J. Numer. Anal. Approx. Theory, vol. 44 (2015) no. 2, pp. 113–126 ictp.acad.ro/jnaat
SEMILOCAL CONVEGENCE OF NEWTON-LIKE METHODS UNDER GENERAL CONDITIONS, WITH APPLICATIONS
IN FRACTIONAL CALCULUS
GEORGE A. ANASTASSIOU1 and IOANNIS K. ARGYROS2
Abstract. We present a semilocal convergence study of Newton-like methods on a generalized Banach space setting to approximate a locally unique zero of an operator. Earlier studies such as [5, 6, 7, 14] require that the operator involved is Fr´echet-differentiable. In the present study we assume that the operator is only continuous. This way we extend the applicability of Newton-like methods to include fractional calculus and problems from other areas. Some applications include fractional calculus involving the Riemann-Liouville fractional integral and the Caputo fractional derivative. Fractional calculus is very important for its applications in many applied sciences.
MSC 2010. 65G99, 65H10, 26A33, 47J25, 47J05.
Keywords. Generalized Banach space, Newton-like method, semilocal conver- gence, Riemann-Liouville fractional integral, Caputo fractional derivative.
1. INTRODUCTION
We present a semilocal convergence analysis for Newton-like methods on a generalized Banach space setting to approximate a zero of an operator. The semilocal convergence is, based on the information around an initial point, to give conditions ensuring the convergence of the method. A generalized norm is defined to be an operator from a linear space into a partially order Banach space (to be precised in section 2). Earlier studies such as [5, 6, 7, 14] for Newton’s method have shown that a more precise convergence analysis is ob- tained when compared to the real norm theory. However, the main assumption is that the operator involved is Fr´echet-differentiable. This hypothesis limits the applicability of Newton’s method. In the present study we only assume the continuity of the operator. This may expand the applicability of these methods.
The rest of the paper is organized as follows: section 2 contains the basic concepts on generalized Banach spaces and auxiliary results on inequalities
1 Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A., e-mail: [email protected].
2 Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505- 6377, USA, e-mail: [email protected].
and fixed points. In section 3 we present the semilocal convergence analysis of Newton-like methods. Finally, in the concluding sections 4-5, we present special cases and applications in fractional calculus.
2. GENERALIZED BANACH SPACES
We present some standard concepts that are needed in what follows to make the paper as self contained as possible. More details on generalized Banach spaces can be found in [5, 6, 7, 14], and the references there in.
LetXbe a linear space. A subset CofX is called a cone ifC+C⊆Cand αC ⊆ C for α > 0. The cone C is proper if C ∩(−C) = {0}. The relation
”≤” defined by
x≤y if and only if y−x∈C
is a partial ordering onC which is compatible with the linear structure of this space. Two elements x and y of X are called comparable if either x ≤ y or y≤xholds. The spaceXendowed with the above relation is called a partially ordered linear space (POL-space). If X has a topology compatible with its linear structure and if the cone C is closed in that topology then X is called a partially ordered topological space (POTL-space).
We remark that in a POTL-space the intervals [a, b] ={x:a≤x ≤b} are closed sets. A stronger connection is considered in the following definitions:
Definition 2.1. A POTL-space is called normal if, given a local base V for the topology, there exists a positive number η so that if 0 ≤ z ∈ U ⊆ V then[0, z]⊂ηU.
Definition 2.2. A POTL-space is called regular if every order bounded increasing sequence has a limit.
If the topology of a POTL-space is given by a norm then this space is called a partially ordered normed space (PON-space). If a PON-space is complete with respect to its topology then it is called a partially ordered Banach space (POB-space). According to Definition 2.1 a PON-space is normal if and only if there exists a positive number α such that
kxk ≤αkyk, for all x, y∈X with 0≤x≤y.
Let us note that any regular POB-space is normal. The reverse is not true.
For example, the spaceC[0,1] of all continuous real functions defined on [0,1], ordered by the cone of nonnegative functions, is normal but is not regular. All finite dimensional POTL-spaces are both normal and regular.
