NABLA DYNAMIC EQUATIONS ON TIME SCALES

NABLA DYNAMIC EQUATIONS ON TIME SCALES DOUGLAS ANDERSON, JOHN BULLOCK AND LYNN ERBE, ALLAN PETERSON, HOAINAM TRAN. preliminaries about time scales The following definitions can be found in Bohner and Peterson [4] and Agarwal and Bohner []. A time scale T is defined to be any closed subset of R. Then the forward and backwards jump operators σ, ρ : T T σ(t) inf{s T : s > t} and ρ(t) sup{s T : s < t} (supplemented by inf : sup T and sup : inf T) are well defined. A point t T is called left-dense if t > inf T and ρ(t) t, left-scattered if ρ(t) < t, right-dense if t < sup T and σ(t) t, right-scattered if σ(t) > t. If T has a right-scattered minimum m, define T κ : T {m}; otherwise, set T κ T. The backwards graininess ν : T κ R + 0 is defined by ν(t) t ρ(t). For f : T R and t T κ, define the nabla derivative [3] of f at t, denoted f (t), to be the number (provided it exists) with the property that given any ε > 0, there is a neighborhood U of t such that f(ρ(t)) f(s) f (t)[ρ(t) s] ε ρ(t) s for all s U. For T R, we have f f, the usual derivative, and for T Z we have the backward difference operator, f (t) f(t) : f(t) f(t ). A function f : T R is left-dense continuous or ld-continuous provided it is continuous at left-dense points in T and its right-sided limits exist (finite) at right-dense points in T. If T R, then f is ld-continuous if and only if f is continuous. If T Z, then any function is ld-continuous. It is known [4] that if f is ld-continuous, then there is a function F (t) such that F (t) f(t). In this case, we define b a f(t) t F (b) F (a). Remark. The following theorems delineate several properties of the nabla derivative; they are found in [3] or [4, p 33-333]. Theorem. Assume f : T R is a function and let t T κ. Then we have the following: (i) If f is nabla differentiable at t, then f is continuous at t. (ii) If f is continuous at t and t is left-scattered, then f is nabla differentiable at t with f (t) f(t) f(ρ(t)). ν(t)

PETERSON AND ANDERSON, ET AL (iii) If t is left-dense, then f is nabla differentiable at t iff the limit f(t) f(s) lim s t t s exists as a finite number. In this case (iv) If f is nabla differentiable at t, then f f(t) f(s) (t) lim. s t t s f ρ (t) f(t) ν(t)f (t). Theorem 3. Assume f, g : T R are nabla differentiable at t T κ. Then: (i) The sum f + g : T R is nabla differentiable at t with (f + g) (t) f (t) + g (t). (ii) The product fg : T R is nabla differentiable at t, and we get the product rules (fg) (t) f (t)g(t) + f ρ (t)g (t) f(t)g (t) + f (t)g ρ (t). (iii) If g(t)g ρ (t) 0, then f g is nabla differentiable at t, and we get the quotient rule (iv) If f and f are continuous, then ( ) f (t) f (t)g(t) f(t)g (t) g g(t)g ρ. (t) ( f(t, s) s) f(ρ(t), t) + a a f (t, s) s.. introduction to nabla equations Definition 4. The function p is ν-regressive if ν(t)p(t) 0 for all t T κ. Define the ν-regressive class of functions on T κ to be R ν {p : T R p is ld continuous and ν-regressive}. If p, q R ν, then for all t T κ. (p ν q)(t) : p(t) + q(t) p(t)q(t)ν(t) Theorem 5. The set {R ν, ν } is an Abelian group.

NABLA DYNAMIC EQUATIONS 3 Proof. Suppose p and q are in R ν. To prove that we have closure under the addition ν, note first that p ν q is also ld-continuous. It only remains to show that p ν q is ν-regressive on T κ, but this follows from ν(p ν q) ν(p + q pqν) νp νq + νpqν ( νp)( νq) 0. Hence R ν is closed under the addition ν. Since 0 R ν and p ν 0 0 ν p p, 0 is the additive identity for ν. To find the additive inverse of p R ν under ν, we must solve for w. Hence we must solve for w. Thus p ν w 0 p + w pwν 0 w : p pν R ν is the additive inverse of p under the addition ν. A straightforward calculation verifies that associativity holds. Hence (R ν, ν ) is a group. Since p ν w p + w pwν w + p wpν w ν p, the commutative law holds, and hence (R ν, ν ) is an Abelian group. Definition 6. For p R ν, define circle minus p by ν p : p pν, and the generalized square of p by Theorem 7. For p R ν, p (t) : p(t)( ν p) p pν. (i) ( ν p) p, (ii) pν p p, (iii) p + ( ν p) p ν, (iv) p ν p p + p.

4 PETERSON AND ANDERSON, ET AL Proof. Throughout this proof assume p R ν. To prove part (i), consider ( ν p) ( ν p) ( ν p)ν p ( pν) + pν pν p ( pν) pν pν + pν p pν p. Formula (ii) follows immediately from the definition of p. To prove part (iii), note that p + ( ν p) p + p pν p p ν p pν p pν ν To prove (iv), we have Definition 8. For h > 0, let and p ν. p ν p p + p p p ν p + p pν p p pν ν p + p p 3 ν pν p + p. 3. The Nabla Exponential Function Z h : { z C : π h C h : < Im(z) < π h { z C : z }. h Define the ν-cylinder transformation ˆξ h : C h Z h by () ˆξh (z) : Log( zh), h }

NABLA DYNAMIC EQUATIONS 5 where Log is the principal logarithm function. z C 0 : C. For h 0, we define ˆξ 0 (z) z for all Definition 9. If p R ν, then we define the nabla exponential function by ( ) () ê p (t, s) : exp ˆξ ν(τ) (p(τ)) τ s where the ν-cylinder transformation ˆξ h is as in (). for s, t T, Lemma 0 (semigroup property). If p R ν, then the semigroup property is satisfied. ê p (t, u)ê p (u, s) ê p (t, s) for all s, t, u T Proof. Suppose p R ν. Let s, t, u T. Then we have by Definition 9 ( ê p (t, u)ê p (u, s) exp u ( exp u ( exp ê p (t, s). s ) ( u ˆξ ν(τ) (p(τ)) τ exp u ˆξ ν(τ) (p(τ)) τ + ) ˆξ ν(τ) (p(τ)) τ s s ) ˆξ ν(τ) (p(τ)) τ ) ˆξ ν(τ) (p(τ)) τ Definition. If p R ν, then the first order linear dynamic equation (3) y p(t)y is called ν-regressive. Theorem. Suppose (3) is ν-regressive and fix T. Then ê p (, ) is a solution of the initial value problem (4) y p(t)y, y( ) on T. Proof. Fix T κ and assume (3) is ν-regressive. First note that ê p (, ). It remains to show that ê p (t, ) satisfies the dynamic equation y p(t)y. Fix t T κ. There are two cases.

6 PETERSON AND ANDERSON, ET AL Case. Assume ρ(t) < t. In this case ( ) ( t ) ρ(t) exp ê t ˆξν(τ) 0 (p(τ)) τ exp t ˆξν(τ) 0 (p(τ)) τ p (t, ) ν(t) ( ) exp ν(t)ˆξ ν(t) (p(t)) ê p (t, ) ν(t) êp(t, ) { exp[log( p(t)ν(t))]} ν(t) p(t) ê p (t, ). Case. Assume ρ(t) t. If y(t) : ê p (t, ), then we want to show that y (t) p(t)y(t). Using Lemma 0 we obtain y(t) y(s) p(t)y(t)(t s) ê p (t, ) ê p (s, ) p(t)ê p (t, )(t s) ê p (t, ) ê p (s, t) p(t)(t s) ê p (t, ) ˆξ ν(τ) (p(τ)) τ ê p (s, t) + ˆξ ν(τ) (p(τ)) τ p(t)(t s) s s ê p (t, ) ˆξ ν(τ) (p(τ)) τ ê p (s, t) s + ê p (t, ) ˆξ ν(τ) (p(τ)) τ p(t)(t s) s ê p (t, ) ˆξ ν(τ) (p(τ)) τ ê p (s, t) s + ê p (t, ) [ˆξ ν(τ) (p(τ)) ˆξ 0 (p(t))] τ. s Let ε > 0 be given. We now show that there is a neighborhood U of t so that the right-hand side of the last inequality is less than ε t s, and the proof will be complete. Since ρ(t) t and p C ld, it follows that (5) lim τ t ˆξν(τ) (p(τ)) ˆξ 0 (p(t)). This implies that there is a neighborhood U of t such that ˆξ ν(τ) (p(τ)) ˆξ ε 0 (p(t)) < for all τ U. 3 ê p (t, ) Let s U. Then (6) ê p (t, ) Next, by L Hôpital s rule s [ˆξ ν(τ) (p(τ)) ˆξ 0 (p(t))] τ < ε t s. 3 z e z lim z 0 z 0,

