On the separation of solutions to fractional differential equations of order α ∈ (1 , 2)

Given the Caputo-type fractional differential equation D α y ( t ) = f ( t, y ( t )) with α ∈ (1 , 2), we consider two distinct solutions y 1 , y 2 ∈ C [0 , T ] to this equation subject to different sets of initial conditions. In this framework, we discuss non-trivial upper and lower bounds for the difference | y 1 ( t ) − y 2 ( t ) | for t ∈ [0 , T ]. The main emphasis is on describing how such bounds are related to the differences of the associated initial values


Introduction
In a recent paper [1], one of the authors contributed to the development of a theory addressing the following question: Given some α ∈ (0, 1] and a fractional initial value problem D α y(t) = f (t, y(t)), y(0) = y 0 , with differential operators of Caputo type and starting point 0 and order α, by how much does the solution change if the initial value y 0 is replaced by, say, y 0 + δ with some δ ̸ = 0?
Upper bounds for the changes in the solution can be easily derived using a Gronwall argument [2], but in many practically relevant cases these can be found to be excessively large [1, p. 170].Therefore it was the primary goal of [1] to provide much more rigorous upper bounds and to complement these with correspondingly strict lower bounds.In this work here, we now extend this theory to the case of differential equations of order α ∈ (1, 2).Thus, we will concentrate on the initial value problems where α ∈ (1, 2), i.e. we look at two functions y 1 and y 2 that solve the same fractional differential equation but satisfy different initial conditions, and derive strict upper and lower bounds for the difference y 1 (t) − y 2 (t) for t ∈ [0, T ] with some T > 0. Throughout our work, D α will denote the Caputo differential operator of order α with starting point 0 [3,Definition 3.2].
Throughout this paper, we use the standard notation to denote the two-parameter Mittag-Leffler function with parameters α > 0 and β > 0 [4].With this convention, we can state our first result.
Theorem 1. Assume that α ∈ (1, 2) and that the function f : [0, T ] × R → R is continuous and satisfies a Lipschitz condition with respect to its second argument whose Lipschitz constant is denoted by L.Then, unique solutions y 1 and y 2 to the two initial value problems given in (1) exist on [0, T ].Moreover, for all t ∈ [0, T ] we have that |y 1 (t) − y 2 (t)| ≤ (|y 1,0 − y 2,0 | + t|y 1,1 − y 2,1 |) E α,1 (Lt α ). ( Proof.The existence and uniqueness of the solutions y 1 and y 2 follows from classical standard results, see, e.g., [5,Theorem 6.4].A careful inspection of the proof of the corresponding Gronwall inequality in this case [3,Theorem 6.20] (see also [2]) then yields the inequality (2).□ The estimate (2) has two fundamental weaknesses: First, it does not provide a nontrivial lower bound for |y 1 (t) − y 2 (t)| at all.And second, by definition, the Lipschitz constant L is always positive (except for the trivial case when f (t, y) depends only on t).This implies that, when the interval [0, T ] is large, the Mittag-Leffler function needs to be evaluated for a large positive argument, and thus in view of its known asymptotic behaviour for t → ∞ (see [4,Theorem 4.3]), the bound on the right-hand side of (2) becomes very large.In Section 2, we will derive refined estimates that address both these weaknesses.
For the case α ∈ (0, 1), it has been shown in [6] that results of the type to be developed below are useful in designing and analyzing numerical methods for solving so-called terminal value problems for fractional differential equations of order α.In the same spirit, we believe that our results described here can be exploited to construct numerical methods for fractional boundary value problems of the form with α ∈ (1, 2) and some T > 0. We intend to elaborate on this topic in a forthcoming paper [7].
It will become evident below that, when transferring the results of [1] from the case α ∈ (0, 1] treated there to α ∈ (1, 2) that we discuss here, some steps of the proofs can be carried over directly, but in some other respects, significant changes are necessary.Specifically, the following issues require major modifications: • In the case α ∈ (0, 1], the initial value problem needs one initial condition to assert the well-posedness; in our case α ∈ (1, 2) here, two such conditions are required.Thus the methods need to be refined in order to allow incorporating the second initial condition.
• When looking at different initial conditions, the case α ∈ (0, 1] necessarily implies that we have two solutions that do not coincide at the initial point.This leads to a substantial simplification of the proof that we do not always have if α ∈ (1, 2) because in this case we may have two solutions with different first derivatives but identical function values at the initial point.
• An inhomogeneous linear fractional differential equation of order α ∈ (0, 1] with constant coefficients has a solution given (via the variationof-constants formula) in terms of the Mittag-Leffler functions E α,1 and E α,α .For α ∈ (0, 1), both these functions are strictly positive on the entire real line, see [11].This property is essential in the proofs of [1].In the case α ∈ (1, 2), the solutions comprise Mittag-Leffler functions E α,1 , E α,2 and E α,α , and for the values of α that are applicable now, some or all (depending on the precise value of α) of these function do have changes of sign on R.This fact demands the proof techniques to be adapted to a significant extent.

