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Volume 6, Number 2 (2015), 63 – 74

SOME INEQUALITIES FOR SECOND ORDER DIFFERENTIAL OPERATORS WITH UNBOUNDED DRIFT

K.N. Ospanov, R.D. Akhmetkaliyeva Communicated by I.N. Parasidis

Key words: second order differential equation, drift, invertibility, separability AMS Mathematics Subject Classification: 34A30.

Abstract. We study coercive estimates for some second-order degenerate and damped differential operators with unbounded coefficients. We also establish the conditions for invertibility of these operators.

1 Introduction

For the Sturm-Liouville operator l0y = −y00 +q(x)y (x ∈ R), coercive estimates and other properties associated which Sobolev spaces are well known (see [1, 3, 4, 15]).

Properties of the operator ly = −y00 +ry0 + qy with the intermediate coefficient r subordinated to the potential q in some sense, are studied in [5, 9].

In this work, we consider the minimal closed differential operator Ly=−ρ(x)(ρ(x)y0)0 +r(x)y0+q(x)y

in L2(R), where ρ, r are continuously differentiable functions, and q is a continuous function. We do not assume that ρ, r, q are bounded in R. The aim of this work is to show that the operator L is continuously invertible when these coefficients satisfy some conditions and to obtain the following estimate for y∈D(L)

k−ρ(ρy0)0k2+kry0k2+kqyk2 ≤CkLyk2, (1.1) where D(L) is the domain of L, k · k2 is the norm in L2(R), and C independent ofy.

Estimate (1.1) already implies that the domain of L coincides with the subspace generated by the norm k−ρ(ρy0)0k2 +kry0k2 + kqyk2. This fact enables us to use the methods of the embedding theory of weighted Sobolev spaces for studying many important properties (for example, regularity, spectral or approximation properties) of L (see [8, 12, 13, 16]).

The operatorL has numerous applications in mathematical physics and stochastic processes. For example, in the theory of Brownian motion the Ornstein - Uhlenbek operator is used (see [10]), which is an operator of type L, and the Fokker - Plank and Kramer differential operators are generalizations of the Ornstein-Uhlenbek operator.

The Ornstein-Uhlenbek operator was studied in works of M. Smoluchowski, A. Fokker,

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M. Plank, H.C. Burger, R. Furth, L. Zernike, S. Goudsmitt, M.C. Wang (see [20] and the references therein). On the other hand, the operator L is used to describe the problem of the propagation of small oscillations in a viscoelastic compressible medium [17, 19]. Also, the operatorLis used in the study of the vibrational motion in mediums with resistance, where the resistance depends on the velocity [18].

Recently in works of J. Pruss, R. Shnaubelt, A. Rhandi, G. Da Prato, V. Vespri, P. Clement, G. Metafune, D. Pallara, M. Hieber, L. Lorenzi and others the following Ornstein-Uhlenbek-type operator

A0u=−div(a∇u) +F · ∇u−V u

was investigated with various properties (see [2] and references therein). In this works are imposed the additional conditions which are sufficient to control the drift term F · ∇u by−div(a∇u) and V u.

The results of the present paper show that if the intermediate coefficientris quickly growing, then the one dimensional operator L is invertible and has regular properties.

Estimate (1.1) is useful for evolutionary partial differential equations associated with the operator L (see [7]).

The paper is organized as follows. In Section 2 we prove several auxiliary statements and the invertibility of the operator

ly =−ρ(ρy0)0+ry0

for a certain class of ρ and r. In Section 3 we prove inequality (1.1) under some additional conditions. We present some examples in Section 4.

Inequality (1.1) for operatorl in the case ρ= 1 was obtained in [11]. The coercive estimate of Lin L1(R)was proved in [14].

We denote by C(R) the class of the continuous functions, and by C(s)(R) (s = 1, 2, ...) the class of all s times continuously differentiable functions and by C0(s)(R) (s= 1, 2, ...) the subset of all compactly supported functions in C(s)(R).

