GLOBAL EXISTENCE AND BLOW-UP OF SOLUTIONS TO A SINGULAR NON-NEWTON POLYTROPIC FILTRATION EQUATION WITH CRITICAL AND SUPERCRITICAL INITIAL ENERGY

. In this paper, we revisit the singular Non-Newton polytropic ﬁltration equation, which was studied extensively in the recent years. However, all the studies are mostly concerned with subcritical initial energy, i.e., E ( u 0 ) < d , where E ( u 0 ) is the initial energy and d is the mountain-pass level. The main purpose of this paper is to study the behaviors of the solution with E ( u 0 ) ≥ d by potential well method and some diﬀerential inequality techniques.

Due to (2), for any u ∈ Q, we have u m ∈ L m+q m (Ω), then the energy functional can be defined by The corresponding Nehari functional and Nehari manifold can be denoted by respectively. Further, we define the potential depth by where M is the optimal constant of the Sobolev embedding for We define the sets related to global existence and blow-up as follows: .
With the above preparations, the main results of [14,15,16,17] can be stated as the following theorem. Theorem 1.2. Let u(t) be the weak solution of problem (1), then we have the following conclusions: (i) If u 0 ∈ Σ 1 , then u(t) exists globally. Moreover, u m (t) decays exponentially in W 1,p 0 (Ω), namely, there exist two positive constants α and C such that lim t→+∞ e αt ∇u m p p ≤ C.
(ii) If u 0 ∈ Σ 2 or E(u 0 ) ≤ 0, then u(t) blows up in finite time. Proof. If E(u(t 0 )) ≤ 0, then taking t 0 as initial time, by Theorem 1.2 (ii), we see the solution blows up at some finite time T .
The assumptions in Theorem 1.2 imply the initial energy is subcritical, i.e., E(u 0 ) < d. For the critical initial case, i.e., E(u 0 ) = d, in [9], the authors get the following result.
then the solution of problem (1) blows up in finite time.
In view of the above results, a natural question is that whether the global solution problem (1) exists when E(u 0 ) = d or E(u 0 ) > d, which is the main task of this paper. More specifically, for any solution of problem (1) with E(u 0 ) = d, we will prove that there exist a sufficient small t 1 such that E(u(t 1 )) < d by some differential inequality techniques, then take t 1 as the initial time, we can deduce our results with the help of Theorem 1.2. For E(u 0 ) > d, we will construct two sets Ψ α and Φ α , and prove that the solution will blow up in finite or infinite time if the initial value belongs to Ψ α , while the solution of problem (1) will exist globally and tends to zero as time t → +∞ when the initial value belongs to Φ α . Furthermore, it worthwhile to point out that the models related to problem (1) were studied extensively in recently years (see for example [4,5,6,7,10,11,12]), and the conditions about global existence and blow-up were got in these papers for the subcritical initial energy case (i.e., E(u 0 ) < d). We remark that the methods used in this paper can also be used to study the solutions of these models with the critical and supercritical initial energy cases (i.e., E(u 0 ) ≥ d).
In order to give the main results of the present paper, we introduce some necessary definitions as follows: where α is a constant. For all α > d, it is easy to see that
Now, we are ready to state the main results of this paper. The first two results are about the case critical initial energy (E(u 0 ) = d).  Finally, we give the results about the solution with E(u 0 ) > d.
Remark 1. If the last inequality of (2) is strict, i.e., the parameters satisfy By (13), we know that λ α admits a positive lower bound and Λ α admits a finite positive upper bound, then the two sets Φ α , Ψ α are well-defined, which ensure the statements in Theorem 1.6 make sense.
The rest of this paper is organized as follows. In Section 2, we give some useful lemmas, which will be used in the proof of the main results. In Section 3, we give the proof of our main results.

2.
Preliminaries. We begin this section with the following lemma.
Then there exists a positive constant C depending on β, n, N and k such that for where C 1 , C 3 are two positive constants given in (14) and (18) respectively.
Proof. For any u ∈ K α , using the Hardy-Sobolev's inequality given in Lemma 2.1, we have where C is a positive constant depending only on N, m, s. Since m ≥ 1, p ≥ 2 and 0 ≤ s ≤ 1 + 1/m, we get Then it follows from the Hölder's inequality and (8) that which, together with the definition of Λ α , leads to On the other hand, since Ω is a bounded domain in R N , there exists a positive constant ρ such that

GUANGYU XU AND JUN ZHOU
Then we have In view of p < N , q > 1 and Then by the Gagliardo-Nirenberg's inequality [2], we have where C 2 is a positive constant depending only on Ω and For any u ∈ K, we get from the definition of K and the above inequality that Since m + q > mp, it follows from the definition of K and (5) that ≥ 0, by the definition of λ α , (15), (16) and (17), for any u ∈ K α ⊂ K, we get < 0, by the definition of λ α , (15), (16) and (8), for any u ∈ K α ⊂ K, we have So for any u ∈ K α , we get λ α ≥ C 3 , where (18) Finally, by the definition of λ α and Λ α , it is easy to see that λ α ≤ Λ α , so Lemma 2.2 is proved.
then it holds Proof. Multiplying the first equation of problem (1) with 1 m (u m ) t and integrating over Ω × (0, t), we get (21), then it is easy to see that E(u(t)) is nonincreasing with respect to t.
