Hyperbolic numbers as operators

Finally, we use the same idea behind the isomorphismΞ to construct an operator algebra iso- morphic to the algebra of polynomials on the hyperbolic numbers.

The hyperbolic numbers†are defined, in a similar way to the complex numbers, as follows, 𝔻 = {𝑥 + 𝑗𝑦 ∶ 𝑥, 𝑦 ∈ ℝ, 𝑗∈ℝ, 𝑗 2= 1}.

The arithmetic in𝔻 is that obtained assuming the commutative, associative and distributive properties for the sum and product. In a parallel fashion to the complex numbers, if𝑤 ∈ 𝔻, with𝑤 = 𝑥 + 𝑗𝑦, we can define

𝑤 ∶= 𝑥 − 𝑗𝑦, ℜ(𝑤) ∶= 𝑥, ℑ(𝑤) ∶= 𝑦, and, since𝑤𝑤 = 𝑥2− 𝑦2∈ ℝ, we set

|𝑤| ∶= √|𝑤𝑤|,

which is called the Minkowski norm. It is clear that|𝑤₁𝑤₂| = |𝑤₁||𝑤₂| for every 𝑤₁, 𝑤₂ ∈ 𝔻 and, if|𝑤| ≠ 0, then 𝑤−1= 𝑤/|𝑤|2. If we add the norm

‖𝑤‖ = √2(𝑥2_{+ 𝑦}2_),

we have that(𝔻, ‖⋅‖) is a Banach algebra, so the exponential and the hyperbolic trigonometric functions are well defined. Although, unlikeℂ, 𝔻 is not a division algebra (not every nonzero element has an inverse), we can derive calculus (differentiation, integration, holomorphic func- tions…) for𝔻 as well [6].

In this setting, we want to derive an operator 𝐽 defined on a suitable space of functions such that satisfies the same algebraic properties as the hyperbolic imaginary unity𝑗. In other words, we want the map

ℝ[𝐷, 𝐽] 𝔻[𝐷]

∑

u�(𝑎u�𝐽 + 𝑏u�)𝐷

u� _∑

u� (𝑎u�𝑗 + 𝑏u�)𝐷 u� Θ

to be an algebra isomorphism. This implies: • 𝐽 is a linear operator,

• 𝐽_∈ℝ[𝐷]._

• 𝐽2= Id, that is, 𝐽 is an involution, • 𝐽𝐷 = 𝐷𝐽.

There is a simple characterization of linear involutions on a vector space: every linear involution 𝐽 is of the form

𝐽 = ±(2𝑃 − Id)

where𝑃 is a projection operator, that is, 𝑃2 = 𝑃. It is clear that ±(2𝑃 − Id) is, indeed a linear operator and an involution. On the other hand, it is simple to check that, if𝐽 is a linear involution,𝑃 ∶= (±𝐽 + Id)/2 is a projection, so 𝐽 = ±(2𝑃 − Id).

Example 5.4.7. Consider the space𝑊 = L2([−𝜋, 𝜋]) and define

𝑃𝑓 (𝑡) ∶= ∑

u�∈ℕ

∫_−u�u� 𝑓 (𝑠) cos(2 𝑛 𝑠) d 𝑠 cos(2 𝑛 𝑡) for every𝑓 ∈ 𝑊,

that is, take only the sum over the even coefficients of the Fourier series of𝑓 . Clearly 𝑃𝐷 = 𝐷𝑃. 𝐽 ∶= 2𝑃 − Id satisfies the aforementioned properties.

The algebraℝ[𝐷, 𝐽], being isomorphic to 𝔻[𝐷], satisfies also very good algebraic prop- erties (see, for instance, [146]). In order to get an analogous theorem to Theorem 5.1.1 for the algebraℝ[𝐷, 𝐽] it is enough to take, as in the case of ℝ[𝐷, 𝐽], 𝑅 = Θ−1(Θ(𝐿)).

𝜑-Laplacian

This chapter is devoted to the study of the existence and periodicity of solutions of initial differ- ential problems, paying special attention to the explicit computation of the period. These prob- lems are also connected with some particular initial and boundary value problems with reflec- tion, which allows us to prove existence of solutions of the latter using the existence of the first.

