The lifespan of classical solutions of semilinear wave equations with spatial weights and compactly supported data in one space dimension

This paper studies initial value problems for semilinear wave equations with spatial weights in one space dimension. The lifespan estimates of classical solutions for compactly supported data are established in all the cases of polynomial weights. The results are classified into two cases according to the total integral of the initial speed.


Introduction
We consider the following initial value problem for semilinear wave equations with spatial weights.
where p > 1, a ∈ R, f and g are given smooth functions of compact support and a parameter ε > 0 is "small enough". When a = −1, (1.1) is well-studied as a model to ensure the optimality of the general theory for nonlinear wave equations. See Introduction in Imai, Kato, Takamura and Wakasa [1] for all the references to this direction including higher dimensions. More precisely, since we have no time decay of the solution of the free wave equation in one space dimension, there is no possibility to construct a global-in-time solution of (1.1) for any p > 1. In fact, we have the finite-time blow-up result by Kato [5]. Therefore we are interested in the so-called lifespan estimates, namely, some kind of a stability of a zero solution because we have an uniqueness of the solution of (1.1). Let T (ε) be, the so-called lifespan, the maximal existence time of the classical solution of (1.1) with arbitrary fixed non-zero data. Due to Zhou [9], we have where T (ε) ∼ A(ε, C) stands for the fact that there are positive constants, C 1 and C 2 , independent of ε satisfying A(ε, C 1 ) ≤ T (ε) ≤ A(ε, C 2 ). We note that p > 1 implies p − 1 2 < p(p − 1) p + 1 , so that the first quantity is smaller than the second one in (1.2). This phenomenon follows from the fact that Huygens' principle holds if the total integral of the initial speed is zero. When a = −1, there are a few results only with the assumption that the data has non-compact support. This kind of the problem was first proposed by Suzuki [7] in which the nonlinearity |u| p is replaced by |u| p−1 u showing the global-in-time existence for odd function data when p > (1 + √ 5)/2 and pa > 1. She also studied the blow-up result of modified integral equations. See the section 6 in [7]. Later, Kubo, Osaka and Yazici [6] extended such a result for all p > 1 and pa > 1. Moreover, they obtained the blow-up in finite-time for (1.1) with some positive data for p > 1 and a ≥ −1. Inspired by some computation of the upper bound of the lifespan in [6], Wakasa [8] obtained the following lifespan estimate for (1.1).
for a > 0, (1.3) where φ −1 is an inverse function of φ defined by φ(s) := s log(2 + s). (1.4) We note that this result is also available even if |u| p is replaced with |u| p−1 u in (1.1). The aim of this paper is to establish the lifespan estimates for compactly supported data in all the cases of a including a < −1. More precisely, our results are the following estimates.
We remark that the quantities in all the cases of (1.5) are larger than those of (1.6). This fact follows from the trivial inequality by p > 1 in the first case of a < 0. For the second case a = 0, one can check it by comparing two functions φ −1 (ξ) and ψ −1 p (ξ p ) with respect to the large variable ξ by making use of differentiation. The third case a > 0 is trivial. We also note that Suzuki obtained T (ε) < ∞ for −1 ≤ a ≤ 1 and g(x) ≥ 0( ≡ 0) in the section 7 in [7]. Her original result is established for the nonlinear term |u| p−1 u, but the proof of u ≥ 0 is missing for compactly supported data.
It is interesting to compare the nonlinear term with time-decaying weights in Kato, Takamura and Wakasa [4] which is closely related to the scaleinvariantly damped wave equations. In such a situation, we have a possibility to obtain the global-in-time existence for the super-critical case, and the exponential type estimate of the lifespan for the critical case.
This paper is organized as follows. In the next section, (1.5) and (1.6) are divided into four theorems, and the preliminaries for their proofs are introduced. Section 3 and 4 are devoted to the proofs of the longtime existence and the blow-up in finite time of the solution, respectively. The main method in this paper is based on point-wise estimates which are originally introduced by John [2] in three space dimensions, and developed by Zhou [9] and Kato, Takamura and Wakasa [4] in one space dimension.

