Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Help solve equations Date: Tue, 24 Jan 2012 02:32:10 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 33 Message-ID: <jfl57a$25$1@newscl01ah.mathworks.com> References: <jf8a82$hu5$1@newscl01ah.mathworks.com> <jfhog2$lac$1@newscl01ah.mathworks.com> <jfim7p$gav$1@newscl01ah.mathworks.com> <jfisnu$53h$1@newscl01ah.mathworks.com> <jfk0um$s39$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-00-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1327372330 69 172.30.248.45 (24 Jan 2012 02:32:10 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Tue, 24 Jan 2012 02:32:10 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:755565 "thinkpadmama " <qutao1125@gmail.com> wrote in message <jfk0um$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)+1-a0-b0) > > 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 = 1.886e-009 > > 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*v1-v2+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 four-parameter 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