([21]). If $u_0\left(x\right)\ge 0$, $u_0\left(x\right)\neg \equiv 0$, then $\left[P_{\alpha }\left(t\right)u_0\right]\left(x\right)>0$, $\left[S_{\alpha }\left(t\right)u_0\right]\left(x\right)>0$ and

See Lemmas 2.1 (a) and 2.2 (a) in [21]. □

([21]). Let $1\le p\le q\le +\infty$ and $1/r=1/p-1/q$.

See Lemmas 2.1 (b) and 2.2 (b) in [21]. □

Let $u_0\in C_0\left({\mathbf{R}}^n\right)$ and $T>0$. We call that $u\in C([0,T],C_0\left({\mathbf{R}}^n\right))$ is a mild solution of the problem (1) if u satisfies the integral equation

Proposition 1 (Theorem 3.2 in [21]). Let $0<\alpha <1$. For given $u_0\in C_0\left({\mathbf{R}}^n\right)$, there exists a maximal existence time $T^{\ast }>0$ such that the problem (1) has a unique mild solution $u\in C([0,T^{\ast }],C_0\left({\mathbf{R}}^n\right))$ and either $T^{\ast }=+\infty$ or $T^{\ast }<+\infty$ and ${\parallel u\parallel }_{L^{\infty }((0,t),C_0\left({\mathbf{R}}^n\right))}\to \infty$. In addition, if $u_0\left(x\right)\ge 0$ and $u_0\left(x\right)\neg \equiv 0$, then $u(x,t)\ge \left[P_{\alpha }\left(t\right)u_0\right]\left(x\right)>0$ for $t\in (0,T^{\ast })$. Moreover, if $u_0\in L^r\left({\mathbf{R}}^n\right)$ for some $r\in [1,\infty )$, then $u\in C([0,T^{\ast }),L^r\left({\mathbf{R}}^n\right))$.

Proposition 2 (Theorem 4.3 in [21]). Let $0<\alpha <1$, $u_0\in C_0\left({\mathbf{R}}^n\right)$ and $u_0\left(x\right)\ge 0$. If

Proposition 3 (Theorem 4.4 in [21]). Let $0<\alpha <1$, $u_0\in C_0\left({\mathbf{R}}^n\right)$, $u_0\left(x\right)\ge 0$ and $u_0\left(x\right)\neg \equiv 0$.

We take the similar strategy as Theorem 1 in [20] and Theorem 3.7 in [22], using the Kaplan’s first eigenvalue method [23].

Let for a sequence ${\left\{x_m\right\}}_{m=1}^{\infty }$ satisfying $|x_m|=m$ for any $m\in \mathbf{N}$.

Let ${\lambda }_m>0$ denote the principal eigenvalue of $-\Delta$ with Dirichlet problem in $B_m$, and let ${\varphi }_m\left(x\right)$ denote the corresponding eigenfunction, normalized by Let $T\in (0,T^{\ast })$ be arbitrarily fixed. We define It follows from (1) that supplemented with the initial condition By integrating by parts, and the fact that ${\varphi }_m\left(x\right)=0$ and $\partial {\varphi }_m/\partial \nu \le 0$ on $\partial B_m$, where $\nu$ denotes the outward unit normal vector to $B_m$ at $x\in \partial B_m$, and applying Green’s formula, we have Since the principal eigenvalue ${\lambda }_m>0$ and the eigenfunction ${\varphi }_m\left(x\right)$ satisfy we obtain By (15), (16) and Hölder’ s inequality, we have So, we obtain Using (16) and (19) in (18), it yields Since $B_m$ is an n-dimensional ball of radius ${\displaystyle \frac{1}{2}}m$, it follows that ${\lambda }_m$ satisfies where $c_1>0$ depends only on the dimension n. Thus, we have Setting $H\left(\zeta \right):=-c_1{m}^{-2}\zeta +{\zeta }^p$, then the function $H\left(\zeta \right)$ is convex in $\zeta >0$ since $H\in C^2(0,\infty )$ and $H^{\prime \prime }\left(\zeta \right)\ge 0$. By (3), writing ${\displaystyle \frac{d}{dt}}(k\ast [F_m-F_m\left(0\right)])\left(t\right)$ instead of ${\partial }_t^{\alpha }{F}_m\left(t\right)$ with $k\left(t\right)={\displaystyle \frac{t^{-\alpha }}{\Gamma (1-\alpha )}}$ in (21), we obtain It is clear that $H\left(\zeta \right)>0$ and $H^{\prime }\left(\zeta \right)>0$ for all $\zeta >{\left(c_1{m}^{-2}\right)}^{{\displaystyle \frac{1}{p-1}}}$.

