Nonexistence of global positive solutions for p-Laplacian equations with non-linear memory

The Cauchy problem in $\mathbb{R}^d,$ $d\geq 1,$ for a non-local in time p-Laplacian equations is considered. The nonexistence of nontrivial global weak solutions by using the test function method is obtained.


Introduction
In this paper, we consider the following problem where I := t 0 (t − s) −γ u(s) q ds, p, q > 1, d ≥ 1, 0 < γ < 1, and u 0 ∈ L 2 loc (R d ). We are interested in the nonexistence of nontrivial global weak solutions.
Weissler in [13] proved that if (1.2) has no global solution u(x, t) satisfying u(· , t) q < ∞ for t > 0 and some q ∈ [1, +∞) whenever if p = 1 + 2/d. Therefore, the value is the limiting exponent of (1.2); it is called Fujita's exponent. Since Fujita's paper, a sizeable number of extensions in many directions have been published. Recently, Cazenave, Dickstein and Weissler [3] extended the results of Fujita to the non-local in time heat equation where w 0 ∈ C 0 (R d ) and ϑ ∈ (0, 1). In this case, the value of the critical Fujita exponent is For the p-Laplacian equation Zhao [15] and Mitidieri and Pohozaev [11] obtained the critical exponent in fact, Zhao [15] proved that if p − 1 < q < q * , then the Cauchy problem (1.6) has no nontrivial global solution, however if q > q * and w 0 (x) is small enough, then (1.6) admits a global solution. Mitidieri and Pohozaev [11] completed the study by proving the nonexistence of nontrivial global solution in the case q ≤ q * and for all p > 2d/(d + 1). Andreucci and Tedeev [1,10] obtained similar results by considering doubly singular parabolic equations. The test function method was used to prove the nonexistence of global solutions. This method was introduced by Baras and Kersner in [2] and developed by Zhang in [14] and Pohozaev and Mitidieri in [11], it was also used by Kirane et al. in [9].
Here, we are concerned with the non-existence of nontrivial global solutions of (1.1); inspired by [11], we choose a suitable test function in the weak formulation of the problem. Our main result is where then no nontrivial global weak solutions exist for problem (1.1).
Remark 1.2 Note that (1.8) can be seen as follows • when p = 2, then q c = p c where p c is defined in (1.5).

Remark 1.3
The same result can be obtained for the problem Carathéodory function, where one assumes the existence of c 1 , c 2 > 0 and p > 1 such that For the study of the non-existence of global solution, the following condition is needed which is a general case of (1.9).
Some words on the structure of the paper: Some definitions and properties on the fractional integrals and derivatives are recalled in Section 2. Section 3, is reserved to the proof of the main result (Theorem 1.1).

Preliminaries
AC[a, b] denotes the space of these functions. Moreover, The Riemann-Liouville left-and right-sided fractional integrals are defined by and where Γ is the Euler gamma function.
The Riemann-Liouville left-and right-sided fractional derivatives are defined by and is satisfied for every f ∈ I α t|d (L p (c, d)), g ∈ I α c|t (L q (c, d)) such that 1 p + 1 q ≤ 1 + α, p, q > 1, where Remark 2.5 A simple sufficient condition for functions f and g to satisfy and are continuous.

7)
where D := d dt . Given T > 0, let us define the function w 1 by The following properties concerning the functions w 1 will be used later on.
Any constant will be denoted by C.

Global nonexistence
In this section, the proof of Theorem 1.1 will be presented.
is said to be a weak solution of (1.1) if u ≥ 0, u | t=0 = u 0 and the following formulation We denote the lifespan for the weak solution by Proof. Let u ≥ 0 be a global weak solution of (1.1), then for all T > 0 and all ϕ ∈ L 2 (0, ∞; where ℓ, η ≫ 1 and Φ ∈ C ∞ (R + ) be Then, using the integration-by-parts formula, we get From (2.5) and Lemma 2.7, we conclude that Moreover, using (2.6), u 0 ≥ 0 and u ≥ 0, it follows u|∂ t ϕ| dx dt =: Next, by introducingφ 1/qφ−1/q in I 2 and applying the following Young's inequality we get In order to obtain a similar estimation on I 1 , let α < 0 be an auxiliary constant such that α > max{−1, 1 − p, 1 − q p−1 } and let By taking ϕ ǫ (x, t) = u α ǫ (x, t)ϕ(x, t) as a test function where ϕ is given in (3.2), and using the fact that u is a weak solution, we obtain Using the integration-by-parts formula, we get Γ(δ) Then, as Whereupon, Similarly, and By J 1 ,J 2 , and J 3 , it follows from (3.5) that where Young's inequality has been used. Consequently, as u ≥ 0, we conclude that Young's inequality and the last inequality, allow to get Applying Fatou's and Lebesgue's theorems, as ǫ → 0, we get Now, we use Young's inequality, ∇(ϕ ℓ 1 ) = ℓϕ ℓ−1 1 ∇ϕ 1 , and the conditions that q > max{1, p − 1}, α < 0 to get estimations of K 1 , K 2 and K 3 ; we have Therefore, we conclude that Using the estimates of I 1 and I 2 into (3.3), we obtain At this stage, we choose B = T θ , θ > 0. Taking α small enough and passing to s = T −1 t, y = T −θ x, we get from (3.7) that If all exponents of T are negative, by taking T −→ +∞ and using the dominated convergence theorem, we conclude that u = 0. In order to ensure the negativity of the exponents of T , it is sufficient to require which is equivalent to q < max θ>0 min{q 1 (θ), q 2 (θ)}. (3.9) To take into consideration q 1 (θ) and q 2 (θ), we first look at (dθ − δ) + and (dθ − θp + 1 − δ) + and try to compare them in terms of θ, i.e, to compare between δ/d and ( using that q 1 (θ) is non-increasing and q 2 (θ) is non-decreasing.
ii) Case 2d/(d + 1) < p ≤ d. In this case, (3.9) can be read as As p 2 (θ) is non-decreasing when p ≥ dδ, and non-increasing when p ≤ dδ, we can see that (3.10) is equivalent to iii) Case of 1 < p ≤ 2d/(d + 1): In this case, (3.9) can be read as Finally, to get similar results in the critical case we choose B = R −θ T θ , where 1 ≪ R < T is such that T and R do not go simultaneously to infinity. Moreover, due to the calculation made above, a positive constant D independent of T exists such that where ∆(B) := {x ∈ R d ; B < |x| < 2B}. Repeating a similar calculation as in the subcritical case, q < max{ p−1 1−δ , q c }, and using Hölder's inequality instead of Young's one in K 1 and K 3 , we get Taking into account that q = max{ p−1 1−δ , q c } and s = T −1 t, y = R θ T −θ x, we get u qφ dx dt p−1 q Taking the limit when T → ∞, and using (3.11), we obtain ∞ 0 R d |u| q dx dt ≤ C R −dθ .
Finally, letting R → ∞, it comes that u = 0. The proof of Theorem 1.1 is complete.