WAVE PROPAGATION IN A 3-D OPTICAL WAVEGUIDE

Transcription

1 WAVE PROPAGATION IN A 3-D OPTICAL WAVEGUIDE OLEG ALEXANDROV Shool of Mathematis, University of Minnesota 26 Churh Str SE, Minneapolis, MN 55455, USA aoleg@math.umn.edu GIULIO CIRAOLO Dipartimento di Matematia U. Dini, University of Firenze viale Morgagni 67/A, 5134 Firenze, Italy iraolo@math.unifi.it Abstrat In this paper we study the problem of eletromagneti wave propagation in a 3-D optial fiber. The goal is to obtain a solution for the time-harmoni field aused by a soure in a ylindrially symmetri waveguide. The geometry of the problem, orresponding to an open waveguide, makes the problem hallenging. To solve it, we onstrut a transform theory whih is a nontrivial generalization of a method for solving a 2-D version of this problem given by Magnanini Santosa.[3] The extension to 3-D is made ompliated by the fat that the resulting eigenvalue problem defining the transform kernel is singular both at the origin at infinity. The singularities require the investigation of the behavior of the solutions of the eigenvalue problem. Moreover, the derivation of the transform formulas needed to solve the wave propagation problem involves nontrivial alulations. The paper provides a omplete desription on how to onstrut the solution to the wave propagation problem in a 3-D optial waveguide with ylindrial symmetry. A follow-up artile will study the partiular ases of a step-index fiber of a oaxial waveguide. In those ases we will obtain onrete formulas for the field numerial examples.

2 1 Introdution In this paper we study the eletromagneti wave propagation in a ylindrial optial fiber. 1 model equation, we use the Helmholtz equation As u + k 2 n(x, y, z) 2 u = f(x, y, z), (x, y, z) R 3, (1) also alled the time-harmoni wave equation. The number k is alled the wavenumber, the funtion f represents a soure of energy. We require that the index of refration n(x, y, z) have the form: { n o( x n(x, y, z) = 2 + y 2 ), if x 2 + y 2 < R, n l, if x 2 + y 2 R, where R is the radius of the waveguide, n o ( ) is an arbitrary, bounded integrable funtion with positive values. The main result of this paper is the onstrution of a representation formula for a solution u of (1) satisfying suitable radiation onditions. To find suh a formula is equivalent to finding a Green s funtion for problem (1), as then the field generated by the soure f an be obtained as a superposition of fields generated by point soures. Our results generalize a similar formula obtained by Magnanini Santosa[3] in the two-dimensional ase. In [3] it is shown that the energy of the eletromagneti field is divided into two parts: a part that propagates without loss inside the waveguide as a finite number of distint guided modes, while the other part either deays exponentially along the fiber or is radiated outside. Our ase reveals a new feature: for speial hoies of the parameters, new kinds of guided modes appear whih, rather than deaying exponentially outside the fiber, vanish as a power of the distane from the fiber s axis. As in,[3] we use the tehnique of separation of variables with a number of important differenes speified below. In our ase the relevant symmetries of the problem suggest that we separate variables in the ylindrial oordinates r, ϑ, z. The equation in the angular variable ϑ is simple to solve for will only introdue a Fourier series in ϑ. The separation in z an be solved as in.[3] More ompliated is the study of the r-oordinate. In the 2-D ase, the authors of [3] start with a domain of the form (x, z) [ t, t] (, ). They impose boundary onditions at x = ±t, then obtain the desired Green s funtion by setting t. They use the fat that the separation of variables yields a differential equation in x whose solutions, in the region where the index n(x) is onstant, are sines osines whose zeros are uniformly distributed. This assumption turns out to play a fundamental role in alulating Green s funtion. In our ase we obtain a differential equation in the variable r whih is somewhat related to Bessel s equation, in partiular, is singular at r =. A further diffiulty is that, unlike the 2-D ase, the distribution of the zeroes of Bessel s funtions is more ompliated. The singularity the non-uniform distribution of zeros make the method used in [3] inappliable. Our approah is to use the theory of singular self-adjoint eigenvalue problems for seond order differential equations as presented in [2]. [8] Applying this theory to our problem doing the expliit alulations took some effort, hiefly for the reason that our self-adjoint eigenvalue problem has, just like in the 2-D ase, a oeffiient whih is, with some restritions, a general funtion, so we annot obtain solutions for our equation in terms of onrete funtions. In spite of the differenes we just mentioned with the work in [3], we wish to remark that qualitatively, the 2-D ase the 3-D ase behave similarly. In both ases one ultimately needs to study the behavior of a differential operator (in the x variable in the 2-D ase in the r variable in the 3-D ase). Both operators exhibit a limit point behavior (see [2] [8]), as 1 We will use the terms optial waveguide optial fiber interhangeably. 2

3 x respetively r. And their spetra have the same struture. They both have a disrete part a ontinuous part. The paper is organized as follows. In Setion 2 we derive the seond order self-adjoint eigenvalue problem assoiated with the Helmholtz equation (1). In Setion 3 we will prove a set of tehnial lemmas aimed at studying the behavior of the solutions of this eigenvalue problem. In Setion 4 we will lassify the solutions of the eigenvalue problem whih are well-behaved (in a sense to be speified there) as r r. The motivation is that the eletromagneti field in the fiber will have a representation in terms of the well-behaved solutions of this eigenvalue problem. In Setion 5 we summarize the theory of self-adjoint eigenvalue problems as exposed in [2]. [8] The funtions defined in Setion 5 are alulated in Setion 6. In Setion 7 the transform defined in Setion 5 is omputed. The obtained transform is used in Setion 8 to find the Green s funtion for the Helmholtz equation (1), in turn, to find the desired eletromagneti field in the fiber given the soure. 2 The Eigenvalue Problem A typial optial fiber is a ylindrial dieletri waveguide, made of silia glass or plasti. Its entral region is alled ore, surrounded by ladding, whih has a slightly lower index of refration. The ladding is surrounded by a protetive jaket. Most of the eletromagneti radiation propagates without loss as a set of guided modes along the fiber axis. The eletromagneti field intensity of the guided modes in the ladding deays exponentially along the radial diretion. This is why, the radius of the ladding, whih is typially several times larger than the radius of the ore, an be onsidered infinite. A typial optial fiber is weakly guided, whih means that the differene in the indexes of refration of the ore ladding is very small. In these onditions the eletromagneti field in the fiber is essentially transverse with eah of the transverse omponents approximately satisfying the Helmholtz equation (1). This is the so-alled weakly guided approximation. For a derivation of (1) from Maxwell s equations for a weakly guided fiber see,[7] hapters Beause of the ylindrial geometry, it will be onvenient to use the ylindrial oordinate system (r, ϑ, z), with z being the axial diretion. Then, the index of refration will depend on the r variable only, n = n(r). In the new oordinates (1) beomes 2 u z r r ( r u r ) + 1 r 2 2 u ϑ 2 + k2 n(r) 2 u = f(r, ϑ, z). (2) This is a linear partial differential equation. The solution of this equation will be determined as soon as we find its Green s funtion. In order to obtain the latter, onsider the homogeneous version of (2), 2 u z r ( r u ) u r r r 2 ϑ + 2 k2 n(r) 2 u =. Look for a solution in separated variables, u(r, ϑ, z) = Z(z) Θ(ϑ) v(r). It is quikly found that we must have u(r, ϑ, z) = e iβkz e imϑ v(r), (3) with β C, m Z, v(r) satisfying the differential equation v + 1 } r v + {k 2 n(r) 2 β 2 m2 v = (the derivative here, in the rest of this paper, will always be with respet to the r variable). r 2 3

