1 Introduction

The Keller–Segel initial-value problem

$$\begin{aligned} \left\{ \begin{array}{ll} u_t = \Delta u - \nabla \cdot (u\nabla v), \quad &{} x\in \mathbb {R}^n, \ t>0, \\ 0 = \Delta v + u, \quad &{} x\in \mathbb {R}^n, \ t>0, \\ u(x,0)=u_0(x), \quad &{} x\in \mathbb {R}^n, \end{array} \right. \end{aligned}$$
(1.1)

has been playing the role of a paradigm in the modeling of self-enhancing aggregation processes both in physics and in biology ([3, 12, 17, 22]; cf. also the detailed discussion in [28]).

In corresponding mathematical studies, a predominant focus is accordingly set on aspects related to spontaneous formation of singularities [3, 10, 14, 20, 25, 27], and in frameworks of finite-mass solutions the literature has created a picture which is quite comprehensive in this regard. Indeed, among the most striking discoveries related to (1.1) seems the observation that if \(n=2\), then nonnegative initial data satisfying \(\int _{\mathbb {R}^n}u_0<8\pi \) lead to globally regular solution behavior, whereas unboundedness phenomena can be observed in the complementary case when \(\int _{\mathbb {R}^n}u_0\ge 8\pi \) [9, 10, 20, 22]. This sharp dichotomy disappears on passing over to three- and higher-dimensional situations, in which yet certain small-data solutions exist globally, but in which at each positive mass level some initial data enforce finite-time blow-up [5, 6, 26].

In the latter case when \(n\ge 3\), some mathematical interest in solutions exhibiting non-integrably slow decay as \(|x|\rightarrow \infty \) arises from the circumstance that then two types of explicit singular solutions exist which satisfy \(u(\cdot ,t)=O(|x|^{-2})\) as \(|x|\rightarrow \infty \): The first of these solutions is the singular steady state determined by letting

$$\begin{aligned} u(x,t):=u_\star (x):= \frac{2(n-2)}{|x|^2}, \quad x\in \mathbb {R}^n{\setminus }\{0\}, \ t>0, \end{aligned}$$
(1.2)

also known as the Chandrasekhar solution [3]; secondly, as already observed in [11], (1.1) possesses a family of explicit backward self-similar solutions given by

$$\begin{aligned} u(x,t):= & {} u^{(\mu )}(x,t):=\frac{1}{2(n-2)} \cdot \frac{8(n-2)^2 |x|^2 + 4n\mu - 8n(n-2)t}{\big (|x|^2 + \mu - 2(n-2)t\big )^2},\nonumber \\{} & {} \quad x\in \mathbb {R}^n, \ 0<t<\frac{\mu }{2(n-2)}, \end{aligned}$$
(1.3)

for \(\mu >0\) (see also [21]). The actual meaning of the latter solutions in the dynamics of (1.1) seems yet widely unexplored; after all, a recent result has revealed a certain strong instability feature thereof, hence surprisingly disproving a conjecture formulated in [11]. Likewise, also the role of the equilibria in (1.2) seems to have been understood only partially so far. For radially symmetric initial data which are less concentrated than \(u_\star \) in the sense that \(\int _{B_r(0)} u_0 \le (1-\varepsilon ) \int _{B_r(0)} u_\star \) for some \(\varepsilon >0\) and all \(r>0\), global bounded solutions exist ([5, 3, Theorem 2.1.1]; cf. also Theorem 1.2 below, and [30] for an analogue in bounded domains). A corresponding complementary implication seems lacking, however: In [2], global existence of solutions which are sufficiently regular for \(t>0\) is ruled out when \(\int _{B_r(0)} u_0 \ge \int _{B_r(0)} u_\star \) for all \(r>0\), which in particular requires already \(u_0\) to be singular itself. Available criteria for blow-up in the presence of bounded initial data appear less explicit, and they seem to mainly focus on local properties, and especially on quantities measuring concentratedness [5, 8]. As a particular consequence, it appears widely open whether the far-field asymptotics of the explicit solutions either in (1.2) or in (1.3) have some threshold character with respect to blow-up in (1.1). More generally, a comprehensive theory of solutions which decay more slowly than these prototypes, or even merely remain bounded as \(|x|\rightarrow \infty \), seems lacking: Precedent approaches addressing non-integrable decay have been operating in the Morrey spaces \(M^\frac{n}{2}(\mathbb {R}^n)\) ([1, 3]; cf. also [4] for a related result in spaces of pseudomeasures), or alternatively in the Lebesgue spaces \(L^p(\mathbb {R}^n)\) with some \(p<n\) [31], hence excluding any function of the form \(\mathbb {R}^n \ni x \mapsto (|x|+1)^{-\alpha }\) with \(\alpha \le 1\), for instance. For initial data with such types of tails larger than those from (1.2) and (1.3), it thus has even remained unclear up to now how far local-in-time solutions can at all be constructed, or if instantaneous blow-up must be expected. Obvious answers to this question seem impeded in view of the considerably slow decay of the Poisson kernel, and hence of unfavorably large taxis gradients potentially arising for large values of |x| if the component v is to be defined via Newtonian potentials. Accordingly, already local theories seem limited to radially symmetric settings [19].

Main results.    The present manuscript now addresses (1.1) for such non-integrable initial data, and intends to discuss the role of the decay seen in (1.2) and (1.3) for the occurrence of blow-up. Here, as in [19] resorting to settings of radially symmetric functions will enable us to rely on comparison-based arguments in essential parts.

In a first and preliminary step, we shall briefly present an approach slightly different from that pursued in [19, Appendix A] to derive the following basic statement on local existence and extensibility of local-in-time classical solutions. In formulating this and throughout the sequel, as usual we shall abbreviate \(B_R:=B_R(0)\) for \(R>0\), and we will call a function \(\varphi :\mathbb {R}^n\rightarrow \mathbb {R}\) radial if \(\varphi (x)=\widetilde{\varphi }(|x|)\) for all \(x\in \mathbb {R}^n\) with some \(\widetilde{\varphi }:[0,\infty )\rightarrow \mathbb {R}\), and then write \(\varphi (r)\) instead of \(\widetilde{\varphi }(r)\) for \(r\ge 0\) whenever this appears convenient; for functions \(\varphi =\varphi (x,t)\) for \(t\in S\subset [0,\infty )\), the term radial will be used in a similar style but referring to the variable x only.

Proposition 1.1

Let \(n\ge 1\) and \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) be radial and nonnegative. Then there exist \(T_{max}=T_{max}(u_0)\in (0,\infty ]\) as well as radial functions

$$\begin{aligned} \left\{ \begin{array}{l}u\in C^0(\mathbb {R}^n\times [0,T_{max})) \cap C^\infty (\mathbb {R}^n\times (0,T_{max})) \quad \hbox {and} \\ v\in C^\infty (\mathbb {R}^n\times (0,T_{max})), \end{array} \right. \end{aligned}$$
(1.4)

with u being uniquely determined by the additional requirements that \(u\ge 0\) and that

$$\begin{aligned} \sup _{t\in (0,T)} \Vert u(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} < \infty \quad \hbox {for all } T\in (0,T_{max}), \end{aligned}$$
(1.5)

such that (1.1) is satisfied in the classical sense in \(\mathbb {R}^n\times (0,T_{max})\), and that

$$\begin{aligned} \hbox {either } T_{max}=\infty , \quad \hbox {or} \quad \Vert u(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} \rightarrow \infty \quad \hbox {as } t\nearrow T_{max}. \end{aligned}$$
(1.6)

Next concerned with the relationship between the profiles in (1.2) and (1.3) and blow-up, by suitably exploiting maximum principles we can first confirm that actually for all initial data which lie below \(u_\star \), and which in addition to the above are assumed to belong to the space \(BUC(\mathbb {R}^n)\) of bounded and uniformly continuous functions on \(\mathbb {R}^n\), the solutions found above indeed exist globally.

Theorem 1.2

Let \(n\ge 3\) and \(u_0\in BUC(\mathbb {R}^n)\) be radial and such that

$$\begin{aligned} 0 \le u_0(x) \le \frac{2(n-2)}{|x|^2} \quad \hbox {for all } x\in \mathbb {R}^n. \end{aligned}$$
(1.7)

Then \(T_{max}(u_0)=\infty \); that is, the problem (1.1) admits a unique global classical solution (uv) in the sense specified in Proposition 1.1. Moreover,

$$\begin{aligned} - \int _{B_r} u(x,t) dx \le \frac{2n}{r^2} \quad \hbox {for all } r>0 \text { and } t>0. \end{aligned}$$
(1.8)

Remark

(i) The proof given below in Sect. 4 will assert global existence under the condition that merely

$$\begin{aligned} - \int _{B_r} u_0(x) dx \le \frac{2n}{r^2} \quad \hbox {for all } r>0, \end{aligned}$$
(1.9)

which is actually slightly weaker than the requirement in (1.7). (ii) A similar observation has already been made in [5]; mainly thanks to an appropriate use of a strong maximum principle, however, no assumption on strictness of the right inequality in (1.7) nor (1.9) is required here. Likewise, by exhausting the entire range of initial data fulfilling (1.7) the above theorem seems to go somewhat beyond [19, Theorem 1.1] where a similar feature has been derived under a comparable condition involving regular relatives of the singular solution in (1.2). (iii) The technical requirement that \(u_0\) belong to \(BUC(\mathbb {R}^n)\) is made here only to guarantee applicability of an extensibility criterion which in precedent literature has been derived under this hypothesis (see Lemma 3.3). (iv) The conclusion of Theorem 1.2 can apparently not be extended so as to include a statement on global boundedness without imposing further restrictions. Indeed, a recent high-dimensional caveat indicates that infinite-time blow-up may occur for some initial data fulfilling (1.7) when \(n\ge 10\) [13].

Our second result in this regard now complements this by revealing that the decay behavior of \(u_\star \) indeed is critical with respect to the occurrence of explosions; in fact, blow-up can be enforced by initial data which are uniformly small in all of \(\mathbb {R}^n\), provided that they lie above \(0<r\mapsto \frac{c}{r^2}\) with some positive c arbitrarily close to \(2(n-2)\):

Theorem 1.3

Let \(n\ge 3\), and let \(a_0>1\) Then for any \(\delta >0\), there exists a nonnegative function \(\psi =\psi ^{(a_0,\delta )} \in W^{1,\infty }_{loc}([0,\infty ))\) fulfilling

$$\begin{aligned} \psi (r) \le \delta \quad \hbox {for all } r>0 \end{aligned}$$
(1.10)

and

$$\begin{aligned} \limsup _{r\rightarrow \infty } r^2\psi (r) \le 2(n-2) a_0, \end{aligned}$$
(1.11)

which has the property that whenever \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is radial and such that

$$\begin{aligned} u_0(x)\ge \psi (|x|) \quad \hbox {for all } x\in \mathbb {R}^n, \end{aligned}$$
(1.12)

the corresponding solution (uv) of (1.1) blows up in finite time; that is, in Proposition 1.1 we have \(T_{max}(u_0)<\infty \) and

$$\begin{aligned} \Vert u(\cdot ,t)\Vert _{L^\infty (\Omega )} \rightarrow \infty \quad \hbox {as } t\nearrow T_{max}(u_0). \end{aligned}$$
(1.13)

We remark that by means of an argument different from ours, Theorem 1.3 could alternatively be derived by combining [19, (1.6), p. 146] with [19, Corollary 1.2 (ii)]. In its form stated here, it intends to explicitly stress the influence of tail behavior, and a short proof independent of the results from [19] will be included below for completeness.

We next supplement this by a criterion on finite-time blow-up which, unlike that in Theorem 1.3 but also that in [19, Theorem 1.1 (ii)], exclusively refers to the asymptotics of the initial data, and hence does not rely on any assumption on the initial behavior within compacta beyond mere positivity. As can be seen by combining [19, Corollary 1.2] with a result on an ordering property of radial solutions \(\Delta \varphi +e^\varphi =0\) [29, Theorem 1.1 (iv)], in the case \(n\ge 10\) only slightly slower decay than that in (1.2) will entail singular behavior. A different reasoning will lead us to the following statement in this regard that does not only include such latter high-dimensional situations, but beyond this also yields information in the apparently less structured case \(3\le n\le 9\) in which steady states intersect.

Theorem 1.4

Let \(n\ge 3\) and \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) be a positive radial function fulfilling

$$\begin{aligned} \liminf _{|x|\rightarrow \infty } \big \{ |x|^2 u_0(x) \big \} > 2(n-2) a_0(n), \end{aligned}$$
(1.14)

where

$$\begin{aligned} a_0(n):=\left\{ \begin{array}{ll}\frac{n+2}{n} - \frac{2(n-2)}{n^2+2n-4} \quad &{} \hbox {if } 3\le n \le 9, \\ 1 &{} \hbox {if } n\ge 10. \end{array} \right. \end{aligned}$$
(1.15)

Then \(T_{max}(u_0)<\infty \), and (1.13) holds.

Remark

(i) In combination, Theorems 1.2, 1.4 and 1.3 provide clear indications for criticality of the asymptotics in (1.2) with respect to the onset of singular behavior. In particular, the slower decay of the initial data associated with (1.3), for which

$$\begin{aligned} |x|^2 u^{(\mu )}(x,0) \rightarrow 4(n-2) \quad \hbox {as } |x|\rightarrow \infty \end{aligned}$$

regardless of the size of \(\mu >0\), seems irrelevant in this regard; in fact, even in the framework of the stronger characterization from Theorem 1.3, the corresponding coefficient \(a_0(n)\) always satisfies \(2(n-2) a_0(n)< \frac{2(n+2)(n-2)}{n} < 4(n-2)\).

(ii) Beyond these quantitative subtleties, especially Theorem 1.3 underlines that finite-time blow-up in (1.1) need not be driven by sufficiently strong concentration near some point in space. In fact, this statement again emphasizes that, alternatively, also large-tail behavior can be responsible for the occurrence of explosions.

Our final result connects to the latter remark by asserting that for radial initial data which diverge to \(+\infty \) as \(|x|\rightarrow \infty \), no solution to (1.1) can be found on any time interval:

Theorem 1.5

Let \(n\ge 3\) and \(u_0\in C^0(\mathbb {R}^n)\) be a nonnegative radial function satisfying

$$\begin{aligned} u_0(x)\rightarrow +\infty \quad \hbox {as } |x|\rightarrow \infty . \end{aligned}$$
(1.16)

Then for any choice of \(T>0\), there do not exist radial functions

$$\begin{aligned} \left\{ \begin{array}{l}u\in C^0(\mathbb {R}^n\times [0,T)) \cap C^{2,1}(\mathbb {R}^n\times (0,T)) \quad \hbox {and} \\ v\in C^{2,0}(\mathbb {R}^n\times (0,T)) \end{array} \right. \end{aligned}$$

such that \(u\ge 0\), and that (1.1) is solved in the classical sense in \(\mathbb {R}^n\times (0,T)\).

2 Comparison principles

This preliminary section provides some comparison principles which form the basis of substantial parts in our subsequent analysis. The first of these is designed for an application in the framework of the original variables from (1.1) in Lemma 3.1 below.

