Kenneth Kuttler

7.4. EXERCISES 185

7.4 Exercises1. Let f (x) = sin(1/x) for x ∈ (0,1] and let f (0) = 0. Show that f is Riemann Darboux

integrable. Is f piecewise continuous?

2. Show that if f is Riemann Darboux integrable and if F (x) =∫ x

a f (t)dt, then F ′ (x) =f (x) at every point x where f is continuous.

3. Show that if fn is Riemann Darboux integrable on [a,b], and fn → f uniformly on[a,b] , then f is also Riemann Darboux integrable and

∫ ba fn (x)dx →

∫ ba f (x)dx.

4. The first order linear initial value problem for the unknown function y is of the formy′ (x)+ p(x)y(x) = q(x) , y(0) = y0 where y0 is given and p(x) ,q(x) are given con-tinuous functions. How do you find y in this problem? This will be discussed in thisproblem. Incidentally, this is the most important equation in differential equationsand properly understood includes almost the entire typical undergraduate differen-tial equations course. Let P(x)≡

∫ x0 p(t)dt. Then multiply both sides by exp(P(x))

and when you do, show that you obtain ddx (exp(P)y)(x) = q(x)exp(P(x)) . Now

explain why, when you take an integral of both sides, you get exp(P(x))y(x)−y0 =∫ x0 q(t)exp(P(t))dt. Then

y(x) = exp(−P(x))y0 + exp(−P(x))∫ x

0q(t)exp(P(t))dt

5. One of the most important inequalities in differential equations is Gronwall’s in-equality. You have u(t) ≤ u0 +K

∫ t0 u(s)ds, t ≥ 0 where t → u(t) is some continu-

ous function usually nonnegative. Then you can conclude that u(t)≤ u0eKt . Explainwhy this is so. Hint: Let w(t) =

∫ t0 u(s)ds and write the inequality in terms of w and

its derivatives. Then use the technique of the previous problem involving integratingfactors.

6. A function f satisfies a Lipschitz condition if | f (x)− f (y)| ≤ K |x− y| . A stan-dard initial value problem is to find a function of t denoted as y such that y′ (t) =f (y(t)) , y(0) = y0 where y0 is a given value called an initial condition. Show thatthis initial value problem has a solution if and only if there is a solution to the integralequation

y(t) = y0 +∫ t

0f (y(s))ds, t ≥ 0 (7.4)

Hint: This is an application of theorems about continuity and the fundamental theo-rem of calculus.

7. Letting f be Lipschitz continuous as in 7.4, use Gronwall’s inequality of Problem 5,to show there is at most one function y which is a solution to the integral equation7.4. Hint: If y, ŷ both work, explain why |y(t)− ŷ(t)| ≤

∫ t0 K |y(s)− ŷ(s)|ds. Also

give a continuous dependence theorem in the case that you have y, ŷ solutions to

y(t) = y0 +∫ t

0f (y(s))ds, t ≥ 0 and ŷ(t) = ŷ0 +

∫ t

0f (ŷ(s))ds, t ≥ 0

respectively. Verify |y0 − ŷ0|eKt ≥ |y(t)− ŷ(t)| .

7.4. EXERCISES 1857.4 Exercises1. Let f (x) =sin(1/x) for x € (0, 1] and let f (0) = 0. Show that f is Riemann Darbouxintegrable. Is f piecewise continuous?2. Show that if f is Riemann Darboux integrable and if F (x) = 7 f (t)dt, then F’ (x) =f (x) at every point x where f is continuous.3. Show that if f, is Riemann Darboux integrable on [a,b], and f, > f uniformly on[a,b], then f is also Riemann Darboux integrable and [? fy (x) dx > f? f (x) dx.4. The first order linear initial value problem for the unknown function y is of the formy’ (x) + p(x) y(x) = ¢g(x), y(0) = yo where yo is given and p(x) ,q(x) are given con-tinuous functions. How do you find y in this problem? This will be discussed in thisproblem. Incidentally, this is the most important equation in differential equationsand properly understood includes almost the entire typical undergraduate differen-tial equations course. Let P(x) = fg p(t)dt. Then multiply both sides by exp (P (x))and when you do, show that you obtain 4 (exp (P) y) (x) = q(x) exp(P(x)). Nowexplain why, when you take an integral of both sides, you get exp (P (x)) y(x) —yo =Jo q(t) exp (P(t)) dt. Theny (x) = exp(—P(x))y0-+exp(—P(x)) | a(e)exp(P(®))at5. One of the most important inequalities in differential equations is Gronwall’s in-equality. You have u(t) <uo+K fj u(s)ds, t > 0 where t — u(t) is some continu-ous function usually nonnegative. Then you can conclude that u(t) < uge*’. Explainwhy this is so. Hint: Let w(t) = 5 u(s)ds and write the inequality in terms of w andits derivatives. Then use the technique of the previous problem involving integratingfactors.6. A function f satisfies a Lipschitz condition if |f(x)—f(y)| < K|x—y|. A stan-dard initial value problem is to find a function of ¢ denoted as y such that y’ (t) =f (v(t)), y(0) = yo where yo is a given value called an initial condition. Show thatthis initial value problem has a solution if and only if there is a solution to the integralequationv= ['Fr())ds, 020 (7.4)Hint: This is an application of theorems about continuity and the fundamental theo-rem of calculus.7. Letting f be Lipschitz continuous as in 7.4, use Gronwall’s inequality of Problem 5,to show there is at most one function y which is a solution to the integral equation7.4. Hint: If y,% both work, explain why |y(t) —¥(t)| < 5K ly(s) —¥(s)|ds. Alsogive a continuous dependence theorem in the case that you have y, f solutions toyi) =y0+ [ £0r(s))ds, 1 > 0.and 9(1) = 0+ [ F(5())ds, 00respectively. Verify |yo — S| e*’ > |y(r) —$(t)|.