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Quadratic functionals for second order matrix equations on time scales
Nonlinear Analysis 33 (1998) 675 – 692
Quadratic functionals for second order matrix equations on time scales Ravi P. Agarwal ∗ , Martin Bohner 1 Department of Mathematics, National University of Singapore, Lower Kent Ridge Road, Singapore 119260, Singapore Received 1 May 1997; accepted 12 May 1997
Keywords: Time scale; Self-adjoint matrix equation; Variational problem; Jacobi’s condition; Sturmian theory
1. Introduction We begin this paper by stating two well-known results concerning quadratic functionals. Result 1 deals with the “continuous” case while Result 2 represents the “discrete” counterpart. We let n ∈ N. Result 1 (Jacobi’s condition). Let R and P be continuous n × n-matrix valued functions on R such that R(t) and P(t) have real entries and are symmetric and such that R(t) is positive de nite for each t ∈ R. Let a; b ∈ R with a¡b. Then the quadratic functional Z b FC (y) = {y˙T Ry˙ − yT Py}(t) dt a
is positive de nite, i.e., FC (y)¿0 for all nontrivial piecewise C 1 -functions y : [a; b] → Rn with y(a) = y(b) = 0, if and only if the solution Y of d [R(t)Y˙ ] + P(t)Y = 0; dt
Y (a) = 0;
Y˙ (a) = R−1 (a)
satis es Y (t) invertible for all t ∈ (a; b]: ∗
(1)
Corresponding author. E-mail: [email protected].
1
The author is grateful to Alexander von Humboldt Foundation for awarding him a Feodor Lynen Research Fellowship to support this work. 0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 6 7 5 - 5
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Result 2 (Discrete version of Jacobi’s condition). Let Rk and Pk be symmetric n × nmatrices with real entries such that Rk is invertible for each k ∈ Z. Let M; N ∈ Z with M ¡N − 1. Then the discrete quadratic functional FD (y) =
N −1 X
T {(yk )T Rk (yk ) − yk+1 Pk yk+1 }
k=M
is positive de nite, i.e., FD (y)¿0 for all nontrivial sequences y : [M; N ] ∩ Z → Rn with yM = yN = 0, if and only if the solution Y of [Rk Yk ] + Pk Yk+1 = 0;
YM = 0;
YM = R−1 M
satis es Yk invertible for all k ∈ {M + 1; : : : ; N }; −1 −1 Rk Yk Yk+1
and
positive de nite for all k ∈ {M + 1; : : : ; N − 1}:
(2)
The proofs of these results and many related topics can be found in recent monographs by Kratz [15] (“continuous” case; see also [11, 17]), and Ahlbrandt and Peterson [3] (“discrete” case; see also [2, 6–9, 16]). Besides featuring minor techniT Pk yk+1 in FD ), the cal di erences (note that yT (t)P(t)y(t) in FC is replaced by yk+1 following two discrepancies occur in the rst view: • In Result 2 the matrices Rk need not be positive de nite, while Result 1 does not hold without this assumption. • The invertibility of the so-called principal solution in Eq. (1) is not enough for Result 2; in fact we need to supplement this condition by an additional condition in Eq. (2). As such it is not clear why these strange di erences occur. In this paper, we will present a theorem on a time scale T which covers both Results 1 and 2 in the sense that they are special cases of our theorem for T = R and T = Z. Such a uni cation by the theory of time scales was initiated by Hilger in [12]. Besides this paper we shall refer to the work of Aulbach and Hilger [4] and the recently published monograph by Kaymakcalan et al. [13]. Some of our results extend the work of Erbe and Hilger [10] to systems. However, the main result of our paper, Theorem 5, is new even in the scalar case. For the convenience of comparing our theorem from Results 1 and 2 above we rst state our version of Jacobi’s condition on time scales. For the notation needed we refer the reader to the next section on preliminaries about time scales. At this moment the only important thing for observing how Results 1 and 2 follow from our theorem is to know that (t) = (t) = t if T = R and (t) = t − 1; (t) = t + 1 if T = Z. Also, the theory has been prepared in such a manner that y = y˙ in the case T = R and y = y in the case T = Z. Result 3 (Jacobi’s condition on time scales). Let R and P be rd-continuous n × nmatrix valued functions on T such that R(t) and P(t) are symmetric and have real entries and such that R(t) is invertible for each t ∈ T. Let a; b ∈ T with a¡(b). Then
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the quadratic functional Z b {(y )T Ry − (y )T Py }(t)t FT (y) = a
1 -functions y : [a; b] ∩T is positive de nite, i.e., FT (y)¿0 for all nontrivial piecewise Crd → Rn with y(a) = y(b) = 0, if and only if the solution Y of
[R(t)Y ] + P(t)Y = 0;
Y (a) = 0;
Y (a) = R−1 (a)
satis es Y (t) Y (t)Y
invertible for all t ∈ (a; b] ∩ T; and −1
−1
((t))R
(t)
positive de nite for all t ∈ (a; (b)] ∩ T:
(3)
The organisation of this paper is as follows: In the next section we introduce time scales and collect some of their basic properties which are needed later. The proofs of these properties and further details can be found in [4, 13]. Section 3 gives a short motivation why it is useful to examine quadratic functionals by considering certain variational problems on time scales. Section 4 contains preliminaries and several fundamental results (e.g. Picone’s identity) for self-adjoined second order matrix equations and corresponding Riccati matrix equations on time scales. Section 5 proves the above mentioned time-scales-version of Jacobi’s condition. We conclude this paper by giving Sturmian separation and comparison results on time scales in Section 6. 2. Preliminaries about time scales A time scale T is de ned to be any closed subset of R. Hence the jump operators ; : T → T (t) = inf {s ∈ T: s¿t}
and
(t) = sup{s ∈ T: s¡t}
(supplemented by inf ∅ := sup T and sup ∅ := inf T) are well-de ned. The point t ∈ T is called left-dense, left-scattered, right-dense, right-scattered if (t) = t; (t)¡t; (t)= t; (t)¿t, respectively. The graininess : T → R+ 0 is de ned by (t) = (t) − t. Throughout we x a; b ∈ T with a¡(b) and put T˜ = [a; b] ∩ T. By T we shall always denote a nonempty subset of T. Also, whenever we write T = [a1 ; b1 ] ∩ T, we mean a1 ; b1 ∈ T with a1 ¡(b1 ). We de ne ( T if T is unbounded above; T = T\((max T); max T] otherwise: We say that a mapping f from T into some real Banach space X is di erentiable at t ∈ T provided (see [10] (Section 2)) f (t) := lim s→t
f((t)) − f(s) ; (t) − s
where s → t; s ∈ T\{(t)};
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exists. The function f is called di erentiable on T if f (t) exists for all t ∈ T. The following lemma contains results for this derivative. We shall write f for f ◦ . Lemma 1. Let f; g : T → X and t ∈ T . Then the following hold: (1) If f (t) exists, then f is continuous in t; (2) If t is right-scattered and f is continuous in t; then f (t) = [f((t))−f(t)]=(t); (3) If f (t) exists, then f((t)) = f(t) + (t)f (t); (4) If f (t); g (t) exist and (fg)(t) is de ned, then (fg) (t) = f((t))g (t) + f (t) g(t); (5) If f exists on T and f is invertible on T; then (f−1 ) = −(f )−1 f f−1 on T ; (6) If |f | ≤ g on T; then |f(s) − f(r)| ≤ g(s) − g(r) for all r; s ∈ T with r ≤ s; (7) If f = 0 on T; then f is constant on T. Proof. For (1) and (2) see [4] (Theorem 3) or [13] (Theorem 1.2.2). While (3) follows (note that (t) = 0 if t is right-dense) from (1) and (2), (4) and (5) are from [4] (Theorem 4) or [13] (Theorem 1.2.3). Finally, for (6) and (7) see [4] (Theorem 5) or [13] (Corollary 1.3.1). Let t1 ; : : : ; tm ∈ T . A function f : T\{t1 ; : : : ; tm } → X is called regulated on T provided f has a left-sided limit at all left-dense points and a right-sided limit at all rightdense points of T. An f is called rd-continuous on T – we write f ∈ Crd (T) = Crd (T; X ) – if it is regulated on T and continuous at all right-dense points of T. We