Proof of Theorem 2.1
As already mentioned, our arguments follow the rescaled test-function argument in [1]. Therefore we implement a proof by contradiction. The idea is to assume that there exists a global weak solution from an appropriate class, and then using a priori asymptotic bounds for this solution, we obtain a contradiction, since there is no nontrivial global weak solution. We combine the relevant identities and inequalities established in the previous section.
(I) Suppose that u is a global weak solution to (1.5). By Definition 2.1 (focusing on (2.2)), for all \(0< T<\infty \) and \(\varphi \in C^{2}(Q_{T})\) satisfying conditions (a) and (b) therein, we have the inequality
$$ \begin{aligned} & \int _{Q_{T}} \vert u \vert ^{p} \varphi \,dx \,dt + \int _{Q_{T}} w(x)\varphi \,dx \,dt+ \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1- \beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad \leq \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta } \Delta \varphi \bigr) \bigr\vert \,dx \,dt+ \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha }\varphi \bigr) \bigr\vert \,dx \,dt \\ &\qquad {} + \int _{Q_{T}} \bigl\vert \operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) \bigr\vert \vert u \vert \,dx \,dt. \end{aligned} $$
(4.1)
On the other hand, using Young’s inequality, we obtain three inequalities related to each term in the right-hand side of (4.1). For the first term, we have
$$ \begin{aligned} & \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta }\Delta \varphi \bigr) \bigr\vert \,dx \,dt \\ &\quad \leq \frac{1}{3} \int _{Q_{T}} \vert u \vert ^{p} \varphi \,dx \,dt+ C \int _{Q_{T}} \varphi ^{\frac{-1}{p-1}} \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta } \Delta \varphi \bigr) \bigr\vert ^{\frac{p}{p-1}} \,dx \,dt. \end{aligned} $$
(4.2)
Similarly, for the second term, we get
$$ \begin{aligned} & \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha } \varphi \bigr) \bigr\vert \,dx \,dt \\ &\quad \leq \frac{1}{3} \int _{Q_{T}} \vert u \vert ^{p} \varphi \,dx \,dt+ C \int _{Q_{T}} \varphi ^{\frac{-1}{p-1}} \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha } \varphi \bigr) \bigr\vert ^{\frac{p}{p-1}} \,dx \,dt \end{aligned} $$
(4.3)
and finally for the third term, we have
$$ \begin{aligned} & \int _{Q_{T}} \bigl\vert \operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) \bigr\vert \vert u \vert \,dx \,dt \\ &\quad \leq \frac{1}{3} \int _{Q_{T}} \vert u \vert ^{p} \varphi \,dx \,dt+ C \int _{Q_{T}} \varphi ^{\frac{-1}{p-1}} \bigl\vert \operatorname{div} \bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) \bigr\vert ^{\frac{p}{p-1}} \,dx \,dt. \end{aligned} $$
(4.4)
Starting from (4.1) and using inequality (4.2) together with the inequalities (4.3) and (4.4), we deduce that
$$ \begin{aligned} & \int _{Q_{T}} w(x)\varphi \,dx \,dt+ \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1-\beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad \leq C \bigl(I_{1}(\varphi )+I_{2}(\varphi )+I_{3}(\varphi ) \bigr), \end{aligned} $$
(4.5)
where
$$\begin{aligned}& I_{1}(\varphi ) = \int _{Q_{T}}\varphi ^{\frac{-1}{p-1}} \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta }\Delta \varphi \bigr) \bigr\vert ^{ \frac{p}{p-1}} \,dx \,dt, \end{aligned}$$
(4.6)
$$\begin{aligned}& I_{2}(\varphi ) = \int _{Q_{T}}\varphi ^{\frac{-1}{p-1}} \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha }\varphi \bigr) \bigr\vert ^{ \frac{p}{p-1}} \,dx \,dt, \\& I_{3}(\varphi ) = \int _{Q_{T}}\varphi ^{\frac{-1}{p-1}} \bigl\vert \operatorname{div} \bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) \bigr\vert ^{\frac{p}{p-1}} \,dx \,dt. \end{aligned}$$
(4.7)
Without loss of generality, consider now the test function
$$ \varphi (t,x)=F(t)G(x),\quad (t,x)\in Q_{T}, $$
(4.8)
where F and G are given, respectively, by (3.1) and (3.5) (with λ, \(L\gg 1\)). We can easily see that \(\varphi \in C^{2}(Q_{T})\) and it satisfies both conditions (a) and (b) of Definition 2.1. Hence the bound (4.5) holds for a function φ given by (4.8).
