c Allerton Press, Inc., 2014. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2014, Vol. 58, No. 1, pp. 60–69. c V.E. Fedorov and E.A. Omel’chenko, 2014, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2014, No. 1, pp. 71–81. Original Russian Text
Linear Equations of the Sobolev Type with Integral Delay Operator V. E. Fedorov1* and E. A. Omel’chenko2** 1
2
Chelyabinsk State University, ul. Br. Kashirinykh 129, Chelyabinsk, 454001 Russia Ural Branch of the Russian Academy of Justice, pr. Pobedy 160, Chelyabinsk, 454084 Russia Received August 23, 2012
Abstract—We establish sufficient conditions of the local and global solvability of initial value problems for a class of linear operator-differential equations of the first order in a Banach space. Equations are assumed to have a degenerate operator at the derivative and an integral delay operator. We apply methods of the theory of degenerate semigroups of operators and the contraction mapping theorem. As examples illustrating the general results we consider the evolution equation for a free surface of a filtered liquid with a delay and a linearized quasistationary system of equations for a phase field with a delay. DOI: 10.3103/S1066369X14010071 Keywords and phrases: delay equation, Sobolev-type equation, integrodifferential equation, contraction mapping theorem, degenerate semigroup of operators.
1. INTRODUCTION Let U and F be Banach spaces. Consider two linear operators L, M : U → F. The Cauchy problem ˙ = M u(t) is of independent interest on one hand, and, on the u(0) = u0 for the evolution equation Lu(t) other hand, gives a scheme of investigation of initial boundary-value problems for partial differential equations (PDE) that describe various processes in nature and technics [1–3]. For continuously invertible operator L the equation is solvable with respect to the derivative and has the form u(t) ˙ = −1 L M u(t) ([1], P. 57). Otherwise, in particular, if the kernel of the linear operator L is nontrivial, we will call it equation of the Sobolev type ([3], P. 12). In mathematical modeling of many real processes, the effect of delay or the aftereffect takes place when previous states have an influence on the contemporary state of the system. Despite of active development of the theory of functional-differential equations that contains equations with delay (see, e.g., [4] and others monographs and papers of the scientific school by Prof. N. V. Azbelev), the equations of the Sobolev type with delay were studied quite a few. We should note here the papers of the authors [5, 6]. In these papers some methods of the theory of operator semigroups were used, they allow to establish global solvability of the problem u(s) = h(s), s ∈ [−r, 0], for equation of the type Lu(t) ˙ = M u(t) + Φut . Here the operators L and M are linear, ker L = {0}, and the delay operator Φ acting on the function ut (s) = u(t + s) s ∈ [−r, 0], is continuous. But, in addition, essential restrictions on the kernel and the image of the operator Φ and on the function h were imposed. In the present paper 0 the delay operator has the integral form Φut = K(s)u(t + s)ds. We managed to waive the restrictions −r
on Φ and h mentioned above. Using the fixed point theorem, let us show that the value of guaranteed interval of existence of solutions to the considered problems for equations of the Sobolev type with delay is defined by the data of the problems, namely, by the operators L and M , and the operator-function K. It could be finite (local existence of solution), and infinite. Note that the considered problem is also a resonance boundary-value problem. Solvability of various classes of resonance boundary-value problems (more often in the nonlinear case) has been investigated * **
E-mail:
[email protected]. E-mail:
[email protected].
