Global Well-Posedness of Compressible Navier–Stokes Equation with $BV\cap L^1$ Initial Data

Abstract

The purpose of this paper is to study the well-posedness problem for weak solutions of Navier–Stokes equations in gas dynamics. We consider rough initial data, in $BV\cap L^1$. The well-posedness theory of Liu and Yu (Commun Pure Appl Math 75(2):223–348, 2022) for the isentropic Navier–Stokes equations is extended to the Navier–Stokes equations with an additional equation for the conservation of energy. A key step is to treat the energy equation as mainly for the dissipation of the temperature. The dissipation is analyzed through the heat kernel with BV variable coefficient constructed in Liu and Yu (2022). This step is natural from the physical point of view, but estimates for the temperature are required to be sufficiently robust for the validity of the conservation of energy in the weak sense; for this, we establish the regularity of the solutions, particularly the estimates of their time derivatives through refined estimates of the heat kernel.

Appendices

Appendix A. Heat Kernel Estimates

The goal of this section is to provide the proofs for the estimates in Section 2. We consider the following heat equation with the coefficient $\rho (x,t)$ being a BV function of x,

$$\begin{aligned} \left\{ \begin{aligned}&\left( \partial _t-\partial _x\rho (x,t)\partial _x\right) H(x,t;y,t_0;\rho )=0,\quad t>t_0,\\&H(x,t_0;y,t_0;\rho )=\delta (x-y). \end{aligned} \right. \end{aligned}$$

Here the BV coefficient $\rho (x,t)$ satisfies the following properties,

$$\begin{aligned} \left\{ \begin{aligned}&\Vert \rho (\cdot )-{\bar{\rho }}\Vert _{L^1}\leqq \delta _*,\quad \Vert \rho (\cdot ,t)\Vert _{BV}\leqq \delta _*,\quad \Vert \rho _t(\cdot ,t)\Vert _{\infty }\leqq \delta _*\max \left( \frac{1}{\sqrt{t}},1\right) ,\quad 0<\delta _*\ll 1,\\&{\mathscr {D}}\equiv \left\{ z\ |\ \rho (z,t)\text { is not continuous at }z \right\} \text{ is } \text{ invariant } \text{ in } t. \end{aligned} \right. \end{aligned}$$

(A.1)

To construct $H(x,t;y,t_0;\rho )$, the strategy is as follows: we first treat the case that $\rho$ is a step function in space variable and independent of time; then we use step function to approximate a general BV function (still time-independent); lastly, we use time-independent solution and time-frozen technique to construct the heat kernel for time-dependent BV coefficient.

1.1 A.1. Step function conductivity coefficient

Consider

$$\begin{aligned} \left\{ \begin{aligned}&\left( \partial _t-\partial _x\mu (x)\partial _x\right) H(x,t;y,t_0;\mu )=0,\quad t>t_0,\\&H(x,t_0;y,t_0;\mu )=\delta (x-y),\\ \end{aligned} \right. \end{aligned}$$

(A.2)

where $\mu (x)$ is a step function.

Proposition A.1

([8], Basic estimates) When step function $\mu$ satisfies that

$$\begin{aligned} \bigl | \mu (x)-\mu _0\bigr |\ll 1, \text{ and } \bigl \Vert \mu \bigr \Vert _{BV}\ll 1, \end{aligned}$$

the heat kernel for equation (A.2) satisfies the following estimates: for all $x,y\in {\mathbb {R}}$

$$\begin{aligned} \left\{ \begin{aligned}&H(x,t;y;\mu ) =\left( 1+ O(1) \Vert \mu \Vert _{BV}\right) \frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{4t}}}{\sqrt{4\pi t}},\\&\left| \partial _x H(x,t;y;\mu )\right| ,\left| \partial _y H(x,t;y;\mu )\right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{Dt}}}{t},\\&\left| \partial _{t} H(x,t;y;\mu ) \right| , \left| \partial _{xy} H(x,t;y;\mu ) \right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{Dt}}}{t^{3/2}},\\&\left| \partial _{tx} H(x,t;y;\mu ) \right| , \left| \partial _{ty} H(x,t;y;\mu ) \right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{Dt}}}{t^{2}}.\\ \end{aligned} \right. \end{aligned}$$

(A.3)

Here for $x,y\notin {\mathscr {D}}$ (the discontinuity set of $\mu (x)$), the partial derivatives are standard ones. While for $y=\alpha \in {\mathscr {D}}$, $H_y(x,t;\alpha +;\mu )$ and $H_y(x,t;\alpha -;\mu )$ both exist and satisfy the estimates.

Moreover, for all $\ell \in {\mathbb {N}}$, when $t>0$,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t^\ell H(x,t;y;\mu ):&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}}, \\&\mu (x)\partial _{t}^{\ell } H_x (x,t;y;\mu ) :&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}},\\&\mu (y) \partial _{t}^{\ell } H_y(x,t;y;\mu ) :&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}}.\\ \end{aligned} \right. \nonumber \\ \end{aligned}$$

(A.4)

Proposition A.2

([8], Comparison estimates) Suppose that the steps function $\mu ^{a}$ and $\mu ^{b}$ satisfy $\Vert \mu ^a \Vert _{BV}+\Vert \mu ^b \Vert _{BV}\ll 1$, $\inf \limits _{z\in {\mathbb {R}}} \mu ^{a}(z),\inf \limits _{z\in {\mathbb {R}}} \mu ^{b}(z)>{\underline{\mu }}>0$. Then for $t\in (0,e^{-1})$, $x,y\in {\mathbb {R}}$,

$$\begin{aligned}&| H(x,t;y;\mu ^{a})-H(x,t;y;\mu ^{b})| \leqq O(1) \Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } \frac{ e^{- \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{\sqrt{t}} , \end{aligned}$$

(A.5)

$$\begin{aligned}&| H_x(x,t;y;\mu ^{a})-H_x(x,t;y;\mu ^{b})| + | H_y(x,t;y;\mu ^{a})-H_y(x,t;y;\mu ^{b})| \nonumber \\&\quad \leqq O(1) \left( |\log t |\Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } + \Vert \mu ^{a}-\mu ^{b} \Vert _{BV} + \sqrt{t}\Vert \mu ^{a}-\mu ^{b} \Vert _{1} \right) \nonumber \\&\quad \times \, \frac{ e^{- \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{t}, \end{aligned}$$

(A.6)

$$\begin{aligned}&| H_{xy}(x,t;y;\mu ^{a})-H_{xy}(x,t;y;\mu ^{b})| \nonumber \\&\quad \leqq O(1) \left( |\log t |\Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } + \Vert \mu ^{a}-\mu ^{b} \Vert _{BV} + \sqrt{t}\Vert \mu ^{a}-\mu ^{b} \Vert _{1} \right) \nonumber \\&\quad \times \, \frac{ e^{- \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{t^{3/2}}. \end{aligned}$$

(A.7)

The first one comes from writing $H(x,t;y;\mu ^b)$ into an integral equation in terms of $H(x,t;y;\mu ^a)$ and direct computations. The derivative comparison are much more subtle. Straightforward differentiating the integral equation will induce non-integrable time singularity. One has to do delicate estimate on the Laplace wave train level, then invert it to physical variable. See [8] for details.

1.2 A.2. Time-independent conductivity coefficient

Now consider conductivity coefficient $\mu (x)$ is a general BV function. The strategy is to construct a sequence of step functions $\{\mu ^{k}(x)\}$ to approximate $\mu (x)$ in the following sense

$$\begin{aligned} \left\{ \begin{aligned}&\Vert \mu ^k \Vert _{BV}\leqq 2 \Vert \mu \Vert _{BV},\\&\Vert \mu ^k-\mu \Vert _{\infty } <\frac{1}{2^k}\rightarrow 0, \text{ as } k\rightarrow \infty . \end{aligned} \right. \end{aligned}$$

For each step function $\mu ^k(x)$, one can construct the heat kernel $H(x,t;y;\mu ^k)$ . Then it is shown that

Proposition A.3

([8], Theorem 3.6) Suppose $\Vert \mu \Vert _{BV}\ll 1$ and $\inf \limits _{z\in {\mathbb {R}}} \mu (z)>{\underline{\mu }}>0$. Let $\mu ^{k}$ be the step functions constructed as above. Then

$$\begin{aligned} H(x,t;y;\mu ) \equiv \lim _{k\rightarrow \infty } H(x,t;y;\mu ^{k}) \text{ exists }. \end{aligned}$$

$H(x,t;y;\mu )$ is a weak solution of

$$\begin{aligned} \left\{ \begin{aligned}&\left( \partial _t-\partial _x\mu (x)\partial _x\right) H(x,t;y;\mu )=0,\quad t>0,\\&H(x,0;y;\mu )=\delta (x-y),\\ \end{aligned} \right. \end{aligned}$$

(A.8)

and satisfies

$$\begin{aligned} \left\{ \begin{aligned}&H(x,t;y;\mu ) =\left( 1+ O(1) \Vert \mu \Vert _{BV}\right) \frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{4t}}}{\sqrt{4\pi t}},\\&\left| \partial _x H(x,t;y;\mu )\right| ,\left| \partial _y H(x,t;y;\mu )\right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{5t}}}{t},\\&\left| \partial _{t} H(x,t;y;\mu ) \right| , \left| \partial _{xy} H(x,t;y;\mu ) \right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{5t}}}{t^{3/2}},\\&\left| \partial _{tx} H(x,t;y;\mu ) \right| , \left| \partial _{ty} H(x,t;y;\mu ) \right| =O(1)\frac{e^{-\frac{\left( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu (z)}}\right) ^2}{5t}}}{t^{2}}.\\ \end{aligned} \right. \end{aligned}$$

(A.9)

Up to a sub-sequence,

$$\begin{aligned} \lim \limits _{k\rightarrow \infty } \partial ^\beta H(x,t;y;\mu ^{k}) = \partial ^\beta H(x,t;y;\mu ) \text{ exits }, \end{aligned}$$

where $\partial ^\beta \in \{\partial _x,\partial _y,\partial _{xy},\partial _t,\partial _{tx},\partial _{ty}\}$. Moreover, for all $\ell \in {\mathbb {N}}$, when $t>0$,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t^\ell H(x,t;y;\mu ):&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}}, \\&\mu (x)\partial _{t}^{\ell } H_x (x,t;y;\mu ) :&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}},\\&\mu (y) \partial _{t}^{\ell } H_y(x,t;y;\mu ) :&\text{ continuous } \text{ in } x\in {\mathbb {R}} \text{ for } \text{ all } y\in {\mathbb {R}},\; \text{ in } y\in {\mathbb {R}} \text{ for } \text{ all } x\in {\mathbb {R}}.\\ \end{aligned} \right. \nonumber \\ \end{aligned}$$

(A.10)

Proposition A.4

([8], Comparison estimates) Suppose that two BV functions $\mu ^{a}$ and $\mu ^{b}$ satisfy $\Vert \mu ^a \Vert _{BV}+\Vert \mu ^b \Vert _{BV}\ll 1$, $\inf \limits _{z\in {\mathbb {R}}} \mu ^{a}(z),\inf \limits _{z\in {\mathbb {R}}} \mu ^{b}(z)>{\underline{\mu }}>0$. Then for $x,y\in {\mathbb {R}}$,

$$\begin{aligned}&| H(x,y;y;\mu ^{a})-H(x,t;y;\mu ^{b})| \leqq O(1) \Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } \frac{ e^{ \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{\sqrt{t}} , \end{aligned}$$

(A.11)

$$\begin{aligned}&| H_x(x,y;y;\mu ^{a})-H_x(x,t;y;\mu ^{b})| + | H_y(x,y;y;\mu ^{a})-H_y(x,t;y;\mu ^{b})| \nonumber \\&\quad \leqq O(1) \left( \big (1+|\log t |\big ) \Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } + \Vert \mu ^{a}-\mu ^{b} \Vert _{BV} + \sqrt{t}\Vert \mu ^{a}-\mu ^{b} \Vert _{1} \right) \nonumber \\&\quad \times \, \frac{ e^{ \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{t}, \end{aligned}$$

(A.12)

$$\begin{aligned}&| H_{xy}(x,y;y;\mu ^{a})-H_{xy}(x,t;y;\mu ^{b})| \nonumber \\&\quad \leqq O(1) \left( \big (1+|\log t |\big ) \Vert \mu ^{a}-\mu ^{b} \Vert _{\infty } + \Vert \mu ^{a}-\mu ^{b} \Vert _{BV} + \sqrt{t}\Vert \mu ^{a}-\mu ^{b} \Vert _{1} \right) \nonumber \\&\qquad \times \, \frac{ e^{ \frac{\big ( \int \nolimits _{y}^{x} \frac{dz}{\sqrt{\mu ^{a}(z)}\vee \sqrt{\mu ^{b}(z)}} \big )^2}{5t} }}{t^{3/2}}. \end{aligned}$$

(A.13)

Propositions A.3 and A.4 are followed from comparison estimates of steps function in Proposition A.2 and a limiting procedure.

