Exact number of local extreme points of curvature function of solutions of secondorder linear differential equations Houssam Chrayteh & Mervan Pašić
Journal of Applied Mathematics and Computing ISSN 1598-5865 J. Appl. Math. Comput. DOI 10.1007/s12190-012-0541-4
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Author's personal copy J Appl Math Comput DOI 10.1007/s12190-012-0541-4
JAMC
O R I G I NA L R E S E A R C H
Exact number of local extreme points of curvature function of solutions of second-order linear differential equations Houssam Chrayteh · Mervan Paši´c
Received: 11 October 2011 © Korean Society for Computational and Applied Mathematics 2012
Abstract We study local properties of the curvature κy (x) of every nontrivial solution y = y(x) of the second-order linear differential equation (P ): (p(x)y ) + q(x)y = 0, x ∈ (a, b) = I , where p(x) and q(x) are smooth enough functions. It especially includes the Euler, Bessel and other important types of second-order linear differential equations. Some sufficient conditions on the coefficients p(x) and q(x) are given such that the curvature κy (x) of every nontrivial solution y of (P ) has exactly one extreme point between each two its consecutive simple zeros. The problem of three local extreme points of κy (x) is also considered but only as an open problem. It seems it is the first paper dealing with this kind of problems. Finally in Appendix, we pay attention to an application of the main results to a study of non-regular points (the cusps) of the ε-parallels of graph Γ (y) of y (the offset curves of Γ (y)). Keywords Curvature · Extreme points · Geometry of solutions · Linear differential equations Mathematics Subject Classification (2000) 26A06 · 34A26 · 34B05 · 53A04
H. Chrayteh () UMR 6086CNRS, Laboratoire de Mathématiques et Applications, Université de Poitiers SP2MI, Boulevard Marie et Pierre Curie, Téléport 2, Futuroscope, BP30179, 86962 Chasseneuil Cedex, France e-mail:
[email protected] M. Paši´c Department of Applied Mathematics, Faculty of Electrical Engineering and Computing-FER, University of Zagreb, 10000 Zagreb, Croatia e-mail:
[email protected]
Author's personal copy H. Chrayteh, M. Paši´c
1 Introduction Let y = y(x) be a C 2 real function defined on the interval I = (a, b) or I = R, where a, b ∈ R, a < b, with the graph Γ (y) = {(t1 , t2 ) : t1 ∈ I, t2 = y(t1 )}. The curvature function of Γ (y), shortly saying: the curvature of y(x), is denoted by κy (x) and defined by formula: κy (x) =
y (x) (1 + y 2 (x))3
,
x ∈ I.
(1)
For more details and elementary properties of κy (x) we refer reader to [1] and [3]. On the one hand, we are aware that the curvature function of a C 2 plane curve can have arbitrary number of local extreme points as well as that the determination of this number is not a simple procedure. For example, we mention the next two essential results: • the famous Four-Vertex Theorem says: the curvature of a simple C 2 closed curve has at least four extreme points—vertices, see [3]; • the conjecture: if y(x) is a polynomial of degree n greater than 1, then the curvature of y(x) has at most n − 1 extreme points, has been solved in [4] provided y (x) admits only real roots. On the other hand, the exact number of local extreme points of y(x) on some interval plays an important role in the theory of nodal solutions of boundary value problems for second-order differential equations. For instance, in [12] it is shown that the spectrum σ (L) of Lu = −u consists of a strictly increasing sequence of eigenvalues λk , k ∈ N, limk→∞ λk = ∞, such that the eigenvector u1 which corresponds to the first eigenvalue λ1 has exactly one extreme point in (0, 1). Moreover, for every k ∈ N, eigenvector uk which corresponds to eigenvalue λk has exactly k local extreme points in (0, 1) as well as uk (0) = 0, uk (0) > 0 and uk (1) = 0. The main problem posed in this paper is different than previous ones, but it is motivated by them, and says: find sufficient conditions on the coefficients of the considered second-order differential equation such that the curvature κy (x) of every nontrivial solution y(x) has exactly one extreme point between each two its consecutive simple zeros. The first main result of the paper solves previous problem for the case of secondorder linear differential equations as follows. Theorem 1 Let J ⊆ I be an open interval and let the function q(x) be such that: q ∈ C 2 (I ),
q 2 (x) − q (x) > 0 in J
and
2 q(x) > 5π/|J | ,
(2)
where |J | denotes the length of J . We consider the linear second-order differential equation: y + q(x)y = 0,
x ∈ I.
(3)
Then the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (3) has exactly one extreme point between each two its consecutive simple zeros in J .
Author's personal copy Geometry of solution near its extreme points Fig. 1 curvature κy (x) (dashed line) of y(x) = sin x has exactly one extreme point between each two its consecutive simple zeros in R
The most simple example of (3) is for q(x) = λ > 0, which obviously satisfies the required condition (2) in J = I = R. The function y(x) = sin x is a solution of such equation and the graphs of y(x) and κy (x) are presented in Fig. 1. Several examples for linear differential equations which satisfy the assumptions of Theorem 1 are given in Examples 1 and 2, see Sect. 2. The proof of Theorem 1 is given in Sect. 3. The same problem is also stated in Theorem 3 of Sect. 2 for the self-adjoint linear differential equation (py ) + qy = 0. As a consequence, we obtain that curvature κy (x) of every solution y(x) of linear differential equation (x μ y ) + x −σ y = 0, μ > 0, σ > 0, has exactly one extreme point between each two its consecutive simple zeros provided μ + σ > 2. Our second main result of the paper solves the following problem: find sufficient conditions on the coefficients of the considered second-order differential equation such that every solution y(x) and its curvature κy (x) have or not have common stationary points between each two consecutive simple zeros of κy (x). Theorem 2 Let q(x) > 0 in I and let y(x) be a solution of the linear equation y + q(x)y = 0, x ∈ I . Let a0 , b0 ∈ I , a0 < b0 be two consecutive simple zeros of κy (x). We have: (i) if q (x) = 0 for all x ∈ I , then y(x) and κy (x) have a common stationary point in (a0 , b0 ); (ii) if q (x) = 0 for all x ∈ I , then y(x) and κy (x) do not have any common stationary point in (a0 , b0 ). Obviously from Fig. 1, we can see that y(x) = sin x and its κy (x) have a common stationary point between each two consecutive simple zeros of κy (x) in R. It is also an immediate consequence of the first statement of previous theorem since y(x) = sin x is a solution of equation y + y = 0 and this equation satisfies the condition (i)Theorem 2 in I = R. However, by previous theorem, the function y(x) = x sin(1/x), x > 0, and its κy (x) do not have any common stationary point between each two consecutive simple zeros of κy (x) in (0, ∞). It is because y(x) = x sin(1/x), x > 0, is a solution of the equation y + x −4 y = 0, x > 0, and this equation satisfies the condition (ii)-Theorem 2 in I = (0, ∞). For more details about the proof of Theorem 2 see Sect. 4. Next, in Sect. 5 as an open problem, we suggest the case when curvature κy (x) of every solution y(x) of considered equation has at least three extreme points between each two its consecutive simple zeros. This property will be established by using Lemma 6 for κy (x) of the function y(x) = sin(sin x), which is a solution of the following second-order linear differential equation: y + (tg x)y + (cos2 x)y = 0. The graphs of such y(x) and κy (x) are given in Fig. 2.
Author's personal copy H. Chrayteh, M. Paši´c Fig. 2 Curvature κy (x) (dashed line) of y(x) = sin(sin x) has exactly three extreme points between each two its consecutive simple zeros
Fig. 3 The number of local extreme points of κy (x) (dashed line) of y(x) = sin4 x alternates between 1 or 3
Remark 1 Regarding to Figs. 1 and 2, we remark that the function z(x) = sin(sin x) passes more slowly through its local extreme points than the function y(x) = sin x. In this situation, one wonders: Does the number of extreme points of curvature κy (x) between each two its consecutive simple zeros influences the growth rate of y(x) near its extreme points? Generally, we do not know. However, from Figs. 1 and 2, we see that |κz (x)| ≤ |κy (x)| near sn = π/2 + nπ , n ∈ Z. It implies that |z (x)| ≤ |y (x)| near sn , where z (sn ) = y (sn ) = 0. Indeed, for arbitrary s0 ∈ R and C 2 functions z(x) and y(x) such that z (s0 ) = y (s0 ) = 0, z (x) ≥ 0 and y (x) ≥ 0 on (s0 , s0 + ε) for some ε > 0 and κz (x) ≤ κy (x) on (s0 , s0 + ε), we have by integrating the last inequality and using (1): s0 +ε s0 +ε dz dy ≤ , s0 s0 (1 + y 2 )3 (1 + z2 )3 which gives
s +ε s +ε tanh Argsinh z s0 ≤ tanh Argsinh y s0 . 0
0
Since z (s0 ) = y (s0 ) = 0, we obtain |z (x)| ≤ |y (x)| on (s0 , s0 + ε). It is interesting to remark that the exact number of extreme points of κy (x) of function y(x) = sin4 x between each two its consecutive simple zeros in R can alternate between 1 or 3, see Fig. 3. In the determination of the exact number of extreme points for the curvature κy (x) between each two its consecutive simple zeros, the computational aspect of the formula (1) is limited whenever the function y(x) is more complex than y(x) = sin x as well as in case where y(x) is not explicitly given, for example, when y(x) is a solution of differential equation (p(x)y ) + q(x)y = 0. Hence in Sect. 2 we give some criteria on y(x) such that the curvature κy (x) of y(x) has exactly one extreme point between each two its simple zeros, see Lemmas 1, 2 and 3. The particular attention is paid to the case of nontrivial solutions y(x) of undamped equation y + q(x)y = 0 (Theorem 1) as well as of self-adjoint equation (p(x)y ) + q(x)y = 0 (Theorem 3) and consequently of damped equation y + g(x)y + f (x)y = 0 (Corollaries 1 and 2).
Author's personal copy Geometry of solution near its extreme points
For more details about qualitative properties of solutions of the second-order linear differential equations we refer reader to [5]. In Sect. 3, we study some local properties of κy (x) (Lemma 4), which are of great interest in the proof of Theorem 1 and the main results stated in Sect. 2. In Appendix of the paper, we consider the exact number of singular points (the cusps) for the ε-parallels of y(x) (the offset curves of y(x)) as a geometric consequence of the main results of the paper, see Theorems 5 and 6. For more details about the cusps we refer reader to [1] and about the offset curves see [8] and [11]. In the applications, the so-called motions by curvature (a kind of evolution of plane curves) appear in the mathematical modelling of various phenomena as crystal growth (see [14]), flame propagation (see [13]), interfaces between phases, image processing, etc. In the last one, it provides an efficient way for smoothing curves detecting the objects boundaries (for an axiomatic approach as well as mathematical results and numerical applications see [2]). The offset curves arises in many industrial applications (for instance in geometric modelling of cutting paths for milling machines, robot-path planning etc., see [11]).
2 The problem of exactly one extreme point of κy (x) When we say that a0 and b0 are two consecutive inflexion points of y(x) in I , it means that a0 < b0 , y (a0 ) = y (b0 ) = 0, y (x) does not change the sign in (a0 , b0 ) and y (x) changes the sign at a0 and b0 . By (1) it also means that a0 and b0 are two consecutive simple zeros of κy (x) in I , that is, κy (a0 ) = κy (b0 ) = 0, κy (x) does not change the sign in (a0 , b0 ) and κy (x) changes the sign at a0 and b0 . Hence, the number of local extreme points of κy (x) is considered in the interval (a0 , b0 ). If I = R, then a0 , b0 ∈ I¯ means that a0 = −∞ and b0 = ∞. In many cases of y(x), the number of local extreme points of κy (x) cannot be directly derived from the formula (1). Therefore, it is very useful to state and prove the following criterion, which is expected from the one-dimensional calculus point of view. Lemma 1 Let y ∈ C 4 (I ) and let a0 , b0 ∈ I¯ be two consecutive inflexion points of y(x), a0 < b0 . (i) Let y (x) ≥ 0 in (a0 , b0 ). The curvature κy (x) has a unique local extreme point in (a0 , b0 ) provided for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0 we have: 2 y I V (ξ0 ) 1 + y 2 (ξ0 ) < 3y 3 (ξ0 ) 1 + 5y 2 (ξ0 ) .
(4)
(ii) Let y (x) ≤ 0 in (a0 , b0 ). The curvature κy (x) has a unique local extreme point in (a0 , b0 ) provided for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0 we have: 2 y I V (ξ0 ) 1 + y 2 (ξ0 ) > 3y 3 (ξ0 ) 1 + 5y 2 (ξ0 ) .
(5)
The result of Lemma 1 is verified also in the case of y(x) has at most one (or does not have any) inflexion point in I¯, as follows.
Author's personal copy H. Chrayteh, M. Paši´c
Lemma 2 Let y ∈ C 4 (I ) and let a0 , b0 ∈ I¯, a0 < b0 . Let I0 ⊆ {a0 , b0 } be a set defined by I0 = {x0 ∈ {a0 , b0 } : x0 is not an inflexion point of y(x)}. If I0 = ∅ and limx→x0 κy (x) = 0 for all x0 ∈ I0 , then the conclusions (i) and (ii) of Lemma 1 are still valid. For instance, the function y(x) = sinh(mx) (m = 0) satisfies all assumptions of Lemma 2 on two intervals in the form (a0 , b0 ), where a0 = −∞, b0 = 0 and I0 = {−∞} (resp. a0 = 0, b0 = ∞ and I0 = {∞}) and limx→±∞ κy (x) = 0. Also, it is true for the function y(x) = x 2 in only one (a0 , b0 ), where a0 = −∞, b0 = ∞, I0 = {−∞, ∞} and limx→±∞ κy (x) = 0. Question 1 In the case of y (x) = 0 for all x ∈ (a0 , b0 ), is it possible to refuse the relations (4) and (5) such that the conclusion of Lemma 1 is still valid? The answer is no and it is given in Theorem 4. With the help of Lemma 1 one can derive the following rather simple sufficient conditions on y(x) such that κy (x) has a unique local extreme point in (a0 , b0 ). Lemma 3 Let y ∈ C 4 (I ) and let a0 , b0 ∈ I¯ be two consecutive inflexion points of y(x), a0 < b0 . We consider two conditions: (i) y (x) ≥ 0 in (a0 , b0 ) and y I V (ξ0 ) < 0 for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0; (ii) y (x) ≤ 0 in (a0 , b0 ) and y I V (ξ0 ) > 0 for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0. If the condition (i) or (ii) is satisfied, then the curvature κy (x) has a unique local extreme point in (a0 , b0 ). The proofs of previous three lemmas are presented in Sect. 3 by using the classic elementary calculus of the functions of one variable. Also, in Sect. 3 we will prove Theorem 1 as a consequence of Lemma 3. Concerning Theorem 1, we have a few remarks. At the first, in the application of Theorem 1, the open interval J ⊆ I can play an essential role because the required condition (2) is often fulfilled near a point x0 ∈ I ; in such a case, J is taken to be a finite or infinite open neighborhood of x0 . Next, the positivity of q(x) and q 2 (x) − q (x) in J ensures that Lemma 3 can be applied to all nontrivial solutions y(x) of (3). Furthermore, the isoperimetric inequality q(x) > (5π/|J |)2 in J provides the existence of at least two inflexion points in J of every nontrivial solution y(x) of (3); when |J | = ∞, this inequality become q(x) > 0 in J , for instance J = R, J = (−∞, T0 ) or J = (T0 , ∞), see the following examples. It is not difficult to check that if q(x) satisfies (2), then λq(x) also satisfies (2) provided λ ≥ 1. Moreover, if q1 (x) and q2 (x) satisfy (2), then λq1 (x) + q2 (x) also satisfies (2). The importance of Theorem 1 is shown in the next examples. Example 1 It is easy to check that the coefficient q(x) in the following linear differential equations satisfies the required conditions from (2) of Theorem 1.
Author's personal copy Geometry of solution near its extreme points
First of all, we consider the fundamental differential equation of oscillations in R, y + λy = 0,
x ∈ I,
(6)
where I = R, λ > 0 and J := R; indeed, q 2 (x) − q (x) = λ2 > 0 in J , q(x) = λ > (5π/∞)2 = 0
in J ,
and thus, for J = R all conditions in (2) are satisfied. Also, we can consider the Euler type equation, y + λx σ y = 0,
x ∈ I,
(7)
where I = (0, ∞), λ > 0, σ > −2 and J1 := (( σ (σλ−1) ) σ +2 , ∞); in fact, 1
q 2 (x) − q (x) = λ λx 2σ − σ (σ − 1)x σ −2 > 0 in J1 , q(x) = λx σ > (5π/∞)2 = 0
in J1 ,
and thus, for J = J1 all conditions in (2) are satisfied. And finally, we consider the equation with exponential coefficient, y + λeαx y = 0,
x ∈ I,
(8)
where I = R, λ > 0, α = 0, and J = J1 := (α −1 ln(α 2 λ−1 ), ∞) if α > 0 and J = J2 := (−∞, α −1 ln(α 2 λ−1 )) if α < 0; in fact, q 2 (x) − q (x) = λeαx λeαx − α 2 > 0 in J , q(x) = λeαx > (5π/∞)2 = 0 in J , and thus, for this choice of J all conditions in (2) are satisfied. In all these examples we have |J | = ∞ and therefore, the last inequality in (2) is trivially fulfilled since q(x) > 0 in J . Remark 2 With the help of previous example, we observe that every function of the type y(x) = c1 sin x + c2 cos x, c1 , c2 ∈ R, its curvature function κy (x) has exactly one extreme point between each two its consecutive simple zeros in J = R, because it is the fundamental system of all solutions of (6). However, it cannot be easily derived directly from the formula (1) since κy (x) is not linear in y(x). In the next example, we consider a linear differential equation with the coefficient q(x) satisfying the required condition (2) in a finite open neighborhood J of x = 0 and hence |J | < ∞. It is a basic difference in respect to Example 1. Example 2 We consider the linear differential equation of Euler type: y + λx −σ y = 0,
x ∈ I,
(9)
Author's personal copy H. Chrayteh, M. Paši´c
where I = (0, ∞), λ > 0 and σ > 2. If T0 > 0 is a real number satisfying, 1 1
σ −2 σ −2 λ λ , , T0 < min 2 σ (σ + 1) 25π then the coefficient q(x) = λx −σ of (9) obviously satisfies: q 2 (x) − q (x) = λ2 x −2σ − λσ (σ + 1)x −σ −2 > 0, 2 q(x) = λx −σ > 5π/|J | , x ∈ (0, T0 ),
x ∈ (0, T0 ),
because of −2σ < −σ − 2. Now, by Theorem 1 it follows that the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (9) has exactly one extreme point between each two its consecutive simple zeros in J = (0, T0 ). Let us remark that every solution y(x) of (9) oscillates near x = 0, which gives infinitely many inflexion points of y(x) near x = 0. Remark 3 In particular for σ = 4, we know that the fundamental system of all solutions of (9) is given by √ √ y(x) = c1 x cos( λ/x) + c2 x sin( λ/x), λ > 0. (10) According to Example 2, we observe that the curvature κy (x) of y(x) given by (10) has exactly one extreme point between each two its consecutive simple zeros in J = (0, T0 ) for some T0 > 0. Open Question 1 Find necessary conditions for the problem of exactly one extreme point of κy (x) of all solutions y(x) of equation y + q(x)y = 0, x ∈ I . That is say, if the curvature κy (x) of every solution y(x) of equation y + q(x)y = 0, x ∈ I , has exactly one extreme point between each two its consecutive simple zeros in I , then what conditions should the function q(x) satisfy? Open Question 2 (i) Does Theorem 1 hold true for the linear differential equation y + q(x)y = 0, x ∈ I , for all q(x) > 0, that is to say, without the condition (2)? (ii) Does Theorem 1 hold true for the linear differential equation y + (q(x) + g(x))y = 0, x ∈ I , where the perturbation g(x) and coefficient q(x) satisfy conditions of different type? (iii) Let q(x) < 0 in J ⊆ I or let q(x) be sign-changing in J . Does it possible to find sufficient conditions on q(x) such that the main conclusion of Theorem 1 remains valid? Theorem 1 could be generalized to the case of self-adjoint linear differential equations as follows. Theorem 3 Let J ⊆ I be an open interval and let the coefficients p = p(x) and q = q(x) be such that: p ∈ C 3 (I ),
0 < p(x) < 1 and
p (x) > 0 in J,
(11)
Author's personal copy Geometry of solution near its extreme points
2 q ∈ C 2 (I ), q(x) > 5π/|J | and q (x) < 0 in J, pp − 2p 2 + pq (x) > 0 in J, 2 −p p + 6pp p − 6p 3 + 4pp q − 2p 2 q (x) > 0 in J, 3pp q − 6p 2 q + 3pp q + pq 2 − p 2 q (x) > 0 in J. We consider the self-adjoint linear second-order differential equation: p(x)y + q(x)y = 0, x ∈ I.
(12) (13) (14) (15)
(16)
Then the curvature κy (x) of each nontrivial solution y ∈ C 4 (I ) of (16) has exactly one extreme point between each two its consecutive simple zeros in J . Remark 4 The conditions of Theorem 3 are not restrictive as they seem, in fact the assumptions (11)–(15) will enable us to apply Lemma 3. For the reader’s convenience, in Example 3 below, a class of p(x) and q(x) is given where all these assumptions are easily verified. Open Question 3 Does it possible to prove previous theorem if we replace the condition “0 < p(x) < 1 in J ” by the weaker one “ 0 < p(x) ≤ 1 in J ”? In the next example we show that in particular for p(x) = x μ and q(x) = x −σ , where μ, σ > 0, the required conditions (11)–(15) are easily fulfilled by supposing only one condition on μ and σ , that is, μ + σ > 2. Example 3 Let I = (0, ∞). Let μ and σ be two positive real numbers such that μ + σ > 2. We consider the Euler type second-order linear equation: μ (17) x y + x −σ y = 0, x ∈ I. In order to show that the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (17) satisfies the problem of exactly one extreme point in J = (0, T0 ) for some T0 > 0, it is enough by Theorem 3 to check that the functions p(x) = x μ and q(x) = x −σ , μ, σ > 0 satisfy the required assumptions (11)–(15) provided μ + σ > 2. Let us remark that the existence of at least two consecutive inflexion points in √ of y(x) / L1 (I ) J follows from the oscillatory behavior of (17) in J . It is because of q/p ∈ which is equivalent to the condition μ + σ > 2. Indeed, (11) and (12) are obviously satisfied near x = 0 since μ and σ are positive real numbers. However, (13), (14) and (15) could be easily verified from the following calculation: (18) P0 (x) = pp − 2p 2 + pq (x) = − μ2 + μ x 2μ−2 + x μ−σ , 2 P1 (x) = −p p + 6pp p − 6p 3 + 4pp q − 2p 2 q (x) = −μ μ2 + 3μ + 2 x 3μ−3 + 2(2μ + σ )x 2μ−σ −1 , (19)
Author's personal copy H. Chrayteh, M. Paši´c
P2 (x) = 3pp q − 6p 2 q + 3pp q + pq 2 − p 2 q (x) = − 3μσ + 3μ2 + 3μ + σ 2 + σ x 2μ−σ −2 + x μ−2σ .
(20)
Let us remark that all of the first terms of previous equalities are negative, that is, for x > 0 we have: − μ2 + μ x 2μ−2 < 0, −μ μ2 + 3μ + 2 x 3μ−3 < 0, − 3μσ + 3μ2 + 3μ + σ 2 + σ x 2μ−σ −2 . Also, the corresponding last terms in P0 (x), P1 (x) and P2 (x) are positive, that is: x μ−σ > 0,
2(2μ + σ )x 2μ−σ −1 > 0 and x μ−2σ > 0 for all x > 0.
Therefore, in order to prove the positiveness of P0 (x), P1 (x) and P2 (x) near x = 0, it is enough to show respectively that: 2μ − 2 > μ − σ,
3μ − 3 > 2μ − σ − 1 and 2μ − σ − 2 > μ − 2σ.
But it is true since μ + σ > 2. Thus, we have shown that P0 (x) > 0, P1 (x) > 0 and P2 (x) > 0 near x = 0, which verifies required (13), (14) and (15) for such p(x) and q(x). Now, by Theorem 3 the main conclusion of this example follows. As a simple consequence of Theorem 3 we obtain the following two important results on the problem of exactly one extreme point of κy (x) of all nontrivial solutions of damped linear second-order differential equations. Corollary 1 Let J ⊆ I be an open interval and assume that the functions g, f ∈ C 3 (I ) and the primitive function G(x) of g(x) satisfy: 2 g(x) > 0, G(x) < 0 and f (x)eG(x) > 5π/|J | in J, (21) 2 (22) gf + f (x) < 0 and g − g + f (x) > 0 in J, (23) −g + 3g g − g 3 + 2g f − 2f (x) > 0 in J, 2g f − g 2 f + g f + f 2 − f (x) > 0 in J. (24) We consider the damped linear second-order differential equation: y + g(x)y + f (x)y = 0,
x ∈ I.
(25)
Then the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (25) has exactly one extreme point between each two its consecutive simple zeros in J . Corollary 2 Let J ⊆ (0, 1) be an open interval. Assume that the function f = f (x) and the real number μ satisfy: 2 f ∈ C 3 (I ), μ > 0 and x μ f (x) > 5π/|J | in J, (26)
Author's personal copy Geometry of solution near its extreme points
μf + xf (x) < 0 and f (x) > μ2 + μ x −2 in J, 2μ f − 2f (x) > μ(μ + 1)(μ + 2)x −3 in J, x μ(μ + 2) μ f2 − (x) > 0 in J. f + − f f x x2
(27) (28) (29)
We consider the damped linear second-order differential equation: y +
μ y + f (x)y = 0, x
x ∈ (0, 1).