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intensively by many authors for over the last few decades (see, e.g., the papers by A. R. Abdullaev and A. B. Burmistrova [7], A. A. Boichuk [8], M. Furi [9], F. Landesman and A. Lazer [10], B. Przeradzki [11], and others). But the method of the present paper differs from those used in the papers listed above and allows us to obtain results of a somewhat different nature. In the second Section we find conditions on the pair of operators L and M and formulate some results and corollaries from them which are used in the main part of the paper. In the third Section we obtain some sufficient conditions of solvability (local and global) of the Cauchy problem and the generalized Showalter problem for equations of the Sobolev type with integral delay operator. In the fourth Section the obtained abstract results are used in investigation of an initial boundary-value problem for the evolution equation of free boundary of filtered liquid with delay. We give an estimate for the guaranteed time of existence of solution to the problem for the case of simply connected domain. The fifth Section is devoted to investigation of a linearized quasi-stationary system of phase field equations with delay. We give a comparison with analogous results from the paper [6] concerning the system. 2. CONDITIONS ON PAIR OF OPERATORS Let us formulate some conditions on operators which will be used further, and recall some statements proved in [12]. Let U and F be Banach spaces, an operator L : U → F be linear and continuous (we write L ∈ L(U; F) for short), and an operator M : dom M → F be linear, closed and densely defined in U (for short, M ∈ Cl(U; F)). Introduce the denotations N0 = {0} ∪ N,
R+ = {0} ∪ R+ ,
ρL (M ) = {μ ∈ C : (μL − M )−1 ∈ L(F; U)},
RμL (M ) = (μL − M )−1 L,
−1 LL μ = L(μL − M ) .
Definition 1. Let p ∈ N0 . An operator M is called strongly (L, p)-radial if (i) ∃a ∈ R (a, +∞) ⊂ ρL (M ); (ii) ∃K ∈ R+ ∀μ ∈ (a, +∞) ∀n ∈ N n(p+1)
L(F) ≤ max (RμL (M ))n(p+1) L(U) , (LL μ (M ))
K ; (μ − a)n(p+1)
◦
(iii) there exists a dense in F lineal F such that ◦ const(f ) ∀f ∈ F; p+2 (μ − a) K
(RμL (M ))p+1 (μL − M )−1 L(F;U) ≤ (μ − a)p+2
p+1 f F ≤
M (μL − M )−1 (LL μ (M ))
for every μ ∈ (a, +∞). Here c(f ) is a constant depending on f . Remark 1. In [13] we prove equivalence of the conditions from Definition 1 to analogous more complicated ones used in [12]. Definition 2. An operator family {U (t) ∈ L(U) : t ∈ R+ } is called resolving semigroup of the equation Lu(t) ˙ = M u(t) if (i) U (s)U (t) = U (s + t) ∀s, t ∈ R+ ; (ii) for every u0 from a dense lineal in U the vector-function u(t) = U (t)u0 is a solution from the class ˙ = M u(t); C 1 (R+ ; U) to the equation Lu(t) RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 58 No. 1 2014
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(iii) for every operator family {V (t) ∈ L(U) : t ∈ R+ } satisfying conditions (i) and (ii) we have im V (0) ⊂ im U (0). p+1 ; let U1 be the closure of the image of the operator Let U0 = ker(RμL (M ))p+1 , F0 = ker(LL μ (M )) L p+1 1 p+1 in the space F. in the space U, and F be the closure of the image im (LL im (Rμ (M )) μ (M )) Denote by Lk (Mk ) the restriction of the operator L (M ) to Uk (dom Mk ≡ dom M ∩ Uk ), k = 0, 1.
Theorem 1 ([12]). Let the operator M be strongly (L, p)-radial. Then (i) U = U0 ⊕ U1 , F = F0 ⊕ F1 ; (ii) Lk ∈ L(Uk ; Fk ), Mk ∈ Cl(Uk ; Fk ), k = 0, 1; 1 1 (iii) there exist operators M0−1 ∈ L(F0 ; U0 ) and L−1 1 ∈ L(F ; U );
