Abstract In this paper, we investigate the existence of integrable solution to a nonlin-ear integral equation, which includes many important integral and functional equa-tions that arise in nonlinear analysis and its applications. Our results are obtainedunder rather general assumptions. In particular, the solvability of the well knownChandrasekhar equation is discussed under appropriate assumptions. Our analysisuses the technique of measures of weak noncompactness and rely on a variant ofSchauder’s fixed point theorem.
Keywords Fixed point theorem · Measure of weak noncompactness · Nonlinearintegral equation · Integrable solution
In this paper we consider the following nonlinear integral equation
x(t) = u(t, x(t)
) + g(t, x(t)
)∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds, t ∈ I = [0,1]. (1.1)
This equation contains as particular cases many important integral and functionalequations in the literature. Specifically, the Chandrasekhar’s integral equation
x(t) = 1 + x(t)
∫ 1
0
t
t + sψ(s)x(s)ds, t ∈ [0,1]. (1.2)
can be considered as a prototype of (1.1). We emphasize that Chandrasekhar’s integralequation was first formulated by Chandrasekhar in 1960. This integral equation playsan important role in radiative transfer theory and neutron transfer theory. For furtherdetails we refer the reader to the book [5].
In our considerations, we look for solutions to (1.1) in the Banach space of realfunctions being integrable on (0,1). We transform our problem into a fixed pointproblem. We choose a strategy which ensures the existence of an integrable solutionto our problem. This approach enables us to develop an existence theory in L1(0,1)
for our integral equation under rather general and nonrestrictive assumptions. In par-ticular, we show that the mere boundedness of the characteristic function ψ with‖ψ‖∞ < 1
4 guarantees the existence of at least one solution to the Chadrasekhar inte-gral equation (1.2) in L1(0,1). The techniques of measures of weak noncompactnessand the concept of ws-compactness play an important role in achieving our goal. Theresults obtained in this paper generalize several ones obtained in [2, 9, 10, 15, 17].Also examples are presented to illustrate our results.
2 Auxiliary facts and results
In this section, we provide some notations, definitions and auxiliary facts which willbe needed for stating our results. Denote by L1(I ) the set of all Lebesgue integrablefunctions on I = [0,1], endowed with the standard norm
‖x‖ =∫
I
∣∣x(t)∣∣dt.
Consider a function f : I × R → R. We say that f satisfies the Carathéodory con-ditions if it is measurable in t for any x ∈ R and continuous in x for almost allt ∈ I . Then to every function x(t) being measurable on I we may assign the func-tion (Fx)(t) = f (t, x(t)), t ∈ I . The operator F defined in such a way is called thesuperposition operator generated by the function f .
We recall the following result due to Krasnosel’skii [12].
Theorem 2.1 The superposition operator F generated by the function f maps con-tinuously the space L1(I ) into itself if and only if |f (t, x)| ≤ a(t) + b|x| for all t ∈ I
and x ∈ R, where a(t) is a function from L1(I ) and b is a nonnegative constant.
Integrable solutions for a nonlinear quadratic integral equation
For our purpose we recall the following theorem of Scorza-Dragoni [16].
Theorem 2.2 Let J be a bounded interval and let f : J × R −→ R be a functionsatisfying Carathéodory conditions. Then, for each ε > 0 there exists a closed subsetDε of the interval J such that meas(J \ Dε) < ε and f |Dε×R is continuous.
Definition 2.3 [11] Let M be a subset of a Banach space X. A continuous mapA : M −→ X is said to be (ws)-compact if for any weakly convergent sequence(xn)n∈N in M the sequence (Axn)n∈N has a strongly convergent subsequence in X.
Recall the following fixed point theorem proved in [14] (see also [13] and [17]).
Theorem 2.4 Let M be a nonempty closed convex subset of a Banach space X.Suppose A : M −→ M satisfies:
(i) A is (ws)-compact.(ii) A(M) is relatively weakly compact.
Then there is an x ∈M such that Ax = x.