Lemma 2.1

Let \(n\ge 1\) and \(T>0\), and let \(h\in C^0([0,\infty )\times [0,T))\) be such that

$$\begin{aligned} \sup _{(r,t)\in (0,\infty )\times (0,\widehat{T})} |h(r,t)| <\infty \quad \hbox {for all } \widehat{T}\in (0,T). \end{aligned}$$
(2.1)

Then whenever \(\underline{z}\) and \(\overline{z}\) belong to \(C^0([0,\infty )\times [0,T)) \cap C^{2,1}([0,\infty )\times (0,T))\) and satisfy

$$\begin{aligned} \sup _{(r,t)\in (0,\infty )\times (0,\widehat{T})} \Big \{ |\underline{z}(r,t)| + |\overline{z}(r,t)| \Big \} <\infty \quad \hbox {for all } \widehat{T}\in (0,T), \end{aligned}$$
(2.2)

assuming that

$$\begin{aligned}{} & {} \underline{z}_t - \underline{z}_{rr} - \frac{n-1}{r} \underline{z}_r - rh(r,t) \underline{z}_r - \underline{z}^2 \le \overline{z}_t - \overline{z}_{rr} - \frac{n-1}{r} \overline{z}_r \nonumber \\{} & {} \quad - rh(r,t) \overline{z}_r - \overline{z}^2 \quad \hbox {for all } r>0 \hbox { and } t\in (0,T) \end{aligned}$$
(2.3)

as well as \(\underline{z}_r(0,t)=\overline{z}_r(0,t)\) for all \(t\in (0,T)\) and

$$\begin{aligned} \underline{z}(r,0)\le \overline{z}(r,0) \quad \hbox {for all } r>0, \end{aligned}$$
(2.4)

one can conclude that

$$\begin{aligned} \underline{z}(r,t)\le \overline{z}(r,t) \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,T). \end{aligned}$$
(2.5)

Proof

We fix \(\widehat{T}\in (0,T)\) and rely on (2.1) and (2.2) to find \(c_1=c_1(\widehat{T})>0\) such that

$$\begin{aligned} |h(r,t)| + |\underline{z}(r,t)| + |\overline{z}(r,t)| \le c_1 \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,\widehat{T}). \end{aligned}$$
(2.6)

Then letting \(\kappa =\kappa (\widehat{T})>0\) be such that

$$\begin{aligned} \kappa >2(n+c_1)+2c_1 \end{aligned}$$
(2.7)

and defining \(\zeta (r):=r^2+1\), \(r\ge 0\), as well as

$$\begin{aligned} d(r,t):=\underline{z}(r,t)-\overline{z}(r,t) - \varepsilon e^{\kappa t} \zeta (r), \quad r\ge 0, \ t\in [0,\widehat{T}], \end{aligned}$$
(2.8)

for \(\varepsilon >0\), from (2.6), (2.4) and the continuity of \(\underline{z}\) and \(\overline{z}\) we obtain that \(t_0:=\sup \big \{ T'\in (0,\widehat{T}) \ \big | \ d(r,t)<0 \hbox { for all } r\ge 0 \hbox { and } t\in [0,T'] \big \}\) is well-defined and positive, and that if \(t_0\) was smaller than \(\widehat{T}\), then there would exist \(r_0\ge 0\) such that \(d(r_0,t_0)=0, d_t(r_0,t_0)\ge 0, d_r(r_0,t_0)=0\) and \(\Delta d(r_0,t_0)\le 0\), where, as usual, for \(\varphi \in C^2([0,\infty ))\) fulfilling \(\varphi _r(0)=0\) we have set \(\Delta \varphi (r):=\varphi _{rr}(r) + \frac{n-1}{r} \varphi (r)\) for \(r>0\), and \(\Delta \varphi (0):=n\varphi _{rr}(0)\). Since \(\zeta _r(r)=2r\) for \(r\ge 0\) and \(\Delta \zeta \equiv 2n\), and since in view of (2.8) and (2.3) we know that

$$\begin{aligned} d_t= & {} \underline{z}_t - \overline{z}_t - \kappa \varepsilon e^{\kappa t} \zeta (r) \\\le & {} \Delta (\underline{z}-\overline{z}) + rh(r,t) (\underline{z}_r-\overline{z}_r) + (\underline{z}+\overline{z})(\underline{z}-\overline{z}) - \kappa \varepsilon e^{\kappa t} \zeta (r) \\= & {} \Delta d + \varepsilon e^{\kappa t} \Delta \zeta (r) + rh(r,t) d_r + rh(r,t) \cdot \varepsilon e^{\kappa t} \zeta _r(r) \\{} & {} + (\underline{z}+\overline{z}) d + (\underline{z}+\overline{z}) \cdot \varepsilon e^{\kappa t} \zeta (r) - \kappa \varepsilon e^{\kappa t} \zeta (r) \quad \hbox { for all } r>0 \text { and } t\in (0,\widehat{T}), \end{aligned}$$

this would imply that

$$\begin{aligned} 0 \le d_t(r_0,t_0)\le & {} \varepsilon e^{\kappa t_0} \cdot 2n + r_0 h(r_0,t_0) \cdot \varepsilon e^{\kappa t_0} \cdot 2r_0 \\{} & {} \quad + (\underline{z}+\overline{z})(r_0,t_0) \cdot \varepsilon e^{\kappa t_0} \cdot (r_0^2+1) - \kappa \varepsilon e^{\kappa t_0} \cdot (r_0^2+1) \\= & {} \varepsilon e^{\kappa t_0} \cdot \Big \{ 2n+ 2r_0^2 h(r_0,t_0) + (r_0^2+1) \cdot (\underline{z}+\overline{z})(r_0,t_0) - \kappa (r_0^2+1)\Big \}. \end{aligned}$$

According to (2.6) and (2.7), however, this would lead to the impossible conclusion that

$$\begin{aligned} 0{} & {} \le 2n + 2r_0^2 c_1 + (r_0^2+1) \cdot 2c_1 - \kappa (r_0^2+1)\\{} & {} \le \big \{ 2(n+c_1)+2c_1\big \} \cdot (r_0^2+1) - \kappa (r_0^2+1) < 0. \end{aligned}$$

Consequently, we must have \(t_0=\widehat{T}\), and thus the claim follows upon taking \(\varepsilon \searrow 0\) and then \(\widehat{T}\nearrow T\). \(\square \)

In order to motivate a second series of comparison principles to be used below, let us recall the following essentially well-known observation concerned with a transformation of (1.1) in its radial version to a scalar parabolic problem.

Lemma 2.2

Let \(n\ge 1\) and \(T>0\), and suppose that \(u_0\in C^0(\mathbb {R}^n)\) as well as

$$\begin{aligned} \left\{ \begin{array}{l}u\in C^0(\mathbb {R}^n\times [0,T)) \cap C^{2,1}(\mathbb {R}^n\times (0,T)) \quad \hbox { and } \\ v\in C^{2,0}(\mathbb {R}^n\times (0,T)) \end{array} \right. \end{aligned}$$

are radial functions such that \(u_0\ge 0\) and \(u\ge 0\), and that (1.1) is solved in \(\mathbb {R}^n\times (0,T)\). Then letting

$$\begin{aligned} U(r,t):=\left\{ \begin{array}{ll}\frac{1}{2r^n} \int _0^r \rho ^{n-1} u(\rho ,t) d\rho , \quad r>0, \ t\in [0,T), \\ \frac{u(0,t)}{2n}, \quad r=0, \ t\in [0,T), \end{array} \right. \end{aligned}$$
(2.9)

defines a nonnegative function \(U\in C^0([0,\infty )\times [0,T)) \cap C^{2,1}([0,\infty )\times (0,T))\) which satisfies

$$\begin{aligned} \left\{ \begin{array}{ll}U_t = U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2, \quad &{} r>0, \ t\in (0,T), \\ U_r(0,t)=0, &{} t\in (0,T), \\ U(r,0) = U_0(r), &{} r>0, \end{array} \right. \end{aligned}$$
(2.10)

where

$$\begin{aligned} U_0(r):=\left\{ \begin{array}{ll}\frac{1}{2r^n} \int _0^r \rho ^{n-1} u_0(\rho ) d\rho , \quad r>0, \\ \frac{u_0(0)}{2n}, \quad r=0. \end{array} \right. \end{aligned}$$
(2.11)

Proof

This can be verified by straightforward computation based on the radial version of (1.1). \(\square \)

Our second group of comparison principles now addresses sub- and supersolutions to the PDE in (2.10) under various assumptions especially on boundary behavior.

Lemma 2.3

Let \(n\ge 1\) and \(T>0\), and for \(i\in \{1,2\}\) let \(R_i: [0,T)\rightarrow [0,\infty ]\) be continuous with \(R_1(t)<R_2(t)\) for all \(t\in [0,T)\), and let

$$\begin{aligned} G_{\widehat{T}}:= \Big \{ (r,t)\in \mathbb {R}^2 \ \Big | \ t\in [0,\widehat{T}), \ r\in (R_1(t),R_2(t)) \Big \}, \quad \widehat{T}\in (0,T]. \end{aligned}$$

Moreover, suppose that \(\underline{U}\) and \(\overline{U}\) are elements of \(C^0(G_T) \cap C^1({\mathop {G}\limits ^{\circ }}_T)\) which are such that

$$\begin{aligned} \underline{U}(\cdot ,t) {\hbox { and }} \overline{U}(\cdot ,t)\text { belong to }W^{2,\infty }_{loc}((R_1(t),R_2(t)))\text { for all } t\in (0,T), \end{aligned}$$
(2.12)

that

$$\begin{aligned}{} & {} \underline{U}_t - \underline{U}_{rr} - \frac{n+1}{r} \underline{U}_r - 2r\underline{U}\underline{U}_r - 2n\underline{U}^2 \nonumber \\{} & {} \quad \le \overline{U}_t - \overline{U}_{rr} - \frac{n+1}{r} \overline{U}_r - 2r\overline{U}\overline{U}_r - 2n\overline{U}^2 \end{aligned}$$
(2.13)
$$\begin{aligned}{} & {} \quad \hbox {for all } t\in (0,T)\text { and a.e. }~r\in (R_1(t),R_2(t)), \end{aligned}$$
(2.14)

and that

$$\begin{aligned} \underline{U}(r,0) \le \overline{U}(r,0) \quad \hbox {for all } r\in (R_1(0),R_2(0)). \end{aligned}$$
(2.15)

Then

$$\begin{aligned} \underline{U}(r,t) \le \overline{U}(r,t) \quad \hbox {for all } (r,t)\in G_T, \end{aligned}$$
(2.16)

provided that one of the following sets of conditions is fulfilled:

(i) We have \(R_2\equiv \infty \) and \(\{\underline{U},\overline{U}\}\subset C^0(\overline{G}_{\widehat{T}})\) for all \(\widehat{T}\in (0,T)\), moreover

$$\begin{aligned} \underline{U}(R_1(t),t) \le \overline{U}(R_1(t),t) \quad \hbox {for all } t\in (0,T), \end{aligned}$$
(2.17)

and

$$\begin{aligned} \sup _{(r,t)\in G_{\widehat{T}}} \Big \{ |\underline{U}(r,t)| + |\overline{U}(r,t)| + |r\underline{U}_r(r,t)| \Big \} < \infty \quad \hbox {for all } \widehat{T}\in (0,T).\nonumber \\ \end{aligned}$$
(2.18)

(ii) The function \(R_2\) satisfies \(R_2\equiv \infty \), the boundedness property (2.18) is valid, and furthermore

$$\begin{aligned} \limsup _{\delta \searrow 0} \sup _{t\in (0,\widehat{T})} \Big \{ \underline{U}(R_1(t)+\delta ,t) - \overline{U}(R_1(t)+\delta ,t)\Big \} <0 \quad \hbox {for all } \widehat{T}\in (0,T).\nonumber \\ \end{aligned}$$
(2.19)

(iii) We have \(R_1\equiv 0\) and \(R_2\equiv \infty \), \(\underline{U}\) and \(\overline{U}\) are members of \(C^1(\overline{G}_{\widehat{T}})\) for all \(\widehat{T}\in (0,T)\), and (2.18) holds.

(iv) The number \(R_2(t)\) is finite for all \(t\in [0,T)\), \(\underline{U}\) and \(\overline{U}\) lie in \(C^1(\overline{G}_{\widehat{T}})\) for all \(\widehat{T}\in (0,T)\), and

$$\begin{aligned} \underline{U}(R_2(t),t) \le \overline{U}(R_2(t),t) \quad \hbox {for all } t\in (0,T), \end{aligned}$$
(2.20)

Proof

We first consider the cases (i) and (iii), in which fixing \(\widehat{T}\in (0,T)\), according to (2.18) we choose \(c_1=c_1(\widehat{T})>0\) such that

$$\begin{aligned} { |\underline{U}(r,t)| + |\overline{U}(r,t)| + |r\underline{U}_r(r,t)| \le c_1 } \quad \hbox {for all } t\in (0,\widehat{T}) \hbox { and } r\in (R_1(t),R_2(t)),\nonumber \\ \end{aligned}$$
(2.21)

and take \(\kappa =\kappa (\widehat{T})>0\) large enough such that

$$\begin{aligned} \kappa >2n { +4 } +(4n+6)c_1. \end{aligned}$$
(2.22)

For each \(\varepsilon >0\), from (2.21) and the respective hypotheses in (i) and (iii) it then readily follows that if we let

$$\begin{aligned} d(r,t):=\underline{U}(r,t)-\overline{U}(r,t) - \varepsilon e^{\kappa t} \zeta (r), \quad t\in [0,\widehat{T}), \ r\ge R_1(t), \end{aligned}$$

again with \(\zeta (r):=r^2+1\), \(r\ge 0\), then \(t_0:=\sup \big \{ T'\in (0,\widehat{T}) \ \big | \ d(r,t)<0 \hbox { for all } t\in (0,T') {\hbox { and }} r>R_1(t) \big \}\) is well-defined and positive, and that if we had \(t_0<\widehat{T}\), then there would exist \(r_0\ge R_1(t_0)\) such that and that

$$\begin{aligned} d(r_0,t_0)=0=\sup _{r\ge R_1(t_0)} d(r,t_0). \end{aligned}$$
(2.23)

Here if (i) holds, then \(d(R_1(t),t)\le -\varepsilon e^{\kappa t} \zeta (R_1(t)) <0\) for all \(t\in (0,\widehat{T})\), so that (2.23) implies that necessarily \(r_0> R_1(t_0)\), that thus the definition of \(t_0\) entails that

$$\begin{aligned} d_t(r_0,t_0) \ge 0, \end{aligned}$$
(2.24)

and that moreover \(d_r(r_0,t_0)=0\) and hence, in particular,

$$\begin{aligned} r_0 d_r(r_0,t_0)=0. \end{aligned}$$
(2.25)

If (iii) is assumed, however, then the inclusion \(\{\underline{U},\overline{U}\} \subset C^1(\overline{G}_{t_0})\) warrants that again (2.24) holds. As before, (2.25) then follows when \(r_0>R_1(t_0)\), while the requirement \(R_1\equiv 0\) ensures that (2.25) trivially holds in the case when \(r_0=R_1(t_0)\). Having thus asserted that (2.25) would be satisfied in any of these situations, we proceed to make sure that under the current hypotheses we could find \((r_j)_{j\in \mathbb {N}} \subset (r_0,\infty )\) such that \(r_j\searrow r_0\) as \(j\rightarrow \infty \), that \(d_r(\cdot ,t_0)\) is differentiable at \(r=r_j\) for all \(j\in \mathbb {N}\), and that

$$\begin{aligned} \limsup _{j\rightarrow \infty } \Bigg \{ d_{rr}(r_j,t_0) + \frac{n+1}{r_j} d_r(r_j,t_0)\Bigg \} \le 0. \end{aligned}$$
(2.26)

Indeed, if this was false, then due to the fact that \(\underline{U}_r(\cdot ,t_0)\) and \(\overline{U}_r(\cdot ,t_0)\) are absolutely continuous we could find \(R_\star >R_1(t_0)\) and \(c_2>0\) such that

$$\begin{aligned} d_{rr}(r,t_0) + \frac{n+1}{r} d_r(r,t_0) \ge c_2 \quad \hbox {for a.e. } r\in (r_0,R_\star ), \end{aligned}$$

that thus

$$\begin{aligned} \big ( r^{n+1} d_r(r,t_0)\big )_r \ge c_2 r^{n+1} \quad \hbox {for a.e. } r\in (r_0,R_\star ), \end{aligned}$$

and that accordingly

$$\begin{aligned} d_r(r,t_0) \ge \frac{c_2\cdot (r^{n+2}-r_0^{n+2})}{(n+2)r^{n+1}} \quad \hbox {for all } r\in (r_0,R_\star ), \end{aligned}$$

because \(r_0^{n+1} d_r(r_0,t_0)=0\) by (2.25). As this would imply that \(d(r,t_0)>d(r_0,t_0)\) for all \(r\in (r_0,R_\star )\), this would contradict (2.23), so that a sequence \((r_j)_{j\in \mathbb {N}}\) with the above properties indeed must exist.

