Eigencurves of non-definite Sturm-Liouville problems for the p-Laplacian ∗ Paul A. Binding † Department of Mathematics and Statistics University of Calgary Calgary, Alberta, Canada T2N 1N4 Patrick J. Browne Department of Mathematics and Statistics University of Saskatchewan Saskatoon, Saskatchewan, Canada S7N 5E6 Bruce A. Watson ‡§ School of Mathematics University of the Witwatersrand Private Bag 3, P O WITS 2050, South Africa email:
[email protected] Tel: +27823769663 Fax: +27117176259 May 17, 2013
Abstract The (λ, µ) eigencurves are studied for a weighted quasi-linear Sturm-Liouville-type problem of the form −∆p y = (p − 1)(λr − q − µs)sgn y|y|p−1 , ∗
on (0, 1)
Keywords: Sturm-Liouville, p-Laplacian, eigencurves, Mathematics subject classification (2000): 34L30, 34B24, 34L15, 47E05. † Research supported in part by NSERC of Canada ‡ Research conducted while visiting University of Calgary. Supported in part by the Centre for Applicable Analysis and Number Theory and by NRF grant number IFR2011032400120. § Corresponding author is Bruce A. Watson.
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with Sturmian-type boundary conditions (∆p being the p-Laplacian). Topics include H¨older classes C k,α and asymptotics (including existence of oblique asymptotes). An example illustrates precision of our H¨older exponents when p > 2.
1
Introduction
The two parameter Sturm-Liouville problem −y 00 + qy = (λr − µs)y,
on
0
where
α0 ∈ [0, π),
(1.2)
0
where
α1 ∈ (0, π],
(1.3)
y (0) sin α0 = y(0) cos α0 , y (1) sin α1 = y(1) cos α1 ,
(0, 1)
(1.1)
has appeared in various contexts, e.g. linearised bifurcation analysis, separation of variables and as a homotopy between indefinite and definite one parameter problems. Under minimal coefficient conditions, specified below, the set of (λ, µ) for which (1.1)-(1.3) has a nontrivial solution turns out to be a countable union of curves known as eigencurves. They can be described variationally or analytically, and the interplay between these descriptions has proved quite fruitful. A review and bibliography of this topic can be found in [7]. Here we shall study the corresponding eigencurves for an extension of (1.1) to the nonlinear but homogeneous equation −∆p y = (p − 1)(λr − q − µs)|y|p−1 sgn y,
on (0, 1)
(1.4)
involving the p-Laplacian ∆p y = (|y 0 |p−1 sgn y 0 )0 . Here 1 < p < ∞, q, r, s ∈ L1 (0, 1), s > 0 and (1.4) is taken a.e. (in the Carath´eodory sense). Note that (1.4) reduces to (1.1) when p = 2. When µ = 0 one obtains a more standard problem (see [4] and references) if r > 0. There is also significant literature when the weight function r is indefinite, although mostly for left definite (see below) or semidefinite cases (e.g. q = 0 and Dirichlet or Neumann boundary conditions [8, 17]). Fully indefinite problems are treated in [3] and [9]. The “principal” eigencurve for the p-Laplacian (where y is of one sign) was examined in [5], and an equivalent construction for “higher” eigencurves, using eigenvalue reciprocals for a left definite case with q = 0, can be found in [10]. Dirichlet boundary conditions were imposed in both these references. There seems to be little known for the general case, and our aim here is to start filling this gap, and in particular to prepare some results for a separate application to a half-line problem. In Section 2 we give some foundations for both the variational and generalised Pr¨ ufer angle approaches. (The corresponding eigencurves coincide although that could fail for
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nonseparated boundary conditions [6]). In particular, (1.2)-(1.4) lead to a relation `[y] = λr[y] − µs[y], (1.5) R1 R1 R1 where r[y] = 0 r|y|p , s[y] = 0 s|y|p and `[y] = 0 (|y 0 |p + q|y|p ) plus (pth power) boundary terms. The problem is “left definite” if `[y] > 0 for all non-zero y in the form domain W defined in Lemma 2.1. In all cases, variational eigencurves can be defined via a minimax operation on a quotient of the form (`[y] − λr[y])/s[y] - see Lemma 2.1. We prepare for a Pr¨ ufer-type angle approach to eigencurves by examining certain modified trigonometric and hyperbolic functions. The best known of these functions is sinp = sinp,p where Z sinp,u x (1 − tu )−1/p dt. x= 0
Equivalently y = sinp x in (1.4) if λ = 1, µ = 0, r = 1, q = 0 and initial conditions y 0 (0) = 1,
y(0) = 0,
are imposed. Indeed Elbert [12] used sinp for a Pr¨ ufer analysis of (1.4) with µ = 0, q = 0 and Dirichlet conditions for (1.2) and (1.3). Many authors have since used sinp (or a ˆ scaled version considered by Otani [19]) to study more general one parameter eigenvalue problems. See [4] for a review. We specify H¨ older classes C k,α for sinp , and corresponding fractional derivatives, where k ∈ {1, 2} and α ∈ [0, 1) depending on where p lies in the intervals√(1, γ], (γ, 2) and (2, ∞) (also sin2 = sin is analytic). Here γ is the golden ratio (1 + 5)/2 and this is the first time that we have seen this constant appear in connection with solutions of p-Laplacian equations. (C 1,α behaviour, with α unspecified, is well known, e.g. from ˆ [11], and C k behaviour was shown by Otani [19].) For a general reference on fractional derivatives, with historical introduction, we cite [18]. We shall also need corresponding results for functions cosp , sinhp , coshp : related functions have been studied by Lindqvist [15]. Most of the literature in this area has employed the definition cosp = sin0p , but here (and in [15]) that is true only if p = 2 – see Lemma 2.5 for details. In our definition cosp is the solution of (1.4) for λ = 1, µ = 0, r = 1, q = 0 with initial conditions y(0) = 1 and y 0 (0) = 0. Already in the 1870’s sinp,u and related twin index functions (with integer p and rational u) were discussed - see [16] for references. ˆ Otani [19] examined a scaled version of sinp,u which is related to (1.4) but with exponent u − 1 on the right hand side, and for recent work on the topic we refer to [13]. In Section 3 we use the above specified H¨older classes to show that the eigencurves, which are graphs of certain functions µn , are at least C 1,α where α depends on p. In Section 4 we study estimates of the form µnλ(λ) → an as |λ| → ∞, giving “asymptotic
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directions” (cf. [20], [21] for the case p = 2). Section 5 gives necessary and sufficient conditions for more precise asymptotics of the form µn (λ) = an λ + bn + o(1), as λ → ∞, giving oblique asymptotes. For related asymptotics when p = 2, we refer to [2], [14]. As a consequence of our eigencurve asymptotics we are able to bound the eigenvalues of the one parameter case of (1.2)-(1.4) under certain conditions when µ = 0. We conclude in Section 6 with an example showing that our eigencurve H¨older class exponents are best possible for p > 2, but we leave the corresponding question open for 1 < p < 2.
2
Preliminaries
We write [y]p−1 = |y|p−1 sgn y and Z πp = 2 0
1
dt 2π = p sin(π/p) (1 − tp )1/p
(2.1)
noting that π2 = π. Define cot∗p ϕ = 0 if ϕ is an integer multiple of πp and cot∗p ϕ = sin0p ϕ/ sinp ϕ otherwise. For y ∈ Wp1 (0, 1) let Z 1 (|y 0 |p + q|y|p ) − |y(1)|p [cot∗ α1 ]p−1 + |y(0)|p [cot∗ α0 ]p−1 (2.2) `q [y] := 0 Z 1 r[y] := r|y|p , (2.3) 0 Z 1 s[y] := s|y|p . (2.4) 0
Combining [4, Theorem 5.1] (where the boundary terms were omitted) and the approach of [1, Corollary 1] (which treats eigenvalue reciprocals for a special left definite case) we obtain the following. Lemma 2.1 The variational eigenvalues of (1.2)-(1.4) are given by −µn (λ) = min
max
A∈Fn y∈A,s[y]=1
`q−λr [y],
(2.5)
where Fn = {F |F an n dimensional subspace of W } and W = {f ∈ W1p (0, 1)|κj f (j) = 0, j = 0, 1} in which κj is 1 if sin αj = 0 and 0 if sin αj 6= 0.
