# A weighted $L_{q}(L_{p})$-theory for fully degenerate second-order evolution
equations with unbounded time-measurable coefficients

Ildoo Kim  Department of mathematics, Korea university, 1 anam-dong sungbuk-
gu, Seoul, south Korea 136-701<EMAIL_ADDRESS>
###### Abstract.

We study the fully degenerate second-order evolution equation

$\displaystyle u_{t}=a^{ij}(t)u_{x^{i}x^{j}}+b^{i}(t)u_{x^{i}}+c(t)u+f,\quad
t>0,x\in\mathbb{R}^{d}$ (0.1)

given with the zero initial data. Here $a^{ij}(t)$, $b^{i}(t)$, $c(t)$ are
merely locally integrable functions, and $(a^{ij}(t))_{d\times d}$ is a
nonnegative symmetric matrix with the smallest eigenvalue $\delta(t)\geq 0$.
We show that there is a positive constant $N$ such that

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left(|u|+|u_{xx}|\right)^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))(\delta(t))^{1-q}dt,$
(0.2)

where $p,q\in(1,\infty)$, $\alpha(t)=\int_{0}^{t}\delta(s)ds$, and $w$ is a
Muckenhoupt’s weight.

###### Key words and phrases:

Degenerate second-order parabolic equations, Weighted $L_{p}$-estimates, zero
initial-value problem

###### 2010 Mathematics Subject Classification:

35K65, 35B65, 35K15

I. Kim has been supported by the National Research Foundation of Korea(NRF)
grant funded by the Korea government(MSIT) (No.NRF-2020R1A2C1A01003959)

##  1\. introduction

Needless to say, the second-order partial differential equations equations
with degenerate or unbounded coefficients have been extensively studied for a
long time. To the best of our knowledge, the starting point of this study was
Keldysh, Fichera, and Oleĭnik’s work (see e.g. [23, 10, 37, 38, 39]).
Moreover, it is very popular to study a (maximal regularity) $L_{p}$-theory
and its generalization to $L_{q}(L_{p})$-theory in harmonic analysis, Fourier
analysis, and partial differential equations after Calderón and Zygmund’s
work. For the historical works and backgrounds of $L_{p}$-theories and their
generalizations, we refer some outstanding books [31, 32, 42, 18, 19, 21, 22].
These days, there are tons of papers handling degenerate and unbounded
coefficients in various prospectives. Among recent works with various
prospectives, we only refer the author to [26, 11, 9, 35, 12, 17, 33, 16, 15,
27, 34, 40, 2, 36, 1, 6, 14, 20, 7, 8, 13, 28, 41, 43]. These results handle
equations having degenerate or unbounded coefficients in Sobolev spaces.

With the degeneracy in the equation, it is hard to expect to obtain full
regularity estimates of solutions unless there are weights involved in
estimates. For instance, by taking the leading coefficients $a^{ij}(t)=0$ for
all $i,j,t$, we see that it is not possible to obtain the unweighted maximal
$L_{p}$-regularity

$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dtdx\leq
N\int_{0}^{T}\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dtdx.$ (1.1)

Hence, weights have been commonly used to controls the degeneracy or
unboundeness (singularity) of the coefficients. However, most results in the
literature focus on degeneracy or singularity near the boundary of a domain.
If we consider the whole space, it is naturally not expected that there is a
regularity gain of a solution in general due to the extreme case such as
$u_{t}=f$, which could be understood as one of equation (0.1) with
coefficients $a^{ij}(t)=0$ for all $t$. Hence when it comes to the solvability
of second-order equations with degeneracy in the whole space, people used to
only prove the existence and uniqueness of a weak solution without considering
regularity gain from the equations.

Nonetheless, there is a way to express an $L_{p}$-norm of second derivatives
of a solution $u$ with a weight which could be singular even in the whole
space. For instance, assume that the degeneracy happens on a time interval
$(a,b)$, then $\delta(t)=0$ for all $t\in(a,b)$. Then we cannot expect the
smoothing gain from the diffusion equations and the Sobolev second derivatives
$u_{xx}$ fails to exist. However, since there is the weight $\delta(t)$ in the
first line of (0.2), the inequality is still true if we understand the second
line of (0.2) as an improper integral. To the best of our knowledge, this type
estimate is firstly introduced by the author and collaborator in [24, 25]. In
this paper, we add Muckenhoupt’s weights in estimates and extend
$L_{p}$-estimates to $L_{q}(L_{p})$-estimates with lower-order terms.

It is well-known that probabilistic methods are very powerfully working for
leading coefficients which are unbounded and have degeneracy (cf. [29, 4]). We
remark that probabilistic tools play very important roles to obtain our
results. Especially, to obtain (0.2), it requires to understand the relation
among the constant $N$, the degeneracy, and the unboundedness of coefficients
$a^{ij}(t)$. Maximal $L_{p}$-regularity estimates such as (1.1) originally
came from $L_{p}$-boundedness of singular integral operators. However, the
exact relation among parameters related to coefficients is hard to obtain from
singular integral theories since all parameters are combined in a complicated
way to control singularities of operators. We found that this relation could
be more clear by applying probabilistic representations of solutions (see
Theorem 4.3).

We believe that our result could initiate various interesting weighted
estimates for degenerate second-order equations with space dependent
coefficients or domain problems.

This paper is organized as follows. In Section 2, we introduce our main
results. A probabilistic solution representation and its application to
estimate a solution $u$ with general weights are given in Section 3 Weighted
estimates for non-degenerate equations are shown in Section 4. Finally, the
proof of the main theorem is specified in Section 5.

We finish the introduction with notation used in the article.

  * •

We use Einstein’s summation convention throughout this paper.

  * •

$\mathbb{N}$ and $\mathbb{Z}$ denote the natural number system and the integer
number system, respectively. As usual $\mathbf{R}^{d}$ stands for the
Euclidean space of points

$x=\begin{pmatrix}x^{1}\\\ x^{2}\\\ \vdots\\\ x^{d}\end{pmatrix}.$

Frequently, the coordinates of the vector $x$ is denoted in a row form, i.e.
$x=(x^{1},\ldots,x^{d})$. We use the notation $(a^{ij})_{d\times d}$ to denote
the $d$ by $d$ matrix whose entry in $i$-th row and $j$-th column is $a^{ij}$.
For $i=1,...,d$, multi-indices $\alpha=(\alpha_{1},...,\alpha_{d})$,
$\alpha_{i}\in\\{0,1,2,...\\}$, and functions $u(x)$ we set

$u_{x^{i}}=\frac{\partial u}{\partial x^{i}}=D_{i}u,\quad
D^{\alpha}u=D_{1}^{\alpha_{1}}\cdot...\cdot D^{\alpha_{d}}_{d}u.$

  * •

$C^{\infty}(\mathbb{R}^{d})$ denotes the space of infinitely differentiable
functions on $\mathbb{R}^{d}$. $\mathcal{S}(\mathbb{R}^{d})$ is the Schwartz
space consisting of infinitely differentiable and rapidly decreasing functions
on $\mathbb{R}^{d}$. By $C_{c}^{\infty}(\mathbb{R}^{d})$, we denote the
subspace of $C^{\infty}(\mathbb{R}^{d})$ with the compact support.

  * •

For $n\in\mathbb{N}$ and $\mathcal{O}\subset\mathbb{R}^{d}$ and a normed space
$F$, by $C(\mathcal{O};F)$, we denote the space of all $F$-valued continuous
functions $u$ on $\mathcal{O}$ having $|u|_{C}:=\sup_{x\in
O}|u(x)|_{F}<\infty$.

  * •

For $p\in[1,\infty)$, a normed space $F$, and a measure space
$(X,\mathcal{M},\mu)$, by $L_{p}(X,\mathcal{M},\mu;F)$, we denote the space of
all $F$-valued $\mathcal{M}^{\mu}$-measurable functions $u$ so that

$\left\|u\right\|_{L_{p}(X,\mathcal{M},\mu;F)}:=\left(\int_{X}\left\|u(x)\right\|_{F}^{p}\mu(dx)\right)^{1/p}<\infty,$

where $\mathcal{M}^{\mu}$ denotes the completion of $\mathcal{M}$ with respect
to the measure $\mu$. If there is no confusion for the given measure and
$\sigma$-algebra, we usually omit them.

  * •

For measurable set $\mathcal{O}\subset\mathbb{R}^{d}$, $|\mathcal{O}|$ denotes
the Lebesgue measure of $\mathcal{O}$.