(iv) the operator H = M0−1 L0 is nilpotent of degree ≤ p; (v) there exists a strongly continuous semigroup of operators {U (t) ∈ L(U) : t ∈ R+ } resolving the equation Lu(t) ˙ = M u(t); (vi) ∀t ∈ R+ U (t) ≤ Keat . Remark 2. In (vi) the constants a and K are the same as in the definition of strongly (L, p)-radial operator. Remark 3. In the case ker L = {0} the unit element U (0) of the resolved semigroup for the equation Lu(t) ˙ = M u(t) is a nontrivial projector, ker L ⊂ ker U (0)=U0 , and im U (0)=U1 . Denote P = U (0), and let Q be the projector onto the subspace F1 along the subspace F0 . Remark 4. In the proof of Theorem 1 (ii) the equalities M P u = QM u, u ∈ dom M , and LP = QL play an essential role. Later they will be used explicitly. Theorem 2 ([12]). Let an operator M be strongly (L, p)-radial, and let a function g ∈ C 1 ([0, T ]; F) be such that (I − Q)g ∈ C p+1 ([0, T ]; F0 ). Then (i) for every u0 ∈ dom M such that (I − P )u0 = −
p
H k M0−1 ((I − Q)g)(k) (0)
(1)
k=0
there exists a unique solution u ∈ C 1 ([0, T ]; U) ∩ C([0, T ]; dom M ) to the Cauchy problem u(0) = u0 for the equation Lu(t) ˙ = M u(t) + g(t),
t ∈ [0, T ],
(2)
and t u(t) = U (t)u0 + 0
U (s)L−1 1 Qg(t
− s)ds −
p
H k M0−1 ((I − Q)g)(k) (t);
(3)
k=0
(ii) for u0 ∈ dom M1 there exists a unique solution u ∈ C 1 ([0, T ]; U) ∩ C([0, T ]; dom M ) to the problem P u(0) = u0 for Eqs. (2). The solution has the form (3). Writing C([0, T ]; dom M ) we mean that the definition domain dom M is provided with the norm of the graph · dom M = · U + M · F of a closed operator M and, consequently, it is Banach space. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 58 No. 1 2014
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3. LOCAL SOLVABILITY OF EQUATION OF THE SOBOLEV TYPE WITH DELAY Let U and F be Banach spaces, L ∈ L(U; F), and M ∈ Cl(U; F). Consider the problem u(t) = h(t),
t ∈ [−r, 0],
for the linear integro-differential equation of the Sobolev type 0 Lu(t) ˙ = M u(t) + K(s)u(t + s)ds, −r
(4)
t ∈ [0, T ]
(5)
where K : [−r, 0] → L(U; F). Solution of problem (4), (5) on the segment [−r, T ) is a function u ∈ C 1 ([0, T ); U) ∩ C([−r, T ); U) satisfying Eq. (5) on [0, T ] and condition (4). Theorem 3. Let an operator M be strongly (L, p)-radial, h ∈ C([−r, 0]; U), h(0) ∈ dom M , K ∈ C p+1 ([−r, 0]; L(U; F)), and K(n) (−r) = K(n) (0) = 0 for n = 0, . . . , p, 0 p k −1 (−H) M0 (I − Q) K(k) (s)h(s)ds. (6) (I − P )h(0) = − −r
k=0
Then for some T0 ∈ (0, +∞] there exists a unique solution u ∈ C 1 ([0, T0 ); U) ∩ C([−r, T0 ); U) of problem (4), (5). Proof. For t ≥ 0, n = 0, 1, . . . , p + 1 we denote 0 0 (n)
K (s) L(U;F) ds ≡ Kn (t), (t + s) K(n) (s) L(U:F) ds ≡ Kn,1 (t). −t
−t
At the origin the values are defined by continuity and, therefore, equal zero. Let hk = H k M0−1 (I − Q) L(F;U) , k = 0, . . . , p, C = L−1 1 Q L(F;U) , K, a ≤ 0; at K(t) = max{K, Ke } = at Ke , a > 0, for t ≥ 0; here K and a are constants from Definition 1. Then, by Theorem 1 (vi),
U (t) L(U) ≤ K(T ) for all t ∈ [0, T ]. Denote tr = min{t, r} and for t ≥ 0 consider the function p+1 p+1 p+1 p+1 Kn,1 (tr ) + h0 Kn (tr ), h1 Kn (tr ), . . . , hp Kn (tr ) . F (t) = max CK(t) n=0
n=0
n=0
n=0
It is obvious that the function is positive, increasing and continuous. Besides, F (0) ≡ lim F (t) = 0. t→0+
If a > 0, then lim F (t) = +∞, except for the case of trivial K. Thus, the function F is unbounded t→+∞
