Higher order evolution inequalities with nonlinear convolution terms

We are concerned with the study of existence and nonexistence of weak solutions to $$ \begin{cases}&\displaystyle \frac{\partial^k u}{\partial t^k}+(-\Delta)^m u\geq (K\ast |u|^p)|u|^q \quad\mbox{ in } \mathbb R^N \times \mathbb R_+,\\[0.1in]&\displaystyle \frac{\partial^i u}{\partial t^i}(x,0) = u_i(x) \,\, \text{ in } \mathbb R^N,\, 0\leq i\leq k-1,\\ \end{cases} $$ where $N,k,m\geq 1$ are positive integers, $p,q>0$ and $u_i\in L^1_{\rm loc}(\mathbb{R}^N)$ for $0\leq i\leq k-1$. We assume that $K$ is a radial positive and continuous function which decreases in a neighbourhood of infinity. In the above problem, $K\ast |u|^p$ denotes the standard convolution operation between $K(|x|)$ and $|u|^p$. We obtain necessary conditions on $N,m,k,p$ and $q$ such that the above problem has solutions. Our analysis emphasizes the role played by the sign of $\displaystyle \frac{\partial^{k-1} u}{\partial t^{k-1}}$.


Introduction and the main results
Let N, k, m ≥ 1 be positive integers. In this work we are concerned with the problem where R + = (0, ∞), N ≥ 1, p, q > 0 and u i ∈ L 1 loc (R N ) for 0 ≤ i ≤ k − 1. We assume that K ∈ C(R + ; R + ) satisfies: K(|x|) ∈ L 1 loc (R N ) and (A) there exists R 0 > 1 such that inf r∈(0,R) K(r) = K(R) for all R > R 0 .
The picture below illustrates some other functions K(r) satisfying the hypothesis (A).
By K * |u| p we denote the standard convolution operator defined by (K * |u| p )(x, t) = R N K(|x − y|)|u(y, t)| p dy for all (x, t) ∈ R N +1 + (2) and we shall require the above integral to be finite for almost all (x, t) ∈ R N +1 + .
We are interested in weak solutions of (1), that is, functions u ∈ L p loc (R N +1 + ) such that (i) (K * |u| p )|u| q ∈ L 1 loc (R N +1 + ); (ii) for any nonnegative test function ϕ ∈ C ∞ c (R N +1 Using a standard integration by parts it is easily seen that any classical solution of (1) is also a weak solution. Let us point out that condition (i) above implies that the convolution integral in (2) is finite for a.a. (x, t) ∈ R N +1 + . This further yields u ∈ L p loc (R N +1 + ). Indeed for R > R 0 and x, y ∈ B R we have |x − y| ≤ 2R, so that by the definition of K and its monotonicity we deduce Since early 1980s many research works have been devoted to the study of the prototype evolution inequalities ∂u ∂t − ∆u ≥ |u| q and ∂ 2 u ∂t 2 − ∆u ≥ |u| q in R N +1 To the best of our knowledge, the first study of nonexistence of solutions to higher order hyperbolic inequalities is due to L. Véron and S.I. Pohozaev [21] related to where and φ is a locally bounded real-valued function such that |φ(u)| ≤ c|u| p , for some c, p > 0. It is obtained in [21] that if q > max{p, 1} and one of the following conditions hold then (5) has no solutions in R N +1 Condition (7) is essential in the study of nonexistence of solutions to (5) and will also play an important role in the study of (1) (see Theorem 1.1 below). Further, if m = 2, it is obtained in [21] that if (6) fails to hold then (5) has a positive solution. Among the results for m = 2, we quote also [10], where a weighted nonlinearity is considered. Another set of results that motivate our present work are due to G.G. Laptev [16,17] where the following problem is studied In the above, Ω ⊂ R N is either the exterior of a ball or an unbounded cone-like domain; for other results on hyperbolic inequalities in exterior domains see [14,15,18]. We observe that solutions of (8) are also required to satisfy (7). It is obtained in [17] that if σ > −2 then the above problem has no solutions provided that 1 < q < q * k := 1 + 2 + σ N − 2 + 2/k .
