The lifespan estimates of classical solutions of one dimensional semilinear wave equations with characteristic weights

In this paper, we study the lifespan estimates of classical solutions for semilinear wave equations with characteristic weights and compactly supported data in one space dimension. The results include those for weights by time-variable, but exclude those for weights by space-variable in some cases. We have interactions of two characteristic directions.


Introduction
In this paper, we focus on the study of a model equation for the purpose of extending the general theory of nonlinear wave equations. Before stating our target, we first overview the general theory for nonlinear wave equations in one space dimension. Consider the following Cauchy problem: where f, g ∈ C ∞ 0 (R) and ε > 0 is a small parameter. Let λ = (λ; (λ i ), i = 0, 1; (λ ij ), i, j, = 0, 1, i + j ≥ 1). Assume that H = H( λ) is a sufficiently smooth function with in a neighborhood of λ = 0, where α ∈ N. Let us define the lifespan T (ε) as the maximal existence time of the classical solution of (1.1) with arbitrary fixed non-zero data. The general theory for the problem (1.1) is to express the lower bound of the lifespan according to α and the initial data. Li, Yu and Zhou [9] have constructed the general theory for this problem: It should be noted that Morisawa, Sasaki and Takamura [8] point out that there is a possibility to improve the above general theory by studying H = |u t | p + |u| q so-called "combined effect" nonlinearity. For the detail, see the end of Section 2 in [8].
We set H = |u| p in (1.1) which is a model to ensure the optimality of the general theory. Note that Kato [5] showed the blow-up result for any p > 1. Zhou [13] has obtained the following lifespan estimates for p > 1: Here, the definition of T (ε) ∼ A(C, ε) is given as follows: there exist positive constants C 1 and C 2 which are independent of ε satisfying A(C 1 , ε) ≤ T (ε) ≤ A(C 2 , ε). The differences between the lifespan estimates of (1.2) come from Huygens' principle which holds if the total integral of the initial speed g vanishes. See Lemma 2.1 below. Our motivation is to extend the general theory to the non-autonomous non-linear terms: H = H(x, t, u, u t , u x , u xx , u xt ).
In order to look for the assumptions on x and t in H, we shall investigate the following model equations in this paper: where p > 1 and F ∈ C 1 (R × (0, ∞)). Our purpose is to get the lifespan estimates for the problem (1.3) with the characteristic weights: where x := (1 + x 2 ) 1/2 for x ∈ R and a, b ∈ R. The meaning of the "characteristic" is originally used as t + |x| and t − |x|. We have changed it to (1.4) by using · , because we need the regularity to get the classical solutions and to avoid the singularity when the power of the exponent is negative. However, this modification is not essential because of Lemma 2.2 below. Let us denote by T (ε) the maximal existence time of the classical solution of (1.3) with arbitrary fixed non-zero data. Then, we have the following main results: T (ε) = ∞ for a + b > 0 and a > 0, (1.5) for a + b = 0 and a > 0, or a = 0 and b > 0, exp(ε −(p−1)/2 ) for a = b = 0, ε −(p−1)/(−a) for a < 0 and b > 0, φ −1 1 (ε −(p−1) ) for a < 0 and b = 0, ε −(p−1)/(−a−b) for a + b < 0 and b < 0 (1.6) if R g(x)dx = 0, where φ −1 1 is an inverse function of φ 1 defined by φ 1 (s) = s −a log(2 + s), and ) for a = 0 and b > 0, exp(ε −p(p−1) ) for a + b = 0 and a > 0, exp(ε −p(p−1)/(p+1) ) for a = b = 0, ε −(p−1)/(−a) for a < 0 and b > 0, ψ −1 1 (ε −p(p−1) ) for a < 0 and b = 0, ε −p(p−1)/(−pa−b) for a < 0 and b < 0, ψ −1 2 (ε −p(p−1) ) for a = 0 and b < 0, ε −p(p−1)/(−a−b) for a + b < 0 and a > 0 where ψ −1 1 and ψ −1 2 are inverse functions of ψ 1 and ψ 2 respectively defined by ψ 1 (s) = s −pa log(2 + s) and ψ 2 (s) = s −b log p−1 (2 + s). (1.9) We explain the background to considering the form of the weight function F in (1.4). Firstly, the pointwise estimates of the wave