Proposition A.5

(Estimates involving time integral) Suppose $\Vert \mu \Vert _{BV}\ll 1$ and $\inf \limits _{z\in {\mathbb {R}}} \mu (z)>{\underline{\mu }}>0$. Let $H(x,t;y;\mu )$ be the associated heat kernel. Then there exists positive constant D such that

$$\begin{aligned} \left\{ \begin{aligned}&\left| \int \nolimits _{0}^{t} H_x(x,\tau ;y;\mu )d\tau \right| \leqq O(1) e^{-\frac{(x-y)^2}{D t}}&\text{ for } x\in {\mathbb {R}}, y\in {\mathbb {R}},\\&\left| \int \nolimits _{0}^{t} H_{xy}(x,\tau ;y;\mu )d\tau - \frac{\delta (x-y)}{\mu (x)} \right| \leqq O(1) \frac{e^{-\frac{(x-y)^2}{D t}}}{\sqrt{t}}&\text{ for } x\in {\mathbb {R}}, y\in {\mathbb {R}},\\&\left| \int \nolimits _{0}^{t} H_{xx}(x,\tau ;y;\mu )d\tau + \partial _x \left( \frac{H(x-y)}{\mu (x)}\right) \right| \leqq O(1) ( \frac{1}{\sqrt{t}} +|\partial _x \mu (x)| ) e^{-\frac{(x-y)^2}{D t}}&\text{ for } x\notin {\mathscr {D}}, y\in {\mathbb {R}},\\&\left| \int \nolimits _{0}^{t} H_{xxy}(x,\tau ;y;\mu )d\tau -\partial _x \left( \frac{\delta (x-y)}{\mu (x)} \right) \right| \leqq O(1) ( \frac{1}{\sqrt{t}} +|\partial _x \mu (x)| ) \frac{e^{-\frac{(x-y)^2}{D t}}}{\sqrt{t}}&\text{ for } x\notin {\mathscr {D}}, y\in {\mathbb {R}}.\\ \end{aligned} \right. \end{aligned}$$

Proof

By Proposition A.3, $H(x,t;y;\mu )$ is a solution to the heat equation (A.8) satisfying (A.9). Integrating the equation (A.8) with respect to time, and switching the differentiation, one has

$$\begin{aligned} H(x,t;y;\mu ) -\delta (x-y)=\partial _x \big (\mu (x) \partial _x \int \nolimits _{0}^{t} H(x,\tau ;y;\mu ) \mathrm{d}\tau \big ). \end{aligned}$$

Integrate against x to yield

$$\begin{aligned}&\int \nolimits _{0}^{t}\mu (x) \partial _x H(x,\tau ;y;\mu )\mathrm{d}\tau = \int \nolimits _{-\infty }^{x} H(z,t;y;\mu )\mathrm{d}z -H(x-y)\\&\quad = {\left\{ \begin{array}{ll} -\int \nolimits _{x}^{+\infty } H(z,t;y;\mu )\mathrm{d}z, &{} \text{ for } x>y,\\ \int \nolimits _{-\infty }^{x} H(z,t;y;\mu ) \mathrm{d}z, &{} \text{ for } x<y. \end{array}\right. } \end{aligned}$$

This then implies the following identities,

$$\begin{aligned} \int \nolimits _{0}^{t} H_x(x,\tau ;y;\mu ) \mathrm{d}\tau&= \frac{1}{\mu (x)} \left( \int \nolimits _{-\infty }^{x} H(z,t;y;\mu )\mathrm{d}z -H(x-y) \right) \nonumber \\&={\left\{ \begin{array}{ll} -\frac{1}{\mu (x)}\int \nolimits _{x}^{+\infty } H(z,t;y)\mathrm{d}z, &{} \text{ for } x>y, \\ \frac{1}{\mu (x)}\int \nolimits _{-\infty }^{x} H(z,t;y) \mathrm{d}z, &{} \text{ for } x<y, \end{array}\right. } \nonumber \\ \int \nolimits _{0}^{t} H_{xy} (x,\tau ;y;\mu ) \mathrm{d}\tau&= \frac{1}{\mu (x)} \left( \int \nolimits _{-\infty }^{x} H_y(z,t;y;\mu ) \mathrm{d}z +\delta (x-y) \right) , \nonumber \\ \int \nolimits _{0}^{t} H_{xx}(x,\tau ;y;\mu ) \mathrm{d}\tau&= \partial _x \left( \frac{1}{\mu (x)}\int \nolimits _{-\infty }^{x} H(z,t;y;\mu )\mathrm{d}z \right) - \partial _x \left( \frac{H(x-y)}{\mu (x)} \right) , \nonumber \\ \int \nolimits _{0}^{t} H_{xxy}(x,\tau ;y;\mu ) \mathrm{d}\tau&= \partial _x \left( \frac{1}{\mu (x)}\int \nolimits _{-\infty }^{x} H_y(z,t;y;\mu )\mathrm{d}z \right) + \partial _x \left( \frac{\delta (x-y)}{\mu (x)} \right) . \end{aligned}$$

(A.14)

By Proposition A.3 and straightforward computations, one completes the proof. $\square$

1.3 A.3. Time-dependent conductivity coefficient

Let $\rho (x,t)$ be a function satisfying (A.1). We are now in the position to consider the Green’s function $H(x,t;y,t_0;\rho )$ to the following equation,

$$\begin{aligned} \left\{ \begin{aligned}&\partial _{t} H=\partial _x \left( \rho (x,t) \partial _x H\right) ,\; t>t_0,\\&H(x,t_0;y,t_0;\rho )=\delta (x-y).\\ \end{aligned} \right. \end{aligned}$$

(A.15)

To establish the estimate for $H(x,t;y,t_0;\rho )$, we shall represent it by an integral equation using heat kernel with time-independent coefficient. We denote $H(x,t;y,t_0;\rho )$ by ${\bar{H}}(x,t;y,t_0)$ for the brevity of notation. In the sequential, we gather all the estimates of $H(x,t;y,t_0;\rho )$ which are needed in this paper.

Theorem A.1

Let $\rho (x,t)$ be a function satisfying (A.1). Then for $\delta _{*}$ sufficiently small and $t_0<t\ll 1$, the following estimates for heat kernel $H(x,y;y,t_0;\rho )$ hold

$$\begin{aligned}&\left| H(x,t;y,t_0;\rho ) \right| \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}}, \end{aligned}$$

(A.16)

$$\begin{aligned}&\left| H_x(x,t;y,t_0;\rho ) \right| , \left| H_y(x,t;y,t_0;\rho ) \right| \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{t-t_0}, \end{aligned}$$

(A.17)

$$\begin{aligned}&\left| H_t(x,t;y,t_0;\rho ) \right| , \left| H_{xy}(x,t;y,t_0;\rho ) \right| \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{(t-t_0)^{3/2}}, \end{aligned}$$

(A.18)

$$\begin{aligned}&\left| H_{ty}(x,t;y,t_0;\rho ) \right| \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{(t-t_0)^2}. \end{aligned}$$

(A.19)

Proof

• (Estimate of (A.16) and (A.17): H, $H_x$, $H_y$) For fixed $T>0$, set $\mu (x)\equiv \rho (x,T)$ and consider

  $$\begin{aligned} \int \nolimits _{t_0}^{t} H(x,t;z,\sigma ;\mu ) \left( \partial _{\sigma } {\bar{H}}(z,\sigma ;y,t_0)- \partial _z \left( \rho (z,\sigma ) \partial _z {\bar{H}}(z,\sigma ;y,t_0) \right) \right) \mathrm{d}z\mathrm{d}\sigma =0. \end{aligned}$$

  By the fact that $H(x,t;z,\sigma ;\mu )$ and $\mu (z) \partial _z H(x,t;z,\sigma ;\mu )$ are continuous in z, one performs integration by parts to get the representation of ${\bar{H}}(x,t;y,t_0)$,

  $$\begin{aligned} {\bar{H}}(x,t;y,t_0)= & {} H(x,t;y,t_0;\mu )\nonumber \\&\quad +\, \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_z(x,t;z,\sigma ) \big ( \rho (z,T)-\rho (z,\sigma )\big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma .\nonumber \\ \end{aligned}$$

  (A.20)

  Differentiate with respect to x to yield the integral equation of ${\bar{H}}_x$,

  $$\begin{aligned} {\bar{H}}_x(x,t;y,t_0)= & {} H_x(x,t;y,t_0;\mu )\nonumber \\&\quad +\, \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx}(x,t;z,\sigma ) \big ( \rho (z,T)-\rho (z,\sigma )\big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma .\nonumber \\ \end{aligned}$$

  (A.21)

  Suppose $\delta _{*}\ll 1$ in (A.1), from Proposition A.3, there exists positive $C_{*}$ such that

  $$\begin{aligned} \left| H_x(x,t;y,\tau ;\mu ) \right|\leqq & {} C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-\tau ) } }}{t-\tau },\quad \left| H_{xy}(x,t;y,\tau ;\mu ) \right| \\\leqq & {} C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-\tau ) } }}{(t-\tau )^{3/2}} \quad \text{ for } t>\tau ,\,x,y\in {\mathbb {R}}. \end{aligned}$$

  One thus makes the following weaker ansatz,

  $$\begin{aligned} \left| {\bar{H}}_x(x,t;y,\tau ) \right| \leqq 2 C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-\tau ) } }}{t-\tau }. \end{aligned}$$

  Setting $T=t$, one then has

  $$\begin{aligned} \left| \rho (z,t)-\rho (z,\sigma ) \right|= & {} \left| \int \nolimits _{\sigma }^{t} \partial _{\tau } \rho (z,\tau )\mathrm{d}\tau \right| \leqq \sup \limits _{ \tau \in [\sigma ,t]} \Vert \sqrt{\tau } \partial _{\tau }\rho (\cdot ,\tau ) \Vert _{\infty }\int \nolimits _{\sigma }^{t} \frac{\mathrm{d}\tau }{\sqrt{\tau }}\\\leqq & {} 2\delta _{*} \frac{t-\sigma }{\sqrt{t}}. \end{aligned}$$

