GEOMETRIC ANALYSIS ON A FAMILY OF PSEUDOCONVEX HYPERSURFACES

(1)

HYPERSURFACES

DER-CHEN CHANG AND STEPHEN S.-T. YAU Abstract. Let Ωk=

n

(z1, . . . , zn, zn+1) ∈ Cn+1 : Im(zn+1) = Pnj=1|zj|2

k

, k ∈ Nobe a family pseudoconvex hypersurfaces. In this article, we shall study geometric and analytical problems related to the Kohn Laplacian ¤b= ¯∂ ¯∂∗+ ¯∂∗∂. We first use analytical method to¯

construct fundamental solution for ¤b. Then we give geometric interpretation of the kernel.

More precisely, the fundamental solution can be expressed as an action integral over the characteristic variety on the cotangent bundle given by H(x, t, ξ, θ) = 0 with respect to the measure EVλ. Here (x, t, ξ, θ) is the Hamiltonian function of ¤b. E is the associated energy

and the volume element Vλis the solution of a second order transport equation.

1. Introduction Let Ω_k= n (z₁, . . . , z_n, z_n+1) ∈ Cn+1: Im(z_n+1) >¡ n X j=1 |z_j|2¢k o = n (z, zn+1) ∈ Cn+1: Im(zn+1) > |z|2k o with k ∈ N. The boundary of Ωk is

∂Ωk=

n

(z, zn+1) ∈ Cn+1: Im(zn+1) = |z|2k

o

.

As usual, we may use new coordinates (z, t) = (z₁, . . . , z_n, t) on the boundary ∂Ω_k to identify it as Cn_{× R. Here z}

j = xj + ixj+n, j = 1, . . . , n and t = Re(zn+1). It is easy to see that the

vector fields Z_j = ∂ ∂zj + ik|z|2(k−1)z¯_j ∂ ∂t = 1 2 ¡ X_j − iX_j+n¢, j = 1, . . . , n where Xj = ∂ ∂xj + 2k ¡Xn j=1 (x2_j + x2_j+n)¢k−1xj+n ∂ ∂t, Xj+n = ∂ ∂xj+n − ¡Xn j=1 (x2_j + x2_j+n)¢k−1xj ∂ ∂t, j = 1, . . . , n, form a basis of the subbundle T(1,0)_(∂Ω

k) of the complex tangent bundle

CT (∂Ωk). When k = 1, ∂Ω1 can be identified with the “Heisenberg group” and the

vec-tor fields {X_j, X_j+n}n

j=1 are left-invariant under the group law

(z, t) ∗ (w, s) =¡z + w, t + s + 2Im(z · ¯w)¢.

2000 Mathematics Subject Classification. 32A20.

Key words and phrases. Kohn Laplacina, fundamental solution, Heisenberg group, Lewy operator,

Hamilton-Jacobi equation, Cauchy-Szeg¨o projection, Whittaker’s function.

The first author is partially supported by a Hong Kong RGC competitive earmarked research grant #600607 and a competitive research grant at Georgetown University. The second author is partially supported by research grants from United State Army Research Office and NSF.

(2)

We define the “sub-Laplacian” on the boundary ∂Ω_k as follows: ∆λ,k= n X j=1 ¡ ZjZ¯j+ ¯ZjZj+ λ[ ¯Zj, Zj] ¢ =1 2 n X j=1 ¡ X_j2+ X_j+n2 + iλ[Xj+n, Xj] ¢ .

Unlike the Casimir operator

L = ∆_λ,k+1 2

∂2

∂t2,

the operator ∆λ,k is a sum of squares of 2n “horizontal” vector fields, and it is therefore not

elliptic, although it is hypoelliptic, see H¨ormander [22], i.e., the solution u of ∆λ,ku = f is

smooth whenever f ∈ C∞(∂Ωk). It is one of the purposes in this paper to discuss the regularity

properties of the fundamental solution for the Kohn Laplacian ∆λ,k. Without loss of generality,

we may assume φ(x) = xk. Then for j = 1, . . . , n, one has

X_j2+ X_j+n2 = ∂ 2 ∂x2 j + ∂ 2 ∂x2 j+n + 4k2(x2_j + x2_j+n)2(k−1) ∂ 2 ∂t2 + 4k(x2_j + x2_j+n)2(k−1) ³ xj+n_∂x∂ j − xj_∂x∂ j+n ´ ∂ ∂t, and [X_j+n, X_j] = 4k2(x2_j + x2_j+n)2(k−2) ∂ ∂t. Therefore, ∆_λ,k =1 2 n X j=1 h ∂2 ∂x2 j + ∂2 ∂x2 j+n + 4k|x|2k−2¡x_j+n ∂ ∂xj − x_j ∂ ∂xj+n ¢ ∂ ∂t i + 2k2|x|4k−2∂2 ∂t2 + 2iλk(n + k − 1)|x|2k−2 ∂ ∂t.

We exploit the partial Fourier transform in the variable t

F(f )(x, τ ) = ef (x, τ ) = Z _∞ −∞ e−iτ tf (t)dt, f (x, t) = Z _∞ −∞ eitτf (x, τ )e dτ 2π, so, F µ ∂f ∂t ¶ (τ ) = iτ ef (τ ), F µ ∂2_f ∂t2 ¶ (τ ) = −τ2f (τ ).e Consequently, F¡∆λ,k ¢ (τ ) =1 2 n X j=1 h ∂2 ∂x2 j + ∂ 2 ∂x2 j+n + 4ikτ |x|2k−2¡xj+n_∂x∂ j − xj ∂ ∂xj+n ¢i − 2k2τ2|x|4k−2− 2λk(n + k − 1)τ |x|2k−2.

In particular, when λ = 0, one has

F¡∆_0,kf¢(τ ) =1 2 n X j=1 h ∂2 ∂x2 j + ∂2 ∂x2 j+n + 4ikτ |x|2k−2¡xj+n ∂ ∂xj − xj ∂ ∂xj+n ¢ − 4k2τ2|x|4k−2 i e f .

Our plan for this article is the following. In section 2, We use a different method to obtain the fundamental solution for the case when k = 1 and general n. This recovers the result obtained

(3)

by Folland-Stein [16] and Greiner-Stein [20] . When λ = 1 and n = 1, the Kohn Laplacian ∆1,1 = −Z1Z¯1 which is not locally solvable because ¯Z1 is the famous Hans Lewy operator.

However, we may construct a relative fundamental solution eK1,1 such that ∆1,1Ke1,1 = δ − S1

in section 3. Here S1 is the Cauchy-Szeg¨o projection from L2(∂Ω1) onto the Hardy space

H2_(∂Ω

1). When λ = 0, n = 1 and k = 2, the result was first discussed in Greiner and Stein

[21]. However, the calculations was just up to certain degree and did not provide a complete answer. Later, a closed form fundamental solution in this case was announced in Greiner [17] but no detailed proof was given. We will give detailed calculation for this case in section 4. In section 5, we will discuss geometric meaning of these formulas. More precisely, the fundamental solution can be expressed as an action integral over the characteristic variety on the cotangent bundle given by H(x, t, ξ, θ) = 0 with respect to the measure EVλ. Here (x, t, ξ, θ) is the

Hamiltonian function of ¤_b. E is the associated energy and the volume element V_λ is the solution of a second order transport equation (see e.g., [2, 3, 4, 5, 9]).

Part of this article is based on a lecture presented by the first author at the International Conference on Geometric Analysis which was held at Taida Institute for Mathematical Sci-ences, June 18-22, 2007. Both authors would like to express their heartfelt gratitude to the organizing committee, especially Professor Ying-Ing Lee for the invitation. Thanks are also due to Professor Chang-Chou Lin, Director of TIMS for his help in providing very pleasant working conditions and for his leadership at TIMS.

The final version of this paper was in part written while the first author visited the National Center for Theoretical Sciences and National Tsing Hua University during May-August of 2007. He would like to thank Professors Jin Yu and Shu-Cheng Chang for their invitation and for the warm hospitality extended to him during his stay in Taiwan.

2. The fundamental solution for ∆_λ,1

This is the case corresponding to k = 1. To simplify the calculation, we may assume n = 1. In this case the basic complex holomorphic tangential vector field is

Z = ∂ ∂z1 + i¯z1 ∂ ∂t and ∆λ,1 = −1₂(Z ¯Z + ¯ZZ) + iλ_∂t∂ = − ∂2 ∂z1∂ ¯z1 + i · z1_∂z∂ 1 − ¯z1 ∂ ∂ ¯z1 ¸ ∂ ∂t− |z1| 2 ∂2 ∂t2 + iλ ∂ ∂t.

