Dedicated to the memory of Professor Dragoslav S. Mitrinovic 1. INTRODUCTION. Let E :[0;+1)!Rbe a nonnegative, non-increasing, locally absolutely

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1 Univ. Beograd. Publ. Elekroehn. Fak. Ser. Ma. 7 (1996), 55{67. DIFFERENTIAL AND INTEGRAL INEQUALITIES Vilmos Komornik Dedicaed o he memory of Professor Dragoslav S. Mirinovic 1. INTRODUCTION Le E :[;)!Rbe a nonnegaive, non-increasing, locally absoluely coninuous funcion. Assume ha here exiss anoher locally absoluely coninuous funcion :[;)!Rand here are hree real numbers a, b and such ha (1) jj ae in [; ) and (2) be E a.e. in [; ): How can we esimae E()? Problems of his ype ofen appear during he sudy of dissipaive linear evoluionary problems where E denoes he energy of he soluion. I is sucien o consider he case where E() = 1. Indeed, if E() =, hen E. On he oher hand, if E() >, hen replacing E,, a and b respecively by E=E(), E() 1, ae() and ae(),we obain a soluion of (1), (2) saisfying E() = 1. We will herefore assume in he sequel ha (3) E()=1: Le us briey recall he Liapunov mehod as usually applied o his problem (see e.g. [1], [4], [5], [1], [11]). Fix a real number d saisfying (4) d>a and d b; 1991 Mahemaics Subjec Classicaion: 26D1 55

2 56 Vilmos Komornik and consider he funcion F := de+. One can readily verify ha F :[;)!R is nonnegaive, non-increasing, locally absoluely coninuous. Furhermore, and (d a)e F (d + a)e in [; ) F (d+a) 1 F a.e. in [; ): Dividing by F and inegraing i follows ha F ()e F () =(d+a) if =; (F() + (d + a) 1 ) if 6= and herefore E() < : d+a d a e =(d+a) if =; if 6= d+a d a d+a+ d+a for all such ha E() >. Nex we minimize he righ-hand side of his esimae wih respec o d saisfying (4). Since (as we shall see a he end of his paper) his mehod does no lead o sharp esimaes, we only consider henceforh he special case where Then we have = and a (5) E() d + a d a e =(d+a) =: f(d) for all and for all d saisfying (4). (Observe ha his inequaliy makes sense and remains valid wihou he assumpion E() >.) An easy compuaion shows ha f (d) =e =(d+a) ( 2a)d ( +2a)a (d a) 2 (d+a) Hence f is decreasing (resp. increasing) if ( 2a)d ( + 2a)a < (resp. > ). If 2a, hen f is decreasing in (a; ) and ends o 1 as!. Therefore we only obain he rivial esimae E() 1. If >2a, hen f decreases in (a; A) and increases in (A; ) where A = +2a 2a a (>a): We disinguish wo cases: If b A, hen choosing d = A in (5) we obain ha E() 2a e ( 2a)=(2a) :

3 Dierenial and inegral inequaliies 57 If b A, hen choosing d = b in (5) we conclude ha E() b + a b a e =(b+a) : If b a, hen b A for all >2a. If b > a, hen b A if and only if 2a <2a b+a b a. We have hus proven he following: Proposiion 1. If E, solve (1) following esimaes : (6) E() (3) wih =and a>, hen we have he 1 if 2a; 2a e(2a )=(2a) if b a and 2a; 2a e(2a )=(2a) if b>aand 2a 2a b+a b+a b a e =(b+a) if b>aand 2a b+a b a. b a ; Despie he very frequen applicaion of his mehod, he above esimaes are no opimal. Applying a dieren mehod we shall prove Theorem 2. a) The problem (1) (3) has no soluion unless > 1,aand a + b>. b) If E, solve (1){(3) wih some >, hen we have he following esimaes : b1) If a<ba, hen (7) E() ( 1 if (a + b); a+b+ (a+b)(1+) if (a + b), and in he second case he inequaliy is sric ; b2) If b>a, hen () E() ( 1 if 2a; a+b+ a+b+2a if 2a. c) If E, solve (1) (3) wih =, hen we have he following esimaes : c1) If a<ba, hen 1 if a + b; (9) E() e (a+b )=(a+b) if a + b, and in he second case he inequaliy is sric ; c2) If b>a, hen 1 if 2a; (1) E() e (2a )=(a+b) if 2a.

