ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXV (1985)
Piotr Ra p p (Poznan)
Existence and uniqueness of special solutions of linear differential equations with deviated arguments
1. Introduction. Let R be the set of real numbers, X — a Banach space and L (X ) — the space of linear bounded operators from X into X.
This paper deals with the problem of special solutions of the equation (1) x(f) = Л (г)х(г)+ £ B i(t)x (t + T t { t ) ) + f ( t ) , t e R ,
i = 1
where x: R -» X is an unknown function and the following continuous functions are given: t,: R -> R, f : R -»■ X , А, В {\ R - * L(X).
A function £: R - > X of class C 1 for which equation (1) and the condition £ (t0) = x 0 are valid for all t e R and given (t0, x 0) e R x X is called a special solution of equation (1) passing through the point (t0, x 0). This term was introduced in paper [8].
Special solutions are of great importance in the theory of differential equations mainly because any solutions of an equation with bounded coefficients and bounded delays of the argument may be asymptotically approximated by means of a certain special solution [1], [3], [8]. The basic results concerning the existence, uniqueness and properties of special solutions can be found in papers [1], [3], [8], [10] and also [2], [4 ]-[7 ],
[9].
This article presents the problems of the existence and uniqueness of special solutions of equation (1). In this respect a number of results more general than the analogous ones quoted in papers [1 ]-[1 0 ] are obtained.
Detailed comparison of the results in case of the equation with small coefficients and small deviations of arguments are presented in Section 5 of this paper.
2. Basic theorem.
Theorem 1. L et H : R —► L (X ) be a continuous function satisfying the equality
П
(2) H (t) = A (t) + X B t (f) N H (t + ti (t), t), t e R, i= 1
where N H(-, •) is the resolvent o f the equation x(t) = H (t)x(t). T hen : Г T he set S o f all special solutions o f (1) passing through an arbitrary point (t0, x 0) e R x X is nonempty i f and only i f the set T o f all solutions o f the equation
„ t + T,(0
(3) y(t) = f { t ) + £ B f t ) J N H(t + X i ( t ) , s)y{s)ds
i = 1 t
defined and continuous on R is nonempty.
2° I f the set S (or T) is nonempty, then the mapping x H- у defined as (4) y(t) = x(t) — H (t)x (t), x e S , t e R ,
is a bijection o f the set S onto T
P r o o f. 1° Let R ^ X be a function of class C 1 satisfying the condition £ (f0) = x 0 . Let <r: R -» X be a continuous function defined by the following formula
(5) <7(i) = f ( t ) - H ( t K ( r ) . We shall show for the function £ that the equality
(
6)
Ш = A (t)Ç (t)+ X B f f ^ t + X iit^ + fit), t e R ,i = 1
is valid if and only if the function о satisfies the equality
(7) a ( t )
= / ( 0 +
Z B i(t) J N H(t + X i ( t ) , s)o (s)d s, t e R .i = 1 t
In fact, formula (5) gives us
(8)
and hence
Ш = N H(t, t0) x o + J N H(t, s)a (s)d s, 'o
t + r,(f)
(9)
£ ( t + T, ( 0 ) = N H ( t + x i (t ) , t ) Ç (t) + J N H (t + X i (t), s) о (s) d s.Let equality (6) be true. Then by substitution of the expressions £
(0
from (5) and Çfa + xfitj) from (9) into (6) we obtain the equalityi = 1
< 7(0= / (0 + Z B i (0 J N H (t + Tf (0, s) <7 (0 ds +
t
+ ( Л ( 0 + Х fii ( 0 W H(t + T i ( 0 ,
t)-H(tj)Z(t), t e R .
i = 1
Hence, taking into account (2) we obtain equality (7). So equalities (6) and (7) are equivalent and it means that ÇgS if and only if o e T .
2° From the first part of the proof it follows that for any o gT the function £ satisfying (5) and the condition f (t0) = x 0 is an element of the set S. In this way we see that the function x h j defined by (4) is onto T This function is also one-to-one. It follows from the uniqueness of the solution of the Cauchy-problem for equation (4) with respect to x.
Co r o l l a r y. L et H : R - * L (X ) be a continuous function satisfying equality (2). A special solution Ç o f equation (1) passing through the given point (t0, x 0) exists and is unique i f and only i f a unique continuous solution о o f equation (3) exists on R. The solution Ç is given by form ula (8).