To find the fundamental solution for the operator ∆_λ,1, we note that the coefficients do not depend on the variable t. It follows that we may assume the fundamental solution has the following form:

K_λ,1(z1, ¯z1; w1, ¯w1) = _2π1

Z _∞

−∞

ei(t−s)τKe_λ,1(z1, ¯z1; w1, ¯w1; τ )dτ.

Applying ∆_λ,1 to K_λ,1, we see that eK_λ,1(z₁, ¯z₁; w₁, ¯w₁; τ ) is a fundamental solution for e∆_λ,1,

i.e., e ∆_λ,1Ke_λ,1(z₁, ¯z₁; w₁, ¯w₁; τ ) = δ(z₁− w₁, ¯z₁− ¯w₁), where e ∆_λ,1 = − ∂2 ∂z1∂ ¯z1 − τ · z1 ∂ ∂z1 − ¯z1 ∂ ∂ ¯z1 ¸ + |z1|2τ2− λτ.

(4)

Lemma 2.1. The fundamental solution, eK_λ,1 of e∆_λ,1 is given by the following formula e Kλ,1 = e τ (z1w¯1−¯z1w1) 4πp2|τ ||z1− w1|2 Γ µ −λ 2sgn(τ ) + 1 2 ¶ Wλ 2sgn(τ ),0(2τ |z1− w1| 2_),

which makes sense

(1) when τ > 0, as long as λ 6= 1, 3, 5, . . . , 2` + 1, . . . ,

(2) when τ < 0, as long as λ 6= −1, −3, −5, . . . , −2` − 1, . . . .

Proof. Here W_α,β(x) denotes Whittaker’s function. More precisely, “Whittaker’s differential equation” (2.1) d2u dx2 + Ã −1 4 + α x + 1 4 − β2 x2 ! u = 0

has the following system of linearly independent solutions

Mα,β(x) = x 1 2+βe−x2 Γ (2β + 1) Γ¡1 2 + α · β ¢ ∞ X m=0 Γ¡1₂ + β − α + m¢ Γ (2β + m + 1·) xm m! Wα,β(x) = Γ (2β) Γ¡1₂ + β − α¢ Mα,−β(x) + Γ (−2β) Γ¡1₂ − β − α¢ Mα,β(x).

In the definition of Mα,β we assumed that 2β is not a negative integer. Mα,β(x) is regular

near x = 0 and W_α,β(x) vanishes exponentially as x → +∞. Readers may consult books [14] and [23] for detailed discussion of Whittaker functions.

To derive eK_λ,1, we introduce polar coordinates z1 = reiθ, w1 = Reiφ. Next we solve for G,

where 4 e∆λ,1G = δ(r − R)δ(θ − φ)_r . We assume that (2.2) G = 1 2π ∞ X m=−∞ e−imθGm.

This implies that Gm solves the following differential equation

(2.3) d 2_G m dr2 + 1 r dGm dr − µ m2 r2 + 4(m − λ)τ + 4τ2r2 ¶ Gm= −δ(r − R)_r eimθ.

We study this equation in different cases.

Case (1) τ > 0. We shall try to solve (2.3) in the form Y(x) = 1

xω(2τ x

2_),

where we replace r by x. A simple calculation shows that ω solves the following equation

d2_{ω(2τ x)} dx2 + " −1 4− m−λ 2 (2τ x2 + 1−m2 4 (2τ x2₎2 # ω(2τ x2) = 0.

as long as x > 0 and τ > 0. According to formula (2.1) and definitions of functions Mα,β,

Wα,β, this yields the following two linearly independent solutions of (2.3) with zero right hand

side Y1(x) = 1_xW₋m−λ 2 , |m| 2 (2τ x2), Y2(x) = _x1M₋m−λ 2 , |m| 2 (2τ x2)

(5)

with Wronskian W (Y₁, Y₂)(x) = 4 τ x Γ ³ 1−λ 2 + |m|+m 2 ´ . Therefore, Gm = Γ ³ 1−λ 2 + |m|+m2 ´ 4τ rR W−m−λ₂ ,|m|₂ (2τ R 2_)M −m−λ₂ ,|m|₂ (2τ r 2_),

as long as r < R. Summing over m in the infinite series (2.2), we obtain e Kλ,1= _{2πτ rR}1 ∞ X m=−∞ eim(φ−θ)Γ µ 1 − λ 2 + |m| + m 2 ¶ W₋m−λ 2 , |m| 2 (2τ R2)M₋m−λ 2 , |m| 2 (2τ r2), as long as r < R. Using a product formula for Whittaker functions (see p. 201 (15) in Vol. 1 of [13, 14]), one has e K_λ,1 = 1 π ∞ X m=−∞ eim(φ−θ) Z _∞ 0 e−τ (r2+R2) cosh νI_|m|(2τ rR sinh ν) ³ cothν 2 ´_−m+λ dν,

which is symmetric in r and R. Here we used one of the modified Bessel functions given by (see [14]) I_`(x) = 2−2` Γ(1 + `)(2x) −1 2M_0,`(2x). Since ∞ X m=−∞ anIm(b) = exp · b 2 µ a + 1 a ¶¸ , we can write e K_λ,1 = 1 π Z _∞ 0 e−τ (r2+R2) cosh ν ³ cothν 2 ´_λ exp ( τ rR sinh ν " ei(φ−θ) cothν₂ + cothν 2 ei(φ−θ) #) dν.

A straight forward computation yields the following simpler form: e Kλ,1 = _π1eτ (z1w¯1−¯z1w1) Z _∞ 0 e−τ |z1−w−1|2cosh ν ³ cothν 2 ´_λ dν = 1 πe τ (z1w¯1−¯z1w1) Z _∞ 1 e−τ |z1−w1|2y µ y + 1 y − 1 ¶λ 2 dy p y2_{− 1}.

The integral exists as long as λ 6= 1, 3, 5, ... . Finally, formula (22) on p. 139 in Vol. 2 of [14] yields e Kλ,1= 1 πΓ µ −λ 2 + 1 2 ¶ eτ (z1w¯1−¯z1w1) p 2τ |z1− w1|2 Wλ 2,0(2τ |z1− w1| 2_).

This completes the proof of (1) of the conclusion of the lemma.

Case (2) τ < 0. In this case, the relevant differential equation is d2_{ω(−2τ x)} dx2 + " −1 4+ m−λ 2 (−2τ x2₎ + 1−m2 4 (−2τ x2₎2 # ω(−2τ x2) = 0. The two linearly independent solutions of (2.3), with zero right hand side, are

Y1(x) = _x1Wm−λ 2 , |m| 2 (−2τ x2), Y2(x) = _x1Mm−λ 2 , |m| 2 (−2τ x2)

(6)

with Wronskian W (Y₁, Y₂)(x) = −4 τ x Γ ³ 1+λ 2 + |m|−m 2 ´ . Therefore, when τ < 0 e Kλ,1 = _π1eτ (z1w¯1−¯z1w1) Z _∞ 1 e−|τ ||z1−w1|2y µ y − 1 y + 1 ¶λ 2 dy p y2_{− 1}.

The integral exists only if λ 6= −1, −3, −5, . . . . In this case, we obtain e Kλ,1 = _π1Γ µ λ 2 + 1 2 ¶ eτ (z1w¯1−¯z1w1) p 2|τ ||z₁− w₁|2W−λ2,0(2|τ ||z1− w1| 2_),

which is the part (2) of the lemma and the proof of the lemma if therefore complete. ¤ Recall that K_λ,1= 1 2π Z _∞ −∞ ei(t−s)Ke_λ,1(z₁, ¯z₁, w₁, ¯w₁; τ )dτ.

It is much easier to understand the idea behind the computation in this case if we introduce the Heisenberg group structure on the surface ∂Ω1. It is easy to see that the vector fields Z, ¯Z and

T = _∂t∂ and the operator ∆λ,1 are left-invariant with respect to the “Heisenberg translation”:

for (z₁, t) = (x₁, x₂, t) and (w₁, s) = (y₁, y₂, s) ∈ R3_, (z1, t) · (w1, s) = ¡ z1+ w1, t + s + Im(z1w¯1) ¢ .

