To obtain the approximation solution of Eqs.(1,2), based on Definition (2.6) we have two cases as follows:
Case (1): is (i)-differentiable, in this case we have,
(3)
Case (2): is (ii)-differentiable, in this case we have,
(4)
Now, we can write successive iterations (by using Picard method) as follows:
Case (1):
(5)
Case (2):
(6)
Remark 1
For we have cases as follows:
Case (1): and be (i)-differentiable and and be (ii)-differentiable
Case (2): is (i)-differentiable and is (ii)-differentiable and is (ii)-differentiable and is (i)-differentiable
Remark 2
We discuss about switching points as follows:
Case (1): is (i)-differentiable
If and and r ∈ [0, 1] then we have and x
0 is a switching point in the form (iv).
If and and r ∈ [0, 1] then we have and x
1 is a switching point in the form (iv).
Case (2): is (ii)-differentiable
If and and r ∈ [0, 1] then we have and x
0 is a switching point in the form (iii).
If and and r ∈ [0, 1] then we have and x
1 is a switching point in the form (iii).
Case (3): is (i)-differentiable
If then we have.
If and and r ∈ [0, 1] then we have and x
0 is a switching point in the form (iv).
Case (4): is (ii)-differentiable
If then we have.
If and and r ∈ [0, 1] then we have and x
1 is a switching point in the form (iii).