Let us introduce the following notations: for any measurable function
$\displaystyle v$ we define

$\begin{array}[]{c}A_{m}(v)=\left\\{(x,t)\in\Omega_{T}:v(x,t)>m\right\\},\text{
and for a.e. }t\in(0,T),\quad
A_{m}^{t}(v)=\\{x\in\Omega:v(x,t)>m\\}.\end{array}$

We use the idea of the classical proof by D.G. Aronson and J. Serrin, which is
to prove a uniform $\displaystyle L^{\infty}$ bound for $\displaystyle u_{k}$
in $\displaystyle\Omega\times(0,\tau)$, for a positive (small)
$\displaystyle\tau$ (to be specified later), and then to iterate such an
estimate. We consider the approximated problem Eq. 2.10 and take
$\displaystyle(u_{k}-m)_{+}\chi_{(0,t)}\in L^{2}(0,T;H^{1}_{0}(\Omega))$, for
$\displaystyle m>0,t\in(0,\tau),\tau<T$, as a test function to obtain

${}\begin{array}[]{c}\int_{\Omega}\varphi_{m}\left(u_{k}(x,t)\right)dx+\int_{0}^{t}\int_{\Omega}\nabla
u_{k}\cdot\nabla(u_{k}-m)_{+}dxd\theta\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
+\frac{1}{2}\int_{0}^{t}\int_{Q}\frac{\left(u_{k}(x,\theta)-u_{k}(y,\theta)\right)\left((u_{k}-m)_{+}(x,\theta)-(u_{k}-m)_{+}(y,\theta)\right)}{|x-y|^{n+2s}}dxdyd\theta\\\
\leq\int_{0}^{\tau}\int_{\Omega}\frac{f_{k}(u_{k}-m)_{+}}{(u_{k}+\frac{1}{k})^{\gamma}}dxd\theta+\int_{\Omega}\varphi_{m}\left(u_{k}(x,0)\right)dx,\end{array}$
(3.16)

where
$\displaystyle\varphi_{m}(\rho)=\int_{0}^{\rho}(\sigma-m)_{+}d\sigma=\frac{(\rho-m)_{+}^{2}}{2}$.
We choose a $\displaystyle m>0$ large enough such that $\displaystyle
m>\operatorname{max}\\{\left\|u_{0}\right\|_{L^{\infty}(\Omega)},1\\}$, in
order to neglect the last term above. Noting that $\displaystyle\nabla
u_{k}\cdot\nabla(u_{k}-m)_{+}=|\nabla(u_{k}-m)_{+}|^{2}$, and

$\left(u_{k}(x,\theta)-u_{k}(y,\theta)\right)\left((u_{k}-m)_{+}(x,\theta)-(u_{k}-m)_{+}(y,\theta)\right)\geq\left((u_{k}-m)_{+}(x,\theta)-(u_{k}-m)_{+}(y,\theta)\right)^{2},$

we take supremum over $\displaystyle t\in(0,\tau]$ in Eq. 3.16, to get

${}\begin{array}[]{c}\|(u_{k}-m)_{+}\|^{2}_{L^{\infty}(0,\tau;L^{2}(\Omega))}+\|(u_{k}-m)_{+}\|^{2}_{L^{2}(0,\tau;X^{s}_{0}(\Omega))}+\|(u_{k}-m)_{+}\|^{2}_{L^{2}(0,\tau;H^{1}_{0}(\Omega))}\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
\leq\int_{0}^{\tau}\int_{A^{t}_{m,k}}f(u_{k}-m)_{+}dxdt.\end{array}$ (3.17)

Note that, in order to deal with the singularity, we have used the fact that
$\displaystyle m\geq 1$ and hence $\displaystyle u_{k}>1$ in $\displaystyle
A^{t}_{m,k}$, here the subscript $\displaystyle k$ in $\displaystyle
A^{t}_{m,k}$ denotes that we are considering the function $\displaystyle
u_{k}$. Now the term of the right-hand side above can be estimated as follows,

${}\int_{0}^{\tau}\int_{A_{m,k}^{t}}f(u_{k}-m)_{+}dxdt\leq\int_{0}^{\tau}\int_{A_{m,k}^{t}}f(u_{k}-m)_{+}^{2}dxdt+\int_{0}^{\tau}\int_{A_{m,k}^{t}}fdxdt.$
(3.18)

We now study each member present in the right-hand side of Eq. 3.18. We first
define the followings,

$\bar{r}=2r^{\prime},\bar{q}=2q^{\prime},\eta=\frac{2\eta_{1}}{n},\hat{r}=\bar{r}(1+\eta),\hat{q}=\bar{q}(1+\eta),$

where $\displaystyle\eta_{1}=1-\frac{1}{r}-\frac{n}{2q}$. Note that
$\displaystyle\eta_{1}\in(0,1)$ as $\displaystyle
0<\frac{1}{r}+\frac{n}{2q}<1$. Further, simple calculation yields that
$\displaystyle\frac{1}{\hat{r}}+\frac{n}{2\hat{q}}=\frac{n}{4}$. Now, applying
Hölder inequality repeatedly, we estimate the first term as

$\begin{array}[]{rcl}\int_{0}^{\tau}\int_{A_{m,k}^{t}}f(u_{k}-m)_{+}^{2}dxdt&\leq&\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}|f(x,t)|^{q}dx\right)^{\frac{1}{q}}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{2q^{\prime}}dx\right)^{\frac{1}{q^{\prime}}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}|f(x,t)|^{q}dx\right)^{\frac{r}{q}}dt\right)^{\frac{1}{r}}\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{2q^{\prime}}dx\right)^{\frac{r^{\prime}}{q^{\prime}}}dt\right)^{\frac{1}{r^{\prime}}}.\end{array}$

As again by Hölder inequality with exponent $\displaystyle(1+\eta)$, we have

$\begin{array}[]{rcl}\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{2q^{\prime}}dx\right)^{\frac{r^{\prime}}{q^{\prime}}}dt\right)^{\frac{1}{r^{\prime}}}&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{2q^{\prime}(1+\eta)}dx\right)^{\frac{r^{\prime}}{q^{\prime}(1+\eta)}}|A^{t}_{m,k}|^{\frac{r^{\prime}\eta}{q^{\prime}(1+\eta)}}dt\right)^{\frac{1}{r^{\prime}}}\\\
&=&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{\hat{q}}dx\right)^{\frac{2r^{\prime}}{\hat{q}}}|A^{t}_{m,k}|^{\frac{2r^{\prime}\eta}{\hat{q}}}dt\right)^{\frac{1}{r^{\prime}}}\\\
&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{\hat{q}}dx\right)^{\frac{\hat{r}}{\hat{q}}}dt\right)^{\frac{1}{r^{\prime}(1+\eta)}}\left(\int_{0}^{\tau}|A^{t}_{m,k}|^{\frac{\hat{r}}{\hat{q}}}\right)^{\frac{\eta}{r^{\prime}(1+\eta)}},\end{array}$

therefore

$\int_{0}^{\tau}\int_{A_{m,k}^{t}}f(u_{k}-m)_{+}^{2}dxdt\leq\|f\|_{L^{r}(0,T;L^{q}(\Omega))}\|(u_{k}-m)_{+}\|^{2}_{L^{\hat{r}}(0,\tau;L^{\hat{q}}(\Omega))}\mu_{k}(m)^{\frac{2\eta}{\hat{r}}},$

where
$\displaystyle\mu_{k}(m)=\int_{0}^{\tau}\left|A_{m,k}^{t}\right|^{\frac{\hat{r}}{\hat{q}}}dt$.
We now use Gagliardo-Nirenberg inequality (Theorem 2.10) for
$\displaystyle(u_{k}-m)_{+}$ to get

${}\begin{array}[]{rcl}\int_{0}^{\tau}\int_{A_{m,k}^{t}}f(u_{k}-m)_{+}^{2}dxdt&\leq&c\|f\|_{L^{r}(0,T;L^{q}(\Omega))}\mu_{k}(m)^{\frac{2\eta}{\hat{r}}}\left(\int_{0}^{\tau}\left\|(u_{k}-m)_{+}\right\|_{L^{2}(\Omega)}^{(1-\theta)\hat{r}}\left\|\nabla(u_{k}-m)_{+}\right\|_{L^{2}\left(\Omega\right)}^{\hat{r}\theta}dt\right)^{\frac{2}{\hat{r}}}\\\
&\leq&c\|f\|\mu_{k}(m)^{\frac{2\eta}{\hat{r}}}\bigg{[}\left\|(u_{k}-m)_{+}\right\|_{L^{\infty}\left(0,\tau;L^{2}(\Omega)\right)}^{2}+\left\|(u_{k}-m)_{+}\right\|_{L^{2}(0,\tau;H^{1}_{0}(\Omega))}^{2}\bigg{]}.\end{array}$
(3.19)

where $\displaystyle\hat{r}\theta=2$, and $\displaystyle c$ is a constant
independent of the choice of $\displaystyle k$ and $\displaystyle m$. Note
that we have used the fact that any function in $\displaystyle
H^{1}_{0}(\Omega)$ can be considered as a function in $\displaystyle
H^{1}(\mathbb{R}^{n})$. Also, we applied Young’s inequality with exponents
$\displaystyle\frac{\hat{r}}{2}$ and its conjugate.
On the other hand, the second term on the right-hand side in Eq. 3.18 can be
estimated by Hölder inequality as

