Strichartz estimates for wave equation with inverse square potential

arXiv:1312.1745v1 [math.AP] 6 Dec 2013

STRICHARTZ ESTIMATES FOR WAVE EQUATION WITH INVERSE SQUARE POTENTIAL CHANGXING MIAO, JUNYONG ZHANG, AND JIQIANG ZHENG

Abstract. In this paper, we study the Strichartz-type estimates of the solution for the linear wave equation with inverse square potential. Assuming the initial data possesses additional angular regularity, especially the radial initial data, the range of admissible pairs is improved. As an application, we show the global well-posedness of the semi-linear wave equation with inverse-square a p−1 u for power p being in some regime potential ∂t2 u − ∆u + |x| 2 u = ±|u| when the initial data are radial. This result extends the well-posedness result in Planchon, Stalker, and Tahvildar-Zadeh.

Contents 1. Introduction and Statement of Main Result 2. Preliminary 2.1. Spherical harmonic expansions and the Bessel functions 2.2. Spectrum of −∆ + |x|a 2 and Hankel transform 3. Estimates of Hankel transforms 4. Proof of Theorem 1.1 5. Proof of Theorem 1.2 6. Appendix: The Proof of Lemma 2.3 References

1 5 5 9 12 14 20 21 22

Mathematics Subject Classification (2000): 35Q40, 35Q55, 47J35. Keywords: Inverse square potential, Strichartz estimate, Spherical harmonics. 1. Introduction and Statement of Main Result The aim of this paper is to study the Lqt (Lrx )-type estimates of the solution for the linear wave equation perturbed by an inverse square potential. More precisely, we shall consider the following wave equation with the inverse square potential ( (t, x) ∈ R × Rn , a ∈ R, ∂t2 u − ∆u + |x|a 2 u = 0, (1.1) u(t, x)|t=0 = u0 (x), ∂t u(t, x)|t=0 = u1 (x). The scale-covariance elliptic operator −∆ + |x|a 2 appearing in (1.1) plays a key role in many problems of physics and geometry. The heat and Schrödinger flows for the elliptic operator −∆ + |x|a 2 have been studied in the theory of combustion [11], and in quantum mechanics [8]. The equation (1.1) arises in the study of the 1

2

CHANGXING MIAO, JUNYONG ZHANG, AND JIQIANG ZHENG

wave propagation on conic manifolds [4]. We refer the readers to [1, 2, 14, 15] and references therein. It is well known that Strichartz-type estimates are crucial in handling local and global well-posedness problems of nonlinear dispersive equations. Along this way, Planchon, Stalker, and Tahvildar-Zadeh [14] first showed a generalized Strichartz estimates for the equation (1.1) with radial initial data. Thereafter, Burq, Planchon, Stalker, and Tahvildar-Zadeh [1] removed the radially symmetric assumption in [14] and then obtained some well-posedness results for the semi-linear wave equation with inverse-square potential. The range of the admissible exponents (q, r) for the Strichartz estimates of (1.1) obtained in [1, 14] is restricted under 2q ≤ (n−1)( 12 − 1r ), which is the same as that of the linear wave equation without potential. Sterbenz and Rodnianski [22] improved the range of the “classical” admissible exponents (q, r) for the linear wave equation with no potential by compensating a small loss of angular regularity. In this paper, we are devoted to study the Strichartz estimates of the solution of the equation (1.1). By employing the asymptotic behavior of the Bessel function and some fine estimates of Hankel transform, we improve the range of the admissible pairs (q, r) in [1, 14] by compensating a small loss of angular regularity. The machinery we employ here is mainly based on the spherical harmonics expansion and some properties of Hankel transform. As an application of the Strichartz estimates, we obtain well-posedness of (1.1) perturbed by nonlinearity |u|p−1 u with power ph < p < pconf (defined below) in the radial case, which extends the wellposedness result in Planchon et al. [14]. Before stating our main theorems, we need some notations. We say the pair (q, r) ∈ Λ, if q, r ≥ 2, and satisfy 1 n−1 1 1 ≥ ( − ) and q 2 2 r

1 1 1 < (n − 1)( − ). q 2 r

Set the infinitesimal generators of the rotations on Euclidean space: Ωj,k := xj ∂k − xk ∂j ,

and define for s ∈ R, ∆θ :=

X j (n−1) 2 − n−2 q n−2 2 For any ǫ > 0 and 0 < s < 1 + min 2 , ( 2 ) + a , • if n ≥ 4, then (1.2) ku(t, x)kLqt Lrx ≤ Cǫ khΩis¯u0 kH˙ s + khΩis¯u1 kH˙ s−1 ,

where (q, r) ∈ Λ, and

2 1 1 s¯ = (1 + ǫ) − (n − 1) − q 2 r

• if n = 3, then (1.3)

and

s=n

1

2

−

1 1 − ; r q

ku(t, x)kLqt Lrx ≤ Cǫ khΩis¯u0 kH˙ s + khΩis¯u1 kH˙ s−1 ,

(n−2)2 . 4

STRICHARTZ ESTIMATES FOR WAVE EQUATION WITH INVERSE SQUARE POTENTIAL 3

where q 6= 2, (q, r) ∈ Λ, and 1 1 1 − − s¯ = (2 + ǫ) q 2 r

and

s=3

1

2

−

1 1 − . r q

In addition, the following estimate holds for r > 4 and s = 3( 12 − r1 ) − 12 , (1.4) ku(t, x)kL2t Lrx ≤ Cǫ khΩis¯(r) u0 kH˙ s + khΩis¯(r) u1 kH˙ s−1 , where s¯(r) = 1 −

2 r

with r 6= ∞.

Remark 1.1. i). We remark that some of admissible pairs (q, r) in Theorem 1.1 are out of the region ACDO or ACO (in the following figures) obtained in [1, 14]. (n−2)2 1 ii). Our restriction a > an := (n−1) is to extend the the range of (q, r) 2 − 4 as widely as possible. We remark that a3 = 0 and an < 0 for n ≥ 4. Therefore, we recover the result of Theorem 1.5 in Sterbenz [22], which considers a = 0 and n ≥ 4. iii). In the extended region Λ (see the below figures), the loss of angular regularity is s¯ = (1 + ǫ)( q2 − (n − 1)( 21 − r1 )). When n = 3, the loss of angular regularity in the line BC is s¯(r) > s¯, since the Strichartz estimate fails at the endpoint (q, r, n) = (2, ∞, 3). It seems that the methods we use here are not available to obtain such estimate at endpoint since Lemma 2.3 and Lemma 2.4 fail at r = ∞. And one might need the wave packet method of Wolff [25] and the argument in Tao [23] to obtain the Strichartz estimate at the endpoint (q, r, n) = (2, ∞, 3) with some loss of angular regularity.

