Equations with Separated Variables on Time Scales

Transcription

1 International Journal of Difference Equations ISSN , Volume 12, Number 1, pp (217) Equations with Separated Variables on Time Scales Antoni Augustynowicz Institute of Mathematics Faculty of Mathematics, Physics and Informatics University of Gdańsk 57 Wit Stwosz str., Gdańsk, Poland antoni.augustynowicz@mat.ug.edu.pl Agata Gołaszewska Faculty of Applied Physics and Mathematics Department of Mathematics Gdańsk University of Technology 11/12 Narutowicz str., Gdańsk, Poland golaszewska@mif.pg.gda.pl, agata@gda.pl Abstract We show that the well-known theory for classical ordinary differential equations with separated variables is not valid in case of equations on time scales. Namely, the uniqueness of solutions does not depend on the convergence of appropriate integrals. AMS Subject Classifications: 26E7, 34N5, 39A1. Keywords: Dynamic equation, time scale, separated variables. 1 Introduction Dynamic equations on time scales have become very interesting for many mathematicians in recent years (see [2, 3, 5], and references for others). The main aim of this paper is to verify how the theory of dynamic equations with separated variables (see [6]) changes if we generalize classical equations to equations on time scales. We show that classical theorems are not true even under some additional assumptions. We introduce some basic definitions (see [2]). Received March 14, 216; Accepted November 22, 216 Communicated by Agnieszka Malinowska

2 2 A. Augustynowicz, A. Gołaszewska Definition 1.1. A time scale is an arbitrary nonempty closed set of the real numbers. Let T be a time scale. We denote I T := I T for an arbitrary I R. Now we define some operators. Definition 1.2. We define the forward jump operator σ : T T by σ(t) = inf {s : s T, s > t}, σ(max T) = max T if sup T <, and the backward jump operator ρ : T T by ρ(min T) = min T if inf T >. ρ(t) = sup {s : s T, s < t}, Definition 1.3. Assume that f : T R and let t T κ = T\ {sup T}. Then we define f (t) to be the number (provided it exists) with the property that for any given ɛ >, there is a neighborhood U of t such that [f(σ(t)) f] f (t)[σ(t) s] ɛ σ(t) s, for all s U T. If f (t) = g(t) for every t [a, b] T, then we denote f(b) = f(a) + b a g(t) t. Let us introduce some special class of time scales that are the most important for this paper. Definition 1.4. If there exists a sequence {t n } n=1 T such that t n and σ(t n ) > t n, then we say that T is a DNC time scale at (dense not classical). In this paper, we assume that T means a DNC time scale and let, for simplicity, min T =. Moreover, we fix the symbol {t n } for a sequence that has the property mentioned in the above definition. We study the uniqueness of solutions of the following initial value problem for equations with separated variables { x (t) = f(t)g(x(t)), t T, (1.1) x() =, where f : T R and g : R R. We say that a function z : T R is a solution of the problem (1.1) if z() = and z (t) = f(t)g(z(t)) for all t T. Rewriting from [6] the uniqueness of the zero solution of (1.1) (if g() = ) in the classical case depends on divergence of the integral ε g(x). Concerning solutions of the problem (1.1) in a local sense, all time scales can be generally divided into three following classes:

3 Equations with Separated Variables on Time Scales 3 (i) [, a] T for some a R; (ii) σ() > ; (iii) T is a DNC time scale at. In the first case (i), we are dealing with the classical theory of ordinary differential equations with separated variables. In the second case (ii), regardless of the time scale, the problem (1.1) has always exactly one solution in a local sense. The DNC time scales are very wide class of different time scales. In this paper, we find different examples of the problem (1.1), where independently from any DNC time scale and from convergence of appropriate integral it may have exactly one solution or it may have more than one solution. In some cases it may even have no solution at all. The following closed sets are some examples of DNC-time scales: {} {τ n : n N}, τ n ; {} n N [ 1 n, 1 n + 1 n 2 ] ; {} { 1 n : } { 1 n N n + 1 nk } : n, k N, k > n ; the Cantor set. In the theory of equations x (t) = F (t, x(t)) on time scales it is natural to consider not necessarily continuous but so called rd-continuous functions F. It means such functions F that for any continuous x : T R the function ϕ(t) = F (t, x(t)) has the following property: ϕ is continuous at t T if σ(t) = t and lim s t ϕ is finite if ρ(t) = t < σ(t) (see [2]). Unfortunately, the Peano type theorem is not valid for equations with rd-continuous right-hand-side (see [1, 4] for counterexample on time scale {} {1, 12, 13 },... ). We present below that for every DNC time scale T there exists a problem (1.1) without any solution, however the right-hand-side of the equation is an rd-continuous function with separated variables. Namely, for any DNC time scale T, we find an rd-continuous function of the form f(t)g(x) such that the problem (1.1) has no solution. Example 1.5. Let us define f : T R and g : R R by the formulas { t, for t = tn, f(t) =, otherwise,