${}\begin{array}[]{rcl}\int_{0}^{\tau}\int_{A_{m,k}^{t}}fdxdt&\leq&\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}|f(x,t)|^{q}dx\right)^{\frac{1}{q}}|A^{t}_{m,k}|^{\frac{1}{q^{\prime}}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}|f(x,t)|^{q}dx\right)^{\frac{r}{q}}dt\right)^{\frac{1}{r}}\left(\int_{0}^{\tau}|A^{t}_{m,k}|^{\frac{r^{\prime}}{q^{\prime}}}dt\right)^{\frac{1}{r^{\prime}}}\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\|f\|_{L^{r}(0,T;L^{q}(\Omega))}\mu_{k}(m)^{\frac{2(1+\eta)}{\hat{r}}}.\end{array}$
(3.20)

Denoting

$|||(u_{k}-m)_{+}|||^{2}=\|(u_{k}-m)_{+}\|_{L^{\infty}(0,\tau;L^{2}(\Omega))}^{2}+\|(u_{k}-m)_{+}\|_{L^{2}(0,\tau;H^{1}_{0}(\Omega))}^{2},$

and using Eq. 3.19, Eq. 3.20, we get from Eq. 3.17,

$|||(u_{k}-m)_{+}|||^{2}\leq
c\|f\|_{L^{r}(0,T;L^{q}(\Omega))}\left[\mu_{k}(m)^{\frac{2\eta}{\hat{r}}}|||(u_{k}-m)_{+}|||^{2}+\mu_{k}(m)^{\frac{2(1+\eta)}{\hat{r}}}\right],$

where $\displaystyle c$ is a constant which does not depend on $\displaystyle
k$ and $\displaystyle m$. Note that
$\displaystyle\mu_{k}(m)\leq\tau|\Omega|^{\frac{\hat{r}}{\hat{q}}}$, for all
$\displaystyle k\in\mathbb{N}$ and $\displaystyle m$, so that we can fix
$\displaystyle\tau$, independent of $\displaystyle u_{k}$ and $\displaystyle
m$, suitable small in such a way that $\displaystyle
c\mu_{k}(m)^{\frac{2\eta}{\hat{r}}}\|f\|_{L^{r}(0,T;L^{q}(\Omega))}=\frac{1}{2}$
and use again the Gagliardo-Nirenberg inequality (see Eq. 3.19), to deduce
that

${}\left\|(u_{k}-m)_{+}\right\|_{L^{\hat{r}}\left(0,\tau;L^{\hat{q}}(\Omega)\right)}^{2}\leq
c|||(u_{k}-m)_{+}|||^{2}\leq
c\|f\|_{L^{r}(0,T;L^{q}(\Omega))}\mu_{k}(m)^{\frac{2(1+\eta)}{\hat{r}}}.$
(3.21)

Consider $\displaystyle
l>m>\operatorname{max}\\{\left\|u_{0}\right\|_{L^{\infty}(\Omega)},1\\}:=m_{0}$.
Then $\displaystyle A^{t}_{l,k}\subset A^{t}_{m,k}$, and using Eq. 3.21, we
get

$\begin{array}[]{rcl}(l-m)\mu_{k}(l)^{\frac{1}{\hat{r}}}=\left(\int_{0}^{\tau}\left((l-m)^{\hat{q}}\left|A_{l,k}^{t}\right|\right)^{\frac{\hat{r}}{\hat{q}}}dt\right)^{\frac{1}{\hat{r}}}&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{l,k}^{t}}(u_{k}-m)_{+}^{\hat{q}}dx\right)^{\frac{\hat{r}}{\hat{q}}}dt\right)^{\frac{1}{\hat{r}}}\\\
&\leq&\left(\int_{0}^{\tau}\left(\int_{A_{m,k}^{t}}(u_{k}-m)_{+}^{\hat{q}}dx\right)^{\frac{\hat{r}}{\hat{q}}}dt\right)^{\frac{1}{\hat{r}}}\leq
c\mu_{k}(m)^{\frac{1+\eta}{\hat{r}}}.\end{array}$

Therefore for all $\displaystyle l>m>m_{0}$, we have

$\mu_{k}(l)\leq\frac{c}{(l-m)^{\hat{r}}}\mu_{k}(m)^{1+\eta},$

where c is a constant independent of $\displaystyle k$. Now applying [27,
Lemma B.$\displaystyle 1$], we conclude that
$\displaystyle\mu_{k}(m_{0}+d_{k})=0,$ where $\displaystyle
d_{k}=c[\mu_{k}(m_{0})]^{\eta}2^{\hat{r}(1+\eta)/\eta}$. As for each
$\displaystyle k$, $\displaystyle d_{k}\leq
c\left(T|\Omega|^{\frac{\hat{r}}{\hat{q}}}\right)^{\eta}$, we get

$\left\|u_{k}\right\|_{L^{\infty}(\Omega\times[0,\tau])}\leq d.$

We can iterate this procedure in the sets
$\displaystyle\Omega\times[\tau,2\tau],\cdots,\Omega\times[j\tau,T]$, where
$\displaystyle T-j\tau\leqslant\tau$ to conclude that

$\left\|u_{k}\right\|_{L^{\infty}\left(\Omega_{T}\right)}\leq C\quad\text{
uniformly in }k\in\mathbb{N}.$

Now, since the sequence $\displaystyle\\{u_{k}\\}_{k}$ is increasing in
$\displaystyle k$, we have $\displaystyle\|u\|_{L^{\infty}(\Omega)}\leq C$.
Further, if $\displaystyle\gamma\leq 1$, testing Eq. 2.10 with the function
$\displaystyle u_{k}$ we can deduce $\displaystyle u\in
L^{2}(0,T;H^{1}_{0}(\Omega))$ (see Theorem 2.26).

### Proof of Theorem 2.36

For $\displaystyle\gamma\geq 1$, we choose
$\displaystyle\delta>\frac{1+\gamma}{2}$ and for $\displaystyle\gamma<1$, we
choose $\displaystyle\delta>1$, and take $\displaystyle
u_{k}^{2\delta-1}\chi_{(0,t)},0<t\leq T$ as test function in Eq. 2.10, with
$\displaystyle u_{0}\equiv 0$. We note that, such a test function is
admissible as $\displaystyle u_{k}\in L^{\infty}(\Omega_{T})$. We get
$\displaystyle\forall t\leq T,$

$\begin{array}[]{c}\quad\frac{1}{2\delta}\int_{\Omega}u_{k}^{2\delta}(x,t)dx+\int_{0}^{t}\int_{\Omega}\nabla
u_{k}\cdot\nabla u_{k}^{2\delta-1}dxd\theta\vskip 3.0pt plus 1.0pt minus
1.0pt\\\
\quad+\frac{1}{2}\int_{0}^{t}\int_{Q}\frac{\left(u_{k}^{2\delta-1}(x,\theta)-u_{k}^{2\delta-1}(y,\theta)\right)\left(u_{k}(x,\theta)-u_{k}(y,\theta)\right)}{|x-y|^{n+2s}}dxdyd\theta\leq\iint_{\Omega_{t}}fu_{k}^{2\delta-1-\gamma}dxd\theta.\end{array}$

Taking supremum over $\displaystyle t\in(0,T]$ and using item
$\displaystyle(i)$ of Lemma 2.11, by Hölder and Sobolev inequalities, we have

${}\begin{array}[]{c}\sup_{t\in[0,T]}\int_{\Omega}u_{k}^{2\delta}(x,t)dx+\frac{2(2\delta-1)}{\delta}\lambda\int_{0}^{T}\left\|u_{k}^{\delta}\right\|_{L^{2^{*}}(\Omega)}^{2}dt+\frac{(2\delta-1)}{\delta}\lambda_{s}\int_{0}^{T}\left\|u_{k}^{\delta}\right\|_{L^{2_{s}^{*}}(\Omega)}^{2}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ \leq
2\delta\left\|f\right\|_{L^{r}\left(0,T;L^{q}(\Omega)\right)}\left(\int_{0}^{T}\left[\int_{\Omega}u_{k}^{(2\delta-1-\gamma)q^{\prime}}dx\right]^{\frac{r^{\prime}}{q^{\prime}}}dt\right)^{\frac{1}{r^{\prime}}}.\end{array}$
(3.22)