Definition2.3. A generalized Banach space is a triplet(X,(E, K,k·k),/·/) such that
(i) X is a linear space over R(C).
(ii) E= (E, K,k·k) is a partially ordered Banach space, i.e.
(ii1) (E,k·k) is a real Banach space,
(ii2) E is partially ordered by a closed convex cone K, (ii3) The norm k·kis monotone on K.
(iii) The operator/·/ :X→K satisfies
/x/ = 0⇔x= 0, /θx/ =|θ|/x/,
/x+y/≤/x/ + /y/, for each x, y∈X, θ∈R(C).
(iv) X is a Banach space with respect to the induced norm k·ki :=k/·/k. Remark2.4. The operator /·/ is called a generalized norm. In view of (iii) and (ii3)k·ki , is a real norm. In the rest of this paper all topological concepts
will be understood with respect to this norm.
Definition 2.5. Let L Xj, Y stand for the space of j-linear symmetric and bounded operators from Xj toY, whereX andY are Banach spaces. For X, Y partially orderedL+ Xj, Ystands for the subset of monotone operators P such that
(2.1) 0≤ai ≤bi ⇒P(a1, ..., aj)≤P(b1, ..., bj).
Definition 2.6. The set of bounds for an operator Q ∈ L(X, X) on a generalized Banach space (X, E,/·/)is defined to be:
(2.2) B(Q) :={P ∈L+(E, E), /Qx/≤P/x/ for each x∈X}. Let D⊂X andT :D→Dbe an operator. If x0 ∈Dthe sequence {xn}given by
(2.3) xn+1:=T(xn) =Tn+1(x0) is well defined. We write in case of convergence (2.4) T∞(x0) := lim (Tn(x0)) = lim
n→∞xn. We need some auxiliary results on inequations.
Lemma2.7. Let (E, K,k·k)be a partially ordered Banach space, ξ∈K and M, N ∈L+(E, E).
(i) Suppose there existsr ∈K such that
(2.5) R(r) := (M+N)r+ξ≤r
and
(2.6) (M+N)kr→0 as k→ ∞.
Then,b:=R∞(0) is well defined, satisfies the equation t=R(t) and is smaller than any solution of the inequalityR(s)≤s.
(ii) Suppose there exists q ∈ K and θ ∈(0,1) such that R(q) ≤ θq, then there existsr ≤q satisfying(i).
Proof. (i) Define sequence {bn} by bn = Rn(0). Then, we have by (2.5) thatb1 =R(0) =ξ ≤r⇒b1 ≤r. Suppose thatbk≤r for each k= 1,2, ..., n.
Then, we have by (2.5) and the inductive hypothesis that bn+1 =Rn+1(0) = R(Rn(0)) = R(bn) = (M+N)bn+ξ ≤ (M+N)r+ξ ≤ r ⇒ bn+1 ≤ r.
Hence, sequence {bn} is bounded above by r. Set Pn =bn+1−bn. We shall show that
(2.7) Pn≤(M+N)nr for each n= 1,2, ...
We have by the definition ofPn and (2.6) that P1 =R2(0)−R(0) =R(R(0))−R(0)
=R(ξ)−R(0) = Z 1
0
R0(tξ)ξdt≤ Z 1
0
R0(ξ)ξdt
≤ Z 1
0
R0(r)rdt≤(M +N)r,
which shows (2.7) for n = 1. Suppose that (2.7) is true for k = 1,2, ..., n.
Then, we have in turn by (2.6) and the inductive hypothesis that Pk+1 =Rk+2(0)−Rk+1(0) =Rk+1(R(0))−Rk+1(0)
=Rk+1(ξ)−Rk+1(0) =RRk(ξ)−RRk(0)
= Z 1
0
R0Rk(0) +tRk(ξ)−Rk(0) Rk(ξ)−Rk(0)dt≤
≤R0Rk(ξ) Rk(ξ)−Rk(0)=R0Rk(ξ) Rk+1(0)−Rk(0)
≤R0(r)Rk+1(0)−Rk(0)≤(M+N) (M+N)kr= (M+N)k+1r, which completes the induction for (2.7). It follows that {bn} is a complete sequence in a Banach space and as such it converges to some b. Notice that R(b) =Rlim
n→∞Rn(0)= lim
n→∞Rn+1(0) =b⇒bsolves the equationR(t) =t.
