Uniqueness for degenerate parabolic equations in weighted $L^1$ spaces

Abstract

We study uniqueness of solutions to degenerate parabolic problems, posed in bounded domains, where no boundary conditions are imposed. Under suitable assumptions on the operator, uniqueness is obtained for solutions that satisfy an appropriate integral condition; in particular, such condition holds for possibly unbounded solutions belonging to a suitable weighted $L^1$ space.

1 Introduction

We investigate uniqueness of solutions to degenerate parabolic problems of the following type

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu=\mathrm{{div}}\{a(x,t)\nabla u\}+f &{} \text{ on } \Omega \times (0,T]=:Q_T\\ u=u_0&{} \text{ on } \Omega \times \{0\}\,, \end{array}\right. } \end{aligned}$$

(1.1)

where $\Omega \subset \mathbb {R}^n$ is an open bounded subset and $T>0$. Note that in (1.1) no boundary conditions are prescribed. Concerning the coefficient a(x, t) and the data f and $u_0$, we always assume that

$$\begin{aligned} a\in C_{x,t}^{1,0}(Q_T), a\ge 0, a\not \equiv 0 \text{ in } Q_T\,, \end{aligned}$$

$f\in C(Q_T), u_0\in C(\Omega )$. Furthermore, we assume that $\partial \Omega$ is a manifold of dimension $n-1$ of class $C^3$.

A wide literature is devoted to degenerate elliptic and parabolic problems, based on both analytical methods (see, e.g., [3, 4]- [7, 12]- [19, 22]) and stochastic calculus (see, e.g., [11, 21]). Under appropriate assumptions on the behavior at the boundary of the coefficients of the operator, in [3] it is shown that uniqueness of solutions can hold without prescribing boundary conditions at some portion of the boundary. Such solutions belong to $C^2(Q_T)\cap C({\bar{Q}}_T)$; therefore, they are bounded.

In [14, 15], by means of appropriate super- and subsolutions, similar uniqueness results have been obtained, also for unbounded solutions. It is assumed that the solutions satisfy suitable pointwise growth conditions near the boundary. Such conditions are related to the constructed super- and subsolutions.

In [17], uniqueness in the weighted Lebesgue space $L^1_{d^{\beta }(x)}(Q_T)\, (\beta >0)$ is shown for degenerate operators in non-divergence form, under appropriate conditions on the coefficients, similar to those in [3]. Here and hereafter,

$$\begin{aligned} d(x):=\mathrm{dist}(x,\partial \Omega )\quad (x\in \Omega )\, \end{aligned}$$

is the function distance from the boundary.

In [20], under suitable hypotheses on the coefficient a, uniqueness results for problem (1.1), in suitable weighted $L^2$ spaces, are established by developing a general idea used, for instance, in [8] and in [9, Theorem 9.2] (see also [10]) for different purposes. Such uniqueness results are obtained as a consequence of suitable integral maximum principles. Note that integral maximum principles in the whole $\mathbb R^n$ for solutions of degenerate parabolic equations are also obtained in [1, 2].

In this paper, we generalize the uniqueness results in [20], since we enlarge the uniqueness class. In fact, we now consider solutions belonging to an appropriate weighted $L^1$ space. The passage from $L^2$ to $L^1$ causes important changes in the proofs. Let us outline the differences between our methods and results, and those in [20]. The line of arguments in [20] is the following: multiply the differential equation in (1.1) by suitable test functions, integrate by parts one time and obtain convenient estimates on the solution. To do this, an important step is to find a function $\xi (x,t)$, depending on the distance function d(x), which is Lipschitz continuous w.r.t. to x and $C^1$ w.r.t. to t, and satisfies

$$\begin{aligned} \partial _t \xi (x,t) + \alpha \, a(x,t) |\nabla \xi (x,t)|^2\le 0 \quad \text {for a.e. } x\in \Omega , \; \text {for any } t\in (0, {\bar{T}}), \end{aligned}$$

(1.2)

for appropriate $\alpha>0, {\bar{T}}>0.$

Now, suppose that, for some $\gamma>1, c_0>{\tilde{c}}_0>0, c_1>0,$ for all $(x,t)\in Q_T$,

$$\begin{aligned}&{\tilde{c}}_0 d^{\gamma }(x) \le a(x,t)\le c_0 d^{\gamma }(x) \nonumber \\&\quad \text{ and } \nonumber \\&\quad |\nabla a(x,t)|\le c_1 d^{\gamma -1}(x)\,. \end{aligned}$$

(1.3)

For every $\varepsilon >0$, let

$$\begin{aligned}\Omega ^{\varepsilon }:=\{x\in \Omega \,:\, d(x)>\varepsilon \}.\end{aligned}$$

In the present paper to obtain uniqueness in a weighted $L^1$ space, we argue as follows: we multiply the differential equation in (1.1) by suitable test functions, then we integrate by parts two times. Hence, to get convenient bounds on the solution, we have to control new terms that appear after the second integration by part. A crucial point in the proof is to exhibit a function $\xi =\xi (x,t)$ with $\xi (\cdot , t)\in C^2(\Omega \setminus \partial \Omega ^{\varepsilon })\cap C^1(\Omega ), \xi (x, \cdot )\in C^1(\Omega )$, which satisfies

$$\begin{aligned} \partial _t\xi +\mathrm{{div}}\big \{a(x,t)\nabla \xi \big \}+\frac{5}{2} a(x,t)|\nabla \xi |^2 \le 0 \quad \text {in } [\Omega \setminus \partial \Omega ^{\varepsilon }]\times (T_1, T_2), \end{aligned}$$

(1.4)

and

$$\begin{aligned} \frac{\partial \xi }{\partial n_{\varepsilon }}=0\quad \text {in }\,\, \partial \Omega ^{\varepsilon }\times (T_1, T_2)\,, \end{aligned}$$

(1.5)

where $n_\varepsilon$ is the unit outward normal vector to $\Omega ^\varepsilon$ at $\partial \Omega ^{\varepsilon }$, for appropriate $0<T_1<T_2.$ Observe that $\xi (x,t)$ is defined in terms of the distance function from the boundary and its behavior as $x\rightarrow \partial \Omega$ is very important, since it influences the integral conditions for the solutions, which guarantees uniqueness. Clearly, the construction of $\xi$ fulfilling (1.4) and (1.5) is more delicate than that verifying only (1.2). The choice of $\xi$ changes according to whether $\gamma >2$ or $\gamma \in [1, 2]$; consequently, in these two cases the proofs present some important differences.

The paper is organized as follows. In Sect. 2, we state our main two uniqueness results, concerning the two cases $\gamma >2$ and $\gamma \in [1, 2]$; in addition, we compare them with some related results in the literature. The uniqueness result for $\gamma >2$ is proved in Sect. 3, while the other one, for $\gamma \in [1, 2]$, in Sect. 4.

2 Statements of the results

Consider the homogeneous problem associated with (1.1), that is

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu=\mathrm{{div}}\{a(x,t)\nabla u\} &{} \text{ in } Q_T\\ u=0&{} \text{ on } \Omega \times \{0\}\,. \end{array}\right. } \end{aligned}$$

(2.1)

The following two uniqueness results are our main contribute in this paper.

Theorem 2.1

Suppose that $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1) and a satisfies (1.3) with $\gamma >2$.

Moreover, suppose that, for some $C>0, \theta>0, \varepsilon _0>0$,

$$\begin{aligned} \int _0^T\int _{\Omega ^{\varepsilon }}|u(x,t)|\, dx\, dt\le Ce^{\theta \, \varepsilon ^{-\gamma +2}} \qquad \text{ for } \text{ every } \varepsilon \in (0,\varepsilon _0). \end{aligned}$$

(2.2)

Then, $u\equiv 0$ in $Q_T$.

Obviously, there exist unbounded functions satisfying condition (2.2). For any $\phi \in C(\Omega ), \phi >0$, $p\ge 1$, let

$$\begin{aligned} L^p_{\phi }(Q_T):=\left\{ u: Q_T \rightarrow \mathbb R \text { measurable} : \int _0^T\int _{\Omega } |u(x,t)|^p \phi (x)\, dx dt<\infty \right\} \,. \end{aligned}$$

Remark 2.2

It is direct to see that if $u\in L^1_{\phi }(Q_T)$ with $\phi (x)=e^{\{-\theta [d(x)]^{2-\gamma }\}}, \theta>0, \gamma >2$, then condition (2.2) holds.

Theorem 2.3

Suppose that $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1) and a satisfies (1.3) with $\gamma \in [1,2]$.

Moreover, suppose that, for some $C>0, \varepsilon _0>0$ and $\mu >-2\gamma +4$,

$$\begin{aligned} \int _0^T\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }} |u(x,t)|\, [d(x)]^{\gamma -2}\, dx\, dt\le C\varepsilon ^{\mu } \qquad \text{ for } \text{ every } \varepsilon \in (0,\varepsilon _0). \end{aligned}$$

(2.3)

Then, $u\equiv 0$ in $Q_T$.

Remark 2.4

Note that, if $u\in L^1_{\phi }(Q_T)$ with $\phi (x)=[d(x)]^{\gamma -2-\mu }$, then (2.3) is valid.

Furthermore, if $\gamma \in \left( \frac{5}{3}, 2\right]$ and

$$\begin{aligned} |u(x,t)|\le {\bar{C}}[d(x)]^{-l}\quad \text {for every}\;\; x\in \Omega , t\in [0, T]\,, \end{aligned}$$

(2.4)

for some ${\bar{C}}>0$ and $0< l<3\gamma -5$, then (2.3) holds with $\mu =\gamma -1-l>-2\gamma +4.$

Remark 2.5

(i) Theorem 2.1 generalizes [20, Theorem 2.1], where (2.2) is replaced by the stronger condition

$$\begin{aligned} \int _0^T\int _{\Omega ^{\varepsilon }}|u(x,t)|^2\, dx\, dt\le Ce^{\theta \, \varepsilon ^{-\gamma +2}} \qquad \text{ for } \text{ every } \varepsilon \in (0,\varepsilon _0). \end{aligned}$$

(2.5)

(ii) Theorem 2.3 generalizes [20, Theorem 2.2], where (2.3) is replaced by the stronger condition

$$\begin{aligned} \int _0^T\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }} |u(x,t)|^2\, [d(x)]^{\gamma -2}\, dx\, dt\le C\varepsilon ^{\mu } \qquad \text{ for } \text{ every } \varepsilon \in (0,\varepsilon _0). \end{aligned}$$

(2.6)

for some $\mu >0$. However, note that in Theorem 2.3 the further request $\mu >-2\gamma +4$ is made.

(iii) We should note that in [20, Theorem 2.1, 22] the hypothesis on the coefficient a is weaker. In fact, instead of (1.3) it is only assumed that

$$\begin{aligned}a(x,t)\le c_0 d^\gamma (x)\quad \text {for all }\,\, (x, t)\in Q_T\,.\end{aligned}$$

Remark 2.6

Let $\gamma \in (1, 2]$ and u be a solution of problem (2.1) satisfying (2.4), for some ${\bar{C}}>0$ and $l>0.$ Observe that [20, Theorem 2.2] yields that if $0<l<\frac{\gamma -1}{2}$, then $u\equiv 0$ in $Q_T$.

