DISCRETE GRONWALL LEMMA AND APPLICATIONS

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DISCRETE GRONWALL LEMMA AND APPLICATIONS JOHN M. HOLTE Variations of Gronwall s Lemma Gronwall s lemma, which solves a certain kind of inequality for a function, is useful in the theory of differential equations. Here is one version of it [1, p, 283]:. Gronwall s inequality. Let y(t, f(t, and g(t be nonnegative functions on [, T] having one-sided limits at every t [, T], and assume that for t T we have Then for t T we also have Proof. y(t f(t + y(t f(t + g(sy(s ds. ( g(sf(s exp g(u du ds. s Let I(t := g(sy(s ds and G(t := g(s ds. Now I (t = g(ty(t g(tf(t + g(ti(t outside a denumerable set, and d dt e G(t I(t = e G(t {I (t g(ti(t} e G(t g(tf(t, so e G(t I(t and y(t f(t + I(t f(t + e G(s g(sf(s ds. Therefore, I(t e G(t G(s g(sf(s ds. e G(t G(s g(sf(s ds Here is our first formulation of a discrete version of this lemma: 1. Preliminary Discrete Gronwall Lemma. If y n, f n, and g n are nonnegative sequences and (1 y n f n + g k y k for n, then (2 y n f n + f k g k exp( k<j<n g j for n. The proof that follows first gives the exact solution for y n when inequality in (1 is replaced by equality. Then it shows that any solution of the inequality is dominated by this exact solution, from which (2 quickly follows. First we prove a lemma giving a solution of the equality when f n 1. 1

2. Lemma. Let g n be a sequence. For n let (3 G n := (1 + g j. j<n Then G n = 1 + j<n g j G j for n, and, more generally, for k n, (4 G n = G k + g j G j. k j<n Proof. An empty product is 1, and an empty sum is, so the formulas hold for n = (and k =. Let m. Assuming the 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, G m+1 = G m + g m G m = G k + g j G j + g m G m = G k + k j<m conclusions follow by mathematical induction. +1 k j<m+1 g j G j. The 3. Proposition. Assume x n, f n, and g n are nonnegative 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. Let χ n := f n + f k g k G n /G k+1. We prove that x n = χ n for n by induction. For n = both sides equal f. Let m >. Assume x n = χ n for n < m. By

(5, x m = f m + g k x k = f m + g k χ k = f m + = f m + = f m + = f m + g k {f k + j<k g k f k + g k f k + g k f k + 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 + f m 1 g m 1 + = f 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 } 1 k<j<m f k g k G m /G k+1 = χ m by (4. The conclusion follows by induction. The following theorem actually gives a sharper result than the preliminary discrete Gronwall lemma. 4. Theorem: Sharp Gronwall Lemma. Assume y n, f n, and g n are nonnegative sequences and (7 y n f n + Then (8 y n f n + for n. g k y k for n. f k g k k<j<n (1 + g j Proof. Let x n be as in the previous proposition. Then an easy induction shows that y n x n for every n. Alternative Proof. For n = inequality (7 implies y f, whence inequality (8 holds for n =. Let m >. Assume that (8 holds for n < m. By hypothesis (7 and this

induction hypothesis, G (m j y m f m + f m + = f m + j<k<m j<m j<i<k g k y k g k ( f j g j G (m j where (by an implicit induction = 1 + g k (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+2 + + g m 1 (1 + g j+2 (1 + g m 2 = = j<i<m (1 + g i. Thus inequality (8 holds for n = m. By mathematical induction, inequality (8 holds for every n. Proof of the Discrete Gronwall Lemma. Use the inequality 1 + g j exp(g j in the previous theorem. 5. Another discrete Gronwall lemma Here is another form of Gronwall s lemma that is sometimes invoked in differential equations [2, pp. 48 49]: Let y and g be nonnegative integrable functions and c a nonnegative constant. If y(t c + g(sy(s ds for t, then ( y(t c exp g(s ds for t. Proposition: Special Gronwall Lemma. Let y n and g n be nonnegative sequences and c a nonnegative constant. If y n c + g k y k for n, then y n c j<n (1 + g j c exp ( j<n g j for n.

