Spaces admissible for the Sturm-Liouville equation

We consider the equation \begin{document}$-{y}''(x)+q(x)y(x)=f(x),\ \ \ \ x\in \mathbb{R}\text{ }\ \ \ \ \ \ \ \ \ \ \left( 1 \right)$ \end{document} where \begin{document}$f∈ L_p^{\text{loc}}(\mathbb R),$\end{document} \begin{document}$p∈[1,∞)$\end{document} and \begin{document}$0≤ q∈ L_1^{\text{loc}}(\mathbb R).$\end{document} By a solution of (1) we mean any function \begin{document}$y,$\end{document} absolutely continuous together with its derivative and satisfying (1) almost everywhere in \begin{document}$\mathbb R.$\end{document} Let positive and continuous functions \begin{document}$μ(x)$\end{document} and \begin{document}$θ(x)$\end{document} for \begin{document}$x∈\mathbb R$\end{document} be given. Let us introduce the spaces \begin{document}$\begin{align} & {{L}_{p}}(\mathbb{R},\mu )=\left\{ f\in L_{p}^{\text{loc}}(\mathbb{R}):\|f\|_{{{L}_{p}}(\mathbb{R},\mu )}^{p}=\int_{-\infty }^{\infty }{|}\mu (x)f(x){{|}^{p}}dx In the present paper, we obtain requirements to the functions \begin{document}$μ,θ$\end{document} and \begin{document}$q$\end{document} under which 1) for every function \begin{document}$f∈ L_p(\mathbb R,θ)$\end{document} there exists a unique solution (1) \begin{document}$y∈ L_p(\mathbb R,μ)$\end{document} of (1); 2) there is an absolute constant \begin{document}$c(p)∈(0,∞)$\end{document} such that regardless of the choice of a function \begin{document}$f∈ L_p(\mathbb R,θ)$\end{document} the solution of (1) satisfies the inequality \begin{document}$\|y\|_{L_p(\mathbb R,μ)}≤ c(p)\|f\|_{L_p(\mathbb R,θ)}.$ \end{document}


1.
Introduction. In the present paper, we consider the equation where f ∈ L loc p (R), p ∈ [1, ∞) and 0 ≤ q ∈ L loc 1 (R). (1.2) Our general goal is to determine a space frame within which equation (1.1) always has a unique stable solution. To state the problem in a more precise way, let us fix two positive continuous functions µ(x) and θ(x), x ∈ R, a number p ∈ [1, ∞), and introduce the spaces L p (R, µ) and L p (R, θ) : For brevity, below we write L p,µ and L p,θ , · p,µ and · p,θ , instead of L p (R, µ), L p (R, θ) and · Lp(R,µ) , · Lp(R,θ) , respectively (for µ = 1 we use the standard notation L p (L p := L p (R)) and · p ( · p := · Lp ). In addition, below by a solution of (1.1) we understand any function y, absolutely continuous together with its derivative and satisfying equality (1.1) almost everywhere on R.
Let us introduce the following main definition (see [12, Ch.5, §50-51]: Definition 1.1. We say that the spaces L p,µ and L p,θ make a pair {L p,µ , L p,θ } admissible for equation (1.1) if the following requirements hold: I) for every function f ∈ L p,θ there exists a unique solution y ∈ L p,µ of (1.1); II) there is a constant c(p) ∈ (0, ∞) such that regardless of the choice of a function f ∈ L p,θ the solution y ∈ L p,µ of (1.1) satisfies the inequality y p,µ ≤ c(p) f p,θ . (1.5) Let us in addition we make the following conventions: For brevity we say "problem I)-II)" or "question on I)-II)" instead of "problem (or question) on conditions for the functions µ and θ under which requirements I)-II) of Definition 1.1 hold." We say "the pair {L p,µ ; L p,θ } admissible for (1.1)" instead of "the pair of spaces {L p,µ ; L p,θ } admissible for equation (1.1)", and we often omit the word "equation" before (1.1). By c, c(·) we denote absolute positive constants which are not essential for exposition and may differ even within a single chain of calculations. Our general requirement (1.2) is assumed to be satisfied throughout the paper, is not referred to, and does not appear in the statements.
