So
Tu(t)=u(t)
2+e−t∗u(t)
3=u∗δ(t)
2+e−t
3,g(t, u(t)) = u(t)
2,
G(t, s, u(s)) = e−(t−s)u(s)
3,a=2,b=3,
where ∗is convolution of uand v; i.e.,
u(t)∗v(t)=t
0
u(t−s)v(s)ds.
We see that
u(t)−Tu(t)=u(t)∗δ(t)
2−e−t
3,v(t)−Tv(t)=v(t)∗δ(t)
2−e−t
3
and
t
0G(t, s, u(s)) −G(t, s, v(s))ds=t
0e−(t−s)u(s)
3−e−(t−s)v(s)
3ds
≤t
0
e−(t−s)∥u−v∥
3ds
≤∥u−v∥
3(1 −e−t)
≤∥u−v∥
3
where δ(t)is Dirichlet function with Laplace transformation L(δ)=1.
|Tu(t)−Tv(t)|=u(t)∗δ(t)
2+e−t
3−v(t)∗δ(t)
2+e−t
3
=(u(t)−v(t)) ∗δ(t)
2+e−t
3
≤5
6|u(t)−v(t)|,
without of loss of generality, let |u(t)|≤|v(t)|for all t∈[0,∞).
u(t)∗δ(t)
2−e−t
3≤v(t)∗δ(t)
2−e−t
3
therefore |Tu(t)−u(t)|≤|Tv(t)−v(t)|.
|u(t)−v(t)|≤|u(t)−Tu(t)|+|Tu(t)−Tv(t)|+|Tv(t)−v(t)|
≤|u(t)−Tu(t)|+5
6|u(t)−v(t)|+|Tv(t)−v(t)|
1
6|u(t)−v(t)|≤|u(t)−Tu(t)|+|Tv(t)−v(t)|
so for some positive number lwe have
∥u−v∥≤4lmin{∥u−Tu∥,∥v−Tv∥}.
All conditions of Example
(3)
with
F
(
r, s
)=4
ls
hold, and integral equation
(12)
has a unique solution
u=0.
Availability of data and material
Not applicable.
https://doi.org/10.17993/3cemp.2022.110250.64-74
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors have read and approved the final manuscript.
ACKNOWLEDGMENT
We would like to thank the referees for their comments and suggestions on the manuscript.
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