3. Lemmas.
Lem m a 1. L et N j( • ,) be a resolvent o f the equation x(t) — H j(t)x(t), where Hj \ R -> L (X ) are continuous functions such that
IIHjII = sup ||flj(r)ll « h, h > 0, j = 1, 2.
t e R
Then, fo r any t, t0 e R , the follow ing inequalities are valid:
(
10
)(И) ||N;(t,(0)|Ke4'- '01.
P r o o f. For the resolvents ЛГД-, •) we have the formulae 00
(12) N j(t, t„ )= £ 7 = 1 ,2 ,
k= 0
where Nf\t, t0) = I,
t„) = J Hj(s)Nf{s, t0)ds.
By the induction we show that the inequalitiesr0
(13)
\\Nf(t,
(0)ll —lol‘ , * = 0 , 1 , 2 , . . . ,IIN f ( t , t0) - N f ( t , r„)|| « НН.-НгН ~~|l f0l^
( k - l )!
are true. Hence
IIJVj(t, t0) - J V2(t, to)ll «
f
||JV?>(r, t0) - N f ( t , f0)ll i= 1«IIH.-HjHIr-tol £ TCprrlt-tol*'1
k = l ( k 1 ) 1
= ||Я
1- Я
2||'|г - г
0|е'‘"_Ч
11 — Prace Matematyczne 25.2
so inequality (10) holds. Inequality (11) is obtained from formulae (12) and (13) .
Let a ^ 0, bh rjt > 0 for i = 1, 2, . . . , n be given real numbers. We denote П
rj = m axrji, b = £ bt.
i i=l
Condition 1. The inequality
(14) eq b eav < 1
holds.
Condition 2. There is a real number y > 0 satisfying the inequality a + t Ь ,е”<у < у .
1 = 1
Condition 3. A real number Ô > 0 exists such that (15) a + £ bi e ib = ô, X b{ rjt е ' Ь < 1.
i = 1 i = 1
Lemma 2.1° I f Condition 1 is satisfied, then Condition 2 is satisfied, too.
V hi = h f or i = 1, 2 then Conditions 1 and 2 are equivalent.
2° Conditions 2 and 3 are equivalent and also Ô < y.
P ro o f. 1° The function z: R -+ R defined as z(t) = a + bent — t is strictly convex on R and has a minimum at the point t = t* defined by the equality z(t*) = 0. Hence we calculate t*:
(16) t* — —- In(brj).
П
From inequality (14) it follows that bq < 1, so t* > 0 and (17) z(t*) = - In (eqbeatl) < 0.
h Now we have
a + £ b i e ^ - t * ^ a + £ Ь{ е т*- t* = z (f*) < o,
i=l i = 1
so Condition 2 is satisfied for y = t*. If r\i = q for i = 1 , 2 , . . . , n, then Condition 2 gives Condition 1. In fact, if z ( y ) < 0 for a у > 0, then z(t*) ^
z(y)
< 0 for t* defined by formula (16). That is why inequality (17) which yields inequality (14) holds.2° The function r : R -> R given by the formula
r(t) = a 4- Y, — t i = 1
is strictly convex on R and has a minimum at the point t — у*, у* > 0
n
defined by the equality r(y*) = 0, i.e., ]T bi rii e t,iy = 1. If Condition 2 is
i
= 1
satisfied, then r(y) < 0, so r(y*) < r(y) < 0. The function r is continuous and strictly decreasing in the interval [0, y*]. Moreover, r(0) > 0. Hence there is a number <5e(0, y*) such that r(0) = 0, so equality (15) holds. From the inequality Ô < y* it follows that
i b,4le * < i = 1
i = 1 i = 1
and hence inequality (15) is satisfied. Conversely, let Condition 3 be satisfied.
From inequality (15) it follows that a number y0 > <5 exists such that
i
m ,v " ° = l,i
- 1
so r(y0) = 0- Thus the function r has a minimum at the point t = y0. Because r(<5) = 0 and Ô < y0, so r(y0) < 0, since the function r is strictly decreasing in the interval [0, y0]- So Condition 2 is satisfied for у = y0.
4. Equations with small coefficients and small deviations. Let Y mean a Banach space. By ^ (R , Y) we denote the space of continuous functions /: R -> Y with the norm ||/|| = sup ||/(f)|| < o o . For given t0 e R , p > 0, by
feR
G ( R , X ; p) we denote the space of continuous functions g: R - + X with the norm
(18) \\g\\p
=
sup (||0 ( f ) l k ~ P l* ~ * 01) < oo.teR
We shall assume that the functions A, B h r, from equation (1) satisfy the following condition:
Condition 4. A, B i e cé ( R , L(X)), xi e (€ (R , R) for i = 1, 2, . . . , n and
(19) I M I K a , \\Bt\\Kbi9 I N K i
By ) we shall denote a resolvent of the equation x(t) — H (t)x(t).
