Periodic solutions of some linear evolution systems of natural differential equations on 2-dimensional tore Dang Khanh Hoi* Department of General Education, Hoa Binh University, Tu Lie
Trang 1Periodic solutions of some linear evolution systems of natural
differential equations on 2-dimensional tore
Dang Khanh Hoi*
Department of General Education, Hoa Binh University, Tu Liem, Hanoi, Vietnam
Received 3 December 2009
Abstract In this paper we study periodic solutions of the equation
with conditions
over a Riemannian manifold _X , where
Gul, 1) = I ale, y)uly)dy
is an integral operator, u(2, ) is a differential form on X, A = 7(d+0) is a natural differential
operator in X We consider the case when X is a tore [I? It is shown that the set of parameters
(v, b) for which the above problem admits a unique solution is a measurable set of complete
measure in x JR”
Keyworks and phrases: Natural differential operators, small denominators, spectrum of compact
operators
1 Introduction
Beside authors, as from A.A Dezin (see, [1]), considered the linear differential equations on manifolds in which includes the external differential operators
At research of such equations appear so named the small denominators, so such equations is incorrect in the classical space
There is extensive literature on the different types of the equations, in which appear small denominators We shall note, in particular, work of B.I Ptashnika (see, [2])
This work further develops part of the authors’ result in [3], on the problem on the periodic solution, to the equation in the space of the smooth functions on the multidimensional tore II” We shall consider one private event, when the considered manifold is 2-dimension tore II? and the considered space is space of the smooth differential forms on II?
* E-mail: dangkhanhhoi@yahoo.com
17
Trang 2We shall note that X-n-dimension Riemannian manifold of the class C’°° is always expected oriented and close Let
€ = OF gf? = OP gAM(T*X) @C
is the complexified cotangent bundle of manifolds X, C°(€) is the space of smooth differential forms and H*(€) is the Sobolev space of differential forms over X (see, [4]) By A we denote operator i(d+0), so-called natural differential operator on manifold X, where d is the exterior differential operator and
6 = d*- his formally relative to the scalar product on C™(€), that inducing by Riemannian structure
on X It is well known (see, [4], [5]) that d+ 6 is an elliptical differential first-order operator on X
From the main result of the elliptical operator theories on close manifolds (see, [4]) there will
be a following theorem
Theorem 1 Jn the Hilbert space H°(£) there is an orthonorm basis of eigenvector {fm},m € Z, of the
operator A = i(d+6) that correspond to the eigenvalues Xp Else Xm = ttm, Lm € Ry A-m = —Am and 1 ~ 0 when m — ov
Proof This theorem was in [5]
The change of variable { = b7 reduces our problem to a problem with a fixed period, but with
a new equation in which the coefficient of the 7-derivative is equal to 1/b:
Ø
(—— + a(d + ð))u(&, br) = G(u(œ, br) — ƒ(œ, br)) ¿bỪr
2 Thus, in II2 = R?/(2Z)? problem (1)(2) turns into the problem on periodic solution of the equation
Ø
Tu = Ly + a(d + ơ))u(œ, t) = vG(u(a,t) — f(a, t)) (3)
with the following conditions:
Here
uo (a, t) uo (a, t)
u(x,t)=( 1 da! dx? dai A da? ) uo(a.t) | = | s0
u3(a, t) u3(a, t)
- complex form with coefficient dependent for t, t € [0,1]; a A 0,v are given numbers,
(ula), v(a)) = uo(a)vo(a) + uy (ar)vy (2) + u2(a)v9(x) + ug (a) v3 (a),
Gulest) =f ale.sulys tty
is an integral operator on the space Lz = L2(H°(€), (0, 1]) with a smooth kernel
defined on IH? x TI? such that
I, ( gool@,y) goil@,y) Ø02(®,) gos(v,y) )dx =0 Vy € LẺ.
