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While solving a research problem, I came across the following set of 8 equations involving 4 complex variables namely, $c2$, $c3$, $c4$, and $c5$ with complex coefficients and 2 real variables namely, $c1$ and $c6$. The complex variables are such that $c2=c3^{*}$ and $c4=c5^{*}$; where, * denotes the complex conjugate.

$2c2 c3+c4 c5=\frac{1}{2\Delta^2} \tag{1}$ $ic1 c4-2i c3 c4-i c1 c5+2ic2 c5=0 \tag{2}$ $c1^2-2c1 c2-2c1c3+4c2c3=0 \tag{3}$ $c6+(i c4-ic5)n\Delta+(c1-c2-c3)n^2\Delta^2=0\tag{4}$ $c5+(ic1-2ic3)n\Delta=-n\tag{5}$ $c4+(-ic1+2ic2)n\Delta=-n\tag{6}$ $c2=\frac{1}{\Delta}(k+il)\tag{7}$ $c3=\frac{1}{\Delta}(k-il)\tag{8}$

Also, $k,l,\Delta,n$ are real constants.

Using equations $(7)$ and $(8)$ and letting $c4=c+id$, $c5=c-id$ ; $\{c,d\in\mathbb{R}\}$, I translated the above-mentioned equations into the following Mathematica code:

Solve[c^2 + d^2 + (2 k^2)/\[CapitalDelta]^2 + (2 l^2)/\[CapitalDelta]^2 == 1/(2 \[CapitalDelta]^2) && -2 c1 d + (4 d k)/\[CapitalDelta] - (4 c l)/\[CapitalDelta] == 0 && c1^2 + (4 k^2)/\[CapitalDelta]^2 + (4 l^2)/\[CapitalDelta]^2 - (4 c1 k)/\[CapitalDelta] == 0 && c - I (d + n (-c1 + (2 k)/\[CapitalDelta] - (2 I l)/\[CapitalDelta]) \[CapitalDelta]) == -n && c + I (d + n (-c1 + (2 k)/\[CapitalDelta] + (2 I l)/\[CapitalDelta]) \[CapitalDelta]) == -n && c6 + n \[CapitalDelta] (-2 d + n (c1 - (2 k)/\[CapitalDelta]) \[CapitalDelta]) == 0, {c, d, c1, c6}]

Now, I need to solve for $c$, $d$, $c_1$, and $c_6$. I have been solving this using Mathematica and also manually by finding ways to reduce the number of equations to match the number of unknowns but without much success. Any guidance on how to approach this would be much appreciated.

Thanks.

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  • $\begingroup$ If you included a paste of your Mathematica code for your system then it is much more likely that someone might try pasting that into their system and seeing if they could think of an idea that would make some progress on this. Please, try to resist the urge to use Subscripts in your code while someone just tries to eliminate c2 and c3 for you, for example. $\endgroup$
    – Bill
    Apr 14, 2023 at 14:54
  • $\begingroup$ @Bill My apologies. Corrected now. $\endgroup$
    – JayanthJ
    Apr 14, 2023 at 15:17
  • $\begingroup$ If you want to solve for n variables you need n equations. You have 6 equations and 4 variables. Therefore, there are no solutions. $\endgroup$ Apr 14, 2023 at 15:22
  • $\begingroup$ Does this help? sys=Simplify[...yoursystem...];Simplify[Reduce[sys,{c,d,c1,c6}]] It instantly gives an answer. If you had any domain information on any of your variables, like perhaps CapitalDelta != 0 or n != 0, then the result might be simpler and easier to understand $\endgroup$
    – Bill
    Apr 14, 2023 at 15:23
  • $\begingroup$ Are you sure k,l are constants? They appear to be real and imaginary parts of (removed) variables c4,c5. $\endgroup$ Apr 15, 2023 at 19:32

1 Answer 1

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Just to have a look for the possibel solutions over the complex domain, is to prepare a Groebner basis of all equations brought to the form expr=0, taking the list of expressions. The GrobnerBasis of a list of polynomials is a list of polynomials with the same setr of zeros, but simplified as a ascending list by the number of variables. The system is overdetermined if seen a s system with four variables, but this is not an obstacle if the the equations are not independent. Here it looks not like this.

