ON THE BOUNDARY VALUE PROBLEMS FOR LOADED DIFFERENTIAL EQUATIONS
MUVASHARKHAN T. DZHENALIEV
ABSTRACT. The equations prescribed innc H:' are called loaded,
if they contain some operations of the traces of desired solution on manifolds (of dimension which is strongly less thann) from closure
O. These equations result from approximations of nonlinear equa- tions by linear ones, in the problems of optimal control when the control actions depends on a part of independent variables, in in- vestigations of the inverse problems and so on. In present work we study the nonlocal boundary value problems for first-order loaded differential operator equations. Criterion of unique solvability is es- tablished. We illustrate the obtained results by examples.
1. Introduction
Boundary value problems for the loaded equations arise and find a wide applications in many applied problems [1], [5], [7], [9]. However they are not always stated correctly. The classical solution of Existence Questions of local boundary value problems for loaded equations are considered in work [1], [2] spectral problems - [3], generalized solvability - [6], [8], etc. Theses equations find a wide applications in many applied problems. In present work we study a nonlocal boundary value problems for loaded differential operator equations of first order. Criterion of unique solvability of considered problems is established. We illustrate the obtained results by examples.
Received July 16, 1999.
2000 Mathematics Subject Classification: 35D05,34K06, 34K10.
Key words and phrases: Evolution equation, loaded differential equation, bound- ary value problem.
2. The Statement of Problem and Main Results
Let (x E) 0 C Rn be a cube with a 21r-ribs; poo be a linear manifold of the smooth all over variables periodic complex-valued functions; H ==
L2(0). Let
be the polynomials with constant coefficients. Further we define the linear differential operators Ak(-iD), k = 0,1, ... , m, i = yCI, Do: =
Dfl .. .D~n, Dj =0/oxj, as the closure inH ofthe differential operations primary-defined onpoo such that
n
Ak(-iD)exp{is· x} = Ak(s)exp{is· x}, s·x =:2::>jXj, k = 0,1, ... , m.
j=l
The operators Ak , k = 0,1, ... , m, will be called by IT-operators [4].
We denote
m
A(s) = LAk(s), k=l
m
B(s) =21rA(s)+(J.L - 1)[1+L Ak(s)tk],
k=l whereJ.L E <C (the complex number).
Let S = {s ={sj}j=llsj =0,±1, ±2, ... ,\fj}, SO ={sls E s, Ao(s)=
O}, lk =degree {Ak(s)}, k=0,1, ... , m, f=max{lk'k =1, ... ,m}, 11=
L2(0, 21r; H), and
IJ.LI < +00, lk ~ 10 , k =1, ... ,m,
(2.1) degree{B(s)} = degree{A(s)} = f, B(s)ls=o-=I 0, 0< tl < ... < tm < 21r, {tk}~l are a fixed points.
We consider the boundary value problem with conditions (2.1):
(2.2)
{
LIU == (Dt +Ao(-iD))u(t) +~Ak(-iD)u(tk) = f(t) in (0, 27r), f Jlu== j.Lu(O) - u(27r) = cp.
The conditions (2.1) rule out the cases j.L = ±oo that correspond to the Cauchy problem [7].
DEFINITION 2.1. The function u(x,t) E 1£ is strong solution of the problem (2.2) if there is a sequence of the functions {Uj(x,t)}~l C C1(O, 271";POO) such that Uj(x,t) -+ u(x,t), L1uj = h -+ f in 1£, and fJluj -+ cp in H.
The following theorem is established.
