Proof. Assuming α = 1, we observe from (4.14) and (2.2) that

(5.1)\begin{equation} u\in C([0,T]; W(\mathbb{R}^{N}))\Longleftrightarrow t\longmapsto \int_{\mathbb R^N}|u(t)|^{2}\log |u(t)|^{2}\,dx\in C([0,T]; \,\mathbb{R}). \end{equation}

In light of (3.4), we have

\begin{equation*}t\mapsto E(u(t))\in C([0,T]; \,\mathbb{R}),\end{equation*}

which implies $t\mapsto \int_{\mathbb R^N}|u(t)|^{2}\log |u(t)|^{2}\,dx\in C([0,T]; \,\mathbb{R}).$ This argument is also valid for the case α = 0 if one uses [Reference Boudjeriou and Alves5, Lemmas 3.4 and 5.1].

Proof. The case where α = 0, the energy identity (3.4) is proved in [Reference Boudjeriou and Alves5, Lemma 3.4]. Here, we focus on proving it for α = 1. Firstly, we observe that the identity (3.1) holds for test functions $v\in H_{loc}^{1}(\mathbb{R}^{N})$, i.e.,

(5.2)\begin{align} \int_{\mathbb{R}^{N}}u_{t}(t)v\,dx+\int_{\mathbb{R}^{N}}\nabla u(t)\cdot\nabla v\,dx+ \int_{\mathbb{R}^N} \nabla u_{t}(t)\cdot\nabla v\,dx=\int_{\mathbb{R}^N} u(t)\log u^2(t)v\,dx,\nonumber\\ \end{align}

for each $v\in H^{1}_{loc}(\mathbb{R}^{N})$ and a.e. $t\in [0,T]$. Let $\phi\in C^{\infty}_{0}(\mathbb{R})$ be a positive function such that $\phi(x)=1$ if $|x|\leq 1$ and $\phi(x)=0$ if $|x|\geq 2$. For $x\in \mathbb{R}^{N}$ and $n\in \mathbb{N}$, define $\phi_{n}(x)=\phi\left(\frac{|x|}{n}\right)$. By definition, $\|\nabla \phi_{n}\|_{\infty} \leq \frac{C}{n}$ for some constant C > 0 independent of n. Taking $v=u_{t}(t)\phi_{n}\in H_{loc}^{1}(\mathbb{R}^{N})$ in (5.2) and integrating over $(0,t)$ for any $t\in [0,T]$, we find

\begin{align*} &\frac{1}{2}\int_{0}^{t}\int_{\mathbb R^N} (u^{2}_{s}(s)\phi_{n}+|\nabla u_{s}(s)|^{2}\phi_{n})\,dxds+\frac{1}{2}\int_{\mathbb R^N} \left(|\nabla u(t)|^{2}+|u(t)|^{2}\right)\phi_{n}\,dx\\ &-\frac{1}{2}\int_{\mathbb R^N} |u(t)|^{2}\log |u(t)|^{2}\phi_{n}\,dx\\ =&-\int_{0}^{t} \int_{\mathbb R^N} \nabla u_{s}(s).\nabla \phi_{n}u_{s}(s)\,dxds-\int_{0}^{t} \int_{\mathbb R^N} \nabla u(s).\nabla \phi_{n}u_{s}(s)\,dxds\\ &+\frac{1}{2}\int_{\mathbb R^N} \left(|\nabla u_{0}|^{2}+|u_{0}|^{2}\right)\phi_{n}\,dx -\frac{1}{2}\int_{\mathbb R^N} |u_{0}|^{2}\log |u_{0}|^{2}\phi_{n}\,dx. \end{align*}

Using the regularity of u as stated in definition 3.1 and the properties of ϕ_n, the dominated convergence theorem allows us to obtain

\begin{equation*} \int_{0}^{t}\|u_{s}(s)\|^{2}_{H^{1}(\mathbb{R}^{N})}\,ds+E(u(t))=E(u_{0}), \;\;\forall t\in [0,T],\end{equation*}

which completes the proof.

Proof. Let us assume first $N\geq 3$, and for k a positive real number, we set

(5.4)\begin{equation} k_{n}=k-\frac{k}{2^{n}},\; n\in \mathbb{N}, \quad \text{and}\quad k\geq 2\|u_{0}\|_{\infty}. \end{equation}

Taking $v=(u-k_{n})_{+}\in W(\mathbb{R}^{N})$ in the identity (3.1) to find

\begin{align*} &\int_{\mathbb{R}^{N}} u_{t}(u-k_{n})_{+}\,dx +\int_{\mathbb R^N} \nabla u.\nabla (u-k_{n})_{+}\,dx+\alpha \int_{\mathbb R^N} \nabla u_{t}.\nabla (u-k_{n})_{+}\,dx\\ &\quad=\int_{\mathbb R^N} u (u-k_{n})_{+}\log|u|^{2}\,dx.\end{align*}

The left-hand side can be written as

\begin{equation*} \frac{1}{2}\partial_{t}\int_{\mathbb{R}^{N}} (u-k_{n})^{2}_{+}\,dx +\int_{\mathbb R^N}|\nabla (u-k_{n})_{+}|^{2}\,dx+\alpha \frac{1}{2}\partial_{t}\int_{\mathbb R^N} |\nabla (u-k_{n})_{+}|^{2}\,dx \end{equation*}

Thus, for any T > 0, integrating this over $(0,t)$ for $t\in (0,T)$, we obtain

(5.5)\begin{align} & \int_{\mathbb{R}^{N}} (u-k_{n})^{2}_{+}\,dx +2\int_{0}^{t}\int_{\mathbb R^N}|\nabla (u-k_{n})_{+}|^{2}\,dxds+\alpha \int_{\mathbb R^N} |\nabla (u-k_{n})_{+}|^{2}\,dx \nonumber\\ &=2\int_{0}^{t}\int_{\mathbb R^N} u (u-k_{n})_{+}\log|u|^{2}\,dxds, \end{align}

where we note that

\begin{equation*} \int_{\mathbb{R}^{N}} (u_{0}-k_{n})^{2}_{+}\,dx= \int_{\mathbb{R}^{N}} |\nabla(u_{0}-k_{n})_{+}|^{2}\,dx=0\end{equation*}

due to $2\|u_{0}\|_{\infty}\leq k.$ Next, we estimate the term that contains the logarithmic nonlinearity. Using the inequality $\log x\leq \frac{1}{e\beta}x^{\beta}$ for any β > 0 and x > 0, we get

\begin{align*} 2\int_{0}^{t}\int_{\mathbb R^N} u (u-k_{n})_{+}\log|u|^{2}\,dxds&=2\int_{0}^{t}\int_{\{u \gt k_{n}\}} u (u-k_{n})_{+}\log|u|^{2}\,dxds\\ &\leq \frac{4}{e\beta}\int_{0}^{t}\int_{\{u \gt k_{n}\}} |u|^{\beta+1} (u-k_{n})_{+}\,dxds, \end{align*}

where

\begin{equation*}\{u \gt k_{n}\}=\{x\in \mathbb{R}^{N}:\; u(x,t) \gt k_{n}\}.\end{equation*}

