CC BY
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MSC 35J20, 35J40 DOI: 10.14529/mmp 170105
NEW RESULTS ON COMPLETE ELLIPTIC EQUATIONS WITH ROBIN BOUNDARY COEFFICIENT-OPERATOR CONDITIONS IN NON COMMUTATIVE CASE
M. Cheggag, Ecole Nationale Polytechnique d'Oran, Oran, Algérie, mustapha.cheggag@enp-oran.dz,
A. Favini, Dipartimento di Matematica, Université degli Studi di Bologna, Bologna, Italia, favini@dm.unibo.it,
R. Labbas, Normandie Univ, Le Havre, France, rabah.labbas@univ-lehavre.fr, S. Maingot, Normandie Univ, Le Havre, France, stephane.maingot@univ-lehavre.fr, Kh. Ould Melha, Laboratoire de Mathématiques Pures et Appliquées, Université Abdelhamid Ibn Badis, Mostaganem, Algérie, ould_melha_khel@yahoo.fr
In this paper, we prove some new results on operational second order differential equations of elliptic type with general Robin boundary conditions in a non-commutative framework. The study is performed when the second member belongs to a Sobolev space. Existence, uniqueness and optimal regularity of the classical solution are proved using interpolation theory and results on the class of operators with bounded imaginary powers. We also give an example to which our theory applies. This paper improves naturally the ones studied in the commutative case by M. Cheggag, A. Favini, R. Labbas, S. Maingot and A. Medeghri: in fact, introducing some operational commutator, we generalize the representation formula of the solution given in the commutative case and prove that this representation has the desired regularity.
Keywords: second-order elliptic differential equations; Robin boundary conditions in non commutative cases; analytic semigroup; maximal regularity.
Dedicated, to Professor Angelo Favini on his 70th birthday.
Introduction and Hypotheses
Let A, B, H be closed linear operators in a complex Banach space X, f E Lp (0,1; X), 1 < p < to and d0, ui E X. A recent paper [1] studies the problem
f u''(x) + 2Bu'(x) + Au(x) = f (x), x E (0,1), m
\ u'(0) - Hu(0) = do, u(1) = ui, ^
A, B, H
The authors have written the above equation in the following form
f u''(x) + (L - M)u'(x) - LMu(x) = f (x), x E (0,1) , , ,
\ u'(0) - Hu(0) = d0,u(1) = uu ^ '
where L = B — VB2 — A and M = —B — VB2 — A under some appropriated hypotheses
LM = ML
In this paper, in order to solve a larger class of boundary value problems, we want to drop the hypothesis LM = ML To this end, we consider some operators , depending on a parameter u > 0 and study the second order operational differential equation
1
u"(x) + (Lu - ) U (x) - ^ (Lu+ ) u(x) = f (x), x E (0,1) together with the abstract boundary conditions of Robin's type
{
u'(0) - Hu(0) = d0, u(1) = Ui.
(3)
(4)
The idea to consider (3) is explained in [2,3]. Here Lu, MM are two closed linear operators in X depending on parameter u.
We seek for a classical solution to (3), (4), that is a function u such that
' u G W2p (0, 1; X) , ж ^ (LJМш) u(x),x ^ (МшЬш) u(x) G Lp (0,1; X) x ^ L - mJ) u' (x) G LP (0,1; X) , u(0) G D(H), u
(5)
Generally, more regularity is required for f to obtain a classical solution, unless X has some particular geometrical properties. This is why we assume in all this study that
X is a UMD space.
Our hypotheses with respect to operators H, L^ and MM are the following:
f 3C0 > 0, 3u0 > 0 : Vu > u0, ]0, +^[C p (Lw) n p M) ,
(6)
ker (Lj) = ker (Мш) = {0} ,R (Lu) = R (Мш) = X,
< C0 and sup IIА (Мш — XI)
(7)
sup || А (Lu — XI)
A>0
i
i
L(X)
A>0
\C(X )
< Co,
(here p(P), ker (P) ,R (P) denote respectively the resolvent set, the kernel and the range of operator P) and
30i,02 G]0, f [, 3C ^ 1 : Vw ^ wo, Vs G R, (—Lj)is, (—MJ)is G L(X),
(—Lj )
L(X )
^ CJeeils and
(—Мш )
L(X )
(8)
Assumptions (7) and (8) involve that, for u > u0, -L^d — MM belong to the class BIP(X, 9) [4, Definition 1, p. 431]. Since 9i,92 G]0, |[, L^ and MM generate bounded analytic semigroups in X, ?>0 , ?>0 [4, p. 437].
For any u > u0, we assume also
d(lj ) = d(mj ),
d((lj + mj)2) с d((mj — lj)2),
(9) (10)
0 G p (Lu + Mu). (11)
Set
Au :=(MW - H) + eL- eM (Lu + H) , (12)
(which will be, in some sense, the abstract determinant of our problem), then we suppose that
and Set
Vw ^ wo, 0 G p (Au), (13)
V£ G DL), Vw > wo, A-£ G D((LU + Mu)2). (14)
Cl„M = MLu - L^Mu) (Lw + Mu)_2 = M; Lu] (Lw + Mu)_2 , (15)
then we will see that, under our hypotheses, CL^>M„ G L(X). We assume in all this work, the following non-commutativity hypothesis
Vw ^ w0, \\CL„,M„\\c{X) ^ X (w) ' (16)
where
X : [w0, R+ is such that lim x (w) = 0. (17)
In many concrete cases, the function x has the following form
C
X (w) = — for w large enough, with C,a > 0.
wa
The commutator defined in (15) was used for the first time by Favini et al. [2]. Remark 1. From assumption (11), we deduce that (Lu + Mu)2 is closed.
Lu Mu
et al. [1,5].
Remark 3. Assume (6)^(16). We then will see in Lemma 1 below, that
D (LuMu) n D (MuLu) = D ((Lu + Mu)2) .
Suppose that Problem (3), (4) has a classical solution u. Then, from above u G W2p (0, 1; X) n Lp (0, 1; D((Lu + Mu)2)) ,
which implies
u (0) , u (1) G (D((Lu + Mu )2),X) = (X,D((Lu + Mu )2))^ , (18)
2p ' p 2p,p
(see [6]).
Remark 4. By the well known reiteration theorem, we have
(X,D((Lu + Mu f))_ x p = (X,D(Lu + Mu ))2_ = (X,D(Lu ))1+1_ P,p
[<P G D(Lu) : Lu0 G (X,D(Lu)): _p^}
and in particular , s , s , N
u (0) ,u (1) G D(LU) = D(MW). (19)
The main result in this work is
Theorem 1. Assume (6)^(16). Let f G Lp (0,1; X) with 1 < p < to. Then, there exists u* > u0 such that, for all u > u*, the two following assertions are equivalent. 1. Problem (3), (4) has a unique classical solution.
2. ui, Л-1 do G {X,D((L„ + Ыш)2)) , . (20)
1 2p
The plan of this paper is as follows. Section 1 contains some technical lemmas. In Section 2, we use the representation formula of the solution given in the commutative case and some heuristic consideration to obtain an integral equation verified by the eventual classical solution u = (Lu + Mu)-2v of Problem (3), (4). This integral equation is written in the form v + (v) = (f) + , where , and depend on and Mu. Section 3 contains the study of (f) + rw and Ru, which will allow us to write
u = L + Mu)-2 ((/ + Ru)-1 F(f) + rw)) , (21)
u.
representation (21). Section 5 is devoted to some comparison with a recent paper [1]. In fact, we obtain, in the case when = Lu, a result concerning Problem (2)
which generalizes the one given in [1] (see Corollary 1). Finally, in Section 6, we give an application to which our results apply.
