Generalized Nonhomogeneous Abstract Degenerate Cauchy Problem

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1 Applie Mathematical Sciences, Vol. 7, 213, no. 49, HIKARI Lt, Generalize Nonhomogeneous Abstract Degenerate Cauchy Problem Susilo Hariyanto Department of Mathematics Gajah Maa University, Yogyakarta, Inonesia Department of Mathematics Diponegoro University, Semarang, Inonesia sus2 Lina Aryati Department of Mathematics Gajah Maa University, Yogyakarta, Inonesia Wioo Department of Mathematics Gajah Maa University, Yogyakarta, Inonesia wioo Copyright c 213 Susilo Hariyanto et al. This is an open access article istribute uner the Creative Commons Attribution License, which permits unrestricte use, istribution, an reprouction in any meium, provie the original work is properly cite. Abstract We will iscuss about how to solve the generalize nonhomogeneous abstract egenerate Cauchy problem by factorization metho. Here, generalization is meant the aition of linear operator B on the nonhomogeneous term s. The problem is formulate in a Hilbert space which can be written as an orthogonal irect sum of KerM an Ran M. In this paper we will solve the problem in which B is a boune or close operator. By using certain assumption, the generalize nonhomogeneous abstract egenerate Cauchy problem can be reuce to

2 2442 Susilo Hariyanto, Lina Aryati an Wioo nonegenerate problem, which is easier, because we can transform it to a common normal form that its solution is unerstoo completely. Finally, by using a specific operator the solution of nonegenerate Cauchy problem can be mappe to the solution of generalization problem. Keywors: Degenerate Cauchy problem, nonegenerate Cauchy problem, factorization metho 1 Introuction We are going to investigate how to solve the generalize nonhomogeneous abstract egenerate Cauchy problem. The generalization is meant the aition of linear operator B on the nonhomogeneous term s. The generalize nonhomogeneous abstract egenerate Cauchy problem have been introuce by Thaller an Thaller [7], but they i not investigate completely. They only iscusse problem (1) in special conition, when function f(t) is a feeback control. In this case the problem (1) becomes homogeneous, where its solution is unerstoo in [7]. In this paper, we will iscuss it without assuming that f(t) is a feeback control. Hence the problem (1) is interesting to iscuss. The generalize egenerate Cauchy problem is Mz(t) =Az(t)+Bf(t), (1) t where M, A are linear operators from Hilbert space H to Hilbert space W. The operator B which is the generalization on nonhomogeneous term s, is a linear operator from Hilbert space U to Hilbert space W. If the operator B is an ientity operator, then we have nonhomogeneous abstract egenerate Cauchy problem: t Mz(t) =Az(t)+f(t), z() = z (2) The operator M is not invertible, so we call it the egenerate problem. If operator M is invertible then we call it the nonegenerate. The solution of nonhomogeneous abstract egenerate Cauchy problem (2) have been iscusse by Thaller an Thalller in [7]. For finite imensional, nonhomogeneous abstract egenerate Cauchy problem (2) is unerstoo completely in the book [2] by Dai, where we can transform matrices M an A to a common normal form. In the infinite imensional, Carrol an Showalter [1] treat the Singular an Degenerate Cauchy Problems. Favini [3], [4] investigates egenerate Cauchy problems in Banach spaces. In

