Stabilization of the solution of a doubly nonlinear parabolic equation

Consider a sequence $\omega _k \in C_0^{\infty }(\Omega)$ of linearly independent functions whose linear hull is dense in $\mathring{W}_{\alpha,p}^1(\Omega)$. We set $I_m=\bigcup _{k=1}^m \operatorname{supp}\omega _k$. Galerkin approximations to the solution will be sought in the form

$$\begin{equation*} u_m(t,x)= \sum _{k=1}^m c_{mk}(t)\omega _k(x), \end{equation*}$$

where the functions $c_{mk}(t)$ are determined from the equations

$$\begin{equation} \begin{aligned} &\int _{\Omega }\biggl (\omega _j\frac{\partial }{\partial t} \biggl (\frac{u_m}{b_m}+|u_m|^{\alpha -2}u_m\biggr) +\sum _{i=1}^n|u_{mx_i}|^{p-2}u_{mx_i}(\omega _j)_{x_i}\biggr)\,dx \\ &\qquad =(f(t),\omega _j)_{\Omega }, \qquad j=1,2,\dots,m. \end{aligned} \end{equation} \tag{ 3.1 }$$

The numbers $b_m>0$ will be chosen later. We claim that equations (3.1) can be solved for the derivatives $c_{mk}^{\prime }$. Clearly, they are as follows:

$$\begin{equation*} A_{jk}(t)c_{mk}^{\prime }=F_j(c_{m1},c_{m2},\dots,c_{mm})+f_j(t). \end{equation*}$$

For each $t$, the coefficient matrix

$$\begin{equation*} A_{jk}(t)=\int _{\Omega }\biggl (\frac{1}{b_m}+(\alpha -1)|u_m|^{\alpha -2}\biggr) \omega _j\omega _k\,dx \end{equation*}$$

is a Gram matrix of the system of linearly independent vectors $\omega _k$, $k=1,2,\dots,m$, and hence is invertible. From equations (3.1) with initial conditions $c_{mk}(0)$, chosen so that $u_m(0,x)\rightarrow u_0(x)$ in $\mathring{W}_{\alpha,p}^1(\Omega)$, we find the functions $c_{mk}(t)$. At first, these functions are determined on a short time interval. Then, the fact that the Galerkin approximations are bounded (to be proved later) lets us define them on an infinite time interval. The numbers $b_m$ will be chosen so as to have $\Vert u_m(0)\Vert _2^2/b_m\rightarrow 0$ as $m\rightarrow \infty$.

Now we proceed to estimate the Galerkin approximations. Multiplying equations (3.1) by $c_{mj}(t)$ and summing, this gives

$$\begin{equation*} \int _{\Omega }\biggl (u_m(t,x)\biggl (\frac{u_m}{b_m}+|u_m|^{\alpha -2}u_m\biggr)_t+ \sum _{i=1}^n|u_{mx_i}|^p\biggr)dx=(f(t),u_m)_{\Omega }. \end{equation*}$$

Integrating with respect to $t$ shows that

$$\begin{equation} \begin{aligned} & \int _{\Omega }\biggl (\frac{u_m^2(t)}{2b_m}+\frac{\alpha -1}{\alpha }|u_m(t)|^{\alpha }\biggr)dx +\Vert \nabla u_m\Vert _{p,D_0^t}^p \\ &\qquad =(f,u_m)_{D_0^t} +\int _{\Omega }\biggl (\frac{u_m^2(0)}{2b_m}+\frac{\alpha -1}{\alpha }|u_m(0)|^{\alpha }\biggr)dx. \end{aligned} \end{equation} \tag{ 3.2 }$$

The last integral on the right is bounded by a constant independent of $m$, thanks to the above convergences. Further, for $t\in (0,T)$ we have

$$\begin{equation} \begin{aligned} |(f,u_m)_{D_0^t}|&\leqslant C(T) \Vert u_m\Vert _{V(D_0^t)} \\ &\leqslant c(\varepsilon,T)+\varepsilon \int _0^t(\Vert u_m(\tau)\Vert _{\alpha }^{\alpha }+\Vert \nabla u_m(\tau)\Vert _p^p)\,d\tau. \end{aligned} \end{equation} \tag{ 3.3 }$$

Hence, from (3.2) and Gronwall's lemma we see that the sequence $u_m$ is bounded in the spaces $C([0,T];L_{\alpha }(\Omega))$ and $V(D^T)$ for all $T>0$ by the constant $C(T,f,\Vert u_0\Vert _\alpha)$, which remains unchanged if the sequence $b_m$ is replaced by a larger one $b_m^{\prime}>b_m$.

Now we multiply equations (3.1) by $c_{mj}^{\prime }(t)$ and sum:

$$\begin{equation*} \int _{\Omega }\biggl (u_m^{\prime }(t,x)\biggl (\frac{u_m}{b_m}+|u_m|^{\alpha -2}u_m\biggr)_t+\sum _{i=1}^n |u_{mx_i}|^{p-2}u_{mx_i}u_{mx_i}^{\prime }\biggr)dx=(f(t),u_m^{\prime })_{\Omega }. \end{equation*}$$

Integrating with respect to $t$ shows that

$$\begin{equation} \begin{aligned} &\int _{D_0^T}\biggl (\frac{1}{b_m}+(\alpha -1)|u_m|^{\alpha -2}\biggr)(u_m^{\prime })^2\,dx\,dt +\frac{1}{p}\Vert \nabla u_m(T)\Vert _p^p \\ &\qquad =\frac{1}{p}\Vert \nabla u_m(0)\Vert _p^p+(f,u_m^{\prime })_{D^T}. \end{aligned} \end{equation} \tag{ 3.4 }$$

Integrating the last term by parts, we obtain

$$\begin{equation*} (f,u_m^{\prime })_{D^T}=(f(T),u_m(T))_{\Omega }-(f(0),u_m(0))_{\Omega }-(f^{\prime },u_m)_{D^T}. \end{equation*}$$

Note that

$$\begin{equation*} \begin{aligned} |(f(T),u_m(T))_{\Omega }|&\leqslant C(T)\Vert u_m(T)\Vert _{W_{\alpha,p}^1} \\ &\leqslant c(\varepsilon,T)+\varepsilon (\Vert u_m(T)\Vert _{\alpha }^{\alpha }+\Vert \nabla u_m(T)\Vert _p^p). \end{aligned} \end{equation*}$$

Further, since $u_m$ is bounded in the space $V(D^T)$, we have

$$\begin{equation*} |(f^{\prime },u_m)_{D^T}|\leqslant C(T). \end{equation*}$$

Now we find from (3.4) that, for any $T>0$, the sequence $|u_m|^{(\alpha -2)/2}u_m^{\prime }$ is bounded in $L_2(D^T)$ and that the sequence $\nabla u_m$ is bounded in $C([0,T]; L_p(\Omega))$. Using the diagonal process, we can now choose a subsequence $u_{m_k}$ which is weakly convergent in these spaces. For brevity of notation we shall omit the subscript $k$:

$$\begin{equation*} \begin{gathered}u_m\rightarrow u \quad \text{weakly in }\ V(D^T); \\ A(u_m)=-\sum _{i=1}^n\frac{\partial }{\partial x_i} (|u_{mx_i}|^{p-2}u_{mx_i})\rightarrow \chi \quad \text{weakly in }\ (V(D^T))^{\prime }; \\ (|u_m|^{(\alpha -2)/2}u_m)^{\prime }\rightarrow \widetilde{u}\quad \text{weakly in }\ L_2(D^T).\end{gathered} \end{equation*}$$

The convergence holds for any $T=1,2,\dots$, and the limit functions agree in their common domain of definition. Consequently, we have convergence for any $T>0$.

