In this section, we present the proofs of main results in this paper.
Proof of Theorem 1.
Before proving Theorem 1, we will mention some results, which will be needed later. We get the initial value problems
(18)-y′′+[ℓ(ℓ+1)x2+q(x)]y=λy, 0<x<1,(19)y(0)=0,(20)-y~′′+[ℓ(ℓ+1)x2 +q~(x)]y~ =λy~, 0<x<1,(21)y~(0)=0.
As known from [18], Bessel's functions of the first kind of order v=ℓ-1/2 are
(22)Jv(x)=∑k=0∞ (-1)kxv+2k2v+2kk!Γ(v+k+1)
and asymptotic formulas for large argument
(23)Jv(x)=2πx{cos[x-vπ2-π4]+O(1x)},Jv′(x)=-2πx{sin[x-vπ2-π4]+O(1x)}.
It can be shown [19] that there exists a kernel H(x,t)(H~(x,t)) continuous in the triangle 0≤t≤x≤1 such that by using the transformation operator every solution of (18), (19) and (20), (21) can be expressed in the form [8, 21],
(24)y(x,λ)=x(λ)vJv(λx)+∫0xH(x,t)t(λ)vJv(λt)dt,(25)y~(x,λ)=x(λ)vJv(λx)+∫0xH~(x,t)t(λ)vJv(λt)dt,
respectively, where the kernel H(x,t) (H~(x,t)) is the solution of the equation
(26)∂2H(x,t)∂x2+ℓ(ℓ+1)x2H(x,t) =∂2H(x,t)∂t2+(ℓ(ℓ+1)t2+q(t))H(x,t)
subject to the boundary conditions
(27)2dH(x,x)dx=q(x),limt→0 H(x,t)tv-1/2=0, [Jv′(t,λ)=O(tv-1/2)].
After the transformations
(28)ξ=14(x+t)2, η=14(x-t)2,H(x,t)=(ξ-η)-v+1/2U(ξ,η),
we obtain the following problem:
(29)∂2U∂ξ∂η-14(ξ-η)∂U∂ξ+14(ξ-η)∂U∂η =14ξηq(ξ+η)U,U(ξ,ξ)=0,∂U∂ξ+αξU=14q(ξ)ξv-1, α=-v+12.
This problem can be solved by using the Riemann method [21].
Multiplying (18) by y~(x,λ) and (20) by y(x,λ), subtracting and integrating from 0 to 1/2, we obtain
(30)∫01/2(q(x)-q~(x))y(x,λ)y~(x,λ)dx =[y~(x,λ)y′(x,λ)-y(x,λ)y~′(x,λ)]|01/2.
The functions y(x,λ) and y~(x,λ) satisfy the same initial conditions (19) and (21), that is,
(31)y~(0,λ)y′(0,λ)-y(0,λ)y~′(0,λ)=0.
Let
(32)Q(x)=q(x)-q~(x),(33)K(λ)=∫01/2Q(x)y(x,λ)y~(x,λ)dx.
If the properties of y(x,λ) and y~(x,λ) are considered, the function K(λ) is an entire function.
Therefore the condition of Theorem 1 implies
(34)y~(12,λn)y′(12,λn)-y(12,λn)y~′(12,λn)=0
and hence
(35)K(λn)=0, n∈ℕ.
In addition, using (24) and (33) for 0<x<1,
(36)|K(λ)|≤M1λv,
where M is constant.
Introduce the function
(37)W(λ)=y′(1,λ)+Hy(1,λ).
By using the asymptotic forms of y and y′, we obtain
(38)W(λ)=λsin(λ-vπ2-π4)+O(1).
The zeros of W(λ) are the eigenvalues of L and hence it has only simple zeros λn because of the seperated boundary conditions. From (38), W(λ) is an entire function of order 1/2 of λ. Since the set of zeros of the entire function W(λ) is contained in the set of zeros of K(λ), we see that the function
(39)Ψ(λ)=K(λ)W(λ)
is an entire function on the parameter λ. From (36), (38), and (39), we get
(40)|Ψ(λ)|=O(1λv+1/2).
So, for all λ, from the Liouville theorem,
(41)Ψ(λ)=0,
or
(42)K(λ)=0.
It was proved in [19] that there exists absolutely continuous function H~~(x,τ) such that we have
(43)y(x,λ)y~(x,λ)=12{∫1+cos2[λx-vπ2-π4] +∫0xH~~(x,τ) ×cos2[λτ-vπ2-π4]dτ},
where
(44)H~~(x,t)=2[H(x,x-2τ)+H~(x,x-2τ)]+2[∫-x+2τxH(x,s)H~(x,s-2τ)ds +∫-xx-2τH(x,s)H~(x,s+2τ)ds].