1 r

1 r

A

1 2

1 2

A

B

n−2 2(n−1)

Λ

Λ

1 6

C

n−3 2(n−1)

B

1 4

C

D O

1 2

n>3

1 q

O

n=3

1 2

As a consequence of Theorem 1.1 and Corollary 3.9 in [14], we have the following Strichartz estimates for radial initial data: Corollary 1.1. Let n ≥ 3 and s < n2 . Suppose (u0 , u1 ) are radial functions, then for q, r ≥ 2, q1 < (n − 1)( 21 − 1r ) and s = n( 21 − 1r ) − q1 , the solution u of the equation

(1.1) with a > (1.5)

1 (n−1)2

−

(n−2)2 4

satisfies ku(t, x)kLqt Lrx ≤ C ku0 kH˙ s + ku1 kH˙ s−1 .

1 q

4


As an application, we obtain some well-posedness result of the following semilinear wave equation, ( a p−1 u, (t, x) ∈ R × Rn , a ∈ R, ∂t2 u − ∆u + |x| 2 u = ±|u| (1.6) u(t, x)|t=0 = u0 (x), ∂t u(t, x)|t=0 = u1 (x). In the case of the semi-linear wave equation without potential (i.e. a = 0), there are many exciting results on the global existence and blow-up. We refer the readers to [9, 16] and references therein. While for the equation (1.6) with 4 and n ≥ 3, Planchon et al. [14] established the global existence p ≥ pconf := 1+ n−1 2 . when the radial initial data is small in H˙ sc × H˙ sc −1 -norm with sc := n2 − p−1 Thereafter, Burq et al. [1] removed the radially symmetric assumption on the initial data. As a consequence of Theorem 1.1, we prove the global existence of the 4n solution to the equation (1.6) with ph := 1 + (n+1)(n−1) < p < pconf for small radial sc −1 sc ˙ ˙ . initial data (u0 , u1 ) ∈ H × H Theorem 1.2. Let n ≥ 3 and ph < p < pconf . Let q0 = (p − 1)(n + 1)/2, r0 = (n + 1)(p − 1)/(2p), and n n n o (n − 2)2 n n 1 − −n+2 , −n −2 . , (1.7) a > max 2 (n − 1) 4 q0 q0 r0 r0

Assume (u0 , u1 ) are radial functions and there is a small constant ǫ(p) such that (1.8)

ku0 kH˙ sc + ku1 kH˙ sc −1 < ǫ(p),

then there exists a unique global solution u to (1.6) satisfying (1.9)

0 (R × Rn ). u ∈ Ct (R; H˙ sc ) ∩ Lqt,x

Remark 1.2. i). The above result extends the well-posedness result in [14] from p ≥ pconf to ph < p < pconf . 0 ii). We remark that the L1 -bound of the operator Kλ,ν defined below is the source of our constraint to p > ph . Inspired by the arguments in Lindblad-Sogge[10, 16] for the usual semi-linear wave equation, if we want to extend the above result to p > pc , one needs to explore new inhomogeneous Strichartz estimates since the 0 operator Kλ,ν is not known as a bounded operator on L1 . Here pc is the positive root of (n − 1)p2c − (n + 1)pc − 2 = 0, and pc is called the Strauss’s index. This paper is organized as follows: In the section 2, we revisit the property of the Bessel functions, harmonic projection operator, and the Hankel transform associated with −∆ + |x|a 2 . Section 3 is devoted to establishing some estimates of the Hankel transform. In Section 4, we use the previous estimates to prove Theorem 1.1. We show Theorem 1.2 in Section 5. In the appendix, we sketch the proof of Lemma 2.3 by using a weak-type (1, 1) estimate for the multiplier operators with respect to the Hankel transform. Finally, we conclude this section by giving some notations which will be used throughout this paper. We use A . B to denote the statement that A ≤ CB for some large constant C which may vary from line to line and depend on various parameters, and similarly use A ≪ B to denote the statement A ≤ C −1 B. We employ A ∼ B to denote the statement that A . B . A. If the constant C depends on a special parameter other than the above, we shall denote it explicitly


by subscripts. We briefly write A + ǫ as A+ for 0 < ǫ ≪ 1. Throughout this paper, pairs of conjugate indices are written as p, p′ , where p1 + p1′ = 1 with 1 ≤ p ≤ ∞. 2. Preliminary In this section, we provide some standard facts about the Hankel transform and the Bessel functions. We use the oscillatory integral argument to show the asymptotic behavior of the derivative of the Bessel function. The Littlewood-Paley theorems associated to the Hankel transform are collected in this section. Finally we prove a Stirchartz estimate for unit frequency by making use of some results in [1]. 2.1. Spherical harmonic expansions and the Bessel functions. We begin with the spherical harmonics expansion formula. For more details, we refer to Stein-Weiss [20]. Let (2.1)

ξ = ρω

and x = rθ

with

ω, θ ∈ Sn−1 .

For any g ∈ L2 (Rn ), we have the expansion formula g(x) =

∞ d(k) X X

ak,ℓ (r)Yk,ℓ (θ)

k=0 ℓ=1

where

{Yk,1 , . . . , Yk,d(k) }

is the orthogonal basis of the spherical harmonic space of degree k on Sn−1 , called Hk , with the dimension 2k + n − 2 k−1 Cn+k−3 ≃ hkin−2 . k We remark that for n = 2, the dimension of Hk is a constant independent of k. We have the orthogonal decomposition d(0) = 1 and d(k) =

L2 (Sn−1 ) =

∞ M k=0

This gives by orthogonality

Hk .

kg(x)kL2θ = kak,ℓ (r)kℓ2k,ℓ .

(2.2)

By Theorem 3.10 in [20], we have the Hankel transforms formula Z ∞ ∞ d(k) X X n k − n−2 2 Jk+ n−2 (2πrρ)ak,ℓ (r)r 2 dr, (2.3) gˆ(ρω) = 2πi Yk,ℓ (ω)ρ 0

k=0 ℓ=1

2

here the Bessel function Jk (r) of order k is defined by Z 1 (r/2)k 1 Jk (r) = eisr (1 − s2 )(2k−1)/2 ds with k > − and r > 0. 2 Γ(k + 12 )Γ(1/2) −1 A simple computation gives the estimates (2.4)

|Jk (r)| ≤

2k Γ(k

1 Crk , 1+ 1 k + 1/2 + 2 )Γ(1/2)

6


and |Jk′ (r)| ≤

(2.5)

