3.3 Split generalized equilibrium problem with multiple output sets
3.3.3 Numerical examples
This section provides some examples to illustrate the implementation of our proposed methods, Algorithm 3.3.5. In our experiment, we let g(x) = x3, αn = 140n+11 , δn =
1
3n+14, µn = 1−αn−δn, for i = 0,1,2, let ri = r = 0.5, , βi,n = 13, γn,0 = 12, γn,j =
1
2j+1, j = 1,2, . . . . Moreover, we consider the effect of varying values of the following parameters θi,n = 2+i1 ,1.0,3+i2 ,1.5,4+i3 ,1.9, ϕi,n = 0.5,4+i3 ,1.0,2+i5 ,2.0,3+i7 on our method.
All numerical computations were carried out using Matlab version R2019(b). We plot the graphs of errors against the number of iterations in each case. The stopping criterion used for our computation is ∥xn+1−xn∥<10−9.The numerical results are reported in Figures 3.4, 3.5, 3.6 and 3.7 and Tables 3.2.11, 3.2.12, 3.2.13 and 3.2.14. In Tables 3.2.11-3.2.14,
“Iter.” means the number of iterations while “CPU” means the CPU time in seconds.
Example 3.3.11. Let H, Hi =R2 for i= 0,1,2, with H =H0. We define the mappings F = F0 : R2 ×R2 → R, F1 : R2 ×R2 → R and F2 : R2 ×R2 → R respectively by F(x, y) = −3x2 +xy+ 2y2, F1(x, y) = −4x2 +xy+ 3y2 and F2(x, y) = −5y2 + 2y+ 5xy−5xy2 for each x = (x1, x2) ∈ R2 and y = (y1, y2) ∈ R2. Also, for i = 0,1,2, let ϕ0 = ϕ : R2 × R2 → R, ϕ1 : R2 × R2 → R and ϕ2 : R2 × R2 → R be defined by ϕ(x, y) = x2 −xy, ϕ1(x, y) = 2x(x−y) and ϕ2(x, y) = 5y2 −2x respectively for each x = (x1, x2) ∈ R2 and y = (y1, y2) ∈ R2. For some r > 0, we obtain by some simple calculation that
v =TrF,ϕu= 1
4r+ 1u, y =TrF1,ϕ1x= 1
1 + 5rx and w=TrF2,ϕ2z = z−2r 1 + 5r.
Let Ai :R2 →R2 be given by Ai(x) = i+1x where x= (x1, x2)∈R2. Let Sj :C →C(B) be defined by
Sjx:= −3
2j x, j= 1,2, ...
It is easy to see that Sj is demicontractive for each j = 1,2, . . . .
Table 3.2.11. Numerical results for Example 3.3.11 (Experiment 1).
Cases θi,n = 2+i1 θi,n= 1 θi,n =
2 3+i
θi,n = 1.5
θi,n =
3 4+i
1 CPU
time(sec)
0.0155 0.0128 0.0096 0.0140 0.0150
No of Iter. 14 14 14 14 14
2 CPU time
(sec)
0.0136 0.0088 0.0128 0.0139 0.0150
No of Iter. 14 14 14 14 14
3 CPU time
(sec)
0.0137 0.0089 0.0145 0.0162 0.0161
No of Iter. 14 14 14 14 14
4 CPU time
(sec)
0.0135 0.0123 0.0097 0.0146 0.0151
No of Iter. 14 14 14 14 14
Table 3.2.12. Numerical results for Example 3.3.11 (Experiment 2).
Cases ϕi,n = 0.5 ϕi,n =
3 4+i
ϕi,n = 1.0
θi,n =
5 2+i
θi,n = 2.0
1 CPU
time(sec)
0.0138 0.0090 0.0132 0.0156 0.0092
No of Iter. 14 14 14 14 14
2 CPU
time(sec)
0.0139 0.0089 0.0128 0.0144 0.0140
No of Iter. 14 14 14 14 14
3 CPU
time(sec)
0.0139 0.0091 0.0148 0.0151 0.0093
No of Iter. 14 14 14 14 14
4 CPU
time(sec)
0.0136 0.0088 0.0140 0.0152 0.0092
No of Iter. 14 14 14 14 14
The next example is in the framework of an infinite dimensional Hilbert spaces.