2 Auxiliary statements and existence of the resolvent for a de- generate operator

Denote by l the closure in L2(R)of the differential expression l0y=−ρ(ρy0)0+ry0

on C0(2)(R), where ρ ∈ C(1)(R), r ∈ C(R). The operator l is a degenerate operator, since it does not have the lower-order term. The domainD(l)is contained in the space L2(R) only in the case when the functions ρ and r satisfy some additional conditions.

In this section, we give some sufficient conditions for bounded invertibility of the operator l. We denote

αg,h(t) = kgkL

2(0, t)

h−1

L2(t,+∞)(t >0), βg,h(τ) =kgkL

2(τ,0)

h−1

L2(−∞, τ)(τ <0),

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γg,h = max

sup

t>0

αg,h(t), sup

τ <0

βg,h(τ)

, where g and h are given functions.

Lemma 2.1. [11]. Let g and h be continuous functions on R and γg,h < ∞. Then for any y∈C0(1)(R) the following inequality holds:

Z

−∞

|g(x)y(x)|2dx≤c1 Z

−∞

|h(x)y0(x)|2dx.

Moreover, the least such constant c1 satisfies γg,h ≤c1 ≤2γg,h.

Lemma 2.2. Let ρ∈C(1)(R) and r∈C(R) satisfy the following conditions

r≥1, γ1,r<∞. (2.1)

Then for y∈D(l) the following estimate holds:

√ry0

2+kyk2 ≤ 1 +p

1,r

√1 rly

2

. (2.2)

Proof. Let y∈C0(2)(R). Integrating by parts, we have (ly, y0) =

Z

R

r(x)(y0)2dx. (2.3)

By H¨older’s inequality,

|(L0y, y0)| ≤

√1 rL0y

2

√ry0

2. (2.4)

Since r ≥1, from (2.3) and (2.4) it follows that

√ry0 2

√1 rL0y

2

. (2.5)

On the other hand, using Lemma 2.1, we get kyk2 ≤2γ1,r

√ry0 2. Then

√ry0

2+kyk2 ≤ 1 + 2γ1,r

√ry0 2. So, using (2.5) we obtain that (2.2) holds for any y∈C0(2)(R).

Let y ∈ D(l). Then there exists a sequence {yn}n=1 ⊂ C0(2)(R) such that kyn−yk2 → 0, klyn−lyk2 → 0 as n → ∞. Since (2.2) holds for all yn (n ∈ N).

Then passing to limit as n→ ∞ we obtain the desired estimate fory ∈D(l).

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Theorem 2.1. Let r∈C(R), ρ∈C(1)(R) be such that

r≥ρ2, γ1,r<∞ (2.6)

and for some N >0 the following inequality holds 1≤ρ(x)≤c2 1 +x2N

. (2.7)

Then the operatorl is invertible and the inverse operatorl−1 is defined on the whole L2(R).

Proof. Inequality (2.2) implies that the inverse l−1 exists. It suffices to show that R(l) = L2(R). Assume that R(l) 6= L2(R). Then there exists a non-zero element v ∈L2(R) such thatv⊥R(l). It follows that

lv ≡ (ρ(ρ v)0)0+ (rv)0 = 0, where l is the adjoint operator of l. Put ρv =z, then

ρz0+ r ρz

0

= 0, or

zexp

Z x

a

r(t) ρ2(t)dt

0

= c ρexp

Z x

a

r(t) ρ2(t)dt

, where cis a constant.

Ifc6= 0, then we can assume that c=−1. Inequalities (2.6), (2.7) imply that

z(x) exp Z x

a

r(t) ρ2(t)dt

0

≤c1 <0, x∈(a, +∞).

Hence (2.6) and (2.7) imply that v /∈L2(R).

Ifc= 0, then we have

v = c2 ρ(x)exp

− Z x

a

r(t) ρ2(t)dt

.

By (2.7), there exists x0 < a such that |v(x)| ≥δ >0 for any x≤x0. So v /∈ L2(R).

Hence, we obtained a contradiction. ThusR(l) = L2(R).

Definition 1. l is called separable in L2(R), if there exists c >0 such that

kρ(ρy0)0k2+kry0k2 ≤c3 klyk2 (2.8) for all y∈D(l).