Note that $\displaystyle\frac{2\delta-1}{\delta}>1$, so we can ignore the
constants in left. Denoting by $\displaystyle
A=\left(\int_{0}^{T}\left\|u_{k}\right\|_{L^{(2\delta-1-\gamma)q^{\prime}}(\Omega)}^{(2\delta-1-\gamma)r^{\prime}}dt\right)^{\frac{1}{r^{\prime}}},$
we get from Eq. 3.22,

${}\sup_{t\in[0,T]}\int_{\Omega}u_{k}^{2\delta}dx\leq
2\delta\left\|f\right\|_{L^{r}\left(0,T;L^{q}(\Omega)\right)}A,$ (3.23)

and

${}\int_{0}^{T}\left[\int_{\Omega}u_{k}^{2_{s}^{*}\delta}dx\right]^{\frac{2}{2_{s}^{*}}}dt\leq
c\int_{0}^{T}\left[\int_{\Omega}u_{k}^{2^{*}\delta}dx\right]^{\frac{2}{2^{*}}}dt\leq\frac{c}{\lambda}2\delta\left\|f\right\|_{L^{r}\left(0,T;L^{q}(\Omega)\right)}A.$
(3.24)

Case 1: $\displaystyle 1<q<\frac{nr}{n+2(r-1)}$ i.e.
$\displaystyle\frac{1}{r}<\frac{n}{n-2}\frac{1}{q}-\frac{2}{n-2}$
Since, $\displaystyle 1<q<\frac{nr}{n+2(r-1)}$ implies $\displaystyle q<r$ and
$\displaystyle\frac{r^{\prime}}{q^{\prime}}<\frac{2}{2^{*}}$, we apply Hölder
inequality with exponent $\displaystyle\frac{2q^{\prime}}{2^{*}r^{\prime}}$ to
get

${}A\leq
T^{\frac{1}{r^{\prime}}-\frac{2^{*}}{2q^{\prime}}}\left[\int_{0}^{T}\left(\int_{\Omega}u_{k}^{(2\delta-1-\gamma)q^{\prime}}dx\right)^{\frac{2}{2^{*}}}dt\right]^{\frac{2^{*}}{2q^{\prime}}}.$
(3.25)

Thus from Eq. 3.24, it follows

${}\int_{0}^{T}\left[\int_{\Omega}u_{k}^{2^{*}\delta}dx\right]^{\frac{2}{2^{*}}}dt\leq
2\delta cA\leq 2\delta
c\left[\int_{0}^{T}\left(\int_{\Omega}u_{k}^{(2\delta-1-\gamma)q^{\prime}}dx\right)^{\frac{2}{2^{*}}}dt\right]^{\frac{2^{*}}{2q^{\prime}}}.\vskip
3.0pt plus 1.0pt minus 1.0pt$

We now choose $\displaystyle 2^{*}\delta=(2\delta-1-\gamma)q^{\prime}$, that
is, $\displaystyle\delta=\frac{1}{2}\cdot\frac{q(n-2)(\gamma+1)}{(n-2q)}$.
Note that if $\displaystyle\gamma\geq 1$, then
$\displaystyle\delta>\frac{\gamma+1}{2}$ if and only if $\displaystyle q>1$,
and for $\displaystyle\gamma<1$, $\displaystyle\delta>1$ if and only if
$\displaystyle q>\left(\frac{2^{*}}{1-\gamma}\right)^{\prime}$. Moreover,
since $\displaystyle q<\frac{n}{2}$ it follows that
$\displaystyle\frac{2^{*}}{2q^{\prime}}<1$. Thus we get

${}\int_{0}^{T}\left(\int_{\Omega}u_{k}^{2^{*}\delta}dx\right)^{\frac{2}{2^{*}}}dt=\int_{0}^{T}\left(\int_{\Omega}u_{k}^{\left(2\delta-1-\gamma\right)q^{\prime}}dx\right)^{\frac{2}{2^{*}}}dt\leq
c,$ (3.26)

and therefore from Eq. 3.23, Eq. 3.25 and Eq. 3.26, it follows

$\left\|u_{k}\right\|_{L^{\infty}\left(0,T;L^{2\sigma}(\Omega)\right)}+\left\|u_{k}\right\|_{L^{2\sigma}\left(0,T;L^{2^{*}\sigma}(\Omega)\right)}\leq
c,\text{ with }\sigma=\frac{q(n-2)(\gamma+1)}{2(n-2q)}>\frac{q}{2}.$

Case 2: $\displaystyle\frac{nr}{n+2(r-1)}\leq q<\frac{n}{2}r^{\prime}$ i.e.
$\displaystyle\frac{1}{r}\geq\frac{n}{n-2}\frac{1}{q}-\frac{2}{n-2}$ and
$\displaystyle\frac{1}{r}+\frac{n}{2q}>1$
In this case we choose
$\displaystyle\delta=\frac{1}{2}\frac{qrn(\gamma+1)}{nr-2q(r-1)}$. Note that
for $\displaystyle\gamma\geq 1$, $\displaystyle\delta>\frac{1+\gamma}{2},$ if
and only if $\displaystyle\frac{1}{r}+\frac{n}{2q}<1+\frac{n}{2}$ which is
always satisfied and for $\displaystyle\gamma<1$, $\displaystyle\delta>1$ if
and only if $\displaystyle q>\frac{2nr}{nr(\gamma+1)+4(r-1)}$ which is
satisfied if $\displaystyle r>\frac{2}{\gamma+1}$. Now by interpolation, we
get that
$\displaystyle\frac{1}{(2\delta-1-\gamma)q^{\prime}}=\frac{1-\theta}{2\delta}+\frac{\theta}{2^{*}\delta}$,
where $\displaystyle\theta=\frac{nq(r-1)}{nr(q-1)+2q(r-1)}\in(0,1]$. Therefore

$\int_{0}^{T}\left\|u_{k}\right\|_{L^{(2\delta-1-\gamma)q^{\prime}}(\Omega)}^{r^{\prime}(2\delta-1-\gamma)}dt\leq
c\left\|u_{k}\right\|_{L^{\infty}\left(0,T;L^{2\delta}(\Omega)\right)}^{2\delta\mu_{1}}\int_{0}^{T}\left(\int_{\Omega}u_{k}^{2^{*}\delta}dx\right)^{\frac{2}{2^{*}}\mu_{2}}dt,$

where
$\displaystyle\mu_{1}=\frac{(1-\theta)r^{\prime}(2\delta-1-\gamma)}{2\delta}$
and $\displaystyle\mu_{2}=\frac{\theta
r^{\prime}(2\delta-1-\gamma)}{2\delta}$. Since $\displaystyle\mu_{2}\leq 1$,
we have that

$\int_{0}^{T}\left\|u_{k}\right\|_{L^{(2\delta-1-\gamma)q^{\prime}}(\Omega)}^{r^{\prime}(2\delta-1-\gamma)}dt\leq
c\left\|u_{k}\right\|_{L^{\infty}\left(0,T;L^{2\delta}(\Omega)\right)}^{2\delta\mu_{1}}\left(\int_{0}^{T}\left[\int_{\Omega}u_{k}^{2^{*}\delta}dx\right]^{\frac{2}{2^{*}}}dt\right)^{\mu_{2}},$

and since $\displaystyle\mu_{1}+\mu_{2}<r^{\prime}$, using Eqs. 3.23 and 3.24,
we can conclude the following which will give our final result

$\int_{0}^{T}\left(\int_{\Omega}u_{k}^{(2\delta-1-\delta)q^{\prime}}dx\right)^{\frac{r^{\prime}}{q^{\prime}}}dt\leq
c.$

### Proof of Theorem 2.39

Consider the approximated problem Eq. 2.11 and corresponding properties of
$\displaystyle\\{u_{k}\\}_{k}$, Theorem 2.22. We denote
$\displaystyle(\omega_{\tilde{T}})_{\delta}=\Omega_{T}\backslash(\Omega_{T})_{\delta}$,
then $\displaystyle u_{k}(x)\geq C((\omega_{\tilde{T}})_{\delta},n,s)>0$ in
$\displaystyle(\omega_{\tilde{T}})_{\delta}$ for all $\displaystyle k$.
Choosing $\displaystyle u_{k}\chi_{(0,\tau)}(t)$ as test function in Eq. 2.11,
we obtain

$\begin{array}[]{l}\quad\frac{1}{2}\int_{\Omega}u_{k}^{2}(x,\tau)dx+\int_{0}^{\tau}\int_{\Omega}|\nabla
u_{k}|^{2}dxdt+\frac{1}{2}\int_{0}^{\tau}\int_{Q}\frac{\left(u_{k}(x,t)-u_{k}(y,t)\right)^{2}}{|x-y|^{n+2s}}dxdydt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
\leq\iint_{\Omega_{T}}\frac{f_{k}u_{k}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\frac{1}{2}\int_{\Omega}u_{0}^{2}(x)dx.\end{array}$