Let us return to Definition 1.1. The question on the admissibility of the pair {L p , L p } for (1.1) was studied in [3,6] (in [3,6] for µ ≡ θ ≡ 1 in the case where I)-II) were valid, we said that equation (1.1) is correctly solvable in L p . We maintain this terminology in the present paper.) Let us quote the main result of [3,6] (in terms of Definition 1.1). (1.6) Below we continue the investigation started in [3,6]. Our goal is as follows: given equation (1.1), to determine requirements to the weights µ and θ under which the pair {L p,µ ; L p,θ }, p ∈ [1, ∞), is admissible for (1.1). Such an approach to the inversion of (1.1) allows to study this equation also in the case where Theorem 1.2 is not applicable, for example, in the following three cases: 1) q 0 (a) > 0 for some a ∈ (0, ∞), f / ∈ L p , p ∈ [1, ∞); 2) q 0 (a) = 0 for all a ∈ (0, ∞), f ∈ L p , p ∈ [1, ∞); 3) q 0 (a) = 0 for all a ∈ (0, ∞), f / ∈ L p , p ∈ [1, ∞). Our main result (see Theorem 2.4 in §2 below) reduces the stated problem to the question on the boundedness of a certain integral operator S : L p → L p (see (2.10) in §2). From this criterion, under additional requirements to the functions µ, θ and q, one can deduce some concrete particular conditions which control the solution of our problem. See §2 for such restrictions.
The structure of the paper is as follows: In §2, we present our main results and relevant comments; all the proofs of these assertions are collected in §4; §3 contains various technical facts and their proofs; see §5 for an example or the main statements.
We thank the anonymous referee for his remarks that allowed us to substantially improve the exposition of the paper.
2. Main results. Throughout the sequel we assume that our standing requirements to the functions q (see (1.2)), and µ and θ (see §1) are satisfied, and we do not mention them in the statements.
Theorem 2.1. Suppose that the function q is nonnegative and continuous at every point of the real axis. Suppose that for a given p ∈ [1, ∞) the following condition holds: To make our a priori requirements independent of the parameter p ∈ [1, ∞), throughout the sequel we assume that together with (1.2), condition (2.2) holds. Similar to (1.2), below this condition is not quoted and does not appear in the statements.
We will also need the following known facts.
). Suppose that (2.2) holds. Then the equation Let us introduce the Green function of equation (1.1): Our main results are the following lemma and theorem.
Lemma 2.3. Suppose that the following condition holds: Then for every p ∈ [1, ∞) equation (2.3) has no solutions z ∈ L p,µ apart from z ≡ 0.
We also note that since the Green function G(x, t) (see (2.8)) is usually not known, Theorem 2.4 seems to be only a theoretical fact that is not applicable to a particular equation ( has a unique finite positive solution. In the sequel, the solution of (2.11) will be denoted by d(x), x ∈ R. This function was introduced in [4] and is one of the versions of auxiliary functions which were first (and then repeatedly) used M.O. Otelbaev (see [13]). Although it is defined in a somewhat exotic way, it is useful to take into account that the function q * (x) def = d −2 (x) (d −2 := 1/d 2 ) can be interpreted as a composed (in the sense of function theory) average of the function q(ξ), ξ ∈ R, at the point ξ = x with step d(x). Indeed, denote Clearly, S x (q)(t) is the Steklov average with step t > 0 of the function q(ξ), ξ ∈ R, at the point ξ = x, and M (f )(η) is the average of the function f (t), t > 0 with step η > 0 at the point t = 0. Now, using Note two additional properties of d(x), x ∈ R. (The proofs of all assertions related to the properties of the function d are given in §3.) Lemma 2.6. The function d(x) is continuously differentiable for all x ∈ R, and the following inequality holds: Here (2.14) Now we can give the following definition.
Definition 2.7. We say that the function q belongs to the class H (and write q ∈ H) if the following equality holds: In the next assertion, we state an important property of the functions q ∈ H.
The next definition is based on Lemma 2.8 and fixes the set of weight functions µ and θ which we study further (using Theorem 2.10 below) for pairs of functions ν and θ generating pairs {L p,µ ; L p,θ } admissible for (1.1). Definition 2.9. Let q ∈ H. We say that a pair of weights (weight functions) {µ, θ} agrees with the function q if for any ε > 0 there is a constant c(ε) ∈ [1, ∞) such that for all t, x ∈ R one has the inequalities (2.17) In the latter case, we say that the pair {L p,µ ; L p,θ }, p ∈ [1, ∞), agrees with equation The next assertions are convenient for the study of concrete equations. They are obvious and are given without proofs.
