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Harmonic functions which vanish on a cylindrical surface
J. Math. Anal. Appl. 433 (2016) 1870–1882
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Harmonic functions which vanish on a cylindrical surface Stephen J. Gardiner ∗ , Hermann Render School of Mathematics and Statistics, University College Dublin, Belfield, Dublin 4, Ireland
a r t i c l e
i n f o
Article history: Received 6 July 2015 Available online 3 September 2015 Submitted by D. Khavinson Keywords: Harmonic continuation Green function Cylindrical harmonics
a b s t r a c t Suppose that a harmonic function h on a finite cylinder vanishes on the curved part of the boundary. This paper answers a question of Khavinson by showing that h then has a harmonic continuation to the infinite strip bounded by the hyperplanes containing the flat parts of the boundary. The existence of this extension is established by an analysis of the convergence properties of a double series expansion of the Green function of an infinite cylinder beyond the domain itself. © 2015 Elsevier Inc. All rights reserved.
1. Introduction The Schwarz reflection principle gives a formula for extending a harmonic function h on a domain Ω ⊂ RN through a relatively open subset E of the boundary ∂Ω on which h vanishes, provided E lies in a hyperplane (and is a relatively open subset thereof). By the Kelvin transformation there is a corresponding result where E lies in a sphere. When N = 2, such a reflection principle holds also when E is contained in an analytic arc (see Chapter 9 of [7]). However, when N ≥ 3 and N is odd, Ebenfelt and Khavinson [4] (see also [6] and Chapter 10 of [7]) have shown that a reflection law can only hold when the containing real analytic surface is either a hyperplane or a sphere. Now let N ≥ 3, let Ωa be the finite cylinder B × (−a, a), where B is the open unit ball in RN −1 and a > 0, and let Ω = B × R. Dima Khavinson raised the following question with the authors: Question. Given a harmonic function h on Ω which vanishes on ∂Ω, does it follow that h must have a harmonic extension to RN ? Although the above results show that there can be no pointwise reflection formula for such an extension, this paper will establish that such an extension does indeed exist. * Corresponding author. E-mail addresses: [email protected] (S.J. Gardiner), [email protected] (H. Render). http://dx.doi.org/10.1016/j.jmaa.2015.08.077 0022-247X/© 2015 Elsevier Inc. All rights reserved.
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We will use the notation x = (x , xN ) to denote a typical point of RN = RN −1 × R. Theorem 1. Let h be a harmonic function on Ωa which continuously vanishes on ∂B × (−a, a). Then h has a harmonic extension h to RN −1 × (−a, a). Further, for any b ∈ (0, a), there is a constant c, depending on a, b, N and h, such that 1−N/2 h(x) ≤ c x
(x ∈ RN −1 \B , |xN | < b).
(1)
It is a classical fact that the Green function for a three-dimensional infinite cylinder can be represented as a double series involving Bessel functions and Chebychev polynomials: see, for example, p. 62 of Dougall [3] or p. 78 of Carslaw [2]. Our approach to proving Theorem 1 involves establishing such a double series representation in N dimensions and analysing its convergence properties outside the cylinder. 2. Preparatory material Let Jν and Yν denote the usual Bessel functions of order ν ≥ 0 of the first and second kinds (see Watson [12]). Thus these functions both satisfy the differential equation z2
dy d2 y + (z 2 − ν 2 )y = 0. +z 2 dz dz
(2)
Further, let (jν,m )m≥1 denote the sequence of positive zeros of Jν , in increasing order. We collect below some facts that we will need. Jν+1 (z) d ν d Jν (z) z Jν (z) = z ν Jν−1 (z) and =− . ν dz dz z zν 2νJν (z) and Jν−1 (z) − Jν+1 (z) = 2Jν (z). (ii) Jν−1 (z) + Jν+1 (z) = z 2 (t > 0). (iii) Jν (t)Yν (t) − Yν (t)Jν (t) = πt −1/2 2 2 2 t2 − ν 2 (iv) {Jν (t)} + {Yν (t)} < (t > ν ≥ 12 ). π ν t 1 (t ≥ 0). (v) |Jν (t)| ≤ 2 Γ(ν + 1) (vi) j0,m ≥ (m + 3/4)π. (vii) jν,m ≥ j0,m + ν. (viii) |Jν (t)| < ν −1/3 (ν > 0, t ≥ 0). (ix) |Jν (t)| ≤ min{1, t−1/3 } (t > 0). 2 2 2 (0 ≤ ν ≤ 12 , t > 0). (x) {Jν (t)} + {Yν (t)} < πt
Lemma 2. (i)
Proof. (i) and (ii). See p. 45 of Watson [12]. (iii) See p. 76, (1) of [12]. (iv) See p. 447, (1) of [12]. (v) See p. 49, (1) of [12]. (vi) See p. 489 of [12]. (vii) See Laforgia and Muldoon [8], (2.4). (viii) See Landau [9]. −1/3 (ix) We know from p. 406, (10) of [12] that |Jν| ≤ 1, and from [9] that . |Jν (t)| ≤ t 2 2 (x) By Section 13.74 of [12] the function t → t {Jν (t)} + {Yν (t)} is non-decreasing when 0 ≤ ν ≤ 12 , and has limit 2/π at ∞. 2
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Some consequences of Lemma 2 are noted below. 2
2 Lemma 3. (i) jν,m {Jν+1 (jν,m )} ≥
2 2 jν,m − ν 2 (m ≥ 1). π
(ii) jν,m ≥ (m + 3/4)π + ν. (iii) |Jν (jν,m s)| ≤ jν,m (1 − s) (0 ≤ s ≤ 1). 1 2 (s ≥ 1). (iv) |Jν (jν,m s)| ≤ π ms
Proof. (i) By the second equation of Lemma 2(i), Jν (jν,m ) = −Jν+1 (jν,m ). If ν ≥ 12 , then, by (iii), (iv) and (vii) of that result, 2 2 jν,m {Jν+1 (jν,m )} = jν,m {Jν (jν,m )} =
4
2
2
π 2 {Yν (jν,m )}
2
>
2 2 jν,m − ν 2 . π
If 0 ≤ ν < 12 , we use part (x) of Lemma 2 in place of part (iv). (ii) This follows from Lemma 2(vi), (vii). (iii) By the mean value theorem, |Jν (jν,m s)| ≤ jν,m (1 − s) sup |Jν (jν,m t)|
(1/2 ≤ s ≤ 1).
1/2≤t≤1
We know from Lemma 2(i), (ii) that Jν = (Jν−1 − Jν+1 )/2 when ν ≥ 1 and Jν (z) = (ν/z)Jν (z) − Jν+1 (z) otherwise. In either case we see from Lemma 2(ix) and part (ii) of this lemma that |Jν (jν,m t)| ≤ 1 when 1/2 ≤ t ≤ 1, so the desired inequality is established when 1/2 ≤ s ≤ 1. It is clearly also valid when 0 ≤ s < 1/2, since |Jν | ≤ 1, and jν,m ≥ 2 by part (ii). (iv) By (ii) we have (jν,m s)2 ≥ (mπs)2 + ν 2 when s ≥ 1. If ν ≥ 1/2, then we see from Lemma 2(iv) that 2
2
{Jν (jν,m s)} <
π 2 ms
(s ≥ 1),
as desired. If 0 ≤ ν < 1/2, we can instead appeal to Lemma 2(x). 2 The following result is taken from Sections 18.24–18.26 of [12]. Proposition 4. Let f : (0, 1] → R be a continuous function of locally bounded variation such that f (1) = 0 1 and 0 t1/2 |f (t)| dt < ∞, and let am =
Then the series
∞
1
2 {Jν+1 (jν,m )}
tf (t)Jν (jν,m t) dt.
