mathscr S(\mathbb R^d)$). Let $u\in \mathscr S'(\mathbb R^d)$ and let us define the …

For $\phi\in C^\infty_c(\mathbb R^m)$ and duality products, we have $$ \langle\frac{\partial f}{\partial x}(x,y),\phi(x)\rangle_{\mathscr D'(\mathbb R^m),\mathscr D(\mathbb R^m)}=- \langle f(x,y),\phi …

Your formula is based on the identity for $s\in\mathbb N$, $$ \partial^s=(1+(-1)+\partial)^s=\sum_{m\ge 0}C_s^m((-1)+\partial)^m=\sum_{m\ge 0\atop 0\le k\le m}C_s^mC_m^k(-1)^{m-k}\partial^k. \tag{$\fl …

Take $A(x)=\vert x\vert$ the Euclidean norm in $\mathbb R^n$, which is obviously homogeneous of degree 1. We have $$ A'(x)=\frac{x}{A(x)},\quad A''(x)=\frac{Id}{A(x)}-\frac{x\otimes x}{A(x)^3}=A(x)^{- …

Let $\Omega$ be an open subset of $\mathbb R^n$ with a $C^1$ boundary and $u\in H^1(\Omega)$. We compute with $D_{x_1}=-i\partial_{x_1}$, $$ 2\Re\langle D_{x_1}u, i x_1u\rangle=-2\Re\int_\Omega x_1(\p …