1. Technical Lemmas
In the following Lemmas, we do not need the dependence of w, so we drop ш in the notations and denote , Ыш, Лш by L, M, Л. We omit some proofs; some are obvious, others can be found in Favini et al. [2, Lemmas 2.1 and 2.2].
Lemma 1. Assume (7)^ (10). Then
1. D(LM) = D(M2) and D(ML) = D(L2).
2. D((L + M)2) = D(L2) П D(M2) = D(LM) П D(ML).
3. D((L - M)2 - (L + M)2) = D(LM) П D(ML) = D((L + M)2).
1. See [2, Lemma 2.2, p. 1499].
2. The proof can be found in [7]. We give the details for the commodity of the readers. Let ф G D((L + M)2). There exists £ G X such that
ф = (L + M )-2
so
Mф = M (L + M)-2 £ = 2 (I - (L + M - 2M) (L + M)-1) (L + M)-1 £ = 2 (I - (L - M) (L + M)-1) (L + M)-1 £ = 1 (L + M)-1 £ - 1 (L - M) (L + M)-2 £,
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but (L + M) 1 e G D(M) and, due to (10),
(L - M) (L + M)_2 e G D(M).
Finally M0 G D(M) that is 0 G D(M2). By exchange of the roles of L and M, we get also 0 G D(L2).
Conversely, let 0 G D(L2) n D(M2); then
L0 G D(L) = D(L + M) and M0 G D(M) = D(L + M),
so (L + M)0 G D(L + M), that is 0 G D((L + M)2). 3. It is enough to combine (10) and statement 2.
Lemma 2. Assume (6)^(15). For any operator S, S G {L, M,L + M,L — M} , we have
1. S (L — I)_1, S (L + M)_1 G L(X),
2. S (L — M) (L + M)_2 ,SS (L + M)_2 gL(X),
3. (L + H) A_1, (M + H) A_1 G L(X).
The proof of the previous lemma is based on assumptions (9)~(11) and the following S X W
G L(X) with W(X) c D(S) then SW G L(X).
Lemma 3. Assume (6)^ (13). Then
1. (L ± M)2 = L2 ± LM ± ML + M2 and for0 G D ((L + M)2) we have
(L ± M)2 0 = L20 ± LM0 ± ML0 + M20,
2. LM + ML = 2 ((L + M)2 — (L — M)2) ,
5. [(L — M) ; (L + M)_1] = —2 [M; (L + M)_1] = 2 [L; (L + M)_1] , I Cl,M = 2 (L + M) [(L — M); (L + M)_1 ] (L + M)_1 ,
5. (L + H) A_1 = (L + M) A_1 + eLeM (L + H) A_1 — I,
(L + M)_1 M — (L + M)_1 Cl , m (L + M) = M (L + M)_1 and (L + M)_1 L — (L + M)_1 Cl ,M (L + M) = L (L + M)_1,
7. M (L + M)_1 M + 2 (L — M) (L + M)_1 CL ,M (L + M)
= 2M — 2 (L — M) M (L + M)_1 and
L (L + M)_1 L + 2 (L — M) (L + M)_1 CL ,M (L + M)
= 2 L + 2(L — M) M (L + M )_1,
Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming & Computer Software (Bulletin SUSU MMCS), 2017, vol. 10, no. 1, pp. 70-96
8. M + (L - M) M (L + M)-1 - (LM + ML) (L + M)-1 = 0 on D (M)
L - (L - M) L (L + M- (LM + ML) (L + M= 0 on D (L).
Proof. For Statements 1 and 2 see the proof of Lemma 2.2 in [2] and for statement 3 see Lemma 2.1 in [2].
Statement 4. For 0 G D ((L + M)2) we write
{
(L + M) (L - M) ф = L20 - LM0 + ML0 - M20, (L - M) (l + M) ф = L^ + LMф - MLф - M^,
then
(L + M) (L - M) 0 - (L - M) (L + M) 0 = 2 (ML - LM) from which we deduce
(L + M) [(L - M) (L + M)_1 - (L + M)_1 (L - M)] (L + M) (L + M)"2 = 2 (ML - LM) (L + M)"2 .
As we have mentioned it in the hypotheses, it follows that our commutator CL,m is well defined in L(X) and can be expressed as
Clmm = 2 (L + M) [(L - M) ; (L + M)"1] (L + M)-1.
Statement 5. It is enough to write
(L + M) A-1 + eLeM (L + H) A-1 - I
= [(L + M) + eLeM (L + H) - (M - H) - eLeM (L + H)] A-1 = (L + H) A-1. Statement 6. Due to statement 4, we have
(L + M)-1 CLmM (L + M) = 2 [(L - M); (L + M)-1] , (22)
then, using statement 3, we get
(L + M)-1 M - (L + M)-1 CL,M (L + M) = (L + M)-1 M - ± [(L - M); (L + M)-1] = (L + M)-1 M + [M; (L + M)-1] = M (L + M)-1.
The second equality is obtained in the same manner. Statement 7. Due to statement 3 and (22), we have
M (L + M)-1 M + 1 (L - M) (L + M)-1 CL, M (L + M)
= M (L + M)-1 M + 4 (L - M) [(L - M); (L + M)-1]
= M (L + M)-1 M - 1 (L - M) [M; (L + M)-1]
= M (L + M)-1 M - 1 (L - M) M (L + M)-1 + 2 (L - M) (L + M)-1 M = M (L + M)-1 M - 1 (L - M) M (L + M)-1 + 1M - M (L + M)-1 M
= 2M - 2 (L - M) M (L + M)-1.
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Similarly, we obtain the second equality.
Statement 8. On the domain D(L) = D(M), we have
M + (L - M) M (L + M)-1 - (LM + ML) (L + M)-1
M + (L - M) M (L + M)-1 + 1 ((L - M)2 - (L + M)2) (L + M)
□
= M +(L - M) M (L + M)-1 + 1(L - M)2 (L + M)-1 - 2 (L + M)
= (L - M) M (L + M)-1 + 2 (L - M)2 (L + M)-1 - 2 (L + M - 2M)
= 1(L - M) [2M + (L - M)] (L + M)-1 - 2 (L - M)
= 1(L - M)[M + L](L + M)-1 - 2(L - M) = 0.
The second equality is similarly obtained.
We need also the following classical results Lemma 4. Assume (6)^(8). Then, for
f e Lp (0,1, X), 1 <p< +TO and Q G {L, M} ,
we have
1.x ^ Q f0x e(x-s)Qf (s)ds G LP (0,1,X),
2. x ^ Q f1 e(s-x)Qf (s)ds e Lp (0,1, X),
3.x ^ Q f01 e(x+s)Qf (s)ds e Lp (0,1, X),
4. fo1 esQf (s)ds, fO e(1-s)Qf (s)ds e (D (Q) , X)p^ .
For statements 1, 2 and 3, see [8,9, p. 167-168] and also [10, (24), (25) and (26)]. Statement 4 is an easy consequence of statements 1 and 2, we proceed as in Remark 3 by using the fact that
x1
x ^ e(x-s)Qf (s)ds, x ^ e(s-x)Qf (s)ds, ox
belong to W1p (0,1,X) n LP (0,1,D (Q)).
Lemma 5. Assume (6)~(14) and let p G (D (Q) ,X) i p with Q G {L,M}. Then
p 'p
1. (L + M)k-1p e (D (Q) ,X) i p .
p''
2. (L + H)k-1p e (D (Q) ,X) i p .
p 'F
Proof. 1. Set T = (L + M) A 1, then T is a linear continuous operator from X to X. But, due to (14), we obtain
T(D(L)) C D(L) and T G L(D(L),D(L))
D(L) property we get
T gl((X,D(L)) 1 p , (X,D(L))Kp),
see [11, p. 19].