Let us consider the problems (3.1.1) and (3.1.2) again for a differentiable involution𝜑. Ob- serve that, from problem (3.1.6), we have that

0 =_{𝑓′(𝑓−1}𝑥″_(𝑥(𝑡)_′(𝑡))) − 𝑓 (𝑥(𝑡))𝜑′_{(𝑡) = (𝑓}−1₎′_(𝑥′_(𝑡))𝑥″_{(𝑡) − 𝑓 (𝑥(𝑡))𝜑}′_(𝑡)

=(𝑓−1_{∘ 𝑥}′₎′_{(𝑡) − 𝑓 (𝑥(𝑡))𝜑}′_(𝑡).

So, clearly, problem (3.1.6) is equivalent to the problem

(𝑓−1_{∘ 𝑥}′₎′_{(𝑡) − 𝜑}′_{(𝑡)𝑓 (𝑥(𝑡)) = 0, 𝑥(𝑎) = 𝑥(𝑏), 𝑥}′_{(𝑎) = 𝑓 (𝑥(𝑎)). (6.0.1)}

Which involves the𝑓−1-Laplacian(𝑓−1∘ 𝑥′)′, although, contrary to most literature, the other term in the equation does not involve𝑓−1but𝑓 . As we will see, this is not more than a further generalization in the line of the𝑝-𝑞-Laplacian.

Problems concerning the𝜑-Laplacian (or, particularly, the 𝑝-Laplacian) have been studied extensively in recent literature. Drábek, Manásevich and others study the eigenvalues of prob- lems with the𝑝-Laplacian in [15, 61, 63, 64, 145] using variational methods. The existence of positive solutions is treated in [62], the existence of an exact number of solutions in [154] and topological existence results can be found in [55]. Anti-maximum principles and sign properties of the solutions are studied in [32, 36]. In [49] the authors study a variant of the𝑝-Laplacian equation with an approach based on variational methods, in [16] they study the eigenvalues of the Dirichlet problem and in [60] they find some oscillation criteria for equations with the 𝑝-Laplacian.

The𝜑-Laplacian is studied from different points of view in several papers, e. g. [2,9–13,33, 38, 48, 53, 54, 86, 110, 127, 136]. Actually, if we consider the problem with the𝑓−1-Laplacian

(𝑓−1_{∘ 𝑥}′

u�)′(𝑡) + 𝑓 (𝑥u�(𝑡)) = 0, 𝑥u�(𝑎) = 𝑐, 𝑥′u�(𝑎) = 𝑓 (𝑐), (6.0.2)

and we assume there exist𝑐₁, 𝑐₂∈ ℝ, 𝑐₁ < 𝑐₂, such that a unique solution of problem (6.0.2) exists for every𝑐 ∈ [𝑐₁, 𝑐₂] and (𝑥_u�

1(𝑏) − 𝑐1)(𝑥u�2(𝑏) − 𝑐2) < 0, then problem (3.1.6) must have at least a solution due to the continuity of𝑥_u�on𝑐 and Bolzano’s Theorem. For this reason we will be interested in studying the properties of problem (6.0.2) and its solutions in this chapter. In the sections to come we study this problem and more general versions of it.

In the following section we will study the existence, uniqueness and periodicity of solutions of problem (6.1.1) and in Section 6.2 we will apply these results to the case of problems with reflection. The results of this chapter can be found in [42].

6.1

General solutions

First, we write in a general way the solutions of equations involving the𝑔-𝑓 -Laplacian.

Let 𝜏_u�, 𝜎_u� ∈ [−∞, ∞], 𝑖 = 1, … , 4, 𝜏₁ < 𝜏₂, 𝜎₁ < 𝜎₂, 𝜏₃ < 𝜏₄, 𝜎₃ < 𝜎₄. Let 𝑓 ∶ (𝜏1, 𝜏2) → (𝜎1, 𝜎2) and 𝑔 ∶ (𝜏3, 𝜏4) → (𝜎3, 𝜎4) be invertible functions such that

𝑓 and 𝑔−1 _{are continuous. Assume there is𝑠}

0 ∈ (𝜏1, 𝜏2) such that 𝑓 (𝑠0) = 0 and define

𝐹(𝑡) ∶= ∫u�

u�0𝑓 (𝑠) d 𝑠. Observe that 𝐹 is 0 at 𝑠0and of constant sign everywhere else. The following Lemma is an straightforward application of the properties of the integral.