Main results and preliminaries
Throughout of this paper, we assume that the initial data (f, g) Our results on (1.5) and (1.6) are splitted into the following four theorems.
Theorem 2.1 Assume the support condition (2.1) and Then, there exists a positive constant ε 1 = ε 1 (f, g, p, a, R) > 0 such that a classical solution u ∈ C 2 (R × [0, T )) of (1.1) exists as far as T satisfies where 0 < ε ≤ ε 1 , c is a positive constant independent of ε and φ is the one in (1.4).
Remark 2.1 In Wakasa [8] for the non-compactly supported data, the assumption on the data is Then, there exists a positive constant ε 2 = ε 2 (f, g, p, a, R) > 0 such that a classical solution u ∈ C 2 (R × [0, T )) of (1.1) exists as far as T satisfies for a > 0, (2.5) where 0 < ε ≤ ε 2 , c is a positive constant independent of ε and ψ p is the one in (1.7).

Theorem 2.3
Assume the support condition (2.1) and Then, there exists a positive constant ε 3 = ε 3 (g, p, a, R) > 0 such that a classical solution u ∈ C 2 (R×[0, T )) of (1.1) cannot exist whenever T satisfies where 0 < ε ≤ ε 3 , C is a positive constant independent of ε and φ is the one in (1.4).

Theorem 2.4
Assume the support condition (2.1) and Then, there exists a positive constant ε 4 = ε 4 (f, p, a, R) > 0 such that a classical solution u ∈ C 2 (R×[0, T )) of (1.1) cannot exist whenever T satisfies for a > 0, (2.9) where 0 < ε ≤ ε 4 , C is a positive constant independent of ε and ψ p is the one in (1.7). All the proofs of above theorems are given in following sections. Here we shall introduce preliminaries. Let u be a classical solution of (1.1) in the time interval [0, T ). Then the support condition of the initial data, For example, see Appendix of John [3] for this fact. It is well-known that u satisfies the following integral equation.
where u 0 is a solution of the free wave equation with the same initial data, and a linear integral operator L a for a function v = v(x, t) in Duhamel's term is defined by (2.13) Proposition 2.1 Assume that (f, g) ∈ C 2 (R) × C 1 (R). Let u be a continuous solution of (2.11). Then, u is a classical solution of (1.1).
Proof. In view of (2.13), the differentiability of L a (v) follows from the continuity of v. Therefore the conclusion follows from the regularity assumption on the initial data. ✷ The following property, namely Huygens' principle, of u 0 will play an essential role in the proofs of Theorems 2.2 and 2.4. (2.14) Proof. For t ≥ R and |x| ≤ t − R, we have Therefore it follows from (2.1), (2.4) and (2.12) that On the other hand, it is trivial that so that (2.14) holds. ✷ Due to Proposition 2.2 as well as (2.10), we shall divide the support of the solution into three pieces, the interior domain the exterior domain and the small domain near the origin We will see that the lifespan is determined by point-wise estimates of the solution in D Int .

Proofs of Theorems 2.1 and 2.2
In this section, we investigate the lower bound of the lifespan. In view of Remark 2.1, only the case of a < −1 should be considered in the proof of Theorem 2.1. But, following the proof of Wakasa [8], all the estimates for the case of −1 ≤ a < 0 hold also for the case of a < −1, so that we can omit its proof here. In fact, we have to show that (4.6) in Wakasa [8] is also established for a < −1. The case of 0 ≤ x ≤ t ≤ T is trivial and another case of x ≥ t follows from |x| ≤ t + R by (2.10). From now on, we shall prove Theorem 2.2 only. To this end, we have to set the following function space which is different from Wakasa [8]. Following Kato, Takamura and Wakasa [4], we shall construct a solution as a limit of the sequence {U n (x, t)} n∈N defined by and a weighted norm of a function U = U(x, t) by We note that Hölder's inequality holds.
Then we have a priori estimates in the following propositions.
Proposition 3.2 Suppose that the assumption of Theorem 2.2 is fulfilled. Let L a be the one in (2.13). Then, for where E(T ) is defiend by First we shall prove the main theorem. The proofs of the propositions above are given later.