The separation of the solutions
Much as in [1], we will break up the analysis into multiple steps.Specifically, we will begin by showing some auxiliary results needed to prove the main theorems.Next, we will discuss the linear case and then finally investigate the general problem involving nonlinear differential equations.In each case, we will keep one of the two initial values (function value and first derivative of the solution to the initial value problem) fixed and estimate the effects of changes of the other one, and afterwards interchange the roles of the two initial values.The overall result then follows by combining the partial results in an appropriate way.

Auxiliary results
Our first result is a comparison statement for solutions to weakly singular Volterra equations.Unlike the other results of our paper, it is not restricted to the case α ∈ (1, 2) but it holds for general α > 0.
Lemma 2. Let J = [0, T ] with some T > 0 or J = [0, ∞).Moreover, let α > 0 and assume that the continuous functions v, v 1 , w, w 1 : J → R and g : respectively, for all t ∈ J. Suppose further that g(t, x) is nondecreasing in x for each fixed t ∈ J and that v 1 (t) < w 1 (t) for all t ∈ J. Then v(t) < w(t) for all t ∈ J.
Proof.This has been claimed and proven for the case α ∈ (0, 1) by Cong and Tuan [8,Lemma 3.4] (see also [9,Theorem 2.1] for a slightly modified version).A close inspection of the proof given in [8] reveals that all arguments used there are valid for α ≥ 1 too.□ Next, we generalize the so-called variation-of-constants formula developed for the case 0 < α ≤ 1 in [8, Lemma 3.1] to the case 1 < α < 2 that is of interest here.
Theorem 3. Let α ∈ (1, 2), y 1,0 , y 1,1 , M ∈ R, and T > 0. Assume that the function g : [0, T ] × R → R is continuous and satisfies a Lipschitz condition with respect to the second variable.Moreover, let the function y 1 : [0, T ] → R be the solution to the initial value problem Then, for all t ∈ [0, T ], we have Proof.The proof is completely analog to the proof of the result for the case α ∈ (0, 1] given in [8,Lemma 3.1].□ Remark 1.In fact, it is easy to see that the result of Theorem 3 can be generalized even more without essential changes in the proof.Specifically, we may admit arbitrary α > 0. Although we do not need this generalization in the context of the present work, we mention it here for the sake of completeness: Let α > 0, m = ⌈α⌉, y 1,ℓ , M ∈ R (ℓ = 0, 1, . . ., m − 1), and T > 0. Assume that the function g : [0, T ] × R → R is continuous and satisfies a Lipschitz condition with respect to the second variable.Moreover, let the function y 1 : [0, T ] → R be the solution to the initial value problem Then, for all t ∈ [0, T ], we have Based on Lemma 2 and Theorem 3, we can now prove a first separation theorem for solutions to the differential equation from (1).This result generalizes the findings of [8,Theorem 3.5] to the case α ∈ (1, 2] on sufficiently small intervals.We shall see in Example 1 below that this restriction is necessary in the sense that-in contrast to the situation encountered for 0 < α ≤ 1-one can construct counterexamples if the intervals are too large. Then, we have y 1 (t) < y 2 (t) for any t ∈ [0, T * ].
Remark 2. Note that Theorem 4 is a qualitative result: Under the given conditions, it asserts that the graphs of two solutions to the same differential equation remain separated from each other and do not touch or intersect.It does, however, not quantify the distance between the solutions.We will develop such quantitative results in Subsections 2.2 and 2.3 below.