3 Measure of weak noncompactness
Let X be a Banach space and let B(X) denote the collocation of all nonemptybounded subsets of X and W(X) the subset of B(X) consisting of all relativelyweakly compact subsets of X. Finally, let Br denote the closed ball centered at 0with radius r . The concept of the axiomatic measure of weak noncompactness isdefined as follows:
Definition 3.1 [3] A function μ : B(X) −→ R+ is said to be a measure of weaknoncompactness if it satisfies the following conditions
(1) The family ker(μ) = {M ∈ B(X) : μ(X) = 0} is nonempty and ker(μ) ⊂ W(X).(2) M1 ⊂ M2 ⇒ μ(M1) ≤ μ(M2).(3) μ(co(M)) = μ(M), where co(M) is the convex hull of M .(4) μ(λM1 + (1 − λ)M2) ≤ λμ(M1) + (1 − λ)μ(M2) for λ ∈ [0,1].(5) If (Mn)n≥1 is a sequence of nonempty, weakly closed subsets of X with M1
bounded and M1 ⊇ M2 ⊇ · · · ⊇ Mn ⊇ · · · . such that limn→∞ μ(Mn) = 0, thenM∞ := ⋂∞
n=1 Mn is nonempty.
The first important example of measure of weak noncompactness was defined by DeBlasi (see [7]). In the space L1(I ), there is a convenient and workable formula for theDe Blasi measure μ which was given by Appel and De Pascale [1] as follows: For anonempty and bounded subset M of the space L1(I )
μ(M) = limε→0
{supx∈M
{sup
[∫
D
∣∣x(t)∣∣dt : D ⊂ I,meas(D) ≤ ε
]}}. (3.1)
The following criterion is crucial for our purpose.
A. Bellour et al.
Theorem 3.2 [8] A bounded set S is relatively weakly compact in L1(I ) if and onlyif for any ε > 0 there exists δ > 0 such that if meas(D) ≤ δ then
∫D
|x(t)| ≤ ε for allx ∈ S.
4 Main result
Equation (1.1) will be studied under the following assumptions:
(i) The function ϕ : I −→ I is continuous.(ii) The function u : I ×R −→ R satisfies the Carathéodory conditions and is Lips-
chitzian with respect to the second variable with a Lipschitz constant α, that is,|u(t, x)−u(t, y)| ≤ α|x −y| for all t ∈ I and all x, y ∈ R. Let β(t) = |u(t,0)| ∈L1(I ).
(iii) The function g : I ×R −→ R satisfies the Carathéodory conditions and is Lip-schitzian with respect to the second variable with a Lipschitz constant b1, thatis, |g(t, x) − g(t, y)| ≤ b1|x − y| for all t ∈ I and all x, y ∈ R. Let a1(t) =|g(t,0)| ∈ L1(I ).
(iv) The function f : I × R −→ R satisfies the Carathéodory conditions and thereexist a constant b2 and a function a2 ∈ L1(I ) such that |f (t, x)| ≤ a2(t) + b2|x|for all t ∈ I and x ∈R.
(v) The function k : I × I −→R is measurable and the linear Volterra operator
Kx(t) =∫ ϕ(t)
0k(t, s)x(s)ds, t ∈ I
transforms the space L1(I ) into L∞(I ). Let ‖K‖ be the norm of this operator.(vi) α + b2‖K‖‖a1‖ + b1‖K‖‖a2‖ + 2
√‖K‖b1b2(‖β‖ + ‖a1‖‖K‖‖a2‖) < 1.
Remark 4.1
(1) We assume that u and g satisfy the Carathéodory conditions. Note that a lessrestrictive assumption of the form (vii) u(., x(.)) and g(., x(.)) are measurablewhenever x(.) is measurable, is only needed here.
(2) It is easy to check that, if the function k is bounded on the set = {(t, s) ∈I × I : 0 ≤ s ≤ ϕ(t)}, then the linear operator K transforms the space L1(I ) intoL∞(I ) and the norm ‖K‖ of this operator is majorized by ‖k‖L∞().