Putρy0 =z. Then

ly =−ρz0+ r ρz.

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Let λ≥0, and ρ be a bounded function. We defineKλ : L2(R)→L2(R)as follows:

Kλz =−z0+ r

ρ2

z, z ∈D(Kλ),

where D(Kλ) is the domain of Kλ. Note that Kλ is separable in L2(R), if for some c4 >0,

kz0k2+

r ρ2

z

2

≤c4kKλzk2 for all z ∈D(Kλ).

Lemma 2.3. Let ρ∈C(1)(R), 1≤ρ≤ s, r∈ C(R) satisfy (2.2). Then l is separable in L2(R) if and only if

Kλz =−z0 + r

ρ2

z is separable in L2(R) for some λ≥0.

Proof. Assume that l is separable in L2(R). Put ρy0 =z. Then k −ρz0k2+

r ρz

2

≤c5−1K0zk2. Hence,

k −z0k2+

r ρ2z

2

≤c5kK0zk2. (2.9)

It is easy to check that for any z∈D(Kλ) the following estimate holds:

r r ρ2 +λ z

2

1 qr

ρ2 +λ Kλz

2

. (2.10)

Therefore,

1 s2

kzk2 ≤ kKλzk2, z ∈D(Kλ). (2.11) By (2.9) and (2.11), we have that

k −z0k2+

r ρ2

z

2

≤c5kK0zk2+λkzk2 ≤(c5+ 2)kKλzk2. (2.12) So, Kλ is separable in L2(R).

LetKλ be separable in L2(R), i.e.

k −z0k2+

r ρ2

z

2

≤c6kKλzk2, z ∈D(Kλ).

By (2.11), we obtain that

kKλzk2 ≤ kK0zk2+ λ

λ+ 1/s2kKλzk2,

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hence

kKλzk2 ≤(s2λ+ 1)kK0zk2. So, it follows that

k −ρz0k2+ r ρz

2

≤s

k −z0k2+

r ρ2z

2

≤c6kKλzk2+λkzk2

≤(c6+ 1)kKλzk2 ≤2c6(s2λ+ 1)kK0zk2. Taking z/ρ=y0,we get that

k −ρ(ρy0)0k2+kry0k2 ≤c7klyk2.

Lemma 2.4. Let ρ∈C(1)(R), 1≤ρ≤s and r ∈C(R). Suppose that sup

|x−η|≤2

r(x)

r(η) <∞ (2.13)

and condition (2.2) hold. Then l is separable in L2(R).

Proof. By Lemma 2.3, it is enough to prove that Kλ is separable in L2(R) for some λ ≥0.

Theorem 2.1 implies that Kλ is continuously invertible on L2(R) for all λ ≥ 0.

Next, we show a useful representation of Kλ−1. Let ∆j = (j −1, j+ 1) (j ∈Z), and {ϕj}+∞j=−∞ be a sequence in C0(∆j) such that

0≤ϕj ≤1, ϕ0j(x)

≤m (j ∈Z),

+∞

X

j=−∞

ϕ2j(x) = 1.

We extend the restriction of r(x)ρ−2(x) to the interval ∆j to R as a piecewise continuous function ψj(x) with period 2. Let Kλ,j be the closure in L2(∆j) of the differential operator −z0 + (ψj(x) +λ) z on C0(1)(∆j). Similarly to (2.10), we obtain that

j+λ z 2,j

1

j +λKλ, jz 2,j

, z ∈C0(1)(∆j), j ∈Z. Hence,

1 s2

kzk2,

j ≤ kKλ,jzk2,

j, z ∈D(Kλ,j), j ∈Z. (2.14) So, Kλ,j−1 exists. On the other hand, by Theorem 2.1, Kλ,j−1 is defined on the whole L2(∆j).

DefineBλ and Mλ as follows:

Bλf =

+∞

X

j=−∞

ϕ0j(x)Kλ,j−1ϕjf, Mλf =

+∞

X

j=−∞

ϕj(x)Kλ,j−1ϕjf, f ∈L2(R).