Now, let $\gamma \in \left( \frac{5}{3},2 \right] .$ From Theorem 2.3 and the subsequent comments, it follows that $u\equiv 0$, provided that $0<l<3\gamma -5$. Since

$$\begin{aligned} 3\gamma -5\in (0,1), \end{aligned}$$

while

$$\begin{aligned} \frac{\gamma -1}{2}\in \left( 0,\frac{1}{2}\right] , \end{aligned}$$

the growth condition for u in Theorem 2.3 is weaker than that in [20, Theorem 2.2]. On the other hand, when $\gamma \in \left( 0, \frac{5}{3}\right)$, [20, Theorem 2.2] can be applied, whereas the hypotheses of Theorem 2.3 are not verified (under the extra condition (2.4)).

Finally, recall that in view of [20, Proposition 3.3], if $\gamma =1, l=0$, then uniqueness holds in $L^\infty (Q_T)$.

By Theorems 2.1 and 2.3, the following uniqueness result immediately follows.

Corollary 2.7

Let $u_1, u_2\in C^{2,1}(Q_T)\cap C(\Omega \times [0, T])$ be two solutions of problem (1.1). Assume that (1.3) holds with $\gamma >2$ and both $u_1$ and $u_2$ satisfy condition (2.2), or that (1.3) holds with $\gamma \in [1, 2]$ and both $u_1$ and $u_2$ satisfy condition (2.3). Then, $u_1\equiv u_2$ in $Q_T$.

Remark 2.8

Assume that, for some $\varepsilon>0, C_0>0$ and $\gamma \in \mathbb R$,

$$\begin{aligned} a(x,t)=C_0 [d(x)]^{\gamma } \quad \text {for any}\;\; x\in \Omega \setminus \Omega ^\varepsilon , t\in [0, T]\,. \end{aligned}$$

(2.7)

If $\gamma \ge 2$, the results in [3] give uniqueness of solutions to problem (2.1) in $C^2(Q_T)\cap C({\bar{Q}}_T)$. So, in particular such solutions are bounded. Hence, our results are in agreement with those in [3] in the special case of bounded solutions, if $\gamma >2$. Instead, when $\gamma <2$, the results in [3] cannot be applied, since the coefficients are not regular enough.

The results in [15] could be applied, once we construct suitable super- and subsolutions; however, we would obtain uniqueness under pointwise growth conditions near $\partial \Omega$. Finally, the results in [14] and in [17] cannot be applied, since our operator does not satisfy the required hypotheses.

From the existence result in [20, Proposition 3.1] and Corollary 2.7, we get the following existence and uniqueness result.

Corollary 2.9

Let $f\equiv 0, \gamma >2$ and $a>0$ in $Q_T$. Suppose that, for some $0<\beta \le \gamma -2, \tau >0,$

$$\begin{aligned} 0\le u_0 \le \exp \left\{ \frac{[d(x)]^{\beta }}{\tau }\right\} \quad \text {for all}\;\; x\in \Omega \,. \end{aligned}$$

(2.8)

Assume that (1.3) holds. Then, there exists a solution $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0, T])$ of problem (1.1) fulfilling

$$\begin{aligned} 0\le u(x,t) \le {\hat{C}}\exp \left\{ \frac{[d(x)]^{\beta }}{\tau -\lambda t}\right\} \quad \text {for all}\;\; x\in \Omega \,, t\in \left[ 0, T\right] , \end{aligned}$$

(2.9)

with $T=\frac{\tau }{2\lambda }$, for suitable $\lambda>0, {\hat{C}}>0$. Furthermore, u is the unique solution of problem (1.1) in $L^1_{\phi }(Q_T)$ with $\phi (x)=e^{\{-\frac{2}{\tau }[d(x)]^{2-\gamma }\}}$.

Observe that Theorems 2.1 and 2.3 imply uniqueness whenever (1.3) holds with $\gamma \ge 1$. Such request on $\gamma$ is indeed optimal. In fact, from [20, Proposition 3.2] it follows that when, for some $\varepsilon>0, \gamma <1, c_2>0, c_3>0, s \in [0, \gamma )$,

$$\begin{aligned} c_2 [d(x)]^{\gamma }\le a(x)\le c_3 [d(x)]^{\gamma -s}\quad \text {for all}\;\; x\in \Omega \setminus \Omega ^{\varepsilon }\,, \end{aligned}$$

(2.10)

problem (1.1) admits infinitely many bounded solutions.

Remark 2.10

We observe that there are important differences between problem (1.1) and the companion problem

$$\begin{aligned} \left\{ \begin{array}{ll} \, \partial _t u = a(x,t)\Delta u + f \, &{}\text {in}\,\,Q_T \\ {} &{}\\ \text { }u \, = u_0&{} \text {in}\,\, \Omega \times \{0\} \,. \end{array} \right. \end{aligned}$$

(2.11)

For example, let

$$\begin{aligned}a(x,t)=[d(x)]^{\gamma }\quad (\gamma >1)\,.\end{aligned}$$

If $\gamma \ge 2$, then there exists a unique bounded solutions to problem (2.11) (see [10, Section 7], [14, Theorem 2.16]). On the other hand, if $\gamma <2$, then nonuniqueness of solutions of problem (2.11) prevails, in the sense that it is possible to prescribe Dirichlet boundary data at $\partial \Omega \times (0, T]$ (see [10, Section 7], [14, Theorem 2.18]). Thus, the change between uniqueness and nonuniqueness occurs for $\gamma =2$. Instead, such change for problem (1.1) occurs for $\gamma =1$.

3 Proof of Theorem 2.1

Observe that

$$\begin{aligned} |\nabla d(x)|\le 1\quad \text {for a.e. }\,\, x\in \Omega \,. \end{aligned}$$

(3.1)

Moreover, (see, e.g., [14]) if $\partial \Omega$ is of class $C^3$, then there exists $\varepsilon _0\in (0,1)$ such that for each $\varepsilon \in (0, \varepsilon _0)$ $d\in C^2(\Omega \setminus \Omega ^\varepsilon ),$ and, for some $k_0>0$,

$$\begin{aligned} |\Delta d(x)|\le k_0\quad \text {in }\,\,\Omega \setminus \Omega ^\varepsilon \,. \end{aligned}$$

(3.2)

In addition, there exists $\nu _0\in (0,1)$ such that

$$\begin{aligned} |\nabla d(x)|\ge \nu _0\quad \text { for any }\,\, x\in \Omega \setminus \Omega ^\varepsilon \,. \end{aligned}$$

(3.3)

For each $\beta >0$, define the function

$$\begin{aligned} \zeta (x,t):={\left\{ \begin{array}{ll} 0 &{} \text{ if } x\in \Omega ^{\varepsilon }\\ {[}d(x)]^{-\beta }-\varepsilon ^{-\beta } &{} \text{ if } x\in \Omega \setminus \Omega ^{\varepsilon }\, \end{array}\right. }\,. \end{aligned}$$

(3.4)

Differentiating the function above, we have

$$\begin{aligned} \nabla \zeta (x,t)=-\beta [d(x)]^{-\beta -1}\nabla d(x)\qquad \text{ for } \text{ any } x\in \Omega \setminus \Omega ^{\varepsilon }\,, \end{aligned}$$

(3.5)

thus

$$\begin{aligned} |\nabla \zeta (x,t)|^2\le \beta ^2[d(x)]^{-2\beta -2}\qquad \text{ for } \text{ any } x\in \Omega \setminus \Omega ^{\varepsilon }\,. \end{aligned}$$

Finally, define the function

$$\begin{aligned} \xi (x,t):=-\frac{\zeta ^2(x)}{2(s-\alpha _1t)} \end{aligned}$$

(3.6)

for any $x\in \Omega , t\ne \frac{s}{\alpha _1}$, where here $\alpha _1>0$ is a parameter to be chosen later. Note that $\xi (\cdot , t)\in C^2(\Omega \setminus \partial \Omega ^{\varepsilon })\cap C^1(\Omega )$ and

$$\begin{aligned} \frac{\partial \xi (x,t)}{\partial n_{\varepsilon }}=0 \qquad \text{ for } \text{ any } x\in \partial \Omega ^{\varepsilon }, \; t\ne \frac{s}{\alpha _1}\,, \end{aligned}$$

(3.7)

where $n_{\varepsilon }$ is the outward normal to $\Omega ^{\varepsilon }$.

Let $\gamma >2$, $c\in \left( 0,\frac{1}{2}\right)$ be such that

$$\begin{aligned} {[}(1-c)^{-\frac{\gamma -2}{2}}-1](c_1+c_0k_0)-\beta \nu _0{\tilde{c}}_0 <0\,, \end{aligned}$$

(3.8)

and define

$$\begin{aligned} \sigma :=1-(1-c)^{\frac{\gamma -2}{2}}\,. \end{aligned}$$

(3.9)

The proof of Theorem 2.1 is based on the combination of the following results.

Proposition 3.1

Under assumption (1.3) with $\gamma >2$, suppose $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1). Suppose that, for some $C>0$ and $\theta >0$, (2.2) holds. Let $\tau \in (0, T), c\in \left( 0,\frac{1}{2}\right)$ be such that (3.8) is satisfied, $\sigma$ be defined by (3.9),

$$\begin{aligned} 0<\delta <\min \left\{ \frac{\sigma ^2}{(\gamma -2)(c_1+c_0)},\; \tau ,\; \frac{\left[ \left( \frac{3}{2} \right) ^{\frac{\gamma -2}{2}}-1\right] ^2}{4\theta \alpha _1}\right\} \, \end{aligned}$$

and

$$\begin{aligned} \alpha _1\ge \max \left\{ \frac{6 c_0 (\gamma -2)^2}{\sigma ^2},\; \frac{5}{4} c_0(\gamma -2)^2 \right\} \,. \end{aligned}$$

Then,

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx\le & {} \int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx + \tilde{C}\varepsilon ^{\gamma -2}\,, \end{array} \end{aligned}$$

(3.10)

where $\tilde{C}>0$ is a suitable constant independent of $\varepsilon$.

Lemma 3.2

Let $u\in C(\Omega \times [0, T])$ with

$$\begin{aligned} u=0\quad \text {in}\;\; \Omega \times \{0\}\,. \end{aligned}$$

(3.11)

Suppose that there exist $c>0, \bar{\varepsilon }>0, \mu>0, \hat{C}>0$ such that for any $\varepsilon \in (0, \bar{\varepsilon })$, $\tau \in (0,T)$ and

$$\begin{aligned} 0<\delta \le \min \{\tau , c\}\,, \end{aligned}$$

(3.12)

there holds

$$\begin{aligned} \int _{\Omega ^{\varepsilon }} |u(x, \tau )|\,dx \le \int _{\Omega ^{\frac{\epsilon }{2}}} |u(x, \tau -\delta )|\, dx + \hat{C} \varepsilon ^{\mu }\,. \end{aligned}$$

(3.13)

Then,

$$\begin{aligned}u\equiv 0\quad \text {in}\;\; \Omega \times (0, T]\,.\end{aligned}$$

Now, we are ready to prove Theorem 2.1.