Example 3.3.12. Let H, Hi = ℓ2 for i = 0,1,2, ... be the linear spaces whose elements consists of 2-summable sequences (x1, x2, ..., xi, ...) of scalars, i.e.,
ℓ2 ={x:x= (x1, x2, ..., xi, ...) and
∞
X
i=1
|xi|2 <∞}, with an inner product ⟨., .⟩:ℓ2×ℓ2 →R defined by
⟨x, y⟩=
∞
X
i=1
xiyi where x={xi}∞i=1, y ={yi}∞i=1 ∈ℓ2.
For i = 0,1,2, let the mapping Ai : ℓ2 → ℓ2 be defined by Aix = x31,x32, ..., x3m, ...
for all x = {xm}∞m=1 ∈ ℓ2 and A∗i : ℓ2 → ℓ2 be defined by A∗iz = z31,z32, ...,z3m, ...
for
0 2 4 6 8 10 12 14 Iteration number (n)
10-10 10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (
i,n = 1) Alg. 3.1 (i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (
i,n = 1) Alg. 3.1 (i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
Figure 3.4: Top left: Case 1; Top right: Case 2; Bottom left: Case 3; Bottom right: Case 4.
all z = {zm}∞m=1 ∈ ℓ2. Define the mapping Fi : ℓ2 × ℓ2 → R by Fi,= F such that F(x, y) = −x2+y2, ∀ x = {xi}∞i=1, y = {yi}∞i=1. and ϕi = 0, for each i = 0,1,2. It is easy to see that
Tr(F,ϕ)x= 1−r 5r+ 1x.
Also, for j = 1,2, ..., we define Sj :C →CB(ℓ2) by Sjx=
0, x 5j
∀j = 1,2, ...
It is easy to see that Sj is 0-demicontractive for each j = 1,2, ... and Fix(Sj) ={0}.
Table 3.2.13. Numerical results for Example 3.3.12 (Experiment 1).
0 2 4 6 8 10 12 14 Iteration number (n)
10-10 10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 0.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (
i,n = 1.0) Alg. 3.1 (i,n = 5/(2 + i)) Alg. 3.1 (i,n = 2.0) Alg. 3.1 (i,n = 7/(3 + i))
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 0.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (
i,n = 1.0) Alg. 3.1 (i,n = 5/(2 + i)) Alg. 3.1 (i,n = 2.0) Alg. 3.1 (i,n = 7/(3 + i))
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 0.5) Alg. 3.1 (
i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.0) Alg. 3.1 (i,n = 5/(2 + i)) Alg. 3.1 (i,n = 2.0) Alg. 3.1 (i,n = 7/(3 + i))
0 2 4 6 8 10 12 14
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 0.5) Alg. 3.1 (
i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.0) Alg. 3.1 (i,n = 5/(2 + i)) Alg. 3.1 (i,n = 2.0) Alg. 3.1 (i,n = 7/(3 + i))
Figure 3.5: Top left: Case 1; Top right: Case 2; Bottom left: Case 3; Bottom right: Case 4.
Cases θi,n = 2+i1 θi,n= 1 θi,n =
2 3+i
θi,n = 1.5
θi,n =
3 4+i
1 CPU
time(sec)
0.0544 0.0112 0.0144 0.0130 0.0158
No of Iter. 33 33 33 33 33
2 CPU
time(sec)
0.0129 0.0093 0.0133 0.0133 0.0138
No of Iter. 34 33 34 33 34
3 CPU
time(sec)
0.0134 0.0124 0.0096 0.0142 0.0137
No of Iter. 33 33 33 33 33
4 CPU
time(sec)
0.0161 0.0174 0.0164 0.0166 0.0133
No of Iter. 33 33 33 34 34
Table 3.2.14. Numerical results for Example 3.3.12 (Experiment 2).
0 5 10 15 20 25 30 35 Iteration number (n)
10-10 10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (
i,n = 1) Alg. 3.1 (i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (
i,n = 1) Alg. 3.1 (i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
Figure 3.6: Top left: Case 1; Top right: Case 2; Bottom left: Case 3; Bottom right: Case 4.