4 4 A. Augustynowicz, A. Gołaszewska g(x) = { 1, for x =,, for x. It is not difficult to verify that the function given by the product of functions f and g as follows F (t, x) = f(t)g(x), is rd-continuous. Suppose that x() =, x (t) = f(t)g(x(t)) for all t [, δ) T and some δ >. Then x(t) on [, δ) T since x (t). Moreover, x(t) > for t >, because if x(t n ) =, then x (t n ) >. Hence for t >, we have x (t) =, and therefore x is a constant positive function on [, δ) T. It means that x() >. We obtain the contradiction to the assumption that x() =. We are able to prove only the following result corresponding to the classical theory of equations with separated variables. Theorem 1.6. Assume that the integral ɛ G(x) is divergent for ɛ, the function g : R R is continuous, f is integrable and nonnegative on T, xg(x) > for x and { max {g(y) : y [, x]}, for x >, G(x) = min {g(y) : y [x, ]}, for x <. Then the problem (1.1) has only a trivial solution. Proof. Suppose that z : [, δ] T R is a nonzero solution of (1.1), then the function z does not change the sign. Assume that z(t ) > for some T (, δ] T, then there exists t [, T ) T such that z(t) > for t (t, T ] T and z(t ) =. Let us assume, that σ(t ) > t. Since g() =, we have z (t ) = f(t )g(z(t )) = f(t )g() =. Consequently z(σ(t )) =. This contradicts the definition of t. If [t, h] T for some h (t, T ], then we obtain z on [t, h] from a well-known theorem for equations with separated variables (see [6], page 45) and from the fact, that ε g(x) ε G(x) = for ε >. The above result means that there exists a sequence {s n } T such that s n t and σ(s n ) > s n (T is DNC at t ). Let us define z : convt R and ρ : convt T by the formulae z(t), for t T, z(t) = z(σ t) + z(σ)(t s) σ s for t (s, σ), s T, ρ(t) = max {s T : s t}.

5 Equations with Separated Variables on Time Scales 5 For each τ T such that σ(τ) > τ, we have σ(τ) τ z σ(τ) g(z) s = τ z (τ) σ(τ) g(z(τ)) s = τ z σ(τ) g(z( ρ)) s = τ z g(z( ρ)) ds, because z is a constant function on the set (τ, σ(τ)). Suppose that y : T R is an integrable function. Let us define a new function ȳ : convt R by ȳ T = y and ȳ = y(t) for s (t, σ(t)), t T. Then Indeed, we put y s = ȳds, t T. t t F (t) = t ȳds, t convt, F (t) = y s t T, t and F = F. T For t T, we have F (t) = F +(t) = ȳ(t) = y(t) = F (t), where F + means the right-hand-side derivative of the function F. Of course F (t ) = = F (t ), so we can conclude that F (t) = F (t), for t T. Now put and Now we have y = z g(z) ȳ = for s T, z + for s convt. g(z( ρ)) t z t g(z) s = t z t + g(z( ρ)) ds = t z g(z( ρ)) ds.