(3) Let f : I ×R → R satisfy (iv) and let F be the superposition operator generatedby f . Let K : L1(I ) → L∞(I ) be the linear operator defined in (v). Note that forall x ∈ L1(I ) we have
Integrable solutions for a nonlinear quadratic integral equation
Thus∣∣∣∣
∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds
∣∣∣∣ = ∣∣KFx(t)∣∣ ≤ ‖K‖‖a2‖ + b2‖K‖‖x‖. (4.3)
Before stating the main result, we need the following lemma:
Lemma 4.2 Assume the assumptions (i)–(vi) hold. Then, there exists r0 > 0 such thatfor each fixed x ∈ Br0 ⊂ L1(I ) there exists a unique ψx ∈ Br0 such that for all t ∈ I ,we have
(i)
ψx(t) = u(t,ψx(t)
) + g(t,ψx(t)
) ∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds.
(ii) For all x, y ∈ L1(I ) we have
∣∣ψx(t) − ψy(t)∣∣ ≤ a1(t) + b1|ψy(t)|
1 − γ
∣∣Hx(t) − Hy(t)∣∣,
where γ = α + b1‖K‖(‖a2‖ + b2r0) ∈ [0,1) and H is defined by
Hx(t) = KFx(t) =∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds.
Proof Let x ∈ L1(I ) be fixed. Let Vx be the map which assigns to y ∈ L1(I ) thefunction Vx(y) defined by
Vx(y)(t) = u(t, y(t)
) + g(t, y(t)
)∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds.
We claim that there exists r0 such that the operator Vx transforms the ball Br0 intoitself for x ∈ Br0 . Assume that x, y ∈ Br . Then
‖Vxy‖ =∫ 1
0
∣∣Vxy(t)∣∣dt
≤∫ 1
0
∣∣u(t, y(t)
)∣∣dt +∫ 1
0
(∣∣g(t, y(t)
)∣∣ ×∣∣∣∣
∫ ϕ(t)
0k(t, s)f
(s, x(s)
)∣∣∣∣ds
)dt
≤∫ 1
0
(β(t) + α
∣∣y(t)∣∣)dt +
∫ 1
0
(a1(t) + b1
∣∣y(t)∣∣)(‖K‖‖a2‖ + b2‖K‖‖x‖)dt
≤ ‖β‖ + α‖y‖ + (‖a1‖ + b1‖y‖)(‖K‖‖a2‖ + b2‖K‖‖x‖)
≤ ‖β‖ + ‖a1‖‖K‖‖a2‖ + (α + b2‖K‖‖a1‖ + b1‖K‖‖a2‖
)r + ‖K‖b1b2r
2
We define the polynomial
f (r) = ‖β‖ + ‖a1‖‖K‖‖a2‖ − ξr + ‖K‖b1b2r2, r > 0
A. Bellour et al.
where ξ = 1 − α − b2‖K‖‖a1‖ − b1‖K‖‖a2‖. Note
= ξ2 − 4(‖K‖b1b2
(‖β‖ + ‖a1‖‖K‖‖a2‖))
is positive from assumption (vi) and for r0 = ξ−√
ξ2−4(‖K‖b1b2(‖β‖+‖a1‖‖K‖‖a2‖))2‖K‖b1b2
theoperator Vx transforms the ball Br0 into itself.
We claim that Vx : Br0 → Br0 is a contraction. Indeed, for t ∈ I we have
∣∣Vxy1(t) − Vxy2(t)∣∣ ≤ ∣∣u
(t, y1(t)
) − u(t, y2(t)
)∣∣ + ∣∣g(t, y1(t)
) − g(t, y2(t)
)∣∣
×∣∣∣∣
∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds
∣∣∣∣
≤ α∣∣y1(t) − y2(t)
∣∣ + b1∣∣y1(t) − y2(t)
∣∣(‖K‖‖a2‖ + b2‖K‖‖x‖)
≤ (α + b1‖K‖(‖a2‖ + b2‖x‖L1(I )
))∣∣y1(t) − y2(t)∣∣
≤ (α + b1‖K‖(‖a2‖ + b2r0
))∣∣y1(t) − y2(t)∣∣
= γ∣∣y1(t) − y2(t)
∣∣
We show that γ < 1. To see this, first notice that from assumption (vi) we knowthat
Thus, Vx : Br0 → Br0 is a contraction mapping, which implies the existence andthe uniqueness of ψx ∈ Br0 so (i) holds.