Remark 3. (a)
In condition (b) of Theorem 4 we demand, among others, that (5a) holds for all t ∈ [0, T * ].By looking concretely at the point t = 0 and taking into consideration that E α,1 (0) = 1, this in particular means that we must have y 1,0 < y 2,0 .For certain applications, in particular the one discussed in [7], this is a little bit too restrictive.Specifically, in that case we only have y 2,0 = y 1,0 and y 2,1 > y 1,1 which means that (5a) holds only for t ∈ (0, T * ] but not for t = 0. To handle this case, we need a slightly different result that, in turn, requires a small modification of the proof, see Theorem 5 below. (b) A sufficient condition for (5a) to hold is that (c) Since E α,1 , E α,α and E α,2 are continuous functions with positive function values at the origin, it is clearly possible to find a value T * > 0 which satisfies the assumptions (5) of Theorem 4 if y 1,0 < y 2,0 .Similarly, we can also see that it is possible to find some T * > 0 that satisfies (5b), (6b) and (6c).We shall discuss this matter in more detail in Section 3.
Proof of Theorem 4. Assume that the statement is not true.Since y 1 and y 2 are continuous functions and y 1 (0) = y 1,0 < y 2,0 = y 2 (0), there exists some t 1 ∈ (0, T * ] such that y 1 (t 1 ) = y 2 (t 1 ) and that y 1 (t) < y 2 (t) for all t ∈ [0, t 1 ).Then, define g(t, y) := f (t, y) + Ly for all t ∈ [0, T ] and y ∈ R. By Theorem 3, we obtain for all these values of t that and Moreover, for all these t and all z, z ∈ R with z > z, we observe in view of the Lipschitz property of f and therefore, by (5b), for these values of t, z and z and for τ ∈ [0, t].In other words, the function •) is nondecreasing for any fixed t ∈ [0, T ].This observation allows us to invoke Lemma 2 which, in combination with (5a) and ( 7), tells us that y 1 (t) < y 2 (t) for all t ∈ [0, t 1 ] which contradicts our basic observation that y 1 (t 1 ) = y 2 (t 1 ).□ The following example demonstrates that the limitation on the magnitude of the quantity T * given in eq. ( 5) is indeed necessary.
Example 1.Consider the differential equation D 1.5 y(t) = −y(t) which clearly satisfies the assumptions of Theorem 4. Evidently, the solution to this equation subject to the initial condition y 1 (0) = y ′ 1 (0) = 0 is y 1 (t) = 0. Also, it is well known [3, Theorem 7.2] that the solution with initial conditions y 2 (0) = 1 and y ′ 2 (0) = 0 is y 2 (t) = E 1.5,1 (−t 1.5 ).Evidently, these two solutions remain separated from each other on any interval [0, T * ] where T * is smaller than the first positive zero of y 2 but not when T * is larger than this zero (which is located at approximately 1.645, see Fig. 1), so even though both solutions exist on [0, ∞), their graphs remain separated from each other only on a subinterval that extends from the initial point to a certain finite value.We conclude this subsection with the modification of Theorem 4 that we had already announced in Remark 3(a).Then, we have y 1 (t) < y 2 (t) for any t ∈ (0, T * ]. As in Remark 3(c), we also emphasize here that a value T * with the properties required in condition (c) exists.
Proof.By [3,Theorem 6.25], we know that y 1 , y 2 ∈ C 1 [0, T ].Our assumption (b) about the initial values asserts that y ′ 1 (0) < y ′ 2 (0), and hence, in view of the continuity of y ′ 1 and y ′ 2 , there exists some δ > 0 such that y ′ 1 (t) < y ′ 2 (t) for all t ∈ [0, δ).Combining this with the information that y 1 (0) = y 2 (0) (which also follows from our assumption about the initial values), we see that Now assume that the statement of the theorem is false.Then, in view of ( 9), there exists some t 1 ≥ δ > 0 such that y 1 (t 1 ) = y 2 (t 1 ) and y 1 (t) < y 2 (t) for t ∈ (0, t 1 ). ( Defining the function g as in the proof of Theorem 4, we notice againas in (8)-that this function is strictly increasing with respect to its second variable.Exploiting this monotonicity, Theorem 3 and conditions (b) and (c) of our Theorem, we thus derive which contradicts (10).Therefore, the hypothesis that our claim be false cannot hold.□