Cases ϕi,n = 0.5 ϕi,n =
3 4+i
ϕi,n = 1.0
θi,n =
5 2+i
θi,n = 2.0
1 CPU
time(sec)
0.0132 0.0100 0.0112 0.0116 0.0118
No of Iter. 33 33 33 33 33
2 CPU
time(sec)
0.0154 0.0122 0.0102 0.0146 0.0138
No of Iter. 33 33 33 33 33
3 CPU
time(sec)
0.0141 0.0100 0.0133 0.0141 0.0140
No of Iter. 33 33 33 33 33
4 CPU
time(sec)
0.0161 0.0169 0.0142 0.0169 0.0160
No of Iter. 33 33 33 34 34
We test these examples under the following experiments:
Experiment 1:
In this experiment, we check the behavior of our method by fixing the other parameters and varying θi,n. We do this to check the effects of the parameter θi,n on our method.
For Example 3.3.11We consider the following cases for the initial value of x0 : Case 1 : x0 = (0.78,1.25);
0 5 10 15 20 25 30 35 Iteration number (n)
10-10 10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-8 10-6 10-4 10-2 100 102
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (
i,n = 1) Alg. 3.1 (i,n = 2/(3 + i)) Alg. 3.1 (i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
0 5 10 15 20 25 30 35
Iteration number (n) 10-10
10-5 100
Errors
Alg. 3.1 (i,n = 1/(2 + i)) Alg. 3.1 (i,n = 1) Alg. 3.1 (
i,n = 2/(3 + i)) Alg. 3.1 (
i,n = 1.5) Alg. 3.1 (i,n = 3/(4 + i)) Alg. 3.1 (i,n = 1.9)
Figure 3.7: Top left: Case 1; Top right: Case 2; Bottom left: Case 3; Bottom right: Case 4.
Case 2 : x0 = (3.78,1.25);
Case 3 : x0 = (4,2);
Case 4 : x0 = (−1,−5).
Also, we consider θi,n ∈ {2+i1 ,1.0,3+i2 ,1.5,4+i3 ,1.9}, which satisfies Assumption (A3). We use Algorithm 3.3.5 for the experiment and report the numerical results in Table 3.2.11 and Figure 3.4.
For Example 3.3.12We consider the following cases for the initial value of x0 : Case 1 : x0 = (2,1,12,· · ·);
Case 2 : x0 = (4,−2,1,· · ·);
Case 3 : x0 = (−3,35,−253,· · ·);
Case 4 : x0 = (6,1,16,· · ·).
Also, we consider θi,n ∈ {2+i1 ,1.0,3+i2 ,1.5,4+i3 ,1.9}, which satisfies Assumption (A3). We use Algorithm 3.3.5 for the experiment and report the numerical results in Table 3.2.13 and Figure 3.6.
Experiment 2:
In this experiment, we check the behavior of our method by fixing the other parameters and varying ϕi,n.We do this to check the effects of the parameter ϕi,n on our method.
For Example 3.3.11: We consider the following cases for the initial value of x0 : Case 1 : x0 = (0.78,1.25);
Case 2 : x0 = (3.78,1.25);
Case 3 : x0 = (4,2);
Case 4 : x0 = (−1,−5).
Also, we consider ϕi,n ∈ {0.5,4+i3 ,1.0,2+i5 ,2.0,3+i7 }, which satisfies Assumption (A4). We use Algorithm 3.3.5 for the experiment and report the numerical results in Table 3.2.12 and Figure 3.5.
For Example 3.3.12: We consider the following cases for the initial value of x0 : Case 1 : x0 = (2,1,12,· · ·);
Case 2 : x0 = (4,−2,1,· · ·);
Case 3 : x0 = (−3,35,−253,· · ·);
Case 4 : x0 = (6,1,16,· · ·).
Also, we consider ϕi,n ∈ {0.5,4+i3 ,1.0,2+i5 ,2.0,3+i7 }, which satisfies Assumption (A4). We use Algorithm 3.3.5 for the experiment and report the numerical results in Table 3.2.14 and Figure 3.7.
Chapter 4
Split Monotone Variational Inclusion Problems and Fixed Point Problems
4.0.1 Introduction
In this chapter, we present our result on relaxed double inertial Tseng’s extragradient method with self-adaptive step sizes for solving split monotone variational inclusion prob- lem (SMVIP) involving non-Lipschitz operators and fixed point problem of strict pseudo- contractive mappings. Furthermore, results on generalized split feasibility problem over a solution set of monotone variational inclusion problem was studied. For each of these problems, we propose iterative algorithm with self-adaptive step size for approximating the solution and prove strong convergence theorem. We also give applications of our results and illustrate our algorithms with numerical examples.