Sufficient conditions of non global solution for fractional damped wave equations with non-linear memory

The focus of the current paper is to prove nonexistence results for the Cauchy problem of a wave equation with fractional damping and non linear memory.  Our method of proof is based on suitable choices of the test functions in the weak formulation of the sought solutions.

2. If γ = 0, p > p * and u 0 ∈ L q * (R N ) (where q * = (p−1)N 4−2γ ) with small data, that is u 0 L q * small enough, then u exists globally. In particular, they proved that the critical exponent in Fujita's sense p * is not the one predicted by scaling, and this is not a surprising result since it is well known that scaling is efficient only for parabolic equations and not for pseudo-parabolic ones. To show this, it is sufficient to note that equation (1.1) can be formally converted into where α = 1−γ and D α 0|t is the fractional derivative operator of order α (α ∈ ]0, 1[) of Riemann-Liouville defined by and I 1−α 0|t is the fractional integral of order 1 − α defined by (2.3) bellow. 3. In the case of γ = 0, Souplet has showed in [18] that nonzero positive solution blows -up in finite time.
After that, the damped wave equation with nonlinear memory was treated by Fino [5] in 2010, when he investigated the global existence and blow-up of solutions for the following equation He used as a main tool in his work for the existence and uniqueness of solution to problem (1.3) the weighted energy method similar as the one introduced by G. Todorova an B. Yardanov [9] in 2001, while he employed the test function method to show the blow-up results. One can found his results in [5]. In particular he found the same p γ and so the same critical exponent p * founded by Cazenave and al in [4]. Our purpose of this work is to generalize some of the above results.
Remark 1.1. Throughout this work, the constants will be denoted C and are different from one place to another one.

Statement of the problem
In this section, we will prove blow-up results of the problem (1)- (2).
Proof. The theorem 3.1 will be demonstrated by absurd. So suppose that u is a global non trivial weak solution for problem (2.1)-(2.2). To prove Theorem 3.1 we need also to some results we will give them in the following section.

Preliminary results
As the principle of the method is the right choice of the test function, we chose it, for some T > 0, as follows: where r > 1 and D α t|T is the right fractional derivative operator of order α in the sense of Riemann-Liouville defined by (3.5) and the functions ϕ 1 and ϕ 2 are given by with β > 1, θ is a positive constant which will be chosen suitably later and φ is a cut-off non increasing function such that We also denote by Ω T for the support of ϕ 1 , that is and by ∆ T for the set containing the support of ∆ϕ 1 which is defined as follows We will also use the fractional version of integration by parts (See [15]) for all f, g ∈ C([0, T ]) such that D α 0|t f (t) and D α t|T g(t) exist and are continuous. The identities (See [15]) and and also the following identity (see [15]) which happens for all u ∈ C n [0, T ] ; T > 0, where ∂ n t is the n−times ordinary derivative with respect to t, will be strongly used in this work.
A simple and immediate calculation, using (3.5) and the identity (3.13) serves to the following proposition: Proposition 3.2. Given β > 1. Let ϕ 2 be the function defined by Therefore, for all α, σ ∈]0, 1[, we have Proof. The proof of the proposition 3.2 is a simple and immediate verification. We have by definition (3.5) 1. We deduce, using formula (3.14) bellow, where B is the famous Beta's function defined by and satisfies in particular (3.14) 2. We apply directly formula (3.13) to show that hence the result is conclude. 3. We apply the identity (3.12) and formula (3.14) we get directly the desired result.

Treatment of the left-hand side
Introducing the test function defined by (3.4), we get using the formula of integration by parts (3.10) and the identity (3.11) For the 2 nd term of the left-hand side of equality (2.4), we use the Proposition 3.2 to obtain For the 3 rd term, noting that using always the Proposition 3.2 we get the following estimate Always by Proposition 3.2, the following estimate will be obtained for the 4 th term of the left hand-side of the weak formulation(2.4)

Treatment of the right-hand side
Taking into account the formula (3.13) we easily get and then (3.20) Using formula (3.12) and (3.13) we show firstly that and then t|T ϕ 2 (t)dtdx.