6 6 A. Augustynowicz, A. Gołaszewska The last equality follows from the fact, that z can be not defined only for s T such that ρ s or s σ. The set of such s is countable. Therefore, the value of the integrand on that set does not affect the value of the Riemann integral. Then z f s = t t g(z) s = z t g(z( ρ)) ds z t G(z( ρ)) ds z z(t) G( z) ds = G(x) = t t for t (t, T ], which contradicts the integrability of function f. Analogously, we obtain a contradiction if z(t ) < for some T >. 2 Results We show that in general there is no correspondence of the uniqueness of solutions of 1 (1.1) with integrability of the function and the sign of xg(x) for x. g(x) Proposition 2.1. There exists a -integrable function f, such that any δ >, f(t), a continuous function g, such that g(x) > for x, ɛ g(x) < for ε, and the problem (1.1) has only trivial solution. δ f(t) t > for Proposition 2.2. There exists a continuous function g such that g(x) > for x > and ɛ g(x) = for ε >, but the problem (1.1) has a positive solution on T\ {} for f(t) 1. Remark 2.3. One may prove an analogous result for negative solutions in a similar manner. Proposition 2.4. There exists a continuous function g such that xg(x) < for x and the integral ɛ g(x) is divergent for ε, but the problem (1.1) for f(t) 1 has a nontrivial solution.

7 Equations with Separated Variables on Time Scales 7 3 Proofs Proof of Proposition 2.1. Let f(t) = where s 1 = 1 2, s k = s 2 k 1. Then, for t t n, s n 1 s n σ(t n ) t n, for t = t n, n = f(t) t = f(t j )(σ(t j ) t j ) = j=n+1 j=n+1 (s j 1 s j ) = s n >. Suppose that x() =, x (t) = f(t) x(t) and x(t) on [, δ) for any δ >. Then x(t) > for t >. Moreover, the function x is constant on any set [σ(t n ), t n 1 ) T, since f(t) = for t [σ(t n ), t n 1 ) T. Let us define the time scale T = {} {s n } n=1 and function z : T R by the formula z() =, z(s n ) = x(t n ). We can see, that hence z (s n ) = x(t n 1) x(t n ) s n 1 s n = x(σ(t n)) x(t n ) s n 1 s n = = x (t n ) f(t n ) = z(s n ), z(s n 1 ) = z(s n ) + (s n 1 s n ) z(s n ). Let us substitute y n = 4z(t n ). Then we obtain y n = (s n 1 s n ) 2 + yn 1 2 (s n 1 s n ) < y n 1, from the fact, that a 2 + b 2 < a + b for a, b >. Moreover y n y n+1 = y n + s n s n+1 (s n s n+1 ) 2 + y 2 n < s n s n+1, hence y n = (y j y j+1 ) < j=n (s n s n+1 ) = s n. j=n

8 8 A. Augustynowicz, A. Gołaszewska Consequently Suppose that z(s n ) < s 2 n. for all natural n and some α 2. Since s n+1 = s 2 n, we have z(s n ) < s α n (3.1) z(s n ) = z(s n+1 ) + (s n s n+1 ) z(s n+1 ) < s α n+1 + (s n s n+1 )s α/2 n+1 = = s 2α n + (s n s 2 n)s α n = s α+1 n (s α 1 n + 1 s n ) s α+1 n. This implies that (3.1) is satisfied for every α 2 and z(s n ) =, so x(t n ) =. We obtain that x(t) on [, δ] T for some δ >, and then we see, that x(t) on T. It is obvious that if T = T, then the function f is continuous, namely f(t) 1. Proof of Proposition 2.2. Let z : T R be an arbitrary -differentiable function such that z() = z () =, and z is a strictly increasing and continuous function, for example z(t) = s s. Let us define Z = z(t) { } { } z(tn ) + z(σ(t n )) x R : x > sup z (t). 2 T n N For x / Z, x >, let us define a(x) = max {y Z : y < x}, b(x) = min {y Z : y > x}. We define function g : R R as follows, for x =, z (t), for x = z(t), t T, z n, for x = z(t n) + z(σ(t n )), 2 g(x) = g(sup z (t)), for x sup z (t), T T g(a(x))(b(x) x) + g(b(x))(x a(x)) for x / Z, x >, b(x) a(x) g( x), for x <. Of course, z (t) = g(z(t)), t T, z() =. Let us suppose that z n >. Then for x, we have xg(x) >, and we can choose {z n } such that the integral is divergent for every ε. ɛ g(x)