4 Let R > be the radius of the fiber ore. Then n(r) = n l for r R. Denote d 2 = k 2 (n 2 n 2 l), l = k 2 (n 2 β 2 ), q(r) = k 2 [n 2 n(r) 2 ]. (4) Then this equation beomes v + 1 } r v + {l q(r) m2 v =. (5) r 2 We will view (5) as an eigenvalue problem in l C. The variable r is in (, ), the number m is an integer, the funtion q is bounded, measurable, real-valued non-negative, with q(r) = d 2 > for r R >. It will be onvenient to make a variable hange. Denote w(r) = rv(r). We get the equation { } w + l q(r) m2 1/4 w =, r (, ). (6) r 2 This will be our self-adjoint eigenvalue problem. d 2 R r The funtion q(r). 4

5 3 A Study of the Solutions of the Eigenvalue Problem Before going further, we will need some information about the behavior of the solutions of the differential equation (6) as funtions of r l m. This will be the subjet of the next four lemmas. Lemma 3.1 There exists a solution j m(r, l) (r >, l C, m Z) of (6) suh that lim r j m (r, l) = 1, r lim m +1/2 r j m(r, l) = 1. (7) ( m + 1/2) r m 1/2 The funtions j m(r, l) j m(r, l) are analyti in l as r is fixed. There exists another solution y m (r, l) of (6) suh that lim r y m (r, l) = 1, r lim m +1/2 r y m(r, l) lim = 1, r r ln r y m(r, l) = 1, if m 1, ( m + 1/2) r (8a) m 1/2 y lim m(r, l) r ln r/(2 = 1, if m =. r) (8b) Proof. We will assume m, then set j m = j m y m = y m. Make the variable hange w = r m+1/2 σ in (6). We obtain σ + 2m + 1 σ + {l q(r)} σ =. r Denote k = 2m + 1, k 1. Multiply this equation by r k. We get (r k σ ) = r k (q(r) l) σ. (9) To prove this lemma we need to find two solutions σ(r, l) τ(r, l) of (9) suh that lim r r2m τ(r, l) = 1, lim σ(r, l) = 1, lim r r σ (r, l) =, (1) lim r 2m+1 τ (r, l) = 2m, if m 1, (11a) r τ(r, l) lim = 1, lim rτ (r, l) = 1, if m =. (11b) r ln r r Also, σ(r, l) must be analyti in l for r fixed. The idea to finding σ τ is to rewrite (9) as an integral equation. Let q = sup r [, ) q(r). If ω : [, ) C is a funtion, bounded integrable on every ompat subset of [, ), p is an integer, then one has r Consider the operator s p ω(s) ds sup ω(t) T ω(r) = t [,r] r t t k r s p ds = sup ω(t) rp+1 t [,r] p + 1. (12) s k (q(s) l)ω(s) ds dt, (13) defined for omplex-valued funtions ω whih are bounded integrable on every ompat subset of [, ). By applying (12) to the inner-most integral in (13) we dedue sup T ω(t) (q + l ) sup ω(t) t [,r] t [,r] 2 (k + 1). r 2 5

6 In partiular, for any r >, T is a bounded linear operator from the spae C([, r]) of ontinuous, omplex-valued funtions defined on [, r] onto itself. By using the same reasoning, one an show by indution that for any integer n T n ω(r) (q + l ) n sup ω(t) t [,r] {2 4 2n} {(k + 1) (k + 3) (k + 2n 1)}. (14) Using (12) it is easy to hek that if funtion σ satisfies (1), then (9) is equivalent to whih in turn is the same as σ (r, λ) = 1 r k r r 2n s k (q(s) λ) σ(s, λ) ds, (15) σ = 1 + T σ. (16) We will find σ by applying the method of suessive approximations. Let σ 1, σ n+1 = σ + T σ n, n. Then, By using (14) we obtain that the series σ n = σ + T σ + + T n σ, n. σ = σ + T σ + + T n σ +..., σ 1, (17) is uniformly onvergent for r l in ompat sets. Sine σ n is ontinuous in r analyti in l, the same property will hold for σ. Using the fat that the operator T is ontinuous on C([, r]) for any r >, it follows from the above series that (16) holds, therefore (9) holds. Now prove the existene of τ satisfying (9) with the boundary onditions (11a) or (11b). That will be done in several steps to be desribed below. For the rest of the lemma, σ = σ(r, l) will be the solution to (9) found above. We are not looking for any properties of τ with respet to l. Therefore, we will fix l C, will onsider τ to be a funtion of r only. At step 1 we will prove that if τ is a solution to (9) whih is linearly independent of σ, then τ is unbounded as r. At step 2, we will show that any solution τ to (9) has the property τ(r) A + Br k, < r r, (18) for some A >, B >, r >. At step 3 we will prove that any solution τ of (9) linearly independent of σ satisfies τ (r) = O(r k ) as r. (19) That will imply (11a) or (11b) depending on whether m 1 or m =, that is, k 3 or k = 1. Step 1. Show that any solution τ to (9) whih is linearly independent of σ is unbounded as r. Assume that τ is suh a solution that it is bounded as r. From (9) we get r k τ (r) = 1 + r s k (q(s) l)τ(r) ds for some onstant 1. We annot have 1, beause then τ (r) = O(r k ) with k 1, so τ annot be bounded as r. Then, if 1 = we an divide by r k integrate again, to obtain τ = 2 + T τ, with T the operator defined by (13) 2 another onstant. Sine the solution σ satisfies σ = 1 + T σ, we dedue that ϕ = τ 2σ will satisfy ϕ = T ϕ. 6

7 Then ϕ = T n ϕ for any n. By using (14) we get ϕ(r) (q + l ) n sup t [,r] r 2n ϕ(t) {2 4 2n} {(k + 1) (k + 3) (k + 2n 1)}, for any r > n, whih implies ϕ =, that is, τ = 2σ. This is a ontradition with the assumption that τ is linearly independent of σ. Step 2. Show that for any τ solution of (9) inequality (18) holds. Let us again write (9) as an integral equation. This time we annot integrate from r =, sine we expet an unbounded solution as r. Let r > be a fixed number. It is easy to show that τ will satisfy (9) if only if τ = τ + Sτ, (2) where τ (r) = + d r r t k dt, with = τ(r ), d = r k τ (r ), the operator S is defined on funtions θ integrable on ompat subsets of (, ) is given by the formula Sθ(r) = r t t k r r s k (q(s) l)θ(s) ds dt. Use one again the method of suessive approximations to express τ as the sum of an infinite r series. Let R, d R, define τ (r) = + d r t k dt, τ n+1 = τ + Sτ n, n. We have τ n = τ + Sτ + + S n τ, n. Notie that if ω is an integrable funtion on (, r ), then for < r r r r s k ω(s) ds s k ω(s) ds r k r r r r ω(s) ds. (21) By applying (21) we get the estimate τ (r) + d r r k for < r r. Then, t r s k (q(s) l)τ (s) ds (q + l )(r k + d r ) ds = C(q + l )(r t), r t where C = r k + d r. Apply (21) to estimate Sτ. Sτ (r) r k With the same reasoning, one obtains by indution r r C(q + l )(r t) dt = C(q + l )r k (r r) 2. 2! S n τ (r) C(q + l ) n r k (r r) 2n, < r r, n 1. (2n)! As it was done for σ, one an prove that the series τ = S n τ n= 7