Below it will be shown that $\widetilde{u}=(|u|^{(\alpha -2)/2}u)^{\prime }$, $\chi =A(u)$ and that the function $u$ is a generalized solution of problem (1.1), (1.2). We will proceed in three steps, which we will also need later in the subsequent analysis.

Step 1. The sequence $u_m(t)$ is bounded in the space $\mathring{W}^1_{\alpha,p}(\Omega)$ on any finite interval:

$$\begin{equation*} \Vert u_m(t)\Vert _{ W_{\alpha,p}^1(\Omega)}\leqslant C(T), \qquad t\in [0,T], \quad m=1,2,\ldots\,. \end{equation*}$$

We fix a countable dense subset $\lbrace t_s\rbrace$ of $[0,\infty]$. We can assume that $t_0=0$. Given any bounded domain $\Omega ^r\subset \Omega$ with smooth boundary, the embedding $\mathring{W}_1^1(\Omega ^r)\subset L_1(\Omega ^r)$ is compact. Hence, using the diagonal process we can choose a subsequence $u_{m_k}(t_s)\rightarrow h_s$ converging strongly in $L_1(\Omega ^r)$ for all natural $s$. Choosing, if necessary, a further subsequence (and dropping the subscripts), we assume that $u_m(t_s,x)\rightarrow h_s(x)$ almost everywhere in $\Omega ^r$ for any $t_s$. In particular, for $t_0=0$ we have $u_m(0,x)\rightarrow u_0(x)$ almost everywhere in $\Omega$.

For the next step we require the following result.

Lemma 1. Let the sequence $v_m(t)\!\in \! C([0,T];L_2(\Omega))$ satisfy the following properties:

• 1)  
  $v_m(t_s,x)$ converges almost everywhere in $\Omega ^r$ for each $t_s$ and some $r>0$;
• 2)  
  the sequence $v_m^{\prime }(t)$ is bounded in $L_2(D^T)$ for each $T>0$.

Then it contains a subsequence $v_{m_k}$ converging to the function $v$ in the space $C([0,T];L_1(\Omega ^r))$, and $v_{m_k}\rightarrow v$ almost everywhere in $(0,T)\times \Omega ^r$.

Proof. The sequence $v_m(t)$ is equicontinuous in $t$ in the space $L_2(\Omega)$:

$$\begin{equation} \begin{aligned} \Vert v_m(t_2)-v_m(t_1)\Vert _2&\leqslant \biggl \Vert \int _{t_1}^{t_2}v_m^{\prime }(t)\,dt\biggr \Vert _2 \\ &\leqslant \biggl (|t_2-t_1|\int _{t_1}^{t_2}\Vert v_m^{\prime }(t)\Vert _2^2\,dt\biggr)^{1/2} \leqslant c|t_2-t_1|^{1/2}. \end{aligned} \end{equation} \tag{ 3.5 }$$

Next, the sequence $v_m(t)$ is bounded in the space $C([0,T];L_2(\Omega))$. Consequently, it contains a subsequence $v_{m_k}(t)$ converging weakly in $L_2(\Omega)$ for the same $t_s$ as above. Along with the almost everywhere convergence in $\Omega ^r$, this implies strong convergence in $L_1(\Omega ^r)$ for any $t_s$ (see [1], Ch. 1, § 12.2). We again write $v_m(t)$ for $v_{m_k}(t)$.

For a bounded domain $\Omega ^r$, it follows easily from (3.5) that $v_m(t)$ is a uniformly Cauchy sequence in the norm $L_1(\Omega ^r)$:

$$\begin{equation*} \begin{aligned} \Vert v_n(t)-v_m(t)\Vert _{1,\Omega ^r} &=\Vert v_n(t)-v_n(t_{s_k})+v_n(t_{s_k})-v_m(t_{s_k})+v_m(t_{s_k})-v_m(t)\Vert _{1,\Omega ^r} \\ &\leqslant C(r)|t-t_{s_k}|^{{1}/{2}}+\Vert v_n(t_{s_k})-v_m(t_{s_k})\Vert _{1,\Omega ^r}. \end{aligned} \end{equation*}$$

Choosing a finite grid $t_{s_k}$ of small spacing and then increasing $n$ and $m$, we ensure that the right-hand side is uniformly small in $t$.

Thus, we see that $v_{m_k}\rightarrow v$ in $C([0,T];L_1(\Omega ^r))$ for any $T=1,2,\dots$ . The convergence also takes place in $L_1((0,T)\times \Omega ^r)$. Hence, one may select a subsequence converging in $(0,T)\times \Omega ^r$ almost everywhere. The lemma is proved.

Step 2. We apply Lemma 1 to the sequence $v_m=|u_m|^{(\alpha -2)/2}u_m$. Since $r>0$ is arbitrary and $T=1,2,\dots$, we can apply the diagonal process to select a subsequence $v_{m_k}$ converging in $D$ almost everywhere. Hence, the sequence $u_{m_k}$ converges almost everywhere in $D$ to $u$ (see [1], Ch. 1, § 1.4, Lemma 1.3). We quote this result.

Lemma 2. Let $Q$ be a domain in ${R}_{n}$ or in ${R}_{n+1}$. Assume that a sequence $g_m$ converges to $g$ almost everywhere in $Q$ and is bounded in $L_q(Q)$. Then $g_m\rightarrow g$ weakly in $L_q(Q)$.

Clearly, this result, which is stated in [1] for bounded domains, also holds for unbounded domains.

From Lemma 2 it also follows that $v_{m_k}\rightarrow v=|u|^{(\alpha -2)/2}u$ weakly in $L_2(D^T)$ for any $T>0$. Given a fixed $T$, the convergence $v_{m_k}(T)\rightarrow v(T)$ in $L_1(\Omega ^r)$ implies that there is a subsequence convergent almost everywhere in $\Omega ^r$ for all $r>0$. Hence,

$$\begin{equation} u_{m_k}(T,x)\rightarrow u(T,x)\quad \text{almost everywhere in }\ \Omega. \end{equation} \tag{ 3.6 }$$

The sequence $u_m(T)$ is bounded in the space $\mathring{W}_{\alpha,p}^1(\Omega)$, and hence we can choose a subsequence such that

$$\begin{equation} {\begin{array}{c}u_{m_k}(T)\rightarrow u(T)\quad \text{weakly in }\ \mathring{W}_{\alpha,p}^1(\Omega), \\ \Vert u(T)\Vert _{W_{\alpha,p}^1}\leqslant C(T,f,\Vert u_0\Vert _{W_{\alpha,p}^1}); \end{array}} \end{equation} \tag{ 3.7 }$$

also, if $f=0$, then the constant $C$ depends only on $\Vert u_0\Vert _\alpha$. The constant $C$ is monotone in $T$, so this proves (2.2).

