4
$\begingroup$

Consider the parabolic equation in $p: \mathbb R^2\to\mathbb R$

$$\partial_t p + b(t)\partial_x p + D(t,x)\partial^2_{xx}p=0,$$

where $b$, $D$ are nice enough functions. I look for the continuity of the derivatives $\partial_t p$, $\partial_x p$ of the solution. It is known by Nash's paper (Continuity of Solutions of Parabolic and Elliptic Equations) that, under very reasonable conditions on $b$, $D$, we have the Hölder-type continuity of $p$. Is there any work concerning such continuity analysis of $\partial_t p$, $\partial_x p$?

PS: My idea is to consider $q\mathrel{:=}\partial_x p$. Then

$$\partial_t q + b(t)\partial_xq + D(t,x)\partial^2_{xx}q=-\partial_x D(t,x)\partial_{xx}p$$

is a similar parabolic equation for $q$ with an additional source. But the term $\partial_x D(t,x)\partial_{xx}p$ contains $\partial_{xx}p$, which makes the estimation even harder….

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The equation for $q = p_x$ can be written in divergence form as $$q_t + b(t)q_x + (D(t,\,x)q_x)_x = 0,$$ so Nash's theorem (which applies to divergence-form equations) implies that $p_x$ is Holder continuous under mild hypotheses on the coefficients (boundedness and measurability).

If the coefficients are more regular (e.g. in Holder classes) then parabolic Schauder estimates (e.g. in the book of Lieberman) give higher regularity of $p$, the general principle being that $p$ has twice as many spatial derivatives as temporal ones (by the scaling of the equation).

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you very kindly for the answer Connor. Do you mean this book worldscientific.com/worldscibooks/10.1142/3302 ? Btw, this post follows from my question at mathoverflow.net/questions/415175/…. Could you please take a look and let me know whether the dependency of fundamental solutions on the coefficients have been studied. Many thanks $\endgroup$
    – GJC20
    Commented Mar 3, 2022 at 18:44
  • $\begingroup$ Thank you so much for your quick reply. While with this transformation, is $v$ still a fundamental solution? Here $v(f^{-1}(s),x)=\delta_{y+h\circ f^{-1}(s)}(x)$. Do you think this matters for the estimation? I do appreciate if you could write an answer (even an outline) to my other question $\endgroup$
    – GJC20
    Commented Mar 3, 2022 at 19:46
  • $\begingroup$ Yes, that is the book. Regarding your other question, I don't know. However, if $u_t + b(t)u_x + \sigma^2/2(1+\alpha(t))^2u_{xx} = 0$ then after the change of variable $u(t,x) = v(f(t),x-h(t))$ with $h' = b$ and $f' = 1/2(1+\alpha)^2$ we see that $v_t + \sigma^2(f^{-1}(t), x+h(f^{-1}(t))v_{xx} = 0$, which simplifies the structure of the equations being considered and reduces the problem to understanding small perturbations of a single Lipschitz coefficient. $\endgroup$
    – Connor Mooney
    Commented Mar 3, 2022 at 19:46
  • $\begingroup$ Here you may assume that $b,\sigma$ are as nice as possible, and $\alpha$ should belong to the class of non-increasing Holder continuous functions taking values in $[0,1]$. The motivation to study this stability is to show the uniqueness at mathoverflow.net/questions/417113/… $\endgroup$
    – GJC20
    Commented Mar 3, 2022 at 19:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .