DISCRETE GRONWALL LEMMA AND APPLICATIONS

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1 DISCRETE GRONWALL LEMMA AND APPLICATIONS JOHN M. HOLTE MAA NORTH CENTRAL SECTION MEETING AT UND 24 OCTOBER 29 Gronwall s lemma saes an inequaliy ha is useful in he heory of differenial equaions. Here is one version of i [1, p, 283]:. Gronwall s lemma. Le y(),f(), and g() be nonnegaive funcions on [,T] having one-sided limis a every [,T], and assume ha for T we have Then for T we also have Proof. y() f()+ y() f()+ g(s)y(s) ds. g(s)f(s)exp g(u) du ds. s Le I() := g(s)y(s) ds and G() := g(s) ds. Now I () = g()y() g()f()+g()i() ouside a denumerable se, and d d e G() I() =e G() {I () g()i()} e G() g()f(), so e G() I() and y() f()+i() f()+ e G(s) g(s)f(s) ds. Therefore, I() e G() G(s) g(s)f(s) ds. e G() G(s) g(s)f(s) ds Here is our firs formulaion of a discree version of his lemma: 1. Discree Gronwall inequaliy. If y n, f n, and g n are nonnegaive sequences and (1) y n f n + g k y k for n, hen (2) y n f n + f k g k exp( k<j<n g j ) for n. The proof ha follows firs gives he exac soluion for y n when inequaliy in (1) is replaced by equaliy. Then i shows ha any soluion of he inequaliy is dominaed by his exac soluion, from which (2) quickly follows. Firs we prove a lemma giving a soluion of he equaliy when f n Lemma. Le g n be a sequence. For n le (3) G n := (1 + g j ). j<n 1

2 Then G n =1+ j<n g j G j for n, and, more generally, for k n, (4) G n = G k + k j<n g j G j. Proof. An empy produc is 1, and an empy sum is, so he formulas hold for n = (and k = ). Lem. Assuming he formulas hold for n = m, we have G m+1 =(1+g m )G m = G m + g m G m =1+ g k G k + g m G m =1+ g k G k, and for k<m k m, G m+1 = G m + g m G m = G k + k j<m conclusions follow by mahemaical inducion. g j G j + g m G m = G k + k<m+1 k j<m+1 g j G j. The 3. Proposiion. Assume x n, f n, and g n are nonnegaive sequences and (5) x n = f n + g k x k for n. Then (6) x n = f n + f k g k (1 + g j )=f n + f k g k G n /G k+1 k<j<n for n. Here G n is defined by (3). Proof. Le χ n := f n + f k g k G n /G k+1. We prove ha x n = χ n for n by inducion. For n = boh sides equal f.lem>. Assume x n = χ n for n<m.by

3 (5), x m g k x k g k χ k k<m k<m k<m k<m k<m g k {f k + j<k g k f k + g k f k + g k f k + k<m k<m j<k f j g j G k /G j+1 } g k f j g j G k /G j+1 j<m 1 j<k<m f m 1 g m 1 + k<m k<m 1 k<j<m g k f j g j G k /G j+1 g j f k g k G j /G k+1 {f k g k (1 + g j G j /G k+1 )} k<m 1 k<j<m f k g k G m /G k+1 = χ m by (4). The conclusion follows by inducion. The following heorem acually gives a sharper resul han he discree Gronwall inequaliy. 4. Theorem: Sharp Gronwall Inequaliy. Assume y n, f n, and g n are nonnegaive sequences and (7) y n f n + g k y k for n. Then (8) y n f n + f k g k (1 + g j ) k<j<n for n. Proof. Le x n be as in he previous proposiion. Then an easy inducion shows ha y n x n for every n. Alernaive Proof. For n = inequaliy (7) implies y f, whence inequaliy (8) holds for n =. Lem>. Assume ha (8) holds for n<m. By hypohesis (7) and his