Solve[(0 == # &) /@
   GroebnerBasis[
    {c^2 + d^2 + (2 k^2)/\[CapitalDelta]^2 + (2 l^2)/\[CapitalDelta]^2 -  1/(2 \[CapitalDelta]^2),
     -2 c1 d + (4 d k)/\[CapitalDelta] - (4 c l)/\[CapitalDelta] ,
     c1^2 + (4 k^2)/\[CapitalDelta]^2 + (4 l^2)/\[CapitalDelta]^2 - (4 c1 k)/\[CapitalDelta], 
     c - I (d + n (-c1 + (2 k)/\[CapitalDelta] -    (2 I l)/\[CapitalDelta])\[CapitalDelta]) + n, 
     c + I (d + n (-c1 + (2 k)/\[CapitalDelta] + (2 I l)/\[CapitalDelta]) 
\[CapitalDelta]) + n, 
     c6 + n \[CapitalDelta] (-2 d + n (c1 - (2 k)/\[CapitalDelta]) \[CapitalDelta])}, {c, d, c1,  c6}]] // TableForm 



c->-n   c1->-(Sqrt[1-2 n^2 \[CapitalDelta]^2]/\[CapitalDelta])  c6->0   d->0    k->-(1/2) Sqrt[1-2 n^2 \[CapitalDelta]^2]   l->0    
c->-n   c1->Sqrt[1-2 n^2 \[CapitalDelta]^2]/\[CapitalDelta] c6->0   d->0    k->1/2 Sqrt[1-2 n^2 \[CapitalDelta]^2]  l->0    
c->0    c1->(2 k \[CapitalDelta]-Sqrt[-\[CapitalDelta]^2+4 k^2 \[CapitalDelta]^2])/\[CapitalDelta]^2    c6->0   d->0    l->-(1/2) Sqrt[1-4 k^2] n->0    
c->0    c1->(2 k \[CapitalDelta]-Sqrt[-\[CapitalDelta]^2+4 k^2 \[CapitalDelta]^2])/\[CapitalDelta]^2    c6->0   d->0    l->1/2 Sqrt[1-4 k^2]    n->0    
c->0    c1->(2 k \[CapitalDelta]+Sqrt[-\[CapitalDelta]^2+4 k^2 \[CapitalDelta]^2])/\[CapitalDelta]^2    c6->0   d->0    l->-(1/2) Sqrt[1-4 k^2] n->0    
c->0    c1->(2 k \[CapitalDelta]+Sqrt[-\[CapitalDelta]^2+4 k^2 \[CapitalDelta]^2])/\[CapitalDelta]^2    c6->0   d->0    l->1/2 Sqrt[1-4 k^2]    n->0    
c->0    c1->-(1/\[CapitalDelta])    c6->0   d->0    k->-(1/2)   l->0    n->0
c->0    c1->1/\[CapitalDelta]   c6->0   d->0    k->1/2  l->0    n->0
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  • $\begingroup$ If you look in the documentation, you read "Possible settings for CoefficientDomain are InexactNumbers, Rationals, RationalFunctions, and Polynomials[x]". There are no complex numbers there. $\endgroup$
    – user64494
    Apr 14, 2023 at 18:41
  • $\begingroup$ Gaussian elimination e.g. for linear systems is a way to get the GroebnerBasis. It is working without any assumptions. As always in Mathematica, symbols implicitely are assumed to be complex valued except 0,oo as default. Thats the reason why no simplification on arbitrary powers, logs and arc functions is attempted without explicit assumptions. $\endgroup$
    – Roland F
    Apr 15, 2023 at 3:41
  • $\begingroup$ Sorry, your statement "Gaussian elimination e.g. for linear systems is a way to get the GroebnerBasis. It is working without any assumptions" does not correspondent to reality (see, e.g. Wiki). In particular, I'd like to quote "Gröbner basis computation can be seen as a multivariate, non-linear generalization of both Euclid's algorithm for computing polynomial greatest common divisors, and Gaussian elimination for linear systems" . $\endgroup$
    – user64494
    Apr 15, 2023 at 7:50
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    $\begingroup$ There are ways to compute Groebner bases using linear algebra. Look up, for example, the F4 method (due to Faugere). $\endgroup$ Apr 15, 2023 at 19:33

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