We have that bn ≤ r ⇒ b ≤ r, where r a solution of R(r) ≤ r. Hence, b is smaller than any solution ofR(s)≤s.
(ii) Define sequences{vn},{wn}by v0 = 0,vn+1=R(vn), w0 =q,wn+1= R(wn). Then, we have that
0≤vn≤vn+1≤wn+1≤wn≤q, (2.8)
wn−vn≤θn(q−vn)
and sequence{vn} is bounded above byq. Hence, it converges to somer with r ≤ q. We also get by (2.8) that wn−vn → 0 as n → ∞ ⇒ wn → r as
n→ ∞.
We also need the auxiliary result for computing solutions of fixed point problems.
Lemma 2.8. Let (X,(E, K,k·k),/·/) be a generalized Banach space, and P ∈ B(Q) be a bound for Q ∈ L(X, X). Suppose there exists y ∈ X and q ∈K such that
(2.9) P q+ /y/≤q and Pkq →0, as k→ ∞.
Then, z =T∞(0), T(x) :=Qx+y is well defined and satisfies: z=Qz+y and/z/≤P/z/ + /y/≤q. Moreover,z is the unique solution in the subspace {x∈X|∃ θ∈R:{x} ≤θq}.
The proof can be found in [14, Lemma 3.2].
3. SEMILOCAL CONVERGENCE
Let (X,(E, K,k·k),/·/) and Y be generalized Banach spaces, D ⊂ X an open subset, G:D →Y a continuous operator and A(·) : D→ L(X, Y). A zero of operator G is to be determined by a Newton-like method starting at a point x0 ∈ D. The results are presented for an operator F = J G, where J ∈L(Y, X). The iterates are determined through a fixed point problem:
xn+1 =xn+yn, A(xn)yn+F(xn) = 0 (3.1)
⇔yn=T(yn) := (I−A(xn))yn−F(xn). LetU(x0, r) stand for the ball defined by
U(x0, r) :={x∈X : /x−x0/≤r}
for somer ∈K.
Next, we present the semilocal convergence analysis of Newton-like method (3.1) using the preceding notation.
Theorem 3.1. Let F :D⊂X →X, A(·) :D →L(X, X) and x0 ∈D be as defined previously. Suppose:
(H1) There exists an operatorM ∈B(I−A(x)) for each x∈D.
(H2) There exists an operatorN ∈L+(E, E) satisfying for eachx, y∈D /F(y)−F(x)−A(x) (y−x)/≤N/y−x/.
(H3) There exists a solution r∈K of
R0(t) := (M+N)t+ /F(x0)/≤t.
(H4) U(x0, r)⊆D.
(H5) (M+N)kr →0 as k→ ∞.
Then, the following hold:
(C1) The sequence {xn} defined by
xn+1 =xn+Tn∞(0),
Tn(y) := (I−A(xn))y−F(xn) (3.2)
is well defined, remains in U(x0, r) for each n = 0,1,2, ... and con- verges to the unique zero of operatorF in U(x0, r).
(C2) An apriori bound is given by the null-sequence{rn} defined by r0:=r and for eachn= 1,2, ...
rn=Pn∞(0), Pn(t) =M t+N rn−1.
(C3) An a posteriori bound is given by the sequence{sn} defined by sn:=R∞n (0), Rn(t) = (M +N)t+N an−1,
bn:= /xn−x0/≤r−rn≤r, where
an−1:= /xn−xn−1/, for each n= 1,2, ...
Proof. Let us define for eachn∈Nthe statement:
(In) xn∈X andrn∈K are well defined and satisfy rn+an−1≤rn−1.