Lemma 2.5. We have E(u) > 0 for any u ∈ K + . Moreover, for any α > 0 and u ∈ E α ∩ K + , it holds Proof. By the definition of K + in (7), for any u ∈ K + , it holds H(u) > 0, i.e., then it follows from the definition of E(u) that E(u) > 0.
Proof. For any u ∈ K − , by the definition of K − we have H(u) < 0, i.e., Since m + q > mp, then it follows from the above inequality that Next, we denote by S(t) the nonlinear semigroup associated to problem (1). Instead of u = u(t) we will also write u = S(t)u 0 for all t < T . If T = +∞, we define the ω-lim set of u 0 as follow: and we have the following lemma, which shows that ω(u 0 ) = ∅ if u m (t) is uniformly bounded in W 1,p 0 (Ω). Lemma 2.7. Let u(t) = S(t)u 0 be a global weak solution of problem (1) such that u m (t) is uniformly bounded in W 1,p 0 (Ω), then ω(u 0 ) contains stationary solutions of problem (1).
Proof. The method in the following proof is similar to [15,Theorem 1.4], for reader's convenience, we give a complete proof here.
We choose a monotone increasing sequence {t n } +∞ n=1 such that t n → +∞ (n → +∞), and let u n = u(t n ). By the uniform boundedness of {u m n } +∞ n=1 in W 1,p 0 (Ω), there exist a subsequence of {u n } +∞ n=1 which is still denoted by {u n } +∞ n=1 and a function ω such that ω m ∈ W 1,p 0 (Ω), u m n ω m weakly in W 1,p 0 (Ω) and u m n → ω m a.e. in Ω.
Next we introduce some suitable test functions. For anyT < +∞, we take two functions ψ and satisfying Multiplying the first term of problem (1) with φ and integrating by parts, then, by the definition of the above functions, we have Hence, for problem (1), it is easy to see that Making a transformations = t − t n in above inequality, we get We claim thatω = ω a.e. in Ω. In fact, since the solution is global, by Corollary 1 we know that E(u(t)) > 0 for all t ∈ [0, +∞), then by (21) we get Let ρ be the positive constant given in (15), by Hölder's inequality and the above inequality, we obtain The proof of the following lemma is very similar to that of [8,13], so we omit the proof here. and E(λu) is increasing on 0 ≤ λ ≤ λ * , decreasing on λ * ≤ λ < +∞ and takes the maximum at λ = λ * ; (iii) H(λu) > 0 for 0 < λ < λ * , H(λu) < 0 for λ * < λ < +∞, and H(λ * u) = 0.
3. Proofs of the main results. In this section, we aim to prove our main results. Firstly, we prove Theorem 1.4.
Proof of Theorem 1.4. The proof is divided into two steps.
Next, we suppose H(u(t)) > 0 for 0 < t < t 0 and H(u(t 0 )) = 0. Let f (t) be the function defined by (19), by Lemma 2.3, we get On the other hand, we have which combining with (29) implies By Lemma 2.4 and E(u 0 ) = d, we have Next, we claim that Arguing by contradiction, if the above inequality does not hold, then we have for a.e. (x, t) ∈ Ω × (0, t 0 ), which contradicts (30). Hence, we get (32), and then it follows from (31) that E(u(t 0 )) < d.
Proof of Theorem 1.5. By E(u 0 ) = d > 0, H(u 0 ) < 0 and the continuity of E(u(t)) and H(u(t)) with respect to t, we know that there exists a sufficiently small t 1 > 0 such that E(u(t 1 )) > 0 and H(u(t)) < 0 for 0 ≤ t ≤ t 1 . Let f (t) be the function defined by (19), by Lemma 2.3, we get On the other hand, we have which combining with the above inequality implies Ω |x| −s |u| m−1 uu t dx > 0, 0 ≤ t ≤ t 1 .
By using the above inequality, similar to the proof of (32), we can get Then it follows from (21) that which combining with H(u(t 1 )) < 0 and E(u(t 1 )) > 0 implies that u(t 1 ) ∈ Σ 2 , so by Theorem 1.2, we know that the weak solution blows up in finite time.
