We first let \(q=\frac{p}{p-1}\) and \(Q_{r}(z,s)=B_{r}(z)\times(-r^{q}+s,r^{q}+s)\) be the cylinder in \(\mathbb{R}^{N+1}\). Write \(Q_{1}=Q_{1}(0,0)\), the unit cylinder. Due to the local character of the results obtained in this paper (Theorem 1.1), we may consider the following local formulation. We say that a function *u* is in \(W^{1,p}(Q_{1})\) belongs to the class \(\mathscr{G}_{a}=\mathscr{G}_{a}(p,\gamma_{0},\gamma_{1})\) if

$$\begin{aligned} &\bigl\| \operatorname{div} a(x,\nabla u)-\partial_{t}u \bigr\| _{\infty,Q_{1}}\leq1; \end{aligned}$$

(2a)

$$\begin{aligned} &0\leq u \leq1, \quad\mbox{a.e. in } Q_{1}; \end{aligned}$$

(2b)

$$\begin{aligned} &u(0,0)=0; \end{aligned}$$

(2c)

$$\begin{aligned} &\partial_{t}u\geq0 \quad\mbox{ a.e. in } Q_{1}. \end{aligned}$$

(2d)

Condition (2a) should be understood in the weak sense, *i.e.*, \(\operatorname{div} a(x,\nabla u)-\partial_{t}u=h\) weakly for \(h\in L^{\infty}(Q_{1})\) with \(\|h\|_{\infty,Q_{1}}\leq1\). Condition (2c) makes sense since (2a) and (2b) provide that \(u\in C^{1,\alpha}_{x}\cap C^{0,\alpha}_{t} (Q_{\frac{1}{2}})\) and \(u\in C^{1,\alpha}(Q_{\frac{1}{2}})\) for some \(\alpha\in(0,1)\) in the case of \(p\geq2\) and \(1< p<2\), respectively (see *e.g.* Chapter IX of [1]).

In this section, we discuss the behavior of solutions to problem (1) and functions in \(\mathscr{G}_{a}\) near the free boundary.

### 2.1 Non-degeneracy of the solution near the free boundary

The following result gives a description of the solution *u* to problem (1) showing that it cannot grow too slowly near the free boundary. This property and the growth rate of the elements in \(\mathscr{G}_{a}\) will pave the way to establish the porosity of the free boundary.

### Lemma 2.1

*Let*
\(u\in W^{1,p}(Q_{1})\)
*be a non*-*negative continuous function in*
\(Q_{1}\), *satisfying*

$$ \operatorname{div} a(x,\nabla u)- \partial_{t}u=f $$

*weakly in*
\(U^{+}=\{u>0\}\). *Then for every*
\((z,s)\in\overline{U^{+}}\)
*and*
\(r>0\)
*with*
\(Q_{r}(z,s)\subset Q_{1}\)

$$ \sup_{(x,t)\in\partial_{p}Q_{r}^{-}(z,s)} u(x,t) \geq c_{0}r^{\frac{p}{p-1}}+u(z,s), $$

*where*
\(Q_{r}^{-}(z,s)=B_{r}(z)\times(s-r^{q},s)\), \(c_{0}\)
*is a positive constant depending only on*
*p*, \(\lambda_{0}\), \(\gamma_{1}\).

### Proof

First suppose that \((z,s)\in U^{+}\), and for small \(\varepsilon>0\) set

$$ u_{\varepsilon}(x,t)=u(x,t)-(1-\varepsilon)u(z,s) $$

and

$$ v(x,t)=C_{1}|x-z|^{\frac{p}{p-1}}-C_{2}(t-s), $$

where \(C_{1}\), \(C_{2}\) are positive constants, depending only on *p*, \(\lambda_{0}\), \(\gamma_{1}\), such that

$$\gamma_{1}C_{1}^{p-1} \biggl(\frac{p}{p-1} \biggr)^{p}+C_{2} \leq\lambda_{0}. $$