We use induction to show (In). The statement (I1) is true: By Lemma 2.7 and (H3), (H5) there exists q≤r such that:
M q+ /F(x0)/ =q and Mkq≤Mkr→0 ask→ ∞.
Hence, by Lemma 2.8x1 is well defined and we havea0≤q. Then, we get the estimate
P1(r−q) =M(r−q) +N r0
≤M r−M q+N r=R0(r)−q
≤R0(r)−q =r−q.
It follows with Lemma 2.7 that r1 is well defined and r1+a0≤r−q+q =r=r0.
Suppose that (Ij) is true for eachj = 1,2, ..., n.We need to show the existence of xn+1 and to obtain a bound q foran. To achieve this notice that:
M rn+N(rn−1−rn) =M rn+N rn−1−N rn=Pn(rn)−N rn≤rn. Then, it follows from Lemma 2.7 that there existsq ≤rn such that (3.3) q =M q+N(rn−1−rn) and (M+N)kq →0, ask→ ∞.
By (Ij) it follows that bn= /xn−x0/≤
n−1
X
j=0
aj ≤
n−1
X
j=0
(rj −rj+1) =r−rn≤r.
Hence,xn∈U(x0, r)⊂Dand by (H1)M is a bound forI−A(xn). We can write by (H2) that
/F(xn)/ = /F(xn)−F(xn−1)−A(xn−1) (xn−xn−1)/
≤N an−1 ≤N(rn−1−rn). (3.4)
It follows from (3.3) and (3.4) that
M q+ /F(xn)/≤q.
By Lemma 2.8,xn+1 is well defined andan≤q ≤rn. In view of the definition of rn+1 we have that
Pn+1(rn−q) =Pn(rn)−q=rn−q, so that by Lemma 2.7, rn+1 is well defined and
rn+1+an≤rn−q+q=rn,
which proves (In+1). The induction for (In) is complete. Let m≥n, then we obtain in turn that
(3.5) /xm+1−xn/≤
m
X
j=n
aj ≤
m
X
j=n
(rj−rj+1) =rn−rm+1 ≤rn. Moreover, we get inductively the estimate
rn+1=Pn+1(rn+1)≤Pn+1(rn)≤(M+N)rn≤...≤(M+N)n+1r.
It follows from (H5) that {rn} is a null-sequence. Hence, {xn} is a complete sequence in a Banach space X by (3.5) and as such it converges to somex∗ ∈ X. By letting m→ ∞in (3.5) we deduce that x∗ ∈U(xn, rn). Furthermore, (3.4) shows that x∗ is a zero ofF. Hence, (C1) and (C2) are proved.
In view of the estimate
Rn(rn)≤Pn(rn)≤rn
the apriori, bound of (C3) is well defined by Lemma 2.7. That issn is smaller in general thanrn. The conditions of Theorem 3.1 are satisfied forxnreplacing x0. A solution of the inequality of (C2) is given by sn (see (3.4)). It follows from (3.5) that the conditions of Theorem 3.1 are easily verified. Then, it follows from (C1) thatx∗ ∈U(xn, sn) which proves (C3).
In general the a posteriori estimate is of interest. Then, condition (H5) can be avoided as follows:
Proposition 3.2. Suppose: condition(H1) of Theorem 3.1 is true.
(H03) There existss∈K, θ∈(0,1)such that
R0(s) = (M+N)s+ /F(x0)/≤θs.
(H04) U(x0, s)⊂D.
Then, there exists r ≤s satisfying the conditions of Theorem 3.1. Moreover, the zero x∗ of F is unique in U(x0, s).
Remark3.3. (i) Notice that by Lemma 2.7R∞n (0) is the smallest solution of Rn(s) ≤s. Hence any solution of this inequality yields on upper estimate forR∞n (0). Similar inequalities appear in (H2) and (H02).