NABLA DYNAMIC EQUATIONS 7 so there is a neighborhood U of t so that if s U, s t, then ˆξ s ν(τ) (p(τ)) τ ê p (s, t) t ˆξ s ν(τ) (p(τ)) τ < ε, where { } ε ε min,. + 3 p(t)ê p (t, ) Let s U : U U. Then ê p (t, ) t ˆξ ν(τ) (p(τ)) τ ê p (s, t) < ê p(t, ) ε ˆξ ν(τ) (p(τ)) τ s s { t } ê p (t, ) ε [ˆξ ν(τ) (p(τ)) ˆξ 0 (p(t))] τ + p(t) t s s ê p (t, ) [ˆξ ν(τ) (p(τ)) ˆξ 0 (p(t))] τ + ê p(t, ) ε p(t) t s using (6). s ε 3 t s + ê p(t, ) ε p(t) t s ε 3 t s + ε t s 3 ε t s, 3 Theorem 3. If (3) is ν-regressive, then ê p (, ) is the unique solution of the IVP (4). Proof. Assume y is a solution of (4). By the definition (), the exponential never vanishes. Using the nabla quotient rule, ( ) y (t) y (t)ê p (t, ) y(t)ê p (t, ) ê p (, ) ê p (t, )ê p (ρ(t), ) p(t)y(t)ê p(t, ) y(t)p(t)ê p (t, ) ê p (t, )ê p (ρ(t), ) 0, so that y is a constant multiple of ê p (, ); the initial condition shows that they are equal. Theorem 4. Let p, q R ν and s, t, u T. Then (i) ê 0 (t, s) and ê p (t, t), (ii) ê p (ρ(t), s) ( ν(t)p(t))ê p (t, s), (iii) ê ê p(t,s) νp(t, s), (iv) ê p (t, s) ê ê p(s,t) νp(s, t), (v) ê p (t, u)ê p (u, s) ê p (t, s), (vi) ê p (t, s)ê q (t, s) ê p νq(t, s), (vii) êp(t,s) ê ê q(t,s) p νq(t, s).

8 PETERSON AND ANDERSON, ET AL (viii) ( ) ê p(t,s) p(t) ê ρ p(t,s). Proof. Part (i). Assume p, q R ν and s, t T. Then ê 0 (t, s) is a solution of the IVP x 0 x, x( ), so x 0 x if and only if x(t) c for some constant c R. But x( ) c, and since the system above has a unique solution, ê 0 (t, s). Moreover, ê p (t, t) follows directly from (). Part (ii). We have ê p (ρ(t), s) ê ρ p(t, s) Part (iii). Let x(t) ê p(t,s). Then (7) x (t) so that ê p (t, s) ν(t)ê p (t, s) ê p (t, s) ν(t)p(t)ê p (t, s) ( ν(t)p(t))ê p (t, s). ê p (t, s) ê p (t, s)ê p (ρ(t), s), x (t) p(t)ê p (t, s) ê p (t, s)( ν(t)p(t))ê p (t, s) p(t) ν(t)p(t) ê p (t, s) ν p(t) x(t). But x(t) is a solution to initial value problem x (t) ν p(t)x(t), x( ). Therefore, ê p(t,s) ê νp(t, s) by uniqueness. Part (iv). This follows from (iii) and (). Part (v). This semigroup property was already shown above in Lemma 0. Part (vi). We know ê p νq(t, s) is the unique solution to the initial value problem (8) x (t) (p ν q)(t)x(t), x(s). If we define (9) x(t) : ê p (t, s)ê q (t, s), then x (t) p(t)ê p (t, s)ê q (t, s) + q(t)ê p (ρ(t), s)ê q (t, s) p(t)ê p (t, s)ê q (t, s) + q(t)ê p (t, s)( ν(t)p(t))ê q (t, s) ê p (t, s)ê q (t, s)(p(t) + q(t)( ν(t)p(t))) (p(t) + q(t) p(t)q(t)ν(t))ê p (t, s)ê q (t, s) (p ν q)(t)ê p (t, s)ê q (t, s) (p ν q)(t)x(t).

It is obvious that x(s) by (). Thus, NABLA DYNAMIC EQUATIONS 9 ê p (t, s)ê q (t, s) ê p νq(t, s). Part (vii). This follows easily using parts (iii) and (vi) of this theorem. Part (viii). We calculate ( ) ê p (t, s) ê p (t, s) ê p (t, s)ê p (ρ(t), s) p(t)ê p (t, s) ê p (t, s)ê p (ρ(t), s) p(t) ê p (ρ(t), s). Lemma 5. Let p R ν. Suppose there exists a sequence of distinct points {t n } n N T κ such that ν(t n )p(t n ) < 0 for all n N. Then lim n t n. In particular, if there exists a bounded set J T κ, then the cardinality of the set of points such that ν(t)p(t) < 0 and t J is finite. Proof. Let J T κ be a bounded set and assume there exists an infinite sequence of distinct points {t n } n N J such that ν(t n )p(t n ) < 0 for all n N. Since the sequence {t n } n N is bounded, it has a convergent subsequence. We assume without loss of generality that the sequence {t n } n N is itself convergent, i.e., lim t n. n Since T is closed, T. Since ν(t n )p(t n ) < 0, we have ν(t n ) > 0 and (0) p(t n ) > ν(t n ) for all n N. There is a subsequence {t nk } k N of {t n } n N with lim k t nk such that {t nk } k N is either strictly decreasing or strictly increasing. If {t nk } k N is strictly decreasing, then lim k ν(t nk ) 0 because of 0 < ν(t nk ) t nk ρ(t nk ) t nk t nk+. If {t nk } k N is strictly increasing, then lim k ν(t nk ) 0 because of 0 < ν(t nk ) t nk ρ(t nk ) t nk t nk. So in either case lim k ν(t nk ) 0 and hence using (0), lim p(t n k ). k But this contradicts the fact that p C ld. Theorem 6. Assume p R ν and T.

0 PETERSON AND ANDERSON, ET AL (i) If ν(t)p(t) > 0 for t T, then ê p (t, ) > 0 for all t T. (ii) If ν(t)p(t) < 0 for t > inf T, then ê p (t, ) ( ) nt α(t, ) for all t T, where ( ) log ν(τ)p(τ) α(t, ) : exp τ > 0 ν(τ) and n t { card (, t] if t card (t, ] if t <. Proof. Part (i) can be shown directly using Definition 9: Since ν(t)p(t) > 0, we have Log[ ν(t)p(t)] R for all t T and therefore ˆξ ν(t) (p(t)) R for all t T. Hence for all t T. ( ) ê p (t, ) exp ˆξν(τ) (p(τ)) τ > 0 Part (ii) follows similarly: Since ν(t)p(t) < 0, for t > inf T we have Log[ ν(t)p(t)] log ν(t)p(t) + iπ for all t > inf T. It follows from Lemma 5 that and By Euler s formula, ( ê p (t, ) exp ( exp ( exp ( exp This leads to the desired result. α(t, ) exp n t < ) ˆξν(τ) (p(τ)) τ ) Log[ ν(τ)p(τ)] τ ν(τ) log ν(τ)p(τ) + iπ ν(τ) { log ν(τ)p(τ) ( iπ ( ) τ exp iπ ν(τ) ν(τ) ) τ. ν(τ) ) τ ( ) τ cos π ν(τ) cos(±n t π) ( ) nt. + iπ } ) τ ν(τ) In view of Theorem 6 (i), we make the following definition.