8 is uniformly onvergent for r in ompat subsets of (, r ] that τ satisfies (2). One an then estimate τ. τ(r) S n τ (r) + Cr e k q + l (r r), < r r, n= whih implies (18). Step 3. Show that any solution τ to (9) whih is linearly independent of σ satisfies (19). From (18) we dedue that r k (q(r) l)τ(r) is bounded as r. Integrate (9) from to r. We get r k τ (r) = 3 + r s k (q(s) l)τ(s) ds, (22) with 3 a onstant. Suppose that 3 =. Then, from (18) (22) it follows that r k τ (r) Mr for < r r some M >. But then, τ (r) M/r k 1 for < r r. Note that k is an odd number, sine k = 2m + 1 with m an integer. In partiular k 2, or k 1 1. Sine τ(r) = τ(r ) + r r τ (s) ds, we obtain τ(r) A 1 + B 1 r k 2, for some A 1 >, B 1 >. This is the same as (18) but with k 2 instead of k. By repeating several times the same reasoning starting with (22) the assumption 3 = we will be able to redue the exponent of r in this inequality by 2 every time, until we get that τ must be bounded as r. As shown at step 1, then τ is linearly dependent of σ, whih is a ontradition. Thus 3. Then, (18) (22) give us τ (r) = O(r k ) as r, or (19). From (19) the equality τ(r) = τ(r ) + r r τ (s) ds it follows that as r, τ(r) = O(ln r) if k = 1, τ(r) = O(r k+1 ) if k > 1. One an then get either (11a) or (11b) by onveniently multiplying τ by a non-zero onstant. This proves the lemma. The following lemma will study the properties of j m (r, l) for l real. Lemma 3.2 Let l = λ R. Then j m (r, λ) j m(r, λ) are real. If λ, then j m (r, λ) r m +1/2 j m(r, λ) >. Proof. We will onsider only the ase m, sine j m(r, λ) is an even funtion of m. As λ is real, we have q(r) λ is real. With the notation from the proof of Lemma 3.1, j m (r, λ) = r m+1/2 σ(r, λ), with σ(r, λ) given by the series (17). By using this series, definition (13) of the operator T we dedue that σ(r, λ) is real, onsequently, j m(r, λ) is real. If λ, then q(r) λ. In this ase, the operator T will map non-negative funtions into non-negative funtions. We onlude from the onstrution (17) of σ that σ(r, λ) 1, so, j m (r, λ) r m+1/2. To show j m(r, λ) > it suffies to prove σ (r, λ), whih is an immediate onsequene of (15). The next lemma will study the properties of j m(r, l) for large absolute value of m l = λ a real number. Lemma 3.3 Let r > be a fixed number Λ [, ) be a bounded set. If m is suffiiently large, then for all λ Λ j m (r, λ) >, j m(r, λ) >. 8

9 Proof. We an assume m. Denote k = 2m We have that j m(r, λ) = r m+1/2 σ(r, λ), with σ(r, λ) defined by the series (17). Here we will prefer to use the notation σ m (r, λ) to emphasize its dependene on m. By using this series estimate (14) it follows that 1 σ m (r, λ) C k + 1, λ Λ, with C = C(Λ, q ) a onstant, where q = sup r [, ) q(r). From (15) we dedue σ m(r, λ) D k + 1, λ Λ, with D = D(Λ, q ) another onstant. These observations imply that for m large enough λ Λ, σ m(r, λ) is lose to 1, while σ m(r, λ) is lose to zero. Then, sine j m (r, λ) = r m+1/2 σ m (r, λ), j m(r, λ) = r m 1/2 {(m + 1/2) σ m(r, λ) + r σ m(r, λ)}, it follows quikly that for m large enough both of these quantities are stritly positive. Lemma 3.4 For l { π/2 < arg(l d 2 ) < 3π/2}, there exist two solutions w m(r, l) x m (r, l), r R, m Z, of (6) whih are analyti in l. If l C + = {l : Im l > }, then as r, w m(r, l) together with its derivative will deay exponentially, while x m(r, l) its derivative will inrease exponentially. Proof. Set Reall that q(r) = d 2 for r R, therefore on [R, ) (6) beomes { } w + l d 2 m2 1/4 w =. r 2 w m (r, l) = rh (1) m ( l d 2 r), x m (r, l) = rh (2) m ( l d 2 r), (23) where H (1) m (ζ) H (2) m (ζ) are the Hankel funtions of m-th order. They are analyti funtions with the domain { π < arg ζ < π}. Consider the funtion z defined on { π/2 < arg z < 3π/2} with values in { π/4 < arg ζ < 3π/4}. We obtain that w m (r, l) x m (r, l) are analyti funtions of l { π/2 < arg(l d 2 ) < 3π/2}, whih is the whole omplex plane exept those l for whih l d 2 has a zero real part a non-positive imaginary part. Aording to formulas (9.2.3) (9.2.4) from,[1] we have the asymptoti expansions as ζ arg ζ < π. Then, w m(r, l) = O ( e i l d 2 r ), H (1) m (ζ) 2/(πζ) e i(ζ mπ/2 π/4) H (2) m (ζ) 2/(πζ) e i(ζ mπ/2 π/4) x m(r, l) = O ( e i l d 2 r ) as r. If l C +, then Im l d 2 > thus, i l d 2 has a stritly negative real part. Therefore, as r, w m (r, l) will deay exponentially, while x m (r, l) will inrease exponentially. By formally differentiating the above equalities (for a rigorous justifiation one needs to use equalities (9.2.13) (9.2.14) from [1]) we dedue that w m(r, l) will deay exponentially, while x m(r, l) will inrease exponentially as r. 9

10 4 Classifiation of the Solutions The main purpose of this paper is to prove that under ertain onditions, the solution to the Helmholtz equation (2) is a superposition of funtions of the form (3). For eah of the funtions in the superposition, v(r) will satisfy (5) with the notations of (4), l will be a real variable, whih we will denote by λ (thus we will reserve the notation l for the omplex variable, the notation λ for its restrition to the real axis). In addition the following properties will hold: v(r) is bounded as r, (24a) r { is in L 2 (R, ), if λ d 2, rv(r) is bounded as r, if λ > d 2. (24b) In this setion we will study lassify the funtions v(r) with the properties (24a) (24b). A solution v(r) of (5) has the form v(r) = w(r)/ r, with w(r) satisfying (6). Lemma 3.1 shows how the solutions of (6) look like. It is lear that in order that v(r) satisfy (24a), we need w(r) = j m(r, λ), or jm(r, λ) v(r) =. (25) r Condition (24b) is then satisfied if only if { is in L 2 (R, ), if λ d 2, r j m (r, λ) is bounded as r, if λ > d 2. (26) Next we will investigate for whih λ ondition (26) holds. We will need to onsider four ases: λ, < λ < d 2, λ = d 2 λ > d 2. In eah of these intervals j m (r, λ) onsequently v(r), will have a different behavior. Case 1. If λ, then aording to Lemma 3.2, j m(r, λ) r m +1/2, therefore j m(r, λ) will be not square integrable on (R, ). Case 2. Assume < λ < d 2. For r R the funtion q(r) defined in (4) is onstant equal to d 2. Then (6) beomes { } w + λ d 2 m2 1/4 w =, r [R, ). (27) r 2 The solutions to this equation are k m(r, λ) = rk m( d 2 λ r), λ < d 2, r R, (28) rim ( d 2 λ r), λ < d 2, r R, where K m I m are the modified Bessel funtions. Formulas (9.7.1) (9.7.2) from [1] give us the expansions skm(s) π/2 e s as s, sim (s) π/2 e s as s. Thus, we have one solution deaying exponentially while another inreasing exponentially. In order that j m(r, λ) be in L 2 (R, ) we need j m(r, λ) = Ck m(r, λ) for r R, 1