Proof of Theorem 2.1

We obtain the thesis, combining Proposition 3.1 and Lemma 3.2 with

$$\begin{aligned} \mu =\gamma -2, \end{aligned}$$

$$\begin{aligned} c=\min \left\{ \frac{\sigma ^2}{(\gamma -2)(c_1+c_0)},\; \frac{\left[ \left( \frac{3}{2} \right) ^{\frac{\gamma -2}{2}}-1\right] ^2}{4\theta \alpha _1}\right\} , \end{aligned}$$

and

$$\begin{aligned} \tilde{C}=\hat{C}\,. \end{aligned}$$

$\square$

3.1 Proofs of Proposition 3.1 and Lemma 3.2

Consider a family of cut-off functions $\{\eta _{\varepsilon }\}\subset C^{\infty }(\Omega )$ such that

$$\begin{aligned} 0\le \eta _{\varepsilon }\le 1\,, \end{aligned}$$

and

$$\begin{aligned} \eta _{\varepsilon }={\left\{ \begin{array}{ll} 1 &{} \text{ in } \Omega ^{\frac{2}{3}\varepsilon }\\ 0 &{} \text{ in } \Omega \setminus \Omega ^{\frac{\varepsilon }{2}}\,. \end{array}\right. } \end{aligned}$$

(3.14)

Notice that

$$\begin{aligned} \begin{array}{rlll} |\nabla \eta _{\varepsilon }|&{}\le &{} \frac{A_1}{\varepsilon } \quad \text{ for } \text{ every } x\in \Omega \,, \\ \\ |\Delta \eta _{\varepsilon }|&{}\le &{} \frac{A_2}{\varepsilon ^2} \quad \text{ for } \text{ every } x\in \Omega \,, \end{array} \end{aligned}$$

(3.15)

where $A_1$ and $A_2$ are two positive constants.

For every $\alpha >0$, consider a function $\psi _{\alpha }:\mathbb {R}\rightarrow \mathbb {R}^{+}$ of class $C^2$ such that

$$\begin{aligned} \psi _{\alpha }''\ge 0 \,. \end{aligned}$$

(3.16)

Then, by the chain rule

$$\begin{aligned} \mathrm{{div}}\{a(x,t)\nabla \psi _{\alpha }(u)\}=\psi _{\alpha }'(u)\mathrm{{div}}\{a(x,t)\nabla u\}+\psi _{\alpha }''(u)a(x,t) |\nabla u|^2 \end{aligned}$$

and, because of (3.16) and the positivity of a, we can estimate the second term on the right-hand side from below, obtaining

$$\begin{aligned} \mathrm{{div}}\{a(x,t)\nabla \psi _{\alpha }(u)\}\ge \psi _{\alpha }'(u)\mathrm{{div}}\{a(x,t)\nabla u\}=\partial _t \psi _{\alpha }(u)\,, \end{aligned}$$

(3.17)

where in the last identity we used equation (2.1). Thus, the composed function $\psi _{\alpha }(u)$ is a subsolution of (1.1).

The main ingredient for the proof of Proposition 3.1 is the following

Lemma 3.3

Under assumption (1.3) with $\gamma >2$, suppose $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1). Let $0<\varepsilon <\varepsilon _0, \tau \in (0, T), c\in \left( 0,\frac{1}{2}\right)$ be such that (3.8) is satisfied, $\sigma$ be defined by (3.9). If

$$\begin{aligned} 0<\delta <\min \left\{ \frac{\sigma ^2}{(\gamma -2)(c_1+c_0)}, \tau \right\} \end{aligned}$$

(3.18)

and

$$\begin{aligned} \alpha _1\ge \max \left\{ \frac{6 c_0 (\gamma -2)^2}{\sigma ^2}, \frac{5}{4} c_0(\gamma -2)^2\right\} \,, \end{aligned}$$

(3.19)

then

$$\begin{aligned}&\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&\quad + C_1\varepsilon ^{ \gamma -2 } \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon } \times (\tau -\delta , \tau )}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} \, dx\, dt\,, \end{aligned}$$

(3.20)

where $\xi$ is defined in (3.6) with $s=\alpha _1(\tau +\delta )$ and $C_1>0$ is a suitable constant independent of $\varepsilon$.

Proof of Lemma 3.3

Define the set $\mathcal {C}=\Omega ^{\frac{\varepsilon }{2}}\times (\tau -\delta ,\tau )$. Testing the time derivative of $\psi _{\alpha }(u)$ with $\eta ^2(x)e^{\xi (x,t)}$, we get

$$\begin{aligned} \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx= & {} \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx\nonumber \\&+\iint _{\mathcal {C}} \partial _t[\psi _{\alpha }(u(x,t))]\eta ^2(x)e^{\xi (x,t)}\, dx\, dt\nonumber \\&+ \iint _{\mathcal {C}} \psi _{\alpha }(u(x,t))\eta ^2(x)\partial _te^{\xi (x,t)}\, dx\, dt\,.\nonumber \\ \end{aligned}$$

(3.21)

We can compute the second term of the right-hand-side of the above equation as

(3.22)

where, in the last inequality we used the positivity of the integrating factor, due to (3.16) and (1.3). We need to estimate the right-hand side of the above inequality further: Integrating by parts a second time, we have

$$\begin{aligned}&\iint _{\mathcal {C}}\mathrm{{div}}\{a\,\nabla \psi _{\alpha }(u)\}\eta ^2e^{\xi }\, dx\,dt\\&= -\iint _{\mathcal {C}}a\,\nabla \psi _{\alpha }(u)\cdot \nabla (\eta ^2e^{\xi })\, dx\, dt\\&=-\iint _{\mathcal {C}}a\,\nabla \psi _{\alpha }(u)\cdot 2\eta \nabla \eta \, e^{\xi }\, dx\, dt - \iint _{\mathcal {C}}a\,\nabla \psi _{\alpha }(u)\cdot \eta ^2 \,e^{\xi }\,\nabla \xi \, dx\, dt\,. \end{aligned}$$

Observe that, since $\xi \in C^2(\Omega \setminus \partial \Omega ^{\varepsilon })$, the last identity is justified by splitting the integral in the set $\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }$ and $\Omega ^{\varepsilon }$, integrating by parts and eliminating the boundary terms thanks to (3.7) and (3.14):

$$ \begin{aligned}&-\iint _{\mathcal {C}}a\,\nabla \psi _{\alpha }(u)\cdot 2\eta \nabla \eta \, e^{\xi }\, dx\, dt - \iint _{\mathcal {C}}a\,\nabla \psi _{\alpha }(u)\cdot \eta ^2 \, e^{\xi }\,\nabla \xi \, dx\, dt\\&\quad =-\iint _{\Omega ^{\varepsilon }}a\,\nabla \psi _{\alpha }(u)\cdot 2\eta \nabla \eta \, e^{\xi }\, dx\, dt - \iint _{\Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}a\,\nabla \psi _{\alpha }(u)\cdot \eta ^2 \,e^{\xi }\,\nabla \xi \, dx\, dt\\&\qquad -\iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}a\,\nabla \psi _{\alpha }(u)\cdot 2\eta \nabla \eta \,e^{\xi }\, dx\, dt\\&\qquad - \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }}a\,\nabla \psi _{\alpha }(u)\cdot \eta ^2 \,e^{\xi }\,\nabla \xi \, dx\, dt\\&\quad =\iint _{\Omega ^{\varepsilon }\times (\tau -\delta ,\tau )} \psi _{\alpha }(u)\mathrm{{div}}(a\, 2\eta \nabla \eta \,e^{\xi }) \, dx\, dt -\int _{\partial \Omega ^{\varepsilon }}a\, \psi _{\alpha }(u) \, 2\eta \frac{\partial \eta }{\partial n_{\varepsilon }} e^{\xi } \, dS_x\, dt\\&\qquad + \iint _{\Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}\psi _{\alpha }(u)\,\mathrm{{div}}(a\, \eta ^2 \,e^{\xi }\,\nabla \xi )\, dx\, dt - \int _{\partial \Omega ^{\varepsilon }}a\, \psi _{\alpha }(u)\,\eta ^2 e^{\xi }\frac{\partial \xi }{\partial n_{\varepsilon }} \, dS_x\, dt\\&\qquad +\iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )} \psi _{\alpha }(u)\,\mathrm{{div}}(a\, 2\eta \nabla \eta \,e^{\xi }) \, dx\, dt\\&\qquad -\int _{\partial \Omega ^{\frac{\varepsilon }{2}}\cup \partial \Omega ^{\varepsilon }\times (\tau -\delta ,\tau ) }a\, \psi _{\alpha }(u)\, 2\eta \frac{\partial \eta }{\partial n_{\varepsilon }} e^{\xi } \, dS_x\, dt\\&\qquad + \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )} \psi _{\alpha }(u)\mathrm{{div}}(a\, \eta ^2 \,e^{\xi }\,\nabla \xi )\, dx\, dt \\&\qquad -\int _{\partial \Omega ^{\frac{\varepsilon }{2}}\cup \partial \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}a\,\psi _{\alpha }(u)\, \eta ^2 e^{\xi } \frac{\partial \xi }{\partial n_{\varepsilon }}\, dS_x\, dt\\&\quad {\mathop {=}\limits ^{(3.7) \& (3.14)}}\iint _{\Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}\psi _{\alpha }(u)\,\mathrm{{div}}(a\, 2\eta \nabla \eta \,e^{\xi }) \, dx\, dt\\&\qquad + \iint _{\Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}\psi _{\alpha }(u)\,\mathrm{{div}}(a\, \eta ^2 \,e^{\xi }\, \nabla \xi )\, dx\, dt\\&\qquad +\iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}\psi _{\alpha }(u)\,\mathrm{{div}}(a\, 2\eta \nabla \eta \,e^{\xi }) \, dx\,dt\\&\qquad + \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\varepsilon }\times (\tau -\delta ,\tau )}\psi _{\alpha }(u)\mathrm{{div}}(a\, \eta ^2 \,e^{\xi }\,\nabla \xi ) dx\, dt\,. \end{aligned}$$

We therefore obtained

$$\begin{aligned}&\iint _{\mathcal {C}}\mathrm{{div}}\{a\,\nabla \psi _{\alpha }(u)\}\eta ^2e^{\xi }\, dx\,dt\\&= 2\iint _{\mathcal {C}}\psi _{\alpha }(u)\left\{ \eta \mathrm{{div}} (a\,\nabla \eta )\,e^{\xi } +a\,|\nabla \eta |^2\,e^{\xi } +a\,\eta \, \nabla \eta \cdot \,e^{\xi }\,\nabla \xi \right\} \, dx\, dt\\&\qquad +\iint _{\mathcal {C}}\psi _{\alpha }(u)\left\{ 2 a\,\eta \nabla \eta \cdot \,e^{\xi }\nabla \xi +a\,\eta ^2 e^{\xi }|\nabla \xi |^2 +\eta ^2 \mathrm{{div}} (a\,\nabla \xi )\,e^{\xi }\right\} \, dx\, dt\,. \end{aligned}$$

and, inserting this new expression in (3.22) and this last one back into inequality (3.21), we get

$$\begin{aligned} \begin{array}{rlll} &{}&{}\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx\\ \\ &{}&{}\quad \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx &{}\\ \\ &{}&{} \qquad +2\iint _{\mathcal {C}} \psi _{\alpha }(u(x,t))\Big \{\eta (x) \mathrm{{div}} (a(x,t)\nabla \eta (x))e^{\xi (x,t)}+a(x,t)|\nabla \eta (x)|^2\,e^{\xi (x,t)}\Big .\\ \\ &{}&{}\qquad \Big .+a(x,t)\, \eta (x) \,\nabla \eta (x)\cdot e^{\xi (x,t)}\nabla \xi (x,t) +a(x,t)\,\eta (x)\nabla \eta (x)\cdot e^{\xi (x,t)}\nabla \xi (x,t)\Big .\\ \\ &{}&{}\qquad \Big .+\frac{1}{2}\,a(x,t)\,\eta ^2(x) |\nabla \xi (x,t)|^2 e^{\xi (x,t)} +\frac{1}{2}\eta ^2(x) \mathrm{{div}} (a(x,t)\nabla \xi (x,t))\,e^{\xi (x,t)}\Big \}\\ \\ &{}&{}\qquad + \iint _{\mathcal {C}} \psi _{\alpha }(u(x,t))\eta ^2(x)\partial _te^{\xi (x,t)}\, dx\, dt. \end{array} \end{aligned}$$

