Sturm–Liouville Differential Equations Involving Kurzweil–Henstock Integrable Functions

In this paper, we give sufficient conditions for the existence and uniqueness of the solution of Sturm–Liouville equations subject to Dirichlet boundary value conditions and involving Kurzweil– Henstock integrable functions on unbounded intervals. We also present a finite element method scheme for Kurzweil–Henstock integrable functions.


Introduction
The Sturm-Liouville equation appears in certain practical areas, such as heat flow and vibration problems, electroencephalography applications, and other areas of physics. It has a relevant role in quantum mechanics, and some of these problems are formulated in unbounded intervals. On occasion, these problems are described by differential equations with highly oscillatory coefficients. A particular characteristic of these coefficients is that they are not square Lebesgue integrable. The study of differential equations involving integrable Henstock-Kurzweil functions has been developed by several authors, for example, [1][2][3][4][5][6][7]. In [8], Pérez et al. introduced the KH-Sobolev space on bounded intervals and guaranteed the existence and uniqueness of the solution to some boundary value problems involving Kurzweil-Henstock integrable functions on [0, 1]. In this paper, particularly in Section 3, we introduce the KH-Sobolev space for unbounded intervals, and then we apply these spaces and the Fredholm alternative theorem to establish the existence and uniqueness of the solution to the Sturm-Liouville differential equation Now we define the Kurzweil-Henstock integral for non-bounded intervals. Let a ∈ R and δ : [a, ∞] → R + be a gauge function. A tagged partition P = {( [s k−1 , s k ], t k )} n+1 k=1 of [a, ∞] is said to be δ-fine, if s n+1 = t n+1 = ∞, 1/δ(t n+1 ) < s n and for each k = 1, 2, ..., n, [s k−1 , s k ] ⊂ (t k − δ(t k ), t k + δ(t k )).

Definition 2.
A function f defined on [a, ∞] is said to be Kurzweil-Henstock integrable (KHintegrable) if there exists a number A ∈ R with the property that for every ε > 0, there exists a gauge δ : [a, ∞] → R + such that The space of functions which are Kurzweil-Henstock integrable on I is denoted by KH(I). The Alexiewicz seminorm for this space is denoted and defined as The Lebesgue space L p (I), for 1 ≤ p < ∞, is defined as the set of Lebesgue-measurable functions f on I for which I | f | p < ∞. The seminorm of this space is given by It is well-known that L 1 (I) ⊂ KH(I). See ( [10], Corollary 4.80). This inclusion is strict, see, for example, ( [11], Example 3.12). In particular, if f ∈ L 1 (I), then f A = f 1 . Moreover, when I is a bounded interval, it follows that L p (I) ⊆ L 1 (I) ⊆ KH(I). Unfortunately, the space (KH(I), · A ) is not a complete space.
The variation of a function h on the interval I is denoted by V I h. If V I h < ∞, then h is of a bounded variation on I, and we write h ∈ BV(I). The functions of bounded variations are the multipliers of the KH-integrable functions. This allows the following Hölder-type theorem to be established: Theorem 1. ([12], Lemma 24) If f ∈ KH(I) and h ∈ BV(I), then f h ∈ KH(I) and A function F is AC * on a set E, if for every ε > 0 there exists δ ε > 0 such that for every collection {J k } n k=1 of non-overlapping closed intervals with endpoints in E, it follows that is a sequence of subsets of J and F is AC * on each E n . In the case of unbounded intervals, we say that f is Theorem 3. ( [12], Lemma 25) Let f : R → R and ω : R 2 → R be functions such that f ∈ KH(R) and for each compact interval J ⊆ R, 1.

The Kurzweil-Henstock-Sobolev Space for Unbounded Intervals
a partition of [a, b] such that for every k = 1, · · · , n, v ∈ C 2 ((s k−1 , s k )), and v (i) ( . It is clear that if v ∈ V I , then v and v belong to BV(I). Throughout this section, we will only consider the interval I = R. We denote by KH loc (R) the space of functions defined on R that are KH-integrable on every compact interval.
Then the function f is zero a.e. on R.
Lemma 2. Let f ∈ KH(R) and define Then, σ ∈ C(R) and Then, for every compact interval J = [a, b], Thus, From (2), ω(·, t) ∈ BV(R) for all t ∈ R. Therefore, by Hake's Theorem, The weak derivative of u ∈ W KH (R) is denoted and defined byu = w, where w is as in Definition 3. Lemma 1 implies the uniqueness ofu, except for sets of measure zero. It is clear that if u 1 , u 2 ∈ W KH (R) and λ is a scalar, then λu 1 + u 2 ∈ W KH (R) and (λu 1 + u 2 )˙= λu 1 +u 2 .
If u 1 = u 2 a.e. on R and u 2 is ACG * on R, then u 1 ∈ W KH (R) andu 1 = u 2 a.e. on R.
Theorem 4. For each u ∈ W KH (R), there is χ u ∈ C(R) for which u = χ u a.e. on R and Therefore, by Corollary 1, there is K ∈ R for which u − σ = K a.e. on R. Putting χ u = σ + K, we obtain the conclusion of the theorem.
By ([6], Corollary 2.4) we have the following integration by a parts formula for the weak derivative. (3)