Definition 7. We define the set R + ν NABLA DYNAMIC EQUATIONS of all positively ν-regressive elements of R ν by R + ν R + ν (T, R) {p R ν : ν(t)p(t) > 0 for all t T}. Exercise 8. Assume p C ld and p(t) 0 for all t T. Show that p R + ν. Lemma 9. R + ν is a subgroup of R ν. Proof. Obviously we have that R + ν R ν and that the constant function 0 R + ν. Now let p, q R + ν. Then νp > 0 and νq > 0 on T. Therefore Hence we have Next, let p R + ν. Then This implies that Hence ν(p ν q) ( νp)( νq) > 0 on T. p ν q R + ν. νp > 0 on T. ν( ν p) + νp νp > 0 on T. νp ν p R + ν. These calculations establish that R + ν is a subgroup of R ν. Theorem 0 (Sign of the Nabla Exponential Function). Let p R ν and T. (i) If p R + ν, then ê p (t, ) > 0 for all t T. (ii) If ν(t)p(t) < 0 for some t T κ, then ê p (ρ(t), )ê p (t, ) < 0. (iii) If ν(t)p(t) < 0 for all t T κ, then ê p (t, ) changes sign at every point t T. (iv) Assume there exist sets T {t i : i N} T κ and S {s i : i N} T κ with < s < s < t < t < such that ν(t)p(t) < 0 for all t S T and ν(t)p(t) > 0 for all t T κ \(S T ). Furthermore if card T, then lim n t n, and if card S, then lim n s n. If T and S, then If card T, then ê p (t, ) > 0 on [s, ρ(t )]. ( ) i ê p (t, ) > 0 on [t i, ρ(t i+ )] for all i N. If card T N N, then and ( ) i ê p (t, ) > 0 on [t i, ρ(t i+ )] for all 0 i N ( ) N ê p (t, ) > 0 on [t N, ).

PETERSON AND ANDERSON, ET AL If T and S, then If card S, then ê p (t, ) > 0 on [s, ). ( ) i ê(t, ) > 0 on [s i+, ρ(s i )] for all i N. If card S M N, then ( ) i ê p (t, ) > 0 on [s i+, ρ(s i )] for all i M and ( ) M ê p (t, ) > 0 on (, ρ(s M )]. If S and T, then ê p (t, ) > 0 on (, ρ(t )]. In particular, the exponential function ê p (, ) is a real-valued function that is never equal to zero but can be negative. Proof. While (i) is just Theorem 6 (i), (ii) follows immediately from Theorem 4 (ii). Part (iii) is clear by Theorem 6 (ii). Now we prove (iv). By Lemma 5, the set of points in T where ν(t)p(t) < 0 is countable. If card T, then lim n t n by Lemma 5. We can assume t < t <. Consider the case where card T with t < t < such that ν(t)p(t) < 0 for all t T and ν(t)p(t) > 0 for all t [, ) \ T. We prove the conclusion of (iv) for this case by mathematical induction with respect to the intervals [, ρ(t )], [t, ρ(t )], [t, ρ(t 3 )],.... First we prove that ê p (t, ) > 0 on [, ρ(t )]. If t, then ê p (t, ) > 0. Hence we can assume t >. Since ν(t )p(t ) < 0, t is left-scattered. Now ν(t)p(t) > 0 on [, ρ(t )], implies that ê p (t, ) e R t ˆξν(s) (p(s)) s > 0, on [t0, ρ(t )]. Assume i 0 and ( ) i ê p (t, ) > 0 on [t i, ρ(t i+ )]. It remains to show that First note that ( ) i+ ê p (t, ) > 0 on [t i+, ρ(t i+ )]. 0 < ( ) i ê p (ρ(t i+ ), ) ( ) i [ ν(t i+ )p(t i+ )]ê p (t i+, ) [ ν(t i+ )p(t i+ )]( ) i+ ê p (t i+, ) implies that ( ) i+ ê p (t i+, ) > 0. Since ν(t i+ )p(t i+ ) < 0, we have that ρ(t i+ ) > t i+. By the semigroup property (Theorem 4, (iv)) ê p (t, ) ê p (t, t i+ )ê p (t i+, ). Since ν(t)p(t) > 0 on (t i+, ρ(t i+ )] it follows that ê p (t, ) > 0 on [t i+, ρ(t i+ )].

NABLA DYNAMIC EQUATIONS 3 The remaining cases are similar and hence are omitted. Exercise. Show that if the constant a > and T Z, then the exponential function ê a (, 0) changes sign at every point in Z. In this case we say the exponential function ê a (, 0) is strongly oscillatory on Z. 4. Examples of Exponential Functions Example. Let T hz for h > 0. Let α R ν be constant, i.e., Then α Ĉh : C \ { }. h () ê α (t, ) ( ) t h αh for all t T. To show this we note that y defined by the right-hand side of () satisfies y( ) and y y(t) y(t h) (t) h ( h αh ) t h ( ) t h h αh { ( αh)}y(t) h αy(t) for all t T. Exercise 3. Show that if T R or T hz, h > 0, and α R ν ê α (t + s, 0) ê α (t, 0)ê α (s, 0) for all s, t T. is a constant, then Example 4. Consider the time scale We claim that T N {n : n N}. () ê (t, ) t ( t)! for t T.

4 PETERSON AND ANDERSON, ET AL Let y be defined by the right-hand side of (). Clearly, y(), and for t T we have y(t) y(ρ(t)) t ( t)! ρ(t) ( ρ(t))! t ( t)! t ( t )! 4 t t ( t)! ν(t) t ( t)! ν(t)y(t). It follows that y (t) y(t). Example 5. Let H 0 : 0, H n n k k, k N be the harmonic numbers [4], and Let α R ν be a constant. We claim that T {t H n : n N 0 }. (3) ê α (H n, 0) where t to the n rising is defined by n! ( α) n, (4) t n t(t + )(t + ) (t + n ) for n N and :. To show this claim, suppose y is defined by the right-hand side of (3) and consider y(t) y(ρ(t)) y(h n ) y(h n ) n! (n )! ( α) n ( α) n n α [ n ]n! ( α) n α n! n ( α) n n! αν(t) ( α) n αν(t)y(t). It follows that y (t) αy(t). Example 6. We consider the time scale T q N 0. Let p R ν. The problem y p(t)y, y()

NABLA DYNAMIC EQUATIONS 5 can be equivalently rewritten as y (q ) q tp(t) yρ, y(). The solution of this problem is (5) ê p (t, ) If α R ν is constant, then we have For q this simplifies as ê α (t, ) s T (,t] s T (,t] (6) ê α (t, ) Now consider the special case of (5) when Using (5), we find p(t) s T (,t] (q ) q sp(s). (q ) q sα. sp(s). q t q t for t T q N 0. ê p (t, ) s T (,t] s T (,t] n k q k s s s q n(n+)/ where t q n. Substituting t q n we finally get that ê p (t, ) te ln (t) ln(q). Exercise 7. Find ê α (t, ) for t, q N 0 with constant α R ν. Exercise 8. Find the exponential function ê λ (, ), where λ R ν is a constant for T t [, ) and T N. Example 9. Let k0 α k L be a convergent series with α 0 R and α k > 0 for k and put { n } T α k : n N {L}. k0

6 PETERSON AND ANDERSON, ET AL Table. Exponential Functions T ê α (t, ) R e α(t ) Z hz n Z q N 0 N 0 s [,t) ( ) t t0 α ( ) (t t0 )/h αh ( ) n(t t0 ) n n α ( s [,t) (q )αs ( αs ) if t ) if t { n k k : n N} (n 0 +) n n 0 (n 0 + α) n n 0 if t n k k Then ê (t, α 0 ) n k n for t α k k0 where we use the convention that 0 k ( ). Therefore lim ê(t, α 0 ) ê (L, α 0 ) t L k In particular, if α k 4k for all k N, then, using that the Wallis product converges, we find that k α k k 4k k lim t L ê(t, α 0 ) ê (L, α 0 ) π. α k, n N, α k. 4k 4k π 5. Nonhomogeneous First Order Linear Equations In this section we first study the first order nonhomogeneous linear equation (7) y p(t)y + f(t) and the corresponding homogeneous equation (8) y p(t)y