Using Young’s inequality in the form

$$\begin{aligned} 2\iint _{\mathcal {C}}\psi _{\alpha }a\, \eta \,\nabla \eta \cdot \, e^{\xi }\,\nabla \xi \le \iint _{\mathcal {C}}\psi _{\alpha }\,a\,\eta ^2\, e^{\xi }|\nabla \xi |^2\, dx\, dt+ \iint _{\mathcal {C}}\psi _{\alpha }a\, |\nabla \eta |^2\,e^{\xi }\, dx\, dt\, \end{aligned}$$

in the second integral of the right-hand side, we get

$$\begin{aligned}&2\iint _{\mathcal {C}}\psi _{\alpha }(u)\Big \{\eta \mathrm{{div}} (a\,\nabla \eta )e^{\xi }+a\,|\nabla \eta |^2\,e^{\xi }+2\,a\,\eta \nabla \eta \cdot e^{\xi }\,\nabla \xi \Big .\\ \\&\qquad \Big . \frac{1}{2} \,a\,\eta ^2 e^{\xi }|\nabla \xi |^2 \,+\frac{1}{2}\,\eta ^2 \mathrm{{div}} (a\,\nabla \xi )e^{\xi }\Big \}\, dx\, dt \\ \\&\le \iint _{\mathcal {C}}\psi _{\alpha }(u) [\eta \mathrm{{div}} (a\,\nabla \eta )+a|\nabla \eta |^2] \,e^{\xi }\, dx\, dt\\ \\&\qquad +\frac{1}{2}\iint _{\mathcal {C}}\psi _{\alpha }(u) [a\,\eta ^2|\nabla \xi |^2+\eta ^2\mathrm{{div}} (a\,\nabla \xi )]e^{\xi }\, dx\, dt\\&\qquad +\iint _{\mathcal {C}}\psi _{\alpha }(u)\, a\,\eta ^2\,e^{\xi }|\nabla \xi |^2\,dx\, dt+\iint _{\mathcal {C}}\psi _{\alpha }(u)\, a\,|\nabla \eta |^2\,e^{\xi }\, dx\, dt. \end{aligned}$$

Putting all the previous estimates together, we have

$$\begin{aligned} \begin{array}{rlll} &{}&{}\displaystyle \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx\\ \\ &{}&{}\displaystyle \quad \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx &{}\\ \\ &{}&{}\displaystyle \qquad +\iint _{\mathcal {C}}\psi _{\alpha }(u(x,t)) [\eta \mathrm{{div}} (a(x,t)\nabla \eta (x))+a(x,t)|\nabla \eta (x)|^2] e^{\xi (x,t)}\, dx\, dt\\ \\ &{}&{}\displaystyle \qquad +\frac{1}{2}\iint _{\mathcal {C}}\psi _{\alpha }(u(x,t))[a(x,t)\eta ^2(x)|\nabla \xi (x,t)|^2\\ \\ &{}&{}\displaystyle \qquad +\eta ^2\mathrm{{div}} (a(x,t)\nabla \xi (x,t))] e^{\xi (x,t)}\, dx\, dt\\ \\ &{}&{}\displaystyle \qquad + \iint _{\mathcal {C}}\psi _{\alpha }(u(x,t))a(x,t)\eta ^2(x)e^{\xi (x,t)}|\nabla \xi (x,t)|^2\, dx\, dt\\ \\ &{}&{}\displaystyle \qquad + \iint _{\mathcal {C}}\psi _{\alpha }(u(x,t))a(x,t)|\nabla \eta (x)|^2e^{\xi (x,t)}\, dx\, dt\\ \\ &{}&{}\displaystyle \qquad + \iint _{\mathcal {C}} \psi _{\alpha }(u(x,t))\eta ^2(x)e^{\xi (x,t)}\partial _t\xi (x,t) \, dx\, dt .\\ \end{array} \end{aligned}$$

Finally, summing up and rearranging the terms we have

$$\begin{aligned}&\displaystyle \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx\nonumber \\&\displaystyle \quad \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&\displaystyle \qquad + \iint _{\mathcal {C}}\psi _{\alpha }(u(x,t))[\eta (x)\mathrm{{div}} (a(x,t)\nabla \eta (x))+2a(x,t)|\nabla \eta (x)|^2] e^{\xi (x,t)} \, dx\, dt\nonumber \\&\displaystyle \qquad +\iint _{\mathcal {C}}\psi _{\alpha }(u)\eta ^2(x)[\partial _t\xi (x,t)+\frac{3}{2}|\nabla \xi (x,t)|^2a(x,t)\nonumber \\&\displaystyle \qquad +\,\mathrm{{div}} (a(x,t)\nabla \xi (x,t))] e^{\xi (x,t)}\, dx\, dt\,. \end{aligned}$$

(3.23)

Our next goal is to show that

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\le 0 \qquad \text{ in } [\Omega \setminus \partial \Omega ^{\varepsilon }]\times (\tau -\delta ,\tau )\,; \end{aligned}$$

(E1)

$$\begin{aligned}&\eta \, \mathrm{{div}} (a\,\nabla \eta )+2|\nabla \eta |^2a\le C_1\varepsilon ^{\gamma -2} \qquad \text{ in } \Omega \times (\tau -\delta ,\tau )\,, \end{aligned}$$

(E2)

for some $C_1>0$ independent of $\varepsilon .$

Claim 1: Condition (E1) holds.

Proof of Claim 1

We start recalling that, by definition of $\zeta$, the function $\xi$ (and all its derivatives in time and space) is supported in $\Omega \setminus \Omega ^{\varepsilon }=\{x\in \Omega \;|\; d(x)\le \varepsilon \}$ so (E1) is trivially verified in $\Omega ^{\varepsilon }$. Now, consider any $x\in \Omega \setminus \overline{\Omega ^\varepsilon }$ and any $t\in (\tau -\delta , \tau ).$

In view of the definition of $\xi$ (3.6), we compute

$$\begin{aligned} \partial _t\xi =-\frac{\alpha _1\zeta ^2}{2(s-\alpha _1t)^2},\qquad |\nabla \xi |^2=\frac{\zeta ^2|\nabla \zeta |^2}{(s-\alpha _1t)^2} \end{aligned}$$

(3.24)

and

$$\begin{aligned} \mathrm{{div}}(a(x,t)\nabla \xi )=-\nabla a\cdot \frac{\zeta \nabla \zeta }{s-\alpha _1t} -a(x,t)\frac{|\nabla \zeta |^2}{s-\alpha _1 t}-a(x,t)\frac{\zeta \Delta \zeta }{s-\alpha _1 t}\,. \end{aligned}$$

(3.25)

Rewriting the right-hand side of (3.25) by inserting the definition of $\nabla \zeta$ and $\Delta \zeta$, we have

$$\begin{aligned} \nabla a\cdot \zeta \nabla \zeta= & {} -\beta \,\zeta \, \nabla a\cdot \nabla d \,d^{-\beta -1} \nonumber \\ a\zeta \Delta \zeta= & {} \beta (\beta +1)\,a\,\zeta \, d^{-\beta -2}\,|\nabla d|^2-\beta \,a\,\zeta \, d^{-\beta -1}\,\Delta d\nonumber \\ a|\nabla \zeta |^2= & {} \beta ^2\,a\, d^{-2\beta -2}\,|\nabla d|^2\,. \end{aligned}$$

(3.26)

Putting all previous terms together, we obtain the expression

$$\begin{aligned}&\partial _t\xi +\frac{5}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a\nabla \xi )\nonumber \\&=\frac{1}{2(s-\alpha _1 t)^2}\left\{ - \alpha _1\zeta ^2+3 a\zeta ^2\beta ^2\,d^{-2(\beta +1)}\,|\nabla d|^2\right. \nonumber \\&\qquad +\left. 2(s-\alpha _1t)\beta \left[ \zeta \,\nabla a\cdot \nabla d \, d^{-\beta -1} +a\, \zeta \, \Delta d \,d^{-\beta -1}\right. \right. \nonumber \\&\qquad \left. \left. -\beta \,a \,d^{-2\beta -2}\,|\nabla d|^2 -(\beta +1)\, a\, \zeta \,d^{-\beta -2}\,|\nabla d|^2 \right] \right\} \,. \end{aligned}$$

(3.27)

$\square$

In order to estimate the right-hand side of the expression above, we decompose the set $\Omega \setminus \overline{\Omega ^{\varepsilon }}$ as

$$\begin{aligned} \Omega \setminus \overline{\Omega ^{\varepsilon }}=(\Omega \setminus \Omega ^{\varepsilon -\varepsilon _1})\cup (\Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }})\,. \end{aligned}$$

Fig. 1

Illustration of the decomposition of the set $\Omega$

Consider first the region

$$\begin{aligned} \Omega \setminus \Omega ^{\varepsilon -\varepsilon _1}=\{x\in \Omega \; |\; d(x)\le \varepsilon -\varepsilon _1\} \quad \text{ with } \quad \varepsilon _1\in (0,\varepsilon /2)\,. \end{aligned}$$

Thanks to (3.3), the last two terms of the right-hand side can be bounded from above by zero, i.e.,

$$\begin{aligned} -\beta \,a\,d^{-2\beta -2}\,|\nabla d|^2\le 0 \end{aligned}$$

and

$$\begin{aligned} -(\beta +1) a\,d^{-\beta -2}\,|\nabla d|^2 \zeta \le 0\,. \end{aligned}$$

Using (3.2) and (1.3)

and $\zeta (x)=d^{-\beta }-\varepsilon ^{-\beta }\le d^{-\beta }$, we can estimate

$$\begin{aligned} a\Delta d\,d^{-\beta -1}\,\zeta \le c_0 k_0d^{-2\beta -1+\gamma }\,, \end{aligned}$$

while using (3.1) and (1.3),

we have

$$\begin{aligned} \nabla a\cdot \nabla d\,d^{-\beta -1}\,\zeta \le c_1 d^{-2\beta -2+\gamma }\,. \end{aligned}$$

Finally,

$$\begin{aligned} 3a\,\zeta ^2\,\beta ^2\,d^{-2(\beta +1)}\,|\nabla d|^2\le 6 \beta ^2 c_0\,d^{-4\beta -2+\gamma }\,, \end{aligned}$$

where we used

$$\begin{aligned} \zeta ^2=d^{-2\beta }+\varepsilon ^{-2\beta }-2d^{-\beta }\varepsilon ^{-\beta }\le 2d^{-2\beta }\,. \end{aligned}$$