The remaining part of the reasoning is now fairly straightforward: using (2.13) along with the identities \(\zeta _r(r)=2r\) and \(\zeta _{rr}(r)=2\) for \(r>0\), we see that for all \(t\in (0,T)\) and \(r>R_1(t)\),

$$\begin{aligned} d_t= & {} \underline{U}_t - \overline{U}_t - \kappa \varepsilon e^{\kappa t} \zeta (r)\\\le & {} (\underline{U}_{rr}-\overline{U}_{rr}) + \frac{n+1}{r} (\underline{U}_r-\overline{U}_r) + 2r(\underline{U}\underline{U}_r - \overline{U}\overline{U}_r) \\{} & {} + 2n(\underline{U}^2-\overline{U}^2) - \kappa \varepsilon e^{\kappa t} \zeta (r)\\= & {} (\underline{U}_{rr}-\overline{U}_{rr}) + \frac{n+1}{r} (\underline{U}_r-\overline{U}_r) + 2r\underline{U}_r (\underline{U}-\overline{U}) + 2r\overline{U}(\underline{U}_r-\overline{U}_r) \\{} & {} + 2n(\underline{U}+\overline{U})(\underline{U}-\overline{U}) - \kappa \varepsilon e^{\kappa t}\zeta (r) \\= & {} d_{rr} + \varepsilon e^{\kappa t} \zeta _{rr}(r) + \frac{n+1}{r}(d_r + \varepsilon e^{\kappa t} \zeta _r(r)) \\{} & {} + 2r\underline{U}_r (d+\varepsilon e^{\kappa t}\zeta (r)) + 2r\overline{U}(d_r + \varepsilon e^{\kappa _t} \zeta _r(r)) \\{} & {} + 2n(\underline{U}+\overline{U}) (d+\varepsilon e^{\kappa t} \zeta (r)) - \kappa \varepsilon e^{\kappa t} \zeta (r) \\= & {} d_{rr} + \frac{n+1}{r} d_r + 2r\underline{U}_r d + 2r\overline{U}d_r + 2n(\underline{U}+\overline{U}) d \\{} & {} + \Big \{ 2n { +4 } +2r\underline{U}_r (r^2+1) + 4r^2 \overline{U}+ 2n(\underline{U}+\overline{U}) (r^2+1) - \kappa (r^2+1)\Big \} \cdot \varepsilon e^{\kappa t}, \end{aligned}$$

which when evaluated at \((r,t)=(r_j,t_0)\) for \(j\in \mathbb {N}\) shows that due to (2.24)–(2.26), in the limit \(j\rightarrow \infty \) we obtain that

$$\begin{aligned} 0\le & {} \Big \{ 2n { +4 } + 2r_0 \underline{U}_r(r_0,t_0) (r_0^2+1) + 4r_0^2 \overline{U}(r_0,t_0) \\{} & {} + 2n \big (\underline{U}(r_0,t_0)+\overline{U}(r_0,t_0)\big ) (r_0^2+1) - \kappa (r_0^2+1)\Big \} \cdot \varepsilon e^{\kappa t_0}. \end{aligned}$$

According to (2.21), however, this entails that

$$\begin{aligned} 0\le & {} 2n { +4 } + 2c_1\cdot (r_0^2+1) + 4r_0^2 c_1 + 2n\cdot 2c_1 \cdot (r_0^2+1) - \kappa (r_0^2+1) \\\le & {} (2n+2c_1+4c_1+4nc_1-\kappa )\cdot (r_0^2+1), \end{aligned}$$

which is impossible due to (2.22) whence actually d(rt) must be negative for all \(t\in [0,\widehat{T})\) and \(r>R_1(t)\). Taking \(\varepsilon \searrow 0\) and then \(\widehat{T}\nearrow T\) we thus infer that (2.16) holds in either of the cases (i) and (iii).

If, alternatively, the conditions in (ii) are presupposed, then an application of the result from (i) again yields (2.16) if for each \(\widehat{T}\in (0,T)\) we replace \(G_{\widehat{T}}\) by \(G_{\widehat{T}}^{(\delta )}:=\big \{ (r,t)\in \mathbb {R}^2 \ \big | \ t\in [0,\widehat{T}), r>R_1(t)+\delta \big \}\) with some suitably small \(\delta =\delta (\widehat{T})>0\), and rely on (2.19) in verifying the corresponding variant of (2.17) with \(R_1\) substituted by \(R_1+\delta \).

The situation in (iv), finally, can be dealt with by a very similar and, in fact, significantly simpler argument than that detailed above for the case (i). \(\square \)

3 A local theory of solvability for bounded continuous radial data

By means of a comparison of certain approximate solutions to (1.1) with spatially flat solutions on the basis of Lemma 2.1, we can now achieve a first and essential step toward our local existence theory. As the only prerequisite that needs to be imported from the literature here (cf. , e.g., [31] or also [3]), we rely on the fact that for compactly supported nonnegative radial data a radial classical solution (uv) can always be found, and that for this solution we have \(v(\cdot ,t)=\Gamma \star u(\cdot ,t)\), where

$$\begin{aligned} \Gamma (z):=\left\{ \begin{array}{ll}-\frac{1}{2\pi } \ln |z| \quad &{} \hbox {if } n=2, \\ \frac{1}{n(n-2)|B_1(0)|} |z|^{2-n} \quad &{} \hbox {if } n\ge 3. \end{array} \right. \end{aligned}$$
(3.1)

denotes the Poisson kernel on \(\mathbb {R}^n\).

Lemma 3.1

Let \(n\ge 1\) and \(M>0\). Then there exists \(T(M)>0\) with the property that if \(u_0\in C^0(\mathbb {R}^n)\) is radial with

$$\begin{aligned} 0 \le u_0 \le M \quad \hbox {in } \mathbb {R}^n, \end{aligned}$$
(3.2)

then one can find radial functions

$$\begin{aligned} \left\{ \begin{array}{l}u\in C^0(\mathbb {R}^n\times [0,T(M)]) \cap C^\infty (\mathbb {R}^n\times (0,T(M))) \quad \hbox {and} \\ v\in C^\infty (\mathbb {R}^n\times (0,T(M))) \end{array} \right. \end{aligned}$$
(3.3)

such that

$$\begin{aligned} 0 \le u\le 2M \quad \hbox {in } \mathbb {R}^n\times (0,T(M)), \end{aligned}$$
(3.4)

and that (uv) solves (1.1) in the classical sense in \(\mathbb {R}^n\times (0,T(M))\).

Proof

For fixed \(M>0\), we let \(T(M):=\frac{1}{2M}\) and

$$\begin{aligned} y(t):=\frac{M}{1-Mt}, \quad t\in [0,T(M)], \end{aligned}$$

noting that then

$$\begin{aligned} \left\{ \begin{array}{l}y'(t)=y^2(t), \quad t\in (0,T(M)), \\ y(0)=M, \end{array} \right. \end{aligned}$$
(3.5)

and that moreover

$$\begin{aligned} 0 \le y(t) \le 2M \quad \hbox {for all } t\in [0,T(M)]. \end{aligned}$$
(3.6)

Given any radial \(u_0\in C^0(\mathbb {R}^n)\) fulfilling (3.2), we now pick a sequence \((u_{0j})_{j\in \mathbb {N}} \subset C^0(\mathbb {R}^n)\) of radial functions \(u_{0j}\) with \(u_{0j}\equiv u_0\) in \(B_j\), \(\textrm{supp} \, u_{0j}\subset B_{j+1}\) and \(0 \le u_{0j} \le u_0\) for all \(j\in \mathbb {N}\). For each fixed \(j\in \mathbb {N}\), thanks to the boundedness of \(\textrm{supp} \, u_{0j}\) we may rely on Proposition 1.1 from [31] to find \(T_j\in (0,T(M)]\) as well as a radial classical solution \((u_j,v_j)\) of (1.1) with \(u_j|_{t=0}=u_{0j}\) and such that with \(\Gamma \) as in (3.1) we have \(v_j(\cdot ,t)=\Gamma \star u_j(\cdot ,t)\) for all \(t\in (0,T_j)\), and that

$$\begin{aligned} \hbox {if } T_j< T(M), \quad \hbox {then} \quad \limsup _{t\nearrow T_j} \Vert u_j(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} = \infty . \end{aligned}$$
(3.7)

Here since \(u_j\) is radial, due to symmetry of \(\Gamma \) it follows that necessarily \(\frac{x}{|x|} \cdot \nabla v_j(x,t)=v_{jr}(|x|,t)\) for all \(x\in \mathbb {R}^n{\setminus }\{0\}\) and \(t\in (0,T_j)\), with

$$\begin{aligned} v_{jr}(r,t)= - r^{1-n} \int _0^r \rho ^{n-1} u_j(\rho ,t) d\rho , \quad r>0, \ t\in (0,T_j), \end{aligned}$$
(3.8)

so that (1.1) implies that

$$\begin{aligned} u_{jt} = u_{jrr} + \frac{n-1}{r} u_{jr} + rh_j(r,t) u_{jr} + u_j^2, \quad r>0, \ t\in (0,T_j), \end{aligned}$$
(3.9)

with

$$\begin{aligned} h_j(r,t):=-\frac{1}{r} v_{jr}(r,t), \quad (r,t)\in (0,\infty )\times (0,T_j), \end{aligned}$$
(3.10)

satisfying

$$\begin{aligned} |h_j(r,t)|{} & {} \le \bigg | r^{-n} \int _0^r \rho ^{n-1} u_j(\rho ,t) d\rho \bigg | \nonumber \\{} & {} \le \frac{1}{n} \Vert u_j(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} \quad \hbox {for all } r>0 \hbox { and } t\in (0,T_j). \end{aligned}$$
(3.11)

Since thus the inclusion \(u_j\in L^\infty (\mathbb {R}^n\times (0,\widehat{T}))\), for each \(\widehat{T}\in (0,T_j)\) asserted by [31, Proposition 1.1], guarantees validity of (2.1) with \(T=T_j\), and since the same token together with (3.6) warrants that for \(\underline{z}(r,t):=u_j(r,t)\) and \(\overline{z}(r,t):=y(t)\), \(r\ge 0, t\in [0,T_j)\), the estimate in (2.2) holds, using that clearly \(\overline{z}_t=\overline{z}^2=\overline{z}_{rr}+\frac{n-1}{r}\overline{z}_r + r h_j(r,t) \overline{z}_r + \overline{z}^2\) in \((0,\infty )\times (0,T_j)\) we may conclude from Lemma 2.1 that \(\underline{z}\le \overline{z}\) in \((0,\infty )\times (0,T_j)\) and hence

$$\begin{aligned} u_j(r,t)\le y(t) \quad \hbox {for all }r>0 \hbox { and } t\in (0,T_j), \end{aligned}$$

because \(u_j(r,0)\le M=y(0)\) for all \(r>0\) by (3.2). In view of (3.6), this implies that

$$\begin{aligned} u_j \le 2M \quad \hbox {in } (0,\infty )\times (0,T_j), \end{aligned}$$
(3.12)

and hence, according to (3.7), ensures that actually \(T_j=T(M)\). As therefore \(|h_j|\le \frac{2M}{n}\) in \((0,\infty )\times (0,T(M))\) due to (3.11), by means of standard parabolic Schauder theory [18] applied to (3.9), (3.10) and (3.8) we readily obtain that \((u_j)_{j\in \mathbb {N}}\) and \((v_{jr})_{j\in \mathbb {N}}\) are bounded in \(C^{2k,k}([0,R] \times [\tau ,T(M)])\) for all \(R>0, \tau \in (0,T(M))\) and \(k\in \mathbb {N}\), whence the Arzelà–Ascoli theorem provides a subsequence, again denoted by \(((u_j,v_j))_{j\in \mathbb {N}}\), such that

$$\begin{aligned} u_j \rightarrow u \ \hbox { and } \ v_{jr} \rightarrow w \quad \hbox {in } C^\infty _{loc}([0,\infty )\times (0,T(M))) \quad \hbox {as } j\rightarrow \infty \end{aligned}$$
(3.13)

with some \(u\in C^\infty ([0,\infty )\times (0,T(M)))\) and \(w\in C^\infty ([0,\infty )\times (0,T(M)))\) fulfilling

$$\begin{aligned} u_t = u_{rr} + \frac{n-1}{r} u_r + w u_r + u^2 \quad \hbox {in } (0,\infty ) \times (0,T(M)) \end{aligned}$$

as well as

$$\begin{aligned} 0 \le u \le 2M \quad \hbox {in } (0,\infty )\times (0,T(M)) \end{aligned}$$

and

$$\begin{aligned} w(r,t)=-r^{1-n} \int _0^r \rho ^{n-1} u(\rho ,t) d\rho \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,T(M)). \end{aligned}$$

To finally confirm that u is continuous at \(t=0\) and satisfies \(u|_{t=0}=u_0\), we return to the original variables and rewrite (3.9) in the form

$$\begin{aligned} u_{jt} = \Delta u_j - \nabla \cdot f_j \quad \hbox {in } \mathbb {R}^n\times (0,T(M)) \quad \text {for } j\in \mathbb {N}, \end{aligned}$$
(3.14)

where thanks to (3.10), (3.11) and (3.12), \(f_j:=u_j\nabla v_j\) satisfies

$$\begin{aligned} |f_j(x,t)| \le 2M\cdot |x| \cdot \frac{2M}{n} \le \frac{4M^2 R}{n} \quad \hbox {for all } x\in B_R, t\in (0,T(M)){\hbox { and }} j\in \mathbb {N}. \end{aligned}$$

Therefore, letting \((e^{t\Delta })_{t\ge 0}\) denote the heat semigroup on \(\mathbb {R}^n\) we see that well-known parabolic Hölder regularity theory applies to the functions given by

$$\begin{aligned} \widehat{u}_j(x,t):=u_j(x,t)-e^{t\Delta } u_0, \quad (x,t)\in \mathbb {R}^n \times (0,T(M)), \ j\in \mathbb {N}, \end{aligned}$$

which due to (3.14), namely, solve

$$\begin{aligned} \left\{ \begin{array}{ll}\widehat{u}_{jt}=\Delta \widehat{u}_j - \nabla \cdot f_j, \quad &{} x\in \mathbb {R}^n, \ t\in (0,T(M)), \\ \widehat{u}_j(x,0)=0, &{} x\in \mathbb {R}^n. \end{array} \right. \end{aligned}$$

In consequence, for any \(R>0\) the local estimates from [23] provide \(\theta =\theta (R)\in (0,1)\) such that

$$\begin{aligned} (\widehat{u}_j)_{j\in \mathbb {N}} \hbox { is bounded in } C^{\theta ,\frac{\theta }{2}}(\overline{B}_R\times [0,T(M)]), \end{aligned}$$

so that thanks to the continuity of \(\mathbb {R}^n\times [0,T(M)] \ni (x,t)\mapsto e^{t\Delta } u_0\) it follows that \((u_j)_{j\in \mathbb {N}}\) is equi-continuous in \(\overline{B}_R\times [0,T(M)]\) for any such R. Again due to the Arzelà–Ascoli theorem, this implies that in addition to (3.13) we have \(u_j\rightarrow u\) in \(C^0_{loc}(\mathbb {R}^n\times [0,T(M)])\) as \(j\rightarrow \infty \), and that hence (1.1) is solved in the claimed classical sense if we define the radial function v by letting \(v(x,t):=\int _0^{|x|} w(\rho ,t)d\rho \) for \(x\in \mathbb {R}^n\) and \(t\in (0,T(M))\), for instance. \(\square \)

Uniqueness in the natural class of classical solutions can now be inferred by means of two comparison arguments acting at the level of solutions to the transformed problem (2.10), hence involving Lemma 2.3.