Actually, the above multiplicative law defines a group structure on R3 which we call the 1-dimensional Heisenberg group H1 with (z1, t)−1 = (−z1, −t). Moreover, the non-isotropic

dilation

(x1, x2, t) 7→ (δx1, δx2, δ2t, ), δ > 0

defines an automorphism on the group H1. The vector fields {Z, ¯Z, T } form a basis of the Lie

algebra of H₁. It is obvious that H₁ is a non-commutative Lie group of step 2, i.e., Z, ¯Z and

their first brackets yield the tangent bundle T H1.

The Heisenberg group and its sub-Laplacian are at the cross-roads of many domains of anal-ysis and geometry: nilpotent Lie group theory, hypoelliptic second order partial differential equations, strongly pseudoconvex domains in complex analysis, probability theory of degen-erate diffusion process, subRiemannian geometry, control theory and semiclassical analysis of quantum mechanics, see e.g., [3], [4], [5], [10], [12], [13] and [18]. The readers can consult the books [6] and [7] for more detailed discussions. We have the following theorem.

Theorem 2.1. ∆λ,1 is invertible as long as λ 6= ±1, ±3, ±5, .... Its inverse is a convolution

operator on the Heisenberg group H1. Its kernel Kλ,1 is induced by

K_λ,1(z1, t) = _2π1₂ Γ¡1+λ₂ ¢Γ¡1−λ₂ ¢ (|z1|2+ it) 1−λ 2 (|z₁|2− it) 1+λ 2 .

Proof. We compute Kλ,1 in two steps. First, p. 216 (16) in Vol. 2 in [14] yields

(2.4) 1 2π Z _∞ 0 ei(t−s)Ke_λ,1(z1, ¯z1, w1, ¯w1; τ )dτ = _2π1₂ µ 1 − λ 2 ¶₋₁ x |x|F µ 1,1 − λ 2 ; 1 − λ 2 + 1; −x 2 ¶ ,

(7)

where we set x = |z1− w1| 2_{+ i(t − s + 2Im(z} 1w¯1)) p |z1− w1|4+ (t − s + 2Im(z1w¯1))2 , and F (a, b; c; y) = Γ(c) Γ(b)Γ(a) ∞ X m=0 Γ(a + m)Γ(b + m) Γ(c + m) ym m!

stands for the hypergeometric function of Gauss. Similarly, (2.5) 1 2π Z ₀ −∞ ei(t−s)Ke_λ,1(z1, ¯z1, w1, ¯w1; τ )dτ = _2π1₂ µ 1 + λ 2 ¶₋₁ 1 x|x|F µ 1,1 + λ 2 ; 1 + λ 2 + 1; − 1 x2 ¶ .

By p. 63(17) in Vol. 2 in [14], one has

F (1, −ν + 1; −ν + 2; −x2) = −1 ν 1 xF µ 1, ν; ν + 1; − 1 x2 ¶ +Γ(2 − ν)Γ(ν) 1 − ν x 2(ν−1)+1_F µ 1 − ν, 0; −ν + 1; − 1 x2 ¶ = −1 ν 1 xF µ 1, ν; ν + 1; − 1 x2 ¶ + Γ(1 − ν)Γ(ν)x2n−1.

If we set ν = 1+λ₂ and add (2.4) and (2.5) together, we obtain

K_λ,1(z1, t) = _2π1₂ Γ¡1+λ₂ ¢Γ¡1−λ₂ ¢ p |z1|4+ t2 " |z1|2+ it p |z1|4+ t2 #_λ .

The conclusion of the theorem follows immediately. ¤

3. Relative fundamental solution for ∆1,1

Now let us turn to a special case: λ = 1. In this case, ∆_1,1= −Z ¯Z.

Hans Lewy (see e.g., [19, 20]) proved that there exists a smooth function f on R3 _{with the}

property that Z(u) = f has no local solution anywhere in R3_{. Let v = − ¯}_{Z(u), then}

∆_1,1(u) = Z[− ¯Z(u)] = Z(v)

has no local solution in general. As we have seen in section 3, we know that the operator ∆1,1

is not invertible because it has an infinite dimensional nullspace. Let S₁ denote the projection operator onto its nullspace. In this section, we shall show that the operator ∆1,1 is invertible

on L2_{\ S}

1L2. Formally, ∆1,1 has no inverse because its “inverse” Kλ,1 is singular at λ = 1.

We shall construct a kerenel, Q_λ when λ 6= 1 such that ∆_λ,1Q_λ= S1. Then we show that the

residue of Qλ and Kλ,1 agree at λ = 1. Hence,

∆1,1 · lim λ→1(Kλ,1− Qλ) ¸ = I − S1.

(8)

Lemma 3.1. The kernel eS1 of the projection onto the nullspace of e∆1,1 is given by (3.1) Se1 = 2τ_π e−τ (|z1| 2_+|w₁_|2₎ X∞ m=0 (2τ z1w¯1)m m! when τ > 0, and eS1 = 0 when τ < 0.

Proof. By Proposition 3.3, the nullspace of e∆1,1 is generated by the functions

eimθe−τ |z1|2_{(2τ |z}

1|2)

m

2, m ∈ Z₊

when τ > 0, and the nullspace of e∆_1,1 is empty set when τ < 0. We now calculate the L2_-norm

of the generating functions, i.e.,

√ 2π ½Z _∞ 0 e−2τ r2(2τ r2)mrdr ¾1 2 = r π τ ½Z _∞ 0 e−s2(s2)msds ¾1 2 = r π 2τ ½Z _∞ 0 e−uum+1du u ¾1 2 = r π 2τ √ m!, m ∈ Z+.

Thus the normalized null solutions of e∆1,1(u) = 0 are

(3.2) Θ−m= r 2τ π 1 √ m!e −τ |z1|2₍√_{2τ z} 1)m, m ∈ Z+

which immediately yields the kernel of the projection, eS1, as given in (3.1). ¤

Now we may take the inverse Fourier transform with respect to the τ -variable to obtain the kernel of the projection onto the nullspace of ∆1,1. It is given by

S1 = 1 2π Z _∞ 0 ei(t−s)τSe1(z1, ¯z1, w1, ¯w1; τ )dτ = 1 π2 Z _∞ 0 τ e−τ (|z1−w1|2−i(t−s+2Im(z1w¯1))_{dτ =} 1 π2_(|z₁_{− w}₁_|2_{− i(t − s + 2Im(z}₁_w_¯₁₎₎2.

Recall from section 2,

A = i 2( ¯w2− z2) = 1 2[|z1− w1| 2_{− i(t − s + 2Im(z} 1w¯1))]

when w2 and z2 are restricted to the boundary ∂Ω1. Therefore, we derived the following

theorem.

Theorem 3.1. On the boundary ∂Ω1 of Ω1= {(z1, z2) ∈ C2 : Im(z2) > |z1|2}, one has

ker(S1) = ker(S) = _π₂_(|z 1

1− w1|2− i(t − s + 2Im(z1w¯1))2.

Here S is the Cauchy-Szeg¨o projection operator from L2_(∂Ω

1) onto H2(∂Ω1).

Next, we shall construct the kernel Qλ such that ∆λ,1Qλ = S1 for λ 6= 1. As in section 2,

we introduce polar coordinates. Set e ∆_λ,1,−m= 1 4 d2 dx2 + 1 4x· d dx − · m2 4x2 − (m + λ)τ + τ 2_r2 ¸ .

(9)

Lemma 3.2. Let m ∈ Z+. Then J−m(r) = √2_π√1 m!(2τ ) m−1 2 r m_e−τ r2 1 − λ satisfies e ∆λ,1,−m(J−m) = Θ−m.

Here Θ−m is defined in formula (3.2).

Proof. To find J−m, we just need to apply the fundamental solution, G−m, of e∆λ,1,−m to Θ−m.

More precisely J_−m(r) =1 2 Z _∞ 0 r 2τ π e−τ x2 √ m!( √ 2τ x)mxdx × Z _∞ 1 e−τ (r2+R2)σI_m ³ 2τ rxpσ2_{− 1}´ µ σ + 1 σ − 1 ¶m+λ 2 dσ √ σ2_{− 1}. (3.3)

We first calculate the x integral which is Z _∞

0

e−τ (1+σ)x2xm+1Im

³

2τ rxpσ2_{− 1}´_dx.

We set x2 _{= s and using page 197 (18) of [14], we obtain}

Z _∞ 0 e−τ (1+σ)x2xm+1Im ³ 2τ rxpσ2_{− 1}´_{dx =} 1 2τr m_eτ r2_(σ−1) (σ + 1)−m2−1(σ−)m2.