The above exponent q * k coincides with the Fujita-Hayakawa critical exponent if k = 1, that is, q * 1 = 1 + 2+σ N and with the Kato critical exponent if k = 2, that is, q * 2 = 1 + 2+σ N −1 . The study of convolution terms in time-dependent partial differential equations goes back to near a century ago. Indeed, the equation was introduced by D.H. Hartree [11,12,13] in 1928 for N = 3 and α = 2 in relation to the Schrödinger equation in quantum physics. Nowadays, (9) bears the name of Choquard equation. Stationary solutions to (9) and its related equations have been extensively investigated from various perspectives: ground states, isolated singularities, symmetry of solutions; see e.g., [3,6,7,8,9,19,20]. In a different direction, G. Whithman [22,Section 6] considered the nonlocal equation to study general dispersion in water waves; for recent results on nonlocal evolution equations we refer the reader to [1,2]. In [4] it is studied a class of quasilinear parabolic inequalities featuring nonlocal convolution terms in the form where Lu is a quasilinear operators that contains as prototype model the m-Laplacian and the generalized mean curvature operator.
In the present work we investigate the existence and nonexistence of weak solutions to (1). The stationary solutions to (1) were discussed recently in [6]. Our main result concerning the inequality (1) reads as follows. (i) If k ≥ 1 is a even integer and q ≥ 1, then (1) admits some positive solutions u ∈ C ∞ (R N +1 then (1) has no nontrivial solutions such that Let us note that condition (10) and the fact that K is decreasing in a neighbourhood of infinity imply that Also, under extra assumptions on u k−1 and strengthening (10), we can handle the case R N u k−1 (x) = 0 in (11) for which the same conclusion as in Theorem 1.1(ii) holds (see Proposition 2.3).
The proof of Theorem 1.1 relies on nonlinear capacity estimates specifically adapted to the nonlocal setting of our problem. More precisely, we derive integral estimates in time for the new quantity where ϕ is a specially constructed test function with compact support (see (26)). Theorem 1.1 shows a sharp contrast in the existence/nonexistence diagram according to whether ∂ k−1 u ∂t k−1 (x, 0) has constant sign (positive or negative) on R N . To better illustrate this fact, let us discuss the case of pure powers in the potential K(r) = r −α , α ∈ (0, N ) and k = 1, 2.
Let us first consider the parabolic problem Note that since k = 1 in (1), condition (11) is satisfied for all nonnegative solutions of (12). (i) If 0 < α < m and then (1) has no nontrivial nonnegative solutions; then (12) has positive solutions.
Part (i) in the above result follows from Theorem 1.1 while part (ii) follows from [6, Theorem 1.4] where the stationary case of (12) was discussed. Let us note that for m = 1, the nonexistence of a nonnegative solution in Corollary 1.2(i) was already observed in [4].
We next take K(r) = r −α , k = 2 and m = 1 in Theorem 1.1. We thus consider Our result on problem (13) is stated below.
then (13) has no nontrivial solutions satisfying then (13) admits some positive solutions which verify The diagram of existence/nonexistence of a weak solution to (13) satisfying (14) in the pqplane is given below. The light shaded region represents the region for which (13) admits solutions satisfying (14) while the dark shaded region describes the pairs (p, q) for which no such solutions exist. Corollary 1.3 leaves open the isue of existence and nonexistence in the white regions of the (p > 0, q > 0) quadrant. Theorem 1.1 also applies to the case where K ≡ 1 for which (1) reads A similar conclusion to Corollary 1.2 and Corollary 1.3 (in which we let α = 0) hold for (16). We can further employ the ideas in the study of (1) to the case of systems of type where N, m, k ≥ 1, p, q, n, s > 0. We assume Our main result concerning (18) is stated below.
then (18) has no nontrivial solutions such that The next sections contain the proofs of the above results.
2 Proof of Theorem 1.1 Proof of Theorem 1 (i) Let γ > 0 be such that pγ > N and define where M > 1 will be specified later. Since K > 0 is continuous in R + and decreasing in a neighbourhood of infinity (by condition (A)) it follows that Furthermore, we have since pγ > N .