NABLA DYNAMIC EQUATIONS 7 Table. Exponential Functions T p(t) ê p (t, ) R 0 e t Z 0 ( ) t hz 0 ( n Z 0 n q N 0 N 0 q N 0 t (q )t N 0 t t ( s [,t) s [,t) ( ) t/h h n ) nt (q )s ( +s te ln (t) ln(q) te ln (t) ln(4) ) if t ) if t N t ( t)! { n k k : n N} 0 n + if t n k k on a time scale T. Using Theorem one can easily prove the following theorem. Theorem 30. Suppose (8) is ν-regressive. Let T and y 0 R. The unique solution of the initial value problem (9) y p(t)y, y( ) y 0 is given by y(t) ê p (t, )y 0. Definition 3. For p R ν we define an operator L : C ld C ld by L y(t) y (t) p(t)y(t), t T κ. Then (8) can be written in the form L y 0 and (7) can be written in the form L y f(t). Since L is a linear operator we say that (7) is a linear dynamic equation. We say y is a solution of (7) on T provided y C ld and L y(t) f(t) for t T κ. Definition 3. The adjoint operator L : C ld C ld is defined by L x(t) x (t) + p(t)x ρ (t), t T κ. Example 33. It is easy to varify that the function x(t) ( αh) t h, t hz

8 PETERSON AND ANDERSON, ET AL is a solution of the adjoint equation where α is a ν-regressive constant. x + αx ρ 0, t hz, Theorem 34 (Lagrange Identity). If x, y C ld, then x ρ L y + yl x (xy) on T κ. Proof. Assume x, y C ld on T κ. and consider (xy) x ρ y + x y x ρ (y py) + y(x + px ρ ) x ρ L y + yl x The next result follows immediately from the Lagrange identity. Corollary 35 (Abel s formula). If x and y are solutions of L y 0 and L x 0, respectively, then x(t)y(t) C for t T, where C is a constant. It follows from this corollary that if a nontrivial y satisfies L y 0, then x : y satisfies the adjoint equation L x 0. Exercise 36. Show directly by substitution into L x 0 that if y is a nontrivial solution of L y 0, then x : y is a nontrivial solution of the adjoint equation L x 0. For later reference we also cite the following existence and uniqueness result for the adjoint initial value problem. Theorem 37. Suppose p R ν. Let T and x 0 R. The unique solution of the initial value problem (0) x p(t)x ρ, x( ) x 0 is given by x(t) ê νp(t, )x 0. Exercise 38. Prove Theorem 37. Show directly that the function x given in Theorem 37 solves the initial value problem (0). We now turn our attention to the nonhomogeneous problem () x p(t)x ρ + f(t), x( ) x 0, where we assume that p R ν and f C ld. Let us assume that x is a solution of (). We multiply both sides of the dynamic equation in () by the so-called integrating factor ê p (t, ) and obtain ê p (t, )x (t) + p(t)ê p (t, )x ρ (t) ê p (t, )f(t).

Using the product rule we get that NABLA DYNAMIC EQUATIONS 9 [ê p (t, )x] ê p (t, )f(t). Taking the nabla integral of both sides we obtain Solving for x(t) we have ê p (t, )x(t) ê p (, )x( ) x(t) ê νp(t, )x 0 + ê p (τ, )f(τ) τ. ê νp(t, τ)f(τ) τ. Before we state the variation of constants formula we give the following definition. Definition 39. The equation (7) is called regressive provided (8) is regressive and f : T R is ld-continuous. Now we give the variation of constants formula for the adjoint equation L x f. Theorem 40 (Variation of Constants). Suppose (7) is regressive. Let T and x 0 R. The unique solution of the initial value problem () x p(t)x ρ + f(t), x( ) x 0 is given by (3) x(t) ê νp(t, )x 0 + ê νp(t, τ)f(τ) τ. Proof. First, it is easily verified that x given by (3) solves the initial value problem () (see Exercise 4 below). Finally, if x is a solution of (), then we have seen above that (3) holds. Remark 4. Because of Theorem 4 (v), an alternative form of the solution of the initial value problem () is given by [ ] x(t) ê νp(t, ) x 0 + ê νp(, τ)f(τ) τ. Exercise 4. Verify directly that the function x given in Theorem 40 solves the IVP (). Next we give the variation of constants formula for L y f. Theorem 43 (Variation of Constants). Suppose (7) is regressive. Let T and y 0 R. The unique solution of the initial value problem (4) y p(t)y + f(t), y( ) y 0 is given by y(t) ê p (t, )y 0 + ê p (t, ρ(τ))f(τ) τ.

0 PETERSON AND ANDERSON, ET AL Proof. We equivalently rewrite y p(t)y + f(t) as i.e., i.e., (use p R ν ) y p(t) [ y ρ + ν(t)y ] + f(t), [ ν(t)p(t)]y p(t)y ρ + f(t), y ( ν p)(t)y ρ + f(t) ν(t)p(t) and apply Theorem 40 to find the solution of (4) as (use ( ν ( ν p))(t) p(t)) For the final calculation y(t) ê p (t, )y 0 + we use Theorem 4 (ii) and (v). f(τ) ê p (t, τ) ν(τ)p(τ) τ. ê p (t, τ) ν(τ)p(τ) êp(t, τ) ê p (ρ(τ), τ) ê p(t, ρ(τ)), Remark 44. Because of Theorem 4 (v), an alternative form of the solution of the initial value problem (4) is given by [ ] y(t) ê p (t, ) y 0 + ê p (, ρ(τ))f(τ) τ. Exercise 45. Use the variation of constants formula from Theorem 43 to solve the following initial value problems on the indicated time scales: (i) y y + t, y(0) 0, where T R; (ii) y y + 3 t, y(0) 0, where T Z; (iii) y p(t)y + ê p (ρ(t), ), y( ) 0, where T is an arbitrary time scale and p R ν. 6. Wronskians In this section we consider the second order linear nabla dynamic equation (5) y + p(t)y + q(t)y f(t), where we assume that p, q, f C ld. If we introduce the operator L : C ld C ld by L y(t) y (t) + p(t)y (t) + q(t)y(t) for t T κ, then (5) can be rewritten as L y f. If y C ld and L y(t) f(t) for all t T κ, then we say y is a solution of L y f on T. The fact that L is a linear operator (see Theorem 46) is why we call equation (5) a linear equation. If f(t) 0 for all t T κ, then we get the homogeneous dynamic equation L y 0. Otherwise we say the equation L y f is nonhomogeneous. The following principle of superposition is easy to prove and is left as an exercise.

NABLA DYNAMIC EQUATIONS Theorem 46. The operator L : C ld C ld is a linear operator, i.e., L (αy + βy ) αl (y ) + βl (y ) for all α, β R and y, y C ld. If y and y solve the homogeneous equation L y 0, then so does y αy + βy, where α and β are any real constants. Exercise 47. Prove Theorem 46. Show also that the sum of a solution of the homogeneous equation L y 0 and the nonhomogeneous equation (5) is a solution of (5). Definition 48. Equation (5) is called ν-regressive provided p, q, f C ld such that the ν-regressivity condition (6) + ν(t)p(t) + ν (t)q(t) 0 for all t T κ holds. Theorem 49. Assume that the dynamic equation (5) is ν-regressive. If T κ, then the initial value problem L y f(t), y( ) y 0, y ( ) y 0, where y 0 and y0 are given constants, has a unique solution, and this solution is defined on the whole time scale T. To motivate the next definition we now try to solve the initial value problem (7) L y 0, y( ) y 0, y ( ) y 0, where T κ and y 0, y0 R. If y and y are two solutions of L y 0, then by Theorem 46 y(t) : αy (t) + βy (t) is a solution of L y 0 for all α, β R. Then we want to see if we can pick α and β so that y 0 y( ) αy ( ) + βy ( ) and y0 y ( ) αy ( ) + βy ( ), i.e., (8) y ( ) y ( ) α y () y () β System (8) has a unique solution provided the occurring -matrix is invertible. This leads to the following definition. Definition 50. For two nabla differentiable functions y and y we define the nabla Wronskian W W (y, y ) by y (t) W (t) det y (t) y 0 y 0 y (t). y (t).