  * •

By $\mathcal{F}$ and $\mathcal{F}^{-1}$ we denote the d-dimensional Fourier
transform and the inverse Fourier transform, respectively. That is,
$\mathcal{F}[f](\xi):=\int_{\mathbf{R}^{d}}e^{-ix\cdot\xi}f(x)dx$ and
$\mathcal{F}^{-1}[f](x):=\frac{1}{(2\pi)^{d}}\int_{\mathbf{R}^{d}}e^{i\xi\cdot
x}f(\xi)d\xi$.

  * •

We write $a\lesssim b$ if there is a positive constant $N$ such that $a\leq
Nb$. The constant $N$ may change from a location to a location, even within a
line. If we write $N=N(a,b,\cdots)$, this means that the constant $N$ depends
only on $a,b,\cdots$. The dependence of the constant $N$ is usually specified
in the statements of theorems, lemmas, and corollaries.

##  2\. Setting and main result

Throughout the paper, we fix $d\in\mathbb{N}$ to denote the dimension of the
space variable and all functions are real-valued if there is no special
comment. We study the following degenerate second-order evolution equation

$\displaystyle
u_{t}(t,x)=a^{ij}(t)u_{x^{i}x^{j}}(t,x)+b^{i}(t)u_{x^{i}}(t,x)+c(t)u(t,x)+f(t,x),$
$\displaystyle
u(0,x)=0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(t,x)\in(0,T)\times\mathbb{R}^{d}.$
(2.1)

We emphasize that our coefficients $a^{ij}(t)$, $b^{i}(t)$, and $c(t)$ do not
satisfy any regularity conditions. More importantly, our coefficients
$a^{ij}(t)$, $b^{i}(t)$, and $c(t)$ can be unbounded and degenerate. Here are
more concrete conditions on the coefficients $a^{ij}(t)$, $b^{i}(t)$, and
$c(t)$.

######  Assumption 2.1.

  1. (i)

Assume that there exists a measurable mapping $\delta(t)$ from $(0,\infty)$ to
$[0,\infty)$ such that

$\displaystyle a^{ij}(t)\xi^{i}\xi^{j}\geq\delta(t)|\xi|^{2}\quad\forall
t\in[0,\infty)~{}\text{and}~{}\xi\in\mathbb{R}^{d}.$

  2. (ii)

Assume that the coefficients $a^{ij}(t)$, $b^{i}(t)$, and $c(t)$ are locally
integrable, i.e.

$\displaystyle\int_{0}^{T}\left(|a^{ij}(t)|+|b^{i}(t)|+|c(t)|\right)dt<\infty\qquad\forall
T\in(0,\infty)~{}\text{and}~{}\forall i,j.$ (2.2)

For $T\in(0,\infty)$ and a measurable function $u$ on
$(0,T)\times\mathbb{R}^{d}$, we say that $u$ is locally integrable if

$\displaystyle\int_{0}^{t}\int_{|x|<c}|u(t,x)|dxdt<\infty\quad\forall
t\in(0,T)~{}\text{and}~{}\forall c>0.$

######  Definition 2.2 (Solution).

Let $T\in(0,\infty)$ and $f$ be a locally integrable function on
$(0,T)\times\mathbb{R}^{d}$. We say that a locally integrable function $u$ is
a solution to (2.1) if for any $\varphi\in C_{c}^{\infty}(\mathbb{R}^{d})$,

$\displaystyle(u(t,\cdot),\varphi)$
$\displaystyle=\int_{0}^{t}\left(u(s,\cdot),a^{ij}(s)\varphi_{x^{i}x^{j}}+b^{i}(s)\varphi_{x^{i}}+c(s)\varphi\right)ds$
$\displaystyle\quad+\int_{0}^{t}\left(f(s,\cdot),\varphi\right)ds\quad\forall
t\in(0,T),$ (2.3)

where $(u(t,\cdot),\varphi)$ denotes the $L_{2}(\mathbb{R}^{d})$-inner
product, i.e.

$(u(t,\cdot),\varphi):=\int_{\mathbb{R}^{d}}u(t,x)\varphi(x)dx.$

######  Remark 2.3.

Due to the definition of a solution, it is obvious that

$a^{ij}(t)u_{x^{i}x^{j}}=\frac{a^{ij}(t)+a^{ji}(t)}{2}u_{x^{i}x^{j}}.$

Thus without loss of generality, we may assume that our coefficient matrix
$(a^{ij}(t))_{d\times d}$ is nonnegative symmetric for all $t$. Additionally,
$\delta(t)$ in Assumption 2.1(i) can be chosen by the smallest eigenvalue of
$(a^{ij}(t))_{d\times d}$.

We recall the definition of Muckenhoupt’s weights.

######  Definition 2.4 (Muckenhoupt’s weight).

For $q\in(1,\infty)$, let $A_{q}(\mathbb{R})$ be the class of all nonnegative
and locally integrable functions $w$ on $\mathbb{R}$ satisfying

$[w]_{A_{q}(\mathbb{R})}:=\sup_{-\infty<a<b<\infty}\left(-\int_{(a,b)}w(t)dt\right)\left(-\int_{(a,b)}w(t)^{-1/(q-1)}dt\right)^{q-1}<\infty,$

where

$-\int_{(a,b)}w(t)dt=\frac{\int_{a}^{b}w(t)dt}{b-a}.$

Finally, we introduce our main result.

######  Theorem 2.5.

Let $T\in(0,\infty)$, $p,q\in(1,\infty)$, and $w\in A_{q}(\mathbb{R})$.
Suppose that Assumption 2.1 holds. Then for any locally integrable function
$f$ on $(0,T)\times\mathbb{R}^{d}$, there is a unique solution $u$ to equation
(2.1) such that

$\displaystyle\sup_{t\in[0,T]}\left[\left(\int_{\mathbb{R}^{d}}\left|u\left(t,x\right)(t,x)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\right]$
$\displaystyle\leq\left[\int_{0}^{\alpha(T)}w(t)^{-\frac{1}{q-1}}dt\right]^{q-1}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt,$
(2.4)

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u\left(t,x\right)(t,x)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq[w]_{A_{q}(\mathbb{R})}[\alpha(T)]^{q}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt,$
(2.5)

and

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u_{xx}\left(t,x\right)(t,x)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt,$
(2.6)

where $\alpha(t)=\int_{0}^{t}\delta(s)ds$ and $N$ is a positive constant
depending only on $d$, $p$, $q$, and $[w]_{A_{q}(\mathbb{R})}$. In particular,
for any $-1<\beta<q-1$,

$\displaystyle\sup_{t\in[0,T]}\left[\left(\int_{\mathbb{R}^{d}}\left|u\left(t,x\right)(t,x)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\right]$
$\displaystyle\leq\left[\frac{q-1}{q-1-\beta}\right]^{q-1}\left[\int_{0}^{T}\delta(t)dt\right]^{q-1-\beta}$
$\displaystyle\quad\times\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\left|\int_{0}^{t}\delta(s)ds\right|^{\beta}(\delta(t))^{1-q}dt,$
(2.7)
$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u\left(t,x\right)(t,x)\right|^{p}\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\left|\int_{0}^{t}\delta(s)ds\right|^{\beta}\delta(t)dt$
$\displaystyle\leq[|t|^{\beta}]_{A_{p}(\mathbb{R})}\left[\int_{0}^{T}\delta(t)dt\right]^{q}$
$\displaystyle\quad\times\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\left|\int_{0}^{t}\delta(s)ds\right|^{\beta}(\delta(t))^{1-q}dt,$
(2.8)

and

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u_{xx}\left(t,x\right)(t,x)\right|^{p}\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\left|\int_{0}^{t}\delta(s)ds\right|^{\beta}\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}\left|\int_{0}^{t}\delta(s)ds\right|^{\beta}(\delta(t))^{1-q}dt,$
(2.9)

where $N$ depends only on $d$, $p$, $q$, and $\beta$.

A proof of Theorem 2.5 is given in Section 5.

######  Remark 2.6.

  1. (i)

For $t\in[0,T]$ so that $\delta(t)=0$, the existence of the Sobolev
derivatives $u_{xx}(t,x)$ is not guaranteed by (2.6). Moreover, $\delta(t)$
can be zero on a set with a positive Lebesgue measure, which is far from
Muckenhoupt’s weight.