Proof of Theorem 2.2.
By virtue of Proposition 2.1, it is sufficient to construct a continuous solution of the integral equation (2.11). Following Kato, Takamura and Wakasa [4], let X be a Banach space defined by which is equipped with the norm (3.3), and its closed subspace Y by The continuity of the sequence is also trivial.
Since we have Propositions 3.1 with m = 0 and 3.2 yield where C is the one in Proposition 3.2. Hence the boundedness in Y of {U n } in (3.1) follows from From now on, we assume (3.9). Since holds with some θ ∈ (0, 1), Propositions 3.1 with m = 1 and 3.2 yield Here we have employed (3.4) as and so on.
We note that (3.9) and (3.10) guarantee the existence of a limit of {U n } in Y .
When a > 0, it is easy to find c and ε 2 in (2.5) because of D(T ) = 1 and E(T ) = T + R. We omit details.
When a = 0, let us look for a sufficient condition on T to (3.9) and (3.10). The definitions of D(T ) and E(T ) in (3.6) and (3.8) respectively yield Assume that T ≥ R.
Then (3.11) follows from Therefore Theorem 2.2 for a = 0 is established with and a number ε 2 is defined to satisfy This is possible. The first inequality is trivial. Setting Hence the second inequality can be valid by taking ε −(p−1) 3 large enough because it is easy to find a point s 0 independent of ε such that The case of a < 0 is almost similar to the above. (3.9) and (3.10) follow from Since holds, it is easy to see that (2.5) for a < 0 is established. Therefore the proof of Theorem 2.2 is now completed. ✷ Proof of Proposition 3.1.
In view of Proposition 2.2 and (2.12), we have and Therefore Proposition 3.1 follows from (3.14) Due to the symmetry of I 0 on x as I 0 (−x, t) = I 0 (x, t), it is sufficient to show (3.14) in case of x ≥ 0.
From now on, all the constants C = C(f, g, a, m, R) > 0 may change from line to line for simplicity. Changing variables by α := s + y, β := s − y (3.15) and making use of we have that First, we shall estimate I 01 in D Ext . Extending the domain of the integral, we have When a > 0, the α-integral is dominated by When a = 0, the α-integral is dominated by log m (t + x + 3R) 2 log 1 + α − β 2 α=t+x α=R ≤ 2 1+m log(t + x + 3R) log m (T + 3R) because of log(t + x + 3R) ≤ log(2t + 4R) ≤ 2 log(T + 3R).
When a < 0, the α-integral is dominated by Hence we obtain On the other hand, it is easy to see that Moreover, similarly to I 02 in D Ext , we also have Therefore, summing up, we obtain (3.14) as desired.  Similarly to I 0 in the proof of Proposition 3.1, it is sufficient to show (3.18) in case of x ≥ 0.
From now on, all the constants C = C(f, g, a, R) > 0 may change from line to line for simplicity. Changing variables by (3.15) again, we have that First, we shall estimate I 11 in D Int . Since the symmetry of the integrand in y = (α − β)/2, we have so that we obtain When a > 0, the α-integral is estimated as so that we have When a = 0, the α-integral is estimated as so that we have When a < 0, the α-integral is estimated as so that we have Next we shall deal with I 12 in D Int . When a > 0, we have so that the estimate is the same as I 01 in the proof of Proposition 3.1 which implies that When a = 0, we have so that follows, which implies When a < 0, we have follows. The α-integral is the same as I 01 in the proof of Proposition 3.1, so that follows.
Similarly to the above, we shall estimate I 13 in D Int . When a > 0, we have follows. When a = 0, we have When a < 0, we have It is easy to estimate I 14 in D Int . Extending the domain of the integral, we have Summing up all the estimates, we obtain Let us step into the estimates in D Ext . When a > 0, we have so that the estimate is completely the same as I 01 in the proof of Proposition 3.1. Hence we obtain When a = 0, we have Hence, similarly to the above, we obtain When a < 0, we have Hence we obtain Moreover, it is easy to estimate I 22 in D Ext . Extending the domain of the integral, we have Summing up all the estimates, we obtain Finally we shall estimate I 3 in D Ori , but this is almost trivial because of Hence we obtain Therefore (3.18) is established as desired. This completes the proof of Proposition 3.2. ✷.