Lemma 6.1.1. If𝑓 is continuous, invertible and increasing (decreasing) then 𝐹₋ ≡ 𝐹|_(−∞,u�

0]is

strictly decreasing (increasing) and𝐹₊ ≡ 𝐹|_[u�

0,+∞)is strictly increasing (decreasing). Further-

more, if𝜏₁ = −∞, 𝐹(−∞) = +∞ (−∞) and if 𝜏₂= +∞, 𝐹(+∞) = +∞ (−∞).

All the same, define𝐺(𝑡) ∶= ∫_u�u�_−1({0})𝑔−1(𝑠) d 𝑠 and consider the problem (𝑔 ∘ 𝑥′₎′_{(𝑡) + 𝑓 (𝑥(𝑡)) = 0, a. e. 𝑡 ∈ ℝ, 𝑥(𝑎) = 𝑐}

1, 𝑥′(𝑎) = 𝑐2, (6.1.1)

for some fixed𝑐₁, 𝑐₂∈ ℝ.

Definition 6.1.2. A solution𝑥 of problem (6.1.1) will be 𝑥 ∈ u�1(𝐼), such that 𝑔∘𝑥′is absolutely continuous on𝐼, where 𝐼 is an open interval with 𝑎 ∈ 𝐼. The solution must further satisfy that the equation in problem (6.1.1) holds a. e. and the initial conditions are satisfied as well.

Theorem 6.1.3. Let 𝑓 ∶ (𝜏₁, 𝜏₂) → (𝜎₁, 𝜎₂) and 𝑔 ∶ (𝜏₃, 𝜏₄) → (𝜎₃, 𝜎₄) be invertible

functions such that𝑓 and 𝑔−1are continuous and assume0 ∈ (𝜏₁, 𝜏₂) ∩ (𝜏₃, 𝜏₄), 𝑓 (0) = 0,

𝑔(0) = 0, 𝑓 and 𝑔 increasing, 𝐹(𝑐1) + 𝐺(𝑔(𝑐2)) < min{𝐺(𝜎3), 𝐺(𝜎4)}. Then there exists

a unique local solution of problem (6.1.1).

Furthermore, if𝐹(𝑐₁) + 𝐺(𝑔(𝑐₂)) < min{𝐹(𝜏₁), 𝐹(𝜏₂)}, then such solution is defined on the whole real line and is periodic of smallest period

𝑇 ∶= ∫u�+−1(u�(u�(u�2))+u�(u�1))

u�₋−1_{(u�(u�(u�2))+u�(u�1))} [_{𝑔−1∘ 𝐺−1} 1

+ (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑟))

−_{𝑔−1∘ 𝐺−1} 1

− (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑟))]d 𝑟.

(6.1.2)

Proof. For the first part of the Theorem and without loss of generality, we will prove the exis-

tence of solution in an interval of the kind[𝑎, 𝑎 + 𝛿), 𝛿 ∈ ℝ+. The proof would be analogous for an interval of the kind(𝑎 − 𝛿, 𝑎].

Let𝑦(𝑡) = 𝑔(𝑥′(𝑡)). Then problem (6.1.1) is equivalent to 𝑥′_{(𝑡) = 𝑔}−1_{(𝑦(𝑡)), 𝑦}′_{(𝑡) = − 𝑓 (𝑥(𝑡)), 𝑡 ∈ ℝ 𝑥(𝑎) = 𝑐}

1, 𝑦(𝑎) = 𝑔(𝑐2).

Hence,

𝑓 (𝑥(𝑡))𝑥′_{(𝑡) + 𝑔}−1_{(𝑦(𝑡))𝑦}′_{(𝑡) = 0, 𝑡 ∈ ℝ,}

so, integrating both sides from𝑎 to 𝑡,

where𝑘 = 𝐹(𝑐₁) + 𝐺(𝑔(𝑐₂)). That is, undoing the change of variables, 𝐺(𝑔(𝑥′_{(𝑡))) = 𝐺(𝑔(𝑐}

2)) + 𝐹(𝑐1) − 𝐹(𝑥(𝑡)), 𝑡 ∈ ℝ. (6.1.3)

If 𝑐₁ = 𝑐₂ = 0 it is clear that the only possible solution is 𝑥 ≡ 0 for, in that case, 𝐺(𝑔(𝑥′_{(𝑡))) + 𝐹(𝑥(𝑡)) = 0 and, since 𝐺 and 𝐹 are nonnegative and increasing, 𝑥}′_{(𝑡) =}

𝑥(𝑡) = 0 for 𝑡 ∈ ℝ. Assume, without loss of generality, that 𝑐2is nonnegative and𝑐1negative

(the other cases are similar). If𝑐₂= 0 then, integrating (6.1.1), 𝑔 ∘ 𝑥′_{(𝑡) = − ∫}u�

u�𝑓 (𝑥(𝑠)) d 𝑠,

which implies𝑥′is positive in some interval[𝑎, 𝑎 + 𝛿).

