to the

We consider the infimum inff max j=1,2,3 k f (j) k L1 (0,T0), where the infimum is taken over every function f which runs through the set KC 3 (0, T0) consisting of all functions f : [0, T0] ! R satisfying the boundary conditions f (j) (0) = aj , f (j) (T0) = 0 for j = 0,1,2, whose derivatives f (j) are continuous for j = 0,1,2 and the third derivative f (3) may have a finite number of discontinuities in the interval (0, T0), and find this infimum explicitly for certain choice of boundary conditions. This problem is motivated by some conditions under which the solution of


Introduction
Let KC 3 (A, B) be a class consisting of all functions f : [A, B] → R satisfying the following conditions: D j f ∈ C(A, B) , i.e. the j th derivative of f is continuous for j = 0, 1, 2 , but D 3 f may have a finite number of discontinuities in the interval (A, B) .(Here, A can be −∞ and B can be ∞ .)Throughout, we denote D = ∂ ∂x and D j f = f (j) for j 0 , so, in particular, D 0 f = f .In this paper we consider the functional where the real valued function f ∈ KC 3 (0, T 0 ) satisfies the boundary conditions D j f (0) = a j and D j f (T 0 ) = 0 for j = 0, 1, 2. ( We find inf M(f ) , where the infimum is taken over each f lying in the class KC 3 (0, T 0 ) with boundary conditions (1) for some choice of a 0 , a 1 , a 2 ∈ R and T 0 > 0 .
The question concerning inf M(f ) arises from certain nonlinear Schrödinger equation, where one needs to estimate the integral in proving that the solution of the nonlinear Schrödinger equation with periodic boundary condition EJQTDE, 2013 No. 8, p. 2 blows up, i.e.
Du L 2 (I) → ∞ as t → t 0 (see [10]).Here I = (−2, 2) .One of the aims of this paper is to investigate the size of the constant M = M(f ) in (2).In general, our results can be applied to all mathematical problems, where the estimates (2) are used for f (x) ∈ KC 3 (R) such that D j f (x) = 0 for x / ∈ I , j = 0, 1, 2 .For example, in some problems of mathematical physics the estimates for derivatives of a truncated function have been used (see [10], [11] and Theorem 1 below).
The blow up problem of the solution given by ( 3) and ( 4) in the whole real line I = R has been studied by many authors; see, for example, [4], [9], [11], [14].Put In the case I = R , the inequality E(u 0 ) < 0 is a sufficient condition for the solution of ( 3) and ( 4) to blow up at finite time t 0 > 0 (see [4]).However, in general the condition E(u 0 ) < 0 is not sufficient for the collapse of (3) − (5) (see [10]).The nonlinear Schrödinger equation with periodic boundary condition have been considered in [2], [5], [6], [10].Some problems related to Schrödinger equation in bounded domain have been studied in [12], [13], etc. Ogawa and Tsutsumi [10] found a sufficient condition for the blow up of the solution of (3) − (5) .Before stating their result let us first give some notation.Assume that φ and Dφ(x) 0 for x 0 φ(y)dy .The sufficient conditions of blow up solution is the following theorem in [10].
The theorem raises the following natural question: how small can the constant M be?In the present note we shall answer this question.Clearly, the functional M = 3 j=1 D j φ L ∞ (I) of Theorem 1 can be replaced by the smaller functional because in the proof of Theorem 1 the authors used the estimate (2) .
The results of the present paper make Theorem 1 applicable in practice.Take the initial function u 0 ∈ H 1 (I) , u 0 (−2) = u 0 (2) .To answer the question on whether the solution of (3)-( 5) blows up one needs verify conditions (7) and (8).However, we cannot verify the conditions (7) and ( 8), if we do not know how small M is, i.e. we cannot use this result in practice.For this one can apply Theorem 1 using the results of Theorem 2 below.
Evaluating the constant M and finding its exact value is complicated, because the function φ(x) is not defined in the interval ( and obtained a sufficient condition for the global existence of the solution of the Schrödinger equation.For this, he solved the Euler-Lagrange equation minimizing the functional because the derivative of this functional does not exist, so we cannot solve the corresponding Euler-Lagrange equation.For minimizing the functional M(f ) we shall use the optimal control problem with a nonfixed termination time.The optimal control problem is one of the cases of Pontryagin's maximum principle and was considered in many papers; see, for example, [1], [7], [8] and the references in those papers.
We solve the following optimal control problem with a nonfixed termination time: i.e. we find the minimal number T for which the conditions (9) are satisfied.We find that the minimal number T is attained at the function ( 13) , ( 14 is satisfied. In the system (10) b 0 , b 1 , b 2 , t 1 , t 2 , a are unknowns and a 0 , a 1 , a 2 , T are parameters.The system (10) can be solved by using resultants.For instance, one can take b 2 from the first equation, b 1 from the third and b 0 from the fifth.Then, since the resulting system with unknowns t 1 , t 2 , a consists of polynomial equations, we can use the elimination of the variables t 1 and t 2 with resultants.(For instance, if P (t, x, y) and Q(t, x, y) are two polynomials in Q[t, x, y] then the resultant of P and Q with respect to t is a polynomial R(x, y) ∈ Q[x, y] which is the determinant of a corresponding Sylvester matrix [3].)The elimination of t 1 and t 2 gives the following equation relating a and T : This was checked with Mathematica (using Eliminate[eqns, vars]) and with Maple.Unfortunately, the resulting equations for other variables (like t 1 and t 2 ) have large degree which leads to many (real and complex) solutions or to no solutions at all.The solution of ( 10) is applicable to our problem in case the following hypothesis holds: Hypothesis (H) The system (10) , where b 0 , b 1 , b 2 , t 1 , t 2 , a are unknowns and a 0 , a 1 , a 2 , T are parameters, has a unique real solution The main result of this paper is the following theorem.EJQTDE, 2013 No. 8, p. 6 Theorem 2 Suppose that the hypothesis (H) holds with T = T 0 , where 0 < T 0 1 , and some a 0 > 0 , a 1 0 , a 2 0 .If a > max{4a 1 /3, −a 2 } then inf M(f ) , where f runs through the class KC 3 (0, T 0 ) with boundary conditions (1), is attained at the function given in (15) with δ = −1 and is equal to a , where a is the positive root of the equation (11).
Finally, we give some numerical calculations and an application of Theorem 2 to Theorem 1.We use these numerical calculations to show that all the conditions of Theorem 2 are satisfied and find inf M(f ) = 562.986 . . .for In particular, we show that Corollary 3 With the conditions of Theorem 1, the smallest possible constant