Multiplying by the appropriate factors from formula (3.3) one has

J−m(r) =√2_π√1 m!(2τ ) m−1 2 rme−τ r2 Z _∞ 1 (σ + 1)λ2−1(σ − 1)−λ2√ dσ σ2_{− 1} =√2 π 1 √ m!(2τ ) m−1 2 rme−τ r2 1 1 − λ.

This proves the lemma. ¤

Theorem 3.2. When λ 6= 1,

Qλ((z1, t) ∗ (w1, s)) = _π1₂_{1 − λ}1 _|z 1

1− w1|2− i(t − s + 2Im(z1w¯1))

solves

∆λ,1Qλ = S1.

Proof. First of all, we know that

e Qλ = ∞ X m=0 e−im(φ−θ)J−m(r)Θ−m(R) and Qλ = _2π1 Z _∞ 0 ei(t−s)τQeλdτ.

(10)

Now, by Lemma 3.2 1 2π Z _∞ 0 ei(t−s)τJ−m(r)Θ−m(R)dτ = 2 m π2_m! (rR)m 1 − λ Z _∞ 0 τme−τ [r2+R2−i(t−s)]dτ = 2m π2[r 2_{+ R}2_{− i(t − s)]}−(m+1)(rR)m 1 − λ. Finally, one has

Qλ = ∞ X m=0 e−im(φ−θ)2m π2[r2+ R2− i(t − s)]−(m+1) (rR)m 1 − λ = 1 π2_{(1 − λ)} 1 [|z1|2+ |w1|2− i(t − s)] ∞ X m=0 · 2z1w¯1 |z1|2+ |w1|2− i(t − s) ¸_m = 1 π2_{(1 − λ)} 1 |z1− w1|2− i(t − s + 2Im(z1w¯1)),

and the proof of the theorem is therefore complete. ¤

Theorem 3.3. 1. Qλ is a convolution operator on the 1-dimensional Heisenberg group

H1. As such, it is induced by the kernel

Qλ(z1, t) = _π₂_{(1 − λ)}1 _|z 1 1|2− it. 2. limλ→1(Kλ,1− Qλ) = K1,1, where K_1,1(z₁, t) = 1 2π2log µ |z₁|2_{− it} |z1|2+ it ¶ × 1 |z1|2− it . Moreover, one has

∆1,1K1,1 = δ − S1.

Proof. The statement (1) is just a restatement of Theorem 3.3. It remains prove the statement

(2). From Theorem 2.1, we know that

Φλ,1(z1, t) = (|z1|2+ it)− 1−λ 2 (|z₁|2− it)− 1+λ 2 satisfies (3.4) −Z ¯Z(Φ_λ,1) = ∆_λ,1Φ_λ,1− i(λ − 1)∂ ∂t(Φλ,1) = Cλδ − i(λ − 1) ∂ ∂t(Φλ,1) where Cλ = 2π 2 Γ¡1+λ₂ ¢Γ¡1−λ₂ ¢ . Formally, we may rewrite Φλ,1(z1, t) as follows:

Φλ,1(z1, t) = µ |z1|2− it |z1|2+ it ¶1−λ 2

(|z1|2− it)−1 = (|z1|2− it)−1exp

½ 1 − λ 2 log µ |z1|2− it |z1|2+ it ¶¾ .

Here the logarithm of the quotient means the difference of the corresponding logarithms. Now, formal differentiation of formula (4.13) with respect to the λ-variable, we obtain

−Z ¯Z · 1 2π2log µ |z1|2− it |z₁|2_{+ it} ¶ × 1 |z₁|2_{− it} ¸ = δ − C(|z1|2− it)−2.

(11)

Set Φε(z1, t) = ε2 + |z1|2 − it. Later, we shall take the limit in the sense of distribution as ε → 0. Now −Z ¯Z µ C0log Φ_Φ ε ε ¶ = 0. On the other hand,

Z ¯Z µ C0log ¯_Φ_nΦε ε ¶ = −4|z1|2+ 2|z1|2+ 2ε2− 2it Φ2 εΦ¯ε = −2|z1|2+ 2ε2− 2it Φ2 εΦ¯ε = 4ε 2_{− 2(ε}2_{+ |z} 1|2+ it) Φ2 εΦ¯ε = 2 µ 2ε2 Φ2 εΦ¯ε − 1 Φ2 ε ¶ .

Here C0= _2π12. Letting ε → 0 one has −Z ¯Z ·µ C0 log ¯Φ_ε Φε ¶ 1 Φε ¸ = 4C0 µZ R3(|z1| 2_{+ 1 − it)}−2_(|z 1|2+ 1 + it)−1dm(z1)dt ¶ δ − 2C0 Φ2 ε .

Now let us look at the integral Z C×R (|z1|2+ 1 − it)−2(|z1|2+ 1 + it)−1dm(z1)dt = Z C (|z1|2+ 1)−2dm(z1) Z _∞ −∞ (1 − it)−2(1 + it)−1dt.

The first integral is easy. Using polar coordinates, one has Z C (|z1|2+ 1)−2dm(z1) = 2π Z _∞ 0 (r2+ 1)−2rdr = π Z _∞ 1 ds s2 = π. (3.5)

For the second integral, we use the idea in pp. 440-441 of [15] to obtain the result. Assume that γ ≥ 1. Then

Z _∞

0

e−xuxγ−1du = Γ(γ)u−γ for Re(u) > 0. Set u = 1 + it, then

(1 + it)−γΓ(γ) = Z _∞

0

e−ixte−xxγ−1dx = ˆf (t),

where f (x) = e−x_xγ−1 _{for x > 0 and f (x) = 0 for x ≤ 0. Here ˆ}_{f is the Fourier transform of}

the function f . Similarly, for β > 1 we have (1 − it)−βΓ(β) = Z _∞ 0 eixte−xxβ−1dx = Z ₀ −∞ e−ixte−|x||x|β−1dx = ˆg(t),

where g(x) = 0 for x ≥ 0 and g(x) = e−|x|_|x|β−1 _{for x < 0. Using Plancherel’s formula in the}

form _Z 0 −∞ ˆ f (t)ˆg(t)dt = 2π Z ₀ −∞ f (x)g(−x)dx

(12)

one may conclude that Γ(β)Γ(γ) Z _∞ −∞ (1 + it)−β(1 − it)−γdt = 2π Z _∞ 0 e−2xxβ+γ−2dx = 2π Z _∞ 0 e−2xxdx = π 2. Therefore, (3.6) Z _∞ −∞ (1 + it)−β(1 − it)−γdt = π 2Γ(β)Γ(γ). Combining formulae (4.8) and (4.9) with β = 1 and γ = 2, one has

Z C×R (|z1|2+ 1 − it)−2(|z1|2+ 1 + it)−1dm(z1)dt = π 2 2 . Hence, ∆1,1K1,1(z1, t) = ∆1,1 · 1 2π2log µ |z₁|2_{− it} |z1|2+ it ¶ × 1 |z1|2− it ¸ = δ − 1 π2 1 (|z1|2− it)2.

This completes the proof of the theorem. ¤

The above theorem tells us that K1,1 inverts the operator ∆1,1 on the subspace orthogonal

to the boundary value of holomorphic functions. Hence, we have the following theorem. Theorem 3.4. ∆1,1(u) = f is solvable in a neighborhood of a point p ∈ H1 if and only if S(f )

is real-analytic in a neighborhood of the point p.

Remark 3.5. For non-isotropic case, the calculation is much more complicated. In general,

it is almost impossible to calculate an exact form of the fundamental solution K0,1(x, t) for

the sub-Laplacian ∆0,1. However, we can compute an exact form of K0,1(x, t) in some special

cases. For example, when a1 = 1 and a2 = 2 on H2, one has

K0(x, t) = 2 π2_(kxk4_{+ t}2₎ Ã 1 + kxk 2 p kxk4_{+ t}2 !1 2 Ã 1 + (x 2 1+ x22)2 p kxk4_{+ t}2 !₋3 2

where kxk2 = (x2₁+ x2₂)2+ 2(x2₃+ x2₄)2. For detailed calculation, see [8].