4 5 Vilmos Komornik d) If E, solve (1) (3) wih some 1 < <, hen we have he following esimaes : d1) If a<ba; hen (11) E() 1 if (a + b); a+b+ (a+b)(1+) if (a + b) <(a+b)=jj; if (a + b)=jj, and in he second case he inequaliy is sric ; d2) If b>a, hen (12) E() 1 if 2a; a+b+ a+b+2a if 2a (a + b)=jj; if (a + b)=jj. The above esimaes are opimal. Remark. Leing! in he formulae corresponding o 6= we nd he formulae for =. For he proof of Theorem 2, we will have o sudy a closely relaed inegral inequaliy, already used in [2], [3], [6]{[9]: (13) Z E(s) ds TE(); : Here we only assume ha E : [;)!R is a nonnegaive, non-increasing (hence measurable) funcion and ha, T are given real numbers. If E() =, hen E. If E() >, hen replacing E by E=E() and T by TE() we obain a soluion of (13) such ha E()=1. Furhermore, in order o avoid he rivial soluion 1 if =; E() = if >, we shall only consider consider soluions of (13) such ha (14) E() = 1 and E 6 in (;1): The following resul, ineresing in iself, complees some earlier heorems of Haraux [2], [3]: Theorem 3. a) The problem (13) (14) has no soluion unless > 1and T >. b) If E solves (13) (14) wih some >, hen we have he following esimaes : ( 1 if T ; (15) E() T + if T. T +T

5 Dierenial and inegral inequaliies 59 Moreover, he second inequaliy is sric if E is righ coninuous. d) If E solves (13) (14) wih =, hen we have he following esimaes : 1 if T ; (16) E() e (T )=T if T. Moreover, he second inequaliy is sric if E is righ coninuous. e) If E solves (13) (14) wih some 1 <<, hen we have he following esimaes : (17) E() 1 if T ; T + T +T if T <T=jj; if T=jj. Moreover, he second inequaliy is sric if E is righ coninuous. These esimaes are opimal. Remark. As in he preceding resuls, leing! in he formulae corresponding o 6= we nd he formulae for =. 2. PROOF Of THEOREM 3 If 1, hen (13) is meaningful only if E() > for all >. However, hen E(s) E() = 1 for all s and herefore he inegral on he lef-hand side of (14) is innie. If T, hen (13) implies a once ha E vanishes in (; ), conradicing (14). Thus par a) of he heorem is proven. Henceforh we may herefore assume ha > 1 and T >. If T, hen he esimaes E() 1 of (15){(17) follow simply from he non-increasingness of E. Also, here is nohing o prove ifbwhere We mayhus assume ha T <<B. The formula B = supfr j E(r) > g: F (r) = Z r E(s) ds denes a nonnegaive, non-increasing and locally absoluely coninuous funcion F :[;1)!R. I follows from (13) ha F T 1 F