4 inducion hypohesis, G (m) j y m f m + k<m f m + k<m j<k<m g k j<m j<i<k g k y k g k f j g j G (m) j where (by an implici inducion) =1+ (1 + g i ) f k + j<k f j g j j<i<k (1 + g i ) =1+g j+1 1+g j+2 (1 + g j+1 )+ + g m 1 (1 + g j+1 ) (1 + g m 2 ) =(1+g j+1 )(1 + g j g m 1 (1 + g j+2 ) (1 + g m 2 )) = = j<i<m (1 + g i ). Thus inequaliy (8) holds for n = m. By mahemaical inducion, inequaliy (8) holds for every n. Proof of he Discree Gronwall inequaliy. he previous heorem. Use he inequaliy 1 + g j exp(g j ) in 5. Anoher discree Gronwall inequaliy Here is anoher form of Gronwall s lemma ha is someimes invoked in differenial equaions [2, pp ]: Le y and g be nonnegaive inegrable funcions and c a nonnegaive consan. If y() c + g(s)y(s) ds for, hen y() c exp g(s) ds for. Proposiion: Special Gronwall Inequaliy. Le y n and g n be nonnegaive sequences and c a nonnegaive consan. If y n c + g k y k for n, hen y n c j<n (1 + g j ) c exp j<n g j for n.

5 Proof. Assume he hypohesis and apply he Sharp Gronwall Inequaliy wih f n c: y n c + cg k (1 + g j ) = c + c = c + c{ = c j<n c exp j<n k<j<n { k j<n (1 + g j ) (1 + g j ) n j<n k+1 j<n (1 + g j ) [empy produc is 1] j<n g j (1 + g j )} (1 + g j )} [elescoping sum] [1 + g j e g j ]. Nonhomogeneous Linear Sysems One area where Gronwall s inequaliy is used is he sudy of he asympoic behavior of nonhomogeneous linear sysems of differenial equaions. We are ineresed in obaining discree analogs. 6. Firs-order nonhomogeneous linear differenial equaions Consider he firs-order nonhomogenous vecor differenial equaion dx = A()x + f() d for wih iniial condiion x() = c, where A() is a coninuous d d marix, f() is a coninuous d 1 vecor, and x() and c are a d 1 vecors. Is soluion may be obained in erms of d linearly independen soluions of he corresponding homogeneous equaion as follows: x() =Y ()c + Y () where Y () is he d d marix soluion of dy d Y (s) 1 f(s) ds = A()Y wih Y () = I. [2, pp ] 7. Firs-order nonhomogeneous linear difference equaions We prove he discree counerpar. For vecor sequences x(n) he forward difference operaor is defined by: x(n) = x(n + 1) x(n). The analog of he differenial equaion is x(n) =A(n)x(n)+f(n), or for n wih x() = c. x(n +1)=(I + A(n))x(n)+f(n) Proposiion. Le x(n) and f(n) be sequences of d 1 vecors, le c be a fixed d 1 vecor, and le A(n) be a sequence of d d marices. Assume ha I + A(n) is inverible

