Abstract

In the first part of this paper sufficient conditions for nonuniqueness of the classical Cauchy problem , are given. As the essential tool serves a method which estimates the “distance” between two solutions with an appropriate Lyapunov function and permits to show that under certain conditions the “distance” between two different solutions vanishes at the initial point. In the second part attention is paid to conditions that are obtained by a formal inversion of uniqueness theorems of Kamke-type but cannot guarantee nonuniqueness because they are incompatible.

1. Introduction

Consider the initial value problem where , with , and .

In the first part (Section 2) we give sufficient conditions for nonuniqueness of the classical -dimensional Cauchy problem (1.1). As the essential tool serves a method which estimates the “distance” between two solutions with an appropriate Lyapunov function and permits to show that under certain conditions the “distance” between two different solutions vanishes at the initial point. In the second part (Section 3) we analyze for the one-dimensional case a set of conditions that takes its origin in an inversion of the uniqueness theorem by Kamke (see, e.g., [1, page 56]) but cannot guarantee nonuniqueness since it contains an inner contradiction. Several attempts were made to get nonuniqueness criteria by using conditions that are (in a certain sense) reverse uniqueness conditions of Kamke type. But this inversion process has to be handled very carefully. It can yield incompatible conditions. This is illustrated by a general set of conditions (in Theorems 3.2, 3.5 and 3.6) that would ensure nonuniqueness, but unfortunately they are inconsistent.

In this paper we study Cauchy problems where is continuous at the initial point. Related results can be found in [15]. In literature there are several investigations for the discontinuous case [1, 613] with different qualitative behaviour.

2. Main Result

In the following let , , and where means the Euclidean norm.

Definition 2.1. We say that the initial value problem (1.1) has at least two different solutions on the interval if there exist solutions , defined on and .

The following notions are used in our paper (see, e.g., [14, pages 136 and 137]).

Definition 2.2. A function is said to belong to the class if it is continuous, strictly increasing on and .

Definition 2.3. A function with is said to be positive definite if there exists a function such that the relation is satisfied for .

For the convenience of the reader we recall the definition of a uniformly Lipschitzian function with respect to a given variable.

Definition 2.4. A function is said to be Lipschitzian uniformly with respect to if for arbitrarily given there exists a constant such that holds for every and for every , within a small neighbourhood of in .

In [1, 15, 16] generalized derivatives of a Lipschitzian function along solutions of an associated differential system are analyzed. A slight modification of Theorem 4.3 [15, Appendix I] is the following lemma.

Lemma 2.5. Let be continuous and let be Lipschitzian uniformly with respect to . Let be any two solutions of where is a continuous function. Then for the upper right Dini derivative the equality holds.

In the proof of Theorem 2.8 we require the following lemmas which are slight adaptations of Theorem 1.4.1 [14, page 15] and Theorem 1.3.1 [1, page 10] for the left side of the initial point.

Lemma 2.6. Let be an open -set in , let be a continuous function, and let be the unique solution of to the left with , . Further, we assume that the scalar continuous function with satisfies and Then holds as far as the solution exists left of in .

Lemma 2.7. Let and be continuous and nondecreasing in for each fixed in . Then, the initial value problem (1.1) has at most one solution in .

Theorem 2.8 (main result). Suppose that (i) is a continuous function such that Let be a solution of problem (1.1) on . Let, moreover, there exist numbers , and continuous functions , such that (ii) is nondecreasing in the second variable, and the problem has a positive solution on ; (iii) is positive definite and is Lipschitzian uniformly with respect to ; (iv)for , , the inequality holds where
Then the set of different solutions of problem (1.1) on interval has the cardinality of the continuum.

Remark 2.9. If condition () is fulfilled then, as it is well known, problem (1.1) is globally solvable and every global solution admits the estimate Moreover, for any local solution of problem (1.1), defined on some interval , there exists a global solution of that problem such that for .

Remark 2.10. For the case the initial value problem is unique and the assumptions of Theorem 2.8 cannot be satisfied. Therefore, without loss of generality, we assume in the proof below.

Proof. At first we show that (1.1) has at least two different solutions on , where , is sufficiently close to . We construct a further solution of (1.1) by finding a point not lying on the solution and starting from this point backwards to the initial point .
First we show that there exist values and , such that holds for the nontrivial solution of . From Lemma 2.7 it follows that is determined uniquely to the left by the initial data . We consider the -tubes for around the solution . There exists such that with is contained in the set For , we define The function is continuous in for . Since , there exists a , , such that . It is clear that inequalities and (due to positive definiteness of ) hold. We define a function continuous on . Taking into account inequalities and we conclude that there exists , , with The value is taken by at a point such that and clearly (in view of the construction) . The above statement is proved and (2.14) is valid for determined above.
Now consider the initial value problem Obviously since and because Peano's theorem implies that there exists a solution of problem (2.22) on . We will show that . Set Note that . Lemma 2.5 and condition (iv) imply for .
Applying Lemma 2.6 we get for . As for and is continuous at , we find . Therefore we have and, as noted above, . Thus problem (1.1) has two different solutions.
According to the well-known Kneser theorem [17, Theorem 4.1, page 15] the set of solutions of problem (1.1) either consists of one element or has the cardinality of the continuum. Consequently, if problem (1.1) has two different solutions on interval and condition () is satisfied, then the set of different solutions of problem (1.1) on interval has the cardinality of the continuum. The proof is completed.