and there exists T0 = min{t ∈ R : F (t) = 1} > 0, F (T ) < 1 for T < T0 . For a ≤ 0 and for all t ≥ r we have F (t) = F (r). Therefore, if F (r) < 1, then F (t) < 1 for all t ≥ 0. In this case we put T0 = +∞. Let us fix a number T ∈ (0, T0 ) and consider the auxiliary equation (2) on the segment [0, T ]. According to Theorem 2 (i), for a function g ∈ C p+1 ([0, T ]; F), satisfying the concordance condition (1) for a given u0 = h(0), the solution u ∈ C 1 ([0, T ]; U) ∩ C([0, T ]; dom M ) to the problem u(0) = h(0) for Eq. (2) exists, it is unique and has the form (3) for t ∈ [0, T ]. Let us define the operator 0 K(s)u(t + s)ds [Φg](t) = −r
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= −t
−
0
[Φg](t) = −r
K(s)U (t + s)h(0)ds +
0 −t
K(s)
p
0
−t
t+s
K(s) 0
U (τ )L−1 1 Qg(t + s − τ )dτ ds
H k M0−1 (I − Q)g(k) (t + s)ds +
k=0
0
K(s − t)h(s)ds
for t ∈ [0, r),
t−r
K(s)u(t + s)ds
0
= −r
K(s)U (t + s)h(0)ds + −
0
−r
0
−r
t+s
K(s) 0
K(s)
p
U (τ )L−1 1 Qg(t + s − τ )dτ ds
H k M0−1 (I − Q)g(k) (t + s)ds
for t ∈ [r, T ].
k=0
Here the function u for t ∈ [0, T ] is defined by the function g according to formula (3), and for t < 0 it coincides with h. Thus, it is a solution u ∈ C 1 ([0, T ]; U) ∩ C([−r, T ]; U) to problem (2), (4). Using the method of mathematical induction we easily make sure that under conditions of the theorem for t ∈ [0, T ], n = 0, 1, . . . , p + 1 0 (n) n K(n) (s)u(t + s)ds, [Φg] (t) = (−1) −r
because K(n) (−r) = K(n) (0) = 0 for all n = 0, . . . , p. Here we use the change of variables from s to s − t before differentiation and the inverse change after that. We have the following obvious equalities 0 K(n) (s)h(s)ds, n = 0, 1, . . . , p + 1. (7) [Φg](n) (0) = (−1)n −r
Consider the Banach space C p+1 ([0, T ]; F) with the norm
g p+1 =
p+1
sup g(k) (t) F,
k=0 t∈[0,T ]
and the metric space
p −1 p+1 k (k) ([0, T ]; F) : − H M0 (I − Q)g (0) = (I − P )u(0) X= g∈C k=0
with the distance d(g, h) = g − h p+1 . The space X is nonempty, because it contains the nonconstant function g ≡ −M h(0). Indeed, by Remark 4, M0−1 (I − Q)M h(0) = M0−1 M (I − P )h(0) = (I − P )h(0). We will show that X is a full metric space. Let
lim gn − gm p+1 = 0 where {gn } ⊂ X. Then, due
n,m→∞
to completeness of C p+1 ([0, T ]; F), there exists g ∈ C p+1 ([0, T ]; F) for which lim gn − g p+1 = 0. n→∞
Therefore, p k −1 (k) H M0 (I − Q)g (0) + (I − P )h(0) k=0
F
p p −1 −1 k (k) k (k) ≤ H M0 (I − Q)g (0) − H M0 (I − Q)gn (0) k=0
k=0
F
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p p −1 k (k) + H M (I − Q)g (0) + (I − P )h(0) ≤
g − g
hk → 0 n p+1 n 0 F
k=0
k=0
as n → ∞. Thus, g ∈ X. Taking into account condition (6) and equalities (7), we have the operator Φ : X → X. Further, for g1 , g2 ∈ X 0 t+s K(s) U (t + s − τ )L−1 [Φg1 ](t) − [Φg2 ](t) = 1 Q(g1 (τ ) − g2 (τ ))dτ ds −tr
0
−
Φg1 − Φg2 p+1 ≤ CK(T )
p+1
0
−tr
K(s)
p
H k M0−1 (I − Q)(g1 (t + s) − g2 (t + s))ds, (k)
(k)
k=0
Kn,1 (Tr ) g1 − g2 0 +
n=0
p+1
Kn (Tr )
n=0
p
hk g1 − g2 k
k=0
≤ F (T ) g1 − g2 p+1 . Since F (T ) < 1, the mapping Φ is contractible in the full metric space X. Consequently, by the fixed point theorem, there exists a unique element g0 ∈ X such that g0 = Φg0 . In this case the function t p (k) U (t − s)L−1 Qg (s)ds − H k M0−1 (I − Q)g0 (t) u(t) = U (t)h(0) + 0 1 0
k=0
is simultaneously a solution to problems (4), (5) and (2), (4), because Lu(t) ˙ − M u(t) = g0 (t) = [Φg0 ](t). Since the conducted arguments are valid for any T < T0 , solution to problem (4), (5) exists on [−r, T0 ). Let u1 , u2 ∈ C 1 ([0, T ); U) ∩ C([−r, T ); U) be two solutions to problem (4), (5) for T < T0 . For 0 i = 1, 2 we denote gi (t) = K(s)ui (t + s)ds. As above, it is easy to show that −r
(n)
gi (0) = (−1)n
0
−r
K(n) (s)h(s)ds,
n = 0, 1, . . . , p + 1,
i = 1, 2.