Choosing $\beta=q-2$ for $2 \lt q \lt 2+\frac{4}{N}$ in the latter inequality and applying Young’s inequality, we find

\begin{align*} 2\int_{0}^{t}\int_{\mathbb R^N} u (u-k_{n})_{+}\log|u|^{2}\,dxds&=2\int_{0}^{t}\int_{\{u \gt k_{n}\}} u (u-k_{n})_{+}\log|u|^{2}\,dxds\\ &\leq \frac{4}{e(q-2)}\int_{0}^{t}\int_{\{u \gt k_{n}\}} |u|^{q-1} (u-k_{n})_{+}\,dxds\\ &\leq \frac{4}{e(q-2)} \iint_{A_{n}} |u|^{q} \,dxds+\frac{4}{e(q-2)}\\&\quad\times \int_{0}^{t}\int_{\mathbb{R}^{N}} (u-k_{n})^{q}_{+} \,dxds, \end{align*}

where

\begin{equation*} A_{n}=\{(x,t)\in \mathbb{R}^{N}\times (0, +\infty):\; u(x,t) \gt k_{n}\}.\end{equation*}

Using the fact that $A_{n+1}\subseteq A_{n}$ for any $n\in \mathbb{N}$, we derive

\begin{align*} \int_{0}^{t}\int_{\mathbb{R}^{N}} (u-k_{n})_{+}^{q}\,dxds&\geq \iint_{A_{n+1}} (u-k_{n})_{+}^{q}\,dxds= \iint_{A_{n+1}}|u|^{q} \left(1-\frac{k_{n}}{k_{n+1}}\right)_{+}^{q}\,dxds\\ &\geq 2^{-q(n+1)}\iint_{A_{n+1}}|u|^{q} \,dxds, \end{align*}

which implies

(5.6)\begin{equation} \iint_{A_{n+1}}|u|^{q} \,dxds\leq 2^{q(n+1)}\int_{0}^{t}\int_{\mathbb{R}^{N}} (u-k_{n})_{+}^{q}\,dxds. \end{equation}

On the other hand, by Hölder’s and Sobolev inequalities, one has

(5.7)\begin{align} &\int_{0}^{t}\int_{\mathbb R^N} (u-k_{n})^{2+\frac{4}{N}}_{+}\,dxds\nonumber\\ &\quad \leq \int_{0}^{t}\left(\int_{\mathbb R^N} (u-k_{n})^{2^{*}}_{+}\,dx\right)^{2/2^{*}}\left(\int_{\mathbb R^N} (u-k_{n})^{2}_{+}\,dx\right)^{2/N}\,ds\nonumber\\ &\quad \leq \|(u-k_{n})_{+} \|^{\frac{4}{N}}_{L^{\infty}(0,T; L^{2}(\mathbb{R}^{N}))}\left(\int_{0}^{t}\int_{\mathbb R^N} (u-k_{n})^{2^{*}}_{+}\,dxds\right)^{2/2^{*}}\nonumber\\ &\quad \leq S^{2}\|\nabla (u-k_{n})_{+}\|^{2}_{L^{2}(0,T; L^{2}(\mathbb{R}^{N}))} \|(u-k_{n})_{+} \|^{\frac{4}{N}}_{L^{\infty}(0,T; L^{2}(\mathbb{R}^{N}))}\nonumber\\ &\quad \leq S^{2}\left(\|\nabla (u-k_{n})_{+}\|^{2}_{L^{2}(0,T; L^{2}(\mathbb{R}^{N}))}+ \|(u-k_{n})_{+} \|^{2}_{L^{\infty}(0,T; L^{2}(\mathbb{R}^{N}))}\right)^{\frac{N+2}{N}}. \end{align}

Substituting k_n by $k_{n+1}$ and combining the latter inequality with (5.5) and (5.6), we get

\begin{align*} &\int_{0}^{t}\int_{\mathbb R^N} (u-k_{n+1})^{2+\frac{4}{N}}_{+}\,dxds \leq S^{2}\left(\|\nabla (u-k_{n+1})_{+}\|^{2}_{L^{2}(0,T; L^{2}(\mathbb{R}^{N}))}\right.\\ &\quad+ \left.\|(u-k_{n+1})_{+} \|^{2}_{L^{\infty}(0,T; L^{2}(\mathbb{R}^{N}))}\right)^{\frac{N+2}{N}}\\ &\leq S^{2}\left(\frac{4}{e(q-2)} \iint_{A_{n+1}} |u|^{q} \,dxds+ \frac{4}{e(q-2)} \int_{0}^{t}\int_{\mathbb{R}^{N}} (u-k_{n+1})^{q}_{+} \,dxds\right)^{\frac{N+2}{N}}\\ &\leq S^{2}\left(\frac{2^{q(n+1)+2}}{e(q-2)}\int_{0}^{t}\!\int_{\mathbb{R}^{N}}\! (u-k_{n})_{+}^{q}\,dxds+\frac{4}{e(q-2)} \int_{0}^{t}\!\int_{\mathbb R^N}\! (u-k_{n+1})^{q}_{+}\,dxds\right)^{\frac{N+2}{N}}. \end{align*}

Since $2 \lt q \lt 2+\frac{4}{N}$, by using Hölder’s inequality, we derive

(5.8)\begin{align} \int_{0}^{t}\int_{\mathbb R^N} (u-k_{n+1})^{q}_{+}\,dxds\leq|A_{n+1}|^{\frac{2N+4-Nq}{2N+4}}\left( \int_{0}^{t}\int_{\mathbb R^N} (u-k_{n+1})^{2+\frac{4}{N}}_{+}\,dxds\right)^{\frac{qN}{2N+4}}\nonumber\\ \end{align}

Letting $Y_{n}=\int_{0}^{t}\int_{\mathbb R^N} (u-k_{n})^{q}_{+}\,dxds$ and noting that $(u-k_{n+1})_{+}\leq (u-k_{n})_{+}$, we infer from the previous inequalities

(5.9)\begin{equation} Y_{n+1}\leq S^{\frac{2qN}{2N+4}}|A_{n+1}|^{\frac{2N+4-Nq}{2N+4}}\left(\frac{2^{2q(n+1)+1}}{e(q-2)}Y_{n}\right)^{\frac{q(N+2)}{2N+4}}. \end{equation}

Through a chain of inequalities, we derive

\begin{equation*} Y_{n}=\int_{0}^{t}\int_{\mathbb R^N} (u-k_{n})^{q}_{+}\,dxds\geq \iint_{A_{n+1}} (u-k_{n})_{+}^{q}\,dxds\geq\left(\frac{k}{2^{n+1}}\right)^{q}|A_{n+1}|.\end{equation*}