write 1 (T) if f is di erentiable on T with f ∈ Crd (T ). If f is continuous on T f ∈ Crd and if it is di erentiable on T except at t1 ; : : : ; tm ∈ T such that f is regulated on T, we abbreviate this piecewise rd-continuous di erentiability by writing f ∈ Cp1 (T). Note that in this case (apply Lemma 1(2)) the points t1 ; : : : ; tm are necessarily leftdense and right-dense at the same time. Moreover, f is said to be regressive on T if I +(t)f(t) is invertible for each t ∈ T, where I denotes the identity in X . Next, for any on T regulated function f : T\{t1 ; : : : ; tm } → X there exists a unique pre-antiderivative F (see [13] (Theorem 1.4.2)), i.e., a continuous function F : T → X that is di erentiable on T \{t1 ; : : : ; tm } and satis esR F (t) = f(t) for all t ∈ T \{t1 ; : : : ; tm }. We then s de ne the Cauchy integral of f by r f(t)t = F(s) − F(r), where r; s ∈ T. If F satis es F (t) = f(t) for all t ∈ T, then F is said to be an antiderivative of f on T. In the following lemma we collect properties of rd-continuous functions that are needed in our work. The terminology “right-sequence for t” is used for strictly decreasing sequences {tm }m∈N ⊂ T with limm→∞ tm = t. Note that due to the de nition of there always exist right-sequences for any right-dense t ∈ T (except if t = max T in case T is bounded above). Lemma 2. Let f ∈ Crd (T). Then the following hold: (1) f has an antiderivative; (2) If f is regressive on T and if t˜ ∈ T; g˜ ∈ X , then any initial value problem g = f(t)g; g(t˜) = g˜ has a unique solution on T; R (t) (3) t f( ) = (t)f(t) hold for all t ∈ T ;
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Rt (4) If t1 ∈ T and f(t1 )¿0, then there exists t2 ∈ T with t12 f(t) t¿0; (5) If a is right-dense and if {am }m∈N ⊂ T = T˜ is a right-sequence for a, then Z b Z b lim f(t) t = f(t) t; m→∞
am
a
(6) If T = [a1 ; b1 ] ∩ T; fm ∈ Crd (T), and limm→∞ fm = f uniformly on T , then Z b1 Z b1 lim fm (t)t = f(t) t: m→∞
a1
a1
Proof. While (1) is from [4] (Theorem 6) or [13] (Theorem 1.4.4), see [4] (Theorem 8) or [13] (Theorem 2.5.1) for (2). Now, (3) follows from (1) and Lemma 1(3). For (4) note the following: Either t1 is right-scattered, then by (3) Z (t1 ) f(t) t = (t1 )f(t1 )¿0; t1
or t1 is right-dense, then f is continuous in t1 and there exists a ¿0 and t2 ∈ (t1 ; ∞) ∩ T such that f(t) ≥ for all t ∈ [t1 ; t2 ] ∩ T. In this case there exist antiderivatives F (according to part (1)) and G (because of lim s→t
G((t)) − G(s) (t) − s = lim = =: g(t) s→t (t) − s (t) − s
actually G(t) = t) of f and g, and an application of Lemma 1(4) yields Z t2 f(t) t = F(t2 ) − F(t1 ) ≥ G(t2 ) − G(t1 ) = (t2 − t1 )¿0: t1
Finally, (5) follows from (1), and (6) is from [12] (Theorem 4.3) or [13] (Theorem 1.4.3). In this paper we shall consider matrix-valued functions. The set of m × n-matrices with real entries will be abbreviated by Rm×n . 3. Variational problems on time scales The motivation of our investigations in Sections 4–6 comes by examining variational problems of the form Z b L(t; y (t); y (t)) t → min; (V) L(y) := a
y(a) = ;
y(b) = ;
where ; ∈ Rn and L : T × R2n → R is a C 2 -function in the last two variables (by Lx and Lu we denote the derivatives of L with respect to the second and the third variables, respectively). For simplicity we allow in this section only functions y : T˜ → Rn that
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˜ A function are di erentiable on T˜ with continuous derivative y , i.e., y ∈ C 1 (T). n 1 ˜ ˜ = , and y(b) ˜ = is called a (weak) local minimum y˜ : T → R with y˜ ∈ C (T), y(a) ˜ with of (V) provided there exists a ¿0 such that L(y) ≥ L(y) ˜ for all y ∈ C 1 (T) y(a) = ; y(b) = , and ky− yk¡; ˜ and the local minimum y˜ is called proper provided L(y) = L(y) ˜ if and only if y = y. ˜ Here we shall consider the (according to Lemma 1(1) well-de ned) norm kfk = max |f (t)| + max |f (t)| t∈T˜
t∈T˜
˜ for f ∈ C 1 (T):
˜ with (a) = (b) = 0. Now let : T → R n be any admissible variation, i.e., ∈ C 1 (T) We put () = (; y; ) = L(y + );
∈ R:
We suppose that L is such that we have Z b ˙ = {Lx (·) (t) + Lu (·) (t)} t; L1 (y; ) := (0) a
and = L2 (y; ) := (0)
Z a
b
{( (t))T Lxx (·) (t) + 2( (t))T Lxu (·) (t) + ( (t))T Luu (·) (t)} t;
where (·) = (t; y (t); y (t)). Then y is said to satisfy the Euler equation and L2 (y) is called positive semide nite if L1 (y; ) = 0 and L2 (y; ) ≥ 0, respectively, for all admissible variations . If L2 (y; )¿0 for all nontrivial admissible variations , then we say that L2 (y) is positive de nite. The following two results provide necessary and sucient conditions for (weak) local minima of (V) in terms of positive de niteness of the quadratic functional L2 (y). Concerning the corresponding results for the discrete case we refer to [1] (Example 1.6.7), [5] (Theorems 1 and 2), and [14] (Theorems 8.1 and 8.9). Theorem 1 (Necessary condition). If y˜ is a local minimum of (V), then y˜ satis es ˜ is positive de nite. the Euler equation and L2 (y) Proof. Let be any admissible variation and de ne : R → R as above, namely ˜ (y˜ +)(a) = ; (y˜ + )(b) = , and, () = (; y; ˜ ). We then have y˜ + ∈ C 1 (T); since there exists a ¿0 such that L(y) ≥ L(y) ˜
˜ for all y ∈ C 1 (T)
with y(a) = ; y(b) = ; ky − yk¡; ˜
we also have () = L(y˜ + ) ≥ L(y) ˜ = (0)
for all ∈ R
with ||¡
: kk
Hence has a local minimum at ˜ = 0, and thus our assertion follows.
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˜ satis es the Euler equation; y(a) Theorem 2 (Sucient condition). If y˜ ∈ C 1 (T) ˜ = ˜ is positive de nite; then y˜ is a proper local minimum of (V). and y(b) ˜ = ; and L2 (y) ˜ is positive de nite, it is also strongly positive (see, e.g., [11] Proof. Since L2 (y) (Footnote 4, p. 100)), i.e., there exists an ¿0 with L2 (y; ˜ ) ≥ 2(b − a)kk2 : Since L is a C 2 -function in x and u, there exists a ¿0 such that |L00 (t; z) − L00 (t; z˜)|¡ ˜ be such that y 6= y; ˜ y(a) = ; whenever z; z˜ ∈ R2n with |z − z˜|¡. Let y ∈ C 1 (T) y(b) = , and ky − yk¡. ˜ We shall show L(y)¿L(y). ˜ To do so, we put = y − y˜ so that is a nontrivial admissible variation. Now, using the above remarks and Taylor’s theorem, we nd that there exists a ∈ (0; 1) with 1 1 ˙ = () L(y) − L(y) ˜ = (1) − (0) = (0) + () 2 2 1 1 1 1 = (0) + { () − (0)} ≥ (0) − | () − (0)| 2 2 2 2 !T ! Z (t) 1 b (t) 1 t {L00 (t; z(t)) − L00 (t; z˜(t))} = (0) − 2 2 a (t) (t) 1 ≥ (b − a)kk − 2 2
= where we put z=
y˜ + y˜ +
Z a
b
1 kkkkt = (b − a)kk2 − (b − a)kk2 2
b−a kk2 ¿0; 2
! and
z˜ =
y˜
!
y˜
˜ . Hence y˜ is indeed a proper so that |z(t) − z˜(t)| ≤ kk¡kk¡ holds for all t ∈ T local minimum of (V). 4. Self-adjoint second order matrix equations on time scales Let R; P ∈ Crd (T; R n×n ) be such that R(t) is symmetric and invertible and P(t) is symmetric for each t ∈ T. Consider the self-adjoint second order matrix equation (E)
[R(t)Y ] + P(t)Y = 0:
1 (T; R n×n ) a solution of (E) provided We call Y ∈ Crd
{[RY ] + PY }(t) = 0
holds for all t ∈ (T ) :
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1 Note that any solution Y of (E) satis es RY ∈ Crd (T ). We shall prove rst the following basic result.
Lemma 3 (Wronskian identity). Let Y and Y1 be solutions of (E). Then Y T RY1 − (Y )T RY1
is a constant function on T :
Proof. We apply the product rule Lemma 1(4) to obtain [Y T RY1 − (Y )T RY1 ] = (Y )T [RY1 ] + (Y )T RY1 −[(Y )T R] Y1 −(Y )T RY1 = −(Y )T PY1 + (Y )T PY1 = 0 on (T ) . Now the claim follows from Lemma 1(7) with T = T . Next, putting Z=
Y
!
RY
on T
and
S=
0
R−1
−P
−PR−1
! on T;
it is easy to see that Y solves (E) if and only if (note that Y = Y + Y by Lemma 1(3)) Z solves Z = S(t)Z:
(4)
Now S ∈ Crd (T; R2n×2n ) is regressive on T because !−1 I R−1 I − 2 R−1 P −1 {I + S} = = P −P I − 2 PR−1
−R−1 I
!