Proof of Theorem 1.6. The idea of the proof comes from [3]. For all t ∈ [0, T ), let u(t) = S(t)u 0 be a weak solution of problem (1). By Lemma 2.3, we have d dt On the other hand, from Lemma 2.4, we know that E(u(t)) is nonincreasing with respect to t. Thus, we get (i). If u 0 ∈ Φ α , then by the definition of Φ α in (11) and the monotonicity property of λ α in (10), we have d < E(u 0 ) ≤ α and We claim that u(t) ∈ K + for all t ∈ [0, T ). Arguing by contradiction, if the claim is not true, then there is a t 0 ∈ (0, T ) such that u(t) ∈ K + for 0 ≤ t < t 0 and u(t 0 ) ∈ K. Then by the definition of K + and (34), we know that Ω |x| −s |u(t)| m+1 dx is strictly decreasing on [0, t 0 ]. So, it follows from (35) and (36) that By u(t 0 ) ∈ K and (37), we get u(x, t 0 ) ∈ K E(u0) . Thus, it follows from the definition of λ E(u0) that which contradicts (38). So the claim is true. Then it follows from (35) that u(t) ∈ E E(u0) ∩ K + . Hence, by (22), we obtain Since the right-hand of (39) is independent of T , then we can extend the solution to infinite, i.e., T = +∞, and we further have (39) holds for 0 ≤ t < +∞. Moreover, by (39) we know that u m is bounded uniformly in W 1,p 0 (Ω), then it follows from Lemma 2.7 that the ω-limit set defined in (23) is not a empty set. Now for any ω ∈ ω(u 0 ), by the above discussions, we get E(ω) ≤ E(u 0 ) and 1 m + 1 Ω |x| −s |ω| m+1 dx < λ E(u0) .
By the first inequality, we know that ω ∈ E E(u0) . By the second inequality and the definition of λ E(u0) , we know that ω ∈ K E(u0) . Since K E(u0) = K ∩ E E(u0) , we then obtain ω ∈ K.
(40) In fact, it follows from u(t) ∈ K + and the definitions of E(u) and K + that i.e., E(u(t)) is bounded below. Then combine the fact that E(u(t)) is nonincreasing with respect to t we know there is a constant c such that lim t→+∞ E(u(t)) = c.
So for any ω ∈ ω(u 0 ), we have E(u ω (t)) = c for all t ≥ 0, where u ω (t) is the solution of problem (1) with initial value ω. Then combining (21) we get u ω (t) ≡ ω, and then it follows from (34) that Combining (41), ω ∈ K and the definition of K we know ω = 0, then by the arbitrariness of ω, we get (40). In other words, the solution u(t) → 0 as t → +∞.
(ii). Similar to the proof of the first part, if u 0 ∈ Ψ α , then by the definition of Ψ α and the monotonicity property of Λ α , we have d < E(u 0 ) ≤ α and We claim that u(t) ∈ K − for all t ∈ [0, T ). By contradiction, if the claim is not true, then there is a t 1 > 0 such that u(t) ∈ K − for 0 ≤ t < t 1 and u(t 1 ) ∈ K. (34) and (35) imply that E(u(t 1 )) ≤ E(u 0 ). Similarly, we get u(x, t 1 ) ∈ K E(u0) . Thus, it follows from the definition of Λ E(u0) that which contradicts (42). So the claim is true. If the solution blows up in infinite time, then the proof is complete. So in the following proof we assume that the solution does not blow up in infinite time and prove the solution blows up in finite by contradiction. Assuming u(t) exists globally, then u(t) ∈ E E(u0) ∩ K − , ∀t ∈ [0, +∞), and Ω |x| −s |u(t)| m+1 dx is strictly increasing on [0, +∞) (by Lemma 2.3). Furthermore, we claim that u m (t) is uniformly bounded in W 1,p 0 (Ω), i.e., there exists a positive constant Θ such that ∇u m (t) p p ≤ Θ, ∀t ∈ [0, +∞). In fact, if the claim is not true, then there exists a monotone increasing sequence {t n } +∞ n=1 such that ∇u m (t n ) p p > n, n = 1, 2, · · · .
By the monotone property of {t n } +∞ n=1 , we know that t n → +∞ (n → +∞) or there exists a positive constant t * such that t n → t * (n → +∞). If the first case happens, by (43), we know that u(t) blows up in infinite time, which contradicts the assumption that solution does not blow up in infinite time; if the second case happen, by (43), we know that u(t) blows up in finite time, which contradicts the assumption that u(t) exists globally. So the claim is true, namely, the assumptions in Lemma 2.7 holds, so we get ω(u 0 ) = ∅.
Since E(u(t)) is nonincreasing with respect to t, then we have following two cases: (a) there is a constant c such that lim t→+∞ E(u(t)) = c; (b) lim t→+∞ E(u(t)) = −∞. Next we will prove both the above cases contradict T = +∞, then we get that the solution u(t) blows up in finite time.
Finally, we consider the case (b). If lim t→+∞ E(u(t)) = −∞, then there must exist a time t 1 such that E(u(t 1 )) ≤ 0. Then take u(t 1 ) as the initial value, by Theorem 1.2 we know that the corresponding solution U (t) = u(t + t 1 ) blows up in finite time, which contradicts T = +∞, thus Theorem 1.6 is proved.