From this and (5.9), it follows that

(5.10)\begin{align} Y_{n+1}&\leq k^{-q\left(\frac{2N+4-Nq}{2N+4}\right)}S^{\frac{2qN}{2N+4}}2^{\frac{2q(N+2)(2q(n+1)+1)}{2N+4}}Y^{1+\frac{2q}{2N+4}}_{n}\nonumber\\&=2^{q+\frac{4q^{2}(N+2)}{2N+4}}k^{-q\left(\frac{2N+4-Nq}{2N+4}\right)}S^{\frac{2qN}{2N+4}}2^{2nq^{2}}Y^{1+\frac{2q}{2N+4}}_{n}. \end{align}

By setting $\vartheta=2^{q+\frac{4q^{2}(N+2)}{2N+4}}k^{-q\left(\frac{2N+4-Nq}{2N+4}\right)}S^{\frac{2qN}{2N+4}}$, (5.10) reduces to

(5.11)\begin{equation} Y_{n+1}\leq \vartheta \left(2^{2q^{2}}\right)^{n}Y^{1+\frac{2q}{2N+4}}_{n}. \end{equation}

Invoking lemma 2.6, we derive $Y_{n}\to 0$ as $n\to \infty$ provided that

(5.12)\begin{equation} Y_{0}\leq \vartheta^{-\frac{2N+4}{2q}}2^{\frac{-(2N+4)^{2}}{2}}\Longleftrightarrow \int_{0}^{t}\int_{\mathbb R^N} u^{q}_{+}\,dx\leq k^{\frac{2N+4-Nq}{2}}\mu^{-\frac{2N+4}{2q}} 2^{\frac{-(2N+4)^{2}}{2}}, \end{equation}

where $\mu=2^{q+\frac{4q^{2}(N+2)}{2N+4}}S^{\frac{2qN}{2N+4}}$. It is clear for any $t\in (0,T)$ and whenever k satisfies

\begin{equation*} \left(\mu^{\frac{2N+4}{2q}} 2^{\frac{(2N+4)^{2}}{2}}\int_{0}^{t}\int_{\mathbb R^N} |u|^{q}\,dxds\right)^{\frac{2}{2N+4-Nq}}\leq k, \end{equation*}

the condition (5.12) holds. Thus, by selecting

\begin{equation*} k=\max\left\{\left(\mu^{\frac{2N+4}{2q}} 2^{\frac{(2N+4)^{2}}{2}}\int_{0}^{t}\int_{\mathbb R^N} |u|^{q}\,dxds\right)^{\frac{2}{2N+4-Nq}}, 2\|u_{0}\|_{\infty}\right\},\end{equation*}

we conclude that $(u-k)_{+}=0$ a.e. $(x,t)\in \mathbb{R}^{N}\times (0, T).$ In other words,

(5.13)\begin{align} \mathop{\text{ess\,sup}}\limits_{(x,t)\mathbb{R}^{N}\times (0,T)}u(x,t)\leq \max\left\{\!\left(\!\mu^{\frac{2N+4}{2q}} 2^{\frac{(2N+4)^{2}}{2}}\int_{0}^{T}\int_{\mathbb R^N} |u|^{q}\,dxdt\!\right)^{\frac{2}{2N+4-Nq}}, 2\|u_{0}\|_{\infty}\!\right\}.\nonumber\\ \end{align}

Arguing similarly as above, we obtain

\begin{align*} &\mathop{\text{ess\,inf}}\limits_{(x,t)\mathbb{R}^{N}\times (0,T)}u(x,t)\\ &\quad\geq - \max\left\{\left(\mu^{\frac{2N+4}{2q}} 2^{\frac{(2N+4)^{2}}{2}}\int_{0}^{T}\int_{\mathbb R^N} |u|^{q}\,dxdt\right)^{\frac{2}{2N+4-Nq}}, 2\|u_{0}\|_{\infty}\right\}. \end{align*}

Consequently, this combined with (5.13) yields (5.3). For N = 2, we adopt a similar approach as for $N\geq 3$, with the difference that in (5.7) we apply the Gagliardo–Nirenberg inequality (see [Reference Zheng35, Theorem 1.3.4]), which we state as follows

(5.14)\begin{equation} \int_{\mathbb R^N}|v|^{4}\,dx \leq S\left(\int_{\mathbb R^N}|\nabla v|^{2}\,dx\right) \left(\int_{\mathbb R^N}|v|^{2}\,dx\right),\;\;\;\forall v\in C^{\infty}_{0}(\mathbb{R}^{N}). \end{equation}

Proof. Assume there are two weak solutions of (1.1), denoted by u and v. Subtracting the identities (3.1) for v and u respectively, and setting $w=v-u$, we obtain

\begin{align*}&\int_{\mathbb{R}^{N}}w_{t}(t)\psi\,dx+\int_{\mathbb{R}^{N}}\nabla w(t)\cdot\nabla \psi\,dx+\alpha \int_{\mathbb{R}^N} \nabla w_{t}(t)\cdot\nabla \psi\,dx\\ &\quad=\int_{\mathbb{R}^N} \big(v(t)\log v^2(t)-u(t)\log u^2(t)\big)\psi\,dx,\end{align*}

for each $\psi\in W(\mathbb{R}^{N})$ and a.e. $t\in [0,T]$. Taking $\psi=w$ in the latter identity yields

(5.15)\begin{equation} \frac{1}{2}\partial_{t}\|w(t)\|^{2}_{H^{1}(\mathbb{R}^{N})}=\int_{\mathbb{R}^N} \big(v(t)\log v^2(t)-u(t)\log u^2(t)\big)w(t)\,dx, \end{equation}

for a.e. $t\in [0,T].$ The right-hand side of this equality can be estimated as follows

\begin{align*} &\int_{\mathbb{R}^N} \big(v(t)\log v^2(t)-u(t)\log u^2(t)\big)w(t)\,dx\\ &\quad=\int_{\mathbb{R}^N} \int_{0}^{1}\frac{d}{ds}\left(s\log |s|^{2}\right)|_{s=\theta v+(1-\theta)u}w^2(t)\,d\theta dx\\ &\quad=2\int_{\mathbb{R}^N} \int_{0}^{1}(\log |\theta v+(1-\theta)u|+1)w^2(t)\,d\theta dx\\ &\quad\leq 2\int_{\mathbb{R}^N} (\log \left(\|u\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}+\|v\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}\right)+1)w^2(t)\, dx. \end{align*}

By remark 3, we know that

\begin{equation*} \|u\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}+\|v\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))} \lt \infty. \end{equation*}

Substituting this into (5.15), we obtain

\begin{align*} \frac{1}{2}\partial_{t}\|w(t)\|^{2}_{H^{1}(\mathbb{R}^{N})} \leq 2(\log \left(\|u\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}+\|v\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}\right)+1)\|w(t)\|^{2}_{H^{1}(\mathbb{R}^{N})}. \end{align*}

Since $w(0)=0$, by Grönwall’s lemma, we conclude that $\|w(t)\|_{H^{1}(\mathbb{R}^{N})}=0$. Therefore, $v(t)=u(t)$ for all $t\in [0,T]$ and T > 0.

