@jalal 

for you have no guarantee that your program will stop, even if the "target determinant" is in the range of reachable determinants. 

Look at this simple example:

restart:
dim := 2:
s := convert({$-2..3} minus {0}, list);
                       [-2, -1, 1, 2, 3]
numelems(s)^(dim^2)
                              625
d := Vector(numelems(s)^(dim^2)): k := 0: for i11 in s do   for i12 in s do
    for i21 in s do
      for i22 in s do         k := k+1:
        d[k] := i11*i22-i12*i21       end do:
    end do:
  end do:
end do:

        
{entries(d, nolist)}
{-15, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 

  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15}

There exis no matrices among the 625 possibles matrices whose determinant is -14 or +14.
In this case your program won't stop if your target determinant is one of these numbers.

With the same set s of values but dim=3, there exist 1 953 125 different matrices.
The extreme values of the determinants are -100 and +100.
But no matrices can have one of those determinants:

{-99, -98, -97, -96, -94, -93, -92, -91, -90, -89, -87, -86, -84, -82, -81, -78, 78, 81, 82, 84, 86, 87, 89, 90, 91, 92, 93, 94, 96, 97, 98, 99}

Raw code here

restart:
dim := 3:
s := convert({$-2..3} minus {0}, list);
                       [-2, -1, 1, 2, 3]
d := {}:
k := 0:
for i11 in s do
for i12 in s do
for i13 in s do
for i21 in s do
for i22 in s do
for i23 in s do
for i31 in s do
for i32 in s do
for i33 in s do
  k := k+1:
  d := d union {LinearAlgebra:-Determinant(Matrix(dim$2, [i11, i12, i13, i21, i22, i23, i31, i32, i33]))}:
end do:
end do:
end do:
end do:
end do:
end do:
end do:
end do:
end do:

MissingDeterminants := {$(min..max)(d)} minus d

@NanaP 

So much the better if it has been useful.

@NanaP 

ALL_Solutions file updated (see below)

Thanks.

A quick reply to your questions:

---------------------------------------------------------------------
1
Let X some random variable anf f its PDF.
Let Y = a+b*X (b assumed <> 0) and g its PDF; the basic formula to keep in mind is

g(t) = f((t-a)/b) / |b|

---------------------------------------------------------------------
2
Concerning the CDF the things are a little bit more complex. See an example here answ_1.mw.
If the scaling factor b is assumed to be strictly positive and if F and G are the CDF of X and Y respectively, then the second basic formula is 

F(t) = F((t-a)/b)

Demonstration of these formulas can be found in any text cours of Probability.

---------------------------------------------------------------------
3
Whatever the value of the scaling parameter Variance(Y) = b^2 * Variance(X).
As Variance is a central moment the shift parameter a doesn't matter.

---------------------------------------------------------------------
4
Skewness and Kurtosis are both defined based on central moments and the shift parameter a doesn't matter.
If you closely look to their formulas (see the Maple help pages for instance), you will see that they are dimensionless
(
Let M_X(r) the central moment of order r of X and  M_Y(r) the central moment of order r of Y.
One has  M_X(r) = b^r * M_Y(r).
Skewness(X) = M_X(3) / M_X(2)^3/2 then Skewness(Y) = b³ *M_X(3) / ( b² *M_X(2)) ^3/2 = M_X(3) / M_X(2)^3/2  = Skewness(X).
For the same reason he same Kurtosis(X) = Kurtosis(Y). 
).


---------------------------------------------------------------------
​​​​​​​5
The Mean of a random variable (more precisely its Expectation), is linear operator over random variables.
Let E this operator.
Then E(Y) = E(a + b*X) = a + b*E(X).
Concerning the Mode: the transformation T : X --> Y being linear one easily deduces that Mode(Y) = a + b*Mode(X).
Concerning the Median, which is by defininition the value such that the PDF of a random variable takes the value 1/2, the demonstration that Median(Y) = a + b*Median(X) is a little bit more complex, see here answ_2.mw

FINALLY:
The following file contains all previous step-by-step demonstrations with Maple ​​​​​​​
 

Download All_demonstrations_b.mw

Here is the demonstration that your problem may have no solution in integer numbers.