holds on T. Hence, by Lemma 2(2), any initial value problem (4);
Z(a) = Za ;
i:e:;
(E);
Y (a) = Ya ;
Y (a) = Ya0
with any choice of a 2n × n-matrix Za , i.e., of n × n-matrices Ya and Ya0 , has a unique solution. We let Y˜ be the unique solution of the initial value problem (E);
Y (a) = 0;
Y (a) = R−1 (a):
This Y˜ is called the principal solution of (E) (at a), and according to the Wronskian identity, Lemma 3, it satis es {Y˜ T RY˜ − (Y˜ )T RY˜ }(t) ≡ {Y˜ T RY˜ − (Y˜ )T RY˜ }(a) = 0 for all t ∈ T , i.e., {Y˜ T RY˜ }(t), and hence (where these are de ned) both ˜ = {RY˜ Y˜ −1 }(t) Q(t)
and
˜ = {Y˜ (Y˜ )−1 R−1 }(t) D(t)
are symmetric matrices for each t ∈ T . Furthermore, we call Y a conjoined solution of (E) provided Y T RY is symmetric on T , and two conjoined solutions Y and Y1
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683
of (E) are called normalized if Y T RY1 − (Y )T RY1 = I on T . Besides the principal solution Y˜ we will also need the associated solution Y˜1 of (E) (at a) which is de ned to be the solution of the initial value problem (E);
Y (a) = −I;
Y (a) = 0:
Of course, Y˜ and Y˜1 are normalized conjoined solutions of (E). We shall now consider the matrix Riccati equation (R)
Q + P(t) + QT {R(t) + (t)Q}−1 Q = 0:
Lemma 4 (Riccati equation). If (E) has a conjoined solution Y such that Y (t) is invertible for all t ∈ T; then Q de ned by Q(t) = R(t)Y (t)Y −1 (t)
for t ∈ T
(5)
is a symmetric solution of (R) on T . Conversely, if (R) has a symmetric solution Q on T , then there exists a conjoined solution Y of (E) such that Y (t) is invertible for all t ∈ T and relation (5) holds. Proof. First we assume that Y solves (E) and is invertible on T. We then de ne Q by Eq. (5) so that on T , Q + P + QT {R + Q}−1 Q = Q + P + QT {I + (Y − Y )Y −1 }−1 R−1 Q = [RY ] (Y )−1 − RY (Y )−1 Y Y −1 + P + RY Y −1 (Y Y −1 )−1 R−1 RY Y −1 = −PY (Y )−1 − RY (Y )−1 Y Y −1 + P + RY (Y )−1 Y Y −1 = 0 holds according to Lemma 1(3)=(4), i.e., Q is a symmetric solution of (R) on T . Conversely, let Q be a symmetric solution of (R) on T , let t0 ∈ T, and let Y be the solution of the initial value problem (note that I + R−1 Q = R−1 (R + Q) is invertible on T and apply Lemma 2(2)) Y = R−1 (t)Q(t)Y;
Y (t0 ) = I:
Then Y is a fundamental solution of Y = R−1 (t)Q(t)Y and hence invertible on T (see [4] (Theorem 9)). Furthermore, it satis es on (T ) , [RY ] = [QY ] = Q Y + QY = −PY − Q(R + Q)−1 QY + QR−1 QY = −PY + Q(R + Q)−1 {(R + Q)R−1 QY − QY } = −PY : We extend Y to a solution of (E) so that, since {Y T RY }(t0 ) = Q(t0 ) is symmetric, Y is indeed a conjoined solution of (E).
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Our next result is the main tool for proving Jacobi’s condition in the next section. Theorem 3 (Picone’s identity). Let ∈ R n and suppose Y and Y1 are normalized conjoined solutions of (E) such that Y is invertible on T. We put Q = RY Y −1
on T
and
D = Y (Y )−1 R−1
on T :
Let t ∈ T and y : T → Rn be di erentiable at t. Then we have at t {yT Qy + 2 T Y −1 y − T Y −1 Y1 } = (y )T Ry − (y )T Py − {Ry − Qy − (Y −1 )T }T D{Ry − Qy − (Y −1 )T }: Proof. According to the di erentiation rules, Lemma 1(4)=(5), we have at t [Y −1 Y1 ] = −(Y )−1 Y Y −1 Y1 + (Y )−1 Y1 = −(Y )−1 R−1 QY1 + (Y )−1 R−1 RY1 = (Y )−1 R−1 (Y −1 )T {Y T RY1 − (Y )T RY1 } = (Y )−1 R−1 (Y −1 )T ; and (also using Lemmas 1(3) and 4), [yT Qy] + {Ry − Qy}T Y (Y )−1 R−1 {Ry − Qy} = (y )T Qy + (y )T Q y + (y )T Qy + (y )T RY (Y )−1 y −yT QY (Y )−1 y − (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Qy − (y )T Py − (y )T QY (Y )−1 R−1 Qy + (y )T Qy +(y )T RY (Y )−1 y − yT QY (Y )−1 y − (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Qy − (y )T Py − (y )T QY (Y )−1 R−1 Qy + (y )T Qy + (y )T R{Y − Y }(Y )−1 y − yT QY (Y )−1 y − (y )T R{Y − Y }(Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy − (y )T RY (Y )−1 y − yT QY (Y )−1 y + (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy − (y )T QY (Y )−1 R−1 Qy
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= (y )T Ry − (y )T Py + (y )T Qy − (y )T QY (Y )−1 R−1 Qy − {y + y}T QY (Y )−1 y + {y + y}T QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy − (y )T QY (Y )−1 R−1 Qy − (y )T QY (Y )−1 y + (y )T QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy −(y )T Q{Y − Y }(Y )−1 y + (y )T QY (Y )−1 R−1 Q{y − y } = (y )T Ry − (y )T Py + (y )T Q{Y (Y )−1 − Y (Y )−1 Y Y −1 }y = (y )T Ry − (y )T Py so that [yT Qy + 2 T Y −1 y − T Y −1 Y1 ] − (y )T Ry + (y )T Py = −{Ry − Qy}T Y (Y )−1 R−1 {Ry − Qy} − 2 T (Y )−1 Y Y −1 y + 2 T (Y )−1 y + T (Y )−1 R−1 (Y −1 )T = −{Ry − Qy − (Y −1 )T }T Y (Y )−1 R−1 {Ry − Qy − (Y −1 )T }; and hence our claimed identity follows. 5. Jacobi’s condition on time scales In this section we will prove the main result of this paper. As it is motivated by the investigations of Section 3, we will consider a quadratic functional of the form Z b F(y) = {(y )T Ry − (y )T Py }(t) t: a
This functional F is called positive de nite – we write F¿0 – provided F(y)¿0
˜ Rn )\{0} for all y ∈ Cp1 (T;
with y(a) = y(b) = 0:
1 Lemma 5 (Quadratic functional). If T = [a1 ; b1 ] ∩ T; y ∈ Crd (T; Rn ); and
{[Ry ] + Py }(t) = 0 for all t ∈ (T ) ; then
Z
(b1 )
a1
{(y )T Ry − (y )T Py }(t) t = {yT Ry }((b1 )) − {yT Ry }(a1 ):
Proof. We have on (T ) [yT Ry ] = (y )T [Ry ] + (y )T Ry = (y )T Ry − (y )T Py so that the assertion follows.
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Our next result, Theorem 4, o ers a sucient condition for the positive de niteness of F. A speical case of this result yields one direction of Jacobi’s condition, which we shall prove in Theorem 5. Theorem 4 also nds application to establish Sturm’s separation theorem, this we shall present in Theorem 6. Theorem 4 (Sucient condition for positive de niteness). A sucient condition for F¿0 is that there exist normalized conjoined solutions Y and Y1 of (E) with Y invertible on (a; b] ∩ T
and
Y (Y )−1 R−1 positive de nite on (a; (b)] ∩ T:
˜ . Pick any y ∈ Proof. We let Y and Y1 as above and put D = Y (Y )−1 R−1 on T 1 ˜ n Cp (T; R ) with y(a) = y(b) = 0. First we shall consider the case that a is rightscattered so that we have (a)¿a and (a)¿0. By Lemma 2(3) and Theorem 3 (with = 0) we have Z b Z (a) {(y )T Ry − (y )T Py }(t) t + {(y )T Ry − (y )T Py }(t)t F(y) = a
(a)
= {(y )T Ry − (y )T Py }(a)(a) Z +
b
(a)
{[yT Qy] + (Ry − Qy)T D(Ry − Qy)}(t) t
= {(y )T Ry −1 − (y )T Py }(a) + {yT Qy}(b) −{yT Qy}((a)) +
Z
b
(a)
{(Ry − Qy)T D(Ry − Qy)}(t) t
≥ {(y )T Ry −1 − (y )T Py }(a) + {yT Qy}(b) − {yT Qy}((a)) = {(y )T Ry −1 − (y )T Py − (y )T [RY Y −1 ] y }(a) = (y )T {R−1 − P − (RY + [RY ] )(Y )−1 }y (a) = (y )T {R−1 − P − (RY −1 − PY )(Y )−1 }y (a) = 0: If however, F(y) = 0; then because of Lemma 2(4) {Ry − Qy}(t) = 0
for all t ∈ [(a); (b)] ∩ T;
i.e., y (t) = {Y Y −1 y}(t) for all t ∈ [(a); (b)] ∩ T. Since I + Y Y −1 = I + [Y − Y ]Y −1 = Y Y −1 ; the map Y Y −1 is regressive on [(a); (b)] ∩ T so that by Lemma 2(2) the initial value problem y = {Y Y −1 }(t)y
on [(a); (b)] ∩ T;
y(b) = 0
˜ so that F¿0 follows. admits only one, namely the trivial solution. Hence y = 0 on T
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
687
˜ for a. With Next, let a be right-dense and pick a right-sequence {am }m∈N ⊂ T T limm→∞ am = a. Let m ∈ N. We put m = −{(Y ) Ry}(am ) and apply Picone’s identity (with = m ) to obtain the following: Z b {(y )T Ry − (y )T Py }(t) t am Z b = [yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m ] (t) t am
Z
+ Z ≥
b
am b
am
{Ry − Qy − (Y −1 )T m }T D{Ry − Qy − (Y −1 )T m }(t) t
[yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m ] (t) t
= −{ mT Y −1 Y1 m }(b) − {yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m }(am ) = −{ mT Y −1 Y1 m }(b) + {yT Qy + yT QY1 (Y )T Ry}(am ) = −{ mT Y −1 Y1 m }(b) + {yT Qy + yT (Y −1 )T [Y T RY1 − I ](Y )T Ry}(am ) = −{ mT Y −1 Y1 m }(b) + {yT RY1 (Y )T Ry}(am ); and hence by Lemma 2(5) Z b F(y) = lim {(y )T Ry − (y )T Py }(t) t m→∞
am
≥ − lim { mT (Y −1 Y1 )(b) m + yT (RY 1 )(am ) m } = 0: m→∞