(30)
Then the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (30) has exactly one extreme point between each two its consecutive simple zeros in J . With the help of Corollary 2 we are able to determine the exact number of local extreme points of κy (x) for all nontrivial solutions of Bessel type equation as follows. Example 4 The well known Bessel equation: t 2 z + t z + t 2 − ν 2 z = 0,
t > 1, ν ∈ R,
could be transformed by the transformation z(t) = y(1/t) into the equation: 1 1 ν2 y = 0, x ∈ (0, 1). y + y + − x x4 x2 We consider previous equation in a more general form: 1 λ ν2 y + y + y = 0, x ∈ (0, 1), − x xσ x2
(31)
(32)
where λ > 0, σ > 3 and ν ∈ R. It is not difficult to see that Bessel type (32) satisfy all required assumptions of Corollary 2 in J = (0, T0 ) for some T0 > 0. Consequently, the curvature κy (x) of every nontrivial solution y ∈ C 4 (I ) of (31) and (32) has exactly one extreme point between each two its consecutive simple zeros in J .
3 Proof of the main results In this section we give the proofs of the main results stated in previous sections. At the first, we present two elementary propositions in order to characterize the stationary and local extreme points of the curvature κy (x). Proposition 1 Let y ∈ C 3 (I ). A point ξ0 ∈ I is a stationary point of κy (x) if and only if y (ξ0 ) 1 + y 2 (ξ0 ) = 3y (ξ0 )y 2 (ξ0 ). (33) Furthermore, if y (ξ0 ) = 0, then y (ξ0 ) and y (ξ0 ) have the same sign.
Author's personal copy H. Chrayteh, M. Paši´c
Proof From the definition of κy (x) given in (1) we derive that κy (x) =
y (x)(1 + y 2 (x)) − 3y (x)y 2 (x) , (1 + y 2 (x))5
(34)
from which clearly follows the proof of this proposition.
Next, the local extreme points of the curvature κy (x) is characterized in the following way. Proposition 2 Let y ∈ C 4 (I ), ξ0 ∈ I and κy (ξ0 ) = 0. Then we have: (i) κy (ξ0 ) < 0 if and only if y I V (ξ0 )(1 + y 2 (ξ0 ))2 < 3y 3 (ξ0 )(1 + 5y 2 (ξ0 )); (ii) κy (ξ0 ) > 0 if and only if y I V (ξ0 )(1 + y 2 (ξ0 ))2 > 3y 3 (ξ0 )(1 + 5y 2 (ξ0 )). Proof Immediately from the formula (34) we derive: −7/2 I V 2 y (x) 1 + y 2 (x) κy (x) = 1 + y 2 (x) − 9y (x)y (x)y (x) 1 + y 2 (x) − 3y 3 (x) + 12y 2 (x)y 3 (x) . (35) With the help of previous equality, the condition κy (ξ0 ) < 0 is equivalent to the following one: 2 y I V (ξ0 ) 1 + y 2 (ξ0 ) − 9y (ξ0 )y (ξ0 )y (ξ0 ) 1 + y 2 (ξ0 ) − 3y 3 (ξ0 ) + 12y 2 (ξ0 )y 3 (ξ0 ) < 0. It together with Proposition 1 yields: 2 0 > y I V (ξ0 ) 1 + y 2 (ξ0 ) − 9y (ξ0 )y (ξ0 )y (ξ0 ) 1 + y 2 (ξ0 ) − 3y 3 (ξ0 ) + 12y 2 (ξ0 )y 3 (ξ0 ) 2 = y I V (ξ0 ) 1 + y 2 (ξ0 ) − 27y 2 (ξ0 )y 3 (ξ0 ) − 3y 3 (ξ0 ) + 12y 2 (ξ0 )y 3 (ξ0 ) 2 = y I V (ξ0 ) 1 + y 2 (ξ0 ) − 15y 2 (ξ0 )y 3 (ξ0 ) − 3y 3 (ξ0 ). It is equivalent to the condition y I V (ξ0 )(1 + y 2 (ξ0 ))2 < 3y 3 (ξ0 )(1 + 5y 2 (ξ0 )), which proves the first statement of this proposition. The second statement analogously follows. Now, we are able to give the proof of Lemma 1. Proof of Lemma 1 Let a0 and b0 be two consecutive inflexion points of y(x). By (1) we also conclude that a0 and b0 are two consecutive simple zeros of κy (x). It ensures the existence of at least one extreme point in (a0 , b0 ). Next, the fact that κy (x) has a unique local extreme point between a0 and b0 is equivalent to that κy (x) does not have any minimum in (a0 , b0 ) when κy (x) ≥ 0 and analogously it does
Author's personal copy Geometry of solution near its extreme points
not have any maximum in (a0 , b0 ) when κy (x) ≤ 0. However, it could be described by the fact that y (x) ≥ 0 in (a0 , b0 ) and κy (ξ0 ) < 0 for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0 and analogously y (x) ≤ 0 in (a0 , b0 ) and κy (ξ0 ) > 0 for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0. Now, both conclusions of this lemma immediately follows from Proposition 2. Proof of Lemma 3 It is clear that y (x) ≥ 0 (resp. y (x) ≤ 0) in (a0 , b0 ) implies y (ξ0 ) ≥ 0 (resp. y (ξ0 ) ≤ 0) which together with y I V (ξ0 ) < 0 (resp. y I V (ξ0 ) > 0) implies that the condition (4) (resp. (5)) is satisfied. Now by Lemma 1 we deduce that the curvature κy (x) has a unique local extreme point in (a0 , b0 ). Proof of Theorem 1 Let y(x) be a nontrivial solution of the equation y + q(x)y = 0, x ∈ I such that 2 q ∈ C 2 (I ), q 2 (x) − q (x) > 0 in J and q(x) > 5π/|J | , where J ⊆ I is such an interval which appears in the assumption (2). We want to show that the curvature κy (x) of y ∈ C 4 (I ) satisfies the problem of exactly one extreme point in J . For this, let a0 , b0 ∈ J , a0 < b0 , be two consecutive inflexion points of y(x). Hence we have: y (x) = −q(x)y(x),
x ∈ I,
(36)
and y (a0 ) = y (b0 ) = 0 and y (x) = 0, x ∈ (a0 , a0 + ε) ∪ (b0 − ε, b0 ),
(37)
for small enough ε > 0. Next, we need to show that κy (x) has a unique extreme point in (a0 , b0 ). To do so, we shall use Lemma 3. In this sense, let us suppose for instance that y (x) ≥ 0,
x ∈ (a0 , b0 )
(38)
(the case y (x) ≤ 0 in (a0 , b0 ) could be analogously considered). From (36), (37), (38) and (2) we obtain y(a0 ) = y(b0 ) = 0
and y(x) < 0,
x ∈ (a0 , a0 + ε) ∪ (b0 − ε, b0 ),
(39)
for small enough ε > 0. Furthermore, we claim that y (x) > 0,
x ∈ (a0 , b0 ).
(40)
Indeed, if (40) does not hold true, then there is an x0 ∈ (a0 , b0 ) such that y (x0 ) = 0 which together with (36), (39) and (2) gives y(a0 ) = y(x0 ) = y(b0 ) = 0 and y(x) < 0,
x ∈ (a0 , a0 + ε) ∪ (b0 − ε, b0 ). (41)
However, since y ∈ C 2 (I ) from (41) we conclude that y (x) has to change the sign in (a0 , b0 ) which contradicts (38). Thus, the assumption y (x0 ) = 0 is not possible and hence, the statement (40) is shown.
Author's personal copy H. Chrayteh, M. Paši´c
Next, by an elementary calculation from (36) we derive: y (x) = − q(x) y (x) − q (x)y(x),
x ∈ J,
(42)
>0
and
y I V (x) = −2q (x)y (x) + q 2 (x) − q (x) y(x),
x ∈ J.