Trang 31,0 10
We assume that operator 7 bar + aA) = BOL + a(d + 6) given in the differential form space i u(a, t) € Œ®(Œ(£), |0, 1|), with these conditions
ult—o = len, | (u(x), 1)da = 0
12
Let L -denote the closure of operation = + a(d +6) in Lo(H°(E), [0, 1]) So, an element i
1
u € Lo(H°(E), [0, 1]) belongs to the domain D(L) of operator L = sứ i + aA), if and only if there
is a sequence {u;} C C®(C™(E), [0, 1]) ujleo = uyleai, fr (uj (x), 1) de = 0 such that limu; = u,
lim Lu; = Lu in Lo(H°(E), [0, 1])
Let H-denote a subspace of space L2(H°(E), [0, 1])
H = {ule,t) € La HC), (0,1) | ff (u(et),1) de =0}
{+iny/k? + k3 = tin|k|;k = (ki, ke) € Z7}
is the set of eigenvalue of operator A = 7(d + 6) on II? and eigenvectors, coresponding to
imnoy/k? + k3,
Sen (@) = it (hie + kox uy,
We note that
are given by the formula:
here win € D> A?(C?),7 = (m,72) € {-1, +1}? is some basic in 4— dimensional space of the complex differential forms with coefficients being constant These coefficients depend on k € Z? and elements of this basic are numbered by parameters 7 We are not show ¿„„ on concrete form (see, [6]) Lemma 1 Zhe forms e€kmy = €?"™ fen(w),k = (ki,k2) A 0, are eigenform operator L that corresponds to the eigenvalues
in the space H These forms form an orthonorm basis in given space The domain of operator L is given by formula
D(L) ={u= S| Ukman€kmn | » [Nkmntkemn | < œ, » lzx„„| < œ}
2m
The spectrum o(L) operator L is the closure of the set \ = {Xkmy}-
We note that the number of dimensions of the eigensubspace is finite and we shall not indicate exactly how many there are of them
Lemma 2 Let g(x,y) € L2(U? x II?) and
1/2
ao = (ff late si Pawar) Then G - linear operator is bounded in H®(€) and his norm ||G|| < Mo
Here ||g(x, y)|| - matric norm g
gla, y) = (9ij(@,y)), 2,7 =0,3
Trang 4lIgll = supf |lgul] | ue R? x R*, ||ul| <1}
Proof If u(a) € H°(E)
Gu? =11 f aeonucnaul?s (f laenutvdiav) <
mm 6h
Iesll#— Ƒ_lIŒa(a)|faz <
[Uf ilatessniPay [inet lPay) ae oul? s ff lig(esø)lBasáy Ƒ_ lio)|24y Miia
|G] < Mo
The lemma is proved
Let B = (—A,)**!,a > 0 Then B is M-operator in H°(€) and B fen = Urfien here
Lp = (r|k|)?*2 are eigenvalues Operator (—A,)°*! is self-conjugate We suppose that kernel g(x,y) of operator G having the following behaviour (—A,,)°*1g;;(z, y) € Le(IP? x II?) (gij(., y)
belongs to space Sobolev wetee for almost every y € 7) Then product operator B o G is integral
operator (—A„)!** o Œ„„ with kernel
(—Az)'*°g(@,y) = (Ax)? 9is)(a, 9), 69 = 0,3
Let M = max{||(—A,)!** 0 GI, ||G]]}-
Lemma 3 Let v = Gu = 3 Vkmn€kmn, then
4A/”||u||Ÿ
Ifk #0 then
2
2< |%ma|
here
Akmn = ((—A,)!** 0 Gu, €kmmn) Lo Proof We have
Akmn = ((—A,,)'t%0 Gu, €kmn) Lo = / ((—A,)!*%0 Gu, €kmn) at = / (Gu, (CAz)**®e„) dt —
/ (Gu, UkCkmn) dt = mm | (Gu, Ckmn) dt = Hr(Gu, Ckmm) Ly = HEUbman-
Then, if Z 0 ( so that || > 1) we have
4lakmn|°
Vkmnl S 7 ng:
_nn,, Thus, by Parseval dentity
Š” la„nl2 = JI(CAz)?!2%e Gall? < MP|IulỦ,
Trang 5So that
4M? ull?
2
Vkm < *
my (uz| + 1)?
In the case 4, = 0 by Parseval dentit y y Ukmnl- = ||Gull? we have n
4M?|\u|l?
Ven |? < GI? ull? < 4G? Jal]? < ———_
[Vmnl” < |G] lull IlœlÍ all (p]+ 1)?