  2. (ii)

Since $\delta(t)$ can be zero on a set with a positive measure, the integral

$\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))(\delta(t))^{1-q}dt$

is understood in an improper sense, i.e.

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))(\delta(t))^{1-q}dt$
$\displaystyle=\lim_{\varepsilon\downarrow
0}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t)+\varepsilon
t)(\delta(t)+\varepsilon)^{1-q}dt.$ (2.10)

  3. (iii)

If

$\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(s,x)|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(s))|\delta(s)|^{1-q}ds<\infty,$

then the local integrability condition on $f$ is not necessary in Theorem 2.5.
In other words, the finiteness condition implies the local integrability of
$f$. To investigate this fact, let $t\in(0,T)$ and $c>0$. Then for any
$\varepsilon\in(0,1)$, applying Hölder’s inequality and the change of variable
$\alpha(t)+\varepsilon t\to t$, we have

$\displaystyle\int_{0}^{t}\int_{|x|<c}|f(t,x)|dxds$
$\displaystyle=\int_{0}^{t}\int_{\mathbb{R}^{d}}|f(t,x)|1_{|x|<c}dx1_{0<s<t}ds$
$\displaystyle\leq
N\int_{0}^{t}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{1/p}1_{0<s<t}ds$
$\displaystyle\leq
N\left[\int_{0}^{t}\left(\int_{\mathbb{R}^{d}}|f(s,x)|^{p}dx\right)^{q/p}e^{-q\int_{0}^{s}c(\rho)d\rho}w(\alpha(s)+\varepsilon
s)|\delta(s)+\varepsilon|^{1-q}ds\right]^{1/q}$
$\displaystyle\quad\times\left[\int_{0}^{t}e^{\frac{q}{q-1}\int_{0}^{s}c(\rho)d\rho}w^{-\frac{1}{q-1}}(\alpha(s)+\varepsilon
s)(\delta(s)+\varepsilon)ds\right]^{(q-1)/q}$ $\displaystyle\leq
N\left[\int_{0}^{t}\left(\int_{\mathbb{R}^{d}}|f(s,x)|^{p}dx\right)^{q/p}w(\alpha(s)+\varepsilon
s)|\delta(s)+\varepsilon|^{1-q}ds\right]^{1/q}$ (2.11)
$\displaystyle\quad\times
e^{\frac{q}{q-1}\int_{0}^{t}|c(s)|ds}\left[\int_{0}^{\alpha(t)+t}w^{-\frac{1}{q-1}}(s)ds\right]^{(q-1)/q}.$

It is obvious that $e^{\frac{q}{q-1}\int_{0}^{t}|c(s)|ds}<\infty$ since the
function $c(t)$ is locally integrable. Moreover, since $w\in
A_{p}(\mathbb{R})$, $\int_{0}^{\alpha(t)+t}w^{-\frac{1}{q-1}}(s)ds$ is finite.
Therefore, taking $\varepsilon\to 0$ in (2.11), (formally) we obtain the local
integrability of $f$.

  4. (iv)

(2.4) and (2.5) hold even for $p=1$ or $p=\infty$ (see Corollary 3.3).
Moreover, it is easy to check that (2.4) is a stronger estimate than (2.5)
with a help of the definition of Muckenhoupt’s weight, i.e. (2.4) implies
(2.5). However, (2.5) can be slightly improved by using the probabilistic
representation of a solution in the sense that it cannot be obtained from
(2.4) directly in general. Indeed, formally using (3.12) with

$h_{1}(t)=w(\alpha(t))\delta(t)$

and

$h_{2}(t)=w(\alpha(t))|\delta(t)|^{1-q},$

we have

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq\int_{0}^{T}\Bigg{[}w(\alpha(t))\delta(t)\left[\int_{0}^{t}|w(\alpha(s))|\delta(s)|^{1-q}|^{-\frac{1}{q-1}}ds\right]^{q-1}$
$\displaystyle\qquad\times\int_{0}^{t}\|f(s,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(s))|\delta(s)|^{1-q}ds\Bigg{]}dt.$

  5. (v)

Obviously, we can obtain

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u_{xx}\left(t,x\right)(t,x)\right|^{p}w_{0}(x)dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}w_{0}(x)dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt$

if $w_{0}(x)$ is bounded both below and above. However, if $w(x)$ has a
degeneracy or a singularity (unboundedness), then we believe that it is
impossible to add $w_{0}\in A_{p}(\mathbb{R}^{d})$ in the estimates. In other
words, generally, it is not expected to find a positive constant $N$ such that

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u_{xx}\left(t,x\right)(t,x)\right|^{p}w_{0}(x)dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}w_{0}(x)dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt.$
(2.12)

To claim it, assume that (2.12) holds with $b(t)=c(t)=0$ for all $t$. Then we
get

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|u_{xx}\left(t,x\right)(t,x)\right|^{p}w_{0}(x)dx\right)^{q/p}w(\alpha(t))\delta(t)dt$
$\displaystyle\leq
N\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{p}w_{0}(x)dx\right)^{q/p}w(\alpha(t))|\delta(t)|^{1-q}dt.$
(2.13)

Then the function $v(t,x)=u\left(t,x+\int_{0}^{t}b(s)ds\right)$ becomes a
solution to

$\displaystyle
v_{t}(t,x)=a^{ij}(t)v_{x^{i}x^{j}}(t,x)+b^{i}(t)v_{x^{i}}(t,x)+f\left(t,x+\int_{0}^{t}b(s)ds\right),$
$\displaystyle
v(0,x)=0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(t,x)\in(0,T)\times\mathbb{R}^{d}.$

Thus by (2.13) with $p=q=2$ and $\delta(t)=1$, we obtain

$\displaystyle\int_{0}^{T}\int_{\mathbb{R}^{d}}\left|v_{xx}\left(t,x\right)(t,x)\right|^{2}w_{0}\left(x+\int_{0}^{t}b(s)ds\right)w(t)dxdt$
$\displaystyle\leq
N\int_{0}^{T}\int_{\mathbb{R}^{d}}\left|f\left(t,x\right)\right|^{2}w_{0}\left(x+\int_{0}^{t}b(s)ds\right)w(t)dxdt.$

Observe that

$(t,x)\mapsto w(t)w_{0}\left(x+\int_{0}^{t}b(s)ds\right)\notin
A_{2}(\mathbb{R}^{d+1})$

unless $\int_{0}^{t}b(s)ds$ is a constant vector uniformly for all $t$ since
$w$ has a singularity or a degeneracy in general. Therefore we cannot expect
(2.12) if there is a non-trivial coefficient $b(t)$ in the equation. Moreover,
our main tool is the probabilistic solution representation such as (3.2). We
use the translation invariant property of $L_{p}$-norms with this
representation in many parts of proofs of the main theorem. Thus (2.12) is
impossible to obtain by our method even for the case $b(t)=0$ for all $t$ (see
Remark 4.4).

  6. (vi)

All constants in estimates in Theorem 2.5 do not depend on the integrals of
coefficients $a^{ij}$, $b^{i}$, and $c$. Thus for a fixed time
$T\in(0,\infty)$, the integrability condition on the coefficients (2.2) can be
relaxed to

$\displaystyle\int_{0}^{t}\left(|a^{ij}(t)|+|b^{i}(t)|+|c(t)|\right)dt<\infty\qquad\forall
t\in(0,T)~{}\text{and}~{}\forall i,j.$

##  3\. Probabilistic solution representations

In this section, we consider equations without lower-order terms first, i.e.

$\displaystyle
u_{t}(t,x)=a^{ij}(t)u_{x^{i}x^{j}}(t,x)+f(t,x)\qquad(t,x)\in(0,T)\times\mathbb{R}^{d}$
$\displaystyle u(0,x)=0.$ (3.1)

Consider a Brownian motion $B_{t}$ in a filtered probability space
$(\Omega,\mathcal{F}_{t},\mathbb{P})$ with the usual condition. It is well-
known that any predictable function $\sigma(t):\Omega\times(0,T)\to\mathbb{R}$
satisfying

$\int_{0}^{t}|\sigma(s)|^{2}ds<\infty\quad(a.s.)\quad\forall t\in[0,t],$

the Itô integral

$X_{t}=\int_{0}^{t}\sigma(s)dB_{s}$

is well-defined and Itô’s formula works for the stochastic process $f(X_{t})$
with a smooth function $f$ (cf. [30, Chapter 5]). Moreover, our solution $u$
to equation (3.1) can be derived from the expectation of a composition of a
function $f$ and the stochastic process $X_{t}$. Here is a more explicit
statement.