Theorem 2. We assume that fo r certain numbers a ^ 0, bh щ > 0 (i = 1, 2, . . . , n) Conditions 3 and 4 are fulfilled. Tnen there is exactly one function H *e4 > (R , L (X )) which satisfies the equality
(20) H *(t)
= A{
t ) + £ B f ^ N ^ t+
x f t f t ) , t e R ,i =
1
and the inequality
(21) ||Я*|| < d,
where Ô is the number appearing in Condition 3.
P ro o f. Let H e c€ (R , L(X)). Let us consider the class of functions D = { H e V ( R , Ц Х )): \\H\\ ^ <5}.
We shall show that the operator Ф defined on D by the formula
<P(H)(t) = A U ) + t B ,(t)N H{t + zt(t),t), t e R ,
i=
1maps the class D into itself and is a contraction map. In fact, for any function H e D we have
||Ф(Я)|| « 1И11+ £ ||В,|| sup ||NH(r + T,(f), ()||.
i = 1 teR
Considering inequalities (19), (11) and equality (15) we obtain
||Ф(Я)|| ^ a+ £ b>e',s = b,
i=
1so the operator Ф maps D into D. Moreover, for any Я 1} H 2eD we have
||Ф(Н1) - Ф ( Н 2)|| ^ £ ||B,.|| 8ир||МЯ1(г + т , ( М - ^ н 2 (* + т , ( 4 #
i = l teR
Now, using Lemma 1, inequalities (15) and (19) we obtain
||ф(н1) - Ф ( н 2)1К||н1- я 2|| £ b i m e" , = q\\Hl - H i \\,
»= 1
where q < 1. Hence, the operator Ф is a contraction within D. On account of the Banach fixed-point principle there is exactly one function H * e D such that
Ф (Я *) = Я *.
Thus the function Я * satisfies equality (20) and inequality (21).
Theorem 3. Assume that
1° В ,е # ( Я , L(X)), т ,е ^ ( Я , R) and ||B,|| ^ bt, ||т;|| ^ rjb where bh rjt > 0 fo r i = 1, 2, . . . , n.
2° F or numbers bh > 0 and fo r a certain number a ^ 0 Condition 2 is satisfied.
3° T he function N R x R -+ L (X ) is continuous and satisfies the inequality
\\N(t, s)|| ^
fo r any t, s e R , where
П
0 < a ^ a + £ bt e h<* < y
i = 1
and y is the number appearing in Condition 2.
By these assumptions, the operator W: у Ч * ( у ) , y e G ( R , X;
y),
defined by the form ulan 1 Tf(*)
(22)
V(y)(t)= Z Bi(0
J N (t + x M ,s ) y ( s ) d s 1 = 1 tmaps the space G (R , X ;
y)
into itself and the follow ing estimation holds(23) im i « —
t
bi (e"‘7- e ' p) < 1 . 1=1P r o o f. Let y e G { R , X ; y). Formula (18) yields the estimation
||у(0Н < 1Ы1,е’"’'01.
so from formula (22) in virtue of assumptions 1° and 3° we have
(24) i =
t btM
1t),
where
t + T , - ( f )
t
In order to the estimate this integral we shall consider six possibilities as follows:
(i) t0 ^ t ^ t + z f t ) , (ii) t + r,(t) ^ t ^ t 0, (iii) t < t + T,(t) ^ t0, (iv) t0 ^ t + x ft ) 5% t, (v) t ^ t0 ^ t + т,-(t), (vi) t + x ft) < t0 < t.
Calculation of each of these integrals yields the following estimation (25) 3 1( 1 ) < 2 _ ( е" » _ еч )<* - « 1
y —a
for any t e R . Joining estimations (25) and (24) we obtain
ll'f'MWIKIWI,— Î
У - а i=i
so it can be deduced from formula (18) that W (y )eG (R , X ;
y)
andn * m < —
У-а i=i i w-e**).
(
26
)By assumptions 2° and 3° we have
£ bi (етУ - е ща) = a + £ h, e ?,'y - (a + £ b{ ещл) < y - a .
i = l i = l ‘ = 1
Consequently, by the above inequality and (26) we obtain estimation (23).
