One of the existence of solutions to fractional differential equations

HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS ——————–o0o——————— NGUYEN THI HAO ON THE EXISTENCE OF SOLUTIONS TO FRACTIONAL DIFFERENTIAL EQUATIONS GRADUATION THESIS Speciality: Analysis Hanoi - 2019 HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS ——————–o0o——————— NGUYEN THI HAO ON THE EXISTENCE OF SOLUTIONS TO FRACTIONAL DIFFERENTIAL EQUATIONS GRADUATION THESIS Speciality: Analysis Supervisor Dr Hoang The Tuan Hanoi - 2019 THESIS ASSURANCE The thesis was written on the basis of my study under the guidance of Dr Hoang The Tuan and my effort I have studied and presented the results from bibliographies The thesis does not coincide others Hanoi, May 2019 Student Nguyen Thi Hao THESIS ACKNOWLEDGMENTS This thesis is the final lesson after years I have learned at Hanoi Pedagogical University I would like to express my sincere gratitude to Dr Hoang The Tuan for his guidance and encouragement He always provides worth opinions and suggestions through all stages of this thesis He guided and supported me by valuable knowledge and explanation I also would like to express my gratitude to ThS Nguyen Phuong Dong for his experience and suggestions that help me to complete my thesis I also would like to thank all lecturers of Department of Mathematics - HPU2, especially Dr Tran Van Bang He inspires and finds the best opportunities for my class to study English It is the most important chance for me to change my mind My deepest gratitude goes to my family for their encouragement every time They always believe and help me to pursue my dreams Due to time, my capacity and conditions are limited, so the thesis can not avoid errors Then, I am looking forward to receiving valuable comments from teachers and friends to complete my thesis Once more times, thank you very much Author Contents Page Thesis Assurance Thesis Acknowledgements List of Symbols Preface Fractional Calculus 1.1 The basic idea 1.2 Riemann-Liouville Integral 1.3 Riemann - Liouville Derivative 15 1.4 Relations Between Riemann-Liouville Integral and Derivative 18 Existence and Uniqueness Results for Riemann-Liouville Fractional Differential Equations 21 2.1 Main result 21 2.2 Existence of Solutions 22 2.3 Uniqueness of Solutions 25 Bibliography 28 LIST OF SYMBOLS N Set of natural numbers R+ Set of strictly positive real numbers An , An [a, b] Set of functions with absolutely continuous derivative of order n − C, C [a, b] Set of continuous functions (cf.Definition 1.1.2) C k , C k [a, b] Set of functions with continuous k th derivative Hµ , Hµ [a, b] Holder Space(cf Definition 1.1.2) Lp , Lp [a, b] Lebesgue space (cf Definition 1.1.2) = supa≤x≤b |f (x)| ∞ Chebyshev norm; f p Lp norm (1 ≤ p < ∞); f ∞ p b |f a = (x)|p dx Ceiling function, x = {z ∈ Z : z ≥ x} Floor function, x = max {z ∈ Z : z ≤ x} 1/p Γ Euler’s Gamma function, (cf Definition 1.1.1) o, O Landau’s symbols D Differential operator, Df (x) = f (x) Dn n ∈ N : n-fold iterate of the differential operator D Dan n ∈ R+ : Riemann- Liouville fractional differential operator I Identity operator Ja Integral operator, Ja f (x) = Jan n ∈ N: n-fold iterate of the integral operator J a n ∈ R+ x f a (t) dt N: Riemann- Liouville fractional integral operator n = 0: identity operator PREFACE In the seventeenth century, Newton and Leibniz developed the foundations of differential and integral calculus In particular, in a letter to de l’Hospital, Leibniz introduced the symbol The question of de l’Hospital ”What does dn dxn f (x) mean if n = 21 ?” was the first occurrence of fractional calculus In the last few decades, mathematicians have studied and published the enormous numbers of very interesting and novel applications of fractional differential equations in physics, chemistry, engineering, finance and psychology