We claim that for \(C_{1}\), \(C_{2}\)

$$ \operatorname{div} a(x,\nabla v)- \partial_{t} v \leq \lambda_{0},\quad \forall (x,t) \in U^{+} \cap Q_{r}^{-}(z,s). $$

(3)

To prove (3), we need to calculate ∇*v* and the divergence of \(a(x,\nabla v)\). Indeed,

$$\nabla v(x,t)=\frac{pC_{1}}{p-1}|x-z|^{\frac{2-p}{p-1}}(x-z),\qquad \bigl|D_{ij}v(x,t)\bigr| \leq \frac{pC_{1}}{(p-1)^{2}}|x-z|^{\frac{2-p}{p-1}}. $$

One may verify that

$$\begin{aligned} \operatorname{div} a(x,\nabla v) -\partial_{t} v =&\sum _{i=1}^{N}\frac{\partial a_{i}}{\partial x_{i}}(x,w)+\sum _{i,j=1}^{N}\frac{\partial a_{i}}{\partial \eta_{j}}(x,w)\frac{\partial w_{j}}{\partial x_{i}}(x)+C_{2} \\ \leq& \gamma_{1}|w|^{p-1}+\gamma_{1}|w|^{p-2} \frac{pC_{1}}{(p-1)^{2}} |x-z|^{\frac{2-p}{p-1}}+C_{2} \\ \leq &\gamma_{1} \biggl(\frac{pC_{1}}{p-1} \biggr)^{p-1} \biggl(|x-z|+\frac {1}{p-1} \biggr)+C_{2} \\ \leq &\gamma_{1} \biggl(\frac{pC_{1}}{p-1} \biggr)^{p-1} \biggl(1+\frac {1}{p-1} \biggr)+C_{2} \\ \leq&\lambda_{0}, \end{aligned}$$

where \(w(x,t)=\nabla v(x,t)=\frac{pC_{1}}{p-1}|x-z|^{\frac{2-p}{p-1}}(x-z)\).

Notice that \(\operatorname{div} a(x,\nabla u)-\partial_{t} u = \operatorname{div} a(x,\nabla u_{\varepsilon})-\partial_{t} u_{\varepsilon}\) in \(U^{+}\cap Q_{r}^{-}(z,s)\). Recall condition (f), it follows that

$$\operatorname{div} a(x,\nabla v) -\partial_{t} v\leq \operatorname{div} a(x,\nabla u_{\varepsilon})-\partial_{t} u_{\varepsilon} \quad\mbox{in } \Omega_{+}\cap Q_{r}^{-}(z,s). $$

It is easy to see \(u_{\varepsilon}(x,t)=-(1-\varepsilon)u(z,s)\leq 0\) on \(\partial U^{+}\) and \(v(x,t)\geq0\) for any \(t\leq s\), thus \(u_{\varepsilon}\leq v\) on \(\partial U^{+}\cap Q_{r}^{-}(z,s)\). If also \(u_{\varepsilon} \leq v\) on \(\partial Q_{r}^{-}(z,s)\cap U^{+} \), then we get by the comparison principle

$$u_{\varepsilon} \leq v \quad\mbox{in } Q_{r}^{-}(z,s)\cap U^{+}. $$

But \(u_{\varepsilon}(z,s)=\varepsilon u(z,s)>0=v(z,s)\), which is a contradiction. Therefore there exists some point \((y,\tau)\in \partial Q_{r}^{-}(z,s)\) such that

$$\begin{aligned} u_{\varepsilon}(y,\tau)\geq v(y,\tau)=c_{0}r^{\frac{p}{p-1}}, \end{aligned}$$

where \(c_{0}=\min\{C_{1},C_{2}\}\). Letting \(\varepsilon\rightarrow 0\) we obtain the desired result for all \((z,s)\in U^{+}\), and by continuity for all \((z,s)\in\overline{U^{+}}\). □

### 2.2 Growth rate of the function *u* in \(\mathscr{G}_{a}\)

In this subsection we prove that every function *u* in \(\mathscr{G}_{a}\) cannot grow too fast near the free boundary but has a growth rate of order \(q=\frac{p}{p-1}\) (Theorem 2.1).