Theorem 2.12. Let q ∈ H, and suppose that the weight function θ(x), x ∈ R, is such that m 0 > 0 where . (2.21) Then for p ∈ [1, ∞) the pair {L p ; L p,θ } is admissible for (1.1). Now we will apply the collection of definitions and assertions introduced above to particular equations (1.1) in order to present methods for checking relations (2.15), (2.17), (2.18), (2.20) and (2.21). All these relations include the implicit function d(x), x ∈ R whose exact values can be found only special rare cases. However, it is easy to see that to check these relations, it is enough to have sharp by order twosided estimates of the function d. Such estimates can easily be found using different methods for different requirements to the function q. Below we present an assertion of such a type.
where q 1 (x), x ∈ R, is positive and absolutely continuous together with its derivative, and q 2 ∈ L loc 1 (R). Denote If we have the condition then the following relations hold: To prove inequalities (2.17), the following lemma can be useful.
Lemma 2.14. Suppose that a function µ(x) is defined, positive and differentiable for all x ∈ R, let q ∈ H, and let d(x), x ∈ R, denote the auxiliary function from Lemma 2.5. Then, if the equality holds, then for any given ε > 0 there is a constant c(ε) ∈ [1, ∞) such that for all t, x ∈ R inequalities (2.17) hold.
3. Auxiliary assertions. In this section, we present the properties of the function d(x), x ∈ R (see Lemma 2.5) and related properties of the FSS {u, v} of equation Recall that conditions (1.2) and (2.2) are assumed to hold, here and in the sequel, and are not mentioned in the statements.
Proof of Lemma 2.6. The existence of the derivative d (x), x ∈ R is a consequence of the theory of implicit functions [7, Ch.III, §1, no.3]. It is proven in the same way as in [5]. The following relations are deduced from (2.11): Now, from (2.11) we obtain the inequality Together with the formula for |d (x)| and (2.14), this implies that Proof. Below we use Lagrange's formula and (2.12): Lemma 3.2. For x ∈ R, we have the inequalities (see Theorem 2.2): Proof. Below we use the following two assertions.
Remark 3.5. Two-sided, sharp by order estimates of the function ρ were first obtained by M. Otelbaev (see [14]), and therefore all such inequalities are referred to by his name. Note that the inequalities given in [14] are expressed in terms of another auxiliary function, more complicated than d(x), x ∈ R, and are proven under auxiliary requirements to the function q.
Let us now consider (3.2). Below we use (3.5) and (3.1): Now we use this together with (3.3) and obtain Inequalities (3.2) for the solution v are checked similarly, and estimates (3.2) for ρ follow from the estimates of u and v and (2.7). Lemma 3.6. For a given x ∈ R, consider the function The function F (η) is differentiable and non-negative, together with its derivative, and In addition, the inequality Proof. To prove that the function F (η) is differentiable and the functions F (η) and F (η) are non-negative for η ≥ 0, we use properties of integral. The last assertion of the lemma follows from Lagrange's formula and the relations Let a function f be defined on R and absolutely continuous together with its derivative. Then for all x ∈ R and t ≥ 0, we have the equality Proof. To obtain (3.8), we use the following simple transformations x+t Proof of Theorem 2.13. Set Hence d(x) ≥ η(x) for |x| 1 by Lemma 3.6. Let now Then by the same arguments we obtain: Hence d(x) ≤ η(x) for |x| 1 by Lemma 3.6, and equality (2.27) is proven. Further, since the function d(x) q 1 (x) is continuous and positive for all x ∈ R, for all x 0 ∈ (0, ∞) we the inequalities: Together with (2.27), this implies (2.28).

Proofs of the main results.
Proof of Theorem 2.1. Assume the contrary. Then (2.2) holds, the pair {L p,µ ; L p,θ } is admissible for (1.1), and there exists x 0 ∈ R such that one of inequalities (2.2), say, the second one, does not hold: Without loss of generality, in what follows we assume x 0 ≥ 1. Let us introduce the functions ϕ and f 0 .