4. Fundamental Solution for the Operator ∆λ,2 when n = 1

In this section, we start with computations in Greiner and Stein [21]. This example gives us a very good motivation for the lengthy calculation for general λ and k. Since the operator ∆_λ,2 is invariant under unitary transformation, without loss of generality, we may ass that n = 1. In this case ∆λ,2 = −1₂(Z ¯Z + ¯ZZ) + 4iλ|z1|2_∂t∂ = − ∂2 ∂z1∂ ¯z1 + 2i|z1| 2 ∂ ∂t µ z1_∂z∂ 1 − ¯z1 ∂ ∂ ¯z1 ¶ − 4|z1|6 ∂ 2 ∂t2 + 4iλ|z1|2 ∂ ∂t where Z = Z1 = _∂z∂ 1 + 2i¯z1|z1|2_∂t∂.

(13)

We are going to apply some techniques developed in section 2 to invert the operator ∆_λ,2. Once again, we may assume a fundamental solution, Kλ, of the form

Kλ(z1, w1, t − s) = 1 2π Z _+∞ −∞ ei(t−s)τKeλ(z1, ¯z1; w1, ¯w1; τ )dτ.

Here eKλ satisfies the following

g ∆_λ,2Ke_λ = δ(z1− w1, ¯z1− ¯w1), where e ∆λ,2 = − ∂ 2 ∂z1∂ ¯z1 − 2|z1| 2_τ µ z1_∂z∂ 1 − ¯z1 ∂ ∂ ¯z1 ¶ + 6|z1|6τ2− 4λ|z1|2τ.

We set z1= reiθ and w1 = Reiφ. Let G solve

4 e∆λ,2G = δ(r − R)δ(θ − φ)_r ,

i.e., eK_λ = 4G. Assume G has the form

G = 1 2π ∞ X `=−∞ e−i`θG`(r),

i.e., G`(r) solves the following differential equation

d2_G ` dr2 + 1 r dG_` dr − µ `2 r2 + 8τ `r 2_{+ 16τ}2_r2_{− 16λτ r}2 ¶ G_` = −δ(r − R) r e i`φ_.

Let us start with the following lemma which was first proved by Greiner and Stein in [21]. Lemma 4.1. The fundamental solution, eK_λ, of e∆_λ,2 can be written in the following form

(4.1) Ke_λ= eK_λ,even+ eK_λ,odd, where (4.2) Ke_λ,even= 1 2πe τ (z2 1w¯21−¯z12w12) Z _∞ 1 e−|τ ||z21−w21|2σ µ σ + 1 σ − 1 ¶λ 2sgn(τ ) dσ √ σ2_{− 1}, and e K_λ,odd = 1 π32 eτ (z12w¯12−¯z21w12) Z _∞ 1 e−|τ ||z21−w21|2σ µ σ + 1 σ − 1 ¶λ 2sgn(τ ) ·Erf h³ z1w¯1 p σ + 1sgn(τ ) + ¯z1w1 p σ − 1sgn(τ ) ´ |τ |12 i _dσ √ σ2_{− 1}. (4.3)

These formulas make sense when λ is in some neighborhood of 1, and when τ > 0 we must also assume that λ 6= 1.

Proof. First, the function Erf(x) stands for

(4.4) Erf(x) = Z _x 0 e−t2dt = ∞ X `=0 (−1)`x2`+1 (2` + 1)`! . To obtain G`, we need to solve the following differential equation

x2d 2_y dx2 + x dy dx − [` 2_{− 8τ (` − 2λ)x}4_{− 16τ}2_x8_{]y = 0} _{for x > 0.}

(14)

There are two cases. Case (1) τ > 0. We set Y(x) = 1 x4ω(2τ x4), then ω is a solution of (4.5) d 2_ω du2 + µ −1 4 + −` + 2λ 4u + 4 − `2 16u2 ¶ ω(u) = 0. Therefore, Y1(x) = 1 x2Wλ 2− ` 4, |`| 4 (2τ x4), Y2(x) = 1 x2Mλ 2− ` 4, |`| 4 (2τ x4) are two linearly independent solutions of (4.5) with the following Wronskian

W (Y1, Y2)(x) = 8τ_x 1 Γ ³ 1−λ 2 + `+|`|4 ´ . Hence, when r < R, one has

e K_λ = 1 4πτ r2_R2 ∞ X `=−∞ ei`(φ−θ)Γ µ 1 − λ 2 + ` + |`| 4 ¶ Wλ 2−`4, |`| 4 (2τ R4)Mλ 2−4`, |`| 4 (2τ r4) = 1 2π ∞ X `=−∞ ei`(φ−θ) Z _∞ 0 e−τ (r4+R4) cosh νI|`| 2 (2τ r2R2sinh ν) ³ cothν 2 ´₋` 2+λ dν.

We note that the second formula is symmetric in r and R. Now let use first sum up even integers `. From results of section 2, one has

e Kλ,even = _2π1 Z _∞ 1 ∞ X m=−∞ eim(2φ−2θ)I_|m|(2τ r2R2sinh ν) ³ cothν 2 ´_−m+λ dν = 1 2πe τ (z2 1w¯12−¯z21w12) Z _∞ 1 e−|z21−w21|2τ σ µ σ + 1 σ − 1 ¶λ 2 dσ √ σ2_{− 1}.

Next, let us sum up odd integers `. In this case e Kλ,odd = _2π1 Z _∞ 0 e−τ (r4+R4) cosh ν ³ cothν 2 ´_λ × ∞ X m=0 I_m+1 2(2τ r 2_R2_{sinh ν)}    " ei(2φ−2θ) cothν₂ #_m+1 2 + · cothν 2 ei(2φ−2θ) ¸_m+1 2   dν. Using the formula

I_m+1 2(x) = 1 √ π ³ x 2 ´_m+1 2 1 m! Z ₁ −1 e−xt(1 − t2)mdt

we obtain the following sum

∞ X m=0 am+12I m+1₂(x) = 1 √ πe 1 2(ax+xa) n Erf¡ r x 2a+ r ax 2 ¢ − Erf¡ r x 2a− r ax 2 ¢o . Therefore, ∞ X m=0 £ am+12 + a−m− 1 2 ¤ I_m+1 2(x) = 2 √ πe 1 2(ax+ x a)Erf ¡r _x 2a + r ax 2 ¢ ,

(15)

which applied to our sum, yields e K_λ,odd = √1 π3e τ (z2 1w¯21−¯z12w12) Z _∞ 1 e−|z21−w12|2τ σ µ σ + 1 σ − 1 ¶λ 2 ×Erf[(z1w¯1 √ σ + 1 + ¯z1w1 √ σ − 1)√τ ]√ dσ σ2_{− 1}.

Thus we derived (4.1), (4.2) and (4.3) if τ > 0 and λ 6= 1 and λ in some neighborhood of 1. Case (2) τ < 0. We set Y(x) = 1 x4ω(−2τ x4), then ω is a solution of d2_ω du2 + µ −1 4+ ` − 2λ 4u + 4 − `2 16u2 ¶ ω(u) = 0 when u > 0. The two linearly independent solutions are given by

Y₁(x) = 1 x2W−λ 2+ ` 4, |`| 4 (−2τ x4), Y₂(x) = 1 x2M−λ 2+ ` 4, |`| 4 (−2τ x4) with the following Wronskian

W (Y1, Y2)(x) = −8τ_x 1 Γ ³ 1+λ 2 +|`|−`4 ´ . Hence e Kλ = _2π1 ∞ X m=−∞ eim(2φ−2θ) Z _∞ 1 e−|τ |(r4+R4) cosh ν ×I|m| 2 (2|τ |r2R2sinh ν) ³ cothν 2 ´m 2−λ dν.

Calculating as in the case τ > 0, we obtain (4.1), (4.2) and (4.3) in the case τ < 0 and λ 6= 1 and λ in some sufficiently small neighborhood of 1. This finish the proof of the lemma. ¤ Let us first calculate the “odd” part of the kernel. Based on Lemma 1.1, we have the following result

Lemma 4.2. The kernel

K_λ,odd+ = 1 2π

Z _∞

0

ei(t−s)τKeλ,odddτ

can be written as an infinite series as follows: K_λ,odd+ = √1 2π52 ∞ X `=0 (−1)`₂` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1 Γ¡` +3₂¢ ` − λ₂ −m₂ + 1 µ X d ¶_`+3 2 F µ ` + 3 2, ` − λ 2 − m 2 + 1; ` − λ 2 − m 2 + 2; −X 2 ¶ ;

and the kernel

K_λ,odd− = 1 2π

Z ₀

−∞

(16)

can be written as an infinite series as follows: K_λ,odd− = √ 2 π52 ∞ X `=0 (−1)` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1 2`_Γ¡_{` +} 3 2 ¢ d`+3₂ 1 m + λ + 1 1 X`+32 F µ ` + 3 2, m 2 + λ 2 + 1 2; m 2 + λ 2 + 3 2; − 1 X2 ¶ , where X = |z12− w21|2+ i[t − s + 2Im(z21w¯21)] £ |z2 1− w12|4+ (t − s + 2Im(z12w¯21))2 ¤1 2 . and d =£|z₁2− w2₁|4+ (t − s + 2Im(z2₁w¯₁2))2¤12 _. Proof. First note that

Erf(x) = ∞ X `=0 (−1)`x2`+1 `!(2` + 1) .

