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Chapter 3 Opial-Type Inequalities 3.1 Introduction In 1960, Z. Opial [231] discovered one of the most fundamental integral inequal- ities involving a function and its derivative, which is now known in the literature as Opial’s inequality. In the same year, C. Olech [230] published a note which addresses a simpler proof of Opial’s inequality under weaker conditions. Starting from the pioneer papers [230,231], the result of Opial has received considerable attention and many papers have appeared, which provides with the simple proofs, various generalizations, extensions and discrete analogues of Opial’s inequality and its generalizations. The importance of Opial’s inequality and its generaliza- tions and extensions lies in successful utilization to many interesting applications in the theory of differential equations. Good surveys of the work on such inequal- ities together with many references are contained in monographs [4,211,215]. In the pastfew years, numerousvariants, generalizationsand extensions of Opial’s inequality which involves functions of one and many independent variables have been found in various directions. This chapter deals with important fundamental results on Opial-type inequalities recently investigated in the literature by various investigators. 3.2 Opial-Type Integral Inequalities In [231] Opial established the following interesting integral inequality h 0 u(t)u (t) dt h 4 h 0 u (t) 2 dt, (3.2.1) 263 264 Chapter 3. Opial-Type Inequalities where u(t) ∈C 1 [0,h], u(t) > 0in(0,h)such that u(0) =u(h) =0. In (3.2.1) the constant h 4 is the best possible. The first simple proof of inequality (3.2.1) is given by Olech [230] in his paper published along with Opial’s paper [231]. In the years thereafter, numerous variants, generalizations and extensions of inequality (3.2.1) have appeared in the literature; see [4,211,215] and the references given therein. In this section we present a weaker form of (3.2.1) and its simplified proof based on Olech [230] as well as variants established by various investigators during the past few years. We begin with the following weaker form of Opial’s inequality (3.2.1) which Olech establishes in [230]. T HEOREM 3.2.1. Let u be an absolutely continuous function on [0,h] and let u(0) = u(h) = 0. Then inequality (3.2.1) holds. Equality holds in (3.2.1) if and only if u(t) = cx for 0 x h 2 , c(h −x) for h 2 x h, (3.2.2) where c is a constant. P ROOF.Lety(t) = t 0 |u (s)|ds and z(t) = h t |u (s)|ds. Then we have the fol- lowing relations y (t) = u (t) =−z (t) (3.2.3) and u(t) y(t), u(t) z(t), (3.2.4) for t ∈[0,h]. From (3.2.3) and (3.2.4), we get h/2 0 u(t)u (t) dt h/2 0 y(t)y (t) dt = 1 2 y 2 h 2 (3.2.5) and h h/2 u(t)u (t) dt − h h/2 z(t)z (t) dt = 1 2 z 2 h 2 . (3.2.6) From (3.2.5) and (3.2.6), we find that h 0 u(t)u (t) dt 1 2 y 2 h 2 +z 2 h 2 . (3.2.7) 3.2. Opial-Type Integral Inequalities 265 On the other hand, using the Cauchy–Schwarz inequality, we have y 2 h 2 = h/2 0 u (t) dt 2 h 2 h/2 0 u (t) 2 dt, (3.2.8) z 2 h 2 = h h/2 u (t) dt 2 h 2 h h/2 u (t) 2 dt. (3.2.9) Now the desired inequality in (3.2.1) follows from (3.2.7)–(3.2.9). Now suppose that the equality holds in (3.2.1), that is, h 0 u(t)u (t) dt = h 4 h 0 u (t) 2 dt. (3.2.10) Then from (3.2.7)–(3.2.9), we get h/2 0 u (t) dt 2 = h 2 h/2 0 u (t) 2 dt, (3.2.11) h h/2 u (t) dt 2 = h 2 h h/2 u (t) 2 dt. (3.2.12) It is easy to see that equalities (3.2.11) and (3.2.12) are possible if and only if |u (t)|=constant almost everywhere in [0, h 2 ] and in [ h 2 ,h]. Hence y(t) and z(t) are linear. Further, it follows from (3.2.10), (3.2.7), (3.2.11) and (3.2.12) that |u(t)|=y(t) for 0 t h 2 and |u(t)|=z(t) for h 2 t h. These facts im- ply (3.2.2). The proof is complete. In [419] Traple has given the inequalities in the following theorem. T HEOREM 3.2.2. Let p be a nonnegative and continuous function on [0,h]. Let u be an absolutely continuous function on [0,h] with u(0) = u(h) =0. Then the following inequalities hold h 0 p(t) u(t) 2 dt h 4 h 0 p(t)dt h 0 u (t) 2 dt , (3.2.13) h 0 p(t) u(t) u (t) dt h 4 h 0 p 2 (t) dt 1/2 h 0 u (t) 2 dt . (3.2.14) P ROOF.Fort ∈[0,h],wehave u(t) t 0 u (s) ds, u(t) h t u (s) ds. (3.2.15) 266 Chapter 3. Opial-Type Inequalities From (3.2.15) we observe that u(t) 1 2 h 0 u (t) dt. From the above inequality and using the Schwarz inequality, we have h 0 p(t) u(t) 2 dt 1 4 h 0 p(t) h 0 u (t) 2 dt 2 h 4 h 0 p(t)dt h 0 u (t) 2 dt , which is the required inequality in (3.2.13). By using inequality (3.2.13) and the Schwarz inequality, we observe that h 0 p(t) u(t) u (t) dt h 0 p 2 (t) u(t) 2 dt 1/2 h 0 u (t) 2 dt 1/2 h 4 h 0 p 2 (t) dt h 0 u (t) 2 dt 1/2 h 0 u (t) 2 dt 1/2 = h 4 h 0 p 2 (t) dt 1/2 h 0 u (t) 2 dt . The proof is complete. In the following two theorems we present the inequalities of the Opial type established by Pachpatte in [348]. T HEOREM 3.2.3. Let p 0, q 1, m 1 be real numbers. If u ∈ C 1 ([0,h], R) satisfies u(0) =u(h) =0, then h 0 u(t) m(p+q) dt (p +q) m I(m) q h 0 u(t) mp u (t) mq dt, (3.2.16) h 0 u(t) m(p+q) dt (p +q) m I(m) p+q h 0 u (t) m(p+q) dt, (3.2.17) 3.2. Opial-Type Integral Inequalities 267 where I(m)= h 0 t 1−m +(h −t) 1−m −1 dt. (3.2.18) P ROOF. From the hypotheses, we have the following identities u p+q (t) = (p +q) t 0 u p+q−1 (s)u (s)ds, (3.2.19) u p+q (t) =−(p +q) h t u p+q−1 (s)u (s)ds, (3.2.20) for t ∈[0,h]. From (3.2.19), (3.2.20) and using Hölder’s inequality with indices m, m/(m −1), we obtain u(t) m(p+q) (p +q) m t m−1 t 0 u(s) p+q−1 u (s) m ds, (3.2.21) u(t) m(p+q) (p +q) m (h −t) m−1 h t u(s) m(p+q−1) u (s) m ds, (3.2.22) for t ∈[0,h]. Multiplying (3.2.21) by t 1−m and (3.2.22) by (h−t) 1−m and adding these inequalities we obtain t 1−m +(h −t) 1−m u(t) m(p+q) (p +q) m h 0 u(s) m(p+q−1) u (s) m ds. (3.2.23) From (3.2.23) we get u(t) m(p+q) (p +q) m t 1−m +(h −t) 1−m −1 × h 0 u(s) mp /q u (s) m u(s) m(p+q−1)−mp/q ds (3.2.24) 268 Chapter 3. Opial-Type Inequalities for t ∈[0,h]. Integrating (3.2.24) on [0,h] and using Hölder’s inequality with indices q,q/(q −1), we obtain h 0 u(t) m(p+q) dt (p +q) m I(m) h 0 u(t) mp /q u (t) m u(t) m(p+q−1)−mp/q dt (p +q) m I(m) × h 0 u(t) mp u (t) mq dt 1/q h 0 u(t) m(p+q) dt (q−1)/q . (3.2.25) If h 0 |u(t)| m(p+q) dt = 0 then (3.2.16) is trivially true, otherwise, dividing both sides of (3.2.25) by ( h 0 |u(t)| m(p+q) dt) (q−1)/q and then taking the qth power on both sides of the resulting inequality we get the required inequality in (3.2.16). By using Hölder’s inequality with indices (p +q)/p, (p + q)/q to the inte- gral on the right-hand side of (3.2.16) and following the arguments as in the last part of