Suppose now that We claim that (22) implies that $F_m\left(t\right)>{\left(c_1{m}^{-2}\right)}^{{\displaystyle \frac{1}{p-1}}}$ for all $t\in (0,T)$. (The fact is stated in the proof of Theorem 3.7 in [22].) Knowing that $F_m\left(t\right)\ge F_m\left(0\right)>{\left(c_1{m}^{-2}\right)}^{{\displaystyle \frac{1}{p-1}}}$ for all $t\in (0,T)$, it follows from (22) that Therefore the function $F_m\left(t\right)$ satisfying (24) is an upper solution of the problem we have by comparison principle $F_m\left(t\right)\ge \zeta \left(t\right)$ (see Theorem 4.10 in [29]).

On the other hand, since $H\left(0\right)=0$, $H\left(\zeta \right)>0$ and $H^{\prime }\left(\zeta \right)>0$ for all $\zeta \ge F_m\left(0\right)>{\left(c_1{m}^{-2}\right)}^{{\displaystyle \frac{1}{p-1}}}$. It then follows from Lemma 3.8 in [22] that $v\left(t\right)=w\left({\displaystyle \frac{t^{\alpha }}{\Gamma (\alpha +1)}}\right)$ is a lower solution for (25), where $v\left(t\right)$ satisfies and $w\left(t\right)$ solves the ordinary differential equation By the comparison principle (see Theorem 4.10 in [29]), we obtain $\zeta \left(t\right)\ge v\left(t\right)$. Solving the initial value problem (26), we have the solution By the comparison principle (see Theorem 4.10 in [29]), we conclude that with $c_2={\displaystyle \frac{c_1{m}^{-2}}{\Gamma (\alpha +1)}}$. Therefore, from (27), we obtain that $v\left(t\right)\to \infty$ as and that $F_m\left(t\right)\to \infty$. This implies that the solution $u(x,t)$ blows up in finite time when (23) holds.

As a result of these arguments, we have the following lemma.

Let $F_m\left(t\right)$ be defined by (16). If $F_m\left(0\right)$ as in (17) satisfies (23) for some $m\in \mathbf{N}$, i.e., where $A=c_1^{{\displaystyle \frac{1}{p-1}}}$ with $c_1$ as in (20), then $u(x,t)$ blows up in finite time.

Here, we shall state the rest of the proof for Theorem 1.

Supposing that $u(x,t)$ is a nontrivial global solution, we prove by reductio ad absurdum.

By Lemma 3, then it follows that for any $m\in \mathbf{N}$

Then, by (17) and $u_0\left(x\right)=\lambda \psi \left(x\right)\ge 0$, we obtain

Here, if we choose $\lambda$ to be large enough for any $m\in \mathbf{N}$, then the left-hand side of (28) is larger than the right-hand side of (28). Thus, we arrive at a contradiction. This completes the proof for the condition (a).

Next, if $\psi \in C_0\left({\mathbf{R}}^n\right)$ satisfies (13), then there is a positive constant L such that $\psi \left(x\right)\ge {L\left|x\right|}^{-a}$ for sufficiently large $\left|x\right|$. Then, we have for sufficiently large m By noting that $\left|x\right|\le {\displaystyle \frac{3}{2}}m$ in $B_m$ by (14), we obtain and then by (15), we have By multiplying both sides of (29) by $m^a$, we obtain Then, if ${\displaystyle 0<a<\frac{2}{p-1}}$ and m is sufficiently large, then the left-hand side of (30) is larger than the right-hand side of (30). Thus, we arrive at a contradiction. This completes the proof for the condition (b). □

Let $n\ge 1$ and $0<\alpha <1$. Assume that the initial data $u_0\left(x\right)=\lambda \psi \left(x\right)\ge 0$, where $\lambda >0$ and $\psi \in C_0\left({\mathbf{R}}^n\right)$. Suppose that $p\ge 1+2/n$, and that $a>2/(p-1)$ and

In what follows, by the letter C we denote generic positive constants, and they may have different values also within the same line.

Since $\psi \in C_0\left({\mathbf{R}}^n\right)$ satisfies (31), there is a constant $C>0$ such that

Let $q_c=n(p-1)/2$. First, if $p>1+2/n$ and $a>2/(p-1)$, then we know $a{q}_c>n$. Hence, it follows from (32) that Next, if $p=1+2/n$ and $a>2/(p-1)$, then we have $q_c=1$ and $a>n=2/(p-1)$. Hence, it follows from (32) that By (33) and (34), if $p\ge 1+2/n$ and $a>2/(p-1)$, then ${\parallel \psi \parallel }_{L^{q_c}\left({\mathbf{R}}^n\right)}\le C$. Since $u_0\left(x\right)=\lambda \psi \left(x\right)$, $\parallel u_0{\parallel }_{L^{q_c}\left({\mathbf{R}}^n\right)}$ is sufficiently small whenever $\lambda >0$ is small enough. Therefore, the mild solution of (1) exists globally by Proposition 3 (b).