First we define the supremum norm of *u* over the cylinder \(Q_{r}^{-}(z,s)\) as [5] by setting

$$\begin{aligned} S(r,u,z,s)=\sup_{x\in Q_{r}^{-}(z,s)}u(x,t),\quad \mbox{and} \quad S(r,u)=\sup _{x\in Q_{r}^{-}(0,0)}u(x,t) . \end{aligned}$$

For each \(u\in\mathscr{G}_{a}\), define the set \(\mathbb{M}_{a}(u,z,s)\) by setting

$$\begin{aligned} \mathbb{M}_{a}(u,z,s)= \bigl\{ j\in\mathbb{N}; AS \bigl(2^{-j-1},u,z,s \bigr)\geq S \bigl(2^{-j},u,z,s \bigr) \bigr\} , \end{aligned}$$

where \(A=2^{q}\max\{1,\frac{1}{c_{0}}\}\) with \(q=\frac{p}{p-1}\), and \(c_{0}\) as in Lemma 2.1. For simplicity, we write \(\mathbb{M}_{a}(u)=\mathbb{M}_{a}(u,0,0)\).

It should be noticed that \(\mathbb{M}_{a}(u)\neq\emptyset\) for all \(u\in \mathscr{G}_{a}\) since \(0\in\mathbb{M}_{a}(u)\). Indeed, it follows from Lemma 2.1 that \(S(1,u) \leq 1=(\frac{1}{c_{0}2^{-q}})c_{0}2^{-q}\leq(\frac {1}{c_{0}2^{-q}})S(2^{-1},u)=AS(2^{-1},u)\).

Now we state the growth property of the elements in the class \(\mathscr{G}_{a}\).

### Theorem 2.1

*There is a positive constant*
\(M_{0}=M_{0}(p,\gamma_{0},\gamma_{1})\)
*such that*, *for every*
\(u\in\mathscr{G}_{a}\),

$$\begin{aligned} \bigl|u(x,t)\bigr|\leq M_{0} \bigl(d(x,t) \bigr)^{q} \quad\forall (x,t)\in Q_{\frac{1}{2}}, \end{aligned}$$

*where*
\(d(x,t)=\sup\{r; Q_{r}(x,t)\subset U^{+}\}\)
*for*
\((x,t)\in U^{+}\), *and*
\(d(x,t)=0\)
*otherwise*.

To prove this theorem we need the following lemma.

### Lemma 2.2

*There is a positive constant*
\(M_{1}=M_{1}(p,\gamma_{0},\gamma_{1})\)
*such that*

$$\begin{aligned} S \bigl(2^{-j-1},u \bigr)\leq M_{1} \bigl(2^{-j} \bigr)^{q}, \end{aligned}$$

*for all*
\(u\in\mathscr{G}_{a}\)
*and*
\(j\in\mathbb{M}_{a}(u)\).

### Proof

Arguing by contradiction, we assume that for every \(k\in \mathbb{N}\), there exists \(u_{k}\in\mathscr{G}_{a}\) and \(j_{k}\in \mathbb{M}_{a}(u_{k})\) such that

$$\begin{aligned} S \bigl(2^{-j_{k}-1},u_{k} \bigr)\geq k \bigl(2^{-j_{k}} \bigr)^{q}. \end{aligned}$$

(4)

Observe that by the uniform boundedness of \(u_{k}\) and (4) it follows that \(j_{k}\rightarrow\infty\) as \(k\rightarrow\infty\).