If𝑐₂is positive, then𝑥′has to be positive at least in some neighborhood of𝑎, so, in a right neighborhood of𝑎, we can solve for 𝑔 ∘ 𝑥′in (6.1.3) as

𝑔 ∘ 𝑥′_{(𝑡) = 𝐺}−1

+ (𝐹(𝑐1) − 𝐹(𝑥(𝑡)) + 𝐺(𝑔(𝑐2))). (6.1.4)

In order to solve for𝑥′in (6.1.4), we need𝐹(𝑐₁) + 𝐺(𝑔(𝑐₂)) < 𝐺(𝜎₄). Then, 𝑥′_{(𝑡) = 𝑔}−1_{∘ 𝐺}−1

+ (𝐹(𝑐1) − 𝐹(𝑥(𝑡)) + 𝐺(𝑔(𝑐2))). (6.1.5)

Integrating between𝑎 and 𝑡,

𝑡 = ∫_u�u� 𝑥′(𝑠) 𝑔−1_{∘ 𝐺}−1 + (𝐹(𝑐1) − 𝐹(𝑥(𝑠)) + 𝐺(𝑔(𝑐2)))d 𝑠 + 𝑎 = 𝐻+(𝑥(𝑡)), where 𝐻+(𝑟) ∶= ∫_u�u�_{1 𝑔−1∘ 𝐺−1} 1 + (𝐹(𝑐1) − 𝐹(𝑠) + 𝐺(𝑔(𝑐2)))d 𝑠 + 𝑎.

𝐻+is strictly increasing in its domain due to the positivity of the denominator in the integrand.

Hence, for𝑡 sufficiently close to 𝑎,

𝑥(𝑡) = 𝐻−1 + (𝑡).

Therefore, a solution of problem (6.1.1) exists and is unique (by construction) on an interval [𝑎, 𝑎 + 𝛿).

If we assume𝐹(𝑐₁) + 𝐺(𝑔(𝑐₂)) < min{𝐹(𝜏₁), 𝐹(𝜏₂)}, 𝑐₂ > 0 (the case 𝑐₂ = 0 is similar),𝐻₊is well defined on

𝐼 ∶= (𝐹−1

− (𝐹(𝑐1) + 𝐺(𝑔(𝑐2))), 𝐹+−1(𝐹(𝑐1) + 𝐺(𝑔(𝑐2)))) .

Now, we study the range of𝐻₊.

𝑔(𝑥′_{(𝑡)) is positive as long as 𝑥}′_{(𝑡) is positive. Hence, consider}

𝑡0∶= sup{𝑡 ∈ [𝑎, +∞) ∶ 𝑥′(𝑠) > 0 for a. e. 𝑠 ∈ [𝑎, 𝑡)} ∈ [𝑎, +∞].

𝐺 is positive on nonzero values, so equation (6.1.3) implies that 𝐹(𝑥(𝑡)) < 𝐺(𝑔(𝑐2)) + 𝐹(𝑐1)

for all𝑡 ∈ (𝑎, 𝑡₀).

Assume𝑡₀= +∞. Now, 𝑥′(𝑡) > 0 a. e. in [𝑎, +∞) so there exists 𝑥(+∞) ∈ (𝑐1, 𝐹+−1(𝐺(𝑔(𝑐2)) + 𝐹(𝑐1))] .

On the other hand, since𝑥 is increasing in [𝑎, +∞) and 𝑐₁ < 0, by equation (6.1.5) we have that𝑥′is increasing as long as𝑥 is negative. This means that, eventually (in finite time), 𝑥 will be positive and therefore, 𝑥′_{is decreasing in[ ̃𝑎, +∞) for ̃𝑎 big enough, so there exists}

𝑥′_{(+∞) ≥ 0. If we assume 𝑥}′_{(+∞) = 𝜖 > 0, this implies that 𝑥(+∞) = +∞, for there}

would exist𝑀 ∈ ℝ such that 𝑥′(𝑡) > 𝜖/2 for every 𝑡 ≥ 𝑀, so 𝑥′(+∞) = 0. Taking the limit 𝑡 → +∞ in equation (6.1.3), 𝑥(+∞) = 𝐹−1

+ (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1)).