Further,

$$\begin{equation*} (v_m^{\prime }, \varphi)_{D^T}=-(v_m,\varphi ^{\prime })_{D^T},\qquad \varphi \in C_0^\infty (D^T). \end{equation*}$$

Passing to the limit, we get

$$\begin{equation*} (\widetilde{u},\varphi)_{D^T}=-(v,\varphi ^{\prime })_{D^T}. \end{equation*}$$

It follows that $\widetilde{u}=v^{\prime }=(|u|^{(\alpha -2)/2}u)^{\prime }$.

We claim that the sequence $|u_m|^{\alpha -2}u_m^{\prime }$ is bounded in $L_{\alpha ^{\prime }}(D^T)$. Indeed,

$$\begin{equation*} \begin{aligned} |(|u_m|^{\alpha -2}u_m^{\prime },\varphi)_{D^T}| &=\bigl |(|u_m|^{(\alpha -2)/2}u_m^{\prime },\varphi |u_m|^{(\alpha -2)/2})_{D^T}\bigr | \\ &\leqslant C\Vert \varphi |u_m|^{(\alpha -2)/2}\Vert _{2,D^T} \\ &\leqslant C\Vert \varphi \Vert _{\alpha,D^T}\Vert u_m\Vert _{\alpha,D^T}^{(\alpha -2)/2}\leqslant C_1\Vert \varphi \Vert _{\alpha,D^T}. \end{aligned} \end{equation*}$$

Hence we can assume that $|u_m|^{\alpha -2}u_m^{\prime }\rightarrow |u|^{\alpha -2}u^{\prime }$ weakly in $L_{\alpha ^{\prime }}(D^T)$. As a result, $|u|^{\alpha -2}u\in C([0,\infty);L_{\alpha ^{\prime }}(\Omega))$. We also have

$$\begin{equation} |u(0)|^{\alpha -2}u(0)=|u_0|^{\alpha -2}u_0. \end{equation} \tag{ 3.8 }$$

In fact, in Step 1 we noted that $u_m(0,x)\rightarrow u_0(x)$ converges almost everywhere in $\Omega$. Now, $v_m(0)\rightarrow v(0)=|u(0)|^{(\alpha -2)/2}u(0)$ in $L_1(\Omega ^r)$, $r>0$, by Lemma 1, thereby proving (3.8).

Step 3. We will now prove the equality $\chi =A(u)$. For this we need some integral relations. We multiply equation (3.1) by the smooth function $d_j(t)$, integrate over $t$, and let $m\rightarrow \infty$, denoting $d_j(t)\omega _j(x)$ by $\varphi$ in the resulting expression:

$$\begin{equation} ((|u|^{\alpha -2}u)^{\prime },\varphi)_{D^T}+ (\chi,\varphi)_{D^T}=(f,\varphi)_{D^T}. \end{equation} \tag{ 3.9 }$$

We note that

$$\begin{equation*} \biggl (\frac{u_m^{\prime }}{b_m},\varphi \biggr)_{D^T} =\frac{1}{b_m}\bigl (-(u_m,\varphi ^{\prime })_{D^T}+(u_m(T),\varphi (T))_{\Omega }-(u_m(0),\varphi)\bigr)\rightarrow 0 \end{equation*}$$

as $u_m$ is bounded in $C([0,T];L_{\alpha }(\Omega))$ and $b_m\rightarrow \infty$. Clearly, any function from $V(D^T)$ can be approximated by linear combinations

$$\begin{equation*} \sum_{j=1}^Nd_j(t)\omega _j(x). \end{equation*}$$

Hence, (3.9) also holds for functions from the space $V(D^T)$. So, in view of (3.8), $u$ will be a generalized solution of problem (1.1), (1.2) if we prove that $\chi =A(u)$.

Note that $v,v^{\prime }\in L_2(D^T)$ implies that $v\in C([0,T];L_2(\Omega))$ and that $\Vert u(t)\Vert _\alpha \in C([0,T])$, $\alpha >1$. Substituting the function $\varphi =u$ into (3.9), this gives

$$\begin{equation} \begin{aligned} (f-\chi,u)_{D^T} &=((|u|^{\alpha -2}u)^{\prime },u)_{D^T}=\frac{\alpha -1}{\alpha } \int _0^T\frac{d}{dt}\Vert v(t)\Vert _2^2\,dt \\ &=\frac{\alpha -1}{\alpha }\bigl (\Vert u(T)\Vert _{\alpha }^{\alpha } -\Vert u(0)\Vert _{\alpha }^{\alpha }\bigr). \end{aligned} \end{equation} \tag{ 3.10 }$$

Our next argument depends on the operator $A$ being monotone. It is readily verified that

$$\begin{equation} X_m=\int _0^T\bigl (A(u_m(t))-A(h(t)), u_m(t)-h(t)\bigr)_{\Omega }\,dt\geqslant 0 \quad \forall \, h\in V(D^T). \end{equation} \tag{ 3.11 }$$

From equations (3.1) it is easily seen that

$$\begin{equation*} \begin{aligned} (A(u_m),u_m)_{D^T} &=(f,u_m)_{D^T}+\frac{\alpha -1}{\alpha } \bigl (\Vert u_m(0)\Vert _{\alpha }^{\alpha }-\Vert u_m(T)\Vert _{\alpha }^{\alpha }\bigr) \\ &\qquad +\frac{1}{2b_m}\bigl (\Vert u_m(0)\Vert _2^2-\Vert u_m(T)\Vert _2^2\bigr). \end{aligned} \end{equation*}$$

Hence,

$$\begin{equation*} \begin{aligned} X_m &=(f,u_m)_{D^T}+\frac{\alpha -1}{\alpha }(\Vert u_m(0)\Vert _{\alpha }^{\alpha } -\Vert u_m(T)\Vert _{\alpha }^{\alpha }) \\ &\qquad +\frac{1}{2b_m}\bigl (\Vert u_m(0)\Vert _2^2-\Vert u_m(T)\Vert _2^2\bigr)-(A(u_m),h)_{D^T}- (A(h),u_m-h)_{D^T}. \end{aligned} \end{equation*}$$

Consequently, since $\liminf \Vert u_m(T)\Vert _{\alpha }\geqslant \Vert u(T)\Vert _{\alpha }$, by (3.6)

$$\begin{equation}\begin{aligned} 0&\leqslant \limsup X_m \\ &\leqslant (f,u)_{D^T} +\frac{\alpha -1}{\alpha }\bigl (\Vert u_0\Vert _{\alpha }^{\alpha }-\Vert u(T)\Vert _{\alpha }^{\alpha }\bigr)- (\chi,h)_{D^T} -(A(h),u-h)_{D^T}. \end{aligned} \end{equation} \tag{ 3.12 }$$

Applying (3.10) shows that

$$\begin{equation*} (\chi -A(h),u-h)\geqslant 0. \end{equation*}$$

Setting $h=u-\lambda \omega$, $\lambda >0$, $\omega \in V(D^T)$, we have

$$\begin{equation*} \lambda (\chi -A(u-\lambda \omega),\omega)_{D^T}\geqslant 0. \end{equation*}$$

Letting $\lambda \rightarrow 0$, we see that $(\chi -A(u),\omega)\geqslant 0$ for any $\omega$. Hence, $\chi =A(u)$.

