CUBO A Mathematical Journal Vol.10, N o ¯ 03, (13–20). October 2008 Poincaré Type Inequalities for Linear Differential Operators George A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152 U.S.A. email: ganastss@memphis.edu ABSTRACT Various Lp form Poincaré type inequalities [1], forward and reverse, are given for a Linear Differential Operator L, involving its related initial value problem solution y, Ly, the associated Green’s function H and initial conditions at point x0 ∈ R. RESUMEN Varias Lp desigualdes de tipo Poincaré [1], hacia adelante o atrás, son dadas para un operador diferencial linear L, envolviendo la solución y de un problema de valor inicial asociado, Ly, la función Green asociada H y las condiciones iniciales en un punto x0 ∈ R. Key words and phrases: Poincaré inequality, linear differential operator. Math. Subj. Class.: 26D10. 14 George A. Anastassiou CUBO 10, 3 (2008) 1. Background Here we follow [2], pp. 145-154. Let [a, b] ⊂ R, ai (x) , i = 0, 1, . . . , n − 1 (n ∈ N) , h (x) be continuous functions on [a, b] and let L = Dn + an−1 (x) D n−1 + . . . + a0 (x) be a fixed linear differential operator on C n ([a, b]) . Let y1 (x) , . . . , yn (x) be a set of linear independent solutions to Ly = 0. Here the associated Green’s functions for L is H (x, t) := ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ y1 (t) . . . yn (t) y′ 1 (t) . . . y′n (t) . . . . . . . . . y (n−2) 1 (t) . . . y (n−2) n (t) y1 (x) . . . yn (x) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ y1 (t) . . . yn (t) y′ 1 (t) . . . y′n (t) . . . . . . . . . y (n−2) 1 (t) . . . y (n−2) n (t) y (n−1) 1 (t) . . . y (n−1) n (t) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ , (1) which is a continuous function on [a, b] 2 . Consider a fixed x0 ∈ [a, b] , then y (x) = ∫ x x0 H (x, t) h (t) dt, ∀x ∈ [a, b] , (2) is the unique solution to the initial value problem Ly = h; y(i) (x0) = 0, i = 0, 1, . . . , n − 1. (3) Next we assume all of the above. 2. Results We present the following Poincaré type inequalities. Theorem 1. Let x0 < b and x ∈ [x0, b] , and p, q > 1 : 1 p + 1 q = 1; ν > 0. CUBO 10, 3 (2008) Poincaré Type Inequalities ... 15 Then 1) ‖y‖ Lν (x0,b) ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| p dt )ν/p dx )1/ν ‖Ly‖ Lq(x0,b) . (4) When ν = q we have 2) ‖y‖ Lq (x0,b) ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| p dt )q/p dx )1/q ‖Ly‖ Lq (x0,b) . (5) When ν = p = q = 2 we get 3) ‖y‖ L2(x0,b) ≤ ( ∫ b x0 ( ∫ x x0 H2 (x, t) dt ) dx )1/2 ‖Ly‖ L2(x0,b) . (6) Proof. From (2) we have |y (x)| ≤ ∫ x x0 |H (x, t)| |h (t)| dt ≤ ( ∫ x x0 |H (x, t)| p dt ) 1/p (∫ x x0 |h (t)| q dt ) 1/q ≤ ( ∫ x x0 |H (x, t)| p dt ) 1/p ( ∫ b x0 |h (t)| q dt ) 1/q . (7) That is |y (x)| ν ≤ ( ∫ x x0 |H (x, t)| p dt )ν/p ‖Ly‖ ν Lq (x0,b) , (8) Therefore ∫ b x0 |y (x)| ν dx ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| p dt )ν/p dx ) ‖Ly‖ ν Lq(x0,b) , (9) proving the claim. 