The lemma 1s proved
We assume that a is real number Then by Lemma 1, the spectrum o(Z) lies on the real axis
The most typical and interesting is the case where the number ab/2 and (ab/2)? are irrational In
this case, 0 A Nkmn Vr € Zk € Z?,k # 0 and the H.Weyl theorem (see, ¢.g., [7]) says that, the set
of the numbers Aj, is everywhere dense on R and o(L) = R Then in the subspace 1 the inverse operator L~! is well defined , but unbounded The expression for this inverse operator involves small
denominators [8]
where the v%m, are the Fourier coefficient of the series
me ,ke 2k40
For positive numbers C’,o let A,(C) denote the set of all positive 6 such that
C
|Akmy| > JkI*”' (8)
for all m € Z,k € Z?,n = (m,n2),m,2 = +L, k # 0
From the definition it follows that the set A,(C) extends as C’ reduces and as o grows There- fore, in what follows, to prove that such a set or its part is nonempty, we require that C > 0 be sufficiently small and o sufficiently large Let A, denote the union of the sets A,(C) over all C > 0
If inequality (8) is fulfilled for some 6 and all m,&, then it is fulfilled for m = 0; this provides a condition necessary for the nonemptiness of A,(C):
Œ <|#|”ˆ”|a|k||'Y k z 0 (9)
We put đ = |a|z and Œ < d/2
Theorem 2 7e se(s A„(C), A¿ are Borel The set Ag has complete measure, i.e., its complement
to the haif-line R* is of zero measure
œ
Proof Obviously, the sets 4„(C) are closed in R* The set Á„ —= U A,(1/r) - is Borel, being
r=1
a countable union of closed sets We show that A, has complete measure in R* Suppose b, | >
0, C < =; we consider the complement (0,/)\A,(C) This set consists of all positive numbers 6,
for which there exist m,k,k A 0, such that
C
|Àkmnl <
Trang 6Solving this inequality for 6, we see that, for m,k,k& 4 0 fixed, the number b forms an interval
Inm = (max, MB), where m = 1, 2,3, ,
san
Ak =
The length of Ip 1s mdz, with
4xŒ|k|~1~#
|a|R||#— GZ|&| 2#
Since C < 5 by assumption, we have
16Œ
For k fixed and m varying, there is only a finite amount of intervals J/;,,, that intersect the given segment (0,1) Such intervals arise for the values of m = 1, 2 , satisfying ma, < I, ie.,
0 <?n < 5, lanl + C|k| 1)
Since Clk] 1? < sI4zlR||; we can write simpler restrictions on m :
The measure of the intervals indicated ( for & zZ 0 fixed ) 1s dominated by ôyỐ„, where
Sz = Sz(L) is the sum of all integers m satisfying (12) Summing an arithmetic progression, we obtain
Considering the union of the intervals in question over k and m, and using (11), we see that
W((0,1)\A(C)) < So ð,5%,<Œ50),
kz0,k€ 2
where
s=80)= ` 8I{/|am||L1 =7
Balk]? lar] kl”
kz0,k€ 2
Observe that the quantity
i|amr|k|| +
z|az |R||
is dominated by a constant J2, therefore, (sinece ø > 0}
We have
(0, \ Ao) < w(0,4) \ Az(G)) SES) VO > 0
It follows that ¿((0,/)\ 4z)=0_ Vĩ > 0 Thus, u((0,00) \ Az) = 0 and A,- has complete measure
The theorem is proved
Trang 7Theorem 3 Suppose g(a, y) € L¿(HZ x II?) such that (—A,)'*%g(a, y) is continuous on I? x TI?
and
I ( goo(a,y) goil@.y) Ø02(,) øØo3(,) ) da =0 Vụ € TỪ
II
Let 0 <o <1, and letb € A,(C) Then in the space H the inverse operator L~' is well defined, and the operator L~' o G is compact
Proof Since b € Ag(C), we have Am, #0 Vm € Zk € Z?, k 4 0 so that in the space H,
||? +22 ((a|k])? 720-4 1)?
|k| — oo because 0 < ao < 1,a > 0 Therefore, given < > 0, we can find an integer kp > 0, such that
Ik”” (eC)
((|k|)#'“^+ 1)” ` M?
E—† is well defined and looks like the expression in (7) Observe that lim =0 as
2 for all |k| > ko We write
L }u(œ,f) = Qkort + Qkogt, v= Gu,
where
Vkmn Vkmn
Qko V = » X Chư QkooV = » X Chư"
0<|ki<ko Oh" Rị>áo “n3
For the operator Q;,, we have
2
[Qkor Ul — » Ầ TT
0<|k|<ko
Observe that if 0 < |k| < ko, then
1
ale [+ a || nal?