9 Equations with Separated Variables on Time Scales 9 Proof of Proposition 2.4. Let a 1 = 1 and a k+1 = 1 k + 1 min {a k, σ(t k ) t k, σ(t k+1 ) t k+1 }, x k = ( 1) k a k. If σ(t k+1 ) = t k, then y k = x k+1, but if σ(t k+1 ) < t k, then y k is close enough to x k+1 in a such way, that A = t k σ(t k+1 ) x k x k+1 y k x k+1 σ(t k ) σ(t k+1 ) 2, and a k+3 < ( 1) k+1 y k < a k+1. We prove that there exists a continuous function z : [, t 1 ] T R, such that z() = z () =, and z(t k ) = y k, z(σ(t k )) = x k for k N. Furthermore, if σ(t k+1 ) < t k, then z is a continuous, monotonic and a sign constant function on [σ(t k+1 ), t k ] T. Let t = st k + (1 s)σ(t k+1 ), y = z(t) z(σ(t k+1)) z(t k ) z(σ(t k+1 )). (3.2) Constructing the definition of the function z on [σ(t k+1 ), t k ] T {σ(t k )}, where σ(t k ) = s t k + (1 s )σ(t k+1 ), comes down to finding a function y : [, 1] T {s } R, such that y() =, y(1) = 1, y (1) = A, where T is the image of T with the above change of variables (3.2). The following conditions for the function y are needed: the function y is continuous, monotonic, and a constant sign function on [, 1] T. If the point 1 is left-scattered in T (it means that t k is left-scattered in T), then y = s for s [, 1], y(s ) = A(s 1) + 1. Suppose that point 1 is left-dense in T. Then lim n 1 τ n τ =. Let n N be, such that 1 τ n τ < 1 A. The function y satisfying the necessary assumptions is the following y = 1 A 1 τ n τ 1 1 τ n τ s + A τ n τ s τ n τ. Let us define Z = {} conv {x k+1, y k } { } xk+3 + y k (, x 1 ] [x 2, ), 2 k N k N

10 1 A. Augustynowicz, A. Gołaszewska and a(x) = max {y Z : y < x}, b(x) = min {y Z : y > x} for x / Z. Now we define g : [x 1, x 2 ] R as g(x) = z k, z (z 1 (x)), for x conv {x k+1, y k },, for x =, g(a(x))(b(x) x) + g(b(x))(x a(x)) b(x) a(x) for x = x k+3 + y k, 2 for x / Z, where z k y k >, lim z k =. Moreover, we put g(x) = g(x 1 ) for x x 1, and k g(x) = g(x 2 ) for x x 2. Of course, the function g is continuous on R\ {} and z (t) = g(z(t)), t T\ {}. We can choose {z n }, such that the following integral ε g(x) is divergent for every ε. Moreover, for all x, we have xg(x) <. It remains to show that z () =, and that z is continuous at. If k, then from the fact, that z (t k ) = x k y k σ(t k ) t k = a k + y k σ(t k ) t k < a k + a k+1 σ(t k ) t k 2 k, we obtain Moreover lim z (t) lim z (t k ) =. t k z () = lim z(t) lim t t k x k σ(t k ) lim a k =. t σ(t k ) t k Consequently, functions z and x are solutions of the problem { x (t) = g(x(t)), t [, t 1 ] T, x() =.

11 Equations with Separated Variables on Time Scales 11 References [1] A. Augustynowicz, A. Gołaszewska, On the Peano Theorem for Some Functional Differential Equation on Time Scale, Functional Differential Equations, Vol. 21, No 3 4, 214, pp ; [2] M. Bohner, A. C. Peterson, Dynamic Equations on Time Scales, Birkhäuser, Boston 21; [3] M. Bohner, A. C. Peterson, Advances in Dynamic Equations on Time Scales, Birkhäuser, Boston 23; [4] J. Jankowski, Istnienie i jednoznaczność rozwia zań równań różniczkowych z wyprzedzonym argumentem, PhD Thesis, University of Gdańsk (29); [5] V. Lakshmikantham, S. Sivasundaram, B. Kaymakcalan, Dynamic Systems on Measure Chains, Kluwer Academic Publishers, Dordrecht 1996; [6] W. F. Trench, Elementary Differential Equations, Brooks/Cole Thomson Learning, 21 (Free Edition 1.1).