(3.28)

We claim that there exists a $\sigma \in (0,1)$ such that

$$\begin{aligned} -\zeta ^2(x)\le - \sigma ^2 d^{-2\beta }(x)\,, \end{aligned}$$

(3.29)

and this will follow if we can show

$$\begin{aligned} d^{-\beta }(x)-\varepsilon ^{-\beta }(x)\ge \sigma d^{-\beta }(x)\,. \end{aligned}$$

The letter is equivalent to

$$\begin{aligned} \sigma \le 1-\left( \frac{d}{\varepsilon }\right) ^{\beta }\,, \end{aligned}$$

which is clearly fulfilled by choosing

$$\begin{aligned} \sigma =1-\left( \frac{\varepsilon -\varepsilon _1}{\varepsilon }\right) ^{\beta }\,. \end{aligned}$$

Let $\varepsilon _1=c\varepsilon$ with $c\in \left( 0,\frac{1}{2}\right)$. Hence,

$$\begin{aligned} \sigma =1-(1-c)^{\beta }. \end{aligned}$$

Now, we can use (3.29) and (3.28) to estimate the right-hand side of (3.27) further:

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\\&\le \frac{1}{2(s-\alpha _1 t)^2} \Big \{-\alpha _1 \sigma ^2 d^{-2\beta } +6 \beta ^2\,c_0\,d^{-4\beta -2+\gamma }\\&\qquad + (s-\alpha _1t)\beta \left[ c_1 d^{-2\beta -2+\gamma }+c_0d^{-2\beta -1+\gamma } \right] \Big \}\\ \\&\le \frac{d^{-4\beta -2+\gamma }}{2(s-\alpha _1 t)^2}\Big \{-\alpha _1\sigma ^2 d^{2\beta +2-\gamma }+6\beta ^2c_0+(s-\alpha _1t)\beta \left[ c_1 d^{2\beta }+c_0d^{2\beta +1}\right] \Big \}\,. \end{aligned}$$

Choose

$$\begin{aligned} \beta =\frac{\gamma -2}{2}\,, \end{aligned}$$

(3.30)

$$\begin{aligned} s=\alpha _1(\tau +\delta )\,. \end{aligned}$$

(3.31)

So, for all $t\in (\tau -\delta , \tau ),$

$$\begin{aligned} \alpha _1\delta<s-\alpha _1t<2\alpha _1\delta \,. \end{aligned}$$

(3.32)

This together with the fact that

$$\begin{aligned} d(x)\le \varepsilon \quad \text{ in } \Omega \setminus \Omega ^{\varepsilon } \end{aligned}$$

yields

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\nonumber \\&\quad \le \frac{d^{-4\beta +\gamma -2}}{2(s-\alpha _1 t)^2}\left\{ -\alpha _1\sigma ^2+6\beta ^2c_0+ 2\alpha _1 \delta \beta \left[ c_1 \varepsilon ^{2\beta }+c_0\varepsilon ^{2\beta +1}\right] \right\} \,. \end{aligned}$$

(3.33)

If we impose that

$$\begin{aligned} 0<\delta \le \frac{\sigma ^2}{ 2 \beta \left[ c_1 \varepsilon ^{2\beta }+c_0\varepsilon ^{2\beta +1}\right] }\,, \end{aligned}$$

(3.34)

then from (3.33) we get

$$\begin{aligned} \partial _t\xi +\frac{3}{2} a|\nabla \xi |^2+\mathrm{{div}}(a\nabla \xi )\le & {} \frac{d^{-4\beta -2+\gamma }}{2(s-\alpha _1 t)^2}\left\{ -\frac{\alpha _1}{2}\sigma ^2+6 c_0\beta ^2\right\} \,. \end{aligned}$$

Now, observe that in view of assumption (3.18), condition (3.34) is true. Finally, if

$$\begin{aligned} \alpha _1\ge \frac{24 c_0 \beta ^2}{\sigma ^2}\,, \end{aligned}$$

(3.35)

then

$$\begin{aligned} \partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\le & {} \frac{d^{-4\beta -2+\gamma }}{2(s-\alpha _1 t)^2}\left\{ -\frac{\alpha _1}{4}\sigma ^2\right\} <0\,. \end{aligned}$$

Now, consider the region

$$\begin{aligned} \Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }}=\{x\in \Omega \;|\; \;\varepsilon -\varepsilon _1< d(x) \le \varepsilon \}\, \end{aligned}$$

For any $x\in \Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }}, t\in (\tau -\delta , \tau )$, thanks to (3.27), (3.2), (3.3), (3.30), (3.32), we have:

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\\&\qquad \le \frac{1}{2(s-\alpha _1 t)^2}\Big \{\zeta ^2(-\alpha _1+5c_0\beta ^2) \\&\qquad \quad +2(s-\alpha _1t)\beta \big (\zeta \,[c_1 d^{\gamma -\beta -2}+ c_0k_0 d^{\gamma -\beta -1}]-\beta {\tilde{c}}_0 \nu _0\big )\Big \}\,. \end{aligned}$$

Observe that for any $x\in \Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }}$,

$$\begin{aligned} \zeta (x)\le (\varepsilon -\varepsilon _1)^{-\beta }-\varepsilon ^{-\beta }=\varepsilon ^{-\beta }[(1-c)^{-\beta }-1], \end{aligned}$$

(3.36)

while

$$\begin{aligned} d(x)<\varepsilon \,. \end{aligned}$$

(3.37)

In view of (3.18), (3.19), (3.36), (3.37), we obtain

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\&\quad \le \frac{\beta }{s-\alpha _1 t}\left\{ [(1-c)^{-\beta }-1](c_1+c_0k_0)-\beta {\tilde{c}}_0\nu _0 \right\} <0, \end{aligned}$$

thanks to (3.8). $\square$

Claim 2: Condition (E2) holds.

Proof for Claim 2

Using (3.14) and (3.15) we have

$$\begin{aligned} \eta \; \mathrm{{div}}(a(x,t)\nabla \eta )+2|\nabla \eta |^2a(x,t)= & {} \eta \nabla a(x,t)\cdot \nabla \eta +\eta a(x,t) \Delta \eta + 2|\nabla \eta |^2a(x,t)\\\le & {} |\nabla a(x,t)||\nabla \eta |+ |a(x,t)| |\Delta \eta |+ 2|\nabla \eta |^2|a(x,t)|\\\le & {} c_0\gamma d^{\gamma -1}\frac{A_1}{\varepsilon }+c_0d^{\gamma }\frac{1}{\varepsilon ^{2}}+2\frac{A_1^2}{\varepsilon ^2}c_0d^{\gamma }\\= & {} \frac{1}{\varepsilon ^2}d^{\gamma }\left( d^{-1}A_1 c_1 \varepsilon +c_0+2 A_1^2 c_0\right) \,. \end{aligned}$$

Because of the support conditions of $\nabla \eta$ and $\Delta \eta$ (contained in the set $\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }=\{x\in \Omega : \frac{\varepsilon }{2}<d\le \frac{2}{3}\varepsilon \}$), the term $d^{-1}A_1 c_1 \varepsilon +c_0+2 A_1^2 c_0$ is bounded by a constant independent of $\varepsilon$, and the claim follows.

Finally, inserting (E1) and (E2) in (3.23) we obtain

$$\begin{aligned}&\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \\&\quad + C_1 \varepsilon ^{\gamma -2}\iint _{\tau -\delta }^{\tau }\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} \, dx\, dt\,, \end{aligned}$$

with $C_1>0$, independent of $\varepsilon$, as in (E2). This completes the proof. $\square$

Proof of Proposition 3.1

For every $\alpha >0$ define

$$\begin{aligned} \psi _{\alpha }(z)=(z^2+\alpha )^{\frac{1}{2}}, \qquad z\in \mathbb {R}\,, \end{aligned}$$

with $\alpha >0$. Since $\psi _{\alpha }''\ge 0$, in view of (3.17) we can infer that $\psi _{\alpha }$ is a subsolution of (1.1). The application of Lemma 3.3 yields

$$\begin{aligned}&\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&\quad + C_1\varepsilon ^{\gamma -2} \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }\times (\tau -\delta , \tau )}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} \, dx\, dt\,; \end{aligned}$$

(3.38)

here, $\xi$ is defined as in (3.6), and conditions (3.30) and (3.31) hold.

Using that $\Omega ^{\varepsilon }\subset \Omega ^{\frac{\varepsilon }{2}}$, $\eta =1$ on $\Omega ^{\frac{2}{3}\varepsilon }$ and the positivity of the integrand we have

$$\begin{aligned}&\int _{\Omega ^{\varepsilon }}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx=\int _{\Omega ^{\varepsilon }}\psi _{\alpha }(u(x,\tau ))e^{\xi (x,\tau )}\, dx\\&\quad \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))e^{\xi (x,\tau )}\, dx\,. \end{aligned}$$

Letting $\alpha \rightarrow 0^+$ in (3.38), applying the Lebesgue’s dominated convergence theorem and observing that $0\le \eta \le 1$, we obtain

$$\begin{aligned} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|e^{\xi (x,\tau )}\, dx\le & {} \int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&+ C_1\varepsilon ^{\gamma -2} \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }\times (\tau -\delta , \tau )}|u(x,t)| e^{\xi (x,t)} \, dx\, dt\,.\nonumber \\ \end{aligned}$$

(3.39)

Recalling (3.4), we first notice that $\xi =0$ in $\Omega ^{\varepsilon }$ for any $t\in [\tau -\delta , \tau ]$. Choose s as in (3.31). Therefore, $\xi (x,t)<0$ for all $x\in \Omega \setminus \overline{\Omega ^{\varepsilon }}$ and $t\in (\tau -\delta ,\tau )$. Since

$$\begin{aligned} \Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }=\{x\in \Omega : \frac{\varepsilon }{2}<d(x)\le \frac{2}{3}\varepsilon \}\subset \Omega \setminus \Omega ^{\varepsilon }, \end{aligned}$$

then

$$\begin{aligned} \zeta (x)=d^{-\beta }-\varepsilon ^{-\beta }= \varepsilon ^{-\beta }\left( \left( \frac{\varepsilon }{d}\right) ^{\beta }-1\right) \quad \text{ in } \Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }\,. \end{aligned}$$

The bound

$$\begin{aligned} \left( \frac{\varepsilon }{d}\right) ^{\beta }\ge \left( \frac{3}{2}\right) ^{\beta } \quad \text{ for } \beta >0\,, \end{aligned}$$

yields

$$\begin{aligned} \zeta (x)\ge \tilde{C}\varepsilon ^{-\beta } \qquad \text{ with } \tilde{C}=\left( \frac{3}{2}\right) ^{\beta }-1\,. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \xi (x,t)=-\frac{\zeta ^2(x)}{2(s-\alpha _1 t)}\le -\frac{\tilde{C}^2 \varepsilon ^{-2\beta }}{4\alpha _1 \delta } \qquad \text{ in } \Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon } \,. \end{aligned}$$

Finally, inserting this bound in (3.39), we obtain

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx &{}\le &{}\int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx &{}\\ \\ &{}&{}+ C_1\varepsilon ^{\gamma -2}e^{-\frac{\tilde{C}^2 \varepsilon ^{-2\beta }}{4\alpha _1 \delta } } \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }\times (\tau -\delta , \tau )}|u(x,t)| \, dx\, dt\,. \end{array} \end{aligned}$$

Due to condition (2.2), it follows that

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx &{}\le &{}\int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx &{}\\ \\ &{}&{}+ C_1 {\text {meas}}(\Omega ) T\varepsilon ^{\gamma -2}e^{-\frac{\tilde{C}^2 \varepsilon ^{-2\beta }}{4\alpha _1 \delta } } e^{\theta \varepsilon ^{(2-\gamma )}}\,. \end{array} \end{aligned}$$

In view of (3.30), since $0<\delta \le \frac{\tilde{C}^2}{4\theta \alpha _1}$, we obtain

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx\le & {} \int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx + C_1 {\text {meas}}(\Omega ) T\varepsilon ^{\gamma -2}\,. \end{array} \end{aligned}$$

This completes the proof. $\square$

Proof Lemma 3.2

The thesis follows by minor variations of the proof of [20, Proposition 4.1]. However, we give the proof for the reader’s convenience.