4
(2.6)
The graph of the function µn is called the nth variational eigencurve. We note that µ0 (λ) > µ1 (λ) > . . . . We now define further p-generalised trigonometric and hyperbolic functions cosp , coshp , sinhp by and
cos0p θ = −(1 − cospp θ)1/p ,
for θ ∈ [0, πp /2],
coshp 0 = 1,
and
for θ ∈ [0, ∞),
sinhp 0 = 0,
and
cosh0p θ = (coshpp θ − 1)1/p , sinh0p θ = (1 + sinhpp θ)1/p ,
cosp 0 = 1,
for θ ∈ [0, ∞).
Thus integration yields
θ θ
R1
dt cosp θ (1−tp )1/p , R cosh θ dt = 1 p (tp −1) 1/p , R sinhp θ dt , = 0 (1+tp )1/p
θ =
for θ ∈ [0, πp /2],
(2.7)
for θ ∈ [0, ∞),
(2.8)
for θ ∈ [0, ∞).
(2.9)
To extend the definition of these functions to R we set sinhp (−θ) := − sinhp θ,
θ ∈ [0, ∞),
coshp (−θ) := coshp (θ),
θ ∈ [0, ∞),
sinp (−θ) := − sinp θ,
θ ∈ [0, πp /2],
cosp (−θ) := cosp (θ),
θ ∈ [0, πp /2],
sinp (θ + nπp ) := (−1)n sinp (θ), n
cosp (θ + nπp ) := (−1) cosp (θ),
θ ∈ [−πp /2, πp /2], n ∈ Z, θ ∈ [−πp /2, πp /2], n ∈ Z.
From the above it follows that | cos0p θ|p | cosh0p θ|p | sin0p θ|p | sinh0p θ|p
= 1 − | cosp θ|p , = | coshp
θ|p
= 1 − | sinp
(2.10)
− 1,
(2.11)
θ|p ,
(2.12)
= 1 + | sinhp
θ|p ,
(2.13)
which shows that cosp and sinp are solutions of −∆p y = (p − 1)[y]p−1 and coshp and sinhp are solutions of ∆p y = (p − 1)[y]p−1 . Proceeding as in [15], one obtains the following analogues of trigonometric identites for
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1 < p < ∞, π p cosp (θ) = sinp −θ , 2 d p d 0 θ p θ = sinp and sind θ = cos0p θ , cosd d p d p d p d p θ θ θ 0 0 coshd = sinhp and sinhd = coshp θ , d p d p p p d d cosd θ + sinp θ = 1 = coshd θ − sinhp θ , d p d p
(2.14) (2.15) (2.16) (2.17)
where d = p/(p − 1). Note that a simple computation shows that pπp = dπd . For 0 < θ < πp /2 the functions sinp and cosp are C ∞ while, away from 0, sinhp and coshp are C ∞ . Thus when considering the smoothness of the p-trigonometric functions it is enough to study them near integer multiples of πp /2, and we need to study the p-hyperbolic functions only near the point 0. Definition 2.2 A function f is in C k,α (Ω), k ≥ 0, α ∈ [0, 1), on an interval Ω if f ∈ C k (Ω) and for each x ∈ Ω there is a constant κx and a relative neighbourhood Nx of x in Ω with |f (k) (ξ) − f (k) (x)| ≤ κx |ξ − x|α , for all ξ ∈ Nx . Note that C k,0 = C k . Definitions of fractional derivatives are explored in [18]. For our purposes it will be sufficient to use the following special case, where Γ is the generalized factorial. Definition 2.3 Let τ ∈ (0, 1) and Nx be a relative neighbourhood of x ∈ Ω. If f ∈ C 0,τ (Nx ) and f (x + h) − f (x) h&0 [h]τ
Γ(τ + 1) lim
(2.18)
τ f (x), of f at x. In this case ∂ τ f (x) exists then it is called the right τ -derivative, ∂+ + satisfies τ f (x + h) = f (x) + [h]τ ∂+ f (x) + o(|h|τ ), as h & 0.
Similarly if (2.18) exists with h & 0 replaced by h % 0, we obtain a left τ -derivative, τ f (x) of f at x. If ∂ τ f (x) = ∂ τ f (x), then we write ∂ τ f (x) for ∂ τ f (x). ∂− − + ± More generally for τ ∈ [0, 1) with f ∈ C n,τ (Nx ) we set ∂ n+τ f (x) = ∂ τ f (n) (x) if ∂ τ f (n) (x) exists, where we take ∂ 0 f (n) (x) = f (n) (x).
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We note that if f ∈ C k [a, b] and f ∈ C k+1 (a, b) with ∂ k+τ f (x) existing at a and b for some τ ∈ [0, 1) then f is C k,τ on [a, b]. The proof of the following Lemmas and Theorem are given in the Appendix.
Lemma 2.4 If d =
p p−1
=1+
1 p−1
then
−∂ d cosp (0) = (p − 1)d−1 Γ(d) = ∂ d coshp (0), −∂ 1+p sinp (0) = Γ(p) = ∂ 1+p sinhp (0). The above derivatives, ∂ τ , can be used to obtain “fractional” Taylor type approximations for the generalised trigonometric and hyperbolic functions and their derivatives near zero.
Lemma 2.5 Let d =
p p−1
=1+
1 p−1
and a =
(p−1)d . p
Then as h → 0 we have
a2 |h|2d + o(|h|2d ), 2(1 + d) a2 |h|2d coshp h = 1 + a|h|d + + o(|h|2d ), 2(1 + d) [h]p+1 p2 − 2p − 1 sinp h = h − − 2 [h]2p+1 + o(|h|2p+1 ), p(p + 1) 2p (p + 1)(2p + 1) [h]p+1 p2 − 2p − 1 sinhp h = h + − 2 [h]2p+1 + o(|h|2p+1 ). p(p + 1) 2p (p + 1)(2p + 1) cosp h = 1 − a|h|d +
The derivatives of these functions have the following expansions near zero a2 d[h]2d−1 + o(|h|2d−1 ), d+1 a2 d[h]2d−1 cosh0p h = ad[h]d−1 + + o(|h|2d−1 ), d+1 |h|p p2 − 2p − 1 2p sin0p h = 1 − − |h| + o(|h|2p ), p 2p2 (p + 1) |h|p p2 − 2p − 1 2p sinh0p h = 1 + − |h| + o(|h|2p ). p 2p2 (p + 1) cos0p h = −ad[h]d−1 +
as h → 0.
We remark that a formal proof of related expansions can be found in [15]. We are now ready to specify H¨older classes for these functions, including the connection with the golden ratio mentioned in Section 1.
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Theorem 2.6 Let 1 < p < ∞, p 6= 2. The functions cosp and sinp are in C p > 2 and are in C 2,ζ (R) for 1 < p < 2 where p − 1, 1 < p < γ, ζ= 1 − 1, γ ≤ p < 2, p−1
1 1, p−1
(R) for
where γ is the golden ratio, see Section 1. The function sinhp is in C 1+bpc,p−bpc (R) and coshp is in C
1 1 1 c, p−1 −b p−1 c 1+b p−1
(R).
The above exponents in the H¨older classes C k,α are sharp, and corresponding (k + α)th derivatives can be obtained from (2.14) - (2.17).
3
Local properties of eigencurves
We next discuss the eigencurves from the viewpoint of the Pr¨ ufer type angle θ (x, λ) as extended by Elbert, [12], where y = ρ sinp θ,
y 0 = ρ sin0p θ.
In this context it is convenient to write (1.2) and (1.3) in the form (sinp β0 ) y 0 (0) = sin0p β0 y(0), β0 ∈ [0, πp ), (sinp β1 ) y 0 (1) = sin0p β1 y(1), β1 ∈ (0, πp ],
(3.1)
(3.2) (3.3)
where cotp βj = cot αj . Here cotp = sin0p / sinp which is not cosp / sinp , except when p = 2, but it is continuous on (0, πp ) and | cotp x|p = 1 +
1 , | sinp x|p
so cotp x decreases strictly from ∞ to 0 as x increases from 0 to πp /2. Similarly cotp x decreases strictly from 0 to −∞ as x increases from πp /2 to πp , giving the existence and uniqueness of β0 , β1 . In particular y = sinp x is the solution of (1.4) where q = 0, λr = 1 and µ = 0 with the initial conditions y(0) = 0, y 0 (0) = 1. Note that sin0p (0) = 1, and sinp (x) = 0 if and only if x = kπp , k ∈ Z. For any non-zero solution y of (1.4), (1.2), θ and ρ are functions of (x, λ, µ) although some or all of these arguments may be suppressed as appropriate. In particular θ0 = 1 − G sinp θ, 0
(3.4) G sin0p θ,
(3.5)
G = [sinp θ]p−1 (1 + q + µs − rλ).
(3.6)
ρ
= ρK,
where
K=
and
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Here θ and ρ are Carath´eodory solutions of the initial value problems given by (3.4) with initial condition θ(0) = β0 , and (3.5) with initial condition ρ(0) = (|y(0)|p + |y 0 (0)|p )1/p . These initial value problems represent (1.4) and (1.2) in p-polar form via (3.1). From [3, Lemma 2.1], for each λ and µ, θ(x, λ, µ) is absolutely continuous, and strictly increasing through non-negative integer multiples of πp , in x. Now as a consequence of (3.4), θ(1, λ, µ) is finite, and so is the oscillation count of each solution y of (1.2), (1.4). From (3.5), ρ(x) is either zero for all x or nonzero for all x.
(3.7)
Lemma 4.2 and Theorem 4.3 of [3] together show that for each λ ∈ R there is precisely one eigenvalue µ = µn (λ) of (1.2)-(1.4) for which θ(1, λ, µn (λ)) = nπp + β1 ,
n = 0, . . . .
(3.8)
By virtue of [4, Theorem 5.1], µn (λ) is as in Lemma 2.1, so the Pr¨ ufer eigencurve via [3, Section 4] coincides with the variational eigencurve of Section 2. The eigenfunctions y corresponding to µn have oscillation count n where n = 0, 1, . . . . In this notation µ0 (λ) > µ1 (λ) > . . . and µn (λ) → −∞ as n → ∞. In addition, from [3, Lemma 2.2] for each x, θ(x, λ, µ) is strictly decreasing in µ. In particular, from [3, Lemma 2.2(iii)], for fixed x, θ(x, λ, µ) is C 1 in λ and µ. The derivatives θλ := ∂θ/∂λ and θµ := ∂θ/∂µ satisfy the initial value problems θλ0 = −pKθλ + r| sinp θ|p ,
θλ (0) = 0.
(3.9)
θµ0
θµ (0) = 0.
(3.10)
p
= −pKθµ − s| sinp θ| ,
Similarly, θ(x, λ, µ) is C 2 in λ and µ Rwhen p < 2, R via Theorem 2.6 and equations (3.9) and (3.10). After multiplication by ep K and e− K respectively, we can integrate (3.9), (3.10) and (3.5) to give Z x Rt θλ (x) = r| sinp θ|p ep x K d t, (3.11) 0 Z x Rt θµ (x) = − s| sinp θ|p ep x K d t, (3.12) 0
ρ(x) = ρ(0)e
Rx 0
K
.
(3.13)
We shall assume for the remainder of this work that y is normalized by |y(0)|p +|y 0 (0)|p = 1. This normalization gives ρ(0) = 1 and the above identities yield Z x ρp θλ (x) = r|y|p , (3.14) 0 Z x p ρ θµ (x) = − s|y|p . (3.15) 0
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For the remainder of this section we also assume that y satisfies (1.4), (3.2) and (3.3). Multiplying (1.4) by y, integrating and using (3.14) and (3.15) we obtain `q+µs [y] = λr[y].
(3.16)
p
r[y] = ρ θλ (1),
(3.17)
p
s[y] = −ρ θµ (1).
(3.18)
From this we obtain λρp (1)θλ (1) = λr[y] = lq+µs [y],
(3.19)
for λ, µ ∈ R and y as above. When p = 2, the eigencurves are real analytic, see [7], while for p 6= 2 we have the following theorem. Theorem 3.1 For all p > 1 and n ≥ 0, µn is C 1 and satisfies R1 r|y|p dµn θλ (1, λ, µn ) , = − = R01 dλ θµ (1, λ, µn ) s|y|p
(3.20)
0
dµn λ dλ
=
`q+µn s [y] `q [y] = µn + , s[y] s[y]
(3.21)
where y is an eigenfunction corresponding to (λ, µn (λ)). Moreover µn is C 1,ω where p − 1, 1 < p < 2, ω= 1 2 < p. p−1 , Proof: Since 0 < s[y] = be applied to give
R1 0
s|y|p = −θµ (1, λ, µn (λ)), the implicit function theorem can
θλ (1, λ, µn (λ)) + µ0n (λ)θµ (1, λ, µn (λ)) = 0,
(3.22)
µn being C 1 . Formulae (3.20), (3.21) and the fact that µn is C 1 follow from (3.22), (3.11), (3.12) and Theorem 2.6. The smoothness of θλ (1) and θµ (1) is determined by the minimum smoothness of sin0p θ sinp−1 θ p p and sinp θ, θ ∈ [0, πp /2], in view of the periodicity and symmetry used in defining sinp . Since (sinpp θ)0 = p sin0p θ sinp−1 θ we need only consider the smoothnesss of p g(θ) = sin0p θ sinpp−1 θ. By virtue of (2.14) it suffices to consider g(θ) and h(θ) = cos0p θ cosp−1 θ p near θ = 0. Now g(0) = 0 = h(0) with g(θ) = O(θp−1 ), h(θ) = O(θ1/(p−1) ), n o 1 as θ & 0, and thus ω = min p − 1, p−1 .
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4
Asymptotic Directions
From Theorem 3.1 we obtain −∞ ≤ r := ess inf
r dµ r ≤ ≤ ess sup =: r ≤ ∞. s dλ s
In this section we investigate the behaviour of µ and
dµ dλ
(4.1)
as λ → ±∞ in terms of r and r.
Note that if r = r ∈ R then it follows directly from (4.1) that for all λ, r=
dµ = r, dλ
so the eigencurves are straight lines of the form µn (λ) = rλ+cn . Hence for the remainder of this section we shall only consider the case of r < r. From [4, Theorem 5.1], if c > µ0 (0) then `q+cs [y] is positive definite. The existence of such a c is critical to the approach used in this section.
Theorem 4.1 Let c be so large that `q+cs [y] is positive definite. For each r˜ ∈ R with r < r˜ < r, the line µ = λ˜ r + c intersects the eigencurve µ = µn (λ) at precisely two points ± ± r) > 0 and r))). Here ±λ± r), µn (λn (˜ (λn (˜ n (˜ dµn − dµn + (λn (˜ r)) < r˜ < (λ (˜ r)). dλ dλ n Proof: We fix r˜ ∈ (r, r), and suppress it as an argument. Note that Z
1
(r − r˜s)± > 0. 0
Let q˜ = q + cs,
(4.2)
µ ˜ = µ − λ˜ r − c,
(4.3)
Then in terms of q˜ and µ ˜, (1.4) becomes −(|y 0 |p−2 y 0 )0 = (p − 1)(λ(r − r˜s) − q˜ − µ ˜s)|y|p−2 y.
(4.4)
Equation (4.4) with boundary conditions (3.2), (3.3) and µ ˜ = 0 meets the conditions for the left definite case of [3, Theorem 5.2]. We can thus conclude that for each n = 0, 1, . . . there are precisely two λ-values λ± ˜ = 0 and the boundary n for which (4.4) with µ + conditions (3.2), (3.3) has a solution, y, with oscillation count n. Moreover λ− n < 0 < λn .