equations have a characteristic factor such as (1.4). A natural question arises how the lifespan estimates change as compared with those for non-weighted case, (1.2). Secondly, our equations include some damped wave equations which were treated by many previous works. For example, let us consider the following Cauchy problem for the nonlinear damped wave equations: (1.10) Then, if we set u(x, t) = (1 + t)v(x, t) (Liouville transform), we have u(x, 0) = εf (x), u t (x, 0) = ε{f (x) + g(x)} for x ∈ R. (1.11) When p > 3, D'Abbicco [2] showed that the energy solution of (1.10) exists globally in time. On the other hand, Wakasugi [12] has obtained the blowup results of the problem (1.10) for 1 < p ≤ 3. We note that they treated more general damping terms µv t /(1 + t) (µ > 0) in (1.10). For the lifespan estimates in (1.11), Wakasa [10] obtained T (ε) ∼ ε −(p−1)/(3−p) for 1 < p < 3, exp(ε −(p−1) ) for p = 3 if R {f (x) + g(x)}dx = 0 (1.12) and Kato, Takamura and Wakasa [6] obtained is a positive number satisfying ε 2 b log(1+b) = 1. Here, we note that the critical exponent 3 which is a threshold between the global-in-time existence and the blow-up in finite time is the so-called Fujita exponent in one space dimension. Remarkably, our lifespan estimates (1.8) with a = p−2 and b = −1 coinside with (1.12) and (1.13). This is because, 1 + t is equivalent to t + x by the finite propagation speeds. See (2.1) and Lemma 2.2 below. We next mention the results to the case where F in (1.3) is a spatial weight only as F = x −(1+b) with b ∈ R. Kitamura, Morisawa and Takamura [7] have obtained the lifespan estimates, and where φ −1 and ψ −1 are inverse functions defined by φ(s) = s log(2 + s) and ψ(s) = s log p (2 + s), respectively. The differences between F = (1 + t) −(p−1) and F = x −(1+b) in the lifespan estimates come from the decay property of the weights near the origin. Indeed, for the time weights, there is a possibility that we get the global existence for p > 3 due to the decay of the nonlinear term even if |x| is small. But there is no such a situation for the spatial weights. Finally, we remark the setting of the weight function F in our problem in (1.3). If F = t + x −(1+a) , we cannot expect to obtain the global existence result, because we have no decay property along t + x = 0 when x < 0. On the other hand, when x > 0, we have the decay property for t + x −(1+a) near the origin. For this reason, we get the lifespan estimates similar to the time weights. Therefore, it is necessary to consider the weighted functions t + |x| −(1+a) and t − |x| −(1+b) separately. In addition, if F ∈ C 1 (R × (0, ∞)) does not hold, we have no chance to obtain the classical solution of (1.3) even locally in time.
Before stating our main results, we assume some conditions for the initial data.
Here, we address our lifespan estimates in the (a, b)-plane for the convenience of explanation. Figure 1 is related to Theorems 1.1, 1.2 and 1.4. Figure 2 is related to Theorems 1.1, 1.3 and 1.5.   Remark 1.1. We compare all the lifespan estimates when a = −1 in our results as for F = t − x −(1+b) with those for F = x −(1+b) by Kitamura, Morisawa and Takamura [7]. If the total integral of g does not vanish, they coincide with each other. However, if the total integral of g vanishes, it does not hold. The reason for this situation can be described as follows. The triangle domain which has a vertex (x, t) in the following figure is the domain of the integral of the Duhamel term, (2.3) below.
When b > −1, we cannot derive any decay of the nonlinear term from the spatial weights y −(b+1) along y = 0. On the other hand, we cannot derive any decay of the nonlinear term from characteristic weights s − |y| −(b+1) along s − |y| = 0. When b < −1, for the weights y −(b+1) , the growth of the solution appears on s − |y| = 0. On the other hands, for the weights s − |y| −(b+1) , the growth of the solution appears on y = 0. Keeping this situation in mind, we shall step into our purpose.