Now taking supremum over $\displaystyle\tau\in(0,T]$, we get

${}\begin{array}[]{l}\quad\sup_{0\leq\tau\leq
T}\int_{\Omega}u_{k}^{2}(x,\tau)dx+2\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}|^{2}dxdt+\int_{0}^{T}\int_{Q}\frac{\left(u_{k}(x,t)-u_{k}(y,t)\right)^{2}}{|x-y|^{n+2s}}dxdydt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ \leq
2\iint_{\Omega_{T}}\frac{f_{k}u_{k}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{2}(\Omega)}\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
=2\iint_{(\Omega_{T})_{\delta}}\frac{f_{k}u_{k}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+2\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{f_{k}u_{k}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{2}(\Omega)}\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ \leq
2\iint_{(\Omega_{T})_{\delta}}fu_{k}^{1-\gamma(x,t)}dxdt+2\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{fu_{k}}{C_{(\omega_{\tilde{T}})_{\delta}}^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{2}(\Omega)}\vskip
3.0pt plus 1.0pt minus 1.0pt\\\ \leq
2\iint_{(\Omega_{T})_{\delta}\cap\left\\{u_{k}\leq
1\right\\}}fu_{k}^{1-\gamma(x,t)}dxdt+2\iint_{(\Omega_{T})_{\delta}\cap\left\\{u_{k}\geq
1\right\\}}fu_{k}^{1-\gamma(x,t)}dxdt\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
\quad+2\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{fu_{k}}{C_{(\omega_{\tilde{T}})_{\delta}}^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{2}(\Omega)}\\\
\leq\left\|u_{0}\right\|_{L^{2}(\Omega)}+2\|f\|_{L^{1}(\Omega_{T})}+2\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right)\iint_{\Omega_{T}}fu_{k}dxdt.\end{array}$
(3.27)

Case 1: $\displaystyle f\in
L^{2}\left(0,T;L^{\left(\frac{2n}{n+2}\right)}(\Omega)\right)$
We note that $\displaystyle(2^{*})^{\prime}=\frac{2n}{n+2}$, then by using
Hölder’s and Sobolev’s inequalities, we have

$\begin{array}[]{rcl}\int_{0}^{T}\int_{\Omega}fu_{k}dxdt&\leq&C\int_{0}^{T}\left(\int_{\Omega}|f(x,t)|^{(2^{*})^{\prime}}dx\right)^{\frac{1}{(2^{*})^{\prime}}}\left(\int_{\Omega}\left|u_{k}(x,t)\right|^{2^{*}}dx\right)^{\frac{1}{2^{*}}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&C\int_{0}^{T}\|f\|_{L^{\left({2^{*}}\right)^{\prime}}{(\Omega)}}\left(\int_{\Omega}\left|\nabla
u_{k}\right|^{2}dx\right)^{\frac{1}{2}}dt\vskip 3.0pt plus 1.0pt minus
1.0pt\\\
&\leq&C\left(\int_{0}^{T}\|f\|_{L^{\left({2^{*}}\right)^{\prime}}{(\Omega)}}^{2}dt\right)^{\frac{1}{2}}\left(\int_{0}^{T}\int_{\Omega}\left|\nabla
u_{k}\right|^{2}dxdt\right)^{\frac{1}{2}}.\end{array}$

In the above, we use Young’s inequality, and then from Eq. 3.27 we get that

$\begin{array}[]{l}\quad\sup_{0\leq\tau\leq
T}\int_{\Omega}u_{k}^{2}(x,\tau)dx+\left\|u_{k}\right\|_{L^{2}(0,T;H_{0}^{1}(\Omega))}^{2}\leq\left\|u_{0}\right\|_{L^{2}(\Omega)}+2\|f\|_{L^{1}(\Omega_{T})}+C\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right).\end{array}$

Therefore the sequence $\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly
bounded in $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))\cap
L^{\infty}(0,T;L^{2}(\Omega))$.
Case 2: $\displaystyle f\in L^{\bar{r}}\left(\Omega_{T}\right)$
For this case, we apply Hölder inequality with exponents
$\displaystyle\bar{r}$ and $\displaystyle\bar{r}^{\prime}$ to get

${}\begin{array}[]{rcl}\int_{0}^{T}\int_{\Omega}fu_{k}dxdt\leq
C\|f\|_{L^{\bar{r}}\left(\Omega_{T}\right)}\left[\iint_{\Omega_{T}}|u_{k}|^{\bar{r}^{\prime}}dxdt\right]^{\frac{1}{\bar{r}^{\prime}}}.\end{array}$
(3.28)

We observe that $\displaystyle\bar{r}^{\prime}=\frac{2(n+2)}{n}$ and hence
using the Hölder inequality with exponents $\displaystyle\frac{n}{n-2}$ and
$\displaystyle\frac{n}{2}$ and by Sobolev embedding (Theorem 2.8), we can
write

${}\begin{array}[]{rcl}\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{2(n+2)}{n}}dxdt=\iint_{\Omega_{T}}|u_{k}|^{2}\left|u_{k}\right|^{\frac{4}{n}}dxdt&\leq&\int_{0}^{T}\left[\int_{\Omega}\left|u_{k}(x,t)\right|^{2}dx\right]^{\frac{2}{n}}\left\|u_{k}\right\|_{L^{2^{*}}(\Omega)}^{2}dt\\\
&\leq&C(n)\left[\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{2}dx\right]^{\frac{2}{n}}\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}|^{2}dxdt.\end{array}$ (3.29)

So using Eq. 3.28 in Eq. 3.27, we get from Eq. 3.29 by convexity argument

$\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{2(n+2)}{n}}dxdt\leq
C(n)2^{\frac{2}{n}}\left(\left(C_{1}C\|f\|_{L^{\bar{r}}\left(\Omega_{T}\right)}\right)^{\frac{n+2}{n}}\left(\iint_{\Omega_{T}}|u_{k}|^{{\bar{r}^{\prime}}}\right)^{\frac{n+2}{n\bar{r}^{\prime}}}+C_{2}^{\frac{n+2}{n}}\right),$

where $\displaystyle
C_{1}=2\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right)$
and $\displaystyle
C_{2}=\left\|u_{0}\right\|_{L^{2}(\Omega)}+2\|f\|_{L^{1}(\Omega_{T})}$. Now
since $\displaystyle\frac{n+2}{n\bar{r}^{\prime}}=\frac{1}{2}$, we use Young’s
inequality to obtain

$\iint_{\Omega_{T}}\left|u_{k}\right|^{\bar{r}^{\prime}}dxdt=\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{2(n+2)}{n}}dxdt\leq
C,$

where $\displaystyle C$ is a positive constant independent of $\displaystyle
k$. Therefore, by Eqs. 3.27 and 3.28 we deduce that
$\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly bounded in
$\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))\cap
L^{\infty}(0,T;L^{2}(\Omega))$.
Since $\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly bounded in the
reflexive Banach space $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))$, there
exist a subsequence of $\displaystyle\left\\{u_{k}\right\\}_{k}$, still
indexed by $\displaystyle k$, and a measurable function $\displaystyle u\in
L^{2}(0,T;H_{0}^{1}(\Omega))$ such that $\displaystyle u_{k}\rightharpoonup u$
weakly in $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))$. Also, since
$\displaystyle\left\\{u_{k}\right\\}$ is increasing in $\displaystyle k$, it
holds by Beppo Levi’s theorem that $\displaystyle u_{k}\rightarrow u$ strongly
in $\displaystyle L^{1}\left(\Omega_{T}\right)$ and hence a.e. in
$\displaystyle\Omega_{T}$. Applying Fatou’s Lemma, we get $\displaystyle u\in
L^{\infty}(0,T;L^{2}(\Omega))$. This pointwise limit is actually defined for
each $\displaystyle t\in[0,T)$ and satisfies $\displaystyle
u(\cdot,0)=u_{0}(\cdot)$ in $\displaystyle L^{1}$ sense (see Remark 2.23). We
show that this $\displaystyle u$ is a very weak solution to Eq. 2.13 in the
sense of Definition 2.25. Choosing arbitrary
$\displaystyle\phi\in\mathcal{C}_{0}^{\infty}(\Omega_{T})$, we have

$\iint_{\Omega_{T}}(\left(u_{k}\right)_{t}-\Delta
u_{k}+(-\Delta)^{s}u_{k})\phi\,dxdt=\iint_{\Omega_{T}}\frac{f_{k}\phi}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt.$

Since $\displaystyle\phi\in\mathcal{C}_{0}^{\infty}(\Omega_{T})$, therefore
$\displaystyle\phi_{t},\Delta\phi$ and $\displaystyle(-\Delta)^{s}\phi$ all
are bounded. As $\displaystyle u_{k}\rightarrow u$ strongly in $\displaystyle
L^{1}\left(\Omega_{T}\right)$, we have