######  Theorem 3.1.

Let $T\in(0,\infty)$ and $f$ be a locally integrable function on
$(0,T)\times\mathbb{R}^{d}$. Assume that the function $t\in(0,\infty)\mapsto
A(t):=\left(a^{ij}(t)\right)_{d\times d}$ is locally integrable, i.e. for each
$i$ and $j$,

$\displaystyle\int_{0}^{T}a^{ij}(t)dt<\infty$

and the coefficients $\left(a^{ij}(t)\right)_{d\times d}$ are nonnegative,
i.e.

$a^{ij}(t)\xi^{i}\xi^{j}\geq
0\quad\forall\xi\in\mathbb{R}^{d}~{}\text{and}~{}\forall t\in(0,T].$

Then there exists a unique solution $u$ to (3.1) and this solution $u$ is
given by

$\displaystyle
u(t,x)=\int_{0}^{t}\mathbb{E}\left[f(s,x+X_{t}-X_{s})\right]ds,$ (3.2)

where

$\displaystyle X_{t}:=\sqrt{2}\int_{0}^{t}\sqrt{A}^{ij}(s)dB_{s}^{j},$

and $B_{t}=(B_{t}^{1},\ldots,B_{t}^{d})$ is a $d$-dimensional Brownian motion
(Wiener process) and the integral is Itô’s stochastic integral. Moreover, for
any $p\in[1,\infty]$, we have

$\displaystyle\|u(t,\cdot)\|_{L_{p}}\leq\int_{0}^{t}\|f(s,\cdot)\|_{L_{p}}ds\quad\forall
t\in[0,T]$ (3.3)

and for any functions $h_{1}$ and $h_{2}$ on $[0,T]$ which are positive
(a.e.), we have

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}h_{1}(t)dt$
$\displaystyle\leq\int_{0}^{T}\left[h_{1}(t)\left[\int_{0}^{t}|h_{2}(s)|^{-\frac{1}{q-1}}ds\right]^{q-1}\int_{0}^{t}\|f(s,\cdot)\|^{q}_{L_{p}}h_{2}(s)ds\right]dt.$
(3.4)

###### Proof.

Part I. (Uniqueness)

Even though the coefficients can be unbounded or degenerate, the uniqueness of
a solution can be easily obtained from a classical Fourier transform method
with Grönwall’s inequality. To give a rigorous detail, choose a $\varphi$
which is a nonnegative function in $C_{c}^{\infty}(\mathbb{R}^{d})$ with a
unit integral. For $\varepsilon\in(0,1)$, denote

$\varphi^{\varepsilon}(x):=\frac{1}{\varepsilon^{d}}\varphi\left(\frac{x}{\varepsilon}\right).$

Let $u$ be a solution to

$\displaystyle u_{t}(t,x)=a^{ij}(t)u_{x^{i}x^{j}}(t,x)+f(t,x)$ $\displaystyle
u(0,x)=0$

and define

$u^{\varepsilon}(t,x)=\int_{\mathbb{R}^{d}}u(t,y)\varphi^{\varepsilon}(x-y)dy.$

It is sufficient to show that for any $\varepsilon\in(0,1)$ and $t\in(0,T)$,
$u(t,x)=0$ for almost every $x\ in\mathbb{R}^{d}$. Fix $\varepsilon\in(0,1)$,
$t\in(0,T)$, and $x\in\mathbb{R}^{d}$. Recalling the definition of a solution
and putting $\varphi^{\varepsilon}(x-\cdot)$ in (2.3), we have

$\displaystyle
u^{\varepsilon}(t,x)=\int_{0}^{t}a^{ij}(s)\left(u^{\varepsilon}\right)_{x^{i}x^{j}}(s,x)ds.$
(3.5)

For each $\varepsilon\in(0,1)$ and $t\in(0,T)$, (3.5) holds for all
$x\in\mathbb{R}^{d}$. Take the $d$-dimensional Fourier transform with respect
to $x$ in (3.5) and absolute value. Then we have

$\displaystyle\left|\mathcal{F}[u^{\varepsilon}(t,\cdot)](\xi)\right|\leq\int_{0}^{t}\left|a^{ij}(s)\xi^{i}\xi^{j}\right|\left|\left[\mathcal{F}[u^{\varepsilon}(s,\cdot)\right](\xi)\right|ds$
(3.6)

for all $\xi\in\mathbb{R}^{d}$. Note that for each $\varepsilon\in(0,1)$ and
$\xi\in\mathbb{R}^{d}$, (3.6) holds for all $t\in(0,T)$. Thus finally applying
Grönwall’s inequality, we have

$|\mathcal{F}[u^{\varepsilon}(t,\cdot)](\xi)|=0$

for all $\varepsilon\in(0,1)$, $t\in(0,T)$, and $\xi\in\mathbb{R}^{d}$, which
completes the uniqueness of a solution $u$.

Part II. (Existence)

The existence of a solution $u$ cannot be shown based on a classical Fourier
transform method since the coefficients can be degenerate. Thus we choose a
probabilistic method to show the existence of a solution. Since it is a well-
known fact if the inhomogeneous term $f$ is smooth (even for more general $f$
in a $L_{p}$-class). However, it is not easy to find an appropriate reference
which exactly fit to our setting (cf. [25, Section 3]), we give a proof with a
detail. Our main tools are Itô’s formula and a smooth approximation. We divide
the proof into three steps.

Step 1. (Smooth case) In this step, we assume that for each $t\in(0,T)$,
$f(t,x)$ is twice continuously differentiable with respect to $x$.

Recall that for each $t$, $\left(a^{ij}(t)\right)_{d\times d}$ is a
nonnegative symmetric matrix. Then there exists a $d\times d$ matrix
$\sqrt{A}(t)$ such that

$A(t)=\sqrt{A}(t)\times\sqrt{A}^{\ast}(t),$

where $\sqrt{A}^{\ast}$ denotes the transpose matrix of $\sqrt{A}$. Recall

$X_{t}=\sqrt{2}\int_{0}^{t}\sqrt{A}(s)dB_{s}.$

We claim that the function

$\displaystyle
u(t,x):=\int_{0}^{t}\mathbb{E}\left[f(s,x+X_{t}-X_{s})\right]ds$ (3.7)

becomes a solution to (2.1). As mentioned before, it is not easy to show that
$u$ defined in (3.2) becomes a solution to (2.1) based on an analytic method
such as the Fourier transform since the degeneracy of $a^{ij}(t)$ makes the
Fourier transform of $u$ lose the integrability. However, it is still possible
to apply Itô’s formula. Fix $s\in(0,T)$ and $x\in\mathbb{R}^{d}$. Apply Itô’s
formula to

$f\left(s,x+\sqrt{2}\int_{s}^{t}\sqrt{A}(r)dB_{r}\right).$

Then we have

$\displaystyle f\left(s,x+\sqrt{2}\int_{s}^{t}\sqrt{A}(r)dB_{r}\right)$
$\displaystyle=f(s,x)+\int_{s}^{t}f_{x^{i}}\left(s,\int_{s}^{\rho}\sqrt{A}(r)dB_{r}\right)dB_{\rho}$
$\displaystyle\quad+\int_{s}^{t}a^{ij}(\rho)f_{x^{i}x^{j}}\left(s,x+\sqrt{2}\int_{s}^{\rho}\sqrt{A}(r)dB_{r}\right)d\rho$
(3.8)

for all $s\leq t<T$ $(a.s)$. Taking the expectations in (3.8), using the
property of the Itô integral that

$\mathbb{E}\left[\int_{s}^{t}f_{x^{i}}\left(s,\int_{s}^{\rho}\sqrt{A}(r)dB_{r}\right)dB_{\rho}\right]=0,$

and recalling the definition of $X_{t}$, we have

$\displaystyle\mathbb{E}\left[f\left(s,x+X_{t}-X_{s}\right)\right]$
$\displaystyle=f(s,x)+\mathbb{E}\left[\int_{s}^{t}a^{ij}(\rho)f_{x^{i}x^{j}}\left(s,x+X_{\rho}-X_{s}\right)d\rho\right]$
(3.9)

for all $0<s\leq t<T$. Taking the integration $\int_{0}^{t}\cdot~{}ds$ to both
sides of (3.9) and applying the Fubini Theorem, we have

$\displaystyle\int_{0}^{t}\mathbb{E}\left[f\left(s,x+X_{t}-X_{s}\right)\right]ds$
$\displaystyle=\int_{0}^{t}f(s,x)ds+\int_{0}^{t}\mathbb{E}\left[\int_{s}^{t}a^{ij}(\rho)f_{x^{i}x^{j}}\left(s,x+X_{\rho}-X_{s}\right)d\rho\right]ds$
$\displaystyle=\int_{0}^{t}f(s,x)ds+\int_{0}^{t}a^{ij}(\rho)\int_{0}^{\rho}\mathbb{E}\left[f_{x^{i}x^{j}}\left(s,x+X_{\rho}-X_{s}\right)\right]dsd\rho.$

Finally due to the definition of $u$ in (3.7), we have

$\displaystyle u_{t}(t,x)=a^{ij}(t)u_{x^{i}x^{j}}(t,x)+f(t,x)$

for all $t\in(0,T)$ and $x\in\mathbb{R}^{d}$.

