Our aim is to obtain an upper/lower bound for the function f = f(x), satisfying the integral inequality

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1 ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII AL.I.CUZA IAŞI Tomul XLVI, s.i a, Matematică, 2, f.2. ON THE INEQUALITY f(x) K + M(s)g(f(s))ds BY ADRIAN CORDUNEANU Our aim is to obtain an upper/lower bound for the function f = f(x), satisfying the integral inequality (1) f(x) K + M(s)g(f(s))ds, x a or a similar one, under hypotheses which will stated in what follows. similar inequality, namely A (2) f(x) K + M(s)g(f(s))ds, x was considered by the author in [1], both in the case of one or several independent variables. Regarding the first inequality, we can state Proposition 1. Let us consider (1), under the following hypotheses (h 1 ) the functions f = f(x) and M = M(x) are continuous nonnegative for x a and K > is a constant; (h 2 ) the function g = g(u) is continuous and nondecreasing for u and g(u) > for u > ; (h 3 ) the primitive G = G(u) of the function 1/g(u) satisfies the condition (3) G(K ) + M(s)ds < G( )

2 296 ADRIAN CORDUNEANU 2 (h 4 ) the function α = is increasing, has a continuous derivative for x a, x for x a and α() =, α(a) = a. Then we have (4) f(x) G 1 (G(K ) + Proof. We denote (5) u(x) = K + M(s)ds), x a M(s)g(f(s))ds, x a which implies f(x) u(x) on the interval [, a]. We also have (6) u (x) = α (x)m()g(f()) on the same interval, so u = u(x) is nonincreasing. Taking into account that g = g(u) is nondecreasing and that x, we finally get (7) u (x) α (x)m()g(u(x)), x a Integrating the inequality (8) u (s)/g(u(s)) α (s)m(α(s)), s a on the interval [x, a] we obtain (9) G(K ) G(u(x)) or, equivalently, (1) G(u(x)) G(K ) + M(s)ds, x a M(s)ds, x a Using condition (3) and the fact that G = G(u) is increasing on (, ) we easily obtain ) a (11) u(x) G (G(K 1 ) + M(s)ds, x a where G 1 stands for the inverse of the function G. This closes the proof, because f(x) u(x) for x a. Remark 1. If G( ) = condition (3) is obviously satisfied and it may be dropped out.

3 3 ON THE INEQUALITY f(x) K + M(s)g(f(s))ds 297 Remark 2. If we suppose α() = α > (α < a) and we replace (3) by the condition G(K ) + M(s)ds < G( ), the other assumptions α remaining unchanged, the earlier bound (4) still remains true. Remark 3. If (3) is not satisfied, the bound (4) is valid only on some interval (a 1, a], on which we have G(K ) + Proposition 2. Let us consider the inequality M(s)ds < G( ). (12) f(x) K + M(s)g(f(s))ds, x a under the hypotheses (h 1 ), (h 2 ) and (h 3 ) of Proposition 1. Assume also that (h 5 ) the function α = is increasing, has continuous derivative for x a, x and α(a) = a. Then we have (13) f(x) G 1 (G(K ) + M(s)ds), x a Proof. Using the notation (5) again, it follows from (12) that f(x) u(x) for x a. From (6), we also have u (x) and thus u = u(x) is nonincreasing on the interval [, a]. Taking into account that g = g(u) is nondecreasing and that x, we have f() u() u(x) and g(f()) (u(x)) for x a. We finally obtain (14) u (x)= α (x)m()g(f()) α (x)m()g(u(x)), x a which implies (15) u (x)/g(u(x))) α (x)m(), x a Integrating (15) on the interval [x, a], it follows (16) G(u(a)) G(u(x)) x α (s)m(α(s))ds = M(s)ds

4 298 ADRIAN CORDUNEANU 4 or, equivalently, (17) G(u(x)) G(K ) + M(s)ds, x a Using (3) again, we obtain (18) u(x) G 1 (G(K ) + M(s)ds), x a which closes the proof, because f(x) u(x) for x a. Remark 4. In the preceding Proposition, the condition x implies α() =. If (3) is not fulfilled, the inequality (13) is valid only on some interval (a 1, a], on which we have G(K ) + Proposition 3. Let us consider the inequality (19) f(x) K + M(s)g(f(s))ds, x b under the following hypotheses M(s)ds) < G( ). (h 6 ) the function f = f(x) and M = M(x) are continuous nonnegative for x b and K > is a constant; (h 7 ) the function α = is decreasing, has a continuous derivative for x b, x for x a < b and α() = b, α(a) = a; (h 8 ) the function g = g(u) is continuous and nondecreasing for u and g(u) > for u > ; (h 9 ) the primitive G = G(u) of the function 1/g(u) satisfies the condition (2) G(K ) + Then we have a M(s)ds < G( ) (21) f(x) G 1 (G(K ) + M(s)ds), x a

5 5 ON THE INEQUALITY f(x) K + M(s)g(f(s))ds 299 Proof. Denoting (22) u(x) = K + M(s)g(f(s))ds, x b we have f(x) u(x) and we remark that u = u(x) is nondecreasing on [, b]. Thus, we can write f() u() u(x) for x a and also g(f()) g(u(x)) for x a. From (23) u (x) = α (x)m()g(f()) α (x)m()g(u(x)), satisfied on the interval [, a], we obtain (24) u (x)/g(u(x)) α (x)m() on the same interval. Integrating this inequality on [, x], with x a, it follows (25) G(u(x)) G(K ) + which easily implies the desired conclusion (21). M(s)ds Example. Taking K >, = b mx with b >, < m < 1 and g(u) = u (the linear case), we obtain from (26) f(x) K + M(s)f(s)ds, x b b mx that ( ) b (2) f(x) K exp M(s)f(s)ds, x a = b/(m + 1) b mx where, of course, f and M are assumed to be continuous nonnegative. Remark 5. In the preceding Proposition, if we suppose G(K ) =, like in [5], i.e. if we take (28) G(u) = u K ds/g(s), u >

6 3 ADRIAN CORDUNEANU 6 then the bounds earlier obtained may be written in a simpler form. example, (21) takes the form ( ) b (29) f(x) G 1 M(s)ds, x a. For REFERENCES 1. CORDUNEANU,A. On the inequality f(x) K + M(s)g(f(s))ds, Bul. Inst. Pol. Iaşi, tom XLV (XLIX), fasc. 3-4, 1999, α(t) 2. CORDUNEANU,A. On the inequality f(t) K(t)+L(t) M(τ)f(τ)dτ, An. Şt. Univ. Al.I. Cuza, Iaşi, s.i-a Mat., t. XLV, fasc.1, 1999, BIHARI, I. A generalization of a lemma of Bellman and its applications to uniqueness problem of differential equations. Acta Math. Acad. Sci. Hungr. 7:1(1956), MARTYNYUK, A.A., GUTOWSKI, R. Integral inequalities and the stability of the motion (Russian). Kiev, Naukova Dumka, 1979, VRABIE, I. Differential equations (Romanian). Bucharest, Matrix Rom., 1999, 37. Received: 14.X.1999 Department of Mathematics Technical University Gh. Asachi of Iaşi ROMANIA