Finally, if F(y) = 0, we know that limm→∞ where
Rb am
{zmT Dzm }(t) exists and equals 0,
zm := Ry − Qy − (Y −1 )T m → Ry − Qy =: z;
m→∞
holds uniformly (observe also Lemma 1(1)) on [ak ; b] ∩ T for each k ∈ N. Now let k ∈ N. We rst note Z b Z b T {zm Dzm }(t) t ≥ {zmT Dzm }(t) t for all m ≥ k: am
ak
Hence, by applying Lemma 2(6) we have Z b Z b lim {zmT Dzm }(t) t ≥ {z T Dz}(t) t: m→∞
am
ak
Now we let k → ∞ and apply Lemma 2(5) to obtain Z b Z b 0 = lim {zmT Dzm }(t) t ≥ {z T Dz}(t) t ≥ 0: m→∞
am
a
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R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
˜ so that as in the rst part of this Hence, again by Lemma 2(4), we have z = 0 on T proof the triviality of y follows. Therefore, F is again positive de nite and the proof of our result is complete. ˜ if the principal solution Y˜ of We say that the equation (E) is disconjugate (on T) (E) satis es Y˜ invertible on (a; b] ∩ T
and
Y˜ (Y˜ )−1 R−1 positive de nite on (a; (b)] ∩ T:
Now we are ready to prove the main result of this paper (see also Result 3 of Section 1). Theorem 5 (Jacobi’s condition). F¿0
if and only if (E) is disconjugate:
Proof. First, if (E) is disconjugate, then we may apply our Theorem 4 with Y˜ and Y˜1 , where Y˜ is the principal solution of (E) and Y˜1 is the associated solution of (E), and hence disconjugacy implies F¿0. Conversely, suppose that (E) is not disconjugate. Then there exists a t0 ∈ T with exactly one of the following two properties: ˜ such that Y˜ (t) is invertible for all t ∈ (a; t0 ) ∩ T and Y˜ (t0 ) is singular, (1) t0 ∈ T\{a} ˜ \{a} such that Y˜ (t) is invertible for all t ∈ (a; b] ∩ T and D(t ˜ 0) = (2) t0 ∈ T {Y˜ (Y˜ )−1 R−1 }(t0 ) is not positive de nite. ˜ Y˜ d) Let d ∈ R n \{0} with Y˜ (t0 )d = 0 in Case (1) and dT (Y˜ T RY˜ )(t0 )d = (RY˜ d)T D(R (t0 ) ≤ 0 in Case (2). Putting ( y(t) =
Y˜ (t)d for t ≤ t0 ; 0 otherwise
˜ except yields y(a) = y(b) = 0, y(t) 6= 0 for all t ∈ (a; t0 ), and, moreover, y ∈ Cp1 (T) for Case (2) with t0 right-dense. Hence we assume for the moment not being in this situation, and we will deal with this special case later. Now it is easy to see that both y (t) and y (t) are zero for all t¿t0 so that {(y )T Ry − (y )T Py }(t) = 0
for all t¿t0 :
In the following we use Lemmas 5 and 2(3), and for ∈ R we put for convenience † = 0 if = 0 and † = −1 if 6= 0. We have Z F(y) =
a
Z =
a
(t0 )
t0
{(y )T Ry − (y )T Py }(t) t
{(y )T Ry − (y )T Py }(t) t +
Z
(t0 )
t0
{(y )T Ry − (y )T Py }(t) t
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
689
= {dT Y˜ T RY˜ d}(t0 ) + {(y )T Ry − (y )T Py }(t0 )(t0 ) = dT {Y˜ T RY˜ + Y˜ T RY˜ † }(t0 )d = dT {† Y˜ T RY˜ + [1 − † ]Y˜ T RY˜ }(t0 )d ≤ 0: Therefore F 6¿ 0. Now we will discuss the remaining Case 2 with right-dense t0 . This situation implies R(t0 ) 6¿ 0, i.e., there exists a d ∈ R n such that dT R(t0 )d¡0. We have to show F 6¿ 0, and for the sake of achieving a contradiction we assume F¿0. First we suppose that ˜ be a right-sequence for t0 . For m ∈ N t0 is left-scattered. Then t0 6= b. Let {tm }m∈N ⊂ T we put √tm − t d if t ∈ (t ; t ] ∩ T; 0 m ym (t) = t −t m0 0 otherwise: √ ˜ Therefore Hence ym (a) = ym (b) = 0, ym (t0 ) = tm − t0 d 6= 0, and y ∈ Cp1 (T). Z 0 ¡ F(ym ) =
(tm )
t0
{(y )T Ry − (y )T Py }(t) t
d tm − (t) tm − (T ) dT R(t) √ − dT √ P(t) √ d t tm − t0 tm − t0 tm − t0 tm − t0 t0 ) ) ( ( Z (tm ) Z (tm ) (tm − (t))2 1 T T =d R(t)t d − d P(t)t d tm − t0 t0 tm − t0 t0 Z
=
(tm )
√
→ dT R(t0 )d¡0;
m → ∞;
which yields our desired contradiction. The remaining case of left-dense t0 can be treated the same way with ym (t) = or
ym (t) =
t−t ∗ √ m ∗ d
t0 −tm
0
if t ∈ [tm∗ ; t0 ) ∩ T; otherwise;
t−t ∗ √ m d if t ∈ [tm∗ ; t0 ) ∩ T; t0 −tm∗ √tm −t d if t ∈ (t0 ; tm ] ∩ T; tm −t0 0 otherwise;
˜ if t0 = b or otherwise, respectively, using a strictly increasing sequence {tm∗ }m∈N ⊂ T with limm→∞ tm∗ = t0 .
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R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
6. Sturmian theory on time scales We call a solution Y of (E) basis whenever Y (a) = n: rank Y (a) A conjoined solution is said to have no focal points in (a; b] provided it satis es ˜ Y invertible on T\{a}
and
˜ \{a}: Y (Y )−1 R−1 positive de nite on T
˜ if and only if the principal Using this terminology, equation (E) is disconjugate (on T) solution of (E) (at a) has no focal points in (a; b]. Concerning conjoined bases of (E) we have the following version of Sturm’s separation result on time scales. Theorem 6 (Sturm’s separation theorem). Suppose there exists a conjoined basis of ˜ (E) with no focal points in (a; b]. Then equation (E) is disconjugate (on T). Proof. Let Y be a conjoined basis of (E) satisfying Y invertible on (a; b] ∩ T
and
Y (Y )−1 R−1 positive de nite on (a; (b)] ∩ T:
Since Y is a basis, we note that the matrix K = Y T (a)Y (a) + (Y )T (a)R2 (a)Y (a)
is invertible:
Let Y1 be the solution of (E);
Y1 (a) = −R(a)Y (a)K −1 ;
Y1 (a) = R−1 (a)Y (a)K −1 :
Then in view of the Wronskian identity, Lemma 3, Y1 satis es {Y1T RY1 − (Y1 )T RY1 }(t) ≡ {Y1T RY1 − (Y1 )T RY1 }(a) = −(K −1 )T (Y )T (a)R(a)Y (a)K −1 + (K −1 )T Y T (a)R(a)Y (a)K −1 = (K −1 )T {Y T RY − (Y )T RY }(a)K −1 = 0; and {Y T RY1 − (Y )T RY1 }(t) ≡ {Y T RY1 − (Y )T RY1 }(a) = Y T (a)Y (a)K −1 + (Y )T (a)R2 (a)Y (a)K −1 = KK −1 = I; and hence Y and Y1 are normalized conjoined solutions of (E). An application of Theorem 4 now yields that F¿0. However, now Jacobi’s condition, Theorem 5, in turn shows that (E) is disconjugate, i.e., the principal solution of (E) has no focal points in (a; b].
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
691
In the nal result of this paper we shall also consider the equation ˜ (E)
˜ ˜ ] + P(t)Y = 0; [R(t)Y
where R˜ and P˜ satisfy the same assumptions as R and P. Theorem 7 (Sturm’s comparison theorem). Suppose we have for all t ∈ T ˜ ≤ R(t) R(t)
and
˜ ≥ P(t): P(t)
˜ is disconjugate, then (E) is also disconjugate. Then, if (E) ˜ is disconjugate. Then by Jacobi’s condition, Theorem 5, Proof. Suppose (E) Z b ˜ ˜ − (y )T Py ˜ }(t) t¿0 {(y )T Ry F(y) = a
˜ with y(a) = y(b) = 0. For such a y we also have for all nontrivial y ∈ Cp1 (T) Z b F(y) = {(y )T Ry − (y )T Py }(t) t a Z b ˜ ˜ − (y )T Py ˜ }(t) t = F(y)¿0: ≥ {(y )T Ry a
Hence F¿0 and thus (E) is disconjugate by Theorem 5. References [1] R.P. Agarwal, Di erence Equations and Inequalities, Marcel Dekker, New York, 1992. [2] C.D. Ahlbrandt, Continued fraction representations of maximal and minimal solutions of a discrete matrix Riccati equation, SIAM J. Math. Anal. 24 (6) (1993) 1597–1621. [3] C.D. Ahlbrandt, A.C. Peterson, Discrete Hamiltonian Systems: Di erence Equations, Continued Fractions, and Riccati Equations, Kluwer Academic Publishers, Boston, 1996. [4] B. Aulbach, S. Hilger, Linear dynamic processes with inhomogeneous time scale, in: Nonlinear Dynamics and Quantum Dynamical Systems, Akademie, Verlag, Berlin, 1990. [5] M. Bohner, Positive de niteness of discrete quadratic functionals, in: C. Bandle (Ed.), Conf. Proc. 7th Int. Conf. on General Inequalities, Oberwolfach, 1995, ISNM, vol. 123, Birkhauser, Basel, 1997, pp. 55–60. [6] M. Bohner, Linear Hamiltonian di erence systems: disconjugacy and Jacobi-type conditions, J. Math. Anal. Appl. 199 (1996) 804 – 826. [7] M. Bohner, On disconjugacy for Sturm–Liouville di erence equations, J. Di erence Equations 2 (2) (1996) 227–237. [8] M. Bohner, O. Dosly, W. Kratz, Inequalities and asymptotics for Riccati matrix di erence operators, J. Math. Anal. Appl., in press. [9] S. Chen, L. Erbe, Oscillation results for second order scalar and matrix di erence equations, Comput. Math. Appl. 28 (1994) 55 – 69. [10] L. Erbe, S. Hilger, Sturmian theory on measure chains, Di erential Equations and Dynamical Systems 1 (3) (1993) 223 –246. [11] I.M. Gelfand, S.V. Fomin, Calculus of Variations, Prentice-Hall, Englewood Cli s, NJ, 1963. [12] S. Hilger, Analysis on measure chains – a uni ed approach to continuous and discrete calculus, Res. Math. 18 (1990) 19 –56.