11 for some onstant C. Sine both j m(r, λ) Ck m(r, λ) satisfy the same seond order differential equation, namely (27), then from the ontinuity of these funtions of their first derivatives at r = R we find that j m (R, λ) Ck m (R, λ) =, j m(r, λ) Ck m(r, λ) =. Thus, C must satisfy simultaneously two different onditions. This is possible if only if Then we will have j m(r, l) j m (R, l) = k m(r, l) k m (R, l), λ < d2. (29) j m (r, λ) = jm(r, λ) k m (R, λ) k m(r, λ), r R. (3) It will be shown later that for eah m, the set of λ suh that (29) holds is finite. Notie that in this ase j m (r, λ) will deay exponentially as r. Its derivative has the same property, this follows from the asymptoti expansion [ sk m (s)] π/2 e s as s (31) (aording to the formulas (9.7.2) (9.7.4) from.[1]) In onlusion, the only λ (, d 2 ) for whih (26) holds, are those satisfying (29). Case 3. Let now λ = d 2. Two linear independent solutions of (27) are in this ase r r 1/2 m, m Z, { r ln r for m =, r r 1/2+ m for m. The seond solution is not bounded for any m Z. With the same reasoning as before (29), one an show that j m (r, λ) will be proportional to the first of these two solutions if only if j m(r, l) j m(r, l) m 1/2 =, λ = d 2, (32) R then, jm(r, λ) j m(r, λ) = R 1/2 m r1/2 m, r R. (33) The funtion j m (r, λ) will be in L 2 (R, ) if only if m 2. A funtion of the form (3), with v(r) given by (25), for whih < λ d 2 either (29), or (32) (with m 2) holds, is alled a guided mode. Note that a guided mode deays in r either exponentially, or as r m ( m 2). Case 4. Let λ > d 2. Two solutions of (27) are then a m (r, λ) = rj m ( λ d2 r ), b m (r, λ) = ry m ( λ d2 r ), λ > d 2, (34) where J m Y m are the Bessel funtions of the first seond kind. Note the formulas s Jm(s) = π/2 os(s mπ/2 π/4) + O(s 1/2 ), s Ym (s) = π/2 sin(s mπ/2 π/4) + O(s 1/2 ), 11

12 as s (they are a partiular ase of formulas (9.2.1) (9.2.2) from.[1]) We infer that a m (r, λ) b m (r, λ) will be bounded as r. By formally differentiating the above formulas it follows that a m(r, λ) b m(r, λ) will also be bounded as r. The funtions a m (r, λ) b m (r, λ) are linearly independent, sine J m Y m are linearly independent. Then, for r R, j m (r, λ) will be a linear ombination of them. To find the oeffiients of the linear ombination, set j m (r, λ) = m (λ)a m (r, λ) + d m (λ)b m (r, λ), r R. By again using the ontinuity of these funtions their derivatives at r = R, we obtain a linear system whih enables us to solve for m d m. Apply the equality Y m(z)j m(z) J m(z)y m(z) = 2/(πz), ((9.1.16) from [1]) to find a value for the determinant of this linear system. We obtain b m(r, λ)a m(r, λ) a m(r, λ)b m(r, λ) = 2/π, (35) therefore, m(λ) = π 2 {b m(r, λ)j m(r, λ) j m(r, λ)b m(r, λ)}, (36a) d m(λ) = π 2 {a m(r, λ)j m(r, λ) j m(r, λ)a m(r, λ)}. (36b) It is easy to see that j m (r, λ) its derivative will be bounded as r, so, ondition (26) will be satisfied for all λ > d 2. Let us look at the expression of (3) for λ > d 2 with v(r) given by (25). Notie that if d 2 < λ < k 2 n 2, then (3) will be osillatory in z (reall that k 2 β 2 = k 2 n 2 λ). In this ase we will say that (3) is a radiation mode. On the other h, if λ > k 2 n 2, then β beomes imaginary. Depending on the sign of Imβ we will have exponential deay in one of the diretions z, z, exponential growth in the other one. For λ > k 2 n 2, (3) will be alled an evanesent mode. 5 The Theory of Eigenvalue Problems Consider the eigenvalue problem w + {l Q(r)}w =, r (, ), (37) where l C. Assume that Q is integrable over any ompat subset of (, ) (in [2] [8] the theory is developed only for ontinuous funtions Q, but it is mentioned in a footnote on page 224 of [2] that it suffies for Q to be as we assume above). Let < R < be an arbitrary but fixed number. Let ϕ(r, l) θ(r, l) be the solutions of (37) with the boundary onditions { ϕ(r, l) =, ϕ (R, l) = 1, (38) θ(r, l) = 1, θ (R, l) =. Sine (37) has an analyti dependene on the parameter l, the solutions θ(r, l) ϕ(r, l) will be analyti funtions of l for r fixed. Any solution to (37) linearly independent of ϕ an be represented, up to a onstant multiple, in the form ψ = θ + Mϕ, (39) with M C. Let τ R < t <. Look for a solution of (37) of the form (39) to satisfy the boundary ondition os τψ(t, l) + sin τψ (t, l) =. 12

13 It is a diret alulation to hek that we need M = M(l) = θ(t, l) os τ + θ (t, l) sin τ ϕ(t, l) os τ + ϕ (t, l) sin τ. (4) Let C + be the open upper omplex-half-plane, C + = {l : Im l > }. As shown in Chapter 2 of [8] the following results hold: as t, M(l) onverges uniformly on ompat subsets of C + to a funtion M (l) analyti on C +. Moreover, the funtion is in L 2 (, R), one has ψ (r, l) = θ(r, l) + M (l) ϕ(r, l), l C +, (41) R ψ (r, l) 2 dr = Im M(l). (42) Im l As t, M(l) onverges uniformly on ompat subsets of C + to an analyti funtion M (l) on C +, if ψ (r, l) = θ(r, l) + M (l) ϕ(r, l), l C +, (43) then ψ (r, l) L 2 (R, ) R ψ (r, l) 2 dr = Im M (l). (44) Im l Note that the obtained M, M, ψ, ψ depend on the parameter τ R. Thus, possibly these quantities, therefore the representation given below, in Theorem 5.1, will not be unique. This might be true in general, but in our onrete ase, given by (6), these quantities will turn out to be unique, as we will see from Lemma 6.1. So then the transform we are looking for (whih is alulated in Theorems ) will be unique. In Setion 9.5 of [2] Chapter 3 of [8] it is proved that for any λ R the following limits exist ξ(λ) = lim δ + η(λ) = lim δ + ζ(λ) = lim δ + λ λ λ 1 Im M (s + iδ) M (s + iδ) ds, M (s + iδ) Im M (s + iδ) M (s + iδ) ds, Im M (s + iδ)m (s + iδ) M (s + iδ) M (s + iδ) ds. (45a) (45b) (45) It is shown there that the funtions ξ ζ are non-dereasing, that η is with bounded variation. In addition, for any λ < λ 1 real numbers, {η(λ 1) η(λ )} 2 {ξ(λ 1) ξ(λ )}{ζ(λ 1) ζ(λ )}, (46) as stated on page 252 of [2] (with a different notation). Then, equalities (3.1.8), (3.1.9), (3.1.1) from [8] give us an expansion formula for a funtion g L 2 (, ) in terms of θ(r, l), ϕ(r, l) the funtions ξ, η ζ. The same result is proved in [2] at Theorem 5.2. In this referene the representation result is stated more rigorously, so we will prefer it over.[8] To state the result we need some notations. 13