2 0
am Jν (jν,m t) converges to f (t) locally uniformly on (0, 1].
m=1
Formula (3) below is stated without proof by Carslaw [2]. Lemma 5. Let 0 < s < 1. (a) If ν > 0, then 4ν
∞ Jν (jν,m s)Jν (jν,m t) 2 m=1 jν,m
{Jν+1 (jν,m )}
and the series converges uniformly for t ∈ [0, 1].
2
=
tν (s−ν − sν ) sν (t−ν − tν )
(0 ≤ t ≤ s) , (s < t ≤ 1)
(3)
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(b) In the case where ν = 0 we have 2
∞ J0 (j0,m s)J0 (j0,m t) m=1
2 j0,m {J1 (j0,m )}
=
2
− log s − log t
(0 ≤ t ≤ s) , (s < t ≤ 1)
(4)
and the series converges uniformly for t ∈ [0, 1]. Proof. (a) Let fs (t) denote the expression on the right hand side of (3). By Proposition 4 it is sufficient to show that
1 tfs (t)Jν (jν,m t) dt = 0
2ν Jν (jν,m s). 2 jν,m
In fact, using parts (i) and (ii) of Lemma 2, we see that
1
tfs (t)Jν (jν,m t) dt = (s−ν − sν )
0
s tν+1 Jν (jν,m t) dt 0
1
(t1−ν − tν+1 )Jν (jν,m t) dt
+ sν s
j ν,m s
uν+1 Jν (u) du ν+2 jν,m 0
jν,m ν−2 jν,m uν+1 ν +s − ν+2 Jν (u) du uν−1 jν,m jν,m s j s ν+1 u Jν+1 (u) ν,m −ν ν = (s − s ) ν+2 jν,m 0 jν,m ν−2 ν+1 jν,m Jν−1 (u) u Jν+1 (u) ν +s − − ν+2 uν−1 jν,m −ν
= (s
−s ) ν
jν,m s
=
s jν,m −
=
{Jν+1 (jν,m s) + Jν−1 (jν,m s)}
sν jν,m
{Jν−1 (jν,m ) + Jν+1 (jν,m )}
2ν Jν (jν,m s). 2 jν,m
To see the uniformity of convergence of the series on all of [0, 1], we note from parts (viii) and (ix) of Lemma 2 that |Jν (jν,m s)Jν (jν,m t)| ≤ (jν,m sν)−1/3 , and thus from Lemma 3(i), (ii) that J (j s)J (j t) π(j sν)−1/3 1 1 ν ν,m ν ν,m ν,m ≤ . ≤
2 1/3 2 jν,m {Jν+1 (jν,m )} 2(πsν) m4/3 2 2 jν,m − ν2
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(b) Now let fs (t) denote the expression on the right hand side of (4). Then
1
s
0
j 0,m s
= (− log s) 0
uJ0 (u) du − 2 j0,m
uJ1 (u) = (− log s) 2 j0,m − =
(log t)tJ0 (j0,m t) dt s
0
1
tJ0 (j0,m t) dt −
tfs (t)J0 (j0,m t) dt = (− log s)
j0,m j0,m s
j0,m s
u uJ0 (u) log du 2 j0,m j0,m
+ log j0,m 0
j0,m
log u uJ1 (u) 2 j0,m
j0,m
+ j0,m s
j0,m s
uJ1 (u) 2 j0,m
j0,m j0,m s
J1 (u) du 2 j0,m
J0 (j0,m s) , 2 j0,m
and (4) follows as before from Proposition 4. The proof of uniform convergence is also similar. 2 We recall (see Section 4.7 of Szegö [11], or Chapter IV of Stein and Weiss [10]) that, when λ > 0, (1 − 2tu + u2 )−λ =
∞
Pn(λ) (t)un
(|t| ≤ 1, |u| < 1),
n=0 (λ)
where Pn
is the usual ultraspherical (Gegenbauer) polynomial. Also, − log(1 − 2tu + u2 ) =
∞ 2 Tn (t)un n n=1
(|t| ≤ 1, |u| < 1),
(5)
where Tn (t) is the Chebychev polynomial given by cos(n cos−1 t) when |t| ≤ 1. In each case, for a given t, the series is locally uniformly convergent in u ∈ (−1, 1). These polynomials are related to each other by the equation Tn (t) = (n/2) lim λ−1 Pn(λ) (t) λ→0+
(n ≥ 1).
The next result summarises properties that we will use. (λ) (λ) Lemma 6. (i) Pn (t) ≤ Pn (1) = (λ)
(ii) Pn
n + 2λ − 1 n
(|t| ≤ 1).
satisfies the differential equation (1 − t2 )
d2 y dy + n(n + 2λ)y = 0. − (2λ + 1)t 2 dt dt
(iii) Tn satisfies the differential equation (1 − t2 )
d2 y dy + n2 y = 0. −t 2 dt dt
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Proof. (i) See Theorem 2.7.1 and Corollary 2.3.8 of [1] for the inequality, and p. 80 of [11] for the equality. (ii) See p. 80 of [11]. (iii) See p. 60 of [11]. 2 3. Expansions in a ball We denote by GB (·, ·) the Green function of the unit ball B of RN −1 (N ≥ 3), and define νn = n +
N −3 2
(n ≥ 0).
Proposition 7. Suppose that y ∈ B \{0 }. (i) Let N ≥ 4. If 0 < x < y , then
GB (x , y ) =
∞
3−N 2
3−N 2
(x y )
Pn
N −3 2
n=0
x , y x y
x
νn
−ν ν y n − y n ;
(6)
y
νn
−ν ν x n − x n .
(7)
and, if y < x < 1, then
GB (x , y ) =
∞
(x y )
Pn
N −3 2
n=0
x , y x y
Further, for any ε > 0, these series converge uniformly on {x : 0 < x ≤ y − ε} and {x : y + ε ≤ x < 1}, respectively. (ii) Let N = 3. If 0 < x < y , then GB (x , y ) = − log y +
∞
x , y 1 n −n n Tn x − y y ; n x y n=1
(8)
∞ 1
x , y n −n n Tn y − x x . n x y n=1
(9)
and, if y < x < 1, then GB (x , y ) = − log x +
Again, for any ε > 0, these series converge uniformly on {x : 0 < x ≤ y − ε} and {x : y + ε ≤ x < 1}, respectively. Proof. (i) Let N ≥ 4. Then (see Chapter 4 of [1]) 3−N y GB (x , y ) = x − y − y x − 2 y 3−N 3−N y y 3−N x = y − y x − y − y y 3−N 3−N 3−N y x − y x − y = x − . x x y
We know that
3−N
3−N
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3−N 3−N 2
x , y 2 2 y x − y x = 1 − 2 y + x y y x y ∞ N −3
x , y 2 n (x y ) , = Pn y x n=0 and the series converges uniformly for x ∈ B \{0 }. If x < y , then 3−N 2 3−N 2 x y x x
x , y − + = 1 − 2 2 y y x y y y n ∞ N −3
x , y x 2 = Pn , x y y n=0 where the series converges uniformly for 0 < x ≤ y − ε, and so
GB (x , y ) =
∞
Pn
N −3 2
n=0
x , y x y
x
n
3−N −n n − y , y
which yields the desired formula. If y < x < 1, then 3−N 2 (3−N )/2 x y y
x , y y + = 1−2 x − x x y x x 2 n ∞ N −3
x , y y 2 = Pn , x y x n=0 where the series converges uniformly for y + ε ≤ x < 1, and so
GB (x , y ) =
∞
Pn
N −3 2
n=0
x , y x y
y
n
3−N −n
x
− x
n
,
which again yields the desired formula. (ii) Let N = 3. Then y GB (x , y ) = − log x − y + log y x − 2 y x y y = − log y − log − + log y x − y y y x y + log y x − y . = − log x − log − x x y
We know from (5), multiplied by the factor 1/2, that ∞ y 1
x , y n = y T (x y ) . − log x − n y n=1 n x y Also,
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n ∞ x
x , y y x 1 = − T − log if x < y , n y y y n x y n=1 and n ∞ x y 1
x , y y = T − log − if y < x < 1. n x x n x y x n=1 Thus we obtained the desired formulae as before. 2 The following result is sufficient for our purposes, but is known to hold under much weaker smoothness assumptions (see [5] and the references therein). Proposition 8. Let f ∈ C ∞ (∂B ) and let ci,j be the Fourier coefficients of f with respect to an orthonor: j = 1, . . . , M (i)} of the spherical harmonics of degree i in RN −1 . Then the series mal basis {H ∞ M (i) i,j i=0 j=1 ci,j Hi,j converges uniformly on ∂B to f , and so the series ∞
M (i)
x
i
i=0
ci,j Hi,j
j=1
x x
converges uniformly on B \{0 } to the Poisson integral of f in B . Let y ∈ B \{0 } and δ ∈ (0, 1), and let Sy be the sphere in RN −1 centred at 0 that contains y . We define μy ,δ to be the probability measure on Sy which has density with respect to surface measure proportional to ⎛ −1 ⎞ 2 z − y ⎠ when z − y < δ y , and 0 otherwise. exp ⎝− 1 − 2 δ 2 y We further define the Green potential
GB (x , z ) dμy ,δ (z )
GB μy ,δ (x ) =
(x ∈ B ),
and the function ( N −3 ) Pn 2 μy ,δ (x )
=
Pn
N −3 2
x , z x z
dμy ,δ (z )
(x ∈ RN −1 \{0 }).