1 C(krk−1 + rk ) , 1+ 1 k + 1/2 + 2 )Γ(1/2)

2k Γ(k

where C is a constant and these estimates will be used when r . 1. Another well known asymptotic expansion about the Bessel function is r kπ π 2 −1/2 Jk (r) = r cos(r − − ) + Ok (r−3/2 ), as r → ∞, π 2 4 but with a constant depending on k (see [20]). As pointed out in [18], if one seeks 1 a uniform bound for large r and k, then the best one can do is |Jk (r)| ≤ Cr− 3 . To investigate the behavior of asymptotic on k and r, we are devoted to Schläfli’s integral representation [24] of the Bessel function: for r ∈ R+ and k > − 12 , Z π Z 1 sin(kπ) ∞ −(r sinh s+ks) Jk (r) = eir sin θ−ikθ dθ − e ds 2π −π π (2.6) 0 := J˜k (r) − Ek (r). We remark that Ek (r) = 0 for k ∈ Z+ . One easily estimates for r > 0 sin(kπ) Z ∞ e−(r sinh s+ks) ds ≤ C(r + k)−1 . (2.7) |Ek (r)| = π 0

Next, we recall the properties of Bessel function Jk (r) in [17, 18], the readers can also refer to [12] for the detailed proof. Lemma 2.1 (Asymptotics of the Bessel function). Assume k ≫ 1. Let Jk (r) be the Bessel function of order k defined as above. Then there exist a large constant C and small constant c independent of k and r such that: • when r ≤ k2 |Jk (r)| ≤ Ce−c(k+r) ;

(2.8) • when

k 2

≤ r ≤ 2k

(2.9) • when r ≥ 2k

1

1

Jk (r) = r− 2

(2.10)

1

1

|Jk (r)| ≤ Ck − 3 (k − 3 |r − k| + 1)− 4 ; X

a± (r)e±ir + E(r),

±

where |a± (r)| ≤ C and |E(r)| ≤ Cr

−1

.

For our purpose, we additionally need the asymptotic behavior of the derivative of the Bessel function Jk′ (r). It is a straightforward elaboration of the argument of proving Lemma 2.1 in [12], but we give the proof for completeness. Lemma 2.2. Assume r, k ≫ 1. Then there exists a constant C independent of k and r such that 1 |Jk′ (r)| ≤ Cr− 2 . k 2

or r ≥ 2k, we apply the recurrence formula [24] 1 Jk′ (r) = Jk−1 (r) − Jk+1 (r) , 2 1 (2.8) and (2.10) to obtaining |Jk′ (r)| ≤ Cr− 2 .

Proof. When r ≤


When

k 2

≤ r ≤ 2k, we have by (2.6)

Jk′ (r) = J˜k′ (r) − Ek′ (r).

A simple computation gives that for r > 0 sin(kπ) Z ∞ ′ |Ek (r)| = e−(r sinh s+ks) sinh s ds ≤ C(r + k)−1 . π 0 Thus we only need to estimate J˜k′ (r). We divide two cases r > k and r ≤ k to estimate it by the stationary phase argument. Let φr,k (θ) = r sin θ − kθ.

Case 1: k < r ≤ 2k. Let θ0 = cos−1 ( kr ), then

φ′r,k (θ0 ) = r cos θ0 − k = 0.

Now we split J˜k (r) into two pieces: Z Z i i ′ ir sin θ−ikθ ˜ Jk (r) = e sin θ dθ + eir sin θ−ikθ sin θ dθ, 2π Ωδ 2π Bδ

where

Ωδ = {θ : |θ ± θ0 | ≤ δ},

Bδ = [−π, π] \ Ωδ

with

δ > 0.

We have by taking absolute values 1 Z eir sin θ−ikθ sin θ dθ ≤ C| sin(θ0 ± δ)|δ. 2π Ωδ

Integrating by parts, we have Z Z eir sin θ−ikθ (r − k cos θ) ei(r sin θ−kθ) sin θ ir sin θ−ikθ dθ, − e sin θ dθ = i(r cos θ − k) ∂Bδ i(r cos θ − k)2 Bδ Bδ

where ∂Bδ = {±π, ±θ0 ± δ}. It is easy to see that ei(r sin θ−kθ) sin θ ≤ c sin(θ0 ± δ)|r cos(θ0 ± δ) − k|−1 . i(r cos θ − k) ∂Bδ

Since r − k cos θ > 0, we obtain Z eir sin θ−ikθ (r − k cos θ) Z |r − k cos θ| sin θ dθ ≤ dθ = 2 2 i(r cos θ − k) (r cos θ − k) (r cos θ − k) ∂Bδ Bδ Bδ ≤ c| sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 .

Therefore,

|J˜k′ (r)| ≤ C| sin(θ0 ± δ)|δ + c| sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 .

We shall choose proper δ such that

| sin(θ0 ± δ)|δ ∼ c| sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 .

Noting that cos(θ0 ± δ) = cos θ0 cos δ ∓ sin θ0 sin δ and the definition of θ0 , we get p r cos(θ0 ± δ) − k = k cos δ ± r2 − k 2 sin δ − k.

Since 1 − cos δ = 2 sin2 2δ , one has

|r cos(θ0 ± δ) − k| ∼ |kδ 2 ± δ

p r2 − k 2 | with small δ.

8


On the other hand, we have by sin(θ0 ± δ) = sin θ0 cos δ ± cos θ0 sin δ, √ r2 − k 2 δ2 k sin(θ0 ± δ) = ± (1 − ) ± δ. r 2 r 1

1

When |r − k| ≤ k 3 , choosing δ = Ck − 3 with large C ≥ 2, we have p 2 2 2 2 | sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 . k − 3 (C 2 − Ck − 3 r2 − k 2 )−1 . k − 3 . r− 3 . 1

1

When |r − k| ≥ k 3 , taking δ = c(r2 − k 2 )− 4 with small c > 0, we obtain

| sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 1 1 1 3 −1 1 . |r − k| 2 r− 2 + r−1 + (r2 − k 2 )− 4 (r2 − k 2 )− 4 c − c2 k(r2 − k 2 )− 4 3

1

1

1

1

.k − 4 |r − k| 4 + k − 2 |r − k|− 2 . r− 2 ,

1

1

1

where we use the fact that (r2 − k 2 )− 4 ≤ (2k)− 4 |r − k|− 4 for k < r. 1 1 Case 2: k2 ≤ r ≤ k. When k − k 3 < r < k, choosing θ0 = 0 and δ = Ck − 3 with large C ≥ 2, it follows from the above argument that |J˜k′ (r)| . | sin(θ0 ± δ)| · |r cos(θ0 ± δ) − k|−1 2

2

. δ(rδ 2 /2 − |r − k|)−1 . k − 3 . r− 3 .

1

When r < k − k 3 , there is no critical point. Hence we obtain 2

|J˜k′ (r)| . |r − k|−2 . r− 3 .