(43)
>0
In order to use Lemma 3, we need to take an arbitrary ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0. Especially from (40) we have y (ξ0 ) > 0 and also y(ξ0 ) < 0 since (36) and (2). Let for a moment q (ξ0 ) ≥ 0. It together with y(ξ0 ) < 0 gives y (ξ0 ) ≥ 0. Indeed, if y (ξ0 ) < 0, then from (42) and (2) we obtain: y (ξ0 ) = −q(ξ0 )y (ξ0 ) − q (ξ0 )y(ξ0 ) > 0, which is not possible since y (ξ0 ) and y (ξ0 ) have the same sign, see Proposition 1. Thus, at this moment we have: q (ξ0 ) ≥ 0, y(ξ0 ) < 0 and y (ξ0 ) ≥ 0. Putting that into (43) we observe: (44) y I V (ξ0 ) = −2q (ξ0 )y (ξ0 ) + q 2 (ξ0 ) − q (ξ0 ) y(ξ0 ) < 0. Let now q (ξ0 ) ≤ 0. Since y(ξ0 ) < 0 we have y (ξ0 ) ≤ 0. Indeed, if y (ξ0 ) > 0, then from (42) and (2) we obtain: y (ξ0 ) = −q(ξ0 )y (ξ0 ) − q (ξ0 )y(ξ0 ) < 0, which is not possible since y (ξ0 ) and y (ξ0 ) have the same sign. Thus, in this case we have: q (ξ0 ) ≤ 0, y(ξ0 ) < 0 and y (ξ0 ) ≤ 0. Putting that into (43) we observe: (45) y I V (ξ0 ) = −2q (ξ0 )y (ξ0 ) + q 2 (ξ0 ) − q (ξ0 ) y(ξ0 ) < 0. Finally, according to (44) and (45), we observe that y I V (ξ0 ) < 0 for all ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0. Now Lemma 3 proves this theorem. Let us mention that the existence of at least two inflexion points of y(x) in J will be shown in the proof of Theorem 3 as a consequence of the inequality q(x) > (5π/|J |)2 stated in assumption (2). Next, we will pay attention to the proof of Theorem 3. At the first, we present some local properties of y(x) near the local extreme points of its curvature κy (x), where y(x) is a nontrivial solution of (16): (p(x)y ) + q(x)y = 0. Unlike (3), the main difficulty with the use of (16) pertains to the fact that the zeros of y(x) do not coincide with its inflexion points. Therefore, the proofs of Theorems 1 and 3 are different. Lemma 4 Let J ⊆ I be an open interval. Let p(x) and q(x) be two functions which satisfy the assumptions (11), (12) and (13). Let y ∈ C 3 (I ) be a nontrivial solution of
Author's personal copy Geometry of solution near its extreme points
(16) and let a0 , b0 ∈ J , a0 < b0 , be two consecutive inflexion points of y(x). For all stationary points ξ0 ∈ (a0 , b0 ) of the curvature κy (x) we have: (i) y (ξ0 ) = 0, (ii) y (ξ0 )y (ξ0 ) < 0, (iii) y (ξ0 )y(ξ0 ) ≥ 0. Proof Let y ∈ C 3 (I ) be a nontrivial solution of (16), that is, y(x) satisfies (p(x)y ) + q(x)y = 0. At the first, we claim: for all stationary points ξ0 ∈ (a0 , b0 ) of the curvature κy (x), we have if y(ξ0 ) > 0,
then y (ξ0 ) > 0 and if y(ξ0 ) < 0,
then y (ξ0 ) < 0.
(46)
Indeed, by (16) and the hypotheses (11), (12) and (13), we derive that p(x)y = −p (x)y − q(x)y and so, 1 pp − 2p 2 + pq (x) y (x) + pq − 2p q (x) y(x) . (47) y (x) = − 2 p (x)
>0
0
>0
It obviously proves (46) in the case of y (ξ0 ) = 0. The second case. Let y (ξ0 ) = 0. Then from Proposition 1 we obviously deduce: if y (ξ0 ) > 0,
then y (ξ0 ) > 0
and if y (ξ0 ) < 0,
then y (ξ0 ) < 0.
(48)
In order to prove (46) let y(ξ0 ) > 0. If we suppose the contrary to the first statement in (46), that is, if y (ξ0 ) ≤ 0, then with the help of (47) we derive y (ξ0 ) > 0. It together with (48) implies y (ξ0 ) > 0. But, it is a contradiction with y (ξ0 ) ≤ 0. Hence y(ξ0 ) > 0 implies y (ξ0 ) > 0. Analogously, one can show that y(ξ0 ) < 0 implies y (ξ0 ) < 0. Thus, both statements in (46) are proved. According to (46), we obviously have: if y (ξ0 ) ≤ 0,
then y(ξ0 ) ≤ 0 and if y (ξ0 ) ≥ 0,
then y(ξ0 ) ≥ 0.
(49)
At the second, we claim: for all stationary points ξ0 ∈ (a0 , b0 ) of the curvature κy (x), we have if y (ξ0 ) > 0,
then y (ξ0 ) < 0 and if y (ξ0 ) < 0,
then y (ξ0 ) > 0.
(50)
Indeed, let for instance y (ξ0 ) > 0. If we suppose the contrary to the first statement in (50), that is, if y (ξ0 ) ≥ 0, then (49) gives y(ξ0 ) ≥ 0. It together with (16) rewritten in the form py = −p y − qy gives y (ξ0 ) ≤ 0. But, it is a contradiction with
Author's personal copy H. Chrayteh, M. Paši´c
y (ξ0 ) > 0 and hence y (ξ0 ) > 0 implies y (ξ0 ) < 0. Analogously, one can show that y (ξ0 ) < 0 implies y (ξ0 ) > 0 and thus, both statements in (50) are proved. Now, with the help of (46) and (50), we are able to prove the first statement (i) of this lemma. In this sense, let us suppose the contrary to (i), that is, there exists a ξ0 ∈ (a0 , b0 ) such that κy (ξ0 ) = 0 and y (ξ0 ) = 0 which by (1) gives κy (ξ0 ) = 0 too. Since a0 and b0 are two consecutive inflexion points of y(x), we may assume y (x) ≥ 0 in (a0 , b0 ) (the opposite case y (x) ≤ 0 in (a0 , b0 ) could be analogously considered). Thus, we have for a moment that: ξ0 ∈ (a0 , b0 ),
y (ξ0 ) = 0 and y (x) ≥ 0 in (a0 , b0 ),
κy (a0 ) = κy (ξ0 ) = κy (b0 ) = 0 and κy (x) ≥ 0 in (a0 , b0 ). It yields that ξ0 is a minimum and zero point of the curvature κy (x), and consequently, there are at least two positive maximum points ξ01 and ξ02 of the curvature κy (x) such that a0 < ξ01 < ξ0 < ξ02 < b0 and κy (ξ0i ) > 0, i = 1, 2. Together with (1), it gives that κy (ξ0i ) = 0 and y (ξ0i ) > 0 for i = 1, 2. It together with (50) and in particular for ξ01 and ξ02 shows that y (ξ01 ) < 0 and y (ξ02 ) < 0.
(51)
On the other hand, from y (ξ0 ) = 0, py = −p y − qy, and (46), we easily observe that y(ξ0 ) = y (ξ0 ) = 0. Since y (x) ≥ 0 in (a0 , b0 ) and a0 , b0 are two consecutive inflexion points of y(x), it gives that ξ0 is a unique stationary-minimum point of the function y(x) in (a0 , b0 ). Hence, we deduce that y (x) ≤ 0 in (a0 , ξ0 ) and y (x) ≥ 0 in (ξ0 , b0 ). Therefore, we obtain y (ξ01 ) ≤ 0 and y (ξ02 ) ≥ 0 which is a contradiction with (51). Hence, the assumption y (ξ0 ) = 0 is not possible and thus, the conclusion (i) of this lemma is shown. Now, the conclusions (ii) and (iii) of this lemma easily follow from the conclusion (i) and the statements (49) and (50). Proof of Theorem 3 Let y ∈ C 4 (I ) be a nontrivial solution of (16): (p(x)y ) + q(x)y = 0, x ∈ I . Then we have: y I V (x) = P1 (x)y (x) + P2 (x)y(x),
(52)
where the functions P1 (x) and P2 (x) are given by: P1 (x) = P2 (x) =
1 p 3 (x)
2 −p p + 6pp p − 6p 3 + 4pp q − 2p 2 q (x),
1 3pp q − 6p 2 q + 3pp q + pq 2 − p 2 q (x). p 3 (x)
From the assumptions (14) and (15), we obviously have: P1 (x) > 0
and P2 (x) > 0 in J.
(53)
In order to prove Theorem 3, we will use Lemma 3. More precisely, we will show that if y (x) ≥ 0 (resp. ≤0) in (a0 , b0 ), then y I V (ξ0 ) < 0 (resp. >0) for all ξ0 ∈ (a0 , b0 )
Author's personal copy Geometry of solution near its extreme points
such that κy (ξ0 ) = 0. Indeed, let ξ0 ∈ (a0 , b0 ) be such that κy (ξ0 ) = 0 and let y (x) ≥ 0 in (a0 , b0 ). Hence by Lemma 4 we observe that y (ξ0 ) > 0 as well as: y (ξ0 ) < 0 and y(ξ0 ) ≤ 0.
(54)
Putting the statements (53) and (54) into (52), we obtain that y I V (ξ0 ) < 0. Now by Lemma 3 follows that the curvature κy (x) has a unique local extreme point in (a0 , b0 ). It is clear that in the case of y (x) ≤ 0 in (a0 , b0 ), this conclusion can be shown in the same way. Finally, we prove the existence of at least two inflexion points of y(x) in J . In this direction, let λ0 = (5π/|J |)2 . It follows that: 2 λ0 > 4π/|J | .