Lemma 3.2

If \(n\ge 1\) and \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is radial and nonnegative, if \(T>0\), and if for \(i\in \{1,2\}\) the functions

$$\begin{aligned} \left\{ \begin{array}{l}u_i\in C^0(\mathbb {R}^n\times [0,T)) \cap C^{2,1}(\mathbb {R}^n\times (0,T)) \quad \hbox {and} \\ v_i\in C^{2,0}(\mathbb {R}^n\times (0,T)) \end{array} \right. \end{aligned}$$
(3.15)

are radial with \(u_i\ge 0\) and such that

$$\begin{aligned} \sup _{t\in (0,\widehat{T})} \Vert u_i(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} < \infty \quad \hbox {for all } \widehat{T}\in (0,T), \end{aligned}$$
(3.16)

and that \((u_i,v_i)\) solves (1.1) classically in \(\mathbb {R}^n\times (0,T)\), then \(u_1\equiv u_2\) in \(\mathbb {R}^n\times (0,T)\).

Proof

Due to (3.15) and our assumptions on nonnegativity and radial symmetry, from Lemma 2.2 we know that the functions \(U_1\) and \(U_2\) accordingly defined as in (2.9) form solutions of (2.10) fulfilling \(U_1(r,0)=U_2(r,0)\) for all \(r>0\). Moreover, for each \(\widehat{T}\in (0,T)\) it follows from (3.16) that with some \(c_1=c_1(\widehat{T})>0\) we have \(u_1(r,t)+u_2(r,t)\le c_1\) for all \(r>0\) and \(t\in (0,\widehat{T})\), so that in line with (2.9),

$$\begin{aligned} 0 \le U_i(r,t) \le \frac{1}{2r^n} \int _0^r \rho ^{n-1} c_1 d\rho = \frac{c_1}{2n} \quad \hbox {for } r>0, t\in (0,\widehat{T}) {\hbox { and }} i\in \{1,2\} \end{aligned}$$

as well as

$$\begin{aligned} \big | rU_{ir}(r,t)\big | \le \frac{1}{2} \cdot c_1 + n \cdot \frac{c_1}{2n} \quad \hbox {for } r>0, t\in (0,\widehat{T}) \hbox { and } i\in \{1,2\}, \end{aligned}$$

because

$$\begin{aligned} U_{ir}(r,t)=\frac{1}{2r} u_i(r,t) - \frac{n}{r} U_i(r,t) \quad \hbox {for } r>0, t\in (0,T) {\hbox { and }} i\in \{1,2\}.\nonumber \\ \end{aligned}$$
(3.17)

We may therefore employ Lemma 2.3 (iii) to conclude that both \(U_1\le U_2\) and \(U_2\le U_1\) in \((0,\infty )\times (0,T)\), meaning that \(U_1\equiv U_2\). Again in view of (3.17), this implies that, indeed, \(u_1\equiv u_2\). \(\square \)

Our local theory of bounded classical radial solutions to (1.1) can thereby be completed:

Proof of Proposition 1.1

The claim directly results upon combining Lemma 3.1 and Lemma 3.2 with a standard extension argument. \(\square \)

For later reference, let us mention already here that thanks to the uniqueness claim in Proposition 1.1, the following additional information on extensibility under an additional assumption on uniform continuity directly follows from [31, Lemma 3.2].

Lemma 3.3

Let \(n\ge 2\), \(p\in [1,n)\), and \(u_0\in BUC(\mathbb {R}^n)\cap L^p(\mathbb {R}^n)\) be radial and nonnegative, and let \(T_{max}=T_{max}(u_0)\in (0,\infty ]\) and (uv) be as in Proposition 1.1. Then it follows that

$$\begin{aligned} \hbox {if } T_{max}<\infty , \quad \hbox {then} \quad \limsup _{t\nearrow T_{max}} \Vert \nabla v(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)}=\infty . \end{aligned}$$

4 Global existence for all data below \(u_\star \). Proof of Theorem 1.2

The following one-parameter family of stationary (super-)solutions to (2.10) has been made use of in the literature already [3, 31].

Lemma 4.1

Let \(n\ge 1\), and for \(b\ge 0\) let

$$\begin{aligned} U(r)\equiv U^{(b)}(r):=\frac{1}{r^2+b}, \quad r>0. \end{aligned}$$
(4.1)

Then

$$\begin{aligned} U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2 = \frac{-4b}{(r^2+b)^3} \quad \hbox {for all } r>0. \end{aligned}$$
(4.2)

Proof

We compute

$$\begin{aligned} U_r(r)=\frac{-2r}{(r^2+b)^2} \quad \hbox { and } \quad U_{rr}(r)=\frac{6r^2-2b}{(r^2+b)^3}, \quad r>0, \end{aligned}$$

and hence obtain that for all \(r>0\),

$$\begin{aligned} U_{rr} + 2rUU_r + \frac{n+1}{r} U_r + 2nU^2= & {} \frac{6r^2-2b}{(r^2+b)^3} - \frac{4r^2}{(r^2+b)^3} \\{} & {} - \frac{2(n+1)}{(r^2+b)^2} + \frac{2n}{(r^2+b)^2} \\= & {} \frac{2r^2-2b}{(r^2+b)^3} - \frac{2}{(r^2+b)^2} \\= & {} \frac{2r^2-2b-2(r^2+b)}{(r^2+b)^3}, \end{aligned}$$

which is equivalent to (4.2). \(\square \)

In the setting of Theorem 1.2, due to the comparison principle from Lemma 2.3 the particular member \(U^{(0)}\) of the latter family forms an upper bound for the cumulated densities from (2.9):

Lemma 4.2

Let \(n\ge 3\) and \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) be radial and such that (1.7) holds. Then the function defined by (2.9) satisfies

$$\begin{aligned} U(r,t) \le \frac{1}{r^2} \quad \hbox {for all } r>0 \hbox { and } t\in (0,T_{max}(u_0)). \end{aligned}$$
(4.3)

Proof

According to (2.9), from (1.7) it follows that

$$\begin{aligned} U(r,0)= & {} \frac{1}{2r^n} \int _0^r \rho ^{n-1} u_0(\rho ) d\rho \nonumber \\\le & {} \frac{n-2}{r^n} \int _0^r \rho ^{n-3} d\rho \nonumber \\= & {} \frac{1}{r^2} \quad \hbox {for all } r>0, \end{aligned}$$
(4.4)

because \(n\ge 3\). Since Lemma 4.1 asserts that \((0,\infty )^2\ni (r,t) \mapsto \overline{U}(r,t):=\frac{1}{r^2}\) solves (2.10) in \((0,\infty )^2\), and since the boundedness of U in \(\mathbb {R}^n\times (0,\widehat{T})\) for \(\widehat{T}\in (0,T_{max}(u_0))\) ensures that for any such \(\widehat{T}\) we have

$$\begin{aligned} \sup _{t\in (0,\widehat{T})} \Big \{ U(\delta ,t)-\overline{U}(\delta ,t) \Big \} \rightarrow -\infty \quad \hbox {as } \delta \searrow 0, \end{aligned}$$

we may draw on Lemma 2.3 (ii) to infer that, indeed, \(U\le \overline{U}\) in \((0,\infty )\times (0,T_{max}(u_0))\). \(\square \)

An argument involving a strong maximum principle now provides \(b>0\) such that U, even lies below \(U^{(b)}\), inter alia, on the parabolic boundary associated with the domain \((0,r_0)\times (\frac{1}{2}T_{max}(u_0),T_{max}(u_0))\).

Lemma 4.3

Let \(n\ge 3\) and \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) be a radial function fulfilling (1.7), and assume that \(T_{max}=T_{max}(u_0)<\infty \). Then there exist \(r_0>0\) and \(b>0\) such that for U as in (2.9) we have

$$\begin{aligned} U(r_0,t) \le \frac{1}{r_0^2+b} \quad \hbox {for all } t\in (0,T_{max}) \end{aligned}$$
(4.5)

and

$$\begin{aligned} U(r,T_0) \le \frac{1}{r^2+b} \quad \hbox {for all } r\in (0,r_0), \end{aligned}$$
(4.6)

where \(T_0:=\frac{1}{2}T_{max}\).

Proof

Since \(T_0<T_{max}\), we can pick \(c_1>0\) such that \(U(r,T_0)\le c_1\) for all \(r>0\), whence writing \(r_0:=\frac{1}{\sqrt{2c_1}}\) we obtain that

$$\begin{aligned} U(r,T_0) \le \frac{1}{2r_0^2} \quad \hbox {for all } r>0. \end{aligned}$$
(4.7)

As Lemma 4.2 guarantees that beyond this we have

$$\begin{aligned} U(r,t)\le \overline{U}(r,t):=\frac{1}{r^2} \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,T_{max}), \end{aligned}$$
(4.8)

it follows that with \(r_0\) as fixed above we have \(U(r,t)\le \frac{64}{r_0^2}\) for all \(r>\frac{r_0}{8}\) and any \(t\in (0,T_{max})\). Based on the \(L^\infty \) bound for U on \((\frac{r_0}{8},8r_0)\times (0,T_{max})\) thereby provided, we may apply standard interior parabolic Schauder theory [18] to (2.10) to find \(\theta \in (0,1)\) and \(c_2>0\) fulfilling

$$\begin{aligned} \Vert U\Vert _{C^{2+\theta ,1+\frac{\theta }{2}}([\frac{r_0}{4},4r_0]\times [\frac{T_0}{2},T])} \le c_2 \quad \hbox {for all }T\in \Bigg (\frac{T_0}{2},T_{max}\Bigg ). \end{aligned}$$

Consequently, U admits an extension to a solution of the PDE in (2.10) in all of \([\frac{r_0}{4},4r_0]\times [\frac{T_0}{2},T_{max}]\), again denoted by U, which satisfies \(U\le \overline{U}\) in \([\frac{r_0}{4},4r_0]\times [\frac{T_0}{2},T_{max}]\) by (4.8), but for which (4.7) warrants that \(U\not \equiv \overline{U}\) in \([\frac{r_0}{4},4r_0]\times [\frac{T_0}{2},T_{max}]\). As \(\overline{U}\) also solves the PDE in (2.10) by Lemma 4.1, we may therefore employ the strong maximum principle [15] in a straightforward manner to conclude that, in fact, \(U<\overline{U}\) throughout \([\frac{r_0}{2},2r_0]\times [T_0,T_{max}]\); in particular, there exists \(c_3>0\) such that

$$\begin{aligned} U(r_0,t) \le \overline{U}(r_0,t) - c_3 \quad \hbox {for all } t\in (T_0,T_{max}). \end{aligned}$$
(4.9)

Thus, if we let \(b:=\min \{c_3 r_0^4 \,, \, r_0^2\}\), then from (4.9) we obtain (4.5) by observing that

$$\begin{aligned} (r_0^2+b) U(r_0,t)\le & {} (r_0^2+b) \cdot \frac{1}{r_0^2} - c_3\cdot (r_0^2+b) \\= & {} \frac{r_0^2+b-c_3 r_0^4 - c_3 b r_0^2}{r_0^2} \\\le & {} 1-c_3 b \le 1 \quad \hbox {for all } t\in (0,T_{max}), \end{aligned}$$

while recalling (4.7) shows that

$$\begin{aligned} U(r,T_0) \le \frac{1}{r_0^2+r_0^2} \le \frac{1}{r_0^2+b} \le \frac{1}{r^2+b} \quad \hbox {for all } r\in (0,r_0), \end{aligned}$$

and that hence also (4.6) holds. \(\square \)

Again thanks to the comparison principle from Lemma 2.3, a combination of the previous three lemmata now shows that under the extra assumption on uniform continuity underlying Theorem 1.2, the extensibility criterion in Lemma 3.3 ensures global existence for any such radial initial data when located below \(u_\star \):

Proof of Theorem 1.2

Supposing that \(T_{max}=T_{max}(u_0)\) be finite, we pick \(p\in (\frac{n}{2},n)\) and then readily obtain, as a particular consequence of (1.7) and the boundedness of \(u_0\), that \(u_0\in L^p(\mathbb {R}^n)\). As we are furthermore assuming that \(u_0\in BUC(\mathbb {R}^n)\), we may therefore rely on Lemma 3.3 to infer that according to our hypothesis,

$$\begin{aligned} \limsup _{t\nearrow T_{max}} \Vert \nabla v(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)}=\infty . \end{aligned}$$
(4.10)

On the other hand, from Lemma 4.3 we obtain \(r_0>0\), \(T_0\in (0,T_{max})\) and \(b>0\) such that \(U\le \overline{U}\) on \(\big ( \{r_0\}\times (T_0,T_{max})\big ) \cup \big ( (0,r_0)\times \{T_0\}\big )\), where \(\overline{U}(r,t):=\frac{1}{r^2+b}\) for \(r\ge 0\) and \(t\ge 0\). Since \(\overline{U}_t-\overline{U}_{rr}-\frac{n+1}{r}\overline{U}_r - 2r\overline{U}\overline{U}_r - 2n\overline{U}^2 \ge 0\) for \(r>0\) and \(t>0\) by Lemma 4.1, part (iv) of Lemma 2.3 shows that \(U\le \overline{U}\) in \([0,r_0] \times [T_0,T_{max})\). As from the second equation in (1.1) it follows that in the considered radial setting we have \(v_r(r,t)=-r^{1-n}\int _0^r \rho ^{n-1} u(\rho ,t)d\rho =-2rU(r,t)\) for all \(r>0\) and \(t\in (0,T_{max})\), this ensures that

$$\begin{aligned} |v_r(r,t)| \le \frac{2r}{r^2+b} \le \frac{r^2+1}{r^2+b} \le \max \Bigg \{ 1 \,, \, \frac{1}{b}\Bigg \} \quad \hbox {for all }r\!\in \! (0,r_0) \hbox { and } t\!\in \! (T_0,T_{max}).\nonumber \\ \end{aligned}$$
(4.11)

But for larger values of r we may recall Lemma 4.2 to simply estimate

$$\begin{aligned} |v_r(r,t)| \le 2r\cdot \frac{1}{r^2} \le \frac{2}{r_0} \quad \hbox {for all } r>r_0 \hbox { and } t\in (0,T_{max}). \end{aligned}$$

In conjunction with (4.11), this contradicts (4.10) and thereby ensures that, indeed, \(T_{max}=\infty \).