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[13] B. Kaymakcalan, V. Lakshmikantham, S. Sivasundaram, Dynamic Systems on Measure Chains, Kluwer Academic Publishers, Boston, 1996. [14] W.G. Kelley, A.C. Peterson, Di erence Equations: An Introduction with Applications, Academic Press, San Diego, 1991. [15] W. Kratz, Quadratic Functionals in Variational Analysis and Control Theory, Akademie Verlag, Berlin, 1995. [16] A. Peterson, J. Ridenhour, Oscillation of second order linear matrix di erence equations, J. Di erential Equations 89 (1991) 69 – 88. [17] W.T. Reid, Sturmian Theory for ordinary Di erential Equations, Springer, New York, 1980.
Quadratic functionals for second order matrix equations on time scales Ravi P. Agarwal ∗ , Martin Bohner 1 Department of Mathematics, National University of Singapore, Lower Kent Ridge Road, Singapore 119260, Singapore Received 1 May 1997; accepted 12 May 1997
Keywords: Time scale; Self-adjoint matrix equation; Variational problem; Jacobi’s condition; Sturmian theory
1. Introduction We begin this paper by stating two well-known results concerning quadratic functionals. Result 1 deals with the “continuous” case while Result 2 represents the “discrete” counterpart. We let n ∈ N. Result 1 (Jacobi’s condition). Let R and P be continuous n × n-matrix valued functions on R such that R(t) and P(t) have real entries and are symmetric and such that R(t) is positive de nite for each t ∈ R. Let a; b ∈ R with a¡b. Then the quadratic functional Z b FC (y) = {y˙T Ry˙ − yT Py}(t) dt a
is positive de nite, i.e., FC (y)¿0 for all nontrivial piecewise C 1 -functions y : [a; b] → Rn with y(a) = y(b) = 0, if and only if the solution Y of d [R(t)Y˙ ] + P(t)Y = 0; dt
Y (a) = 0;
Y˙ (a) = R−1 (a)
satis es Y (t) invertible for all t ∈ (a; b]: ∗
(1)
Corresponding author. E-mail: [email protected].
1
The author is grateful to Alexander von Humboldt Foundation for awarding him a Feodor Lynen Research Fellowship to support this work. 0362-546X/98/$19.00 ? 1998 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 7 ) 0 0 6 7 5 - 5
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Result 2 (Discrete version of Jacobi’s condition). Let Rk and Pk be symmetric n × nmatrices with real entries such that Rk is invertible for each k ∈ Z. Let M; N ∈ Z with M ¡N − 1. Then the discrete quadratic functional FD (y) =
N −1 X
T {(yk )T Rk (yk ) − yk+1 Pk yk+1 }
k=M
is positive de nite, i.e., FD (y)¿0 for all nontrivial sequences y : [M; N ] ∩ Z → Rn with yM = yN = 0, if and only if the solution Y of [Rk Yk ] + Pk Yk+1 = 0;
YM = 0;
YM = R−1 M
satis es Yk invertible for all k ∈ {M + 1; : : : ; N }; −1 −1 Rk Yk Yk+1
and
positive de nite for all k ∈ {M + 1; : : : ; N − 1}:
(2)
The proofs of these results and many related topics can be found in recent monographs by Kratz [15] (“continuous” case; see also [11, 17]), and Ahlbrandt and Peterson [3] (“discrete” case; see also [2, 6–9, 16]). Besides featuring minor techniT Pk yk+1 in FD ), the cal di erences (note that yT (t)P(t)y(t) in FC is replaced by yk+1 following two discrepancies occur in the rst view: • In Result 2 the matrices Rk need not be positive de nite, while Result 1 does not hold without this assumption. • The invertibility of the so-called principal solution in Eq. (1) is not enough for Result 2; in fact we need to supplement this condition by an additional condition in Eq. (2). As such it is not clear why these strange di erences occur. In this paper, we will present a theorem on a time scale T which covers both Results 1 and 2 in the sense that they are special cases of our theorem for T = R and T = Z. Such a uni cation by the theory of time scales was initiated by Hilger in [12]. Besides this paper we shall refer to the work of Aulbach and Hilger [4] and the recently published monograph by Kaymakcalan et al. [13]. Some of our results extend the work of Erbe and Hilger [10] to systems. However, the main result of our paper, Theorem 5, is new even in the scalar case. For the convenience of comparing our theorem from Results 1 and 2 above we rst state our version of Jacobi’s condition on time scales. For the notation needed we refer the reader to the next section on preliminaries about time scales. At this moment the only important thing for observing how Results 1 and 2 follow from our theorem is to know that (t) = (t) = t if T = R and (t) = t − 1; (t) = t + 1 if T = Z. Also, the theory has been prepared in such a manner that y = y˙ in the case T = R and y = y in the case T = Z. Result 3 (Jacobi’s condition on time scales). Let R and P be rd-continuous n × nmatrix valued functions on T such that R(t) and P(t) are symmetric and have real entries and such that R(t) is invertible for each t ∈ T. Let a; b ∈ T with a¡(b). Then
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
677
the quadratic functional Z b {(y )T Ry − (y )T Py }(t)t FT (y) = a
1 -functions y : [a; b] ∩T is positive de nite, i.e., FT (y)¿0 for all nontrivial piecewise Crd → Rn with y(a) = y(b) = 0, if and only if the solution Y of
[R(t)Y ] + P(t)Y = 0;
Y (a) = 0;
Y (a) = R−1 (a)
satis es Y (t) Y (t)Y
invertible for all t ∈ (a; b] ∩ T; and −1
−1
((t))R
(t)
positive de nite for all t ∈ (a; (b)] ∩ T:
(3)
The organisation of this paper is as follows: In the next section we introduce time scales and collect some of their basic properties which are needed later. The proofs of these properties and further details can be found in [4, 13]. Section 3 gives a short motivation why it is useful to examine quadratic functionals by considering certain variational problems on time scales. Section 4 contains preliminaries and several fundamental results (e.g. Picone’s identity) for self-adjoined second order matrix equations and corresponding Riccati matrix equations on time scales. Section 5 proves the above mentioned time-scales-version of Jacobi’s condition. We conclude this paper by giving Sturmian separation and comparison results on time scales in Section 6. 2. Preliminaries about time scales A time scale T is de ned to be any closed subset of R. Hence the jump operators ; : T → T (t) = inf {s ∈ T: s¿t}
and
(t) = sup{s ∈ T: s¡t}
(supplemented by inf ∅ := sup T and sup ∅ := inf T) are well-de ned. The point t ∈ T is called left-dense, left-scattered, right-dense, right-scattered if (t) = t; (t)¡t; (t)= t; (t)¿t, respectively. The graininess : T → R+ 0 is de ned by (t) = (t) − t. Throughout we x a; b ∈ T with a¡(b) and put T˜ = [a; b] ∩ T. By T we shall always denote a nonempty subset of T. Also, whenever we write T = [a1 ; b1 ] ∩ T, we mean a1 ; b1 ∈ T with a1 ¡(b1 ). We de ne ( T if T is unbounded above; T = T\((max T); max T] otherwise: We say that a mapping f from T into some real Banach space X is di erentiable at t ∈ T provided (see [10] (Section 2)) f (t) := lim s→t
f((t)) − f(s) ; (t) − s
where s → t; s ∈ T\{(t)};