When N = 3 the function Tn μy ,δ is defined from Tn analogously.
x ,y Remark 9. By Proposition 7 we obtain formulae for G μ (x ) if we replace Pn by x y ( N 2−3 ) x ,y Pn μy ,δ (x ) in (6) and (7), and Tn x y by Tn μy ,δ (x ) in (8) and (9). Further, the series in (6) B
y ,δ
N −3 2
and (8) would now converge uniformly on {x : 0 < x ≤ y }, by Proposition 8, because the restriction of GB μy ,δ to Sy is C ∞ (cf. Theorem 3.3.3 of [1]). Also, the series in (7) and (9) would converge uniformly on {x : y ≤ x < 1}. To see this in the case of (9) we write the series as the difference of ∞ 1 n −n Tn μy ,δ (x ) y x n n=1
and
∞ 1 n n Tn μy ,δ (x ) y x . n n=1
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The second of these series clearly converges uniformly on B , and we can use inversion together with −1 −1 Proposition 8 to see that the first series converges uniformly on {x : x ≤ y }. A similar argument, based on the Kelvin transformation, applies to the series in (7). Let fn,m (s, t) =
Jνn (jνn ,m s)Jνn (jνn ,m t) jνn ,m {Jνn +1 (jνn ,m )}
2
(s ≥ 0, t ≥ 0, n ≥ 0, m ≥ 1).
The above remark and Lemma 5 combine to yield the following. Proposition 10. Suppose that x , y ∈ B \{0 } and δ ∈ (0, 1). (i) If N ≥ 4, then GB μy ,δ (x ) =
∞
(x y )
3−N 2
( N 2−3 )
Pn
μy ,δ (x )4νn
n=0
∞ fn,m (x , y ) . jνn ,m m=1
(ii) If N = 3, then GB μy ,δ (x ) = 2
∞ ∞ ∞ f0,m (x , y ) fn,m (x , y ) +4 Tn μy ,δ (x ) . j0,m jνn ,m m=1 n=1 m=1
4. Proof of Theorem 1 We define aN = σN (N − 2) when N ≥ 3, and a2 = σ2 , where σN denotes the surface area of the unit sphere in RN . Lemma 11. For any n ≥ 0, m ≥ 1 and any y ∈ (B \{0 }) × R, let un,m,y be the function defined by
x , y fn,m (x , y )e−jνn ,m |xN −yN | x → x Pn (N ≥ 4), x y
x , y x → Tn fn,m (x , y )e−jνn ,m |xN −yN | (N = 3). x y
3−N 2
N −3 2
Then un,m,y (i) is harmonic on RN −1 \{0 } × (R\{yN }); (ii) has a harmonic continuation to RN −1 × (R\{yN }); (iii) continuously vanishes on ∂Ω; (iv) is, in Ω, the Green potential of the signed measure given by 2a−1 N jνn ,m un,m,y (x , yN )dx on B × {yN };
(10)
(v) satisfies |un,m,y (x)| ≤
1 − y N/2−1
x
n+N −4 n
e−jνn ,m ε {Jνn +1 (jνn ,m )}
2
on RN −1 × ((−∞, yN − ε) ∪ (yN + ε, ∞)), where ε > 0 and the binomial coefficient is interpreted as 1 when N = 3.
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Proof. (i) If N ≥ 4, we know from Theorem 2.7.1 of [1] that the function
n
x → x Pn
N −3 2
x , y x y
,
interpreted as 0 at 0 , is harmonic on RN −1 , so the function
x → Pn
N −3 2
x , y x y
is an eigenfunction of the Laplace–Beltrami operator on ∂B with eigenvalue −n(n + N − 3). Next, the Laplacian of the radial function x → x
3−N 2
Jνn (jνn ,m x )
is, by direct computation, r
−(N +1)/2
jν2n ,m r2 Jνn (jνn ,m r) + jνn ,m rJν n (jνn ,m r) −
N −3 2
2 Jνn (jνn ,m r) ,
where r = x . Hence, by (2), the function
x → x
3−N 2
Pn
N −3 2
x , y x y
Jνn (jνn ,m x )
is an eigenfunction of the Laplacian on RN −1 \{0 }, with eigenvalue −jν2n ,m . The harmonicity of un,m,y in N −1 \{0 } × (R\{yN }) is now clear. R The case where N = 3 is similar (and simpler). (ii) This follows from Lemma 2(v) and the fact that any line segment, being polar, is a removable singularity for bounded harmonic functions in RN (N ≥ 3). (See Corollary 5.2.3 of [1].) (iii) This is obvious. (iv) Let Ψ be a C ∞ function on Ω with compact support. This support is contained in Ωa for some a > |yN |. By applying Green’s theorem to the cylinders B × (−a, yN − ε) and B × (yN + ε, a), we obtain (in the sense of distributions)
(Δun,m,y ) (Ψ) =
un,m,y ΔΨ dx Ω
⎧ ⎫ ∂un,m,y ∂Ψ ⎬ ⎨ − u Ψ dx n,m,y ∂xN ∂xN B ×{yN +ε} = lim ∂un,m,y ∂Ψ ⎭ ε→0+ ⎩ − − u Ψ n,m,y ∂xN dx ∂xN B ×{yN −ε}
=− 2jνn ,m un,m,y Ψdx . B ×{yN }
Thus, if v denotes the Green potential in Ω of the signed measure given by (10), we see from Theorems 4.3.8(i) and 4.3.5 of [1] that un,m,y − v is harmonic on Ω. Since this difference vanishes on ∂Ω and at infinity, we conclude that un,m,y = v on Ω. −1/2 (v) Lemma 3(iv) shows that |Jνn (jνn ,m x )| ≤ x when x ≥ 1, and this inequality remains valid for x < 1 by Lemma 2(ix). Also, Lemma 3(iii) shows that |Jνn (jνn ,m y )| ≤ jνn ,m (1 − y ). On RN −1 × ((−∞, yN − ε) ∪ (yN + ε, ∞)) we thus have the stated estimate, in view of Lemma 6(i). 2
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Remark 12. Now let δ ∈ (0, 1) and let uδn,m,y have the same definition as un,m,y , except that we use N −3 ( N 2−3 ) 2 x ,y x ,y Pn μy ,δ (x ) (respectively, Tn μy ,δ (x )) in place of Pn T (respectively, n x y x y ). Then Lemma 11 clearly remains true if we replace un,m,y by uδn,m,y throughout. Ω (·, y) to RN −1 × (R\{yN }). Theorem 13. Let y ∈ (B \{0 }) × R. Then GΩ (·, y) has a harmonic extension G On this set, ∞ ∞ 3−N Ω (x, y) = aN y 2 2νn un,m,y (x) G aN −1 n=0 m=1
Ω (x, y) = 2 G
∞
u0,m,y (x) + 4
m=1
∞ ∞
(N ≥ 4),
(11)
(N = 3).