I consider the elementary case of 2x2 random matrices of integer generated this way:

RandomMatrix(2$2, generator=-10..10);

All these matrices have a determinant in the range -200..200.
Nevertheless one can proove these determinants can only take 333 values out of the 401 possible a priori.
This means, for instance, that there exist no such matrix whose determinant is 173.

In more general terms your problem doesn't always has a solution: this will depend on the value of the determinant you give.
(note that it has always an infinity of solutions in real numbers, see my answer).

Download add_on.mw

@Axel Vogt @one man

You're right, it's pretty common, but one might expect all Mapleprimes users to be a step above.

The plot with scale x = 1850 .. 2100 is not a zoom of the plot with scale x = 1800 .. 2100 : the date are not the same given the way you buid them.

@one man

What's disturbing is that you sometimes put a lot of effort into providing a complete answer, but after a few back-and-forths, the OP becomes silent, so you never know whether what you've done has been helpful or not.
Sometimes you feel like a sucker.

@delvin 

Thanks.

Were you able to go forward from file ddd_2_mmcdara.mw ?

@Carl Love 

You're right Carl, I should have open a post.
Sorry for the inconvenience.

@delvin 

Here is a correction up to the evaluation of fin1.
Check it aand try to a step ahead.

ddd_2_mmcdara.mw

BTW, given the time I have already spent to help you I would be pleased you vote up :-)

@ 

Updated June 20, GMT 7pm, have you suddenly gone mute?

Can't you run my code after modifying the parameter values as described in the last few lines?
Do you still need help?

In which case you have to provide more informations

• Do you have any idea of alpha AP and aAlpha S upper bounds?
   
• What were the initial conditions for the experimental data you provide?

@delvin 
I propose you a rewritting of your ddd.mw file.
I modified it up to the command 

fin1 := ...

I have pointed out an inconsistency in the equations (a function depends on xi[n] and you take its derivative wrt t).
I corrected this in a way which seems correct to me, but I ask you to validate what i did.

If it is ok, one could go further following a step by step way.

ddd_1_mmcdara.mw

@ 

for I will not answer it.
Keep doing exchanges within this thread.

TIA

@sursumCorda 

As the new parameters a and b are of very different orders of magnitude, a second transformation b -> 1/beta helps finding values closer to three (error about 1e-6 to 3e-6).
With this transformation  a and b then have relatively closed values and a rand()() inititialpoint seems to become more "robust".

This transformation comes from this observation: if one chooses an initial point such that a=r, b=1/r, then the solution is extremely close to a=r, b=1/r with a value of the objective function closed to 3.
 

Download 2D_parameters_2.mw

What initial value of a can we use as a good guess?
The answer is given by the following result

J := eval(fun, [y=a*x, z=x/a]):
K := simplify(normal(J), size) assuming x::real, x > 0:
asympt(K, a, 3)
                             (1/2)       
                          /1\         /1\
                      3 - |-|      + O|-|
                          \a/         \a/

which suggests that the higher the value of a (together with beta=1/b=a) the smaller the error.

A few numerical investigations lead to infer this result:

Led D the value of Digits, (D "large") and set the initial value of a to 10^D/4, then the error is 10^D/8 :
 

Digits:=400:

simplify(eval(K, b=1/beta), size):
simplify(eval(%, beta=a), size) assuming a > 0:
opt := Optimization:-Maximize(%, initialpoint={a=10^100}, iterationlimit=10^3):
printf("%1.3e", 3-opt[1]);
      1.000e-50
    

restart
interface(version)
Standard Worksheet Interface, Maple 2015.2, Mac OS X, ...
solve({a > 0, ln(a*(1 + a)) >= 0}, a);

allvalues(%);
             /      /  2                    \     \ 
            { RootOf\_Z  + _Z - 1, index = 1/ <= a }
             \                                    / 
                      /1  (1/2)   1     \ 
                     { - 5      - - <= a }
                      \2          2     /