Set a = −1 as F = t − |x| −(1+b) . If we assume (1.17), then the Huygens' principle (Lemma 2.1) does not hold. Hence, we have that u ∼ O(ε) in its support. See the construction of the solution in the function space Y 1 below at the end of section 3. This fact helps us understand that the lifespan estimates in Theorem 1.2 and Theorem 1.4 are the same as that of (1.14).
In contrast, if we assume (1.18) which implies that Huygens' principle holds, we have that u ∼ O(ε p ) in the interior domain {t + |x| ≥ R and t − |x| ≥ R} while we have that u ∼ O(ε) in the exterior domain {t + |x| ≥ R and |t − |x|| ≤ R}. See the construction of the solution in the function space Y 2 below at the end of section 4. For F = x −(b+1) (b > −1), the solution u does not decay along y = 0 which is located in the interior domain. On the other hands, for F = t − |x| −(b+1) , the solution u does not decay along s − |y| = 0 which is located in the exterior domain. This fact helps us understand that the lifespan estimates in Theorems 1.3 and 1.5 are smaller than those of (1.15) when b > −1. Conversely, in view of the considerations above, when b < −1, the lifespan estimates in Theorems 1.3 and 1.5 are larger than those of (1.15). Remark 1.2. We remark that the weight function F of t + x and t − x can cover a large class of power-type weights. For example, let us consider F = t + x 2 −(a+1) . When a > −1, we may have the same results as for F = t −(a+1) because x 2 in F has a minor effect. On the other hands, when a < −1, we may have the same results as for F = x 2 −(1+a) because x 2 in F has a major effect. It is sufficient to consider the linear combinations of x and t in the weighted functions.
Let us consider F = t − C|x| −(b+1) , where C is a positive constant. If 0 < C < 1 and t − |x| > 0, then we have that so that the lifespan estimates may coincide with that for the time weights. On the other hands, if C > 1, the line s − C|y| = 0 in (s, y)-domain of the integral of the Duhamel term is located in the interior domain in Remark 1.1. Therefore, the lifespan estimates may coincide with that for the spatial weights.
This work was almost completed during the period when the first author was in the master course of Mathematical Institute of Tohoku University and the second author had the second affiliation of Research Alliance Center of Mathematical Sciences, Tohoku University. This paper is organized as follows. In the next section, we set the notation and prove some lemmas needed later. The proofs of Theorem 1.1 with (1.17) and Theorem 1.2 are established in Section 3. In Section 4, we prove Theorem 1.1 with (1.18) and Theorem 1.3. Finaliy, we prove Theorem 1.4 and Theorem 1.5 in Section 5. All the proofs in this paper are based on the point-wise estimates of solutions which are originally introduced by John [4].

Preliminaries
In this section, we set the notation and prove some useful lemmas. Assume that u ∈ C 2 (R × [0, T ]) is a solution to the Cauchy problem (1.3). Then the following finite propagation speeds holds: For the proof, see Appendix of John [3]. We define and is the solution to the Cauchy problem (1.3).
The following lemma is so-called Huygens' principle which is essential for the proof of main theorems.
For the proof, see Proposition 2.2 in [7]. and Proof. We only prove Others are trivial by taking the difference between the squares of right-hand side and left-hand side. It is easy to see that Then, we see ✷ Finally, we introduce the following domains to estimate the solutions: Note that the lifespan is determined by the point-wise estimates of the solution in D Int .

Proofs of Theorem 1.1 and Theorem 1.2
We define L ∞ -norm of u by The following a priori estimate plays a key role in the proofs of Theorem 1.1 and Theorem 1.2.
where E 1 (T ) is defined by Proof. We denote by C a positive constant which depends only on p, a, b, R and j may change from line to line. The definition of (2.3) gives us Then, the inequality (3.2) follows from Thus, we show (3.6) in the following. Without loss of generality, we may assume x ≥ 0 due to the symmetry of I(x, t) as I(x, t) = I(−x, t) holds by (3.5). Changing variables by in the integral of (3.5), we get Here we set We first state the following lemma which is useful to estimate the above integrals.