Theorem 4. Assume that fo r certain numbers a ^ 0, frf, rji > 0, Conditions 2 and 4 are fulfilled. Then fo r every function f e G ( R , X ;
y)
tfcere is exactly one special solution o f equation (1) in the space G (R , X ;y ) passing through an arbitrary point (t0, x 0) e R x X .P ro o f. By Condition 2 and Lemma 2 it follows that Condition 3 is fulfilled. So Theorem 2 implies the existence of function H : R - * L { X | satisfying equality (2). Moreover, since ||Я|| ^ Ô we have ||iVH(i, t0)|| ^ *°
and hence the assumptions of Theorem 3 are fulfilled. Thus, by Theorem 3 it follows that there is exactly one function о in the space G (R , X ; y) satisfying equation (3). Consequently, in virtue of Theorem 1, the function R - + X defined as
t
Ç (t) = N H(t, t0) x o+ { N H{t, s)a (s)d s хо
belongs to the set S. This means that ^ is a special solution of equation (1) passing through the point (t0, x 0) e R x X . Moreover, we have the estimation
НШ11 « e4|' " ol||Xoll+IH U } e‘,,' ,| + r|,- 'ol* | .
l0
Recalling that 0 < ô < y (from Lemma 2) we can calculate
If / l t_sl + yls_tolds| = —— (ev|t_,ol- / |t_I()l).
I J 1 y —
S
lo '
The last and the previous inequality imply £ e G (R , X ;
y).
The solution £ is unique in the space G (R , X ;y).
Indeed, let £ be another solution of equation (1) in the space G (R , X ;y).
Then from the equalityf(0 = л ( ') Ш + t BMUt+zAtïl+fV)
i= 1
it follows that t e G ( R , X ; y), so there is also â e G ( R , X ; y), where a(t)
= Ç(t) — H(t)Ç(t). From Theorem 1 it follows that the function ô satisfies equation (3) and next, from Theorem 3 it follows that a = cr. Hence ^
5.
Discussion. In papers [1], [4 ] there can be found theorems analogous to Theorem 4. They assume that instead of Condition 2 the inequality(27) erj(a + b) < 1
is satisfied, where b = £ bf, r\ = max for i = 1, 2, . . n and instead of the
i = 1 »
space G (R , X ; y) the space G (R , X ; 1/rj) is considered.
We shall compare Condition 2 and inequality (27). There are four possibilities:
C a s e 1. a = 0, rji = t] for i = 1, 2, . . . , n. Lemma 2 implies the equivalence of Condition 2 and inequality (27).
C a s e 2. я = 0, г]{ < r\ for some i e {1, 2, . .. , n}. From Lemma 2 it follows that inequality (27) is valid, if and only if a number у > 0 exists such that
b eny < y.
For у > 0 we have the inequality
£ b{ еПгУ — y < b em — у
i= 1
from which one can deduce that if Condition 2 is satisfied, then inequality (27) need not be satisfied.
C a s e 3. a > 0, щ = г] for i = 1, 2, . .. , и. Lemma 2 implies the equivalence of Conditions 1 and 2. Denoting и = rjb, v = rja one can state that Condition 1 is satisfied in the region
{(и, v): и, v > 0, и < e~ v~1}
and inequality (27) is satisfied in the region
{(m, t>): u, v > 0, u + v < l/e ] (see the figure)
It can be seen that Condition 1 and consequently Condition 2 are evidently weaker than inequality (27).
C a s e 4. a > 0, r\t < rj for some ie {1, 2, . . . , n}. For у > 0 we have
It
a + Y, btе 1У — у < a + bem — y.
i = 1
Hence, Condition 2 is weaker than Condition 1 and therefore much weaker than condition (27), according to case 3.
Let us compare numbers
y
and\/rj.
If Condition 2 is valid, then numbersô lt ô 2, à3
exist such that 0 < <b2
<S3
and the following equalities are valid:a + Y, 01 = S1, a A- Y bie41** —
£ = 1 i= 1
i b , 4 , e * 2 - 1.
»= 1
Condition 2 is satisfied for any ye(<5l5 <53). For
y e(S 2,
<53) we have Ybi rii e liy
> 1 and a + Y^ е 117
<у.
i = 1 i= l
Then
a + - < a + - Y Ь щ е * 17 < a + Y ^ е ' 7 < y,
П 4 i= 1 ,-=i
hence
a + l/rj < y.
It can be seen that the space G (R , X ; 1 /rj) is a proper subspace of the space G ( R ,X ;y ) .
A ck n ow led g m en t. The author wishes to express his gratitude to Professor Marian Kwapisz for his valuable conversations and suggestions.
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