Step 2. (Bounded case) In this step, we assume that $f$ is bounded.

We use Sobolev’s mollifiers. For $\varepsilon\in(0,1)$, denote

$f^{\varepsilon}(t,x)=\int_{\mathbb{R}^{d}}f(x-\varepsilon y)\varphi(y)dy,$

and

$\displaystyle
u^{\varepsilon}(t,x)=\int_{0}^{t}\mathbb{E}\left[f^{\varepsilon}(s,x+X_{t}-X_{s})\right]ds$

for all $t\in(0,T)$ and $x\in\mathbb{R}^{d}$. Then by the result in Step 1, we
have

$\displaystyle
u^{\varepsilon}_{t}(t,x)=a^{ij}(t)u^{\varepsilon}_{x^{i}x^{j}}(t,x)+f^{\varepsilon}(t,x).$

In particular, applying the integration by parts, for any $\phi\in
C_{c}^{\infty}(\mathbb{R}^{d})$, we have

$\displaystyle(u^{\varepsilon}(t,\cdot),\phi)=\int_{0}^{t}\left(u^{\varepsilon}(s,\cdot),a^{ij}(s)\phi_{x^{i}x^{j}}\right)ds+\int_{0}^{t}\left(f^{\varepsilon}(s,\cdot),\phi\right)ds\quad\forall
t\in(0,T).$

Since $f$ is bounded, applying the dominate convergence theorem one can easily
check that

$u(t,x):=\int_{0}^{t}\mathbb{E}\left[f(s,x+X_{t}-X_{s})\right]ds=\limsup_{\varepsilon\downarrow
0}u^{\varepsilon}(t,x)~{}(a.e.)$

and it becomes a solution to (3.1).

Step 3. (General case)

It suffices to remove the condition that $f$ is bounded. Due to the linearity
of equation (3.1) and the trivial decomposition
$f(t,x)=f^{+}(t,x)-f^{-}(t,x)$, we may assume that $f$ is nonnegative, where
$f^{+}(t,x)=\frac{|f(t,x)|+f(t,x)}{2}$ and
$f^{-}(t,x)=\frac{|f(t,x)|-f(t,x)}{2}$. For $M>0$, define
$f^{M}(t,x):=f(t,x)\wedge M:=\min\\{f(t,x),M\\}$ and denote

$\displaystyle
u^{M}(t,x)=\int_{0}^{t}\mathbb{E}\left[f^{M}(s,x+X_{t}-X_{s})\right]ds.$

Then by the result of step 2, for any $M>0$, we have

$\displaystyle(u^{M}(t,\cdot),\phi)=\int_{0}^{t}\left(u^{M}(s,\cdot),a^{ij}(s)\phi_{x^{i}x^{j}}\right)ds+\int_{0}^{t}\left(f^{M}(s,\cdot),\phi\right)ds\quad\forall
t\in(0,T).$ (3.10)

It is obvious that $u^{M}(t,x)\to u(t,x)$ for all $t\in(0,T)$ and
$x\in\mathbb{R}^{d}$ as $M\to\infty$. Finally, taking $M\to\infty$ and
applying the monotone and dominate convergence theorems in (3.10), we show
that $u$ is a solution to (3.1).

Part III. (Estimate)

We prove (3.3) and (3.4). By (3.2), the generalized Minkowski inequality, and
the translation invariant property of the $L_{p}$-space,

$\displaystyle\|u(t,\cdot)\|_{L_{p}}\leq\int_{0}^{t}\|f(s,\cdot)\|_{L_{p}}ds.$

Moreover, applying Hölder’s inequality, we have

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}h_{1}(t)dt$
$\displaystyle\leq\int_{0}^{T}h_{1}(t)\int_{0}^{t}\|f(s,\cdot)\|^{q}_{L_{p}}h_{2}(s)ds\left[\int_{0}^{t}|h_{2}(s)|^{-\frac{1}{q-1}}ds\right]^{q-1}dt.$

$\Box$

######  Remark 3.2.

Assume that

$\int_{0}^{T}\|f(s,\cdot)\|_{L_{p}}dt<\infty.$

Then due to (3.3) and the linearity of (3.1), one can easily find a continuous
modification of $u$ so that

$\displaystyle\sup_{t\in[0,T]}\|u(t,\cdot)\|_{L_{p}}\leq\int_{0}^{T}\|f(s,\cdot)\|_{L_{p}}ds\quad\forall
t\in[0,T].$

######  Corollary 3.3.

Let $T\in(0,\infty)$, $p\in[1,\infty]$, and $q\in(1,\infty)$. Suppose that
Assumption 2.1 holds. Additionally, assume that $h_{1}$ and $h_{2}$ are
functions on $[0,T]$ which are positive (a.e.). Then for any locally
integrable function $f$ on $(0,T)\times\mathbb{R}^{d}$, there is a unique
solution $u$ to equation (2.1) such that

$\displaystyle\sup_{t\in[0,T]}\left[\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}\right]$
$\displaystyle\leq\left[\int_{0}^{T}|h_{2}(t)|^{-\frac{1}{q-1}}dt\right]^{q-1}\int_{0}^{T}e^{-q\int_{0}^{t}c(s)ds}\|f(t,\cdot)\|^{q}_{L_{p}}h_{2}(t)dt.$
(3.11)

and

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}h_{1}(t)dt$
$\displaystyle\leq\int_{0}^{T}\left[h_{1}(t)\left[\int_{0}^{t}|h_{2}(s)|^{-\frac{1}{q-1}}ds\right]^{q-1}\int_{0}^{t}e^{-q\int_{0}^{s}c(\rho)d\rho}\|f(s,\cdot)\|^{q}_{L_{p}}h_{2}(s)ds\right]dt.$
(3.12)

###### Proof.

Let $v$ be a solution to the equation

$\displaystyle
v_{t}(t,x)=a^{ij}(t)v_{x^{i}x^{j}}(t,x)+e^{-\int_{0}^{t}c(s)ds}f\left(t,x-\int_{0}^{t}b(s)ds\right),$
$\displaystyle
v(0,x)=0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(t,x)\in(0,T)\times\mathbb{R}^{d}.$

Define $U(t,x)=e^{\int_{0}^{t}c(s)ds}v\left(t,x+\int_{0}^{t}b(s)ds\right)$,
where $b(t)=(b^{1}(t),\ldots,b^{d}(t))$. Then

$\displaystyle U_{t}(t,x)$
$\displaystyle=c(t)U(t,x)+e^{\int_{0}^{t}c(s)ds}\left(v_{t}\left(t,x+\int_{0}^{t}b(s)ds\right)+b^{i}(t)v_{x^{i}}\left(t,x+\int_{0}^{t}b(s)ds\right)\right)$
$\displaystyle=c(t)U(t,x)$
$\displaystyle\quad+e^{\int_{0}^{t}c(s)ds}\left(a^{ij}(t)v_{x^{i}x^{j}}\left(t,x+\int_{0}^{t}b(s)ds\right)+e^{-\int_{0}^{t}c(s)ds}f(t,x)\right)$
$\displaystyle\quad+e^{\int_{0}^{t}c(s)ds}\left(b^{i}(t)v_{x^{i}}\left(t,x+\int_{0}^{t}b(s)ds\right)\right)$
$\displaystyle=a^{ij}(t)U_{x^{i}x^{j}}(t,x)+b^{i}(t)U_{x^{i}}(t,x)+c(t)U(t,x)+f(t,x)$

and

$\displaystyle U(0,x)=0.$

Thus by the uniqueness of a solution, the solution $u$ to (2.1) is given by

$u(t,x)=e^{\int_{0}^{t}c(s)ds}v\left(t,x+\int_{0}^{t}b(s)ds\right)$

and obviously

$v(t,x)=e^{-\int_{0}^{t}c(s)ds}u\left(t,x-\int_{0}^{t}b(s)ds\right).$

Applying (3.4) to $v$ and using the translation invariant property of
$L_{p}$-norms, we obtain (3.12). Moreover, by (3.3) and Hölder’s inequality,
for any $0\leq t\leq T$, we have