In view of (2.9), the inequality in (1.8) thereupon becomes equivalent to (4.3). \(\square \)

5 Upper estimates for existence times via Kaplan-type arguments

As a preliminary for our subsequent nonexistence analysis underlying Theorems 1.4, 1.3 and 1.5, this section briefly records the outcome of a Kaplan-type blow-up argument [16] when applied to (2.10) (cf. also [6] for a precedent in the context of (1.1)). Here and in what follows, given \(n\ge 1\) we let \(\Delta _{n+2}\) represent the \((n+2)\)-dimensional Laplacian, and let \(\Theta ^{(1)}=\Theta ^{(1)}(r)\) denote the principal Dirichlet eigenfunction of \(-\Delta _{n+2}\) in \(B_1=B_1(0) \subset \mathbb {R}^{n+2}\), normalized such that \(- \int _{B_1} \Theta ^{(1)}=1\), and corresponding to the first eigenvalue \(\lambda _1>0\). Then for arbitrary \(R>0\),

$$\begin{aligned} \Theta ^{(R)}:=\Theta ^{(1)}\Bigg (\frac{\cdot }{R}\Bigg ) \end{aligned}$$
(5.1)

defines a positive eigenfunction of \(-\Delta _{n+2}\) in \(B_R\), hence satisfying \(-\Delta _{n+2} \Theta ^{(R)}= \frac{\lambda _1}{R^2} \Theta ^{(R)}\) in \(B_R\) and \(\Theta ^{(R)}|_{\partial B_R}=0\).

Testing the PDE in (2.10) against \(\Theta ^{(R)}\) now yields the following quantitative statement on nonexistence.

Lemma 5.1

Let \(n\ge 3\), and suppose that for some \(T>0\), \(U\in C^0([0,\infty )\times [0,T)) \cap C^{2,1}([0,\infty )\times (0,T))\) is a nonnegative solution of

$$\begin{aligned} U_t = U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2, \quad r>0, \ t\in (0,T), \end{aligned}$$
(5.2)

for which there exist \(t_0\in [0,T)\) and \(R>0\) such that

$$\begin{aligned} \int _0^R r^{n+1} U(r,t_0) \Theta ^{(R)}(r) dr \ge \frac{2\lambda _1 R^n}{ { (n+2) } (n-2)}. \end{aligned}$$
(5.3)

Then

$$\begin{aligned} T \le t_0 + \frac{2 R^{n+2}}{ { (n+2) } (n-2)} \cdot \Bigg \{ \int _0^R r^{n+1} U(r,t_0) \Theta ^{(R)}(r) dr \Bigg \}^{-1}. \end{aligned}$$
(5.4)

Proof

According to (5.2) and several integrations by parts, \(y(t):=\int _0^R r^{n+1} U(r,t) \)\( \Theta (r) dr\), \(t\in [0,T)\), with \(\Theta =\Theta ^{(R)}\), satisfies

$$\begin{aligned} y'(t)= & {} \int _0^R r^{n+1} \cdot r^{-n-1} (r^{n+1} U_r)_r \Theta dr + \int _0^R r^{n+1} \cdot r(U^2)_r \Theta dr \nonumber \\{} & {} + 2n \int _0^R r^{n+1} U^2 \Theta dr \nonumber \\= & {} \int _0^R U (r^{n+1}\Theta _r)_r dr + \big [ r^{n+1} U_r \Theta - r^{n+1} U\Theta _r \big ]_{r=0}^{r=R} \nonumber \\{} & {} - (n+2) \int _0^R r^{n+1} U^2 \Theta dr + \big [ r^{n+2} U^2 \Theta \big ]_{r=0}^{r=R} - \int _0^R r^{n+2}U^2\Theta _rdr\nonumber \\{} & {} +2n \int _0^R r^{n+1} U^2 \Theta dr \nonumber \\= & {} - \frac{\lambda _1}{R^2} \int _0^R r^{n+1} U\Theta dr + (n-2) \int _0^R r^{n+1} U^2\Theta dr \nonumber \\{} & {} - R^{n+1} U(R,t) \Theta _r(R) - \int _0^R r^{n+2}U^2\Theta _rdr \nonumber \\\ge & {} - \frac{\lambda _1}{R^2} y(t) + \frac{ { (n+2) } (n-2)}{R^{n+2}} y^2(t) \quad \hbox {for all } t\in (t_0,T), \end{aligned}$$
(5.5)

because \(\Theta (R)=0\ge \Theta _r(R)\), and because

$$\begin{aligned} y^2(t)\le & {} \bigg \{ \int _0^R r^{n+1} \Theta dr \bigg \} \cdot \int _0^R r^{n+1} U^2 \Theta dr \\= & {} R^{n+2} \cdot \bigg \{ \int _0^1 \rho ^{n+1} \Theta ^{(1)}(\rho ) { d\rho } \bigg \} \cdot \int _0^R r^{n+1} U^2\Theta dr \\= & {} \frac{R^{n+2}}{ { n+2 } } \int _0^R r^{n+1} U^2\Theta dr \quad \hbox {for all } t\in (t_0,T) \end{aligned}$$

according to the Cauchy–Schwarz inequality and our normalization of \(\Theta ^{(1)}\). But since (5.3) ensures that

$$\begin{aligned} -\frac{\lambda _1}{R^2} y(t_0) + \frac{ { (n+2) } (n-2)}{R^{n+2}} y^2(t_0) \ge \frac{ { (n+2) } (n-2)}{2R^{n+2}} y^2(t_0), \end{aligned}$$

an ODE comparison argument applied to (5.5) shows that

$$\begin{aligned} y'(t) \ge \frac{ { (n+2) } (n-2)}{2R^{n+2}} y^2(t) \quad \hbox {for all } t\in (t_0,T), \end{aligned}$$

which upon an integration leads to the inequality

$$\begin{aligned} \frac{1}{y(t_0)} \ge - \frac{1}{y(t)} + \frac{1}{y(t_0)} \ge \frac{ { (n+2) } (n-2)}{2R^{n+2}} \cdot (t-t_0) \quad \hbox {for all } t\in (t_0,T), \end{aligned}$$

yielding (5.4) in the limit \(t\nearrow T\). \(\square \)

Although in this form essentially made in [6] already, for completeness let us derive the following lemma, to be used in Theorem 1.4, as a particular consequence of Lemma 5.1.

Lemma 5.2

Let \(n\ge 3\) and \(r_0>0\). Then there exists \(K(r_0)>0\) with the property that if \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is nonnegative and radial and such that with U as in (2.9) we have

$$\begin{aligned} U(r_0,t_0)\ge K(r_0) \quad \hbox {for some } t_0\in [0,T_{max}(u_0)), \end{aligned}$$
(5.6)

then

$$\begin{aligned} T_{max}(u_0) < \infty . \end{aligned}$$
(5.7)

Proof

Given \(r_0>0\), we let \(R\equiv R(r_0):=2r_0\) and \(c_1\equiv c_1(r_0):=\int _{r_0}^{2r_0} r\Theta ^{(R)}(r) dr\) as well as

$$\begin{aligned} K(r_0):= \frac{2^{n+1} \lambda _1}{ { (n+2) } (n-2) c_1}. \end{aligned}$$
(5.8)

Then assuming that \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is nonnegative and radial and such that (5.6) holds for some \(t_0\in [0,T_{max}(u_0))\), we observe that since \((r^n U)_r = \frac{1}{2} r^{n-1} u \ge 0\) for all \(r>0\) and \(t\in (0,T_{max}(u_0))\) by (2.9), from (5.6) it follows that

$$\begin{aligned} r^n U(r,t_0) \ge r_0^n U(r_0,t_0) \ge r_0^n K(r_0) \quad \hbox {for all } r>r_0. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _0^R r^{n+1} U(r,t_0) \Theta ^{(R)}(r) dr \ge \int _{r_0}^{2r_0} r^{n+1} U(r,t_0) \Theta ^{(R)}(r) dr \ge c_1 r_0^n K(r_0), \end{aligned}$$

so that since

$$\begin{aligned} \frac{c_1 r_0^n K(r_0)}{\frac{2\lambda _1 \cdot (2r_0)^n}{ { (n+2) } (n-2)}} = \frac{ { (n+2) } (n-2) c_1 K(r_0)}{2^{n+1}\lambda _1} =1 \end{aligned}$$

by (5.8), from Lemma 5.1 we infer that, in fact, \(T_{max}(u_0)<\infty \). \(\square \)

6 Blow-up for some special initial data decaying more slowly than \(u_\star \)

6.1 Conservation of spatial decay. A necessary blow-up condition involving U

A further comparison argument independent from the above, now addressing a framework of time-dependent spatial domains in the style of Lemma 2.3 (ii), will be of importance in making sure that within finite time intervals, for initial data satisfying a mild assumption on spatial decay the corresponding taxis gradients remain uniformly bounded (see Lemma 6.2).

Lemma 6.1

Let \(n\ge 1\), and suppose that \(0\le u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is radial and such that the function \(U_0\) from (2.11) satisfies

$$\begin{aligned} U_0(r) \le \frac{A}{r-r_\star } \quad \hbox {for all } r>r_\star \end{aligned}$$
(6.1)

with some \(A>0\) and \(r_\star \ge \frac{1}{A}\). Then

$$\begin{aligned} U(r,t) \le \frac{A}{r{-}r_\star {-}2nAt} \quad \hbox {for all } t\in (0,T_{max}(u_0)) \hbox { and } r>r_\star {+}2nAt. \end{aligned}$$
(6.2)

Proof

We let

$$\begin{aligned} \overline{U}(r,t):=\frac{A}{r-r_\star -2nAt}, \quad t\ge 0, \ r>r_\star +2nAt, \end{aligned}$$

and compute

$$\begin{aligned} \overline{U}_r\!=\!-\frac{A}{(r-r_\star -2nAt)^2}, \quad \overline{U}_{rr} \!=\! \frac{2A}{(r\!-\!r_\star \!-\!2nAt)^3} \quad \hbox {and} \quad \overline{U}_t \!=\! \frac{2nA^2}{(r\!-\!r_\star \!-\!2nAt)^2} \end{aligned}$$

for \(t>0\) and \(r>r_\star +2nAt\). Therefore,

$$\begin{aligned}{} & {} \overline{U}_t - \overline{U}_{rr} - \frac{n+1}{r} \overline{U}_r - 2r\overline{U}\overline{U}_r - 2n\overline{U}^2 \\{} & {} \qquad \frac{2nA^2}{(r-r_\star -2nAt)^2} - \frac{2A}{(r-r_\star -2nAt)^3} + \frac{(n+1)A}{r(r-r_\star -2nAt)^2} \\{} & {} \qquad + \frac{2rA^2}{(r-r_\star -2nAt)^3} - \frac{2nA^2}{(r-r_\star -2nAt)^2} \\{} & {} \quad = \frac{2A(rA-1)}{(r-r_\star -2nAt)^3} + \frac{(n+1)A}{r(r-r_\star -2nAt)^2} \\{} & {} \quad \ge 0 \quad \hbox {for all } t>0 \hbox { and } r>r_\star +2nAt, \end{aligned}$$

because \(r_\star A-1\ge 0\) according to our hypothesis. Since (6.1) ensures that \(U(r,0)\le \overline{U}(r,0)\) for all \(r>r_\star \), since

$$\begin{aligned} \inf _{t\in (0,\widehat{T})} \overline{U}(r_\star +2nAt+\delta ,t)=\frac{A}{\delta } \rightarrow + \infty \quad \hbox {as } \delta \searrow 0 \quad \hbox {for all } \widehat{T}>0, \end{aligned}$$

and since clearly

$$\begin{aligned} \sup _{t\in (0,\widehat{T})} \sup _{r>r_\star +2nAt+\delta } \overline{U}(r,t)=\frac{A}{\delta } < \infty \quad \hbox {for all }\widehat{T}>0 {\hbox { and }} \delta >0, \end{aligned}$$

an application of Lemma 2.3 (ii) shows that, as claimed, \(U(r,t)\le \overline{U}(r,t)\) for all \(t\in (0,T_{max}(u_0))\) and each \(r>r_\star +2nAt\). \(\square \)

In fact, the latter provides some favorable information on spatial decay under a comparably mild assumption on decay of the initial data.

Lemma 6.2

Let \(n\ge 1\), and assume that \(0\le u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is radial and such that for \(U_0\) as in (2.11) we have

$$\begin{aligned} \limsup _{r\rightarrow \infty } rU_0(r) < \infty , \end{aligned}$$
(6.3)

and that moreover \(T_{max}(u_0)<\infty \). Then there exist \(r_0>0\) and \(C>0\) such that

$$\begin{aligned} rU(r,t) \le C \quad \hbox {for all } r>r_0{\hbox { and }} t\in (0,T_{max}(u_0)). \end{aligned}$$
(6.4)

Proof

In line with (6.3), we fix \(r_1>0\) and \(A>0\) such that

$$\begin{aligned} U_0(r) \le \frac{A}{r} \quad \hbox {for all } r>r_1, \end{aligned}$$

whence letting \(r_\star :=\max \big \{\frac{1}{A},r_1\big \}\) we particularly obtain that the assumptions of Lemma 6.1 are met. We therefore may rely on (6.2) to infer upon defining \(r_0:=2(r_\star +2nAT_{max}(u_0))\) that

$$\begin{aligned} rU(r,t) \le \frac{Ar}{r-r_\star -2nAt} = \frac{A}{1-\frac{r_\star +2nAt}{r}} \le \frac{A}{1-\frac{r_\star +2nAT_{max}(u_0)}{r_0}} = 2A \end{aligned}$$

for all \(r>r_0\) and \(t\in (0,T_{max}(u_0))\). \(\square \)

Under a condition slightly stronger than that from Lemma 6.2, the above can be used in the context of an application of Lemma 3.3 to ensure that finite-time blow-up in (1.1) is possible only when U becomes unbounded inside some fixed compact spatial domain.

Lemma 6.3

If \(n\ge 1\), \(p\in [1,n)\) and \(u_0\in BUC(\mathbb {R}^n)\) is nonnegative and radial with

$$\begin{aligned} \limsup _{|x|\rightarrow \infty } |x|^\beta u_0(x) < \infty \end{aligned}$$
(6.5)

for some \(\beta >1\), and if furthermore \(T_{max}(u_0)<\infty \), then one can find \(r_0>0\) such that

$$\begin{aligned} \limsup _{t\nearrow T_{max}(u_0)} \sup _{r\in [0,r_0]} U(r,t)=\infty . \end{aligned}$$
(6.6)

Proof

Since \(\beta >1\), we can fix \(p\in [1,n)\) such that \(p\beta >n\), whence observing that then \(\int _ {\mathbb {R}^n{\setminus }B_1} |x|^{-p\beta } dx <\infty \), from (6.5) we obtain that \(u_0\) also belongs to \(L^p(\mathbb {R}^n)\). We may therefore employ Lemma 3.3 to infer that due to the assumed finiteness of \(T_{max}=T_{max}(u_0)\),

$$\begin{aligned} \limsup _{t\nearrow T_{max}} \Vert \nabla v(\cdot ,t)\Vert _{L^\infty (\mathbb {R}^n)} = \infty . \end{aligned}$$
(6.7)

To derive (6.6) from this, we recall that by (1.1) and (2.9), the radial function v satisfies

$$\begin{aligned} v_r(r,t)=-2rU(r,t) \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,T_{max}). \end{aligned}$$
(6.8)

As Lemma 6.2 provides \(r_0>0\) and \(c_1>0\) such that \(U(r,t) \le \frac{c_1}{r}\) for all \(r>r_0\) and \(t\in (0,T_{max})\), namely, this firstly ensires that

$$\begin{aligned} \big |v_r(r,t)\big | \le 2c_1 \quad \hbox {for all } r>r_0 \hbox { and } t\in (0,T_{max}). \end{aligned}$$
(6.9)

Secondly, if (6.6) was false and hence \(U(r,t) \le c_2\) for all \(r\in [0,r_0]\), each \(t\in (0,T_{max})\) and some \(c_2>0\), then from (6.8) we could conclude that

$$\begin{aligned} \big |v_r(r,t)\big | \le 2c_2 r_0 \quad \hbox {for all } r\in [0,r_0] \hbox { and } t\in (0,T_{max}). \end{aligned}$$

In combination with (6.9), this would contradict (6.7). \(\square \)

6.2 Some general properties of stationary (sub-)solutions to (2.10)

The following statement on asymptotic behavior of stationary solutions to (2.10) is essentially well-known, actually even in a somewhat stronger form [2]; as our subsequent application will involve solutions which are possibly singular at the origin and hence do not exactly meet the regularity assumptions made in the literature, we include a brief proof for completeness.