Main results for the linear case
We are now in a position to derive the desired quantitative statements about the separation of different solutions to the same fractional differential equation for the linear case, thus extending the theory developed for α ∈ (0, 1] in [1, Theorems 4 and 5] to the case α ∈ (1, 2).Theorem 6. Assume the hypotheses of Theorem 4 or Theorem 5 and choose T * in the corresponding way.Under the condition that f (t, y) = a(t)y with some continuous function a : Proof.For t ∈ [0, T * ], we define u(t) := y 2 (t) − y 1 (t).We then see that u is the unique solution to the initial value problem on [0, T * ], and Theorem 4 or 5 asserts that u(t) > 0 for all t ∈ (0, T * ].Now we pick an arbitrary value T ∈ (0, T * ].Since [0, T ] ⊆ [0, T * ], we can see that ( 12) is satisfied on [0, T ].For t in this interval, we then rewrite the problem (12) as By Theorem 3, it follows that The definition of the function a * implies that a(τ In combination with the positivity of u in the relevant interval that we had observed at the beginning of the proof and the choice of T * which implies that the Mittag-Leffler function inside the integral on the right-hand side of (13) has a positive function value, we conclude that this integral is nonnegative for all t ∈ [0, T ] and hence, in particular, for t = T .Therefore, we obtain Since T was an arbitrary element of (0, T * ], the upper bound for |y 2 (t) − y 1 (t)| claimed in eq. ( 11) follows from the given assumption (5a) upon the change of variable T → t.
To prove the lower bound, we also start with the problem (12) and again pick some arbitrary T ∈ (0, T * ].Then we consider the initial value problem According to [3, Theorem 7.2], the unique solution to this problem is Then, for t ∈ [0, T ], we set h(t) := u(t) − v(t) and see that the function h is the unique solution of the initial value problem and so Theorem 3 implies that From this relation, we can see (using the same arguments as above) that h(t) ≥ 0 for all t ∈ [0, T * ], and so u(t) ≥ v(t) for all t.Hence, by definition of u(t) and ( 14), the lower bound for |y 2 (t) − y 1 (t)| stated in (11)  Remark 5.The observations of Remark 3 apply to Theorem 6 as well.