Galerkin approximations to the solution will be sought in the previous form, but now the functions $c_{mk}(t)$ are determined from the equations

$$\begin{equation} \begin{aligned} &\int _{\Omega }\biggl (\omega _j\frac{\partial }{\partial t} (v_m^{{\alpha }/{2}-1 }(t,x)u_m(t,x)) +\sum _{i=1}^n|u_{mx_i}|^{p-2}u_{mx_i}(\omega _j)_{x_i}\biggr)dx \\ &\qquad =(f(t),\omega _j)_{\Omega }, \qquad j=1,2,\dots,m. \end{aligned} \end{equation} \tag{ 3.13 }$$

Here, the functions $v_m=u_m^2+\varepsilon _m$ are introduced for regularization; the numbers $\varepsilon _m>0$ will be chosen later. We claim that equations (3.13) can be solved with respect to the derivatives $c_{mk}^{\prime }$. Clearly, they are as follows

$$\begin{equation*} A_{jk}(t)c_{mk}^{\prime }=F_j(c_{m_1},c_{m_2},\dots,c_{m_m})+f_j(t). \end{equation*}$$

For each $t$, the coefficient matrix

$$\begin{equation*} A_{jk}(t)=\int _{\Omega }((\alpha -1)u_m^2+\varepsilon _m)v_m^{{\alpha }/{2}-2} \omega _j\omega _k\,dx \end{equation*}$$

is a Gram matrix of the system of linearly independent vectors $\omega _k$, $k=1,2,\dots$, and hence is invertible. From equations (3.13) with initial conditions $c_{mk}(0)$, chosen so that $u_m(0,x)\rightarrow u_0(x)$ in the space $\mathring{W}_{\alpha,p}^1$, we find the functions $c_{mk}(t)$.

Now we estimate the Galerkin approximations. Multiplying equations (3.13) by $c_{mj}(t)$ and summing, we find that

$$\begin{equation*} \begin{aligned} &\int _{\Omega }\biggl (u_m(t,x)((\alpha -1)v_m^{{\alpha }/{2}-1}- \varepsilon _m(\alpha -2)v_m^{{\alpha }/{2}-2})u_m^{\prime } +\sum _{i=1}^n|u_{mx_i}|^p\biggr)dx \\ &\qquad =(f(t),u_m(t))_{\Omega }. \end{aligned} \end{equation*}$$

We have $u_mu_m^{\prime }=v_m^{\prime }/2$ and $\operatorname{supp} u_m \subset I_m$, and hence, integrating with respect to $t$ shows that

$$\begin{equation} \begin{aligned} &\int _{I_m}\biggl (\frac{\alpha -1}{\alpha }v_m(t)^{\alpha /2} -\varepsilon _mv_m(t)^{\alpha /2-1}\biggr)dx +\Vert \nabla u_m\Vert _{p,D_0^t}^p \\ &\qquad =(f,u_m)_{D_0^t} +\int _{I_m}\biggl (\frac{\alpha -1}{\alpha }v_m(0)^{\alpha /2} -\varepsilon _mv_m(0)^{\alpha /2-1}\biggr)dx. \end{aligned} \end{equation} \tag{ 3.14 }$$

Since the sets $I_m$ are bounded, we may choose $\varepsilon _m$ to satisfy

$$\begin{equation} \int _{I_m}\varepsilon _mv_m(t)^{\alpha /2-1}\,dx\leqslant \int _{I_m}\varepsilon _m^{\alpha /2}\,dx\leqslant \varepsilon _m^{{\alpha }/{4}}. \end{equation} \tag{ 3.15 }$$

Now the last integral on the right of (3.14) is bounded by a constant independent of $m$, thanks to the above convergences. As before, inequality (3.3) holds. Hence, using (3.14) and Gronwall's lemma it follows that the integrals $\displaystyle \int _{I_m}v_m(t)^{\alpha /2}\,dx$ are uniformly bounded in $t$ and $m$, and hence so is the sequence $u_m$ in the spaces $C([0,T];L_{\alpha }(\Omega))$ and $V(D^T)$ for all $T>0$.

We now multiply equations (3.13) by $c_{mj}^{\prime }(t)$ and sum:

$$\begin{equation*} \begin{aligned} &\int _{\Omega }\biggl ((u_m^{\prime })^2((\alpha -1)u_m^2+\varepsilon _m)v_m^{{\alpha }/{2}-2}+\sum _{i=1}^n |u_{mx_i}|^{p-2}u_{mx_i}u_{mx_i}^{\prime }\biggr)dx \\ &\qquad =(f(t),u_m^{\prime }(t))_{\Omega }. \end{aligned} \end{equation*}$$

Integrating over $t$ gives

$$\begin{equation} \begin{aligned} &\int _{D_0^T}(u_m^{\prime })^2((\alpha -1)u_m^2+\varepsilon _m)v_m^{\alpha /2-2}\,dx\,dt +\frac{1}{p}\Vert \nabla u_m(T)\Vert _p^p \\ &\qquad =\frac{1}{p}\Vert \nabla u_m(0)\Vert _p^p+(f,u_m^{\prime })_{D^T}. \end{aligned} \end{equation} \tag{ 3.16 }$$

As before, $|(f,u_m^{\prime })_{D^T}|\leqslant C(T)$. In addition, $(\alpha -1)u_m^2+\varepsilon _m\geqslant (\alpha -1)v_m$. Setting

$$\begin{equation*} g_m(u)=\int _0^u(t^2+\varepsilon _m)^{(\alpha -2)/4}\,dt, \end{equation*}$$

it follows from (3.16) that the sequence $v_m^{(\alpha -2)/4}u_m^{\prime }=(g_m(u_m))^{\prime }$ is bounded in $L_2(D^T)$ and the sequence $\nabla u_m$ is bounded in $C([0,T];L_p(\Omega))$. In view of the above results, we can choose a subsequence $u_{m_k}$ that is weakly convergent in the spaces specified below. For brevity, the subscript $k$ will be dropped:

$$\begin{equation*} \begin{gathered}u_m\rightarrow u\quad \text{weakly in }\ {V(D^T)}; \\ A(u_m)\rightarrow \chi \quad \text{weakly in }\ (V(D^T))^{\prime }; \\ (g_m(u_m))^{\prime }\rightarrow \widetilde{u}\quad \text{weakly in }\ L_2(D^T).\end{gathered} \end{equation*}$$

Repeating Steps 1–3, as adapted to the new setting, we will be able to prove that $\widetilde{u}=|u|^{(\alpha -2)/2}u^{\prime }$.

Proceeding as in Step 1, we can assume (dropping the subscripts) that the sequence $u_m(t_s,x)$ converges almost everywhere in $\Omega ^r$ for each $t_s$.