Now let us estimate the integrals \(I_{i}(\varphi )\), \(i=1,2,3\), always in the case that φ is given by (4.8). We have
$$ I_{1}(\varphi )= \biggl( \int _{0}^{T} F(t)^{\frac{-1}{p-1}} \bigl\vert \bigl(I_{T}^{1-\beta }F \bigr)'(t) \bigr\vert ^{\frac{p}{p-1}} \,dt \biggr) \biggl( \int _{\mathbb{R}^{N}} G(x)^{\frac{-1}{p-1}} \bigl\vert \Delta G(x) \bigr\vert ^{ \frac{p}{p-1}} \,dx \biggr). $$
(4.9)
Using (3.13) with \(\rho =1-\beta \), we obtain
$$ \int _{0}^{T} F(t)^{\frac{-1}{p-1}} \bigl\vert \bigl(I_{T}^{1-\beta }F \bigr)'(t) \bigr\vert ^{\frac{p}{p-1}} \,dt \leq C T^{ \frac{(1-\beta )p-1}{p-1}}. $$
(4.10)
Next, using Lemma 3.3 with \(\theta =0\), we have that
$$ \int _{\mathbb{R}^{N}} G(x)^{\frac{-1}{p-1}} \bigl\vert \Delta G(x) \bigr\vert ^{ \frac{p}{p-1}} \,dx\leq C T^{\xi (N-\frac{2p}{p-1} )}. $$
(4.11)
Therefore (4.9), (4.10), and (4.11) yield the estimate
$$ I_{1}(\varphi )\leq C T^{\xi N+\frac{(1-\beta )p-1-2\xi p}{p-1}}. $$
(4.12)
Proceeding in a similar way, we now estimate \(I_{2}(\varphi )\). Indeed, considering (4.8), we get the identity
$$ I_{2}(\varphi )= \biggl( \int _{0}^{T} F(t)^{\frac{-1}{p-1}} \bigl\vert \bigl(I_{T}^{1-\alpha }F \bigr)'(t) \bigr\vert ^{\frac{p}{p-1}} \,dt \biggr) \biggl( \int _{\mathbb{R}^{N}} G(x) \,dx \biggr). $$
(4.13)
Using (3.13) with \(\rho =1-\alpha \), we obtain
$$ \int _{0}^{T} F(t)^{\frac{-1}{p-1}} \bigl\vert \bigl(I_{T}^{1-\alpha }F \bigr)'(t) \bigr\vert ^{\frac{p}{p-1}} \,dt \leq C T^{ \frac{(1-\alpha )p-1}{p-1}}, $$
(4.14)
and hence using (4.13) and (4.14) together with Lemma 3.2, we deduce that
$$ I_{2}(\varphi )\leq C T^{\xi N+\frac{(1-\alpha )p-1}{p-1}}. $$
(4.15)
Employing a similar argument as above, by (4.8) we retrieve the identity
$$ I_{3}(\varphi )= \biggl( \int _{0}^{T} F(t) \,dt \biggr) \biggl( \int _{ \mathbb{R}^{N}} G(x)^{\frac{-1}{p-1}} \bigl\vert \operatorname{div} \bigl( \vert x \vert ^{\theta }\nabla G\bigr) \bigr\vert ^{\frac{p}{p-1}} \,dx \biggr). $$
(4.16)
Then by Lemma 3.3 and identity (3.14), we deduce for (4.16) the estimate
$$ I_{3}(\varphi )\leq C T^{\xi (N+\frac{(\theta -2)p}{p-1} )+1}. $$
(4.17)
Therefore it follows from (4.5), (4.12), (4.15), and (4.17) that
$$ \begin{aligned} & \int _{Q_{T}} w(x)\varphi \,dx \,dt+ \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1-\beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad \leq C \bigl( T^{\xi N+\frac{(1-\beta )p-1-2\xi p}{p-1}}+T^{\xi N+ \frac{(1-\alpha )p-1}{p-1}}+T^{\xi (N+\frac{(\theta -2)p}{p-1} )+1} \bigr). \end{aligned} $$
(4.18)
On the other hand, by (3.5), (4.8), and (3.14) we have
$$ \begin{aligned} \int _{Q_{T}} w(x)\varphi \,dx \,dt &= \biggl( \int _{0}^{T}F(t) \,dt \biggr) \biggl( \int _{\mathbb{R}^{N}}w(x)G(x) \,dx \biggr) \\ &= C T \int _{\mathbb{R}^{N}}w(x) \psi \biggl( \frac{ \vert x \vert ^{2}}{T^{2\xi }} \biggr)^{L} \,dx. \end{aligned} $$
(4.19)
Notice that since \(w\in L^{1}(\mathbb{R}^{N})\), by the dominated convergence theorem and properties of the cut-off function in (3.4), we have the asymptotic behavior