From 1)-2) we obtain the equality According to (4.5), we conclude that f 0 ∈ L p,θ : Since the pair {L p,µ ; L p,θ } is admissible for (1.1), we conclude that (1.1) for f = f 0 has a unique solution y 0 ∈ L p,µ . Then (see (4.4) and (4.5)) where z(x), x ∈ R, is some soluton of (2.3). From (2.3) and (4.1), we obtain the equality Let us show that c 2 = 0. Assume to the contrary that c 2 = 0. Choose x 1 so that to have the inequality Then (see (4.7)) and we get a contradiction. Hence c 2 = 0. Let us check that also c 1 = 0. Assume that c 1 = 0. Since ϕ ∈ C ∞ (R), from (4.2) it follows that ϕ(x 0 ) = ϕ (x 0 ) = 0 and therefore (see (4.7)): In addition, ϕ(x) ≡ 0 for x ≤ x 0 , and therefore from (4.5) and (4.6) it follows that the function z is a solution of the Cauchy problem Further, without loss of generality, we assume that c 1 = 1. Let us check that then we have the inequality z(x) ≥ 1 for x ≤ x 0 .
Hence z (x) = 0. But then the function z(x) is a solution of the Cauchy problem We get a contradiction because z(x 0 ) = 1. Thus z(x) > 0 for x ≤ x 0 . Then for We get a contradiction. Hence c 1 = 0, and we obtain the equality We get a contradiction. Hence (4.1) does not hold.
Proof of Lemma 2.3. Let us show that in the case of (2.9) for all p ∈ [1, ∞) we have the equalities We only consider the second equality because the first one can be proved in the same way. For p = 1 equality (4.12) follows from Theorem 2.2 and (2.9) is a straightforward manner. Let p ∈ (1, ∞), p = p(p − 1) −1 . The following relations rely only on Theorem 2.2: (4.13) Let A > 0. Below we use Hölder's inequality and (4.13):

SPACES ADMISSIBLE FOR THE STURM-LIOUVILLE EQUATION 1035
Now, to obtain (4.12), in the last inequality we let A tend to infinity. Let us now go to the proof of the lemma. By Theorem 2.2, the general solution of (2.3) is of the form x ∈ R. Let z ∈ L p,µ . Then c 2 = 0. Indeed, if c 2 = 0, then denote x 1 1, a number such that for all x ≥ x 1 we have the inequality (see (2.6)): (4.14) Now from (4.12), (4.14) and Theorem 2.2 it follows that We get a contradiction. Hence c 2 = 0. The equality c 1 = 0 now follows from (2.4) and (4.12).
Remark 4.1. Below we use the following known theorems. It is convenient to formulate them separately from the main statements of the present paper.
In addition, For p ∈ (1, ∞) the operatorH : L p → L p is bounded if and only ifH p < ∞. Herẽ In addition,H Let us introduce the Green operator  1)) if and only if the operator G : L p → L p is bounded. In the latter case, for f ∈ L p , the solution y ∈ L p of (1.1) is of the form y = Gf. Note that the functiond(x) was introduced by M. Otelbaev (see [13]). The following inequalities hold (see [4] and (2.11)): Proof of Theorem 2.4 for p ∈ (1, ∞). Necessity. We need the following lemma. Proof. Below we only consider the case p ∈ (1, ∞) (for p = 1 the arguments are similar). Let us continue the function f by zero beyond the segment [x 1 , x 2 ] and maintain the original notation. From the obvious inequalities it follows that f ∈ L p,θ . Set (see (2.8), (4.21)) Let us estimate the integrals in (4.26): From (4.27) and (4.28) it follows that the functionỹ(x), s ∈ R, is well-defined. It is also easy to see that the functionỹ(x), x ∈ R is a particular solution of (1.1). But, since f ∈ L p,θ , (1.1) has a unique solution y ∈ L p,θ . This means that we have the equality x ∈ R. Let us check that c 1 = c 2 = 0. Assume, say, that c 2 = 0. Then for x ≥ x 2 , we get From (2.6) and (4.27) it follows that there exists x 3 ≥ max{1, x 2 } such that We get a contradiction. Hence c 2 = 0. Similarly, we prove that also c 1 = 0, and therefore y =ỹ (see (4.26)).
Let [x 1 , x 2 ] be any finite segment. Set (4.30) Therefore, since the pair {L p,µ ; L p,θ } is admissible for (1.1), in the case of (4.29) equation (1.1) has a solution y ∈ L p,µ . This solution is of the form (4.21) (see Lemma 4.6). This implies that Now, using (4.31), (4.30) and (1.5), we obtain Since in this inequality x 1 and x 2 (x 1 ≤ x 2 ) are arbitrary numbers, we conclude that This inequality means that the operator S 2 : L p → L p , is bounded (see Theorem 4.2). Similarly, we use Theorem 4.3 to conclude that the operator S 1 : L p → L p , is bounded. Since we have the equality (see (2.8) and (2.10)) our assertion now follows from the triangle inequality for norms.