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exists. The function f is called di erentiable on T if f (t) exists for all t ∈ T. The following lemma contains results for this derivative. We shall write f for f ◦ . Lemma 1. Let f; g : T → X and t ∈ T . Then the following hold: (1) If f (t) exists, then f is continuous in t; (2) If t is right-scattered and f is continuous in t; then f (t) = [f((t))−f(t)]=(t); (3) If f (t) exists, then f((t)) = f(t) + (t)f (t); (4) If f (t); g (t) exist and (fg)(t) is de ned, then (fg) (t) = f((t))g (t) + f (t) g(t); (5) If f exists on T and f is invertible on T; then (f−1 ) = −(f )−1 f f−1 on T ; (6) If |f | ≤ g on T; then |f(s) − f(r)| ≤ g(s) − g(r) for all r; s ∈ T with r ≤ s; (7) If f = 0 on T; then f is constant on T. Proof. For (1) and (2) see [4] (Theorem 3) or [13] (Theorem 1.2.2). While (3) follows (note that (t) = 0 if t is right-dense) from (1) and (2), (4) and (5) are from [4] (Theorem 4) or [13] (Theorem 1.2.3). Finally, for (6) and (7) see [4] (Theorem 5) or [13] (Corollary 1.3.1). Let t1 ; : : : ; tm ∈ T . A function f : T\{t1 ; : : : ; tm } → X is called regulated on T provided f has a left-sided limit at all left-dense points and a right-sided limit at all rightdense points of T. An f is called rd-continuous on T – we write f ∈ Crd (T) = Crd (T; X ) – if it is regulated on T and continuous at all right-dense points of T. We write 1 (T) if f is di erentiable on T with f ∈ Crd (T ). If f is continuous on T f ∈ Crd and if it is di erentiable on T except at t1 ; : : : ; tm ∈ T such that f is regulated on T, we abbreviate this piecewise rd-continuous di erentiability by writing f ∈ Cp1 (T). Note that in this case (apply Lemma 1(2)) the points t1 ; : : : ; tm are necessarily leftdense and right-dense at the same time. Moreover, f is said to be regressive on T if I +(t)f(t) is invertible for each t ∈ T, where I denotes the identity in X . Next, for any on T regulated function f : T\{t1 ; : : : ; tm } → X there exists a unique pre-antiderivative F (see [13] (Theorem 1.4.2)), i.e., a continuous function F : T → X that is di erentiable on T \{t1 ; : : : ; tm } and satis esR F (t) = f(t) for all t ∈ T \{t1 ; : : : ; tm }. We then s de ne the Cauchy integral of f by r f(t)t = F(s) − F(r), where r; s ∈ T. If F satis es F (t) = f(t) for all t ∈ T, then F is said to be an antiderivative of f on T. In the following lemma we collect properties of rd-continuous functions that are needed in our work. The terminology “right-sequence for t” is used for strictly decreasing sequences {tm }m∈N ⊂ T with limm→∞ tm = t. Note that due to the de nition of there always exist right-sequences for any right-dense t ∈ T (except if t = max T in case T is bounded above). Lemma 2. Let f ∈ Crd (T). Then the following hold: (1) f has an antiderivative; (2) If f is regressive on T and if t˜ ∈ T; g˜ ∈ X , then any initial value problem g = f(t)g; g(t˜) = g˜ has a unique solution on T; R (t) (3) t f( ) = (t)f(t) hold for all t ∈ T ;
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Rt (4) If t1 ∈ T and f(t1 )¿0, then there exists t2 ∈ T with t12 f(t) t¿0; (5) If a is right-dense and if {am }m∈N ⊂ T = T˜ is a right-sequence for a, then Z b Z b lim f(t) t = f(t) t; m→∞
am
a
(6) If T = [a1 ; b1 ] ∩ T; fm ∈ Crd (T), and limm→∞ fm = f uniformly on T , then Z b1 Z b1 lim fm (t)t = f(t) t: m→∞
a1
a1
Proof. While (1) is from [4] (Theorem 6) or [13] (Theorem 1.4.4), see [4] (Theorem 8) or [13] (Theorem 2.5.1) for (2). Now, (3) follows from (1) and Lemma 1(3). For (4) note the following: Either t1 is right-scattered, then by (3) Z (t1 ) f(t) t = (t1 )f(t1 )¿0; t1
or t1 is right-dense, then f is continuous in t1 and there exists a ¿0 and t2 ∈ (t1 ; ∞) ∩ T such that f(t) ≥ for all t ∈ [t1 ; t2 ] ∩ T. In this case there exist antiderivatives F (according to part (1)) and G (because of lim s→t
G((t)) − G(s) (t) − s = lim = =: g(t) s→t (t) − s (t) − s
actually G(t) = t) of f and g, and an application of Lemma 1(4) yields Z t2 f(t) t = F(t2 ) − F(t1 ) ≥ G(t2 ) − G(t1 ) = (t2 − t1 )¿0: t1
Finally, (5) follows from (1), and (6) is from [12] (Theorem 4.3) or [13] (Theorem 1.4.3). In this paper we shall consider matrix-valued functions. The set of m × n-matrices with real entries will be abbreviated by Rm×n . 3. Variational problems on time scales The motivation of our investigations in Sections 4–6 comes by examining variational problems of the form Z b L(t; y (t); y (t)) t → min; (V) L(y) := a
y(a) = ;
y(b) = ;
where ; ∈ Rn and L : T × R2n → R is a C 2 -function in the last two variables (by Lx and Lu we denote the derivatives of L with respect to the second and the third variables, respectively). For simplicity we allow in this section only functions y : T˜ → Rn that
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˜ A function are di erentiable on T˜ with continuous derivative y , i.e., y ∈ C 1 (T). n 1 ˜ ˜ = , and y(b) ˜ = is called a (weak) local minimum y˜ : T → R with y˜ ∈ C (T), y(a) ˜ with of (V) provided there exists a ¿0 such that L(y) ≥ L(y) ˜ for all y ∈ C 1 (T) y(a) = ; y(b) = , and ky− yk¡; ˜ and the local minimum y˜ is called proper provided L(y) = L(y) ˜ if and only if y = y. ˜ Here we shall consider the (according to Lemma 1(1) well-de ned) norm kfk = max |f (t)| + max |f (t)| t∈T˜
t∈T˜
˜ for f ∈ C 1 (T):
˜ with (a) = (b) = 0. Now let : T → R n be any admissible variation, i.e., ∈ C 1 (T) We put () = (; y; ) = L(y + );
∈ R:
We suppose that L is such that we have Z b ˙ = {Lx (·) (t) + Lu (·) (t)} t; L1 (y; ) := (0) a
and = L2 (y; ) := (0)
Z a
b
{( (t))T Lxx (·) (t) + 2( (t))T Lxu (·) (t) + ( (t))T Luu (·) (t)} t;
where (·) = (t; y (t); y (t)). Then y is said to satisfy the Euler equation and L2 (y) is called positive semide nite if L1 (y; ) = 0 and L2 (y; ) ≥ 0, respectively, for all admissible variations . If L2 (y; )¿0 for all nontrivial admissible variations , then we say that L2 (y) is positive de nite. The following two results provide necessary and sucient conditions for (weak) local minima of (V) in terms of positive de niteness of the quadratic functional L2 (y). Concerning the corresponding results for the discrete case we refer to [1] (Example 1.6.7), [5] (Theorems 1 and 2), and [14] (Theorems 8.1 and 8.9). Theorem 1 (Necessary condition). If y˜ is a local minimum of (V), then y˜ satis es ˜ is positive de nite. the Euler equation and L2 (y) Proof. Let be any admissible variation and de ne : R → R as above, namely ˜ (y˜ +)(a) = ; (y˜ + )(b) = , and, () = (; y; ˜ ). We then have y˜ + ∈ C 1 (T); since there exists a ¿0 such that L(y) ≥ L(y) ˜
˜ for all y ∈ C 1 (T)
with y(a) = ; y(b) = ; ky − yk¡; ˜
we also have () = L(y˜ + ) ≥ L(y) ˜ = (0)
for all ∈ R
with ||¡
: kk
Hence has a local minimum at ˜ = 0, and thus our assertion follows.
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681
˜ satis es the Euler equation; y(a) Theorem 2 (Sucient condition). If y˜ ∈ C 1 (T) ˜ = ˜ is positive de nite; then y˜ is a proper local minimum of (V). and y(b) ˜ = ; and L2 (y) ˜ is positive de nite, it is also strongly positive (see, e.g., [11] Proof. Since L2 (y) (Footnote 4, p. 100)), i.e., there exists an ¿0 with L2 (y; ˜ ) ≥ 2(b − a)kk2 : Since L is a C 2 -function in x and u, there exists a ¿0 such that |L00 (t; z) − L00 (t; z˜)|¡ ˜ be such that y 6= y; ˜ y(a) = ; whenever z; z˜ ∈ R2n with |z − z˜|¡. Let y ∈ C 1 (T) y(b) = , and ky − yk¡. ˜ We shall show L(y)¿L(y). ˜ To do so, we put = y − y˜ so that is a nontrivial admissible variation. Now, using the above remarks and Taylor’s theorem, we nd that there exists a ∈ (0; 1) with 1 1 ˙ = () L(y) − L(y) ˜ = (1) − (0) = (0) + () 2 2 1 1 1 1 = (0) + { () − (0)} ≥ (0) − | () − (0)| 2 2 2 2 !T ! Z (t) 1 b (t) 1 t {L00 (t; z(t)) − L00 (t; z˜(t))} = (0) − 2 2 a (t) (t) 1 ≥ (b − a)kk − 2 2
= where we put z=
y˜ + y˜ +
Z a
b
1 kkkkt = (b − a)kk2 − (b − a)kk2 2
b−a kk2 ¿0; 2
! and
z˜ =
y˜
!
y˜
˜ . Hence y˜ is indeed a proper so that |z(t) − z˜(t)| ≤ kk¡kk¡ holds for all t ∈ T local minimum of (V). 4. Self-adjoint second order matrix equations on time scales Let R; P ∈ Crd (T; R n×n ) be such that R(t) is symmetric and invertible and P(t) is symmetric for each t ∈ T. Consider the self-adjoint second order matrix equation (E)
[R(t)Y ] + P(t)Y = 0:
1 (T; R n×n ) a solution of (E) provided We call Y ∈ Crd
{[RY ] + PY }(t) = 0
holds for all t ∈ (T ) :
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1 Note that any solution Y of (E) satis es RY ∈ Crd (T ). We shall prove rst the following basic result.
Lemma 3 (Wronskian identity). Let Y and Y1 be solutions of (E). Then Y T RY1 − (Y )T RY1
is a constant function on T :
Proof. We apply the product rule Lemma 1(4) to obtain [Y T RY1 − (Y )T RY1 ] = (Y )T [RY1 ] + (Y )T RY1 −[(Y )T R] Y1 −(Y )T RY1 = −(Y )T PY1 + (Y )T PY1 = 0 on (T ) . Now the claim follows from Lemma 1(7) with T = T . Next, putting Z=
Y
!