2 We continue with Theorem 2. Let x0 > a and x ∈ [a, x0] , and p, q > 1 : 1 p + 1 q = 1; ν > 0. Then 1) ‖y‖ Lν (a,x0) ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| p dt )ν/p dx ) 1/ν ‖Ly‖ Lq (a,x0) . (10) When ν = q we have 2) ‖y‖ Lq(a,x0) ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| p dt )q/p dx ) 1/q ‖Ly‖ Lq (a,x0) . (11) 16 George A. Anastassiou CUBO 10, 3 (2008) When ν = p = q = 2 we get 3) ‖y‖ L2(a,x0) ≤ ( ∫ x0 a ( ∫ x0 x H 2 (x, t) dt ) dx ) 1/2 ‖Ly‖ L2(a,x0) . (12) Proof. From (2) we have |y (x)| ≤ ∫ x0 x |H (x, t)| |h (t)| dt ≤ ( ∫ x0 x |H (x, t)| p dt )1/p (∫ x0 x |h (t)| q dt )1/q ≤ ( ∫ x0 x |H (x, t)| p dt ) 1/p (∫ x0 a |h (t)| q dt ) 1/q . (13) That is |y (x)| ν ≤ ( ∫ x0 x |H (x, t)| p dt )ν/p ‖Ly‖ ν Lq (a,x0) , (14) Therefore ∫ x0 a |y (x)| ν dx ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| p dt )ν/p dx ) ‖Ly‖ ν Lq (a,x0) , (15) proving the claim. 2 Extreme cases follow Proposition 3. Here x0 < b, x ∈ [x0, b] , and p = 1, q = ∞. Then 1) ‖y‖ Lν (x0,b) ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| dt )ν dx )1/ν ‖Ly‖ L∞(x0,b) . (16) When ν = 1 we have 2) ‖y‖ L1(x0,b) ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| dt ) dx ) ‖Ly‖ L∞(x0,b) . (17) Proof. From (2) we have |y (x)| ≤ ∫ x x0 |H (x, t)| |h (t)| dt ≤ ( ∫ x x0 |H (x, t)| dt ) ‖h‖ L∞(x0,b) . (18) CUBO 10, 3 (2008) Poincaré Type Inequalities ... 17 That is |y (x)| ν ≤ ( ∫ x x0 |H (x, t)| dt )ν ‖Ly‖ ν L∞(x0,b) , (19) and ∫ b x0 |y (x)| ν dx ≤ ( ∫ b x0 ( ∫ x x0 |H (x, t)| dt )ν dx ) ‖Ly‖ ν L∞(x0,b) , (20) proving the claim. 2 We continue with Proposition 4. Here x0 > a, x ∈ [a, x0] , and p = 1, q = ∞. Then 1) ‖y‖ Lν (a,x0) ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| dt )ν dx )1/ν ‖Ly‖ L∞(a,x0) . (21) When ν = 1 we get 2) ‖y‖ L1(a,x0) ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| dt ) dx ) ‖Ly‖ L∞(a,x0) . (22) Proof. From (2) we have |y (x)| ≤ ∫ x0 x |H (x, t)| |h (t)| dt ≤ ( ∫ x0 x |H (x, t)| dt ) ‖h‖ L∞(a,x0) . (23) That is |y (x)| ν ≤ ( ∫ x0 x |H (x, t)| dt )ν ‖Ly‖ ν L∞(a,x0) , (24) and ∫ x0 a |y (x)| ν dx ≤ ( ∫ x0 a ( ∫ x0 x |H (x, t)| dt )ν dx ) ‖Ly‖ ν L∞(a,x0) , (25) proving the claim. 