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In particular the line µ ˜ = 0 and the graph of µ ˜n : λ 7→ µn (λ) − r˜λ − c intersect at precisely two points (λ± ˜)-plane. The eigencurve µ = µn (λ) n , 0) in the (λ, µ ± intersects the line µ = c + λ˜ r at precisely the two points (λ± n , µn (λn )) for each n = 0, 1, . . . . As in (2.3) and (3.19), λ
`q˜+˜µs [y] dµ d˜ µ − r˜λ = λ = . dλ dλ s[y]
(4.5)
Thus, with λ = λ± ˜n (λ± n, µ n ) = 0, and λ± n
`q˜[y] dµn − r˜λ± > 0. n = dλ s[y]
Hence dµn − dµn + (λn ) > r˜ > (λ ). dλ dλ n
Corollary 4.2 Let r < r and c be as in Theorem 4.1 and r˜ ∈ (r, r). For λ > λ+ r) we n (˜ have λ˜ r + c < µn (λ) < λr + c while for λ < λ− (˜ r ), λ˜ r + c < µ (λ) < λr + c. n n Proof: We shall only consider the case of λ > λ+ r). The other case follows from this n (˜ by replacing λ by −λ, r by −r and r˜ by −˜ r. By Theorem 4.1, at λ = λ+ r) the eigencurve µ = µn (λ) and the line µ = r˜λ + c n (˜ intersect. Here the slope of the eigencurve is strictly greater than that of the line. Also r). Thus, by continuity of the the eigencurve and line do not intersect for λ > λ+ n (˜ eigencurve and line, for λ > λ+ (˜ r ) we have µ (λ) > λ˜ r + c. n n If r = ∞ then µn (λ) < ∞ = λr + c so we now assume r < ∞. As the slope of the eigencurve does not exceed r it follows for λ > λ+ r) that n (˜ µn (λ) ≤ r[λ − λ+ r)] + µn (λ+ r)) n (˜ n (˜ = r[λ − λ+ r)] + λ+ r)˜ r+c n (˜ n (˜ = rλ − λ+ r)[r − r˜] + c n (˜ < rλ + c, since λ+ r) > 0 and r − r˜ > 0. n (˜ We have thus shown that r and r are asymptotic directions for the eigencurve as λ → ±∞ respectively. We now refine these results to limits for the slope of the eigencurve.
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Theorem 4.3 If r < r then as λ → +∞ we have
dµn dµn → r, while → r as λ → −∞. dλ dλ
Proof: We prove the result for λ → +∞; the case λ → −∞ can be proved similarly. We already know that dµn /dλ ≤ r. From Theorem 4.1, for each r˜ ∈ I := (r, r), there exists precisely one λ > 0, (i.e. λ+ r)) at which the eigencurve µ = µn (λ) and the line n (˜ µ = λ˜ r + c intersect, where c is as in Theorem 4.1. Here µn is a C 1 function of λ. Let R ∈ I; we show that dµn (λ)/dλ > R for λ ≥ λ+ ˜≥R n (R). From Theorem 4.1, if r then dµn + (λ (˜ r)) > r˜ ≥ R. dλ n It thus suffices to show that for each λ ≥ λ+ ˜ ≥ R, r˜ ∈ I, with λ = λ+ r). n (R) there is r n (˜ Note that µn (λ+ r)) = λ+ r)˜ r+c n (˜ n (˜ for all r˜ ∈ I. More precisely, the map M (λ, r˜) := µn (λ) − λ˜ r−c is at least C 1 in λ and r˜ and
∂M + r), r˜) > 0, (λ (˜ ∂λ n so by the implicit function theorem, for each λ in a neighbourhood λ+ r) there exists a n (˜ ∗ unique r∗ (λ) in a neighbourhood of r˜ so that M (λ, r∗ (λ)) = 0. In particular λ+ n (r (λ)) = λ so by Theorem 4.1, dµn (λ) > r∗ (λ). dλ Moreover, in this neighbourhood, r∗ (λ) is a C 1 function with r∗ (λ) + λ
dr∗ dµn = > r∗ (λ) dλ dλ
so
dr∗ > 0. dλ Hence r∗ (λ) is an increasing function, making λ+ r) a continuous strictly increasing n (˜ function. To conclude the proof we need only show that λ+ r) → ∞ as r˜ % r. For n (˜ r < ∞, λ
λ+ r)˜ r + c = µn (λ+ r)) n (˜ n (˜ + r ) − λ+ ≤ r[λ+ n (˜ n (R)] + µn (λn (R))
= rλ+ r) − λ+ n (˜ n (R)[r − R] + c,
13
giving λ+ n (R)
r−R ≤ λ+ r). n (˜ r − r˜
Hence λ+ r) % ∞ as r˜ % r. n (˜ For r = ∞, if λ+ r) % Λ < ∞ as r˜ → ∞ then taking λ > Λ and n (˜ r˜ =
µn (λ) − c λ
we have λ+ r) < Λ. However the uniqueness of the intersection of the line µ = r˜λ + c n (˜ and the eigencurve µ = µn (λ) for λ > 0 gives that λ+ r) = λ > Λ, a contradiction. n (˜ Hence Λ = ∞. Following directly from the above we have: ± th Corollary 4.4 If µn (0) < 0, let λ± n , where ±λn > 0, denote any λ abscissa of the n ± eigencurve (i.e. µn (λn ) = 0). Then for n = 0, 1, . . . ,
µn (0) ≤ λ+ n, r µn (0) ≥ λ− − n, r
∞ > r > 0,
−
−∞ < r < 0.
Remark 4.5 In the right definite, i.e. r > 0 a.e. (resp. left definite) case, λk (resp. − λ+ k and λk ) are unique by (3.20) and (3.21). For the final result of this section, we recall that, by assumption, r < r.
Theorem 4.6 The eigencurve function µ0 is strictly convex. Proof: From [4] −µ0 (λ) = min y6=0
Thus for
λ†
<
`q−λr [y] . s[y]
λ× −µ0
λ† + λ× 2
`q−λ† r [y] `q−λ× r [y] 1 = inf + 2 y6=0 s[y] s[y] `q−λ† r [y] `q−λ× r [y] 1 ≥ inf + inf y6=0 2 y6=0 s[y] s[y] † × µ0 (λ ) + µ0 (λ ) = − 2
14
proving convexity. It remains to show that µ0 is strictly convex. If µ0 is convex but not strictly convex then it has an affine restriction of the form µ0 (λ) = µ0 (λ† ) + (λ − λ† )˜ r where r˜ = µ00 (λ) for all λ ∈ [λ† , λ× ]. Let q˜ and µ ˜ be as in (4.2)-(4.3) with c = µ0 (λ† ) − λ† r˜. Equation (1.4) transforms to (4.4) which with (1.2), (1.3) has 0th eigencurve µ ˜(λ) = 0 for λ ∈ [λ† , λ× ]. Now from (3.21) we have 0=λ
`q˜[y] `q˜[y] d˜ µ =µ ˜+ = dλ s[y] s[y]
for λ ∈ [λ† , λ× ] and y an eigenfunction of (1.2)-(1.4) for (λ, µ0 (λ)). Moreover y is (up to `q˜[y] rescaling) the unique minimizer of s[y] which is non-negative. Thus y is independent of λ ∈ [λ† , λ× ] and from (4.4) −(|y 0 |p−2 y 0 )0 = (p − 1)(λ(r − r˜s) − q˜)|y|p−2 y.
(4.6)
Taking the difference between (4.6) at λ = λ† and at λ = λ× we have 0 = (p − 1)(λ× − λ† )(r − r˜s)|y|p−2 y. As r 6= r, there is a set of positive measure on which r − r˜s 6= 0 making y1 = 0 on a set of positive measure. In particular y has a zero in (0, 1), which is not possible as y is a principal eigenfunction.
5
Existence of Asymptotes
In this section we assume r and r of (4.1) to be finite. If µ ˆ := µ − rλ,
(5.1)
rˆ := r − rs,
(5.2)
then equation (1.4) becomes −∆p y = (p − 1)(λˆ r−q−µ ˆs)[y]p−1 ,
on (0, 1).