(ii) The weak assumptions of Theorem 3.1 do not imply the existence of A(xn)−1. In practice the computation of Tn∞(0) as a solution of a linear
equation is no problem and the computation of the expensive or impossible to compute in general A(xn)−1 is not needed.
(iii) We can use the following result for the computation of the a posteriori estimates. The proof can be found in [14, Lemma 4.2] by simply exchanging
the definitions ofR.
Lemma 3.4. Suppose that the conditions of Theorem 3.1 are satisfied. If s∈Kis a solution ofRn(s)≤s, thenq :=s−an∈Kand solvesRn+1(q)≤q.
This solution might be improved by Rkn+1(q)≤q for each k= 1,2, ....
4. SPECIAL CASES AND APPLICATIONS
Application4.1. The results obtained in earlier studies such as [5, 6, 7, 14]
require that operatorF (i.e. G) is Fr´echet-differentiable. This assumption lim- its the applicability of the earlier results. In the present study we only require that F is a continuous operator. Hence, we have extended the applicability of Newton-like methods to classes of operators that are only continuous. If A(x) =F0(x) Newton-like method (3.1) reduces to Newton’s method consid-
ered in [14].
Example 4.2. The j-dimensional space Rj is a classical example of a gen- eralized Banach space. The generalized norm is defined by componentwise absolute values. Then, as ordered Banach space we set E =Rj with compo- nentwise ordering with e.g. the maximum norm. A bound for a linear operator (a matrix) is given by the corresponding matrix with absolute values. Simi- larly, we can define the ”N” operators. Let E =R. That is we consider the case of a real normed space with norm denoted by k·k. Let us see how the
conditions of Theorem 3.1 look like.
Theorem 4.3. Assume:
(H1) kI−A(x)k ≤M for some M ≥0.
(H2) kF(y)−F(x)−A(x) (y−x)k ≤Nky−xk for some N ≥0.
(H3) M+N <1,
(4.1) r= kF(x0)k
1−(M+N). (H4) U(x0, r)⊆D.
(H5) (M+N)kr →0 as k→ ∞, where r is given by(4.1).
Then, the conclusions of Theorem 3.1hold.
5. APPLICATION TO FRACTIONAL CALCULUS
Our presented earlier semilocal convergence Newton-type general methods, see Theorem 4.3, apply in the next two fractional settings given that the following inequalities are fulfilled:
(5.1) k1−A(x)k∞≤γ0 ∈(0,1),
and
(5.2) |F(y)−F(x)−A(x) (y−x)| ≤γ1|y−x|, whereγ0, γ1∈(0,1), furthermore
(5.3) γ =γ0+γ1∈(0,1),
for all x, y∈[a, b∗].
Here we consider a < b∗ < b.
The specific functionsA(x),F(x) will be described next.
I) Let α >0 and f ∈ L∞([a, b]). The right Riemann-Liouville integral [4, pp. 333–354] is given by
(5.4) (Jbαf) (x) := Γ(α)1 Z b
x
(t−x)α−1f(t)dt, x∈[a, b]. Then
|(Jbαf) (x)| ≤ Γ(α)1 Z b
x
(t−x)α−1|f(t)|dt
≤ Γ(α)1 Z b
x
(t−x)α−1dtkfk∞= Γ(α)1 (b−x)α α kfk∞ (5.5)
= Γ(α+1)(b−x)α kfk∞= (ξ1). Clearly
(5.6) (Jbαf) (b) = 0.
(5.7) (ξ1)≤ Γ(α+1)(b−a)α kfk∞. That is
(5.8) kJbαfk∞,[a,b]≤ Γ(α+1)(b−a)α kfk∞<∞, i.e. Jbα is a bounded linear operator.
By [3] we get that (Jbαf) is a continuous function over [a, b] and in particular over [a, b∗]. Thus there existx1, x2 ∈[a, b∗] such that
(Jbαf) (x1) = min (Jbαf) (x), (5.9)
(Jbαf) (x2) = max (Jbαf) (x) , x∈[a, b∗]. We assume that
(5.10) (Jbαf) (x1)>0.