2. Due to statement 5 of Lemma 3, we have
(L + H) A-1 = (L + M) A-1 + eLeM (L + H) A-1 - I,
then, from statement 1, we deduce that (L + H)A-1p G (D (Q) ,X) i p for any p G
p
(D (Q) ,X) i,p .
p
□
2. Representation Formula
Again, in this section, we drop u in the notations. Assume that there exists a classical solution u to (3), (4). We want to find an integral equation satisfied by
v(.) := (L + M)2u(.). (23)
To this end, we begin par recalling that the solution in the commutative case writes in the following form
$ + ^ + d,
where, for x G [0,1]
r-x f'1
$ (x) = (L + M)-1 / e(x-s)M f (s)ds + (L + M)-1 / e(s-x)Lf (s)ds, (24)
Ф(ж) = - (L + M)-1 exM (L + H)A-1eL /'e(1-s)M f (s)ds
Jo
+ (L + M)-1 exM (L + H)A-1 / esL f (s)ds (25)
Jo
- (L + M)-1 e(1-x)LeM (L + H) Л-1 i esL f (s)ds
o
i-1
\l-eM (L + HU-1eLl I ee
(L + M)-1 e(1-x)L [I - eM (L + H) Л-^] / e(1-s)M f (s)ds,
o
and
d (x) = exMЛ-1do - e(1-x)LeMЛ-1do + exMeL (L + H)Л-1щ (26)
+e(1-x)Lu1 - e(1-x)LeMeL (L + H) Л-1 щ.
See Cheggag et al. in [1, p. 63].
o
Now, in our non commutative case, since m satisfies (3), for almost all x E (0,1), we have
i'x 1
$(x) = (L + M)-1 e(x-s)M[u"(s) + (L - M) u'(s) -- (LM + ML) u(s)]ds Jo 2
r i 1
+ (L + M)-1 e(s-x)L[u''(s) + (L - M) u'(s) - - (LM + ML) u(s)]ds.
J x 2
6
Set $ = $i where
i=1
$1(x) = x (L + M )-1 e(x-Jo ~s)M u' '(s)ds,
$2(x) = (L + M)-1 f e(s- J x x)Lu'' (s)ds,
$3(x) = x (L + M )-1 e(x-Jo ~s)M (L - M) u'(s)ds,
$4(x) = (L + M)-1 i e(s- J x ■x)L (L - M) u'(s)ds,
$5(x) = 1x - -2{L + MI e(x-s)M (LM + ML) u(s)ds
$6(x) = 1 r 1 - + M rll e(s-x)L (LM + ML) u(s)ds,
then the main idea is to carry out integration by parts in view to deduce the integral equation satisfied by $. Since u satisfies (5), all the above calculations are justified for almost all x E (0,1). Then
$i(x) = (L + M)-1 f0x e(x-s)Mu''(s)ds = (L + M)-1 u'(x) - (L + M)-1 exMu'(0) + (L + M)-1 Mu(x) - (L + M)-1 MexMu(0) + (L + M)-1 f0x M2e(x-s)Mu(s)ds,
here we have used the fact that, for any g E W1 'p (0,1; X)
x
g(x) - g (0)= g'(s)ds-
Jo
Similarly, we have
$2(x) = (L + M)-1 fx e(s-x)Lu''(s)ds = (L + M)-1 e(1-x)Lu'(1) - (L + M)-1 u'(x) - (L + M)-1 Le(1-x)LU1 + (L + M)-1 Lu(x) + (L + M)-1 £ L2e(s-x)Lu(s)ds.
Due to (19), u (0), U1 E D(L) = D(M); so
(L + M)-1MexMu (0) = (L + M)-1exMMu (0), (L + M )-1Le(1-x)Lu1 = (L + M )-1e(1-x)LLu1.
Jg Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming
& Computer Software (Bulletin SUSU MMCS), 2017, vol. 10, no. 1, pp. 70-96
Since v(.) = (L + M)2u(.) and U (0) = do + Hu (0), we get
(L + M)2 ($i(x) + &2(x)) = v(x) - (L + M) exM (H + M) u (0)
+ (L + M) e(1-x)L (u'(1) - Lui) - (L + M) exMdo (27)
+ (L + M) f0x M2e(x-s)M (L + M)-2 v(s)ds + (L + M) £ L2e(s-x)L (L + M)-2 v(s)ds.
In the same manner, for $3(x) + $4(x), the integration by parts gives
$s(x) + $4(x) = - (L + M)-1 exM (L - M) u(0) + (L + M)-1 e(1-x)L (L - M) u (1) + (L + M)-1 f0x Me(x-s)M (L - M) u(s)ds - (L + M)-1 f1 Le(s-x)L (L - M) u(s)ds,
but
M(L - M)u(x) = M(L - M)(L + M)-2(L + M)2u(x), L(L - M)u(x) = L(L - M)(L + M)-2(L + M)2u(x), so
(L + M) ($3(x) + $4(x)) = -exM (L - M) u (0) + e(1-x)L (L - M) ui + f0x e(x-s)MM (L - M)(L + M)-2v(s)ds - £ e(s-x)LL (L - M) (L + M)-2v(s)ds.
Therefore, applying again L + M, we get
(L + M)2 ($3(x) + $4(x)) - (L + M) exM (L - M) u (0) + (L + M) e(1-x)L (L - M) ui + (L + M) f0x e(x-s)MML (L + M)-2 v(s)ds (28) + (L + M) f1 e(s-x)LLM (L + M)-2 v(s)ds
+ M) f0x e(x-s)M
Finally
- (L + M) f0x e(x-s)MM2 (L + M)-2 v(s)ds - (L + M) ¡1 e(s-x)LL2 (L + M)-2 v(s)ds.
(L + M) (Ф5(x) + Фа(ж)) = -!f0x e(x-s)MLM (L + M)-2 v(s)ds -110x e(x-s)MML (L + M)-2 v(s)ds - 1 ¡1 e(s-x)LLM (L + M)-2 v(s)ds
-1 ¡1 e(s-x)LML (L + M)-2 v(s)ds,
it follows that
(L + M)2 (Ф5(x) + Фб(х)) = -2 (L + M) ¡0x e(x-s)MLM (L + M)-2 v(s)ds
-1 (L + M) ¡0x e(x-s)MML (L + M)-2 v(s)ds (29)
v(
-2 (L + M) x e(s-x)LLM (L + M)-2 v(s)ds
2 ^ 1 ^) ¡0
1 (L + M )fx
2 ( 1 + m 11 ev~ i\/I L ( L + i\/I x 2
-2 (L + M) j'1 e(s-x)LML (L + M)-2 v(s)ds.
Then, using (27), (28) and (29), we have
(L + M)2 Ф(х) = v(x) - (L + M) exM ((H + L) u (0) + do) + (L + M) e(1-x)L (u'(1) - Миг) + 2 (L + M) ¡0x e(x-s)MML (L + M)-2 v(s)ds -1 (L + M) ¡0x e(x-s)MLM (L + M)-2 v(s)ds +1 (L + M) ¡1 e(s-x)LLM (L + M)-2 v(s)ds - \ (L + M) ¡J e(s-x)LML (L + M)-2 v(s)ds.
