Hyers Ulam stability of first-order linear dynamic equations on time scales

Hyers Ulam stability of first-order linear dynamic equations on time scales Douglas R. Anderson Concordia College, Moorhead, Minnesota USA April 21, 2018, MAA-NCS@Mankato

Introduction In 1940, Stan Ulam posed the following problem concerning the stability of functional equations:

Introduction In 1940, Stan Ulam posed the following problem concerning the stability of functional equations: Give conditions in order for a linear mapping near an approximately linear mapping to exist.

Introduction In 1940, Stan Ulam posed the following problem concerning the stability of functional equations: Give conditions in order for a linear mapping near an approximately linear mapping to exist. The problem for the case of approximately additive mappings was solved by Hyers in 1941, who proved that the Cauchy equation is stable in Banach spaces.

Introduction In 1940, Stan Ulam posed the following problem concerning the stability of functional equations: Give conditions in order for a linear mapping near an approximately linear mapping to exist. The problem for the case of approximately additive mappings was solved by Hyers in 1941, who proved that the Cauchy equation is stable in Banach spaces. The result of Hyers was generalized by Rassias in 1978.

Introduction In 1940, Stan Ulam posed the following problem concerning the stability of functional equations: Give conditions in order for a linear mapping near an approximately linear mapping to exist. The problem for the case of approximately additive mappings was solved by Hyers in 1941, who proved that the Cauchy equation is stable in Banach spaces. The result of Hyers was generalized by Rassias in 1978. In 1993 Obloza was the first author who investigated the Hyers-Ulam stability of a differential equation.

Motivation Onitsuka and Shoji (2017) proved the Hyers Ulam stability of both homogeneous and non-homogeneous linear differential equations of first order, while Onitsuka (2017) proved the Hyers Ulam stability of the corresponding linear difference equation with step size h.

Motivation Onitsuka and Shoji (2017) proved the Hyers Ulam stability of both homogeneous and non-homogeneous linear differential equations of first order, while Onitsuka (2017) proved the Hyers Ulam stability of the corresponding linear difference equation with step size h. I emailed Onitsuka in Japan a month ago and wondered whether we could generalize his approach to handle all time scales, including the continuous and the discrete, simultaneously.

Motivation Onitsuka and Shoji (2017) proved the Hyers Ulam stability of both homogeneous and non-homogeneous linear differential equations of first order, while Onitsuka (2017) proved the Hyers Ulam stability of the corresponding linear difference equation with step size h. I emailed Onitsuka in Japan a month ago and wondered whether we could generalize his approach to handle all time scales, including the continuous and the discrete, simultaneously. In what follows we give some of our preliminary results on the Hyers Ulam stability of homogeneous linear dynamic equations of first order on all time scales.

Define Hyers-Ulam Stability (HUS) on T = R Consider on the time scale T = R, the interval I R, and the differential equation x (t) ax(t) = 0, t I. (1)

Define Hyers-Ulam Stability (HUS) on T = R Consider on the time scale T = R, the interval I R, and the differential equation x (t) ax(t) = 0, t I. (1) Let ε > 0 be given. If whenever φ C 1 (I ) satisfies φ (t) aφ(t) ε, t I there exists a solution x C 1 (I ) of (1) such that φ x Kε on I for some constant K > 0,

Define Hyers-Ulam Stability (HUS) on T = R Consider on the time scale T = R, the interval I R, and the differential equation x (t) ax(t) = 0, t I. (1) Let ε > 0 be given. If whenever φ C 1 (I ) satisfies φ (t) aφ(t) ε, t I there exists a solution x C 1 (I ) of (1) such that φ x Kε on I for some constant K > 0, then (1) has Hyers-Ulam stability I. The contant K is an HUS constant.

No HUS on R if a = 0 Example Let a = 0, and let ε > 0 be given. Note that the function φ(t) = εt, t R satisfies φ (t) = ε, t T. As x(t) c is the general solution to x (t) = 0, we see that lim φ(t) x(t), t so that x (t) ax(t) = 0 does not have Hyers Ulam stability on R when a = 0.

Onitsuka-Shoji Theorem for T = R Theorem Let a R\{0}. Then equation x (t) ax(t) = 0, t R has HUS with HUS constant 1/ a on R.