6 6 Vilmos Komornik almos everywhere in (; 1). Dividing by F and inegraing in (;s), we obain for every <s<bhe following inequaliies: F (s) (F () + T 1 s) if 6= ; F()e s=t if =. Since F () T by (13) (14); hese inequaliies remain valid if we replace F () by T.Furhermore, we have F(s) Z T+()s s E(r) dr (T + s)e(t +()s) : Therefore, we deduce from he preceding inequaliies he esimaes (T + s)e(t +()s) (T + T 1 s) if 6= ; Te s=t if =, or equivalenly, E(T +()s) ( T+s if 6= ; T e s=t if =, for all <s<b. If, hen hese esimaes obviously remain valid for all s>. Choosing s = T hence (15) (16) follow. If 1 <<, hen he righ-hand side of he above esimae is meaningless for s T=jj. Hence E() = for all T=jj, proving he hird inequaliy in (17). Furhermore, he above esimae obviously remains valid for all <s<t=jj. Since T <<Bimplies ha < T <T=jj,wemaychoose s = T in he above esimae, and he second inequaliy of (17) follows. Now assume ha E is righ coninuous and prove ha he second inequaliies of (15) (17) are sric. Assume on he conrary ha we have equaliy in he second inequaliy of one of he formulae (15) (17) for some T : (1) E( )= ( T+ T +T if 6= ; e (T )=T if =. Using he righ coninuiy ofein, here is a consan <<1 such ha I follows ha he funcion Z Z E ds E ds: G() =n E() if ; if >

7 Dierenial and inegral inequaliies 61 also saises (13) (14); even if we replace he consan T in (13) by T. Applying he already proved (weak) esimaes (15) (17); we have G( ) ( T+ T+T if 6= ; e (T )=(T) if =. (Noe ha he hird case in (17) canno occur because G( ) > by assumpion.) Using (1) and he equaliy G( )=E( )>, i follows ha ( T + T +T T+ T+T if 6= ; e (T )=T e (T )=(T) if =. Bu boh inequaliies conradic he propery <1. Le us now urn o he proof of he opimaliy of he esimaes (15) (17): Fix > 1, T > and arbirarily. If <T, hen we have o consruc a soluion of (13) (14) such ha E() = E( ) = 1. Choose simply E() =n 1 if T ; if >T. The vericaion of (13) is immediae: he case >T is rivial, while for T we have Z Z T E(s) ds 1dsT=TE(): Wemayeven consruc coninuous examples, e.g., ( 1 if ; E() = (T )=(T ) if T ; if >T. If T (for ) or T <T=jj(for consruc a soluion of (13) (14) such ha If =, hen le us choose E( )= E() = If 6=, hen le us choose E() = ( T+ T +T if 6= ; e (T )=T if =. ( e =T if T ; e ( T )=T if T ; if >. T+ T T + T +T if T ; if T ; if >. 1 <<), hen we haveo

8 62 Vilmos Komornik (Noe ha hese funcions are no coninuous.) The only nonrivial propery overify is (13) for T. Since E = TE in (; T ) in all cases, we have in fac equaliy: Z E(s) ds = Z ( T )=() = TE() TE = TE(): E(s) ds + + T The proof of Theorem 3 is compleed. Z ( T )=() E T E(s) ds T 3. PROOF OF THEOREM 2 We begin wih a lemma relaing he problem (1) (13) (14): (3) o he inegral inequaliy Lemma 4. If E, solve (1) (3) wih some a, b and, hen E also solves (13) (14) wih he same and wih T = a + b. Proof. Since he soluions E of (1) (3) are coninuous, (3) implies (14). I follows from (1) (2) and from he non-increasingness of E ha (19) Z E(s) ds [be + ] 2(jaj + jbj)e() for all < <. Leing! hence we conclude ha Z E(s) ds 2(jaj + jbj)e() for all. Applying Theorem 3 i follows ha E( )! as!1. Using (1) we also obain ha ( )! as!. Hence, leing!1in he rs inequaliy of (19), we conclude ha Z Applying (1) again, hence (13) follows. 2 E(s) ds be()+(): I follows a once from (1) and (3) ha a. The res of par a and pars b1, c1, d1 Theorem 2 follow a once from Lemma 4 and Theorem 3, including he sric inequaliies. I remains o prove he esimaes (), (1) and (12). Since he inequaliy E() 1isobvious, we haveoprove for > 1, b > aand >2ahe