Lemma 6.4

Let \(n\ge 3\) and \(r_0\ge 0\), and let \(U\in C^2((r_0,\infty ))\) be nonnegative and such that

$$\begin{aligned} rU_r + nU \ge 0 \quad \hbox {for all } r>r_0 \end{aligned}$$
(6.10)

and

$$\begin{aligned} U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2 =0 \quad \hbox {for all } r>r_0. \end{aligned}$$
(6.11)

Then

$$\begin{aligned} \liminf _{r\rightarrow \infty } r^2 U(r) \le 1. \end{aligned}$$
(6.12)

Proof

If the claim was false, then we could find \(a>1\) and \(r_1>r_0\) such that \(U(r)\ge \frac{a}{r^2}\) for all \(r>r_1\), where since \(n\ge 3\), we may assume that \(2a<n\). Together with (6.10), this would entail that

$$\begin{aligned} 2rUU_r + 2nU^2 \ge \frac{2a}{r^2} \cdot (rU_r+nU) = 2ar^{-n-1} (r^n U)_r \quad \hbox {for all } r>r_1. \end{aligned}$$

From (6.11) we would therefore obtain that

$$\begin{aligned} (r^{n+1} U_r + 2ar^n U)_r= & {} r^{n+1} \Bigg (U_{rr}+\frac{n+1}{r}U_r\Bigg ) + 2a(r^n U)_r \\\le & {} r^{n+1}\Bigg (U_{rr}+\frac{n+1}{r}U_r\Bigg ) + r^{n+1} (2rUU_r + 2nU^2) =0 \end{aligned}$$

for all \(r>r_1\), and that hence \(r^{n+1} U_r + 2ar^n U \le c_1:=r_1^{n+1} U_r(r_1)+ 2ar_1^n U(r_1)\) for all \(r>r_1\). Consequently, \(U_r \le - \frac{2a}{r} U + c_1 r^{-n-1}\) for all \(r>r_1\), so that one further integration relying on our restriction \(n<2a\) would show that

$$\begin{aligned} U(r)\le & {} U(r_1) e^{-2a\int _{r_1}^r \rho ^{-1} d\rho } + c_1 \int _{r_1}^r e^{-2a\int _{\rho }^r \sigma ^{-1} d\sigma } \cdot \rho ^{-n-1} d\rho \\= & {} U(r_1) \cdot \Bigg (\frac{r}{r_1}\Bigg )^{-2a} + c_1 \int _{r_1}^r \Bigg (\frac{r}{\rho }\Bigg )^{-2a} \cdot \rho ^{-n-1} d\rho \\= & {} U(r_1) \cdot \Bigg (\frac{r}{r_1}\Bigg )^{-2a} + \frac{c_1}{n-2a} r^{-2a} (r_1^{2a-n} - r^{2a-n}) \\\le & {} \Bigg \{ r_1^{2a} U(r_1) + \frac{c_1 r_1^{2a-n}}{n-2a}\Bigg \} \cdot r^{-2a} \quad \hbox {for all } r>r_1, \end{aligned}$$

which due to the inequality \(2a>2\) is incompatible with our hypothesis. \(\square \)

We add a basic observation on monotonicity of \(0\le r\mapsto r^n U\) for suitably regular steady-state subsolutions to (2.10). This will be used both in Theorem 1.4 and in Theorem 1.3 below.

Lemma 6.5

Let \(n\ge 1\), and suppose that \(U\in W^{2,\infty }_{loc}([0,\infty ))\) is nonnegative and such that

$$\begin{aligned} U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2 \ge 0 \quad \hbox {for a.e. } r>0. \end{aligned}$$
(6.13)

Then

$$\begin{aligned} rU_r + nU \ge 0 \quad \hbox {for all } r>0. \end{aligned}$$
(6.14)

Proof

Letting \(z(r):=rU_r(r)+nU(r)\), \(r\ge 0\), we see that \(z\in W^{1,\infty }_{loc}([0,\infty ))\) with \(z(0)\ge 0\) and

$$\begin{aligned} z_r{} & {} = rU_{rr} + (n+1)U_r \ge r\cdot \Big \{ - \frac{n+1}{r} U_r - 2rUU_r - 2nU^2\Big \} + (n+1) U_r \\{} & {} = - 2r^2 UU_r - 2nrU^2 \end{aligned}$$

for a.e. \(r>0\). Therefore, \(z_r \ge - 2rUz\) and hence \(\partial _r \big \{ \exp \big ( 2\int _0^r \rho U(\rho ) d\rho \big ) \cdot z(r)\big \} \ge 0\) for a.e. \(r>0\), so that z must indeed remain nonnegative throughout \((0,\infty )\). \(\square \)

6.3 Blow-up for temporally nondecreasing solutions to (2.10) with slow decay

The following lemma now contains the main result of this auxiliary section: By combining Lemma 5.2 with Lemma 6.3, by means of a further maximum principle type argument based on Lemma 2.3 we can make sure that in the particular situation when the initial data in (2.10) are stationary subsolutions decaying more slowly than the functions addressed in (6.12), the corresponding solution must blow up in the following sense.

Lemma 6.6

Let \(n\ge 3\) and \(p\in [1,n)\), and suppose that \(u_0\in BUC(\mathbb {R}^n)\cap L^p(\mathbb {R}^n)\) is nonnegative and radial, fulfilling (6.5) for some \(\beta >1\) and such that the function \(U_0\) defined in (2.11) is a member of \(W^{2,\infty }_{loc}([0,\infty ))\) which satisfies

$$\begin{aligned} U_{0rr} + \frac{n+1}{r} U_{0r} + 2rU_0 U_{0r} + 2nU_0^2 \ge 0 \quad \hbox {a.e. in } (0,\infty ) \end{aligned}$$
(6.15)

as well as

$$\begin{aligned} \liminf _{r\rightarrow \infty } r^2 U_0(r) >1. \end{aligned}$$
(6.16)

Then \(T_{max}=T_{max}(u_0)<\infty \), and there exists \(r_\star >0\) such that

$$\begin{aligned} U(r_\star ,t) \rightarrow +\infty \quad \hbox {as } t\nearrow T_{max}. \end{aligned}$$
(6.17)

Proof

In view of a standard argument based on Lemma 2.3 [24, Proposition 52.19], the subsolution property expressed in (6.15) ensures that

$$\begin{aligned} U_t(r,t)\ge 0 \quad \hbox {for all } r\ge 0 \hbox { and } t\in (0,T_{max}), \end{aligned}$$
(6.18)

and that thus there exists \(U_\infty : [0,\infty )\rightarrow [0,\infty ]\) such that

$$\begin{aligned} U(r,t)\nearrow U_\infty (r) \quad \hbox {for all } r\ge 0 \quad \hbox {as } t\nearrow T_{max}. \end{aligned}$$
(6.19)

Now assuming that \(T_{max}=\infty \), upon employing Lemma 5.2 we would infer that necessarily \(U_\infty (r)<\infty \) for each individual \(r>0\), whence again relying on the fact that

$$\begin{aligned} (r^n U)_r = \frac{1}{2} \cdot r^{n-1} u \ge 0 \quad \hbox {for all } r>0 \hbox { and } t\in (0,T_{max}) \end{aligned}$$
(6.20)

by (2.9), once more due to (6.19) we readily obtain that for all \(r_1>0\) and \(r_2>r_1\) we could find \(c_1=c_1(r_1,r_2)>0\) fulfilling

$$\begin{aligned} U(r,t)\le c_1 \quad \hbox {for all } r\in (r_1,r_2) {\hbox { and }} t>0. \end{aligned}$$
(6.21)

Based on this \(L^\infty \) bound, we could therefore apply interior parabolic Schauder estimates [18] to (2.10) to infer that for any such \(r_1\) and \(r_2\) there would exist \(\theta =\theta (r_1,r_2)\in (0,1)\) and \(c_2=c_2(r_1,r_2)>0\) such that

$$\begin{aligned} \Vert U\Vert _{C^{2+\theta ,1+\frac{\theta }{2}}([r_1,r_2]\times [t,t+1])} \le c_2 \quad \hbox {for all } t>1. \end{aligned}$$

In particular, due to the Arzelà–Ascoli theorem this implies that actually \(U_\infty \in C^2([0,\infty ))\), and that in (6.19) we would have

$$\begin{aligned} U(\cdot ,t)\rightarrow U_\infty \quad \hbox { in } C^2_{loc}((0,\infty )) \quad \hbox {as } t\rightarrow \infty , \end{aligned}$$
(6.22)

and that moreover also \(U_t(\cdot ,t)\rightarrow 0\) in \(C^0_{loc}((0,\infty ))\) as \(t\rightarrow \infty \), because from (6.18) and (6.21) we know that \(\int _0^\infty |U_t(r,t)|dt <\infty \) for each fixed \(r>0\). Consequently, taking \(t\rightarrow \infty \) in (2.10) we would infer that \(0=U_{\infty rr} + \frac{n+1}{r} U_{\infty r} + 2r U_\infty U_{\infty r} + 2nU_\infty ^2\) for all \(r>0\), whence noting that \(r^n U_{\infty r} + nr^{n-1} U_\infty \ge 0\) for all \(r>0\) by (6.20) and (6.22), we could employ Lemma 6.4 to conclude that

$$\begin{aligned} \liminf _{r\rightarrow \infty } r^2 U_\infty (r) \le 1. \end{aligned}$$

In light of (6.18), however, this contradicts (6.16) and thereby confirms that, indeed, \(T_{max}<\infty \).

As our assumptions on \(u_0\) guarantee applicability of Lemma 6.3, the divergence feature in (6.17), finally, is implied by (6.6) when combined with (6.19). \(\square \)

7 A blow-up criterion exclusively involving tail behavior. Proof of Theorem 1.4

7.1 A family of favorably small stationary subsolutions to (2.10)

The key for our derivation of the sufficient condition for blow-up provided by Theorem 1.4 can be found in the following obsevation on subsolution properties in a second family of stationary profiles with decay of quadratic nature similar to that in (4.1). The corresponding subclass obtained on letting \(\alpha =1\) in (7.2) has already been used for an analysis in bounded domains [7], but admitting more general \(\alpha \ge 1\) will enable us to achieve significantly smaller choices of the parameter a which measures asymptotic distance to \(u_\star \).

Lemma 7.1

Let \(n\ge 3\), \(\alpha \ge 1\) and \(a>1\) be such that

$$\begin{aligned} a\ge g(\alpha ):= & {} \min \Bigg \{ \frac{n\alpha }{2}-\frac{3n}{2}+4+\frac{n}{\alpha }-\frac{3}{\alpha }+\frac{2}{n\alpha } \,, \, \nonumber \\{} & {} \frac{n\alpha }{2}-\frac{n}{2}+2-\frac{1}{\alpha }+\frac{2}{n\alpha } - \Bigg (1-\frac{n-2}{n\alpha }\Bigg ) \cdot \sqrt{n(n-2)(\alpha -1)} \Bigg \}.\nonumber \\ \end{aligned}$$
(7.1)

Then for any choice of \(b>0\),

$$\begin{aligned} U(r)\equiv U^{(\alpha ,a,b)}(r):=\frac{ar^{n\alpha -n}}{r^{n\alpha -n+2}+b}, \quad r\ge 0, \end{aligned}$$
(7.2)

satisfies

$$\begin{aligned} U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2 \ge 0 \quad \hbox {for all } r>0. \end{aligned}$$
(7.3)

Proof

Since for \(r>0\) we have

$$\begin{aligned} U_r(r)=\frac{a}{(r^{n\alpha -n+2}+b)^2} \cdot \Big \{ -2r^{2n\alpha -2n+1} + (n\alpha -n) b r^{n\alpha -n-1}\Big \} \end{aligned}$$

and

$$\begin{aligned} U_{rr}(r)= & {} \frac{a}{(r^{n\alpha -n+2}+b)^3} \cdot \Big \{ 6r^{3n\alpha -3n+2} \\{} & {} + (-n^2 \alpha ^2 + 2n^2 \alpha - 9n\alpha - n^2 + 9n-2) b r^{2n\alpha -2n} \\{} & {} + (n^2\alpha ^2 - 2n^2 \alpha - n\alpha + n^2+n) b^2 r^{n\alpha -n-2}\Big \}, \end{aligned}$$

we obtain that for all \(r>0\),

$$\begin{aligned}{} & {} \frac{(r^{n\alpha -n+2}+b)^3}{a} \cdot \Bigg \{ U_{rr} + \frac{n+1}{r} U_r + 2rUU_r + 2nU^2\Bigg \} \nonumber \\{} & {} \quad = 6r^{3n\alpha -3n+2} \nonumber \\{} & {} \qquad + (-n^2 \alpha ^2 + 2n^2 \alpha - 9n\alpha - n^2 +9n -2) br^{2n\alpha -2n} \nonumber \\{} & {} \qquad +(n^2 \alpha ^2 - 2n^2 \alpha - n\alpha + n^2 +n) b^2 r^{n\alpha -n-2} \nonumber \\{} & {} \qquad + \frac{n+1}{r} \cdot \big \{ -2r^{2n\alpha -2n+1} + (n\alpha -n) br^{n\alpha -n-1}\big \}\cdot \big \{ r^{n\alpha -n+2}+b\big \} \nonumber \\{} & {} \qquad +2ar^{n\alpha -n+1} \cdot \big \{ -2r^{2n\alpha -2n+1} + (n\alpha -n) b r^{n\alpha -n-1}\big \} \nonumber \\{} & {} \qquad +2na r^{2n\alpha -2n} \cdot \big \{ r^{n\alpha -n+2}+b\big \} \nonumber \\{} & {} \quad = (6-2n-2-4a+2na) r^{3n\alpha -3n+2} \nonumber \\{} & {} \qquad + (-n^2 \alpha ^2 + 2n^2 \alpha -9n\alpha - n^2+9n-2 - 2n-2+n^2 \alpha - n^2 +n\alpha -n \nonumber \\{} & {} \qquad +2n\alpha a -2na +2na) br^{2n\alpha -2n} \nonumber \\{} & {} \qquad + (n^2 \alpha ^2 - 2n^2 \alpha -n\alpha +n^2+n+n^2 \alpha - n^2 +n\alpha -n) b^2 r^{n\alpha -n-2} \nonumber \\{} & {} \quad = I(r):= c_1(a-1) r^{3n\alpha -3n+2} + 2n\alpha (a-c_2) b r^{2n\alpha -2n} + n^2 \alpha (\alpha -1) b^2 r^{n\alpha -n-2}, \nonumber \\ \end{aligned}$$
(7.4)

where

$$\begin{aligned} c_1:=2(n-2) \quad \hbox { and } \quad c_2\equiv c_2(\alpha ):=\frac{n\alpha }{2}-\frac{3n}{2}+4+\frac{n}{\alpha } - \frac{3}{\alpha } + \frac{2}{n\alpha }. \end{aligned}$$