Now, take 𝜖 ∈ (0, 𝑓 (𝑥(+∞))). Since 𝑔 ∘ 𝑥′(+∞) = 0 and 𝑔 ∘ 𝑥′ is continuous and decreasing in[ ̃𝑎, +∞), there exists 𝑀 ∈ ℝ+ such that|𝑔(𝑥′(𝑀₂)) − 𝑔(𝑥′(𝑀₁))| < 𝜖 for every𝑀₁, 𝑀₂ > 𝑀. Since 𝑓 is continuous, there exits ̃𝑀 > 𝑀 such that 𝑓 (𝑥(𝑀₃)) > 𝜖 for every𝑀₃ > ̃𝑀. Take 𝑀₃in such a way. Then, integrating equation (6.1.1) between𝑀₃and 𝑀3+ 1,

(𝑔 ∘ 𝑥′_)(𝑀

3+ 1) − (𝑔 ∘ 𝑥′)(𝑀3) = ∫_u�u�₃3+1𝑓 (𝑥(𝑠)) d 𝑠 > 𝜖,

a contradiction. Therefore,𝑡₀ ∈ ℝ.

Observe that𝑥′(𝑡₀) = 0, so 𝑥 attains its maximum at 𝑡₀and𝑥(𝑡₀) = 𝐹₊−1(𝐺(𝑔(𝑐₂)) + 𝐹(𝑐1)) by equation (6.1.3), that is, 𝑥(𝑡0) = sup 𝐼. In order for this value to be well defined it

is necessary that𝐺(𝑔(𝑐₂)) + 𝐹(𝑐₁) ≤ 𝐹(𝜏₂).

Now, we have that𝐻₊ is well defined atsup 𝐼 (assuming it is defined continuous at that point). Indeed,

𝑡0 = lim_u�→u�

0𝐻+(𝑥(𝑡)) = 𝐻+(𝐹

−1

+ (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1))).

We prove now that there is a neighborhood(𝑡₀, 𝑡₀+𝜖) where 𝑥′is negative, which means that we can take

𝑡1∶= sup{𝑡 ∈ [𝑡0, +∞) ∶ 𝑥′(𝑠) < 0 for a. e. 𝑠 ∈ [𝑡0, 𝑡)}.

Fix𝜉 such that 0 < 𝜉 < 𝑓 (𝑥(𝑡₀)) and take 𝜖 such that 𝑓 (𝑥(𝑡)) > 𝜉 in (𝑡₀, 𝑡₀+ 𝜖). Take 𝑡 ∈ (𝑡0, 𝑡0+ 𝜖), then, integrating equation (6.1.1) between 𝑡0and𝑡,

𝑔(𝑥′_{(𝑡)) = − ∫}u�

u�0𝑓 (𝑥(𝑠)) d 𝑠 < −𝜉(𝑡 − 𝑡0) < 0.

We deduce that𝑡₁< +∞ by the same kind of reasoning we used to prove 𝑡₀< +∞. Observe that𝑥′(𝑡₁) = 0 and 𝑥(𝑡₁) = 𝐹₋−1(𝐺(𝑔(𝑐₂)) + 𝐹(𝑐₁)). This last equality comes from eval- uating equation (6.1.3) at𝑡₁and Rolle’s Theorem as we show now: the other possibility would be𝑥(𝑡₁) = 𝐹₊−1(𝐺(𝑔(𝑐₂)) + 𝐹(𝑐₁)). Observe that, by equation (6.1.5), 𝑥′is continuous, so 𝑥 ∈ u�1_{([𝑎, 𝑡}

1)). Since 𝑥(𝑡0) = 𝑥(𝑡1), there would exist ̃𝑡 ∈ (𝑡0, 𝑡1) such that 𝑥′( ̃𝑡) = 0, a

contradiction.