$\begin{array}[]{rcl}\iint_{\Omega_{T}}(\left(u_{k}\right)_{t}-\Delta
u_{k}+(-\Delta)^{s}u_{k})\phi\,dxdt&=&\iint_{\Omega_{T}}u_{k}(-(\phi)_{t}-\Delta\phi+(-\Delta)^{s}\phi)dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\rightarrow&\iint_{\Omega_{T}}u(-(\phi)_{t}-\Delta\phi+(-\Delta)^{s}\phi)dxdt.\end{array}$

Now since for all $\displaystyle\omega\subset\subset\Omega$ and for all
$\displaystyle t_{0}>0,$ $\displaystyle u_{k}(x,t)\geq
C\left(\omega,t_{0},n,s\right)$ in
$\displaystyle\omega\times\left[t_{0},T\right)$, we reach that $\displaystyle
u_{k}(x,t)\geq C$ in $\displaystyle\operatorname{Supp}\phi$ (say
$\displaystyle\omega\times[t_{1},t_{2}]$) for all $\displaystyle k\geq 1$.
Therefore

$0\leq\left|\frac{f_{k}\phi}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}\right|\leq\|\phi
C_{\omega,t_{1},t_{2}}^{-\gamma(x,t)}\|_{L^{\infty}(\Omega_{T})}f.$

Hence by the dominated convergence theorem

$\iint_{\Omega_{T}}\frac{f_{k}\phi}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt\rightarrow\iint_{\Omega_{T}}\frac{f\phi}{u^{\gamma(x,t)}}dxdt\text{
as }k\rightarrow\infty.$

Thus, we get the following and conclude

$\iint_{\Omega_{T}}(u_{t}-\Delta
u+(-\Delta)^{s}u)\phi\,dxdt=\iint_{\Omega_{T}}\frac{f\phi}{u^{\gamma(x,t)}}dxdt.$

### Proof of Theorem 2.41

Consider $\displaystyle u_{k}$ to be the unique nonnegative solution to Eq.
2.11. Since $\displaystyle\gamma^{*}>1$, so we can use $\displaystyle
u_{k}^{\gamma^{*}}\chi_{(0,\tau)}(t)$ as a test function in Eq. 2.11 to get
for all $\displaystyle\tau\leq T$,

$\begin{array}[]{c}\frac{1}{\gamma^{*}+1}\int_{\Omega}u_{k}^{\gamma^{*}+1}(x,\tau)dx+\int_{0}^{\tau}\int_{\Omega}\nabla
u_{k}\cdot\nabla u_{k}^{\gamma^{*}}dxdt\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
+\frac{1}{2}\int_{0}^{\tau}\int_{Q}\frac{(u_{k}^{\gamma^{*}}(x,t)-u_{k}^{\gamma^{*}}(y,t))\left(u_{k}(x,t)-u_{k}(y,t)\right)}{|x-y|^{n+2s}}dxdydt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
\leq\iint_{\Omega_{T}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\frac{1}{\gamma^{*}+1}\int_{\Omega}u_{0}^{\gamma^{*}+1}(x)dx.\end{array}$

Now taking supremum over $\displaystyle\tau\in(0,T]$ and using item
$\displaystyle i)$ of Lemma 2.11, we get

$\begin{array}[]{c}\quad\frac{({\gamma^{*}}+1)}{2\gamma^{*}}\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{{\gamma^{*}}+1}dx+\frac{({\gamma^{*}}+1)^{2}}{2}\int_{0}^{T}\int_{\Omega}u_{k}^{{\gamma^{*}}-1}|\nabla
u_{k}|^{2}dxdt\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
\quad+\int_{0}^{T}\int_{Q}\frac{|u_{k}^{\frac{{\gamma^{*}}+1}{2}}(x,t)-u_{k}^{\frac{{\gamma^{*}}+1}{2}}(y,t)|^{2}}{|x-y|^{n+2s}}dxdydt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
\leq\frac{({\gamma^{*}}+1)^{2}}{2\gamma^{*}}\left(\iint_{\Omega_{T}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{\gamma^{*}+1}(\Omega)}\right).\end{array}$

We note that $\displaystyle 1<\frac{\gamma^{*}+1}{\gamma^{*}}<2$ and hence it
follows

${}\begin{array}[]{c}\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{{\gamma^{*}}+1}dx+4\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt\\\
+2\int_{0}^{T}\int_{Q}\frac{|u_{k}^{\frac{{\gamma^{*}}+1}{2}}(x,t)-u_{k}^{\frac{{\gamma^{*}}+1}{2}}(y,t)|^{2}}{|x-y|^{n+2s}}dxdydt\leq
4\gamma^{*}\left(\iint_{\Omega_{T}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\left\|u_{0}\right\|_{L^{\gamma^{*}+1}(\Omega)}\right).\end{array}$
(3.30)

Denoting
$\displaystyle(\omega_{\tilde{T}})_{\delta}=\Omega_{T}\backslash(\Omega_{T})_{\delta}$,
we have $\displaystyle u_{k}(x,t)\geq C((\omega_{\tilde{T}})_{\delta},n,s)>0$
in $\displaystyle(\omega_{\tilde{T}})_{\delta}$ for all $\displaystyle k$. We
estimate like Theorem 2.39 as

$\begin{array}[]{rcl}\iint_{\Omega_{T}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt&=&\iint_{(\Omega_{T})_{\delta}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt+\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{f_{k}u_{k}^{\gamma^{*}}}{\left(u_{k}+\frac{1}{k}\right)^{\gamma(x,t)}}dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\iint_{(\Omega_{T})_{\delta}}fu_{k}^{{\gamma^{*}}-\gamma(x,t)}dxdt+\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{fu_{k}^{\gamma^{*}}}{C_{(\omega_{\tilde{T}})_{\delta}}^{\gamma(x,t)}}dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\iint_{(\Omega_{T})_{\delta}\cap\left\\{u_{k}\leq
1\right\\}}fu_{k}^{{\gamma^{*}}-\gamma(x,t)}dxdt+\iint_{(\Omega_{T})_{\delta}\cap\left\\{u_{k}\geq
1\right\\}}fu_{k}^{\gamma^{*}-\gamma(x,t)}dxdt\vskip 3.0pt plus 1.0pt minus
1.0pt\\\
&&+\iint_{(\omega_{\tilde{T}})_{\delta}}\frac{fu_{k}^{\gamma^{*}}}{C_{(\omega_{\tilde{T}})_{\delta}}^{\gamma(x,t)}}dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\|f\|_{L^{1}(\Omega_{T})}+\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right)\iint_{\Omega_{T}}fu_{k}^{\gamma^{*}}dxdt.\end{array}$

Using the above in Eq. 3.30 we get

${}\begin{array}[]{l}\quad\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{{\gamma^{*}}+1}dx+2\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt\vskip 3.0pt plus 1.0pt minus
1.0pt\\\ \leq
4\gamma^{*}\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right)\iint_{\Omega_{T}}fu_{k}^{\gamma^{*}}dxdt+4\gamma^{*}\left(\|f\|_{L^{1}(\Omega_{T})}+\left\|u_{0}\right\|_{L^{\gamma^{*}+1}(\Omega)}\right).\end{array}$
(3.31)

For convenience we take $\displaystyle
C_{1}=4\gamma^{*}\left(1+\left\|C_{(\omega_{\tilde{T}})_{\delta}}^{-\gamma(\cdot)}\right\|_{L^{\infty}(\Omega_{T})}\right)$
and $\displaystyle
C_{2}=4\gamma^{*}\left(\|f\|_{L^{1}(\Omega_{T})}+\left\|u_{0}\right\|_{L^{\gamma^{*}+1}(\Omega)}\right)$.
Case 1: $\displaystyle f\in
L^{\gamma^{*}+1}\left(0,T;L^{\left(\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}\right)}(\Omega)\right)$
In this case using Hölder inequality first in the space variable and then by
Sobolev inequality and another application of Hölder inequality (in time
variable) we get

$\begin{array}[]{rcl}\int_{0}^{T}\int_{\Omega}fu_{k}^{\gamma^{*}}dxdt&=&\int_{0}^{T}\int_{\Omega}f\left(u_{k}^{\frac{\gamma^{*}+1}{2}}\right)^{\frac{2\gamma^{*}}{\gamma^{*}+1}}dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&C\int_{0}^{T}\left(\int_{\Omega}|f(x,t)|^{\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}}dx\right)^{\frac{n+2\gamma^{*}}{n(\gamma^{*}+1)}}\left(\int_{\Omega}|u_{k}^{\frac{\gamma^{*}+1}{2}}(x,t)|^{2^{*}}dx\right)^{\frac{(n-2)\gamma^{*}}{n(\gamma^{*}+1)}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&C\int_{0}^{T}\|f(\cdot,t)\|_{L^{\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}}{(\Omega)}}\left(\int_{\Omega}|\nabla
u_{k}^{\frac{\gamma^{*}+1}{2}}|^{2}dx\right)^{\frac{\gamma^{*}}{\gamma^{*}+1}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&C\left(\int_{0}^{T}\|f(\cdot,t)\|_{L^{\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}}{(\Omega)}}^{\gamma^{*}+1}dt\right)^{\frac{1}{\gamma^{*}+1}}\times\left(\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{\gamma^{*}+1}{2}}|^{2}dxdt\right)^{\frac{\gamma^{*}}{\gamma^{*}+1}}.\end{array}$