$\displaystyle e^{-q\int_{0}^{t}c(s)ds}\|u(t,\cdot)\|^{q}_{L_{p}}$
$\displaystyle=\|v(t,\cdot)\|^{q}_{L_{p}}$
$\displaystyle\leq\int_{0}^{t}e^{-q\int_{0}^{s}c(\rho)d\rho}\|f(s,\cdot)\|^{q}_{L_{p}}h_{2}(s)ds\left[\int_{0}^{t}|h_{2}(s)|^{-\frac{1}{q-1}}ds\right]^{q-1}$
$\displaystyle\leq\int_{0}^{T}e^{-q\int_{0}^{t}c(s)ds}\|f(t,\cdot)\|^{q}_{L_{p}}h_{2}(t)dt\left[\int_{0}^{T}|h_{2}(s)|^{-\frac{1}{q-1}}ds\right]^{q-1},$

which obviously implies (3.11). $\Box$

##  4\. Estimates for non-degenerate equations

We start the section by reviewing previous weighted estimates with uniform
elliptic and bounded coefficients and apply these estimates to our model
equation (3.1). We denote

$\displaystyle\|f\|_{L_{p,q}(T,w)}=\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q}.$

As usual, $L_{p,q}(T,w)$ denote the spaces of all locally integrable functions
$f$ on $(0,T)\times\mathbb{R}^{d}$ such that $\|f\|_{L_{p,q}(T,w)}<\infty$.

######  Theorem 4.1.

Let $T\in(0,\infty)$, $p,q\in(1,\infty)$, and $w\in A_{q}(\mathbb{R})$. Assume
that the coefficients $a^{ij}(t)$ are uniformly bounded and elliptic, i.e.
there exist positive constants $M$ and $\delta$ such that

$\displaystyle M|\xi|^{2}\geq
a^{ij}(t)\xi^{i}\xi^{j}\geq\delta|\xi|^{2}\quad\forall\xi\in\mathbb{R}^{d}.$
(4.1)

Then for any $f\in L_{p,q}(T,w)$, there exists a unique solution $u$ to (3.1)
such that

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q}$
$\displaystyle\qquad\leq
N\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q},$
(4.2)

where

$N=N\left(p,q,M,\delta,[w]_{A_{q}(\mathbb{R})}\right).$

###### Proof.

It is a well-known result which could be easily obtained by combining some
classical results. However, it is not easy to find a paper covering the result
directly. Thus we refer two recent papers [5, Theorem 2.2] handling more
general coefficients and [3, Theorem 2.14] studying time measurable pseudo-
differential operators. $\Box$

######  Remark 4.2.

Theorem 4.1 is enough for our application. However, as shown in [5, Theorem
2.2] and [3, Theorem 2.14], $w_{0}(x)\in A_{p}(\mathbb{R}^{d})$ can be inside
(4.2) if (4.1) holds. In other words, we can find a positive constant $N$ such
that such that

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}w_{0}(x)dx\right)^{q/p}w(t)dt\right)^{1/q}$
$\displaystyle\qquad\leq
N\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}w_{0}(x)dx\right)^{q/p}w(t)dt\right)^{1/q},$

where

$N=N\left(p,q,M,\delta,[w]_{A_{q}(\mathbb{R})},[w_{0}]_{A_{p}(\mathbb{R}^{d})}\right).$

Next we want to enhance Theorem 4.1. Specifically, we show the constant $N$ in
(4.2) is independent of the upper bound $M$ of the coefficients $a^{ij}(t)$
and more precise relation between the constant $N$ and the elliptic constant
$\delta$. However, it seems to be almost impossible to prove it with only
analytic tools. Thus we recall probabilistic representations of solutions to
upgrade Theorem 4.1.

######  Theorem 4.3.

Let $T\in(0,\infty)$, $p,q\in(1,\infty)$, and $w\in A_{q}(\mathbb{R})$. Assume
that the coefficients $a^{ij}(t)$ are uniformly elliptic, i.e. there exists a
positive constant $\delta$ such that

$\displaystyle
a^{ij}(t)\xi^{i}\xi^{j}\geq\delta|\xi|^{2}\quad\forall\xi\in\mathbb{R}^{d}.$
(4.3)

Additionally, we assume that the coefficients $a^{ij}(t)$ are locally
integrable, i.e.

$\displaystyle\int_{0}^{t}a^{ij}(s)ds<\infty\qquad\forall t\in(0,T).$

Then for any $f\in L_{p,q}(T,w)$, there exists a unique solution $u$ to (3.1)
such that

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\leq\frac{N}{\delta^{q}}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(t)dt,$
(4.4)

where

$N=N\left(p,q,[w]_{A_{q}(\mathbb{R})}\right).$

###### Proof.

(Step 1) $a^{ij}(t)u_{x^{i}x^{j}}=\delta\Delta u$.

For this simple case, we use a basic scaling property of the equation. Put
$v(t,x)=u(t,\sqrt{\delta}x)$. Since $u$ is the solution to

$\displaystyle u_{t}(t,x)=\delta\Delta u(t,x)+f(t,x)$ $\displaystyle
u(0,x)=0,$

we have

$\displaystyle v_{t}(t,x)=\Delta v(t,x)+f(t,\sqrt{\delta}x)$ $\displaystyle
v(0,x)=0.$

Thus applying (4.2), we have

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|v_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q}$
$\displaystyle\qquad\leq
N\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,\sqrt{\delta}x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q},$

where

$N=N\left(p,q,[w]_{A_{q}(\mathbb{R})}\right).$

Finally, we obtain (4.4) by the simple change of the variable
$\sqrt{\delta}x\to x$.

(Step 2) General $a^{ij}(t)u_{x^{i}x^{j}}$.

To prove a general case, we use probabilistic solution representations. We may
assume that

$\int_{0}^{T}a^{ij}(t)dt<\infty$

since the constant $N$ in (4.4) is independent of $T$. Additionally, due to
the trivial constant extension $a^{ij}(t)1_{t\in(0,T)}+a^{ij}(T)1_{t\geq T}$,
we may assume that $a^{ij}(t)$ is defined on $(0,\infty)$. Consider two
independent $d$-dimensional Brownian motions $B_{t}$ and $W_{t}$ in a
probability space $(\Omega,\mathcal{F}_{t},\mathbb{P})$. Set

$\left(a^{ij}(t)\right)_{d\times d}=A(t)=\sqrt{A}(t)\times\sqrt{A}^{\ast}(t),$
$\displaystyle X_{t}:=\sqrt{2}\int_{0}^{t}\sqrt{A}^{ij}(s)dB_{s}^{j},$
$\displaystyle
X^{2}_{t}:=\sqrt{2}\int_{0}^{t}\left(\sqrt{A(s)-\delta{I}}^{ij}\right)dB_{s}^{j},$
$\displaystyle X^{1}_{t}:=\sqrt{2}\sqrt{\delta}{I}^{ij}W^{j}_{t},$

where $I=(I^{ij})_{d\times d}$ denotes the $d$ by $d$ identity matrix whose
diagonal entries are 1 and the other entries are zero and
$\sqrt{{A}(s)-\delta{I}}$ is a matrix so that

$\sqrt{{A}(s)-\delta{I}}\sqrt{{A}(s)-\delta{I}}=A(s)-\delta I,$

which exists due to (4.3), i.e. $A(s)-\delta I$ is a nonnegative symmetric
matrix. Then due to (3.2), the solution $u$ is given by

