On the Cauchy Problem for Nondegenerate Parabolic Integro-Differential Equations in the Scale of Generalized Hölder Spaces

Abstract

Parabolic integro-differential nondegenerate Cauchy problem is considered in the scale of Hölder spaces of functions whose regularity is defined by a radially O-regularly varying Lévy measure. Existence and uniqueness and the estimates of the solution are derived.

Appendix

We simply state a few results that were used in this paper. Let $\nu \in \mathfrak {A}^{\alpha }$, and

$$ \begin{array}{@{}rcl@{}} \delta \left( r\right) &=&\delta_{\nu }\left( r\right) =\nu \left( \left\{ \left\vert y\right\vert >r\right\} \right) >0,r>0, \\ w &=&w_{\nu }\left( r\right) =\delta \left( r\right) ^{-1},r>0,\lim_{r\rightarrow 0}w\left( r\right) =0. \end{array} $$

We assume that w = w_ν is an O-RV function at zero, i.e.,

$$ r_{1}\left( \varepsilon \right) =\overline{\lim_{x\rightarrow 0}}\frac{ \delta \left( \varepsilon x\right)^{-1}}{\delta \left( x\right)^{-1}} <\infty ,\varepsilon >0. $$

By Theorem 2 in [2], the following limits exist:

$$ p_{1}=\lim_{\varepsilon \rightarrow 0}\frac{\log r_{1}\left( \varepsilon \right) }{\log \varepsilon }\leq q_{1}=\lim_{\varepsilon \rightarrow \infty } \frac{\log r_{1}\left( \varepsilon \right) }{\log \varepsilon }. $$

(1)

Lemma 8

Assume w = w_ν is an O-RV function at zero.

• a) Let β > 0 and τ > −βp₁. There is C > 0 so that

  $$ {{\int}_{0}^{x}}t^{\tau }w\left( t\right)^{\beta }\frac{dt}{t}\leq Cx^{\tau }w\left( x\right)^{\beta },x\in (0,1], $$

  and $\lim _{x\rightarrow 0}x^{\tau }w\left (x\right )^{\beta }=0.$

• b) Let β > 0 and τ < −βq₁. There is C > 0 so that

  $$ {{\int}_{x}^{1}}t^{\tau }w\left( t\right)^{\beta }\frac{dt}{t}\leq Cx^{\tau }w\left( x\right)^{\beta },x\in (0,1],\text{ } $$

  and $\lim _{x\rightarrow 0}x^{\tau }w\left (x\right )^{\beta }=\infty .$

• c) Let β < 0 and τ > −βq₁. There is C > 0 so that

  $$ {{\int}_{0}^{x}}t^{\tau }w\left( t\right)^{\beta }\frac{dt}{t}\leq Cx^{\tau }w\left( x\right)^{\beta },x\in (0,1], $$

  and $\lim _{x\rightarrow 0}x^{\tau }w\left (x\right )^{\beta }=0.$

• d) Let β < 0 and τ < −βp₁. There is C > 0 so that

  $$ {{\int}_{x}^{1}}t^{\tau }w\left( t\right)^{\beta }\frac{dt}{t} ={\int}_{1}^{x^{-1}}t^{-\tau }w\left( \frac{1}{t}\right)^{\beta }\frac{dt}{t} \leq Cx^{\tau }w\left( x\right)^{\beta },x\in (0,1], $$

  and $\lim _{x\rightarrow 0}x^{\tau }w\left (x\right )^{\beta }=\infty .$

Proof

The claims follow easily by Theorems 3, 4 in [2]. Because of the similarities, we will prove c) only. Let β < 0 and τ > −βq₁ . Then

$$ \begin{array}{@{}rcl@{}} \overline{\lim_{t\rightarrow \infty }}\frac{w\left( \frac{1}{\varepsilon t} \right)^{\beta }}{w\left( \frac{1}{t}\right)^{\beta }} &=&\overline{ \lim_{x\rightarrow 0}}\frac{w\left( x\right)^{-\beta }}{w\left( \varepsilon^{-1}x\right)^{-\beta }}=\overline{\lim_{x\rightarrow 0}}\frac{w\left( \varepsilon \varepsilon^{-1}x\right)^{-\beta }}{w\left( \varepsilon^{-1}x\right)^{-\beta }} \\ &=&\overline{\lim_{x\rightarrow 0}}\frac{w\left( \varepsilon x\right)^{-\beta }}{w\left( x\right)^{-\beta }}=r_{1}\left( \varepsilon \right)^{-\beta }<\infty ,\varepsilon >0. \end{array} $$