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Since suppϕj ⊂ ∆j−1S

jS

j+1 (j ∈ Z), at each point x ∈ R the sums of the right-hand side ofBλ and Mλ contain no more than two summands, soBλ and Mλ are well-defined on the whole L2(R). Moreover, it is clear that

KλMλ =E−Bλ. (2.15)

Notice that in (j, j+ 1) (j ∈Z)only the functions ϕj and ϕj+1 are not equal to zero.

So, we have that kBλfk22 =

+∞

X

j=−∞

ϕ0j(x)Kλ,j−1ϕjf

2

2

= Z

−∞

+∞

X

j=−∞

ϕ0j(x)Kλ,j−1ϕjf

2

dx

=

+∞

X

i=−∞

Z i+1

i

+∞

X

j=−∞

ϕ0j(x)

Kλ,j−1jf) (x)

!2

dx

=

+∞

X

i=−∞

Z i+1

i

0i|

Kλ,i−1kf) +

ϕ0i+1

Kλ,i+1−1i+1f)

2

dx

≤2

+∞

X

i=−∞

Z

i

0i|2

Kλ,i−1if)

2dx+ Z

i+1

ϕ0i+1

2

Kλ,i+1−1i+1f)

2dx

= 4

+∞

X

i=−∞

Z

i

0i(x)|2

Kλ,i−1if) (x)

2dx.

Furthermore

kBλfk22 ≤4m2

+∞

X

j=−∞

Kλ,j−1

2

L2(∆j)→L2(∆j)jfk22,

j

≤8m2 sup

j∈Z

Kλ,j−1

2

L2(∆j)→L2(∆j)

Z

R

X

j

ϕ2j

!

|f|2dx

= 8m2sup

j∈Z

Kλ,j−1

2

L2(∆j)→L2(∆j)kfk22. By inequality (2.14),

Kλ,j−1

L2(∆j)→L2(∆j)≤ s2 1 +s2λ. Thus kBλfk22

2ms2

1+s2λ kfk2, f ∈L2(R). Let λ0 = (4√

2ms2−1)s−2. Then kBλkL

2(R)→L2(R) ≤ 1 2

holds for any λ ≥λ0. So, E+Bλ (λ≥λ0) is invertible. By (2.15), we get

Kλ−1 =Mλ(E−Bλ)−1, λ≥λ0. (2.16)

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Now, we can prove (2.8). Let m1 = sup

|x−η|≤2 r(x)

r(η). By (2.16) and the properties of ϕj(j ∈Z), we obtain that

r ρ2

Kλ−1f

2

≤4√

2 (m1s2+ 1)kfk2.

Then, for λ≥(4√

2ms2−1)s−2, we have that kz0k2+

r ρ2

z

2

≤(1 + 4√

2 + 4√

2m1s2)kKλzk2. (2.17) Put z =ρy0. By (2.17), we get that

ρ(ρy0)0

2 +kry0k2 ≤8√

2ms(1 + 4√

2 + 4√

2m1s2)klyk2, y ∈D(l), (2.18) hence l is separable.

3 Separability of the damped differential operator

Denote by L the closure in L2(R) of the differential expression Ly˜ =−ρ(ρy0)0 +ry0+qy

on C0(2)(R), where ρ is a continuously differentiable function, r and q are continuous functions.

Theorem 3.1. Let ρ be a bounded continuously differentiable function, r and q be continuous functions. Suppose that ρ≥1, r and q satisfy conditions (2.2), (2.17) and γq,r < ∞. Then L is continuously invertible, and L−1 is defined on the whole L2(R).

Furthermore, there exists c8 such that

k−ρ(ρy0)0k2+kry0k2+kqyk2 ≤c8kLyk2, (3.1) for any y∈D(L).

Proof. We consider the equation

Ly=f. (3.2)

A function y ∈ L2(R) is called a solution to (3.2), if there is a sequence {yn}+∞n=1 ⊂ C0(2)(R) such that kyn−yk2 → 0, kLyn−fk2 → 0 (n → +∞). It is clear that L is continuously invertible if and only if there exists a unique solution y to (3.2) for each f ∈L2(R). Putting x=at (a >0), we rewrite (3.2) in the following form:

−˜ρ(t)( ˜ρ(t)˜yt0)0t+ 1/α˜r(t)˜yt0 + 1/α2q(t)˜˜ y= ˜f , (3.3) where

˜

y(t) =y(at),ρ(t) =˜ ρ(at),r(t) =˜ r(at),q(t) =˜ q(at),f(t) =˜ f(at)/a2.