  Substitute it and the ansatz into the integral in (A.21) to find

  $$\begin{aligned}&\left| \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx}(x,t;z,\sigma ) \big ( \rho (z,T)-\rho (z,\sigma )\big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \right| \\&\quad \leqq O(1) \delta _{*} C_{*}^2 \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \frac{e^{-\frac{(x-z)^2}{C_{*} (t-\sigma ) }}}{(t-\sigma )^{3/2}} \frac{t-\sigma }{\sqrt{t}} \frac{e^{-\frac{(z-y)^2}{C_{*} (\sigma -t_0) }}}{\sigma -t_0} \mathrm{d}z \mathrm{d}\sigma \\&\quad \leqq O(1) \delta _{*} C_{*}^2 \int \nolimits _{t_0}^{t} \frac{e^{-\frac{(x-y)^2}{C_{*} (t-t_0) }}}{\sqrt{t} \sqrt{t-t_0}\sqrt{\sigma -t_0}} \mathrm{d}\sigma \\&\quad \leqq O(1) \delta _{*} C_{*}^2\frac{e^{-\frac{(x-y)^2}{C_{*} (t-t_0) }}}{\sqrt{t}}. \end{aligned}$$

  Then one gets

  $$\begin{aligned} \left| {\bar{H}}_x(x,t;y,t_0) \right| \leqq \left( C_{*} + O(1) \delta _{*} C_{*}^2 \frac{t-t_0}{\sqrt{t}}\right) \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } }}{t-t_0}. \end{aligned}$$

  When $t-t_0<1$, and $\delta _{*}$ is sufficiently small, the ansatz is justified. This also follows that

  $$\begin{aligned} \left| {\bar{H}}(x,t;y,t_0) \right| \leqq 2 C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } }}{\sqrt{t-t_0}}. \end{aligned}$$

  If setting $T=t_0$ in (A.20) instead of $T=t$, taking derivative with respect to y, and following the similar argument, one can get the estimate for ${\bar{H}}_y(x,t;y,t_0)$ as well,

  $$\begin{aligned} \left| {\bar{H}}_y(x,t;y,t_0) \right| \leqq 2 C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } }}{t-t_0}. \end{aligned}$$

• (Estimate of the first term in (A.18): $H_{t}$) Next we estimate ${\bar{H}}_t(x,t;y,t_0)$ by difference estimate. By (A.20), we consider

  $$\begin{aligned}&{\bar{H}}(x,t+h;y,t_0)- {\bar{H}}(x,t;y,t_0)\\&\quad =H(x,t+h;y,t_0)-H(x,t;y,t_0)\\&\qquad +\, \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \left[ H_z(x,t+h;z,\sigma )-H_z(x,t;z,\sigma )\right] \\&\qquad \times \, \big ( \rho (z,T)-\rho (z,\sigma )\big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \\&\qquad +\, \int \nolimits _{t}^{t+h} \int \nolimits _{{\mathbb {R}}} H_z(x,t+h;z,\sigma )\big ( \rho (z,T)-\rho (z,\sigma )\big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma . \end{aligned}$$

  Taking $T=t$ and using Proposition A.3, one can estimate each term on the right-hand-side and get

  $$\begin{aligned} \bigl |{\bar{H}}(x,t+h;y,t_0)- {\bar{H}}(x,t;y,t_0) \bigr | \leqq O(1) |h| \frac{e^{-\frac{(x-y)^2}{C (t-t_0) }}}{(t-t_0)^{3/2}}. \end{aligned}$$

  Therefore one concludes that there exists positive $C_{*}$ such that

  $$\begin{aligned} \left| {\bar{H}}_t(x,t;y,t_0;\rho )\right| \leqq C_{*} \frac{e^{-\frac{(x-y)^2}{C_{*} (t-t_0) }}}{(t-t_0)^{3/2}}\quad \text{ for } 0<t-t_0<1,\, x,y\in {\mathbb {R}}. \end{aligned}$$

• (Estimate of the second term in (A.18): $H_{xy}$) The representation (A.20) is insufficient to get the higher order estimate ${\bar{H}}_{xy}$ due to high singularity in time integral. We take advantage of the estimates for time-independent coefficient problem and interpolate the heat kernels of time-independent coefficients frozen at t and $t_0$ to approximate the heat kernel for time-dependent coefficient, and prove it is indeed a good approximation when $t-t_0\ll 1$. Introduce a smooth non-increasing cutoff function $\chi (s)$ with the property

  $$\begin{aligned} \chi (s)= {\left\{ \begin{array}{ll} 1, &{}\quad \text{ for } 0\leqq s<\frac{1}{3},\\ 0, &{}\quad \text{ for } s>\frac{2}{3}. \end{array}\right. } \end{aligned}$$

  For fixed $t_0$ and t, set

  $$\begin{aligned} \left\{ \begin{aligned}&\mu ^t(x)\equiv \rho (x,t),\; \mu ^{t_0}(x)\equiv \rho (x,t_0),\\&{\tilde{H}}(x,t;y,\sigma )\equiv \chi \big ( \frac{\sigma -t_0}{t-t_0} \big ) H(x,t;y,\sigma ;\mu ^{t_0}) + \left( 1- \chi \big (\frac{\sigma -t_0}{t-t_0}\big ) \right) H(x,t;y,\sigma ;\mu ^{t}) \text{ for } \sigma \in [t_0,t]. \end{aligned} \right. \end{aligned}$$

  (A.22)

  Consider

  $$\begin{aligned}&\int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} {\tilde{H}}(x,t;z,\sigma ) \left( \partial _{\sigma } H(z,\sigma ;y,t_0;\rho )\right. \\&\quad \left. - \partial _z \big ( \rho (z,\sigma ) \partial _z H(z,\sigma ;y,t_0;\rho )\big ) \right) \mathrm{d}z \mathrm{d}\sigma =0. \end{aligned}$$

  Using integration by parts, and in view of the facts that $H(x,t;z,\sigma ;\mu ^t)$, $\mu ^t (z)H(x,t;z,\sigma ;\mu ^t)$, $H(x,t;z,\sigma ;\mu ^{t_0})$ and $\mu ^{t_0}(z)H_z(x,t;z,\sigma ;\mu ^{t_0})$ are all continuous in z, one obtains the representation of $H(x,t;y,t_0;\rho )$ for $t>t_0$,

  $$\begin{aligned} \begin{aligned} H(x,t;y,t_0;\rho )&= {\tilde{H}} (x,t;y,t_0) + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \frac{\chi ^{\prime }\big ( \frac{\sigma -t_0}{t-t_0} \big )}{t-t_0} \\&\quad \times \,\big ( H(x,t;z,\sigma ;\mu ^{t_0}) -H(x,t;z,\sigma ;\mu ^t)\big ) H(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma \\&\quad +\int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \chi \big ( \frac{\sigma -t_0}{t-t_0} \big ) H_z(x,t;z,\sigma ;\mu ^{t_0}) \\&\quad \times \,\big ( \rho (z,\sigma )-\mu ^{t_0}(z) \big ) H_z(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma \\&\quad + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \left( 1- \chi \big (\frac{\sigma -t_0}{t-t_0}\big ) \right) \\&\quad \times \,H_z(x,t;z,\sigma ; \mu ^t) \big ( \rho (z,\sigma )-\mu ^{t}(z) \big ) H_z(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma . \end{aligned} \end{aligned}$$

  (A.23)

  Differentiate (A.23) with respect to x and y to yield

  $$\begin{aligned} \begin{aligned} H_{xy}(x,t;y,t_0;\rho )&= {\tilde{H}}_{xy} (x,t;y,t_0) + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \frac{\chi ^{\prime }\big ( \frac{\sigma -t_0}{t-t_0} \big )}{t-t_0} \big ( H_x(x,t;z,\sigma ;\mu ^{t_0}) \\&\quad -H_x(x,t;z,\sigma ;\mu ^t)\big ) H_y(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma \\&\quad +\int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \chi \big ( \frac{\sigma -t_0}{t-t_0} \big ) H_{zx}(x,t;z,\sigma ;\mu ^{t_0}) \\&\quad \times \,\big ( \rho (z,\sigma )-\mu ^{t_0}(z) \big ) H_{zy}(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma \\&\quad + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \left( 1- \chi \big (\frac{\sigma -t_0}{t-t_0}\big ) \right) H_{zx}(x,t;z,\sigma ; \mu ^t) \\&\quad \times \,\big ( \rho (z,\sigma )-\mu ^{t}(z) \big ) H_{zy}(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma . \end{aligned} \end{aligned}$$

  (A.24)

  This gives rise to an integral equation for $H_{xy}(\cdot ,\cdot ;y,t_0;\rho )$. In the $\sigma$ integral, there are two possible singularities, that is, when $\sigma =t_0$ and $\sigma =t$. The advantage of this representation is that in each integral on the right-hand side of (A.24), only one singularity shows up thanks to cutoff function, and it can be controlled by either $\rho (\cdot ,\sigma )-\mu ^{t_0}$ or $\rho (\cdot ,\sigma )-\mu ^{t}$. By Propositions A.3 and A.4, following the similar arguments as in the estimate of $H_x(x,t;y,t_0;\rho )$, i.e., making weaker ansatz and proving a stronger one, we can conclude the estimate

  $$\begin{aligned} \left| H_{xy}(x,t;y,t_0;\rho ) \right| \leqq 2 C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } }}{(t-t_0)^{3/2}}. \end{aligned}$$

• (Estimate of (A.19): $H_{ty}$) In a similar way, one differentiates (A.23) in y to find the integral representation of $H_y(x,t;y,t_0;\rho )$. Taking the difference between $H_y(x,t+h;y,t_0;\rho )$ and $H_y(x,t;y,t_0;\rho )$, and by lengthy computations, we obtain that

  $$\begin{aligned}&\left| H_{y}(x,t+h;y,t_0;\rho )-H_{y}(x,t;y,t_0;\rho )\right| \\&\quad \leqq O(1) |h| \frac{e^{ \frac{(x-y)^2}{C(t-t_0)} }}{(t-t_0)^2} \qquad \text{ when } |h|<(t-t_0)/10. \end{aligned}$$

  Therefore we arrive at the conclusion that there exists positive $C_{*}$ such that

  $$\begin{aligned} \left| H_{ty}(x,t;y,t_0;\rho ) \right| \leqq C_{*} \frac{e^{ \frac{(x-y)^2}{C_{*}(t-t_0)} }}{(t-t_0)^2}. \end{aligned}$$

  $\square$

Theorem A.2

(Hölder continuity in time) When $t_0<s<t\ll 1$, one has Hölder continuity in time estimates

$$\begin{aligned}&\left\| H_x(\cdot ,t;y,t_0;\rho )-H_x(\cdot ,s;y,t_0;\rho ) \right\| _{\infty } \leqq C_{*} \frac{(t-s)\left| \log (t-s)\right| }{(s-t_0)(t-t_0)}, \end{aligned}$$

(A.25)

$$\begin{aligned}&\left\| H_x(\cdot ,t;y,t_0;\rho )-H_x(\cdot ,s;y,t_0;\rho ) \right\| _{1} \leqq C_{*} \frac{(t-s)\left| \log (t-s)\right| }{(t-t_0)\sqrt{s-t_0}}, \end{aligned}$$

(A.26)

$$\begin{aligned}&\left\| H_{xy}(\cdot ,t;y,t_0;\rho )-H_{xy}(\cdot ,s;y,t_0;\rho ) \right\| _{\infty } \leqq C_{*} \frac{(t-s)\left| \log (t-s)\right| }{(s-t_0)^{3/2}(t-t_0)}, \end{aligned}$$

(A.27)

$$\begin{aligned}&\left\| H_{xy}(\cdot ,t;y,t_0;\rho )-H_{xy}(\cdot ,s;y,t_0;\rho ) \right\| _{1} \leqq C_{*} \frac{(t-s)\left| \log (t-s)\right| }{(s-t_0)(t-t_0)}. \end{aligned}$$

(A.28)

