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2 On Second Order Hyperbolic Equations with Coefficients Degenerating at Infinity and the Loss of Derivatives and Decays Tamotu Kinoshita Institute of Mathematics, University of Tsukuba Abstract In this paper, we study well-posedness issues in the weighted L space for the Cauchy problem for the wave equation of the form t u a(t, x xu = 0. We give the condition a(t, x > 0 for all (t, x [0, T ] R x which is between the strictly hyperbolic condition and weakly hyperbolic one, and allows the decaying coefficient a(t, x such that lim x a(t, x = 0 for all t [0, T ]. Our concerns are the loss of derivatives and decays of the solutions. Keywords Hyperbolic equations, Decaying coefficients, Weighted L space. AMS Mathematics Subject Classification(010 35B30, 35L15. 1 Introduction We consider the Cauchy problem on [0, T ] R x { t u a(t, x xu = 0, u(0, x = u 0 (x, t u(0, x = u 1 (x, (1.1 where the real coefficient a(t, x satisfies a(t, x > 0 for all (t, x [0, T ] R x. (1. A standard hyperbolic condition is usually given as follows: a(t, x c T for all (t, x [0, T ] R x. (1.3 1

3 For the cases c T > 0 and c T = 0, we call strictly hyperbolic and weakly hyperbolic respectively. So, the condition (1. is between the strictly hyperbolic condition and weakly hyperbolic one, and allows the decaying coefficient a(t, x such that lim x a(t, x = 0 for all t [0, T ] like an integrable function on R x. Then we shall assume that a(t, x C 1( [0, T ] C (R x. (1.4 In the strictly hyperbolic case, more relaxed assumptions can be considered (see [], [7]. But, the assumption (1.4 is not enough to show the wellposedness in the weakly hyperbolic case. Therefore, we also assume that t a(t, x M(ta(t, x with M(t L 1 (0, T for (t, x [0, T ] R x. (1.5 Remark 1.1 The assumption (1.5 under the (standard weakly hyperbolic condition (1.3 with c T = 0 would be strong (see [10], but the assumption (1.5 (under the condition (1. is not so restrictive. Indeed, when the coefficient has the form a(t, x = a 1 (ta (x (including the case independent of t, thanks to (1. there exists a constant M( M(t such that t a(t, x = t a 1 (t a (x Ca (x Ma 1 (ta (x = Ma(t, x, here we used the fact that a 1 (t must satisfies a 1 (t c T > 0 on the compact set [0, T ] (since a 1 (t 0 = 0 for some t 0 [0, T ] contradicts (1.. The assumption (1.5 plays a similar role as the regularity assumption C k under the (standard weakly hyperbolic condition (1.3 with c T = 0 (see [5], [6], [8], [9], [1], etc.. Actually, defining the energy E(t := t u L (R x + a(t, x 1/ x u + L (R x δu L (R x, (1.6 a standard argument gives the energy inequality E(t E(0 which can be translated into the following estimate of the solution: t u L (R x + δu L (R x E(t E(0 u 1 L (R x + (A + δu 0 H 1 (R x, where A := sup (t,x [0,T ] Rx a(t, x. The usual loss of derivatives means the decreasing weight functions in the Fourier space (see [11], etc.. Though this estimate does not include a weight function, it can be regarded as the second type of loss of derivatives, since u 0 H 1 (R x in the righthand side corresponds to not u H 1 (R x but u L (R x in the lefthand side. In this paper we shall propose a new kind of the energy to avoid such a loss of derivatives.