At Step 2, Lemma 1 is applied to the sequence $v_m=g_m(u_m)$ to produce a subsequence converging almost everywhere in $(0,T)\times \Omega ^r$. Since $r>0$, $T>0$ are arbitrary, the diagonal process can be applied to extract a subsequence $g_m(u_{m_k})$ converging almost everywhere in $D$. The sequence of inverse functions $g_m^{-1}(v)$ converges pointwise. Now, Lemma 2 shows that the sequence $u_{m_k}$ converges almost everywhere to $u$ in $D$. Hence, we have proved that

$$\begin{equation*} g_m(u_{m_k})\rightarrow v=\frac{2}{\alpha }|u|^{(\alpha -2)/2}u. \end{equation*}$$

Moreover, $\widetilde{u}=|u|^{(\alpha -2)/2}u^{\prime }$.

As in the case $\alpha \geqslant 2$, (3.6) and (3.7) are satisfied for a fixed $T$.

For $\alpha <2$, the sequence $u_m^{\prime }$ is bounded in $L_{\alpha }(D^T)$. Indeed, using (3.16),

$$\begin{equation*} \begin{gathered}\begin{aligned} |(u_m^{\prime },\varphi)_{D^T}| &=|(v_m^{(\alpha -2)/4}u_m^{\prime },\varphi v_m^{(2-\alpha)/4})_{D^T}|\leqslant C\Vert \varphi v_m^{(2-\alpha)/4}\Vert _{2,D^T(m)} \\ & \leqslant C\Vert \varphi \Vert _{\alpha ^{\prime },D^T}\Vert v_m\Vert _{{\alpha }/{2},D^T(m)}^{(2-\alpha)/4}\leqslant C_1\Vert \varphi \Vert _{\alpha ^{\prime },D^T}, \end{aligned} \\ D^T(m)=(0,T)\times I_m.\end{gathered} \end{equation*}$$

Hence, we can assume that $u_m^{\prime }\rightarrow u^{\prime }$ weakly in $L_{\alpha }(D^T)$, $T>0$, and so $u\in C([0,\infty);L_{\alpha }(\Omega))$. This proves (2.4).

At Step 3, our aim is to show that $\chi =A(u)$. For this purpose we shall need some integral relations. We multiply equations (3.13) by a smooth function $d_j(t)$, integrate with respect to $t\in (0,T)$, and then integrate the first term by parts. Now, writing $\varphi$ for $d_j(t)\omega _j(x)$, we have

$$\begin{equation*} \begin{aligned} &(v_m^{\alpha /2-1}(T)u_m(T),\varphi (T))_{\Omega }- (v_m^{\alpha /2-1}u_m,\varphi _t)_{D^T}+ (A(u_m),\varphi)_{D^T} \\ &\qquad =(f,\varphi)_{D^T}+(v_m^{\alpha /2-1}(0)u_m(0),\varphi (0))_{\Omega }. \end{aligned} \end{equation*}$$

We note that

$$\begin{equation*} |v_m^{\alpha /2-1}u_m|\leqslant v_m^{\alpha /2-1}v_m^{{1}/{2}}\in C([0,T],L_{\alpha ^{\prime }}(\Omega ^r)),\qquad r>0, \end{equation*}$$

since $(v_m^{(\alpha -1)/2})^{\alpha ^{\prime }}=v_m^{\alpha /2}$ is a bounded sequence in $C([0,T];L_1(\Omega ^r))$. Hence, we can choose a subsequence to ensure that $v_m^{\alpha /2-1}u_m\rightarrow |u|^{\alpha -2}u$ weakly in $L_{\alpha ^{\prime },{\text{loc }}}(D^T)$ and $v_m^{\alpha /2-1}(T)u_m(T)\rightarrow |u|^{\alpha -2}u(T)$ weakly in $L_{\alpha ^{\prime },{\text{loc}}}(\Omega)$. That the limit functions have exactly this form is justified by the fact that, as we said above, the subsequence $u_m$ converges almost everywhere in $D^T$ and almost everywhere in $\Omega$ for $t=T$ (see (3.6)). Now making $m\rightarrow \infty$, in view of (3.8) it follows that

$$\begin{equation} \begin{aligned} & (|u|^{\alpha -2}(T)u(T),\varphi (T))_{\Omega }- (|u|^{\alpha -2}u,\varphi _t)_{D^T}+ (\chi,\varphi)_{D^T} \\ &\qquad =(f,\varphi)_{D^T}+(|u_0|^{\alpha -2}u_0,\varphi (0))_{\Omega }. \end{aligned} \end{equation} \tag{ 3.17 }$$

As in the case $\alpha \geqslant 2$, we show that this equality is satisfied for all functions $\varphi \in C_0^\infty (D_{-1}^\infty)$. Relation (3.17) also means that $u$ is a generalized solution of problem (1.1), (1.2), provided that $\chi =A(u)$.

Substituting $\varphi =u$ into (3.17), we see that

$$\begin{equation} (f-\chi,u)_{D^T}=\frac{\alpha -1}{\alpha } \bigl (\Vert u(T)\Vert _{\alpha }^{\alpha }-\Vert u(0)\Vert _{\alpha }^{\alpha }\bigr). \end{equation} \tag{ 3.18 }$$

Our next argument depends on $A$ being monotone. Using equations (3.14), it is easily seen that

$$\begin{equation}\begin{aligned} (A(u_m),u_m)_{D^T} &=(f,u_m)_{D^T}+\int _{I_m}\biggl (\frac{\alpha -1}{\alpha }v_m^{\alpha /2}(0) -\varepsilon _mv_m^{\alpha /2-1}(0)\biggr)dx \\ &\qquad -\int _{I_m}\biggl (\frac{\alpha -1}{\alpha }v_m^{\alpha /2}(T) -\varepsilon _mv_m^{\alpha /2-1}(T)\biggr)dx. \end{aligned} \end{equation} \tag{ 3.19 }$$

Employing inequality (3.15), by (3.19) we have

$$\begin{equation*} \begin{aligned} X_m &\leqslant (f,u_m)_{D^T}+\frac{\alpha -1}{\alpha } \int _{I_m}\bigl (v_m^{\alpha /2}(0)-v_m^{\alpha /2}(T)\bigr)\,dx +2\varepsilon _m^{\alpha /4} \\ &\qquad -(A(u_m),h)_{D^T}-(A(h),u_m-h)_{D^T}. \end{aligned} \end{equation*}$$

Now (3.12) follows because $\liminf \Vert v_m^{\alpha /4}(T)\Vert _{2}\geqslant \Vert u(T)\Vert _{\alpha }$ and $\varepsilon _m\rightarrow 0$. Finally, a similar argument as in the case $\alpha \geqslant 2$ shows that $\chi =A(u)$.

The proof of Theorem 1 is complete.