Onitsuka-Shoji Theorem for T = R Theorem Let a R\{0}. Then equation x (t) ax(t) = 0, t R has HUS with HUS constant 1/ a on R. Moreover, given the approximate solution φ, the solution x satisfying φ(t) x(t) ε/ a, t R is unique, and is given by ( lim φ(t)e at) e at : a > 0, (t x(t) = ) lim t φ(t)e at e at : a < 0.

Onitsuka Theorem for T = hz Theorem Let a R\{0} with a > 1/h. Then equation h x(t) ax(t) = 0, t hz, h x(t) = has HUS with HUS constant 1/ a on hz. x(t + h) x(t) h

Onitsuka Theorem for T = hz Theorem Let a R\{0} with a > 1/h. Then equation h x(t) ax(t) = 0, t hz, h x(t) = x(t + h) x(t) h has HUS with HUS constant 1/ a on hz. Moreover, given the approximate solution φ, the solution x satisfying φ(t) x(t) ε/ a, t hz is unique, and is given by ( lim φ(t)(1 + ah) t/h) (1 + ah) t/h : a > 0, (t x(t) = ) lim φ(t)(1 + ah) t/h (1 + ah) t/h : 1/h < a < 0. t

No HUS on hz if a = 0 or a = 2/h Example If a = 0, consider the function φ(t) = εt satisfying h φ(t) = ε for all t hz. Since x c is the general solution of h x(t) = 0, we have φ(t) x(t) as t.

No HUS on hz if a = 0 or a = 2/h Example If a = 0, consider the function φ(t) = εt satisfying h φ(t) = ε for all t hz. Since x c is the general solution of h x(t) = 0, we have φ(t) x(t) as t. If a = 2/h, the function φ(t) = εt( 1) t/h satisfies h φ(t) + 2φ(t)/h = ε, t hz. Since the general solution of h x(t) + 2x(t)/h = 0 is x(t) = c( 1) t/h, we see that φ(t) x(t) as t.

Comparing T = R and T = hz Remark Note that and for t hz, d dt eat = ae at, t R h (1 + ah) t/h = (1 + ah)(t+h)/h (1 + ah) t/h h = a(1 + ah) t/h, so that exponential functions with respect to the derivative seem to be key. Also,

Comparing T = R and T = hz Remark Note that and for t hz, d dt eat = ae at, t R h (1 + ah) t/h = (1 + ah)(t+h)/h (1 + ah) t/h h = a(1 + ah) t/h, so that exponential functions with respect to the derivative seem to be key. Also, On R, no HUS if a = 0.

Comparing T = R and T = hz Remark Note that and for t hz, d dt eat = ae at, t R h (1 + ah) t/h = (1 + ah)(t+h)/h (1 + ah) t/h h = a(1 + ah) t/h, so that exponential functions with respect to the derivative seem to be key. Also, On R, no HUS if a = 0. On hz, no HUS if a = 0 and a = 2/h;

Comparing T = R and T = hz Remark Note that and for t hz, d dt eat = ae at, t R h (1 + ah) t/h = (1 + ah)(t+h)/h (1 + ah) t/h h = a(1 + ah) t/h, so that exponential functions with respect to the derivative seem to be key. Also, On R, no HUS if a = 0. On hz, no HUS if a = 0 and a = 2/h; On hz, exponential DNE if a = 1/h.

Time Scales A time scale T is any nonempty closed subset of R, with jump operator σ : T T σ(t) = inf{s T : s > t} and forward graininess µ(t) := σ(t) t.

Time Scales A time scale T is any nonempty closed subset of R, with jump operator σ : T T σ(t) = inf{s T : s > t} and forward graininess µ(t) := σ(t) t. For f : T R and t T, the delta derivative of f at t is provided the limit exists. x x(σ(t)) x(s) (t) := lim, s t σ(t) s

Time Scales A time scale T is any nonempty closed subset of R, with jump operator σ : T T σ(t) = inf{s T : s > t} and forward graininess µ(t) := σ(t) t. For f : T R and t T, the delta derivative of f at t is x x(σ(t)) x(s) (t) := lim, s t σ(t) s provided the limit exists. For T = R, we have f = f, the usual derivative, and for T = hz we have the forward difference operator, f (t) = h f (t) = [f (t + h) f (t)]/h.

Definition A function p : T R is regressive provided 1 + µ(t)p(t) 0, t T. Let R := {p C rd (T; R) : 1 + µ(t)p(t) 0, t T}. Also, p R + iff 1 + µ(t)p(t) > 0, t T.