We first assume that τ > 0. Then one has ¡ z1w¯1 √ σ + 1√τ + ¯z1w1 √ σ − 1√τ¢2`+1 = 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1 √ τ )m(¯z1w1 √ τ )2`−m+1(σ + 1)m2(σ − 1) (2`−m+1) 2 . Thus Erf£¡z1w¯1 √ σ + 1 + ¯z1w1 √ σ − 1¢ √τ¤ = ∞ X `=0 (−1)` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1τ 2`+1 2 (σ + 1) m 2(σ − 1)`− m 2+ 1 2.

First we integrate out the σ-variable with respect to dσ. Denote Σ(σ) = τ`+12(σ + 1) (α+m−1) 2 (σ − 1)`− m 2− λ 2.

From page 139(22) in Vol. 1 of [14], with a = b = 1, 2µ − 1 = (λ+m−1)₂ and 2ν − 1 = ` −m 2 −λ2, one has µ = (λ + m + 1) 4 , ν = ` 2 − λ 4 − m 4 + 1 2. Hence, Z _∞ 1 e−τ |z12−w21|2σΣ(σ)dσ = Γ µ ` −λ 2 − m 2 + 1 ¶ 2`2−14¡τ |z2 1− w21|2 ¢₋` 2−34 _W λ 2+ m 2− ` 2− 1 2, ` 2+ 1 4 ¡ τ |z2₁− w₁2|2¢· τ`+12.

Here W_κ,µ is the Whittaker functions (see [13,14] and [23]). Next we need to calculate the following: Z _∞ 0 eitτ +τ (z12w¯12−¯z21w12)τ`2−14W_λ 2+ m 2− ` 2− 1 2, ` 2+ 1 4 ¡ τ |z₁2− w₁2|2¢dτ = Z _∞ 0 e−τ {−i[t+2Im(z21w¯12)]}τ2`−14W_λ 2+m2−`2−12,2`+14 ¡ τ |z₁2− w₁2|2¢dτ. (4.6)

(17)

Using page 216(16) in Vol. 1 of [14] with ν = ` 2+ 3 4, κ = λ 2 + m 2 − ` 2− 1 4, µ = ` 2 + 1 4, the equation (4.6) equals to

Γ¡` + 3₂¢Γ(1) Γ¡` − λ₂ − m₂ + 2¢ ¡ 2|z₁2− w2₁|2¢2`+34 [|z2 1− w12|2− i(t + 2Im(z12w¯12))]`+ 3 2 F ³ `+3 2, `− λ 2− m 2 +1; `− λ 2− m 2+2; − X ¯ X ´ where X = _£ |z12− w21|2+ i[t − s + 2Im(z12w¯21)] |z2 1− w12|4+ (t − s + 2Im(z12w¯21))2 ¤1 2 .

Note that |X| = 1 and

X ¯ X = X2 |X|2 = X2, 1 ¯ X = X. We set d = |(z₁2, t)(w2₁, s)−1| =£|z2₁− w₁2|4+ (t − s + 2Im(z₁2w¯2₁))2¤12_.

Then we get from the representation of Σ(σ): 2`+12Γ ¡ ` + 3₂¢ d`+32 X`+32 ` − λ₂ − m₂ + 1F ³ ` + 3 2, ` − λ 2 − m 2 + 1; ` − λ 2 − m 2 + 2; −X 2´_.

Therefore, for τ > 0, we have

K_λ,odd+ = 1 2π52 ∞ X `=0 (−1)` `!(2` + 1) 2`+1_X m=0 2`+1₂_{(2` + 1)!} m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1 Γ¡` + 3₂¢ d`+32 X`+32 ` − λ₂ −m₂ + 1F ³ ` + 3 2, ` − λ 2 − m 2 + 1; ` − λ 2 − m 2 + 2; −X 2´_.

Now let us turn to the case τ < 0. In this case, one has Erf¡z1w¯1 √ σ − 1|τ |12 + ¯z₁w₁√σ + 1|τ |12¢ = ∞ X `=0 (−1)` `!(2` + 1) ¡ z₁w¯₁√σ − 1|τ |12 + ¯z₁w₁ √ σ + 1|τ |12 ¢_2`+1 = ∞ X `=0 (−1)` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)! × (z₁w¯₁)m(¯z₁w₁)2`−m+1|τ |`+12(σ − 1) m 2(σ + 1) (2`−m+1) 2 . (4.7)

Thus we need to calculate (4.8) |τ |`+12 Z _∞ 1 e−|τ ||z12−w21|2σ(σ − 1) (m+λ−1) 2 (σ + 1) (2`−m−λ) 2 dσ.

Again, from page 139(22) in Volume I of [14] with a = b = 1, 2µ − 1 = ` − m₂ − λ₂ and 2ν − 1 = m 2 +λ2 −12, we get µ = ` 2 − m 4 − λ 4 + 1 2, ν = m 4 + λ 4 + 1 4.

(18)

Therefore, the equation (4.8) equals |τ |`+12Γ³ m 2 + λ 2 + 1 2 ´ 2`2− 1 4 ¡ |τ ||z2₁− w₁2|2¢−`2− 3 4_W_` 2−m2−λ2+14,2`+14 ¡ 2|τ ||z₁2− w2₁|2¢.

Next, note that Z ₀ −∞ e−itτ +τ (z21w¯21−¯z12w21)f (|τ |)dτ = Z _∞ 0 e−s(it+z21w¯21−¯z21w21)f (s)ds,

thus, we have to evaluate the following integral:

(4.9) Z _∞ 0 e−s(it+z21w¯21−¯z21w21)s2`−14W_` 2− m 2− λ 2+ 1 4, ` 2+ 1 4 ¡ 2s|z2₁− w2₁|2¢ds.

Now we use equation (10) in page 216, Volume I of [14] with ν = `₂ +3₄, κ = `₂ −m₂ −λ₂ + 1₄ and µ = `₂+ 1₄. Then integral (4.9) becomes

Γ¡` + 3₂¢Γ(1) Γ¡m₂ +λ₂ +3₂¢ ¡ 2|z2 1− w12|2 ¢` 2+ 3 4 [|z2 1− w12|2− i(t + 2Im(z12w¯21))]`+ 3 2 F ³ ` +3 2, m 2 + λ 2 + 1 2; m 2 + λ 2 + 3 2; − 1 X2 ´ where X = _£ |z12− w21|2+ i[t − s + 2Im(z12w¯21)] |z2 1− w12|4+ (t − s + 2Im(z12w¯21))2 ¤1 2 . Thus we have, if τ < 0, 2`+12Γ ¡ ` +3 2 ¢ d−`−32 2 m + λ + 1 1 X`+32 F ³ ` +3 2, m 2 + λ 2 + 1 2; m 2 + λ 2 + 3 2; − 1 X2 ´ .

Plugging this into (4.7), we have, if τ < 0: 1 2π52 ∞ X `=0 (−1)` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1 2`+12Γ ¡ ` +3₂¢ d`+32 2 m + λ + 1 1 X`+32 F µ ` + 3 2, m 2 + λ 2 + 1 2; m 2 + λ 2 + 3 2; − 1 X2 ¶ .

This completes the proof of the lemma. ¤

Next we simplify the sum of τ > 0 and of τ < 0. Set

a = ` + 3 2, b = ` − λ 2 − m 2 + 1, c = ` − λ 2 − m 2 + 2.