In addition, some applications of fractional calculus within various fields of mathematics itself It turns out that many of these applications gave rise to a type of equations, but they had not been covered in the standard mathematical literature Leibniz was not able to find the answer of de l’Hospital’s question, except for the special case f (x) = x There are many possible generalizations of dn dxn f (x) to the case n ∈ / N, while Riemann- Liuoville derivative and Caputo’derivative are popular In the first half of the nineteenth centery, in works of Abel, Riemann and Liouville, the former concepts of fractional calculus was established Athough it leads to be difficult when applying it to ”real world” problems, the theories were studied and presented carefully The Riemann-Liouville idea is closed to others and is the basis of other results Based on the monograph by Kai Diethelm, in this thesis, we focuse on equations with Riemann-Liouville differential operators The structure of the thesis is arranged in the following way: • Chapter 1: Fractional Calculus, we introduce the fundamental concepts and definitions of fractional Riemann-Liouville differential and integral operators; • Chapter 2: Existence and Uniqueness results for Riemann-Liouville fractional differential equations Chapter Fractional Calculus In this chapter, the basic idea behind fractional calculus is related to a classical standard result from differential and integral calculus, the fundamental theorem We also recall definitions of fractional integral and differential operators 1.1 The basic idea Theorem 1.1.1 (Fundamental Theorem of Classical Calculus) Let f : [a, b] → R be a continuous function, and let F : [a, b] → R be defined by x f (t) dt F (x) := a Then, F is differentiable and F = f Therefore, we have a closed relation between differential operator and integral operator It is one of the goals of fractional calculus to retain this relation in a suitably generalized sense Hence there is also a need to deal with fractional integral operators and actually it turns out to be useful to discuss these first before coming to fractional differential operators Thus, it is an answer of de l’Hospital’s question It has proven to be convenient to use the notational conventions introduced in the following definition Definition 1.1.1 (i) By D, we denote the operator that maps a differentiable function onto its derivative, i.e Df (x) := f (x) (ii) By Ja , we denote the operator that maps a function f , assumed to be Riemann integrable on the compact interval [a, b], onto its primitive centered at a, i.e x Ja f (x) := f (t) dt a for a ≤ x ≤ b (iii) For n ∈ N we use the symbols Dn and Jan to denote the n-fold iterates of D and Ja , respectively, i.e we set D1 := D, Ja1 := Ja , Dn := DDn−1 and Jan := Ja Jan−1 for n ≥ Note 1.1.1 The key question is ”How can we extend the concept of Definition 1.1.1 to n∈ / N?” Once we will have provided such an extension, we need to ask for the mapping properties of the resulting operators and in particular, this includes the question for their domains and ranges Note that, from the Fundamental Theorem of Classical Calculus, we have a notation DJa f = f which implies that Dn Jan f = f for n ∈ N, i.e Dn is the left inverse of Jan in a suitable space of functions We wish to retain this property However, the conditions of the theorem are not straightforward to the fractional case n ∈ / N We give a generalization of the concepts in the definition 1.1.1 in next part Following the above outline, we begin with the integral operator Jan for n ∈ N Two following lemmas are well known in [4, Theorem 2.16] Lemma 1.1.2 Let f be Riemann integrable on [a, b] Then, for a ≤ x ≤ b and n ∈ N, we have Jan f (x) = (n − 1)! x (x − t)n−1 f (t) dt a Moreover, it is an immediate consequence of the fundamental theorem