THEOREM 2.1. Tbe problem (2.2) for"'If E 1£ and cp E H admits a unique strong solution in1£ ifI
(2.3) B(s) =f
°
VsE So,(2.4) C(s)=j.L-exp{-Ao(s)27r}=fO VSES\So,
(2.5)
m
D(s) == 1+[AO(s)t1LAk(s){l
k=l
- [C(s)t1(j.L - 1)exp{-AO(s)tk}} =f
°
VsES \ So.For j.L= 1 we obtain from Theorem 2.1
COROLLARY 2.1. Tbe problem (2.2) for"'If E 1£ and cp E H admits a unique strong solution in1£ ifI
(2.6) (2.7)
A(s)+Ao(s) =f 0 Vs E S,
Ao(s) =f iq Vs E S\So, q= ±1,±2, ....
REMARK 2.1. The condition (2.6) follows from (2.3) and (2.5), the condition (2.7) follows from (2.4).
3. Auxiliary results
We shall find a solution of the problem (2.2) by Fourier method:
(3.1) U= L Us (t)eis.x, 1=L Is (t)eis.X•
sES SES
For the coefficients of the series (3.1) we have
According to the ([4], p.ll8) we have:
LEMMA 3.1. Boundaryvalue problem (2.2) hasaunique strong solu- tion in the space1-£ forVI E 1-£,'\/rp EH ifI each of the problems(3.2), (3.3) admit a unique solution and there is a constant C >
°
that doesn't depend on s and such that(3.4) lIus(t)IIL.!(0,21r) :S ClII/s(t)IIL2(0,21r)+l!Psl] Vs ES.
By the Lemma 3.1 Theorem 2.1 will follow immediately from the following Lemmas.
LEMMA 3.2. For allsE SO and each Is E C(O,27l') the Problem (3.2) has a uniqueUs E C1(O,27l')nC[O,27l'] and the estimate (3.4) is true ifI the condition (2.3) is held.
LEMMA 3.3. For all s E S \ SO and each Is E C(O,27l') the Problem (3.3) has a unique Us E C1(O,27l')nC[O,27l'] and the estimate (3.4) is true ifI the conditions (2.4) and (2.5) are held.
Proof of the Lemma 3.2. At the conditions (2.3) only the Problem (3.2) has a unique solution
(3.5)
t m
us(t) = / fs(r) dr+[21rA(s)t1[1+L Ak(s)(tk - t)]
o k=1
211" m tk
. [ / fs(r)dr +'Ps] - [A(s)t1LAk(s) / fs(r)dr, if ft = 1,
o k=l 0
(3.6)
t m ~
us(t) = / fs(r) dr - [B(s)t1[t(ft - 1)+21r] L Ak(s)
J
fs(r) dro k=l 0
+{(ft - 1)-1 - [B(s)t1A(s) [t +(ft - 1)-121T] }
211"
. ['Ps +
J
fs(r)dr], if ft i= 1,o
where A(s), B(s) are defined in section 2.
These representations (3.5), (3.6) are obtained by the following way.
Firstly we integrate the equation (3.2) and obtain
t m
(3.7) us(t) = us(O) + / fs(r)dr - t LAk(s)Us(tk).
o k=l
Secondly we have for t = 21r from (3.7)
211" m
(3.8) (ft -l)usCO)='Ps +
J
fs(r)dr - 21r LAk(s)us(tk)'o k=l
Further we have for t = tk,k = 1, , m, from (3.7) (3.9)
[1+
tAk(S)!k] [~A'(S)U,(tk)]~A(s)u,(O)+~Ak(S)I
j,(r)dr.Specifically, from here we have for f-L = 1
(3.10)
(3.11)
u,(O) ~ [211"A(8)t' [I+t,Ak(S)tk]
. [I'" +
1
f'(T)dT] - [A(sJr' t,Ak(S)1
f,(T)dT;for f-L =I 1 correspondingly
(3.12)
t,Ak(S)",(tk)~ [B(s)t'A(s) [I'" +
1
f'(T)dT]m tk
+[B(s)tl(f-L -1) LAk(S)
J
fs(7)d7;k=l °
",(0) ~ (I-' - I)-'{I - 211"[B(8)t'A(8)} [10,+
1
f,(T)dT](3.13)
m tk
- 21r[B(s)t1LAk(s)
J
fs(T)d7.k=l °
Thus from equalities (3.10), (3.11), (3.12), (3.13) and (3.7) we obtain the representations (3.5), (3.6) of unique solution of problem (3.2).