We deduce the following integral equation
(L + Mf $(x) = v(x) - (L + M) exM [(H + L) u (0) + do]
+ (L + M) e(1 -x)L (u'(1) - Mui) (30)
1 Cx
+ -(L + M) e(x-s)M Cl , m v(s)ds
2 J o
1 1
- -(L + M) e(s-x)LCL, m v(s)ds,
where
Cl,M := [M; L] (L + M)-2 = (ML - LM) (L + M)-2
(L + M)exM (L + H) u (0) - (L + M)e(l-x)Lu'(1)
Concerning by the same method, we obtain the following integral equation (L + M )2^(x)
xM
-(L + M)exM (L + H) A-1eL (L + M) ux -(L + M)exM (L + H) A-1 (I - eLeM) do + (L + M)e(1-x)M (L + H) A-1eL (L + M) u)
+ (L + M)e(1-x)LeM (L + H) A-1 (I - eLeM) do (31)
-(L + M)e(1-x)LLui + (L + M J e(1-x)LeM do - fi --(L + M)exM (L + H)A-1eL e(l-s)MCl,mv(s)ds 2 Jo
-(L + M)exM (L + H)A-1 j esLCL,mv(s)ds
o
1 i'i
+-(L + M)e(1-x)LeM (L + H) A-1 esLCL,mv(s)ds 2o
--(L + M)e(1-x)L [I - eM (L + H) A-1eL] i' e(1-s)MCl,mv(s)ds. 2o
This last equation, together with (30), leads us to set
v + R(v) = F (f ) + r,
where for all x e]0,1[ R(v)(x)
1 l'x 1 I'1
= -(L + M) e(x-s)MCl,mv(s)ds - -(L + M) e(s-x)LCL,mv(s)ds
ox
1 r1
--(L + M)exM (L + H)A-1 esLC1L,mv(s)ds
2 Jo
1 f1 --(L + M)exM (L + H)A-1eL e(l-s)MCl,mv(s)ds (32)
2 Jo
- 1(L + M)e(1-x)L [I - eM (L + H) A-1 eL] f e(1-s)MCl,mv(s)ds 2 Jo
1 i'1 + -(L + M)e(1-x)LeM (L + H) A-1 esLCL,mv(s)ds, 2 Jo
F(f) (х)
i>x r-1
(L + M) e(x-s)M f (s)ds + (L + M) e(s-x)Lf (s)ds
J0 Jx
- (L + M) exM (L + H) k-leL f e(1-s)M f (s)ds
Jo
+ (L + M) exM (L + H)A-1 f esLf (s)ds (33)
Jo
¡л
-1 i „sL .
- (L + M) e(1-x)LeM (L + H) A-1 esLf (s)ds
o
- (L + M) eS1-x)L [I - eM (L + H)A-1eL] f e(l-s)M f (s)ds,
and
Г(х) = (L + M)exM (L + H)Л-1 (I - eLeM) do
xM
+ (L + M) exMdo - (L + M)e{1-x)LeMdo
-(L + M)e(1-x)LeM (L + H) A-1 - eLeM) do (34)
+ (L + M)exM (L + H) A-1eL (L + M) ui
-(L + M)e(l-x)LeM (L + H) A-1eL (L + M) u1
+ (L + M)e(1-x)L (L + M) u1.
Finally, if u is a classical solution of (3), (4), then v(.) := (L + M)2 u(.) satisfies the integral equation
v + R(v) = F (f) + r, (35)
with R, F(f) and r are defined by (32), (33) and (34) respectively. Remark 5. Since
(L + H) A-1 = (L + M) A-1 + eLeM (L + H) A-1 - I, we can write T(x) = R(x) + S(x), where
R(x) = (L + M)exMeLeM (L + H) A-1dt
-(L + M -(L + M + (L + M -(L + M + (L + M -(L + M -(L + M + (L + M
o
exM (L + M)A-1eLeMdo exMeLeM (L + H)A-1eLeMdo exM eLeM do e(1-x)LeM [I + (L + H) A-1 (I - eLeM)] d( exMeLeM (L + H) A-1eL (L + M) щ exMeL (L + M) U1
e(1-x)LeM (L + H) A-1eL (L + M) u1 exM (L + M) A-1 eL (L + M) щ,
o
and
5(x) = (L + M)exM (L + M) A-1 do + (L + M)e(1-x)L (L + M) uv
R will be regular since it corresponds to the terms of r containing eL or eM. Then, to obtain a uniqueness and existence result for our problem, it will be sufficient to study the regularity of R, F(f) and S.
3. Study of F^(f), S^ Ru and Rw
Suppose that u1 E D (L + M) = D (L).
Let us recall the following result (see [12, p. 96]):
Proposition 1. Let Q the infinitesimal generator of an analytic semigroup in X, (exQ)x>0> P e X, 1 < p < to and m E N \ {0} . Then
1. eQp E LP (0,1; X) ,
2. QmeQp E LP (0,1; X) & e;Qp E Wmp (0,1; X) & p E (X, D (Qm)) p .
mp
Set, for a given function g fern (0,1) to X and for x E (0,1)
x1
G(g)(x) = e(x-s)Mg(s)ds, K (g)(x) = e(s-x)Lg(s)ds.
0x
Applying [4], a remark in [13, p. 25] and using the closed graph theorem, we obtain Proposition 2. Assume (6)~(9). Let g E Lp (0,1; X), 1 <p < to. Then
K(g),G(g) E W1 p (0,1; X) n Lp (0,1; D(M)),
and this implies that
G(g) (1), K(g)(0) E (D (M), X)^ = (D (L), X)^ . (36)
pp
Moreover, for a.e. x E (0,1)
G'(g)(x) = MG(g)(x) + g(x) and K'(g)(x) = -LK(g)(x) - g(x),
and there exists C > 0 such that
\\G(g)\\ LP(0,1 ; X) ^ C \\g\\ Lp(0,1;X) , \\K (g)\\ LP(0,1;X) ^ C \\g\\ Lp(0,1;X) , \\x ^ MG(g)(x)\\LP(0,1;X) < C \\g\\LP(0,1;X) , \\x m- LK(g)(x)\\Lp(0,1;X) ^ C \\g\\ LP(0,1;X) ■
Proposition 3. Assume (6)~(9) and (13)~(14). Let g E Lp (0,1; X) with 1 < p < to. Then the following functions
(g) := eM (L + H)A-1K(g)(0),
(g) := eM (L + H) A-1 eLG(g)(1),
(g) := e(1-)LeM (L + H) A-1eLG(g)(1),
(g) : = e(1-)LeM (L + H) A-1K(g)(0),
(g) : = e(1~)LG(g)(1),
belong to Lp (0,1; D(L)) and there exists C > 0, such that
\№i(9)\\LP(0,1;X) < C \\9\\lp(0,1;x) ,i = h 5 ||x ^ MVi(g)(x)\\LP(0,1;X) < C \\9\\lp(0,1;x) ,i = 1 \\x ^ LVi(g)(x)\\LP(0,1;X) < C \\g\\LP(0,1;X) ,i = 4> 5
Proof. We study only V1(g) (the other functions are similarly treated). In the following, C denotes various constants which can depend on p but do not depend on g). Clearly V1(g) E Lp (0,1; X) Mid 1 (g)\LP(0 1 ■ x) < C Mlp{0 1 ■ xy Let us show that ^(g) E LP (0,1; D (M)). We have
(g)(x) = exM (L + H) Л-1 I esLg(s)ds
-1 „sb.
x1
e(x-s)MesM (L + H)Л-1esbg(s)ds + exM (L + H) Л-1 esbg(s)ds,
that is ^ (g) (x) = G (g) (x) + g 1 (g) (x), where
g (s) = e (L + H) Л-1esbg(s) and Ф1 (g) = exM (L + H) Л-1 esbg(s)ds.