Therefore, the quantity War dominated by a constant C'(ko) Then
IF + an |k|n2|?
|I9,,9llƒ < À ˆ loems|?C(Œo) < C06) |Io|Ì
which means that @;,, is a bounded operator
Consider the operator Q;,, 0G By Lemma 3 and (8), we have
Viren?
||đQz.„+|l” — Il€)xo; ° Gul? — » am 2 <
|k|>ko [Aenea
lœx„„lÊ 15 219g 12 ,£Œ 3 9 2 5
Consequently, ||Qz„„ o G|| < £
Since G' is compact and @;,, is bounded, @;,, 0 G' is compact Next, we have
ID oG|| — l|€z›; oGil| SE
Thus, we see that the operator L~! 0G is the limit of sequence of compact operators Therefore, it is compact itself The theorem is proved We denote K = Ky = L7! 0G
Theorem 4 Suppose b € A,(C) Then problem (1)(2) has the unique periodic solution with period
b for allv € C, except, possibly, an at most countable discrete set of values of v
Trang 8Proof Equation (3) reduces to
We write ÌoŒ—-—= K—-
Since K = L~!oG is a compact operator, is spectram ø(#€) is at most countable, and the
1
limit point of o( A’) (if any ) can only be zero Therefore, the set S = {vy 4 0 | — € o(K)} is at most Vv
1 countable and discrete, and for all vy £ 0, v ¢ S the operator (K — —) is invertible, i.e., equation (3)
Ụ
is uniquely solvable The theorem is proved
We pass to the question about the solvability of problem (1)(2) for fixed v We need to study
1
the structure of the set FH C C x R™, that consists of all pairs (v,b), such that 0 and — # ø(Kp),
Ụ where / = L !oŒ
Theorem 5 F is a measurable set of complete measure in C x R*
For the proof, we need several auxiliary statements
Lemma 4 For any < > 0 there exists an integer ko such that ||Ky — Ky|| < < for allb € Ag(-), 0<
r
o <1, where r= 1,2, ,
Kyu = Lp Tạ — » x omn(Ö) TF Ckmn, Kyu = sa, AemiÐ) x TF Ckmn:
||? +22 8
Proof Observe that for any ¢ > 0 there is an integer ko such that aI ((z|k|)?!2?+1)2 — tạ) ff x <1
for all |k| > kọ, 0 <ø <1,œ>0 We have
Niemn (0)
|k|>ko
2.2 2420
||UX¿, — K¿)w||Ÿ = |LKz„+||ˆ = "` ` | ol < mm
2, —— = \2 2 <r2(-—~ 2, = \2y72 2 = 2 2
(A)? > Immal2 <+2(CT)?MP||a|l2 = z?|la|
|k|>ko
Thus ||, — ¿|| = |LKx+„„|| < © as required
1 Lemma 5 The operator-valued function b > Ky is continuous for b € Ag(-)
r
Proof Suppose b,b+ Ab € AC) and « > 0 By Lemma 4 there exists an integer kg (independent
of b, b+ Ab) such that || — Kol| = || Kgoal| < € and || Kyy.a5 — Ko+asl| = || Keoo+ao)|| < & Next
EyAp — Ấy = (ẤwcAw + Egy@wAp)) — (Ấy + Ki), whence we obtain
|| Ko+as — Koll < ||Ko+ae — Koll + |W go(o+aoy ll + || all:
Trang 9Considering the operators Koso, Ki, we have
(Koray — Kyu = fe, Nemn(b + Ab) (———————x — xx)Đkmn€km Apmg(b) 1
|| Kyu — Kyyavul|? = p=, Se BOF ADP, 24 ]Aamulb + AOVP Din) 5 5 (14)
1
Ifb+ Abe A¿(—), 0< |š| < kọ, 0< ø <1, then
+
l„„l? < | |? 2) p|? +22 < 2 2420 |?
The relation lim ———, = b? and the condition 0 < |k| < ko imply that the quantity ————_ —
4
me is dominated by a constant C'(ko) depending on ko Therefore
” +az|klip|°
BO FADE , 4, DMan(b + A9)|Z]Asny(Đ|P —
PT" 0<|k|<ko <<
|A»l ø
brane” Rơ 7 C(ko) » Uk |?