Consider the function

$$\begin{aligned} \widetilde{u}_{k}(x,t)=\frac{u_{k}(2^{-j_{k}}x,\alpha _{k}t)}{S(2^{-j_{k}-1},u_{k})} \end{aligned}$$

defined in the unit cylinder, where \(\alpha_{k}=(2^{-j_{k}})^{p}(S(2^{-j_{k}-1},u_{k}))^{2-p}\). Note that by (4) we have

$$\begin{aligned} \alpha_{k} \leq& \frac{1}{k^{p-1}} \bigl(S \bigl(2^{-j_{k}-1},u_{k} \bigr) \bigr)^{p-1}\cdot \bigl(S \bigl(2^{-j_{k}-1},u_{k} \bigr) \bigr)^{2-p} \\ =&\frac{1}{k^{p-1}}S \bigl(2^{-j_{k}-1},u_{k} \bigr) \\ \leq&\frac{1}{k^{p-1}}\rightarrow0 \quad\mbox{as } k\rightarrow \infty. \end{aligned}$$

By the definition of \(\mathbb{M}_{a}(u_{k})\) and \(\mathscr{G}_{a}\) it follows that

$$\begin{aligned}& 0\leq\widetilde{u}_{k} \leq A \quad\mbox{in } Q_{1}^{-}, \\& \sup_{Q_{\frac{1}{2}}^{-}} \widetilde{u}_{k}\geq1 \quad \bigl( \mbox{by } (2d) \mbox{ and } \bigl(2^{-1} \bigr)^{q} \alpha_{k}\geq \bigl(2^{-j_{k}-1} \bigr)^{q} \bigr), \\& \widetilde{u}_{k}(0,0)=0, \\& \partial_{t}\widetilde{u}_{k}\geq0\quad \mbox{in } Q_{1}^{-}. \end{aligned}$$

Now, define for \((x,\eta)\in B_{1}\times\mathbb{R}^{N}\)

$$\begin{aligned} a^{k}(x,\eta) =& \biggl(\frac{2^{-j_{k}}}{S(2^{-j_{k}-1},u_{k})} \biggr)^{p-1} \cdot a \biggl(2^{-j_{k}}x, \frac{S(2^{-j_{k}-1},u_{k})}{2^{-j_{k}}}\eta \biggr). \end{aligned}$$

We claim that \(a^{k}(x,\eta)\) satisfies the same structural conditions as \(a(x,\eta)\) for large *k*. Indeed, letting \(s_{k}=\frac{2^{-j_{k}}}{S(2^{-j_{k}-1},u_{k})}\), one may verify directly that

$$ \begin{aligned} \sum^{N}_{i,j=1} \frac{\partial a^{k}_{i}}{\partial\eta _{j}}(x,\eta) \xi_{i}\xi_{j}&= \sum ^{N}_{i,j=1}s_{k}^{p-2} \frac{\partial a_{i}}{\partial\eta_{j}} \bigl(2^{-j_{k}}x, s_{k}^{-1}\eta \bigr)\xi_{i}\xi_{j} \\ &\geq\gamma_{0}s_{k}^{p-2} \bigl|s_{k}^{-1} \eta\bigr|^{p-2}|\xi|^{2} \\ &=\gamma_{0}|\eta|^{p-2}|\xi|^{2}, \\ \sum^{N}_{i,j=1}\biggl| \frac{\partial a^{k}_{i}}{\partial\eta _{j}}(x, \eta)\biggr|&=\sum^{N}_{i,j=1}s_{k}^{p-2} \biggl\vert \frac{\partial a_{i}}{\partial\eta_{j}} \bigl(2^{-j_{k}}x, s_{k}^{-1} \eta \bigr) \biggr\vert \\ &\leq\gamma_{1}s_{k}^{p-2} \bigl|s_{k}^{-1} \eta\bigr|^{p-2} \\ &=\gamma_{1}|\eta|^{p-2}, \\ \sum^{N}_{i,j=1}\biggl| \frac{\partial a^{k}_{i}}{\partial x _{j}}(x, \eta)\biggr|&=\sum^{N}_{i,j=1} s_{k}^{p-1} 2^{-j_{k}}\biggl|\frac{\partial a_{i}}{\partial x _{j}} \bigl( 2^{-j_{k}}x, s_{k}^{-1} \eta \bigr)\biggr| \\ &\leq 2^{-j_{k}}\gamma_{1}|\eta|^{p-1} \\ &\leq\gamma_{1}|\eta|^{p-1}. \end{aligned} $$