3 Nonhomogeneous abstract egenerate Cauchy problem 2443 his articles, the problem is treate uner the assumption that the Banach space of the system can be written as a irect sum of suitable subspaces. Thaller an Thaller [7], [8] emphasize on the possibility of factorization an the relation of the factorize problem with the original egenerate system-without assuming parabolicity. Making use of the ecomposition of the Hilbert space into a irect sum of Ker M an Ran M (resp. Ker M an Ran M) they formulate the conitions which allow us to obtain an equivalent but nonegenerate Cauchy problem in the factor space H/Ker M an give the explicit form of generators of the factorize problems. The crucial assumptions is that the restriction of A to a mapping from Ker M to Ker M is well efine an invertible. This allows to efine a factorization operator Z A which maps the solutions of the factorize system to solutions of the original egenerate system. Hariyanto et. al. [5] have been iscusse the solution of problem (2) by the alternative form on factorization metho. The generalization of nonhomogeneous abstract egenerate Cauchy problem with boune operator on nonhomogeneous term s was solve in [6]. Accoring the first paragraph, problem (1) have been introuce by Thaller an Thaller in their article [7], but they only iscusse problem (1) in special conition when f(t) is a feeback control. In this article we will iscuss problem (1) without assuming that f(t) is a feeback control. Here we will try to solve the generalization problem with boune operator on nonhomogeneous term s by the alternative form on factorization metho. An then accoring the earlier research on the topics, we will also look for sufficient conitions, so the problem with B close operator can be solve by factorization metho in [7] an alternative form on factorization metho in [5]. Moreover, uner the several assumptions we will give an example problem that can be solve by the metho. 2 Main Results We will investigate how to solve the generalize nonhomogeneous abstract egenerate Cauchy problem by factorization metho which was iscusse by Thaler an Thaler [7]. Hence, we use again several notations an assumptions in [7]. Here, D(T ) is represente omain of T, where T is an operator. Accoring of [4], in this article the problem is treate also uner the assumption that the Hilbert spaces of the system can be written as a irect sum of the kernel of M (Ker M ) an the range of ajoint M (Ran M ). In orer to write Hilbert space H as an orthogonal irect sum of Ker M an Ran M, they have assume that operator M : D(M) H Wan A : D(A) H Ware close

4 2444 Susilo Hariyanto, Lina Aryati an Wioo linear operators an ensely efine. So, we efine the orthogonal projection operator P on Hilbert space H onto Ker M an operator Q on Hilbert space W onto Ker M. In orer to transform the problem (2) to nonegenerate problem, they restricte D(M) tod(m r ). Hence, operator M can be reuce to invertible operator M r = M D(Mr) where D(M r )=(Ker M) D(M) an (Ker M) enotes orthogonal complement of Kernel M. Before we iscuss assumptions, lemmas, an theorems for solving the problem (1), we introuce the efinition strict solution of problem (1). Definition 2.1. Function z :[, ) His a strict solution of nonhomogeneous abstract egenerate Cauchy problem with the linear operator on the nonhomogeneous term if z(t) D(A) D(M) for all t, Mz is continuously ifferentiable, an (1) hols. Any strict solution z(t) of the egenerate Cauchy problem (1) clearly satisfies z(t) D A,B for all t, where D A,B = { z D(A) Az + Bf(t) Ran M }, (3) an Ran M enotes the closure of range of M. Next, we efine an operator A which is restriction A on (Ker M), { (P T A x(t) =A ) } 1 {x(t)} DA,B for every x(t) D(A ) { where D(A )= x(t) (KerM) ( P ) } T 1 {x(t)} DA,B φ. In this iscussion, ( P ) T 1 {x(t)} is inverse image of x(t) uner projection P T,sowe efine it by ( P ) T 1 {x(t)} = {x(t)+y(t) y(t) Ker M}, for every x(t) (Ker M). The { operator A, however, will become a multi value operator A on (KerM). (P T If ) } 1 {x(t)} D A,B is singleton, then operator A on (Ker M) become a single value operator. So, we nee the following Assumption 2.2, an 2.3. Assumption 2.2. The operator B : D(B) U W is a linear operator which is Ran B Ran M. Assumption 2.3. P D A,B D(A) an operator (QAP ) P D A,B has a boune inverse. Lemma 2.4. Uner the Assumption 2.2, an 2.3 a vector z(t) His in the subspace D A,B if an only if z(t) D(A) an Pz(t) = (QAP ) 1 QAP T z(t).