Hence
(5.11) kJbαfk∞,[a,b∗]= (Jbαf) (x2)>0.
Here it is
(5.12) J(x) =mx, m6= 0.
Therefore the equation
(5.13) J f(x) = 0, x∈[a, b∗], has the same solutions as the equation
(5.14) F(x) := J f(x)
2 Jbαf(x2) = 0, x∈[a, b∗]. Notice that
(5.15) Jbα f
2 Jbαf(x2)
!
(x) = (Jbαf) (x) 2 Jbαf(x2) ≤ 1
2 <1, x∈[a, b∗]. Call
(5.16) A(x) := (Jbαf) (x)
2 Jbαf(x2), ∀ x∈[a, b∗]. We notice that
(5.17) 0< (Jbαf) (x1)
2 Jbαf(x2) ≤A(x)≤ 1
2, ∀x∈[a, b∗]. Hence the first condition (5.1) is fulfilled
(5.18) |1−A(x)|= 1−A(x)≤1− (Jbαf) (x1)
2 Jbαf(x2) =:γ0, ∀ x∈[a, b∗]. Clearly γ0 ∈(0,1).
Next we assume thatF(x) is a contraction, i.e.
(5.19) |F(x)−F(y)| ≤λ|x−y|; all x, y∈[a, b∗], and 0< λ < 12.
Equivalently we have
(5.20) |J f(x)−J f(y)| ≤2λ(Jbαf) (x2)|x−y|, all x, y∈[a, b∗]. We observe that
|F(y)−F(x)−A(x) (y−x)| ≤ |F(y)−F(x)|+|A(x)| |y−x|
≤λ|y−x|+|A(x)| |y−x|
= (λ+|A(x)|)|y−x|
=: (ψ1) , ∀x, y∈[a, b∗]. (5.21)
We have that
(5.22) |(Jbαf) (x)| ≤ (b−a)α
Γ (α+ 1)kfk∞<∞, ∀x∈[a, b∗]. Hence
(5.23)
|A(x)|= |(Jbαf) (x)|
2 Jbαf(x2) ≤ (b−a)αkfk∞
2Γ (α+ 1) Jbαf(x2) <∞, ∀ x∈[a, b∗].
Therefore we get
(5.24) (ψ1)≤ λ+ (b−a)akfk∞ 2Γ (α+ 1) Jbαf(x2)
!
|y−x|, ∀ x, y∈[a, b∗]. Call
(5.25) 0< γ1 :=λ+ (b−a)akfk∞ 2Γ (α+ 1) Jbαf(x2),
choosing (b−a) small enough we can makeγ1 ∈(0,1), fulfilling (5.2).
Next we call and we need that
(5.26) 0< γ:=γ0+γ1= 1− (Jbαf) (x1)
2 Jbαf(x2)+λ+ (b−a)akfk∞
2Γ (α+ 1) Jbαf(x2) <1, equivalently,
(5.27) λ+ (b−a)akfk∞
2Γ (α+ 1) Jbαf(x2) < (Jbαf) (x1) 2 Jbαf(x2), equivalently,
(5.28) 2λ(Jbαf) (x2) +(b−a)akfk∞
Γ (α+ 1) <(Jbαf) (x1),
which is possible for small λ, (b−a). That is γ ∈ (0,1), fulfilling (5.3). So our numerical method converges and solves (5.13).
II) Let again a < b∗ < b, α > 0, m = dαe (d·e ceiling function), α /∈ N, G∈Cm−1([a, b]), 06=G(m) ∈L∞([a, b]). Here we consider the right Caputo fractional derivative (see [4, p. 337]),
(5.29) Dαb−G(x) = (−1)m Γ (m−α)
Z b x
(t−x)m−α−1G(m)(t)dt.
By [3] Dαb−G is a continuous function over [a, b] and in particular continuous over [a, b∗]. Notice that by [4, p. 358], we have that Db−α G(b) = 0.