We define

$$\begin{equation*} \begin{gathered}E_m(t)=E(t)=\int _{\Omega }\biggl (\frac{u_m^2(t,x)}{2b_m} +\frac{\alpha -1}{\alpha }|u_m(t,x)|^{\alpha }\biggr)dx, \\ H(t)= \Vert \nabla u_m(t)\Vert _p^p,\end{gathered} \end{equation*}$$

dropping the subscripts where possible. After differentiating with respect to $t$, for $f=0$ formula (3.2) assumes the form

$$\begin{equation} E^{\prime }+H=0. \end{equation} \tag{ 4.1 }$$

Differentiating (3.4), we obtain

$$\begin{equation} \int _{\Omega }\biggl (\frac{1}{b_m}+(\alpha -1)|u_m(t)|^{\alpha -2}\biggr) u_m^{\prime 2}(t)\,dx+\frac{1}{p}H^{\prime }(t)=0. \end{equation} \tag{ 4.2 }$$

The following estimates hold:

$$\begin{equation*} \begin{aligned} (E^{\prime })^2 &=\biggl (\int _{\Omega }\biggl (\frac{u_mu_m^{\prime }(t)}{b_m} +(\alpha -1)|u_m(t)|^{\alpha -2}u_m(t)u_m^{\prime }(t)\biggr)dx\biggr)^2 \\ &\leqslant \biggl (\frac{\Vert u_m(t)\Vert _2\Vert u_m^{\prime }(t)\Vert _2}{b_m}+ (\alpha -1)\Vert u_m^{\alpha /2}\Vert _2\Vert u_m^{\prime }(t)|u_m|^{\alpha /2-1}\Vert _2\biggr)^2. \end{aligned} \end{equation*}$$

Applying the Cauchy-Schwarz inequality for the inner product in $\mathbb {R}_2$, this gives

$$\begin{equation*} \begin{aligned} &(E^{\prime })^2\leqslant \biggl (\int _{\Omega }\frac{u_m^{\prime 2}(t)}{b_m}\,dx+(\alpha -1) \int _{\Omega }|u_m|^{\alpha -2}u_m^{\prime 2}(t)\,dx\biggr) \\ &\qquad \qquad \times \biggl (\int _{\Omega }\frac{u_m^2(t)}{b_m}\,dx+(\alpha -1) \int _{\Omega }|u_m|^{\alpha }\,dx\biggr). \end{aligned} \end{equation*}$$

Hence,

$$\begin{equation} (E^{\prime })^2\leqslant -\frac{\alpha }{p}H^{\prime }(t) E(t). \end{equation} \tag{ 4.3 }$$

Using (4.1), we rewrite this as follows:

$$\begin{equation*} -HE^{\prime }\leqslant -\frac{1}{\gamma }EH^{\prime }, \qquad \gamma =\frac{p}{\alpha }. \end{equation*}$$

As a result, $\gamma {E^{\prime }}/{E}\geqslant {H^{\prime }}/{H}$, or, on integrating,

$$\begin{equation*} H(t)\leqslant \frac{H(0)E^{\gamma }(t)}{E^{\gamma }(0)}. \end{equation*}$$

Consequently,

$$\begin{equation*} E^{\prime }(t)=-H(t)\geqslant -\frac{H(0)E^{\gamma }(t)}{E^{\gamma }(0)}, \end{equation*}$$

or

$$\begin{equation*} \frac{E^{\prime }}{E^{\gamma }}\geqslant -\frac{H(0)}{E^{\gamma }(0)}. \end{equation*}$$

It follows that

$$\begin{equation*} E^{1-\gamma }(t)-E^{1-\gamma }(0)\leqslant (\gamma -1)\frac{H(0)t}{E^{\gamma }(0)}. \end{equation*}$$

Thus,

$$\begin{equation} E(t)\geqslant E(0)\biggl (1+(\gamma -1)\frac{H(0)t}{E(0)}\biggr)^{{1}/(1-\gamma)}. \end{equation} \tag{ 4.4 }$$

In view of (3.7), for fixed $t>0$ and $\alpha \leqslant p$ if the domain is bounded one may choose a subsequence $u_{m_k}(t)$ which converges strongly in the space $L_\alpha (\Omega)$. The functions

$$\begin{equation*} u_m(t)=\sum _{k=1}^nc_{mk}(t)\omega _k \end{equation*}$$

lie in the linear hull of the functions $\omega _1,\omega _2,\dots,\omega _m$. In a finite-dimensional space all the norms are equivalent, and hence

$$\begin{equation*} \int _{\Omega }u_m^2(t)\,dx\leqslant c_m\Vert u_m(t)\Vert _{\alpha }^2\leqslant \widetilde{c_m}\quad \forall \, t>0. \end{equation*}$$

We choose numbers $b_m$ so as to have $\widetilde{c_m}\leqslant b_m/m$. Now

$$\begin{equation*} E_m(t)\rightarrow \frac{\alpha -1}{\alpha }\Vert u(t)\Vert _{\alpha }^{\alpha }. \end{equation*}$$

Passing to the limit in (4.4) as $m\rightarrow \infty$, we obtain

$$\begin{equation} \Vert u(t)\Vert _{\alpha }^{\alpha }\geqslant \Vert u(0)\Vert _{\alpha }^{\alpha } (1+C(u_0)t)^{-\alpha /(p-\alpha)}. \end{equation} \tag{ 4.5 }$$

In this case we set

$$\begin{equation*} E(t)=\int _{I_m}\biggl (\frac{\alpha -1}{\alpha }v_m^{\alpha /2}(t,x) -\varepsilon _mv_m^{(\alpha -2)/2}(t,x)\biggr)dx+2\varepsilon _m^{\alpha /4}. \end{equation*}$$

Note that (3.15) implies that $E(t)\geqslant \varepsilon _m^{\alpha /4}$.

Differentiating (3.14) with respect to $t$ and rewriting it for $f=0$, this establishes

$$\begin{equation} E^{\prime }+H=0. \end{equation} \tag{ 4.6 }$$

From (3.16) it follows that

$$\begin{equation} \int _{I_m}((\alpha -1)u_m^2+\varepsilon)v_m^{(\alpha -4)/2}u_m^{\prime 2}\,dx=-\frac{1}{p}H^{\prime }(t). \end{equation} \tag{ 4.7 }$$

For any $\mu >0$, clearly

$$\begin{equation*} \begin{aligned} (E^{\prime }(t))^2 &=\biggl (\int _{I_m}\biggl (\frac{\alpha -1}{2}v_m^{(\alpha -2)/2}(2u_mu_m^{\prime }) -\varepsilon _m\frac{\alpha -2}{2}v_m^{(\alpha -4)/2}(2u_mu_m^{\prime })\biggr)dx\biggr)^2 \\ &\leqslant \frac{1}{4}\biggl ((\alpha -1)\int _{I_m} v_m^{(\alpha -2)/2}\biggl (\mu u_m^2+\frac{u_m^{\prime 2}}{\mu }\biggr)dx \\ &\qquad +\varepsilon _m(2-\alpha)\int _{I_m}v_m^{(\alpha -4)/2}\biggl (\mu u_m^2 +\frac{u_m^{\prime 2}}{\mu }\biggr)dx\biggr)^2 \\ &\leqslant \frac{1}{4}\biggl (-\frac{1}{p\mu }H^{\prime }+\alpha E(t)\mu +\varepsilon _m(3-\alpha)\mu \int _{I_m}v_m^{(\alpha -2)/2}\,dx- 2\alpha \mu \varepsilon _m^{\alpha /4}\biggr)^2. \end{aligned} \end{equation*}$$

In view of (3.15),

$$\begin{equation*} (E^{\prime }(t))^2\leqslant \frac{1}{4}\biggl (-\frac{1}{p\mu }H^{\prime }(t)+\alpha E(t)\mu \biggr)^2. \end{equation*}$$

Now (4.3) follows by minimizing the right-hand side over $\mu$.