(5)

Now by (2a) and (4) we obtain

$$\begin{aligned} \bigl\| \operatorname{div} a^{k} \bigl(x,\nabla \widetilde{u}_{k}(x,t) \bigr)-\partial_{t}\widetilde{u}_{k}(x,t) \bigr\| _{\infty} =& 2^{-j_{k}}s_{k}^{p-1}\bigl\| (Au_{k}-\partial _{t}u_{k}) \bigl(2^{-j_{k}}x, \alpha_{k}t \bigr)\bigr\| _{\infty} \\ \leq &2^{-j_{k}} \biggl(\frac{2^{-j_{k}}}{S(2^{-j_{k}-1},u_{k})} \biggr) ^{p-1} \\ \leq&\frac{1}{k^{p-1}}\rightarrow0 \quad\mbox{as } k \rightarrow\infty, \end{aligned}$$

where \((Au)(x,t)\) is defined by \((Au)(x,t)=\operatorname{div} a(x,\nabla u(x,t))\).

Observe that by (5), for any \(M>0\) we have

$$\begin{aligned} \biggl|\frac{\partial a^{k}_{i}}{\partial x _{j}}(x,\eta)\biggr|\rightarrow0 \quad \mbox{as } k \rightarrow\infty, \end{aligned}$$

uniformly in \((x,\eta)\in B_{1}\times B_{M}\). Therefore the pointwise limit of \(a^{k}(x,\eta)\) does not depend on *x*:

$$\begin{aligned} a^{k}(x,\eta) \rightarrow\widetilde{a}(\eta), \end{aligned}$$

with *ã* satisfying the same structural conditions as (a_{1})-(a_{4}). Then invoking compactness arguments (see Lemma 14.1 on p.75 and 14-(iii) on p.115 of [1]), we deduce that, up to a subsequence, \(\widetilde{u}_{k}\) converges locally uniformly in \(Q_{1}^{-}\) to a function *u*. Moreover, the limit function *u* satisfies

$$ \begin{aligned} &\operatorname{div} \widetilde{a}(\nabla u)-\partial_{t} u =0,\qquad u \geq0,\qquad u(0,0)=0,\\ &\sup_{Q_{\frac{1}{2}}^{-}}u\geq1,\qquad \partial_{t}u\geq0 \quad\mbox{in } Q_{1}^{-}. \end{aligned} $$

(6)

To get a contradiction, we divide our problem into two cases.

-*Case* 1 (\(1< p\leq2\)). In this case, we need the following lemma originating from [6], where the authors stated it for *p*-parabolic equations (\(1< p<2\)). One should pay attention to the fact that the proof of the following lemma can be repeated as in [6] with slight modifications. Moreover, the result is valid for \(p=2\) since the process is ‘stable’ as \(p\nearrow2\) so that one may recover the regularity results by letting \(p\nearrow2\) (see the proofs of Theorems 1 and 2, or the remarks in 1-(iii) on p.323 of [6]).

### Lemma 2.3

(Theorem 2 [6])

*Let* Ω *be a region of*
\(\mathbb {R}^{N}\), \(\Omega_{\infty}=\Omega\times(0,\infty)\)
*and*
\(u\in C(0,T; L^{2}(\Omega))\cap L^{2}(0,T; W^{1,p}(\Omega))\)
*be any non*-*negative local solution of*

$$\begin{aligned} \operatorname{div} \widetilde{a}(\nabla u)-\partial_{t} u =0 \quad \textit{in } \Omega_{\infty}. \end{aligned}$$

*Suppose*
\(u(x_{0},t_{0})>0\)
*for some*
\((x_{0},t_{0})\in \Omega_{\infty}\). *Then*, *for any ball*
\(B_{\rho}(x_{0})\subset\Omega\),

$$\begin{aligned} u(x,t_{0})>0 \quad\forall x\in B_{\rho}(x_{0}). \end{aligned}$$