Proof. The proof of first part in theorem 3.2

For the case α = 0, where the initial energy is subcritical, this was already proved in [Reference Boudjeriou and Alves5, Theorem 2.7]. When α = 1, and the initial energy is subcritical, a similar argument as in [Reference Boudjeriou and Alves5, Theorem 2.7] can be applied to establish that

(6.1)\begin{align} \int_{0}^{t}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^N)}\,ds \lt d, \quad \|u(t)\|^2_{2}\leq 2E(u_{0}) \lt 2d, \quad \text{and}\quad & 0 \lt I(u(t)) \lt 2d, \nonumber\\ &\quad\;\;\forall t\geq 0. \end{align}

This implies that $u\in L^{2}(0, +\infty; H^{1}(\mathbb{R}^{N}))$. Next, by using (2.9), we have

\begin{align*} \|\nabla u(t)\|^2_{2}&=I(u(t))+\int_{\mathbb{R}^{N}} |u(t)|^2\log|u(t)|\,dx\\ &\leq I(u(t))+\frac{a^2}{\pi}\|\nabla u(t)\|^2_{2} +\|u(t)\|^2_{2}\log \big(e^{-N(1+\log a)}\|u(t)\|^2_{2}\big), \;\;\forall a \gt 0 \\ &\leq I(u(t))+\frac{a^2}{\pi}\|\nabla u(t)\|^2_{2} +\|u(t)\|^2_{2}\log \big(1+e^{-N(1+\log a)}\|u(t)\|^2_{2}\big), \;\;\forall a \gt 0. \end{align*}

Choosing $a=\sqrt{\frac{\pi}{2}}$, and combining this with (6.1), we obtain

(6.2)\begin{equation} \|\nabla u(t)\|^2_{2}\leq 4d+4d\log \big(1+e^{-N(1+\log\sqrt{\frac{\pi}{2}} )}2d\big), \;\;\;\forall t\geq 0. \end{equation}

From this, along with (2.9) and (2.2), we deduce

\begin{equation*} \sup_{t\in (0, +\infty)}\int_{\mathbb R^N} |u(t)|^2\log|u(t)|^2\,dx \lt \infty \quad \text{and} \quad \sup_{t\in (0, +\infty)} \int_{\mathbb{R}^N} B(|u(t)|)\,dx \lt \infty.\end{equation*}

Since $B(|u|)=|u|^2\log|u|^2+A(|u|)$, it follows that

\begin{equation*}\sup_{t\in (0, +\infty)} \int_{\mathbb{R}^N} A(|u(t)|)\,dx \lt \infty.\end{equation*}

Consequently, combining these results with lemma 5.1 yields $u\in C_{b}(0, +\infty; W(\mathbb{R}^N)).$

For the case where the initial energy is critical, the proof proceeds similarly once the following lemma is established.

Proof. Assume by contradiction that there exists $t_{*}\in (0, +\infty)$ such that

(6.3)\begin{equation} I(u(t_{*}))=0 \quad \text{and}\quad I(u(t)) \gt 0, \;\;\forall t\in [0, t_{*}). \end{equation}

Substituting v = u in (3.1), we obtain

(6.4)\begin{equation} -\langle u_{t}(t), u(t)\rangle=I(u(t)) \gt 0, \;\;\;\forall t\in [0, t_{*}). \end{equation}

This implies that

\begin{equation*} \|u(t)\|_{H^{1}(\mathbb{R}^{N})} \gt 0 \quad \text{and} \quad \|u_{t}(t)\|_{H^{1}(\mathbb{R}^{N})} \gt 0, \;\;\;\forall t\in [0, t_{*}).\end{equation*}

Hence, $\int_{0}^{t}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds$ is strictly positive for any $t\in [0, t_{*}]$. From (3.4), we derive

(6.5)\begin{equation} E(u(t))=E(u_{0})-\int_{0}^{t}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds \lt E(u_{0})=d, \;\;\;\forall t\in [0, t_{*}]. \end{equation}

On the other hand, by proposition 2.3, there exists $\lambda_{*}=\lambda_{*}(u) \gt 1$ such that

\begin{equation*} \lambda_{*}(u)=\exp\left(\frac{\|u\|^2_{H^{1}(\mathbb{R}^{N})}-\|u\|^{2}_{2}-\int_{\mathbb R^N} |u|^2\log |u|^2\,dx}{2\|u\|^2_{2}}\right) \quad \text{and}\quad I(\lambda_{*}(u)u)=0.\end{equation*}

Using this and the definition of d, we obtain

\begin{align*} 0 \lt d\leq E(\lambda_{*}(u(t))u(t))=E(\lambda_{*}(u(t))u(t))-\frac{1}{2}I(\lambda_{*}(u(t))u(t))&=\frac{\lambda^2_{*}(u(t))}{2}\|u(t)\|^2_{2},\\ &\quad \;\;\forall t\in [0, t_{*}).\end{align*}

From this, we infer that $u(t_{*})\neq 0$. This, combined with (6.3), implies that $d\leq E(u(t_{*}))$, which contradicts (6.5). This concludes the proof.

Proof of the third part of theorem 3.2

Taking v = u in (3.1) and integrating over (t, T), we obtain

(6.6)\begin{equation} \frac{1}{2} \|u(T)\|^2_{H^{1}(\mathbb{R}^{N})}-\frac{1}{2} \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}=-\int_{t}^{T}I(u(s))\,ds. \end{equation}

From above, we have $I(u(t)) \gt 0$ for any $t\geq 0$, which implies that $t\mapsto \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}$ is a decreasing function on $\mathbb{R}_{+}$. Next, using the logarithmic Sobolev inequality (2.9), we find

\begin{align*} I(u)&\geq \left(1-\frac{a^2}{\pi}\right) \|\nabla u\|^{2}_{2}+\left(N(1+\log a)-\log \|u\|^2_{2}\right)\|u\|^2_{2}\\ &\geq \left(1-\frac{a^2}{\pi}\right) \|\nabla u\|^{2}_{2}+\left(N(1+\log a)-\log\left( 2E(u_{0})\right)\right)\|u\|^2_{2}\\ &\geq \beta\|u\|^2_{H^{1}(\mathbb{R}^{N})}, \end{align*}

where $\beta =\min\left\{\left(1-\frac{a^2}{\pi}\right), \left(N(1+\log a)-\log\left( 2E(u_{0})\right)\right) \right\}$ and $\frac{(2E(u_{0}))^{1/N}}{e} \lt a \lt \sqrt{\pi}.$ Substituting this into (6.6) yields

\begin{equation*} \int_{t}^{T}\|u(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds\leq \frac{1}{2\beta}\|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}, \;\;\forall t\geq 0.\end{equation*}

Letting $T\to +\infty$, we get

(6.7)\begin{equation} \int_{t}^{+\infty}\|u(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds\leq \frac{1}{2\beta}\|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}, \;\;\forall t\geq 0. \end{equation}

Applying lemma 2.7, we derive

\begin{equation*} \|u(t)\|^{2}_{H^{1}(\mathbb{R}^{N})}\leq \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})}e^{1-2\beta t}, \;\;\;\forall t\geq 0. \end{equation*}