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Let

ˆl0y˜=−˜ρ(t)( ˜ρ(t)˜yt0)0t+ ˜r/a y˜t0, then from (3.3) we obtain

ˆl0y˜+ ˜q(t)/a2y˜= ˜f(t). (3.4) Note that r/a˜ satisfies the conditions of Lemma 2.3, so the operatorˆl0 is continuously invertible. By (2.18),

−˜ρ(t)( ˜ρ(t)˜yt0)0t 2

+ ˜r/a y˜0t

2

≤T

ˆl02

,∀y˜∈D(ˆl0), (3.5) where T = 8√

2ms(1 + 4√

2 + 4√

2m1s2).

It is clear thatγ˜q,˜r = 1/a γq,r. By Lemma 2.1 and (3.5),

1 a2q˜˜y

2

≤ 2γq,˜˜r

a2 k˜ry˜0k2 ≤2γq,ra−2 1 ar˜y˜0

2

≤ 2T γq,r a2

ˆl02. Choose a= 2p

T γq,r, then

1 a2q˜˜y

2

≤ 1 2

ˆl02

. (3.6)

By Theorem 1.16 in Chapter IV of [6], ˆl0+a121(t)E is invertible and R

ˆl0+a121E

= L2(R). Let y˜ be a solution to (3.4). Then, by (3.5) and (3.6), we get that

−ρ(t)( ˜˜ ρ(t)˜y0)0 2

+ 1 ar˜y˜0

2

+

1 a2q˜y˜

2

T

1 + 2γq,r a2

ˆl0

2. (3.7)

On the other hand,

ˆl02

ˆl0 + 1

a2qE˜

˜ y 2

+

1 a2q˜˜y

2

. (3.8)

Using (3.4) and (3.6), we obtain that

1 a2q˜˜y

2

ˆl0+ 1

a2qE˜

˜ y 2

, and

ˆl02

ˆl0+ 1

a2qE˜

˜ y 2

+

1 a2q˜y˜

2

≤2

ˆl0+ 1

a2qE˜

˜ y 2

. (3.9)

So, (3.7) and (3.9) imply that the inequality

−ρ( ˜˜ρy˜0)0 2+

1 ar˜y˜0

2

+

1 a2q˜y˜

2

≤2

T

1 + 2γq,r

a2

2

holds for any solution y˜to (3.4). Lett =x/a. Rewriting the above formula, we obtain (3.1).

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4 Examples

1. Let L0y = (1 +x2) ((1 +x2)y0)0+ (5 +x4)y0. Then all conditions of Theorem 2.1 are satisfied. Hence, L0 is invertible, and L−10 is continuous.

2. We consider

Ly =−y00+ (1 +x2)ωy0+|x|σy,

whereω >0,σ ≥0. Ifω ≥σ/2 + 3/4, then the conditions of Theorem 3.1 are satisfied.

So Lhas a bounded inverse L−1, and there exists c9 >0such that ky00k2+

1 +x2ω

y0

2+k|x|σyk2 ≤c9kLyk2 for all y∈D(L).

3. By Theorem 3.1, Ly˜ =−y00+ exp(1 +x2)y0+ exp|x|y is continuously invertible on L2(R). Moreover, for all y∈D( ˜L),

ky00k2+

exp(1 +x2)y0

2+kexp|x|yk2 ≤c10

Ly˜ 2, where c10 is independent of y.

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Kordan Nauryzkhanovich Ospanov, Raya Duisenbekovna Akhmetkaliyeva Department of Mechanics and Mathematics

L.N. Gumilyov Eurasian National University 13 Kazhymukan St,

010008 Astana, Kazakhstan

E-mails: [email protected], [email protected]

Received: 22.04.2015

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