PETERSON AND ANDERSON, ET AL We say that two solutions y and y of L y 0 form a fundamental set of solutions (or a fundamental system) for L y 0 provided W (y, y )(t) 0 for all t T κ. Theorem 5. If the pair of functions y, y forms a fundamental system of solutions for L y 0, then y(t) αy (t) + βy (t), where α and β are constants, is a general solution of L y 0. By a general solution we mean every function of this form is a solution and every solution is in this form. Proof. Assume that the pair of functions y, y is a fundamental system of solutions for L y 0. By Theorem 46, any function of the form y(t) αy (t) + βy (t), where α and β are constants, is a solution of L y 0. Now we show that any solution of L y 0 is of this form. Let y 0 (t) be a fixed but arbitrary solution of L y 0. Fix T κ and let y 0 : y 0 ( ), y 0 : y 0 ( ). Let We want to pick α and β so that and u(t) αy (t) + βy (t). u( ) y 0 αy ( ) + βy ( ) u ( ) y 0 αy ( ) + βy ( ). Hence we want to pick α and β so that (8) is satisfied. Since the determinant of the coefficient matrix in (8) is nonzero, there is a unique set of constants α 0, β 0 so that (8) is satisfied. It follows that u(t) α 0 y (t) + β 0 y (t) solves the initial value problem (7). By the uniqueness theorem (Theorem 49), y 0 (t) u(t) α 0 y (t) + β 0 y (t). It is an important fact that the nabla Wronskian of two solutions of (5) is nonzero at a single point if and only if it is nonzero for all t. This is a consequence of the subsequent Abel s formula. Before we prove Abel s theorem we prove the following lemma concerning nabla Wronskians. Lemma 5. Let y and y be twice nabla differentiable. Then y ρ y ρ (i) W (y, y ) det ; y y

y ρ (ii) W y ρ (y, y ) det NABLA DYNAMIC EQUATIONS 3 y y ; y ρ (iii) W y ρ (y, y ) det + ( p νq)w (y, y ). Ly Ly Proof. By Definition 50 we have y y W (y, y ) det y y y ρ det + νy y ρ + νy y y y ρ y ρ det. y y This proves part (i). For part (ii), we use the product formula to calculate [W (y, y )] (y y y y ) Finally we employ part (ii) to obtain [W (y, y )] y ρ y ρ det y y y ρ y + y y y ρ y y y y ρ y y ρ y y ρ y ρ det. y y y ρ y ρ det Ly py qy Ly py qy

4 PETERSON AND ANDERSON, ET AL y ρ y ρ det + det y ρ y ρ Ly Ly py qy py qy y ρ y ρ det + det y ρ y ρ Ly Ly py qyρ qνy py qyρ qνy y ρ y ρ det + det y ρ y ρ Ly Ly py qνy py qνy y ρ y ρ y ρ y ρ det + ( p νq) det. Ly Ly y y From here, part (iii) follows by applying part (i). Theorem 53 (Abel s Theorem). Let T κ and assume that L y 0 is ν-regressive. Suppose that y and y are two solutions of L y 0. Then their Wronskian W W (y, y ) satisfies W (t) ê p νq (t, )W ( ) for t T κ. Proof. Using the fact that y and y are solutions of L y 0 and Lemma 5 (iii), we see that W is a solution of the initial value problem (9) W [ p(t) ν(t)q(t)]w, W ( ) W 0, where we put W 0 W ( ). Using condition (6) we get that the coefficient function p νq is ν-regressive. Since p, q, ν C ld, we have that p νq C ld. Altogether p νq R ν. By Theorem 30, the unique solution of (9) is W (t) ê p νq (t, )W 0 for t T κ. Alternatively, we may consider a linear dynamic equation of the form (30) x + p(t)x ρ + q(t)x ρ 0. Definition 54. We say that (30) is ν-regressive provided p R ν and q C ld. Theorem 55. If (30) is ν-regressive, then it is equivalent to a ν-regressive equation of the form L y 0. Conversely, if L y 0 is ν-regressive, then it is equivalent to a ν-regressive equation of the form (30).

NABLA DYNAMIC EQUATIONS 5 Proof. Assume (30) is ν-regressive. Then p, q C ld and ν(t)p(t) 0 for t T κ. We write x + px ρ + qx ρ x + p(x νx ) + q(x νx ) Hence equation (30) is equivalent to where p : p νq νp Note that p, q C ld, and since x νpx + (p νq)x + qx ( νp)x + (p νq)x + qx [ [ νp] x + p νq νp x + q ] νp x. y + p (t)y + q (t)y 0, and q : q νp. + νp + ν q + ν p νq νp + q ν νp νp + νp ν q + ν q νp νp 0, we get that (30) is equivalent to a ν-regressive equation of the form (5). Next assume that (5) is ν-regressive. Then + ν(t)p(t) + ν (t)q(t) 0 for t T κ and p, q C ld. Consider y + py + qy y + p(y ρ + νy ) + q(y ρ + νy ) ( + νp)y + py ρ + qy ρ + νq(y ρ + νy ) ( + νp + ν q)y + (p + νq)y ρ + qy ρ ( ( + νp + ν q) y + Hence L y 0 is equivalent to the equation (3) p + νq + νp + ν q y ρ + x + p (t)x ρ + q (t)x ρ 0, ) q + νp + ν q yρ. where p + νq p : + νp + ν q and q q : + νp + ν q. Note that p, q C ld, and since νp p + νq ν + νp + ν q + νp + ν q 0,

6 PETERSON AND ANDERSON, ET AL we have p R ν so that (3) is ν-regressive. Theorem 56 (Abel s Theorem for (30)). Assume that (30) is ν-regressive and let T κ. Suppose that x and x are two solutions of equation (30). Then their Wronskian W W (x, x ) satisfies W (t) ê νp(t, )W ( ) for t T κ. Proof. Let x and x be two solutions of (30). We use Theorem 5 (ii) to find W x ρ x ρ det x x x ρ x ρ det px ρ qx ρ px ρ qx ρ x ρ x ρ det px ρ px ρ x ρ x ρ p det x ρ x ρ p det x x x x pw ρ. Hence W is a solution of the initial value problem W p(t)w ρ, W ( ) W 0, where we put W 0 W ( ). Since p C ld we get from Theorem 37 that the unique solution of this initial value problem is given by W (t) ê νp(t, )W 0. Corollary 57. Assume q C ld. Then the Wronskian of any two solutions of is independent of t. x + q(t)x ρ 0 ρ

NABLA DYNAMIC EQUATIONS 7 Proof. Let T. The above assumptions ensure that (30) with p 0 is ν-regressive. Since ν p 0, we have ê νp(t, ), and by Theorem 56, W (x, x )(t) W (x, x )( ) follows, where x and x can be any two solutions of (30). 7. Nabla Hyperbolic and Trigonometric Functions Here and in the following section we consider the second order linear dynamic homogeneous equation with constant coefficients (3) y + αy + βy 0 with α, β R, on a time scale T. We assume throughout that the dynamic equation (3) is ν-regressive, i.e., + αν(t) + βν (t) 0 for t T κ, i.e., βν α R ν. We try to find numbers λ C with λν(t) 0 for t T κ such that y(t) ê λ (t, ) is a solution of (3). Note that if y(t) ê λ (t, ), then y (t) + αy (t) + βy(t) λ ê λ (t, ) + αλê λ (t, ) + βê λ (t, ) ( λ + αλ + β ) ê λ (t, ). Since ê λ (t, ) does not vanish, y(t) ê λ (t, ) is a solution of (3) iff λ satisfies the so-called characteristic equation (33) λ + αλ + β 0. The solutions λ and λ of the characteristic equation (33) are given by (34) λ α α 4β and λ α + α 4β. Exercise 58. Let λ and λ be defined as in (34). Show that (3) is regressive iff λ, λ R ν. Theorem 59. Suppose α 4β 0. If βν α R ν, then a general solution of (3) is given by y(t) c ê λ (t, ) + c ê λ (t, ), where T κ and λ, λ are given in (34). The solution y of the initial value problem (35) y + αy + βy 0, y( ) y 0, y ( ) y 0 is given by y(t) y 0 ê λ (t, ) + ê λ (t, ) + αy 0 + y0 α 4β ê λ (t, ) ê λ (t, ).