5 Nonhomogeneous abstract egenerate Cauchy problem 2445 Proof. By efinition (3) an Assumption 2.2, z(t) D A,B if an only if z(t) D(A) an Q(Az(t) + Bf(t)) = QAz(t) =. By Assumption 2.3, the vector Pz(t) an P T z(t) are in D(A). Writing z(t) =Pz(t) +P T z(t) gives QAP z(t) +QAP T z(t) = from which the result follows immeiately, since QAP is invertible on the range of the projection Q. Therefore any x(t) P T D A,B (KerM) uniquely etermines z(t) D A,B such that x(t) =P T z(t) an z(t) =(1 (QAP ) 1 QA)x(t). Hence the set (P ) 1 x(t) D A,B contains pricesly one element. Accoring to Lemma 2.4, we can efine the operator ZA by Z A = P T (QAP ) 1 QAP T which is efine on D(Z A ) P T D A,B. The restriction Z A P T D A,B is 1 (QAP ) 1 QA on P T D A,B. This is the inverse of the projection P T D A,B. Hence, the operator A can be etermine by A = AZ A, on D(A )=P T D A,B. For every z(t) D A,B, we can write A x(t) =Az(t) where x(t) =P T z(t). Since Az(t) =Q T Az(t) for all z(t) D A,B, the operator A can be written in the more symetric form: A = Q T AP T Q T AP (QAP ) 1 QAP T. To factorize A, we efine an operator Y A = Q T Q T AP (QAP ) 1 Q. Therefore Y A AP =, then Y A AP T = Y A A an A = Y A A on D(A )=P T D A,B. Assumption 2.5. The operator A has a boune inverse an D A,B D(M). This Assumption implies that A D A,B has a boune inverse. A D A,B : D A,B Q T W an (A D A,B ) 1 : Q T W D A,B. Hence, the operator A 1 = P T (AZ A ) 1 = P T A 1 Q T W is boune an ensely efine on Q T W. Accoring to Assumption 2.2, we have: Bf(t) =(Q T Q T AP (QAP ) 1 Q)Bf(t) =Y A Bf(t), (4) for every f(t) D(B). Finally by Assumptions 2.2, 2.3, 2.5 an for every z(t) D A,B D(M), the egenerate Cauchy problem (1) can be reuce to nonegenerate problem: t M rx(t) =A x(t)+y A Bf(t), x(t) =P T z (5)

6 2446 Susilo Hariyanto, Lina Aryati an Wioo where M r is invertible. The nonegenerate Cauchy problem (5) can be written in two types of normal form: t x(t) =A 1x(t)+(M r ) 1 Y A Bf(t), where A 1 =(M r ) 1 A. (6) t y(t) =A 2y(t)+Y A Bf(t), where A 2 = A (M r ) 1. (7) The next step epens on the assumption of operator B an M. Assumption 2.6. Operator B is a boune operator. Solution of problem (1) with B boune operator via normal form (6) has been iscusse completely in [6]. The Assumption 2.7.a. bellow has given in [6], when we solve problem (1) via (6). In solving the problem (1) via normal form (7), we nee the following Assumption 2.7.b. Assumption 2.7. a. Operator M has a close omain. b. Operator M has a close range. Operator M is close an ensely efine, then by using Close Graph Theorem, (M r ) 1 is boune an efine on all of Q T W. If the Assumption 2.7.b. is satisfie, operator A 2 = A (M r ) 1 is close on the omain, D(A 2 )={y Q T W (M r ) 1 y D(A )} = M r P T D A,B = MD A,B. It is close because the prouct of a close operator A an a boune operator (M r ) 1. It is also ensely efine in the Hilbert space W = (MD A,B ). By Assumption 2.2, 2.3, 2.5, we can efine: k(t) =M r P T A 1 Bf(t) =MA 1 Bf(t), an then (7) can be written in the following form: t y(t) =A 2(y(t)+k(t)). Assumption 2.8. A 2 generates a strongly continuous semigroup in W. By Assumptions 2.2, 2.3, 2.5, 2.6, 2.7.b, 2.8, an whenever k(t) isind(a 2 )=