$$ \lim_{T\to +\infty } \int _{\mathbb{R}^{N}}w(x) \psi \biggl( \frac{ \vert x \vert ^{2}}{T^{2\xi }} \biggr)^{L} \,dx= \int _{\mathbb{R}^{N}}w(x) \,dx. $$
Since \(\int _{\mathbb{R}^{N}}w(x) \,dx>0\), we deduce that for sufficiently large T,
$$ \int _{\mathbb{R}^{N}}w(x) \psi \biggl(\frac{ \vert x \vert ^{2}}{T^{2\xi }} \biggr)^{L} \,dx\geq C \int _{\mathbb{R}^{N}}w(x) \,dx. $$
(4.20)
Again for sufficiently large T, combining (4.19) and (4.20), it follows that
$$ \int _{Q_{T}} w(x)\varphi \,dx \,dt\geq CT \int _{\mathbb{R}^{N}}w(x) \,dx. $$
(4.21)
On the other hand, using (4.8) and (3.2) with \(\rho \in \{1-\alpha ,1-\beta \}\), we obtain
$$ \begin{aligned} & \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1- \beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad = \int _{\mathbb{R}^{N}} u_{0}(x) \bigl( \bigl(I_{T}^{1-\alpha }F\bigr) (0) G(x)-\bigl(I_{T}^{1- \beta }F \bigr) (0)\Delta G(x) \bigr) \,dx \\ &\quad = \int _{\mathbb{R}^{N}} u_{0}(x) \biggl( \frac{\Gamma (\lambda +1)}{\Gamma (2-\alpha +\lambda )}T^{1-\alpha } G(x)- \frac{\Gamma (\lambda +1)}{\Gamma (2-\beta +\lambda )}T^{1-\beta } \Delta G(x) \biggr) \,dx \\ &\quad =\mu _{T}(u_{0}) T, \end{aligned} $$
(4.22)
where
$$ \mu _{T}(u_{0})= \Gamma (\lambda +1) \int _{\mathbb{R}^{N}} u_{0}(x) \biggl( \frac{1}{\Gamma (2-\alpha +\lambda )}T^{-\alpha } G(x)- \frac{1}{\Gamma (2-\beta +\lambda )}T^{-\beta } \Delta G(x) \biggr) \,dx. $$
We claim that
$$ \lim_{T\to +\infty } \mu _{T}(u_{0})=0. $$
(4.23)
Indeed, by (3.4) and (3.5), since \(u_{0}\in L^{1}(\mathbb{R}^{N})\), we have
$$ \bigl\vert \mu _{T}(u_{0}) \bigr\vert \leq C \biggl(T^{-\alpha } \int _{\mathbb{R}^{N}} \bigl\vert u_{0}(x) \bigr\vert \,dx+ T^{-\beta } \int _{T^{\xi }< \vert x \vert < \sqrt{2} T^{\xi }} \bigl\vert u_{0}(x) \bigr\vert \bigl\vert \Delta G(x) \bigr\vert \,dx \biggr). $$
On the other hand, using (3.4) and (3.11) with \(\theta =0\), we obtain
$$ \bigl\vert \Delta G(x) \bigr\vert \leq C T^{-2\xi } \psi \biggl( \frac{ \vert x \vert ^{2}}{T^{2\xi }} \biggr)^{L-2}\leq C T^{-2\xi },\quad T^{\xi }< \vert x \vert < \sqrt{2} T^{\xi} $$
(recall that \(L\gg 1\)). Hence we have the estimate
$$ \bigl\vert \mu _{T}(u_{0}) \bigr\vert \leq C \biggl(T^{-\alpha } \int _{\mathbb{R}^{N}} \bigl\vert u_{0}(x) \bigr\vert \,dx+ T^{-\beta -2\xi } \int _{T^{\xi }< \vert x \vert < \sqrt{2} T^{\xi }} \bigl\vert u_{0}(x) \bigr\vert \,dx \biggr). $$
Therefore, passing to the limit as \(T\to +\infty \) in the above inequality, the result in (4.23) is established. If we combine appropriately inequalities (4.18), (4.21), and (4.22), we get
$$ \begin{aligned} &T \biggl( \int _{\mathbb{R}^{N}}w(x) \,dx+\mu _{T}(u_{0}) \biggr) \\ & \quad \leq C \bigl( T^{\xi N+\frac{(1-\beta )p-1-2\xi p}{p-1}}+T^{\xi N+ \frac{(1-\alpha )p-1}{p-1}}+T^{\xi (N+\frac{(\theta -2)p}{p-1} )+1} \bigr), \end{aligned} $$
and hence