Now since $\displaystyle\frac{\gamma^{*}}{\gamma^{*}+1}<1$, so using Young’s
inequality in the above estimate, we get from Eq. 3.31 that

$\begin{array}[]{l}\quad\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{{\gamma^{*}}+1}dx+2\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt\vskip 3.0pt plus 1.0pt minus
1.0pt\\\ \leq
C\left\|f\right\|_{L^{\gamma^{*}+1}\left(0,T;L^{\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}}(\Omega)\right)}\times\left(\varepsilon\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt+C(\varepsilon)\right)+C_{2}.\end{array}$

Choosing $\displaystyle\varepsilon$ small enough such that
$\displaystyle\varepsilon
C\left\|f\right\|_{L^{\gamma^{*}+1}\left(0,T;L^{\frac{n(\gamma^{*}+1)}{n+2\gamma^{*}}}(\Omega)\right)}=1,$
we get

$\sup_{0\leq\tau\leq
T}\int_{\Omega}\left|u_{k}(x,\tau)\right|^{{\gamma^{*}}+1}dx+\int_{0}^{T}\int_{\Omega}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt\leq C,$

which implies that the sequence
$\displaystyle\left\\{u_{k}^{\frac{\gamma^{*}+1}{2}}\right\\}_{k}$ is
uniformly bounded in $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))\cap
L^{\infty}(0,T;L^{2}(\Omega))$ and $\displaystyle\left\\{u_{k}\right\\}_{k}$
is uniformly bounded in $\displaystyle
L^{\infty}(0,T;L^{\gamma^{*}+1}(\Omega))$.
Case 2: $\displaystyle f\in L^{\tilde{r}}\left(\Omega_{T}\right)$
For this case, we apply Hölder inequality with exponents
$\displaystyle\tilde{r}$ and $\displaystyle\tilde{r}^{\prime}$ to get

${}\begin{array}[]{rcl}\int_{0}^{T}\int_{\Omega}fu_{k}^{\gamma^{*}}dxdt\leq
C\|f\|_{L^{\tilde{r}}\left(\Omega_{T}\right)}\left[\iint_{\Omega_{T}}|u_{k}|^{\gamma^{*}\tilde{r}^{\prime}}dxdt\right]^{\frac{1}{\tilde{r}^{\prime}}}.\end{array}$
(3.32)

We note that
$\displaystyle\gamma^{*}\tilde{r}^{\prime}=\frac{(n+2)(\gamma^{*}+1)}{n}$.
Again, we use the same technique as that of Theorem 2.30, Theorem 2.32 and
Theorem 2.39. Using the Hölder inequality with exponent
$\displaystyle\frac{n}{n-2}$ and $\displaystyle\frac{n}{2}$ we get

$\begin{array}[]{rcl}\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{({\gamma^{*}}+1)(n+2)}{n}}dxdt&=&\iint_{\Omega_{T}}\left|u_{k}\right|^{2\frac{{\gamma^{*}}+1}{2}}\left|u_{k}\right|^{\frac{4}{n}\frac{{\gamma^{*}}+1}{2}}dxdt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&\leq&\int_{0}^{T}\left(\int_{\Omega}\left|u_{k}(x,t)\right|^{2^{*}\frac{{\gamma^{*}}+1}{2}}dx\right)^{\frac{n-2}{n}}\left(\int_{\Omega}\left|u_{k}(x,t)\right|^{{\gamma^{*}}+1}dx\right)^{\frac{2}{n}}dt\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
&=&\int_{0}^{T}\|u_{k}^{\frac{{\gamma^{*}}+1}{2}}\|_{L^{2^{*}}(\Omega)}^{2}\left(\int_{\Omega}\left|u_{k}(x,t)\right|^{{\gamma^{*}}+1}dx\right)^{\frac{2}{n}}dt.\end{array}$

Now taking supremum over $\displaystyle t\in[0,T]$ and applying the Sobolev
embedding as that of Theorem 2.8, we can write

$\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{({\gamma^{*}}+1)(n+2)}{n}}dxdt\leq
C(n)\left(\sup_{0\leq t\leq
T}\int_{\Omega}\left|u_{k}(x,t)\right|^{{\gamma^{*}}+1}dx\right)^{\frac{2}{n}}\left(\iint_{\Omega_{T}}|\nabla
u_{k}^{\frac{{\gamma^{*}}+1}{2}}|^{2}dxdt\right).$

Using Eq. 3.31 and Eq. 3.32, from the above inequality we get by convexity
argument

$\begin{array}[]{rcl}\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{({\gamma^{*}}+1)(n+2)}{n}}dxdt&\leq&C(n)\left(C_{1}\iint_{\Omega_{T}}fu_{k}^{{\gamma^{*}}}dxdt+C_{2}\right)^{\frac{n+2}{n}}\\\
&\leq&C(n)2^{\frac{2}{n}}\left(\left(C_{1}\iint_{\Omega_{T}}fu_{k}^{{\gamma^{*}}}dxdt\right)^{\frac{n+2}{n}}+C_{2}^{\frac{n+2}{n}}\right)\\\
&\leq&C(n)2^{\frac{2}{n}}\left(\left(C_{1}C\|f\|_{L^{\tilde{r}}(\Omega_{T})}\right)^{\frac{n+2}{n}}\left(\iint_{\Omega_{T}}|u_{k}|^{\gamma^{*}\tilde{r}^{\prime}}dxdt\right)^{\frac{n+2}{n\tilde{r}^{\prime}}}+C_{2}^{\frac{n+2}{n}}\right).\end{array}$

Now as
$\displaystyle\frac{n+2}{n\tilde{r}^{\prime}}=\frac{\gamma^{*}}{\gamma^{*}+1}<1$,
therefore using Young’s inequality we get

$\iint_{\Omega_{T}}\left|u_{k}\right|^{\gamma^{*}\tilde{r}^{\prime}}dxdt=\iint_{\Omega_{T}}\left|u_{k}\right|^{\frac{({\gamma^{*}}+1)(n+2)}{n}}dxdt\leq
C,$

where $\displaystyle C$ is a positive constant independent of $\displaystyle
k$. The above boundedness along with Eqs. 3.32 and 3.31 gives that the
sequence $\displaystyle\left\\{u_{k}^{\frac{\gamma^{*}+1}{2}}\right\\}_{k}$ is
uniformly bounded in $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))\cap
L^{\infty}(0,T;L^{2}(\Omega))$ and $\displaystyle\left\\{u_{k}\right\\}_{k}$
is uniformly bounded in $\displaystyle
L^{\infty}(0,T;L^{\gamma^{*}+1}(\Omega))$. We now show
$\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly bounded in
$\displaystyle L^{2}(t_{0},T;H_{loc}^{1}(\Omega))$ for each $\displaystyle
t_{0}>0$. Since $\displaystyle\gamma^{*}>1$, and $\displaystyle\Omega_{T}$ is
bounded and $\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly bounded in
$\displaystyle L^{\infty}(0,T;L^{\gamma^{*}+1}(\Omega))$, we thus have
$\displaystyle\left\\{u_{k}\right\\}_{k}$ is uniformly bounded in
$\displaystyle L^{2}(0,T;L^{2}(\Omega))=L^{2}(\Omega_{T})$, in particular in
$\displaystyle L^{2}(K\times(t_{0},T))$, for every $\displaystyle
K\subset\subset\Omega$ and for each $\displaystyle t_{0}>0$. Again as
$\displaystyle\left\\{u_{k}^{\frac{\gamma^{*}+1}{2}}\right\\}_{k}$ is
uniformly bounded in $\displaystyle L^{2}(0,T;H_{0}^{1}(\Omega))$, so we have
by nonnegativity of $\displaystyle u_{k}$

$\begin{array}[]{c}\int_{t_{0}}^{T}\int_{K}u_{k}^{\gamma^{*}-1}|\nabla
u_{k}|^{2}dxdt\leq\int_{0}^{T}\int_{\Omega}u_{k}^{\gamma^{*}-1}|\nabla
u_{k}|^{2}dxdt\leq\frac{4C}{(\gamma^{*}+1)^{2}},\end{array}$

for every $\displaystyle K\subset\subset\Omega$ and for each $\displaystyle
t_{0}>0$. Using the positivity of $\displaystyle u_{k}$ in
$\displaystyle\omega\times[t_{0},T)$ for all $\displaystyle k$, we conclude