(12)
un,m,y (x)
n=1 m=1
Further, if ε > 0, then there is a positive constant c, depending on ε and N , such that 1 − y GΩ (x, y) ≤ c N/2−1 x
on {(x , xN ) : |xN − yN | ≥ ε}.
(13)
Proof. Let δ ∈ (0, 1). If |xN − yN | ≥ ε > 0, then Lemma 11(v), Remark 12 and Lemma 3(i) together show that δ un,m,y (x) ≤ π 1 − y N/2−1 2 x
n+N −4 n
jν2n ,m e−jνn ,m ε
. jν2n ,m − νn2
(14)
Since
1 jν2n ,m − νn2
≤
1 1 ≤ j0,m π
by Lemma 2(vii), and jνn ,m ≥ (m + 3/4)π + n by Lemma 3(ii), we see that uδ 1 − y n+N −4 n,m,y (x) e−(πm+n)ε 2 ≤ jνn ,m x N/2−1 n
(n ≥ 0, m ≥ 1).
It follows that the series ∞ ∞ 3−N uδn,m,y (x) aN y 2 2νn aN −1 jν2n ,m n=0 m=1
2
∞ ∞ ∞ uδ0,m,y (x) uδn,m,y (x) + 4 2 2 j0,m jn,m m=1 n=1 m=1
(N ≥ 4),
(N = 3)
converge absolutely and locally uniformly on (RN −1 \{0 }) × (R\{yN }), and indeed (by the volume mean value property of harmonic functions) on RN −1 × (R\{yN }), to a harmonic function g satisfying the bound in (13). We know from Remark 12 and Lemma 11(iv) that (aN /aN −1 )jν−2 uδ is, in Ω, the potential of n ,m n,m,y the signed measure given by −1 δ 2a−1 N −1 jνn ,m un,m,y (x , yN )dx on B × {yN }.
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(When N = 2, we have aN /aN −1 = 2 and νn = n.) Thus, by Proposition 10, and the uniformity of convergence in Lemma 5 and Remark 9, the function g is, in Ω, the potential GΩ μδ1 of the measure given by dμδ1 = a−1 N −1 GB μy ,δ (x )dx on B × {yN }.
By the invariance of GΩ (·, ·) under translation in the xN -direction, and harmonicity, ∂ 2 GΩ μδ1 = ∂x2N
GΩ (·, (z , yN )) dμδ2 (z )
on Ω,
B
where dμδ2 = −a−1 N −1 ΔGB μy ,δ on B × {yN }
in the sense of distributions. Hence ∂ 2 GΩ μδ1 = ∂x2N
GΩ (·, (z , yN )) dμy ,δ (z )
on Ω.
(15)
B
The left hand side of (15) equals ∞ ∞ 3−N aN y 2 2νn uδn,m,y (x) aN −1 n=0 m=1
2
∞
uδ0,m,y (x) + 4
m=1
∞ ∞
uδn,m,y (x)
(N ≥ 4), (N = 3)
n=1 m=1
on Ω\(RN −1 × {yN }) by the local uniform convergence of the above series there. Further, as δ → 0+, the functions defined by these series converge locally uniformly on the same set to the expressions on the right hand side of (11) and (12). Since the right hand side of (15) converges to GΩ (·, y) as δ → 0+, we have established (11) and (12). Further, (13) follows easily from (14). 2 Remark 14. The proof of the above results can be simplified when N = 3: the use of the smoothing measure μy ,δ can be avoided, since the partial sums of the series in Proposition 7(ii) are then dominated by a min{x , y } on B . multiple of the integrable function x → − log 1 − max{x , y } Proof of Theorem 1. Let h be a harmonic function on Ωa which vanishes on ∂B × (−a, a). There is no loss of generality in assuming that h is continuous on Ωa . Further, since h (being the integral of h|∂Ωa against harmonic measure) can be written as the difference of two such functions which are positive, there is no loss of generality in assuming that h > 0. Next, let 0 < b < a, and let h∗ be defined as h on B × [−b, b], as 0 on B × (−∞, −a] and B × [a, ∞) and ∂Ω, and extended to Ω by solving the Dirichlet problem in B × (−a, −b) and in B × (b, a). Then h∗ is subharmonic on B × ((−∞, −b) ∪ (b, ∞)) and superharmonic on B × (−a, a), and continuously vanishes on ∂Ω. It can be written as GΩ μ, where μ is a signed measure on B × {±a, ±b} satisfying
(1 − |y |)d |μ| (y) < ∞.
It now follows from Theorem 13 that h can be extended to a harmonic function h on RN −1 × (−b, b). The estimate (1) is a consequence of (13). Since b can be arbitrarily close to a, the result follows. 2
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Remark 15. The rate of decay in (1) is sharp. This follows from the observations that each of the functions un,m,y in Lemma 11 satisfies the hypotheses of Theorem 1 (with a = |yN |), and that (for a fixed ν) Jν (t) =
νπ π 2 cos t − − + O(t−3/2 ) πt 2 4
(t → ∞)
(see [11], equation (1.71.7)). References [1] D.H. Armitage, S.J. Gardiner, Classical Potential Theory, Springer, London, 2001. [2] H.S. Carslaw, Integral equations and the determination of Green’s functions in the theory of potential, Proc. Edinb. Math. Soc. 31 (1913) 71–89. [3] J. Dougall, The determination of Green’s function by means of cylindrical or spherical harmonics, Proc. Edinb. Math. Soc. 18 (1900) 33–83. [4] P. Ebenfelt, D. Khavinson, On point to point reflection of harmonic functions across real-analytic hypersurfaces in Rn , J. Anal. Math. 68 (1996) 145–182. [5] H. Kalf, On the expansion of a function in terms of spherical harmonics in arbitrary dimensions, Bull. Belg. Math. Soc. Simon Stevin 2 (1995) 361–380. [6] D. Khavinson, H.S. Shapiro, Remarks on the reflection principle for harmonic functions, J. Anal. Math. 54 (1990) 60–76. [7] D. Khavinson, Holomorphic Partial Differential Equations and Classical Potential Theory, Universidad de La Laguna, Departamento de Análisis Matemático, La Laguna, 1996. [8] A. Laforgia, M.E. Muldoon, Inequalities and approximations for zeros of Bessel functions of small order, SIAM J. Math. Anal. 14 (1983) 383–388. [9] L.J. Landau, Bessel functions: monotonicity and bounds, J. Lond. Math. Soc. (2) 61 (2000) 197–215. [10] E.M. Stein, G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Univ. Press, Princeton, 1971. [11] G. Szegö, Orthogonal Polynomials, fourth edition, Amer. Math. Soc., Providence, RI, 1975. [12] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Univ. Press, 1922.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Harmonic functions which vanish on a cylindrical surface Stephen J. Gardiner ∗ , Hermann Render School of Mathematics and Statistics, University College Dublin, Belfield, Dublin 4, Ireland
a r t i c l e
i n f o
Article history: Received 6 July 2015 Available online 3 September 2015 Submitted by D. Khavinson Keywords: Harmonic continuation Green function Cylindrical harmonics
a b s t r a c t Suppose that a harmonic function h on a finite cylinder vanishes on the curved part of the boundary. This paper answers a question of Khavinson by showing that h then has a harmonic continuation to the infinite strip bounded by the hyperplanes containing the flat parts of the boundary. The existence of this extension is established by an analysis of the convergence properties of a double series expansion of the Green function of an infinite cylinder beyond the domain itself. © 2015 Elsevier Inc. All rights reserved.