Take any $\varepsilon >0, \tau \in (0, T)$. Define

$$\begin{aligned}\varepsilon _k:=2^{-k} \varepsilon \quad \text {for all}\;\; k\in \mathbb N\,,\end{aligned}$$

Furthermore, let $\{\delta _k\}_{k\in \mathbb N}\subset (0, \infty )$ be a sequence fulfilling (3.12), that is

$$\begin{aligned} 0<\delta _k\le \min \{\tau , C\} \quad \text {for all}\;\; k\in \mathbb N\,. \end{aligned}$$

(3.40)

Also, let $\{\tau _k\}_{k\in \mathbb N}$ be a sequence defined inductively as follows

$$\begin{aligned} \tau _0:= & {} \tau ,\\ \tau _{k+1}:= & {} \tau _k-\delta _k\quad \text {for every}\;\; k\in \mathbb N\setminus \{0\}\,. \end{aligned}$$

Observe that

$$\begin{aligned} \tau -\tau _{k+1}=\delta _0+\delta _1+\ldots + \delta _k \quad \text {for every}\;\; k\in \mathbb N\,. \end{aligned}$$

(3.41)

From (3.13), it follows that for every $k\in \mathbb N$

$$\begin{aligned} \int _{\Omega ^{\varepsilon _k}} |u(x, \tau _k)|\, dx \le \int _{\Omega _{\varepsilon _{k+1}}} |u(x, \tau _{k+1})| dx + C \varepsilon _k^\mu \,. \end{aligned}$$

(3.42)

Claim: there exists $k_0\in \mathbb N$ such that $\tau _{k_0+1}=0$.

In fact, in view of (3.41), $\tau _{k_0+1}=0$ if and only if

$$\begin{aligned} \tau =\delta _0+\delta _1+\ldots +\delta _{k_0}\,. \end{aligned}$$

(3.43)

Clearly, we can select the sequence $\{\delta _k\}$ so that (3.40) and (3.43) hold, for some $k_0\in \mathbb N$. So, the Claim has been shown.

The Claim combined with (3.11) yields

$$\begin{aligned} \int _{\Omega ^{\varepsilon _{k_0+1}}} |u(x, \tau _{k_0+1})|\, dx = \int _{\Omega ^{\varepsilon _{k_0+1}}} |u(x, 0)|\, dx = 0\,. \end{aligned}$$

(3.44)

By iterating (3.42) up to $k=k_0$, in view of (4.28), we get

$$\begin{aligned} \int _{\Omega ^{\varepsilon }}| u(x, \tau ) |dx \le \int _{\Omega ^{\varepsilon _{k_0+1}}} |u(x, \tau _{k_0+1})| + C \sum _{k=0}^{k_0} \varepsilon _k^{\mu } \le C \sum _{k=0}^{k_0} \varepsilon _k^{\mu }\,. \end{aligned}$$

(3.45)

Observe that

$$\begin{aligned}\sum _{k=0}^{k_0} \varepsilon _k^{\mu } \le \sum _{k=0}^{k_0} 2^{-k\mu }\varepsilon ^{\mu } \le \varepsilon ^{\mu } \sum _{k=0}^{+\infty } 2^{-\mu k}\le \frac{1}{1-2^{-\mu }}\varepsilon ^{\mu } \mathop {\longrightarrow }_{\varepsilon \rightarrow 0}0\,.\end{aligned}$$

Hence, by letting $\varepsilon \rightarrow 0^+$ in (3.45), we obtain

$$\begin{aligned}\int _{\Omega } |u(x, \tau )| dx = 0\,. \end{aligned}$$

Since $\tau \in (0, T)$ was arbitrary, the conclusion follows.

$\square$

4 Proof of Theorem 2.3

Let $\beta =-\gamma +2$ whenever $\gamma \in (1,2);$ let $\beta =b>0$ whenever $\gamma =2$, with $b>0$ arbitrary. Consider $\ell \in \left( 0,\frac{1}{2}\right)$ such that

$$\begin{aligned}{}[1-(1-\ell )^{\beta }](c_1+c_0k_0)-\beta \nu _0{\tilde{c}}_0 <0\,, \end{aligned}$$

(4.1)

and define

$$\begin{aligned} \bar{\sigma }:=1-(1-\ell )^{\beta }\,. \end{aligned}$$

(4.2)

Define the functions

$$\begin{aligned} \zeta (x)={\left\{ \begin{array}{ll} 0 &{} \text{ for } x\in \Omega ^{\varepsilon }\,\\ \varepsilon ^{\beta }-d^{\beta }(x) &{} \text{ for } x\in \Omega \setminus \Omega ^{\varepsilon }\,, \end{array}\right. } \end{aligned}$$

(4.3)

and

$$\begin{aligned} \xi (x)=-\frac{\zeta ^2(x)}{2(s-\alpha _1 t)}\quad \text {for all }\,\, x\in \Omega , t\ne \frac{s}{\alpha _1}\,. \end{aligned}$$

(4.4)

The proof of Theorem 2.3 is based on the combination of the following results.

Proposition 4.1

Under assumption (1.3) with $\gamma \in [1,2]$, suppose $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1). Let $\tau>0, b>0,$ $\ell \in \left( 0, \frac{1}{2} \right)$ be such that (4.1) is true, $\bar{\sigma }$ be defined by (4.2). Suppose that

$$\begin{aligned} 0<\delta < {\left\{ \begin{array}{ll}\min \left\{ \frac{\bar{\sigma }^2}{16(-\gamma +2)(c_1 +c_0)}\varepsilon ^{-2\gamma +4}, \tau \right\} &{} \text { for }\gamma \in [1,2) \\ \min \left\{ \frac{\bar{\sigma }^2}{16b[c_1 +(b-1)^++c_0]}, \tau \right\} &{} \text { for }\gamma =2, \end{array}\right. } \end{aligned}$$

$$\begin{aligned} \alpha _1 \ge {\left\{ \begin{array}{ll}\max \left\{ \frac{24\,c_0(-\gamma +2)^2}{\bar{\sigma }^2}, 5c_0(2-\gamma )^2\right\} &{} \text { for }\gamma \in [1,2)\\ \max \left\{ \frac{24\,c_0b^2}{\bar{\sigma }^2}, 5c_0 b^2\right\} &{} \text { for }\gamma =2\,, \end{array}\right. } \end{aligned}$$

and that, for some $C>0, \mu >0$,

$$\begin{aligned} \int _0^T\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }}|u(x,t)|d(x)^{\gamma -2}\,dx\, dt\le C\varepsilon ^{\mu } \text{ for } \text{ every } \varepsilon \in (0,\varepsilon _0)\,. \end{aligned}$$

(4.5)

Then,

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx\le & {} \int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx + \tilde{C}\varepsilon ^{\mu }\,, \end{array} \end{aligned}$$

(4.6)

for some constant ${\tilde{C}}>0$ independent of $\varepsilon$.

Lemma 4.2

Let $u\in C(\Omega \times [0, T])$ with

$$\begin{aligned} u=0\quad \text {in}\;\; \Omega \times \{0\}\,. \end{aligned}$$

(4.7)

Suppose that there exist $c>0, {\hat{C}}>0, \varepsilon _0>0, \mu _2>\mu _1>0$ such that for any $\varepsilon \in (0, \varepsilon _0)$, $\tau \in (0,T)$,

$$\begin{aligned} 0<\delta < \min \{\tau , c \varepsilon ^{\mu _1}\}\,, \end{aligned}$$

(4.8)

there holds

$$\begin{aligned} \int _{\Omega ^{\varepsilon }} u(x, \tau )\,\mathrm {d}x \le \int _{\Omega ^{\frac{\epsilon }{2}}} u(x, \tau -\delta ) \mathrm {d}x + \hat{C} \varepsilon ^{\mu _2}\,. \end{aligned}$$

(4.9)

Then,

$$\begin{aligned}u\equiv 0\quad \text {in}\;\; \Omega \times (0, T]\,.\end{aligned}$$

Lemma 4.2 is an extension of Lemma 3.2. Differently from Lemma 3.2, in Lemma 4.2 the bound on $\delta$ goes to zero as $\varepsilon \rightarrow 0^+$. To manage this situation, the condition $\mu _2>\mu _1$ will be expedient.

Proof of Theorem 2.3

The thesis follows, combining Proposition 4.1 and Lemma 4.2, with

$$\begin{aligned} \mu _1=-2\gamma +4, \quad \mu _2=\mu >\mu _1, \\ c={\left\{ \begin{array}{ll} \frac{\bar{\sigma }^2}{16(-\gamma +2)(c_1+c_0)} &{} \text{ for } \gamma \in [1,2)\\ \frac{\bar{\sigma }^2}{16b[c_1+(b-1)^++c_0]} &{} \text{ for } \gamma =2\,, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \tilde{C}=\hat{C}\,. \end{aligned}$$

$\square$

4.1 Proofs of Proposition 4.1 and Lemma 4.2

The proof of Proposition 4.1 is based on the following crucial lemma.

Lemma 4.3

Under assumption (1.3) with $\gamma \in [1,2]$, suppose $u\in C^{2,1}(Q_T)\cap C(\Omega \times [0,T])$ solves (2.1). Let $\varepsilon \in (0, \varepsilon _0), \tau>0, b>0,$ $\ell \in \left( 0, \frac{1}{2} \right)$ be such that (4.1) is true, $\bar{\sigma }$ be defined by (4.2). If

$$\begin{aligned} 0<\delta < {\left\{ \begin{array}{ll}\min \left\{ \frac{\bar{\sigma }^2}{16(-\gamma +2)(c_1 +c_0)}\varepsilon ^{-2\gamma +4}, \tau \right\} &{} \text { for }\gamma \in [1,2) \\ \min \left\{ \frac{\bar{\sigma }^2}{16b[c_1 +(b-1)^++c_0]}, \tau \right\} &{} \text { for }\gamma =2, \end{array}\right. } \end{aligned}$$

(4.10)

$$\begin{aligned} \alpha _1 \ge {\left\{ \begin{array}{ll}\max \left\{ \frac{24\,c_0(-\gamma +2)^2}{\sigma ^2}, 5c_0(2-\gamma )^2\right\} &{} \text { for }\gamma \in [1,2)\\ \max \left\{ \frac{24\,c_0b^2}{\bar{\sigma }^2}, 5c_0 b^2\right\} &{} \text { for }\gamma =2\,, \end{array}\right. } \end{aligned}$$

(4.11)

then

$$\begin{aligned}&\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&\quad + C_1\int _{\tau -\delta }^{\tau }\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} d^{\gamma -2}(x)\, dx\, dt\,, \end{aligned}$$

(4.12)

where $\xi$ is defined as in (4.4) with $s=\alpha _1(\tau +\delta )$, for a suitable $C_1>0$ independent of $\varepsilon$.