x
But g £ Lp (0,1; X), so due to Proposition 2, G(g) £ Lp (0,1; D(M)) and ||x м- MG(g)(x)\\bP(01 ;X) ^ C ||g|bp(01 x) ^ C \\gll bP(0,1 ;X) ■
Now, from Lemma 3, statement 5, we have
(L + H) Л-1 = (L + M) Л-1 + ebeM (L + H) Л-1 - I
and then
(37)
Ф1 (g) (x) = exM (L + M) Л-1 esbg(s)ds
(38)
+ exMebeM (L + H) Л-1 / esbg(s)ds - exM esbg(s)ds
xx
= I1 (x) + I2 (x) + I3 (x). Since eb(X) С D(M) we have MebeM (L + H) Л-1 £ L(X) and we can write
fl\MI2 (x)\\p dx = f 00
exMMebeM (L + H)Л-1exb e(s-x)bg(s)ds
dx
^ C
As-x)b
g(s)ds
dx ^ C \\K(g)\pbp(01.x) < C \g\PbP(01; X)
similarly
¡0 \\MI3 (x) \\p dx = f01 exMM (L - I)-1 (L - I) ¡1 esbg(s)ds
dx
1 1 p / \p
^ C£ (L - I)Ц esbg(s)ds dx ^ C l\\LK(g)\\ bP(0,1 ;X) + \\K(g)\\ ^ C g
LP (0,1; X)
1
0
0
1
1
11
p
1
p
1
0
For h, we use (14) which shows that M (L + M) A 1 (L — I) 1 G L(X) and then
¡0 \\MIi (x)\\p dx = ¡J exMM (L + M) A-1 (L — I)-1 (L — I) ¡1 esLg(s)ds
1 1 p ^ C tf (L — I)H esLg(s)ds dx ^ C Ml^x) .
p
dx
Finally, the previous estimates concerning I1,I^d I3 together with (37) and (38) prove that V1 (g) G Lp (0,1; D(M)) and
\\X ^ MV1(g)(X)\Lp(0,1;X) < C Mlp(0,1;x) •
□
Proposition 4. Assume (6)~(14). L et f G Lp (0,1; X ) , 1 < p < œ. Then
F(f ) G Lp (0,1; X) .
Proof. Using Propositions 2 and 3, we can write
F(f)(.) = (L + M)(M — I)-1 (M — I) G(f ) + (L + M)(L — I)-1 (L — I)[K(f)(.)]
— (L + M)(M — I)-1 (M — I) (f) + (L + M)(M — I)-1 (M — I) [V1 (f) (.)]
— (L + M)(L — I)-1 (L — I) V4 (f) + (L + M)(L — I)-1 (L — I) [V3 (f) (.)]
— (L + M)(L — I)-1 (L — I) [V5 (f)(.)].
But (L + M )(M — I )-1 , (L + M )(L — I )-1 G L(X ). So F (f ) G LP (0,1; X ).
□
Proposition 5. Assume (6)~(14). L et 1 <p< œ. Then R G Lp (0,1; X ). Proof. Set
U :=(L + M) (L — I)-1 G L(X) and V := (L + M) (M — I)-1 G L(X). It is well known that, for any n G N* and any £ G X, we have
eLi G D(Ln)rnià eM£ G D(Mn), thus (M — I)eL, (L — I)eM G L(X) and also, using hypothesis (14), (M — I)(L + M) A-1eL = M (L + M) A-1 (L — I)-1 (L — I) eL — (L + M) A-1eL G L(X). For x G (0,1), we can write
Ri(x) = VexM(M — I)eLeM (L + H)A-1do — VexM(M — I)(L + M)A-1eLeMdo
— VexM(M — I)eLeM (L + H) A-1eLeMdo + VexM]M — I)eLeMdo
— Ue(1-x)L(L — I)eM [I +(L + H) A-1 (I — eLeM)] do
+ VexM(M — I)eLeM (L + H) A-1eL (L + M) m — VexM(M — I)eL (L + M) m
— Ue(1-x)L(L — I)eM (L + H) A-1eL (L + M) m +VexM(M — I)(L + M) A-1eL (L + M) u1.
Then, from Proposition 1, statement 1, we obtain R G Lp (0,1; X). □
Proposition 6. Assume (6)~(13). Let 1 < p < œ. Then S e Lp (0,1; X) if and only if
U\, A-1da e (X,D((L + M)2))^^
2p'
Proof. For x e (0,1), we have
S (x) = (L + M )exM (L + M )A-1da + (L + M )e(i-x)L (L + M ) ui = (L + M)(M - I)-l(M - I)exM (L + M)A-ld0 + (L + M)(L - I)-1 (L - I)e(l-x)L (L + M) ui = Si(x) + S2(x). Since (L + M)(M - I)-1 te invertible in L(X), then Si e Lp (0,1; X) if and only if
x ^ (M - I)exM (L + M) A-ida e Lp (0,1; X) , and this, from Proposition 1, is equivalent to
(L + M) A-ido e (X, D(M))i_i p = (X, D(L + Mi p .
p if pif
Similarly, S2 e Lp (0,1; X) if and only if
(L + M) ui e (X, D(L))i-1 p = (X, D(L + M))-p p .
pp
We have Si e Lp (2, 1; X) and S2 e Lp (0, i; X); then
S e Lp (0,1; X) ^ S e Lp (0, 2; X) and S e Lp (1,1; X) ^ Si e Lp (0,2; X) and S2 e Lp (|, 1; X)
{
(L + M]A-ldo e (X, D(L + M1 p and
(L + M) m e (.X, D(L + M1 p,
p <r
and, by using the reiteration property, we then get
5 e Lp (0,1; X) ^ ui, A-1do e (X, D((L + M)2)) 1-_L p .
Now, we need the dependence of u. Operators R, R, F(f) and S become RM, RM, FM (fj and SM where M and L are replaced by Mw and Lu.
We must estimate \\Ru|| in order to invert I + R^ for u large enough and obtain the
u
u(.) = L + Мш )-2 (I + Ru )-1 ( Fu (f ) + Иш + Бш) (.)
(39)
Proposition 7. Assume (6)~(16). Let 1 < p < Then Ru e L (Lp (0,1; X)) and there exits u* ^ u0 such that, for each u ^ u*, I + Ru is invertible in L (Lp (0,1; X)).
Proof. Let v e Lp (0,1; X) .Since , m„ e L(X ) then CLut m„ v e Lp (0,1; X). Using the notation of Propositions 2 and 3, we can write
2Ru (v)(.) = L + Мш ) Мш - I) -1 ( Мш - I) G (C^,mыv(.))(.)
- L + мш ) LM - I)~ 1 ( Lu - I) K (CL^,mыv(.))(.)
- L + мш ) Мш - I) -1 (Мш - I)Ф1 (clы,m^v(.))(.)
- L + мш ) Мш - I) -1 (Мш - I)^2 (clm^v(.))(.)
+ L + мш ) LM - I)- 1 (Lu - I(clы,m^v(.))(.)
+ L + мш ) - I)- 1 (Lu - I(clмшv(.))(.)
- L + мш ) L„- - I)~ 1 (Lu - I(clмшv(.))(.).
Recall that (Lw + Mw) (Mw - I) 1, - (Lw + Mw) (Lw - I) 1 £ L(X). Then, due to Propositions 2 and 3, we have Rw (v) £ Lp (0,1; X) and there exists a constant b > 0 such that, for every u ^ u0
\\Rw (v)\\lp(0,1;X) < b \\CL„ M\\c{X) M LP(0,1;X) , and by hypothesis (16), we have
\R
w(v)\lp(0,1;x) ^ Cx (u) \\V\lp(0,1;X) '
with lim x (u) = 0. Therefore, there exists u* ^ u0 such that for every u ^ u*
\\Rw\c(LP(0,1;X)) < 1-
So I + Rw is invertible in L(Lp (0,1; X)) for any u ^ u*.
□
4. Proof of Theorem 1
We consider the fixed number u* built above and let u ^ u*.