(3.21)
Finally for the third term of the right-hand side of the formulation (2.4), using the following identity (3.23) The facts that ϕ 1 ≤ 1 and allows us to deduce from the formula (3.23) the following inequality for some constant C > 0. Next, applying the following ε− Young inequality AB ≤ εA p + C(ε)B q , pq = p + q, to the terms of the right-hand side of inequality (3.24) we get For the third term of the right-hand side we obtain  and noting that they are null outside Ω T (defined by (3.8)), then, using Fubini's theorem, we get, for I 1 (3.30) We have and using Proposition 3.2, we get Combining (3.31) and (3.32) into (3.30) we obtain then By the same way we have for I 2 So, if we replace q by its value p p−1 we get For Then we find for J 31 and as usual Finally, we replace (3.33),(3.37) and (3.41) into (3.28) we obtain (3.42) Now, since θ is arbitrary and it must only be positive, we choose it as follows This choice of θ allows us to have with this choice of θ, and by (3.43) we get from (3.42) At this stage, to prove the first result in Theorem.3.1, we distinguish two cases.

Case of p ≤ p γ (σ)
This case itself is divided into two subcases as follows 1. i. Subcase of p < p γ (σ). In this case, one can remark that the condition p < p γ (σ) is equivalent to δ < 0, then we pass to the limit as T → ∞ in (3.44), we get and this implies that u = 0, which is a contradiction because we have supposed that the solution u is not trivial.
1. ii. Subcase of p = p γ (σ). Firstly, we remark that the condition p = p γ (σ) is equivalent to δ = 0. Next, taking the limit as T → ∞ in (3.44) with the consideration δ = 0 we get from which we can deduce that where ∆ T is defined by (3.9). Fixing arbitrarily R in ]0, T [ for some T > 0 such that when T → ∞ we don't have R → ∞ at the same time and choosing ϕ 1 as with θ is an arbitrary positive constant and φ is the cut-off function defined by (3.7). Using the following Hölder's inequality instead of the ε−Young's one to estimate integral I 2 in (3.28) on the set Recalling Integrals I 1 , I 3 in page 231 andĨ 2 such that To estimate them, we use at this stage the change of variables x = T θ 2 R − θ 2 y, and t = T τ on the set Ω T R −1 . We have firstly and using the hypothesis δ = 0 we conclude from (3.50) Calculating the integralĨ 2 using the same change of variables and the same form of function ϕ 1 and using (3.51) we obtain from (3.28) (3.52) Now taking the limit as T → +∞ in (3.52) and using (3.47) and the fact that lim which means that necessarily R → +∞ and this is a contradiction. Case of p ≤ 1 γ . Even this case is divided into two subcases as follows 2. i. Subcase of p < 1 γ . In this case we recall (3.28), we take ϕ 1 (x) = φ |x| 2 R θ where φ is the function defined by (3.7) and R is a fixed positive number. Trying to calculate generalized integrals I 1 , I 2 and I 3 (page 231) with respect to x on the set Employing the scaled variables x = R θ 2 y and t = T τ, for the first integral we have By the same way, we obtain (3.54) Finally, for the third integral, we have Using formula (3.53), (3.54) and (3.55) we get (3.56) Firstly, we note that p < 1 γ implies that 1 − α p p−1 < 0. So, the facts that (α + 2) p p−1 > α p p−1 and (σ + α + 1) p p−1 > α p p−1 allow us the fact that lim T →+∞ ψ(t, x) = ϕ r 1 (x), (3.57) hence, by taking the limit as T → +∞ in (3.56), we arrive at Next, taking the limit in (3.58) as R → +∞ and taking into account the fact that lim This implies that u = 0 which is a contradiction. 2. ii. Subcase of p = 1 γ . In this case, we assume furthermore that p < N N − 1 . (3.59) First, we observe that (3.59) implies that N 2 − p p − 1 < 0.  Finally, one can remark easily that if N = 1, 2 then N 2 − p p−1 < 0 for all p > 1, then by taking the limit as R → ∞ in (3.62) and using the facts that θ > 0 and lim which implies that u = 0 and this is a contradiction.
If N ≥ 3 then N 2 − p p−1 will be negative and we can get (3.63) by letting R → ∞ in (3.62), if we assume furthermore that (3.59) or equivalently (3.60) is satisfied. This achieved the proof of Theorem 3.1. Remark 3.3. We remark that the condition (3.59) is needed only in the case of p = 1 γ and N ≥ 3 and not otherwise. We, also, point out that the condition (3.59) is equivalent to N − 2 N < γ < 1.

Conclusion
First, one can show that if σ → 0 then p γ (σ) → p γ , and we find the same critical exponent obtained by Fino ([5]), and this is reasonable because if σ = 0 then D σ 0|t u t = u t . In other word, our result is a generalization of the result of ( [5]). Also, thanks to the presence of the term u t in the model of ( [5]), one can show using Fourier transform, for example, or by scaling argument, that this model is parabolic like and then it tends to the model of Cazenave and al ( [4]) as t → +∞. For this reason, the two problems have the same critical exponent p * .