4 Remark 1. If we suppose that the initial data have (fixed compact supports, from the finite propagation property, we may use just the energy E(t with δ = 0 under the strictly hyperbolic condition (1.3 with c T > 0. Then, usual loss of derivatives and the second type of loss of derivatives never occur for the strictly hyperbolic case, since there exists c T > 0 such that t u L (R x + c T u H 1 (R x E(t E(0 u 1 L (R x + Au 0 H 1 (R x. We have to investigate how the supports of the initial data can be enlarged. So, our strategy is to consider the weight functions which control the solution degenerating at infinity. Now, let us consider the following example: Example Let χ R C (R be a cut off function such that 0 χ R (x 1 and χ R (x = 0 for x 0, = 1 for x R(> 0 and the coefficient independent of t a R (t, x = { 1 χ R ( x } 1 + χ R ( x (x 1 1. Put T = 1, u 0 (x = u 1 (x = e x / and a(t, x = a 1 (t, x (with R = 1 satisfying (1., (1.4, (1.5, that is, the Cauchy problem on [0, 1] R x { t u a 1 (t, x xu = 0, u(0, x = e x /, t u(0, x = e x /. (1.7 Then, we find the following (i and (ii: (i The coefficient a 1 (t, x satisfies A 1 := sup (t,x [0,1] Rx a 1 (t, x 1. More precisely, a 1 (t, x has a polynomial decay such as c x a 1 (t, x C x ( 1 for all (t, x [0, 1] R x. (ii For the initial data u 0, u 1 having a super-exponential decay, there exists a unique solution u having a super-exponential decay such that the loss of derivatives does not occur. (i is trivial. In order to show (ii, we prepare the auxiliary Cauchy problems (CP 1, (CP and (CP 3. (CP 1: u (1 (t, x and u (1 0 (x = u (1 (CP : u ( (t, x and u ( 0 (x = u ( (CP 3: u (3 (t, x and u (3 0 (x = u (3 1 (x = {1 χ( x }e x /, 1 (x = χ( x e x /, 1 (x = χ( x {1 χ( x 5}e x /. 3

5 We immediately see that superposition principle gives the solution of (1.7 u(t, x u (1 (t, x + u ( (t, x. By Remark 1. we obtain the solutions u (1 (t, x and u (3 (t, x of (CP 1 and (CP 3 such that the loss of derivatives does not occur. So, our concern is the loss of derivatives of u ( (t, x. By uniqueness and finite propagation speed, we know that u ( e t x / for t [0, 1] and x + R + A (t, x 1, u (3 (t, x for t [0, 1] and x 5 A 1, noting that u ( 0 (x u (3 0 (x for all x [ 5, 5]. Since R = 1 and A 1 1, u ( (t, x is determined for all (t, x [0, 1] R x. Thus, we find that u ( (t, x has a super-exponential decay such that the loss of derivatives does not occur. This also means (ii for u(t, x. From the above example, the following questions arise when a(t, x has a polynomial decay and the initial data u 0, u 1 have a (sub- or super-exponential decay: Question 1: Would not the (second type of loss of derivatives for the solution of the Cauchy problem (1.1 under (1. occur in general? Question : Can we expect that the solution keeps to have a (sub- or super- exponential decay same as the initial data? We shall give answers for these questions in the next section. Energy Estimate We may suppose that A := sup (t,x [0,T ] Rx a(t, x 1 by considering u(t, x = U(t, x/ A without loss of generality. Now we define the energy E(t := e ρ(ta(t,x γ t u + a(t, x 1/ e ρ(ta(t,x γ x u + δ e ρ(ta(t,x γ u L L L, (.1 where γ > 0 and ρ(t is a differentiable function such that ρ (t < 0. Remark.1 Under our condition (1., we remark that a(t, x 1/ e ρ(ta(t,x γ c ε e (ρ(t εa(t,x γ. Then, we get a(t, x 1/ e ρ(ta(t,x γ x u c L εe (ρ(t εa(t,x γ x u L, (. 4

6 even when lim x a(t, x = 0. Consequently, this enables us to avoid a second type of loss of derivatives. In the below proof our task is to check carefully whether the use of the weight e ρ(ta(t,x γ is suitable for the energy computations. By differentiating E(t in t, we have t {E(t } = t {( e ρa γ u t, e ρa γ u t + ( ae ρa γ u x, e ρa γ u x + δ ( e ρa γ u, e ρa γ u } ρ a γ/ e ρa γ u t + γ ρ at 1/ a (γ+1/ e ρa γ u t +R ( e ρa γ u tt, e ρa γ u t +ρ a (1 γ/ e ρa γ at u x + γ ρ 1/ a γ/ e ρa γ u x + a t 1/ e ρa γ ( u x + R ae ρa γ u tx, e ρa γ u x +δρ a γ/ e ρa γ u + δγ ρ a t 1/ a (γ+1/ e ρa γ u +δr ( e ρa γ u t, e ρa γ u = ρ a γ/ e ρa γ u t + γ ρ at 1/ a (γ+1/ e ρa γ u t +R ( e ρa γ au xx, e ρa γ u t +ρ a (1 γ/ e ρa γ at u x + γ ρ 1/ a γ/ e ρa γ u x + a t 1/ e ρa γ ( u x R ut, x ae ρa γ u x +δρ a γ/ e ρa γ u + δγ ρ a t 1/ a (γ+1/ e ρa γ u +δr ( e ρa γ u t, e ρa γ u = ρ a γ/ e ρa γ u t + ρ a (1 γ/ e ρa γ u x + δρ a γ/ e ρa γ u at +γ ρ 1/ a (γ+1/ e ρa γ at u t + δγ ρ 1/ a (γ+1/ e ρa γ u +γ ρ a t 1/ a γ/ e ρa γ at u x + 1/ e ρa γ u x R ( ( ( u t, a x e ρa γ u x + 4γρR ut, a γ a x e ρa γ u x + δr e ρa γ u t, e ρa γ u =: I + II + III + IV + V + V I + V II + V III + IX + X. By (1.5 we also get IV + V Mγ ρ a γ/ e ρa γ a u t + Mδγ ρ γ/ e ρa γ u, 5