7 Nonhomogeneous abstract egenerate Cauchy problem 2447 MD A,B, the solution of (7) is given by: y(t) =e A 2t y() + A 2 e A 2(t s) k(s)s, where k(t) =MA 1 Bf(t). (8) Lemma 2.9. If z(t) is a solution of (1), then z(t) =Z A M 1 r y(t) (QAP ) 1 QBf(t), for all t Proof. We will prove this Lemma by reuctio a absurum. If Z A Mr 1 y(t) (QAP ) 1 QBf(t) =r(t) an z(t) is a solution, then we have: t Mr(t) = t M [ Z A M 1 r y(t) (QAP ) 1 QBf(t) ] = [ MZ A M 1 r y(t) M(QAP ) 1 QBf(t) ] = t MZ Ax(t) = Mz(t) =Az(t)+Bf(t), for z(t) =r(t). t Moreover, if we assume z(t) r(t), we have: t Mz(t) Mr(t) Az(t)+Bf(t). t It is a contraiction, thefore z(t) =Z A M 1 r y(t) (QAP ) 1 QBf(t). Theorem 2.1. Let f(t) be a continuosly ifferentiabel function with values in Ran M. Uner Assumption 2.2, 2.3, 2.5, 2.6, 2.7.b, an 2.8, the egenerate Cauchy problem (1) for each initial value z D A,B has a unique strict solution: z(t) =Z A Mr 1 y(t) (QAP ) 1 QBf(t), where y(t) =e A 2t y() + A 2 e A 2 (t s) k(s)s, an k(t) =MA 1 Bf(t). Proof. We know that y(t) =e A 2t y() + A 2 e A 2 (t s) k(s)s, where k(t) =M r P T A 1 Bf(t) is a solution of nonhomogeneous abstract egenerate Cauchy problem with boune operator on the nonhomogeneous term (7). By Assumption 2.8, the function t y(t) is continuosly ifferentiable an containe in D(A 2 ) for all t. Therefore z(t) =Z A M 1 r efine. Using the continuity of M 1 r y(t) is well we fin Mz(t) =M r x(t) continuosly

8 2448 Susilo Hariyanto, Lina Aryati an Wioo ifferentiable with t Mz(t) = t M rx(t) =M r t x(t) =M r[a 1 x(t)+(m r ) 1 Y A Bf(t)] = A x(t)+y A Bf(t) =Az(t)+f(t). Hence z(t) is a strict solution of egenerate Cauchy problem (1) with B boune operator. Next, we will investigate problem (1) with B close operator. In orer to solve of problem (1) with B close operator on the nonhomogeneous term, we will replace Assumption 2.6 by Assumption 2.11: Assumption Operator B is a close operator. Operator B is close, so we have a close subspace Ker B in W. Let R be orthogonal projection on Ker B, then R T =1 R is also orthogonal projection on (KerB). Hence we have : RW = Ker B,R T W =(KerB), (9) an the restriction operator B to (KerM) D(B) is efine by: B r = B D(Br), engan D(Br) =(KerB) D(B). (1) Lemma Operator B D(Br) = B r is boune. Assumption Function f(t) U is containe in (KerB). Accoring to Assumption 2.13, we have: Bf(t) =Q T BR T f(t) =(Q T Q T AP (QAP ) 1 Q)BR T f(t) =Y A B r f(t), (11) for every f(t) (Ker B) D(B). An then by using (11) an for every z(t) D A,B, the right han sie on the original problem can be reuce to Az(t)+Bf(t) =A x(t)+y A B r f(t) where x(t) =P T z(t). (12) Hence the generalization of nonhomogeneous abstract egenerate Cauchy problem on nonhomogeneous term s can be reuce to t M rx(t) =A x(t)+y A B r f(t), x() = P T z (13)