1. Introduction The Schwarz reflection principle gives a formula for extending a harmonic function h on a domain Ω ⊂ RN through a relatively open subset E of the boundary ∂Ω on which h vanishes, provided E lies in a hyperplane (and is a relatively open subset thereof). By the Kelvin transformation there is a corresponding result where E lies in a sphere. When N = 2, such a reflection principle holds also when E is contained in an analytic arc (see Chapter 9 of [7]). However, when N ≥ 3 and N is odd, Ebenfelt and Khavinson [4] (see also [6] and Chapter 10 of [7]) have shown that a reflection law can only hold when the containing real analytic surface is either a hyperplane or a sphere. Now let N ≥ 3, let Ωa be the finite cylinder B × (−a, a), where B is the open unit ball in RN −1 and a > 0, and let Ω = B × R. Dima Khavinson raised the following question with the authors: Question. Given a harmonic function h on Ω which vanishes on ∂Ω, does it follow that h must have a harmonic extension to RN ? Although the above results show that there can be no pointwise reflection formula for such an extension, this paper will establish that such an extension does indeed exist. * Corresponding author. E-mail addresses: [email protected] (S.J. Gardiner), [email protected] (H. Render). http://dx.doi.org/10.1016/j.jmaa.2015.08.077 0022-247X/© 2015 Elsevier Inc. All rights reserved.
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We will use the notation x = (x , xN ) to denote a typical point of RN = RN −1 × R. Theorem 1. Let h be a harmonic function on Ωa which continuously vanishes on ∂B × (−a, a). Then h has a harmonic extension h to RN −1 × (−a, a). Further, for any b ∈ (0, a), there is a constant c, depending on a, b, N and h, such that 1−N/2 h(x) ≤ c x
(x ∈ RN −1 \B , |xN | < b).
(1)
It is a classical fact that the Green function for a three-dimensional infinite cylinder can be represented as a double series involving Bessel functions and Chebychev polynomials: see, for example, p. 62 of Dougall [3] or p. 78 of Carslaw [2]. Our approach to proving Theorem 1 involves establishing such a double series representation in N dimensions and analysing its convergence properties outside the cylinder. 2. Preparatory material Let Jν and Yν denote the usual Bessel functions of order ν ≥ 0 of the first and second kinds (see Watson [12]). Thus these functions both satisfy the differential equation z2
dy d2 y + (z 2 − ν 2 )y = 0. +z 2 dz dz
(2)
Further, let (jν,m )m≥1 denote the sequence of positive zeros of Jν , in increasing order. We collect below some facts that we will need. Jν+1 (z) d ν d Jν (z) z Jν (z) = z ν Jν−1 (z) and =− . ν dz dz z zν 2νJν (z) and Jν−1 (z) − Jν+1 (z) = 2Jν (z). (ii) Jν−1 (z) + Jν+1 (z) = z 2 (t > 0). (iii) Jν (t)Yν (t) − Yν (t)Jν (t) = πt −1/2 2 2 2 t2 − ν 2 (iv) {Jν (t)} + {Yν (t)} < (t > ν ≥ 12 ). π ν t 1 (t ≥ 0). (v) |Jν (t)| ≤ 2 Γ(ν + 1) (vi) j0,m ≥ (m + 3/4)π. (vii) jν,m ≥ j0,m + ν. (viii) |Jν (t)| < ν −1/3 (ν > 0, t ≥ 0). (ix) |Jν (t)| ≤ min{1, t−1/3 } (t > 0). 2 2 2 (0 ≤ ν ≤ 12 , t > 0). (x) {Jν (t)} + {Yν (t)} < πt
Lemma 2. (i)
Proof. (i) and (ii). See p. 45 of Watson [12]. (iii) See p. 76, (1) of [12]. (iv) See p. 447, (1) of [12]. (v) See p. 49, (1) of [12]. (vi) See p. 489 of [12]. (vii) See Laforgia and Muldoon [8], (2.4). (viii) See Landau [9]. −1/3 (ix) We know from p. 406, (10) of [12] that |Jν| ≤ 1, and from [9] that . |Jν (t)| ≤ t 2 2 (x) By Section 13.74 of [12] the function t → t {Jν (t)} + {Yν (t)} is non-decreasing when 0 ≤ ν ≤ 12 , and has limit 2/π at ∞. 2
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Some consequences of Lemma 2 are noted below. 2
2 Lemma 3. (i) jν,m {Jν+1 (jν,m )} ≥
2 2 jν,m − ν 2 (m ≥ 1). π
(ii) jν,m ≥ (m + 3/4)π + ν. (iii) |Jν (jν,m s)| ≤ jν,m (1 − s) (0 ≤ s ≤ 1). 1 2 (s ≥ 1). (iv) |Jν (jν,m s)| ≤ π ms
Proof. (i) By the second equation of Lemma 2(i), Jν (jν,m ) = −Jν+1 (jν,m ). If ν ≥ 12 , then, by (iii), (iv) and (vii) of that result, 2 2 jν,m {Jν+1 (jν,m )} = jν,m {Jν (jν,m )} =
4
2
2
π 2 {Yν (jν,m )}
2
>
2 2 jν,m − ν 2 . π
If 0 ≤ ν < 12 , we use part (x) of Lemma 2 in place of part (iv). (ii) This follows from Lemma 2(vi), (vii). (iii) By the mean value theorem, |Jν (jν,m s)| ≤ jν,m (1 − s) sup |Jν (jν,m t)|
(1/2 ≤ s ≤ 1).
1/2≤t≤1
We know from Lemma 2(i), (ii) that Jν = (Jν−1 − Jν+1 )/2 when ν ≥ 1 and Jν (z) = (ν/z)Jν (z) − Jν+1 (z) otherwise. In either case we see from Lemma 2(ix) and part (ii) of this lemma that |Jν (jν,m t)| ≤ 1 when 1/2 ≤ t ≤ 1, so the desired inequality is established when 1/2 ≤ s ≤ 1. It is clearly also valid when 0 ≤ s < 1/2, since |Jν | ≤ 1, and jν,m ≥ 2 by part (ii). (iv) By (ii) we have (jν,m s)2 ≥ (mπs)2 + ν 2 when s ≥ 1. If ν ≥ 1/2, then we see from Lemma 2(iv) that 2
2
{Jν (jν,m s)} <
π 2 ms
(s ≥ 1),
as desired. If 0 ≤ ν < 1/2, we can instead appeal to Lemma 2(x). 2 The following result is taken from Sections 18.24–18.26 of [12]. Proposition 4. Let f : (0, 1] → R be a continuous function of locally bounded variation such that f (1) = 0 1 and 0 t1/2 |f (t)| dt < ∞, and let am =
Then the series
∞
1
2 {Jν+1 (jν,m )}
tf (t)Jν (jν,m t) dt.