8 PETERSON AND ANDERSON, ET AL Proof. Since λ and λ given in (34) are solutions of the characteristic equation (33), we find that (note λ, λ R ν according to Exercise (58)) both ê λ (, ) and ê λ (, ) solve the dynamic equation (3). Moreover, the Wronskian of these two solutions is found to be ê λ (t, ) ê λ (t, ) det (λ λ )ê λ (t, )ê λ (t, ) λ ê λ (t, ) λ ê λ (t, ) α 4βê λ νλ (t, ), which does not vanish since α 4β 0. Hence the pair of functions ê λ (, ) and ê λ (, ), is a fundamental system of (3). The proof of the last statement in this theorem is straight forward and so will be omitted. When α 0 and β < 0, we are interested in the nabla hyperbolic functions defined as follows. Definition 60 (Nabla Hyperbolic Functions). If p C ld and νp R ν, then we define the nabla hyperbolic functions cosh p (, ) and ŝinh p(, ) by for t T. cosh p (t, ) : êp(t, ) + ê p (t, ) and ŝinh p (t, ) : êp(t, ) ê p (t, ) Note that the ν-regressivity condition on νp is equivalent to 0 ν p ( νp)( + νp) and hence is equivalent to both p and p being ν-regressive. In the following, if f is a function in two variables, then by f we mean the nabla derivative with respect to the first variable. Lemma 6. Let p C ld. If νp R ν and T κ, then for t T we have (i) cosh p (t, ) p(t)ŝinh p(t, ); (ii) ŝinh p (t, ) p(t) cosh p (t, ); (iii) cosh p(t, ) ŝinh p(t, ) ê νp (t, ); (iv) cosh p (t, ) + ŝinh p(t, ) ê p (t, ); (v) cosh p (t, ) ŝinh p(t, ) ê p (t, ); (vi) W [ cosh p (t, ), ŝinh p(t, )] p(t)ê νp (t, ).

NABLA DYNAMIC EQUATIONS 9 Proof. Using Definition 60, the first two formulas are easily verified, while the formula in (iii) follows from ) cosh (êp + ê ) p (êp ê p p ŝinh p ê p + ê p ê p + ê p 4 ê p ê p ê p ê p ê p + ê p 4 ê p νp ê νp, where we have used Theorem 4 (vi). The Euler type formulas (iv) and (v) follow immediately from Definition 60. Finally note that W ( ) coshp (t, ), ŝinh p(t, ) cosh p (, ) ŝinh p (, ) det p(t)ŝinh p(t, ) p(t) cosh p (t, ) [ ] p(t) cosh p(t, ) ŝinh p(t, ) p(t)ê νp (t, ) so that (vi) holds. Exercise 6. Show that if γ > 0 with γ ν R ν, then cosh γ (, ) and ŝinh γ(, ) are solutions of (36) y γ y 0. Theorem 63. If γ > 0 with γ ν R ν, then a general solution of (36) is given by where c and c are constants. y(t) c coshγ (t, ) + c ŝinh γ (t, ), Proof. From Exercise 6, cosh γ (, ) and ŝinh γ(, ) are solutions of (36). From Theorem 6, (vi) we get that these solutions are a fundamental set of solutions and hence the conclusion of this theorem follows from Theorem 5. Exercise 64. Show that if γ > 0 and γ ν R ν, then the solution of the IVP y γ y 0, y( ) y 0, y ( ) y 0 is given by y(t) y 0 coshγ (t, ) + y 0 γ ŝinh γ(t, ).

30 PETERSON AND ANDERSON, ET AL Theorem 65. Suppose α 4β > 0. Define p α α and q 4β. If p and βν α are ν-regressive and T κ, then a general solution of (3) is given by y(t) c ê p (t, ) cosh q/( νp) (t, ) + c ê p (t, )ŝinh q/( νp)(t, ), where T, and the Wronskian W [ê p (t, ) cosh q/( νp) (t, ), ê p (t, )ŝinh q/( νp)(t, ] qê νβ α (t, ) for t T κ. The solution of the IVP (35) is given by [ y(t) ê p (t, ) y 0coshq/( νp) (t, ) + y 0 py ] 0 ŝinh q q/( νp) (t, ). Proof. In this proof we use the convention that ê p ê p (, ) and similarly for cosh p and ŝinh p. We apply Theorem 59 to find two solutions of (3) as ê p+q and ê p q. By Theorem 46 we can construct two other solutions of (3) by We use the formulas y êp+q + ê p q ( ) q p ν νp and ( p ν q ) νp to obtain, by using Theorem 4 (vi), and y êp+q ê p q. p + q νp νpq νp p + q p q νp + νpq νp p q and similarly y êp+q + ê p q êp ν(q/( νp)) + ê p ν( q/( νp)) êpê q/( νp) + ê p ê q/( νp) ê q/( νp) + ê q/( νp) ê p ê pcoshq/( νp) y êp+q ê p q ê p ŝinh q/( νp).

NABLA DYNAMIC EQUATIONS 3 Instead of using Abel s Theorem (Theorem 53) we will calculate this Wronskian directly. First we find, by using Theorem 4 (ii) and Lemma 6, that y ê p cosh q/( νp) + ê ρ pcosh q/( νp) pê pcoshq/( νp) + ê ρ q p νpŝinh q/( νp) pê p coshq/( νp) + qê p ŝinh q/( νp) and similarly that y pê p ŝinh q/( νp) + qê pcoshq/( νp). Then the Wronskian of y and y is given by ê pcoshq/( νp) ê p ŝinh q/( νp) det pê pcoshq/( νp) + qê p ŝinh q/( νp) pê p ŝinh q/( νp) + qê p ê pcoshq/( νp) det qê p ŝinh q/( νp) [ qê p cosh q/( νp) ŝinh q/( νp) qê pê νq /( νp) qê p( νp) ν(νq /( νp) ) ê p ŝinh q/( νp) qê pcoshq/( νp) ] coshq/( νp) qê p ν(p q ) qê νβ α, where we have used Lemma 6. When α 0 and β > 0, we are interested in the nabla trigonometric functions defined as follows. Definition 66 (Nabla Trigonometric Functions). If p C ld and νp R ν, then we define the nabla trigonometric functions ĉos p (, ) and ŝin p (, ) by ĉos p (t, ) : êip(t, ) + ê ip (t, ) for t T. and ŝin p (t, ) : êip(t, ) ê ip (t, ) i Note that νp is ν-regressive iff both ip and ip are ν-regressive, so ĉos p (, ) and ŝin p (, ) in Definition 66 are well defined. The proofs of Lemma 67 and Theorem 70 are similar to the proofs of Lemma 6 and Theorem 65 and hence will left as exercises (Exercises 68 and 7). Lemma 67. Let p C ld. If νp R ν and T κ, then we have for t T κ (i) ĉos p (t, ) p(t)ŝin p (t, );

3 PETERSON AND ANDERSON, ET AL (ii) ŝin p (t, ) p(t)ĉos p (t, ); (iii) ĉos p(t, ) + ŝin p(t, ) ê νp (t, ); (iv) ê ip (t, ) ĉos p (t, ) + iŝin p (t, ); (v) ê ip (t, ) ĉos p (t, ) iŝin p (t, ); (vi) W [ĉos p (t, ), ŝin p (t, )] p(t)ê νp (t, ). We remark that for p R, the regressivity condition on νp is always satisfied. Exercise 68. Prove Lemma 67. Theorem 69. Assume γ > 0 and T κ. Then is a general solution of y(t) c ĉos γ (t, ) + c ŝin γ (t, ) (37) y + γ y 0. Proof. Note that the equation (37) is ν-regressive because + γ ν (t) 0 for t T κ. Using Lemma 67,(i), (ii), we can easily show that ĉos γ (t, ) and ŝin γ (t, ) are solutions of (37). Using Lemma 67,(vi) we get that ĉos γ (, ) and ŝin γ (, ) form a fundamental set of solutions of (37). It follows that y(t) c ĉos γ (t, ) + c ŝin γ (t, ) is a general solution of (37). Theorem 70. Suppose α 4β < 0. Define p α and q 4β α. If p and νβ α are regressive, then a general solution of (3) is given by y(t) c ê p (t, )ĉos q/( νp) (t, ) + c ê p (t, )ŝin q/( νp) (t, ), where T, and the Wronskian W [ê p (t, )ĉos q/( νp) (t, ), ê p (t, )ŝin q/( νp) (t, )] qê νβ α (t, ) for t T κ. The solution of the initial value problem (35) is given by y(t) y 0 ê p (t, )ĉos q/( νp) (t, ) + y 0 py 0 ê p (t, )ŝin q q/( νp) (t, ). Exercise 7. Prove Theorem 70.