that the following relation holds for the operators D and Ja : Lemma 1.1.3 Let m, n ∈ N such that m > n, and let f be a function having the continuous n-th derivative on the interval [a, b] Then, Dn f = Dm Jam−n f Proof We have f = Dm−n Jam−n f Applying the operator Dn to both sides of this relation and using the fact that Dn Dm−n = Dm , the statement follows We recall here the definition of Gamma function Definition 1.1.2 The function Γ : (0, ∞) → R, defined by ∞ tx−1 e−t dt Γ (x) := is called Euler’s Gamma function (or Euler’s integral of the second kind) We have some properties of Gamma function Theorem 1.1.4 (i) The functional equation Γ (x + 1) = xΓ (x) holds for any x ∈ (0, ∞) (ii) For n ∈ N, Γ (n + 1) = n! The proof of this theorem is presented in [2, P 192] Before we start the main work, we introduce some function spaces in which we are going to discuss matters Definition 1.1.3 Let < µ ≤ 1, k ∈ N and p ≥ We define b f : [a, b] → R, f is measurable on [a, b] and Lp [a, b] := |f (x)|p dx < ∞ a L∞ [a, b] := {f : [a, b] → R, f is measurable and bounded on [a, b]} Hµ [a, b] := f : [a, b] → R, ∃c > such that ∀x, y ∈ [a, b] : |f (x) − f (y)| ≤ c|x − y|µ C k [a, b] := f : [a, b] → R, f has a continuous k th derivative C [a, b] := C [a, b] and H0 [a, b] := C [a, b] Definition 1.1.4 By H ∗ or H ∗ [a, b], we denote the set of functions f : [a, b] → R with the property that there exists some constants L > such that |f (x + h) − f (x)| ≤ L |h| ln |h|−1 whenever |h| < and x, x + h ∈ [a, b] When working in a Lebesgue space rather than in a space of continuous functions, we can still retain the main part of the statement of the fundamental theorem A proof of this theorem can be found in [3] for a ≤ x < a + h/2, and ∞ Jan f (x) = for a ≤ x < a + h In particular, (x − a)k+n k D f (a) Γ (k + + n) k=0 n Ja f is analytic in (a, a + h) Example 1.2.2 Let f (x) = exp (λx) with some λ > Compute J0n f (x) for n > In the case n ∈ N, we obviously have J0n f (x) = λ−n exp (λx) For the case n ∈ / N, we find ∞ J0n f (x) = J0n k=0 ∞ = k=0 (λ·)k (x) = k! ∞ k=0 λk n J (.)k (x) k! λk xk+n = λ−n Γ (k + n + 1) ∞ k=0 (λx)k+n Γ (k + n + 1) 1.3 Riemann - Liouville Derivative Having established these fundamental properties of Riemann- Liouville integral operators, we come to the corresponding differential operators Definition 1.3.1 (See [1]) Let n ∈ R+ and m = n The operator Dan defined by Dan f := Dm Jam−n f is called the Riemann - Liouville fractional differential operator of order n For n = 0, we set Da0 := I, the identity operator Lemma 1.3.1 Let n ∈ R+ and let m ∈ N such that m > n Then, Dan = Dm Jam−n Proof Our assumptions on m imply that m ≥ n Thus, Dm Jam−n = D n Dm− n m− n Ja Ja n −n = Dan in the view of the semigroup property of fractional integration Next, we present a very simple sufficient condition for the existence of Dan f Lemma 1.3.2 Let f ∈ A1 [a, b] and < n < Then, Dan f exists almost everywhere in [a, b] Moreover, Dan f ∈ Lp [a, b] for ≤ p < and Dan f (x) = Γ (1 − n) f (a) + (x − a)n 15 x a f (t) (x − t)−n dt Proof We use the definition of the Rienmann - Liouville differential operator and the fact that f ∈ A1 It follows that Dan f (x) = = = = x d Γ (1 − n) dx a x d Γ (1 − n) dx t f (a) + a f (u) du (x − t)−n dt a x d Γ (1 − n) dx Γ (1 − n) f (t) (x − t)−n dt f (a) a x dt + (x − t)n f (a) d n + dx (x − a) x a t a t f (u)(x − t)−n dudt a f (u)(x − t)−n dudt a By Fubini’s theorem, we may interchange the order of integration in the double integral This implies that Dan f (x) = Γ (1 − n) f (a) d n + dx (x − a) x f (u) a (x − u)1−n du 1−n The standard rules on the differentiation of parameter integrals give the desired representation The integrability statement is an immediate consequence of the representation using classical results from Lebesgue integration theory Example 1.3.1 Let f (x) = (x − a)β for some β > −1 and n > Then, Dan f (x) = D n n Ja−n f (x) = Γ (β + 1) D ( − a) Γ ( n − n + β + 1) n −n−β (x) Specifically, if n − β ∈ N, the right-hand side is the n -th derivative of a classical polynomial of degree n −(n − β) ∈ {0, 1, , n − 1}, and so the expression vanishes, i.e, Dan ( − a)n−m (x) = for all n > 0, m ∈ {1, 2, , n } On the other hand, if n − β ∈ / N, we find Dan ( − a)β (x) = Γ (β + 1) (x − a)β−n Γ (β + − n) Both these relations are straightforward generalization of what we know for integerorder derivatives Note that in the last expression the argument of Gamma function may be negative Theorem 1.3.3 Assume that n1 , n2 ≥ Moreover let φ ∈ L1 [a, b] and f = Jan1 +n2 φ, Dan1 Dan2 f = Dan1 +n2 f 16 Proof By our assumption on f and the definition of the Riemann-Liouville differential operator, we have Dan1 Dan2 f = Dan1 Dan2 Jan1 +n2 φ = D n1 J n1 −n1 D n2 Ja n2 −n2 n1 +n2 φ Ja The semigroup property of the integral operators allows us to rewrite this expression as Dan1 Dan2 f = D n1 J n1 −n1 D n2 Ja n2 +n1 φ = D n1 J n1 −n1 D n2 Ja J n1 φ The orders of the integral and differential operators involved are natural numbers, we n2 find that this is equivalent to Dan1 Dan2 f = D n1 J n1 −n1 n1 Ja φ =D n1 Ja n1 φ where we have once again the semigroup property of fractional integration We imply that Dan1 Dan2 f = φ The proof that Dan1 +n2 f = φ goes along similar lines The smoothness and zero condition in this theorem is not just a technicality The following examples show some cases where the condition is not satisfied They prove that an unconditional semigroup property of fractional differentiation in the RiemannLiouville sense does not hold Example 1.3.2 Consider example 1.3.1, let f (x) = x−1/2 and n1 = n2 = 1/2 Then, D0n1 f (x) = D0n2 f (x) = and D0n1 D0n2 f (x) = However, D0n1 +n2 f (x) = D1 f (x) = − 2x2/3 −1 Example 1.3.3 Let f (x) = x1/2 , n1 = 1/2 and n2 = 3/2 Consider Example 1.3.1 again, we have D0n1 f (x) = √ π and D0n2 f (x) = This implies D0n1 D0n2 f (x) = 0, but D0n2 D0n1 f (x) = −x−3/2 /4 = D2 f (x) = Dn1 +n2 f (x) In other words, the first of these two examples shows that it is possible to have D0n1 D0n2 = D0n2 D0n1 f (x) = Dn1 +n2 f (x) , 17 and in the second example, the case D0n1 D0n2 = D0n2 D0n1 f (x) = Dn1 +n2 f (x) holds 1.4 Relations Between Riemann-Liouville Integral and Derivative In this section we discuss on the relationship between the Riemann-Liouvill integral and derivative Theorem 1.4.1 Let n ≥ Then for f ∈ L1 [a, b], the following statement holds Dan Jan f = f almost everywhere Proof See [1, Theorem 2.14] By Theorem 1.4.1, we imply a very important result that Dan is the left inverse of Jan However, we can not claim that it is the right inverse We consider the following situation Theorem 1.4.2 Let n > If there exists some φ ∈ L1 [a, b] such that f = Jan φ, then Jan Dan f = f almost everywhere Proof This is an immediate consequence of Theorem 1.4.1, that Jan Dan f = Jan [Dan Jan φ] = Jan φ = f If f is not required in Theorem 1.4.2, then we consider one of different representation for Jan Dan f Proof The case n = is trivial for then Dan and Jan are both the identity operator For n > 0, let m = n Dan Jan f (x) = Dm Jam−n Jan f (x) = Dm Jam f (x) = f (x) 18 Theorem 1.4.3 Let n > and m = n + Assume that f is satisfied Jam−n f ∈ Am [a, b] Then, m−1 Jan Dan f (x) = f (x) − k=0 (x − a)n−k−1 lim Dm−k−1 Jam−n f (z) Γ (n − k) z→a+ Specifically, for < n < 1, we have Jan Dan f (x) = f (x) − (x − a)n−1 lim J 1−n f (z) Γ (n) z→a+ a Proof It is easy to imply that the right hand side exists (since assumption on f and the continuity of Dm−a Jam−n f ) Moreover, because of this assumption, there exists some φ ∈ L1 such that Dm−a Jam−n f = Dm−1 Jam−n f (a) + Ja1 φ This is a classical differential equation of order m − for Jam−n f Its solution is in the form m−1 Jam−n f (x) = k=0 (x − a)k lim Dk Jam−n f (z) + Jam φ (x) z→a+ k! Thus, by definition of Dan , Jan Dan f (x) = Jan Dm Jam−n f (x) = Jan Dm m−1 (.