14 Denote ρ = (ξ, η, ζ). For any vetor Γ = (Γ 1, Γ 2), where Γ 1, Γ 2 : R C, let Γ 2 = Γ 1 (λ) 2 dξ + 2Re{Γ 1 (λ) Γ 2 (λ)} dη + Γ 2 (λ) 2 dζ. (47) The fat that ξ η are non-dereasing, together with (46), gives us that Γ 2. It is easy to hek that is a semi-norm. Denote by L 2 (ρ) the spae of all Γ = (Γ 1, Γ 2 ) suh that Γ <. This is then the statement of Theorem 5.2 from.[2] Theorem 5.1 If g L 2 (, ), the vetor Γ = (Γ 1, Γ 2 ), where Γ 1(λ) = θ(r, λ)g(r) dr, Γ 2(λ) = onverges in L 2 (ρ), that is, there exists Γ L 2 (ρ) suh that where for < < d < Γ Γ d as, d, ϕ(r, λ)g(r) dr, The expansion Γ d 1 (λ) = d θ(r, λ)g(r) dr, Γ d 2 (λ) = d ϕ(r, λ)g(r) dr. (48) g(r) = 1 π { θ(r, λ)γ1(λ) dξ(λ) + θ(r, λ)γ 2(λ) dη(λ) + ϕ(r, λ)γ 1(λ) dη(λ) + ϕ(r, λ)γ 2(λ) dζ(λ) } holds, with the latter integral onvergent in L 2 (, ), that is, g στ, τ, where for < τ < σ < g in L 2 (, ) as σ g στ (r) = 1 π τ σ { θ(r, λ)γ1(λ) dξ(λ) + θ(r, λ)γ 2(λ) dη(λ) + ϕ(r, λ)γ 1(λ) dη(λ) + ϕ(r, λ)γ 2(λ) dζ(λ) }. (49) We have the Parseval identity g(r) 2 dr = 1 π Γ 2. (5) 6 Computing the Measures In the setions to follow we will apply the results of Setion 5 to our partiular eigenvalue equation, given by (6). Thus, for the funtion Q(r) in (37) we will have the expression Q(r) = q(r) m2 1/4 r 2, r (, ). In this setion we will alulate the measures given in (45a), (45b) (45). 14

15 Lemma 6.1 Let M m (l), M (l), m ψ m (r, l), ψ (r, m l) (l C + = {l : Im l > }) be the quantities defined in Setion 5 for equation (6) (we use the supersript m to emphasize their dependene on m Z). Let j m(r, l) w m(r, l) be the solutions of (6) defined respetively in Lemma 3.1 by (23). Then, M m (l) = j m(r, l) j m(r, l), M (l) m = w m(r, l) w m(r, l), (51) ψ m (r, l) = j m(r, l) j m(r, l), ψm (r, l) = w m(r, l) w m(r, l). (52) Proof. It is easy to note that j m (r, l) y m (r, l) defined in Lemma 3.1 are linearly independent solutions of (6), thus any other solution will be a linear ombination of these two. Then θ(r, l) φ(r, l) defined in Setion 5 an be represented as { θ(r, l) = α(l)j m(r, l) + β(l)y m(r, l), (53) ϕ(r, l) = γ(l)j m(r, l) + δ(l)y m(r, l), for some oeffiients α, β, γ, δ. Let (l) = j m(r, l)y m(r, l) j m(r, l)y m(r, l). (54) Being the Wronskian of two linearly independent solutions, (l) is non-zero. Using the boundary onditions (38) it is easy to find that α = y m(r, l) (l) From (53) (4) it follows that, β = j m(r, l), γ = (l) ym(r, l), δ = (l) M(l) = {αjm(t, l) + βym(t, l)} os τ + {αj m(t, l) + βy m(t, l)} sin τ {γj m (t, l) + δy m (t, l)} os τ + {γj m(t, l) + δy m(t, l)} sin τ. jm(r, l). (55) (l) For t the formulas (7), (8a) (8b) tell us that the term y m (t, l) will dominate j m (t, l) j m(t, l), while y m(t, l) will dominate y m(t, l). Then, M m (l) = lim t M(l) = β δ. By applying (55) we obtain the first equality in (51). To get the first equality in (52) use definition (41) of ψ m (r, l) equalities (53) to (55). The quantities M (l) m ψ (r, m l) are alulated in exatly the same way. One needs to express ϕ(r, l) θ(r, l) in terms of w m (r, l) x m (r, l), then put t in (4) use the properties of w m(r, l), x m(r, l), their derivatives shown in Lemma 3.4. The lemma is proved. The next lemma will study the properties of the funtions M m (l), M (l), m M m (l) M (l). m Here we will set some notation write some formulas whih will be used in the lemma. By λ we will denote a real variable. Let w m(r, λ), k m(r, λ), a m(r, λ), b m(r, λ) be the funtions defined by (23), (28) (34). The following equalities hold, H (1) m (iz) = 2 iπ e im/2 K m (z), π < arg z π/2 H (1) m (z) = J m (z) + iy m (z), π < arg z < π, (56) 15

16 (formulas (9.6.4) (9.1.3) from.[1]) We dedue w m (r, λ) = 2 iπ e im/2 k m (r, λ), λ < d 2, (57a) w m (r, λ) = a m (r, λ) + ib m (r, λ) λ > d 2. (57b) Lemma 6.2 M m (l) is meromorphi aross all the omplex plane, while M m (l) (l C +) extends ontinuously to the real axis. M m (λ) M m (λ) is real or infinite for λ < d 2, it has a finite number of zeros on the real axis, all in the interval (, d 2 ]. Proof. As stated in Lemma 3.1, j m(r, l) j m(r, l) will be analyti funtions of l C. Then M m (l), being obtained as their ratio, will be a meromorphi funtion. The funtion w m (r, l) was defined in Lemma 3.4. It was proved there that w m (r, l) is defined analyti for all l { π/2 < arg(l d 2 ) < 3π/2}. In partiular w m(r, λ) is defined for all real λ d 2. To prove that M (l) m extends ontinuously to the real axis, it suffies to show that its denominator, w m(r, λ), is not zero for λ R\{d 2 } that lim l d 2 M (l) m exists. If λ < d 2, then w m (r, λ) beause of (28) (57a), sine the Bessel funtion K m (s) takes real stritly positive values for s > (as it follows from setion (9.6.1) of.[1]) If λ > d 2, then w m (r, λ) beause of (34) (57b), sine for s > the Bessel funtions J m (s) Y m (s) take real values annot be zero at the same time. Show that lim l d 2 M (l) m exists. If m, then aording to (9.1.8) (9.1.9) from, [1] we have that for z C, z H (1) m (z) (1/π)(m 1)!(z/2) m, for m 1, H (1) m (z) ( 2i/π) ln z, for m =. We an extend these to m <, by using (56) together with J m (z) = ( 1) m J m (z), Y m (z) = ( 1) m Y m (z), m Z (formula (9.1.5) from.[1]) One an obtain the behavior of the derivative of H (1) m (z) as z by formally differentiating the above. Set z = R l d 2. We dedue that for all m Z lim M (l) m = l d 2 m 1/2. (58) R Note that M m (λ) is real or infinite for λ < d 2, being the quotient of j m(r, λ) j m (R, λ) both of whih are real aording to Lemma 3.1. We have that M m (λ) is real for λ < d 2, that follows from (57a) by using again the properties of the funtion K m (s). Then, M m (λ) M m (λ) is real or infinite for λ < d 2. Let us show the last part of this lemma, the fat that M m (λ) M m (λ) finite number of zeros on the real axis, all in the interval (, d 2 ]. Let first prove that this funtion an have no zeros for λ. From (57a) we have where M m (λ) M m (λ) = j m(r, λ) j m (R, λ) + k m(r, λ) k m (R, λ) = D m(r, λ) j m (R, λ)k m (R, λ), D m (r, λ) = j m(r, λ)k m (r, λ) j m (r, λ)k m(r, λ). Assume for some λ, M m (λ) M m (λ) =. Then D m (R, λ) =. We will prove that is false. First note that D m(r, λ) = D m(r, λ) for r R. Indeed, j m(r, λ) k m(r, λ) will satisfy (6) (for j m (r, λ) this follows by its definition, for k m (r, λ) it follows from the observation that 16