4.1. Statement 1 Implies Statement 2
Assume statement 1, that is u, defined by (39), is the classical solution of (3), (4). Then
x ^ (Lw + Mw f u(x) = (I + Rw)-1 (f„(f )(x) + Rw(x) + Sw(x)) £ Lp (0,1; X) ,
and we get Fw (f) + Rw + Sw £ Lp (0,1; X). So, by Propositions 4, 5 and 6, statement 2 is satisfied.
4.2. Statement 2 Implies Statement 1
u
of (3), (4). Set
v(x) = (I + Rw)-1 (Fw(f )(x) + Rw(x) + Sw(x)) = (Lw + Mw) u(x).
• First step: u £ Lp (0,1; D((Lw - Mw)2)) . From Proposition 4-7, we get that v £ Lp (0,1; X). Since
(Lw - Mw) (Lw + Mw)-2 £ L(X),
we have
x ^ (Lw - Mw f u(x) = (Lw - Mw f (Lw + Mw)-2 v(x) £ Lp (0,1; X).
• Second step: (Lw - Mw) u' £ Lp (0,1; X). Since v = Fw (f) + rw - Rw (v), we have
u(.) := (Lw + Mw )-2 (Fw (f)(.) + rw (.)) - (Lw + Mw )-2 Rw (v)(.), (40)
gg Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming
& Computer Software (Bulletin SUSU MMCS), 2017, vol. 10, no. 1, pp. 70-96
so, using (32), (33) and (34), we can write и in the following form u = u + и + u where
f u(x) = (Ьш + Ыш)-lG(f)(x) + (Ьш + Ыш)-lK(f)(x)
-2(Lш + мш)-lG (Clu,mиv(.)) (x) + 1(ЬШ + Ыш)-lK (Clu,mиv(.)) (x),
-l„ xMи
u(x) = (Ьш + мш )-lexMu fo,
k u(x) = (Lu + мш )-le(l-x)Lu f
with
fo = - L + H)Л-leLuG(f )(1) + L + H)Л-ш1 K (f )(0)
+ L + H) л-1 [(I - eLueMu) do + eLu (Lu + Мш) ui] + do
+2 L + H) Л-1К (Clu,Muv(.)) (0) + 2(Lu + H) K-leLuG (Clmиv(.)) (1),
and
fi = -eMu L + H)Л-1К (f )(0) + (-eMudo + (Мш + Lu) щ) - [I - eMu L + H)A-leLu] G(f)(1) -eMu L + H) Л-1 )(I - eLueMu) do + eLu (Мш + LM) щ]
+ 1 [I - eMu L + H) k-leLu] G (Clu,Muv(.)) (1) -1 eMu L + H) л-1 К (Clu,Muv(.)) (0).
We have
L + Мш) u(x) = МшG(f )(x) - LMК(f )(x) - 2мшG (Clu,muv(.)) (x)
- 2 Lш K (Clu , mu v(-)) (x) - Clu , mu v (x) ,
but f E Lp (0,1; X) and also v (see first step) so, in virtue of Proposition 2 and Lemma 2, statement 5, we obtain (Lu + Мш) u'(.) E Lp (0,1; X). Since
(Lu - Мш) u'(.) = (Lu - Мш) L + Мш)~l L + Мш) u'(.),
we get
x ^ L - Мш) u!(x) E Lp (0,1; X). (41)
Now in f\, putting together the terms сontaining eMu; we write
fl = (Мш + Lш) ui - G(f )(1) + 2G (Clu,muv(.)) (1) + eMu^, where fil E X. So, from (20) and (36), we deduce that
fl E (В(Мш),X )pp ,
pn
and thus
x ^ (Lw - Mw) u'(x) £ Lp (0,1; X). (42)
It remains to study u. For f0, we use Lemma 3, statement 5; then, we can write
fo = (Lw + Mw)A-1wo + eL-eM- (Lw + H)A-1 (wo + do) + eL-eM-do
- eL- (Lw + Mw) u1 + eL-G(f )(1) - 2 eL-G (Cl-,m-v(.)) (1)
- K (f) (0) - 1K (Cl-,m-v(.)) (0) + (Lw + Mw) A-1 do,
where
wo = -eL-eM-do + eL (Lw + Mw) u1 - eL"G(f )(1)
+K (f) (0) + 1K (Cl-v(.)) (0) + 2eL-G (Cl-,muv(.)) (1).
So
fo = (Lw + Mw) A-1wo - K (f) (0) - 2K (Cl-,m-v(.)) (0) + (Lw + Mw) A-1do + eL-/io,
where ¡io £ X. It's easily seen that wo £ (D (Mw) ,X) 1 p so that
p 'p
fo £ (D(Mw),X ) ip ,
p
and
x ^ (Lw - Mw) u'(x) £ Lp (0,1; X). (43)
Finally, (41), (42) and (43) prove that x ^ (Lw - Mw) u'(x) £ LP (0,1; X). •u
u £ W2'p (0,1; X) since
x ^ u''(x) = - (Lw - Mw) u' (x) + 2 (Lw Mw + Mw Lw) u(x) + f (x) £ Lp (0,1; X).
Moreover, we have for almost all x £ (0,1)
v(x) = (Lw + Mw) G(f )(x) + (Lw + Mw) K(f )(x) (44)
- 2Lw + Mw) G (Cl- ,m- v(.)) (x) + + Mw) K (Cl-,m- v(.)) (x)
+ (Lw + Mw) exM- fo + (Lw + Mw) e(1-x)L- fu
and
(Lw + Mw) u'(x) = MwG(f )(x) - LwK(f )(x) - 2MwG (Cl-,m-v(.)) (x)
-2LwK (Cl-,m-v(.)) (x) + MwexM- fo - Lwe(1-X)L-h - Cl-m-v (x).
Therefore, inserting (44) in (Lw + Mw) u'(x) , and using Lemma 3 , statement 6, we have for almost all x £ (0,1)
u'(x) = Mw (Lw + Mw)-1 G(f)(x) - Lw (Lw + Mw)-1 K(f )(x) (45)
- 2Mw (Lw + Mw)-1 G (Cl-,m-v(.)) (x) - 2Lw (Lw + Mw)-1 K (Cl-,m-v(.)) (x) + Mw (Lw + Mw)-1 exM- fo - Lw (Lw + Mw)-1 e(1-x)L- fv
Then for almost all x £ (0,1)
u''(x) = f (x) + Mw (Lw + Mw)-1 MwG(f)(x) + Lw (Lw + Mw)-1 LwK(f)(x) (46)
- 2Mw (Lw + Mw)-1MwG (Cl-,m-v(.))(x) + 2Lw (Lw + Mw)-1LwK (Cl-,m- v(.))(x) + 2(Lw - Mw) (Lw + Mw)-1 Cl-,M-v(x)
+ Mw (Lw + Mw)-1 MwexM- fo + Lw (Lw + Mw)-1 Lwe(1-x)L- f\, we insert (44) in (46), so we obtain for almost all x £ (0,1)
u''(x) = f (x) + EG(f )(x)+ T1K(f )(x) - 2E1G (Cl-,m-v)(x) (47)
+2T1K (Cl-,m-v) (x) + E1 exM fo + T1e(1-x)Lfh where, on the domain D(Lw + Mw)
E1 = Mw (Lw + Mw)-1 Mw + 2 (Lw - Mw) (Lw + Mw)-1 Cl-,m- (Lw + Mw) = 2Mw - 2(Lw - Mw) Mw (Lw + Mw)-1,
and
T1 = Lw (Lw + Mw)-1 Lw + 2(Lw - Mw) (Lw + Mw)-1 Cl-,m- (Lw + Mw)
= 2Lw + 2 (Lw - Mw) Lw (Lw + Mw)-1,
in virtue of Lemma 3, statement 6. Using (40), (45) and (47), we have for almost all x £ (0,1)
u''(x) + (Lw - Mw )u' (x) - 2 (Lw Mw + Mw Lw) u(x)f (x) + E2G(f )(x) + T2K (f )(x)
2E2G (Cl-,m-v(.))(x) + 2
where
-1E2G (CLыv(.)) (x) + -T2K (CLыv(.)) (x) + E2exM" f0 + T2e(1-x'L fu
E2 = -Мш - 2(Lu - мш) Ыш (Lu + Ыш+ (ЬШ - Ыш) Ыш (Lu + Ыш -2 Lмш + мшLM) (Lu + мш,
and
T2 = 1Lu + - M) Lu (Lu + Mu)-1 - (Lu - Mu) Lu (Lu + Mu)-1
-2 LMu + MMLu) (Lu + Mu)-1. But in virtue of Lemma 3, statement 8, we obtain
E2G(f )(.)= T2K(f )(.) = 0 in Lp (0,1; X) .