The rest of the proof is similar to the case $\alpha \geqslant 2$.

We claim that the estimate (4.5), in the case of a bounded domain $\Omega$, is sharp. To prove this, we need the following Friedrichs-Steklov-type inequality

$$\begin{equation*} \Vert \varphi \Vert _p\leqslant C\Vert \nabla \varphi \Vert _p\quad \forall \, \varphi \in C_0^\infty (\Omega). \end{equation*}$$

We note in passing that the following Friedrichs-Steklov inequality holds for an unbounded domain located inside a cone:

$$\begin{equation} \Vert \chi ^b{v}\Vert _{p}\leqslant Cb\Vert \chi ^b \nabla {v}\Vert _{p}. \end{equation} \tag{ 4.8 }$$

For $p>\alpha$, we have

$$\begin{equation} \Vert \varphi \Vert _\alpha \leqslant C_1\Vert \varphi \Vert _p\leqslant C_2\Vert \nabla \varphi \Vert _p. \end{equation} \tag{ 4.9 }$$

Differentiating (3.10) or (3.18) with respect to $T$ and employing (4.9) for $u$, we see that

$$\begin{equation*} \frac{\alpha -1}{\alpha }\,\frac{d}{dt}\Vert u(t)\Vert _\alpha ^\alpha =-(A(u),u(t))_\Omega =-\Vert \nabla u(t)\Vert _p^p\leqslant -C_2^{-1}\Vert u(t)\Vert _\alpha ^p. \end{equation*}$$

Solving this differential inequality, we obtain the estimate

$$\begin{equation*} \Vert u(t)\Vert _{\alpha }^{\alpha }\leqslant \Vert u(0)\Vert _{\alpha }^{\alpha } (1+c(u_0)t)^{-\alpha /(p-\alpha)}, \end{equation*}$$

which proves that (4.5) is sharp.

Our aim here is to derive (4.5) for the solution of problem (1.3), (1.2) in the case when $\Omega$ is unbounded.

We let $u^{l}$, $l=1,2,\dots$, denote the solutions in $[0,\infty)\times \Omega ^{l}$ with fixed initial function $u_0$. We can assume that these solutions are extended by zero outside $[0,\infty)\times \Omega ^{l}$. Using (3.7), we have the estimate

$$\begin{equation*} \Vert u^{l}(t)\Vert _{W_{\alpha,p}^1(\Omega)}\leqslant C, \qquad t>0,\quad l=1,\dots,\infty. \end{equation*}$$

Using the properties of the solutions $u^{l}$ listed in Theorem 1, we can choose a subsequence $l_i$ such that $u^{l_i}(t)\rightarrow u(t)$ weakly in $V(D^T)$ as $i\rightarrow \infty$ for all $T>0$, and then, applying Steps 1–3, prove that the limit function $u(t)$ is a generalized solution of problem (1.3), (1.2).

Further, given a fixed $t>0$ we can assume that $u^{l_i}(t)\rightarrow u(t)$ weakly in $\mathring{W}_{\alpha,p}^1(\Omega)$ as $i\rightarrow \infty$. As the embedding $\mathring{W}_{\alpha,p}^1(\Omega ^{r})\subset L_\alpha (\Omega ^{r})$ is compact, $u^{l_i}(t)\rightarrow u(t)$ strongly in $L_\alpha (\Omega ^{r})$ as $i\rightarrow \infty$ for any $r>0$.

By (2.11), for any $\varepsilon$ there exists $r$ such that

$$\begin{equation*} \Vert u^{l}(t)\Vert _{\alpha,\Omega \setminus \Omega ^{r}}^{\alpha }\leqslant \varepsilon. \end{equation*}$$

Estimate (4.5) holds for $u^l$. Now

$$\begin{equation*} \Vert u^l(t)\Vert _{\alpha,\Omega ^{r}}^{\alpha }\geqslant \Vert u_0\Vert _{\alpha }^{\alpha } (1+C(u_0)t)^{-\alpha /(p-\alpha)}-\varepsilon. \end{equation*}$$

Using the strong convergence in $L^{\alpha }(\Omega ^{r})$, we pass to the limit as $l\rightarrow \infty$ and then as $r\rightarrow \infty \ (\varepsilon \rightarrow 0)$. Thus, we have now proved (2.10) for an unbounded domain $\Omega$ as well.

The proof of Theorem 2 is complete.

Let $\theta (\rho)$, $\rho >0$, be an absolutely continuous function equal to 1 for $\rho \geqslant r$, to 0 for $\rho \leqslant R_0$, linear for $\rho \in [R_0,2R_0]$ and satisfying the equation

$$\begin{equation} \theta ^{\prime }(\rho)=\delta {\nu (\rho)}\theta (\rho),\qquad \rho \in (2R_0,r) \end{equation} \tag{ 6.1 }$$

(the constant $\delta$ will be specified below). Solving this equation, we find, in particular, that

$$\begin{equation} \theta ^{\prime }(\rho)=\theta ^{\prime }(R_0) =\frac{\theta (2R_0)}{R_0}=\frac{1}{R_0}\exp \biggl (-\delta \int _{2R_0}^r {\nu (\rho)}\,d\rho \biggr)\quad \text{for }\ \rho \in (R_0,2R_0). \end{equation} \tag{ 6.2 }$$

For any function $v(x)\in C_0^{\infty }(\Omega)$ it follows from the definition of $\nu (\rho)$ that

$$\begin{equation*} \nu (\rho)\Vert {v}\Vert _{p,\gamma _{\rho }}\leqslant \Vert \nabla {v}\Vert _{p,\gamma _{\rho }},\qquad \rho >0. \end{equation*}$$

Multiplying by $\theta ^p(\rho)$ and integrating over $\rho \in [a,b]$, we derive the inequality

$$\begin{equation*} \int _{a}^b\nu ^{p}(\rho)\theta ^p(\rho)\Vert {v}\Vert _{p,\gamma _{\rho }}^{p}\,d\rho \leqslant \int _{a}^b\theta ^p(\rho)\Vert \nabla {v} \Vert ^{p}_{p,\gamma _{\rho }}\,d\rho, \end{equation*}$$

which now holds for any function ${v}\in \mathring{W}_{\alpha,p}^1(\Omega)$. For $a=2R_0$ and $b=r$, we rewrite the last inequality as follows

$$\begin{equation} \Vert \chi _{2R_0}^r\nu (|x|)\theta (|x|){v}\Vert _p \leqslant \Vert \chi _{2R_0}^r\theta (|x|)\nabla {v}\Vert _{p}. \end{equation} \tag{ 6.3 }$$