Proof. Assume the hypothesis and apply the Sharp Gronwall Lemma with 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 [empty product is 1] ( j<n g j (1 + g j } (1 + g j } [telescoping sum] [1 + g j e g j ]. Applications One area where Gronwall s inequality is used is the study of the asymptotic behavior of nonhomogeneous linear systems of differential equations. We are interested in obtaining discrete analogs. 6. First-order differential equations The special Gronwall lemma in the continuous case can be used to establish uniqueness of solutions of dy dt = f(t, y for t with y = y when t =. It can also be used to bound the difference between two solutions with different initial values. Details... 7. First-order difference equations The special discrete Gronwall lemma can similarly be applied to solutions of the general first-order difference equation y n = f(n, y n for n. Details... 8. A vector differential equation application of Gronwall s lemma One application of the less general Gronwall lemma is in the proof of the following theorem. [2, pp. 119-12] Let denote a norm for d-vectors and the corresponding operator norm for d d matrices. For example, x and A could be defined as the sums of the absolute values of their components. Theorem. Assume lim t A(t = A, a constant matrix. If all the solutions of the limiting homogeneous equation dx dt = Ax remain bounded as t, then the same is true of all solutions of the nonhomogeneous equation dx = A(tx + f(t dt

provided that and A(t A dt < f(t td <. 9. A vector difference equation application of the discrete Gronwall lemma Our analogous theorem is the following. Theorem. Assume that lim n A(n = A, a constant matrix. If all the solutions of the limiting homogeneous equation x(n = Ax(n for n remain bounded as n, then the same is true of all solutions of the nonhomogeneous equation (9 x(n = A(nx(n + f(n for n provided that and A(n A < n= f(n <. n= Proof. First consider x(n = Ax(n, i.e., x(n + 1 x(n = Ax(n. Since x(n + 1 = (I + Ax(n for n, we have x(n = (I + A n x( for n. By hypothesis, all these solutions remain bounded as n, so there must be a constant C 1 such that for n, Let B(n = A(n A and rewrite (9 as (I + A n C 1. (1 x(n = Ax(n + B(nx(n + f(n. If B(n =, then by backtracking we get x(n = (I + Ax(n 1 + f(n 1 = (I + A[(I + Ax(n 2 + f(n 2] + f(n 1 = (I + A 2 x(n 2 + (I + Af(n 2 + f(n 1 = (I + A 2 [(I + Ax(n 3 + f(n 3] + (I + Af(n 2 + f(n 1 = (I + A 3 x(n 3 + (I + A 2 f(n 3 + (I + Af(n 2 + f(n 1 =... = (I + A n x( + (I + A n 1 f( + (I + A n 2 f(1 + + f(n 1. Therefore, if x(n = Ax(n + f(n, then x(n = (I + A n x( + (I + A n 1 k f(k.

Now apply this result to (1 with B(nx(n + f(n playing the role of f(n : x(n = (I + A n x( + (I + A n 1 k [B(kx(k + f(k]. Taking norms, we get x(n (I + A n x( + whence, Therefore, where x(n C 1 x( + x(n f n + (I + A n 1 k { B(k x(k + f(k }, C 1 { B(k x(k + f(k }, f n = C 1 x( + g k x(k C 1 f(k and g n = C 1 B(k = C 1 A(k A. By hypothesis, there are constants C 2 and C 3 such that f(k C 2 and A(k A C 3, k= so f n C 1 x( +C 1 C 2 and, by Gronwall s lemma (the special case, ( x(n (C 1 x( +C 1 C 2 exp g j (C 1 x( +C 1 C 2 exp(c 2 C 3 <, showing that the solutions are bounded. j= k= I still need to find a nice discrete application that requires the strong version of the discrete Gronwall Lemma. References [1] Dieudonné, J., Foundations of Modern Analysis, Academic Press, 196. [2] Struble, Raimond A., Nonlinear Differential Equations, McGraw-Hill, New York, 1962. E-mail address: holte@gustavus.edu

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