7 V I Mγ ρ a (1 γ/ e ρa γ u x, V II M a 1/ e ρa γ u x. By (1.4 from the Glaeser inequality with respect to x, i.e., x a(t, x Ka(t, x 1/ for (t, x R, we get V III = R ( K 1/ a γ/ e ρa γ u t, K 1/ a γ/ a x e ρa γ u x K a γ/ e ρa γ u t + K 1 a γ/ a x e ρa γ u x Ka e ρa γ a u t + K (1+γ/ e ρa γ u x, IX = 4γρR ( K 1/ a γ/ e ρa γ u t, K 1/ a γ/ a x e ρa γ u x Kγ ρ a γ/ e ρa γ u t + K 1 γ ρ a γ/ a x e ρa γ u x Kγ ρ a γ/ e ρa γ a u t + Kγ ρ (1 γ/ e ρa γ u x, X = δr ( e ρa γ u t, e ρa γ u δe e u t + δ ρa γ u. Summing up these terms and taking δ = 1, we have t {E(t } { ρ + γ(m + K ρ } { a γ/ e ρa γ a u t + (1 γ/ e ρa γ a u x + δ γ/ e ρa γ u } +(M + K { a 1/ e ρa γ a u x + (1+γ/ e ρa γ u } x +δ { e ρa γ e u t + ρa γ u } { ρ + γ(m + K ρ + 1 (M + K + 1} { a γ/ e ρa γ u t + a (1 γ/ e ρa γ u x + a γ/ e ρa γ u }, (.3 here we used that a 1/ a (1 γ/, a (1+γ/ a (1 γ/ and a γ/ 1 for γ > 0, thanks to 0 < a 1. Now we shall solve the ordinary differential equation ρ + γ(m + K ρ + 1 (M + K + 1 = 0. (.4 When ρ 0 > 0, we shall put the one point t = t 0 > 0 such that ρ 0 = 1 t0 0 e γ s 0 (M+Kdτ (M + K + 1ds 6

8 and choose ρ(t = e γ t { (M+Kdτ 0 ρ 0 1 s t 0 e γ (M+Kdτ 0 (M + K + 1ds } if ρ 0 0 and 0 t T, e γ t 0 (M+Kdτ { ρ 0 1 t 0 eγ s 0 (M+Kdτ (M + K + 1ds } if ρ 0 > 0 and 0 t t 0, e γ t (M+Kdτ { t 0 ρ(t 0 1 t t 0 e γ s } (M+Kdτ t 0 (M + K + 1ds if ρ 0 > 0 and t 0 t T, (.5 where we remark that ρ(t 0 = 0. Then, we see that E(t E(0. Hence, it follows that for γ > 0 e ρ(ta(t,x γ t u + a(t, x 1/ e ρ(ta(t,x γ x u + e ρ(ta(t,x γ u L L L e ρ 0a(0,x γ a(0, u 1 + x 1/ e ρ 0a(0,x γ e x u 0 + ρ 0 a(0,x γ u 0 L L. (.6 Thus, noting (., we can avoid the (second type of loss of derivatives. This is an answer of the Question 1 in 1. In particular, when there exist c T > 0, C T > 0 and α 0 such that c T x α a(t, x C T x α for all (t, x [0, T ] R x, (.7 noting the signature of ρ(t and ρ 0, we have for γ > 0 e min{ρ(tc T, ρ(tc T } x αγ t u + c L T x α/ e min{ρ(tc T, ρ(tc T } x αγ x u + e min{ρ(tc T, ρ(tc T } x αγ u L e max{ρ 0c T, ρ 0 C T } x αγ u 1 + C L T x α/ e max{ρ 0c T, ρ 0 C T } x αγ x u 0 L + e max{ρ 0c T, ρ 0 C T } x αγ u 0. (.8 So, the solution u also has a (sub- or super-exponential decay. Remark. In order to derive (.8, we have not directly used the energy as E(t := e ρ(t x γ t u L + a(t, x 1/ e ρ(t x γ x u L + e ρ(t x γ u L for the energy computations. This would make the energy computations complicated in our case. After getting the desired energy inequality with the energy (.1, we have replaced a(t, x γ by the power of x in the exponent as the above. 7 L L L