Now, we have that𝑥′(𝑡) = 𝑔−1∘ 𝐺−1₋ (𝐺(𝑔(𝑐₂)) + 𝐹(𝑐₁) − 𝐹(𝑥(𝑡))), that is, 1 = 𝑥′_(𝑡)/(𝑔−1_{∘ 𝐺}−1

Thus,

𝑡1− 𝑡0= ∫_u�u�₀1 𝑥

′_{(𝑠) d 𝑠}

𝑔−1_{∘ 𝐺}−1

− (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑥(𝑠)))

= ∫u�−−1(u�(u�(u�2))+u�(u�1))

u�−1 + (u�(u�(u�2))+u�(u�1)) d 𝑟 𝑔−1_{∘ 𝐺}−1_{− (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑟))}. If we define 𝐻−(𝑠) ∶= ∫_u�u�−1 + (u�(u�(u�2))+u�(u�1)) d 𝑟 𝑔−1_{∘ 𝐺}−1 − (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑟))+ 𝑡0,

𝐻− is strictly decreasing in its domain and𝑥(𝑡) = 𝐻−−1(𝑡) for 𝑡 ∈ [𝑡0, 𝑡1].

We can again deduce that

𝑡2∶= sup{𝑡 ∈ [𝑡1, +∞) ∶ 𝑥′(𝑠) > 0 for a. e. 𝑠 ∈ [𝑡1, 𝑡)} < +∞.

Using the positivity and growth conditions of the functions involved, it is easy to check that 𝑥(𝑡1) = 𝐹−−1(𝐺(𝑔(𝑐2)) + 𝐹(𝑐1)) < 𝑐1 < 𝐹+−1(𝐺(𝑔(𝑐2)) + 𝐹(𝑐1)) = 𝑥(𝑡2), so there

exists a unique𝑏 ∈ (𝑡₁, 𝑡₂) such that 𝑥(𝑏) = 𝑐₁. Now, 𝑏 − 𝑡1 = ∫_u�u�₁ 𝑥

′_{(𝑠) d 𝑠}

𝑔−1_{∘ 𝐺}−1

+ (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑥(𝑠)))

= ∫u�1

u�−1_{− (u�(u�(u�2))+u�(u�1))𝑔−1∘ 𝐺−1} d 𝑟

+ (𝐺(𝑔(𝑐2)) + 𝐹(𝑐1) − 𝐹(𝑟)).

Defining𝑇 ∶= 𝑏 − 𝑎 and extending 𝑥 periodically in the following way (we have 𝑥 already defined in[𝑎, 𝑎 + 𝑇]),

𝑥(𝑡) = 𝑥 (𝑡 − ⌊𝑡 − 𝑎_{𝑇 ⌋ 𝑇) ,}

where⌊𝑡⌋ ∶= sup{𝑘 ∈ ℤ ∶ 𝑘 ≤ 𝑡}, it is easy to check that 𝑥, extended in such a way, is a global periodic solution of problem (6.1.1).

Take𝑧(𝑡) ∶= 𝑥(𝑡 − 𝑇), 𝑡 ∈ ℝ, we show that 𝑧 is a solution of the problem in [𝑎 + 𝑇, 𝑎 + 2𝑇]. 0 = (𝑔 ∘ 𝑥′₎′_{(𝑡) + 𝑓 (𝑥(𝑡)) = (𝑔 ∘ 𝑧}′₎′_{(𝑡 + 𝑇) + 𝑓 (𝑧(𝑡 + 𝑇)) for a. e. 𝑡 ∈ ℝ} This is equivalent to (𝑔 ∘ 𝑧′₎′_{(𝑡) + 𝑓 (𝑧(𝑡)) = 0 for a. e 𝑡 ∈ ℝ.} Also, 𝑧(𝑎 + 𝑇) = 𝑥(𝑎) = 𝑐1, 𝑧′_{(𝑎 + 𝑇) = 𝑥}′_{(𝑎) = 𝑐} 2. 

Remark 6.1.4. A similar argument can be done for the case𝑓 and 𝑔 have different growth type

(e. g. 𝑓 increasing and 𝑔 decreasing), but taking the negative branch of the inverse function 𝐺−1_{in (6.1.5).}

Remark 6.1.5. In the hypotheses of Theorem 6.1.3, if instead of𝑔(0) = 𝑓 (0) = 0 we have that

𝑔(𝑠0) = 𝑓 (𝑠0) = 0, define ̃𝑓(𝑥) ∶= 𝑓 (𝑥 + 𝑠0), ̃𝑔(𝑥) ∶= 𝑔(𝑥 + 𝑠0). Then ̃𝑓(0) = ̃𝑔(0) = 0

and problem (6.1.1) is equivalent to

( ̃𝑔 ∘ 𝑣′₎′_{(𝑡) + ̃𝑓(𝑣(𝑡)) = 0, 𝑣(𝑎) = 𝑐}

1− 𝑠0, 𝑣(𝑎) = 𝑐2,

with𝑣(𝑡) = 𝑥(𝑡) − 𝑠₀. Hence, we can apply Theorem 6.1.3 to this case.