Main results for the nonlinear case
Based on Theorem 6, we can now establish similar results for the general nonlinear case.In this situation, we temporarily impose a restriction on the given initial value problem (1), namely we assume that the function f on the right-hand side of the differential equation has the property f (t, 0) = 0 for all t ∈ [0, T ].Our first results in this subsection will require this limitation; later on, we will remove it.
In complete analogy with the notions introduced in Theorem 6, we start our considerations in the nonlinear case (under the restriction mentioned above) by defining Clearly, in the linear case f (t, y) = a(t)y discussed in Subsection 2.2, these definitions coincide with those introduced in the formulation of Theorem 6.Moreover, in the same way as in [1, Proof of Lemma 2], we can see that these quantities are well defined and that the functions a * and a * are continuous on [0, T ] if the function f satisfies a Lipschitz condition.
Our first result then exploits the fact that, under the assumption f (t, 0) = 0 for all t, the function given by y 1 (t) = 0 for all t is a solution to the differential equation given in (1).
Proof.If y 2,0 = y 2,1 = 0 then all statements are trivial, so we only need to explicitly consider the case that at least one of these values is nonzero.Furthermore, the statements are also trivial for t = 0. Thus we only have to show the inequalities for t ∈ (0, T * ].
For the proof of (a), we note that, as indicated above, our assumptions imply that the function y 1 (t) = 0 is the unique solution to the given differential equation subject to the initial conditions y 1 (0) = y ′ 1 (0) = 0. Thus, y 2 (t) > y 1 (t) = 0 for all t ∈ (0, T * ] by Theorem 4 if y 2,0 > 0 and by Theorem 5 if y 2,0 = 0. Now, as in the proof of Theorem 6, choose an arbitrary T ∈ (0, T * ].Then, we can see that, for t ∈ [0, T ], the function y 2 satisfies the differential equation and by definition of a * ( T ), the expression in square brackets on the right-hand side of the definition of g * (t, y 2 ) is nonnegative.Thus, by Theorem 3, and in view of our considerations above-in particular also taking into account that y 2 (t) > 0 for all t ∈ (0, T * ]-we see that the integral in this equation is nonnegative.Thus, for t = T we find for arbitrary T ∈ (0, T * ] (and also, trivially, for T = 0).This is equivalent to the lower bound in the claim of part (a).
To show the upper bound in part (a), we note that y 2 also satisfies the differential equation and by definition of a * ( T ), the expression in square brackets on the right-hand side of the definition of g * (t, y 2 ) is nonpositive.Thus, here we have by Theorem 3 that where now the integral is nonpositive, so by an argument as above we also establish the upper bound.Statement (b) can be shown in a completely analog manner.□ As announced above, we now finish our quantitative discussion of upper and lower bounds for the difference between two distinct solutions to the same fractional differential equation of order α ∈ (1, 2) by a look at the general nonlinear case without assuming the hypothesis that f (t, 0) = 0 for all t.Note that this final theorem of this section is the only theorem that requires knowledge of one specific solution to the differential equation in question.All other theorems proved so far in this paper were exclusively formulated in terms of the given function f on the right-hand side of the differential equation and the given initial values.
Theorem 8. Consider the initial value problems (1) for t ∈ [0, T ] with 1 < α < 2 where f is assumed to be continuous and to satisfy a Lipschitz condition in the second variable with Lipschitz constant L. Assume that y 1,0 ≤ y 2,0 and y 1,1 ≤ y 2,1 .Moreover, let T * ∈ (0, T ] satisfy the conditions (5b), (6b) and (6c).Then, defining we have we see the following facts: (a) The function u := y 2 − y 1 satisfies the differential equation and the initial conditions u(0) = y 2,0 − y 1,0 and u ′ (0) = y 2,1 − y 1,1 .(b) f (t, 0) = 0 for all t.(c) f satisfies a Lipschitz condition with the same Lipschitz constant L as f .Therefore, we may apply Theorem 7(a) to the solution u of the differential equation ( 16) and derive the desired result.□

Zeros of Mittag-Leffler functions
The conditions (5b), (6b) and (6c) play a significant role in deciding when the results of Section 2 can be applied.To obtain a thorough understanding of the meaning of these conditions, it is essential to have some knowledge of the locations of the smallest positive zeros of the three functions While a large amount of information is available on the zeros with large modulus of these functions (see, e.g., [4, Section 4.6]), we have unfortunately been unable to find any concrete statements about the zeros with small modulus.Therefore, to allow the reader to at least obtain a reasonable impression of the situation, we now present some numerical results.To this end, we define Throughout this section, whenever we had to evaluate some Mittag-Leffler function, we have used Garrappa's MATLAB implementation of his algorithm described in [10].We hope that our paper will initiate further analytical investigations on the question for the exact values of Z β (α).