Remark 2.11. Note that in the scalar case with condition (2.11) has the form

Example 2.12. Consider for , , and the scalar differential equation with the initial condition . Let us show that the set of different solutions of this problem on interval has the cardinality of . Obviously we can set . Put Conditions (i), (ii), and (iii) are satisfied. Let us verify that the last condition (iv) is valid, too. We get Thus, all conditions of Theorem 2.8 hold and, consequently, the set of different solutions on of given problem has the cardinality of .

3. Incompatible Conditions

In this section we show that the formulation of condition (iv) in Theorem 2.8 without knowledge of a solution of the Cauchy problem can lead to an incompatible set of conditions. In the proof of Theorem 3.2 for the one-dimensional case we use the following result given by Nekvinda [18, page 1].

Lemma 3.1. Let and let be a continuous function in . Let equation has the property of left uniqueness. For any let be the set of all such that and, for some , the initial-value problem (1.1) has more than one solution in the interval . Then is at most countable.

Theorem 3.2. The set of conditions ()–(): (i) with is continuous; (ii) is continuous, nondecreasing in the second variable, and has the following property: there exists a continuous function on , which satisfies the differential equation for with and does not vanish for ; (iii) is continuous, positive definite, and Lipschitzian uniformly with respect to ; (iv)for , where we define contains a contradiction.

Proof. Any initial value problem with has at least two different solutions due to Theorem 2.8. Thus we have an uncountable set of nonuniqueness points. We show that solutions passing through different initial points are left unique. Suppose that it does not hold. Let be a solution starting from , and let be a solution starting from with . If we assume that these solutions cross at a point and if we set then , . Therefore there exists a point such that (we apply Lemma 2.5) in contradiction to (3.3). Thus we obtain left uniqueness. From Lemma 3.1 we conclude in contrast to the above conclusion that the set of nonuniqueness points can be at most countable.

In [1, Theorem 1.24.1, page 99] the following nonuniqueness result (see [14, Theorem 2.2.7, page 55], too) is given which uses an inverse Kamke's condition (condition (3.9) below).

Theorem 3.3. Let be continuous on , , , and for . Suppose that, for each , is a differentiable function on , and continuous on for which exists, Let , where , and, for , Then, the scalar problem , has at least two solutions on .

Remark 3.4. In the proof of Theorem 3.3 at first is assumed. Putting in (3.9) leads to the inequality As is continuous and for it follows that must have constant sign for each of the half planes and . For the upper half plane this implies that For the first inequality nonuniqueness is shown in [1]. But a similar argumentation cannot be used for the second inequality as the following example in [5] shows. We consider the initial value problem , with and . Thus inequality holds. In the upper half-plane we have . The function is a nontrivial solution of the comparison equation. Therefore all assumptions are fulfilled, but the initial value problem has at most one solution because of Theorem 1.3.1 [1, page 10].

The next theorem analyzes in the scalar case (for ) that even fulfilling a rather general condition (see condition (3.14) in the following theorem) cannot ensure nonuniqueness since the set of all conditions contains an inner contradiction. The proof was motivated by the paper [5].

Theorem 3.5. There exists no system of three functions , , and satisfying the following suppositions: (i) with is a continuous function; (ii)the continuous function , if , has the following property: there exists a continuously differentiable function on , satisfying the differential equation for such that and for ;(iii)the continuous function is positive definite, and for all , continuously differentiable; (iv)for , , where we define and subscript indices denote the derivative with respect to the first and second argument, respectively; (v)there exist a positive constant and a function such that for and (vi)for and , with the inequality holds.

Proof. Let us show that the above properties are not compatible. For fixed numbers , with consider the auxiliary function Clearly, is continuous and assumes a (positive) maximum. Set If the function fulfills the inequality with a positive constant in a domain , , , then the initial value problem has the unique trivial solution . Really, since for , by integrating inequality with limits , we get and for which contradicts positivity of . Therefore problem (3.21) has only the trivial solution. Hence, there exist a sequence with , , and a sequence , , with such that the inequality holds for every . Consider now the relation Due to the properties of we conclude that for all sufficiently small positive numbers , (i.e., for all sufficiently large ) there exists a (sufficiently small and positive) number such that the equation has the solution . Thus a sequence with and corresponds to the sequence . For every define a number as where is the ceiling function. Without loss of generality we can suppose that Obviously, Moreover, without loss of generality we can suppose that for every sufficiently large the inequality holds. Set Consider for all sufficiently large n the expression Then Estimating the expression from above we get (see (3.32)) These two above estimations yield in contrast to . Since the initially taken points and , , can be chosen arbitrarily close to zero, the theorem is proved.

The following result is a consequence of Theorem 3.5 if , and . Condition (3.38) below was discussed previously in [5].

Theorem 3.6. There exists no system of two functions and satisfying the following suppositions: (i) with is a continuous function; (ii)the continuous function , if , has the following property: there exists a continuously differentiable function on , satisfying the differential equation for such that and for ;(iii) for , (iv)for the inequality holds.

Remark 3.7. Let us note that in the singular case, that is, when we permit that the function is not continuous at , the given sets of conditions in Theorems 3.5 and 3.6 can be compatible. This can be seen from the proof where the continuity of is substantial. Such singular case was considered in [13].

Acknowledgments

This research was supported by the Grants P201/11/0768 and P201/10/1032 of the Czech Grant Agency (Prague), by the Project FEKT-S-11-2(921) and by the Council of Czech Government MSM 00216 30529.