Therefore there exist x1, x2 ∈[a, b∗] such thatDαb−G(x1) = minDαb−G(x), and Db−α G(x2) = maxDb−α G(x), for x∈[a, b∗].
We assume that
Dαb−G(x1)>0.
(i.e. Db−α G(x)>0,∀x∈[a, b∗]).
Furthermore
Dαb−G∞,[a,b∗]=Dαb−G(x2). Here it is
J(x) =mx,m6= 0.
The equation
J G(x) = 0, x∈[a, b∗],
has the same set of solutions as the equation F(x) := J G(x)
2Dαb−G(x2) = 0, x∈[a, b∗]. Notice that
Dαb− G(x) 2Dαb−G(x2)
!
= Db−α G(x) 2Dαb−G(x2) ≤ 1
2 <1, ∀ x∈[a, b∗]. We call
A(x) := Db−α G(x)
2Dαb−G(x2), ∀x∈[a, b∗]. We notice that
0< Db−α G(x1)
2Db−α G(x2) ≤A(x)≤ 1 2. Hence the first condition (5.1) is fulfilled
|1−A(x)|= 1−A(x)≤1− Db−α G(x1)
2Db−α G(x2) =:γ0, ∀ x∈[a, b∗]. Clearly γ0 ∈(0,1).
Next we assume thatF(x) is a contraction over [a, b∗], i.e.
|F(x)−F(y)| ≤λ|x−y|; ∀ x, y∈[a, b∗], and 0< λ < 12.
Equivalently we have
|J G(x)−J G(y)| ≤2λ Dαb−G(x2)|x−y|, ∀ x, y∈[a, b∗]. We observe that
|F(y)−F(x)−A(x) (y−x)| ≤ |F(y)−F(x)|+|A(x)| |y−x|
≤λ|y−x|+|A(x)| |y−x|
= (λ+|A(x)|)|y−x|
=: (ξ2) , ∀ x, y∈[a, b∗]. We observe that
Db−α G(x)≤ Γ(m−α)1 Z b
x
(t−x)m−α−1G(m)(t)dt
≤ Γ(m−α)1 Z b
x
(t−x)m−α−1dtG(m)∞
= Γ(m−α)1 (b−x)(m−α)m−αG(m)∞
= Γ(m−α+1)1 (b−x)m−αG(m)∞≤ Γ(m−α+1)(b−a)m−αG(m)∞. That is
Dαb−G(x)≤ Γ(m−α+1)(b−a)m−αkG(m)k∞<∞, ∀ x∈[a, b].
Hence,∀ x∈[a, b∗] we get that
|A(x)|=
Dαb−G(x)
2Db−α G(x2) ≤ (b−a)m−α 2Γ (m−α+ 1)
G(m)∞ Dαb−G(x2) <∞.
Consequently we observe (ξ2)≤λ+ (b−a)m−α
2Γ (m−α+ 1)
kG(m)k∞ Db−α G(x2)
|y−x|, ∀x, y∈[a, b∗]. Call
0< γ1:=λ+ (b−a)m−α 2Γ (m−α+ 1)
kG(m)k∞ Dαb−G(x2),
choosing (b−a) small enough we can makeγ1 ∈(0,1). So (5.2) is fulfilled.
Next we call and need
0< γ :=γ0+γ1 = 1− Dαb−G(x1)
2Db−α G(x2)+λ+ (b−a)m−α 2Γ (m−α+ 1)
kG(m)k∞ Db−α G(x2) <1, equivalently we find,
λ+ (b−a)m−α 2Γ (m−α+ 1)
kG(m)k∞
Dαb−G(x2) < Db−α G(x1) 2Dαb−G(x2), equivalently,
2λDb−α G(x2) + (b−a)m−α
Γ (m−α+ 1)kG(m)k∞< Dαb−G(x1), which is possible for small λ, (b−a).
That is γ ∈ (0,1), fulfilling (5.3). Hence equation (5) can be solved with our presented numerical methods.
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Received by the editors: July 9, 2015.