(55)
√ Next, let z(x) = sin( λ0 x − c), c ∈ R. Since z(x) is a solution of equation (P0 ): z + λ0 z = 0, the sequence zn of all consecutive zeros of z(x) satisfy: nπ + c zn = √ λ0
4π and zn+4 − zn = √ < |J |, λ0
(56)
where in the last inequality we use (55). Now, (56) ensures that there is k ∈ R such that {zk , zk+1 , zk+2 , zk+3 , zk+4 } ⊂ J . On the other hand, from (12) and (55) follows that 2 q(x) > 5π/|J | = λ0 in J, and therefore, by Sturm comparison theorem applied to equations (P0 ) and (16), it follows that there is xk+j ∈ (zk+j , zk+j +1 ) for j = 0, 1, 2, 3 such that y(xk+j ) = 0 for all j = 0, 1, 2, 3. Hence, y(x) has at least two inflexion points in (xk , xk+3 ) ⊂ J . Thus, all statements of Theorem 3 are proved.
4 Proof of Theorem 2 In this section, we study the problem of common stationary points of y(x) and κy (x). Regarding to Fig. 1, we see that y(x) = sin x and its curvature κy (x) have a common stationary point between each two consecutive simple zeros of κy (x) (inflexion points of y(x)). But this does not occur always. For this, we give sufficient conditions to prove the existence or non-existence of common stationary points of y(x) and κy (x) in a more general setting, as follows. Before we give the proof of Theorem 2, we present some simple sufficient conditions such that y(x) and κy (x) have or do not have common stationary points. Proposition 3 Let y ∈ C 3 (I ) and let a0 < b0 be two consecutive inflexion points of y(x). If y (x)y (x) ≤ 0 in (a0 , b0 ), then y(x) and κy (x) have a common stationary point in (a0 , b0 ).
Author's personal copy H. Chrayteh, M. Paši´c
Proof Since y (x)y (x) ≤ 0 in (a0 , b0 ) and also a0 and b0 are two consecutive inflexion points, there is an s0 ∈ (a0 , b0 ) such that y (s0 ) = 0. Hence the assumption y (x)y (x) ≤ 0 in (a0 , b0 ) gives the following two cases: either
y (x) ≤ 0
in (a0 , s0 )
y (x) ≥ 0 in (a0 , s0 ) or
y (x) ≥ 0
in (a0 , s0 )
y (x) ≤ 0 in (a0 , s0 )
and y (x) ≥ 0 in (s0 , b0 ), and y (x) ≤ 0 in (s0 , b0 ) and y (x) ≤ 0 in (s0 , b0 ), and y (x) ≥ 0 in (s0 , b0 ).
(57)
(58)
Putting (57) and (58) separately into the formula (34) we conclude: either κy (x) ≥ 0 in (a0 , s0 )
and κy (x) ≤ 0
κy (x) ≤ 0 in (a0 , s0 )
and κy (x) ≥ 0 in (s0 , b0 ).
in (s0 , b0 )
or
Hence, κy (s0 ) = 0. It proves this proposition.
Proposition 4 Let s0 ∈ R be a stationary point of y(x), that is y (s0 ) = 0. Then κy (s0 ) = 0 if and only if y (s0 ) = 0. Proof Putting x = s0 into (34) and since y (s0 ) = 0, we deduce that κy (s0 ) = y (s0 ). The proof is achieved. Now, we are able to complete this section with the following proof. Proof of Theorem 2 Let y(x) be a solution of y + q(x)y = 0, x ∈ I and q(x) > 0 in I . (i) We first suppose q (x) = 0 for all x ∈ I . Since q(x) > 0 in I , there is a constant λ > 0 such that q(x) = λ for all x ∈ I . Hence, y (x) = −λy(x) as well as y (x) = −λy (x). Therefore y (x)y (x) = −λy 2 (x) ≤ 0 in (a0 , b0 ). Now by Proposition 3 follows that y(x) and κy (x) have a common stationary point in (a0 , b0 ). It proves (i). (ii) Let now q (x) = 0 for all x ∈ I and let s0 ∈ (a0 , b0 ) be such that y (s0 ) = 0. Let us remark that because of y (x) = −q(x)y(x) and q(x) > 0 in I we have for all x ∈ (a0 , b0 ) that: y (x) > 0 when y(x) < 0, and y (x) < 0 when y(x) > 0. Therefore y(s0 ) = 0. Next, since y (x) = −q (x)y(x) − q(x)y (x), we also have y (s0 ) = −q (s0 )y(s0 ) − q(s0 )y (s0 ) = −q (s0 )y(s0 ) = 0. Now, by Proposition 4 follows that κy (s0 ) = 0 and therefore, y(x) and κy (x) do not have any common extremum point. Thus, both assertions of Theorem 2 are proved.
Author's personal copy Geometry of solution near its extreme points
5 The problem of three or at least three extreme points of κy (x) The functions y(x) = x 2m (m ≥ 2), y(x) = cosh(ωx) (0 < ω2 < 1/3) and y(x) = sin(sin x) give us the main motivation to consider the problems of exactly three and of at least three extreme points of κy (x). At the first, we start this section by giving the existence of a real function y(x) with κy (x) having exactly three extreme points between each two its consecutive simple zeros in R. It solves the Question 1. The method for proving this result describes a classic procedure for the reconstruction of y(x) from its curvature κy (x). For instance see [6]. Theorem 4 Let I = [T , ∞) for some T > 0. There are a function y ∈ C 3 (I ) and an increasing sequence an ∈ I of the consecutive inflexion points of y(x) such that an → ∞ when n → ∞, and κy (x) has exactly three local extreme points in (an , an+1 ), and y (x) = 0 in (an , an+1 ) for all n ∈ N. Before giving the proof of this theorem, we present the following lemma. Lemma 5 Let κ(x) be a real continuous function defined on an interval J ⊆ R. Let x0 ∈ J and let G(x) be a function given by G(x) =
x
κ(t) dt,
x ∈ J,
x0
such that |G(x)| < 1 for all x ∈ J . Let x1 ∈ J and let y(x) be a function defined by y(x) =
x x1
G(t) dt, 1 − G2 (t)
x ∈ J.
(59)
Then y ∈ C 2 (J ), κy (x) = κ(x) in J , and y (x) = 0
for all x ∈ J where κ(x) = 0.
(60)
Proof Let y(x) be given by (59). It is elementary to check that G(x) y (x) = 1 − G2 (x)
and y (x) =
κ(x) (1 − G2 (x))3
,
x ∈ J,
(61)
from which follows that y ∈ C 2 (J ) and κy (x) =
y (x) (1 + y 2 (x))3
√ =
κ(x) (1−G2 (x))3
(1 +
Finally, from (61) immediately follows (60).
G2 (x) 3 ) 1−G2 (x)
= κ(x),
x ∈ J.
Author's personal copy H. Chrayteh, M. Paši´c Fig. 4 The graph of periodic step function s(x) is shown by a dashed line and the graph of F3 (s(x)) by a thick line; κy (x) = π5 F3 (s(x))
Proof of Theorem 4 Let F3 (s(x)) be a Fourierov polynomial of third order of the step function s(x) given by: 1 if x ∈ (2kπ, (2k + 1)π), k ∈ Z, s(x) = −1 if x ∈ ((2k + 1)π, (2k + 2)π), k ∈ Z. Since s(x) is an odd 2π -periodic function, it is not difficult to see that π π 2 2 F3 s(x) = s(t) sin(t) dt sin x + s(t) sin(3t) dt sin 3x 2π −π 2π −π =
4 4 sin x + sin 3x, π 3π
see Fig. 4. Next, let κ(x) and G(x) be two functions defined by: 4 π 4 F3 s(x) = sin x + sin 3x, 5 5 15 x 4 4 G(x) = cos 3x. κ(t) dt = − cos x − 5 45 −π/2 κ(x) =
It is clear that |G(x)| < 40/45 < 1 for all x ∈ J = R. Now, let y(x) be defined by: x G(t) y(x) = dt, x ∈ J. −π 1 − G2 (t) Now by Lemma 5 we obtain κy (x) = κ(x) = π5 F3 (s(x)). Hence, the main conclusions of this theorem follows from the fact that such defined κ(x) has exactly three local extreme points between each two its consecutive simple zeros in R. Let us remark that each of inflexion points of y(x) makes at the same time a simple zero of the curvature κy (x) and conversely. Now we state and prove the following lemma, which could help us to continue our study of the problem of at least three extreme points of κy (x). Lemma 6 Let a0 , b0 ∈ I be two consecutive inflexion points of y ∈ C 3 (I ) and let s0 ∈ I be a unique stationary point of y(x) in (a0 , b0 ). Let y (x), y (x) and y (x)
Author's personal copy Geometry of solution near its extreme points
satisfy near x = s0 the following extra conditions: y (x) > 1, y (x)
(62)
√ 3 near x = s0 . 3 Then κy (x) has at least three local extreme points in (a0 , b0 ).