Using this result alongside the logarithmic Sobolev inequality (2.9), we obtain

\begin{equation*} \int_{\mathbb R^N} |u(t)|^{2}\log|u(t)|^{2}\,dx \leq \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})} \left(1+\log(1+\|u_0\|^2e^{1-2\beta t})\right)e^{1-2\beta t}. \end{equation*}

Moreover, setting $\epsilon=\frac{4}{N} $ and applying Hölder’s and Sobolev’s inequalities in connection with (2.2), we find

\begin{equation*} \int_{\mathbb{R}^{N}} B(|u(t)|)\,dx \precsim \|u_{0}\|^{2+\frac{4}{N}}_{H^{1}(\mathbb{R}^{N})}e^{\frac{N+2}{N}(1-2\beta t)}, \;\;\;\forall t\geq 0.\end{equation*}

By combining these results with the fact that $B(|u|)=A(|u|)+F(|u|)$, for all $t\geq 0$, we conclude

\begin{align*} \int_{\mathbb{R}^{N}} A(|u(t)|)\,dx &\precsim \|u_{0}\|^{2+\frac{4}{N}}_{H^{1}(\mathbb{R}^{N})}e^{\frac{2N+4}{N}(1-2\beta t)}+ \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})} \\ &\quad\times\left(1+\log(1+\|u_0\|^2e^{1-2\beta t})\right)e^{1-2\beta t}. \end{align*}

Proof of the fourth part of theorem 3.1

In light of lemma 6.1 and inequality (6.4), there exists $t_{\delta}\in (0, +\infty)$ such that $\int_{0}^{t_{\delta}}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds \gt 0$. Therefore, using (3.1), we obtain

\begin{align*} E(u(t))&\leq E(u(t_{\delta}))=E(u_{0})-\int_{0}^{t_{\delta}}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds\\ &= d-\int_{0}^{t_{\delta}}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds \lt d, \;\;\;\forall t\geq t_{\delta}.\end{align*}

Letting $\gamma=d-\int_{0}^{t_{\delta}}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds$ and by applying a similar reasoning as in the subcritical initial energy case, we obtain

\begin{equation*} \|u(t)\|^{2}_{H^{1}(\mathbb{R}^{N})}\leq \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})}e^{1-2\beta_{*} t}, \;\;\;\forall t\geq t_{\delta},\end{equation*}

where $\beta_{*} =\min\left\{\left(1-\frac{a^2}{\pi}\right), \left(N(1+\log a)-\log\left( 2\gamma \right)\right) \right\}$ and $\frac{(2 \gamma)^{1/N}}{e} \lt a \lt \sqrt{\pi}.$ Arguing similary as done for (3.14), for all $ t\geq t_{\delta}$, we have

\begin{align*} \int_{\mathbb{R}^{N}} A(|u(t)|)\,dx &\precsim \|u_{0}\|^{2+\frac{4}{N}}_{H^{1}(\mathbb{R}^{N})}e^{\frac{2N+4}{N}(1-2\beta_* t)}+ \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})} \\ &\quad\times\left(1+\log(1+\|u_0\|^2e^{1-2\beta_* t})\right)e^{1-2\beta_* t}. \end{align*}

Proof of the third part of theorem 3.1

According to lemma 5.3, it suffices to show that

\begin{equation*} \int_{0}^{+\infty}\int_{\mathbb R^N} |u(x,t)|^q\,dxdt \lt +\infty,\end{equation*}

where $2 \lt q \lt 2+\frac{4}{N}$ for $N\geq 3$ and $2 \lt q \lt 4$ for N = 2. We will focus on the case $N\geq 3$, as the case N = 2 can be handled in a similar manner. Following the approach outlined in remark 3, we obtain

(6.10)\begin{equation} \|u(t)\|^{q}_{q}\leq \|u(t)\|^{(1-\theta)q}_{2}\|u(t)\|^{\theta q}_{2+\frac{4}{N}}\precsim \|u(t)\|^q_{H^{1}(\mathbb{R}^N)}. \end{equation}

Integrating this over $(0, +\infty)$, and using (6.8) along with the fact that $u\in C_{b}(0, +\infty; W(\mathbb{R}^{N}))$, we obtain

\begin{align*} \int_{0}^{+\infty} \|u(t)\|^{q}_{q}\,dt&\precsim \int_{0}^{+\infty} \|u(t)\|^{q}_{H^1(\mathbb{R}^N)}\,dt=\int_{0}^{+\infty}\|u(t)\|^{2}_{H^1(\mathbb{R}^N)} \|u(t)\|^{q-2}_{H^1(\mathbb{R}^N)}\,dt\\ &\precsim \frac{\|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})}}{2\beta}\sup_{t\in (0, +\infty)}\|u(t)\|^{q-2}_{H^1(\mathbb{R}^N)} \lt +\infty. \end{align*}

From (6.10), we see that $u\in C([0,t_{\delta}]; L^{q}(\mathbb{R}^{N}))$. Proceeding as before, we obtain

\begin{align*} \int_{0}^{+\infty} \|u(t)\|^{q}_{q}\,dt&= \int_{0}^{t_{\delta}} \|u(t)\|^{q}_{q}\,dt+\int_{t_{\delta}}^{+\infty} \|u(t)\|^{q}_{q}\,dt\\ &\precsim \sup_{t\in [0,t_{\delta}]}\|u(t)\|_{q}^{q}+\frac{\|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})}}{2\beta_{*}}\sup_{t\in (0, +\infty)}\|u(t)\|^{q-2}_{H^1(\mathbb{R}^N)} \lt +\infty. \end{align*}

Combining these results, we conclude the proof of (3.10).

Proof of the case $E(u_{0}) \lt 0$

Substituting v = u into (3.1) and integrating over $(0,t)$, we obtain

(7.1)\begin{align} \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}-\|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})}&=-2\int_{0}^{t}I(u(s))\,ds=-2\int_{0}^{t}I(u(s))\,ds\\ &=-2\int_{0}^{t}\left(I(u(s))+\|u(s)\|^2_{2}\right)\,ds+2\int_{0}^{t}\|u(s)\|^2_{2}\,ds.\nonumber \end{align}

Since $t\mapsto E(u(t))=\frac{1}{2}\|u(t)\|^2_{2}+\frac{1}{2}I(u(t))$ is nonincreasing, we have

\begin{equation*} \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}\geq \|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})} -2t\left(I(u_{0})+\|u_{0}\|^2_{2}\right),\end{equation*}

which implies

\begin{equation*} \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}\geq \|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})} -2t\left(I(u_{0})+\|u_{0}\|^2_{2}\right), \;\;\;\;\forall t\geq 0.\end{equation*}

The upper bound can be obtained using (4.16) and taking the limit as $n\to \infty$ in (4.7) resulting in

\begin{align*}\|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})} -2t\left(I(u_{0})+\|u_{0}\|^2_{2}\right)&\leq\|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}\leq e^{N(1+\log \sqrt{\pi})}\\ &\quad\times\left(1+e^{-N(1+\log \sqrt{\pi})} \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})}\right)^{e^{2t}}, \;\forall t\geq 0.\end{align*}

From this, we infer that $\lim\limits_{t\to +\infty}\|u(t)\|_{H^{1}(\mathbb{R}^{N})}=+\infty$, and this concludes the proof.