2 0
am Jν (jν,m t) converges to f (t) locally uniformly on (0, 1].
m=1
Formula (3) below is stated without proof by Carslaw [2]. Lemma 5. Let 0 < s < 1. (a) If ν > 0, then 4ν
∞ Jν (jν,m s)Jν (jν,m t) 2 m=1 jν,m
{Jν+1 (jν,m )}
and the series converges uniformly for t ∈ [0, 1].
2
=
tν (s−ν − sν ) sν (t−ν − tν )
(0 ≤ t ≤ s) , (s < t ≤ 1)
(3)
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(b) In the case where ν = 0 we have 2
∞ J0 (j0,m s)J0 (j0,m t) m=1
2 j0,m {J1 (j0,m )}
=
2
− log s − log t
(0 ≤ t ≤ s) , (s < t ≤ 1)
(4)
and the series converges uniformly for t ∈ [0, 1]. Proof. (a) Let fs (t) denote the expression on the right hand side of (3). By Proposition 4 it is sufficient to show that
1 tfs (t)Jν (jν,m t) dt = 0
2ν Jν (jν,m s). 2 jν,m
In fact, using parts (i) and (ii) of Lemma 2, we see that
1
tfs (t)Jν (jν,m t) dt = (s−ν − sν )
0
s tν+1 Jν (jν,m t) dt 0
1
(t1−ν − tν+1 )Jν (jν,m t) dt
+ sν s
j ν,m s
uν+1 Jν (u) du ν+2 jν,m 0
jν,m ν−2 jν,m uν+1 ν +s − ν+2 Jν (u) du uν−1 jν,m jν,m s j s ν+1 u Jν+1 (u) ν,m −ν ν = (s − s ) ν+2 jν,m 0 jν,m ν−2 ν+1 jν,m Jν−1 (u) u Jν+1 (u) ν +s − − ν+2 uν−1 jν,m −ν
= (s
−s ) ν
jν,m s
=
s jν,m −
=
{Jν+1 (jν,m s) + Jν−1 (jν,m s)}
sν jν,m
{Jν−1 (jν,m ) + Jν+1 (jν,m )}
2ν Jν (jν,m s). 2 jν,m
To see the uniformity of convergence of the series on all of [0, 1], we note from parts (viii) and (ix) of Lemma 2 that |Jν (jν,m s)Jν (jν,m t)| ≤ (jν,m sν)−1/3 , and thus from Lemma 3(i), (ii) that J (j s)J (j t) π(j sν)−1/3 1 1 ν ν,m ν ν,m ν,m ≤ . ≤
2 1/3 2 jν,m {Jν+1 (jν,m )} 2(πsν) m4/3 2 2 jν,m − ν2
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(b) Now let fs (t) denote the expression on the right hand side of (4). Then
1
s
0
j 0,m s
= (− log s) 0
uJ0 (u) du − 2 j0,m
uJ1 (u) = (− log s) 2 j0,m − =
(log t)tJ0 (j0,m t) dt s
0
1
tJ0 (j0,m t) dt −
tfs (t)J0 (j0,m t) dt = (− log s)
j0,m j0,m s
j0,m s
u uJ0 (u) log du 2 j0,m j0,m
+ log j0,m 0
j0,m
log u uJ1 (u) 2 j0,m
j0,m
+ j0,m s
j0,m s
uJ1 (u) 2 j0,m
j0,m j0,m s
J1 (u) du 2 j0,m
J0 (j0,m s) , 2 j0,m
and (4) follows as before from Proposition 4. The proof of uniform convergence is also similar. 2 We recall (see Section 4.7 of Szegö [11], or Chapter IV of Stein and Weiss [10]) that, when λ > 0, (1 − 2tu + u2 )−λ =
∞
Pn(λ) (t)un
(|t| ≤ 1, |u| < 1),
n=0 (λ)
where Pn
is the usual ultraspherical (Gegenbauer) polynomial. Also, − log(1 − 2tu + u2 ) =
∞ 2 Tn (t)un n n=1
(|t| ≤ 1, |u| < 1),
(5)
where Tn (t) is the Chebychev polynomial given by cos(n cos−1 t) when |t| ≤ 1. In each case, for a given t, the series is locally uniformly convergent in u ∈ (−1, 1). These polynomials are related to each other by the equation Tn (t) = (n/2) lim λ−1 Pn(λ) (t) λ→0+
(n ≥ 1).
The next result summarises properties that we will use. (λ) (λ) Lemma 6. (i) Pn (t) ≤ Pn (1) = (λ)
(ii) Pn
n + 2λ − 1 n
(|t| ≤ 1).
satisfies the differential equation (1 − t2 )
d2 y dy + n(n + 2λ)y = 0. − (2λ + 1)t 2 dt dt
(iii) Tn satisfies the differential equation (1 − t2 )
d2 y dy + n2 y = 0. −t 2 dt dt
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Proof. (i) See Theorem 2.7.1 and Corollary 2.3.8 of [1] for the inequality, and p. 80 of [11] for the equality. (ii) See p. 80 of [11]. (iii) See p. 60 of [11]. 2 3. Expansions in a ball We denote by GB (·, ·) the Green function of the unit ball B of RN −1 (N ≥ 3), and define νn = n +
N −3 2
(n ≥ 0).
Proposition 7. Suppose that y ∈ B \{0 }. (i) Let N ≥ 4. If 0 < x < y , then
GB (x , y ) =
∞
3−N 2
3−N 2
(x y )
Pn
N −3 2
n=0
x , y x y
x
νn
−ν ν y n − y n ;
(6)
y
νn
−ν ν x n − x n .
(7)
and, if y < x < 1, then
GB (x , y ) =
∞
(x y )
Pn
N −3 2
n=0
x , y x y
Further, for any ε > 0, these series converge uniformly on {x : 0 < x ≤ y − ε} and {x : y + ε ≤ x < 1}, respectively. (ii) Let N = 3. If 0 < x < y , then GB (x , y ) = − log y +
∞
x , y 1 n −n n Tn x − y y ; n x y n=1
(8)
∞ 1
x , y n −n n Tn y − x x . n x y n=1
(9)
and, if y < x < 1, then GB (x , y ) = − log x +
Again, for any ε > 0, these series converge uniformly on {x : 0 < x ≤ y − ε} and {x : y + ε ≤ x < 1}, respectively. Proof. (i) Let N ≥ 4. Then (see Chapter 4 of [1]) 3−N y GB (x , y ) = x − y − y x − 2 y 3−N 3−N y y 3−N x = y − y x − y − y y 3−N 3−N 3−N y x − y x − y = x − . x x y
We know that
3−N
3−N
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3−N 3−N 2
x , y 2 2 y x − y x = 1 − 2 y + x y y x y ∞ N −3
x , y 2 n (x y ) , = Pn y x n=0 and the series converges uniformly for x ∈ B \{0 }. If x < y , then 3−N 2 3−N 2 x y x x
x , y − + = 1 − 2 2 y y x y y y n ∞ N −3
x , y x 2 = Pn , x y y n=0 where the series converges uniformly for 0 < x ≤ y − ε, and so
GB (x , y ) =
∞
Pn
N −3 2
n=0
x , y x y
x
n
3−N −n n − y , y
which yields the desired formula. If y < x < 1, then 3−N 2 (3−N )/2 x y y
x , y y + = 1−2 x − x x y x x 2 n ∞ N −3
x , y y 2 = Pn , x y x n=0 where the series converges uniformly for y + ε ≤ x < 1, and so
GB (x , y ) =
∞
Pn
N −3 2
n=0
x , y x y
y
n
3−N −n
x
− x
n
,
which again yields the desired formula. (ii) Let N = 3. Then y GB (x , y ) = − log x − y + log y x − 2 y x y y = − log y − log − + log y x − y y y x y + log y x − y . = − log x − log − x x y
We know from (5), multiplied by the factor 1/2, that ∞ y 1
x , y n = y T (x y ) . − log x − n y n=1 n x y Also,
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n ∞ x
x , y y x 1 = − T − log if x < y , n y y y n x y n=1 and n ∞ x y 1
x , y y = T − log − if y < x < 1. n x x n x y x n=1 Thus we obtained the desired formulae as before. 2 The following result is sufficient for our purposes, but is known to hold under much weaker smoothness assumptions (see [5] and the references therein). Proposition 8. Let f ∈ C ∞ (∂B ) and let ci,j be the Fourier coefficients of f with respect to an orthonor: j = 1, . . . , M (i)} of the spherical harmonics of degree i in RN −1 . Then the series mal basis {H ∞ M (i) i,j i=0 j=1 ci,j Hi,j converges uniformly on ∂B to f , and so the series ∞
M (i)
x
i
i=0
ci,j Hi,j
j=1
x x
converges uniformly on B \{0 } to the Poisson integral of f in B . Let y ∈ B \{0 } and δ ∈ (0, 1), and let Sy be the sphere in RN −1 centred at 0 that contains y . We define μy ,δ to be the probability measure on Sy which has density with respect to surface measure proportional to ⎛ −1 ⎞ 2 z − y ⎠ when z − y < δ y , and 0 otherwise. exp ⎝− 1 − 2 δ 2 y We further define the Green potential
GB (x , z ) dμy ,δ (z )
GB μy ,δ (x ) =
(x ∈ B ),
and the function ( N −3 ) Pn 2 μy ,δ (x )
=
Pn
N −3 2
x , z x z
dμy ,δ (z )
(x ∈ RN −1 \{0 }).
