Thread Subject: Help solve equations

Hi,
I have some difficulties here. How can I solve equations like this?

1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
-0.0168/E1*(-1+v1^2)-5.684e-5=0;
1/2/y/(1+v2)*E2-0.04791=0;
1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;

I want the solution of E1, E2, v1, v2 and y.
E1>0, E2>0, 0<v1<0.5, 0<v2<0.5, y>0. I tried SOLVE and FSOLVE, but not work.
Any suggestions?

Thanks a lot!

On 19 Jan., 06:38, "thinkpadmama " <qutao1...@gmail.com> wrote:
> Hi,
> I have some difficulties here. How can I solve equations like this?
>
> 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=­0;
> -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> 1/2/y/(1+v2)*E2-0.04791=0;
> 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
>
> I want the solution of E1, E2, v1, v2 and y.
> E1>0, E2>0, 0<v1<0.5, 0<v2<0.5, y>0. I tried SOLVE and FSOLVE, but not work.
> Any suggestions?
>
> Thanks a lot!

What makes you think you can impose the constraints
E1>0, E2>0, 0<v1<0.5, 0<v2<0.5, y>0
to the variables ?
You have 5 equations to determine 5 unknowns - usually there exists
only one
solution - so imposing constraints to it is not possible.
Or do you want to solve the equations in the least-squares sense with
the
above conditions being satisfied ? Then try MATLAB's lsqnonlin.

Best wishes
Torsten.

"thinkpadmama " <qutao1125@gmail.com> wrote in message <jf8a82$hu5$1@newscl01ah.mathworks.com>...
> Hi,
> I have some difficulties here. How can I solve equations like this?
>
> 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
> -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> 1/2/y/(1+v2)*E2-0.04791=0;
> 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
>
> I want the solution of E1, E2, v1, v2 and y.
> E1>0, E2>0, 0<v1<0.5, 0<v2<0.5, y>0. I tried SOLVE and FSOLVE, but not work.
> Any suggestions?
>
> Thanks a lot!
- - - - - - - - - -
  I am surprised that 'solve' can't handle this problem. A cursory examination of these equations indicates to me that you can eliminate first 'y', then 'E1' and 'E2', and finally 'v2' to get a single polynomial equation in v1. Accordingly this polynomial can be solved, if not by 'solve', then by using matlab's 'roots' function to numerically determine a number of roots for v1, depending on the order of the polynomial, of which some may be complex-valued. For each of these v1 values, the other four unknowns would then be uniquely determined. You could determine if any of these sets of solutions would fit within your required bounds.

  If you have to do the above manipulations by hand, it will admittedly require some careful algebra on your part to avoid errors. You can probably use your Symbolic Toolbox to assist in this, both to carry out some of the manipulations and to check your work.

Roger Stafford

"Roger Stafford" wrote in message <jf8k4u$g8r$1@newscl01ah.mathworks.com>...
> "thinkpadmama " <qutao1125@gmail.com> wrote in message <jf8a82$hu5$1@newscl01ah.mathworks.com>...
> > .......
> > 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> > 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
> > -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> > 1/2/y/(1+v2)*E2-0.04791=0;
> > 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
> > .......
> - - - - - - - - - -
> I am surprised that 'solve' can't handle this problem. A cursory examination ....
- - - - - - - -
  I now believe the reason 'solve' did not work for you is that these five equations are such that they have an interdependency. If your six coefficients within the equations are just right, then v1 is uniquely determined, but v2 can be any arbitrary value, which would result in an infinite continuum of possible solutions. If the coefficients are not just right, then no solutions exist.

  An analogy in three dimensions would be three planes, each represented by an equation in x, y, and z. However the three lines of intersection from the three possible plane pairings happen to all be parallel to one another. If the additive constants in the planes' equations are just right, these lines will all coincide giving infinitely many points of common intersection of the three planes. Otherwise there will be no intersection of the three planes at all. That would be analogous to the situation with your five equations in five unknowns. They unfortunately have this same kind of fatal property, (though of course they aren't really planes.)

  What it all amounts to is that the problem you have described has not been well formulated so as to as to yield proper solutions. You need to reduce these equations to four independent ones and add another which is independent of them. In other words there just isn't enough information present to allow for unique solutions to your problem.

