Appendix A
To resolve equation (6), we used the same method developed in previous study [
53] by transforming equation (6) into difference between two perfect squares and then to transform the expression
into perfect square, it has to have double solution and then his discriminant has to be cancelled.
Now, considering the following equation:
And cancelling the discriminant
this second-degree equation function in r, one obtains:
By choosing
, one obtains:
Equation (A4) can be written as general form:
With and Putting and taking or one obtains:
And
et
will be the solutions of the following second-degree equation A6:
With the following discriminant:
or
Two cases have to be considered:
The limit value of
u is given by:
In this case, one obtains:
This leads to the solution of equation (A5):
And the solution of equation (104):
This value of
µ will cancel the discriminant of equation (A1). The solution
r is then given by:
Knowing that
, one obtains:
And in complex development:
One deduced that in this case, the four solutions will be complex. The solutions of equation (A18) are equivalent to the solutions of the two following equations:
The discriminants of equations (A19) and (A20) are given by the respective following expressions:
The two discriminants can be given as:
The cases studied here are obtained for
. The four complex solutions are then given by:
Now, the solution in this case for
, is given by:
The following boundary conditions
Allowed to deduce the following parameters:
And the final solution is therefore given by:
Equation A27 can be written in general form as:
Where , , , and are respectively given by Eqs. A9, A12 and A14.
The three first derivatives of the function
are given by the following equations:
In this case, one obtains:
Where
, and
and finally,
With:
And the solution of equation (A1):
This value of
µ will cancel the discriminant of equation (A1). The solution
r is given by:
The expression
can be given by:
The discriminants of equations (A35) are given by the respective following expressions:
Two cases can be studied:
First case:
Where
is positive for
Second case:
Where is positive for
The two other solutions are then given by
Now, the final solution for
, is given by:
One obtains for
the following solution
In conclusion, for
The three first derivatives of
are given below:
And the function g and its different derivatives are given by the following boundary conditions at point 0:
The expressions of the function
and derivatives in 0 are given below as a function of the different parameters
Aij
Taking:
;
and
; one obtains the different derivatives of
and
:
One obtains the derivatives of
and
:
The same procedure was used with
and
:
Values in 0 for
and all derivatives
Values of different parameters at point 0 and derivatives
Comparison between parameters in 0
Calculation of the different derivatives of
Knowing that
;
and
, one obtains:
;
;
and then the first derivative
:
; let’s put
; therefore,
Knowing that:
, one obtains:
By using
, one obtains:
and therefore:
Using
, one obtains:
Values of the successive derivatives of
in u=0