L 1 -maximal Regularity for Quasilinear Second Order Differential Equation with Damped Term

We investigated a quasilinear second order equation with damped term on the real axis. We gave some suitable conditions for existence of the L 1-maximal regular solutions of this equation.

By C (k) 0 (R) (k = 1, 2, . ..) we denote the set of k times continuously differentiable functions with compact support.Let C (j) Then y is called a solution of (1.1).
The purpose of this work is to find some conditions for r and q such that for every f ∈ L 1 , the equation (1.1) has a solution y which satisfies y 1 + r(•, y)y 1 + q(•, y)y 1 < ∞.
B Email: ospanov_kn@enu.kz The separability of differential operators introduced by Everitt and Giertz in [7,8] plays an important role in the study of second order differential equations.Recall that the Sturm-Liouville operator Ly := −y + q 1 (x)y, Everitt and Giertz [7,8] proved that if q 1 and its derivatives satisfy some conditions, then L is separable in L 2 (R).In the case q 1 is not differentiable function, the separability of L in L 2 (R) was discussed in [3,23].In [9], Everitt, Giertz and Weidmann give an example of non-separable Sturm-Liouville operator in L 2 (R) with strongly oscillating and infinitely smooth coefficient q 1 .The separability of linear partial differential operators was studied in [4,15,17,21,24].Some sufficient conditions of separability of operators on Riemann manifolds are obtained in [1,2,12,13].
The separability is also an important tool when dealing with quasilinear equations.In [16], Muratbekov and Otelbaev used the separability to discuss the solvability of the nonlinear equation where f ∈ L 2 (R).Grinshpun and Otelbaev showed that the solvability of the equation (1.2) in L 1 implies q 0 ≥ 1 (see [6]).This method is useful for the multidimensional (Schrödinger) equation −∆u + q(x, u)u = F(x), x ∈ R n (see [17,20] for details).
In general, the expression (1.1) can be converted neither to (1.2) nor to the form − p 2 (x, y)y + q 2 (x, y)y = f (x).
In [22], we considered the equation −y + r(x, y)y = f (x), f ∈ L 2 (R), and found some conditions for r such that this equation is solvable.In the present paper, we discuss the more general equation (1.1), in the case f ∈ L 1 .Under weaker conditions on r than in [22] the existence and regularity of solutions of (1.1) are established.
Schauder's fixed-point theorem is used to prove our main result (see [10]).
Let g and h be some functions on R and let The main result of this paper is the following.
Theorem 1.2.Let r be a continuously differentiable function and q a continuous function satisfying and sup Then for any f ∈ L 1 , the equation (1.1) has a solution y such that Example 1.3.Let r = 10 + x 10 + 5y 4 , q = x 3 + cos 4 x + 2y.Then r and q satisfy the conditions of Theorem 1.2.
Lemma 2.1.Let g and h be continuous functions on R such that Moreover, γ g,h is the smallest constant which satisfies (2.1).
Let r be a continuously differentiable function and Denote by l the closure of l 0 in L 1 .
From Lemma 2.1, (1.3) and (2.4) follows that So the inverse l −1 of l exists.
Next, we show that R(l) = L 1 .Let R(l) = L 1 .Since l is closed, (2.5) implies R(l) is closed.Hence there exists a nonzero element z 0 ∈ ⊥ R(l) such that l * z 0 = −z 0 − (r(x)z 0 ) = 0, where l * is adjoint operator of l.Then Let c 2 = 0. Without loss of generality, we can assume that c 2 = −1.Then We consider the following linear equation The function y ∈ L 1 is called a solution of (2.6), if there exists a sequence {y n } ∞ n=1 ⊂ C (2) Lemma 2.3.Let r 1 be a continuously differentiable function such that r 1 ≥ δ 1 Assume q 1 is a continuous function and γ q 1 ,r 1 < ∞.Then for every f ∈ L 1 , the equation (2.6) has a unique solution y such that where c 4 depends only on γ q 1 ,r 1 .
L will denote the closure in L 1 of the operator L 0 y := −y + r 1 y + q 1 y, D(L 0 ) = C (2) If the conditions of Lemma 2.3 hold, then the operator L is separable in L 1 .

Proof of the main theorem
Let C(R) be the space of bounded continuous functions on R with the norm y C(R) = sup x∈R |y(x)|.Let ε and A be positive numbers.Set Let v ∈ S A .L v,ε denote the closure in L 1 of the following linear differential expression We consider the equation L v,ε y = f (x).
(3.1) r1,ε,v (x) := r(x, v(x)) + ε 1 + x 2 and qv (x) := q(x, v(x)) satisfy all of the conditions of Lemma 2.3.Indeed, by (1.3), r1,ε,v (x) Therefore, for any f ∈ L 1 , the equation (3.1) has a unique solution y and where C 2 does not depend on A. By 2.1, we have that By using (3.2), (3.3) and Theorem 1 given in Chapter 3 of [18], we obtain that where C 5 also does not depend on A. (3.4), the operator P ε maps S A into itself.Moreover, the operator P ε maps S A to the set . Indeed, let γ > 0, then by (3.4) there exists l ∈ N such that for any < γ/2, (3.5) and sup x:|x|≥l We denote T l = {ϕ l z : z ∈ Q A }.By (3.5) and (3.6), T l is a γ-net of Q A .On the other hand, T l is a subset of the Sobolev space Notice that the embedding of [19,27]).So T l is a compact γ-net of Q A .By Hausdorff's theorem (see [10, Next, we show that the operator P ε is continuous on S A .Let {v n } ∞ n=1 ⊂ S A be a sequence such that sup x∈R |v n (x) − v(x)| → 0 as n → +∞.If y n (n = 1, 2, . ..) and y satisfy Since functions v and v n (n = 1, 2, . ..) are continuous, we see that r(x, v n (x)) − r(x, v(x)) and q(x, v n (x)) − q(x, v(x)) are continuous functions.Therefore, from (3.8) it follows that as n → ∞, for every a > 0. On the other hand, by (3.4), we have Since the operator L −1 v,ε is closed, by (3.9) and (3.10), we obtain that z = y.Thus P ε is continuous.
So P ε is a completely continuous operator in C(R) and it maps the ball S A into itself.By Schauder's theorem (see [10, Chapter XVI]), P ε has a fixed point y in S A , i.e.P ε (y) = y.And y satisfies the equality −y + r(x, y) + ε(1 + x 2 ) y + q(x, y)y = f (x).

y j 1 ≤Remark 3 . 1 .
C 5 f 1 .(3.11)    Let (a, b) be an arbitrary finite interval.It is known that the space W 2 1 (a, b) is compactly embedded to L 1 (a, b).Therefore, by virtue of (3.11), we can select a subsequence ỹj∞ j=1 of y j ∞ j=1 ⊂ W 2 1 (a, b) such that ỹj − y L 1 (a, b) → 0 as j → ∞.By Definition 1.1, y is a solution of equation(1.1).By Lemma 2.3, we obtain that for y the estimate (1.5) holds.The condition (1.3) is natural.If (1.3) does not hold, from Lemma 2.1 it follows that the domain D(L) of L is not included in L 1 .