Proof

• (Hölder continuity in time of $H_x(x,t;y,t_0;\rho )$) Assume $t>s>t_0$ and $t-s<1$, by (A.21),

  $$\begin{aligned}&H_{x} (x,t;y,t_0;\rho ) = H_{x}(x,t;y,t_0;\mu ^T) \\&\quad + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz} (x,t;z,\sigma ;\mu ^T) \big ( \rho (z,T)-\rho (z,\sigma ) \big ) H_{z}(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma , \end{aligned}$$

  where $\mu ^T=\rho (\cdot ,T)$. Set $T=s$ and denote

  $$\begin{aligned} {\bar{H}}(x,t;y,t_0) \equiv H(x,t;y,t_0;\rho ),\quad H(x,t;y,t_0)\equiv H(x,t;y,t_0;\mu ^s) \end{aligned}$$

  for simplicity of notations. Replacing t by s in the above representation and taking the difference, one gets

  $$\begin{aligned}&{\bar{H}}_x(x,t;y,t_0) -{\bar{H}}_x (x,s;y,t_0) \nonumber \\&\quad = H_x(x,t;y,t_0)- H_x(x,s;y,t_0) \nonumber \\&\qquad +\, \int \nolimits _{t_0}^{s} \int \nolimits _{{\mathbb {R}}} \big ( H_{zx}(x,t;z,\sigma )-H_{zx}(x,s;z,\sigma ) \big )\nonumber \\&\qquad \times \, \big ( \rho (z,s)-\rho (z,\sigma ) \big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&\qquad +\, \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx} (x,t;z,\sigma ) \big ( \rho (z,s)-\rho (z,\sigma ) \big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&\quad \equiv {\mathcal {I}}_1 + {\mathcal {I}}_2 +{\mathcal {I}}_3. \end{aligned}$$

  (A.29)

  Furthermore, one rewrites ${\mathcal {I}}_3$ as

  $$\begin{aligned} {\mathcal {I}}_3&= \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz}(x,t;z,\sigma ) \big ( \rho (z,s)-\rho (z,\sigma ) \big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&=- \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx} (x,t;z,\sigma ) \big ( \rho (z,\sigma ) -\rho (z,t) \big ) {\bar{H}}_z(z,\sigma ;y,t_0) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&\qquad - \, \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx} (x,t;z,\sigma ) \big ( \rho (z,t) -\rho (z,s) \big ) {\bar{H}}_z(z,t;y,t_0) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&\qquad - \, \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{zx} (x,t;z,\sigma ) \big ( \rho (z,t) -\rho (z,s) \big ) \nonumber \\&\qquad \times \, \big ( {\bar{H}}_z(z,\sigma ;y,t_0)- {\bar{H}}_z(z,t;y,t_0)\big ) \mathrm{d}z \mathrm{d}\sigma \nonumber \\&\quad \equiv {\mathcal {I}}_{31} +{\mathcal {I}}_{32}+{\mathcal {I}}_{33}. \end{aligned}$$

  (A.30)

  We first consider $L^\infty$ estimate. By Proposition A.3,

  $$\begin{aligned} \left| {\mathcal {I}}_1\right|= & {} \left| \int \nolimits _{s}^{t} H_{x \sigma } (x,\sigma ;y,t_0) \mathrm{d}\sigma \right| \\\leqq & {} O(1) \int \nolimits _{s}^{t} \frac{e^{-\frac{(x-y)^2}{C_{*} (\sigma -t_0)}}}{(\sigma -t_0)^{2}} \mathrm{d}\sigma \leqq O(1) \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } }}{t-t_0} \frac{t-s}{s-t_0}. \end{aligned}$$

  For ${\mathcal {I}}_2$, one has

  $$\begin{aligned} \begin{aligned} |{\mathcal {I}}_2 |&= \left| \int \nolimits _{t_0}^{s} \int \nolimits _{{\mathbb {R}}} \int \nolimits _{s}^{t} \int \nolimits _{\sigma }^{s} H_{\tau x z} (x,\tau ;z,\sigma ) \rho _{\chi }(z,\chi ) {\bar{H}}_z(z,\sigma ;y,t_0) d\chi \mathrm{d}\tau \mathrm{d}z \mathrm{d}\sigma \right| \\&\leqq O(1) C_{*}^2 \int \nolimits _{t_0}^{s} \int \nolimits _{{\mathbb {R}}} \int \nolimits _{s}^{t} \frac{ e^{-\frac{(x-z)^2}{C_{*} (\tau -\sigma ) }} }{(\tau -\sigma )^{5/2}} \delta _{*} \frac{s-\sigma }{\sqrt{s}} \frac{ e^{-\frac{(z-y)^2}{C_{*} (\sigma -t_0) }} }{\sigma -t_0} \mathrm{d}\tau \mathrm{d}z \mathrm{d}\sigma \\&\leqq O(1) \delta _{*} C_{*}^2 \int \nolimits _{t_0}^{s} \int \nolimits _{s}^{t} \frac{e^{-\frac{(x-y)^2}{C_{*} (\tau -t_0) }}}{\sqrt{s}} \frac{1}{\sqrt{\tau -t_0}} \frac{s-\sigma }{(\tau -\sigma )^2\sqrt{\sigma -t_0}} \mathrm{d}\tau \mathrm{d}\sigma . \end{aligned} \end{aligned}$$

  Carrying out the $\sigma$ integral,

  $$\begin{aligned} \begin{aligned}&\int \nolimits _{t_0}^{s} \frac{s-\sigma }{(\tau -\sigma )^2\sqrt{\sigma -t_0}} \mathrm{d}\sigma \\&\quad \leqq O(1) \frac{s-t_0}{(\tau -t_0)^2} \int \nolimits _{t_0}^{(t_0+s)/2} \frac{\mathrm{d}\sigma }{\sqrt{\sigma -t_0}} \\&\qquad +\, O(1) \frac{1}{\sqrt{s-t_0}} \int \nolimits _{(t_0+s)/2}^{s} \frac{s-\sigma }{(\tau -s +s-\sigma )^2} \mathrm{d}\sigma \\&\quad \leqq O(1) \left( \frac{(s-t_0)^{3/2}}{(\tau -t_0)^2} + \frac{\log \left( 1 + \frac{s-t_0}{2(\tau -s)}\right) }{\sqrt{s-t_0}}- \frac{\sqrt{s-t_0}}{\tau -s+\tau -t_0} \right) . \end{aligned} \end{aligned}$$

  Now one needs to calculate

  $$\begin{aligned} \int \nolimits _{s}^{t} \left( \frac{(s-t_0)^{3/2}}{(\tau -t_0)^{5/2}} + \frac{\log \left( 1 + \frac{s-t_0}{2(\tau -s)}\right) }{\sqrt{s-t_0}\sqrt{\tau -t_0}}- \frac{\sqrt{s-t_0}}{(\tau -s+\tau -t_0)\sqrt{\tau -t_0}} \right) \mathrm{d}\tau . \end{aligned}$$

  Straightforward computations show

  $$\begin{aligned} \int \nolimits _{s}^{t} \frac{(s-t_0)^{3/2}}{(\tau -t_0)^{5/2}} \mathrm{d}\tau + \int \nolimits _{s}^{t} \frac{\sqrt{s-t_0}}{(\tau -s+\tau -t_0)\sqrt{\tau -t_0}}\mathrm{d}\tau \leqq O(1) \frac{t-s}{t-t_0}. \end{aligned}$$

  To calculate

  $$\begin{aligned} \int \nolimits _{s}^{t} \frac{\log \left( 1 + \frac{s-t_0}{2(\tau -s)}\right) }{\sqrt{s-t_0}\sqrt{\tau -t_0}} \mathrm{d}\tau , \end{aligned}$$

  one consider two cases: (i). $t-s<\frac{s-t_0}{10}$ and (ii). $t-s>\frac{s-t_0}{10}$. For case (i),

  $$\begin{aligned} \begin{aligned}&\int \nolimits _{s}^{t} \frac{\log \left( 1 + \frac{s-t_0}{2(\tau -s)}\right) }{\sqrt{s-t_0}\sqrt{\tau -t_0}} \mathrm{d}\tau \leqq O(1) \frac{1}{s-t_0} \int \nolimits _{s}^{t} \log \left( \frac{s-t_0}{\tau -s}\right) \mathrm{d}\tau \\&\quad \leqq O(1) \frac{t-s}{s-t_0} \log \left( \frac{s-t_0}{t-s}\right) \leqq O(1) \frac{(t-s)\left| \log (t-s)\right| }{t-t_0}. \end{aligned} \end{aligned}$$

  For case (ii), one splits the integral to two parts,

  $$\begin{aligned}&\int \nolimits _{s}^{t} \frac{\log \left( 1 + \frac{s-t_0}{2(\tau -s)}\right) }{\sqrt{s-t_0}\sqrt{\tau -t_0}} \mathrm{d}\tau \\&\quad = \frac{1}{\sqrt{s-t_0}} \int \nolimits _{0}^{t-s} \frac{\log \left( 1 +\frac{s-t_0}{2\sigma } \right) }{\sqrt{\sigma +s-t_0}}\mathrm{d}\sigma \\&\quad \lesssim \int \nolimits _{0}^{(s-t_0)/10} \frac{\log \left( 1+ \frac{s-t_0}{2\sigma }\right) }{s-t_0} \mathrm{d}\sigma + \int \nolimits _{(s-t_0)/10}^{t-s} \frac{1}{\sqrt{s-t_0}} \frac{1}{\sqrt{\sigma }} \frac{s-t_0}{\sigma } \mathrm{d}\sigma \\&\quad \lesssim 1 + \left( 1-\frac{s-t_0}{10(t-s)}\right) . \end{aligned}$$

  Combining the above estimates then follows that

  $$\begin{aligned} |{\mathcal {I}}_2|\leqq O(1) \delta _{*} C_{*}^2 \frac{e^{-\frac{(x-y)^2}{C_{*} (t-t_0) }}}{\sqrt{s}} \frac{(t-s)\left| \log (t-s)\right| }{t-t_0} . \end{aligned}$$

  Now we estimate the first two terms in ${\mathcal {I}}_3$.

  $$\begin{aligned} |{\mathcal {I}}_{31}|&\leqq O(1) \delta _{*} C_{*}^2 \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} \frac{e^{-\frac{(x-z)^2}{C_{*} (t-\sigma ) }}}{(t-\sigma )^{3/2}} \frac{t-\sigma }{\sqrt{t}} \frac{e^{-\frac{(z-y)^2}{C_{*} (\sigma -t_0) }}}{\sigma -t_0} \mathrm{d}z \mathrm{d}\sigma \\&\leqq O(1)\delta _{*} C_{*}^2 \frac{e^{-\frac{(x-y)^2}{C_{*} (t-t_0) }}}{t-t_0} \frac{t-s}{\sqrt{t}}, \end{aligned}$$

  and

  $$\begin{aligned} |{\mathcal {I}}_{32}|&= \left| \int \nolimits _{{\mathbb {R}}\setminus {\mathscr {D}}} \left( \int \nolimits _{s}^{t} H_{xz}(x,t;z,\sigma ) \mathrm{d}\sigma \right) \big ( \rho (z,t) -\rho (z,s) \big ) {\bar{H}}_z(z,t;y,t_0) \mathrm{d}z \right| \\&\leqq \left| \int \nolimits _{{\mathbb {R}}\setminus {\mathscr {D}}} \frac{\delta (x-z)}{\mu (x)} \big ( \rho (z,t) -\rho (z,z) \big ) {\bar{H}}_z(z,t;y,t_0) \mathrm{d}z \right| \\&\quad +O(1) \delta _{*} C_{*}^2\int \nolimits _{{\mathbb {R}}\setminus {\mathscr {D}}} \frac{e^{- \frac{(x-z)^2}{C_{*}(t-s)} }}{\sqrt{t-s}} \frac{t-s}{\sqrt{t}} \frac{e^{- \frac{(z-y)^2}{C_{*}(t-t_0)} }}{t-t_0} \mathrm{d}z\\&\leqq O(1) \delta _{*} \frac{e^{- \frac{(x-y)^2}{C_{*} (t-t_0) } } }{t-t_0} \frac{t-s}{\sqrt{t}} + O(1) \delta _{*} C_{*}^2 \frac{e^{- \frac{(x-y)^2}{ C_{*} (t-t_0+t-s) } } }{t-t_0} \frac{t-s}{\sqrt{t}}\\&\leqq O(1) \delta _{*} \frac{e^{- \frac{(x-y)^2}{2 C_{*} (t-t_0) } } }{t-t_0} \frac{t-s}{\sqrt{t}}. \end{aligned}$$