Lemma 3.2. Let q ∈ R and µ, ν ≥ 0. Then there exists a positive constant The proof is easy so that we shall omit it. and Proof. We show the inequality (3.9) only as (3.10) is trivial. The assumptions give us Case 1. a > 0 and a + b > 0, or a > 0 and a + b = 0, or a > 0 and a + b < 0.
Making use of (3.8) with q = a(< 0) and (3.10) to the α-integral, we obtain It follows from (3.8) with q = b and (3.10) that Summing up all the estimates, we get Next, we shall estimate I 12 in D Int . Noticing that β ≤ t − x ≤ t + x, the estimates of this integral are the same as the estimates for I 11 , which is replaced by α and β. So, we have Next, we shall estimate I 13 and I 14 in D Int . We note that both the βintegral in I 13 and α-integral in I 14 are bounded by C. Thus, we get Making use of (3.8) with q = a, (3.9) and (3.10) to the above integrals, we obtain It follows from T + R ≥ 1 and log(T + 3R) ≥ 1, (3.12) (3.11) and the definition of Next, we shall estimate I 21 and I 22 in D Ext ∪ D Ori . First, we investigate them in D Ext . Since |t − x| ≤ R holds for (x, t) ∈ D Ext , we get For t − x > 0, we have Making use of (3.12) for the estimates of I 22 , we get where we have used β ≥ −R. It follows from (3.9), (3.10) and (3.12) that Finally, we shall estimate I 21 and I 22 in D Ori . Noticing that |t − x| ≤ R and t + x ≤ R hold for (x, t) ∈ D Ori , we obtain Making use of (3.12), we get Summing up all the estimates, we obtain (3.6). Therefore the proof of (3.2) is established. ✷ Proofs of Theorem 1.1 and Theorem 1.2. Let us define a Banach space which is equipped with the norm (3.1). Define a sequence of functions {u n } n∈N by where L and u 0 are defined in (2.3) and (2.2) respectively.
We shall construct a solution of the integral equation (2.4) in a closed subspace Because it follows from (2.2) and the assumption for (f, g) that |u 0 (x, t)| ≤ εC 0 . Then, analogously to the proof of Theorem 1.2 in [11] with M : 4 Proofs of Theorem 1.1 and Theorem 1.3 We define the following weight function: Denote a weighted L ∞ -norm of U by Then we have the following a priori estimates.
where D(T ) is defined by Suppose that the assumptions of Theorem 1.1 and Theorem if a + b < 0 and a > 0.
Proof of Lemma 4.1. We denote by C a positive constant which depends only on p, a, b, R and j may change from line to line. Making use of (2.5), we have Then, the inequality (4.3) follows from Thus, we show the above inequality in the following.
Without loss of generality, we may assume x ≥ 0 due to the symmetry of J(x, t) in x. Changing variables by (3.7), we obtain Here we set First, we shall estimate J 11 in D Int . Noticing that by α ≤ t + x, (3.9) and (3.10), we obtain Next, we shall estimates J 12 in D Int . Because of β ≤ t − x, we get where we have used (3.9) and (3.10). Hence we obtain Making use of (3.13), we obtain We next estimate J 21 and J 22 in D Ext . Because of |t − x| ≤ R, we get Thus, the desired estimates for J 21 and J 22 are established. Finally, we shall estimate J 21 and J 22 in D Ori . Noticing that |t − x| ≤ R and t + x ≤ R hold for (x, t) ∈ D Ori , we obtain for t − x > 0. It follows from w(|x|, t) ≥ 1 and (3.12) that Therefore, we get (4.3). ✷ Proof of Lemma 4.2. We denote by C a positive constant which depends only on p, a, b, R and j may change from line to line. The definition of (2.3) gives us Then, the inequaltiy (4.4) follows from (4.5) Thus, we show (4.5) in the following. Without loss of generality, we may assume x ≥ 0 due to the symmetry of J (x, t) in x. Changing variables by (3.7), we get Here we set Case 1. a > 0 and a + b > 0, or a > 0 and a + b = 0, or a > 0 and a + b < 0.