$$ \int _{\mathbb{R}^{N}}w(x) \,dx+\mu _{T}(u_{0}) \leq C \bigl(T^{A_{1}( \xi )}+T^{A_{2}(\xi )}+T^{A_{3}(\xi )} \bigr), $$
(4.24)
where we set
$$\begin{aligned}& A_{1}(\xi ) = \xi N+\frac{(1-\beta )p-1-2\xi p}{p-1}-1, \end{aligned}$$
(4.25)
$$\begin{aligned}& A_{2}(\xi ) =\xi N+\frac{(1-\alpha )p-1}{p-1}-1, \\& A_{3}(\xi ) =\xi \biggl(N+\frac{(\theta -2)p}{p-1} \biggr). \end{aligned}$$
(4.26)
Now we take \(\xi >0\) such that
$$ \xi \biggl(N-\frac{2p}{p-1} \biggr)< \frac{\beta p}{p-1} \quad \mbox{and} \quad \xi < \frac{\alpha p}{N(p-1)}. $$
(4.27)
Notice that under the above conditions, we have \(A_{i}(\xi )<0\), \(i=1,2\). Moreover, if \(\theta \leq 2-N\), then \(A_{3}(\xi )<0\) for all \(p>1\). Hence, passing to the limit as \(T\to +\infty \) in (4.24) and using (4.23), we obtain a contradiction to the positivity condition \(\int _{\mathbb{R}^{N}}w(x) \,dx>0\). Therefore we deduce that for all \(p>1\), problem (1.5) admits no global weak solution. This proves part (I)-(i) of Theorem 2.1.
Next, if \(\theta >2-N\), then we observe that \(A_{3}(\xi )<0\) for all \(1< p<\frac{N}{\theta -2+N}\). So, proceeding as in the previous case, we arrive again at contradiction to \(\int _{\mathbb{R}^{N}}w(x) \,dx>0\). Thus part (I)-(ii) of Theorem 2.1 is also established.
Now we focus on the second part of the theorem.
(II) We assume the restrictions
$$ \frac{1}{p-1}< \delta < \frac{N-2+\theta }{2-\theta } $$
(4.28)
and
$$ 0< \varepsilon < \bigl[(2-\theta )\tau \bigr]^{\frac{1}{p-1}}, $$
(4.29)
where
$$ \tau =N-(2-\theta ) (\delta +1). $$
(4.30)
Notice that since \(0\leq \theta <2\), \(\theta >2-N\), and \(p>\frac{N}{\theta -2+N}\), the set of values δ satisfying (4.28) and the set of values ε satisfying (4.29) are nonempty. Without loss of generality, consider the function
$$ u(x)=\varepsilon \bigl(1+r^{2-\theta } \bigr)^{-\delta }, \quad x\in \mathbb{R}^{N}, \vert x \vert =r. $$
(4.31)
An elementary calculation shows that
$$ \begin{aligned} &{-}\operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla u\bigr) \\ &\quad = -r^{1-N} \partial _{r} \bigl(r^{N-1+\theta }\partial _{r} u \bigr) \\ &\quad = \varepsilon (2-\theta ) \bigl[N \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}-(2-\theta ) (\delta +1)r^{2-\theta } \bigl(1+r^{2-\theta } \bigr)^{-\delta -2} \bigr] . \end{aligned} $$
(4.32)
Let
$$ w(x)=-\operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla u\bigr)- \bigl\vert u(x) \bigr\vert ^{p},\quad x\in \mathbb{R}^{N}, $$
(4.33)
so that by (4.31) and (4.32) we obtain
$$ \begin{aligned} w(x)&= \varepsilon (2-\theta ) \bigl[N \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}-(2-\theta ) (\delta +1)r^{2-\theta } \bigl(1+r^{2-\theta } \bigr)^{-\delta -2} \bigr] \\ &\quad {} -\varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p} \\ &\geq \varepsilon (2-\theta ) \bigl[N \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}-(2-\theta ) (\delta +1) \bigl(1+r^{2-\theta } \bigr)^{- \delta -1} \bigr] \\ & \quad {}-\varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p} \\ &=\varepsilon (2-\theta ) \tau \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}- \varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p}. \end{aligned} $$