9 Thus, we conclude the following theorem: Theorem.3 Let ρ be defined by (.5. Assume that a(t, x satisfies (1., (1.4 and (1.5. Then the Cauchy problem (1.1 has a unique solution u satisfying (.6. In particular, if a(t, x has a polynomial decay as (.7 and the initial data u 0, u 1 have (sub- or super-exponential decay, the Cauchy problem (1.1 has a unique solution u having a (sub- or super-exponential decay and satisfying (.8. Remark.4 It is not necessary that the decreasing function ρ(t is positive on [0, T ] for the given T > 0. If we take a sufficiently large ρ 0 > 0 such that ρ(t > 0 for t [0, T ], the (second type of loss of derivatives does not occur, since the weight functions in the nd terms in the left hand sides of (.6 and (.8 do not disappear. Remark.5 In general, weakly hyperbolic equations are concerned with the study of the Levi condition ([3], [5], etc. If we consider the equation with the lower order terms, i.e., t u a(t, x xu + b(t, x x u + d(t, x t u + c(t, xu = 0, we also need Levi conditions b(t, x M(ta(t, x 1/ with M(t L 1 (0, T for (t, x [0, T ] R x, (.9 d(t, x + c(t, x M(t L 1 (0, T for (t, x [0, T ] R x, (.10 where M(t is arranged to be a positive function (M(t 1/, M(t 1 exist. Indeed, we have the additional terms in the energy computations, which can be estimated as follows: R ( e ρa γ bu x, e ρa γ u t = R ( M 1/ a γ/ be ρa γ u x, M 1/ a γ/ e ρa γ u t R ( e ρa γ du t, e ρa γ u t M e ρa γ u t, M 1 a γ/ be ρa γ a u x + M γ/ e ρa γ u t M a (1+γ/ e ρa γ a u x + M γ/ e ρa γ u t, R ( e ρa γ cu, e ρa γ u t M e ρa γ u x + M e ρa γ u t. (.11 When c(t, x > 0, we may replace δe ρ(ta(t,x γ u L by c(t, x 1/ e ρ(ta(t,x γ u L in the energy (.1. Then the term (.11 is canceled in the energy computations. 8

10 The energy (.1 includes the parameter γ > 0. We can also consider the case corresponding to γ = 0 with the following energy instead of (.1: E(t ρ(t log a(t,x 1 := e ρ(t log a(t,x 1 t u a(t, x L + 1/ e ( a(t, = x ρ(t t u L + a(t, x 1/ ρ(t x u x u e L +δ ρ(t log a(t,x 1 u L + δ a(t, x ρ(t u L L. (.1 In the energy computations, we replace γ by 0 in the inside of the norms or the inner products (but replace γ by 1 in the outside of them. Similarly we have the estimate instead of (.3 t {E(t } { ρ + (M + K ρ + 1 (M + K + 1}{ (log a 1 1/ e ρ log a 1 u t + (a log a 1 1/ e ρ log a 1 u x + (log a 1 1/ e ρ log a 1 u }, (.13 and solve the ordinary differential equation ρ + (M + K ρ + 1 (M + K + 1 = 0. When ρ 0 > 0, we shall put the one point t = t 0 > 0 such that and choose ρ(t = ρ 0 = 1 t0 0 s e (M+Kdτ 0 (M + K + 1ds t e (M+Kds{ 0 ρ 0 1 s t 0 e (M+Kdτ 0 (M + K + 1ds } if ρ 0 0 and 0 t T, e t 0 (M+Kds{ 0 ρ 0 1 t 0 e s (M+Kdτ 0 (M + K + 1ds } if ρ 0 > 0 and 0 t t 0, t (M+Kds t e { 0 ρ(t 0 1 t t 0 e s } (M+Kdτ t 0 (M + K + 1ds if ρ 0 > 0 and t 0 t T, (.14 where we remark that ρ(t 0 = 0. Then, we see that t {E(t } 0 and E(t E(0. Hence, by it follows that a(t, x ρ(t t u L + a(t, x 1/ ρ(t x u L + a(t, x ρ(t u a(0, x ρ 0 u 1 L + a(0, x 1/ ρ 0 x u 0 9 L L + a(0, x ρ 0 u 0 L,(.15