We are now going to show that Q(x)=0 a.e. on (0,1/2]. From (33), (43) we have
(45)12∫01/2Q(x){∫1+cos2[λx-vπ2-π4] +∫0xH~~(x,τ) ×cos2[λτ-vπ2-π4]dτ}dx=0.
This can be written as
(46)∫01/2Q(x)dx+∫01/2cos2[λτ-vπ2-π4] ×[Q(τ)+∫01/2Q(x) ×H~~(x,τ)dx∫01/2]dτ=0.
Let λ→∞ along the real axis, by the Riemann-Lebesgue lemma, one should have
(47)∫01/2Q(x)dx=0,∫01/2cos2[λτ-vπ2-π4] ×[Q(τ)+∫τ1/2Q(x)H~~(x,τ)dx]dτ=0.
Thus from the completeness of the functions cos, it follows that
(48)Q(τ)+∫τ1/2Q(x)H~~(x,τ)dx=0, 0<x<12.
But this equation is a homogeneous Volterra integral equation and has only the zero solution. Thus we have obtained
(49)Q(x)=q(x)-q~(x)=0,
or
(50)q~(x)=q(x)
almost everywhere on (0,1/2]. Therefore Theorem 1 is proved.
Proof of Lemma 2.
As in the proof of Theorem 1 we can show that
(54)G(ρ)=∫0bQ(x)y(x,λ)y~(x,λ)dx=[y~(x,λ)y′(x,λ)-y(x,λ)y~′(x,λ)]|x=b,
where ρ=λ=reiθ and Q(x)=q(x) -q~(x). From the assumption
(55)ym(n)′(b)ym(n)(b)=y~m(n)′(b)y~m(n)(b)
together with the initial condition at 0 it follows that,
(56)G(ρm(n))=0, n∈ℕ.
Next, we will show that G(ρ)=0 on the whole ρ plane. The asymptotics (23) imply that the entire function G(ρ) is a function of exponential type ≤2b.
Define the indicator of function G(ρ) by
(57)h(θ)=limr→∞supln|G(reiθ)|r.
Since |Imλ|=r|sinθ|, θ=argλ from (23) it follows that
(58)h(θ)≤2b|sinθ|.
Let us denote by n(r) the number of zeros of G(ρ) in the disk {|ρ|≤r}. According to [22] set of zeros of every entire function of the exponential type, not identically zero, satisfies the inequality
(59)limr→∞infn(r)r≤12π∫02πh(θ)dθ,
where n(r) is the number of zeros of G(ρ) in the disk |ρ|≤r. By (58),
(60)12π∫02πh(θ)dθ≤bπ∫02π|sinθ|dθ=4bπ.
From the assumption and the known asymptotic expression (7) of the eigenvalues λn we obtain
(61)n(r)≥2∑(πn/σ)[1+O(1/n)]<r1=2πσr(1+o(1)), r⟶∞.
For the case σ>2b,
(62)limr→∞n(r)r≥2πσ>4bπ=2b∫02π|sinθ|dθ≥12π∫02πh(θ)dθ.
The inequalities (59) and (62) imply that G(ρ)=0 on the whole ρ plane.
Similar to the proof of Theorem 1, we have
(63)q(x)= q~(x) a.e on the interval (0,b].
This completes the proof of Lemma 2.
Proof of Theorem 3.
From
(64)λr(n)=λ~r(n), yr(n)′(b)yr(n)(b)=y~r(n)′(b)y~r(n)(b),
where {r(n)}n∈ℕ satisfies (14) and σ2>2-2b. Similar to the proof of Lemma 2, we get
(65)q(x)= q~(x) a.e on [b,1).
Thus, it needs to be proved that q(x)= q~(x) a.e on (0,b]. The eigenfunctions yn(x,λn) and y~n(x,λn) satisfy the same boundary condition at 1. It means that
(66)yn(x,λn)=ξny~n(x,λn)
on [b,1] for any n∈ℕ where ξn are constants.
Let ρn=λn, sn=μn. From (54) and (66) we obtain
(67)G(ρn)=0, n∈ℕ,G(sln)=0, n∈ℕ.
We are going to show that inequality (59) fails and consequently, the entire function of exponential type G(ρ) vanishes on the whole ρ-plane. The ρn and sn have the same asymptotics (7). Counting the number of ρn and sn located inside the disc of radius r, we have
(68)1+2πr[1+O(1n)]
of ρn's and
(69)1+2πrσ1[1+O(1n)].
of sn's.
This means that
(70)n(r)=2+2π[r(σ1+1)+O(1n)],limr→∞n(r)r=2π(σ1+1).
Repeating the last part of the proof of Lemma 2, and considering the condition σ1>2b-1, we can show that G(ρ)=0 identically on the whole ρ-plane which implies that
(71)q(x)= q~(x) a.e on (0,b]
and consequently
(72)q(x)= q~(x) a.e on (0,1).
Hence the proof of Theorem 3 is completed.