Hence $w\left (\frac {1}{t}\right )^{\beta },t\geq 1$, is an O-RV function at infinity with

$$ p=\lim_{\varepsilon \rightarrow 0}\frac{\log r_{1}\left( \varepsilon \right)^{-\beta }}{\log \varepsilon }=-\beta p_{1}\leq -\beta q_{1}=\lim_{\varepsilon \rightarrow \infty }\frac{\log r_{1}\left( \varepsilon \right)^{-\beta }}{\log \varepsilon }=q. $$

Then for x ∈ (0, 1],

$$ {{\int}_{0}^{x}}t^{\tau }w\left( t\right)^{\beta }\frac{dt}{t} ={\int}_{x^{-1}}^{\infty }t^{-\tau }w\left( \frac{1}{t}\right)^{\beta }\frac{ dt}{t}\leq Cx^{\tau }w\left( x\right)^{\beta } $$

by Theorem 3 in [2], and $\lim _{x\rightarrow 0}x^{\tau }w\left (x\right )^{\beta }=0$ according to Theorem 4 in [2]. □

Corollary 5

Assume w = w_ν is an O-RV function at zero and p₁ > 0. Let N > 1,β > 0. Then

$$ \sum\limits_{j=0}^{\infty }w\left( N^{-j}\right)^{\beta }<\infty . $$

Proof

Indeed,

$$ \sum\limits_{j=0}^{\infty }w\left( N^{-j}\right)^{\beta }\leq {\int}_{0}^{\infty }w\left( N^{-x}\right)^{\beta }dx\leq C{{\int}_{0}^{1}}w\left( t\right)^{\beta }\frac{dt}{t}<\infty , $$

because, by Lemma 8a),

$$ {{\int}_{0}^{x}}w\left( t\right)^{\beta }\frac{dt}{t}\leq Cw\left( x\right)^{\beta },x\in \left[ 0,1\right] . $$

□

We will need some Lévy measure moment estimates.

Lemma 9

Let $\nu \in \mathfrak {A}^{\alpha },$andw = w_ν be an O-RV function at zero with p₁,q₁ defined in Eq. 1. Assume

$$ \begin{array}{@{}rcl@{}} 0 &<&p_{1}\leq q_{1}<1\text{ if }\alpha \in \left( 0,1\right) , \\ 0 &<&p_{1}\leq 1\leq q_{1}<2\text{ if }\alpha =1, \\ 1 &<&p_{1}\leq q_{1}<2\text{ if }\alpha \in \left( 1,2\right) . \end{array} $$

Then

• (i)
  $$ \begin{array}{@{}rcl@{}} \sup_{R\in (0,1]}{\int}_{\mathbf{R}^{d}}\left( \left\vert y\right\vert \wedge 1\right) \tilde{\nu}_{R}\left( dy\right) &<&\infty \text{ if }\alpha \in \left( 0,1\right) , \\ \sup_{R\in (0,1]}{\int}_{\mathbf{R}^{d}}\left( \left\vert y\right\vert^{2}\wedge 1\right) \tilde{\nu}_{R}\left( dy\right) &<&\infty \text{ if } \alpha =1, \\ \sup_{R\in (0,1]}{\int}_{\mathbf{R}^{d}}\left( \left\vert y\right\vert^{2}\wedge \left\vert y\right\vert \right) \tilde{\nu}_{R}\left( dy\right) &<&\infty \text{ if }\alpha \in (1,2). \end{array} $$
• (ii)
  $$ \inf_{R\in (0,1]}{\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\tilde{\nu}_{R}\left( dy\right) \geq c_{1}, $$

  for some c₁ > 0.

Proof

• (i) Let α ∈ (0, 1) . Then by Lemma 8 (recall $\nu _{R}\left (dy\right ) =\nu \left (Rdy\right ) ,\tilde {\nu }_{R}\left (dy\right ) =w\left (R\right ) \nu _{R}\left (dy\right ) ,\delta \left (R\right ) =w\left (R\right )^{-1}$),

  $$ \begin{array}{@{}rcl@{}} {\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert \nu_{R}\left( dy\right) &=&R^{-1}{\int}_{\left\vert y\right\vert \leq R}\left\vert y\right\vert \nu \left( dy\right) \\ &=&R^{-1}{{\int}_{0}^{R}}[\delta \left( s\right) -\delta \left( R\right) ]ds, \end{array} $$

  and

  $$ {\int}_{\mathbf{R}_{0}^{d}}\left( \left\vert y\right\vert \wedge 1\right) \tilde{\nu}_{R}\left( dy\right) =R^{-1}w\left( R\right) {{\int}_{0}^{R}}w\left( s\right)^{-1}ds\leq C,~R\in (0,1]. $$