$\displaystyle u(t,x)$
$\displaystyle=\int_{0}^{t}\mathbb{E}\left[f(s,x+X_{t}-X_{s})\right]ds$
$\displaystyle=\int_{0}^{t}\mathbb{E}\left[f(s,x+X^{1}_{t}-X^{1}_{s}+X^{2}_{t}-X^{2}_{s})\right]ds,$
(4.5)

where the last equality is due to the fact that two probabilistic
distributions of $X_{t}-X_{s}$ and $X^{1}_{t}-X^{1}_{s}+X^{2}_{t}-X^{2}_{s}$
are equal for all $0<s<t$. Moreover, due to the independence of two Brownian
motions $B_{t}$ and $W_{t}$, we can split the random parameters in (4.5).
Additionally, applying Fubini’s theorem we have

$\displaystyle u(t,x)$
$\displaystyle=\int_{0}^{t}\mathbb{E}\left[f(s,x+X^{1}_{t}-X^{1}_{s}+X^{2}_{t}-X^{2}_{s})\right]ds$
$\displaystyle=\int_{0}^{t}\mathbb{E}^{\prime}\left[\mathbb{E}\left[f(s,x+X^{1}_{t}(\omega)-X^{1}_{s}(\omega)+X^{2}_{t}(\omega^{\prime})-X^{2}_{s}(\omega^{\prime}))\right]\right]ds$
$\displaystyle=\mathbb{E}^{\prime}\left[\int_{0}^{t}\mathbb{E}\left[f(s,x+X^{1}_{t}(\omega)-X^{1}_{s}(\omega)+X^{2}_{t}(\omega^{\prime})-X^{2}_{s}(\omega^{\prime}))\right]ds\right].$
(4.6)

For each fixing $\omega^{\prime}$, the function

$\displaystyle
v^{\omega^{\prime}}(t,x):=\int_{0}^{t}\mathbb{E}\left[f(s,x+X^{1}_{t}(\omega)-X^{1}_{s}(\omega)-X^{2}_{s}(\omega^{\prime}))\right]ds$

becomes a solution to the equation

$\displaystyle v^{\omega^{\prime}}_{t}(t,x)=\delta\Delta
v^{\omega^{\prime}}(t,x)+f(t,x-X^{2}_{t}(\omega^{\prime}))$ $\displaystyle
v^{\omega^{\prime}}(0,x)=0.$

Thus by the result in Step 1,

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|v^{\omega^{\prime}}_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\leq\frac{N}{\delta^{q}}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x-X^{2}_{t}(\omega^{\prime}))|^{p}dx\right)^{q/p}w(t)dt,$
(4.7)

where $N$ depends only on $p$, $q$, $[w]_{A_{q}(\mathbb{R})}$, and $\kappa$.
Moreover, by (4.6),

$\displaystyle
u_{xx}(t,x)=\mathbb{E}^{\prime}\left[v_{xx}^{\omega^{\prime}}\left(t,x+X_{t}^{2}(\omega^{\prime})\right)\right].$
(4.8)

Finally applying (4.8), (4.7), the generalized Minkowski’s inequality, and
Jensen’s inequality, we have

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt$
$\displaystyle\leq
N\mathbb{E}^{\prime}\left[\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|v^{\omega^{\prime}}_{xx}(t,x+X_{t}^{2}(\omega^{\prime}))|^{p}dx\right)^{q/p}w(t)dt\right]$
$\displaystyle\leq\frac{N}{\delta^{q}}\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}w\left(x+k(t)\right)dx\right)^{q/p}w(t)dt.$

$\Box$

######  Remark 4.4.

We hope that there is a positive constant $N$ such that such that

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q}$
$\displaystyle\qquad\leq\frac{N}{\delta^{q}}\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q},$

where

$N=N\left(p,q,[w]_{A_{q}(\mathbb{R})},[w_{0}]_{A_{p}(\mathbb{R}^{d})}\right).$

However, it cannot be obtained by following the proof of Theorem 4.3 since

$\int_{\mathbb{R}^{d}}|f(t,x-X^{2}_{t}(\omega^{\prime}))|^{p}dx=\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\quad\forall\omega^{\prime}~{}\text{and}~{}\forall
t$

is used in the proof.

##  5\. Proof of the main theorem

Proof of Theorem 2.5

Due to Theorem 3.1, the existence and uniqueness of a solution $u$ is obvious.
Moreover, (2.7), (2.8), and (2.9) can be easily obtained from (2.4), (2.5),
and (2.6) since $|t|^{\beta}\in A_{q}(\mathbb{R})$ for any $-1<\beta_{1}<q-1$
(see [18, Example 7.1.7]). Thus it suffices to show (2.4), (2.5) and (2.6).
Let $u$ be the solution to (2.1). First we show (2.4) and (2.5). For each
$\varepsilon\in(0,1)$, we denote

$h_{1,\varepsilon}(t)=w(\alpha(t)+\varepsilon
t)\left(\delta(t)+\varepsilon\right)$

and

$h_{2,\varepsilon}(t)=w(\alpha(t)+\varepsilon
t)|\delta(t)+\varepsilon|^{1-q}.$

Then by (3.11) and (3.12) with a simple change of variable,

$\displaystyle\sup_{t\in[0,T]}\left[\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}\right]$
$\displaystyle\leq\left[\int_{0}^{T}\left|w(\alpha(t)+\varepsilon
t)|\delta(t)+\varepsilon|^{1-q}\right|^{-\frac{1}{q-1}}dt\right]^{q-1}$
$\displaystyle\quad\times\int_{0}^{T}\|f(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t)+\varepsilon
t)|\delta(t)+\varepsilon|^{1-q}(t)dt$
$\displaystyle\leq\left[\int_{0}^{\alpha(T)+\varepsilon
T}|w(t)|^{-\frac{1}{q-1}}dt\right]^{q-1}$
$\displaystyle\quad\times\int_{0}^{T}\|f(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t)+\varepsilon
t)|\delta(t)+\varepsilon|^{1-q}(t)dt$

and

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t)+\varepsilon
t)\left(\delta(t)+\varepsilon\right)dt$
$\displaystyle\leq\left[\int_{0}^{T}w(\alpha(t)+\varepsilon
t)\left(\delta(t)+\varepsilon\right)\left[\int_{0}^{t}|w(\alpha(s)+\varepsilon
s)|\delta(s+\varepsilon)|^{1-q}|^{-\frac{1}{q-1}}ds\right]^{q-1}dt\right]$
$\displaystyle\quad\times\int_{0}^{T}\|f(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t)+\varepsilon
t)|\delta(t+\varepsilon)|^{1-q}dt.$

Moreover, by taking $\varepsilon\to 0$, we have

$\displaystyle\sup_{t\in[0,T]}\left[\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}\right]$
$\displaystyle\leq\left[\int_{0}^{\alpha(T)}|w(t)|^{-\frac{1}{q-1}}dt\right]^{q-1}\int_{0}^{T}\|f(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}(t)dt$

and

$\displaystyle\int_{0}^{T}\|u(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))\left(\delta(t)\right)dt$
$\displaystyle\leq\left[\int_{0}^{T}w(\alpha(t))\left(\delta(t)\right)\left[\int_{0}^{t}|w(\alpha(s))|\delta(s)|^{1-q}|^{-\frac{1}{q-1}}ds\right]^{q-1}dt\right]$
$\displaystyle\quad\times\int_{0}^{T}\|f(t,\cdot)\|^{q}_{L_{p}}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))|\delta(t)|^{1-q}dt.$
(5.1)

One may think that this limit procedure does not seem to be clear. However, it
is clear if our weight $w$ is continuous. Moreover, if $w$ is bounded, then
$w$ can be approximated by a sequence of continuous functions with a uniform
upper bound. Finally, considering $w\wedge M$ for any positive constant $M>0$,
we can complete the limit procedure due to the monotone convergence theorem as
$M\to\infty$.

We keep going to estimate the term in the middle of (5.1). Recalling the
definition of $[w]_{A_{p}(\mathbb{R})}$ and applying the change of variable
$\alpha(t):=\int_{0}^{t}\delta(s)ds\to t$, we have

$\displaystyle\int_{0}^{T}w(\alpha(t))\delta(t)\left[\int_{0}^{t}|w(\alpha(s))|\delta(s)|^{1-q}|^{-\frac{1}{q-1}}ds\right]^{q-1}dt$
$\displaystyle\leq\int_{0}^{T}w(\alpha(t))\delta(t)dt\left[\int_{0}^{T}|w(\alpha(t))|^{-\frac{1}{q-1}}\delta(t)ds\right]^{q-1}$
$\displaystyle\leq\int_{0}^{\alpha(T)}w(t)dt\left[\int_{0}^{\alpha(T)}|w(t)|^{-\frac{1}{q-1}}ds\right]^{q-1}$
$\displaystyle\leq[w]_{A_{p}(\mathbb{R})}\left[\alpha(T)\right]^{q}.$

By putting the above computations in (5.1), we obtain (2.5).