Definition A function p : T R is regressive provided 1 + µ(t)p(t) 0, t T. Let R := {p C rd (T; R) : 1 + µ(t)p(t) 0, t T}. Also, p R + iff 1 + µ(t)p(t) > 0, t T. Definition If p R, t 0 T, define the generalized exponential function e p (t, t 0 ) to be the unique solution of the initial value problem x = p(t)x, x(t 0 ) = 1.

Many of the properties of this generalized exponential function e p (t, t 0 ) listed below are employed throughout this work. Theorem If p, q R and s, t T, then 1. e 0 (t, s) 1 and e p (t, t) 1; 2. e p (σ(t), s) = (1 + µ(t)p(t))e p (t, s); 3. 1 e p(t,s) = e p(t, s), where p := 4. e p (t, s) = 1 e p(s,t) = e p(s, t); 5. e p (t, s)e p (s, r) = e p (t, r); p 1+µp ; 6. If p R +, then e p (t, s) > 0 for all t T.

New Theorem for T Theorem Let a R\{0} with 1 + aµ(t) > 0. Then equation x (t) ax(t) = 0, t T, (2) has HUS with HUS constant 1/ a on T.

New Theorem for T Theorem Let a R\{0} with 1 + aµ(t) > 0. Then equation x (t) ax(t) = 0, t T, (2) has HUS with HUS constant 1/ a on T. Moreover, given the approximate solution φ, the solution x satisfying φ(t) x(t) ε/ a, t T is unique, and is given by ( ) lim φ(t)e a(t) e a (t) : a > 0, t x(t) = ( ) lim φ(t)e a(t) e a (t) : 1/µ(t) < a < 0. t

Example: T = N 0 = {0, 1, 2, 3, } For n T we have µ ( n ) = n + 1 n,

Example: T = N 0 = {0, 1, 2, 3, } For n T we have µ ( n ) = n + 1 n, and the exponential function e a ( n ) is given by e a ( n ) = n k=1 ( ( k )) 1 + aµ 1, n N 0.

Example: T = N 0 = {0, 1, 2, 3, } For n T we have µ ( n ) = n + 1 n, and the exponential function e a ( n ) is given by e a ( n ) = n k=1 ( ( k )) 1 + aµ 1, n N 0. Let ε > 0 be given and let a > 0. Suppose φ ( n ) aφ ( n ) ε, n N 0.

Example: T = N 0 = {0, 1, 2, 3, } For n T we have µ ( n ) = n + 1 n, and the exponential function e a ( n ) is given by e a ( n ) = n k=1 ( ( k )) 1 + aµ 1, n N 0. Let ε > 0 be given and let a > 0. Suppose φ ( n ) aφ ( n ) ε, n N 0. Then by the theorem we know that there exists a unique solution x ( n ) { := lim φ ( n ) ( ) } ( ) e a n e a n n such that φ ( n ) x ( n ) ε/a for all n N 0.

Negative Exponential Function ( Let k 0 N and a 1 µ( k 0), 1 µ( k 0 1) ). Then ( ) ( ) 1+aµ(0) < < 1+aµ k0 1 < 0 < 1+aµ k0 < < 1,

Negative Exponential Function ( Let k 0 N and a 1 µ( k 0), 1 µ( k 0 1) ). Then ( ) ( ) 1+aµ(0) < < 1+aµ k0 1 < 0 < 1+aµ k0 < < 1, and we have e a ( n ) = n k=1 ( ( k )) { < 0 : k 0 odd, 1 + aµ 1 > 0 : k 0 even, so that the theorem does not apply, as a R +.

T = N 0 Example Continued If we can show that is bounded, namely that s=0 t 0 e a (t, σ(s)) s n 1 ( ) n s 1 1 s + 1 + s 1 + a <, k + k 1 k=1 then we speculate HUS for this time scale. Numerical evidence in the next slide with a = 1.25 suggests this indeed holds, and we would have HUS for T = N for these a < 0 values.