Proof of Theorem 2.4. Sufficiency. Proof. The upper estimate in (4.35) follows from (4.34). To prove the lower estimate in (4.35), we use the following obvious relations: This implies that S 1 p→p ≤ S p→p . Similarly, we check that S 2 p→p ≤ S p→p . These inequalities imply the lower estimate in (4.35).
Let us now go to the proof of the theorem. Since (2.2) holds, equation (2.3) has a FSS {u, v} with the properties from Theorem 2.2. Since the operator S : L p → L p is bounded, so are also the operators S i : L p → L p , i = 1, 2 (see (4.35)). Then, by Theorems 4.2 and 4.3, we obtain the inequalities These inequalities imply that the function is well-defined because the integrals in (4.38) converge: Further, one can check in a straightforward manner (see Theorem 2.2) that the function y(x), x ∈ R is a solution of (1.1). In addition, i.e., (1.5) holds. It only remains to refer to Lemma 2.3.
Proof of Theorem 2.4 for p = 1. Necessity.
Let [x 1 , x 2 ] be an arbitrary finite segment, and let f ∈ L 1 be such that supp f = [x 1 , x 2 ]. Then (see (4.25)) f ∈ L 1,θ and therefore equation (1.1) with such a righthand side has a unique solution y ∈ L 1,µ . By Lemma 4.6, this solution is given by formula y = Gf (see (4.21)) and satisfies (1.5). Let us introduce the operatorS : and the function g given on the segment [x 1 , x 2 ] by the formula Then we have This implies that S L1(x1,x2)→L1(x1,x2) ≤ c(1).  ∈ (a, b), and let K be an integral operator ∈ (a, b).
In the last inequality, x 1 and x 2 are arbitrary numbers. Hence But then by Theorem 4.8 we obtain that S L1→L2 ≤ c(1) < ∞, as required.
From (2.2) it follows that equation (2.3) has a FSS {u, v} (see Theorem 2.2), the Green function and the operator S are defined (see (2.8) and (2.10)). Further, the operators S i , i = 1, 2 (see (4.32), (4.33)) are bounded because so is the operator S : L 1 → L 1 (see Lemma 4.7). Let now f ∈ L 1,θ and g = θ · |f |. Then 0 ≤ g ∈ L 1 , S i g ∈ L 1 , i = 1, 2, and one has the inequalities 0 ≤ (S i g)(x) < ∞, ∀x ∈ R, i = 1, 2. (4.42) We will prove (4.42) for i = 1 (the case i = 2 is considered in a similar way). Assume to the contrary that there exists x 1 ∈ R such that (S 1 g)(x 1 ) = ∞. Let x 2 > x 1 . Then, since the functions µ and u are continuous, we have We get a contradiction. Hence, inequalities (4.42) hold. From (4.42) and the definition of g we obtain Thus, if f ∈ L 1,θ , then by (4.43) the following integrals converge: and therefore, for x ∈ R, the function is well-defined. This immediately implies that y(x) is a solution of (1.1). In addition, (1.5) holds: It remains to note that by Lemma 2.3 this solution is unique in the class L 1,µ .

Cases 1.1 and 3.3.
Both cases are treated in the same way. Let us introduce the standing notation for the whole proof: Consider, say, Case 3.3. The following implications are obvious:

Cases 1.3 and 3.1.
Both cases are treated in the same way. For instance, in Case 1.3 we have

Cases 2.3 and 3.2.
Both cases are treated in the same way. For instance, in Case 2.3 we have Proof of Theorem 2.10 for p ∈ (1, ∞). Necessity. We need some auxiliary assertions.
For the reader's convenience, we enumerate the main steps of the proof of assertions A) and B). Note that since the functions in (5.1) and (5.2) are even, all proofs are only given for x ∈ [0, ∞) or for x ∈ [x 0 , ∞), x 0 1. 1) Checking condition (2.2).
Let us check that in the case of (5.1) condition (2.2) holds. Assume to the contrary that there is x 0 ∈ R such that ∞ x0 q(t)dt = 0. (5. 3) The function q in (5.1) is continuous and non-negative. Therefore, from (5.3) it follows that q(t) ≡ 0 for t ∈ [x 0 , ∞) which is obviously false. This contradiction implies (2.2).