Next, using (4.28) and (4.29), we get the positivity condition
$$ \begin{aligned} \varepsilon ^{-1}w(x) &\geq (2-\theta ) \tau \bigl(1+r^{2-\theta } \bigr)^{-\delta -1}-\varepsilon ^{p-1} \bigl(1+r^{2-\theta } \bigr)^{- \delta p} \\ &\geq (2-\theta ) \tau \bigl[ \bigl(1+r^{2-\theta } \bigr)^{-\delta -1}- \bigl(1+r^{2-\theta } \bigr)^{-\delta p} \bigr] \\ &> 0. \end{aligned} $$
This shows that for all δ and ε satisfying, respectively, (4.28) and (4.29), the function u defined by (4.31) is a stationary solution (and hence a global solution) to (1.5), where \(u_{0}(x)=\varepsilon (1+r^{2-\theta } )^{-\delta }>0\), and \(w>0\) is given by (4.33). This proves part (II) of Theorem 2.1. □
Next, we give the proof of the second main result of this manuscript (i.e., we consider (1.1) when \(\iota >0\)).
Proof of Theorem 2.2
We construct the proof following a similar strategy to the previous proof, and hence we aim to obtain a contradiction to the assumption that there exists a global weak solution to problem (2.4). We provide the precise details as follows.
(I) Suppose that u is a global weak solution to (2.4). We first consider the case
$$ 0\leq \theta < 2 \quad \mbox{and} \quad 1< p< p^{*}(N,\theta ). $$
From (2.5) we deduce that u solves (4.1) for all \(0< T<\infty \) and \(\varphi \in C^{2}(Q_{T})\), \(\varphi \geq 0\), satisfying conditions (a) and (b) of Definition 2.1. Hence in this case the proof is the same as that of part (I) of Theorem 2.1. Consider now the case
$$ 0\leq \theta < 1 \quad \mbox{and}\quad p>1,\quad 1< q< q^{*}(N,\theta ). $$
(4.34)
From (2.5), for all \(0< T<\infty \), we have
$$\begin{aligned} &\int _{Q_{T}} \bigl( \vert u \vert ^{p}+ \vert \nabla u \vert ^{q} \bigr) \varphi \,dx \,dt + \int _{Q_{T}} w(x)\varphi \,dx \,dt \\ &\qquad {} + \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1-\beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad = \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta } \Delta \varphi \bigr) \bigr\vert \,dx \,dt+ \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha }\varphi \bigr) \bigr\vert \,dx \,dt \\ &\qquad {} - \int _{Q_{T}} \operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) u \,dx \,dt, \end{aligned}$$
(4.35)
where φ is the test function given by (4.8). On the other hand, integrating by parts, we obtain
$$ - \int _{Q_{T}} \operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) u \,dx \,dt= \int _{Q_{T}} \vert x \vert ^{\theta }\nabla \varphi \cdot \nabla u \,dx \,dt, $$
which yields (by the Cauchy–Schwarz inequality)
$$ - \int _{Q_{T}} \operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla \varphi \bigr) u \,dx \,dt\leq \int _{Q_{T}} \vert x \vert ^{\theta } \vert \nabla \varphi \vert \vert \nabla u \vert \,dx \,dt. $$