Numerical Evidence of HUS Plot of n 1 ( ) n s 1 1 s + 1 + s 1 + a k + k 1 s=0 for a = 1.25. 1.0 y k=1 y(n)= 0 n ea (n,σ(s)) Δs 0.8 0.6 0.4 0.2 n 5 10 15

Example: T = P 1 β,β for 0 < β < 1 Simple electric circuit with capacitor discharge: T = P 1 β,β = k Z[k, k + 1 β], has graininess µ(t) = { 0 : k t < k + 1 β β : t = k + 1 β,

Example: T = P 1 β,β for 0 < β < 1 Simple electric circuit with capacitor discharge: T = P 1 β,β = k Z[k, k + 1 β], has graininess µ(t) = { 0 : k t < k + 1 β β : t = k + 1 β, and for a R\{0, 1/β} the exponential function e a (t) is given by e a (t) = (aβ + 1) k e a(t kβ).

Example: T = P 1 β,β for 0 < β < 1 Simple electric circuit with capacitor discharge: T = P 1 β,β = k Z[k, k + 1 β], has graininess µ(t) = { 0 : k t < k + 1 β β : t = k + 1 β, and for a R\{0, 1/β} the exponential function e a (t) is given by e a (t) = (aβ + 1) k e a(t kβ). If a > 1/β,

Example: T = P 1 β,β for 0 < β < 1 Simple electric circuit with capacitor discharge: T = P 1 β,β = k Z[k, k + 1 β], has graininess µ(t) = { 0 : k t < k + 1 β β : t = k + 1 β, and for a R\{0, 1/β} the exponential function e a (t) is given by e a (t) = (aβ + 1) k e a(t kβ). If a > 1/β, then x ax = 0 has HUS with HUS constant 1/ a by the theorem.

T = k Z[k, k + 1 β] e a (t) = (aβ + 1) k e a(t kβ).

T = k Z[k, k + 1 β] Note that if a = 1 (β 1)β, then e a (k) = e a (t) = (aβ + 1) k e a(t kβ). ( ) β k β 1 e 1/β, k Z.

T = k Z[k, k + 1 β] Note that if a = 1 (β 1)β, then e a (k) = e a (t) = (aβ + 1) k e a(t kβ). ( ) β k β 1 e 1/β, k Z. Thus if β = 0.7821882943 and a = 5.869586019, then e a (k, 0) = ( 1) k, k Z.

T = k Z[k, k + 1 β] Note that if a = 1 (β 1)β, then e a (k) = e a (t) = (aβ + 1) k e a(t kβ). ( ) β k β 1 e 1/β, k Z. Thus if β = 0.7821882943 and a = 5.869586019, then e a (k, 0) = ( 1) k, k Z. For ε > 0, fixed k Z and t [k, k + 1 β], ( ) β k e a (t) = e t kβ (β 1)β, k t k + 1 β, β 1 is bounded and alternates sign with k.

T = P 1 β,β Example Concluded For k N 0, satisfies φ (t) aφ(t) = φ(t) := εte a (t), t [k, k + 1 β] {ε ( 1) k e t k β(β 1) ε ( 1) k : t = k, : t (k, k + 1 β], so that φ (t) aφ(t) ε for all t [k, k + 1 β].

T = P 1 β,β Example Concluded For k N 0, satisfies φ (t) aφ(t) = φ(t) := εte a (t), t [k, k + 1 β] {ε ( 1) k e t k β(β 1) ε ( 1) k : t = k, : t (k, k + 1 β], so that φ (t) aφ(t) ε for all t [k, k + 1 β]. But the general solution of x ax = 0 is x(t) = ce a (t), we have that φ(k) x(k) = εk c as k, so no HUS in this case.

Theorem Let a R but a R +, and assume T is a time scale such that τ := min T exists and T is unbounded above. 1. If t K := sup e a (t, σ(s)) s <, t T τ then (2) has Hyers Ulam stability on [τ, ) T with HUS constant K. 2. If there exist constants 0 < m < M such that 0 < m e a (t, τ ) M, t T, then (2) does not have HUS.

Thanks to my co-author Masakazu Onitsuka Okayama University of Science Japan.

Selected Bibliography [1] S. Hilger, Analysis on measure chains a unified approach to continuous and discrete calculus, Results Math., 18 (1990) 18 56. [2] D. H. Hyers, On the stability of the linear functional equation, Proc. Nat. Acad. Sci. U.S.A. 27 (1941) 222 224. [3] Th. M. Rassias, On the stability of linear mapping in Banach spaces, Proc. Amer. Math. Soc. 72 (1978) 297 300. [4] S. M. Ulam, A Collection of the Mathematical Problems, Interscience, New York, 1960.

Thank you for your attention.