Also, for t ∈ I and x, y ∈ Br0 we have
∣∣ψx(t) − ψy(t)∣∣ ≤ ∣∣u
(t,ψx(t)
) − u(t,ψy(t)
)∣∣ + ∣∣Hx(t)∣∣∣∣g
(t,ψx(t)
) − g(t,ψy(t)
)∣∣
+ ∣∣g(t,ψy(t)
)∣∣∣∣Hx(t) − Hy(t)∣∣
≤ α∣∣ψx(t) − ψy(t)
∣∣ + (‖K‖‖a2‖ + b2‖K‖‖x‖)b1∣∣ψx(t) − ψy(t)
∣∣
+ (a1(t) + b1
∣∣ψy(t)∣∣)∣∣Hx(t) − Hy(t)
∣∣
Integrable solutions for a nonlinear quadratic integral equation
≤ (α + (b1‖K‖‖a2‖ + b1b2‖K‖r0)
)∣∣ψx(t) − ψy(t)∣∣
+ (a1(t) + b1
∣∣ψy(t)∣∣)∣∣Hx(t) − Hy(t)
∣∣.
Hence,
∣∣ψx(t) − ψy(t)∣∣ ≤ a1(t) + b1|ψy(t)|
1 − γ
∣∣Hx(t) − Hy(t)∣∣. �
Now we are in a position to state our main result.
Theorem 4.3 Under the assumptions (i)–(vi) the quadratic integral equation (1.1)has at least one solution x ∈ L1(I ).
Proof The problem of solving Eq. (1.1) is equivalent to finding a fixed point of theoperator A which assigns to x ∈ Br0 the function ψx . We will show that A satisfies theconditions of Theorem 2.4. First notice that Ax belongs to Br0 for any x ∈ Br0 . Thisfollows from Lemma 4.2. Now we show that A(Br0) is relatively weakly compact.To see this, take an arbitrary number ε > 0 and a nonempty subset D of I such thatD is measurable and meas(D) ≤ ε. Then for any x ∈ Br0 we have,
∫
D
∣∣Ax(t)∣∣dt ≤
∫
D
∣∣u(t,Ax(t)
)∣∣dt
+∫
D
(∣∣g
(t,Ax(t)
)∣∣ ×∣∣∣∣
∫ ϕ(t)
0k(t, s)f
(s, x(s)
)ds
∣∣∣∣
)dt
≤∫
D
β(t)dt + α
∫
D
∣∣Ax(t)∣∣dt
+ (‖a2‖‖K‖ + b2‖K‖r0)(∫
D
∣∣a1(t)∣∣dt + b1
∫
D
∣∣Ax(t)∣∣dt
)
≤∫
D
β(t)dt + γ
∫
D
∣∣Ax(t)∣∣dt + (‖a2‖ + b2r0
)‖K‖∫
D
∣∣a1(t)∣∣dt.
This implies
(1 − γ )
∫
D
∣∣Ax(t)∣∣dt ≤
∫
D
β(t)dt + (‖a2‖ + b2r0)‖K‖
∫
D
∣∣a1(t)∣∣dt.
Now using (3.1) and the fact that a set consisting of one element is weakly compact,we get
limε→0
{sup
[∫
D
β(t)dt : D ⊂ R+,meas(D) ≤ ε
]}= 0,
and
limε→0
{sup
[∫
D
a1(t)dt : D ⊂ R+,meas(D) ≤ ε
]}= 0.
A. Bellour et al.
Thus, from the above estimate, we obtain μ(A(Br0)) = 0 and therefore A(Br0) isrelatively weakly compact. Now we show that A is (ws)-compact. To see this, weuse some ideas in [2]. Let (yn) be a weakly convergent sequence in Br0 . The set{yn,n ∈ N} is relatively weakly compact. Since A(Br0) is relatively weakly compactthen the set {ψyn = Ayn, n ∈ N} is relatively weakly compact. By Theorem 3.2, weinfer that the set
{a1(t) + b1ψyn(t), n ∈N
}
is also relatively weakly compact.Now take an arbitrary number ε > 0. In view of Theorem 3.2 there exists δ(ε) > 0
such that if S ⊂ I and meas(S) ≤ δ(ε), we have for all n ∈ N
∫
S
(a1(t) + b1
∣∣ψyn(t)∣∣)dt ≤ ε(1 − γ )
4‖K‖(‖a2‖ + b2r0). (4.4)
Therefore, in view of Theorem 2.2, we can find a closed subset Dε ⊂ I such thatmeas(I \ Dε) ≤ δ(ε) and the function k |Dε×I is uniformly continuous.