17 aording to (57a), k m(r, λ) is a onstant times w m(r, λ) whih, as defined by (23), is a solution of (6) for r R). Then it is easy to hek that D m(r, λ) = j m(r, λ)k m (r, λ) j m (r, λ)k m(r, λ) =, so D m(r, λ) is onstant in r. Seond, note that for r large enough D m(r, λ) <. Indeed, for r >, j m (r, λ) >, j m(r, λ) > by Lemma 3.2, k m (r, λ) > by the properties of Bessel funtions. Also, for r suffiiently large we have k m(r, λ) <, this is a onsequene of (31). We infer that for r large enough D m (r, λ) <. These two observations imply D m (R, λ) <, therefore D m(r, λ) is non-zero, M m (λ) M m (λ). Now let < λ d 2. We showed that M m (l) is meromorphi on the whole omplex plane. The funtion w m (r, l), as defined by (23), is analyti on { π/2 < arg(l d 2 ) < 3π/2}, in partiular it is analyti on the set of l suh that Re l < d 2. Then M m (l) = w m(r, l)/w m(r, l) is meromorphi on the same region, so is M m (l) M m (l). Therefore, it an have only a disrete number of zeros on the interval (, d 2 ). Assuming that the number of zeros is infinite, their only possible aumulation point is λ = d 2. So, there would exist a sequene λ n < d 2, n 1, with lim λ n = d 2 M m (λ n ) M m (λ n) = for all n 1. Aording to formulas (9.1.3), (9.1.1) (9.1.11) from, [1] the Hankel funtion H (1) (z) will have the representation H (1) m (z) = f 1(z) + f 2(z) ln z, with f 1 (z) f 2 (z) meromorphi funtions of z C. Then, if we reall definition (23) of w m(r, l), one an alulate that we will have the representation M m (l) M m (l) = g1(t) + g2(t) ln t g 3 (t) + g 4 (t) ln t, with g 1, g 2, g 3 g 4 meromorphi funtions on the whole omplex plane, t = l d 2 C. We infer g 1(t n) + g 2(t n) ln t n =, for all n 1, where t n = λ n d 2. Two ases are possible. If g 2 (t n ) is zero for infinitely many n, this implies g 1 (t n ) = at the same points. Then the meromorphi funtions g 1 (t) g 2 (t) are identially zero, so is M m (l) M m (l), obtaining a ontradition. Otherwise, if g 2(t n) is zero only for finitely many n, we an write ln t n = g 1(t n ) g 2 (t n ), n n. On the right-h side we have a meromorphi funtion. As n we have t n, thus, for some integer p, there follows ln t n = O( t n p ), whih is learly impossible. This ontradition shows that M m (λ) M (λ) m an have only finitely many zeros on [, d 2 ). Another possible zero ould be at λ = d 2. Lastly, show that M m (λ) M (λ) m is never zero on (d 2, ). Aording to (51) (57b), so, M m (λ) = a m(r, λ) + ib m(r, λ) a m (R, λ) + ib m (R, λ), M m (λ) M m (λ) = j m j m + a m + ib m a m + ib m = (j ma m a mj m ) + i(j mb m b mj m ) j m (a m + ib m ) (59) 17

18 d 2 The funtion χ m (λ). λ (the arguments (R, λ) were omitted for simpliity). If we assume that for some λ > d 2, M m (λ) M m (λ) =, this implies j m(r, λ)a m(r, λ) a m(r, λ)j m(r, λ) =, j m(r, λ)b m (R, λ) b m(r, λ)j m (R, λ) = (sine these quantities are real, as it follows from Lemma 3.2 (34)). The numbers j m (R, λ) j m(r, λ) annot be both zero, sine j m(r, λ) is a non-zero solution of the seond order differential equation (6). In this ase the vetors (a m (R, λ), a m(r, λ)) (b m (R, λ), b m(r, λ)) must be linearly dependent. That annot be sine the funtions a m (r, λ) b m (r, λ) (r R) are also solutions of (6), they are linearly independent. Thus M m (λ) M (λ) m. This finishes the lemma. Theorem 6.3 Let ξ m, η m ζ m be the funtions defined by (45a), (45b) (45). There exists a non-dereasing funtion χ m : R R suh that the following measures are equal dξ m(λ) = j m(r, λ) 2 dχ m(λ), dη m (λ) = j m(r, λ)j m (R, λ) dχ m (λ), dζ m(λ) = j m(r, λ) 2 dχ m(λ). (6) The funtion χ m is identially zero for λ (, ], is pieewise onstant for λ (, d 2 ) where it has a finite number of disontinuities, is ontinuous for λ (d 2, ). Proof. Denote 1 M(l) = M m(l) M (l), m Aording to (45a), for any λ < λ 1 real numbers l C+. ξ m (λ 1 ) ξ m (λ ) = lim δ + λ 1 λ Im M(s + iδ) ds. 18

19 In partiular, if M(l) extends ontinuously to the interval [λ, λ 1], then by using Lebesgue s theorem of dominant onvergene it is easy to show that ξ m (λ 1 ) ξ m (λ ) = λ 1 λ Im M(s) ds. (61) The same kind of reasoning learly holds for η m ζ m. As it follows from Lemma 6.2, if λ then M m (λ) M (λ) m is real or infinite, non-zero. By applying (61) we find that for any λ 1 < λ 2 <, ξ m (λ 1 ) ξ m (λ ) =. From (45a) we have that ξ m() =, thus ξ m(λ) = for all λ. In the same fashion one obtains that for λ, η m (λ) = ζ m (λ) =. Set χ m (λ) = for λ, (6) will hold. Let λ m 1 < λ m 2 < < λ m P m be the points in the interval (, d 2 ] where, aording to Lemma 6.2, M m (λ) M (λ) m =. We an use Lemma 6.2 (61) to dedue that ξ m, η m ζ m are onstant on eah of the intervals making up (, d 2 ]\{λ m 1, λ m 2,..., λ m P m }. Set χ m to be onstant on eah of these intervals. At eah of the points λ m 1, λ m 2,..., λ m P m the funtions ξ m, η m ζ m ould have a jump. To show (6) on (, d 2 ] we need to find a relationship between the jumps of these funtions. We will return to this shortly. The remaining ase, λ > d 2, is treated similarly. Equation (59) in the proof of Lemma 6.2 gives an expression for M m (λ) M (λ) m on this interval (with the notation from (34)). By using the fat that the quantities a m, b m j m together with their derivatives are real, we an alulate Im M(λ) = Im = { j m (a m + ib m ) (j ma m a mj m) + i(j mb m b mj m) j 2 m(b ma m a mb m) (j ma m a mj m ) 2 + (j mb m b mj m ) 2. Use (35) to simplify the numerator of this fration. Apply (61). We get that for any d 2 < λ < λ 1, ξ m (λ 1 ) ξ m (λ ) = π 2 λ 1 λ j m(r, λ) 2 m (λ) 2 + d m (λ) 2 dλ, where m(λ) d m(λ) are defined by (36a) (36b). So we have dξ m(λ) = π 2 j m(r, λ) 2 dλ m (λ) 2 + d m (λ) 2. We want (6) to hold. Define χ m(λ) for λ > d 2 suh that dχ m (λ) = π 2 dλ m (λ) 2 + d m (λ) 2, then the first of the three identities (6) is valid. It is easy to repeat the same alulation for η m ζ m show that for λ > d 2 the other two identities in (6) hold. Now we will return to what is the longest part of the proof, the study of what happens at the points λ (, d 2 ] where M m (λ) M (λ) m =. Let λ be suh a point. An immediate observation is that M m (λ ) is finite j m (R, λ ). Indeed, it was proved in Lemma 6.2 that M (λ) m is finite for λ real. So, M m (λ ) whih equals M (λ m ) is finite. Then, sine M m (λ ) = j m(r, λ )/j m (R, λ ) beause j m(r, λ ) annot beome zero simultaneously with j m(r, λ ) (j m(r, l) is a solution of (6)), we dedue j m(r, λ ). } 19