When N = 3 the function Tn μy ,δ is defined from Tn analogously.
x ,y Remark 9. By Proposition 7 we obtain formulae for G μ (x ) if we replace Pn by x y ( N 2−3 ) x ,y Pn μy ,δ (x ) in (6) and (7), and Tn x y by Tn μy ,δ (x ) in (8) and (9). Further, the series in (6) B
y ,δ
N −3 2
and (8) would now converge uniformly on {x : 0 < x ≤ y }, by Proposition 8, because the restriction of GB μy ,δ to Sy is C ∞ (cf. Theorem 3.3.3 of [1]). Also, the series in (7) and (9) would converge uniformly on {x : y ≤ x < 1}. To see this in the case of (9) we write the series as the difference of ∞ 1 n −n Tn μy ,δ (x ) y x n n=1
and
∞ 1 n n Tn μy ,δ (x ) y x . n n=1
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The second of these series clearly converges uniformly on B , and we can use inversion together with −1 −1 Proposition 8 to see that the first series converges uniformly on {x : x ≤ y }. A similar argument, based on the Kelvin transformation, applies to the series in (7). Let fn,m (s, t) =
Jνn (jνn ,m s)Jνn (jνn ,m t) jνn ,m {Jνn +1 (jνn ,m )}
2
(s ≥ 0, t ≥ 0, n ≥ 0, m ≥ 1).
The above remark and Lemma 5 combine to yield the following. Proposition 10. Suppose that x , y ∈ B \{0 } and δ ∈ (0, 1). (i) If N ≥ 4, then GB μy ,δ (x ) =
∞
(x y )
3−N 2
( N 2−3 )
Pn
μy ,δ (x )4νn
n=0
∞ fn,m (x , y ) . jνn ,m m=1
(ii) If N = 3, then GB μy ,δ (x ) = 2
∞ ∞ ∞ f0,m (x , y ) fn,m (x , y ) +4 Tn μy ,δ (x ) . j0,m jνn ,m m=1 n=1 m=1
4. Proof of Theorem 1 We define aN = σN (N − 2) when N ≥ 3, and a2 = σ2 , where σN denotes the surface area of the unit sphere in RN . Lemma 11. For any n ≥ 0, m ≥ 1 and any y ∈ (B \{0 }) × R, let un,m,y be the function defined by
x , y fn,m (x , y )e−jνn ,m |xN −yN | x → x Pn (N ≥ 4), x y
x , y x → Tn fn,m (x , y )e−jνn ,m |xN −yN | (N = 3). x y
3−N 2
N −3 2
Then un,m,y (i) is harmonic on RN −1 \{0 } × (R\{yN }); (ii) has a harmonic continuation to RN −1 × (R\{yN }); (iii) continuously vanishes on ∂Ω; (iv) is, in Ω, the Green potential of the signed measure given by 2a−1 N jνn ,m un,m,y (x , yN )dx on B × {yN };
(10)
(v) satisfies |un,m,y (x)| ≤
1 − y N/2−1
x
n+N −4 n
e−jνn ,m ε {Jνn +1 (jνn ,m )}
2
on RN −1 × ((−∞, yN − ε) ∪ (yN + ε, ∞)), where ε > 0 and the binomial coefficient is interpreted as 1 when N = 3.
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Proof. (i) If N ≥ 4, we know from Theorem 2.7.1 of [1] that the function
n
x → x Pn
N −3 2
x , y x y
,
interpreted as 0 at 0 , is harmonic on RN −1 , so the function
x → Pn
N −3 2
x , y x y
is an eigenfunction of the Laplace–Beltrami operator on ∂B with eigenvalue −n(n + N − 3). Next, the Laplacian of the radial function x → x
3−N 2
Jνn (jνn ,m x )
is, by direct computation, r
−(N +1)/2
jν2n ,m r2 Jνn (jνn ,m r) + jνn ,m rJν n (jνn ,m r) −
N −3 2
2 Jνn (jνn ,m r) ,
where r = x . Hence, by (2), the function
x → x
3−N 2
Pn
N −3 2
x , y x y
Jνn (jνn ,m x )
is an eigenfunction of the Laplacian on RN −1 \{0 }, with eigenvalue −jν2n ,m . The harmonicity of un,m,y in N −1 \{0 } × (R\{yN }) is now clear. R The case where N = 3 is similar (and simpler). (ii) This follows from Lemma 2(v) and the fact that any line segment, being polar, is a removable singularity for bounded harmonic functions in RN (N ≥ 3). (See Corollary 5.2.3 of [1].) (iii) This is obvious. (iv) Let Ψ be a C ∞ function on Ω with compact support. This support is contained in Ωa for some a > |yN |. By applying Green’s theorem to the cylinders B × (−a, yN − ε) and B × (yN + ε, a), we obtain (in the sense of distributions)
(Δun,m,y ) (Ψ) =
un,m,y ΔΨ dx Ω
⎧ ⎫ ∂un,m,y ∂Ψ ⎬ ⎨ − u Ψ dx n,m,y ∂xN ∂xN B ×{yN +ε} = lim ∂un,m,y ∂Ψ ⎭ ε→0+ ⎩ − − u Ψ n,m,y ∂xN dx ∂xN B ×{yN −ε}
=− 2jνn ,m un,m,y Ψdx . B ×{yN }
Thus, if v denotes the Green potential in Ω of the signed measure given by (10), we see from Theorems 4.3.8(i) and 4.3.5 of [1] that un,m,y − v is harmonic on Ω. Since this difference vanishes on ∂Ω and at infinity, we conclude that un,m,y = v on Ω. −1/2 (v) Lemma 3(iv) shows that |Jνn (jνn ,m x )| ≤ x when x ≥ 1, and this inequality remains valid for x < 1 by Lemma 2(ix). Also, Lemma 3(iii) shows that |Jνn (jνn ,m y )| ≤ jνn ,m (1 − y ). On RN −1 × ((−∞, yN − ε) ∪ (yN + ε, ∞)) we thus have the stated estimate, in view of Lemma 6(i). 2
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Remark 12. Now let δ ∈ (0, 1) and let uδn,m,y have the same definition as un,m,y , except that we use N −3 ( N 2−3 ) 2 x ,y x ,y Pn μy ,δ (x ) (respectively, Tn μy ,δ (x )) in place of Pn T (respectively, n x y x y ). Then Lemma 11 clearly remains true if we replace un,m,y by uδn,m,y throughout. Ω (·, y) to RN −1 × (R\{yN }). Theorem 13. Let y ∈ (B \{0 }) × R. Then GΩ (·, y) has a harmonic extension G On this set, ∞ ∞ 3−N Ω (x, y) = aN y 2 2νn un,m,y (x) G aN −1 n=0 m=1
Ω (x, y) = 2 G
∞
u0,m,y (x) + 4
m=1
∞ ∞
(N ≥ 4),
(11)
(N = 3).