11 and in particular under (.7 with c T > 0 and C T > 0 x αρ(t t u + c L T x α{ρ(t 1/} x u + x αρ(t u L L x αρ 0 u 1 + C L T x α{ρ0 1/} x u 0 + x αρ 0 u 0 L L. (.16 Thus, we get the following: Theorem.6 Let ρ be defined by (.14. Assume that a(t, x satisfies (1., (1.4 and (1.5. Then the Cauchy problem (1.1 has a unique solution u satisfying (.15. In particular, if a(t, x has a polynomial decay as (.7 and the initial data u 0, u 1 have polynomial decays, the Cauchy problem (1.1 has a unique solution u having a polynomial decay and satisfying (.16. A result when the initial data having less regularities can be given, is better in the study of the wellposedness of Cauchy problems. So, one will also think that Theorem.6 is a better result than Theorem.3 in sense that the initial data having less decays can be given. Similarly as the technical term finite loss of derivatives, we may call finite loss of decays for (.16. The energy inequality (.16 corresponding to γ = 0 with α = (and especially ρ 0 = 1/ becomes x ρ(t t u L + c T x ρ(t 1 x u L + x ρ(t u L x u 1 L + C T x u 0 L + x u 0 L. (.17 But, if we consider the Cauchy problem (1.7, the energy inequality (.17 does not show a positive answer for the Question in 1. On the other hand, the energy inequality (.8 with γ = 1 and α = in Theorem.3 just becomes e min{ρ(tc T, ρ(tc T } x t u + c L T x 1 e min{ρ(tc T, ρ(tc T } x x u L + e min{ρ(tc T, ρ(tc T } x u L e max{ρ 0 c T, ρ 0 C T } x u 1 + C L T x 1 e max{ρ 0c T, ρ 0 C T } x x u 0 L + e max{ρ 0 c T, ρ 0 C T } x u 0. (.18 (.18 means that the solution keeps to have a super-exponential decay same as the initial data (see the example in 1. This is an answer of the Question. 10 L

12 Remark.7 If one considers the standard energy (1.6, that is (.1 with ρ = 0, one would obtain the following instead of (.13: t {E(t } (M + K + 1 { u t + a 1/ u x + u } = (M + K + 1E(t, t and E(t e (M+K+1dτ 0 E(0 and t u +a 1/ x u +u t e { (M+K+1dτ 0 u 1 +a(0, x 1/ x u 0 +u 0 }. Under the condition (1. the left hand side should be changed into t u + u. So, the second type of loss of derivatives occurs for the standard energy (1.6. From this inequality with the standard energy, we do not know any information about the loss of decay. Remark.8 Thanks to the use of the loss of decays, we can avoid the second type of loss of derivatives for the decaying coefficient a(t, x, but can not expect the time decay of solutions. [13] and [14] developed the theory of Fourier integral operators which help us to construct the parametrix, and gave weighted estimates for hyperbolic equations, where the decay in x is translated into the time decay of solutions. In the above, we have proved the global results with respect to x by the global condition (1.. If we give compactly supported initial data, by the finite propagation property we may consider the problem locally with respect to x. Instead of (1., we shall assume (A a(t, x > 0 for all (t, x [0, T ] R x \{0}, (.19 which is regarded as a local condition around x = 0. Without changes of other conditions and the proof, we can also obtain the following: Corollary.9 Let ρ be defined by (.5 (resp. (.14. Assume that a(t, x satisfies (1.4, (1.5 and (.19. Then the Cauchy problem (1.1 has a unique solution u satisfying (.6 (resp. (.15. In particular, if there exist c T > 0, C T > 0 and α 0 such that c T x α a(t, x C T x α for all (t, x [0, T ] R x, (.0 11

