"Roger Stafford" wrote in message <jfl57a$25$1@newscl01ah.mathworks.com>...
> "thinkpadmama " <qutao1125@gmail.com> wrote in message <jfk0um$s39$1@newscl01ah.mathworks.com>...
> > I am doing some fitting for some experimental data plot using the model
> > y=d0*(a0*exp(a*x)+b0*exp(b*x)+1a0b0)
> >
> > Here d0 = 3/4*p/R^(1/2)/E1*(1 + v1^2)
> > a0=1/2*E1*(1 + v2)/E2/(1 + v1^2)
> > b0=1/2*(1 + 2*v1)*(1 + 2*v1  v2 + 2*v2*v1)*E1/(2*E1 + 2*E1*v2 + 3*E2)/(1 + v1^2)
> > a=1/2/y/(1 + v2)*E2
> > b=1/6*(2*E1 + 2*E1*v2 + 3*E2)/y/(1 + v2)
> >
> > p and R are known. Now I got a good curve fitting for the plot and found the parameters for this model, such as
> >
> > a = 0.04445
> > a0 = 1.162
> > b = 0.4979
> > b0 = 0.3821
> > d0 = 1.886e009
> >
> > I need to solve these equations to find E1, E2, v1, v2 and y. Here I got the a0, b0, a, b, d0 first and then solve the E1, E2, v1, v2, and y. As you said I checked the equations and found there cannot be a solution for this one. So I am trying to get E1, E2, v1, v2 and y directly from the curve fitting without getting the a0,b0,a,b and d0 first, but it takes too long to calculate. Do you have any suggestions? Thanks.
>          
> I do not know how you obtained the following equations in connection with fitting a five parameter double exponential curve to your data:
>
> a0 = 1/2*E1*(1+v2)/E2/(1+v1^2)
> b0 = 1/2*(1+2*v1)*(1+2*v1v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(1+v1^2)
> d0 = 3/4*p/R^(1/2)/E1*(1+v1^2)
> a = 1/2/y/(1+v2)*E2
> b = 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)
>
> However I do know that in the process you lost one degree of freedom in your parameters. Even though there are five parameters here, v1, v2, E1, E2, and y, the five expressions have a mutual interdependency which would place a constraint on the values of a0, b0, d0, a, and b. It is the same kind of constraint I already told you about, since these expressions are just those you described in your original posting. If they have to satisfy these five equations, then they cannot be arbitrarily assigned values. In particular the five values you specified (a = .04445, etc.) may not be possible if they satisfy those equations. That means your curve fit would really be only a fourparameter fit.
>
> So the question is, where did you get those five equations? What is the mathematical significance of those five variables, v1, v2, E1, E2, and y? In what way are they related to your data or to your model? Why choose expressions where one of them can be derived from the others? Why not do the curve fitting using a0, b0, d0, a, and b directly where you still have five independent parameters?
>
> Roger Stafford
The purpose of my work is to find the values of E1, E2, v1, v2 and y from the curve fitting. It is convenient and easier to solve for a0, b0, d0, a and b first which can be described by E1,E2,v1,v2 and y as the parameters of the curve fitting model. To get the curve fitting result, I give the initial guess for the five parameters and that works and I got a good curve fitting result with the values of a0,b0,d0,a and b that I posted above. So the work now is to solve for E1,E2,v1,v2 and y using the five equations which you thought not right. Now my question is that whether I can get E1, E2, v1, v2 and y directly from the curve fitting problem by using some "curve fitting" functions or methods in MATLAB without getting the a0, b0, d0 , a and b first.
Thanks for your patience and illustration.