Substituting $\xi (x)=\theta (|x|)$ into (5.5), we have

$$\begin{equation} \int _{ \Omega }u^{\alpha }(t,{x}) \theta ^{p}(|x|) \,d{x} + \int _{D^{t}} \theta ^{p} |\nabla u|^{p}\,d{x}\,d \tau \leqslant C(\alpha,p)\int _{D^{t}} |u|^{p} (\theta ^{\prime }(|x|))^{p}\,d{x}\,d\tau. \end{equation} \tag{ 6.4 }$$

Using (6.1), (6.2), one easily reduces (6.4) to the form

$$\begin{equation}\begin{aligned} &\int _{ \Omega }u^{\alpha }(t,{x}) \theta ^{p}(x_1)\,d{x} + \int _{D^{t}} \theta ^{p} |\nabla u|^{p}\,d{x}\,d\tau \\ &\qquad \leqslant C (\theta ^{\prime }(R_0))^p\int _0^t\int _{\Omega } \chi _{R_0}^{2R_0}|u|^{p}\,d{x}\,d\tau +C\delta ^p\int _0^t\int _{\Omega }\chi _{2R_0}^{r}|u|^{p}\nu ^{p}(|x|)\theta ^{p}(|x|)\,d{x}\, d\tau \\ &\qquad =I_1^t+I_2^t. \end{aligned} \end{equation} \tag{ 6.5 }$$

Applying (4.8) with $b=2R_0$ and using (3.18), it follows that

$$\begin{equation} I_1^t\leqslant C_1(R_0)(\theta ^{\prime }(R_0))^p\int _0^t\Vert \nabla u(\tau)\Vert ^{p}_{p}\,d\tau \leqslant C_2\exp \biggl (-\delta p\int _{2R_0}^r{\nu (\rho)}\,d\rho \biggr)\Vert u_0\Vert ^{\alpha }_{\alpha }. \end{equation} \tag{ 6.6 }$$

Using (6.3) shows that

$$\begin{equation} I_2^t\leqslant C\delta ^{p}\int _0^t\Vert \chi _{2R_0}^r\theta \nabla u \Vert ^{p}_p\,d\tau. \end{equation} \tag{ 6.7 }$$

Taking $\delta =C^{-1/p}$ and combining (6.5)–(6.7), we obtain

$$\begin{equation*} \Vert \chi _{r}u(t)\Vert ^{\alpha }_{\alpha } +\int _0^t \Vert \chi _{r}\nabla u(\tau)\Vert ^{p}_{p}\,d \tau \leqslant C_2\exp \biggl (-\delta p \int _{1}^r{\nu (\rho)}\,d\rho \biggr)\Vert u_0\Vert _\alpha ^\alpha, \end{equation*}$$

thereby proving inequality (2.14).

The proof of Theorem 4 is complete.

We choose a positive number $r\geqslant 2R_0$. Setting

$$\begin{equation*} \varepsilon (r)=\mathscr{M}^{\alpha }\exp \biggl (-\alpha \kappa \int _{1}^r \nu (\rho)\,d\rho \biggr)\Vert u_0 \Vert _\alpha ^{\alpha }, \end{equation*}$$

in view of (2.14), we have

$$\begin{equation*} \Vert u(t)\Vert _\alpha^{\alpha }\leqslant \Vert \chi ^ru(t)\Vert _{\alpha }^{\alpha }+\varepsilon (r),\qquad t\geqslant 0. \end{equation*}$$

By (4.8),

$$\begin{equation*} \begin{aligned} \Vert \chi ^ru(t)\Vert _{\alpha }^{\alpha } & =\int _\Omega \chi ^ru(t)^\alpha \,dx \leqslant \biggl (\int _\Omega \chi ^rdx\biggr)^{(p-\alpha)/p}\Vert \chi ^ru(t)\Vert ^{\alpha }_p \\ & \leqslant Cr^{n(p-\alpha)/p}r^\alpha \Vert \chi ^r\nabla u(t)\Vert _p^\alpha. \end{aligned} \end{equation*}$$

Setting $\mu (r)=C_1r^{1+n(p-\alpha)/(p\alpha)}$ with an appropriate factor $C_1$, we have

$$\begin{equation} \Vert u(t)\Vert _\alpha ^{\alpha }\leqslant \mu ^{\alpha }(r) \Vert \chi ^r\nabla u(t)\Vert _p^{\alpha }+\varepsilon (r), \qquad t\geqslant 0. \end{equation} \tag{ 6.8 }$$

Let $t_r$ denote a point in the interval $[0,\infty)$ such that $E(t_r)=\Vert u(t_r)\Vert _\alpha ^{\alpha }=\varepsilon (r)$. If $E(t)>\varepsilon (r)$ for any $t\geqslant 0$, we let $t_r=\infty$. Since the function $E(t)$ is monotone nonincreasing for $t\in [0,t_r)$, we have $E(t)>\varepsilon (r)$. Now (6.8) implies that

$$\begin{equation} (E(t)-\varepsilon (r))^{p/\alpha }\leqslant \mu ^{p}(r)\Vert \nabla u(t)\Vert ^{p}_{p}, \qquad t\in [0,t_r). \end{equation} \tag{ 6.9 }$$

Differentiating (3.18) with respect to $T$, we see that

$$\begin{equation*} \frac{dE(t)}{d t}=-\frac{\alpha }{\alpha -1}\Vert \nabla u(t)\Vert _p^p. \end{equation*}$$

Next, in view of (6.9),

$$\begin{equation} \frac{dE(t)}{d t}\leqslant -\frac{\alpha }{\alpha -1}\mu ^{-p}(r)(E(t)-\varepsilon (r))^{p/\alpha }, \qquad t\in (0,t_r). \end{equation} \tag{ 6.10 }$$

Solving this differential inequality, we obtain

$$\begin{equation*} (E(t)-\varepsilon (r))^{(p-\alpha)/\alpha }\leqslant \frac{\alpha -1}{(p-\alpha)t\mu ^{-p}(r)}, \qquad t\in (0,t_r). \end{equation*}$$

The substitution $\varepsilon (r)$ into the last inequality shows that, for $t\in (0,t_r)$,

$$\begin{equation} E(t)\leqslant C_2(t\mu ^{-p}(r))^{-\alpha /(p-\alpha)}+\mathscr{M}^\alpha \exp \biggl (-\alpha \kappa \int _{1}^r \nu (\rho)\,d\rho \biggr)\Vert u_0\Vert _\alpha ^{\alpha }. \end{equation} \tag{ 6.11 }$$

Note that, for $t\in [t_r,\infty)$, the inequality $E(t)\leqslant \varepsilon (r)$ holds and estimate (6.11) is also valid.

In (6.11) we set $r=r(t)$ and use (2.15):

$$\begin{equation*} \begin{aligned} E(t) &\leqslant C_2 t^{-\alpha /(p-\alpha)}r(t)^{p\alpha /(p-\alpha)+n}+ \mathscr{M}^\alpha \exp \biggl (-\alpha \kappa \int _{1}^{r(t)} \nu (\rho)\,d\rho \biggr)\Vert u_0\Vert _\alpha ^{\alpha } \\ &\leqslant C_3 t^{-\alpha /(p-\alpha)}r(t)^{p\alpha /(p-\alpha)+n},\qquad t>0. \end{aligned} \end{equation*}$$

This proves inequality (2.16).

The proof of Theorem 5 is complete.