6 for each n. If x(n) =A(n)x(n)+f(n) for n and x() = c, and if Y (n) is he d d soluion of Y (n) =A(n)Y (n) wih Y () = I, hen (9) x(n) =Y (n)c + Y (n) Y (k +1) 1 f(k) for n. Proof. Assuming I + A(n) is inverible for n, we le Z(n) be he marix soluion of he adjoin equaion Z(n) = Z(n +1)A(n) for n wih Z() = I. Then ( Z(n))Y (n) = Z(n +1)A(n)Y (n) and Z(n +1) Y (n) =Z(n +1)A(n)Y (n), so ( Z(n))Y (n)+z(n +1)A(n)Y (n) =, he zero marix, i.e., (Z(n)Y (n)) =. Therefore Z(n)Y (n) =C, a consan marix. Because Z() = I and Y () = I, we have Z(n)Y (n) =I, whence Z(n) =Y (n) 1 for n. Now muliply x(n) =A(n)x(n)+f(n) on he lef by Z(n +1)oge Z(n +1) x(n) =Z(n +1)A(n)x(n)+Z(n +1)f(n), muliply Z(n) = Z(n +1)A(n) on he righ by x(n) oge and add o ge for n, i.e., Therefore, so ( Z(n))x(n) = Z(n +1)A(n)x(n), Z(n +1) x(n)+( Z(n))x(n) =Z(n +1)f(n) (Z(n)x(n)) = Z(n +1)f(n). (Z(k)x(k)) = Z(n)x(n) Z()x() = Since Z() = I and each Z(n) =Y (n) 1, x(n) =Y (n)x() + Y (n) Z(k +1)f(k), Z(k +1)f(k). Y (k +1) 1 f(k). 8. Marix differenial equaions and linearly independen soluions In he case of he marix differenial equaion dx d X = A()X() one can show ha = d d (race A()) X(), where X() = de X(), whence X() = X() exp race A(s) ds. [2, p. 88] From his i follows ha a se of d linearly independen vecor soluions of dx = A()x will remain linearly independen as ime goes on. d 9. Marix difference equaions and linearly independen soluions The discree analog does no play ou in he same nea way, bu i is even simpler. If he sequence X(n) of d d marices saisfies X(n) =A(n)X(n) forn, hen X(n +1)= (I +A(n))X(n) forn, so ha X(n) =(I +A(n 1))(I +A(n 2)) (I +A())X() for

7 n>. Thus, if each I + A(k) is inverible, hen X(n) is nonsingular whenever X() is, and so a se of d linearly independen vecor soluions of x(n) = A(n)x(n) will remain linearly independen as ime n goes on. Le denoe a norm for d-vecors and he corresponding operaor norm for d d marices. For example, x and A could be defined as he sums of he absolue values of heir componens. 1. A differenial equaion applicaion of Gronwall s inequaliy One applicaion of Gronwall s inequaliy is in he proof of he following heorem. [2, pp ] Theorem. Assume lim A() =A, a consan marix. If all he soluions of he limiing homogeneous equaion dx d = Ax remain bounded as, hen he same is rue of all soluions of he nonhomogeneous equaion provided ha and dx d = A()x + f() A() A d < f() d <. 11. A difference equaion applicaion of Gronwall s inequaliy Our analogous heorem is he following. Theorem. Assume ha lim n A(n) =A, a consan marix. If all he soluions of he limiing homogeneous equaion x(n) =Ax(n) for n remain bounded as n, hen he same is rue of all soluions of he nonhomogeneous equaion (1) x(n) =A(n)x(n)+f(n) for n provided ha and Proof. A(n) A < n= f(n) <. n= Le B(n) =A(n) A and rewrie (1) as x(n) =Ax(n)+B(n)x(n)+f(n).

8 By (9) wih f(n) replaced by B(n)x(n)+f(n), we find x(n) =Y (n)x() + Y (n) Y (k +1) 1 (B(k)x(k)+f(k)) where Y (n) =AY (n) forn and Y () = I. Therefore Y (n) =(I + A) n for n and Y (n)y (k +1) 1 =(I + A) n (I + A) (k+1) =(I + A) n k 1 = Y (n k 1) and x(n) =Y (n)x() + Y (n k 1)(B(k)x(k)+f(k)). By hypohesis, every Y (n) c 1, a consan, so, upon aking norms, we ge x(n) Y (n) x() + Y (n k 1) ( B(k) x(k) + f(k) ) c 1 x() +c 1 = f n + g k x(k) f(k) + c 1 B(k) x(k) where f n is he sum of he firs wo erms and g k = c 1 B(k). Here f n = c 1 x() +c 1 f(k) so c 1 x() +c 1 x(n) c 2 + k< f(k) =: c 2 <, c 1 B(k) x(k). By he Special Gronwall Inequaliy, x(n) c 2 exp c 1 B(k) c 2 exp showing ha he soluions are bounded. References k< c 1 A(k) A [1] Dieudonné, J., Foundaions of Modern Analysis, Academic Press, 196. [2] Sruble, Raimond A., Nonlinear Differenial Equaions, McGraw-Hill, New York, address: hole@gusavus.edu <,