Small positive zeros of
We begin our discussion of the quantity Z β (α) defined in (17) with the case β = α.From [11,Section 7] we obtain that the number of zeros of this function is positive and finite.A limit consideration using the power series representation of the Mittag-Leffler function yields From (18b) we deduce that the smallest positive z for which E 2,2 (−z 2 ) = 0 is z = π, and so we have   Specifically, we find from our numerical results that min α∈(1,2] Z α (α) ≈ 2.9378538 and that this minimum is attained for α ≈ 1.586.In particular, Z α (α) does not depend on α in a monotonic way.
Also, we can analyze the asymptotic behaviour of Z α (α) for α → 1+ and α → 2−, respectively, based on the numerical results.The observations obtained in this way give rise to the following conjecture.To support items (b) and (c) of the conjecture, the suspected asymptotics are plotted as the dotted lines in the two parts of Figure 3.Note that coefficients with values approximately equal to −0.81 occur both in the asymptotic expansion for α → 1+ and in the expansion for α → 2−.We do not know whether or not these two coefficients are exactly identical.

Small positive zeros of
Next, we look at the zeros of E α,β (−z α ) for β = 2. Here, we observe a substantially different behaviour: The function has real zeros if and only if α is above or equal to a certain threshold α 0 whose value is not known exactly [11,Sections 6 and 7].The approximation provided in [11, eq. (8)] indicates that α 0 ≈ 1.59927; our numerical observations indicate that the correct value is α 0 ≈ 1.599115206 (see the right part of Figure 4).We can then actually proceed in the way indicated for the case β = α in Subsection 3.1 for α ∈ [α 0 , 2] while for α ∈ (1, α 0 ) the search for zeros is pointless.We can summarize our findings as follows (see, in particular, Figures 4 and 5 for a motivation of these formulations): Conjecture 2. There exists some α 0 ∈ (1.59911, 1.59912) such that the following statements hold: (a) For each α ∈ [α 0 , 2], the equation E α,2 (−z α ) = 0 possesses at least one solution in (0, ∞).Denoting (as in eq.(17) above) the smallest of these solutions by Z 2 (α), we have:

Small positive zeros of
As far as the function E α,1 (−z α ) is concerned, we once again obtain from [11] that it has a positive and finite number of zeros for any α ∈ (1, 2).Since arguments similar to those used above yield Z 1 (2) = π/2.On the other hand, by (19a), the equation E 1,1 (−z) = 0 has no positive solution at all, and thus we may expect lim α→1+ Z 1 (α) = ∞.The graphs of these functions for some values of α that cover a wide range of our interval of interest are shown in  The graphs in Figure 7 show the locations Z 1 (α) of the smallest positive zero of E α,1 (−z α ) vs. α obtained in this way.In particular, we once agan see (like in the case β = α) that Z 1 (α) does not depend on α monotonically.More precisely, the smallest value of Z 1 (α) is attained for α ≈ 1.833, with min α∈(1,2] Z 1 (α) ≈ 1.559.More precisely, our numerical calculations give rise to the following conjecture (see also Figures 6 and 7

Conclusion and outlook
As indicated in Section 1, we believe that our findings can be useful to obtain an improved understanding of fractional boundary value problems of the form  (3) and in the development and analysis of associated numerical solution techniques, in particular for algorithms based on the principle of shooting methods [7,12], thus extending the developments for terminal value problems of order α ∈ (0, 1) described in [6].
To fully exploit the results presented here, it will be necessary to obtain further, and in particular analytic, information about the location of the negative real zeros of the Mittag-Leffler functions used above.We hope that our work will motivate the initiation of such investigations.