(8)
By (8) and (6), we have Lu˙ i − M ui = gi ∈ X, and, by the definition of the mapping Φ, Φgi = gi , Due to uniqueness of the fixed point of the mapping Φ : X → X we obtain the equality g1 ≡ g2 . Denote ˙ − M w(t) ≡ 0, w(0) = 0. By Theorem 2 (i), w ≡ 0, w(t) = u1 (t) − u2 (t), t ∈ [0, T ), then we have Lw(t) therefore, solution to problem (4), (5) is unique. Remark 5. The restrictions on T0 are defined by the operators L and M , the operator function K, and the number r > 0, only, they do not depend on h. From the conducted arguments we obtain Corollary 1. Let an operator M be strongly (L, p)-radial, h ∈ C([−r, 0]; U), h(0) ∈ dom M , K ∈ C p+1 ([−r, 0]; L(U; F)), K(n) (−r) = K(n) (0) = 0 for n = 0, . . . , p, 0 p (−H)k M0−1 (I − Q) K(k) (s)h(s)ds, (I − P )h(0) = − k=0
−r
and F (r) < 1. Then there exists a unique solution u ∈ C 1 ([0, +∞); U) ∩ C([−r, +∞); U) to problem (4), (5). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 58 No. 1 2014
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In addition to problem (4), we consider an analog of the generalized Showalter problem [14, 15] P u(t) = h(t),
t ∈ [−r, 0].
(9)
Theorem 4. Let an operator M be strongly (L, p)-radial, h ∈ C([−r, 0]; U1 ), h(0) ∈ dom M1 , K ∈ C p+1 ([−r, 0]; L(U; F)), and K(n) (−r) = K(n) (0) = 0 for n = 0, . . . , p. Then for some T0 ∈ (0, +∞] there exists a unique solution u ∈ C 1 ([0, T0 ); U) ∩ C([−r, T0 ); U) to problem (5), (9). Proof differs from that in the previous theorem only by the fact that we use the metric space X = C p+1 ([0, T ]; F). This is stipulated by the fact that in Theorem 2 (ii) on solvability of the problem P u(0) = u0 for Eq. (2) we do not use condition (1) of matching the initial value with the function in the right-hand side of the equation; the condition is necessary for solvability of the Cauchy problem. Corollary 2. Let an operator M be strongly (L, p)-radial, h ∈ C([−r, 0]; U1 ), h(0) ∈ dom M1 , K ∈ C p+1 ([−r, 0]; L(U; F)), and K(n) (−r) = K(n) (0) = 0 for n = 0, . . . , p, F (r) < 1. Then there exists a unique solution u ∈ C 1 ([0, +∞); U) ∩ C([−r, +∞); U) to problem (5), (9). 4. DZEKTSER EQUATION WITH DELAY As an example of application of Theorem 3, we consider the initial boundary-value problem z(x, t) = z0 (x, t), z(x, t) = Δz(x, t) = 0,
(x, t) ∈ Ω × [−1, 0], (x, t) ∈ ∂Ω × [0, T ),
(10) (11)
for the Dzektser equation; it describes evolution of free boundary of a filtered liquid [16] with memory 0 k(s)z(x, t + s)ds, (x, t) ∈ Ω × [0, T ). (12) (λ − Δ)zt (x, t) = z(x, t) − βΔ2 z(x, t) + −1
Here a bounded domain Ω ⊂
Rn
has a smooth boundary, λ ∈ R, and β ∈ R+ .