Sturm-Liouville Differential Equations for Unbounded Intervals
We denote by W 2 KH (I) the space of functions u ∈ W KH (I) such thatu ∈ W KH (I), and by Ω 0 (I) the space of functions u ∈ W KH (I) such thatu ∈ L ∞ (I), u(q 1 ) = u(q 1 +) = 0 and u(q 2 ) = u(q 2 −) = 0, where q 1 is the left end of the interval I and q 2 is the right end.
Observe that if u ∈ Ω 0 (I), then by Theorem 4, there is χ u ∈ C(I) so that u = χ u a.e. on I, therefore u ∈ L ∞ (I). In this way, we can equip the space Ω 0 (I) with the seminorm In this section, we will only consider I = [a, ∞); however, the results are also true for intervals of the form (−∞, b] or (−∞, ∞).
We solve the following boundary value problem: This problem is equivalent to the following variational problem: We will provide a solution to this variational problem. Define where and where Then Then, Ψ is a bilinear operator and Γ is a linear operator that satisfies ∞) . Therefore, equality (5) is represented by the equation If there were u ∈ Ω 0 ([a, ∞)) such that then the equality Ψ(u, ϕ) + Ψ(Γu, ϕ) = Ψ(x f , ϕ) would be satisfied, for every ϕ ∈ V [a,∞) . Therefore, u would be a solution to the variational problem (5). We will show, using Fredholm's alternative theorem, that under certain conditions there is indeed a solution to Equation (10).

Theorem 6.
Suppose that Y is a compact Hausdorff topological space. A subset H of C(Y, R) is relatively compact in the topology induced by the uniform norm if, and only if: (ii) For every y * ∈ Y and > 0, there exists a neighborhood U y * of y * such that for all y ∈ U y * and v ∈ H.
Proof. Let E ⊆ Ω 0 ([a, ∞)) be such that u W ≤ K 1 for all u ∈ E and some K 1 > 0.
Theorem 8 (Fredholm's alternative theorem). Let X be a normed space and consider a compact linear operator T : X → X. Then, the transformation T + I is injective if only if T + I is surjective. Therefore, (T + I) −1 : X → X is bounded, when T + I is injective. Proof. Let u ∈ W 2 KH ([a, ∞)) be a solution of the homogeneous problem. Then [ρu]˙= qu, a.e. on (a, ∞). Since ρ > 0 and q > 0, we have that u = 0 a.e. on [a, ∞).
Theorem 9. Let f ∈ KH([a, ∞)), q ∈ L 1 ([a, ∞)) and ρ ∈ L 1 ([a, ∞)) be such that ρ ∈ BV(J) for all compact interval J ⊆ [a, ∞), and 1 ρ is ACG * on [a, ∞]. Suppose that either of the two following conditions hold: Proof. By Theorem 7, the operator Γ is compact, and by Proposition 1, Γ + I is injective. Thus, by Fredholm's alternative theorem, the transformation Γ + I is surjective. Then, there exists u ∈ Ω 0 ([a, ∞)) such that (Γ + I)u = x f . Hence, Ψ(u + Γu − x f , ϕ) = 0 for all ϕ ∈ V [a,∞) , consequently, u is a solution to the variational problem (5), and so u is a solution to the boundary problem (4). It is clear that if w is another solution to the boundary problem (4), then u − w is a solution the homogeneous problem, which implies that u = w a.e. on [a, ∞).
Now, let ( f n ) be a sequence in KH([a, ∞)) such that f n − f A → 0, and for each n ∈ N, let u n be a solution to the boundary problem Observe that 1 Therefore, On the other hand, by Fredholm's alternative theorem, (Γ + I) −1 is a bounded operator. Then Consequently, u n − u W → 0 when f n − f A → 0.