NABLA DYNAMIC EQUATIONS 33 8. Reduction of Order Now we consider the nabla dynamic equation (3) in the case that α 4β 0. From (34) we have λ λ p, where p α. In this case α p, β p, and so the dynamic equation (3) is of the form (38) Hence one solution y of (3) is given by y py + p y 0. y (t) ê p (t, ), where T κ. We will now find a second linearly independent solution of (38) using the so-called method of reduction of order. We will look for a second linearly independent solution of the form (39) By the product rule and Theorem 4 (ii), (40) y(t) u(t)ê p (t, ). y (t) u (t)ê ρ p(t, ) + u(t)ê p (t, ) u (t)[ ν(t)p]ê p (t, ) + pê p (t, )u(t). Now we must be careful because there are many time scales where the graininess function ν is not nabla differentiable. We assume u is a function such that u ( νp) is nabla differentiable. Then from (40) we get using the product rule that (4) y (t) { u (t)[ ν(t)p] } êρ p (t, ) + u (t)[ ν(t)p]ê p (t, ) +pê ρ p(t, )u (t) + pê p (t, )u(t) { u (t)[ ν(t)p] } êρ p (t, ) + u (t)[ ν(t)p]pê p (t, ) +p[ ν(t)p]ê p (t, )u (t) + p ê p (t, )u(t) { u (t)[ ν(t)p] } êρ p (t, ) Using (39), (40), and (4) we obtain +p[ ν(t)p]ê p (t, )u (t) + p ê p (t, )u(t). y (t) py (t) + p y(t) { u (t)[ ν(t)p] } êρ p (t, ) +p[ ν(t)p]ê p (t, )u (t) + p ê p (t, )u(t) pu (t)[ ν(t)p]ê p (t, ) p ê p (t, )u(t) +p u(t)ê p (t, ) { u (t)[ ν(t)p] } êρ p (t, ). Hence for y(t) u(t)ê p (t, ) to be a solution we want to choose u so that { u (t)[ ν(t)p] } 0.

34 PETERSON AND ANDERSON, ET AL We will get that the above equation is true if we choose u so that u (t)[ ν(t)p]. Recall in the steps above that we assumed u (t)[ ν(t)p] to be differentiable, and that is true in this case. We then solve for u (t) to get Hence if we take u (t) ν(t)p. u(t) ν(τ)p τ, we can check that all of the above steps are valid and so (4) y(t) ê p (t, ) ν(τ)p τ is a solution of (3). This leads to the following theorem. Theorem 7. Suppose α 4β 0. Define p α. If p R ν, then a general solution of (3) is given by y(t) c ê p (t, ) + c ê p (t, ) where T, and the Wronskian [ W ê p (t, ), ê p (t, ) ] pν(τ) τ The solution of the initial value problem (35) is given by [ ê p (t, ) y 0 + (y0 py 0 ) pν(τ) τ, ê α να /4(t, ). τ pν(τ) In the next exercise we use a slight variation of the method of reduction of order to find a second linearly independent solution of (38). We also call this the method of reduction of order and we will use this method several times later in this chapter. Exercise 73. In this exercise we outline another method for finding the solution (4) of the dynamic equation (38). The reader should fill in the missing steps. Suppose y is the solution of (38) satisfying the initial conditions and consider y( ) 0 and y ( ) ]. W (ê p (, ), y)(t) ê p (t, )y (t) ê p (t, )y(t) ê p (t, )y (t) pê p (t, )y(t) (y (t) py(t))ê p (t, ).

In particular, NABLA DYNAMIC EQUATIONS 35 W (ê p (, ), y)( ) y ( ) py( ). On the other hand, by Abel s theorem (Theorem 53), we have W (ê p (, ), y)(t) W (ê p (, ), y)( )ê νβ α (t, ) ê νβ α (t, ). Hence y is a solution of the first order linear equation (y py)ê p (t, ) ê νβ α (t, ) or, equivalently according to Theorem 4 (vii), y py ê ( νβ α) νp(t, ) ê p (t, ). Since y( ) 0, we have by the variation of constants formula given in Theorem 43 that y(t) ê p (t, ρ(τ))ê p (τ, ) τ ê p (t, ) 9. Nabla Riccati Equation ν(τ)p τ. In this section we will introduce and briefly study the nabla Riccati equation associated with the second order linear equation M x x + p(t)x ρ + q(t)x ρ 0. We assume throughout that p and q be in C ld. For z C ld R ν we define the nabla Riccati operator R by R z z + q + pz ρ + z, where the nabla generalized square of z R ν is defined by z : ( z)( ν z) z νz. We say z is a solution of R z 0 on T κ provided z C ld R ν and R z(t) 0 for t T κ. First we prove the following factorization theorem relating the second order linear operator M and the first order nonlinear Riccati operator R. Theorem 74. Assume x C ld and x(t) 0 for t T. Then z : x x for t T κ. x ρ (t)r z(t) M x(t) R ν and Proof. Assume x C ld and x(t) 0 for t T and let z : x x. First note that so z R ν. Next consider ν(t)z(t) ν(t) x (t) x(t) x(t) ν(t)x (t) xρ (t) x(t) x(t) 0

36 PETERSON AND ANDERSON, ET AL x ρ R z x ρ [z + q + pz ρ + z ] { (x ) x ρ + q + p x ρ x x ρ x ρ {( xx (x ) xx ρ + z } ) + q + p x ρ x ρ + (x ) x[x νx ] x (x ) + qx ρ + px ρ + (x ) x x M x. Corollary 75. Assume x Cld, x(t) 0 for t T, and z : x x. Then x is a solution of M x 0 on T iff z is a solution of R z 0 on T κ. In particular if z is a solution of R z 0 on T κ, then x ê z (, ) is a solution of M x 0 on T. Proof. The first statement follows from Theorem 74. Now assume z is a solution of R z 0, then z R ν and it follows from Theorem 74 that the solution x ê z (, 0) of the IVP is a solution M x 0. x z(t)x, x( ) 0. ν-regressive Matrices Definition 76. Let A be an m n-matrix-valued function on T. We say that A is ldcontinuous on T if each entry of A is ld-continuous on T, and the class of all such ldcontinuous m n-matrix-valued functions on T is denoted by C ld C ld (T) C ld (T, R m n ). We say that A is nabla differentiable on T provided each entry of A is nabla differentiable on T, and in this case we put A (a ij) i m, j n, where A (a ij ) i m, j n. Theorem 77. If A is nabla differentiable at t T κ, then A ρ (t) A(t) ν(t)a (t). Proof. We obtain at t. A ρ (a ρ ij ) (a ij νa ij) (a ij ) ν(a ij) A νa Theorem 78. Suppose A and B are nabla differentiable n n-matrix-valued functions. Then }