Further, a direct calculation shows that where c 1 , c 2 are real coefficients depending on N and γ. Hence, an induction argument yields Thus, the function u given by (22) satisfies where c j ∈ R are independent of M . Using the fact that k is an even integer, by taking M > 1 large, we may ensure that where C(K, p, γ) > 0 is the constant from (23). Now, combining (23)-(25) and using that u p e M pt = v −pγ/2 , we deduce being q ≥ 1, which concludes the proof of part (i). Since k is even, one has (ii) Assume that p + q > 2, (10) and (11) hold and that (1) admits a weak solution u. We shall first construct a suitable test function ϕ for (3). To do so, take a standard cut off function ∈ C ∞ c (R) such that: Now take R > 0, γ > 0 to be precised later, κ ≥ 1 sufficiently large and consider the function Clearly and where C is a positive constant changing from line to line.
Observe that by (11) we have u k−1 ≥ 0 or u k−1 ∈ L 1 (R N ) and R N u k−1 (x)dx > 0. In the latter case, from u k−1 ∈ L 1 loc (R N ), we deduce, using (26), Thus, from (11) we deduce that for large R > 0 we have case in which (31) yields provided R > 0 in the definition of ϕ (see (26)) is large enough. An important tool of our approach is the following result. loc (R N ) and where Proof. First note that for x, y ∈ B 2R one has |x − y| ≤ 4R, so that, thanks to the monotonicity of K, where we have used that ϕ ≤ 1 and that ϕ(·, t) ≡ 0 outside of B 2R for all t.
Furthermore, by Hölder's inequality we have which, by (37) and part (i) in the definition of a solution, yields u(·, t) ∈ L p+q 2 loc (R N ) and (35). Inserting (35) into (34) we find We next estimate the integral term on the right hand side of (38). Using Hölder's inequality, we have where in the last inequality we have used (29) with ς = p+q p+q−2 , i = k and thanks to (28). Note that By Hölder's inequality we find Similarly, by Hölder's inequality, (30) with ς = p+q p+q−2 and (28), we find Since a new application of Hölder's inequality in the above estimate yields Comparing the powers of R in (41) and (44) we are led to choose γ > 0 so that kγ = 2m, that is γ = 2m/k. Also, Thus, (38) together with (41) and (44) yield Using (36), the above estimate implies Let {R j } j be a divergent sequence that achieves the limsup in (10), namely Passing to a subsequence we may assume R j > 2R j−1 for all j > 1.
If = ∞, we can pass to the limit in (47), by replacing R with R j , to raise If ∈ (0, ∞), then (47) shows that J ∈ L 2 (0, ∞). Using this fact we infer that as j → ∞. Indeed, the first limit in (48) follows immediately by the fact that J ∈ L 2 (0, ∞). To check the second limit in (48) we observe that The convergence of the last series in the above estimate implies the second part of (48). Using this fact in (45) we find Again, by letting R j → ∞, we can conclude, also when ∈ (0, ∞), that A similar argument allows us to treat the case R N u k−1 (x)dx = 0. In this case we need to be more precise on the behavior of u k−1 .
Proof. With the help of the above we construct the test function ϕ as defined in (26). Thus, (33) is no more in force and (31) changes to Consequently using (41) and (44), with γ = 2k/m, and (35), then the above inequality gives Let σ = 2N +2m/k p+q − N + 2m(1 − 1/k). By Young's inequality we have Using this last inequality into (51) we find Now, in virtue of (50) we may let R → ∞ in the above estimate to deduce J = 0 and then u ≡ 0.

Proof of Corollary 1.3
Part (i) and (ii) in Corollary 1.3 follow directly from Theorem 1.1.
Thus, (60), (32) and (21) yield for R > 0 large. We next employ the estimates (39) and (42) with γ = 2m/k. We find , applying Hölder's inequality we derive In particular, Similarly and in particular, From (62), (64) and (66) we derive Assume now that (19) holds and let {R j } be an increasing sequence of positive real numbers such that R j > 2R j−1 for all j ≥ 2 and K(R j )L(R j ) Assume first = ∞. Then, using the second estimate of (67) into the first we find as j → ∞. This implies that J(t) ≡ 0 so that v ≡ 0. From the second estimate of (67) it follows now that I(t) ≡ 0, hence u ≡ 0. Assume now ∈ (0, ∞). From (69) it follows that J 2 ∈ L 1 (0, ∞) and arguing as in (48), this further implies  Using this fact in (62) along with (64) we deduce Using the second estimate of (71) into the first one obtains