Proof of the case $E(u_{0}) \lt d$ and $I(u_{0}) \lt 0$

The proof is based on the following lemmas that we establish.

Lemma 7.1. We have

\begin{equation*}d=\inf\left\{\frac{1}{2}\|u\|^2_{2}:\; u\in W(\mathbb{R}^N)\backslash \{0\}, \; I(u)\leq 0\right\}=d_{*}\end{equation*}

Proof. It is clear that $d_{*}\leq d$. To show $d\leq d_{*}$, it is sufficient to prove that

\begin{equation*} d\leq \frac{1}{2}\|u\|^2_{2}, \;\;\;\forall u\in W(\mathbb{R}^N)\backslash \{0\}, \; I(u) \lt 0.\end{equation*}

Indeed, let $u\in W(\mathbb{R}^N)\backslash \{0\}$ and $I(u) \lt 0$. Then, by proposition 2.3, there exists $\lambda_{*}\in (0,1)$ such that $I(\lambda_{*}u)=0$. In view of the definition of d, we have

\begin{equation*} d\leq E(\lambda_{*}u)=E(\lambda_{*}u)-\frac{1}{2}I(\lambda_{*}u)=\frac{\lambda^2_{*}}{2}\|u\|^2_{2} \lt \frac{1}{2}\|u\|^2_{2}.\end{equation*}

Hence, the proof is complete.

Proof. According to lemma 7.1, for any η > 0 we have

\begin{align*} d_{\eta}&=\inf\left\{E(u):\; u\in W(\mathbb{R}^N)\backslash \{0\}, \; I(u)=-\eta\right\}\\ &= \inf\left\{\frac{1}{2}\|u\|^2_{2}+\frac{-\eta}{2}:\; u\in W(\mathbb{R}^N)\backslash \{0\}, \; I(u)=-\eta\right\}\\ &= \frac{-\eta}{2}+\inf\left\{\frac{1}{2}\|u\|^2_{2}:\; u\in W(\mathbb{R}^N)\backslash \{0\}, \; I(u)=-\eta\right\}\\ &\geq \frac{-\eta}{2}+d. \end{align*}

Since $\min\left\{d-E(u_{0}), -I(u_{0})\right\} \gt 0$, there exists η > 0 such that $I(u_{0}) \lt -\eta$ and $d-E(u_{0}) \gt \eta$. This implies

(7.2)\begin{equation} E(u(t))\leq E(u_{0}) \lt \frac{-\eta}{2}+d \lt d_{\eta}, \;\;\forall t \gt 0. \end{equation}

To finish the proof, it’s enough to show that $I(u(t))\neq 0$ for all t > 0. Indeed, arguing by contradiction, suppose there exists $t_{0}\in (0, +\infty)$ such that $I(u(t_{0}))=-\beta$. This would imply $d_{\beta}\leq E(u(t_{0})) $. However, this contradicts (7.2), and thus completes the proof.

From (7.1) and (3.17), and using lemma 7.2, we derive for some β > 0,

\begin{align*} \|u_{0}\|^2_{H^1(\mathbb{R}^N)}+2\beta t&\leq \|u(t)\|^2_{H^1(\mathbb{R}^N)}\leq e^{N(1+\log \sqrt{\pi})}\\ &\quad\times\left(1+e^{-N(1+\log \sqrt{\pi})} \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})}\right)^{e^{2t}}, \;\forall t\geq 0.\end{align*}

Proof of the case $E(u_{0})=d$ and $I(u_{0}) \lt 0$

We first need to prepare the following lemma.

Proof. To show the first part, we argue by contradiction. Assume there exists $t_{*}$ such that $I(u(t_{*}))=0$ and $I(u(t)) \lt 0$ for all $t\in [0, t_{*})$. This implies

\begin{equation*} \|u_{t}(t)\|_{H^{1}(\mathbb{R}^{N})}\|u(t)\|_{H^{1}(\mathbb{R}^{N})}\geq \langle u_{t}(t), u(t)\rangle \gt 0\;\;\;\;\forall t\in [0, t_{*}).\end{equation*}

From this, we obtain $\int_{0}^{t}\|u_{s}(s)\|^2_{H^{1}(\mathbb{R}^{N})}\,ds \gt 0$ for all $t\in [0,t_{*}]$. Therefore, by (3.4), we derive

(7.3)\begin{equation} E(u(t)) \lt d, \;\;\;\forall t\in [0, t_{*}]. \end{equation}

By proposition 2.3, there exists $\lambda_{*}=\lambda_{*}(u) \lt 1$ such that

\begin{equation*} \lambda_{*}(u)=\exp\left(\frac{\|u\|^2_{H^{1}(\mathbb{R}^{N})}-\|u\|^{2}_{2}-\int_{\mathbb R^N} |u|^2\log |u|^2\,dx}{2\|u\|^2_{2}}\right) \quad \text{and}\quad I(\lambda_{*}(u)u)=0.\end{equation*}

Using the definition of d, we find

\begin{align*} 0 \lt d\leq E(\lambda_{*}(u(t))u(t))=E(\lambda_{*}(u(t))u(t))-\frac{1}{2}I(\lambda_{*}(u(t))u(t))&=\frac{1}{2}\|u(t)\|^2_{2},\\ & \;\;\forall t\in [0, t_{*}).\end{align*}

From this, it follows that $u(t_{*})\neq 0$. By the definition of d, we have $d\leq E(u(t_{*}))$, which contradicts (7.3). Hence, $I(u(t)) \lt 0$ for all $t\geq 0$. Further, arguing as we did previously, there exists $t_{0} \gt 0$ such that $E(u(t_{0})) \lt d$ and $I(u(t_{0})) \lt 0$. Invoking lemma 7.2, we deduce the existence of τ > 0 such that $I(u(t)) \lt -\tau$ for all $t\geq t_{0}.$

In view of (7.1) and lemma 7.3, we have

\begin{align*} \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}-\|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})}&=-2\int_{0}^{t}I(u(s))\,ds=-2\int_{0}^{t_{0}}I(u(s))\,ds\\ &\quad-2\int_{t_{0}}^{t}I(u(s))\,ds\\ &\geq \tau(t-t_{0}), \;\;\;\forall t\geq t_{0}. \end{align*}

Consequently, this completes the proof of (3.19).

Proof of the case $E(u_{0}) \lt \frac{\min\left\{\|u_{0}\|_{2}, \|u_{0}\|^2_{2}\right\}}{4\sqrt{e}}-\frac{1}{2}$

The proof will be complete upon establishing the following lemma.