Let α = 1. Then similarly using Lemma 8, we have

$$ \int \left( \left\vert y\right\vert^{2}\wedge 1\right) \tilde{\nu}_{R}\left( dy\right) =2R^{-2}w\left( R\right) {{\int}_{0}^{R}}s^{2}w\left( s\right)^{-1}\frac{ds}{s}\leq C,~R\in (0,1]. $$

Let α ∈ (1, 2). Then similarly,

$$ \begin{array}{@{}rcl@{}} &&R^{-1}{\int}_{\left\vert y\right\vert >R}\left\vert y\right\vert \nu \left( dy\right) =R^{-1}{\int}_{0}^{\infty }\delta \left( s\vee R\right) ds \\ &=&\delta \left( R\right) +R^{-1}{\int}_{R}^{\infty }\delta \left( s\right) ds=\delta \left( R\right) +R^{-1}{\int}_{R}^{\infty }w\left( s\right)^{-1}ds \end{array} $$

and with R ∈ (0, 1],

$$ \begin{array}{@{}rcl@{}} R^{-2}{\int}_{\left\vert y\right\vert \leq R}\left\vert y\right\vert^{2}\nu \left( dy\right) &=&2R^{-2}{{\int}_{0}^{R}}s^{2}[w\left( s\right)^{-1}-w\left( R\right)^{-1}]\frac{ds}{s} \\ &=&2R^{-2}{{\int}_{0}^{R}}s^{2}w\left( s\right)^{-1}\frac{ds}{s}-w\left( R\right)^{-1}. \end{array} $$

(2)

Hence, by Lemma 8,

$$ \begin{array}{@{}rcl@{}} &&\int \left( \left\vert y\right\vert^{2}\wedge \left\vert y\right\vert \right) \nu_{R}\left( dy\right) \\ &\leq &2R^{-2}{{\int}_{0}^{R}}s^{2}w\left( s\right)^{-1}\frac{ds}{s} +R^{-1}{{\int}_{R}^{1}}w\left( s\right)^{-1}ds+R^{-1}{\int}_{1}^{\infty }w\left( s\right)^{-1}ds \\ &=&2R^{-2}{{\int}_{0}^{R}}s^{2}w\left( s\right)^{-1}\frac{ds}{s} +R^{-1}{{\int}_{R}^{1}}w\left( s\right)^{-1}ds+R^{-1}{\int}_{\left\vert y\right\vert >1}\left\vert y\right\vert \nu \left( dy\right) \\ &\leq &Cw\left( R\right)^{-1},~R\in (0,1]. \end{array} $$

• (ii) By Eq. 2, for R ∈ (0, 1],

  $$ \begin{array}{@{}rcl@{}} {\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\tilde{\nu} _{R}\left( dy\right) &=&w\left( R\right) {\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\nu_{R}\left( dy\right) \\ &=&2R^{-2}{{\int}_{0}^{R}}s^{2}\left[\frac{w\left( R\right) }{w\left( s\right) }-1\right] \frac{ds}{s}=2{{\int}_{0}^{1}}s^{2}\left[\frac{w\left( R\right) }{w\left( Rs\right) } -1\right]\frac{ds}{s}. \end{array} $$

  Hence, by Fatou’s lemma,

  $$ \underline{\lim }_{R\rightarrow 0}{\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\tilde{\nu}_{R}\left( dy\right) \geq 2{{\int}_{0}^{1}}s^{2}\left[\frac{1}{r_{1}\left( s\right) }-1\right]\frac{ds}{s}=c_{1}>0 $$

  if |{s ∈ [0, 1] : r₁ (s) < 1}| > 0, because

  $$ \lim \inf_{R\rightarrow 0}\frac{w\left( R\right) }{w\left( Rs\right) }=\frac{ 1}{\lim \sup_{R\rightarrow 0}\frac{w\left( Rs\right) }{w\left( R\right) }}= \frac{1}{r_{1}\left( s\right) },~s\in (0,1]. $$