Example 1.
In order to derive the steady-state heat conduction model, consider a non-uniform bar of infinite length with cross-sectional area Λ. Let u(t) be the temperature, φ(t) the heat flux and f (t) the source term that models the generation or loss of heat at each point of the cross-section of the bar at position t, where 0 < a ≤ t < ∞. If [t, t + dt] is a small and arbitrary portion of the bar, then by the law of conservation of energy, we have, Dividing by Λdt and taking the limit as dt → 0, we have If ρ(t) is the thermal conductivity of the bar, then by Fourier's heat conduction law, φ(t) = −ρ(t)u (t), we obtain the steady-state heat conduction equation As a particular example, let us consider a non-uniform bar such that its property of conducting heat is greater as the position increases, then the thermal conductivity can be modeled by the function Furthermore, if we assume that heat is continuously lost in certain portions, and in others it is gained due to some source with null effect at distant locations, then one way to represent this behavior is by the function Setting the boundary conditions u(a) = u(a+) = 0 y u(∞) = u(∞−) = 0, we obtain the problem with boundary values for the temperature u(t): − ρu = f a.e. on [a, ∞); (14) u(a) = u(a+) = 0, u(∞) = u(∞−) = 0.

Sturm-Liouville Differential Equations for Bounded Intervals
Let us begin this section by showing that when I is a compact interval on R we have where H 1 (I) = u ∈ L 2 (I) : ∃g ∈ L 2 (I) such that and I • is the interior of I. Let a be the left end of the interval I and b be the right end, and take u ∈ H 1 (I), then there exists g ∈ L 2 (I) such that I uϕ = I gϕ for all ϕ ∈ C 1 c (a, b). Take an arbitrary ϕ ∈ V I , then ϕ ∈ L 1 (I).
Then (ψ n ) is a sequence in C ∞ c (a, b). We prove that 1.
For the first convergence, observe that Now, since it follows that Thus, (16) tends to zero. The second convergence is deduced by the following. From Theorem 1, Finally, since each ψ n ∈ C ∞ c (a, b), it follows that However, it is also true that − I gψ n → − I gϕ.
Consequently, due to the uniqueness of limits, it follows that Therefore, u ∈ W KH (I).
From (15), we obtain the following sequence of inclusions: As in Section 4, we will consider Ω 0 (I) with the seminorm z W = z ∞ + ż ∞ + ż A . The form of this semi-norm is required in the following section. Based on ( [8], Theorem 4.3) and the results of the previous section, we state the following theorems: Then, the following assertions are equivalent: 1. The

Finite Element Method
In this section, we give a finite element method scheme for KH-integrable functions. We consider ρ and f as in Let N ∈ N and a = s 0 < s 1 < · · · < s N < x N+1 = b be a partition of [a, b]. We set h = max{s i − s i−1 : i = 1, 2, · · · , N + 1} and we consider the finite element space V h given by Let r h f be an interpolate of f on [a, b], that is, r h f (s i ) = f (s i ) for all i = 0, · · · , N + 1. Then from Theorem 11,there existsũ Now, we will find u h ,ũ h ∈ V h such that they satisfy A basis for the space V h is given by the functions ϕ i , i = 1, · · · , N, defined as Observe that if u h ,ũ h are defined by then the Equations (20) and (21) are equivalent to Then M is symmetric and tridiagonal. Additionally, M is positive-definitive, because if η = (η 1 , η 2 , · · · , η N ) ∈ R N \ {0} then is a contradiction. Thus, η T Mη > 0. Consequently, M is invertible. Thus, there exist (u 1 , u 2 , · · · , u N ) and (ũ 1 ,ũ 2 , · · · ,ũ N ) unique solutions of the systems Mz = y and Mz =ỹ, where (24) and (25) We will now estimate the error committed. First, observe that for every Indeed, equality (27) is deduced by the following and equality (29) is deduced by From (17) we have that u,ũ ∈ H 1 ([a, b]). Then by (18) Observe that u h is the optimal approximation for u, that is, for all v ∈ V h , and so the inequality (31) is satisfied. Now, take an interpolate r h u ∈ V h of u on [a, s N+1 ], that is, r h u(s i ) = u(s i ) for all i = 0, 1, · · · , N + 1. We take z i ∈ (s i , s i+1 ) such thatu(z i ) = (r h u)˙(z i ), for all i = 0, · · · , N. Then, for every i = 0, · · · , N, there exists c i between s and z i such that Consequently, by (30), (32), and (33) we have that