Proof. We begin by proving that $I(u_{0}) \lt 0$. Suppose $\|u_{0}\|_{2}\leq 1$. Then $\frac{1}{2}\|u_{2}\|^2_{2}+\frac{1}{2}I(u_{0})=E(u_{0}) \lt \frac{\|u_{0}\|^2_{2}}{4\sqrt{e}}.$ Given $\frac{1}{4\sqrt{e}} \lt \frac{1}{2}$ it follows that $I(u_{0}) \lt 0$. Now, if $\|u_{0}\|_{2} \gt 1$, we have

\begin{equation*}\frac{1}{2}\|u_{2}\|^2_{2}+\frac{1}{2}I(u_{0})=E(u_{0}) \lt \frac{\|u_{0}\|_{2}}{4\sqrt{e}} \lt \frac{\|u_{0}\|^2_{2}}{4\sqrt{e}}.\end{equation*}

This implies

\begin{equation*}\left(\frac{1}{2}-\frac{1}{4\sqrt{e}}\right)\|u_{2}\|^2_{2}+\frac{1}{2}I(u_{0}) \lt 0.\end{equation*}

which leads to the conclusion that $I(u_{0}) \lt 0$. Since

\begin{equation*}\min\left\{-I(u_{0}), \frac{\min\left\{\|u_{0}\|_{2}, \|u_{0}\|^2_{2}\right\}}{4\sqrt{e}}-E(u_{0})-\frac{1}{2} \right\} \gt 0,\end{equation*}

there exists κ > 0 such that $I(u_{0}) \lt -\kappa$ and

(7.4)\begin{equation} \frac{1}{4}\|u_{0}\|^2_{2} \gt e\left(2E(u_{0})+1+\kappa\right)^2. \end{equation}

Next, we show that $I(u(t)) \lt -\kappa$ for all $t\geq 0$. We argue by contradiction. Assume there exists $t_{*} \gt 0$ such that $I(u(t_{*}))=-\kappa$ and $I(u(t)) \lt -\kappa$ for all $t\in [0, t_{*})$. From this, it follows that $t\mapsto \|u(t)\|^2_{H^{1}(\mathbb{R}^{N})}$ is increasing on $[0,t_{*}]$. Thus, (7.4) implies that

(7.5)\begin{equation} \|u(t_{*})\|^2_{H^{1}(\mathbb{R}^{N})}\geq \|u_{0}\|^2_{H^{1}(\mathbb{R}^{N})}\geq 4e\left(2E(u_{0})+1+\kappa\right)^2. \end{equation}

On the other hand, by the fact that $t\mapsto E(u(t))$ is non-increasing, we obtain

\begin{equation*} E(u_{0})\geq E(u(t_{*}))=\frac{\|u(t_{*})\|^2_{2}}{2}+\frac{I(u(t_{*}))}{2} =\frac{\|u(t_{*})\|^2_{2}}{2}-\frac{\kappa}{2}.\end{equation*}

This is equivalent to

\begin{equation*} \|u(t_{*})\|^2_{2}\leq 2E(u_{0})+ \kappa.\end{equation*}

Hence,

(7.6)\begin{equation} \|u(t_{*})\|^2_{2} \log \|u(t_{*}\|^2_{2})\leq \left(2E(u_{0})+1+ \kappa\right)\log \left(2E(u_{0})+1+ \kappa\right). \end{equation}

Using the logarithmic Sobolev inequality (2.9) and again using the fact that $t\mapsto E(u(t))$ is non-increasing, we find

\begin{align*} 2E(u_{0})+ 1+\kappa&\geq 2E(u(t_{*}))+ 1+\kappa\\ &\geq \frac{\|u(t_{*})\|^2_{H^{1}(\mathbb{R}^{N})}}{2} -\frac{a^2\|u(t_{*})\|^2_{H^{1}(\mathbb{R}^{N})}}{\pi }-\|u(t_{*})\|^2_{2}\log\|u(t_{*})\|^2_{2}\\ &\quad +N(1+\log(a))\|u(t_{*})\|^2_{2}\\ &\geq \frac{1}{4}\|u(t_{*})\|^2_{H^1(\mathbb{R}^N)}-\left(2E(u_{0})+ 1+\kappa\right)\log\left(2E(u_{0})+ 1+\kappa\right), \end{align*}

where we have chosen $a=\frac{\sqrt{\pi}}{2} $, and used (7.6). Combining this with (7.5), we find

\begin{equation*} 2E(u_{0})+1+ \kappa\geq e\left(2E(u_{0})+1+ \kappa\right)^2-\left(2E(u_{0})+1+ \kappa\right)\log\left(2E(u_{0})+1+ \kappa\right),\end{equation*}

This leads to

\begin{equation*} \log \left(2E(u_{0})+1+ \kappa\right)+1\geq e\left(2E(u_{0})+1+ \kappa\right),\end{equation*}

which equivalent to

\begin{equation*}\log \left(e\left(2E(u_{0})+1+ \kappa\right)\right)\geq e\left(2E(u_{0})+1+\kappa\right). \end{equation*}

However, this is a contradiction since $\log b \lt b$ for all b > 0. This completes the proof.

From (7.1), and using lemma 7.4, for some κ > 0, we derive

\begin{align*} \|u_{0}\|^2_{H^1(\mathbb{R}^N)}+2\kappa t&\leq \|u(t)\|^2_{H^1(\mathbb{R}^N)}\geq e^{N(1+\log \sqrt{\pi})}\\ &\quad\times\left(1+e^{-N(1+\log \sqrt{\pi})} \|u_{0}\|^{2}_{H^{1}(\mathbb{R}^{N})}\right)^{e^{2t}}, \;\;\forall t\geq 0.\end{align*}

Proof. Subtracting the identities in (8.1) and taking $w=(u-v)_{+}$, we obtain

(8.3)\begin{align} &\frac{1}{2}\partial_{t}\left(\int_{\mathbb R^N} (u(t)-v(t))^2_{+}\,dx+\alpha \int_{\mathbb R^N} |\nabla (u(t)-v(t))_{+}|^2\,dx\right)\nonumber\\&\quad+\int_{\mathbb R^N} |\nabla (u(t)-v(t))_{+}|^2\,dx\nonumber\\ &\leq \int_{\mathbb R^N} \left(u(t)\log |u(t)|^2-v(t)\log |v(t)|^2\right)(u(t)-v(t))_{+}\,dx. \end{align}

Note that the term on the right-hand side can be estimated as follows

\begin{align*} &\int_{\mathbb R^N} \left(u(t)\log |u(t)|^2-v(t)\log |v(t)|^2\right)(u(t)-v(t))_{+}\,dx\\ & =\int_{\mathbb R^N} 2(u(t)-v(t))\log |u(t)|+2v(t)\log\left(\left|1+\frac{u(t)-v(t)}{v(t)}\right|\right) (u(t)-v(t))_{+}\,dx\\ & \leq \int_{\mathbb R^N} 2(u(t)-v(t))\log |u(t)|+2|v(t)|\log\left(1+\left|\frac{u(t)-v(t)}{v(t)}\right|\right) (u(t)-v(t))_{+}\,dx\\ & \leq 2\int_{\mathbb R^N} (u(t)-v(t))_{+}\left(\log |u(t)|+1\right)\,dx\\ & \leq 2 \left(\log \|u\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}+1\right)\int_{\mathbb R^N} (u(t)-v(t))^2_{+}\,dx. \end{align*}

Combining this with (8.3) gives

\begin{align*} &\partial_{t}\left(\int_{\mathbb R^N} (u(t)-v(t))^2_{+}\,dx+\alpha \int_{\mathbb R^N} |\nabla (u(t)-v(t))_{+}|^2\,dx\right)\\ &\leq 4 \left(\log \|u\|_{L^{\infty}(\mathbb{R}^{N}\times (0,T))}+1\right)\int_{\mathbb R^N} ((u(t)-v(t))^2_{+}+|\nabla (u(t)-v(t))_{+}|^2)\,dx. \end{align*}

By applying Gronwall’s lemma and noting that $(u(0)-v(0))_{+}=0$, we infer the desired conclusion.