the proof of inequality (3.2.16) with suitable changes, we get the required inequality in (3.2.17). The proof is complete. T HEOREM 3.2.4. Let p 0, q 1, r 0, m 1 be real numbers. If u ∈ C 1 ([0,h], R) satisfies u(0) =u(h) =0, then h 0 u(t) m(p+q) u (t) mr dt (p +q +r) m I(m) q h 0 u(t) mp u (t) m(q+r) dt, (3.2.26) h 0 u(t) m(p+q) u (t) mr dt (p +q +r) m I(m) p+q h 0 u (t) m(p+q+r) dt, (3.2.27) where I(m) is defined by (3.2.18). P ROOF. Rewriting the integral on the left-hand side of (3.2.26) and using Hölder’s inequality with indices (q + r)/r, (q + r)/q and inequality (3.2.16), 3.2. Opial-Type Integral Inequalities 269 we observe that h 0 u(t) m(p+q) u (t) mr dt = h 0 u(t) m(pr/(q+r)) u (t) mr u(t) m(p+q)−m(pr/(q+r)) dt h 0 u(t) mp u (t) m(q+r) dt r/(q+r) h 0 u(t) m(p+q+r) dt q/(q+r) h 0 u(t) mp u (t) m(q+r) dt r/(q+r) × (p +q +r) m I(m) q+r h 0 u(t) mp u (t) m(q+r) dt q/(q+r) = (p +q +r) m I(m) q h 0 u(t) mp u (t) m(q+r) dt. This result is the required inequality in (3.2.26). From (3.2.26) and using Hölder’s inequality with indices (p + q)/p, (p +q)/q, we observe that h 0 u(t) m(p+q) u (t) mr dt (p +q +r) m I(m) q × h 0 u(t) mp u (t) m(rp/(p+q)) u (t) m(q+r)−m(rp/(p+q)) dt (p +q +r) m I(m) q × h 0 u(t) m(p+q) u (t) mr dt p/(p+q) h 0 u (t) m(p+q+r) dt q/(p+q) . Now, by following the arguments as in the last part of the proof of inequality (3.2.16) with suitable modifications, we get the required inequality in (3.2.27). The proof is complete. R EMARK 3.2.1. We note that the inequalities in (3.2.16) and (3.2.27) are similar to that of Opial’s inequality given in (3.2.1) which in turn yield respectively the lower and upper bounds on the integral of the form involved on the left-hand side 270 Chapter 3. Opial-Type Inequalities of (3.2.1) while the inequalities obtained in (3.2.17) and (3.2.26) are different from those of (3.2.1). In [304] Pachpatte has established the inequalities in the following theorems which can be considered as their origin to well-known Weyl’s inequality [423], see also [141, p. 165] and Opial’s inequality in (3.2.1). T HEOREM 3.2.5. Let α 0, p 0, q 1 be real constants and f be a real- valued continuously differentiable function defined on (0,b)for a fixed real num- ber b>0. Then the following inequalities hold b 0 t α f(t) p+q dt M b 0 t α+1 f(t) p |f(t)| b + f (t) q dt 1/q × b 0 t α+1 f(t) p+q dt (q−1)/q , (3.2.28) b 0 t α f(t) p+q dt M b 0 t (α+1)q f(t) p |f(t)| b + f (t) q dt 1/q × b 0 f(t) p+q dt (q−1)/q , (3.2.29) where M ={ α+2 α+1 , 2(p+q) α+1 }. R EMARK 3.2.2. It is interesting to note that, if the function f is continuously differentiable on (0, ∞), then letting b →∞in (3.2.28) and (3.2.29) we get re- spectively the following inequalities ∞ 0 t α f(t) p+q dt M ∞ 0 t α+1 f(t) p f (t) q dt 1/q × ∞ 0 t α+1 f(t) p+q dt (q−1)/q , (3.2.30) ∞ 0 t α f(t) p+q dt M ∞ 0 t (α+1)q f(t) p f (t) q dt 1/q × ∞ 0 f(t) p+q dt (q−1)/q . (3.2.31) 3.2. Opial-Type Integral Inequalities 271 In particular, if we take α = 0, p = 0, q = 2, then the inequalities obtained in (3.2.30), (3.2.31) reduce to the slight variants of Weyl’s inequality given in [141, p. 165]. T HEOREM 3.2.6. Let α, p, q, f be as defined in Theorem 3.2.5. Then the follow- ing inequalities hold b 0 t α f(t) p+q dt M q b 0 t α+q f(t) p |f(t)| b + f (t) q dt, (3.2.32) b 0 t α f(t) p+q dt M p+q b 0 t α+p+q |f(t)| b + f (t) p+q dt, (3.2.33) where M is as defined in Theorem 3.2.5. R EMARK 3.2.3. If the function f is continuously differentiable on (0, ∞), then letting b →∞in (3.2.32) and (3.2.33) we get respectively the following inequal- ities ∞ 0 t α f(t) p+q dt M q ∞ 0 t α+q f(t) p f (t) q dt, (3.2.34) ∞ 0 t α f(t) p+q dt M p+q ∞ 0 t α+p+q f (t) p+q dt. (3.2.35) Here we note that the inequalities obtained in (3.2.34) and (3.2.35) are similar to that of the Opial-type inequality (3.2.1) and those of well-known Hardy’s inequal- ity (2.4.1). T HEOREM 3.2.7. Let α, p, q be as defined in Theorem 3.2.5 and let f be a real- valued continuously differentiable function defined on (a, b) for fixed real num- bers a<b. Then the following inequalities hold b a |t| α f(t) p+q dt |H |+ p +q α + 1 b a |t| α+1 f(t) p f (t) q dt 1/q × b a |t| α+1 f(t) p+q dt (q−1)/q , (3.2.36) 272 Chapter 3. Opial-Type Inequalities b a |t| α f(t) p+q dt |H |+ p +q α + 1 b a |t| (α+1)q f(t) p f (t) q dt 1/q × b a f(t) p+q dt (q−1)/q , (3.2.37) where H = 1 α + 1 b α+1 (sgnb) α f(b) p+q −a α+1 (sgna) α f(a) p+q . (3.2.38) R EMARK 3.2.4. We note that, in the special case when q = 1, the inequalities obtained in (3.2.36) and (3.2.37) reduces to the following inequality b a |t| α f(t) p+1 dt |H 0 |+ p +1 α + 1 b a |t| α+1 f(t) p f (t) dt, (3.2.39) where H 0 is defined by the right-hand side of (3.2.38) taking q =1. P ROOFS OF THEOREMS 3.2.5–3.5.7. Integrating bypartswehavethe following identity b 0 t α+1 − 1 b t α+2 f(t) p+q−1 f (t) sgnf(t)dt =− b 0 (α + 1)t α − 1 b (α + 2)t α+1 |f(t)| p+q p +q dt. (3.2.40) From (3.2.40) we observe that (α + 1) b 0 t α f(t) p+q dt =(α +2) b 0 t α+1 1 b f(t) p+q dt −(p +q) b 0 t α+1 1 − 1 b t f(t) p+q−1 f (t) sgnf(t)dt [...]... From (3. 3 .38 ) and (3. 3 .39 ), we have zr (t) = ur (t) for a t c and r = 1, , m and vr (t) = − ur (t) for c t t ur (s) ds a t (3. 3.41) b and r = 1, , m We note that ur (t) = for a (3. 3.40) (3. 3.42) c and r = 1, , m and b ur (t) = − t ur (s) ds (3. 3. 43) for c t b and r = 1, , m From the hypotheses, the arithmetic mean– geometric mean inequalities (3. 3.26), (3. 3.42), (3. 3.40) and (3. 3 .38 ) we... integral inequalities P ROOFS OF T HEOREMS 3. 3.1 3. 3 .3 From the hypotheses, for every t ∈ I and i = 1, , n, we have t D i−1 f (t) = D i f (s) ds, a a D i f (s) ds, (3. 3.12) t t D i−1 g(t) = b D i−1 f (t) = − D i g(s) ds, b D i−1 g(t) = − t D i g(s) ds (3. 3. 13) 279 3. 3 Wirtinger–Opial-Type Integral Inequalities From (3. 3.12) and (3. 3. 13) , we observe that b 1 2 D i−1 f (t) (3. 3.14) D i g(t) dt, (3. 3.15)... (t) dt, (3. 3.18) a b 2 D i g(t) dt, (3. 3.19) a for i = 1, , n Now, from (3. 3.18) and (3. 3.19), and following exactly the same arguments as in the proof of inequality (3. 3.1) in Theorem 3. 3.1, we obtain the desired inequality in (3. 3.6) The details of the proofs of inequalities (3. 3.7) and (3. 3.8) follow from (3. 3.18) and (3. 3.19), and following exactly the same arguments as in the proofs of inequalities. .. inequalities in (3. 3.6)– (3. 3.8) reduce to the new inequalities In this special case, it is easy to observe from inequalities (3. 3.6) and (3. 3.8) that the following inequality b a 4 pi−1 (t) D i−1 f (t) dt 1 4 2 (b − a )3 278 Chapter 3 Opial-Type Inequalities b × b pi−1 (t) dt a 4 D i f (t) dt (3. 3.9) a holds for i = 1, , n T HEOREM 3. 