(63)
y (s0 ) = 0
and
lim
x→s0
y (x)
0 (resp. y (x) < 0) in (a0 , b0 ). By assumption y (s0 ) = y (s0 ) = 0 and (34), we obtain κy (s0 ) = 0. At first, we will show that s0 is a local minima (resp. maxima) for both two functions κy (x) and y(x) because it together with κy (a0 ) = κy (b0 ) = 0 and κy (x) > 0 (resp. κy (x) < 0) in (a0 , b0 ) obviously implies that κy (x) has at least three extreme points in (a0 , b0 ). However, in order to show that s0 is a local minima (resp. maxima) for both two functions κy (x) and y(x), it is enough to show that y (x) and κy (x) have the same sign near x = s0 . It is formally written as: y (x)κy (x) > 0 near x = s0 , x = s0 . Because of (34), the statement (64) can be rewritten in the form: y (x) y (x) 1 + y 2 (x) − 3y (x)y 2 (x) > 0 near x = s0 , x = s0 .
(64)
(65)
Since s0 is unique stationary point of y(x), we have that y (x) = 0 for all x ∈ (a0 , b0 ), x = s0 . Hence, (65) is equivalent to y (x) 1 + y 2 (x) − 3y 2 (x) > 0 near x = s0 , x = s0 . y (x)
(66)
Now, according to (62) we have: y (x) ≥ 1 near x = s0 , x = s0 . y (x)
(67)
Also, from y (s0 ) = 0 and (63) immediately follows 1 + y 2 (x) > 3y 2 (x)
near x = s0 .
(68)
With the help of (67) and (68) we observe: y (x) 1 + y 2 (x) ≥ 1 + y 2 (x) > 3y 2 (x) y (x)
near x = s0 ,
which proves (66). Hence, the desired statement (64) is proved and thus, κy (x) has at least three local extreme points in (a0 , b0 ). With the help of Lemma 6, we are able now to determine the exact number of local extreme points of the curvature κy (x) of y(x) = sin(sin x), see Fig. 2 in Sect. 1.
Author's personal copy H. Chrayteh, M. Paši´c
Example 5 Let y(x) = sin(sin x). We will show that κy (x) has at least three extreme points between each two consecutive inflexion points of y(x) in R. Since y(x) is a 2π -periodic function, it is enough to show previous statement on the intervals (0, π) and (π, 2π). At the first, by an elementary calculation we observe that: y (x) = cos x cos(sin x) and y (x) = − cos2 x sin(sin x) − sin x cos(sin x), y (x) = − cos3 x − cos x cos(sin x) + 3 sin x cos x sin(sin x). Next, since y (x) > 0 in (0, π/2) and y (x) < 0 in (π/2, π), it is clear that s0 = π/2 is a unique extreme point of y(x) in (0, π) as well as a0 = 0 and b0 = π are two consecutive inflexion points of y(x). Next, we have: y (x) 0 (− cos2 x − 1) cos(sin x) + 3 sin x sin(sin x) = = lim x→π/2 y (x) 0 x→π/2 cos(sin x) lim
=
− cos 1 + 3 sin 1 > 1, cos 1
√ because of sin 1 > cos 1 > (2/3) cos 1. Also, since cos 1 < 3/3, we obtain: √ 2 y (x) = cos x sin(sin x) + sin x cos(sin x) < 3 near x = π/2. 3 Thus, y(x) satisfies all required conditions of Lemma 6 and therefore, κy (x) of y(x) = sin(sin x) has at least three extreme points in (0, π). The same could be analogously shown in (π, 2π). In the same way as in Example 5, one can show that all statements of Lemma 6 are fulfilled for the function y(x) = sin(θ sin(ωx)), where ω > 0 and for some θ ∈ (0, π/2) such that: √ 3 ω2 + 1 and cos θ < . tg θ > 2 3θ ω 3θ ω2 Example 6 Let y(x) be a function defined by: y(x) =
1 1 1 1 cos(4x) − cos(2x) − = sin4 x − , 8 2 8 2
see Fig. 3 in Sect. 1. Let a0 = −π/2, b0 = π/2 and s0 = 0. It is not difficult to check that such defined a0 and b0 are two consecutive inflexion points of y(x) and also, s0 is a unique stationary point of y(x) in (a0 , b0 ). Moreover, y (0) = 0 and y (x) 8 sin(4x) − 4 sin(2x) 16 cos(2x) − 4 = lim 1 = ∞ > 1. = lim x→0 y (x) x→0 − sin(4x) + sin(2x) x→0 − cos(2x) + 1 2 lim
√ Next, since y (0) = 0 it is clear that |y (x)| < 3/3 near x = 0. Thus, for such choice of a0 , b0 and s0 , the function y(x) satisfies all required conditions of Lemma 6 and therefore, κy (x) has at least three local extreme points in (a0 , b0 ).
Author's personal copy Geometry of solution near its extreme points
Example 7 Consider y(x) = A cos(4x) + B cos(2x) + C, where A, B, C ∈ R. We leave the reader to find sufficient conditions on A and B and to give the corresponding constants a0 , b0 and s0 which enable us to apply Lemma 6. Remark 5 Let y(x) be a nontrivial solution of equation (P ): y + q(x)y = 0, x ∈ I , where q (x) = 0 in I . According to Theorem 2, we have shown that y(x) and κy (x) do not have common stationary points in the whole interval I . Hence, Lemma 6 could not be applied in this case. We complete this section with the main question of the paper. Open Question 4 Does exist a class of linear equation (P ): y + g(x)y + f (x)y = 0 such that the curvature κy (x) of every nontrivial solution y(x) of (P ) has exactly three extreme points between each two its consecutive simple zeros in J ? As yet, we only know that the curvature κy (x) of y(x) = cosh(ωx) (ω2 < 1/3) which is a solution of equation (Pω ): y − ω2 y = 0 has exactly three extreme points between each two its consecutive simple zeros in R. But at the same time, the curvature κy (x) of y(x) = sinh(ωx) (ω = 0) which is also a solution of equation (Pω ) has exactly one extreme point between each two its consecutive simple zeros in R. Therefore, the equation (Pω ) does not give the answer to previous question.
Appendix: Non-regular points (the cusps) of ε-parallels (the offset curves) In this section, we pay attention to an application of the main results of the paper to a study of exact number of all non-regular points of the ε-parallels of all nontrivial solutions y(x) of (16), where ε > 0. Definition 1 Let γ (t) be a C 1 -plane parametric curve and let N (t) be its normal, that is, let there be two C 1 functions X(t) and Y (t) such that γ (t) = (X(t), Y (t)) and (−Y (t), X (t)) N (t) = . X 2 (t) + Y 2 (t) For any ε > 0 let Γlow,ε,γ (t) and Γup,ε,γ (t) denote the lower and upper ε-parallels of γ (t) respectively, which are defined as two plane curves by: Γlow,ε,γ (t) = γ (t) − εN (t)
and Γup,ε,γ (t) = γ (t) + εN (t).
Let us remark that in the case of γ (t) is a C 1 -parametrization of the graph Γ (y) = {(x, y(x)) : x ∈ I¯} ⊆ R2 of a differentiable function y(x), the lower and upper ε-parallels of the graph Γ (y), often said the offset curves of Γ (y), are denoted respectively by Γlow,ε,y (t) and Γup,ε,y (t), see Fig. 5. Definition 2 Let γ (t) = (X(t), Y (t)) be a C 1 -parametric plane curve. A point dY (X(t∗ ), Y (t∗ )) is said to be a non-regular point of γ (t) if dX dt (t∗ ) = dt (t∗ ) = 0.
Author's personal copy H. Chrayteh, M. Paši´c Fig. 5 The graph of y(x) is shown by a thick line, and the ε-parallels Γlow,ε,y (t) and Γup,ε,y (t) are shown by corresponding lower and upper thin lines
Fig. 6 There exist one or two non-regular points of the ε-parallels between any two consecutive inflexion points a0 < b0 of y(x), because the curvature κy (x) has a unique local extreme point in (a0 , b0 ) and hence κy (x) is one time or twice intersected by the horizontal line y = 1/ε (or y = −1/ε respectively)
Fig. 7 There exist two or four non-regular points of the upper ε-parallel Γup,ε,y (t) between any two consecutive inflexion points a0 < b0 of y(x), because the curvature κy (x) has exactly three local extreme points in (a0 , b0 ) and hence κy (x) is two or four times intersected by the horizontal line y = 1/ε
Let a0 and b0 be two consecutive inflexion points of y(x), a0 < b0 . By regarding to Figs. 6 and 7 below, we can observe the following geometric facts: • a non-regular point of an ε-parallel is a point in which the corresponding ε-parallel has a cusp; also, it is a point of intersection between the curvature κy (x) and horizontal line y = ±1/ε; • if the curvature κy (x) is intersected by horizontal line y = 1/ε (resp. y = −1/ε) and κy (x) ≥ 0 (resp. ≤0) in [a0 , b0 ], then the number of non-regular points of the ε-parallel Γup,ε,y (t) (resp. Γlow,ε,y (t)) depends on the number of the local extreme points of the curvature κy (x) in the following way: if κy (x) has a unique local extreme point in (a0 , b0 ), then Γup,ε,y (t) (resp. Γlow,ε,y (t)) has either one or two non-regular points, see Fig. 6 below and if κy (x) has exactly three local extreme points in (a0 , b0 ), then Γup,ε,y (t) (resp. Γlow,ε,y (t)) has either two or four nonregular points, see Fig. 7.