Remark 6.1.6. Using the notation of Theorem 6.1.3, the explicit form of the solution of problem

(6.1.1) is given by 𝑥(𝑡) =⎧{{ ⎨{{ ⎩ 𝐻−1 + (𝑡 − ⌊𝑡 − 𝑎_{𝑇 ⌋ 𝑇) , 𝑡 ∈ [𝑎 + 2𝑇𝑘, 𝑎 + (2𝑘 + 1)𝑇], 𝑘 ∈ ℤ,} 𝐻−1 − (𝑡 − ⌊𝑡 − 𝑎_{𝑇 ⌋ 𝑇) , 𝑡 ∈ [𝑎 + (2𝑘 − 1)𝑇, 𝑎 + 2𝑘𝑇], 𝑘 ∈ ℤ,}

Remark 6.1.7. Consider the following particular case of problem (6.1.1) with𝑓 (0) = 0, 𝑔(0) =

0, 𝑓 and 𝑔 increasing and the hypothesis for a unique global solution of the following problem are satisfied in Theorem 6.1.3.

(𝑔 ∘ 𝑥′₎′_{(𝑡) + 𝑓 (𝑥(𝑡)) = 0, 𝑥(0) = 0, 𝑥}′_{(0) = 1. (6.1.6)}

It is clear that, in the case𝑔(𝑥) = 𝑓 (𝑥) = 𝑥, the unique solution of problem (6.1.6) is sin(𝑡), which suggests the definition of thesin_u�,u� function as the unique solution of problem (6.1.6) for general𝑔 and 𝑓 . Correspondingly,

arcsin+

u�,u�(𝑟) ∶= 𝐻+(𝑟).

This function, defined as such, coincides with the arcsin_u� function defined in [24, 115] for the𝑝-Laplacian 𝑓 (𝑥) = 𝑔(𝑥) = |𝑥|u�−2𝑥, the function arcsin_u�,u�defined in [14, 65, 108] for the 𝑝-𝑞-Laplacian 𝑓 (𝑥) = |𝑥|u�−2_{𝑥, 𝑔(𝑥) = |𝑥|}u�−2_{𝑥, which first appeared with a slightly different de-}

finition in [64], and the hyperbolic version of this function, also in [14, 108], which corresponds to the case𝑓 (𝑥) = |𝑥|u�−2𝑥, 𝑔(𝑥) = −|𝑥|u�−2𝑥. [164] derives generalized Jacobian functions in a similar way, defining

arcsnu�,u�(𝑡, 𝑘) ∶= ∫₀u� u� 1

√(1 − 𝑠u�_{)(1 − 𝑘}u�_𝑠u�₎d 𝑠,

of which the inverse (see [164, Proposition 3.2]) is precisely a solution of (𝑓u�∘ 𝑥′(𝑡))′+ 𝑞_𝑝∗𝑓u�(𝑥(𝑡))(1 + 𝑘u�− 2𝑘u�|𝑥(𝑡)|u�) = 0,

where 𝑓_u� is the 𝑟-Laplacian for 𝑟 = 𝑝, 𝑞 and 𝑝∗𝑝 = 𝑝∗ + 𝑝. Observe that this case is also covered by our definition.

In all of the aforementioned works they are interested on the inverse of thearcsin_u�,u� func- tion, thesin_u�,u� function, which they extend to the whole real line by symmetry and periodicity. Observe that in our case𝑓 and 𝑔 need not to be odd functions, contrary to the above examples, but we can still give the definition of thesin_u�,u� function in the whole real line. Also, this lack of symmetry gives rise to a richer set of right inverses ofsin_u�,u�, for instance,

arcsin−

u�,u�(𝑟) ∶= 𝐻−(𝑟).

In general, if we have a problem of the kind

Φ((𝑔 ∘ 𝑥′₎′_{, 𝑥(𝑡)) = 0; 𝑥(0) = 0, 𝑥}′_{(0) = 1,}

and we know it has a unique solution in a neighborhood of0, then we can define sin_u�,Φ as such unique solution and its inverse, in a neighborhood of0, arcsin_u�,Φ.

In document Existence and Multiplicity of Solutions of Functional Differential Equations (Page 119-129)