(5.3)
For fixed λ we denote nth µ ˆ-eigenvalue of (5.3) with boundary conditions (1.2), (1.3) by µ ˆn (λ) = µn (λ) − rλ.
Theorem 5.1 If there is an open interval (a, b) = J ⊆ (0, 1), J 6= φ, with r(x) = rs(x) a.e. on J, then each eigencurve has an asymptote as λ → +∞ with slope r.
15
Proof: With ˆ and rˆ as in (5.1) and (5.2) we have rˆ ≤ 0. By assumption rˆ = 0 a.e. on R1 µ J, and 0 rˆ ≤ 0. From this and Theorem 3.1 it follows that R1 rˆ|y|p dˆ µn = R01 ≤ 0, dλ s|y|p 0
y being an eigenfunction for (λ, µn (λ)). Note that r(x) = rs(x) a.e. on (0, 1) contradicts our assumption that r < r. Thus we assume r(x) < rs(x) on a set of positive measure making rˆ(x) < 0 on a set of positive measure. R1 If 0 rˆ|y|p = 0 then y = 0 on a set of positive measure, making y identically zero, which R1 µn contradicts y being an eigenfunction. Hence 0 rˆ|y|p < 0 and dˆ ˆn is a dλ < 0, and µ strictly decreasing function. In order to show that µ ˆn has a horizontal asymptote it remains only to show that µ ˆn (λ) is bounded below as λ → ∞. Suppose instead that µ ˆn (λ) → −∞ as λ → ∞. Then for large λ > 0 we have µ ˆn (λ) < 0. 0 Let ρˆ = (−ˆ µn )1/p and ϕ be the continuously defined Pr¨ ufer-type angle with cotp ϕ = ρˆyy . Note that y(x) = 0 = y 0 (x) is impossible for any x by (3.13), since y is an eigenfunction. Then, by [4], ϕ0 = (ˆ ρ)1−p (λˆ r − q − sˆ µn )| sinp ϕ|p + ρˆ| sin0p ϕ|p q − λˆ r = ρˆ + s | sinp ϕ|p + | sin0p ϕ|p . µ ˆn Now on J, rˆ = 0 a.e. so a.e. on J we have q 0 p ϕ ≥ ρˆ | sinp ϕ| + min{1, s} , µ ˆn giving Rb
q| sinp ϕ|p ρˆp−1 a Z b kqk1 ≥ ρˆ min{1, s(x)} dx − p−1 . ρ ˆ a Z
ϕ(b) − ϕ(a) ≥ ρˆ
b
min{1, s(x)} dx −
a
Consequently b kqk1 ϕ(1) − ϕ(0) ≥ ϕ(1) − ϕ(b) + ρˆ min{1, s(x)} dx − p−1 + ϕ(a) − ϕ(0) ρ ˆ a Z b kqk1 ≥ −2πp + ρˆ min{1, s(x)} dx − p−1 . ρ ˆ a
Z
Now by assumption µn (λ) → −∞ as λ → ∞, giving ρˆ → ∞ and thus ϕ(1) − ϕ(0) → ∞. But nπp < ϕ(1) ≤ (n + 1)πp and 0 ≤ ϕ(0) < πp , giving ϕ(1) − ϕ(0) ≤ (n + 1)πp , a contradiction. Thus µ ˆn (λ) is bounded below as λ → ∞ and hence has a horizontal asymptote.
16
Remark 5.2 If r in Theorem 5.1 is replaced by r, then the eigencurves have asymptotes as λ → −∞ with slope r. Similar remarks hold for each result in this section.
The Pr¨ ufer angle θ introduced in (3.4) when written in terms of µ ˆ, rˆ of (5.1), (5.2) and the parameters x, λ gives rise to a Pr¨ ufer angle ˆ λ, µ θ(x, ˆ) = θ(x, λ, µ ˆ + λr) ˆ λ, µ associated with equation (5.3) and the initial condition θ(0, ˆ) = β0 . If rˆ is replaced by −ˆ r, λ by −λ and µ ˆ by −ˆ µ then we are in the setting of [3, Section 4]. ˆ λ, µ Hence by [3, Lemma 4.1], θ(1, ˆ) is strictly decreasing in λ and µ ˆ with θˆλ (1, λ, µ ˆn (λ)) < 0
and θˆµˆ (1, λ, µ ˆn (λ)) < 0.
(5.4)
We can thus define ˆ λ, µ L(ˆ µ) := lim θ(1, ˆ) = lim θ(1, λ, µ ˆ + λr) λ→∞
λ→∞
which is also a decreasing function of µ ˆ. Here L(ˆ µ) can be computed via [3, Lemma 4.2] as follows. Let S be the union of the maximal open intervals in (0, 1) where rˆ = 0 a.e., i.e., S is the essential interior of rˆ−1 ({0}).
(5.5)
Then we have the finite sum L(ˆ µ) =
X
Θ(I, µ ˆ),
(5.6)
I∈S ∗ (ˆ µ)
where S ∗ (ˆ µ) is the set of maximal open intervals I ⊂ S with Θ(I, µ ˆ) ≥ πp if 1 ∈ / I¯ and ¯ Θ(I, µ ˆ) ≥ β1 if 1 ∈ I (this should be corrected in [3]). On each such interval I = (a, b), Θ(x, µ ˆ) is the solution to the equation Θ0 = 1 − | sinp Θ|p (1 + q + µ ˆs),
(5.7)
with initial condition Θ(a, µ ˆ) = β0 if a = 0 and Θ(a, µ ˆ) = 0 if a > 0. Now Θ(I, µ ˆ) is assigned the value Θ(1, µ ˆ) if b = 1 and kπp if b < 1 where k ∈ Z is such that Θ(b, µ ˆ) ∈ [kπp , (k + 1)πp ). Note that if µ ˆ, λ > 0, then θ0 = 1 − | sinp θ|p G where G = 1 + q + µs − λr = 1 + q + µ ˆs − λˆ r.
17
ˆ =1+q+µ ˆ so ψ ≥ θ = θˆ where Setting G ˆs we see that G ≥ G, ˆ λ, µ ψ(0, µ ˆ) = θ(0, λ, µ ˆ + λ¯ r) = θ(0, ˆ), and ˆ ψ 0 = 1 − | sinp ψ|p G. ˆ λ, µ Here from [4], ψ(1, µ ˆ) → 0 as µ ˆ → ∞, thus giving θ(1, ˆ) → 0 as µ ˆ → ∞ uniformly in λ for λ > 0. Hence L(ˆ µ) → 0 as µ ˆ → ∞. Under the assumptions of Theorem 5.1, L(ˆ µ) → ∞ as µ ˆ → −∞, so the set {ˆ µ|L(ˆ µ) < nπp + β1 } is bounded below and non-empty. We now give an explicit form for the asymptotes of the nth eigencurve when the condition of Theorem 5.1 are met.
Theorem 5.3 Under the conditions of Theorem 5.1, the nth eigencurve has an (oblique) asymptote µ = λr + hn as λ → ∞, where hn = inf{ˆ µ|L(ˆ µ) < nπp + β1 }, and L is as in (5.6).