NABLA DYNAMIC EQUATIONS 37 (i) (A + B) A + B ; (ii) (αa) αa if α is constant; (iii) (AB) A B ρ + AB A ρ B + A B; (iv) (A ) (A ρ ) A A A A (A ρ ) if AA ρ is invertible; (v) (AB ) (A AB B )(B ρ ) (A (AB ) ρ B )B if BB ρ is invertible. Proof. We only show the first parts of (iii) and (iv) and leave the remainder of the proof as an exercise. Let i, j n. We calculate the ijth entry of (AB) : ( n ) n a ik b kj (a ik b kj ) k k n k ( ) a ik bρ kj + a ikb kj n n a ik bρ kj + a ik b kj, k and this is the ijth entry of the matrix A B ρ + AB (see Definition 76). Next, we use part (iii) to differentiate I AA : 0 A A + A ρ (A ), and solving for (A ) yields the first formula in (iv). We consider the linear system of dynamic equations (43) y A(t)y, where A is an n n-matrix-valued function on T. A vector-valued y : T R n is said to be a solution of (43) on T provided y (t) A(t)y(t) holds for each t T κ. In order to state the main theorem on solvability of initial value problems involving equation (43), we make the following definition. Definition 79 (Regressivity). An n n-matrix-valued function A on a time scale T is called ν-regressive (with respect to T) provided (44) I ν(t)a(t) is invertible for all t T κ, and the class of all such ν-regressive and ld-continuous matrix functions is denoted by k R ν R ν (T) R ν (T, R n n ). We say that the system (43) is ν-regressive provided A R ν. Exercise 80. Show that the n n-matrix-valued function A is ν-regressive iff the eigenvalues λ i (t) of A(t) are ν-regressive for all i n. Exercise 8. Show that a -matrix-valued function A is ν-regressive iff the scalar-valued function tr A ν det A is ν-regressive (where tr A denotes the trace of the matrix A, i.e., the sum of the diagonal elements of A). Derive similar characterizations for the 3 3-case, the 4 4-case, and the general n n-case.

38 PETERSON AND ANDERSON, ET AL Now the existence and uniqueness theorem for initial value problems for equation (43) reads as follows. Theorem 8 (Existence and Uniqueness Theorem). Let A R ν be an n n-matrix-valued function on T and suppose that f : T R n is ld-continuous. Let T and y 0 R n. Then the initial value problem y A(t)y + f(t), y( ) y 0 has a unique solution y : T R n. It follows from Theorem 8 that the matrix initial value problem (45) Y A(t)Y, Y ( ) Y 0, where Y 0 is a constant n n-matrix, has a unique (matrix-valued) solution Y. Before we define and give some properties of the important nabla exponential matrix function we define the ν circle plus addition ν and the ν circle minus subtraction ν. Definition 83. Assume A and B are ν-regressive n n-matrix-valued functions on T. Then we define A ν B by and we define ν A by (A ν B)(t) : A(t) + B(t) ν(t)a(t)b(t) for all t T κ, ( ν A)(t) : [I ν(t)a(t)] A(t) for all t T κ. Exercise 84. Show that if A is ν-regressive on T, then ( ν A)(t) A(t)[I ν(t)a(t)] for all t T κ. Lemma 85. (R ν (T, R n n ), ν ) is a group. Proof. Let A, B R ν. Then I ν(t)a(t) and I ν(t)b(t) are invertible for all t T κ, and therefore I ν(t)(a ν B)(t) I ν(t)[a(t) + B(t) ν(t)a(t)b(t)] I ν(t)a(t) ν(t)b(t) + ν (t)a(t)b(t) [I ν(t)a(t)][i ν(t)b(t)] is also invertible for each t T κ, being the product of two invertible matrices. Hence (note that A ν B C ld ) A ν B R ν. Next, if A R ν, then ν A defined in Definition 83 satisfies (A ν ( ν A))(t) 0 for all t T κ. It remains to show that ν A R ν. But this follows since because of Exercise 84 I ν(t)( ν A)(t) I + ν(t)a(t)[i ν(t)a(t)] [I ν(t)a(t)][i ν(t)a(t)] + ν(t)a(t)[i ν(t)a(t)] [I ν(t)a(t) + ν(t)a(t)][i ν(t)a(t)] [I ν(t)a(t)]

NABLA DYNAMIC EQUATIONS 39 is invertible for all t T κ. In Exercise 86, the reader is asked to check that ν satisfies the associative law. Exercise 86. Show that ν satisfies the associative law, i.e., A ν (B ν C) (A ν B) ν C for all A, B, C R ν. Definition 87. If the matrix-valued functions A and B are ν-regressive on T, then we define A ν B by (A ν B)(t) (A ν ( ν B))(t) for all t T κ. If A is a matrix, then we let A denote its conjugate transpose. Exercise 88. Show that (A ) (A ) holds for any nabla differentiable matrix-valued function A. Exercise 89. Suppose A and B are ν-regressive matrix-valued functions. Show that (i) A is ν-regressive; (ii) ν A ( ν A) ; (iii) (A ν B) B ν A. Definition 90 (Nabla Matrix Exponential Function). Let T and assume that A R ν is an n n-matrix-valued function. The unique matrix-valued solution of the IVP Y A(t)Y, Y ( ) I, where I denotes as usual the n n-identity matrix, is called the nabla matrix exponential function (at ), and it is denoted by ê A (, ). In the following theorem we give some properties of the nabla matrix exponential function. Theorem 9. If A, B R ν are matrix-valued functions on T, then (i) ê 0 (t, s) I and ê A (t, t) I; (ii) ê A (ρ(t), s) (I ν(t)a(t))ê A (t, s); (iii) ê A (t, s) ê νa (t, s); (iv) ê A (t, s) ê A (s, t) ê νa (s, t); (v) ê A (t, τ)ê A (τ, s) ê A (t, s); (vi) ê A (t, s)ê B (t, s) ê A νb(t, s) if ê A (t, s) and B(t) commute. Proof. Part (i). Consider the initial value problem Y 0, Y (s) I, which has exactly one solution by Theorem 8. Since Y (t) I is a solution of this IVP, we have ê 0 (t, s) Y (t) I. Clearly ê A (t, t) I. Part (ii). According to Theorem 77 we have ê A (ρ(t), s) ê A (t, s) ν(t)ê A(t, s) ê A (t, s) ν(t)a(t)ê A (t, s) (I ν(t)a(t))ê A (t, s).

40 PETERSON AND ANDERSON, ET AL Part (iii). Let Y (t) (ê A (t, s)). Then Y (t) ( ê A (ρ(t), s)ê A(t, s)ê A (t, s)) ( ê A (ρ(t), s)a(t)ê A(t, s)ê A (t, s)) ( ê A (ρ(t), s)a(t)) ( ê A (t, s)(i ν(t)a(t)) A(t) ) ( ê A (t, s)( νa)(t) ) ( ν A )(t)(ê A (t, s)) ( ν A )(t)y (t), where we have used (ii) and Exercise 89 (ii). Also, Y (s) (ê A (s, s)) (I ) I. Hence Y solves the initial value problem Y ( ν A )(t)y, Y (s) I, which has exactly one solution according to Theorem 8, and therefore we have ê νa (t, s) Y (t) (ê A (t, s)) so that ê A (t, s) ê νa (t, s). Part (iv). This is Exercise 9. Part (v). Consider Y (t) ê A (t, τ)ê A (τ, s). Then Y (t) ê A(t, τ)ê A (τ, s) A(t)ê A (t, τ)ê A (τ, s) A(t)Y (t) and Y (s) ê A (s, s)ê A (s, s) I according to (iv). Hence Y solves the IVP Y A(t)Y, Y (s) I, which has exactly one solution according to Theorem 8, and therefore we have ê A (t, s) Y (t) ê A (t, τ)ê A (τ, s). Part (vi). Let Y (t) ê A (t, s)ê B (t, s) and suppose that ê A (t, s) and B(t) commute. We use Theorem 77 and Theorem 78 (iii) to calculate Y (t) ê A(t, s)ê ρ B (t, s) + ê A(t, s)ê B(t, s) A(t)ê A (t, s)(i ν(t)b(t))ê B (t, s) + ê A (t, s)b(t)ê B (t, s) A(t)(I ν(t)b(t))ê A (t, s)ê B (t, s) + B(t)ê A (t, s)ê B (t, s) [A(t)(I ν(t)b(t)) + B(t)] ê A (t, s)ê B (t, s) (A ν B)(t)ê A (t, s)ê B (t, s) (A ν B)(t)Y (t). Also Y (s) ê A (s, s)ê B (s, s) I I I. So Y solves the initial value problem Y (A ν B)(t)Y, Y (s) I, which has exactly one solution according to Theorem 8, ê A νb(t, s) Y (t) ê A (t, s)ê B (t, s). Exercise 9. Prove part (iv) of Theorem 9. and therefore we have