Theorem 4 .
Consider the two initial value problems stated in (1) on some interval [0, T ] with α ∈ (1, 2) and the associated solutions y 1 and y 2 subject to the following assumptions: (a) The function f is continuous and satisfies a Lipschitz condition with respect to the second variable with Lipschitz constant L. (b) The quantity T

Theorem 5 .
Consider the two initial value problems stated in (1) on some interval [0, T ] with α ∈ (1, 2) and the associated solutions y 1 and y 2 subject to the following assumptions: (a) The function f is continuous and satisfies a Lipschitz condition with respect to the second variable with Lipschitz constant L. (b) y 1,0 = y 2,0 and y 1,1 < y 2,1 .(c) The quantity T * ∈ (0, T ] is chosen such that the inequalities (5b) and (6c) hold for all t ∈ [0, T * ].
Z 2 (2) = π.On the other hand, by (18a), the equation E 1,1 (−z) = 0 has no positive solution at all, and thus we may expect lim α→1+ Z α (α) = ∞.For α ∈ (1, 2), it is a relatively easy matter to compute the values of Z α (α) numerically: The graphs of the function (see the examples given in Figure2) indicate that we may use the standard Newton iteration algorithm with starting value z 0 = π/2; indeed the numerical experiments reveal that this always leads to a rapid convergence against the correct solution.To be precise, the number of iterations required to compute the first positive zero of E α,α (−z α ) with an absolute error of 10 −12 was 19 for α = 1.001, decreased monotonically to 10 as α increased to 1.045, further to 7 as α reached the value 1.303, and stayed at or below 6 for α ≥ 1.615.The required derivative of the function E α,α (−z α ) can be computed using the chain rule and a classical formula for derivatives of Mittag-Leffler functions [4, eq.(4.3.2)],thus obtaining

Figure 2 :
Figure 2: Plots of Eα,α(−z α ) vs. z for α ∈ {1.1, 1.5, 1.9}.The algorithm then yields the numerical results shown in the continuous line in the graphs of Figure 3.To provide a clearer picture, each of the two graphs in this figure covers only a part of the interval α ∈ (1, 2].

Figure 4 :
Figure 4: Plots of E α,2 (−z α ) vs. z for α = 1.599 (dashed line) and α = 1.6 (continuous line).Left: Plot range z ∈ [4, 150] with logarithmically scaled horizontal axis showing that both functions (visually hard to distinguish from each other) seem to remain positive for large z; Right: Zoom into z ∈ [5.2106, 5.21075] with linear scale on horizontal axis reveals that the function for α = 1.59911520635 has a zero at z ≈ 5.210663 whereas the function for α = 1.5991152064 has a local minimum with a strictly positive function value and hence does not have a zero.

Figure 6 .
Figure 6.The shapes of these graphs indicate that we may find the smallest zeros of each of these functions with the help of a Newton iteration with starting value π/2, and our numerical experiments confirm this expectation.Indeed, for all α ∈ [1.001, 2] the method converged up to an accuracy of 10 −12 in at most 13 iterations.The graphs in Figure7show the locations Z 1 (α) of the smallest positive zero of E α,1 (−z α ) vs. α obtained in this way.In particular, we once agan see (like in the case β = α) that Z 1 (α) does not depend on α monotonically.More precisely, the smallest value of Z 1 (α) is attained for α ≈ 1.833, with min α∈(1,2] Z 1 (α) ≈ 1.559.More precisely, our numerical calculations give rise to the following conjecture (see also Figures6 and 7):
follows as well.□ Remark 4. Much as in [1, Remark 2], we can see that the upper bound in (11) also holds if the differential equation under consideration is vector valued, i.e. if f : [0, T ] × R d → R d and y : [0, T ] → R d with arbitrary d ∈ N, whereas the lower bound holds only in the scalar case (d = 1).