Now notice that \(\sup_{B_{\frac{1}{2}}}u(x,0)\geq1\) by \(\partial_{t} u \geq0\) and (6). One may find \(x_{0}\in B_{\frac{1}{2}}(0)\) such that \(u(x_{0},0)\geq\frac{1}{2}\). On the other hand, since, for any \(\overline{Q'}\subset Q_{1}\), \(u\in C^{1,\alpha}(Q')\) for some \(\alpha\in(0,1)\) (see Chapter IX of [1]), and then Lemma 2.3 gives \(u(0,0)\geq\frac{1}{2}\), which is a contradiction. Indeed, in Lemma 2.3, one may let \(\rho=\frac{1}{2}\in(|x_{0}|,1-|x_{0}|)\) and \(\Omega=B_{r}\subset B_{1}\) with \(r=\frac{\frac{1}{2}+|x_{0}|+1}{2}\). Therefore \(B_{\rho}(x_{0})\subset\Omega\) and \(0\in B_{\rho}(x_{0})\).

-*Case* 2 (\(2< p<\infty\)). In this case, due to the lack of a strong minimum principle, we need further discussion to get a contradiction. At this point, we show that *u* is time independent, *i.e.*,

$$\partial_{t} u =0 \quad\mbox{in } Q_{1}^{-}. $$

Then *u* is a nonzero, non-negative harmonic function in the unit ball and it vanishes at the origin. Indeed, this is a contradiction to the strong minimum principle; see [10] for instance. To this end, choosing \((x,t),(x',t')\in Q_{\frac{1}{2}}^{-} \) and using the definition of \(\mathbb{M}_{a}(u_{k})\) and Hölder’s estimates for solutions with \(G_{T}=Q_{2^{-j_{k}}}^{-}\) and \(K=Q_{2^{-j_{k}-1}}^{-}\) (see Theorem 1.1 on p.41 of [1]), we arrive at

$$\begin{aligned} \bigl|\widetilde{u}_{k}(x,t)-\widetilde{u}_{k} \bigl(x,t' \bigr)\bigr| =& \frac{|u_{k}(2^{-j_{k}}x, \alpha_{k}t)-u_{k}(2^{-j_{k}}x,\alpha _{k}t')|}{S(2^{-j_{k}-1},u_{k})} \\ \leq &A\frac{|u_{k}(2^{-j_{k}}x,\alpha_{k}t)-u_{k}(2^{-j_{k}}x,\alpha _{k}t')|}{S(2^{-j_{k}},u_{k})} \\ \leq& A \gamma \frac{\|u_{k}\|_{\infty,G_{T}}}{S(2^{-j_{k}},u_{k})} \biggl(\frac{\|u_{k}\|_{\infty,G_{T}}^{\frac{p-2}{p}}\alpha _{k}^{\frac{1}{p}}|t-t'|^{\frac{1}{p}}}{ \operatorname{dist}_{p}(K, \partial_{p}G_{T};p)} \biggr)^{\alpha}, \end{aligned}$$

where \(\operatorname{dist}_{p}(K, \partial_{p}G_{T};p)=\inf_{(x,t)\in K, (y,s)\in \partial_{p}G_{T}} (|x-y|+\|u\|_{\infty, G_{T}}^{\frac{2-p}{p}}|t-s|^{\frac{1}{p}} )\), \(\alpha\in(0,1)\) is the Hölder exponent, the constant *γ* does not depend on \(\|u_{k}\|_{\infty,G_{T}}\).