2 Next we give reverse Poincaré type inequalities. Theorem 5. Let x0 < b, x ∈ [x0, b] , and 0 < p < 1, q < 0 : 1 p + 1 q = 1, ν > 0. Assume H (x, t) ≥ 0 for x0 ≤ t ≤ x, and Ly = h is of fixed sign and nowhere zero. Then 1) ‖y‖ Lν (x0,b) ≥ ( ∫ b x0 ( ∫ x x0 (H (x, t)) p dt )ν/p dx ) 1/ν ‖Ly‖ Lq (x0,b) . (26) 18 George A. Anastassiou CUBO 10, 3 (2008) When ν = p we get 2) ‖y‖ Lp(x0,b) ≥ ( ∫ b x0 ( ∫ x x0 (H (x, t)) p dt ) dx )1/p ‖Ly‖ Lq(x0,b) . (27) When ν = 1 we obtain 3) ‖y‖ L1(x0,b) ≥ ( ∫ b x0 ( ∫ x x0 (H (x, t)) p dt )1/p dx ) ‖Ly‖ Lq(x0,b) . (28) Proof. By (2) we have |y (x)| = ∫ x x0 H (x, t) |h (t)| dt, for all x0 ≤ x ≤ b. (29) From (29) by reverse Hölder’s inequality we obtain |y (x)| ≥ ( ∫ x x0 (H (x, t)) p dt )1/p (∫ x x0 |h (t)| q dt )1/q ≥ ( ∫ x x0 (H (x, t)) p dt )1/p ( ∫ b x0 |h (t)| q dt ) 1/q , (30) for all x0 < x ≤ b. I.e. it holds |y (x)| ν ≥ ( ∫ x x0 (H (x, t)) p dt )ν/p ‖h‖ ν Lq (x0,b) , (31) for all x0 ≤ x ≤ b, and ∫ b x0 |y (x)| ν dx ≥ ( ∫ b x0 ( ∫ x x0 (H (x, t)) p dt )ν/p dx ) ‖h‖ ν Lq(x0,b) , (32) proving the claim. 2 We continue with Theorem 6. Let x0 > a, x ∈ [a, x0] , and 0 < p < 1, q < 0 : 1 p + 1 q = 1, ν > 0. Assume H (x, t) ≤ 0 for x ≤ t ≤ x0, and Ly = h is of fixed sign and nowhere zero. Then 1) ‖y‖ Lν (a,x0) ≥ ( ∫ x0 a ( ∫ x0 x (−H (x, t)) p dt )ν/p dx ) 1/ν ‖Ly‖ Lq(a,x0) . (33) CUBO 10, 3 (2008) Poincaré Type Inequalities ... 19 When ν = p we get 2) ‖y‖ Lp(a,x0) ≥ ( ∫ x0 a ( ∫ x0 x (−H (x, t)) p dt ) dx )1/p ‖Ly‖ Lq (a,x0) . (34) When ν = 1 we have 3) ‖y‖ L1(a,x0) ≥ ( ∫ x0 a ( ∫ x0 x (−H (x, t)) p dt )1/p dx ) ‖Ly‖ Lq(a,x0) . (35) Proof. From (2) we have |y (x)| = ∣ ∣ ∣ ∣ ∫ x x0 H (x, t) h (t) dt ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∫ x0 x H (x, t) h (t) dt ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∫ x0 x (−H (x, t)) h (t) dt ∣ ∣ ∣ ∣ = ∫ x0 x (−H (x, t)) |h (t)| dt. (36) From (36) by reverse Hölder’s inequality we obtain |y (x)| ≥ ( ∫ x0 x (−H (x, t)) p dt )1/p (∫ x0 x |h (t)| q dt )1/q ≥ ( ∫ x0 x (−H (x, t)) p dt )1/p (∫ x0 a |h (t)| q dt )1/q . (37) for all a ≤ x < x0. I.e. it holds |y (x)| ν ≥ ( ∫ x0 x (−H (x, t)) p dt )ν/p ‖Ly‖ ν Lq(a,x0) , (38) for all a ≤ x ≤ x0, and ∫ x0 a |y (x)| ν dx ≥ ( ∫ x0 a ( ∫ x0 x (−H (x, t)) p dt )ν/p dx ) ‖Ly‖ ν Lq (a,x0) , (39) proving the claim. 2 Received: December 2007. Revised: April 2008. 20 George A. Anastassiou CUBO 10, 3 (2008) References [1] G. Acosta and R. G. Durán, An optimal Poincaré inequality in L1 for convex domains, Proc. AMS, 132 (1)2003, pp.195–202. [2] D. Kreider, R. Kuller, D. Ostberg and F. Perkins, An Introduction to Linear Anal- ysis, Addison-Wesley Publishing Company, Inc., Reading, Mass., USA, 1966. N02