Proof of Lemma 4.3

Let $\zeta$ be defined by (4.3). Then,

$$\begin{aligned} \nabla \zeta =-\beta d^{\beta -1}\nabla d \quad \text{ and } \quad \Delta \zeta =-\beta (\beta -1)d^{\beta -2}(\nabla d)^2-\beta d^{\beta -1}\Delta d\,.\qquad \end{aligned}$$

(4.13)

Let $\xi$ be defined as in (4.4). Imitating the arguments in Proposition 3.3, we derive

$$\begin{aligned}&\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx\nonumber \\&\quad \le \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx \nonumber \\&\qquad + \iint _{\mathcal {C}}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} [\eta (x)\mathrm{{div}} (a(x,t)\nabla \eta (x))+2|\nabla \eta (x)|^2a(x,t)]\, dx\, dt\nonumber \\&\qquad +\iint _{\mathcal {C}}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)}\eta ^2(x)[\partial _t\xi +\frac{3}{2}|\nabla \xi (x,t)|^2a(x,t)\nonumber \\&\qquad +\,\mathrm{{div}} (a(x,t)\nabla \xi (x,t))]\, dx\, dt\,, \end{aligned}$$

(4.14)

which is exactly (3.23). Our next goal is to ensure that the following two conditions

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a\,|\nabla \xi |^2+\mathrm{{div}}(a\,\nabla \xi )\le 0 \quad \text{ in } [\Omega \setminus \partial \Omega ^{\varepsilon }]\times (\tau -\delta ,\tau )\, ; \end{aligned}$$

(D1)

$$\begin{aligned}&\eta \, \mathrm{{div}} (a\,\nabla \eta )+2|\nabla \eta |^2a\,\le C_1d^{\gamma -2} \quad \text{ in } \Omega \times (\tau -\delta ,\tau ) \end{aligned}$$

(D2)

are simultaneously satisfied. $\square$

Claim 3: Condition (D1) holds.

Proof of for Claim 3

By the same arguments used to obtain (3.27), we deduce that

$$\begin{aligned} \begin{array}{rlll} &{}&{}\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\ \\ &{}&{}\quad =\frac{1}{2(s-\alpha _1 t)^2}\left\{ - \alpha _1\zeta ^2(x)+3 \beta ^2 a(x,t)\,\zeta ^2(x)\,d^{2\beta -2}(x)\,|\nabla d(x)|^2\right. \\ \\ &{}&{}\left. \qquad +\,2(s-\alpha _1 t)\beta \left[ \zeta (x)\nabla a(x,t)\cdot \nabla d(x) d^{\beta -1}(x)-\beta a(x,t) d^{2\beta -2}(x)|\nabla d(x)|^2\right. \right. \\ \\ &{}&{}\left. \left. \qquad +\,(\beta -1)a(x,t)\zeta (x) d^{\beta -2}(x)|\nabla d(x)|^2+a(x,t)\zeta (x) d^{\beta -1}(x)\Delta d(x)\right] \right\} . \end{array}\nonumber \\ \end{aligned}$$

(4.15)

Also here, because of (3.3) and the non-negativity of a(x, t), we have

$$\begin{aligned} -\beta a(x,t) d^{2\beta -2}(x)|\nabla d(x)|^2\le 0\,. \end{aligned}$$

We now analyze all the other terms on the right-hand-side singularly, using the fact that in $\Omega \setminus \Omega ^{\varepsilon }$ we have $d(x)\le \varepsilon$. We start with the second term:

$$\begin{aligned} 3\beta ^2 a(x,t)\zeta ^2(x)\,d^{2\beta -2}(x)\,|\nabla d(x)|^2\le 6 \beta ^2 c_0\varepsilon ^{2\beta } d^{2\beta -2+\gamma }(x) \le 6\beta ^2c_0 \varepsilon ^{4\beta +\gamma -2}\,, \end{aligned}$$

where we used (1.3), (3.1) and that $\zeta ^2\le 2\varepsilon ^{2\beta }$, together with the hypothesis that $\gamma +2\beta -2\ge 0$ for the last inequality.

For the third term, we use again (1.3) and (3.1) to obtain

$$\begin{aligned} \zeta (x)\nabla a(x,t)\cdot \nabla d(x) d^{\beta -1}(x)\le \varepsilon ^{\beta }c_1d^{\gamma -1}(x)|\nabla d(x)|d^{\beta -1}(x)\le c_1 \varepsilon ^{\beta +\gamma -2}\,, \end{aligned}$$

where the last inequality holds if $\gamma +\beta -2\ge 0$.

Again, if $\gamma +\beta -2\ge 0$, the fourth term is estimated easily as

$$\begin{aligned} (\beta -1)a(x,t)\zeta (x)d^{\beta -2}(x)|\nabla d(x)|^2\le c_0(\beta -1)\varepsilon ^{\beta }d^{\beta +\gamma -2}(x)\le (\beta -1)c_0\varepsilon ^{2\beta +\gamma -2}\,. \end{aligned}$$

Finally, we estimate the last term as

$$\begin{aligned} a(x,t)\zeta (x) d^{\beta -1}(x)\Delta d(x)\le c_0 \varepsilon ^{\beta }d^{\beta +\gamma -1}(x)k_0\le c_0 \varepsilon ^{2\beta +\gamma -1} \end{aligned}$$

if $\gamma +\beta -1\ge 0$. Collecting these estimates in the range $1\le \gamma \le 2$ we choose $\beta \ge -\gamma +2$, so that all the previous conditions are satisfied. In particular, we set

$$\begin{aligned} \beta = {\left\{ \begin{array}{ll} -\gamma +2 &{} \text{ for } 1\le \gamma <2\\ b &{} \text{ for } \gamma =2 \end{array}\right. } \end{aligned}$$

where b is any positive number.

Finally, choose

$$\begin{aligned} s=\alpha _1(\tau +\delta )\,. \end{aligned}$$

(4.16)

so, for all $t\in (\tau -\delta , \tau ),$

$$\begin{aligned} \alpha _1\delta<s-\alpha _1t<2\alpha _1\delta \,. \end{aligned}$$

(4.17)

We now write the set $\Omega \setminus \overline{\Omega ^{\varepsilon }}$ as a union of two disjoint sets

$$\begin{aligned} \Omega \setminus \overline{\Omega ^{\varepsilon }}=(\Omega \setminus \Omega ^{\varepsilon -\varepsilon _2})\cup (\Omega ^{\varepsilon -\varepsilon _2}\setminus \overline{\Omega ^{\varepsilon }})\,, \end{aligned}$$

and analyze the validity of condition (D1) separately in the two domains. First, let us consider the set $\Omega \setminus \Omega ^{\varepsilon -\varepsilon _2}=\{x\in \Omega : d(x)\le \varepsilon -\varepsilon _2\}$ and look at the case $\gamma \in [1,2)$ and $\gamma =2$ separately.

• For the case $\gamma \in [1,2)$ and $\beta =-\gamma +2$, we have

  $$\begin{aligned} \begin{array}{rlll} &{} &{} \partial _t \xi + \frac{3}{2} a(x,t) |\nabla \xi |^2 + \mathrm{{div}} (a(x,t) \nabla \xi ) &{}\\ \\ &{} &{}\quad \le \frac{1}{2(s-\alpha _1 t)^2} \left\{ - \alpha _1\zeta ^2(x) + 6 c_0\beta ^2 \varepsilon ^{-3\gamma +6} \right. &{}\\ \\ &{} &{}\qquad + \left. 2 (s-\alpha _1t) (-\gamma +2) \left[ c_1 +c_0(-\gamma +1)\varepsilon ^{-\gamma +2}+c_0\varepsilon ^{-\gamma +3} \right] \right\} \,.&{} \end{array} \end{aligned}$$

  For the first term on the right-hand-side, we claim the existence of a $\bar{\sigma }\in (0,1]$ such that

  $$\begin{aligned} -\zeta ^2\le -\bar{\sigma }^2 \varepsilon ^{2\beta }\, \end{aligned}$$

  and this is equivalent to the condition

  $$\begin{aligned} \bar{\sigma }\le 1-\left( \frac{d}{\varepsilon }\right) ^{\beta }\,. \end{aligned}$$

  (4.18)

  We set $\varepsilon _2=\ell \varepsilon$ with $\ell \in (0,\frac{1}{2})$ and choose $\bar{\sigma }=1-(1-\ell )^{\beta }$ so that (4.18) is trivially satisfied. Thus,

  $$\begin{aligned} \begin{array}{rlll} &{}&{}\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\ \\ &{}&{}\quad \le \frac{1}{2(s-\alpha _1 t)^2}\left\{ - \alpha _1\bar{\sigma }^2 \varepsilon ^{-2\gamma +4}+ 6 c_0(-\gamma +2)^2 \varepsilon ^{-3\gamma +6} \right. \\ \\ &{}&{} \qquad \left. +2(s-\alpha _1t)(-\gamma +2) \left[ c_1 +c_0(-\gamma +1)\varepsilon ^{-\gamma +2}+c_0\varepsilon ^{-\gamma +3} \right] \right\} \,, \end{array} \end{aligned}$$

  and, using that $0<s-\alpha _1 t<2\alpha _1\delta$, we obtain

  $$\begin{aligned} \begin{array}{rlll} &{}&{}\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\ \\ &{}&{}\quad \le \frac{1}{2(s-\alpha _1 t)^2}\left\{ - \alpha _1\bar{\sigma }^2 \varepsilon ^{-2\gamma +4}+ 6 c_0(-\gamma +2)^2 \varepsilon ^{-3\gamma +6}\right. \\ \\ &{}&{}\qquad \left. +4\alpha _1\delta (-\gamma +2) \left[ c_1+c_0(-\gamma +1)\varepsilon ^{-\gamma +2}+c_0\varepsilon ^{-\gamma +3} \right] \right\} \\ \\ &{}&{}\quad \le \frac{\varepsilon ^{-2\gamma +4}}{2(s-\alpha _1t)^2}\left\{ -\alpha _1\bar{\sigma }^2+6 c_0(-\gamma +2)^2\varepsilon ^{-\gamma +2}\right. \\ \\ &{}&{}\qquad \left. +4\alpha _1\delta (-\gamma +2)\varepsilon ^{2\gamma -4}[c_1 +c_0\varepsilon ^{-\gamma +3}]\right\} , \end{array} \end{aligned}$$

  where in the last inequality we used $(-\gamma +1)\varepsilon ^{-\gamma +2}\le 0$ since $\gamma \ge 1$. Comparing the three terms (the first with the third and then the first with the second), we obtain

  $$\begin{aligned} \partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\le \frac{\varepsilon ^{-3\gamma +6}}{2(s-\alpha _1 t)^2}\left\{ - \frac{\alpha _1}{4}\bar{\sigma }^2 \right\} <0 \, \end{aligned}$$

  if the following two conditions are satisfied:

  $$\begin{aligned} 0<\delta \le \frac{\bar{\sigma }^2\varepsilon ^{-2\gamma +4}}{16(-\gamma +2)[c_1 +c_0\varepsilon ^{-\gamma +3}]}\qquad \text{ and } \qquad \alpha _1\ge \frac{24 c_0(-\gamma +2)^2\varepsilon ^{-\gamma +2}}{\bar{\sigma }^2}\,. \end{aligned}$$