−a)k lim Dk Jam−n f k! z→a+ k=0 = Jan Dm Jam φ (x) + (z) + Jam φ (x) m−1 Jan Dm (.−a)k (x) k! k=0 lim Dk Jam−n f (z) k→a+ = Jan φ (x) Next, we apply the operator Dam−n to both side and find m−1 Jan Dan f (x) = Jan Dm ( − a)k (x) k! k=0 m−1 Jan Dm ( − a)k (x) = k=0 lim Dk Jam−n f (z) + Dm−n Jam φ (x) k→a+ k! lim Dk Jam−n f (z) + Da1 Ja1−m+n Jam φ (x) k→a+ Moreover, we have m−1 f (x) = k=0 (x − a)k+n−m lim Dk Jam−n f (z) + Jan φ (x) Γ (k + n − m + 1) k→a+ 19 Finally, we substitute k in the sum by m − k − 1, solve for Jan φ (x) and from above results, we imply that m−1 Jan Dan f (x) = Jan φ (x) = f (x) − k=0 (x − a)n−k−1 lim Dm−k−1 Jam−n f (z) Γ (n − k) k→a+ 20 Chapter Existence and Uniqueness Results for Riemann-Liouville Fractional Differential Equations The question of existence and uniqueness of solutions, i.e, the classical questions concerning ordinary differential equations involving fractional derivative is discussed by many mathematicians under other types of conditions The thesis focuses on equations with Riemann-Liouville differential operators The fundamental result presented and proved in this chapter is an existence and uniqueness theorem Without loss of generality, we assume in this result and in the ensuing developments that the fractional derivatives are developed at the point 2.1 Main result Our task in this chapter is to prove the following theorem Theorem 2.1.1 Let n > 0, n ∈ / N and m = n Moreover let K > 0, h∗ > and b1 , , bm ∈ R Define G :=    (x; y) ∈ R2 : ≤ x ≤ h∗ , y ∈ R x = and m xm−n y −   k=1    bk xm−k /Γ (n − k + 1) < K   , and assume that the function f : G −→ R is continuous and bounded in G It fulfills a Lipschitz condition with respect to the second variable, i.e there exists a constant L > such that for all (x, y1 ) and (x, y2 ) ∈ G, |f (x, y1 ) − f (x, y2 )| < L |y1 − y2 | Then, the differential equation D0n y (x) = f (x, y (x)) equipped with the initial conditions D0n−k y (0) = bk (k = 1, 2, , m − 1) , 21 lim J0m−n y (z) = bm z→0+ has a uniquely solution y ∈ C (0, h] where Γ (n + 1) K M ∗ h := h , h, 1/m with M := sup(x,z)∈G |f (x, z)| and h being an arbitrary positive number satisfying the constraint Γ (2n − m + 1) Γ (n − m + 1) L h< 1/n The result is very similar to the known classical results for first-order equations We shall first transform the initial value problem into an equivalent Volterra integral equation Then we are going to prove the existence and uniqueness of the solution of the fractional differential equation by a Picard -type iteration process The proof of Theorem 2.1.1 is the consequence of two lemmas below 2.2 Existence of Solutions Lemma 2.2.1 Assume the hypotheses of Theorem 2.1.1 and let h > The function y ∈ C (0, h] is a solution of the differential equation D0n y (x) = f (x, y (x)) , equipped with the initial conditions D0n−k y (0) = bk (k = 1, 2, , m − 1) , lim J0m−n y (z) = bm , z→0+ if and only id it is a solution of the Volterra integral equation m y (x) = k=1 x bk xn−k + Γ (n − k + 1) Γ (n) (x − t)n−1 f (t, y (t)) dt Proof If M = then f (x, y) = for all (x, y) ∈ G In this case, it is evident that m−1 the function y : [0, h] −→ R with y (x) = k=0 (k) y0 xk k! is a solution of the initial value problem Hence we conclude, as required, that is a solution exists in this case Otherwise, we consider the Volterra equation We introduce the polynomial T that satisfies the initial conditions, as follow m−1 T (x) := k=0 and the set U := {y ∈ C [0, h] : y − T ∞ xk (k) y k! ≤ K} It implies that U is closed and convex subset of the Banach space of all continuous functions on [0, h], equipped with the 22 Chebyshev norm Hence, U is a Banach space too Since the polynomial T is an element