Now observe that \(\operatorname{dist}_{p}(K, \partial_{p}G_{T};p)\geq(2^{-j_{k}-1})^{\frac{q}{p}}\|u\|_{\infty, G_{T}}^{\frac{2-p}{p}}\). It follows that

$$\begin{aligned} \bigl|\widetilde{u}_{k}(x,t)-\widetilde{u}_{k} \bigl(x,t' \bigr)\bigr| \leq& A \gamma\bigl|t-t'\bigr|^{\frac{\alpha}{p}} \bigl(\alpha_{k}\cdot 2^{(j_{k}+1)q} \bigr)^{\frac{\alpha}{p}} \\ =&A \gamma\bigl|t-t'\bigr|^{\frac{\alpha}{p}} \bigl[ \bigl(2^{-j_{k}} \bigr)^{p}\cdot \bigl(S \bigl(2^{-j_{k}-1},u_{k} \bigr) \bigr)^{2-p}\cdot 2^{(j_{k}+1)q} \bigr]^{\frac{\alpha}{p}} \\ \leq& A \gamma\bigl|t-t'\bigr|^{\frac{\alpha}{p}} \bigl[ \bigl(2^{-j_{k}} \bigr)^{p-1}\cdot k^{2-p}\cdot \bigl(2^{-j_{k}} \bigr)^{q(2-p)} \cdot \bigl(2^{j_{k}} \bigr)^{q}\cdot 2^{q} \bigr]^{\frac{\alpha}{p}} \\ =& 2^{\frac{\alpha}{p-1}}A \gamma\bigl|t-t'\bigr|^{\frac{\alpha}{p}}k^{2-p} \rightarrow 0 \quad\mbox{as } k\rightarrow\infty. \end{aligned}$$

Hence *u* is *t*-independent, and the proof is completed. □

### Proof of Theorem 2.1

The proof of this theorem is standard (see [5]). For convenience, we recover the process. Let us take *j* for which

$$\begin{aligned} S \bigl(2^{-j},u \bigr)>2^{q}M_{1}2^{-qj}. \end{aligned}$$

It follows that

$$\begin{aligned} S \bigl(2^{-j+1},u \bigr)\leq2^{q}M_{1}2^{-q(j-1)}< 2^{q}S \bigl(2^{-j},u \bigr)\leq AS \bigl(2^{-j},u \bigr), \end{aligned}$$

(7)

*i.e.*
\(j-1\in\mathbb{M}_{a}(u)\), so Lemma 2.2 holds for \(j-1\). Now we arrive at the following obvious contradiction to (7):

$$\begin{aligned} S \bigl(2^{-j},u \bigr)\leq S \bigl(2^{-j+1},u \bigr)\leq M_{1}2^{-q(j-1)}=2^{q}M_{1}2^{-qj}. \end{aligned}$$

Therefore

$$\begin{aligned} S \bigl(2^{-j},u \bigr)\leq2^{q}M_{1}2^{-qj},\quad \forall j, \end{aligned}$$

which implies

$$\begin{aligned} \sup_{Q_{r}^{-}(0,0)}u\leq2^{q}M_{1}2^{-qj},\quad \forall r\leq1. \end{aligned}$$

To obtain a similar estimate for *u* over the whole cylinder (and not only over the lower half part) we use an upper barrier. Define \(w(x,t)=C_{3}|x|^{q}+C_{4}t\) where \(C_{4}=1+\gamma_{1}(qC_{3})^{p-1} (1+\frac{1}{p-1} )\) and \(C_{3}>0\). Let now \(Q^{+}_{1}=B_{1}(0)\times(0,1)\). Then proceeding as Lemma 2.1, we deduce

$$\begin{aligned} \operatorname{div} a(x,\nabla w) -\partial_{t} w \leq& \gamma_{1}(qC_{3})^{p-1} \biggl(1+\frac{1}{p-1} \biggr)-C_{4} \\ =& -1\leq\operatorname{div} a(x,\nabla u) -\partial_{t} u \quad \mbox{in } Q^{+}_{1}. \end{aligned}$$

Since by choosing \(C_{3}\) large, we will have \(w\geq u\) on \(\partial_{p}Q^{+}_{1}\), where for the estimate on \(\{t=0\}\) we have sued the previous discussion, *i.e.*, \(S(r,u)\leq Cr^{q}\). Hence by the comparison principle we have \(w\geq u\) in \(Q^{+}_{1}\). Therefore

$$\begin{aligned} \sup_{Q_{r}(0,0)}u\leq M_{2}r^{q}. \end{aligned}$$

The proof is completed. □