Because of w(|x|, t) = 1, the desired estimates are established by the same manner as that of Case 1 of Lemma 3.1. We shall omit the detalis. We shall estimate J 11 in D Int . The definition of (4.1) yields Let a = 0 and b > 0, or a = 0 and b = 0. Making use of (3.8), we get It follows from (3.9) and (4.6) that For the case of a = 0 and b < 0, we employ the same argument as in the proof of Lemma 3.1. By virtue of (3.8) and (3.10), we get Noticing that log p (t + x + 3R) ≤ log p−1 (T + 3R)w −1 (x, t) by (3.9), we obtain Therefore, we get Next, we shall estimate J 12 . It follows from (4.1) that The estimates of this integral are the same as those of J 11 , in which α and β are replaced with each other. Hence, we get Next, we shall estimate J 13 and J 14 . The definition of (4.1) and (3.9) gives us It follows from (3.13), (3.12) and Next, we shall estimate J 21 and J 22 in D Ext ∪ D Ori . Since |t − x| ≤ R holds for (x, t) ∈ D Ext , we have Thus, the estimates of above integrals are the same as those of J 13 and J 14 . Moreover, since (x, t) ∈ D Ori is bounded, we have Making use of log(t + |x| + 3R) ≥ 1 and (3.12), we get the desired estimates. Hence, we obtain Case 3. a < 0 and b > 0, or a < 0 and b = 0, or a < 0 and b < 0.
We shall show the estimate for J 11 in D Int . The definition of w in (4.1) and the trivial inequality (3.10) yield Making use of (3.8) for the above integral, we get Thus, (4.7) implies Next, we shall estimate J 12 . It follows from (4.1) that The estimates of this integral are the same as those of J 11 , in which α and β are replaced with each other. Hence, we get Next, we shall estimate J 13 and J 14 . It follows from (4.1) and (3.9) that Hence, because of (3.13) and (3.12), we get Next, we shall estimate J 21 and J 22 in D Ext ∪ D Ori . Since |t − x| ≤ R holds for (x, t) ∈ D Ext , we have for t − x > 0 in D Ext . Thus, the estimates of above integrals are the same as those of J 13 and J 14 . Moreover, since (x, t) ∈ D Ori is bounded, we have Making use of t + |x| + 3R ≥ 1 and (3.12), we get the desired estimates. Hence, we obtain Summing up all the estimates, we obtain (4.5). Therefore the proof of (4.4) is now established. ✷ Proofs of Theorem 1.1 and Theorem 1.3. We shall employ the same argument as in the proof of Theorem 2.2 of [7]. We consider an integral equation: where L and u 0 are defined in (2.3) and (2.2) respectively. Let us define a Banach space which is equipped with a norm (4.2). We shall construct a solution of the integral equation (4.8) in a closed subspace Analogously to the proof of Theorem 1.2 in [7] with M := C 2 , we can see that U n ∈ Y 2 (n ∈ N) provided the inequality holds, where C 3 , E 2 (T ) are defined in (4.4). Moreover, {U n } is a Cauchy sequence in Y 2 provided the inequality holds, where D(T ) is the one in (4.3). When a > 0, we can easily find ε 1 in Theorem 1.1, or c and ε 3 in Theorem 1.3, because of D(T ) = 1. When a = 0 and b > 0, the conditions (4.9) and (4.10) follow from where we set When a = b = 0, since p(p − 1)/(p + 1) < p − 1, the conditions (4.9) and (4.10) follow from where we set When a = 0 and b < 0, the estimate in Theorem 1.3 is derived by Here ε 3 has to satisfy When a < 0 and b > 0, the conditions (4.9) and (4.10) follow from where we set When a < 0 and b = 0, the estimate in Theorem 1.3 is derived by Here ε 3 has to satisfy Finally, when a < 0 and b < 0, since p(p − 1)/(−pa − b) ≤ (p − 1)/(−a), the conditions (4.9) and (4.10) follow from where we set In this section, we shall investigate the upper bounds of the lifespan. We note that they are determined by point-wise estimates of the solution in the interior domain, D Int . In fact, it follows from (1.16) and (2.2) that In this section, we assume that Making use of Lemma 2.2 and introducing the characteristic coordinate by (3.7), we have that Employing this integral inequality, we shall estimate the lifespan from above. We also use the following lemma. Then, Proof. The definition of M n yields log M n+1 = log C − n log(ηµ) + p log M n .