Finally, we collect all the estimates to get |J˜k′ (r)| . r− 2 . 1

Next, we record the two basic results about the modified square function expressions. ∞ + Lemma 2.3 (A modified Littlewood-Paley P theorem [18]). Let β ∈ C0 (R ) be 1 −j supported in [ 2 , 2], βj (ρ) = β(2 ρ) and βj = 1. Then for any ν(k) > 0 and j

1 < p < ∞, we have

XZ ∞

0 n−1 − n−2

2 J (rρ) cos(tρ)b (ρ)ρ β (ρ)dρ (rρ) j ν(k) k,ℓ

(2.11)

j∈Z

0

Z

X

∼ j∈Z

0

∞

− n−2 2

(rρ)

Lpn−1 r

2 12

Jν(k) (rρ) cos(tρ)b0k,ℓ (ρ)ρn−1 βj (ρ)dρ

.

Lpn−1 r

For the sake of the completeness, we will prove Lemma 2.3 in the appendix by using a weak-type (1, 1) estimate for the multiplier operators with respect to the Hankel transform. Lemma 2.4 (Littlewood-Paley-Stein theorem for the sphere, [19, 21, 22]). Let β ∈ C0∞ (R+ ) be supported in [ 12 , 4] and β(ρ) = 1 when ρ ∈ [1, 2]. Assume βj (ρ) = β(2−j ρ). Then for any 1 < p < ∞ and any test function f (θ) defined on Sn−1 , we have (2.12) ∞ X d(k)

X 2 12 2 X

βj (k)ak,ℓ Yk,ℓ (θ) kf (θ)kLpθ (Sn−1 ) ∼ a0,1 Y0,1 (θ) +

p n−1 , j=0

k

ℓ=1

Lθ (S

)


where f =

∞ d(k) P P

ak,ℓ Yk,ℓ (θ).

k=0 ℓ=1

We conclude this subsection by showing the “Bernstein” inequality on sphere j+1

(2.13)

j+1

21 2X d(k) X X

2X d(k) 1 1

ak,ℓ Yk,ℓ (θ) Lq (Sn−1 ) ≤ Cq,n 2j(n−1)( 2 − q ) |ak,ℓ |2 k=2j ℓ=1

k=2j ℓ=1

for q ≥ 2, j = 0, 1, 2 · · · . d(k) P |Yk,ℓ (θ)|2 = d(k)|Sn−1 |−1 , ∀θ ∈ Sn−1 (see Stein-Weiss [20]), one In fact, since ℓ=1

has

j+1 d(k)

2

2X X

ak,ℓ Yk,ℓ (θ)

∞

L

k=2j ℓ=1

(Sn−1 )

≤C ≤C

j+1 d(k) 2X X

k=2j ℓ=1

j+1 d(k) 2X X

k=2j ℓ=1

≤ C2j(n−1)

j+1 d(k)

2X 21 2 X

|ak,ℓ | |Yk,ℓ (θ)|2 ∞

2

|ak,ℓ |2 j+1

2X

k=2j

k=2j ℓ=1

j+1 2X

L

(Sn−1 )

k n−2

k=2j

d(k)

X ℓ=1

|ak,ℓ |2 .

Interpolating this with j+1 d(k)

2

2X X

ak,ℓ Yk,ℓ (θ)

2

k=2j

yields (2.13).

L (Sn−1 )

ℓ=1

≤C

j+1 d(k) 2X X

k=2j

ℓ=1

|ak,ℓ |2

2.2. Spectrum of −∆ + |x|a 2 and Hankel transform. Let us first consider the eigenvalue problem associated with the operator −∆ + |x|a 2 : ( −∆u + |x|a 2 u = ρ2 u x ∈ B = {x : |x| ≤ 1}, x ∈ Sn−1 .

u(x) = 0,

If u(x) = f (r)Yk (θ), we have f ′′ (r) +

n−1 ′ k(k + n − 2) + a ]f (r) = 0. f (r) + [ρ2 − r r2

Let λ = ρr and f (r) = λ− (2.14)

g ′′ (λ) +

n−2 2

g(λ), we obtain

2 (k + n−2 1 ′ 2 ) +a ]g(λ) = 0. g (λ) + [1 − λ λ2

Define p n−2 + k, and ν(k) = µ2 (k) + a with a > −(n − 2)2 /4. 2 The Bessel function Jν(k) (λ) solves the Bessel equation (2.14). And the eigenfunctions corresponding to the spectrum ρ2 can be expressed by (2.15) µ(k) =

(2.16)

φρ (x) = (ρr)−

n−2 2

Jν(k) (ρr)Yk (θ)

with

x = rθ,

10


where

(2.17)

−∆+

a φρ = ρ2 φρ . |x|2

We define the following elliptic operator

k(k + n − 2) + a n−1 ∂r + r r2 2 (2.18) 2 ν (k) − n−2 n−1 2 2 = −∂r − , ∂r + r r2 then Aν(k) φρ = ρ2 φρ . Define the Hankel transform of order ν: Z ∞ n−2 (2.19) (Hν f )(ξ) = (rρ)− 2 Jν (rρ)f (rω)rn−1 dr, Aν(k) : = −∂r2 −

0

where ρ = |ξ|, ω = ξ/|ξ| and Jν is the Bessel function of order ν. Specially, if the function f is radial, then Z ∞ n−2 (2.20) (Hν f )(ρ) = (rρ)− 2 Jν (rρ)f (r)rn−1 dr. 0

If f (x) =

∞ d(k) P P

ak,ℓ (r)Yk,ℓ (θ), then we obtain by (2.3)

k=0 ℓ=1

(2.21)

fˆ(ξ) =

∞ d(k) X X

k=0 ℓ=1

2πik Yk,ℓ (ω) Hµ(k) ak,ℓ (ρ).

We will also make use of the following properties of the Hankel transform, which appears in [1, 14]. Lemma 2.5. Let Hν and Aν be defined as above. Then (i) Hν = Hν−1 , (ii) Hν is self-adjoint, i.e. Hν = Hν∗ , (iii) Hν is an L2 isometry, i.e. kHν φkL2ξ = kφkL2x , (iv) Hν (Aν φ)(ξ) = |ξ|2 (Hν φ)(ξ), for φ ∈ L2 . 0 Let Kµ,ν = Hµ Hν , then as well as in [14] one has

(2.22)

0 0 Aµ Kµ,ν = Kµ,ν Aν .