(19)

Then we have F µ ` +3 2, ` − λ 2 − m 2 + 1; ` − λ 2 − m 2 + 2; −X 2 ¶ = Γ ¡ ` − λ₂ −m₂ + 2¢Γ¡−λ₂ −m₂ −1₂¢ Γ¡` − λ₂ −m₂ + 1¢Γ¡−λ₂ −m₂ +1₂¢ (X 2₎−`−3₂_F µ ` +3 2, λ 2 + m 2 + 1 2; λ 2 + m 2 + 3 2; − 1 X2 ¶ +Γ ¡ ` − λ₂ −m₂ + 2¢Γ¡λ₂ + m₂ +1₂¢ Γ¡` +3₂¢Γ(1) (X 2₎−`+λ₂+m₂−1_F µ ` −λ 2 − m 2 + 1, 0; − λ 2 − m 2 + 1 2; − 1 X2 ¶ = ` − λ 2 −m2 + 1 −λ₂ −m₂ −1₂ (X 2₎−`−3 2F µ ` + 3 2, λ 2 + m 2 + 1 2; λ 2 + m 2 + 3 2; − 1 X2 ¶ +Γ ¡ ` − λ₂ −m₂ + 2¢Γ¡λ₂ + m₂ +1₂¢ Γ¡` +3₂¢ (X 2₎−`+λ₂+m₂−1_. This gives us X`+32 ` −λ₂ −m₂ + 1F ³ ` + 3 2, ` − λ 2 − m 2 + 1; ` − λ 2 − m 2 + 2; −X 2´ + _λ 1 2 +m2 +12 X−`−32F ³ ` + 3 2, λ 2 + m 2 + 1 2; λ 2 + m 2 + 3 2; − 1 X2 ´ =Γ ¡ ` − λ₂ −m₂ + 1¢Γ¡λ₂ +m₂ +1₂¢ Γ¡` +3 2 ¢ X−`+λ+m−12.

Therefore, one has

K_λ,odd = √1 2π52 ∞ X `=0 (−1)`₂` `!(2` + 1) 2`+1_X m=0 (2` + 1)! m!(2` − m + 1)!(z1w¯1) m_(¯_z 1w1)2`−m+1 ·Γ µ ` − λ 2 − m 2 + 1 ¶ Γ µ λ 2 + m 2 + 1 2 ¶ d−`−32X−`+λ+m−12.

We may state the above result as the following lemma.

Lemma 4.3. The “odd” part K_λ,odd of the kernel K_λ can be written as the following infinite series:

Kλ,odd = K_λ,odd+ + K_λ,odd−

= √1 2π52 ∞ X `=0 (−2)` `!(2` + 1) 2`+1_X m=0 Γ(2` + 2) Γ(m + 1)Γ(2` − m + 2)(z1w¯1) m_(¯_z 1w1)2`−m+1 ·Γ µ ` − λ 2 − m 2 + 1 ¶ Γ µ λ 2 + m 2 + 1 2 ¶ d−`−32X−`+λ+m−12.

Now let us calculate the kernel eKλ,even. As we know

e Kλ,even = _2π1 eτ (z 2 1w¯21−¯z21w21) Z _∞ 1 e−|τ ||z12−w21|2σ µ σ + 1 σ − 1 ¶λ 2sgn(τ ) dσ √ σ2_{− 1} = Γ ¡₁ 2− λ2sgn(τ ) ¢ 4π eτ (z2 1w¯21−¯z12w12) (2τ |z2 1 − w21|2) 1 2 Wλ 2sgn(τ ),0(2τ |z 2 1 − w21|2).

(20)

We have the following lemma.

Lemma 4.4. The “even” part of the kernel Kλ can be written in the following form:

Kλ,even = 1 4π2 Γ¡1+λ 2 ¢ £ |z2 1 − w21|2− i(t − s + 2Im(z21w¯12)) ¤1+λ 2 · Γ ¡_1−λ 2 ¢ £ |z2 1 − w21|2+ i(t − s + 2Im(z12w¯21)) ¤1−λ 2 .

Proof. We compute the kernel Kλ,even in two steps. Applying page 216(16) in [14], one has

(4.10) 1 2π Z _∞ 0 ei(t−s)τK_λ,even(τ ) dτ = 1 2π2 2 1 − λ X |X|F µ 1,1 − λ 2 ; 1 − λ 2 + 1; −X 2 ¶ , where X = _£|z12− w21|2+ i(t − s + 2Im(z12w¯21)) |z2 1− w12|4+ (t − s + 2Im(z12w¯21))2 ¤1 2 , and F (a, b; c; z) = Γ(c) Γ(a)Γ(b) ∞ X `=0 Γ(a + `)Γ(b + `) Γ(c + `) z` Γ(` + 1) stands for the hypergeometric function of Gauss.

Similarly, one has

(4.11) 1 2π Z ₀ −∞ ei(t−s)τK_λ,even(τ ) dτ = 1 2π2 2 1 + λ 1 X|X|F µ 1,1 + λ 2 ; 1 + λ 2 + 1; − 1 X2 ¶ .

Now, page 63(17) in [14] yields

F (1, 1 − ν; 2 − ν; −z2) = −1 ν 1 zF ³ 1, ν; ν + 1; − 1 z2 ´ +Γ(2 − ν)Γ(ν) 1 − ν z 2(ν−1)+1_F³_{1 − ν, 0; 1 − ν; −}1 z2 ´ = −1 ν 1 zF ³ 1, ν; ν + 1; − 1 z2 ´ + Γ(1 − ν)Γ(ν)z2ν−1. If we set ν = 1+λ

2 and add (4.10) and (4.11), we obtain

Kλ,even =_4π1₂ Γ¡1+λ₂ ¢Γ¡1−λ₂ ¢ £ |z2 1− w12|4+ (t − s + 2Im(z12w¯21))2 ¤1 2 ×n |z 2 1 − w21|2+ i(t − s + 2Im(z12w¯21)) £ |z2 1− w12|4+ (t − s + 2Im(z12w¯12))2 ¤1 2 o_λ . (4.12)

This completes the proof of the lemma. ¤

Theorem 4.1. The fundamental solution, Kλ,2, of ∆λ,2 can be written in the following form

Kλ,2 = Kλ,even+ Kλ,odd= Kλ,even+ K_λ,odd+ + K_λ,odd− ,

where K_λ,even is given by (4.12) and K_λ,odd is given in Lemma 4.3.

Now let us turn to a special case, λ = 0. One has (4.13) K0,even= _4π1₂ Γ¡1₂¢Γ¡1₂¢ £ |z2 1− w12|4+ (t − s + 2Im(z12w¯21)) ¤1 2 = 1 4πd.

(21)

Next, we calculate K_0,odd. K_0,odd = √1 2π52 ∞ X `=0 (−1)`₂` `!(2` + 1) 2`+1_X m=0 Γ(2` + 2) Γ(m + 1)Γ(2` − m + 2)(z1w¯1) m_(¯_z 1w1)2`−m+1 ·Γ µ m 2 + 1 2 ¶ Γ ³ ` − m 2 + 1 ´ d−`−32X−`+m−12.

Let us simplify the above equation. Using the formula Γ(2z) = π−1222z−1Γ(z)Γ µ z +1 2 ¶ , we conclude that Γ(2` + 2) Γ(` + 1) = π −1 222`+1Γ µ ` +3 2 ¶ , and Γ¡m₂ +1₂¢Γ¡` −m₂ + 1¢ Γ(m + 1)Γ(2` − m + 2) = π12 2m_Γ¡m 2 + 12 ¢ · π 1 2 22`−m+1_Γ¡_{` −} m 2 +32 ¢ . Therefore, Γ(2` + 2)Γ¡m₂ +1₂¢Γ¡` − m₂ + 1¢ `!Γ(m + 1)Γ(2` − m + 2) = π12Γ ¡ ` + 3₂¢ Γ¡m₂ + 1¢Γ¡` −m₂ +3₂¢ . Note that Γ¡` + 3 2 ¢ 2` + 1 = Γ¡` + 1 2 ¢ 2 . Hence we have K0,odd = √ 2 4π2 ∞ X `=0 (−2)`Γ µ ` + 1 2 ¶2`+1_X m=0 1 Γ¡m 2 + 1 ¢ Γ¡` −m 2 +32 ¢ ·d−`−32(z₁w¯₁)m(¯z₁w₁)2`−m+1X−`+m− 1 2. Denote A = 1 2 ¡ |z₁|4+ |w₁|4+ it¢. Then we have d = £2A − 2¯z₁2w₁2¤12£_{2 ¯}_{A − 2z}2 1w¯21 ¤1 2 = 2A12A¯12 Ã 1 − · ¯ z1w1 A12 ¸₂!1 2 Ã 1 − · z1w¯1 ¯ A12 ¸₂!1 2 , i.e., d = 2A12A¯ 1 2(1 − P2) 1 2(1 − ¯P2) 1 2 where we set P = z¯1w1 A12 .