Hence by (4.35) we obtain the identity
$$ \begin{aligned} & \int _{Q_{T}} \bigl( \vert u \vert ^{p}+ \vert \nabla u \vert ^{q} \bigr) \varphi \,dx \,dt + \int _{Q_{T}} w(x)\varphi \,dx \,dt \\ &\qquad {} + \int _{\mathbb{R}^{N}} u_{0}(x) \bigl(I_{T}^{1-\alpha } \varphi (0,x)-I_{T}^{1-\beta }\Delta \varphi (0,x) \bigr) \,dx \\ &\quad = \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\beta } \Delta \varphi \bigr) \bigr\vert \,dx \,dt+ \int _{Q_{T}} \vert u \vert \bigl\vert \partial _{t} \bigl(I_{T}^{1-\alpha }\varphi \bigr) \bigr\vert \,dx \,dt \\ &\qquad {} + \int _{Q_{T}} \vert x \vert ^{\theta } \vert \nabla \varphi \vert \vert \nabla u \vert \,dx \,dt. \end{aligned} $$
(4.36)
On the other hand, by Young’s inequality we have
$$ \begin{aligned} & \int _{Q_{T}} \vert x \vert ^{\theta } \vert \nabla \varphi \vert \vert \nabla u \vert \,dx \,dt \\ &\quad \leq \varepsilon \int _{Q_{T}} \vert \nabla u \vert ^{q} \varphi \,dx \,dt +C \int _{Q_{T}} \vert x \vert ^{\frac{\theta q}{q-1}}\varphi ^{\frac{-1}{q-1}} \vert \nabla \varphi \vert ^{\frac{q}{q-1}} \,dx \,dt, \end{aligned} $$
(4.37)
where \(0<\varepsilon <1\). Therefore, using appropriately (4.36), (4.37), (3.10), (4.3), (4.21), and (4.22), we deduce that
$$ T \biggl( \int _{\mathbb{R}^{N}}w(x) \,dx+\mu _{T}(u_{0}) \biggr)\leq C \bigl(I_{1}(\varphi )+I_{2}(\varphi )+J( \varphi ) \bigr), $$
(4.38)
where \(I_{1}(\varphi )\) and \(I_{2}(\varphi )\) are given, respectively, by (4.6) and (4.7), and
$$ J(\varphi )= \int _{Q_{T}} \vert x \vert ^{\frac{\theta q}{q-1}}\varphi ^{ \frac{-1}{q-1}} \vert \nabla \varphi \vert ^{\frac{q}{q-1}} \,dx \,dt. $$
Referring to (3.14) and (4.8), by Lemma 3.4 we deduce that
$$\begin{aligned} J(\varphi ) &= \biggl( \int _{0}^{T} F(t) \,dt \biggr) \biggl( \int _{ \mathbb{R}^{N}} \vert x \vert ^{\frac{\theta q}{q-1}}G(x)^{\frac{-1}{q-1}} \bigl\vert \nabla G(x) \bigr\vert ^{\frac{q}{q-1}} \,dx \biggr) \\ &\leq C T^{1+\xi (\frac{q}{q-1}(\theta -1)+N )}, \end{aligned}$$
(4.39)
Therefore it follows from (4.38), (4.39), (4.12), and (4.15) that
$$ \begin{aligned} &T \biggl( \int _{\mathbb{R}^{N}}w(x) \,dx+\mu _{T}(u_{0}) \biggr) \\ &\quad \leq C \bigl(T^{\xi N+\frac{(1-\beta )p-1-2\xi p}{p-1}}+T^{\xi N+ \frac{(1-\alpha )p-1}{p-1}}+T^{1+\xi (\frac{q}{q-1}(\theta -1)+N )} \bigr), \end{aligned} $$
and hence
$$ \int _{\mathbb{R}^{N}}w(x) \,dx+\mu _{T}(u_{0}) \leq C \bigl(T^{A_{1}( \xi )}+T^{A_{2}(\xi )}+T^{B(\xi )} \bigr), $$
(4.40)
where \(A_{1}(\xi )\) and \(A_{2}(\xi )\) are given respectively by (4.25) and (4.26), and
$$ B(\xi )=\xi \biggl(\frac{q}{q-1}(\theta -1)+N \biggr). $$
Now we take \(\xi >0\) such that (4.27) is satisfied, which guarantees that \(A_{i}(\xi )<0\), \(i=1,2\). On the other hand, (4.34) guarantees that \(B(\xi )<0\). Hence, passing to the limit as \(T\to +\infty \) in (4.40) and using (4.23), we obtain a contradiction to the positivity condition \(\int _{\mathbb{R}^{N}}w(x) \,dx>0\). This proves part (I) of Theorem 2.2.