(12)
un,m,y (x)
n=1 m=1
Further, if ε > 0, then there is a positive constant c, depending on ε and N , such that 1 − y GΩ (x, y) ≤ c N/2−1 x
on {(x , xN ) : |xN − yN | ≥ ε}.
(13)
Proof. Let δ ∈ (0, 1). If |xN − yN | ≥ ε > 0, then Lemma 11(v), Remark 12 and Lemma 3(i) together show that δ un,m,y (x) ≤ π 1 − y N/2−1 2 x
n+N −4 n
jν2n ,m e−jνn ,m ε
. jν2n ,m − νn2
(14)
Since
1 jν2n ,m − νn2
≤
1 1 ≤ j0,m π
by Lemma 2(vii), and jνn ,m ≥ (m + 3/4)π + n by Lemma 3(ii), we see that uδ 1 − y n+N −4 n,m,y (x) e−(πm+n)ε 2 ≤ jνn ,m x N/2−1 n
(n ≥ 0, m ≥ 1).
It follows that the series ∞ ∞ 3−N uδn,m,y (x) aN y 2 2νn aN −1 jν2n ,m n=0 m=1
2
∞ ∞ ∞ uδ0,m,y (x) uδn,m,y (x) + 4 2 2 j0,m jn,m m=1 n=1 m=1
(N ≥ 4),
(N = 3)
converge absolutely and locally uniformly on (RN −1 \{0 }) × (R\{yN }), and indeed (by the volume mean value property of harmonic functions) on RN −1 × (R\{yN }), to a harmonic function g satisfying the bound in (13). We know from Remark 12 and Lemma 11(iv) that (aN /aN −1 )jν−2 uδ is, in Ω, the potential of n ,m n,m,y the signed measure given by −1 δ 2a−1 N −1 jνn ,m un,m,y (x , yN )dx on B × {yN }.
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(When N = 2, we have aN /aN −1 = 2 and νn = n.) Thus, by Proposition 10, and the uniformity of convergence in Lemma 5 and Remark 9, the function g is, in Ω, the potential GΩ μδ1 of the measure given by dμδ1 = a−1 N −1 GB μy ,δ (x )dx on B × {yN }.
By the invariance of GΩ (·, ·) under translation in the xN -direction, and harmonicity, ∂ 2 GΩ μδ1 = ∂x2N
GΩ (·, (z , yN )) dμδ2 (z )
on Ω,
B
where dμδ2 = −a−1 N −1 ΔGB μy ,δ on B × {yN }
in the sense of distributions. Hence ∂ 2 GΩ μδ1 = ∂x2N
GΩ (·, (z , yN )) dμy ,δ (z )
on Ω.
(15)
B
The left hand side of (15) equals ∞ ∞ 3−N aN y 2 2νn uδn,m,y (x) aN −1 n=0 m=1
2
∞
uδ0,m,y (x) + 4
m=1
∞ ∞
uδn,m,y (x)
(N ≥ 4), (N = 3)
n=1 m=1
on Ω\(RN −1 × {yN }) by the local uniform convergence of the above series there. Further, as δ → 0+, the functions defined by these series converge locally uniformly on the same set to the expressions on the right hand side of (11) and (12). Since the right hand side of (15) converges to GΩ (·, y) as δ → 0+, we have established (11) and (12). Further, (13) follows easily from (14). 2 Remark 14. The proof of the above results can be simplified when N = 3: the use of the smoothing measure μy ,δ can be avoided, since the partial sums of the series in Proposition 7(ii) are then dominated by a min{x , y } on B . multiple of the integrable function x → − log 1 − max{x , y } Proof of Theorem 1. Let h be a harmonic function on Ωa which vanishes on ∂B × (−a, a). There is no loss of generality in assuming that h is continuous on Ωa . Further, since h (being the integral of h|∂Ωa against harmonic measure) can be written as the difference of two such functions which are positive, there is no loss of generality in assuming that h > 0. Next, let 0 < b < a, and let h∗ be defined as h on B × [−b, b], as 0 on B × (−∞, −a] and B × [a, ∞) and ∂Ω, and extended to Ω by solving the Dirichlet problem in B × (−a, −b) and in B × (b, a). Then h∗ is subharmonic on B × ((−∞, −b) ∪ (b, ∞)) and superharmonic on B × (−a, a), and continuously vanishes on ∂Ω. It can be written as GΩ μ, where μ is a signed measure on B × {±a, ±b} satisfying
(1 − |y |)d |μ| (y) < ∞.
It now follows from Theorem 13 that h can be extended to a harmonic function h on RN −1 × (−b, b). The estimate (1) is a consequence of (13). Since b can be arbitrarily close to a, the result follows. 2
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Remark 15. The rate of decay in (1) is sharp. This follows from the observations that each of the functions un,m,y in Lemma 11 satisfies the hypotheses of Theorem 1 (with a = |yN |), and that (for a fixed ν) Jν (t) =
νπ π 2 cos t − − + O(t−3/2 ) πt 2 4
(t → ∞)
(see [11], equation (1.71.7)). References [1] D.H. Armitage, S.J. Gardiner, Classical Potential Theory, Springer, London, 2001. [2] H.S. Carslaw, Integral equations and the determination of Green’s functions in the theory of potential, Proc. Edinb. Math. Soc. 31 (1913) 71–89. [3] J. Dougall, The determination of Green’s function by means of cylindrical or spherical harmonics, Proc. Edinb. Math. Soc. 18 (1900) 33–83. [4] P. Ebenfelt, D. Khavinson, On point to point reflection of harmonic functions across real-analytic hypersurfaces in Rn , J. Anal. Math. 68 (1996) 145–182. [5] H. Kalf, On the expansion of a function in terms of spherical harmonics in arbitrary dimensions, Bull. Belg. Math. Soc. Simon Stevin 2 (1995) 361–380. [6] D. Khavinson, H.S. Shapiro, Remarks on the reflection principle for harmonic functions, J. Anal. Math. 54 (1990) 60–76. [7] D. Khavinson, Holomorphic Partial Differential Equations and Classical Potential Theory, Universidad de La Laguna, Departamento de Análisis Matemático, La Laguna, 1996. [8] A. Laforgia, M.E. Muldoon, Inequalities and approximations for zeros of Bessel functions of small order, SIAM J. Math. Anal. 14 (1983) 383–388. [9] L.J. Landau, Bessel functions: monotonicity and bounds, J. Lond. Math. Soc. (2) 61 (2000) 197–215. [10] E.M. Stein, G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Univ. Press, Princeton, 1971. [11] G. Szegö, Orthogonal Polynomials, fourth edition, Amer. Math. Soc., Providence, RI, 1975. [12] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Univ. Press, 1922.