On 06/09/2011 02:39 AM, Sourabh Bajaj wrote:
>
> I have simplified some of the equations using a divide and conquer
> kind of strategy.
>
> And reduced my problem to a two variable one.
>
> Now the equations are :
>
> exp1=(2*beta4^2+2*beta3^2+2*beta2^2)*d^2*z1^2+(
>
> (4*alpha4*beta4+4*alpha3*beta3+4*alpha2*beta2)*d^2*y1+4*beta4*d^2*gamma4+4*beta3*d^2*gamma3+4*
>
> beta2*d^2*gamma2)*z1+(2*alpha4^2+2*alpha3^2+2*alpha2^2)*d^2*y1^2+
>
> (4*alpha4*d^2*gamma4+4*alpha3*d^2*gamma3+4*alpha2*d^2*gamma2)*y1+l2^2-l1^2+2*d^2*gamma4^2+2*d^2*
>
> gamma3^2+2*d^2*gamma2^2
>
> exp2=zp^2-2*z1*zp+(beta1^2+1)*z1^2+(2*alpha1*beta1*y1-2*beta1*xp+2*beta1*gamma1)*z1+yp^2
>
> -2*y1*yp+(alpha1^2+1)*y1^2+(2*alpha1*gamma1-2*alpha1*xp)*y1+xp^2-2*gamma1*xp-l1^2+gamma1^2
>
> exp3=(-4*beta4*d*z1-4*alpha4*d*y1-4*d*gamma4)*zp+(4*beta4+4*beta1*beta2)*d*z1^2+(-4*
>
> beta3*d*yp+(4*beta3+4*alpha1*beta2+4*alpha2*beta1+4*alpha4)*d*y1-4*beta2*d*xp+4*d*gamma4+4*
>
> beta1*d*gamma2+4*beta2*d*gamma1)*z1+(-4*alpha3*d*y1-4*d*gamma3)*yp+(4*alpha3+4*alpha1*alpha2)*
>
> d*y1^2+(-4*alpha2*d*xp+4*d*gamma3+4*alpha1*d*gamma2+4*alpha2*d*gamma1)*y1-4*d*gamma2*xp-3*l2^2+
>
> 3*l1^2+4*d*gamma1*gamma2
>
> I just want to find, y1 & z1 and everything else is a constant.
>
> Can you please help me how to solve these. I have tried running the
> solution a lot of times but it gives the same error I stated earlier.
>
>
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I'm not much of a mathematician, but this caught my eye so I did the
following as an exercise to see what maxima could do. I broke up your
lengthy expressions into something I thought was more manageable because
you stated everything except y1 and z1 were considered constant. If I
didn't make a typo (or some other error) this is what maxima did:
A=(2*beta4^2+2*beta3^2+2*beta2^2)*d^2;
B=(4*alpha4*beta4+4*alpha3*beta3+4*alpha2*beta2)*d^2;
C=+4*beta4*d^2*gamma4+4*beta3*d^2*gamma3+4*beta2*d^2*gamma2;
D=(2*alpha4^2+2*alpha3^2+2*alpha2^2)*d^2;
E=(4*alpha4*d^2*gamma4+4*alpha3*d^2*gamma3+4*alpha2*d^2*gamma2);
F=l2^2-l1^2+2*d^2*gamma4^2+2*d^2*gamma3^2+2*d^2*gamma2^2;
/*Then exp1 becomes exp1a*/
exp1a=A *z1^2 +( B *y1 + C ) *z1 + D *y1^2 + E *y1 +F;
G=zp^2-2;
H=zp+(beta1^2+1);
I=2*alpha1*beta1;
J=2*beta1*xp+2*beta1*gamma1;
K=yp^2-2;
L=yp+(alpha1^2+1);
M=(2*alpha1*gamma1-2*alpha1*xp);
N=xp^2-2*gamma1*xp-l1^2+gamma1^2;
/*Then exp2 becomes exp2a*/
exp2a=G *z1 *H *z1^2 +(I *y1 -J) *z1 +K *y1 *L *y1^2 +M *y1 +N;
O=-4*beta4*d;
P=4*alpha4*d;
Q=4*d*gamma4;
R=zp+(4*beta4+4*beta1*beta2)*d;
S=-4*beta3*d*yp+(4*beta3+4*alpha1*beta2+4*alpha2*beta1+4*alpha4)*d;
T=4*beta2*d*xp+4*d*gamma4+4*beta1*d*gamma2+4*beta2*d*gamma1;
U=-4*alpha3*d;
V=4*d*gamma3;
W=yp+(4*alpha3+4*alpha1*alpha2)*d;
X=(-4*alpha2*d*xp+4*d*gamma3+4*alpha1*d*gamma2+4*alpha2*d*gamma1);
Y=4*d*gamma2*xp-3*l2^2+3*l1^2+4*d*gamma1*gamma2;
/*Then exp3 becomes exp3a*/
exp3a=(O *z1 -P *y1 -Q) *R *z1^2 +(S *y1 -T) *z1 +(U *y1 -V) *W *y1^2 +X
*y1 -Y;
solve([exp1a,exp2a,exp3a],[y1,z1]);
[]
As you can see by the empty solution from solve, there appears to be no
solution, but since I'm not much of a mathematician, there might be
better ways to approach this problem.
I also tried doing:
solve([exp1,exp2,exp3],[y1,z1]);
[]
Which maxima also could not solve, but I got no error messages from the
attempt.
build_info ();
gives the following:
Maxima version: 5.24post
Maxima build date: 13:38 5/16/2011
Host type: i686-pc-linux-gnu
Lisp implementation type: CMU Common Lisp
Lisp implementation version: Snapshot 2011-02 (20B Unicode)
Paul