0 For our purpose, we need another crucial properties of Kµ(k),ν(k) with k = 0: 0 Lemma 2.6 (The boundness of Kλ,ν , [1, 14]). Let ν, α, β ∈ R, ν > −1, λ = µ(0) = n−2 , −n < α < 2(ν + 1) and −2(ν + 1) < β < n. Then the conjugation operator 2 β 0 n ˙ Kλ,ν is continuous on Hp,rad (R ) provided that nλ + ν + 2 λ + ν + 2 + β o n λ − ν βo 1 < < min , , ,1 max 0, n n p n n

0 α while the inverse operator Kν,λ is continuous on H˙ q,rad (Rn ) provided that n λ − ν λ − ν + αo 1 nλ + ν + 2 α o max 0, < < min , ,1 + ,1 . n n q n n

We also need the Strichartz estimates for (1.1) in [1]:

STRICHARTZ ESTIMATES FOR WAVE EQUATION WITH INVERSE SQUARE POTENTIAL11

Lemma 2.7 (Strichartz estimates). For n ≥ 2, let 2 ≤ r < ∞ and q, r, γ, σ satisfy n 1 n − 1 1 1 o 1 1 1 1 , σ =γ+ −n . ≤ min , − − (2.23) q 2 2 2 r q 2 r

There exists a positive constant C depending on n, a, q, r, γ such that the solution u of (1.1) satisfies

(−∆) σ2 u q (2.24) ≤ C ku0 kH˙ γ + ku1 kH˙ γ−1 L (R;Lr (Rn )) t

provided that when n = 2, 3 o nn + 1 nn − 1 1 1 1o , ν(1) − , 1 + ν(0) < γ < min , ν(1) + , 1 + ν(0) − − min 2 2 2 2 q

and when n ≥ 4 nn o nn + 1 n+3 n+3 1 1o − min − , ν(1)− , 1+ν(0) < γ < min , ν(1)+ , 1+ν(0)− . 2 2(n − 1) 2(n − 1) 2 2 q Next, define the projectors Mjj ′ = Pj P˜j ′ and Njj ′ = P˜j Pj ′ , where Pj is the usual dyadic frequency localization at |ξ| ∼ 2j and P˜j is the localization with respect to 21 a . More precisely, let f be in the k-th harmonic subspace, then − ∆ + |x| 2 Pj f = Hµ(k) βj Hµ(k) f

and P˜j f = Hν(k) βj Hν(k) f,

where βj (ξ) = β(2−j |ξ|) with β ∈ C0∞ (R+ ) supported in [ 14 , 2]. Then, we have the almost orthogonality estimate which is proved in [1]. Lemma 2.8 (Almost orthogonality estimate, [1]). There exists a positive constant C independent of j, j ′ , and k such that the following inequalities hold for all positive 1 (n−2)2 ǫ1 < 1 + min{ n−2 + a) 2 } 2 ,( 4 ′

kMjj ′ f kL2 (Rn ) , kNjj ′ f kL2 (Rn ) ≤ C2−ǫ1 |j−j | kf kL2(Rn ) , where f is in the k-th harmonic subspace. As a consequence of Lemma 2.7 and Lemma 2.8, we have Lemma 2.9 (Strichartz estimates for unit frequency). Let n ≥ 3, k ∈ N. Let u solve ( (∂t2 − ∆ + |x|a 2 )u = 0, where u0 ∈ L2 (Rn ) and

u|t=0 = u0 (x), ut |t=0 = 0,

u0 =

∞ d(k) X X

ak,ℓ (r)Yk,ℓ (θ).

k=0 ℓ=1

Assume that for all k, ℓ ∈ N, supp Hν(k) ak,ℓ ⊂ [1, 2]. Then the following estimate

holds for a >

1 (n−1)2

(n−2)2 4

ku(t, x)kLq (R;Lr (Rn )) ≤ Cku0 kL2 (Rn ) ,

(2.25) where q ≥ 2,

−

1 q

=

n−1 1 2 (2

− r1 ) and (q, r, n) 6= (2, ∞, 3).

12


n+1 , we Proof. By making use of Lemma 2.7 with σ = 0, γ = n( 21 − 1r ) − 1q = q(n−1) obtain that kukLqt (R;Lr (Rn )) ≤ Cku0 kH˙ γ . Since 0 < γ ≤ 1, we have by Lemma 2.8 with ǫ1 = 1+,

kuk

Lqt (R;Lr (Rn ))

≤C =C

X j∈Z

X j∈Z

≤C

2

2jγ

X j∈Z

kPj u0 k2L2

12

=C

X j∈Z

2

∞ d(k)

X 1 X

2 2 Pj ak,ℓ (r)Yk,ℓ (θ) 2

2jγ

L

k=0 ℓ=1

∞ d(k)

X 1 X

2 2 22jγ Pj P˜1 ak,ℓ (r)Yk,ℓ (θ) 2 L

k=0 ℓ=1

22jγ−2ǫ1 |j−1| ku0 k2L2

1 2

≤ Cku0 kL2 .

This completes the proof of Lemma 2.9.

3. Estimates of Hankel transforms In this section, we prove some estimates for the Hankel transforms of order ν(k). These estimates will be utilized to prove the main results in the next section. Proposition 3.1. Let k ∈ N, 1 ≤ ℓ ≤ d(k) and let ϕ be a smooth function supported in the interval I := [ 21 , 2]. Then (3.1)

Z ∞

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)ϕ(ρ)dρ

0

L2t (R;L2r ([R,2R]))

1 ≤ C min R 2 , 1 kb0k,ℓ (ρ)kL2ρ (I) ,

where R ∈ 2Z and C is a constant independent of R, k, and ℓ. Proof. Using the Plancherel theorem in t, we have

L.H.S of (3.1) .

Jν(k) (rρ)b0k,ℓ (ρ)ϕ(ρ) L2 (3.2) ρ

L2r ([R,2R])

.

We first consider the case R . 1. Since ν(k) > 0, one has by (2.4) 2 21 Z 2R

rν(k) L.H.S of (3.1) . b0k,ℓ (ρ) L2 (I) ν(k) 1 1 dr ρ 2 Γ(ν(k) + )Γ( ) R (3.3) 2 2

1 . R 2 b0k,ℓ (ρ) L2 (I) . ρ

Next we consider the case R ≫ 1. It follows from (3.2) that (3.1) can be reduced to show Z 2R (3.4) |Jk (r)|2 dr ≤ C, R ≫ 1, R

where the constant C is independent of k and R. To prove (3.4), we write Z Z Z 2R Z |Jk (r)|2 dr |Jk (r)|2 dr + |Jk (r)|2 dr + |Jk (r)|2 dr = R

I1

I2

I3

where k I1 = [R, 2R] ∩ [0, ], 2

k I2 = [R, 2R] ∩ [ , 2k] and I3 = [R, 2R] ∩ [2k, ∞]. 2


Using (2.8) and (2.10) in Lemma 2.1, we have Z Z 2 e−cr dr ≤ Ce−cR , |J (r)| dr ≤ C (3.5) k I1

I1

and

Z

(3.6)

I3

|Jk (r)|2 dr ≤ C.