This implies, by (4.4), that for all n ∈ N we have
∫
I\Dε
(a1(t) + b1
∣∣ψyn(t)∣∣)dt ≤ ε(1 − γ )
4‖K‖(‖a2‖ + b2r0). (4.5)
Let t1, t2 ∈ Dε . Without loss of generality we can assume that ϕ(t1) ≤ ϕ(t2). Then,for an arbitrary n ∈ N we have
∣∣Hyn(t2) − Hyn(t1)∣∣
=∣∣∣∣
∫ ϕ(t2)
0k(t2, s)f
(s, yn(s)
)ds −
∫ ϕ(t1)
0k(t1, s)f
(s, yn(s)
)ds
∣∣∣∣
≤∣∣∣∣
∫ ϕ(t2)
ϕ(t1)
k(t2, s)f(s, yn(s)
)ds
∣∣∣∣ +∣∣∣∣
∫ ϕ(t1)
0k(t2, s)f
(s, yn(s)
)ds
−∫ ϕ(t1)
0k(t1, s)f
(s, yn(s)
)ds
∣∣∣∣
≤∫ ϕ(t2)
ϕ(t1)
∣∣k(t2, s)∣∣[a2(s) + b2
∣∣yn(s)∣∣]ds
+∫ ϕ(t1)
0
∣∣k(t1, s) − k(t2, s)∣∣[a2(s) + b2
∣∣yn(s)∣∣]ds
≤ ‖k‖L∞(Dε×I )
∫ ϕ(t2)
ϕ(t1)
[a2(s) + b2
∣∣yn(s)∣∣]ds
+ ω(k, |t1 − t2|
) ∫ ϕ(t1)
0
[a2(s) + b2
∣∣yn(s)∣∣]ds
Integrable solutions for a nonlinear quadratic integral equation
≤ ‖k‖L∞(Dε×I )
∫ ϕ(t2)
ϕ(t1)
a2(s)ds + ‖k‖L∞(Dε×I )b2
∫ ϕ(t2)
ϕ(t1)
∣∣yn(s)∣∣ds
+ ω(k, |t1 − t2|
)(‖a2‖ + b2r0),
where ω(k, .) denotes the modulus of continuity of the function k on the set Dε × I .Now, let ε > 0 be fixed. Since M := {yn : n ∈ N} is relatively weakly compact,
then from Theorem 3.2 there exists a η > 0 such that∫D
|yn(t)|dt < ε for any D ⊂ I
with meas(D) < η. Then, since ϕ is continuous, there exists a η0 > 0 such that|ϕ(t1)−ϕ(t2)| < η whenever |t1 − t2| < η0. Thus, for t1, t2 ∈ I with |t1 − t2| < η0 wehave
∫ ϕ(t2)
ϕ(t1)|yn(t)|dt < ε.
Consequently, for all n ∈ N, we have∫ ϕ(t2)
ϕ(t1)|yn(t)|dt → 0 as t1 → t2. Similarly,
we can show that∫ ϕ(t2)
ϕ(t1)|a1(s)|ds is arbitrarily small provided the number |t2 − t1| is
small enough. This means that the sequence (Hyn) is a sequence of equicontinuousfunctions on Dε . Moreover, for an arbitrary t ∈ Dε and for n ∈N, we have
∣∣Hyn(t)∣∣ =
∣∣∣∣
∫ ϕ(t)
0k(t, s)f
(s, yn(s)
)ds
∣∣∣∣
≤∫ ϕ(t)
0
∣∣k(t, s)∣∣[a2(s) + b2
∣∣yn(s)∣∣]ds
≤ ‖k‖L∞(Dε×I )
∫ ϕ(t)
0a2(s)ds + ‖k‖L∞(Dε×I )b2
∫ ϕ(t)
0
∣∣yn(s)∣∣ds
≤ ‖k‖L∞(Dε×I )
(‖a2‖ + b2r0).