We finally get
u''(x) + (LM - Mu)u'(x) - 1 (LMMu + MuLu) u(x) = f (x). We conclude that u, determined by (39), is the classical solution of (3), (4).
5. Go Back to the Commutative Case
This section is devoted to some comparison with the recent paper [1]. We will show that this work improves the results contained in [1]. In fact, instead of considering families (Lu)u^u0 , (Mu)u^u0 of linear operators, we consider L, M two closed linear operators in X such that
D(M) = D(L)
{
LM = ML,
as assumptions (11) and (12) in [1, p. 59]. But we do not assume the commutativities between L and H and between M and H, as (15) and (16) in [1, p. 59]. Note that here, in virtue of (48), we have
A=(M - H) + eLeM(L + H) = (M - H) + eL+M(L + H),
and since CL,M = 0 we have R = 0. Then, by formula (21), the solution of Problem
(49)
{
u''(x) + (L - M)u'(x) - LMu(x) = f (x), x E (0,1) , u'(0) - Hu(0) = d0, u(1) = u1,
is
\-2
u(.) = (L + M) (F(f)(.) + T(.)), (here (I + R)-1 = I). Moreover, for x E [0,1] (L + M)-2F(f) (x)
i>x r-1
= (L + M )-1 e(x-s)M f (s)ds + (L + M )-1 e(s-x)Lf (s)ds
'0
1
- (L + M)-1 exM (L + H) K-1eL e(1-s)M f (s)ds
0
+ (L + M)-1 exM (L + H)A-1 f esLf (s)ds
0
- (L + M)-1 e(1-x)LeM (L + H) A-1 i esLf (s)ds
0
- (L + M)-1 e(1-x)L [I - eM (L + H) A-1eL] i' e(1-s)M f (s)ds,
Jo
and
(L + M)-2r (x) = (L + M)-1exM [(L + H)A-1 + I - (L + H) A-1eL+M] do -(L + M)e(1-x)LeM [(L + H) A-1 + I - (L + H) A-1eL+M] do + (L + M)-1exM (L + H) A-1eL (L + M) u] + (L + M)-1e(1-x)L [I - eM (L + H) A-1 eL] (L + M) u1.
Now, using the fact that
(L + H) A-1 = (L + M) A-1 + eL+M (L + H) A-1 - I,
(see Lemma 3, statement 5), we deduce
eMdo + eM[L + H) A-1do - eM (L + H) A-1eL+Mdo = eMdo + eM [(L + M) A-1 + eL+M (L + H) A-1 - I] do
-eM (L + H)A-leL+Mdo
eM (L + M) A-1do + eMeL+M (L + H) A-1do
-eM (L + H)A-1eL+Md0
eM (L + M) A-1do + eM [eL+M; (L + H) Л-1] do,
and since
(L + H) Л-Чо + do - (L + H) Л-\^мdo
o
= [(L + H) + (M - H) + eL+M (L + H)] Л-^о - (L + H) do
= [(L + M) + eL+M (L + H)] Л-1do - (L + H) Л-1 eL+Mdo = (L + M)Л-1do + [eL+M; (L + H)Л-1] do,
then we obtain
(L + M )-2Г(х) = exMЛ-1do + (L + M)-1exM [eL+M; (L + H) Л-1] do
-e(1-x)LeMЛ-1do - (L + M)-1e(1-x)LeM [eL+M; (L + H) Л-1] do + (L + M)-1exM (L + H) Л-1eL (L + M) щ + (L + M)-1e(1-x)L [I - eM (L + H) Л-1eL] (L + M) u1.
Вестник ЮУрГУ. Серия «Математическое моделирование Q]^
и программирование» (Вестник ЮУрГУ ММП). 2017. Т. 10, № 1. С. 70-96
Finally, the representation solution of Problem (49), for a.e x E (0,1), is u(x)
= exMA-1 do + (L + M)-1 exM (L + H) A-leL (L + M) ui
- (L + M)-1 exM (L + H)A-1eL e(1-s)M f (s)ds
1
ee o
1
-1 xM -1 sL
+ (L + M)-1 e (L + H) A-1 esLf (s)ds
Jo
-e(1-x)LeMA-1d0 + (L + M)-1e(1-x)L [I - eM (L + H) A-1eL] (L + M) u1
- (L + M)-1 e(1-x)LeM (L + H) A-1 f* esLf (s)ds
Jo
- (L + M)-1 e(1-x)L [I - eM (L + H) A-1eL] i' e(1-s)M f (s)ds
Jo
x1
+ (L + M )-1 e(x-s)M f (s)ds + (L + M )-1 e(s-x)Lf (s)ds
(0 Jx
+ (L + M)-1 (exM - e(1-x)LeM) [eL+M; (L + H) A-1]do.
This representation generalizes the one used in [1, p. 63]. Note that, the last term
(L + M)-1 (exM - e(1-x)LeM) [eL+M; (L + H) A-1]do,
vanishes when H and operators L and M commute in the sense of the resolvent operators and then, our solution coincides with the one used in [1].
Corollary 1. Let L, M be closed linear operators in X satisfying (48). Assume (6)^(14) where Lu = L, MM = M for any u ^ u0. Let f E Lp (0,1; X) with 1 < p < to. Then, the two following assertions are equivalent
1. Problem (49) has a unique classical solution.
2. ui, A-1do E (X,D((L + M)2))^ , .
2p
Remark 6. Note that, instead of commutativity between H and operators L and M, we have assumed the following hypothesis
E D(L) = D(M) = D(L + M), A-1£ E D((L + M)2),
which is obviously verified when the operator H commutes with L and M in the sense of the resolvent operators. Notice also that (48) implies (10).
6. Application
Let X = L2 (R). Define operators LM, Mw and H by
D(LU) = D(MU) = H2 (R), D(H) = H1
Lwv(v) = v"(y) + a(vW(y) - Mv (y) = v"(y) -
Hv(v) = v' (v),
where a > 0,u > 0 and a E Cb (R) , a = 0 It is not difficult to prove that Lu, Mu generate analytic semigroups and that
Lu + Mu is boundedly invertible.
On the other hand, we have the following properties:
{
D(L + Мш f) = H4 (R) С D(L - Мш f) = H3 D(Mm LM) = D(LU Мш) = H4
(МшLM)ф (y) = ф4 (y) + a (y) ф(3 (y) + 2(a' (y) - иа) ф (y) + (a" (y) - uaa (y)) ф (y) + и2аф (y),
LМш)ф (y) = ф(4 (y) + a (y) ф(3 (y) - 2шаф" (y) - иаф' (y) + и2аф (y) note that Мш = МшLu. We have
{
DM - H) = H2 (Мш - H)ф(у) = Ф'(у) - иаф(у) - ф'(y) = ф(у).