Now in the case when \(a\ge c_2\), our assumptions that \(a>1\) and \(\alpha \ge 1\) directly ensure that (7.4) implies (7.3). If, conversely, \(a<c_2\), then we use Young’s inequality to estimate

$$\begin{aligned} \big |2n\alpha (a-c_2)\big | \cdot br^{2n\alpha -2n}{} & {} \le c_1(a-1) r^{3n\alpha -3n+2} + \frac{n^2 \alpha ^2 (a-c_2)^2}{c_1(a-1)} \cdot b^2 r^{n\alpha -n-2} \\{} & {} \quad \hbox {for all } r>0, \end{aligned}$$

so that our hypothesis in (7.1) guarantees that (7.3) also holds in this case, because

$$\begin{aligned} \frac{n^2 \alpha ^2 (a-c_2)^2}{c_1(a-1)} \le n^2 \alpha (\alpha -1). \end{aligned}$$
(7.5)

In fact, writing

$$\begin{aligned} a_{\pm } \equiv a_{\pm }(\alpha ):=c_2 + \frac{(\alpha -1) c_1}{2\alpha } \pm \sqrt{\frac{(\alpha -1) c_1}{\alpha } (c_2-1) + \frac{(\alpha -1)^2 c_1^2}{4\alpha ^2}}, \end{aligned}$$

we readily verify that

$$\begin{aligned} \alpha (a-c_2)^2 - (\alpha -1) c_1(a-1)= & {} \alpha \cdot \Bigg \{ a^2-2c_2 a {-} \frac{(\alpha -1)c_1}{\alpha } \, a + c_2^2 {+} \frac{(\alpha -1)c_1}{\alpha } \Bigg \} \nonumber \\= & {} \alpha \cdot (a-a_+)(a-a_-), \end{aligned}$$
(7.6)

and that by definition of \(c_1\) and \(c_2\), both numbers \(a_+\) and \(a_-\) are real with

$$\begin{aligned} a_\pm= & {} \frac{n\alpha }{2} - \frac{3n}{2} + 4 + \frac{n}{\alpha } - \frac{3}{\alpha } + \frac{2}{n\alpha } + \frac{(n-2)(\alpha -1)}{\alpha } \\{} & {} \pm \sqrt{\frac{(n-2)(\alpha -1)}{\alpha } \cdot \Bigg (n\alpha -3n+6+\frac{2n}{\alpha }-\frac{6}{\alpha }+\frac{4}{n\alpha }\Bigg ) + \frac{(n-2)^2 (\alpha -1)^2}{\alpha ^2}} \\= & {} \frac{n\alpha }{2} - \frac{n}{2} + 2 - \frac{1}{\alpha } + \frac{2}{n\alpha } \pm \Bigg (1-\frac{n-2}{n\alpha }\Bigg ) \cdot \sqrt{n(n-2)(\alpha -1)}. \end{aligned}$$

Since the inequality \(a<c_2\) thus clearly implies that \(a\le a_+\), and since (7.1) in this case precisely states that \(a\ge a_-\), from (7.6) we infer that indeed (7.5) holds.\(\square \)

Indeed, elementary calculus shows that for the subsolution property in question it is sufficient to choose the parameter a in (7.1) above a certain threshold which always lies below the number \(\frac{n+2}{n}\) seen in the bounded domain analysis in [7], and which in high-dimensional settings even coincides with the coefficient 1 that corresponds to the asymptotics in (1.2).

Lemma 7.2

Let \(n\ge 3\), and let g be as in Lemma 7.1. Then

$$\begin{aligned} \min _{\alpha \ge 1} g(\alpha ) \left\{ \begin{array}{ll}<\frac{n+2}{n} - \frac{2(n-2)}{n^2+2n-4} \quad &{} \hbox {if } 3\le n\le 9, \\ =1 &{} \hbox {if } n\ge 10. \end{array} \right. \end{aligned}$$
(7.7)

Proof

We first consider the case when \(n\ge 10\), in which we use that according to (7.1) we have

$$\begin{aligned} g(\alpha )\le g_1(\alpha ):=\frac{n\alpha }{2}-\frac{3n}{2} + 4 + \frac{n}{\alpha } - \frac{3}{\alpha } + \frac{2}{n\alpha }, \quad \alpha \ge 1. \end{aligned}$$

Since

$$\begin{aligned} g_1'(\alpha )=\frac{n}{2}-\Bigg (n-3+\frac{2}{n}\Bigg ) \cdot \frac{1}{\alpha ^2} \quad \hbox {for all } \alpha >1, \end{aligned}$$

it follows that if we let \(\alpha _1:=\sqrt{\frac{2}{n}(n-3+\frac{2}{n})} \equiv \sqrt{2-\frac{6}{n}+\frac{4}{n^2}}\), then

$$\begin{aligned} \min _{\alpha \ge 1} g_1(\alpha ) = g_1(\alpha _1)= & {} \frac{n}{2}\sqrt{2-\frac{6}{n}+\frac{4}{n^2}} - \frac{3n}{2} + 4 + \Bigg (n-3+\frac{2}{n}\Bigg ) \cdot \frac{1}{\sqrt{2-\frac{6}{n}+\frac{4}{n^2}}} \\= & {} \frac{n}{2} \sqrt{2-\frac{6}{n}+\frac{4}{n^2}} - \frac{3n}{2} + 4 + \frac{n}{2} \sqrt{2-\frac{6}{n}+\frac{4}{n^2}} \\= & {} \sqrt{2n^2-6n+4} - \frac{3n}{2} +4, \end{aligned}$$

because our current assumption \(n\ge 10\) particularly guarantees that \(\alpha _1 \ge \sqrt{2-\frac{6}{n}} \ge \sqrt{2-\frac{6}{10}}>1\). As the inequality \(n\ge 10\) furthermore entails that

$$\begin{aligned} \big ( \sqrt{2n^2-6n+4} \big )^2 - \Bigg (\frac{3n}{2}-4+1\Bigg )^2= & {} 2n^2-6n+4-\frac{9n^2}{4}+9n-9 \\= & {} - \frac{n^2}{4}+3n-5 = -\frac{(n-2)(n-10)}{4} \le 0, \end{aligned}$$

this implies that

$$\begin{aligned} \min _{\alpha \ge 1} g_1(\alpha )\le 1, \end{aligned}$$

as claimed.

If \(n\le 9\), however, then making use of the second option offered in (7.1) we estimate

$$\begin{aligned} g(\alpha ) \le g_2(\alpha ):=\frac{n\alpha }{2} - \frac{n}{2} + 2 - \frac{1}{\alpha } + \frac{2}{n\alpha } -\Bigg (1-\frac{n-2}{n\alpha }\Bigg ) \sqrt{n(n-2)(\alpha -1)}, \quad \alpha \ge 1, \end{aligned}$$

where we observe that since \(1-\xi \le \frac{1}{1+\xi }\le 1\) for all \(\xi \ge 0\), we have

$$\begin{aligned} g_2(1+\xi )= & {} \frac{n\xi }{2} + 2 - \frac{n-2}{n} \cdot \frac{1}{1+\xi } - \Bigg (1-\frac{n-2}{n(1+\xi )} \Bigg ) \sqrt{n(n-2)\xi } \nonumber \\\le & {} \frac{n\xi }{2} + 2 - \frac{n-2}{n} (1-\xi ) - \Bigg (1-\frac{n-2}{n}\Bigg )\sqrt{n(n-2)\xi } \nonumber \\= & {} h(\xi ):=\frac{n^2+2n-4}{2n} \cdot \xi + \frac{n+2}{n} - 2\sqrt{\frac{n-2}{n} \cdot \xi } \quad \hbox {for all } \xi \ge 0.\nonumber \\ \end{aligned}$$
(7.8)

Minimizing here shows that since

$$\begin{aligned} h'(\xi )=\frac{n^2+2n-4}{2n} - \sqrt{\frac{n-2}{n\xi }} \quad \hbox {for } \xi >0, \end{aligned}$$

we have

$$\begin{aligned} \min _{\xi >0} h(\xi )= & {} h\Bigg ( \frac{4n(n-2)}{(n^2+2n-4)^2}\Bigg ) \\= & {} \frac{n^2+2n-4}{2n} \cdot \frac{4n(n-2)}{(n^2+2n-4)^2} + \frac{n+2}{n} - 2\sqrt{\frac{n-2}{n} \cdot \frac{4n(n-2)}{(n^2+2n-4)^2}} \\= & {} \frac{2(n-2)}{n^2+2n-4} + \frac{n+2}{n} - \frac{4(n-2)}{n^2+2n-4} \\= & {} \frac{n+2}{n} - \frac{2(n-2)}{n^2+2n-4}. \end{aligned}$$

Inserting this into (7.8) completes the proof of (7.7). \(\square \)

7.2 Proof of Theorem 1.4

To conclude as planned, we now only need to apply Lemma 6.6 to initial data suitably chosen from the family in Lemma 7.1, and use the accordingly obtained exploding solutions as comparison functions:

Proof of Theorem 1.4

According to (1.14), we can fix \(a>a_0(n)\) and \(a_1>a\) such that with some \(r_0>0\) we have

$$\begin{aligned} u_0(r) \ge \frac{2(n-2)a_1}{r^2} \quad \hbox {for all } r>r_0. \end{aligned}$$
(7.9)

As \(a_1>a\), we may then pick \(r_1>r_0\) suitably large such that

$$\begin{aligned} a_1\cdot \Bigg (1-\Bigg (\frac{r_0}{r_1}\Bigg )^{n-2}\Bigg ) \ge a, \end{aligned}$$
(7.10)

and relying on the positivity of \(u_0\) we can thereupon choose \(c_1>0\) small enough such that the function \(U_0\) from (2.11) satisfies

$$\begin{aligned} U_0(r) \ge c_1 \quad \hbox {for all } r\in [0,r_1]. \end{aligned}$$
(7.11)

We finally make use of (1.15) to infer from Lemma 7.1 and Lemma 7.2 that we can find \(\alpha \ge 1\) with the property that the functions \(U^{(\alpha ,a,b)}\) in (7.2) satisfy (7.3) for arbitrary \(b>0\), and specifying this parameter by letting

$$\begin{aligned} b:=\frac{a r_1^{n\alpha -n}}{c_1}, \end{aligned}$$
(7.12)

we set

$$\begin{aligned} \underline{U}_0(r):=U^{(\alpha ,a,b)}(r), \quad r\ge 0. \end{aligned}$$
(7.13)

Then since (7.3) together with Lemma 6.5 guarantees that \((r^n \underline{U}_0)_r \ge 0\) on \((0,\infty )\), we readily see that

$$\begin{aligned} \underline{u}_0(r):=2r\underline{U}_{0r}(r) + 2n\underline{U}_0(r), \quad r\ge 0, \end{aligned}$$

defines a nonnegative radial element of \(BUC(\mathbb {R}^n)\) which due to the evident fact that

$$\begin{aligned} r^2 \underline{U}_0(r)\rightarrow a \quad \hbox {as } r\rightarrow \infty \end{aligned}$$
(7.14)

clearly satisfies (6.5) with \(\beta =2>1\), and hence particularly belongs to \(L^p(\mathbb {R}^n)\) for any \(p\in (\frac{n}{2},n)\). But since Lemma 7.1 asserts that \(\underline{U}_0\) satisfies (6.15), and since the fact that \(a>a_0(n)\ge 1\) ensures that (7.14) implies (6.16), we may employ Lemma 6.6 to infer that \(T:=T_{max}(\underline{u}_0)\) is finite, and that letting \((\underline{u},\underline{v})\) denote the solution of (1.1) with \(\underline{u}|_{t=0}=\underline{u}_0\) from Proposition 1.1, for the corresponding solution \(\underline{U}\) of (2.10) accordingly defined by (2.9) we have

$$\begin{aligned} \underline{U}(r_\star ,t)\rightarrow + \infty \quad \hbox {as } t\nearrow T \end{aligned}$$
(7.15)

with some \(r_\star >0\).

To conclude as intended from this, we note that due to (7.13), (7.2), (7.12) and (7.11),

$$\begin{aligned} \underline{U}_0(r) = \frac{ar^{n\alpha -n}}{r^{n\alpha -n+2} +b} \le \frac{ar^{n\alpha -n}}{b} \le c_1 \le U_0(r) \quad \hbox {for all } r\in [0,r_1], \end{aligned}$$

while thanks to (7.10),

$$\begin{aligned} \underline{U}_0(r) \le \frac{ar^{n\alpha -n}}{r^{n\alpha -n+2}}= & {} \frac{a}{r^2} \le \frac{a_1}{r^2} \cdot \Bigg (1-\Bigg (\frac{r_0}{r_1}\Bigg )^{n-2}\Bigg ) \le \frac{a_1}{r^2} \cdot \Bigg (1-\Bigg (\frac{r_0}{r}\Bigg )^{n-2}\Bigg ) \\{} & {} \hbox {for all } r>r_1. \end{aligned}$$

Since (2.11) along with (7.9) ensures that

$$\begin{aligned} U_0(r)= & {} \frac{1}{2r^n} \int _0^r \rho ^{n-1} u_0(\rho )d\rho \\\ge & {} \frac{1}{2r^n} \int _{r_0}^r \rho ^{n-1} u_0(\rho )d\rho \\\ge & {} \frac{(n-2) a_1}{r^n} \int _{r_0}^r \rho ^{n-3} d\rho \\= & {} \frac{a_1}{r^n} \cdot (r^{n-2}-r_0^{n-2}) \\= & {} \frac{a_1}{r^2} \cdot \Bigg (1-\Bigg (\frac{r_0}{r}\Bigg )^{n-2}\Bigg ) \quad \hbox {for all } r>r_0, \end{aligned}$$

this shows that

$$\begin{aligned} \underline{U}_0(r)\le U_0(r) \quad \hbox {for all } r\ge 0. \end{aligned}$$

Now if (uv) existed globally, then due to the latter ordering, the accordingly transformed variable U from (2.9) would satisfy

$$\begin{aligned} U(r,t)\ge \underline{U}(r,t) \quad \hbox {for all } r>0 \hbox { and } t\in (0,T) \end{aligned}$$

by the comparison principle from Lemma 2.3. In light of (7.15), however, this is impossible. \(\square \)

8 Blow-up for small initial data with decay close that of \(u_\star \). Proof of Theorem 1.3

Alternatively to the above, in the essentially remaining case when \(3\le n\le 9\) our proof of Theorem 1.3 will rely on a different construction of steady-state subsolutions to (2.10) on the basis of the following result on corresponding equilibria which in its major part can already be found in [2].