Roger Stafford

"Roger Stafford" wrote in message <jf8k4u$g8r$1@newscl01ah.mathworks.com>...
> "thinkpadmama " <qutao1125@gmail.com> wrote in message <jf8a82$hu5$1@newscl01ah.mathworks.com>...
> > .......
> > 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> > 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
> > -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> > 1/2/y/(1+v2)*E2-0.04791=0;
> > 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
> >
> > I want the solution of E1, E2, v1, v2 and y.
> > E1>0, E2>0, 0<v1<0.5, 0<v2<0.5, y>0. I tried SOLVE and FSOLVE, but not work.
> > .......
> - - - - - - - - - -
> I am surprised that 'solve' can't handle this problem. ......
- - - - - - - - - -
  I now believe the reason 'solve' did not work for you is that these five equations are such that they have an interdependency. If your six coefficients within the equations are just right, then v1 is uniquely determined, but v2 can be any arbitrary value, which would result in an infinite continuum of possible solutions. If the coefficients are not just right, then no solutions exist.

  An analogy in three dimensions would be three planes, each represented by an equation in x, y, and z. However the three lines of intersection from the three possible plane pairings happen to all be parallel to one another. If the additive constants in the planes' equations are just right, these lines will all coincide giving infinitely many points of common intersection of the three planes. Otherwise there will be no intersection of the three planes at all. That would be analogous to the situation with your five equations in five unknowns. They unfortunately have this same kind of fatal property, (though of course they aren't really planes.)

  What it all amounts to is that the problem you have described has not been well formulated so as to as to yield proper solutions. You need to reduce these equations to four independent ones and add another which is independent of them. In other words there just isn't enough information present to allow for unique solutions to your problem.

Roger Stafford

"thinkpadmama " <qutao1125@gmail.com> wrote in message <jf8a82$hu5$1@newscl01ah.mathworks.com>...
> 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
> -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> 1/2/y/(1+v2)*E2-0.04791=0;
> 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
>
> I want the solution of E1, E2, v1, v2 and y.
- - - - - - - - - -
  The following reasoning will let you demonstrate for yourself what the difficulty is with your equations.

  Your five equations can be rewritten as:

 k1 = -E1*(1+v2)/E2/(-1+v1^2)
 k2 = -(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)
 k3 = -1/E1*(-1+v1^2)
 k4 = 1/y/(1+v2)*E2
 k5 = (2*E1+2*E1*v2+3*E2)/y/(1+v2)

where k1 = 2*1.229, k2 = 2*0.4108, k3 = (5.684e-5)/0.0168, k4 = 2*0.04791, and k5 = 6*0.697 .

  After working on your equations it turns out that the following expression exhibits the interdependence that exists between four of these equations:

 F(k1,k2,k4,k5) = (3*(k1+2)*k4-(2+k2)*k5)^2-4*k1*k2*k4*k5

It is shown this way. If you use the Symbolic Toolbox to substitute into this expression the right hand sides of equations 1, 2, 4, and 5 for k1, k2, k4, and k5, and apply 'simplify' or 'simple' to the result, you will see that it becomes identically zero. In other words, any simultaneous solution in terms of v1, v2, E1, E2, and y must yield a zero result for F. Yet if you substitute in the values of k1, k2, k4, and k5 which you have given, F is decidedly not zero. The only way this could be true is that there can be no simultaneous solutions to those four equations, and consequently no solutions to all five. You have therefore posed an impossible problem.

  It is not necessary to know how the above F was derived to validate the above reasoning, but it arises from the process of eliminating one at a time the variables y, E1, E2, and v2. In this process it turns out that the remaining variable v1 needs to satisfy two different equations, whereas v2 can be any arbitrary value. This double requirement on v1 in turn puts a constraint on the parameters k1, k2, k4, and k5, the constraint being that F(k1,k2,k4,k5) must be zero. If your requirement on these k's did happen to satisfy that constraint, there would be infinitely many possible solutions corresponding to the arbitrary values of v2. Thus this situation is indeed analogous to the intersecting planes problem I mentioned in an earlier post.