Proof: As noted above, under the conditions of Theorem 5.1, hn exists. For each component I of S, Θ(I, µ ˆ) is a decreasing function of µ ˆ (see (5.6) et seq. for notation). Let M (ˆ µ) be the least integer for which L(ˆ µ) < M (ˆ µ)πp + β1 . Then by [3, Theorem 4.3], M (ˆ µ) is the minimal oscillation number for solutions of (5.3) obeying (3.2)-(3.3), for λ ∈ R. Observe that from the definition of the infimum, if µ ˆ < hn then L(ˆ µ) ≥ nπp + β1 . For µ ˆ > hn , from the definition of hn there is a µ ˆ1 with µ ˆ>µ ˆ1 ≥ hn with L(ˆ µ1 ) < nπp + β1 . From the monotonicity of L it follows that L(ˆ µ) ≤ L(ˆ µ1 ) < nπp + β1 so L(ˆ µ) < nπp + β1 . From the previous section we obtain that µ ˆ(λ) is a decreasing function of λ, so by Theorem 5.1, µ ˆn (λ) & Υ as λ → ∞, for some Υ ∈ R. Since µ = µ ˆn (λ) is the nth eigencurve of (5.3) with boundary conditions (3.2)-(3.3), ˆ λ, µ θ(1, ˆn (λ)) = nπp + β1 , for all λ ∈ R. Using (5.4) we get ˆ λ, Υ) ≥ θ(1, ˆ λ, µ L(Υ) ← θ(1, ˆn (λ)) = nπp + β1 ,
18
as λ → ∞. Hence Υ ≤ hn . If Υ < hn then take k with Υ < k < hn . Now L(k) ≥ nπp + β1 and ˆ λ, k) = L(k) ≥ nπp + β1 . lim θ(1,
λ→∞
ˆ λ, k) is strictly decreasing in λ, so for each λ ∈ R, But θ(1, ˆ λ, k) > nπp + β1 . θ(1, ˆ λ, k) is strictly decreasing in k, k < µ Again as θ(1, ˆn (λ) for all λ. In particular k ≤ Υ = lim µ ˆn (λ). λ→∞
This contradicts how k was chosen. Hence Υ 6< hn , i.e. hn = Υ. The intercept hn can also be found by the following construction, which is useful in concrete examples; see the next section. Let the components of S in (5.5) be labelled Sj = (aj , bj ).
Theorem 5.4 Consider the one parameter family of eigenvalue problems (Pj ) given by the differential equation −∆p y = −(p − 1)(q + σs)[y]p−1 ,
on Sj ,
with Dirichlet boundary conditions, unless aj = 0 (resp. bj = 1) in which case (1.2) (resp. (1.3)) is imposed. If σ0j > σ1j > . . . are the eigenvalues of (Pj ) then the double sequence Σ = (σij )i,j has no finite accumulation point. If Σ is reordered as σ0 ≥ σ1 ≥ . . . , then hn = σn , n = 0, 1, . . . . Proof: We begin by showing that the double sequence (σij ) has no finite accumulation point. Suppose the contrary. As each of the sequences (σij )i does not have a finite accumulation point, j = 1, 2, . . . there is an infinite increasing sequence of distinct indices j1 , j2 , j3 , . . . with corresponding i indices i1 , i2 , . . . so that (σijkk )k is a Cauchy sequence. We assume none of the intervals S¯jk contains the points 0, 1, other cases being similar. If η = −1 + inf k σijkk then, in the notation of (5.7), jk πp ≤ Θ(Sjk , η) = Θ(bjk , η) − Θ(ajk , η), but |Θ0 | ≤ 2 + |q| + |η|s
19
giving Z jk πp ≤ Θ(Sjk , η) ≤ 2|Sjk | +
Z |q| + |η|
Sjk
Z
1
|q| + |η|
s≤2+ Sjk
Z
0
1
s, 0
which is a contradiction if k is large. Thus the double sequence (σij ) has no finite accumulation point and consequently the double sequence (σij ) can be reordered as σ0 ≥ σ1 ≥ . . . . Let L1 (η) = Θ(SJ , η) where J ∈ N satisfies 1 ∈ S¯J . If there is no such J take L1 = 0 = ˆ 1 . If L ˆ 1 (η) = b L1 (η)−β1 +πp cπp , then at each eigenvalue of the problem (Pj ) on SJ the L πp ˆ 1 (η) has a jump discontinuity of size πp . If L(η) ˆ ˆ 1 (η). It is function L := L(η) − L1 (η) + L ˆ is constant on each interval (σn+1 , σn ) and that if clear from the above definition that L ˆ 2 )− L(η ˆ 1 ) = (k+1)πp η1 > η2 with σn−1 > η1 > σn ≥ · · · ≥ σn+k > η2 > σn+k+1 then L(η ˆ where we take σ−1 = ∞. As L(η) → 0 as η → ∞ it follows that L(η) = 0 for all η > σ0 . ˆ n ) = (n + k)πp and L(η) ˆ Hence if σn−1 > σn = · · · = σn+k > σn+k+1 then L(σ < nπp for all η > σn , giving ˆ µ) < (n + k)πp } = inf{ˆ inf{ˆ µ|L(ˆ µ|ˆ µ > σn } = σj for j = n, . . . , n + k. ˆ µ) < (n + k)πp } = σj for j = n, . . . , n + k, but L(ˆ ˆ µ) < (n + k)πp Hence hn+k = inf{ˆ µ|L(ˆ if and only if L(η) < β1 + (n + k)πp . Theorem 5.5 If S in Theorem 5.3 is empty, i.e., there is no interval (a, b) 6= φ on which rˆ = 0 a.e., then no eigencurve has an (oblique) asymptote as λ → ∞. Proof: As noted above L(ˆ µ) → 0 as µ ˆ → ∞. The determination of L(ˆ µ) via Θ0 shows that in this case L(ˆ µ) is independent of µ ˆ as is S = φ. Thus L(ˆ µ) = 0 for all µ ˆ. Now if µn (λ) were to have an oblique asymptote, it would be of the form µn = λ¯ r + h. If c ≤ h, then ˆ λ, c) ≥ θ(1, ˆ λ, h) = θ(1, ˆ λ, µn − λ¯ θ(1, r) = θ(1, λ, µn (λ)) = nπp + β1 > 0, ˆ λ, c) → L(c) = 0 as λ → ∞, and we have a for n = 0, 1, . . . . On the other hand θ(1, contradiction. Corollary 5.6 Suppose r > 0 and that the nth eigencurve has asymptote µ = λr + hn . If µn (0) < 0 then hn + λr ≤ µn (λ) ≤ µn (0) + λr, so −µn (0) −hn ≤ λ+ n ≤ r r + for any λ+ n > 0 with µn (λn ) = 0.
20
Note that this holds for all n in the left definite case since then µ0 (0) < 0.
Remark 5.7 In the right semi-definite case λk is unique for all k satisfying hk < 0 if the asymptote exists (Theorem 5.1) and for all k ≥ 0 if no asymptotes exists (Theorem 5.5). This follows from (3.20) R 1 and the fact that y has only finitely many zeros, so, as in the proof of Theorem 5.1, 0 r|y|p > 0 for each eigenfunction y.
6
Example
In this section, we consider an equation (discussed by Richardson for the case of p = 2, cf. [7]) of the form −([y 0 ]p−1 )0 = (p − 1)(λr − µ)[y]p−1 ,
on (−1, 1),
(6.1)
and r(x) = sgn (x). Here r = −1 and r = 1. The above differential equation will be subjected to boundary conditions y 0 (−1) = 0 = y 0 (1).
(6.2)
We take y(x; λ, µ) to be the solution of (6.1) with initial conditions y(−1) = 1 and y 0 (−1) = 0.
Theorem 6.1 For λ = 0 the eigenvalues of (6.1)-(6.2) are nπ p p µn (0) = − , n = 0, 1, . . . . 2 In addition the eigencurve µ = µn (λ) satisfies µ0n (0) = 0 and µn (λ) = µn (0) + o(λ) for n = 0, 1, . . . as λ → 0.