RY
on T
and
S=
0
R−1
−P
−PR−1
! on T;
it is easy to see that Y solves (E) if and only if (note that Y = Y + Y by Lemma 1(3)) Z solves Z = S(t)Z:
(4)
Now S ∈ Crd (T; R2n×2n ) is regressive on T because !−1 I R−1 I − 2 R−1 P −1 {I + S} = = P −P I − 2 PR−1
−R−1 I
!
holds on T. Hence, by Lemma 2(2), any initial value problem (4);
Z(a) = Za ;
i:e:;
(E);
Y (a) = Ya ;
Y (a) = Ya0
with any choice of a 2n × n-matrix Za , i.e., of n × n-matrices Ya and Ya0 , has a unique solution. We let Y˜ be the unique solution of the initial value problem (E);
Y (a) = 0;
Y (a) = R−1 (a):
This Y˜ is called the principal solution of (E) (at a), and according to the Wronskian identity, Lemma 3, it satis es {Y˜ T RY˜ − (Y˜ )T RY˜ }(t) ≡ {Y˜ T RY˜ − (Y˜ )T RY˜ }(a) = 0 for all t ∈ T , i.e., {Y˜ T RY˜ }(t), and hence (where these are de ned) both ˜ = {RY˜ Y˜ −1 }(t) Q(t)
and
˜ = {Y˜ (Y˜ )−1 R−1 }(t) D(t)
are symmetric matrices for each t ∈ T . Furthermore, we call Y a conjoined solution of (E) provided Y T RY is symmetric on T , and two conjoined solutions Y and Y1
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of (E) are called normalized if Y T RY1 − (Y )T RY1 = I on T . Besides the principal solution Y˜ we will also need the associated solution Y˜1 of (E) (at a) which is de ned to be the solution of the initial value problem (E);
Y (a) = −I;
Y (a) = 0:
Of course, Y˜ and Y˜1 are normalized conjoined solutions of (E). We shall now consider the matrix Riccati equation (R)
Q + P(t) + QT {R(t) + (t)Q}−1 Q = 0:
Lemma 4 (Riccati equation). If (E) has a conjoined solution Y such that Y (t) is invertible for all t ∈ T; then Q de ned by Q(t) = R(t)Y (t)Y −1 (t)
for t ∈ T
(5)
is a symmetric solution of (R) on T . Conversely, if (R) has a symmetric solution Q on T , then there exists a conjoined solution Y of (E) such that Y (t) is invertible for all t ∈ T and relation (5) holds. Proof. First we assume that Y solves (E) and is invertible on T. We then de ne Q by Eq. (5) so that on T , Q + P + QT {R + Q}−1 Q = Q + P + QT {I + (Y − Y )Y −1 }−1 R−1 Q = [RY ] (Y )−1 − RY (Y )−1 Y Y −1 + P + RY Y −1 (Y Y −1 )−1 R−1 RY Y −1 = −PY (Y )−1 − RY (Y )−1 Y Y −1 + P + RY (Y )−1 Y Y −1 = 0 holds according to Lemma 1(3)=(4), i.e., Q is a symmetric solution of (R) on T . Conversely, let Q be a symmetric solution of (R) on T , let t0 ∈ T, and let Y be the solution of the initial value problem (note that I + R−1 Q = R−1 (R + Q) is invertible on T and apply Lemma 2(2)) Y = R−1 (t)Q(t)Y;
Y (t0 ) = I:
Then Y is a fundamental solution of Y = R−1 (t)Q(t)Y and hence invertible on T (see [4] (Theorem 9)). Furthermore, it satis es on (T ) , [RY ] = [QY ] = Q Y + QY = −PY − Q(R + Q)−1 QY + QR−1 QY = −PY + Q(R + Q)−1 {(R + Q)R−1 QY − QY } = −PY : We extend Y to a solution of (E) so that, since {Y T RY }(t0 ) = Q(t0 ) is symmetric, Y is indeed a conjoined solution of (E).
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Our next result is the main tool for proving Jacobi’s condition in the next section. Theorem 3 (Picone’s identity). Let ∈ R n and suppose Y and Y1 are normalized conjoined solutions of (E) such that Y is invertible on T. We put Q = RY Y −1
on T
and
D = Y (Y )−1 R−1
on T :
Let t ∈ T and y : T → Rn be di erentiable at t. Then we have at t {yT Qy + 2 T Y −1 y − T Y −1 Y1 } = (y )T Ry − (y )T Py − {Ry − Qy − (Y −1 )T }T D{Ry − Qy − (Y −1 )T }: Proof. According to the di erentiation rules, Lemma 1(4)=(5), we have at t [Y −1 Y1 ] = −(Y )−1 Y Y −1 Y1 + (Y )−1 Y1 = −(Y )−1 R−1 QY1 + (Y )−1 R−1 RY1 = (Y )−1 R−1 (Y −1 )T {Y T RY1 − (Y )T RY1 } = (Y )−1 R−1 (Y −1 )T ; and (also using Lemmas 1(3) and 4), [yT Qy] + {Ry − Qy}T Y (Y )−1 R−1 {Ry − Qy} = (y )T Qy + (y )T Q y + (y )T Qy + (y )T RY (Y )−1 y −yT QY (Y )−1 y − (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Qy − (y )T Py − (y )T QY (Y )−1 R−1 Qy + (y )T Qy +(y )T RY (Y )−1 y − yT QY (Y )−1 y − (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Qy − (y )T Py − (y )T QY (Y )−1 R−1 Qy + (y )T Qy + (y )T R{Y − Y }(Y )−1 y − yT QY (Y )−1 y − (y )T R{Y − Y }(Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy − (y )T RY (Y )−1 y − yT QY (Y )−1 y + (y )T RY (Y )−1 R−1 Qy + yT QY (Y )−1 R−1 Qy − (y )T QY (Y )−1 R−1 Qy
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685
= (y )T Ry − (y )T Py + (y )T Qy − (y )T QY (Y )−1 R−1 Qy − {y + y}T QY (Y )−1 y + {y + y}T QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy − (y )T QY (Y )−1 R−1 Qy − (y )T QY (Y )−1 y + (y )T QY (Y )−1 R−1 Qy = (y )T Ry − (y )T Py + (y )T Qy −(y )T Q{Y − Y }(Y )−1 y + (y )T QY (Y )−1 R−1 Q{y − y } = (y )T Ry − (y )T Py + (y )T Q{Y (Y )−1 − Y (Y )−1 Y Y −1 }y = (y )T Ry − (y )T Py so that [yT Qy + 2 T Y −1 y − T Y −1 Y1 ] − (y )T Ry + (y )T Py = −{Ry − Qy}T Y (Y )−1 R−1 {Ry − Qy} − 2 T (Y )−1 Y Y −1 y + 2 T (Y )−1 y + T (Y )−1 R−1 (Y −1 )T = −{Ry − Qy − (Y −1 )T }T Y (Y )−1 R−1 {Ry − Qy − (Y −1 )T }; and hence our claimed identity follows. 5. Jacobi’s condition on time scales In this section we will prove the main result of this paper. As it is motivated by the investigations of Section 3, we will consider a quadratic functional of the form Z b F(y) = {(y )T Ry − (y )T Py }(t) t: a
This functional F is called positive de nite – we write F¿0 – provided F(y)¿0
˜ Rn )\{0} for all y ∈ Cp1 (T;
with y(a) = y(b) = 0:
1 Lemma 5 (Quadratic functional). If T = [a1 ; b1 ] ∩ T; y ∈ Crd (T; Rn ); and
{[Ry ] + Py }(t) = 0 for all t ∈ (T ) ; then
Z
(b1 )
a1
{(y )T Ry − (y )T Py }(t) t = {yT Ry }((b1 )) − {yT Ry }(a1 ):
Proof. We have on (T ) [yT Ry ] = (y )T [Ry ] + (y )T Ry = (y )T Ry − (y )T Py so that the assertion follows.