(II) We consider the ranges
$$ \max \biggl\{ \frac{1}{p-1},\frac{(1-\theta )q}{(2-\theta )(q-1)}-1 \biggr\} < \delta < \frac{N+\theta -2}{2-\theta } $$
(4.41)
and
$$ 0< \varepsilon < \min \biggl\{ \biggl[\frac{(2-\theta )\tau }{2} \biggr]^{ \frac{1}{p-1}}, \biggl[\frac{(2-\theta )^{1-q}\delta ^{-q}\tau }{2} \biggr]^{\frac{1}{q-1}} \biggr\} , $$
(4.42)
where τ is given by (4.30). Notice that since \(0\leq \theta <1\), \(\theta >2-N\), \(p> \frac{N}{\theta -2+N}\), and \(q> \frac{N}{\theta +N-1}\), the set of values δ satisfying (4.41) and the set of values ε satisfying (4.42) are nonempty. Let u be the function defined by (4.31) and set
$$ w(x)=-\operatorname{div}\bigl( \vert x \vert ^{\theta }\nabla u\bigr)- \bigl\vert u(x) \bigr\vert ^{p}- \vert \nabla u \vert ^{q},\quad x \in \mathbb{R}^{N}. $$
(4.43)
An elementary calculation shows that
$$ \vert \nabla u \vert ^{q}=\varepsilon ^{q} \delta ^{q}(2-\theta )^{q} r^{(1- \theta )q} \bigl(1+r^{2-\theta }\bigr)^{-(\delta +1)q},\quad r= \vert x \vert . $$
Hence, using (4.31) and (4.32), we obtain
$$\begin{aligned} w(x)&= \varepsilon (2-\theta ) \bigl[N \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}-(2-\theta ) (\delta +1)r^{2-\theta } \bigl(1+r^{2-\theta } \bigr)^{-\delta -2} \bigr] \\ &\quad {} -\varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p}- \varepsilon ^{q} \delta ^{q}(2-\theta )^{q} r^{(1-\theta )q} \bigl(1+r^{2- \theta }\bigr)^{-(\delta +1)q} \\ &\geq \varepsilon (2-\theta ) \bigl[N \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}-(2-\theta ) (\delta +1) \bigl(1+r^{2-\theta } \bigr)^{- \delta -1} \bigr] \\ &\quad {} -\varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p}- \varepsilon ^{q} \delta ^{q}(2-\theta )^{q} \bigl(1+r^{2-\theta }\bigr)^{- ( \frac{1}{2-\theta }+\delta )q} \\ &= \varepsilon (2-\theta ) \tau \bigl(1+r^{2-\theta } \bigr)^{- \delta -1}- \varepsilon ^{p} \bigl(1+r^{2-\theta } \bigr)^{-\delta p} \\ &\quad {} -\varepsilon ^{q} \delta ^{q}(2-\theta )^{q} \bigl(1+r^{2-\theta }\bigr)^{- (\frac{1}{2-\theta }+\delta )q}. \end{aligned}$$
Next, using (4.41) and (4.42), we deduce that
$$ \begin{aligned} \varepsilon ^{-1}w(x)&\geq (2-\theta ) \tau \bigl(1+r^{2-\theta } \bigr)^{-\delta -1}-\varepsilon ^{p-1} \bigl(1+r^{2-\theta } \bigr)^{- \delta p} \\ &\quad {} -\varepsilon ^{q-1} \delta ^{q}(2-\theta )^{q} \bigl(1+r^{2-\theta }\bigr)^{- (\frac{1}{2-\theta }+\delta )q} \\ &> (2-\theta ) \tau \bigl(1+r^{2-\theta } \bigr)^{-\delta -1}- \frac{(2-\theta )\tau }{2} \bigl(1+r^{2-\theta } \bigr)^{-\delta -1} \\ &\quad {} -\frac{(2-\theta )\tau }{2}\bigl(1+r^{2-\theta }\bigr)^{-\delta -1} \\ &=0. \end{aligned} $$
This shows that for all δ and ε satisfying, respectively, (4.41) and (4.42), the function u defined by (4.31) is a stationary solution (and hence a global solution) to (2.4), where \(u_{0}(x)=\varepsilon (1+r^{2-\theta } )^{-\delta }>0\), and \(w>0\) is given by (4.43). This proves part (II) of Theorem 2.2. □