Author's personal copy Geometry of solution near its extreme points
Now, according to Theorems 1 and 3 and previous geometric observation, we are able to state the second main result of the paper. Theorem 5 Let y(x) be a nontrivial solution of (16) (resp. (3)), where the coefficient p(x) and q(x) satisfy required assumptions (11)–(15) (resp. (2)). Let a0 and b0 be two consecutive inflexion points of y(x) in J such that a0 < b0 . (i) If {x ∈ (a0 , b0 ): κy (x) ≥ 1/ε} = ∅ for some ε > 0, then Γup,ε,y (t) has either one or two non-regular points in (a0 , b0 ). (ii) If {x ∈ (a0 , b0 ): κy (x) ≥ 1/ε} = ∅ for some ε > 0, then Γup,ε,y (t) has no any non-regular point in (a0 , b0 ). (iii) If {x ∈ (a0 , b0 ): κy (x) ≤ −1/ε} = ∅ for some ε > 0, then Γlow,ε,y (t) has either one or two non-regular points in (a0 , b0 ). (iv) If {x ∈ (a0 , b0 ): κy (x) ≤ −1/ε} = ∅ for some ε > 0, then Γlow,ε,y (t) has no any non-regular point in (a0 , b0 ). The proof of Theorem 5 will be sketched at the end of this section. Remark 6 It is clear that if κy (x) is a bounded function on an interval J , then we can choose small enough ε0 > 0 such that |κy (x)| < 1/ε for all x ∈ J and ε ∈ (0, ε0 ). Indeed, if there is an M > 0 such that |κy (x)| < M for all x ∈ J , then for ε0 = 1/M we have: |κy (x)| < M = 1/ε0 < 1/ε for all ε ∈ (0, ε0 ) and x ∈ J . In this sense, the cases (i) and (ii) of Theorem 5 do make a sense exactly if κy (x) is not a bounded function on an interval J . According to previous remark, we give a class of linear differential equations y + q(x)y = 0, x ∈ J , such that for all its solutions y(x) the curvature κy (x) is not bounded in J . Theorem 6 Let y(x) be a nontrivial solution of (3), where I = (0, T ]. Assume that the coefficient q(x) satisfies: q(x) > 0 in I , q(0+) = ∞, q(x) ∼ x −σ near x = 0 and σ > 2, and the following Hartman-Wintner type condition: 1 1 q − 4 q − 4 ∈ L1 (I ).
(69)
Then κy (x) is an unbounded function on I . Moreover, there is a decreasing sequence sn ∈ I of stationary points of y(x) such that sn 0, |κy (sn )| → ∞ when n → ∞, and κy (sn ) ∼ n 2σ3σ−4 as n → ∞. (70) The proof of Theorem 6 will be given at the end of this section. Let us remark that in [7] author gives a new approach for demonstrating the global stability of ordinary differential equations by means of the boundedness of the curvature κy (x) of solutions y(x) on some set. Next, in [9] author derives the fractal dimension of the graph Γ (y) of all nontrivial solutions y(x) of a nonlinear p-Laplace equation, by calculating the Lebesgue measure of the ε-neighborhood of Γε (y) of graph Γ (y), where Γε (y) is bounded with the ε-parallels Γlow,ε,y (t) and Γup,ε,y (t).
Author's personal copy H. Chrayteh, M. Paši´c
In the sequel, we give a short observation on the non-regular points of ε-parallels Γlow,ε,y (t) and Γup,ε,y (t) associated to the graph Γ (y) of a function y(x). We will show that Γlow,ε,y (t) and Γup,ε,y (t) allow two kinds of singularities, which are influenced by the inequality κy (x) ≥ 1/ε and two kinds of the profiles of the curvature κy (x) studied in the previous section. That is, Theorem 5 will be proved. Also, at the end of this section, we establish an observation from Remark 6 presented in the introduction of the paper. For many real functions y(x) and values of the positive parameter ε, its ε-parallels Γlow,ε,y (t) and Γup,ε,y (t) are similar to the graph Γ (y) as in Fig. 3 above. It is because in such a case, Γlow,ε,y (t) and Γup,ε,y (t) do not possess non-regular points which have been defined in Sect. 1 of the paper. Next, we recall a basic fact from the differential geometry which says that the non-regular points of ε-parallels of the function y(x) are determined by the equality |κy (x)| = 1/ε. Proposition 5 Let y ∈ C 1 (I ) and let graph Γ (y) be parametrized by (t, y(t)). Let Γlow,ε,y (t) = (Xlow,ε (t), Ylow,ε (t)) and Γup,ε,y (t) = (Xup,ε (t), Yup,ε (t)). Then we have: ⎧ y (t) ⎪ ⎪ , ⎨Xlow,ε (t) = t + ε 1 + y 2 (t) (71) 1 ⎪ ⎪ , ⎩Ylow,ε (t) = y(t) − ε 1 + y 2 (t) and
⎧ y (t) ⎪ ⎪ , ⎨Xup,ε (t) = t − ε 1 + y 2 (t) ⎪ ⎪Yup,ε (t) = y(t) + ε 1 . ⎩ 1 + y 2 (t)
(72)
Proof Directly from Definition 1 we obtain: (−y (t), 1) Γlow,ε,y (t) = γ (t) − εN (t) = t, y(t) − ε 1 + y 2 (t) y (t) 1 = t +ε , y(t) − ε , 1 + y 2 (t) 1 + y 2 (t) which proves this proposition.
Proposition 6 Let y ∈ C 1 (I ) and let graph Γ (y) be parametrized by (t, y(t)). Let Γlow,ε,y (t) = (Xlow,ε (t), Ylow,ε (t)) and Γup,ε,y (t) = (Xup,ε (t), Yup,ε (t)). Then we have: ⎧ dXlow,ε ⎪ ⎪ = 1 + εκy (t), ⎨ dt (73) ⎪ dY ⎪ ⎩ low,ε = y (t)(1 + εκy (t)), dt
Author's personal copy Geometry of solution near its extreme points
and
⎧ dXup,ε ⎪ ⎪ = 1 − εκy (t), ⎨ dt ⎪ dY ⎪ ⎩ up,ε = y (t)(1 − εκy (t)). dt
(74)
Proof Taking the derivative in (71) and (72), we get (73) and (74).
Proof of Theorem 5 We leave this proof to the reader by suggesting to combine Theorem 3 and Proposition 6. Proof of Theorem 6 In the famous Hartman’s book [5], one can find the asymptotic formula for the behavior near x = ∞ of all solution of the linear second order differential equations. With the help of it, the corresponding asymptotic formula near x = 0 of solutions of equation (p(x)y ) + q(x)y = 0 can be found for instance in [10] provided q(x) satisfies q(x) > 0, q(0+) = ∞ and the Hartman-Winter condition (69). Moreover, when q(x) ∼ x −σ near x = 0 and σ > 2, it implies that: if y(x) is a solution of (3), then there is a decreasing sequence sn ∈ I of stationary points of y(x) such that sn 0,
2
sn ∼ n− σ −2
σ/4 and y(sn ) ∼ sn
as n → ∞.
(75)
Now, from (1), (3) and (75) we obtain: κy (sn ) = |y (sn )| = y (sn ) (1 + y 2 (sn ))3 3σ σ/4 −3σ/4 = q(sn )y(sn ) ∼ sn−σ sn ∼ sn ∼ n 2σ −4 ,
which proves this theorem.
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Author's personal copy H. Chrayteh, M. Paši´c 9. Paši´c, M.: Minkowski–Bouligand dimension of solutions of the one-dimensional p-Laplacian. J. Differ. Equ. 190, 268–305 (2003) 10. Paši´c, M., Tanaka, S.: Rectifiable oscillations of self-adjoint and damped linear differential equations of second-order. J. Math. Anal. Appl. 381, 27–42 (2011) 11. Pham, B.: Offset curves and surfaces: a brief survey. Comput. Aided Des. 24(4), 223–229 (1992) 12. Rynne, B.P.: Spectral properties and nodal solutions for second-order, m-point, boundary value problems. Nonlinear Anal. 67, 3318–3327 (2007) 13. Sethian, J.A.: Curvature and the evolution of fronts. Commun. Math. Phys. 101, 487–499 (1985) 14. Taylor, J.E., Cahn, J.W., Handwerker, C.A.: Geometric models of crystal growth. Acta Metall. Mater. 40, 1443–1474 (1992)