3 .3 Let pi−1 , f, g be as in Theorem 3. 3.1 Then the following inequalities. .. inequality in (3. 3.20) From the hypotheses, we have the following identities D r−1 f (t) 2 t =2 D r−1 f (s)D r f (s) ds, (3. 3.29) 0 D r−1 f (t) 2 b = −2 D r−1 f (s)D r f (s) ds, (3. 3 .30 ) t for t ∈ I and r = 1, , n From (3. 3.29) and (3. 3 .30 ), we obtain D r−1 f (t) b 2 D r−1 f (t) D r f (t) dt (3. 3 .31 ) 0 for t ∈ I and r = 1, , n From (3. 3 .31 ) and using inequalities (3. 3.26), (3. 3.27) and Schwarz... to Wirtinger- and Opial-type inequalities, see [241] P ROOF OF T HEOREM 3. 3.4 From the hypotheses we have the following identities t D r−1 f (t) = D r f (s) ds, (3. 3. 23) 0 b D r−1 f (t) = − D r f (s) ds, (3. 3.24) t for t ∈ I and r = 1, , n From (3. 3. 23) and (3. 3.24), we obtain b 1 2 D r−1 f (t) D r f (t) dt (3. 3.25) 0 for t ∈ I and r = 1, , n From (3. 3.25) and using the elementary inequalities. .. m b Gr r=1 c ur (t) dt (3. 3.45) The desired inequality in (3. 3 .37 ) follows from (3. 3.44) and (3. 3.45) and the proof is complete R EMARK 3. 3.7 In the special case when m = 1, inequality (3. 3 .37 ) reduces to b a c g1 u1 (t) u1 (t) dt G1 a u1 (t) dt + G1 b c u1 (t) dt (3. 3.46) for c ∈ [a, b] Inequality (3. 3.46) is a variant of the inequality due to Calvert given u in Theorem 3. 3.5 On taking g1 (t) = t... Inequalities b t a a b = u (t) dt u (s) ds f f z(t) z (t) dt = F z(b) (3. 3 .34 ) a Now, using Hölder’s inequality with indices p, q, we get b z(b) = u (t) dt a b = r −1/p (t)r 1/p (t)z (t) dt a b 1/q b r 1−q (t) dt a 1/p r(t)z (t)p dt (3. 3 .35 ) a Inequality (3. 3 .33 ) now follows from (3. 3 .34 ) and (3. 3 .35 ) and the fact that F is nondecreasing R EMARK 3. 3.5 Let f (t) = t p−1 , p > 1, u(a) = 0, −∞ b up−1 (t)u (t) dt... dt 0 (3. 3 .32 ) Multiplying both sides of (3. 3 .32 ) by p(t) and integrating the resulting inequality from 0 to b we obtain the desired inequality in (3. 3.21) By using Schwarz inequality and inequalities (3. 3.20) and (3. 3.27), we observe that 1/n n b D p(t) 0 r−1 D r f (t) f (t) r=1 b dt r=1 2/n n 2 D p (t) 0 n r=1 r−1 f (t) 1/2 b 2 n r D f (t) dt 0 r=1 1/2 dt 285 3. 3 Wirtinger–Opial-Type Integral Inequalities. .. (3. 3.10) a b n pi−1 (t) D i−1 f (t) 2 2 D i f (t) 2 dt a i=1 n 2 1 4 i=1 b (b − a )3 a b × 4 1/2 2 pi−1 (t) dt D i f (t) + D i g(t) 4 dt (3. 3.11) a R EMARK 3. 3 .3 We note that, for n = 1, the inequality established in (3. 3.10) reduces to the inequality established by Pachpatte in [2 43] In the special case when D i f = D i g, the inequalities established in (3. 3.10) and (3. 3.11) reduce to the Opial-type . of inequality (3. 3.1) in Theorem 3. 3.1, we obtain the desired inequality in (3. 3.6). The details of the proofs of inequalities (3. 3.7) and (3. 3.8) follow from (3. 3.18) and (3. 3.19),and following. dt b a D i f(t) 2 + D i g(t) 2 dt , which is the desired inequality in (3. 3 .3) and the proof of Theorem 3. 3.1 is com- plete. 3. 3. Wirtinger–Opial-Type Integral Inequalities 281 From (3. 3.14), (3. 3.15) and using Schwarz inequality,. D i−1 f(t)=− b t D i f(s)ds, (3. 3.12) D i−1 g(t) = t a D i g(s)ds, D i−1 g(t) =− b t D i g(s)ds. (3. 3. 13) 3. 3. Wirtinger–Opial-Type Integral Inequalities 279 From (3. 3.12) and (3. 3. 13) , we observe that D i−1 f(t) 1 2 b a D i f(t) dt,
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