  Combine the estimates of ${\mathcal {I}}_1$, ${\mathcal {I}}_2$, ${\mathcal {I}}_{31}$ and ${\mathcal {I}}_{32}$ to yield an integral equation

  $$\begin{aligned} \begin{aligned}&{\bar{H}}_{x}(x,t;y,t_0) - {\bar{H}}_{x}(x,s;y,t_0) \\&\quad = \Phi (x,t;y,t_0) - \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz}(x,t;z,\sigma ) \big ( \rho (z,t)-\rho (z,s) \big ) \\&\qquad \times \, \big ( {\bar{H}}_z(z,\sigma ;y,t_0)- {\bar{H}}_z(z,t;y,t_0)\big ) \mathrm{d}z \mathrm{d}\sigma . \end{aligned} \end{aligned}$$

  (A.31)

  Here

  $$\begin{aligned} \left| \Phi (x,t;y,t_0) \right| \leqq O(1) e^{-\frac{(x-y)^2}{2C_{*} (t-t_0) }} \frac{(t-s)\left| \log (t-s) \right| }{(t-t_0)(s-t_0)}. \end{aligned}$$

  Thus there exists positive $C_1$ such that

  $$\begin{aligned} \left\| \Phi (\cdot ,t;y,t_0) \right\| _{\infty } \leqq C_{1} \frac{(t-s)\left| \log (t-s) \right| }{(t-t_0)(s-t_0)}. \end{aligned}$$

  We make the following ansatz:

  $$\begin{aligned} \left\| {\bar{H}}_{x}(\cdot ,t;y,t_0)-{\bar{H}}_{x}(\cdot ,s;y,t_0) \right\| _{\infty } \leqq 2 C_1 \frac{(t-s)\left| \log (t-s) \right| }{(t-t_0)(s-t_0)}. \end{aligned}$$

  Substituting the ansatz into ${\mathcal {I}}_{33}$, one gets

  $$\begin{aligned} \left| {\mathcal {I}}_{33}\right|\leqq & {} O(1) \int \nolimits _{s}^{t} \int \nolimits _{{\mathbb {R}}} \frac{e^{- \frac{(x-z)^2}{C_{*}(t-\sigma )}}}{(t-\sigma )^{3/2}} \delta _{*} \frac{t-s}{\sqrt{t}} \frac{(t-\sigma )\left| \log (t-\sigma )\right| }{(t-t_0)(\sigma -t_0)} \mathrm{d}z \mathrm{d}\sigma \nonumber \\\leqq & {} O(1) \delta _{*} \int \nolimits _{s}^{t} \frac{t-s}{\sqrt{t} (t-t_0) } \frac{\left| \log (t-\sigma )\right| }{\sigma -t_0} \mathrm{d}\sigma \nonumber \\\leqq & {} O(1) \delta _{*} \frac{t-s}{t-t_0} \frac{1}{\sqrt{t}} \left( \int \nolimits _{s}^{(t+s)/2} \frac{\left| \log (t-s)\right| }{\sigma -t_0}\mathrm{d}\sigma + \int \nolimits _{(t+s)/2}^{t} \frac{\left| \log (t-\sigma )\right| }{t-t_0} \mathrm{d}\sigma \right) \nonumber \\\leqq & {} O(1) \delta _{*}\frac{t-s}{t-t_0} \frac{\left| \log (t-s)\right| }{\sqrt{t}} \left[ \log \left( 1+\frac{t-s}{s-t_0}\right) + \frac{t-s}{t-t_0} \right] \nonumber \\\leqq & {} O(1) \delta _{*} \frac{(t-s)\left| \log (t-s)\right| }{(t-t_0)(s-t_0)} \frac{s-t_0}{\sqrt{t}} \left( \frac{t-s}{s-t_0} + \frac{t-s}{t-t_0}\right) \nonumber \\\leqq & {} O(1) \delta _{*} \frac{(t-s)\left| \log (t-s) \right| }{(t-t_0)(s-t_0)}. \end{aligned}$$

  (A.32)

  Therefore the ansatz is justified provided $\delta _{*}$ is sufficiently small. The estimates for $L^1_x$ norm is even simpler. Actually, integrating the magnitude of (A.29) with respect to x and using (A.30), we obtain the representation of

  $$\begin{aligned} \int \nolimits _{{\mathbb {R}}} \left| {\bar{H}}_x(x,t;y,t_0) -{\bar{H}}_x (x,s;y,t_0) \right| \mathrm{d}x. \end{aligned}$$

  By calculating known terms, making suitable ansatz and justifying it provided $\delta _{*}$ sufficiently, we conclude

  $$\begin{aligned} \left\| {\bar{H}}_{x}(\cdot ,t;y,t_0)-{\bar{H}}_{x}(\cdot ,s;y,t_0) \right\| _{1} \leqq O(1) \frac{(t-s)\left| \log (t-s) \right| }{(t-t_0)\sqrt{s-t_0}}. \end{aligned}$$

• (Hölder continuity in time of $H_{xy}(x,t;y,t_0;\rho )$) In this case, due to the high singularity, (A.21) is not appropriate to use, one has to resort to expression (A.24). By applying similar arguments as for Hölder estimates of $H_x$, we can conclude the proof. $\quad \square$

Theorem A.3

(Estimates involving time integral for time-dependent coefficient) Let $\rho (x,t)$ be a function satisfying (A.1). Then for $\delta _{*}$ sufficiently small and $t_0<t\ll 1$. The following estimates for heat kernel $H(x,y;y,t_0;\rho )$ hold

$$\begin{aligned}&\left| \int \nolimits _{t_0}^tH_x(x,\tau ;y,t_0;\rho )d\tau \right| \leqq C_*e^{-\frac{(x-y)^2}{C_*(t-t_0)}}, \end{aligned}$$

(A.33)

$$\begin{aligned}&\left| \int \nolimits _{t_0}^t H_x(x,t;y,s;\rho )\mathrm{d}s\right| \leqq C_*e^{-\frac{(x-y)^2}{C_*(t-t_0)}}, \end{aligned}$$

(A.34)

$$\begin{aligned}&\left| \int \nolimits _{t_0}^tH_{xy}(x,\tau ;y,t_0;\rho )d\tau -\frac{\delta (x-y)}{\rho (x,t_0)}\nonumber \right. \\&\quad \left. -\, \int \nolimits _{t_0}^{t} \frac{\rho (x,t_0)-\rho (x,\tau )}{\rho (x,t_0)} H_{xy}(x,\tau ;y,t_0;\rho ) d\tau \right| \leqq C_{*} \frac{e^{-\frac{(x-y)^2}{C_*(t-t_0)}}}{\sqrt{t-t_0}}, \end{aligned}$$

(A.35)

$$\begin{aligned}&\left| \int \nolimits _{t_0}^t H_{xy}(x,t;y,s;\rho )\mathrm{d}s +\frac{\delta (x-y)}{\rho (y,t)}\nonumber \right. \\&\left. \quad -\, \int \nolimits _{t_0}^{t} \frac{\rho (y,t)-\rho (y,s)}{\rho (y,t)} H_{xy}(x,t;y,s;\rho ) \mathrm{d}s \right| \leqq C_{*} \frac{e^{-\frac{(x-y)^2}{C_*(t-t_0)}}}{\sqrt{t-t_0}}, \end{aligned}$$

(A.36)

$$\begin{aligned}&\int \nolimits _{t_0}^{t} H_{xx}(x,\tau ;y,t_0;\rho ) d\tau = -\frac{\delta (x-y)}{\rho (x,t_0)}-\frac{1}{\rho (x,t_0)} \partial _{x}\nonumber \\&\quad \times \, \left[ \int \nolimits _{t_0}^{t} \left( \rho (x,\tau )-\rho (x,t_0)\right) H_x(x,\tau ;y,t_0) d\tau \right] \nonumber \\&\quad +\, O(1) \left( |\partial _{x} \rho (x,t_0)| e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} } + \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}}\right) , \text{ for } x\notin {\mathscr {D}}, \, \end{aligned}$$

(A.37)

$$\begin{aligned}&\int \nolimits _{t_0}^{t} {\bar{H}}_{xxy}(x,\tau ;y,t_0) d\tau = \frac{1}{\rho (x,t_0)} \nonumber \\&\quad \times \, \left[ \delta ^{\prime }(x-y) - \int \nolimits _{t_0}^{t} \partial _{x}\Big [ \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_{xy}(x,\tau ;y,t_0) \Big ] d\tau \right] \nonumber \\&\quad -\, \frac{\partial _{x}\rho (x,t_0)}{\rho ^2(x,t_0)} \left[ \delta (x-y) - \int \nolimits _{t_0}^{t} \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_{xy}(x,\tau ;y,t_0) d\tau \right] \nonumber \\&\quad +\, O(1) \left( |\partial _{x} \rho (x,t_0)| \frac{ e^{ -\frac{(x-y)^2}{C_{*}(t-t_0)} } }{\sqrt{t-t_0}}+ \frac{e^{ -\frac{(x-y)^2}{C_{*}(t-t_0)} }}{t-t_0} \right) \text{ for } x\notin {\mathscr {D}}, \end{aligned}$$

(A.38)

$$\begin{aligned}&\int \nolimits _{t_0}^tH_{t}(x,t;y,s;\rho )\mathrm{d}s = H(x,t-t_0;y;\mu ^t)-\delta (x-y) + O(1) \delta _{*} e^{-\frac{(x-y)^2}{C_*(t-t_0)}}. \end{aligned}$$

(A.39)

Proof

$\bullet$ (Estimates of (A.33)–(A.38))

We begin with

$$\begin{aligned} \partial _{\tau } {\bar{H}}(x,\tau ;y,t_0) = \partial _{x} \left[ \rho (x,\tau ) \partial _{x} {\bar{H}}(x,\tau ;y,t_0)\right] . \end{aligned}$$

Integrate with respect to $\tau$ from $t_0$ to t to yield

$$\begin{aligned} {\bar{H}}(x,t;y,t_0)-\delta (x-y) = \partial _{x} \left[ \int \nolimits _{t_0}^{t} \rho (x,\tau ) \partial _{x} {\bar{H}} (x,\tau ;y,t_0) \mathrm{d}\tau \right] . \end{aligned}$$

Integrate against x to get

$$\begin{aligned} \int \nolimits _{-\infty }^{x} {\bar{H}}(z,t;y,t_0) \mathrm{d}z - H(x-y) = \int \nolimits _{t_0}^{t} \rho (x,\tau ) {\bar{H}}_{x} (x,\tau ;y,t_0) \mathrm{d}\tau . \end{aligned}$$

Using of this, one can write

$$\begin{aligned} \begin{aligned}&\int \nolimits _{t_0}^{t} {\bar{H}}_{x} (x,\tau ;y,t_0) \mathrm{d}\tau \\&\quad = \int \nolimits _{t_0}^{t} \left( \frac{\rho (x,\tau )}{\rho (x,t_0)} -\frac{ \rho (x,\tau )-\rho (x,t_0)}{\rho (x,t_0)} \right) {\bar{H}}_x(x,\tau ;y,t_0) \mathrm{d}\tau \\&\quad = \frac{1}{\rho (x,t_0)} \left[ \int \nolimits _{-\infty }^{x} {\bar{H}}(z,t;y,t_0) \mathrm{d}z - H(x-y) \right. \\&\qquad \left. -\, \int \nolimits _{t_0}^{t} \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_x(x,\tau ;y,t_0) \mathrm{d}\tau \right] . \end{aligned} \end{aligned}$$

