Regularity estimates of the Fokker-Planck equation on the torus - MathOverflow most recent 30 from mathoverflow.net 2022-01-23T03:43:56Z https://mathoverflow.net/feeds/question/327336 https://creativecommons.org/licenses/by-sa/4.0/rdf https://mathoverflow.net/q/327336 1 Regularity estimates of the Fokker-Planck equation on the torus Richard https://mathoverflow.net/users/62049 2019-04-06T17:25:45Z 2019-04-07T00:41:09Z <p>The following comes from Pages 37-41 of the following paper:</p> <p><a href="https://arxiv.org/abs/1509.02505" rel="nofollow noreferrer">https://arxiv.org/abs/1509.02505</a></p> <blockquote> <p>Let <span class="math-container">$\mathbb{T}^d$</span> be a <span class="math-container">$d$</span>-dimensional torus and let <span class="math-container">$C^{n+ \alpha}$</span> be the <span class="math-container">$n$</span>th order <span class="math-container">$\alpha$</span>-Holder space defined in the usual way. We define functions <span class="math-container">$V \in C([0,T], (C^{n+ 1+\alpha}(\mathbb{T}^d))^d)$</span> and <span class="math-container">$c \in L^{\infty} \big([0,T], \big( C^{n+ \alpha}(\mathbb{T}^d) \big)' \big)$</span>. Let <span class="math-container">$\rho$</span> be a measure on <span class="math-container">$\mathbb{T}^d$</span>. We consider, in the distribution sense, the PDE given by <span class="math-container">$$\partial_t \rho + \Delta \rho - \text{div} (\rho V)- \text{div} (c) = 0, \quad \quad \text{ on } [0,T],$$</span> with initial condition <span class="math-container">$\rho(0)=0$</span>. Lemma 3.5 of the paper states that there exists a unique solution which satisfies <span class="math-container">$$\sup_{t \in [0,T]} \| \rho (t) \|_{( C^{n+ 1+\alpha}(\mathbb{T}^d) )'} \leq C \sup_{t \in [0,T]} \| c (t) \|_{( C^{n+ \alpha}(\mathbb{T}^d) )'} &lt; \infty.$$</span> </p> </blockquote> <p>My question is whether or not this holds for <span class="math-container">$\mathbb{R}^d$</span> instead of <span class="math-container">$\mathbb{T}^d$</span>. I am not able to check in the paper as most of the technical details are hidden. Thanks.</p>