Data fitting leastsq vs. lsqrsolve, don't success to use lsqrsolve

Hi,
I am beginner using scilab and I would like to fit experimental data point.
I have a set of data :
j = [0.02 0.045 0.13 0.351 0.7 0.998 1.226 1.411 1.574]';
U = [0.9 0.85 0.8 0.75 0.7 0.65 0.6 0.55 0.5]';
The model that links j to U is based on the following equation:
function Etheo = CdP2fit(j, Param)
Etheo = Param(1) - Param(2).*j - Param(3)*log10(j./(Param(4))) + 0.0152*log(Param(5)./(Param(5)- j));
endfunction

So the question is how to find Param vector that fit as best Etheo with U.

Therefore I defined an error function that has to be minimized:

function e = err(Param,j, U,m)
e = abs(CdP2fit(j, Param) - U)
endfunction

Then, defining a set of initial value and using the function leastsq as below:

Param_init = [1.18; 0.2; 0.06; 3e-8; 2];
[f, ParaOpt] = leastsq(list(err,j,U),Param_init);

I can find a good fitting. I 've obtianed :
ParaOpt =
1.1794511
0.1808364
0.0521901
4.745D-08
2.0009558
f =
0.0025729
But if I want to use lsqrsolve function, I can't fit, I don't see why !

[ParaOpt,f] = lsqrsolve(Param_init,err,size(j,1))
disp ("ParaOpt =")
disp(ParaOpt)
disp ("f =")
disp(f)

I got
ParaOpt =

2.5424888
0.2046764
0.0000009
- 0.0000021
- 342.05439

f =

0.2
0.15
0.1
0.0500000
0.0000013
0.0500000
0.1000000
0.15
0.2

The reasult is meaningless.
Why the result is so different, it seems as well it is highly dependent from the initial value.

In which case should I use lsqrsolve or leastsq ,

Thank you so much for your help,

Best regards,
Ph

ps: Is scilab 5.5.2 is more robust than 6.0.2 ?

Scilab 18-02-20, 4:39 p.m. phoudha