(A.40)

This immediately follows that

$$\begin{aligned} \begin{aligned}&\int \nolimits _{t_0}^{t} {\bar{H}}_{xx} (x,\tau ;y,t_0) \mathrm{d}\tau \\&\quad = \partial _{x} \left( \frac{1}{\rho (x,t_0)}\right) \left[ \int \nolimits _{-\infty }^{x} {\bar{H}}(z,t;y,t_0) \mathrm{d}z - H(x-y) \right. \\&\left. \qquad -\, \int \nolimits _{t_0}^{t} \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_x(x,\tau ;y,t_0) \mathrm{d}\tau \right] \\&\qquad +\, \frac{1}{\rho (x,t_0)} \left[ {\bar{H}}(x,t;y,t_0) -\delta (x-y) -\int \nolimits _{t_0}^{t} \partial _{x}\right. \\&\qquad \times \,, \left. \Big [ \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_x(x,\tau ;y,t_0) \Big ] \mathrm{d}\tau \right] , \end{aligned} \end{aligned}$$

which in turn gives

$$\begin{aligned} \begin{aligned}&\int \nolimits _{t_0}^{t} {\bar{H}}_{xxy} (x,\tau ;y,t_0) \mathrm{d}\tau \\&\quad = \partial _{x} \left( \frac{1}{\rho (x,t_0)}\right) \left[ \int \nolimits _{-\infty }^{x} {\bar{H}}_{y}(z,t;y,t_0) \mathrm{d}z +\delta (x-y)\right. \\&\left. \quad -\, \int \nolimits _{t_0}^{t} \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_{xy}(x,\tau ;y,t_0) \mathrm{d}\tau \right] \\&\qquad +\, \frac{1}{\rho (x,t_0)} \left[ {\bar{H}}_{y}(x,t;y,t_0) +\delta ^{\prime }(x-y) \right. \\&\left. \qquad -\, \int \nolimits _{t_0}^{t} \partial _{x}\Big [ \big (\rho (x,\tau )-\rho (x,t_0) \big ) {\bar{H}}_{xy}(x,\tau ;y,t_0) \Big ] \mathrm{d}\tau \right] . \end{aligned} \end{aligned}$$

From above expressions, and the following identities (which hold for $x\notin {\mathscr {D}}$),

$$\begin{aligned} \begin{aligned}&{\bar{H}}_{xx}(x,\tau ;y,t_0) = \frac{1}{\rho (x,\tau )}{\bar{H}}_{\tau }(x,\tau ;y,t_0) - \frac{\partial _{x} \rho (x,\tau )}{\rho (x,\tau )} {\bar{H}}_{x}(x,\tau ;y,t_0),\\&{\bar{H}}_{xxy}(x,\tau ;y,t_0) = \frac{1}{\rho (x,\tau )}{\bar{H}}_{\tau y}(x,\tau ;y,t_0) - \frac{\partial _{x} \rho (x,\tau )}{\rho (x,\tau )} {\bar{H}}_{xy}(x,\tau ;y,t_0), \end{aligned} \end{aligned}$$

We conclude the proof.

The estimate for (A.39) is of more technical difficulties. We first write down the integral equation for ${\bar{H}}(x,t;y,s)$ in terms of $H(x,t;y,s;\mu ^t)$, then represent the time derivative in t, perform the time integral in s, and use comparison estimates in Theorem (A.4) to conclude the proof. $\quad \square$

Theorem A.4

(Comparison Estimates for time-dependent coefficient) Let $\rho ^a(x,t)$ and $\rho ^b(x,t)$ be two functions satisfying (A.1). Suppose $t_0<t\ll 1$. Then the following comparison estimates hold:

$$\begin{aligned}&\left| H(x,t;y,t_0;\rho ^b) - H(x,t;y,t_0;\rho ^a) \right| \nonumber \\&\quad \leqq C_{*}\frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}} {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }, \end{aligned}$$

(A.41)

$$\begin{aligned}&\left| H_x(x,t;y,t_0;\rho ^a) - H_x(x,t;y,t_0;\rho ^b) \right| ,\;\nonumber \\&\quad \left| H_y(x,t;y,t_0;\rho ^a) - H_y(x,t;y,t_0;\rho ^b) \right| \nonumber \\&\quad \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{t-t_0} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\nonumber \right. \\&\left. \qquad +\, \sqrt{t-t_0} \Big ({\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1}+\left| \log t\right| {\left| \left| \left| \frac{\sqrt{\tau }}{\left| \log \tau \right| } \partial _{\tau }\left[ \rho ^a-\rho ^b\right] \right| \right| \right| }_{\infty } \Big ) \right] , \end{aligned}$$

(A.42)

$$\begin{aligned}&\left| H_{xy}(x,t;y,t_0;\rho ^a) - H_{xy}(x,t;y,t_0;\rho ^b) \right| ,\; \nonumber \\&\quad \left| H_{t}(x,t;y,t_0;\rho ^a) - H_{t}(x,t;y,t_0;\rho ^b) \right| \nonumber \\&\quad \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{(t-t_0)^{3/2}} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\nonumber \right. \\&\left. \qquad +\, \sqrt{t-t_0} \Big ({\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1}+\left| \log t\right| {\left| \left| \left| \frac{\sqrt{\tau }}{\left| \log \tau \right| } \partial _{\tau }\left[ \rho ^a-\rho ^b\right] \right| \right| \right| }_{\infty } \Big ) \right] . \end{aligned}$$

(A.43)

Proof

• (Comparison estimates of H) Consider

  $$\begin{aligned}&0=\int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H(x,t;z,\sigma ;\rho ^a)\\&\quad \times \left[ \partial _{\sigma } H(z,\sigma ;y,t_0;\rho ^b) -\partial _{z} \big (\rho ^b(z,\sigma ) \partial _{z} H(z,\sigma ;y,t_0;\rho ^b) \big ) \right] \mathrm{d}z \mathrm{d}\sigma . \end{aligned}$$

  Use integration by parts to find

  $$\begin{aligned}&H(x,t;y,t_0;\rho ^b) - H(x,t;y,t_0;\rho ^a) = -\int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_z(x,t;z,\sigma ;\rho ^a)\\&\quad \times \, \big [\rho ^b(z,\sigma )-\rho ^a(z,\sigma )\big ] H_z (z,\sigma ;y,t_0;\rho ^b) \mathrm{d}z \mathrm{d}\sigma . \end{aligned}$$

  It follows from this that

  $$\begin{aligned} \begin{aligned}&\left| H(x,t;y,t_0;\rho ^b) - H(x,t;y,t_0;\rho ^a) \right| \\&\quad \leqq O(1) {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty } \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \frac{e^{- \frac{(x-z)^2}{C_{*}(t-\sigma )} }}{t-\sigma } \frac{e^{- \frac{(z-y)^2}{C_{*}(\sigma -t_0)} }}{\sigma -t_0} \mathrm{d}z \mathrm{d}\sigma \\&\quad \leqq O(1) {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty } \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}}. \end{aligned} \end{aligned}$$

• (Comparison estimates of $H_x$) By (A.21) and setting $T=t$,

  $$\begin{aligned}&H_x(x,t;y,t_0;\rho ) = H_x (x,t;y,t_0;\mu ^t) \\&\quad + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz} (x,t;z,\sigma ;\mu ^t) \left[ \rho (z,t)-\rho (z,\sigma ) \right] H_z(z,\sigma ;y,t_0;\rho ) \mathrm{d}z \mathrm{d}\sigma , \end{aligned}$$

  where $\mu ^t(z)\equiv \rho (z,t)$. Substituting $\rho =\rho ^a,\rho ^b$ and taking difference, one has

  $$\begin{aligned}&H_x (x,t;y,t_0;\rho ^a)-H_x (x,t;y,t_0;\rho ^b)\\&\quad = H_x (x,t;y,t_0;\mu _a^t) - H_x (x,t;y,t_0;\mu _b^t)\\&\qquad +\, \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \left[ H_{xz} (x,t;z,\sigma ;\mu _a^t)-H_{xz} (x,t;z,\sigma ;\mu _b^t) \right] \\&\qquad \times \,\left[ \rho ^a (z,t)-\rho ^a(z,\sigma ) \right] H_z(z,\sigma ;y,t_0;\rho ^a) \mathrm{d}z \mathrm{d}\sigma \\&\qquad +\, \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz} (x,t;z,\sigma ;\mu _b^t)\\&\qquad \times \, \left[ \big (\rho ^a (z,t)-\rho ^a(z,\sigma )\big )-\big (\rho ^b(z,t)-\rho ^b(z,\sigma )\big ) \right] H_z(z,\sigma ;y,t_0;\rho ^a) \mathrm{d}z \mathrm{d}\sigma \\&\qquad + \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_{xz} (x,t;z,\sigma ;\mu _b^t)\\&\qquad \times \, \left[ \rho ^b (z,t)-\rho ^b(z,\sigma ) \right] \left[ H_z(z,\sigma ;y,t_0;\rho ^a)-H_z(z,\sigma ;y,t_0;\rho ^b) \right] \mathrm{d}z \mathrm{d}\sigma \\&\quad \equiv T_1 + T_2 + T_3 + T_4. \end{aligned}$$

  Calculating out the terms $T_1$, $T_2$ and $T_3$, one has

  $$\begin{aligned} \begin{aligned}&\left| T_1\right| + \left| T_2\right| + \left| T_3\right| \\&\quad \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{t-t_0} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \qquad +\, \sqrt{t-t_0} \Big ({\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1}+ \left| \log t\right| {\left| \left| \left| \frac{\sqrt{\tau }}{\left| \log \tau \right| } \partial _{\tau }\left[ \rho ^a-\rho ^b\right] \right| \right| \right| }_{\infty } \Big ) \right] . \end{aligned} \end{aligned}$$

  Note that to estimate $T_4$, one needs to bound

  $$\begin{aligned} H_z(z,\sigma ;y,t_0;\rho ^a)-H_z(z,\sigma ;y,t_0;\rho ^b), \end{aligned}$$

  which is exactly the estimates we are seeking for. In other words, we have an integral equation for function

  $$\begin{aligned} H_x (x,t;y,t_0;\rho ^a)-H_x (x,t;y,t_0;\rho ^b). \end{aligned}$$

  Making ansatz and substituting into $T_4$, one can justify the ansatz provided $O(1)\delta _{*} \sqrt{t-t_0}<C_{*}/2$.