Roger Stafford

"Roger Stafford" wrote in message <jfhog2$lac$1@newscl01ah.mathworks.com>...
> "thinkpadmama " <qutao1125@gmail.com> wrote in message <jf8a82$hu5$1@newscl01ah.mathworks.com>...
> > 1/2*E1*(1+v2)/E2/(-1+v1^2)+1.229=0;
> > 1/2*(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)+0.4108=0;
> > -0.0168/E1*(-1+v1^2)-5.684e-5=0;
> > 1/2/y/(1+v2)*E2-0.04791=0;
> > 1/6*(2*E1+2*E1*v2+3*E2)/y/(1+v2)-0.697=0;
> >
> > I want the solution of E1, E2, v1, v2 and y.
> - - - - - - - - - -
> The following reasoning will let you demonstrate for yourself what the difficulty is with your equations.
>
> Your five equations can be rewritten as:
>
> k1 = -E1*(1+v2)/E2/(-1+v1^2)
> k2 = -(-1+2*v1)*(-1+2*v1-v2+2*v2*v1)*E1/(2*E1+2*E1*v2+3*E2)/(-1+v1^2)
> k3 = -1/E1*(-1+v1^2)
> k4 = 1/y/(1+v2)*E2
> k5 = (2*E1+2*E1*v2+3*E2)/y/(1+v2)
>
> where k1 = 2*1.229, k2 = 2*0.4108, k3 = (5.684e-5)/0.0168, k4 = 2*0.04791, and k5 = 6*0.697 .
>
> After working on your equations it turns out that the following expression exhibits the interdependence that exists between four of these equations:
>
> F(k1,k2,k4,k5) = (3*(k1+2)*k4-(2+k2)*k5)^2-4*k1*k2*k4*k5
>
> It is shown this way. If you use the Symbolic Toolbox to substitute into this expression the right hand sides of equations 1, 2, 4, and 5 for k1, k2, k4, and k5, and apply 'simplify' or 'simple' to the result, you will see that it becomes identically zero. In other words, any simultaneous solution in terms of v1, v2, E1, E2, and y must yield a zero result for F. Yet if you substitute in the values of k1, k2, k4, and k5 which you have given, F is decidedly not zero. The only way this could be true is that there can be no simultaneous solutions to those four equations, and consequently no solutions to all five. You have therefore posed an impossible problem.
>
> It is not necessary to know how the above F was derived to validate the above reasoning, but it arises from the process of eliminating one at a time the variables y, E1, E2, and v2. In this process it turns out that the remaining variable v1 needs to satisfy two different equations, whereas v2 can be any arbitrary value. This double requirement on v1 in turn puts a constraint on the parameters k1, k2, k4, and k5, the constraint being that F(k1,k2,k4,k5) must be zero. If your requirement on these k's did happen to satisfy that constraint, there would be infinitely many possible solutions corresponding to the arbitrary values of v2. Thus this situation is indeed analogous to the intersecting planes problem I mentioned in an earlier post.
>
> Roger Stafford

Thank you so much. You mean my equations are not enough to find the solutions, or there may be something wrong with one of these equations?

"thinkpadmama " <qutao1125@gmail.com> wrote in message <jfim7p$gav$1@newscl01ah.mathworks.com>...
> Thank you so much. You mean my equations are not enough to find the solutions, or there may be something wrong with one of these equations?
- - - - - - - - - -
  The only thing I can be sure of is that, as they stand, your five equations can have no simultaneous solution. The reasoning with F(k1,k2,k4,k5) I showed you earlier demonstrates conclusively why a solution is quite impossible.

  What you need to find the five unknowns is a set of five equations that are mutually independent, that is that none of them can be derived from the others. Whether you can correct things by making a minor change in one or more of the equations, or whether you need to replace some of them by totally different equations I cannot tell.

  What I suspect is that both assertions may be true in this case but that is only a guess. Perhaps if you could explain the underlying problem that led you to these equations, an better answer to your question might be forthcoming.

Roger Stafford

"Roger Stafford" wrote in message <jfisnu$53h$1@newscl01ah.mathworks.com>...
> "thinkpadmama " <qutao1125@gmail.com> wrote in message <jfim7p$gav$1@newscl01ah.mathworks.com>...
> > Thank you so much. You mean my equations are not enough to find the solutions, or there may be something wrong with one of these equations?
> - - - - - - - - - -
> The only thing I can be sure of is that, as they stand, your five equations can have no simultaneous solution. The reasoning with F(k1,k2,k4,k5) I showed you earlier demonstrates conclusively why a solution is quite impossible.
>
> What you need to find the five unknowns is a set of five equations that are mutually independent, that is that none of them can be derived from the others. Whether you can correct things by making a minor change in one or more of the equations, or whether you need to replace some of them by totally different equations I cannot tell.
>
> What I suspect is that both assertions may be true in this case but that is only a guess. Perhaps if you could explain the underlying problem that led you to these equations, an better answer to your question might be forthcoming.
>
> Roger Stafford

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.

"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

"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)+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

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.