Mu - H
r P2%nyi J,(p)
M - H)- ф -L и+П2? +U) *
So
лш = (Мш - H) + e eM L + H)
= (I + eL-eM (Lu + H) (Мш - H)-1) (Мш - H),
and
L + h) [(Мш - hф] (y) = Lu [(Мш - Hф] (y) + H [(Мш - Hф] (y)
^ ) _dC - Ш + 1)Г f ^fZl
Jr (иа + 4п2е2 + 2гпе) Л (иа + 4п2е2 + 2гпе)
+ « [ e2iny^ ф(0
+U Jr (иа + 4п2е2 + 2гпе)de
then
|(LW + H)[(МЫ - Hф] || ^ С
Ф
+ Си0
L2(R)
Ф
L2(R)
С (1 + иа)
L2(R)
and
lleL-eM (Lu + H)(МШ - HфЦ < С (1 + иа) ||eL- ||||еМ-|ц^u2{r) _< С (1 + иа) ml2(r) ._
Вестник ЮУрГУ. Серия «Математическое моделирование Q2
и программирование» (Вестник ЮУрГУ ММП). 2017. Т. 10, № 1. С. 70-96
Therefore, for a large u, we deduce that Aw is boundedly invertible. Now, notice that
Vv E D(Lw) = H2 (R) , A-1v E D((LU + M)2) = H4
Set A-1v = then
v = Aw^ = [(M - H) + eLeM(Lw + H)] ^,
thus
(Mu - H)^ = - u*^ - tf E D(Lw) = H2 (R) ,
which implies
A~-1v = ^ E H4 (R). Moreover, there exists C > 0 such that, for any u > 0,
2,, C
\\M,Lw] (Lw + Mw)-2\\l{x) < —,
where 7 = 2a if 0 < u < 1 and 7 = a if u ^ 1.
u > 0
problem:
f d2u . . ^ d3u . . , . d2u . . d4u . .
dx (x,v) +2 dV2dX(x,v + a(v) dVdX(x,v) - W(x,v)
d 3u d 2u 1 du du
-a(v)dk? (x,v) - a'(v)dv? (x,v) - 2a''(v)dv (x,v) + uaa(v)dv (x,v)
+uadv (x,v) + 2uad^u (x,v) - u2au(x,v) = f (x,v), x E (0,1) , v E R, (50) du (0,v) - duu(0,v) = do, v E R,
dx dv
K u(1,v) = ui, v E R.
Proposition 8. Let f E Lp (0,1; L2 (R)) with 1 < p < Then, for a large u > 0, the two following assertions are equivalent.
u.
2. ui, A-1 do E (H4 (R) ,L2 (R)) ± p ,
2p
where (H4 (R) ,L2 (R)) p = B\(lp-1/2p) (R) (see [6, p. 680681]).
2p ' p
Let us explicit A-1 do. Setting Vw = eL"eM (Lw + H) (Mw - H)-1 and noting that
(I + Vw )-1 = I - Vw (I + Vw )-1
we have
A-1do = (Mw - H)-1 (I + Vw)-1 do
= (Mw - H)-1 do - (Mw - H)-1 Vw (I + Vw)-1 do.
But Vw contains the regularizing operator eL", then A-1do E B4p 1/2p) (R) if and only if (Mw - H)-1 do E B%-1/2p) (R), that is
v /*_^ do(&_B4(1-1/2P)
v ^ Jr u + 4n2e + 2in£) ^ E B2,p
References
1. Cheggag M., Favini A., Labbas R., Maingot S., Medeghri A. Elliptic Problems with Robin Boundary Coefficient-Operator Conditions in General Lp Sobolev Spaces and Applications. Bulletin of the South Ural State University. Series: Mathematical Modelling, Programming and Computer Software, 2015, vol. 8, no. 3, pp. 56-77. DOI: 10.14529/mmpl50304
2. Favini A., Labbas R., Maingot S., Meisner M. Study of Complete Abstract Elliptic Differential Equations in Non Commutative Cases. Applicable Analysis, 2012, vol. 91, issue 8, pp. 1495-1510. DOI: 10.10801.00036811.2011.635652
3. Favini A., Labbas R., Maingot S., Meisner M. Boundary Value Problem For Elliptic Differential Equations in Non-Commutative Cases. Discrete and Continuous Dynamics System. Series A, 2013, vol. 33, issue 11-12, pp. 4967-4990. DOI: 10.3934/dcds.2013.33.4967
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6. Grisvard P. Spazi di tracce ed applicazioni. Rendiconti di matematica. Serie VI, 1972, vol. 5, pp. 657-729. (in Italian)
7. Meisner M. Etude unifiée d'équations aux dérivées partielles de type elliptique régies par des équations différentielles à coefficients opérateurs dans un cadre non commutatif: applications concrètes dans les espaces de Holder et les espaces Iß. Thèse de doctorat de l'Université du Havre, 2012.
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9. Favini A., Labbas R., Maingot S., Tanabe H., Yagi A. A Simplified Approach in the Study of Elliptic Differential Equation in UMD Space and New Applications. Funkcialaj Ekvacioj, 2008, vol. 51, no. 2, pp. 165-187.
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11. Lunardi A., Analytic Semigroups and Optimal Regularity in Parabolic Problems. Basel, Birkhäuser, 1995.
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Received November 9, 2016
УДК 517.9 DOI: 10.14529/mmpl70105
НОВЫЕ РЕЗУЛЬТАТЫ О ПОЛНЫХ ЭЛЛИПТИЧЕСКИХ УРАВНЕНИЯХ С КОЭФФИЦИЕНТНО-ОПЕРАТОРНЫМИ ГРАНИЧНЫМИ УСЛОВИЯМИ ТИПА РОБИН В НЕКОММУТАТИВНОМ СЛУЧАЕ
М. Чеггаг, А. Фавини, Р. Лаббас, С. Менго, Х.У. Мелха
В статье доказываются некоторые новые результаты для дифференциальных уравнений эллиптического типа второго порядка с общими граничными условиями типа Робина с некоммутативными структурами. Исследование выполнено при условии, что второй член принадлежит пространствам Соболева. Существование, единственность и оптимальная регулярность классического решения доказаны с использованием теории интерполяции, и получены результаты для класса операторов с ограниченными мнимыми степенями. В работе приводится пример применения данной теории. Результаты, полученные в этой работе, улучшают исследования в коммутативном случае М. Чегага, А. Фавини, Р. Лаббаса, С. Менго и А. Медегри. Вводя в рассмотрение некоторые операторные коммутаторы, обобщено представление решения задачи в коммутативном случае и доказано, что это представление обладает свойством регулярности.
Ключевые слова: эллиптическое дифференциальное уравнение второго порядка; граничные условия Робин в случаях коммутативности; аналитические полугруппы; максимальная регулярность.
Мустафа Чеггаг, Национальная политехническая школа (г. Оран, Алжир), mustapha.cheggag@enp-oran.dz
Анджело Фавини, кафедра математики, Болонский университет (г. Болонья, Италия ), favini@dm.unibo.it.
Раба Лаббас, Нормандский университет; лаборатория прикладной математики, Университет Гавра (г. Гавр, Франция), rabah.labbas@univ-lehavre.fr.
Стефан Менго, Нормандский университет; лаборатория прикладной математики, Университет Гавра (г. Гавр, Франция), stephane.maingot@univ-lehavre.fr.
Хеллаф Ульд Мелха, лаборатория чистой и прикладной математики, Университет имени Абдул Хамид Ибн-Бадиса (г. Мостаганем, Алжир), ould_melha_khel@yahoo.fr.
Поступила в редакцию 9 ноября 2016 г.