of U, we also see that U is non-empty On this set U, we define the operator A by x (Ay) (x) := T (x) + Γ (n) (x − t)n−1 f (t, y (t)) dt Using this operator, the equation whose solvability we need to prove, viz the Volterra equation, can be rewritten as y = Ay and thus, in order to prove our desired existence result, we have to show that A has a fixed point We therefore proceed by investigating the properties of the operator more closed Our first goal in this context is to show that Ay ∈ U for y ∈ U To this end we begin by notice that, for ≤ x1 ≤ x2 ≤ h, |(Ay) (x1 ) − (Ay) (x2 )| = Γ(n) x1 = Γ(n) x1 ≤ M Γ(n) (x1 − t)n−1 f (t, y (t)) dt − x1 x2 (x2 − t)n−1 f (t, y (t)) dt (x1 − t)n−1 − (x2 − t)n−1 f (t, y (t)) dt − (x1 − t)n−1 − (x2 − t)n−1 dt − x2 x1 x2 x1 (x2 − t)n−1 f (t, y (t)) dt (x2 − t)n−1 dt The second integral in the right hand side of the above equation has the value (x2 − x1 )n /n For the first integral, we look at the three cases n = 1, n < and n > separately In the first case n = 1, the integrand vanishes identically and hence the integral has the value zero Secondly, for n < 1, we have n−1 < and hence (x1 − t)n−1 ≥ (x2 − t)n−1 Thus, x1 x1 (x1 − t)n−1 − (x2 − t)n−1 dt = (x1 − t)n−1 − (x2 − t)n−1 dt = n x − xn2 + (x2 − x1 )n n ≤ (x2 − x1 )n n 23 Finally, if n > then (x1 − t)n−1 ≤ (x2 − t)n−1 , hence x1 (x1 − t)n−1 − (x2 − t)n−1 dt = = x1 (x1 − t)n−1 − (x2 − t)n−1 dt n x1 − xn2 + (x2 − x1 )n ≤ (x2 n − x1 n ) n n A combination of these results yields 2M (x − x1 )n Γ(n+1) M (x2 − x1 )n Γ(n+1) |(Ay) (x1 ) − (Ay) (x2 )| ≤ if n ≤ 1, + xn2 − xn1 if n > In either case, the expression on the right-hand side of the above inequality converges to as x2 −→ x1 that means Ay is a continuous function Moreover, for y ∈ U and x ∈ [0, h] we find |(Ay) (x) − T (x)| = Γ (n) ≤ x (x − t)n−1 f (t, y (t)) dt ≤ M xn Γ (n) 1 KΓ (n + 1) M xn ≤ M = K Γ (n + 1) Γ (n + 1) M Thus, we have shown that Ay ∈ U if y ∈ U , i.e, A maps the set U to itself Since we want to apply Schauder’s Fixed Point Theorem, we need to show that A (U ) := {Au : u ∈ U } is relatively compact set This can be done by means of the Arzela- Ascoli Theorem For z ∈ A (U ) we find that, for all x ∈ [0, h], |z (x)| = |(Ay) (x)| ≤ T ≤ T ∞+ ∞ + Γ (n) x (x − t)n−1 |f (t, y (t))| dt M hn ≤ T Γ (n + 1) ∞ +K which is the required boundedness property Moreover, the equicontinuity property can be derived from the inequality Specifically, for ≥ x1 ≥ x2 ≥ h, we have found in the case n ≥ that |Ay (x1 ) − Ay (x2 )| ≤ 2M (x2 − x1 )n Γ (n + 1) Thus, if |x2 − x1 | < δ, then |Ay (x1 ) − Ay (x2 )| ≤ 24 2M δn Γ (n + 1) Nothing that the expression on the right - hand side is independent of y, x1 and x2 , we see that the set A (U ) is equicontinuous Similarly, in the case n > we may use the Mean Value Theorem to conclude that |Ay (x1 ) − Ay (x2 )| ≤ M (x2 − x1 )n + xn2 − xn1 Γ (n + 1) M (x2 − x1 )n + n (x2 − x1 ) ξ n−1 Γ (n + 1) M ≤ (x2 − x1 )n + n (x2 − x1 ) hn−1 , Γ (n + 1) with some ξ ∈ [x1 , x2 ] ⊆ [0, h] Hence, if once again |x2 − x1 | < δ, then = |(Ay) (x1 ) − (Ay) (x2 )| ≤ M δ n + nδhn−1 Γ (n + 1) and the right-hand side is one more independent of y, x1 , x2 , proving the equicontinuity In either case Arzela- Ascoli Theorem implies that A (U ) is relatively compact, and hence Schauder’s Fixed Point Theorem asserts that A has a fixed point By construction, a fixed point of A is a solution of our initial value problem We note two important special case: the first of these states that under certain assumptions, the solution exists on the entire interval [0, h∗ ] 2.3 Uniqueness of Solutions Lemma 2.3.1 Under the assumptions of Theorem 2.1.1, the Volterra equation m bk xn−k + Γ (n − k + 1) Γ (n) y (x) = k=1 x (x − t)n−1 f (t, y (t)) dt possesses a unique determined solution y ∈ C (0, h] Proof We define the set m B := y ∈ C (0, h] : sup x m−n y (x) − 0