9 following esimaes: (2) E() Dierenial and inegral inequaliies 63 a+b+ a+b+2a if >; e (2a )=(a+b) if =; a+b+ a+b+2a if < and (a + b)=jj; if < and >(a+b)=jj: Clearly, wemayalso assume ha <B:= supfr j E(r) > g: Dividing he inequaliy (2) by E, hen inegraing in (;) and using (1), we obain ha whence Z be 1 E ds (b + a + a) Z 1 E 1 ds Z =[ E 1 ] + 1 ( )E 2 E ds ae() + ae() ( )a Z Compuing he inegral, i follows easily ha E() E 1 E ds ae() + ae() : a+b+ a+b+2a if >; e (2a )=(a+b) if =; a+b+ a+b+2a if <. Z E 1 E ds; Comparing wih (2), i only remains o show ha E() =if < and > (a+b)=jj. Le us observe ha for < he righ-hand side of he las inequaliy vanishes for =(a+b)=jj. I canno occur if E() >, herefore E((a+b)=jj) = and our claim follows. Now we are going o prove he opimaliy of our esimaes (7) (12): Fix > 1, a, b> aarbirarily. Furhermore, x arbirarily if and x < (a + b)=jj arbirarily if 1 <<. Le us dene a number R in he following way: se R = < : if b a and <a+b; if b>aand < 2a; (a+b)( 2a) if b>aand 2a. a+b+2a

10 64 Vilmos Komornik Furhermore, choose an arbirary number (21) a b 1+ <R if b a and a + b; is value will be precised laer. These deniions are correc and R in all cases. Nex we dene he funcion E. For >we se where E() = For =we dene E() = a+b+ if R; a+b E(R) if R< ; E(R) 1+ ( )E(R) a+b ( R)E(R) if >. < : e =(a+b) if R; E(R) if R< ; E(R)e ( )=(a+b+r ) if >. Finally, for 1 <<we se a+b+ if R; a+b E(R) if R< E() = ; E(R) 1+ ( )E(R) a+b ( R)E(R) if << ; if, = + a + b ( R)E(R) jje(r) : If, hen a + b + > ; hence E() is correcly dened and sricly posiive. In paricular, E(R) >. Le us show ha (22) ( R)E(R) <a+b and (23) ( R)E(R) 2a: Indeed, if b a and <a+b, hen If b>aand < 2a, hen ( R)E(R) = <a+b2a: ( R)E(R) = < 2a <a+b:

11 If b>aand 2a, hen Dierenial and inegral inequaliies 65 ( R)E(R) =2a<a+b by a simple compuaion. Finally, ifbaand a + b, hen ( R)E(R) = ( R)(a + b) a + b + R because R>( a b)=(1 + ) (see (21)). <a+b2a Using (22) one can readily verify ha E is a correcly dened, nonnegaive, non-increasing, locally absoluely coninuous funcion for all, and E() = 1. Le us assume for he momen he exisence of a locally absoluely coninuous funcion saisfying (1) (2); and prove he opimaliy of he esimaes of Theorem 2. Le us compue E( )=E(R). If b>a, hen E( )= 1 if < 2a; a+b+ a+b+2a if 2a and 6= ; e (2a )=(a+b) if 2a and =. This proves he opimaliy of he esimaes (), (1), (12). If b a, hen E( )= 1 if <a+b; if a + b and 6= ; a+b+r a+b e R=(a+b) if a + b and =. Leing R! ( a b)=(1 + ) (see (21)) hence he opimaliy of he esimaes (7), (9), (11) follows. I remains o consruc a locally absoluely coninuous funcion :[;)! Rsaisfying (1) and (2). Dene () = < : ae() if R; ae(r) ( R)E(R) if R ; (a ( R)E(R) )E() if. Then is locally absoluely coninuous. R; for >Ri follows easily using (23): The propery (1) is obvious for ae() () (a ( R)E(R) )E() ae(): Nex we claim ha = be E a.e. in [; ); in paricular, (2) is saised. Indeed, in (;R) wehave (be + )() =(a+b)e ()= E() :