The optimal control problem
We first solve the following optimal control problem with a nonfixed termination time (9) .The simplest optimal control problem with a nonfixed termination time was solved in [1].Some other problems with a nonfixed termination time have been considered in [7], [8].The following lemma is a necessary condition in our optimal control problem.
Lemma 5 Suppose that the solution T of the optimal control problem ( 9) is attained at a function f (x) , a > 0 , and suppose that δ ∈ {−1, 1} .Then the function f (x) can only be one of the following: or where 0 < t 1 < t 2 < T and the constants b j , j = 0, 1, 2 , are such that f (x) ∈ KC 3 (0, T ) .
Proof: In this lemma, we shall use the usual notation Df = f ′ .Let us reduce our problem to the standard problem of Pontryagin's maximum principle [1], by changing the variables The Lagrange function for this problem is If the solution exists, then there exist some constants λ j , j = 0, 1, . . ., 6 , the functions p k (x) , k = 1, 2, 3 , do not vanish simultaneously and satisfy the following conditions (a), (b), (c), (d) given below.
The following lemma is a sufficient condition for the optimal control problem.
Lemma 6 Suppose hypothesis (H) holds and a 0 > 0 .Then the solution of the optimal control problem is attained at the function f (x) as defined (15) Proof: By hypothesis (H), there exists a unique real positive number a = a(T 0 ) such that (b 0 , b 1 , b 2 , t 1 , t 2 , a) is a solution of the system (10) satisfying 0 < t 1 < t 2 < T 0 .Hence there exists at least one function f (x) (as in (15) with δ = −1 ) satisfying f ∈ KC 3 , D j f (T 0 ) = 0 , j = 0, 1, 2 .Lemma 5 shows that the solution of the optimal control problem (9) with a = a(T 0 ) is equal to T 0 if this solution exists.We shall prove that the solution of the optimal control problem is attained at this function f (x) given in (15) with δ = −1 .
(20) Indeed, using boundary conditions and integration by parts, we obtain By the same argument, Similarly, by integrating over the interval (t 2 , T 0 ) , we find that and in the same way Df ) .This completes the proof of (20).
To complete the proof of (19) assume that D j f 1 (t 1 ) = D j f 2 (t 1 ) for some j ∈ {0, 1, 2} .Let S be a finite set of points in (0, t 1 ) , where the derivative D 3 f 2 (x) does not exist.Then from the expression of the difference D j f 1 (t 1 )− D j f 2 (t 1 ) by a corresponding integral we see that a+D 3 f 2 (x) must be zero in the set (0, t 1 ) \ S .Thus D 3 f 2 (x) = −a = D 3 f 1 (x) for each x ∈ (0, t 1 ) \ S .This equality and the boundary conditions D j f 1 (0) = D j f 2 (0) = a j for j = 0, 1, 2 give us D j f 1 (t 1 ) = D j f 2 (t 1 ) for each j = 0, 1, 2 .Now, by EJQTDE, 2013 No. 8, p. 11 integrating by parts over the interval (t 1 , t 2 ) , we deduce that This contradicts to D 2 f 1 (t 2 ) < D 2 f 2 (t 2 ) (which is already proved in (20)), and so completes the proof of (19) .
We next prove that there exist τ 1 and τ 2 satisfying Since the function f 1 (x) − f 2 (x) and its derivative in the interval (t 1 , t 2 ) are continuous, the inequalities f 1 (t 1 ) < f 2 (t 1 ) and s) for some point s ∈ (t 1 , θ) .This proves the first inequality in (21) .In the same way the second inequality of (21) with Now, by (21) , we find that EJQTDE, 2013 No. 8, p. 12 If the derivative D 3 f 2 exists in the interval (τ 1 , τ 2 ) then, by ( 22) and the Lagrange theorem, we conclude that there exists θ 2 ∈ (τ 1 , τ 2 ) for which which is a contradiction with |D 3 f 2 (x)| a for each x , where D 3 f 2 (x) exists.