□

According to [9], Chapter 3, 70-74, any Lévy measure $\nu \in \mathfrak {A}^{\alpha }$ can be disintegrated as

$$ \nu \left( {\Gamma} \right) =-{\int}_{0}^{\infty }{\int}_{S_{d-1}}\chi_{\Gamma }\left( rw\right) {\Pi} \left( r,dw\right) d\delta \left( r\right) ,{\Gamma} \in \mathcal{B}\left( \mathbf{R}_{0}^{d}\right) , $$

where δ = δ_ν, and π (r,dw),r > 0, is a measurable family of measures on the unit sphere S_d− 1 with π (r,S_d− 1) = 1,r > 0. The following is a straightforward consequence of Lemma 9(ii).

Corollary 6

Let $\nu \in \mathfrak {A}^{\alpha },$

$$ \nu \left( {\Gamma} \right) =-{\int}_{0}^{\infty }{\int}_{S_{d-1}}\chi_{\Gamma }\left( rw\right) {\Pi} \left( r,dw\right) d\delta \left( r\right) ,{\Gamma} \in \mathcal{B}\left( \mathbf{R}_{0}^{d}\right) , $$

where δ = δ_ν,π (r,dw), r > 0, is a measurable family of measures on S_d− 1 with π (r,S_d− 1) = 1, r > 0. Assume $w=w_{\nu }=\delta _{\nu }^{-1}$ is an O-RV function at zero satisfying assumptions of Lemma 9, and

$$ \inf_{\left\vert \hat{\xi}\right\vert =1}{\int}_{S_{d-1}}\left\vert \hat{\xi} \cdot w\right\vert^{2}{\Pi} \left( r,dw\right) \geq c_{0}>0. $$

(3)

Then assumption B holds.

Proof

Indeed, for $\left \vert \hat {\xi }\right \vert =1,R\in (0,1],$ with C > 0,

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\left\vert y\right\vert \leq 1}\left\vert \hat{\xi}\cdot y\right\vert^{2}\nu_{R}\left( dy\right) \\ &=&R^{-2}{\int}_{\left\vert y\right\vert \leq R}\left\vert \hat{\xi}\cdot y\right\vert^{2}\nu \left( dy\right) =-R^{-2}{{\int}_{0}^{R}}{\int}_{S_{d-1}}\left\vert \hat{\xi}\cdot w\right\vert^{2}{\Pi} \left( r,dw\right) r^{2}d\delta \left( r\right) \\ &\geq &-c_{0}R^{-2}{{\int}_{0}^{R}}r^{2}d\delta \left( r\right) =c_{0}R^{-2}{\int}_{\left\vert y\right\vert \leq R}\left\vert y\right\vert^{2}\nu \left( dy\right) =c_{0}{\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\nu_{R}\left( dy\right) . \end{array} $$

Hence by Lemma 9(ii),

$$ \begin{array}{@{}rcl@{}} \inf_{R\in (0,1]}\inf_{\left\vert \hat{\xi}\right\vert =1}{\int}_{\left\vert y\right\vert \leq 1}\left\vert \hat{\xi}\cdot y\right\vert^{2}\tilde{\nu}_{R}\left( dy\right) &\geq &c_{0}~\inf_{R\in (0,1]}{\int}_{\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{2}\tilde{\nu}_{R}\left( dy\right) \\ &\geq &c_{0}c_{1}>0. \end{array} $$

□

Remark 1

Let α ∈ (0, 2) , $\nu \in \mathfrak {A}^{\alpha }$, and w_ν be an O-RV function at zero, p₁ > 0. By Theorems 3 and 4 in [2], for any σ ∈ (0,p₁),

$$ \begin{array}{@{}rcl@{}} {\int}_{r<\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{\sigma }\nu \left( dy\right) &=&\sigma {{\int}_{r}^{1}}t^{\sigma }w\left( t\right)^{-1} \frac{dt}{t}-\delta \left( 1\right) \\ &\geq &cr^{\sigma }w\left( r\right)^{-1}-\delta \left( 1\right) \rightarrow \infty \end{array} $$

as r → 0. Hence p₁ ≤ α. On the other hand for any σ > q₁, by Lemma 9,

$$ {\int}_{0<\left\vert y\right\vert \leq 1}\left\vert y\right\vert^{\sigma }\nu \left( dy\right) \leq \sigma {{\int}_{0}^{1}}t^{\sigma }w\left( t\right)^{-1} \frac{dt}{t}<\infty , $$

and α ≤ q₁.