According to (2.1) there is the constantC such that IB(s)I-1
IAk(s)1 ::; C for all k= 1, ... ,m, and sESo.
Here the constant C is independent on k,s. For f-L = 1 we note this condition is identical to the following
IA(s)!-1 IAk(s)1 ::; C for all k= 1, ... ,m, and sESo.
(3.14)
The desired estimate (3.4) follows from here and (3.5). This finishes
the proof of the Lemma 3.2. 0
Proof of the Lemma 3.3. At the. conditions (2.4), (2.5) only the Problem (3.3) has a unique solution
211"
us(t) =
J
Gs(t, T)fs(T)dT+[C(s)tlcpsexp{-Ao(s)t}o
27r m 27r
- [D(s)tl
J
Gs(t, T)dTL Ak(s)J
Gs(tk' T)fs(r)dT,o k=l °
whereGs(t, r) is the Green function (3.15)
{
Jl[C(S)]_l exp{-Ao(s)(t - TH
Gs(t, T) =
[C(S)]-l exp{-Ao(s)(271'+t - TH
and C(s), D(s) are defined in (2.4), (2.5).
We note that the following equality
if O::S: T ::; t ::; 271', if O::S: t ::;r ::; 271',
27r
(3.16)
J
Gs(t, T)dT = [Ao(s)tl{1 - [C(s)tl(Jl - 1)exp{-Ao(s)t}}is hold. °
Considering (3.15) the representation (3.14) is obtained similarly to Lemma 3.2.
Itonly remains to establish the estimates (3.4) for the solutions (3.14).
The first item of the solution (3.14) estimate similar to ([4], p.12Q-- 121)
27r
(3.17)
J
Gs(t,T)fs(r)dr ::;Kl llfs(T)IIL2(o,27r) '° L2(O,27r)
where the constant Kl doesn't depend on s.
By the equality (3.16) the estimate of the third item of the solution (3.14) follows from the inequalities
• because of (3.17) we obtain
211"
J
Gs(tk, 7)fs(7)d7 :S K2"fS(7)"~(0,211") Vs E S \ SO, k = 1, ... ,m;°
• because of D(s) ¥= 0 (2.5) there is the constant 8> 0 such that ID(s)I-1:S8-1< +00 VsE S \ SO;
• by definition C(s) (2.4) we have
11- [C(s)t1(jL - 1) exp{-Ao(s)t}I:s K3Vs E S \ Sa, t E (0, 21TJ;
• because of (2.1) and (2.3) we have
IAo(s)I-1IAk(s)1 :sK4 VsE S \ SO k = 1, ... ,m;
where the constants 8,K2 ,K3 and K4 don't depend on s.
By (2.3) the second item of the solution (3.14) yields a estimate
I[C(s)t 1exp{-Ao(s)t}CPsl :s K51CPsl Vs ES \ So, t E (0,21T].
Thus the desired estimates (3.4) follows by the established inequali- ties.
This finishes the proof of the Lemma 3.3. 0
4. Proof of the Theorem 2.1
To finish the proof of the Theorem 2.1 it is necessary to prove the closure of our problem operator (2.2) in the space 1i., that is to check the correctness offollowing proposition ([10], p.209).
PROPOSITION 4.1. The operator T +E is closed ifT is closed and E is bounded in theV(E), VeT) C V(E).
HereVeX) is the domain of a operator X.
Let I :::; 10/2. Then we define T as the operator £1 of problem (2.2) without item ~~=1Ak(-iD)u(tk)' and E =~~=1 Ak(-iD)u(tk)' By the trace theorem ([11]' p.33) we obtain
V(T)={vlvEk! (0,27r;V(Ao»), DIvE1/.}CC([0,27r];[V(Ao),Hh/2)~V(E),
where [X, Y]o is the interpolating space ([11]' p.23), X and Y are the Hilbert spaces, the embeddingX c Yis dense and continuous,eE (0,1).