Proof: Taking µ as the eigenparameter, for λ = 0, this is a left semi-definite and right definite problem. In the terminology Rof [3] if yn is an eigenfunction with λ = 0 and 1 µ = µn (0) then 0 ≤ `0 [yn ] = −µn (0) −1 |yn |p . Thus µn (0) ≤ 0, moreover it is clear that for λ = 0 = µ the constant function obeys both the differential equation and the boundary conditions. Thus µ0 (0) = 0 with eigenfunction say y0 (x, 0, 0) = 1. For λ = 0 and µ < 0, y(x) = cosp (−µ)1/p (x + 1) satisfies (6.1) on [−1, 0), with initial conditions y 0 (−1) = 0 and y(−1) = 1, while y(x) = cosp (−µ)1/p (1 − x) satisfies (6.1)
21
on (0, 1], with terminal conditions y 0 (1) = 0 and y(1) = 1. Hence the eigen-condition becomes cos0p (−µ)1/p cosp (−µ)1/p + cosp (−µ)1/p cos0p (−µ)1/p = 0, i.e. 2 cos0p (−µ)1/p cosp (−µ)1/p = 0. Now cosp (ϕ) = 0 for ϕ = (n + 21 )πp , n ∈ Z, while cos0p (ϕ) = 0 when | cosp (ϕ)| = 1, i.e. for ϕ = nπp , n ∈ Z. Thus the eigenvalues are given nπ by (−µn (0))1/p = 2 p , n = 0, 1, 2, . . . , making µn (0) = −
nπ p p
2
,
n = 0, 1, 2, . . . ,
as claimed. Finally observe that if y(x; λ, µ) solves (6.1) and (6.2) then so does y(−x; −λ, µ). Hence µn (−λ) = µn (λ). Now as µn ∈ C 1 we have µ0n (0) = 0 giving µn (λ) = µn (0) + o(λ). We are now in a position to study the nature of the eigencurves µ = µn (λ), n = 0, 1, . . . , near λ = 0. As will be seen, µn is approximately quadratic there for n ≥ 1, while the case of n = 0 (which illustrates tightness of the H¨older exponent ω in Theorem 3.1 when p > 2) requires more care.
Theorem 6.2 For 1 < p < ∞ and λ near 0, µ0 (λ) =
ap |λ|d (1 + o(1)), d+1
while for n ≥ 1, µn (λ) = µn (0) + λ
2
2 nπp
p
(−1)n p − 1 (1 + o(1)), 2p
where a and d are as in Lemma 2.5.
Proof: From the previous theorem for small λ > 0 and x 6= 0: for n ≥ 1 we have λr(x) − µn (λ) > 0 while for n = 0, (λr(x) − µ0 (λ))sgn (x) > 0. By the symmetry of the eigencurves the case λ < 0 can be treated similarly. We begin with the case n = 0. For x ≤ 0 solutions obeying the left hand boundary condition are multiples of y− (x) = coshp (λ + µ)1/p (x + 1) while for x ≥ 0 solutions obeying the right hand boundary condition are multiples of y+ (x) = cosp (λ−µ)1/p (1−x). 0 (0)y (0) − y (0)y 0 (0) = 0. Hence for µ = µ (λ) we have Thus the eigencondition is y+ − + 0 − −u1/p cos0p u1/p coshp v 1/p = v 1/p cosp u1/p cosh0p v 1/p .
(6.3)
where, for ease of notation, we have set u = λ − µ and v = λ + µ. Note that u = λ + o(λ) ≥ 0 and v = λ + o(λ) ≥ 0. Here and below all order estimates are as λ & 0.
22
With a and d as in Lemma 2.5, where d/p = d − 1, (6.3) gives ! ! aud/p a2 v 2d/p d/p d/p d/p 2d/p adu 1− + o(u ) 1 + av + + o(v ) d+1 2(d + 1) ! ! av d/p a2 u2d/p 2d/p d/p d/p d/p + o(u ) 1+ + o(v ) . = adv 1 − au + 2(d + 1) d+1 This, when raised to the power p − 1 = p/d, gives u−
apud apuv d−1 apv d apvud−1 + =v+ − + o(ud + v d ). d(d + 1) d d(d + 1) d
Recalling that u = λ + o(λ) and v = λ + o(λ) with 2µ = v − u we obtain µ=
ap d λ + o(λd ). d+1
nπ
We now fix n ≥ 1 and write ξ = 2 p . Let µn (λ) = µn (0)+k(λ). Here k(λ) = o(λ) and 0 < λr(x) − µn (λ) = λr(x) − µn (0) − k(λ) for λ > 0 and small. Thus taking µ = µn (0) + k(λ) we have for x ≤ 0 solutions obeying the left hand boundary condition are multiples of y− (x) = cosp (−λ − µn (0) − k(λ))1/p (x + 1) while for x ≥ 0 solutions obeying the right hand boundary condition are multiples of y+ (x) = cosp (λ − µn (0) − k(λ))1/p (1 − x). As 0 (0)y (0) − y (0)y 0 (0) = 0. Hence we have before the eigencondition is y+ − + − (ξ p + λ − k(λ))1/p cos0p (ξ p + λ − k(λ))1/p cosp (ξ p − λ − k(λ))1/p = − (ξ p − λ − k(λ))1/p cosp (ξ p + λ − k(λ))1/p cos0p (ξ p − λ − k(λ))1/p . For ease of notation set σ p = ξ p − λ − k(λ), δ p = ξ p + λ − k(λ). Now δ cos0p δ cosp σ = −σ cosp δ cos0p σ. Here 1/p σ = ξ 1 − ξ −p v 1 p − 1 1−2p 2 = ξ − ξ 1−p v − ξ v (1 + o(1)), p 2p2 1/p δ = ξ 1 + ξ −p u , 1 p − 1 1−2p 2 = ξ + ξ 1−p u − ξ u (1 + o(1)), p 2p2
23
where u = λ − k(λ) and v = λ + k(λ). Thus for n even cos0p cos0p 1 1−p p − 1 −p (σ) = − ξ v 1+ ξ v(1 + o(1)) , cosp cosp p 2p cos0p 1 1−p cos0p p − 1 −p (δ) = ξ u 1− ξ u(1 + o(1)) , cosp cosp p 2p and for n odd cosp sinp 1 1−p p − 1 −p (σ) = − 0 ξ v 1+ ξ v(1 + o(1)) , cos0p sinp p 2p sinp 1 1−p cosp p − 1 −p (δ) = ξ u 1 − ξ u(1 + o(1)) , cos0p sin0p p 2p Combining these results and using Lemma 2.5 we see that for even n, cos0p σ (σ) = cosp
1/(p−1)
v 1− (1 + o(1)) , 2pξ p cos0p u(p − 1) 1/(p−1) u (δ) = 1+ (1 + o(1)) . −δ cosp p 2pξ p v(p − 1) p
Raising the above two equations to the power p − 1 and equating them, we find that the eigencondition for even n ≥ 1 becomes v − v2
p−1 p−1 (1 + o(1)) = u + u2 (1 + o(1)). 2pξ p 2pξ p
Thus k(λ) = λ2
p−1 (1 + o(1)). 2pξ p
Finally for odd n 1 cosp − (σ) = σ cos0p 1 cosp (δ) = δ cos0p
1 p+1 v 1+ v(1 + o(1)) , pξ p 2pξ p 1 p+1 u 1 − u(1 + o(1)) . pξ p 2pξ p
Thus the eigencondition gives v+
p+1 2 p+1 2 v (1 + o(1)) = u − u (1 + o(1)), 2pξ p 2pξ p
24
and so k(λ) = −λ2
p+1 (1 + o(1)), 2pξ p
which completes the study of the eigencurves near λ = 0. Now using Theorems 5.3 and 5.4 we give the behaviour of the eigencurves for large λ.
Theorem 6.3 For |λ| large the eigencurve µ = µn (λ) approaches the oblique asymptote µ = |λ| + hn where (2n + 1)πp p hn = − , n = 0, 1, . . . . 2
Proof: By symmetry we assume λ > 0. In the notation of (1.4), s(x) = 1 for all x ∈ [−1, 1] while r(x) = r¯ = 1 on (0, 1], so by Theorem 5.3 the eigencurve µ = µn (λ) has an asymptote and it is of the form µ = λ + hn . From Theorem 5.4, hn = σn where σ0 > σ1 > . . . are the eigenvalues of −([y 0 ]p−1 )0 = (p − 1)(−σ)[y]p−1 ,
on (0, 1),
with boundary conditions y(0) = 0 = y 0 (1), giving the value of hn , above.