Next we prove (2.6). We may assume that $f$ has a compact support in
$[0,T]\times\mathbb{R}^{d}$. We divide the proof into several steps.

(Step 1) $\delta(t)\geq\varepsilon$ and $b^{i}(t)=c(t)=0$ for all $i$ and $t$.

We first assume that there exists a positive constant $\varepsilon\in(0,1)$
such that $\delta(t)\geq\varepsilon$ for all $t$. Additionally, suppose that
$b^{i}(t)=0$ and $c(t)=0$ for all $t$ and $i$ in this first step. Denote

$\alpha(t)=\int_{0}^{t}\delta(s)ds.$

Then $\beta(t)$ becomes a strictly increasing function and it has the inverse
$\beta(t):[0,\infty)\to[0,\infty)$ such that

$\displaystyle\beta^{\prime}(t)=\frac{1}{\alpha^{\prime}(\beta(t))}=\frac{1}{\delta(\beta(t))}\quad\forall
t\in[0,\infty).$ (5.2)

Define $v(t,x)=u(\beta(t),x)$. Then since $u$ is a solution to (2.1),

$\displaystyle
v_{t}(t,x)=u_{t}(\beta(t),x)\beta^{\prime}(t)=\frac{a^{ij}(\beta(t))}{\delta(\beta(t))}v_{x^{i}x^{j}}(t,x)+\frac{f(\beta(t),x)}{\delta(\beta(t))}$

and $v(0,x)=0$. Note that

$\frac{a^{ij}(\beta(t))}{\delta(\beta(t))}\xi^{i}\xi^{j}\geq|\xi|^{2}\quad\forall\xi\in\mathbb{R}^{d}.$

In other words, $v$ becomes the solution to

$\displaystyle
v_{t}(t,x)=\tilde{a}^{ij}(t)v_{x^{i}x^{j}}(t,x)+\frac{f(\beta(t),x)}{\delta(\beta(t))}\qquad(t,x)\in(0,T)\times\mathbb{R}^{d},$
$\displaystyle u(0,x)=0,$ (5.3)

with the coefficients
$\tilde{a}^{ij}(t)=\frac{a^{ij}(\beta(t))}{\delta(\beta(t))}$ whose elliptic
constant is $1$. Moreover, it is obvious that $\tilde{a}^{ij}(t)$ is locally
integrable. Indeed, by the change of the variable $\beta(t)\to t$ and (5.2),

$\displaystyle\int_{0}^{T}\tilde{a}^{ij}(t)dt=\int_{0}^{\beta(T)}a^{ij}(t)dt<\infty.$

Thus applying (4.4), we have

$\displaystyle\left(\int_{0}^{T_{0}}\left(\int_{\mathbb{R}^{d}}|v_{xx}(t,x)|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q}$
$\displaystyle\qquad\leq
N\left(\int_{0}^{T_{0}}\left(\int_{\mathbb{R}^{d}}\left|\frac{f(\beta(t),x)}{\delta(\beta(t))}\right|^{p}dx\right)^{q/p}w(t)dt\right)^{1/q},$
(5.4)

where

$N=N\left(p,q,[w_{0}]_{A_{q}(\mathbb{R})},\kappa\right)$

and $T_{0}$ is a constant so that $\beta(T_{0})=T$. By considering the change
of variables $\beta(t)\to t$ in (5.4), we finally obtain

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(\alpha(t))\delta(t)dt\right)^{1/q}$
$\displaystyle\qquad\lesssim\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(\alpha(t))(\delta(t))^{1-q}dt\right)^{1/q}.$
(5.5)

(Step 2) $b^{i}(t)=c(t)=0$ for all $i$ and $t$.

In this step, we remove the condition $\delta(t)\geq\varepsilon$. For any
$\varepsilon\in(0,1)$, we can rewrite (2.1) as

$\displaystyle u_{t}(t,x)=(a^{ij}(t)+\varepsilon I_{d\times
d})u_{x^{i}x^{j}}(t,x)+f(t,x)-\varepsilon\Delta u,$ $\displaystyle
u(0,x)=0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(t,x)\in(0,T)\times\mathbb{R}^{d},$

where $I_{d\times d}$ denotes the $d$ by $d$ identity matrix whose diagonal
entries are 1 and the other entries are zero. Thus applying (5.5), we have

$\displaystyle\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w(\alpha_{\varepsilon}(t))(\delta(t)+\varepsilon)dt\right)^{1/q}$
$\displaystyle\qquad\lesssim\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w(\alpha_{\varepsilon}(t))(\delta(t)+\varepsilon)^{1-q}dt\right)^{1/q}$
$\displaystyle\qquad\quad+\left(\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|\varepsilon\Delta
u(t,x)|^{p}dx\right)^{q/p}w(\alpha_{\varepsilon}(t))(\delta(t)+\varepsilon)^{1-q}dt\right)^{1/q},$
(5.6)

where $\alpha_{\varepsilon}(t)=\int_{0}^{t}(\delta(s)+\varepsilon)ds$. Observe
that

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|\varepsilon\Delta
u(t,x)|^{p}w(x+k(t))dx\right)^{q/p}w_{0}(\alpha_{\varepsilon}(t))(\delta(t)+\varepsilon)^{1-q}dt$
$\displaystyle=\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|\Delta
u(t,x)|^{p}w(x+k(t))dx\right)^{q/p}w_{0}(\alpha_{\varepsilon}(t))(\delta(t)+\varepsilon)\left(\frac{\varepsilon}{\delta(t)+\varepsilon}\right)^{q}dt,$
$\displaystyle(\delta(t)+\varepsilon)\left(\frac{\varepsilon}{\delta(t)+\varepsilon}\right)^{q}\leq(\delta(t))^{1-q}$

and

$\displaystyle(\delta(t)+\varepsilon)\left(\frac{\varepsilon}{\delta(t)+\varepsilon}\right)^{q}\to
0~{}\text{as}~{}\varepsilon\to 0,$

where $0^{1-q}:=\infty$. Thus due to the dominate convergence theorem and the
definition of the integral in (2.10), taking $\varepsilon\to 0$ in (5.6), we
have

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|u_{xx}(t,x)|^{p}dx\right)^{q/p}w_{0}(\alpha(t))\delta(t)dt$
$\displaystyle\lesssim\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}|f(t,x)|^{p}dx\right)^{q/p}w_{0}(\alpha(t))(\delta(t))^{1-q}dt.$
(5.7)

(Step 3) (General case).

Let $v$ be a solution to the equation

$\displaystyle
v_{t}(t,x)=a^{ij}(t)v_{x^{i}x^{j}}(t,x)+e^{-\int_{0}^{t}c(s)ds}f\left(t,x-\int_{0}^{t}b(s)ds\right),$
$\displaystyle
v(0,x)=0,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(t,x)\in(0,T)\times\mathbb{R}^{d}.$

The as shown in the proof of Corollary 3.3, the solution $u$ is given by

$u(t,x)=e^{\int_{0}^{t}c(s)ds}v\left(t,x+\int_{0}^{t}b(s)ds\right)$

and obviously

$v(t,x)=e^{-\int_{0}^{t}c(s)ds}u\left(t,x-\int_{0}^{t}b(s)ds\right).$

Applying (5) to $v$, we have

$\displaystyle\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|e^{-\int_{0}^{t}c(s)ds}u_{xx}\left(t,x-\int_{0}^{t}b(s)ds\right)(t,x)\right|^{p}dx\right)^{q/p}$
$\displaystyle\quad\times w(\alpha(t))\delta(t)dt$
$\displaystyle\lesssim\int_{0}^{T}\left(\int_{\mathbb{R}^{d}}\left|f\left(t,x-\int_{0}^{t}b(s)ds\right)\right|^{p}dx\right)^{q/p}e^{-q\int_{0}^{t}c(s)ds}w(\alpha(t))(\delta(t))^{1-q}dt.$

Finally, the translation $x\to x+\int_{0}^{t}b(s)ds$ leads us to (2.6).

$\Box$

##  6\. Acknowledgement

I would like to thank prof. Kyeong-Hun Kim for careful reading and suggesting
valuable comments.

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