12 66 Vilmos Komornik In (R; )wehave In ( ; ) wehave (be + )() = E(R) = E() : (be + )() =(a+b ( R)E(R) )E ()= E() by anoher simple compuaion. The proof of Theorem 2 is compleed. 4. COMPARISON OF PROPOSITION 1 AND THEOREM 2 We are going o show ha he esimaes of Proposiion 1 are opimal only in rivial cases. As in Proposiion 1, assume ha = and a>. a) If b a, hen (1) (3) has no soluion; his was no revealed by he Liapunov mehod: we only obained in his case he esimae 1 if 2a; E() 2a e(2a )=(2a) if 2a (cf. (6)). b) If a<ba, hen we have o compare he esimaes (6) and (9). For a + b hey boh give E() 1. For a + b<2ahe esimae (9) is beer because e (a+b )=(a+b) < 1: Finally, for >2ahe esimae (9) is beer again because Indeed, we have e (a+b )=(a+b) < 2a e(2a )=(2a) : e (a+b )=(a+b) e (2a )=(2a) < 2a e(2a )=(2a) : c) If b>a, hen we have o compare he esimaes (6) and (1). For 2a hey boh give E() 1. In order o show ha for 2a b+a he esimae (1) is beer han (6), we b a have o prove ha (24) e (2a )=(a+b) < b + a b a e =(a+b) : Puing x =2a=(a + b) wehave < x <1, and he inequaliy akes he form e x < (1 x). This inequaliy is rivially saised: e x = 1X i=1 x i i! < 1X i=1 x i =(1 x):

13 Dierenial and inegral inequaliies 67 Finally, in order o show ha for 2a <2a b+a he esimae (1) is beer b a han (6), we haveoprove he inequaliy e (2a )=(a+b) < 2a e(2a )=(2a) : Keeping a and xed, le us increase b unil =2a b+a (hen he lef-hand side b a of he inequaliy increases). Then our inequaliy coincides wih (24) and he claim follows. REFERENCES 1. F. Conrad, B. Rao: Decay of soluions of wave equaions in a sar-shaped domain wih nonlinear boundary feedback. Asympoic Anal., 7 (1993), 159{ A. Haraux: Oscillaions forcees pour cerains sysemes dissipaifs non lineaires. Publicaion du Laboraoire d'analyse Numerique No. 71, Universie Pierre e Marie Curie, Paris, A. Haraux:,Semi-groupes lineaires e equaions d'evoluion lineaires periodiques. Publicaion du Laboraoire d'analyse Numerique No. 711, Universie Pierre e Marie Curie, Paris, A. Haraux, E. Zuazua: Decay esimaes for some semilinear damped hyperbolic problems. Arch. Ra. Mech. Anal. (19), 191{ V. Komornik, E. Zuazua: A direc mehod for he boundary sabilizaion of he wave equaion. J. Mah. Pures Appl., 69 (199), 33{ V. Komornik: Rapid boundary sabilizaion of he wave equaion. SIAM J. Conrol Op., 29 (1991), 197{2. 7. V. Komornik: On he nonlinear boundary sabilizaion of he wave equaion. Chin. Ann. of Mah. 14B:2 (1993), 153{164.. V. Komornik: On he nonlinear boundary sabilizaion of Kirchho plaes. Nonlinear Di. Equaions and Appl. (NoDEA) 1 (1994), 323{ V. Komornik: Boundary sabilizaion, observaion and conrol of Maxwell's equaions. PanAmerican Mah. J., 4 (1994) No. 4, 47{ J. Lagnese: Boundary Sabilizaion of Thin Plaes. SIAM Sudies in Appl. Mah., Philadelphia, E. Zuazua: Uniform sabilizaion of he wave equaion by nonlinear boundary feedback. SIAM J. Conrol Op. 2 (199), 265{26. Isiu de Recherche Mahemaique Avancee, (Received Sepember 15, 1995) Universie Louis Paseur e C.N.R.S, 7, rue Rene Descares, 674 Srasbourg Cedex, France