Clearly, |P| ≤ 1. Moreover, |P| = 1 if and only if t = 0 and z1 = eiθw1. In fact, we have the

following lemma.

Lemma 4.5. Assume w1 6= 0. Then P is close to ±1 if and only if (z1, t) is close to (±w1, s),

(22)

Proof. Since |P| ≤ 1, |1 − P| ≥ 1 2|1 − P 2_{| ≥} 1 2(1 − |P| 2_{) =} 1 2 Ã 1 − 2|z1|2|w1|2 |z₁|4_{+ |w} 1|4 ·p 1 1 + γ2 ! , where γ = t − s |z1|4+ |w1|4. Note that 2|z1|2|w1|2 |z1|4+ |w1|4 ≤ 1 and 1 p 1 + γ2 ≤ 1.

It is easy to see that the first inequality in the above implies 1 − 2 (|z1|/|w1|) 2 1 + (|z1|/|w1|)4 ≤ 2|1 − P|, 1 − (1 + γ2)−12 ≤ 2|1 − P|. If 2|1 − P| < δ < 1, then ||z1|2− |w1|2| < 3|w1|2· √ δ(1 − δ)−1 and |t − s| < 35|w1|2· √ δ(1 − δ)−2,

i.e., (|z1|, t) must be arbitrarily close to (|w1|, s) by choosing |P| sufficiently close to 1. The

converse is clear, i.e.,s (|z1|, t) is near (|w1|, s) implies that |P| is near 1. Finally set z1= reiθ,

w1 = Reiφ. Then

1 − P = 1 − rR

|A|12

ei(θ−φ).

Therefore, P is close to 1 if and only if rR

|A|12 is close to 1 and φ is close to θ, i.e., P is close to

1 if and only if (z₁, t) is close to (±w₁, s). This completes the proof of the lemma. ¤ From the definition of A, d, P and X, we also know that

Xd = 2A(1 − P2) i.e., X = Xd d = A12 ¯ A12 (1 − P2₎1₂ (1 − ¯P2₎12 .

Hence one has

d−`−32(z1w¯1) m (¯z1w1)m Xm(¯z₁w₁)2`+1X−`−12 = 1 d (¯z₁w₁)2`+1 (Xd)`+12 (z₁w¯₁)m (¯z1w1)m Xm = 1 d (¯z1w1)2`+1 (2A)`+12(1 − P2)`+12 Ã z1w¯1 ¯ z₁w₁ A12 ¯ A12 (1 − P2₎1₂ (1 − ¯P2₎12 !_m = 1 d2 −`−1₂ µ P2 1 − P2 ¶`+1 2 Ã ¯ P (1 − ¯P2₎12 ·(1 − P 2₎1 2 P !_m . Denote Q = P (1 − P2₎12 .

(23)

Then d−`−32(z1w¯1) m (¯z1w1)mX m_(¯_z 1w1)2`+1X−`− 1 2 = 1 d2 −`−1₂ µ P2 1 − P2 ¶`+1 2 Ã ¯ P (1 − ¯P2₎12 ·(1 − P 2₎1₂ P !_m = 1 d2 −`−1 2Q2`+1 µ ¯_Q Q ¶_m . It follows that K_0,odd = √ 2 4π2_d ∞ X `=0 (−1)`Γ µ ` +1 2 ¶ Q2`+1 2`+1_X m=0 1 Γ¡m₂ + 1¢Γ¡` − m₂ +3₂¢ µ ¯_Q Q ¶_m = K(1)+ K(2),

where K(1) _{is the sum over m = 2k + 1 and K}(2) _{is the sum over m = 2k.}

Case 1. m = 2k + 1. In this case, one has

K(1) = Q¯ 4π2_d ∞ X `=0 (−Q2)`Γ µ ` +1 2 ¶_X` k=0 1 Γ¡k +3 2 ¢ Γ (` − k + 1) · ¯_Q Q ¸_2k = Q¯ 4π2_d ∞ X k=0 · ¯_Q Q ¸_2k 1 Γ¡k + 3 2 ¢ ∞ X `=k (−Q2₎`_Γ¡_{` +} 1 2 ¢ Γ (` − k + 1) = Q¯ 4π2_d ∞ X k=0 (− ¯Q2₎k Γ¡k + 3₂¢ ∞ X `=0 (−Q2₎`_Γ¡_{` + k +} 1 2 ¢ Γ (` + 1) = Q¯ 4π2_d ∞ X k=0 (− ¯Q2₎k k +1₂ F µ k + 1 2, 1; 1; −Q 2 ¶ = Q¯ 4π2_d ∞ X k=0 (− ¯Q2₎k k +1₂ (1 + Q 2₎−k−1₂_.

But we know that

1 + Q2= 1 + P2 1 − P2 = 1 1 − P2, hence, K(1) = Q¯ 4π2_d(1 − P2) 1 2 ∞ X k=0 Γ¡k + 1₂¢ Γ¡k + 3 2 ¢Γ(k + 1) k! µ − ¯Q2 1 + Q2 ¶_k = Q¯ 4π2_d(1 − P 2₎1 2Γ ¡₁ 2 ¢ Γ¡3₂¢ Γ¡3₂¢ Γ¡1₂¢ ∞ X k=0 Γ¡k + 1₂¢ Γ¡k + 3₂¢ Γ(k + 1) k! µ − ¯Q2 1 + Q2 ¶_k = 2 ¯Q 4π2_d(1 − P2) 1 2F µ 1 2, 1; 3 2; − ¯ Q2 1 + Q2 ¶ . Since ¯ Q(1 − P2)12 = ¯ P (1 − ¯P2₎12 (1 − P2)12 = ¯ P(1 − P2)12 (1 − ¯P2₎12 and ¯ Q 1 + ¯Q2 = ¯Q2(1 − ¯P2) = ¯ P2(1 − P2) 1 − ¯P2 ,

(24)

it follows that K(1)= P¯ 4π2_d (1 − P2)12 (1 − ¯P2₎12 F µ 1 2, 1; 3 2; − ¯ P2(1 − P2) 1 − ¯P2 ¶ . By a formula in p.38 of [23], 2zF µ 1 2, 1; 3 2; z 2 ¶ = log µ 1 + z 1 − z ¶ ,

setting z = iw, one has

2iwF µ 1 2, 1; 3 2; −w 2 ¶ = log µ 1 + iw 1 − iw ¶ . Therefore, (4.14) K(1)= −i 4π2_dlog        1 + iP(1−P¯ 2) 1 2 (1− ¯P2₎12 1 − iP(1−P¯ 2) 1 2 (1− ¯P2₎12        .

Case 2. m = 2k. In this case, one has

K(2) = Q 4π2_d ∞ X `=0 (−Q2)`Γ µ ` +1 2 ¶_X_` k=0 1 Γ¡` − k +3₂¢Γ (k + 1) · ¯_Q Q ¸_2k = Q 4π2_d ∞ X k=0 · ¯_Q Q ¸_2k 1 k! ∞ X `=k (−Q2)`Γ¡` + 1₂¢ Γ¡` − k +3₂¢ = Q 4π2_d ∞ X k=0 (− ¯Q2₎k k! ∞ X `=0 (−Q2₎`_Γ¡_{` + k +} 1 2 ¢ Γ¡` +3 2 ¢ = Q 4π2_d ∞ X k=0 (− ¯Q2₎k k! Γ¡k + 1₂¢ Γ¡3 2 ¢ F µ k + 1 2, 1; 3 2; −Q 2 ¶ = Q 4π2_d ∞ X k=0 (− ¯Q2₎k k! Γ¡k + 1 2 ¢ √ π Z ₁ 0 (1 − ξ)−12(1 + Q2ξ)−k− 1 2dξ = Q 4π2_d Z ₁ 0 (1 − ξ)−12(1 + Q2ξ)−12 ( 1 √ π ∞ X k=0 Γ¡k +1₂¢ k! µ − ¯Q2 1 + Q2_ξ ¶_k) dξ = Q 4π2_d Z ₁ 0 (1 − ξ)−12(1 + Q2ξ)− 1 2F µ 1 2, 1; 1; − ¯ Q2 1 + Q2_ξ ¶ dξ = Q 4π2_d Z ₁ 0 (1 − ξ)−12(1 + Q2ξ)− 1 2 µ 1 + Q¯2 1 + Q2_ξ ¶₋1 2 dξ = Q 4π2_d Z ₁ 0 (1 − ξ)−12(1 + ¯Q2+ Q2ξ)−12dξ.