9 Nonhomogeneous abstract egenerate Cauchy problem 2449 where M r is invertible. Simillary with (6) an (7), we can transform (13) to normal form: t x(t) =A 1x(t)+(M r ) 1 Y A B r f(t), or (14) t y(t) =A 2y(t)+Y A B r f(t), (15) where A 1 =(M r ) 1 A or A 2 = A (M r ) 1 is a generator of strongly continuous semigroup in H or W. Uner the Assumption 2.5, we can efine l(t) = P T A 1 B r f(t) =P T A 1 BR T f(t) =P T A 1 Bf(t), an then (14) can be written in the following form: t x(t) = A 1x(t)+(M r ) 1 Y A B r f(t) ( = A 1 x(t)+(a1 ) 1 (M r ) 1 Y A B r f(t) ) ( = A 1 x(t)+(a ) 1 Y A B r f(t) ) ( = A 1 x(t)+p T A 1 B r f(t) ) = A 1 {x(t)+l(t)}. If Assumption 2.7.a. is satisfie, the operator A 1 =(M r ) 1 A is close on the omain D(A 1 )= { x P T D A,B A x Ran M } = A 1 Ran M = P T D A,B because it is the prouct of a bounely invertible operator (M r ) 1 an a close operator A. It is also ensely efine in the Hilbert space H = (P T D A,B ). Let l(t) D(A 1 ) an be continuosly ifferentiable, the solution of (14) is: x(t) =e A 1t P T z + t e A 1t(t s) A 1 l(s)s t = e A1t P T z + A 1 e A1t(t s) l(s)s. (16) Solving problem (1) via (15) is analogue, we can efine m(t) =M r P T A 1 B r f(t) =M r P T A 1 BR T f(t) =M r P T A 1 Bf(t), an (16) can be written in the following form : t y(t) =A 2(y(t)+m(t)). (17)

10 245 Susilo Hariyanto, Lina Aryati an Wioo Let m(t) D(A 2 ) an be continuosly ifferentiable, the solution of (17) is : t y(t) =e A21t P T z + e A2t(t s) A 2 m(s)s t = e A2t P T z + A 2 e A2t(t s) m(s)s. (18) Finally, we have the following theorem for generalization of nonhomogeneous abstract egenerate Cauchy problem with close operator on nonhomogeneous term. Theorem i. Let f(t) be a continuously ifferentiable function with value Bf(t) in Ran M. The problem (1) which is satisfie Assumption 2.2, 2.3, 2.5, 2.6, 2.7.a, 2.11, an 2.13, has a unique strict solution, z(t) =Z A x(t) (QAP ) 1 QBf(t) for each initial value z D A,B, where x(t) =e A1t P T z + t e A1t(t s) l(s)s an l(t) =P T A 1 B r f(t). A 1 ii. Let f(t) be a continuously ifferentiable function with value Bf(t) in Ran M. The problem (1) which is satisfie Assumptions 2.2, 2.3, 2.5, 2.6, 2.7.b, 2.11, an 2.13, has a unique strict solution, z(t) =Z A Mr 1 y(t) (QAP ) 1 QBf(t) for each initial value z D A,B, where y(t) =e A2t P T t z +A 2 e A2t(t s) m(s)s an m(t) =M r P T A 1 B r f(t). Proof. i. For each initial value z (t) D A,P T z (t) D(A 1 )=P T D A,B, the function t x(t) is continuously ifferentiable, an x(t) D(A 1 )=P T D A,B for all t. Therefore z(t) =Z A x(t) is well efine. Using the continuity of M r we fin Mz(t) =M r x(t) continuously ifferentiable with: t Mz(t) = M r t x(t) = M r (A 1 x(t)+mr 1 Y A Bf(t)) = A x(t)+y A Bf(t) = Az(t)+Bf(t).