• For the case $\gamma =2$ and $\beta =b>0$, we have

  $$\begin{aligned} \begin{array}{rlll} &{} &{} \partial _t \xi + \frac{3}{2} a(x,t) |\nabla \xi |^2 + \mathrm{{div}} (a(x,t) \nabla \xi ) &{}\\ \\ &{} &{}\quad \le \frac{1}{2(s-\alpha _1 t)^2} \left\{ - \alpha _1\zeta ^2(x) + 6 c_0b^2 \varepsilon \right. &{}\\ \\ &{} &{} \qquad + \left. 2 (s-\alpha _1t) \beta \left[ c_1 +(b-1)^++c_0\varepsilon \right] \right\} \,. \end{array} \end{aligned}$$

  Proceeding as above, we deduce easily

  $$\begin{aligned} \partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\le \frac{1}{2(s-\alpha _1 t)^2}\left\{ - \frac{\alpha _1}{4}\bar{\sigma }^2 \right\} <0 \, \end{aligned}$$

  if the following two conditions are satisfied:

  $$\begin{aligned} 0<\delta \le \frac{\bar{\sigma }^2}{16b[c_1 +(b-1)^++c_0\varepsilon ]}\qquad \text{ and } \qquad \alpha _1\ge \frac{24 c_0b^2}{\bar{\sigma }^2}\,. \end{aligned}$$

Now, consider the region

$$\begin{aligned} \Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }}=\{x\in \Omega \;|\; \;\varepsilon -\varepsilon _1< d(x)\le \varepsilon \}\, \end{aligned}$$

For any $x\in \Omega ^{\varepsilon -\varepsilon _1}\setminus \Omega ^{\varepsilon }, t\in (\tau -\delta , \tau )$, thanks to (4.15), (3.2), (3.3), (4.17) we have:

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\&\quad =\frac{1}{2(s-\alpha _1 t)^2}\Big \{\zeta ^2(-\alpha _1+3c_0\beta ^2)\Big .\\&\Big .\qquad +2(s-\alpha _1t)\beta \Big [ \zeta \left( c_1 d^{\gamma +\beta -2}(x)+ c_0k_0 d^{\gamma +\beta -1}(x)\right) -d^{2\beta -2+\gamma }(x)\beta {\tilde{c}}_0 \nu _0\Big ]\Big \}. \end{aligned}$$

Observe that for any $x\in \Omega ^{\varepsilon -\varepsilon _1}\setminus \overline{\Omega ^{\varepsilon }}$,

$$\begin{aligned} \zeta (x)\le \varepsilon ^{-\beta }- (\varepsilon -\varepsilon _1)^{-\beta }=\varepsilon ^{-\beta }[1-(1-\ell )^{\beta }], \end{aligned}$$

(4.19)

while

$$\begin{aligned} d(x)<\varepsilon \,. \end{aligned}$$

(4.20)

In view of (3.18), (3.27), (4.19), (4.20), we obtain

$$\begin{aligned}&\partial _t\xi +\frac{3}{2} a(x,t)|\nabla \xi |^2+\mathrm{{div}}(a(x,t)\nabla \xi )\\&\quad \le \frac{\beta \varepsilon ^{2\beta -2+\gamma }}{s-\alpha _1 t}\left\{ [(1-\ell )^{\beta }-1](c_1+c_0k_0)-\beta \nu _0{\tilde{c}}_0 \right\} <0, \end{aligned}$$

thanks to (4.1). $\square$

Claim 4: Condition (D2) holds.

Proof of Claim 4

Using the properties of $\eta$ in (3.14) and (3.15) and the assumption on a(x, t) in (1.3), we have

$$\begin{aligned} \eta \mathrm{{div}} (a(x,t)\nabla \eta )+2|\nabla \eta |^2a(x,t)\le & {} c_1d^{\gamma -1}\frac{A_1}{\varepsilon }+c_0d^{\gamma }\frac{A_2}{\varepsilon ^2}+2 \frac{A_1^2}{\varepsilon ^2}c_0 d^{\gamma }\\\le & {} C_1 d^{\gamma -2}\, \end{aligned}$$

in $\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }$.

Finally, inserting the estimates in (D1) and (D2) in (4.14) we obtain

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau ))\eta ^2(x)e^{\xi (x,\tau )}\, dx &{}=&{}\int _{\Omega ^{\frac{\varepsilon }{2}}}\psi _{\alpha }(u(x,\tau -\delta ))\eta ^2(x)e^{\xi (x,\tau -\delta )}\, dx &{}\\ \\ &{}&{}+ C_1\int _{\tau -\delta }^{\tau }\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }}\psi _{\alpha }(u(x,t)) e^{\xi (x,t)} d^{\gamma -2}(x)\, dx\, dt\,, \end{array} \end{aligned}$$

which coincides with (4.12). $\square$

Proof of Proposition 4.1

Using the same arguments as in Proposition 3.1 and the Lebesgue’s dominated convergence theorem, from (4.12), we have

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|e^{\xi (x,\tau )}\, dx &{}\le &{}\int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|e^{\xi (x,\tau -\delta )}\, dx &{}\\ \\ &{}&{}+ C_1\int _{\tau -\delta }^{\tau }\int _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3}\varepsilon }}|u(x,t)| e^{\xi (x,t)} d^{\gamma -2}(x)\, dx\, dt\,. \end{array} \end{aligned}$$

(4.21)

By the definition of (4.3), $\xi =0$ in $\Omega ^{\varepsilon }$ for any $t\in [\tau -\delta , \tau ]$. Choose s as in (4.16). So, $\xi (x,t)<0$ for all $x\in \Omega \setminus \overline{\Omega ^{\varepsilon }}$ and $t\in [\tau -\delta , \tau ]$, so $e^{\xi (x,t)} \le 1$.

Therefore, from (4.21) we obtain

$$\begin{aligned} \begin{array}{rlll} \int _{\Omega ^{\varepsilon }}|u(x,\tau )|\, dx &{}\le &{}\int _{\Omega ^{\frac{\varepsilon }{2}}}|u(x,\tau -\delta )|\, dx &{}\\ \\ &{}&{}+ C_1 \iint _{\Omega ^{\frac{\varepsilon }{2}}\setminus \Omega ^{\frac{2}{3} \varepsilon }\times (\tau -\delta , \tau )}|u(x,t)| d^{\gamma -2}(x) \, dx\, dt\,. \end{array} \end{aligned}$$

Finally, we use the assumption (4.5) to get (4.6).

$\square$

Proof of Lemma 4.2

Take any $\varepsilon >0, \tau \in (0, T)$. Define

$$\begin{aligned}\varepsilon _k:=\frac{\varepsilon }{k^{\frac{1}{\mu _1}}}\quad \text {for all}\;\; k\in \mathbb N\,.\end{aligned}$$

Note that

$$\begin{aligned} \sum _{k=1}^{+\infty } \varepsilon _k^{\mu _1}=+\infty \,, \end{aligned}$$

(4.22)

while, since $\mu _2>\mu _1$,

$$\begin{aligned} \sum _{k=1}^{+\infty } \frac{1}{k^{\frac{\mu _2}{\mu _1}}}=:S<+\infty \,. \end{aligned}$$

(4.23)

Furthermore, let $\{\delta _k\}_{k\in \mathbb N}\subset [0, \infty )$ and $\{\tau _k\}_{k\in \mathbb N}\subset [0, \tau ]$ be two sequences with $\{\tau _k\}$ defined inductively as follows

$$\begin{aligned}&\tau _1:=\tau ,\\&\tau _{k+1}:=\tau _k-\delta _k\quad \text {for every}\;\; k\in \mathbb N, k\ge 2\,, \end{aligned}$$

and

$$\begin{aligned} 0\le \delta _k\le \min \{\tau _k, C_1\varepsilon _k^{\mu _1}\} \quad \text {for all}\;\; k\in \mathbb N\,. \end{aligned}$$

(4.24)

Observe that

$$\begin{aligned} \tau -\tau _{k+1}=\delta _1+\delta _2+\ldots + \delta _k \quad \text {for every}\;\; k\in \mathbb N\,. \end{aligned}$$

(4.25)

We can choose $\{\tau _k\}$ so that there exists $k_0\in \mathbb N$ with $\tau _{k_0+1}=0$. In fact, in view of (4.25), $\tau _{k_0+1}=0$ if and only if

$$\begin{aligned} \tau =\delta _0+\delta _1+\ldots +\delta _{k_0}\,. \end{aligned}$$

(4.26)

Due to (4.22), we can select the sequence $\{\delta _k\}$, and thus $\{\tau _k\}$, so that (4.24) and (4.26) hold, for some $k_0\in \mathbb N$.

From (4.9), it follows that for every $k=1\ldots k_0$

$$\begin{aligned} \int _{\Omega ^{\varepsilon _k}} u(x, \tau _k)\, \mathrm {d}x \le \int _{\Omega _{\varepsilon _{k+1}}} u(x, \tau _{k+1}) \mathrm {d}x + C_2 \varepsilon _k^{\mu _2}\,. \end{aligned}$$

(4.27)

Since $\tau _{k_0+1}=0$, thanks to (3.11) we get

$$\begin{aligned} \int _{\Omega ^{\varepsilon _{k_0+1}}} u(x, \tau _{k_0+1})\, \mathrm {d}x = \int _{\Omega ^{\varepsilon _{k_0+1}}} u(x, 0)\, \mathrm {d}x = 0\,. \end{aligned}$$

(4.28)

By iterating (4.27) up to $k=k_0$, in view of (4.28), we get

$$\begin{aligned} \int _{\Omega ^{\varepsilon }} u(x, \tau ) \mathrm {d}x \le \int _{\Omega ^{\varepsilon _{k_0+1}}} u(x, \tau _{k_0+1}) + C_2 \sum _{k=1}^{k_0} \varepsilon _k^{\mu _2} = C_2 \sum _{k=1}^{k_0} \varepsilon _k^{\mu _2}\,. \end{aligned}$$

(4.29)

Thanks to (4.23),

$$\begin{aligned}\sum _{k=1}^{k_0} \varepsilon _k^{\mu _2} = \sum _{k=1}^{k_0} \frac{\varepsilon ^{\mu _2}}{k^{\frac{\mu _2}{\mu _1}}} \le \varepsilon ^{\mu _2} \sum _{k=1}^{+\infty } \frac{1}{k^{\frac{\mu _2}{\mu _1}}}= S \varepsilon ^{\mu _2} \mathop {\longrightarrow }_{\varepsilon \rightarrow 0}0\,.\end{aligned}$$

Hence, by letting $\varepsilon \rightarrow 0^+$ in (4.29), we obtain

$$\begin{aligned}\int _{\Omega } u(x, \tau ) \mathrm {d}x = 0\,. \end{aligned}$$

Since $\tau \in (0, T)$ was arbitrary, the conclusion follows. $\square$