7
Appendix
We collect here the proofs of Theorem 2.6 and Lemmas 2.4 and 2.5. Proof: [of Lemma 2.4] The results for sinp and sinhp follow directly from their definitions. We now prove the result for the case of cosp . From the definition of cosp , cosp (0) = 1 and cos0p (0) = 0 with cosp θ continuously approaching 1 from below as θ tends down to 0. Thus setting cosp (h) = 1 − δ(h) we have 0 < δ(h) → 0 as h & 0. In the integral (2.7) the integration variable t lies in (cosp (h), 1). Setting t = 1 − η, we have η ∈ (0, δ(h)) and η → 0 as h & 0, so tp = 1 − pη + o(η)
25
and (1 − tp )1/p = (pη + o(η))1/p = p1/p η 1/p (1 + o(1)). Returning to the integral (2.7) we have h=
1
1−cosp h
Z
p1/p
0
1− 1
dη p p 1− 1 (1 + o(1)) = (δ(h)) p (1 + o(1)), 1/p p−1 η
whence 1 − cosp (h) = δ(h) =
(p − 1)d hd (1 + o(1)), p
giving cosp (h) − cosp (0) (p − 1)d → − = −a, p hd as h & 0. A similar calculation holds as h % 0. For coshp , coshp (0) = 1 and cosh0p (0) = 0 with coshp θ & 1 as θ & 0. Thus we can set coshp (h) = 1 + δ(h) with 0 < δ(h) → 0 as h & 0. In the integral (2.8) the integration variable t lies in (1, cosp (h)), and we take t = 1 + η, where η ∈ (0, δ(h)) and η → 0 as h & 0. Thus tp = 1 + pη + o(η) and (tp − 1)1/p = (pη + o(η))1/p = p1/p η 1/p (1 + o(1)),
as
h & 0.
Returning to the integral (2.8) we have h=
1 p1/p
Z 0
coshp h−1
1− 1
dη p p 1− 1 (1 + o(1)) = (δ(h)) p (1 + o(1)), 1/p p−1 η
so coshp (h) − 1 = δ(h) =
(p − 1)d hd (1 + o(1)), p
giving coshp (h) − coshp (0) (p − 1)d → = a, p hd as h & 0. Again the calculation as h % 0 is similar.
26
Proof: [of Lemma 2.5] We consider first the cases of cosp . Let g(t) := cosp t − 1 + atd . From Lemma 2.4 we know that g(t)/td → 0 as t & 0 with g(0) = 0. It remains only to show that a2 (p − 1) ∂ 2d g(0) = Γ(2d + 1) . 2 To do this let δ(h) > 0 be such that g(h) = hd δ(h) for h > 0. Then we know from Lemma 2.4 that δ(h) → 0 as h & 0. Now for 1 > η > 0 and t = 1 − η we have that tp = 1 − pη +
p(p − 1) 2 η + o(η 2 ) 2
as η → 0, so p −1/p
(1 − t )
−1/p p(p − 1) 2 2 pη − η + o(η ) 2 −1/p p−1 1 1 − η + o(η) 2 p1/p η 1/p 1 p−1 1+ η + o(η) 2p p1/p η 1/p
= = = =
η −1/p p−1 + (p+1)/p η (p−1)/p + o(η (p−1)/p ). 1/p p 2p
Now returning to the definition of cosp we obtain Z
hd (a−δ(h))
h= 0
! η −1/p p − 1 (p−1)/p + (p+1)/p η (1 + o(1)) dη, p1/p 2p
giving #hd (a−δ(h)) η 1−(1/p) p−1 2−(1/p) h = + η (1 + o(1)) p1/p (1 − (1/p)) 2p(p+1)/p (2 − (1/p)) 0 " #hd (a−δ(h)) η (p−1)/p p−1 = + η 1+(p−1)/p (1 + o(1)) p1/p ((p − 1)/p) 2p(p+1)/p (1 + (p − 1)/p) 0 " #hd (a−δ(h)) dη 1/d 1 = + 1/p η (d+1)/d (1 + o(1)) 1/p p 2p (1 + d) "
0
=
δ(h))1/d
hd(a − p1/p
+
1 2p1/p (1
+ d)
(a − δ(h))(d+1)/d hd+1 (1 + o(1)).
27
Hence 2p1/p (1 + d)(a − δ(h))−1/d − 2d(1 + d) = (a − δ(h))hd (1 + o(1)) from which 2p1/p (1 + d)a−1/d (1 +
δ(h) + o(δ(h))) − 2d(1 + d) = (a − δ(h))hd (1 + o(1)). ad
As p1/p a−1/d = d, the above simplifies to δ(h) =
hd a2 (1 + o(1)). 2(1 + d)
Thus cosp h = 1 − ahd +
h2d a2 + o(h2d ). 2(1 + d)
The case of coshp follows in similar manner. To obtain an approximation to the derivative of cosp h for h > 0 small, observe cos0p h
p 1/p h2d a2 d 2d = − 1 − 1 − ah + + o(h ) 2(1 + d) 1/p (p − 1)aphd d 1/p = −(aph ) 1− (1 + o(1)) 2p − 1 (p − 1)ahd = −(aphd )1/p 1 − (1 + o(1)) 2p − 1
and similarly for cosh0p . In contrast to the above, the expansion of sinp can be derived simply from Lemma 2.4 and the definition of sinp which give sin0p h = (1 − hp (1 + o(1)))1/p = 1 −
hp (1 + o(1)). p
Integrating this and noting that sinp (0) = 0 we come to sinp h = h −
hp+1 (1 + o(1)), p(p + 1)
28
as
h & 0.
From the definition of sinp we now obtain sin0p h
p 1/p hp+1 (1 + o(1)) p(p + 1) 1/p hp p (1 + o(1)) 1−h 1− p+1 2 hp hp 1 1 hp p 1− (1 + o(1)) + −1 h 1− (1 + o(1)) 1− + o(h2p ) p p+1 2p p p+1 hp 1 p−1 h2p + o(h2p ) 1− + − p p(p + 1) 2p2 hp p2 − 2p − 1 2p 1− − h + o(h2p ). p 2p2 (p + 1)
= = = = =
1− h−
Integration gives p2 − 2p − 1 hp+1 − 2 h2p+1 + o(h2p+1 ). p(p + 1) 2p (p + 1)(2p + 1)
sinp h = h −
Proceeding in a similar manner for sinhp one obtains sinhp h = h +
p2 − 2p − 1 hp+1 − 2 h2p+1 + o(h2p+1 ). p(p + 1) 2p (p + 1)(2p + 1)
Proof: [of Theorem 2.6] From (2.14), the functions sinp and cosp when considered of the whole interval [0, πp ] are in the same H¨ older class. The functions cosp and sinp are in C ∞ (0, πp /2) thus in proving the result we need only consider behaviour at and near the points 0 and πp /2. In the light of (2.14), to obtain the behaviour of sinp near πp /2 it suffices to consider cosp at 0. It follows from (2.10) and (2.12) that cosp and sinp are at least C 1 on [0, πp /2] with cos0p (0) = 0 and sin0p (0) = 1. Appealing to Lemma 2.5, we have 1
cos0p (θ) − cos0p (0) = O(θ p−1 ), sin0p (θ) − sin0p (0) = O(θp ), as θ & 0. Now for p > 2, 1 > For 1 < p < 2, both p and
1 p−1
1 p−1
giving sinp , cosp ∈ C
1 1, p−1
.
are greater than 1, so cos00p and sin00p exist at 0 and are
29
both 0. Meanwhile (2.14)-(2.17) and direct computation give, that for θ ∈ (0, πp /2), 0 1/(p−1) cos00p (θ) = − sind ((p − 1)θ) 1
= − sin0d ((p − 1)θ) sindp−1 = −((p − 1)θ)
1 −1 p−1
−1
((p − 1)θ)
(1 + o(1)),
and similarly 1
sin00p (θ) = cos0d ((p − 1)θ) cosdp−1
−1
((p − 1)θ)
= ((p − 1)θ)p−1 (1 + o(1)), 2 as θ & 0. As can be seen from the above n two displays, o sinp , cosp ∈ C [0, πp /2] and are 1 in fact in C 2,ζ [0, πp /2] where ζ = min p − 1, p−1 −1 .
For the p-hyperbolic functions the proof is similar but simpler as they are C ∞ on R\{0}.
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