Proof of Theorem 1.4
Let u = u(x, t) ∈ C 2 (R × [0, T )) be a solution of (1.3). It follows from (1.19), (5.2) and (5.5) that for (x, t) ∈ D, where Case 1. a > 0 and a + b < 0, or a ≤ 0 and b < 0. First, we consider only the case where a > 0 and a + b < 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 (5.9) and (5.10) that and b < 0, we get Therefore, if {a n } is defined by a n+1 = pa n + 1, a 1 = 0, (5.13) then (5.10) holds for all n ∈ N as far as M n satisfies In view of (5.8), we note that (5.10) holds for n = 1 with Therefore, it follows from (5.13) that According to a n < p n−1 /(p − 1), (5.6) and (5.15), we define {M n } by Hence, making use of Lemma 5.1, we reach to Therefore, we obtain If there exists a point (x 0 , t 0 ) ∈ D such that we have u(x 0 , t 0 ) = ∞ by letting n → ∞, so that T < t 0 . Let us set 2x 0 = t 0 and 6R < t 0 . Then because the inequality holds for 6R < t 0 . The condition (5.17) follows from The proof for a > 0 and a + b < 0 is now completed. Next, we consider the case where a ≤ 0 and b < 0. Assume that holds. Then, similarly to the above computations with (5.9), we have Making use of (5.11), we get This estimate is almost same as (5.12). Therefore, the same lifespan estimate is obtained except for the included constant independent of ε. ✷ Case 2. a < 0 and b = 0. Assume that an estimate holds, where a n is the one in (5.16). In this case, M n is defined by (5.6) with C = C 2 := C 0 (p − 1) 2 (1 − a) −1 > 0 and η = µ = p. We note that the same argument as in Case 1 can be applicable to also this case. It follows from (5.8) and (5.18) that For the α-integral, we obtain Replacing β by (t − x) in the quantity above, we get Therefore, we obtain and S p is the one in (5.7). If there exists (x 0 , t 0 ) ∈ D such that K 2 (x 0 , t 0 ) > 0, then we have u(x 0 , t 0 ) = ∞ as before. Set 2x 0 = t 0 and 2 + t 0 ≥ 16R 2 . Then for 0 < ε ≤ ε 4 , where ε 4 is defined by The proof for a < 0 and b = 0 is now completed. ✷ Case 3. a < 0 and b > 0.
Hence K 3 (x 0 , t 0 ) > 0 follows from This inequality provides us the desired estimate as before. The proof for a < 0 and b > 0 is now completed. ✷ Case 4. a = 0 and b > 0.
Assume that an estimate holds, where a n is the one in (5.16). In this case M n is defined by (5.6) with C = C 4 := C 0 (p − 1) 2 > 0 and η = µ = p. Making use of (5.8) and (5.21), we have .
For the α-integral, we have that t+x β log pan 1 + α 1 + β which gives us Therefore, we have and S p is the one in (5.7). As in Case 3, let us fix (x 0 , t 0 ) ∈ D such that t 0 − x 0 = R + 1 and t 0 > (R + 2) 2 .
Then we have Hence, K 4 (x 0 , t 0 ) > 0 follows from This inequality provides us with the desired estimate as before. The proof for a < 0 and b > 0 is now completed. ✷ Case 5. a = 0 and b = 0.
Then we have which is the desired estimate as before. The proof for a = 0 and b = 0 is now completed. ✷ Case 6. a > 0 and a + b = 0.
In this case, we employ so-called "slicing method" which was introduced by Agemi, Kurokawa and Takamura [1]. Let us set We note that D n+1 ⊂ D n for all n ∈ N. Assume that an estimate u(x, t) ≥ M n log 1 + t − x 1 + l n R an in D n (5.24) holds, where a n is the one in (5.16). The sequence {M n } with M n > 0 is defined later. Then (5.8) and (5.24) imply that Let (x, t) ∈ D n+1 . Then, we get Note that l n R < λ n (α) < α for all n ∈ N holds, where λ n (α) := (1 + α)(1 + l n R) 1 + l n+1 R − 1.