On the other hand, one has by (2.9) Z Z |Jk (r)|2 dr ≤ C

[k 2 ,2k]

[k 2 ,2k]

1

2

1

k − 3 (1 + k − 3 |r − k|)− 2 dr ≤ C.

Observing [R, 2R] ∩ [ k2 , 2k] = ∅ unless R ∼ k, we obtain Z |Jk (r)|2 dr ≤ C. (3.7) I2

This together with (3.5) and (3.6) yields (3.1).

Proposition 3.2. Let γ ≥ 2 and let k ∈ N, 1 ≤ ℓ ≤ d(k). Suppose supp b0k,ℓ (ρ) ⊂ I := [1, 2]. Then

Hν(k) cos(tρ)b0k,ℓ (ρ) (r) 2 Lt (R;Lγn−1 ([R,2R])) dr (3.8) o n (n+1)+(γ−2)ν(k)r n−1 n−1 n−2 − 2 γ ≤ C min R , R γ − 2 kb0k,ℓ (ρ)kL2ρ (I) , where R ∈ 2Z and C is a constant independent of R, k and ℓ.

Proof. We first consider the case R ≫ 1. Using the definition of Hankel transform and the interpolation, we only need to prove

Z ∞

n−2

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)(rρ)− 2 ρn−1 dρ 2

Lt (R;L2n−1 ([R,2R])) 0 (3.9) r dr 1 . R 2 kb0k,ℓ (ρ)kL2ρ , and (3.10)

Z

∞

0

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)(rρ)−

n−2 2

ρn−1 dρ

L2t (R;L∞n−1

.R

− n−2 2

r

dr

([R,2R]))

kb0k,ℓ (ρ)kL2ρ .

(3.9) follows from Proposition 3.1. To prove (3.10), it is enough to show that there exists a constant C independent of k, ℓ such that

Z ∞

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)ϕ(ρ)dρ 2 ≤ Ckb0k,ℓ (ρ)kL2ρ (I) . (3.11)

∞ Lt (R;Lr ([R,2R]))

0

1

∞

By the Sobolev embedding H (Ω) ֒→ L (Ω) with Ω = [R, 2R], it suffices to show

Z ∞

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)ϕ(ρ)dρ 2 ≤ Ckb0k,ℓ (ρ)kL2ρ (I) , (3.12)

0

and

(3.13)

Z

0

∞

Lt (R;L2r ([R,2R]))

′ (rρ)b0k,ℓ (ρ)ρϕ(ρ)dρ e−itρ Jν(k)

L2t (R;L2r ([R,2R]))

≤ Ckb0k,ℓ (ρ)kL2ρ (I) .

14


In fact, (3.12) follows from Proposition 3.1, we apply the Plancherel theorem in t and Lemma 2.2 to showing (3.13). Secondly, we consider the case R . 1. From the definition of Hankel transform, we need to prove

Z ∞ n−2


Lt (R;Lγ r ([R,2R])) 0 (3.14) 2+(γ−2)ν(k) n−1 − 2 0 γ .R kbk,ℓ (ρ)kL2ρ . On the other hand, we have by Proposition 3.1

Z ∞ n−2


Lt (R;L2r ([R,2R])) 0 (3.15) − n−3 0 . R 2 kbk,ℓ (ρ)kL2ρ . By interpolation, it suffices to prove the estimate

Z ∞ n−2


Lt (R;L∞ r ([R,2R])) 0 (3.16) n−1 − 2 +ν(k) 0 .R kbk,ℓ (ρ)kL2ρ . Indeed, using Sobolev embedding, we can prove (3.16) by showing

Z ∞

1

e−itρ Jν(k) (rρ)b0k,ℓ (ρ)ϕ(ρ)dρ 2 ≤ CR 2 +ν(k) kb0k,ℓ (ρ)kL2ρ (I) ,

2 Lt (R;Lr ([R,2R]))

0

and

Z

∞

0

′ e−itρ Jν(k) (rρ)b0k,ℓ (ρ)ρϕ(ρ)dρ

1

L2t (R;L2r ([R,2R]))

≤ CRν(k)− 2 kb0k,ℓ (ρ)kL2ρ (I) .

These two estimates are implied by (2.4) and (2.5). Therefore, we conclude this proposition. 4. Proof of Theorem 1.1

In this section, we use Proposition 3.1 and Proposition 3.2 to prove Theorem 1.1. We first consider the Cauchy problem: ( (∂tt − ∆ + |x|a 2 )u(x, t) = 0, (4.1) u(x, 0) = u0 (x), ∂t u(x, 0) = 0. We use the spherical harmonic expansion to write (4.2)

u0 (x) =

∞ d(k) X X

a0k,ℓ (r)Yk,ℓ (θ).

k=0 ℓ=1

Then we have the following proposition:

0 Proposition 4.1. Let γ = 2(n−1) n−2 + and suppose supp Hν ak,ℓ ⊂ [1, 2] for all k, ℓ ∈ N and 1 ≤ ℓ ≤ d(k). Then (4.3)

ku(x, t)kL2t Lγn−1 r

dr

L2 (Sn−1 )

≤ Cku0 kL2x .


Proof. Let us consider the equation (4.1) in polar coordinates. Write v(t, r, θ) = u(t, rθ) and g(r, θ) = u0 (rθ). Then v(t, r, θ) satisfies that ( 1 a ∂tt v − ∂rr v − n−1 r ∂r v − r 2 ∆θ v + r 2 v = 0, (4.4) v(0, r, θ) = g(r, θ), ∂t v(0, r, θ) = 0. By (4.2), we also have g(r, θ) =

∞ d(k) X X

a0k,ℓ (r)Yk,ℓ (θ).

∞ d(k) X X

vk,ℓ (t, r)Yk,ℓ (θ),

k=0 ℓ=1

Using separation of variables, we can write v as a superposition (4.5)

v(t, r, θ) =

k=0 ℓ=1

where vk,ℓ satisfies the following equation ( k(k+n−2)+a vk,ℓ = 0 ∂tt vk,ℓ − ∂rr vk,ℓ − n−1 r ∂r vk,ℓ + r2 0 vk,ℓ (0, r) = ak,ℓ (r), ∂t vk,ℓ (0, r) = 0 for each k, ℓ ∈ N, and 1 ≤ ℓ ≤ d(k). From the definition of Aν , it becomes ( ∂tt vk,ℓ + Aν(k) vk,ℓ = 0, (4.6) vk,ℓ (0, r) = a0k,ℓ (r), ∂t vk,ℓ (0, r) = 0. Applying the Hankel transform to the equation (4.6), we have by Lemma 2.5 ( ∂tt v˜k,ℓ + ρ2 v˜k,ℓ = 0, (4.7) v˜k,ℓ (0, ρ) = b0k,ℓ (ρ), ∂t v˜k,ℓ (0, ρ) = 0, where (4.8)

v˜k,ℓ (t, ρ) = (Hν vk,ℓ )(t, ρ),

b0k,ℓ (ρ) = (Hν a0k,ℓ )(ρ).