Proof of theorem 3.4

Set $v(x,t)=\varphi(x)g(t)$, where $\varphi(x)=\exp\left(\frac{N}{2}-\frac{|x|^2}{2}\right)$ and $g\in C^{1}(\mathbb{R}_{+})$. It is straightforward to check that v satisfies (1.1) for α = 0 if and only if

\begin{equation*}g'(t)= 2g(t)\log|g(t)|. \end{equation*}

Solving this (ODE) with the initial condition $g(0)=1+\delta$, we obtain $g(t)=e^{e^{2t}\log (1+\delta)}$. Thus,

\begin{equation*} v(x,t)=\exp\left(\frac{N}{2}+\frac{-|x|^2}{2}\right)e^{e^{2t}\log (1+\delta)}\end{equation*}

is a weak solution of (1.1) with the initial condition $v(x,0)=(1+\delta)\varphi(x)$. Assume,

\begin{equation*} v(x,0)\leq u(x,0), \;\;\;x\in \mathbb{R}^{N}.\end{equation*}

By applying lemma 8.1, we find

\begin{equation*} u(x,t)\geq \exp\left(\frac{N}{2}+\frac{-|x|^2}{2}\right)e^{e^{2t}\log (1+\delta)}, \;\;\text{a.e. }\;\;(x,t)\in \mathbb{R}^{N}\times (0,+\infty)\end{equation*}

This implies, for all t > 0

\begin{equation*}\|u(t)\|_{\infty}\geq e^{\frac{N}{2}} e^{e^{2t}\log (1+\delta)}, \end{equation*}

and

\begin{equation*}\|u(t)\|_{2}\geq e^{\frac{N}{2}}\pi^{\frac{N}{2}} e^{e^{2t}\log (1+\delta)}. \end{equation*}

Next, we suppose that

\begin{equation*} u(x,0)\leq (1-\delta)\varphi(x),\end{equation*}

for some δ > 0. Then, following a similar argument, we get

\begin{equation*} u(x,t)\leq \exp\left(\frac{N}{2}+\frac{-|x|^2}{2}\right)e^{e^{2t}\log (1-\delta)}, \;\;\text{a.e. }\;\;(x,t)\in \mathbb{R}^{N}\times (0,+\infty).\end{equation*}

This implies, for all t > 0

\begin{equation*}\|u(t)\|_{\infty}\leq e^{\frac{N}{2}} e^{e^{2t}\log (1-\delta)}, \end{equation*}

and

\begin{equation*}\|u(t)\|_{2}\leq e^{\frac{N}{2}}\pi^{\frac{N}{2}} e^{e^{2t}\log (1-\delta)}. \end{equation*}

To conclude the proof of theorem 3.4, we choose $u^{0}_{1}(x)=(1+\delta)\varphi(x)$, which satisfies for small $0 \lt \delta \lt 1$ that $ \|u^{0}_{1}-\varphi\|_{\infty}+\|u^{0}_{1}-\varphi\|_{2} \lt \epsilon$ for all ϵ > 0. Similarly, we choose $u^{0}_{2}=(1-\delta)\varphi(x)$. Thus, the proof is complete.

Proof of theorem 3.5

Let α = 1. We seek a subsolution of (1.1) in the form $v(x,t)=\varphi(x)h(t)$, where $h\in C^1(\mathbb{R}_{+})$ and $h, h' \gt 0$. Substituting this into Eq. (1.1) yields that $v(x,t)$ is a subsolution if and only if

(8.4)\begin{equation} \left(\varphi-\Delta\varphi \right)h'(t)\leq 2\varphi h(t) \log h(t). \end{equation}

Since $-\Delta \varphi(x)=(N-|x|^2)\varphi(x)$, it follows that

\begin{equation*} \left(\varphi+(N-|x|^2)\varphi\right)h'(t)\leq 2\varphi h(t) \log h(t), \end{equation*}

which simplifies to

\begin{equation*} \left(1+N-|x|^2)\right)\varphi h'(t)\leq 2\varphi h(t) \log h(t). \end{equation*}

To find h, it is enough to solve

\begin{equation*} (N+1)h'(t)=2 h(t) \log h(t).\end{equation*}

with $h(0)=1+\delta$. Solving this ODE, we obtain $h(t)=\exp\left(e^{\frac{2t}{N+1}\log(1+\delta)}\right)$. Suppose that

\begin{equation*} v(x,0)=(1+\delta)\varphi(x)\leq u(x,0).\end{equation*}

Then, it follows from lemma 8.1 that

\begin{equation*} u(x,t)\geq \exp\left(\frac{N}{2}+\frac{-|x|^2}{2}\right) \exp\left(e^{\frac{2t}{N+1}\log(1+\delta)}\right),\end{equation*}

which implies for all t > 0

\begin{equation*} \|u(t)\|_{\infty}\geq e^{\frac{N}{2}} \exp\left(e^{\frac{2t}{N+1}\log(1+\delta)}\right),\end{equation*}

and

\begin{equation*} \|u(t)\|_{2}\geq e^{\frac{N}{2}}\pi^{\frac{N}{2}} \exp\left(e^{\frac{2t}{N+1}\log(1+\delta)}\right).\end{equation*}

To finish the proof, we take $u_{0}(x)=(1+\delta)\varphi(x)$ for $0 \lt \delta \lt 1$ sufficiently small. This choice satisfies $ \|u_{0}-\varphi\|_{\infty}+\|u_{0}-\varphi\|_{2} \lt \epsilon$ for all ϵ > 0.

On the other hand, if $u_{0}\geq 1+\delta $, for some δ > 0, then $v(t)=e^{e^{2t}\log (1+\delta)}$ is a subsolution of (1.1). Therefore,

\begin{equation*} u(x,t)\geq e^{e^{2t}\log (1+\delta)}.\end{equation*}

Conversely, if $u_{0}\leq 1-\delta$, for some δ > 0 then $v(t)=e^{e^{2t}\log (1-\delta)}$ is a supersolution of (1.1). Thus,

\begin{equation*} u(x,t)\leq e^{e^{2t}\log (1-\delta)}.\end{equation*}