$\int_{t_{0}}^{T}\int_{K}|\nabla
u_{k}|^{2}dxdydt\leq\frac{C}{\gamma^{*}c_{(K,t_{0})}^{\gamma^{*}-1}}.$

Since $\displaystyle\left\\{u_{k}\right\\}_{k}$ is increasing in
$\displaystyle k$ and is uniformly bounded in $\displaystyle
L^{\infty}(0,T;L^{\gamma^{*}+1}(\Omega))$, we get by Beppo Levi’s Lemma
$\displaystyle u_{k}\uparrow u$ strongly in $\displaystyle
L^{\gamma^{*}+1}\left(\Omega_{T}\right)$ and hence in $\displaystyle
L^{1}(\Omega_{T})$, where $\displaystyle u$ is the pointwise limit of
$\displaystyle u_{k}$ (possibly infinite). Then employing Fatou’s lemma we
will get $\displaystyle u^{\frac{\gamma^{*}+1}{2}}\in
L^{\infty}(0,T;L^{2}(\Omega))\cap L^{2}(0,T;H_{0}^{1}(\Omega))$ and
$\displaystyle u\in L^{2}(t_{0},T;H_{loc}^{1}(\Omega))$ for each
$\displaystyle t_{0}>0$. Remark 2.23 implies that $\displaystyle u$ satisfies
$\displaystyle u(\cdot,0)=u_{0}(\cdot)$ in $\displaystyle L^{1}$ sense. Also,
we have $\displaystyle u(x,t)\geq$ $\displaystyle
C\left(\omega,t_{0},n,s\right)$ in
$\displaystyle\omega\times\left[t_{0},T\right)$ for all
$\displaystyle\omega\subset\subset\Omega$ and for each $\displaystyle
t_{0}>0$. This $\displaystyle u$ will be a very weak solution to Eq. 2.13 in
the sense of Definition 2.25, and the proof follows exactly similar to Theorem
2.39.

######  Remark 3.3.

In the above existence results, the cases where $\displaystyle u\in
L^{2}(0,T;H^{1}_{0}(\Omega))$, we can show that the weak solution
$\displaystyle u$ is unique, if it has finite energy. For this we assume that
$\displaystyle v$ is another finite energy solution to Eq. 2.12 or Eq. 2.13.
Then, by the construction of $\displaystyle u$ in each existence result, we
get $\displaystyle v$ is a supersolution to all the approximating problems.
Hence by the comparison principle in Lemma 2.20, we deduce that $\displaystyle
v\geq u_{k}$ for all $\displaystyle k$. Then $\displaystyle v\geq u$. To prove
the inverse inequality, let $\displaystyle\phi=v-u$, then
$\displaystyle\phi\geq 0$ and $\displaystyle\phi\in
L^{2}(0,T;H_{0}^{1}(\Omega))$ and hence we can use both
$\displaystyle\phi_{+}$ or $\displaystyle\phi_{-}$ as test functions. Also, we
note that $\displaystyle\phi$ solves the problem

$\displaystyle\displaystyle\begin{cases}\phi_{t}-\Delta\phi+(-\Delta)^{s}\phi=f\left(\frac{1}{v^{\gamma(x,t)}}-\frac{1}{u^{\gamma(x,t)}}\right)&\text{
in }\Omega_{T},\\\ \phi(x,t)=0&\text{ in
}\left(\mathbb{R}^{n}\backslash\Omega\right)\times(0,T),\\\ \phi(x,0)=0&\text{
in }\Omega.\end{cases}$

Since $\displaystyle
f\left(\frac{1}{v^{\gamma(x,t)}}-\frac{1}{u^{\gamma(x,t)}}\right)\leq 0$ in
$\displaystyle\Omega_{T}$, by comparison principle, we get
$\displaystyle\phi\leq 0$ in $\displaystyle\Omega_{T}$. Thus
$\displaystyle\phi=0$ i.e. $\displaystyle u=v$.

######  Remark 3.4.

As we can notice, in each of the existence results, we have $\displaystyle
u\in L^{2}(t_{0},T;W^{1,1}_{\operatorname{loc}}(\Omega))$ for each
$\displaystyle t_{0}>0$ and this gives that our definition of a weak solution
is well motivated in the sense that we have derived them integrating by parts
twice upon multiplying by a test function.

##  4 Asymptotic behaviour

This section is devoted to the study of the asymptotic behaviour of finite
energy solution to the problem Eq. 2.12, as $\displaystyle
t\rightarrow\infty$, for the particular case where $\displaystyle f$ depends
only on $\displaystyle x$ and that $\displaystyle f>0$ with $\displaystyle
f\in L^{q}(\Omega)$ and $\displaystyle q$ will be specified later. We first
state the following existence and uniqueness result to the corresponding mixed
local-nonlocal elliptic problem, which can be done as a direct application of
Theorem 2.27. Consider the problem

${}\begin{array}[]{lcr}\begin{cases}-\Delta
w+(-\Delta)^{s}w=\frac{f}{w^{\gamma}}&\mbox{ in }\Omega,\\\ w=0&\mbox{ in
}(\mathbb{R}^{n}\backslash\Omega).\end{cases}\end{array}$ (4.1)

We define the weak solution to the above problem as:

######  Definition 4.1.

Suppose that $\displaystyle f\in L^{1}(\Omega)$ is a nonnegative function and
$\displaystyle\gamma>0$. We say that $\displaystyle w$ is a very weak solution
to problem Eq. 4.1 if $\displaystyle w\in L^{1}(\Omega)$ satisfies the
boundary conditions and $\displaystyle\forall\omega\subset\subset\Omega,$
$\displaystyle\exists c_{\omega}>0$ such that $\displaystyle w(x)\geq
c_{\omega}>0,\text{ a.e. in }\omega,$ and for all
$\displaystyle\varphi\in\mathcal{C}_{0}^{\infty}(\Omega)$, we have

$\int_{\Omega}w(-\Delta\varphi+(-\Delta)^{s}\varphi)dx=\int_{\Omega}\frac{f\varphi}{w^{\gamma}}dx.$

######  Theorem 4.2.

Let $\displaystyle f\in L^{1}(\Omega)$ be a non-negative function. Then for
all $\displaystyle\gamma>0$, the problem Eq. 4.1 has a nonnegative very weak
solution $\displaystyle w$ such that $\displaystyle w^{\frac{\gamma+1}{2}}\in
H_{0}^{1}(\Omega)$ in the sense of Definition 4.1.

###### Proof.

Let $\displaystyle w_{k}$ be the unique positive bounded solution to the
approximating problem

${}\begin{array}[]{lcr}\begin{cases}-\Delta
w_{k}+(-\Delta)^{s}w_{k}=\frac{f_{k}}{(w_{k}+\frac{1}{k})^{\gamma}}&\mbox{ in
}\Omega,\\\ w_{k}>0&\mbox{ in }\Omega,\\\ w_{k}=0&\mbox{ in
}(\mathbb{R}^{n}\backslash\Omega);\end{cases}\end{array}$ (4.2)

with $\displaystyle f_{k}:=\min(k,f)$. We note that the existence of
$\displaystyle w_{k}$ follows using a simple monotonicity argument. More
precisely, one can prove a similar version of existence results and comparison
principle as of Theorem 2.13 for the elliptic case also, and using that, one
can find the elliptic version of Lemma 2.20 and Corollary 2.21. Again, by the
comparison principle, we deduce that the sequence
$\displaystyle\left\\{w_{k}\right\\}_{k}$ is increasing in $\displaystyle k$.
By the Harnack inequality, see [22, Theorem 8.3], it holds that for any set
$\displaystyle\omega\subset\subset\Omega$, for all $\displaystyle k$ we have,
$\displaystyle w_{k}(x)\geq w_{1}(x)\geq c(\omega,n,s)\text{ in }\omega.$
Also, the details of existence, uniqueness, positivity and monotonicity of
$\displaystyle w_{k}$’s can be found in [26, Lemma 3.2].
Now taking
$\displaystyle((w_{k}+\varepsilon)^{\gamma}-\varepsilon^{\gamma}),0<\varepsilon<1/k,$
as a test function in Eq. 4.2 and letting $\displaystyle\varepsilon\to 0$, it
follows that

$\int_{\Omega}\nabla w_{k}\cdot\nabla
w_{k}^{\gamma}+\frac{1}{2}\iint_{\mathbb{R}^{2n}}\frac{\left(w_{k}(x)-w_{k}(y)\right)\left(w_{k}^{\gamma}(x)-w_{k}^{\gamma}(y)\right)}{|x-y|^{n+2s}}dxdy\leq\int_{\Omega}f(x)dx.$