Now let10/2 < [:::; 10 , [= degree{Ak(s)},wherek= arg{max{lk'k =
1, ... , m}}. Then we define T and E as the operators Tu = Diu +
m
Ak(-iD)u(tk), Eu= Ao(-iD)u+ ~ Ak(-iD)u(tk)' k=l,k#/C
By the trace theorem ([11], p.35-36, 55-56) we obtain V(T) = {vlv E £2(0,271";V(AkAk)), Djv E 1l}
c {C([O,271"];V(Ak )) n£2(0,271";V(Ao))} ~ V(E).
It only remains to apply Proposition 4.1. Proof of Theorem 2.1 is completed.
5. Examples
Let Q= {x, tlO< x, t < 271"}.
(i). For the boundary value problem:
m
(5.1) (Dj - D;)u+'L(ak+f3kD~)u(x, tk)=f, {x, t} E Q, k=l
(5.2) D;u(O, t) =D;u(271", t), j =0,1; J1u(x,O) =u(x, 271"), the following corollaries take place:
COROLLARY 5.1. The problem (5.1)-(5.2) for Vf E L2(Q) admits a unique strong solution u E L2 (Q) iff
(5.3) (5.4)
(5.5)
m
L ak[27r+(j.t - l)tk] +j.t - 1i= 0 k=1
j.t - exp{-s227r} i= 0 Vs E S \ {O},
m
1+S-2L(ak +is.Bk){l - [j.t - exp{-s227r}t1
k=1
. (j.t - 1) exp{-s2tk }} i= 0 Vs E S \ {O}.
COROLLARY 5.2. Let j.t = 1. Then the problem (5.1)-(5.2) for Vf E
L2(Q) admits a unique strong solutionU E L2(Q) iff
(5.6)
m
L(ak+is· .Bk)+82 i=0 Vs E S.
k=1
COROLLARY 5.3. Let j.t = 1 and.Bk = 0,k = 1, ... ,m. Then the problem (5.1)-(5.2) for Vf E L2(Q) admits a unique strong solution
U E L2(Q) iff
(5.7) ( m)1/2
-{;ak tj:S.
REMARK 5.1. For j.t =1 the correctness of the boundary value prob- lem(5.1)-(5.2) doesn't depend on the pointstk. This follows from corol- laries 5.2, 5.3.
(ii). Leta =const. For the boundary value problem:
m
(5.8) (DJ - D;+a)u+L aku(x, tk) =f, {x, t} E Q,
k=1
(5.9) D~u(O,t) =D~u(27r,t), j =0,1,2; j.tu(x,O) =u(x,27r), the following corollaries take place:
COROLLARY 5.4. The problem (5.8)-(5.9) for "If E L2(Q) admits a unique strong solution u E L2(Q) if[
m m
(5.10) (5.11)
(5.12)
21fL D:k+(/1-1)[1+L D:ktkJ :f=0
k=l k=l
j.t :f=exp {- 21fa},
m
is3+a+L D:k{1 - [j.t - exp {- 21fa}r1
k=l
.(j.t - 1) exp{-(is3+a)tk}} :f= 0 Vs ES \ So.
(5.13)
COROLLARY 5.5. Let a:f= 0, j.t =1.Then the problem (5.8)-(5.9) for
"If EL2(Q) admits a unique strong solution u E L2 (Q) if[
~ ( m )~
sinh(1f'Y) :f=0, ~'Y rJ.S, ~here l'= a+
f;
D:kThese corollaries follow from Theorem 2.1 and Corollary 2.1. The conditions (5.3)-(5.12) of the corollaries 5.1-5.4 follow from (2.3)-(2.5) and (2.6)-(2.7) correspondingly.
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