13 the Cauchy problem (1.1 has a unique solution u satisfying e min{ρ(tc T, ρ(tc T } x αγ t u + c L T x α/ e min{ρ(tc T, ρ(tc T } x αγ x u L + e min{ρ(tc T, ρ(tc T } x αγ u L e max{ρ 0 c T, ρ 0 C T } x αγ u 1 + C L T x α/ e max{ρ 0c T, ρ 0 C T } x αγ x u 0 L ( resp. x αρ(t t u x αρ 0 u 1 + e max{ρ 0 c T, ρ 0 C T } x αγ u 0 L + c L T x α{ρ(t 1/} x u x + αρ(t u L L + C L T x α{ρ0 1/} x x u 0 + αρ 0 u 0. L L Remark.10 In order that a(t, x belongs to C with respect to x as (1.4, actually α should satisfy α. The last energy inequality corresponding to γ = 0 with α = (and especially ρ 0 = 1/ becomes x ρ(t t u L + c T x ρ(t+1 x u L + x ρ(t u L x 1 u 1 L + C T x u 0 L + x 1 u 0 L. If the initial data u 0 and u 1 have degeneration of polynomial order at x = 0, the right hand side is finite. So, we find that the solution u must degenerate at x = 0 for the degenerating coefficient a(t, x (propagation speed 0 at x = 0. Remark.11 The case when the coefficient is just a(t, x x, satisfies (.0 with c T = C T = 1 and α =, but not (.19 with A <. If we give compactly supported initial data, the finite propagation property allows the coefficient a(t, x x with some modifications to satisfy (.19. Indeed, [1] shows the following representation formula of the exact solution u with a(t, x x and u 0 0: 1 u(t, x = t 1 u 1 (xe yt e yt ( t 1 y J 0 dy, where J 0 (x is the Bessel function of order 0, i.e., J 0 (x = n=0 ( x 4 n /(n!. Hence, we really find that the solution u degenerates at x = 0 if u 1 (0 = 0. 1

14 Acknowledgments This work was supported by Grant-in-Aid for Scientific Research (C (No. 16K053, Japan Society for the Promotion of Science. The author warmly thanks the referee for her/his comments: they have been very useful to improve the quality of the paper. References [1] A. Galstian and T. Kinoshita, Representation of solutions for nd order one-dimensional model hyperbolic equations, to appear in Journal D Analyse Mathématique. [] M. Cicognani, Strictly hyperbolic equations with non regular coefficients with respect to time, Ann. Univ. Ferrara Sez. VII (N.S. 45, (000. [3] C. Garetto and R. Ruzhansky, Weakly hyperbolic equations with nonanalytic coefficients and lower order terms, Math. Ann. 357, (013. [4] F. Colombini, E. De Giorgi and S. Spagnolo, Sur les équations hyperboliques avec des coefficients qui ne dépendent que du temps, Ann. Scuola Norm Sup. Pisa 6, ( [5] F. Colombini, E. Jannelli and S. Spagnolo, Wellposedness in the Gevrey classes of the Cauchy problem for a non strictly hyperbolic equation with coefficients depending on time, Ann. Scuola Norm Sup. Pisa 10, (1983. [6] P. D Ancona, Gevrey well posedness of an abstract Cauchy problem of weakly hyperbolic type, Publ. RIMS Kyoto Univ. 4, (1988. [7] D. Del Santo, T. Kinoshita and M. Reissig, Energy estimates for strictly hyperbolic equations with low regularity in coefficients, Differential Integral Equations 0, (007. [8] N. Fukuda and T. Kinoshita, On a coefficient concerning an ill-posed Cauchy problem and the singularity detection with the wavelet transform, Rend. Istit. Mat. Univ. Trieste 45, (

15 [9] E. Jannelli, Gevrey well-posedness for a class of weakly hyperbolic equations, J. Math. Kyoto. Univ. 4, (1984. [10] T. Kinoshita, On second order weakly hyperbolic equations with oscillating coefficients, Differential Integral Equations 8, (015. [11] T. Kinoshita and M. Reissig, About the loss of derivatives for strictly hyperbolic equations with non-lipschitz coefficients, Adv. Differential Equations 10, 191 (005. [1] T. Nishitani, Sur les équations hyperboliques à coefficients hölderiens en t et de classes de Gevrey en x, Bull. Sci. Math. 107, (1983. [13] M. Ruzhansky and M. Sugimoto, Global calculus of Fourier integral operators, weighted estimates, and applications to global analysis of hyperbolic equations. Pseudo-differential operators and related topics, Oper. Theory Adv. Appl. 164, Birkhäluser, Basel, 006. [14] M. Ruzhansky and M. Sugimoto, Weighted Sobolev L estimates for a class of Fourier integral operators, Math. Nachr. 84, (011. Addresses of the authors Tamotu Kinoshita, Institute of Mathematics, Tsukuba University, Tsukuba Ibaraki , Japan kinosita@math.tsukuba.ac.jp 14