By Lemma 2.11, we have

$\left(w_{k}(x)-w_{k}(y)\right)\left(w_{k}^{\gamma}(x)-w_{k}^{\gamma}(y)\right)\geq
C\left(w_{k}^{\frac{\gamma+1}{2}}(x)-w_{k}^{\frac{\gamma+1}{2}}(y)\right)^{2},$

and on the other hand

$\nabla w_{k}\cdot\nabla w_{k}^{\gamma}=\gamma w_{k}^{\gamma-1}|\nabla
w_{k}|^{2},$

we deduce that the sequence
$\displaystyle\left\\{w_{k}^{\frac{\gamma+1}{2}}\right\\}_{k}$ is bounded in
the reflexive Banach space $\displaystyle X_{0}^{s}(\Omega)\cap
H_{0}^{1}(\Omega)$. Now by the monotonicity of
$\displaystyle\left\\{w_{k}\right\\}_{k}$, using Beppo-Levi’s theorem, we get
the existence of a measurable function $\displaystyle w$ such that
$\displaystyle w_{k}\uparrow w$ a.e. in $\displaystyle\mathbb{R}^{n},$
$\displaystyle w_{k}^{\frac{\gamma+1}{2}}\rightharpoonup
w^{\frac{\gamma+1}{2}}$ weakly in $\displaystyle X_{0}^{s}(\Omega)\cap
H_{0}^{1}(\Omega)$. Clearly $\displaystyle w\geq c(\omega)$ for any set
$\displaystyle\omega\subset\subset\Omega$. Then, letting $\displaystyle
k\rightarrow\infty$ in the approximating problems and using the dominated
convergence theorem, we can show that $\displaystyle w$ is a very weak
solution to problem Eq. 4.1 with $\displaystyle w^{\frac{\gamma+1}{2}}\in
H_{0}^{1}(\Omega)$. We refer [26, Theorem $\displaystyle 2.16$] for the
boundedness of $\displaystyle w$. One can find conditions when $\displaystyle
w\in H^{1}_{0}(\Omega)$ in [26]. Further, [26, Corollary $\displaystyle 5.2$]
gives such solution is unique. ∎

We now write the asymptotic behaviour of solutions to problem Eq. 2.12 under
suitable conditions on $\displaystyle f$ and $\displaystyle u_{0}$.

######  Theorem 4.3.

Suppose that $\displaystyle f\in L^{q}(\Omega)$ is a non-negative function
depending only on $\displaystyle x$, and $\displaystyle q$ is such that
solution to the problem Eq. 4.1 is in $\displaystyle H^{1}_{0}(\Omega)$ and is
unique. Let $\displaystyle u$ be a finite energy solution to problem Eq. 2.12
with $\displaystyle 0\leq u_{0}\leq w$, where $\displaystyle w$ is the
solution to Eq. 4.1. Then $\displaystyle u(\cdot,t)\rightarrow w$ as
$\displaystyle t\rightarrow\infty$, in $\displaystyle L^{\sigma}(\Omega)$,
where $\displaystyle\sigma<2^{*}$.

###### Proof.

We divide the proof into two cases according to the value of the initial
condition $\displaystyle u_{0}$.
The first case $\displaystyle u_{0}=0$: In this case using the comparison
principle as in Lemma 2.20, we get that $\displaystyle u\leq w$ in
$\displaystyle\Omega\times(0,T)$ for all $\displaystyle T>0$. Hence,
$\displaystyle u$ is globally defined. We now show that $\displaystyle
u(x,\cdot)$ is increasing in $\displaystyle t$. Since $\displaystyle u$ has
finite energy, we fix $\displaystyle t_{1}>0$ and define $\displaystyle
v(x,t)=u(x,t_{1}+t),$ then $\displaystyle v$ satisfies the problem

$\displaystyle\displaystyle\begin{cases}v_{t}-\Delta
v+(-\Delta)^{s}v=\frac{f(x)}{v^{\gamma}(x,t)}&\text{ in }\Omega_{T},\\\
v(x,t)=0&\text{ in }\left(\mathbb{R}^{n}\backslash\Omega\right)\times(0,T),\\\
v(x,0)=u\left(x,t_{1}\right)&\text{ in }\Omega.\end{cases}$

Now as $\displaystyle u\left(x,t_{1}\right)\geq u_{0}\equiv 0$ in
$\displaystyle\Omega$, using the comparison principle as in Lemma 2.20 again,
we have $\displaystyle u\leq v$ in $\displaystyle\Omega_{T}$. Therefore for
all $\displaystyle t_{1}>0$, we have $\displaystyle u(x,t)\leq
u\left(x,t_{1}+t\right)$ for all $\displaystyle x\in\mathbb{R}^{n}$. Therefore
$\displaystyle u(\cdot,t)$ is increasing in $\displaystyle t$. Since
$\displaystyle u\leq w$, so there exists a measurable function
$\displaystyle\hat{u}(\cdot)=\underset{t\rightarrow\infty}{\operatorname{lim}\operatorname{sup}}~{}u(\cdot,t)$.
By dominated convergence theorem, $\displaystyle
u(\cdot,t)\rightarrow\hat{u}(\cdot)$ as $\displaystyle t\rightarrow\infty$ in
$\displaystyle L^{\sigma}(\Omega)$ for $\displaystyle\sigma<2^{*}$. Also, we
have $\displaystyle\hat{u}\leq w$. It is now sufficient to show that
$\displaystyle\hat{u}$ is a very weak solution to problem Eq. 4.1.
Let $\displaystyle\varphi\in\mathcal{C}_{0}^{\infty}(\Omega)$, then using
$\displaystyle\varphi$ as a test function in Eq. 2.12 and integrating in
$\displaystyle\Omega\times(k,k+1)$, we get

$\int_{\Omega}(u(x,k+1)-u(x,k))\varphi(x)dx+\int_{k}^{k+1}\int_{\Omega}u(x,t)(-\Delta\varphi+(-\Delta)^{s}\varphi)dxdt=\int_{k}^{k+1}\int_{\Omega}\frac{f(x)}{u^{\gamma}(x,t)}\varphi(x)dxdt.$

Now clearly we have

$\int_{\Omega}(u(x,k+1)-u(x,k))|\varphi(x)|dx\rightarrow 0\text{ as
}k\rightarrow\infty.$

On the other hand, in the set $\displaystyle\Omega\times(k,k+1)$ using the
monotonicity of $\displaystyle u$ in the variable $\displaystyle t$, we have

$\frac{f(x)}{u^{\gamma}(x,k+1)}|\varphi(x)|\leq\frac{f(x)}{u^{\gamma}(x,t)}|\varphi(x)|\leq\frac{f(x)}{u^{\gamma}(x,k)}|\varphi(x)|,\quad\forall
t\in(k,k+1),$ $u(x,k)\left|-\Delta\varphi\right|\leq
u(x,t)\left|-\Delta\varphi\right|\leq
u(x,k+1)\left|-\Delta\varphi\right|,\quad\forall t\in(k,k+1),$

and

$u(x,k)\left|(-\Delta)^{s}\varphi\right|\leq
u(x,t)\left|(-\Delta)^{s}\varphi\right|\leq
u(x,k+1)\left|(-\Delta)^{s}\varphi\right|,\quad\forall t\in(k,k+1).$

Then, since $\displaystyle\phi,\Delta\phi,(-\Delta)^{s}\phi$ are bounded, so
by the dominated convergence theorem, we will get that, as $\displaystyle
k\rightarrow\infty$,

$\int_{k}^{k+1}\int_{\Omega}u(x,t)(-\Delta\varphi+(-\Delta)^{s}\varphi)dxdt\rightarrow\int_{\Omega}\hat{u}(x)(-\Delta\varphi+(-\Delta)^{s}\varphi)dx,$

and

$\int_{k}^{k+1}\int_{\Omega}\frac{f(x)}{u^{\gamma}(x,t)}\varphi(x)dxdt\rightarrow\int_{\Omega}\frac{f(x)}{\hat{u}^{\gamma}(x)}\varphi(x)dx.$

Hence, $\displaystyle\hat{u}$ is a very weak solution to problem Eq. 4.1. By
the uniqueness we then get $\displaystyle\hat{u}=w$.
The second case $\displaystyle 0<u_{0}\leq w$: Let us denote by
$\displaystyle\hat{u}$ a finite energy solution to problem Eq. 2.12 in this
case. Then by the comparison principle in Lemma 2.20, we deduce that
$\displaystyle u\leq\hat{u}\leq w$, where $\displaystyle u$ is the weak
solution to problem Eq. 2.12 with $\displaystyle u_{0}=0$. Hence by the
convergence result of the first case, we get that
$\displaystyle\hat{u}(\cdot,t)\rightarrow w$, as $\displaystyle
t\rightarrow\infty$, strongly in $\displaystyle L^{\sigma}(\Omega)$ for all
$\displaystyle\sigma<2^{*}$. ∎

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