Solving this ODE and using the inverse Hankel transform, we obtain Z ∞ n−2 vk,ℓ (t, r) = (rρ)− 2 Jν(k) (rρ)˜ vk,ℓ (t, ρ)ρn−1 dρ 0 Z n−2 1 ∞ = (rρ)− 2 Jν(k) (rρ) eitρ + e−itρ b0k,ℓ (ρ)ρn−1 dρ. 2 0

Therefore, we get

u(x, t) = v(t, r, θ)

(4.9)

=

∞ d(k) X X k=0 ℓ=1

=

∞ d(k) X X k=0 ℓ=1

Yk,ℓ (θ)

Z

∞

(rρ)− 0

n−2 2

Jν(k) (rρ) cos(tρ)b0k,ℓ (ρ)ρn−1 dρ

Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (ρ) (r).

To prove (4.3), it suffices to show (4.10)

∞ d(k)

X X 1

Hν(k) cos(tρ)b0k,ℓ (ρ) (r) 2 2

k=0 ℓ=1

L2t (R;Lγn−1 r

dr

(R+ ))

≤ Cku0 kL2x .

16


Using the dyadic decomposition, we have by ℓ2 ֒→ ℓγ (γ > 2) (4.11) ∞ d(k)

X X 1 2

Hν(k) cos(tρ)b0 (ρ) (r) 2 2

2

k,ℓ

Lt (R;Lγn−1

k=0 ℓ=1

r

dr

(R+ ))

∞ d(k)

X X X 1 γ

Hν(k) cos(tρ)b0k,ℓ (ρ) (r) 2 2 =

γ

L

k=0 ℓ=1

R∈2Z

∞ d(k) X X X

2 .

Hν(k) cos(tρ)b0k,ℓ (ρ) (r) 2

Lt (R;Lγn−1

R∈2Z k=0 ℓ=1

r

dr

rn−1 dr

([R,2R])

([R,2R]))

γ1 2

2

Lt (R)

.

By Proposition 3.2, we obtain

∞ d(k)

X X 1 2

Hν(k) cos(tρ)b0k,ℓ (ρ) (r) 2 2

2

Lt (R;Lγn−1

k=0 ℓ=1

(4.12)

.

∞ d(k) X X X

R∈2Z k=0 ℓ=1

.

∞ d(k) X X

k=0 ℓ=1

r

dr

(R+ ))

n (n+1)+(γ−2)ν(k) n−1 n−1 n−2 o2 − 2 γ min R ,R γ − 2 kb0k,ℓ (ρ)k2L2ρ

kb0k,ℓ (ρ)k2L2ρ .

Since supp b0k,ℓ (ρ) ⊂ [1, 2], we have ∞ d(k) X X

k=0 ℓ=1

kb0k,ℓ (ρ)k2L2ρ .

∞ d(k) X X k=0 ℓ=1

k Hν(k) a0k,ℓ (ρ)k2L2

ρn−1 dρ

.

It follows from Lemma 2.5 that ∞ d(k) X X k=0 ℓ=1

k

Hν(k) a0k,ℓ

(ρ)k2L2

ρn−1 dρ

=

∞ d(k) X X

k=0 ℓ=1

ka0k,ℓ (r)k2L2

rn−1 dr

= ku0 (x)k2L2x (Rn ) .

Therefore, we complete the proof of (4.3). Now we turn to prove Theorem 1.1. We choose β ∈ C0∞ (R+ ) supported in [ 21 , 2] P β( Nρ ) = 1 for all ρ ∈ R+ . Let βN (ρ) = β( Nρ ) and βÑ be similar to such that N ∈2Z

βN . For simplicity, we assume u1 = 0. Then we can write (4.13) u(x, t) =

X n

M∈2Z

Y0,1 (θ)Hν(0) cos(tρ)b00,1 (ρ)βM (ρ) (r) +

X X

N ∈2N

d(k)

βÑ (k)

k

:=u< (x, t) + u≥ (x, t).

X ℓ=1

o Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (ρ)βM (ρ) (r)


Without loss of the generality, it suffices to estiamte u≥ (x, t). By Lemma 2.3, Lemma 2.4 and the scaling argument, we show that for 2 ≤ q, r and r < ∞ (4.14) ku≥ (t, x)k2Lq (R;Lr (Rn )) t

x

d(k) X X X

X ˜

2 . βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (ρ)βM (ρ) (r) q

k

M∈2Z N ∈2N

.

X

M

Lt (R;Lrx (Rn ))

ℓ=1

2(n− q1 − n r)

d(k) X X

X ˜

2 βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) q

N ∈2N

M∈2Z

k

Lt (R;Lrx (Rn ))

ℓ=1

• Case 1: n ≥ 4. we have by interpolation

(4.15) d(k)

X X

βÑ (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r)

k

ℓ=1

Lqt (R;Lrx (Rn ))

d(k)

X X

λ . βÑ (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) 2 k

d(k)

X X

1−λ × βÑ (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) ∞ k

where

γ

Lt (R;Lx0 (Rn ))

ℓ=1

1 λ 1−λ = + , q 2 ∞

(4.16)

Lt (R;L2x (Rn ))

ℓ=1

1−λ 1 λ + = , r γ0 2

Since (q, r) ∈ Λ, one has 2(n−1) n−2 < γ0 ≤ proving Proposition 4.1, one has

2(n−1) n−3 .

q 1 1 1 1 = ( + − ). γ0 2 r q 2 By (2.13) and the argument in

(4.17) d(k)

X X


k

ℓ=1

2(n−1)

L2t (R;Lx n−2

X d(k) X 1 1

βÑ (k)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) 2 2 . N 2 +

k

.N

1 2+

k

+

(Rn ))

L2t (R;L

ℓ=1

X d(k) X 1

βÑ (k)b0k,ℓ (M ρ)β(ρ) 2 2

L2ρ

ℓ=1

,

2(n−1) + n−2 (R+ )) rn−1 dr

.

On the other hand, using the endpoint Strichartz estimate in Lemma 2.9, we have

(4.18)

d(k)

X X

˜ βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r)

k

ℓ=1

X d(k) X 1

βÑ (k)b0k,ℓ (M ρ)β(ρ) 2 2 .

k

ℓ=1

L2ρ

.

2(n−1)

L2t (R;Lx n−3 (Rn ))

.