• (Comparison estimates of $H_y$) Next, we shall estimate

  $$\begin{aligned} H_y(x,t;y,t_0;\rho ^a) - H_y(x,t;y,t_0;\rho ^b). \end{aligned}$$

  From

  $$\begin{aligned}&0= \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} \Big [ \partial _{\sigma } H(x,t;z,\sigma ;\rho ) + \partial _z \left[ \rho (z,\sigma ) \partial _z H(x,t;z,\sigma ;\rho ) \right] \Big ] \\&\quad \times \, H(z,\sigma ;y,t_0;\mu ^{t_0}) \mathrm{d}z \mathrm{d}\sigma , \end{aligned}$$

  one can find another representation of $H(x,t;y,t_0;\rho )$ in terms of the time-independent heat kernel,

  $$\begin{aligned}&H(x,t;y,t_0;\rho )= H(x,t;y,t_0;\mu ^{t_0}) - \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_z(x,t;z,\sigma ;\rho )\\&\quad \times \, \big [ \rho (z,\sigma )-\rho (z,t_0)\big ] H_z(z,\sigma ;y,t_0;\mu ^{t_0}) \mathrm{d}z \mathrm{d}\sigma . \end{aligned}$$

  Differentiate it with respect to y to obtain

  $$\begin{aligned}&H_y(x,t;y,t_0;\rho )= H_y(x,t;y,t_0;\mu ^{t_0}) - \int \nolimits _{t_0}^{t} \int \nolimits _{{\mathbb {R}}} H_z(x,t;z,\sigma ;\rho ) \\&\quad \times \, \big [ \rho (z,\sigma )-\rho (z,t_0)\big ] H_{zy}(z,\sigma ;y,t_0;\mu ^{t_0}) \mathrm{d}z \mathrm{d}\sigma . \end{aligned}$$

  Then by substituting $\rho =\rho ^a,\rho ^b$, taking difference and following the similar arguments as above, one can conclude

  $$\begin{aligned}&\left| H_y(x,t;y,t_0;\rho ^a) - H_y(x,t;y,t_0;\rho ^b) \right| \\&\quad \leqq C_{*} \frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{t-t_0} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \quad +\, \sqrt{t-t_0} \Big ({\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1}+\left| \log t\right| {\left| \left| \left| \frac{\sqrt{\tau }}{\left| \log \tau \right| } \partial _{\tau }\left[ \rho ^a-\rho ^b\right] \right| \right| \right| }_{\infty } \Big ) \right] . \end{aligned}$$

• (Comparison estimates of $H_{xy}$) As for

  $$\begin{aligned} H_{xy}(x,t;y,t_0;\rho ^a) - H_{xy}(x,t;y,t_0;\rho ^b), \end{aligned}$$

  thanks to (A.24), one can do much more tedious but similar in spirit estimates as above to complete the estimate.

• (Comparison estimates of $H_t$) For comparison of $H_t$, we shall use

  $$\begin{aligned} \begin{aligned}&H_t(x,t;y,t_0;\rho ^a) - H_t (x,t;y,t_0;\rho ^b)\\&\quad =\lim \limits _{h\rightarrow 0}\frac{1}{h} \left[ \big (H(x,t+h;y,t_0;\rho ^a)-H(x,t;y,t_0;\rho ^a)\big )\right. \\&\left. \qquad - \big (H(x,t+h;y,t_0;\rho ^b)-H(x,t;y,t_0;\rho ^b)\big ) \right] \end{aligned} \end{aligned}$$

  and prove the uniform estimate of difference quotient in h. We may assume $0<h<(t-t_0)/10$. As the estimates are of the same spirit, but lengthy, we omit the details. $\quad \square$

Taking advantage of the equation itself or representing the time-dependent coefficient heat kernel in terms of time-independent heat kernel as in Theorem A.3, we have the following comparison estimates involving time integral.

Theorem A.5

(Comparison estimates involving time integral) Let $\rho ^a(x,t)$ and $\rho ^b(x,t)$ be two functions satisfying (A.1). Suppose $t_0<t\ll 1$. Then the following comparison estimates hold:

$$\begin{aligned}&\left| \int \nolimits _{t_0}^{t} \left[ H_{x} (x,\tau ;y,t_0;\rho ^a)- H_{x} (x,\tau ;y,t_0;\rho ^b) \right] d\tau \right| ,\;\\&\left| \int \nolimits _{t_0}^{t} \left[ H_{y} (x,t;y,s;\rho ^a)- H_{y} (x,t;y,s;\rho ^b) \right] \mathrm{d}s \right| \\&\quad \leqq C_{*} e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} } \left[ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \qquad +\, {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1} +{\left| \left| \left| \frac{\sqrt{\tau }}{|\log \tau |}\partial _{\tau }(\rho ^a-\rho ^b) \right| \right| \right| }_{\infty } \right] ,\\&\left| \int \nolimits _{t_0}^{t} \left[ H_{x} (x,t;y,s;\rho ^a)- H_{x} (x,t;y,s;\rho ^b) \right] \mathrm{d}s \right| ,\; \\&\left| \int \nolimits _{t_0}^{t} \left[ H_{y} (x,\tau ;y,t_0;\rho ^a)- H_{y} (x,\tau ;y,t_0;\rho ^b) \right] d\tau \right| \\&\quad \leqq C_{*} e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} } \left[ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty }+ {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \qquad +\, {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1} +{\left| \left| \left| \frac{\sqrt{\tau }}{|\log \tau |}\partial _{\tau }(\rho ^a-\rho ^b) \right| \right| \right| }_{\infty } \right] ,\\&\int \nolimits _{t_0}^{t} \left[ H_{xy} (x,\tau ;y,t_0;\rho ^a)- H_{xy} (x,\tau ;y,t_0;\rho ^b) \right] d\tau \\&\quad = \left[ \frac{1}{\rho ^a(x,t_0)}-\frac{1}{\rho ^b(x,t_0)}\right] \delta (x-y)\\&\qquad -\, \int \nolimits _{t_0}^{t} \left[ \frac{\rho ^a(x,\tau )-\rho ^a(x,t_0)}{\rho ^a(x,t_0)} H_{xy}(x,\tau ;y,t_0;\rho ^a)\right. \\&\left. \qquad -\, \frac{\rho ^b(x,\tau )-\rho ^b(x,t_0)}{\rho ^b(x,t_0)} H_{xy}(x,\tau ;y,t_0;\rho ^b) \right] d\tau \\&\qquad + O(1)\frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty } + {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \qquad +\, {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1} + {\left| \left| \left| \frac{\sqrt{\tau }}{|\log \tau |}\partial _{\tau }(\rho ^a-\rho ^b) \right| \right| \right| }_{\infty } \right] ,\\&\int \nolimits _{t_0}^{t} \left[ H_{xy} (x,t;y,s;\rho ^a)- H_{xy} (x,t;y,s;\rho ^b) \right] \mathrm{d}s\\&\quad = \left[ \frac{1}{\rho ^a(y,t)}-\frac{1}{\rho ^b(y,t)}\right] \delta (x-y)\\&\qquad +\, \int \nolimits _{t_0}^{t} \left[ \frac{\rho ^a(y,t)-\rho ^a(y,s)}{\rho ^a(y,t)} H_{xy}(x,t;y,s;\rho ^a)\right. \\&\left. \qquad - \, \frac{\rho ^b(y,t)-\rho ^b(y,s)}{\rho ^b(y,t)} H_{xy}(x,t;y,s;\rho ^b) \right] d\tau \\&\qquad +\, O(1)\frac{e^{- \frac{(x-y)^2}{C_{*}(t-t_0)} }}{\sqrt{t-t_0}} \left[ \left| \log (t-t_0)\right| {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{\infty } + {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{BV}\right. \\&\left. \qquad +\, {\left| \left| \left| \rho ^a-\rho ^b \right| \right| \right| }_{1} + {\left| \left| \left| \frac{\sqrt{\tau }}{|\log \tau |}\partial _{\tau }(\rho ^a-\rho ^b) \right| \right| \right| }_{\infty } \right] . \end{aligned}$$

Appendix B. Explicit Expressions of Constants in Green’s Function

1.1 B.1. Constants in Lemmas 6.1 and 6.2

The following give the explicit expressions of the constants appeared in Lemmas 6.1 and 6.2.

$$\begin{aligned}&\alpha ^*_2 =\frac{\mu }{v},\quad \alpha ^*_3 =\frac{\kappa \theta _e}{v},\quad \beta ^*_2=\frac{v \left( \mu p p_e+\kappa \theta _e p_v-\mu p_v\right) }{\mu \left( \mu -\kappa \theta _e\right) },\quad \beta ^*_3=\frac{p v p_e}{\kappa \theta _e-\mu },\\&A_{1,1}= -\frac{v^3 \left( \kappa \theta _e p_v^2+\mu p p_e p_v\right) }{ \kappa \mu ^3 \theta _e},\\&A_{2,1}=\frac{v^3 \left( \mu ^3 p^2 p_e^2-\mu p p_e p_v \left( \kappa ^2 \theta _e^2-3 \kappa \mu \theta _e+2 \mu ^2\right) +p_v^2 \left( \mu -\kappa \theta _e\right) {}^3\right) }{\mu ^3 \left( \mu -\kappa \theta _e\right) {}^3},\\&A_{3,1}= \frac{p v^3 p_e \left( \kappa p \theta _e p_e+p_v \left( \kappa \theta _e-\mu \right) \right) }{\kappa \theta _e \left( \kappa \theta _e-\mu \right) {}^3}. \end{aligned}$$

$$\begin{aligned}&M_1^{*,0}=\left( \begin{array}{ccc} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array}\right) ,\quad M_1^{*,1}=\left( \begin{array}{ccc} 0 &{} \frac{v}{\mu } &{} 0 \\ -\frac{v p_v}{\mu } &{} 0 &{} 0 \\ -\frac{u v p_v}{\mu } &{} 0 &{} 0 \\ \end{array} \right) ,\quad M_1^{*,2}=\left( \begin{array}{ccc} -\frac{v^2 p_v}{\mu ^2} &{} -\frac{u v^2 p_e}{\kappa \mu \theta _e} &{} \frac{v^2 p_e}{\kappa \mu \theta _e} \\ 0 &{} \frac{v^2 p_v}{\mu ^2} &{} 0 \\ -\frac{p v^2 p_v}{\kappa \mu \theta _e} &{} \frac{u v^2 p_v}{\mu ^2} &{} 0 \\ \end{array} \right) , \end{aligned}$$

$$\begin{aligned}&M_1^{*,3}=\left( \begin{array}{ccc} 0 &{} -\frac{ v^3 \left( p \mu p_e+2 \kappa p_v \theta _e\right) }{\kappa \mu ^3 \theta _e} &{} 0 \\ \frac{ v^3 p_v \left( p \mu p_e+2 \kappa p_v \theta _e\right) }{\kappa \mu ^3 \theta _e} &{} \frac{ u v^3 p_e p_v}{\kappa \mu ^2 \theta _e} &{} -\frac{ v^3 p_e p_v}{\kappa \mu ^2 \theta _e} \\ \frac{ u v^3 p_v \left( p \mu p_e+2 \kappa p_v \theta _e\right) }{\kappa \mu ^3 \theta _e} &{} -\frac{ v^3 \left( p p_v-u^2 p_e p_v\right) }{\kappa \mu ^2 \theta _e} &{} -\frac{ u v^3 p_e p_v}{\kappa \mu ^2 \theta _e} \end{array} \right) .\\&M_2^{*,0}=\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} 1 &{} 0 \\ 0 &{} u &{} 0 \\ \end{array} \right) ,\quad M_2^{*,1}=\left( \begin{array}{ccc} 0 &{} -\frac{ v}{\mu } &{} 0 \\ \frac{ v p_v}{\mu } &{} -\frac{u v p_e}{\mu -\kappa \theta _e} &{} \frac{v p_e}{\mu -\kappa \theta _e} \\ \frac{u v p_v}{\mu } &{} \frac{v \left( p-u^2 p_e\right) }{\mu -\kappa \theta _e} &{} \frac{ u v p_e}{\mu -\kappa \theta _e} \\ \end{array} \right) ,\\&M_2^{*,2}=\left( \begin{array}{ccc} \frac{v^2 p_v}{\mu ^2} &{} -\frac{u v^2 p_e}{\mu ^2-\kappa \mu \theta _e} &{} \frac{v^2 p_e}{\mu ^2-\kappa \mu \theta _e} \\ 0 &{} \frac{v^2 \left( p \mu ^2 p_e-p_v \left( \mu -\kappa \theta _e\right) {}^2\right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^2} &{} 0 \\ -\frac{p v^2 p_v}{\mu ^2-\kappa \mu \theta _e} &{} \frac{u v^2 \left( 2 p \mu ^2 p_e-p_v \left( \mu -\kappa \theta _e\right) {}^2\right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^2} &{} -\frac{p v^2 p_e}{\left( \mu -\kappa \theta _e\right) {}^2} \\ \end{array} \right) , \end{aligned}$$