11 Nonhomogeneous abstract egenerate Cauchy problem 2451 This shows that Az(t) + Bf(t) is continuous. Hence z(t) is a strict solution generalization of egenerate Cauchy problem with close operator on nonhomogeneous term. ii. For each initial value z (t) D A,B,Mz (t) D(A 2 ), the function t y(t) is continuously ifferentiable, an y(t) D(A 2 )=MD A,B for all t. Therefore (M r ) 1 is boune, x(t) =Mr 1 y(t) is continuously ifferentiable an z(t) =Z A Mr 1 y(t) is well efine. Hence, Mz(t) = M r x(t) =y(t) is continuously ifferentiable, so: t y(t) = t M rx(t) = Mz(t) =Az(t)+Bf(t). t This shows that Az(t)+Bf(t) is continuous, so z(t) is continuous. Hence function z(t) is a strict solution generalization of egenerate Cauchy problem with close operator on nonhomogeneous term. In the following we give an application of our approach on escriptor system. Example Consier the escriptor system on the Sobolev space W 2,2 (R) 3 : t Mz(t) =Az(t)+Bf(t), z() = z (19) y(t) =Cz(t) (2) where M = 1, A = x x x a 2 x 2 x, B = x, an C = ( 1 ). In this problem we assume that z D A,B an f(t) continuosly ifferentiable.

12 2452 Susilo Hariyanto, Lina Aryati an Wioo We choose orthogonal projection operators: P = Q = R = 1 1, then P T = Q T = R T = Therefore the problem (19) can be reuce to: 1 t M rx(t) =A x(t)+y A BR T f(t), x() = P T z on D(A )=P T D A,B (21) where M r = MP T = 1, an A = Y A A = ( a 1 ) 2 a x 2 Accoring to (14), the problem (21) can be reuce to normal form: t x(t) =A 1x(t)+(M r ) 1 Y A B r f(t), x() = P T z.. on D(A 1 )=P T D A,B, where A 1 = ( a 1 ) 2 a x 2. The assumptions in Theorem 2.14.i. are satisfie then we get a solution of problem (19) in the following z(t) =Z A x(t) (QAP ) 1 QBf(t) for each initial value z D A,B, where x(t) =e A1t P T t z + A 1 an Z A = e A 1t(t s) l(s)s, 1. a x x l(s) =P T A 1 BR T f(s), Hence, the solution of escriptor system (19-2) on P T D A,B =(Ker M) D A,B is y(t) = ( 1 ) z(t).

13 Nonhomogeneous abstract egenerate Cauchy problem 2453 References [1] R.W. Carrol an R.E. Showalter, Singular an Degenerate Cauchy Problems:Math. Sci. Engrg., vol 127, Acaemic Press, New York-San Fransisco-Lonon, [2] L. Dai, A Singular Control Systems, Lecture Notes in Control an Inform, Sci., vol.118, Springer-Verlag, Berlin-Heielberg-New York, [3] A. Favini, Laplace Tranform Metho for a Class of Degenerate Evolution Problems, Ren. Mat. Appl. 12 (1979), [4] A. Favini, Abstract Potential Operator an Spectral Metho for a Class of Degenerate Evolution Problems, J. Differential Equations, 39 (1981), [5] S. Hariyanto, L. Aryati an Wioo, Solving of Degenerate Cauchy Problem Via The Alternative Form on Factorization Problem, Proceeing of InoMS International Conference on Mathematics an Its Applications. (21), [6] S. Hariyanto, L. Aryati an Wioo, Nonhomogeneous Abstract Degenerate Cauchy Problem: The Boune Operator on The Nonhomogen Term, Proceeing of The 6th SEAMS-UGM Conference 211 (212), [7] B. Thaller an S. Thaller, Factorization of Degenerate Cauchy Problems: The Linear Case, J. Operator Theory, 36 (1996), [8] B. Thaller an S. Thaller, Approximation of Degenerate Cauchy Problems. SFB F3 Optimierung un Kontrolle 76 University of Graz. [9] J. Weiman Linear Operators in Hilbert Spaces, Springer-Verlag, Berlin- Heielberg- New York, 198. [1] E. Zeiler, Nonlinear Functional Analysis an Its Applications II/A, Springer-Verlag, Berlin-Heielberg- New York, 199. Receive: February 5, 213