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R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
Our next result, Theorem 4, o ers a sucient condition for the positive de niteness of F. A speical case of this result yields one direction of Jacobi’s condition, which we shall prove in Theorem 5. Theorem 4 also nds application to establish Sturm’s separation theorem, this we shall present in Theorem 6. Theorem 4 (Sucient condition for positive de niteness). A sucient condition for F¿0 is that there exist normalized conjoined solutions Y and Y1 of (E) with Y invertible on (a; b] ∩ T
and
Y (Y )−1 R−1 positive de nite on (a; (b)] ∩ T:
˜ . Pick any y ∈ Proof. We let Y and Y1 as above and put D = Y (Y )−1 R−1 on T 1 ˜ n Cp (T; R ) with y(a) = y(b) = 0. First we shall consider the case that a is rightscattered so that we have (a)¿a and (a)¿0. By Lemma 2(3) and Theorem 3 (with = 0) we have Z b Z (a) {(y )T Ry − (y )T Py }(t) t + {(y )T Ry − (y )T Py }(t)t F(y) = a
(a)
= {(y )T Ry − (y )T Py }(a)(a) Z +
b
(a)
{[yT Qy] + (Ry − Qy)T D(Ry − Qy)}(t) t
= {(y )T Ry −1 − (y )T Py }(a) + {yT Qy}(b) −{yT Qy}((a)) +
Z
b
(a)
{(Ry − Qy)T D(Ry − Qy)}(t) t
≥ {(y )T Ry −1 − (y )T Py }(a) + {yT Qy}(b) − {yT Qy}((a)) = {(y )T Ry −1 − (y )T Py − (y )T [RY Y −1 ] y }(a) = (y )T {R−1 − P − (RY + [RY ] )(Y )−1 }y (a) = (y )T {R−1 − P − (RY −1 − PY )(Y )−1 }y (a) = 0: If however, F(y) = 0; then because of Lemma 2(4) {Ry − Qy}(t) = 0
for all t ∈ [(a); (b)] ∩ T;
i.e., y (t) = {Y Y −1 y}(t) for all t ∈ [(a); (b)] ∩ T. Since I + Y Y −1 = I + [Y − Y ]Y −1 = Y Y −1 ; the map Y Y −1 is regressive on [(a); (b)] ∩ T so that by Lemma 2(2) the initial value problem y = {Y Y −1 }(t)y
on [(a); (b)] ∩ T;
y(b) = 0
˜ so that F¿0 follows. admits only one, namely the trivial solution. Hence y = 0 on T
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687
˜ for a. With Next, let a be right-dense and pick a right-sequence {am }m∈N ⊂ T T limm→∞ am = a. Let m ∈ N. We put m = −{(Y ) Ry}(am ) and apply Picone’s identity (with = m ) to obtain the following: Z b {(y )T Ry − (y )T Py }(t) t am Z b = [yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m ] (t) t am
Z
+ Z ≥
b
am b
am
{Ry − Qy − (Y −1 )T m }T D{Ry − Qy − (Y −1 )T m }(t) t
[yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m ] (t) t
= −{ mT Y −1 Y1 m }(b) − {yT Qy + 2 mT Y −1 y − mT Y −1 Y1 m }(am ) = −{ mT Y −1 Y1 m }(b) + {yT Qy + yT QY1 (Y )T Ry}(am ) = −{ mT Y −1 Y1 m }(b) + {yT Qy + yT (Y −1 )T [Y T RY1 − I ](Y )T Ry}(am ) = −{ mT Y −1 Y1 m }(b) + {yT RY1 (Y )T Ry}(am ); and hence by Lemma 2(5) Z b F(y) = lim {(y )T Ry − (y )T Py }(t) t m→∞
am
≥ − lim { mT (Y −1 Y1 )(b) m + yT (RY 1 )(am ) m } = 0: m→∞
Finally, if F(y) = 0, we know that limm→∞ where
Rb am
{zmT Dzm }(t) exists and equals 0,
zm := Ry − Qy − (Y −1 )T m → Ry − Qy =: z;
m→∞
holds uniformly (observe also Lemma 1(1)) on [ak ; b] ∩ T for each k ∈ N. Now let k ∈ N. We rst note Z b Z b T {zm Dzm }(t) t ≥ {zmT Dzm }(t) t for all m ≥ k: am
ak
Hence, by applying Lemma 2(6) we have Z b Z b lim {zmT Dzm }(t) t ≥ {z T Dz}(t) t: m→∞
am
ak
Now we let k → ∞ and apply Lemma 2(5) to obtain Z b Z b 0 = lim {zmT Dzm }(t) t ≥ {z T Dz}(t) t ≥ 0: m→∞
am
a
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R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
˜ so that as in the rst part of this Hence, again by Lemma 2(4), we have z = 0 on T proof the triviality of y follows. Therefore, F is again positive de nite and the proof of our result is complete. ˜ if the principal solution Y˜ of We say that the equation (E) is disconjugate (on T) (E) satis es Y˜ invertible on (a; b] ∩ T
and
Y˜ (Y˜ )−1 R−1 positive de nite on (a; (b)] ∩ T:
Now we are ready to prove the main result of this paper (see also Result 3 of Section 1). Theorem 5 (Jacobi’s condition). F¿0
if and only if (E) is disconjugate:
Proof. First, if (E) is disconjugate, then we may apply our Theorem 4 with Y˜ and Y˜1 , where Y˜ is the principal solution of (E) and Y˜1 is the associated solution of (E), and hence disconjugacy implies F¿0. Conversely, suppose that (E) is not disconjugate. Then there exists a t0 ∈ T with exactly one of the following two properties: ˜ such that Y˜ (t) is invertible for all t ∈ (a; t0 ) ∩ T and Y˜ (t0 ) is singular, (1) t0 ∈ T\{a} ˜ \{a} such that Y˜ (t) is invertible for all t ∈ (a; b] ∩ T and D(t ˜ 0) = (2) t0 ∈ T {Y˜ (Y˜ )−1 R−1 }(t0 ) is not positive de nite. ˜ Y˜ d) Let d ∈ R n \{0} with Y˜ (t0 )d = 0 in Case (1) and dT (Y˜ T RY˜ )(t0 )d = (RY˜ d)T D(R (t0 ) ≤ 0 in Case (2). Putting ( y(t) =
Y˜ (t)d for t ≤ t0 ; 0 otherwise
˜ except yields y(a) = y(b) = 0, y(t) 6= 0 for all t ∈ (a; t0 ), and, moreover, y ∈ Cp1 (T) for Case (2) with t0 right-dense. Hence we assume for the moment not being in this situation, and we will deal with this special case later. Now it is easy to see that both y (t) and y (t) are zero for all t¿t0 so that {(y )T Ry − (y )T Py }(t) = 0
for all t¿t0 :
In the following we use Lemmas 5 and 2(3), and for ∈ R we put for convenience † = 0 if = 0 and † = −1 if 6= 0. We have Z F(y) =
a
Z =
a
(t0 )
t0
{(y )T Ry − (y )T Py }(t) t
{(y )T Ry − (y )T Py }(t) t +
Z
(t0 )
t0
{(y )T Ry − (y )T Py }(t) t
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
689
= {dT Y˜ T RY˜ d}(t0 ) + {(y )T Ry − (y )T Py }(t0 )(t0 ) = dT {Y˜ T RY˜ + Y˜ T RY˜ † }(t0 )d = dT {† Y˜ T RY˜ + [1 − † ]Y˜ T RY˜ }(t0 )d ≤ 0: Therefore F 6¿ 0. Now we will discuss the remaining Case 2 with right-dense t0 . This situation implies R(t0 ) 6¿ 0, i.e., there exists a d ∈ R n such that dT R(t0 )d¡0. We have to show F 6¿ 0, and for the sake of achieving a contradiction we assume F¿0. First we suppose that ˜ be a right-sequence for t0 . For m ∈ N t0 is left-scattered. Then t0 6= b. Let {tm }m∈N ⊂ T we put √tm − t d if t ∈ (t ; t ] ∩ T; 0 m ym (t) = t −t m0 0 otherwise: √ ˜ Therefore Hence ym (a) = ym (b) = 0, ym (t0 ) = tm − t0 d 6= 0, and y ∈ Cp1 (T). Z 0 ¡ F(ym ) =
(tm )
t0
{(y )T Ry − (y )T Py }(t) t
d tm − (t) tm − (T ) dT R(t) √ − dT √ P(t) √ d t tm − t0 tm − t0 tm − t0 tm − t0 t0 ) ) ( ( Z (tm ) Z (tm ) (tm − (t))2 1 T T =d R(t)t d − d P(t)t d tm − t0 t0 tm − t0 t0 Z
=
(tm )
√
→ dT R(t0 )d¡0;
m → ∞;
which yields our desired contradiction. The remaining case of left-dense t0 can be treated the same way with ym (t) = or
ym (t) =
t−t ∗ √ m ∗ d
t0 −tm
0
if t ∈ [tm∗ ; t0 ) ∩ T; otherwise;
t−t ∗ √ m d if t ∈ [tm∗ ; t0 ) ∩ T; t0 −tm∗ √tm −t d if t ∈ (t0 ; tm ] ∩ T; tm −t0 0 otherwise;
˜ if t0 = b or otherwise, respectively, using a strictly increasing sequence {tm∗ }m∈N ⊂ T with limm→∞ tm∗ = t0 .
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R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
6. Sturmian theory on time scales We call a solution Y of (E) basis whenever Y (a) = n: rank Y (a) A conjoined solution is said to have no focal points in (a; b] provided it satis es ˜ Y invertible on T\{a}
and
˜ \{a}: Y (Y )−1 R−1 positive de nite on T
˜ if and only if the principal Using this terminology, equation (E) is disconjugate (on T) solution of (E) (at a) has no focal points in (a; b]. Concerning conjoined bases of (E) we have the following version of Sturm’s separation result on time scales. Theorem 6 (Sturm’s separation theorem). Suppose there exists a conjoined basis of ˜ (E) with no focal points in (a; b]. Then equation (E) is disconjugate (on T). Proof. Let Y be a conjoined basis of (E) satisfying Y invertible on (a; b] ∩ T
and
Y (Y )−1 R−1 positive de nite on (a; (b)] ∩ T:
Since Y is a basis, we note that the matrix K = Y T (a)Y (a) + (Y )T (a)R2 (a)Y (a)
is invertible:
Let Y1 be the solution of (E);
Y1 (a) = −R(a)Y (a)K −1 ;
Y1 (a) = R−1 (a)Y (a)K −1 :
Then in view of the Wronskian identity, Lemma 3, Y1 satis es {Y1T RY1 − (Y1 )T RY1 }(t) ≡ {Y1T RY1 − (Y1 )T RY1 }(a) = −(K −1 )T (Y )T (a)R(a)Y (a)K −1 + (K −1 )T Y T (a)R(a)Y (a)K −1 = (K −1 )T {Y T RY − (Y )T RY }(a)K −1 = 0; and {Y T RY1 − (Y )T RY1 }(t) ≡ {Y T RY1 − (Y )T RY1 }(a) = Y T (a)Y (a)K −1 + (Y )T (a)R2 (a)Y (a)K −1 = KK −1 = I; and hence Y and Y1 are normalized conjoined solutions of (E). An application of Theorem 4 now yields that F¿0. However, now Jacobi’s condition, Theorem 5, in turn shows that (E) is disconjugate, i.e., the principal solution of (E) has no focal points in (a; b].
R.P. Agarwal, M. Bohner / Nonlinear Analysis 33 (1998) 675 – 692
691
In the nal result of this paper we shall also consider the equation ˜ (E)
˜ ˜ ] + P(t)Y = 0; [R(t)Y
where R˜ and P˜ satisfy the same assumptions as R and P. Theorem 7 (Sturm’s comparison theorem). Suppose we have for all t ∈ T ˜ ≤ R(t) R(t)
and
˜ ≥ P(t): P(t)
˜ is disconjugate, then (E) is also disconjugate. Then, if (E) ˜ is disconjugate. Then by Jacobi’s condition, Theorem 5, Proof. Suppose (E) Z b ˜ ˜ − (y )T Py ˜ }(t) t¿0 {(y )T Ry F(y) = a
˜ with y(a) = y(b) = 0. For such a y we also have for all nontrivial y ∈ Cp1 (T) Z b F(y) = {(y )T Ry − (y )T Py }(t) t a Z b ˜ ˜ − (y )T Py ˜ }(t) t = F(y)¿0: ≥ {(y )T Ry a
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