Denote by λm , m ∈ N, the eigenvalues of the Laplace operator defined on H02 (Ω) = {u ∈ H 2 (Ω) : u(x) = 0, x ∈ ∂Ω} and acting in L2 (Ω); the numeration of λm is made according to decrease of their values and with account taken of their multiplicities. Let {ϕm : m ∈ N} be an orthonormal system of eigenfunctions of the operator. We assume that λm = λ for some m, i.e., Eq. (12) is not solvable with respect to zt . Theorem 5. Let λ = 0, λ = β −1 , z0 ∈ C([−1, 0]; H02 (Ω)), z0 (·, 0) ∈ H 4 (Ω), z0 (x, 0) = Δz0 (x, 0) = 0 for x ∈ ∂Ω, k ∈ C 1 ([−1, 0]; R), k(−1) = k(0) = 0, and 0
1 k(s)z (·, s)ds, ϕ (13) z0 (·, 0), ϕm = − 0 m λ − βλ2 −1 for m ∈ N such that λm = λ. Then for some T0 ∈ (0, +∞] there exists a unique solution u ∈ C 1 ([0, T0 ); H02 (Ω)) ∩ C([−1, T0 ); H02 (Ω)) to problem (10)–(12). Proof. Consider the spaces U = H02 (Ω), F = L2 (Ω), and the operators L = λ − , M = Δ − βΔ2 with domain of definition dom M = {u ∈ H 4 (Ω) : u(x) = Δu(x) = 0, x ∈ ∂Ω}. In [15] (theorem 5.1) it is proved that for the case λ = 0, λ = β −1 the operator M is strongly (L, 0)-radial. It is obvious that the function z0 (·, t) matches to h(t) from condition (4) for t ∈ [−1, 0]. For all s ∈ [−1, 0] the operators K(s) : H02 (Ω) → L2 (Ω), s ∈ [−1, 0] have the form [K(s)u](·) = k(s)u(·), and K(s) L(H02 (Ω);L2 (Ω)) ≤ |k(s)|, K ∈ C 1 ([−1, 0]; L(H02 (Ω); L2 (Ω))). Condition (13) is exactly coincides with condition (6), because in this case we have I − P = I − Q = ·, ϕm ϕm (see [15]). Application of Theorem 3 completes the proof. λm =λ
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Now we will estimate the value T0 for problem (10)–(12) letting n = 1, Ω = (0, π), λ = −1, and β = 2, and applying the results of the paper [15]. In this case λm = −m2 , ϕm = sin mx for m ∈ N, L−1 1 Q
∞ ·, sin mx sin mx , = m2 − 1 m=2
therefore, for v ∈ L2 (Ω) ∞ (1 + m4 )|v, sin mx|2 , (m2 − 1)2 m=2 √ √ 1 + m4 17 −1 C = L1 Q L(L2 (Ω);H 2 (Ω)) = sup = . 2 3 m=2,3,... m − 1 2
L−1 1 Qv H 2 (Ω) =
Analogously, ·, sin x sin x 1 = − ·, sin x sin x, 2 3 λ1 − 2λ1 √ 2 −1 , h0 = M0 (I − Q) L(L2 (Ω);H02 (Ω)) = 3 √
√ −m2 − 2m4 17 17 = −12, K(t) ≡ K = sup 1, = . a = sup 2−1 m 3 3 m=2,3,... M0−1 (I − Q) =
By Theorem 3, we have T0 = min{T > 0 : F (T ) = 1} where √ 0 2 17 0 (T + s)(|k(s)| + |k (s)|)ds + (|k(s)| + |k (s)|)ds, F (T ) = 9 −T1 3 −T1 If k ≡ C1