Hence, the sequence (Hyn) is uniformly bounded in C(Dε). Applying the Arzela-Ascoli Theorem we obtain that {Hyn,n ∈ N} is a relatively compact subset of C(Dε).Thus {Hyn} has a convergent subsequence in C(Dε) say (Hynk
). This subsequenceis clearly a Cauchy sequence in C(Dε). Thus for a given ε > 0, there exists k0 suchthat for all m,k ≥ k0 the following inequality holds
∣∣Hynm(t) − Hynk(t)
∣∣ ≤ ε(1 − γ )
2(‖a1‖ + b1r0)(4.6)
for any t ∈ Dε .Now we prove that the subsequence (Aynk
) is convergent in L1(I ). Since L1(I ) isa complete Banach space, it suffices to prove that the subsequence (Aynk
) is a Cauchysequence in L1(I ). From (4.5) and (4.6), we have for all m,k ≥ k0 that
∫ 1
0
∣∣Aynm(t) − Aynk
(t)∣∣dt ≤
∫ 1
0
a1(t) + b1|ψynk(t)|
1 − γ
∣∣Hynm(t) − Hynk
(t)∣∣dt
=∫
Dε
a1(t) + b1|ψynk(t)|
1 − γ
∣∣Hynm(t) − Hynk(t)
∣∣dt
+∫
I\Dε
a1(t) + b1|ψynk(t)|
1 − γ
∣∣Hynm(t) − Hynk
(t)∣∣dt
A. Bellour et al.
≤ ε
2+ 2‖K‖(‖a2‖ + b2r0)
1 − γ
∫
I\Dε
(a1(t) + b1
∣∣ψynk(t)
∣∣)dt
≤ ε
2+ ε
2= ε,
which implies that (Aynk) is a Cauchy sequence in L1(I ) and as a result the oper-
ator A is (ws)-compact. Theorem 2.4 guarantees a fixed point for A and hence anintegrable solution to Eq. (1.1). �
5 Examples
Example 5.1 Consider the Chandrasekhar’s integral equation (1.2). We assume thatψ ∈ L∞(I ). Notice that Eq. (1.2) is a particular case of Eq. (1.1) with
u(t, x(t)
) = 1, g(t, x(t)
) = x(t), f(t, x(t)
) = ψ(t)x(t),
k(t, s) = t
t + s, ϕ(t) = 1,
β(t) = 1, α = 0, a1(t) = 0, b1 = 1,
a2(t) = 0, b2 = ‖ψ‖L∞(I ).
Now inequality (vi) in Theorem 4.3 becomes
2√‖ψ‖L∞(I ) < 1 ⇐⇒ ‖ψ‖L∞(I ) <
1
4
Consequently, the Chandrasekhar’s equation has a solution x ∈ L1(I ) whenever‖ψ‖L∞(I ) < 1
4 .
Example 5.2 Consider the quadratic integral equation
x(t) = 1 + t2 + 1
5x(t) + (
t2 + x(t))∫ t
0
1
ts + λln
(1 + x2(s)
)ds (5.1)
where t ∈ I and λ is a positive number. Eq. (5.1) is a particular case of Eq. (1.1) with
u(t, x(t)
) = 1 + t2 + 1
5x, g
(t, x(t)
) = t2 + x(t), k(t, s) = 1
ts + λ,
f(t, x(t)
) = ln(1 + x(t)2), ϕ(t) = t, β(t) = 1 + t2, α = 1
5,
a1(t) = t2, b1 = 1, a2(t) = 0, b2 = 1.
Now inequality (vi) becomes
1
5+ 1
3λ+ 2
√4
3λ< 1 ⇐⇒ 1
3+ 4√
3
√λ <
4
5λ ⇐⇒ λ >
55 + 10√
30
12
Integrable solutions for a nonlinear quadratic integral equation
From Theorem 4.3 we conclude that Eq. (5.1) has a solution x ∈ L1(I ) whenever
λ > 55+10√
3012 .
6 Final remarks
(1) The continuity of the characteristic function ψ is not required in our considera-tions (in [4] note ψ is assumed to be continuous). However, we should mentionthat the solution obtained in [4] is continuous.
(2) In [6], the existence of integrable solution to a more general quadratic integralequation has been discussed. However, the conditions required on the nonlinear-ity are strong and do not cover the Chandrasekhar integral equation.
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