20 With this observation in h, in order to prove that (6) holds at λ one needs to show that dη m(λ ) = M m (λ ) dξ m(λ ), dζ m(λ ) = M m (λ ) 2 dξ m(λ ), (62) then define dχ m(λ ) = dη m(λ )/j m(r, λ ) 2. Let r be the jump of ξ m at λ. Reall, ξ m was defined by (45a), so, r = lim lim λ +ε ε + δ + λ ε 1 Im M m(s + iδ) M (s m + iδ) ds. By using (45b), the first equality in (62) an be written as lim lim λ +ε ε + δ + λ ε M m (s + iδ) Im M m(s + iδ) M (s m + iδ) ds = M m (λ ) r. M m (l) is analyti around λ. Therefore, in a neighborhood of λ M m (s + iδ) = M m (λ ) + {M m (s) M m (λ )} + iδh(s + iδ), with H a ontinuous funtion around λ. Substitute this above. By using the fat that M m (λ ) is real the definition of r, we get the equivalent equality M m (λ ) r + lim lim λ +ε ε + δ + λ ε Im {M m (s) M m (λ )} + iδh(s + iδ) M m (s + iδ) M m (s + iδ) ds = M m (λ ) r, (63) so we have to prove that the limit of the integral on the left-h side of (63) is zero. To do this, we will need additional information about M m M. m Let l = s + iδ (δ > ). Reall that formulas (42) (44) hold, where ψ m ψ m are defined in (41) (43) an expression for them is given by (52). Add equalities (42) (44). We obtain 1 δ Im{M m (l) M m (l)} = R jm(r, l) 2 j m (R, l) dr + R wm(r, l) 2 w m (R, l) dr. This equality gives us two things. First, that Im{M m (l) M m (l)} >, therefore, 1 Im M m(l) M >. (64) (l) m Seond, for l lose to λ, the quantity δ 1 Im{M m (l) M (l)} m is bounded from below by a stritly positive number, say ω 1. Then δ 1 M m (l) M (l) m is bounded below by the same number therefore, δ M m(l) M (l) ω. (65) m 2

21 The integral in (63) an be written as λ +ε λ ε M m (s) M m (λ ) Im M m(s + iδ) M (s m + iδ) ds + λ +ε λ ε iδh(s + iδ) Im M m(s + iδ) M ds. (66) (s m + iδ) Denote ε = sup s λ ε M m (s) M m (λ ). Sine M m is ontinuous in a neighborhood of λ, we will have ε as ε. Let us estimate the first integral from (66). λ +ε M m (s) M m λ+ε (λ ) Im M m(s + iδ) M (s m + iδ) ds Im M m (s) M m (λ ) M m(s + iδ) M (s m + iδ) ds λ ε = λ +ε λ ε λ ε M m (s) M m (λ ) Im 1 M m(s + iδ) M (s m + iδ) ds ε = ε λ +ε λ ε λ +ε λ ε Im 1 M m(s + iδ) M (s m + iδ) ds 1 Im M m(s + iδ) M (s m + iδ) ds = ε [ξ m (λ + ε) ξ m (λ ε)]. In deriving this we used the fat that M m (s) M m (λ ) is real, together with (64) (45a). Clearly as ε, the integral goes to zero. Now estimate the seond integral in the sum (66). Let H be an upper bound of H(s + iδ) for l = s + iδ in a neighborhood of λ. Inequality (65) implies that as ε, λ +ε λ+ε iδh(s + iδ) Im M m(s + iδ) M (s m + iδ) ds H ω ds = 2εH ω. λ ε This proves the first equality in (62). Let us prove the seond equality in (62). Reall that ζ m is given by (45). The above approah does not apply immediately, sine unlike M m (l) (the numerator in (45b)), the funtion M m (l)m (l) m (the numerator in (45)) will not be analyti around λ if λ = d 2 (sine as seen from Lemma 3.4, w m (r, l) is not defined in a neighborhood of l = d 2 ). The idea is then to use the equality xy x y = λ ε x2 x y x, to write the jump of ζ m at λ as λ +ε lim lim M m (s + iδ) 2 Im ε + δ + M m(s + iδ) M ds + lim (s m + iδ) λ ε lim λ +ε ε + δ + λ ε Im M m (s + iδ) ds. The limit of the seond integral in the sum will be zero, sine as argued above, M m will be finite at λ ( therefore, around λ ) thus the quantity inside the integral is bounded. For the 21

22 first integral in the sum we an proeed in the same way we alulated the jump of η m. This finishes the proof of (6). Finally, we need to justify the laim that χ m is a non-dereasing funtion. From (6) we have dξ m (λ) + dζ m (λ) = {j m (R, λ) 2 + j m(r, λ) 2 }dχ m (λ). The left-h side of this is non-negative measure, sine by theory ξ m ζ m are non-dereasing. The number j m (R, λ) 2 + j m(r, λ) 2 is stritly positive, as j m (R, λ) j m(r, λ) annot be both zero, j m(r, λ) being a non-zero solution to the seond order differential equation (6). Then we get that dχ m (λ) is a non-negative measure, whih shows that χ m is non-dereasing. Corollary 6.4 Let λ (, d 2 ] be a disontinuity point for χ m. Then, if λ < d 2, the following hold: j m(r, λ) j m (R, λ) = k m(r, λ) k m (R, λ), (67a) While for λ = d 2, j m(r, λ) = j m(r, λ) km(r, λ), r R. (67b) k m(r, λ) j m(r, λ) j m (R, λ) m 1/2 =, (68a) R j m (r, λ) = j m(r, λ) R 1/2 m r1/2 m, r R. (68b) In partiular, for λ (, d 2 ] a disontinuity point of χ m, j m (r, λ) deays exponentially as r if λ < d 2, j m(r, λ) r 1/2 m as r if λ = d 2. Proof. As it follows from the proof of Theorem 6.3, for suh a λ we will have M m (λ ) M (λ m ) =. With the help (51), (57a), (58), we dedue (67a) (68a). Notie that these equalities are exatly the onditions (29) (32). Then (67b) (68b) follow from (3) (33). 7 Computing the Transform Denote by L 2 (χ m) the spae of all funtions G : R C suh that G(λ) 2 dχ m(λ) <, where χ m is the non-dereasing funtion defined in Theorem 6.3. Theorem 7.1 Let g L 2 (, ). The integral G m (λ) = j m (r, λ)g(r) dr (69) is onvergent in L 2 (χ m), in the sense that there exists G m L 2 (χ m) suh that G d m G m in L 2 (χ m ) as d, where G d m (λ) = d j m (r, λ)g(r) dr, < < d <. (7) 22

