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We fix $T>0$ and let $\mathbb T := [0, T]$. Let $\sigma : \mathbb R \to \mathbb R$ belong to the Hölder space $C^{1, \alpha}_b (\mathbb R)$ for some $\alpha \in (0, 1)$. Let $u : \mathbb T \times \mathbb R^d \to \mathbb R_+$ be the distributional solution to the non-linear PDE $$ \partial_t u (t, x) = \Delta_x \{ |\sigma (u (t, x))|^2 u(t, x) \}. $$

For brevity, let $u_t := u(t, \cdot)$. We assume that

  • $u_0 \in C^\infty_b (\mathbb R^d)$ is a probability density function.
  • there is $\lambda >0$ such that $\frac{1}{\lambda} \le \sigma \le \lambda$.

Are there some estimates of $\| u_t \|_{C^{0, \beta}_b}$ in terms of $u_0$ and $\sigma$?

Any reference is greatly appreciated! Thank you so much for your help!

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    $\begingroup$ Could you clarify? 1. Did you mean $\sigma \in C_b^{1,\alpha}(\mathbb{R})$? 2. By $\sigma^2(u(t,x))$, do you mean $\sigma(\sigma(u(t,x)))$ or $(\sigma(u(t,x)))^2$? $\endgroup$
    – Fiktor
    Commented Feb 11 at 20:54
  • $\begingroup$ @Fiktor Thank you for your comment! It is a typo and yess it should be $\sigma \in C_b^{1,\alpha}(\mathbb{R})$. By $\sigma^2(u(t,x))$, I meant $|\sigma(u(t,x))|^2$. $\endgroup$
    – Akira
    Commented Feb 13 at 10:04
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    $\begingroup$ If you let $H(s) := s\sigma^2(s)$ then the equation takes the appealing form $\partial_t(H(u)) = H'(u)\Delta(H(u))$. If $H' > 0$ then standard uniformly parabolic theory (e.g. from the book of Lieberman) can be applied. $\endgroup$
    – Connor Mooney
    Commented Feb 21 at 21:16

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