Lemma 8.1

Let \(3\le n \le 9\) and \(A>0\). Then there exists a solution \(\widetilde{U}\in C^2([0,\infty ))\) of

$$\begin{aligned} \left\{ \begin{array}{l}\widetilde{U}_{rr} + \frac{n+1}{r} \widetilde{U}_r + 2r\widetilde{U}\widetilde{U}_r + 2n\widetilde{U}^2 =0, \quad r>0, \\ \widetilde{U}_r(0)=0, \\ \widetilde{U}(0)=A, \end{array} \right. \end{aligned}$$
(8.1)

which satisfies \(\widetilde{U}(r)>0\) and \(\widetilde{U}_r(r)<0\) for all \(r>0\), and which is such that

$$\begin{aligned} r^2 \widetilde{U}(r) \rightarrow 1 \quad \hbox {as } r\rightarrow \infty . \end{aligned}$$
(8.2)

Moreover,

$$\begin{aligned} \widetilde{U}(r) \le \frac{2}{r^2+\frac{2}{A}} \quad \hbox {for all } r>0 \end{aligned}$$
(8.3)

and

$$\begin{aligned} \widetilde{U}_r(r) \ge - \frac{8r}{(r^2+\frac{2}{A})^2} \quad \hbox {for all } r>0, \end{aligned}$$
(8.4)

and one can find a strictly increasing unbounded sequence \((\widetilde{r}_k)_{k\in \mathbb {N}} \subset (0,\infty )\) such that whenever \(l\ge 1\), we have

$$\begin{aligned} r^2 \widetilde{U}(r)<1 \quad \hbox {for all } r\in (\widetilde{r}_{2l-1}, \widetilde{r}_{2l}) \quad \hbox { and} \quad r^2 \widetilde{U}(r)>1 \quad \hbox {for all } r\in (\widetilde{r}_{2l}, \widetilde{r}_{2l+1}).\nonumber \\ \end{aligned}$$
(8.5)

Proof

The existence of a solution \(\widetilde{U}\) of (8.1) fulfilling \(\widetilde{U}>0\) and \(\widetilde{U}_r<0\) on \((0,\infty )\) as well as (8.2) and (8.5) was already asserted in [2, pp.181 ff.]. Since an integration in (8.1) shows that

$$\begin{aligned} r^{n+1} \widetilde{U}_r(r)= & {} \int _0^r (\rho ^{n+1} \widetilde{U}_r)_r d\rho \nonumber \\= & {} - \int _0^r \rho ^{n+2} (\widetilde{U}^2)_r d\rho - 2n\int _0^r \rho ^{n+1} \widetilde{U}^2 d\rho \nonumber \\= & {} -(n-2) \int _0^r \rho ^{n+1} \widetilde{U}^2 d\rho - r^{n+2} \widetilde{U}^2(r) \quad \hbox {for all } r>0, \end{aligned}$$
(8.6)

we particularly obtain that \(\widetilde{U}_r(r) \le -r\widetilde{U}^2(r)\) for all \(r>0\), so that

$$\begin{aligned} - \frac{1}{\widetilde{U}(r)} + \frac{1}{A} \le - \frac{r^2}{2} \quad \hbox {for all } r>0, \end{aligned}$$

which is equivalent to (8.3). In (8.6), we can therefore estimate

$$\begin{aligned} r^{n+2} \widetilde{U}^2(r) \le \frac{4r^{n+2}}{(r^2+\frac{2}{A})^2} \quad \hbox {for all } r>0 \end{aligned}$$

and

$$\begin{aligned} (n-2)\int _0^r \rho ^{n+1} \widetilde{U}^2 d\rho \le 4(n-2) \int _0^r \frac{\rho ^{n+1}}{(\rho ^2+\frac{2}{A})^2} d\rho \quad \hbox {for all } r>0, \end{aligned}$$

where an integration by parts shows that

$$\begin{aligned} 4(n-2) \int _0^r \frac{\rho ^{n+1}}{(\rho ^2+\frac{2}{A})^2} d\rho= & {} 4\int _0^r \partial _\rho \rho ^{n-2} \cdot \frac{1}{(1+\frac{2}{A\rho ^2})^2} d\rho \\= & {} 4r^{n-2} \cdot \frac{1}{(1+\frac{2}{Ar^2})^2} - 4\int _0^r r^{n-2} \cdot \partial _\rho \frac{1}{(1+\frac{2}{A\rho ^2})^2} d\rho \\\le & {} 4r^{n-2} \cdot \frac{1}{(1+\frac{2}{Ar^2})^2} = \frac{4r^{n+2}}{(r^2+\frac{2}{A})^2} \quad \hbox {for all } r>0, \end{aligned}$$

because \(0<\rho \mapsto 1+\frac{2}{A\rho ^2}\) is nonincreasing. A second application of (8.6) thus confirms (8.4). \(\square \)

Thanks to the osciallatory behavior described in (8.5), some favorably small stationary subsolutions can be constructed by simply extending these equilibria by exact power-type functions outside suitably large annuli:

Lemma 8.2

Let \(3\le n \le 9\). Then for any choice of \(\eta >0\), one can find \(r_0=r_0^{(\eta )}>0\) and a positive function \(\widehat{U}=\widehat{U}^{(\eta )} \in C^2([0,\infty )) \cap W^{2,\infty }_{loc}([0,\infty ))\) such that

$$\begin{aligned} \left\{ \begin{array}{l}\widehat{U}_{rr} + \frac{n+1}{r} \widehat{U}_r + 2r\widehat{U}\widehat{U}_r + 2n\widehat{U}^2 \ge 0, \quad r\in (0,\infty ){\setminus }\{r_0\}, \\ \widehat{U}_r(0)=0, \end{array} \right. \end{aligned}$$
(8.7)

that

$$\begin{aligned} \widehat{U}(r) + \big | r\widehat{U}_r(r)\big | \le \eta \quad \hbox {for all } r>0, \end{aligned}$$
(8.8)

and that

$$\begin{aligned} r^2\widehat{U}(r) \rightarrow a \quad \hbox {and} \quad r^3 \widehat{U}_r(r) \rightarrow -2a \quad \hbox {as } r\rightarrow \infty \end{aligned}$$
(8.9)

with some \(a=a^{(\eta )} \in (1,1+\eta )\).

Proof

For fixed \(\eta >0\), choosing \(A\equiv A^{(\eta )}:=\frac{\eta }{5}\) we obtain from (8.3) and (8.4) that the solution \(\widetilde{U}=\widetilde{U}^{(A)}\) of (8.1) satisfies

$$\begin{aligned} \widetilde{U}(r) + \big | r\widetilde{U}_r(r)\big |\le & {} \frac{2}{r^2+\frac{2}{A}} + \frac{8r^2}{(r^2+\frac{2}{A})^2} \end{aligned}$$
(8.10)
$$\begin{aligned}\le & {} \frac{2}{\frac{2}{A}} + \frac{8}{\frac{2}{A}} = 5A=\eta \quad \hbox {for all } r>0. \end{aligned}$$
(8.11)

Moreover, according to (8.2) and (8.5) the function \(\varphi \in C^2([0,\infty ))\) defined by letting \(\varphi (r):=r^2 \widetilde{U}(r)\), \(r\ge 0\), has the property that \(\varphi (r)\rightarrow 1\) as \(r\rightarrow \infty \), and that this convergence is oscillatory in the sense that with \((\widetilde{r}_k)_{k\in \mathbb {N}}\) as in Lemma 8.1 we have \(\varphi <1\) on \((\widetilde{r}_{2l-1},\widetilde{r}_{2l})\) and \(\varphi >1\) on \((\widetilde{r}_{2\,l},\widetilde{r}_{2\,l+1})\) for all \(l\ge 1\). Consequently, if we let \(l_0=l_0^{(\eta )}\ge 1\) be large enough such that \(\varphi (r) <1+\eta \) for all \(r>\widetilde{r}_{2l_0}\), then we can choose \(r_0=r_0^{(\eta )}\in (\widetilde{r}_{2l_0},\widetilde{r}_{2l_0+1})\) in such a way that \(\varphi _r(r_0)=0\), and that \(a\equiv a^{(\eta )}:=\varphi (r_0)\) satisfies \(a\in (1,1+\eta )\). We now set

$$\begin{aligned} \widehat{U}(r)\equiv \widehat{U}^{(\eta )}(r):= \left\{ \begin{array}{ll}\widetilde{U}(r), \quad &{} r\in [0,r_0], \\ \frac{a}{r^2}, \quad r>r_0, \end{array} \right. \end{aligned}$$

and note that, by construction, \(\widehat{U}\) indeed is an element of \(C^2([0,\infty ){\setminus }\{r_0\}) \cap W^{2,\infty }_{loc}([0,\infty ))\) which clearly satisfies (8.9), while (8.8) follows from (8.10). To verify (8.7), we only need to combine (8.1) with the observation that for \(r>r_0\),

$$\begin{aligned} \widehat{U}_{rr} + \frac{n+1}{r} \widehat{U}_r + 2r\widehat{U}\widehat{U}_r + 2n\widehat{U}^2= & {} \frac{6a}{r^4} + \frac{n+1}{r} \cdot \Bigg ( - \frac{2a}{r^3}\Bigg ) \\{} & {} + 2r \cdot \frac{a}{r^2} \cdot \frac{-2a}{r^3} + 2n\cdot \Bigg (\frac{a}{r^2}\Bigg )^2 \\= & {} \frac{2(n-2) a(a-1)}{r^4} >0 \end{aligned}$$

due to the fact that \(a>1\). \(\square \)

In its essence, our derivation of Theorem 1.3 now reduces to a second application of the general blow-up result from Lemma 6.6.

Proof of Theorem 1.3

We first consider the case when \(3\le n\le 9\), in which we note that since \(a_0>1\), we can fix \(\eta =\eta (a_0,\delta )>0\) such that besides

$$\begin{aligned} 2(n+1)\eta \le \delta , \end{aligned}$$
(8.12)

we have

$$\begin{aligned} (1+\eta )\le a_0. \end{aligned}$$
(8.13)

An application of Lemma 8.2 then yields a positive function \(\widehat{U}=\widehat{U}^{(\eta )}\) with the properties listed there, whence letting

$$\begin{aligned} \psi (r)\equiv \psi ^{(a_0,\delta )}(r):=2r\widehat{U}_r(r) + 2n\widehat{U}(r), \quad r\ge 0, \end{aligned}$$
(8.14)

we particularly see using (8.7) and recalling Lemma 6.5 that \(\psi \) is a nonnegative function from \(W^{1,\infty }_{loc}([0,\infty ))\) which, by (8.8) and (8.12), satisfies

$$\begin{aligned} \psi (r) \le 2\big | r\widehat{U}_r(r)\big | + 2n\widehat{U}(r) \le 2\eta + 2n\eta \le \delta \quad \hbox {for all } r>0. \end{aligned}$$

Moreover, according to (8.9) and (8.14) there exists \(a\in (1,1+\eta )\) such that

$$\begin{aligned} r^2 \psi (r) = 2r^3 \widehat{U}_r(r) + 2nr^2 \widehat{U}(r) \rightarrow - 4a + 2na = 2(n-2)a \quad \hbox {as } r\rightarrow \infty , \end{aligned}$$

so that (1.11) becomes a consequence of (8.13).

Now if \(u_0\in C^0(\mathbb {R}^n)\cap L^\infty (\mathbb {R}^n)\) is radial and such that (1.12) holds, then in view of (8.7) and the comparison principle from Lemma 2.3, the function U from (2.9) satisfies

$$\begin{aligned} U(r,t)\ge \underline{U}(r,t) \quad \hbox {for all } r>0 {\hbox { and }} t\in (0,T), \end{aligned}$$
(8.15)

where \(T:=\min \{T_{max}(u_0), T_{max}(\psi (|\cdot |))\}\), and where \(\underline{U}\) is as defined through (2.9) with u replaced by \(\underline{u}\), where \((\underline{u},\underline{v})\) solves (1.1) with \(\underline{u}(r,0)=\psi (r)\), \(r\ge 0\). But since \(\underline{U}(\cdot ,0)=\widehat{U}\) satisfies \(\liminf _{r\rightarrow \infty } r^2 \underline{U}(r,0)>1\) according to (8.9), Lemma 6.6 guarantees that \(T_{max}(\psi (|\cdot |))\) is finite, and that there exists \(r_\star >0\) such that \(\underline{U}(r_\star ,t)\rightarrow +\infty \) as \(t\nearrow T_{max}(\psi (|\cdot |))\). In light of (8.15), this is possible only if \(T_{max}(u_0) \le T_{max}(\psi (|\cdot |))\).

In the case when \(n\ge 10\), the claim immediately results from Theorem 1.4 upon letting

$$\begin{aligned} \psi (r)\equiv \psi ^{(a_0,\delta )}(r):=\min \Bigg \{ \delta \,, \, \frac{2(n-2)a_0}{r^2} \Bigg \} \end{aligned}$$

for \(r\ge 0\). \(\square \)

9 Instantaneous blow-up. Proof of Theorem 1.5

The detailed quantitative design of our blow-up argument from Lemma 5.1 finally enables us to infer the announced local-in-time nonexistence result in a fairly straightforward style:

Proof of Theorem 1.5

Supposing on the contrary that (uv) be a solution of (1.1) in \(\mathbb {R}^n\times (0,T)\) with the indicated properties, through (2.9) we would obtain a nonnegative classical solution \(U\in C^0([0,\infty )\times [0,T)) \cap C^{2,1}([0,\infty )\times (0,T))\) of (5.2). To see that this is impossible, with \((\Theta ^{(R)})_{R>0}\) taken from (5.1) we abbreviate

$$\begin{aligned} c_1:=\int _{\frac{1}{2}}^1 \rho ^{n+1} \Theta ^{(1)}(\rho ) d\rho , \end{aligned}$$
(9.1)

and note that \(c_1\) is positive according to the strict positivity of \(\Theta ^{(1)}\) on [0, 1). We can therefore fix \(M>0\) such that

$$\begin{aligned} M> { \frac{8n}{(n+2)(n-2)c_1 T}, } \end{aligned}$$
(9.2)

and draw on (1.16) in choosing \(R>0\) large enough such that apart from the inequality

$$\begin{aligned} R^2 \ge { \frac{8n\lambda _1}{(n+2)(n-2) c_1 M}, } \end{aligned}$$
(9.3)

we also have

$$\begin{aligned} u_0(r) \ge M \quad \hbox {for all } r>2^{-1-\frac{1}{n}} R. \end{aligned}$$

Then in line with (2.9), for each \(r>\frac{R}{2}\) we can estimate

$$\begin{aligned} U(r,0)= & {} \frac{1}{2r^n} \int _0^r \rho ^{n-1} u_0(\rho ) d\rho \\\ge & {} \frac{1}{2r^n} \int _{2^{-1-\frac{1}{n}}R}^r \rho ^{n-1} \cdot M d\rho \\= & {} \frac{M}{2r^n} \cdot \frac{r^n - (2^{-1-\frac{1}{n}} R)^n}{n} \\= & {} \frac{M}{2n} \cdot \Bigg (1-2^{-n-1} \cdot \Bigg (\frac{R}{r}\Bigg )^n\Bigg ) \\\ge & {} \frac{M}{2n} \cdot \big (1-2^{-n-1} \cdot 2^n\big ) = \frac{M}{4n}, \end{aligned}$$

whence by nonnegativity of \(\Theta ^{(R)}\),

$$\begin{aligned} y_0:=\int _0^R r^{n+1} U(r,0)\Theta ^{(R)}(r) dr \end{aligned}$$

satisfies

$$\begin{aligned} y_0 \ge \frac{M}{4n} \int _{\frac{R}{2}}^R r^{n+1} \Theta ^{(R)}(r) dr = \frac{M}{4n} \int _{\frac{R}{2}}^R r^{n+1} \Theta ^{(1)}\Bigg (\frac{r}{R}\Bigg ) dr = \frac{c_1 M R^{n+2}}{4n}\nonumber \\ \end{aligned}$$
(9.4)

according to (1.16). Now thanks to (9.3), this firstly implies that

$$\begin{aligned} \frac{y_0}{\frac{2\lambda _1 R^n}{ { (n+2) } (n-2)}} \ge \frac{\frac{c_1 M R^{n+2}}{4n}}{\frac{2\lambda _1 R^n}{ { (n+2) } (n-2)}} = \frac{ { (n+2) } (n-2) c_1 M R^2}{8 { n } \lambda _1} \ge 1, \end{aligned}$$

and that thus (5.3) holds. We may therefore invoke Lemma 5.1 to see that, again by (9.4),

$$\begin{aligned} T\le \frac{2 R^{n+2}}{ { (n+2) } (n-2)} \cdot \frac{1}{y_0} \le \frac{2R^{n+2}}{n { (n+2) } (n-2)} \cdot \frac{4n}{c_1 M R^{n+2}} = \frac{8 { n } }{ { (n+2) } (n-2) c_1 M}. \end{aligned}$$

In view of (9.2), however, this leads to the absurd conclusion that \(T<T\), meaning that such a solution actually cannot exist. \(\square \)