23 The equality g(r) = 1 π j m(r, λ)g m(λ) dχ m(λ) (71) holds, in the sense that g στ g in L 2 (, ) as τ, σ, where g τσ (r) = 1 π We have the Parseval identity σ τ j m (r, λ)g m (λ) dχ m (λ), < σ < τ <. (72) g(r) 2 dr = 1 π G m (λ) 2 dχ m (λ). (73) Proof. We will apply Theorem 5.1. First note that if Γ = (Γ 1, Γ 2) with Γ 1, Γ 2 : R C, then beause of (6), the norm as defined in (47) an be written in the form Γ 2 = j m(r, λ)γ 1(λ) j m(r, λ)γ 2(λ) 2 dχ m(λ). (74) Reall the identities (53) to (55) for expressing θ(r, λ) ϕ(r, λ) in terms of j m(r, λ) y m (r, λ). Note that from (54) (55) we get the following equalities involving the oeffiients α, β, γ, δ α(λ)j m(r, λ) γ(λ)j m(r, λ) = 1, (75) β(λ)j m (R, λ) δ(λ)j m(r, λ) =. (76) Then, for < < d <, the funtions Γ d 1 Γ d 2 defined by (48) beome Γ d 1 (λ) = Γ d 2 (λ) = We denoted Γ d = (Γ d 1, Γ d 2 ). Let d d { α(λ)jm (r, λ) + β(λ)y m (r, λ) } g(r) dr, { γ(λ)jm (r, λ) + δ(λ)y m (r, λ) } g(r) dr. Γ d 1 (λ) = d α(λ)j m (r, λ)g(r) dr, Γd 2 (λ) = d γ(λ)j m (r, λ)g(r) dr. (77) set Γ d = ( Γ d d 1, Γ 2 ). We have Γ d Γ d =. (78) That follows from (74) (76). Theorem 5.1 guarantees the existene of Γ = (Γ 1, Γ 2 ) L 2 (ρ) suh that Γ Γ d as, d. 23

24 Then, equality (78) says that we have Γ Γ d as, d. But, aording to (74), Γ Γ d 2 = Note that {j m (R, λ)γ 1 (λ) j m(r, λ)γ 2 (λ)} j m(r, λ) Γ d 1 (λ) j m(r, λ) Γ d 2 (λ) = That follows from (75) (77). Thus, if we denote we obtain from (79) that {j m(r, λ) Γ d 1 (λ) j m(r, λ) Γ d 2 (λ)} 2 dχ m. (79) d j m(r, λ)g(r) dr. G m(λ) = j m(r, λ)γ 1(λ) j m(r, λ)γ 2(λ), λ R, (8) G m(λ) G d m (λ) 2 dχ m as, d (G d m was defined by (7)). This shows the first part of Theorem 7.1. Next we need to show that representation (71) holds. It suffies to prove that g τσ as defined by (49) in Theorem 5.1 is the same as g τσ defined in (72). And they are. To hek this one needs to start with g τσ as given in (49), substitute θ(r, λ) ϕ(r, λ) from (53), use (6) to express ξ m, η m ζ m in terms of χ m, use the equalities (75) (76), finally use definition (8) for G m (λ). Lastly, the Parseval identity (73) follows from (5), (74) (8). The theorem is proved. Theorem 7.2 Let g L 2 (, ). Let χ m be the non-dereasing funtion defined in Theorem 6.3. Let < λ m 1 < < λ m P m d 2 (P m ) be the points where χ m is disontinuous. Let r1 m,..., rp m m be the orresponding jumps. Let a m (r, λ) b m (r, λ) be the funtions defined by (34), m(λ) d m(λ) be defined by (36a) (36b). Then, r m k We have the representation g(r) = 1 π { 1 = π j m (r, λ m k ) dr} 2, k = 1,..., P m, (81) dχ m(λ) = π 2 P m k=1 dλ m(λ) 2 + d m(λ) 2, λ (d2, ). (82) r m k j m(r, λ m k )G m(λ m k ) d 2 j m (r, λ)g m (λ) dλ. (83) m(λ) 2 + d m(λ) 2 Proof. In Theorem 6.3 we proved the existene of the funtion χ m along the way we found its ontinuous part, that is, equality (82). We need to find its disrete part, that is, the value of all the jumps of χ m. Then (83) will follow by applying (71). 24

25 Let us notie the following observation. If λ 1 λ 2, w 1(r) w 2(r) satisfy (6) with λ = λ 1, λ = λ 2 respetively, then for any < < d < d Indeed, we an write d d w 1(r)w 2(r) dr = (λ 1 λ 2) 1 [w 1(r)w 2(r) w 1(r)w 2(r)]. (84) w 1 (r)w 2(r) dr = w 1(r)w 2 (r) dr = d d { } λ 1 q(r) m2 1/4 w 1(r)w 2(r) dr, r 2 { } w 1(r) λ 2 q(r) m2 1/4 w 2(r) dr. r 2 If we integrate by parts the left-h sides of these two equalities, then subtrat from first the seond, we get exatly (84). Before we prove (81), reall the behavior of j m (r, λ) as r. For λ d 2 a disontinuity point of χ m it is desribed in Corollary 6.4, while for λ > d 2 see the disussion in Setion 4. Let λ d 2 be one of the disontinuity points of χ m. Let r be the orresponding jump. We will onsider two ases: when j m(r, λ ) is square integrable, when it is not. The first ase splits into two subases: we an have either λ < d 2, or λ = d 2 with m 2. The seond ase happens for λ = d 2 m {, 1}. Let us start with the first ase. Denote g(r) = j m(r, λ ). Sine g(r) is square integrable, we an apply Theorem 7.1 for this partiular funtion. We will show that the orresponding G m (λ) as defined by (69) is suh that d G m (λ) = { j m(r, λ ) 2 dr, if λ = λ,, if λ λ. (85) for all λ suh that dχ m (λ). Then (81) will follow promptly; one needs to apply the Parseval identity (73) notie that sine r is the jump of χ m at λ, then dχ m(λ ) = r δ(λ λ ) with δ being Dira s funtion. Consider first the subase λ < d 2. That (85) is true for λ = λ follows immediately from (69). Assume now λ λ. Let us ompute G d m (λ) for < < d < as defined in (7). Use (84) with w 1 (r) = j m (r, λ ), w 2 (r) = j m (r, λ). Put d. From (7) we dedue that j m(, λ )j m (, λ) j m (, λ )j m(, λ) as. Also, both j m(d, λ )j m (d, λ) j m (d, λ )j m(d, λ) go to zero as d if dχ m (λ), sine on one h, j m (r, λ ) its derivative derease exponentially, on the other h, j m(r, λ) its derivative either derease exponentially for λ < d 2, or behave like a power of r for λ = d 2, or are bounded for λ > d 2. In any ase we get that G d m (λ) as d, so G m(λ) =. The subase λ = d 2 with m 2 follows in the same way. The statement that for λ λ dχ m (λ) both j m(d, λ )j m (d, λ) j m (d, λ )j m(d, λ) go to zero as d is argued in a little different way. We have that j m(r, λ ) its derivative deay as a negative power of r for r, while j m (r, λ) its derivative either deay exponentially for λ < λ, or stay bounded for λ > λ. But the onlusion is the same, G d m (λ) as d, thus (85) holds in this ase too. 25