$$\begin{aligned}&M_2^{*,3}=\left( \begin{array}{ccc} 0 &{} \frac{v^3 \left( 2 p_v+\frac{p \mu p_e \left( \kappa \theta _e-2 \mu \right) }{\left( \mu -\kappa \theta _e\right) {}^2}\right) }{\mu ^3} &{} 0 \\ -\frac{ v^3 p_v \left( 2 p_v+\frac{p \mu p_e \left( \kappa \theta _e-2 \mu \right) }{\left( \mu -\kappa \theta _e\right) {}^2}\right) }{\mu ^3} &{} -\frac{ u v^3 p_e \left( 2 p \mu ^2 p_e-p_v \left( 2 \mu ^2-3 \kappa \theta _e \mu +\kappa ^2 \theta _e^2\right) \right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^3} &{} \frac{ v^3 p_e \left( 2 p \mu ^2 p_e-p_v \left( 2 \mu ^2-3 \kappa \theta _e \mu +\kappa ^2 \theta _e^2\right) \right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^3} \\ -\frac{ u v^3 p_v \left( 2 p_v \left( \mu -\kappa \theta _e\right) {}^2+p \mu p_e \left( \kappa \theta _e-2 \mu \right) \right) }{\mu ^3 \left( \mu -\kappa \theta _e\right) {}^2} &{} \frac{ v^3 \left( p-u^2 p_e\right) \left( 2 p \mu ^2 p_e-p_v \left( 2 \mu ^2-3 \kappa \theta _e \mu +\kappa ^2 \theta _e^2\right) \right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^3} &{} \frac{ u v^3 p_e \left( 2 p \mu ^2 p_e-p_v \left( 2 \mu ^2-3 \kappa \theta _e \mu +\kappa ^2 \theta _e^2\right) \right) }{\mu ^2 \left( \mu -\kappa \theta _e\right) {}^3} \end{array} \right) . \end{aligned}$$

$$\begin{aligned}&M_3^{*,0}=\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} -u &{} 1 \\ \end{array} \right) ,\quad M_3^{*,1}=\left( \begin{array}{ccc} 0 &{} 0 &{} 0 \\ 0 &{} \frac{u v p_e}{\mu -\kappa \theta _e} &{} -\frac{v p_e}{\mu -\kappa \theta _e} \\ 0 &{} -\frac{v \left( p-u^2 p_e\right) }{\mu -\kappa \theta _e} &{} -\frac{u v p_e}{\mu -\kappa \theta _e} \\ \end{array} \right) ,\\&M_3^{*,2}=\left( \begin{array}{ccc} 0 &{} \frac{u v^2 p_e}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) } &{} -\frac{v^2 p_e}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) } \\ 0 &{} -\frac{p v^2 p_e}{\left( \mu -\kappa \theta _e\right) {}^2} &{} 0 \\ \frac{p v^2 p_v}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) } &{} -\frac{2 p u v^2 p_e}{\left( \mu -\kappa \theta _e\right) {}^2} &{} \frac{p v^2 p_e}{\left( \mu -\kappa \theta _e\right) {}^2} \\ \end{array} \right) ,\\&M_3^{*,3}=\left( \begin{array}{ccc} 0 &{} \frac{p v^3 p_e}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) {}^2} &{} 0 \\ -\frac{p v^3 p_e p_v}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) {}^2} &{} \frac{u v^3 p_e \left( p_v \left( \mu -\kappa \theta _e\right) -2 p \kappa p_e \theta _e\right) }{\kappa \theta _e \left( \kappa \theta _e-\mu \right) {}^3} &{} \frac{v^3 p_e \left( 2 p \kappa p_e \theta _e+p_v \left( \kappa \theta _e-\mu \right) \right) }{\kappa \theta _e \left( \kappa \theta _e-\mu \right) {}^3} \\ -\frac{p u v^3 p_e p_v}{\kappa \theta _e \left( \mu -\kappa \theta _e\right) {}^2} &{} \frac{v^3 \left( p-u^2 p_e\right) \left( 2 p \kappa p_e \theta _e+p_v \left( \kappa \theta _e-\mu \right) \right) }{\kappa \theta _e \left( \kappa \theta _e-\mu \right) {}^3} &{} \frac{u v^3 p_e \left( 2 p \kappa p_e \theta _e+p_v \left( \kappa \theta _e-\mu \right) \right) }{\kappa \theta _e \left( \kappa \theta _e-\mu \right) {}^3} \end{array} \right) . \end{aligned}$$

1.2 B.2. Constants in Lemma 6.3

The constant matrices $M_j^k$ in Lemma 6.3 are given as follows.

$$\begin{aligned}&M_1^0=\left( \begin{array}{ccc} \frac{p p_e}{p p_e-p_v} &{} -\frac{u p_e}{p p_e-p_v} &{} \frac{p_e}{p p_e-p_v} \\ 0 &{} 0 &{} 0 \\ -\frac{p p_v}{p p_e-p_v} &{} \frac{u p_v}{p p_e-p_v} &{} -\frac{p_v}{p p_e-p_v} \\ \end{array} \right) ,\\&M_2^0=\left( \begin{array}{ccc} -\frac{p_v}{2 p p_e-2 p_v} &{} \frac{u p_e-\sqrt{p p_e-p_v}}{2 p p_e-2 p_v} &{} -\frac{p_e}{2 p p_e-2 p_v} \\ \frac{p_v}{2 \sqrt{p p_e-p_v}} &{} \frac{1}{2}-\frac{u p_e}{2 \sqrt{p p_e-p_v}} &{} \frac{p_e}{2 \sqrt{p p_e-p_v}} \\ \frac{\left( p+u \sqrt{p p_e-p_v}\right) p_v}{2 p p_e-2 p_v} &{} \frac{-p_e \sqrt{p p_e-p_v} u^2-p_v u+p \sqrt{p p_e-p_v}}{2 p p_e-2 p_v} &{} \frac{p_e \left( p+u \sqrt{p p_e-p_v}\right) }{2 p p_e-2 p_v} \\ \end{array} \right) ,\\&M_3^0=\left( \begin{array}{ccc} -\frac{p_v}{2 p p_e-2 p_v} &{} \frac{u p_e+\sqrt{p p_e-p_v}}{2 p p_e-2 p_v} &{} -\frac{p_e}{2 p p_e-2 p_v} \\ -\frac{p_v}{2 \sqrt{p p_e-p_v}} &{} \frac{1}{2} \left( \frac{u p_e}{\sqrt{p p_e-p_v}}+1\right) &{} -\frac{p_e}{2 \sqrt{p p_e-p_v}} \\ \frac{\left( p-u \sqrt{p p_e-p_v}\right) p_v}{2 p p_e-2 p_v} &{} -\frac{-p_e \sqrt{p p_e-p_v} u^2+p_v u+p \sqrt{p p_e-p_v}}{2 p p_e-2 p_v} &{} \frac{p_e \left( p-u \sqrt{p p_e-p_v}\right) }{2 p p_e-2 p_v} \\ \end{array} \right) .\\&M_1^1=\left( \begin{array}{ccc} 0 &{} \frac{ p \kappa p_e \theta _e}{v \left( p_v-p p_e\right) {}^2} &{} 0 \\ -\frac{p \kappa p_e p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} &{} \frac{u \kappa p_e p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} &{} -\frac{\kappa p_e p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} \\ -\frac{p u \kappa p_e p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} &{} -\frac{\kappa \left( p-u^2 p_e\right) p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} &{} -\frac{u \kappa p_e p_v \theta _e}{v \left( p_v-p p_e\right) {}^2} \\ \end{array} \right) ,\\&M_2^1=(\xi _1,\xi _2,\xi _3),\\&\xi _1=\left( \begin{array}{ccc} \frac{p_v \left( \mu p_v-p p_e \left( \mu +3 \kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \frac{p \kappa p_e p_v \theta _e}{2 v \left( p_v-p p_e\right) {}^2} \\ \frac{p p_v \left( p_e \left( p \mu +\kappa \left( p+2 u \sqrt{p p_e-p_v}\right) \theta _e\right) -p_v \left( \mu -2 \kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \end{array} \right) ,\\&\xi _2=\left( \begin{array}{ccc} \frac{p_e \left( -2 p \kappa \sqrt{p p_e-p_v} \theta _e-u p_v \left( \mu -2 \kappa \theta _e\right) +p u p_e \left( \mu +\kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ -\frac{ \left( p_e \left( p \sqrt{p p_e-p_v} \mu +\left( 2 u \kappa p_v-p \kappa \sqrt{p p_e-p_v}\right) \theta _e\right) -\mu \sqrt{p p_e-p_v} p_v\right) }{4 v \left( p_v-p p_e\right) {}^2} \\ -\frac{\left( 2 p^2 u \left( \mu -\kappa \theta _e\right) p_e^2+u p_v \left( \kappa \left( 5 p+2 u \sqrt{p p_e-p_v}\right) \theta _e-3 p \mu \right) p_e+p_v \left( u \mu p_v-2 p \kappa \sqrt{p p_e-p_v} \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \end{array} \right) ,\\&\xi _3=\left( \begin{array}{ccc} -\frac{p_e \left( p p_e \left( \mu +\kappa \theta _e\right) -p_v \left( \mu -2 \kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \frac{\kappa p_e p_v \theta _e}{2 v \left( p_v-p p_e\right) {}^2}\\ \frac{p_e \left( p_e \left( \mu -\kappa \theta _e\right) p^2+p_v \left( 2 \kappa \left( 2 p+u \sqrt{p p_e-p_v}\right) \theta _e-p \mu \right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \end{array} \right) .\\&M_3^1=(\zeta _1,\zeta _2,\zeta _3),\\&\zeta _1=\left( \begin{array}{ccc} \frac{p_v \left( p p_e \left( \mu +3 \kappa \theta _e\right) -\mu p_v\right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \frac{p \kappa p_e p_v \theta _e}{2 v \left( p_v-p p_e\right) {}^2} \\ \frac{p_v \left( p p_v \left( \mu -2 \kappa \theta _e\right) -p p_e \left( p \mu +\kappa \left( p-2 u \sqrt{p p_e-p_v}\right) \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \end{array} \right) ,\\&\zeta _2=\left( \begin{array}{ccc} -\frac{p_e \left( 2 p \kappa \sqrt{p p_e-p_v} \theta _e-u p_v \left( \mu -2 \kappa \theta _e\right) +p u p_e \left( \mu +\kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \frac{\left( p_e \left( p \mu \sqrt{p p_e-p_v}-\kappa \left( \sqrt{p p_e-p_v} p+2 u p_v\right) \theta _e\right) -\mu \sqrt{p p_e-p_v} p_v\right) }{4 v \left( p_v-p p_e\right) {}^2} \\ \frac{\left( 2 p^2 u \left( \mu -\kappa \theta _e\right) p_e^2+u p_v \left( \kappa \left( 5 p-2 u \sqrt{p p_e-p_v}\right) \theta _e-3 p \mu \right) p_e+p_v \left( u \mu p_v+2 p \kappa \sqrt{p p_e-p_v} \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \end{array} \right) ,\\&\zeta _3=\left( \begin{array}{ccc} \frac{p_e \left( p p_e \left( \mu +\kappa \theta _e\right) -p_v \left( \mu -2 \kappa \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \frac{ \kappa p_e p_v \theta _e}{2 v \left( p_v-p p_e\right) {}^2} \\ \frac{p_e \left( p_e \left( \kappa \theta _e-\mu \right) p^2+p_v \left( p \mu +\left( 2 u \kappa \sqrt{p p_e-p_v}-4 p \kappa \right) \theta _e\right) \right) }{4 v \left( p p_e-p_v\right) {}^{5/2}} \\ \end{array} \right) . \end{aligned}$$