18


Therefore, we obtain by interpolation d(k)

X X


k

(4.19)

ℓ=1

γ

L2t (R;Lx0 (Rn ))

X d(k) X 1 n−3 1

βÑ (k)b0k,ℓ (M ρ)β(ρ) 2 2 . N (n−1)( γ0 − 2(n−1) )+

k

By Lemma 2.5, we have

[Hν(k) ak,ℓ ](r)

L2ρ

ℓ=1

L2n−1 r

dr

.

= kak,ℓ (ρ)kL2n−1 . dρ

ρ

We are in sprit of energy estimate to obtain

(4.20) d(k)

X X


k

ℓ=1

2 L∞ t (R;L n−1 r

X d(k) X 1

βÑ (k)Hν(k) cos(tρ)b0 (M ρ)β(ρ) (r) 2 2 .

k,ℓ k

.

k

ℓ=1

(R+ ;L2 (Sn−1 )))

2 L∞ t (R;L n−1

ℓ=1

X d(k) X

dr

r

dr

(R+ ))

2 21 βÑ (k) b0k,ℓ (M ρ)β(ρ) L2 . ρ

Combining (4.14),(4.15), (4.19), and (4.20), we have (4.21) ku≥ (t, x)k2Lq (R;Lr (Rn )) t

.

X

x

1

n

M 2(n− q − r )

X

n−3

1

N 2λ(n−1)( γ0 − 2(n−1) )+

1

n

M 2(n− q − r )

M∈2Z

X

N ∈2N

X d(k) X k

N ∈2N

M∈2Z

.

X

2

1

1

N 2[ q +(n−1)( r − 2 )]+

ℓ=1

X d(k) X k

ℓ=1

2 βÑ (k) b0k,ℓ (M ρ)β(ρ) L2

ρ

2 βÑ (k) b0k,ℓ (M ρ)β(ρ) L2 . ρ

By making use of Lemma 2.8, s = n( 12 − 1r ) − q1 , and s¯ = (1 + ǫ)( 2q − (n − 1)( 12 − r1 )), we get ku≥ (t, x)kLqt (R;Lrx (Rn )) . khΩis¯u0 kH˙ s .

(4.22)

• Case 2: n = 3. Since the endpoint Strichartz estimate fails, the above argument breaks down. By the interpolation, we have (4.23) d(k)

X X

˜ β (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r)

N k

ℓ=1

Lqt (R;Lrx (R3 ))

d(k)

X X

λ ˜ . βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) 2 k

3 Lt (R;L4+ x (R ))

ℓ=1

d(k)

X X

1−λ ˜ × βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) q0 k

ℓ=1

r

Lt (R;Lx0 (R3 ))

,


where λ 1−λ 1 , = + q 2 q0

(4.24)

1 λ 1−λ , = + r 4+ r0

1 1 1 + , = 2 q0 r0

r0 6= ∞.

By (2.13) and the argument in proving Proposition 4.1, one has (4.25) d(k)

X X


k

ℓ=1

3 L2t (R;L4+ x (R ))

X d(k) X 1 1

βÑ (k)Hν(k) cos(tρ)b0 (M ρ)β(ρ) (r) 2 2 . N 2 +

k,ℓ k

.N

1 2+

L2t (R;L4+ n−1

ℓ=1

X d(k) X 1

βÑ (k)b0k,ℓ (M ρ)β(ρ) 2 2

k

L2ρ

ℓ=1

r

dr

(R+ ))

.

On the other hand, by the Strichartz estimate with (q0 , r0 ) in Lemma 2.9, we have d(k)

X X


k

(4.26)

ℓ=1

X d(k) X 1

βÑ (k)b0k,ℓ (M ρ)β(ρ) 2 2 .

k

L2ρ

ℓ=1

q

r

Lt 0 (R;Lx0 (R3 ))

.

This together with (4.14), (4.23) and (4.25) yields that ku≥ (t, x)k2Lq (R;Lr (Rn )) t

. (4.27)

X

x

1

.

N λ+

M

2(n− q1 − n r)

M∈2Z

X

X d(k) X k

N ∈2N

M∈2Z

X

X

n

M 2(n− q − r )

N

ℓ=1

2 βÑ (k) b0k,ℓ (M ρ)β(ρ) L2

(4+ǫ)[ q1 + r1 − 21 ]

ρ

X d(k) X k

N ∈2N

ℓ=1

2 βÑ (k) b0k,ℓ (M ρ)β(ρ) L2 . ρ

Since s = n( 12 − r1 ) − 1q , by the scaling argument, Lemma 2.8, we get ku≥ (t, x)kLqt (R;Lrx (Rn )) . khΩis¯u0 kH˙ s .

(4.28)

Moreover, for q = 2, 4 < r < ∞, (4.14) and the “Bernstein” inequality (2.13) imply that ku≥ (t, x)k2L2 (R;Lr (Rn )) t

.

X

x

1

n

M 2(n− 2 − r )

N ∈2N

M∈2Z

.

X

d(k) X X

X ˜

2 βN (k) Yk,ℓ (θ)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) 2

1

n

M 2(n− 2 − r )

M∈2Z

.khΩis¯(r) u0 k2H˙ s .

X

N ∈2N

k

ℓ=1

Lt (R;Lrx (Rn ))

X d(k) X 1 2

βÑ (k)Hν(k) cos(tρ)b0k,ℓ (M ρ)β(ρ) (r) 2 2 N 2¯s(r)

2 k

ℓ=1

Combining this with (4.22) and (4.28), we complete the proof of Theorem 1.1.

Lt (R;Lrn−1 r

dr

(R+ ))

20


5. Proof of Theorem 1.2 To prove Theorem 1.2, we first use the inhomogeneous Strichartz estimates for the wave equation without potential in [7, 13] and the arguments in [14] to prove an inhomogeneous Strichartz estimates for the wave equation with inverse-square potential. e = ∂t2 + Aν and let Proposition 5.1 (Inhomogeneous Strichartz estimates). Let e = h in R × Rn with zero initial data. v solve the inhomogeneous wave equation v n n n−2 − − 2}, then , − If ν > max{ n−2 2 q0 r0 2 (5.1)

r0 kvkLqt,x 0 (R×Rn ) . khkLt,x (R×Rn ) ,

where q0 = (p − 1)(n + 1)/2 and r0 = (n + 1)(p − 1)/(2p) with ph < p < pconf .

0 Proof. By the continuity property of Kν,λ in Lemma 2.6, it follows that

(5.2)

0 0 q0 kvkLqt,x 0 (R×Rn ) . (R×Rn ) ≤ kKν,λ kq0 →q0 kKλ,ν vkLt,x

0 0 e 0 Noting that (2.22) with k = 0, one has Kλ,ν h = Kλ,ν v = Kλ,ν v. We recall the

inhomogeneous Strichartz estimates, for

1 r

−

1 q

=

2 n+1

and

2n n−1