Then we have that Due to 1 < l n < 2 for all n ∈ N and R > 1, the β-integral is estimated as follows: Therefore, we can employ the same argument as in Case 1 with a n in (5.16) and M n defined by M n+1 = C 6 2 −n p −n M p n , M 1 = C g ε, where C 6 := C 0 (p − 1)/2 · 3 1−b > 0. Here, we assume t − x > 2R = lim n→∞ l n R. Making use of Lemma 5.1 with C = C 6 , η = 2 and µ = p, we obtain and S p is the one in (5.7). As in Case 1, let us fix (x 0 , t 0 ) ∈ D ∞ such that t 0 = 2x 0 and t 0 > 4(1 + 2R) 2 .
Hence K 6 (x 0 , t 0 ) > 0 follows from which is the desired estimate as before. The proof for a > 0 and a + b = 0 is now completed. Therefore the proof of Theorem 1.4 finishes. ✷

Proof of Theorem 1.5
The proof is almost similar to the one of Theorem 1.4. Let u = u(x, t) ∈ C 2 (R × [0, T )) be a solution of (1.3). Since the assumption on the initial data in (1.20) yields It follows from (5.2), (5.3), (5.5) and for (x, t) ∈ D, where D and C 0 are defined in (5.1) and (5.4) respectively, and (5.28) For (5.28), we may assume that without loss of generality. Because, if not, we have to assume f (z) ≡ 0 in z ∈ (0, R) and to change the definition of D in which x > 0 is replaced with x < 0. For such a case, we obtain all the estimates below with −x instead of x. In fact, taking f (x + t) instead of f (x − t) in (5.25), we have, instead of (5.9), that Case 1. a > 0 and a + b < 0.
Since J ′ in (5.28) is estimated from below by it follows from (5.27) and (5.29) that Therefore, we can employ the same argument as Case 1 of Theorem 1.4 in which the constant C g ε in (5.9) is simply replaced with C f,1 ε p . ✷ Case 2. a = 0 and b < 0.
Since J ′ in (5.28) is estimated from below by it follows from (5.26) and (5.29) that We employ the "slicing method" again. Assume that an estimate holds, where a n ≥ 0, b n > 0 and M n > 0. Here, D n and l n are defined in (5.23 Let (x, t) ∈ D n+1 . Note that l n R < λ n < t − x for all n ∈ N holds, where Then we have It follows from Hence, (5.33) holds for all n ∈ N provided a n+1 = pa n + 1, a 1 = 0 b n+1 = pb n , b 1 = 1 and M n+1 ≤ C 7 2 −2n M p n , M 1 = C f,2 ε p , where C 7 := C 0 /72 > 0. Therefore, we define a n is the one in (5.16), b n = p n−1 and M n as above. Making use of Lemma 5.1 with C = C 7 and η = µ = 2, for (x, t) ∈ D ∞ , we obtain and S p is the one in (5.7). Let us fix (x 0 , t 0 ) ∈ D ∞ such that Then, we obtain , which is the desired estimate as before. This completes the proof for a = 0 and b < 0. ✷ Case 3. a < 0 and b < 0.
Then we obtain which is the desired estimate as before. This completes the proof for a < 0 and b < 0. ✷ Case 4. a < 0 and b = 0. Hence we define a n by the one in (5.16) and set b n = a n+1 . Moreover, M n is defined by M n+1 = C 9 p −2n M p n , M 1 = C f,4 ε p , where C 9 := C 0 (p − 1) 2 /(1 − a)p 2 > 0. Therefore, making use of Lemma 5.1 with C = C 9 and η = µ = p, we obtain, by (5.39), that Let us fix (x 0 , t 0 ) ∈ D such that t 0 = 2x 0 and t 0 ≥ 16R 2 .
Therefore, we obtain the desired estimate as before. This completes the proof for a < 0 and b > 0. ✷ Case 6. a = 0 and b > 0.