0<|k|<ko
Since
S- |vemml? < |le|lf < MỸ|lu|Ứ,
0<|k|<ko
we arrive at the estimate
Ab/?
= 5 M?r7ko? 7° C(ko)
I| Koray — Koll — |b(b + AĐ)|
We choose Ab so as to satisfy the condition
|Ao|?
|b(b + Ab)|?
Then ||Ky1ay — ¿|| < 32 This shows that the operator-valued function b — Ky is continuous on
M?r2ko?*??C'(ko) < £
Az(—) The Lemma 1s proved
r
Lemma 6 The spectrum o(K) of the compact operator K depends continuously on K in the space Comp(Ho) of compact operators on Ho, in the sense that for any « there exists 6 > 0 such that for all compact ( and even bounded ) operators B with ||B — K|| <6 we have
Here V.(0) = {\ € C | |A| < €} is the e-neighborhood of the point 0 in C
Trang 10Proof Let k be a compact operator; we fix ¢ > 0 The structure of the spectrum of a compact operator
shows that there exists ¢, < ¢/2 such that <, 4 |A| for all X € o(K) Let S = {\j, , Ax} be the set
of all spectrum points A with |A| > <1 and let V = U V.,(A) Then V is neighborhood of o(K)
AcsU{0}
and V Cc o(K) + V.(0) By the well-known property of spectra ( see, e.g.,[9], Theorem 10.20) there
exists 6 > 0 such that o(B) C V for any bounded operator B with ||B—K'|| < 6 Moreover (see, ¢.g.,
[9], p.293, Exercise 20), the number 6 > 0 can be chosen so that o(B) NV.,(A) 4 0 VA € SU {0}
Then for all bounded operators B with ||B—K'|| < 6 the required inclusion o(K) C ơ() + V2z, (0) C o(B) + V.(0) and o(B) CV C o(K) + V.(0) are fulfilled The lemma is proved
From Lemma 6 we have the following statement
Proposition 1 The function p(A, K) = dist(A, o(K)) is continuous on C x Comp(Ho)
Proof Suppose 4 € C, K € Comp(Ho) and ¢ > 0 By Lemma 6 there exists 6 > 0 such that for any operator H lying in the 6-neighborhood of K, ||H — K]|| < ở, the inclusions (15) are fulfilled; these
inclusions directly imply the estimate |p(A, K) — p(A, H)| < « Then for all « € C with |u — À| < £ and all H with ||H — K|| < 6 we have
|ø(w, ) — ø(A, H)| < |o(u, K) — pQ, K)| + |eQ, 1) — pQ, H)| < Ju —Al +2 < 2¢,
Since < > 0 is arbitrary, the function p(\, A’) is continuous The proposition is proved
Combining Proposition 1 and Lemma 5 we obtain the following fact
1
Corollary 1 The function p(A, b) = dist(A, o(Ky)) is continuous on (A, b) € C x Ag(-)
r Now we are ready to prove Theorem 5
Proof of Theorem 5 By Corollary 1, the function p(1/v, b) is continuous with respect to the variable
(v, b) € (C \ {O}) x Ag( ty, Consequently, the set
B, = {(v,b) | p(/v,b) 49, be AC]
is measurable, and is so the set B = U,B, Clearly, B C EF and = BU Đụ, where Øọ = E \ B
Obviously, Bo lies in the set C x (R* \ A,) of zero measure ( recall that, by Theorem 3, A, has complete measure in R™ ) Since the Lebesgue measure is complete, Bo is measurable Thus, the set
F is measurable, being the union of two measurable sets Next, by Theorem 4, for 6 € A, the section
Be = {v €C | (v,b) € E} has complete measure, because its complement {1/1 | € ø(K§p)} is at
most countable Therefore, the set EF is of full plane Lebesgue measure The Theorem is proved The following important statement is a consequence of Theorem S
Corollary 1 For a.e v € C, problem (1)(2) has a unique periodic solution with almost every period b€R'
Proof Since the set Fis measurable and has complete measure, for a.e € C the section ?2„ = {b €
R* | (vb) € E}= {bE R* | 1/v ¢ ø(K§)} has complete measure, and for such b’s problem (1)(2)
has an unique periodic solution with period b The Corollary is proved
Aknowledgments The work is partially supported by the study QG-10-03