Proofs of Theorems 2.3 and 2.4
In this section, we shall investigate the upper bounds of the lifespan. As stated at the end of Section 2, the upper bounds of the lifespan are also determined by point-wise estimates of the solution in the interior domain, D Int in (2.15). In fact, it follows from (2.1) and (2.12) that Throughout this section, we assume that Making use of (3.16) and introducing the characteristic coordinate by (3.15), we have that where and Employing this integral inequality, we shall estimate the lifespan from above.

Proof of Theorem 2.3
Let u = u(x, t) ∈ C 2 (R × [0, T )) be a solution of (1.1). It follows from (2.6), (4.2) and (4.4) that for (x, t) ∈ D, where Assume that an estimate holds, where a n ≥ 0 and M n > 0. The sequences {a n } and {M n } are defined later. Then it follows from (4.5) and (4.7) that Note that the domain of the integral is included in D R , that is, Therefore, if {a n } is defined by a n+1 = pa n + 1, a 1 = 0, (4.9) then (4.7) holds for all n ∈ N as far as M n satisfies In view of (4.5), we note that (4.7) holds for n = 1 with Let us fix {M n }. It follows from (4.9) that which implies pa n + 1 = a n+1 ≤ p n p − 1 .
In view of (4.10) and (4.11), one of the choice of the definition of {M n } is where Hence we obtain that M n > 0 for all n ∈ N and log M n+1 = log C 2 − 2n log p + p log M n which implies log M n+1 = (1 + p + · · · + p n−1 ) log C 2 −2{n + p(n − 1) + · · · + p n−1 (n − n + 1)} log p + p n log M 1 because of d'Alembert's criterion. Therefore it follows from (4.7) that where and (4.16) If there exists a point (x 0 , t 0 ) ∈ D R such that we have a contradiction u(x 0 , t 0 ) = ∞ by letting n → ∞, so that T < t 0 . Let us set x 0 = R and t 0 ≥ 4R. (4.17) This condition follows from We note that (4.18) is stronger than t 0 ≥ 4R for where ε 3 is defined by It is easy to see that (R, t 0 ) ∈ D R with t 0 satisfying (4.18). The proof for a > 0 is now completed.

Case 2. a = 0.
Assume that an estimate holds, where a n ≥ 0 and M n > 0. The sequences {a n } and {M n } are defined later. Then it follows from (4.5) and (4.19) that Note that the domain of the integral is included in D, that is, Hence we can employ the same definitions of {M n } and {a n } as Case 1 in which C 1 is replaced with 2C 0 , so that we have where and where S p is the one in (4.14). Therefore the same argument as Case 1 is valid. The difference appears only in finding (x 0 , t 0 ) ∈ D with K 2 (x 0 , t 0 ) > 0. Let t 0 = 2x 0 and t 0 ≥ 4R.
Then, since we have . This completes the proof for a = 0.
This case is almost similar to Case 2. Assume that an estimate holds, where a n ≥ 0 and M n > 0. The sequences {a n } and {M n } are defined later. Then it follows from (4.5) and (4.21) that Note that the domain of the integral is included in D, that is, Hence we can employ the same definitions of {M n } and {a n } as Case 1 in which C 1 is replaced with 2C 0 /(1 − a), so that we have u(x, t) ≥ C 6 (t − x − R) x 1−a 1 + t + x −1/(p−1) exp K 3 (x, t)p n−1 for (x, t) ∈ D, where C 6 := exp − 1 p − 1 log C 7 > 0, C 7 := 2(p − 1) 2 C 0 1 − a > 0 (4.22) and where S p is the one in (4.14). Therefore the same argument as Case 1 is valid. The difference appears only in finding (x 0 , t 0 ) ∈ D with K 3 (x 0 , t 0 ) > 0. Let t 0 = 2x 0 and t 0 ≥ 4R.
Then, since we have This completes the proof for a < 0. ✷

Proof of Theorem 2.4
The proof is almost similar to the one of Theorem 2.3. Let u = u(x, t) ∈ C 2 (R × [0, T )) be a solution of (1.1). Since the assumption on the initial data in (2.8) yields Because, if not, we have to assume that f (x) ≡ 0 for x ∈ (0, R).
Therefore we obtain all the estimates below for x < 0 by replacing x with −x. Because, taking f (x + t) instead of f (x − t) in (4.23), we have, in stead of (4.24), that This implies the symmetry of the domain as well as the estimates. Case 1. a > 0.