I have four nonlinear equations I want to find the points of intersection of these equations, and I used the software Mathematica, unfortunately after many hours of waiting it does not give me any result Do you have an idea how to solve this kind of problem?.

My equations are the following

1) $(52 \alpha ^2+\alpha (104 \beta +440 \sqrt{3}-1071)+\beta (65 \beta +110 \sqrt{3}-752)-2 (52 \gamma +55 \sqrt{3}-376) \delta +(-52 \gamma -440 \sqrt{3}+1071) \gamma -65 \delta ^2)=0$

2) $52 \alpha ^2+\alpha (-52 \beta +80 \sqrt{3}-450)+2 \beta (13 \beta +10 \sqrt{3}+38)-52 \gamma ^2+\gamma (52 \delta -80 \sqrt{3}+450)-2 \delta (13 \delta +10 \sqrt{3}+38)=0$

3) $\beta ^2-\alpha ^3=0$

4) $\delta ^2-\gamma ^3=0$

  • $\begingroup$ Where do those equations come from $-$ i.e. why do you care about this particular system with its complicated coefficients? $$ $$ (3) and (4) mean that $(\alpha,\beta) = (x^2, x^3)$ and $(\gamma,\delta) = (y^2, y^3)$ for some $x,y$. So now you have two equations in two variables and can use a resultant to eliminate one of them. Then you still have to deal with the $\sqrt 3$'s in the coefficients, though some packages know how to do this. $\endgroup$ – Noam D. Elkies Dec 25 '18 at 15:59
  • $\begingroup$ I need to solve this system of equation to prove that it has only two solutions, I used Mathematica software but it's very complicated $\endgroup$ – Sara yaqob Dec 25 '18 at 16:03
  • $\begingroup$ Again I ask: why is this particular system of equations of interest, and in what context do you "need to solve" it? Knowing where a problem comes from (and why you expect that it has only two (real? complex?) solutions) is often useful in finding a solution $-$ and also in motivating others to help you with it. $\endgroup$ – Noam D. Elkies Dec 25 '18 at 17:21
  • $\begingroup$ I have two differential linear systems with centers the first one has a first integral H1(x,y)=k, the second one has the first integral H2(x,y)=h, I am looking for the number of intersection points between H1(x,y)=k, H2(x,y)=h, and the curve y^2-x^3=0......I suppose that there are two pairs (\alpha, \beta) and (\gamma, \delta) satisfying the four following equations .(it is necessary)....................1) H1(\alpha, \beta) =H1(\gamma, \delta) ....2) H2(\alpha, \beta) =H2(\gamma, \delta) ...3) \beta^2-\alpha^3=0 ............4) \delta^2-\gamma^3=0. $\endgroup$ – Sara yaqob Dec 25 '18 at 17:55
  • $\begingroup$ The most straightforward way is by computing resultants. If $P, Q$ are polynomials in $x, y, z, \ldots$, then Resultant[P, Q, x] gives a polynomial in $y, z, \ldots$, which vanishes at $(y_0, z_0, \ldots)$ if and only if there exists some $x_0$, such that $(x_0, y_0, \ldots)$ is a common zero of $P$ and $Q$. Thus you can pass from a system of $n$ equations in $n$ variables to a system of $n-1$ equations in $n-1$ variables. Unfortunately the degree increases, so in this case it would be wise to exploit (3) and (4) as Noam Elkies described first. $\endgroup$ – Jan-Christoph Schlage-Puchta Dec 25 '18 at 19:49

This is an expansion of the comment by Noam D. Elkies. Indeed, we can rewrite your equations 3) and 4) as $(\alpha,\beta)=(x^2,x^3)$ and $(\gamma,\delta)=(y^2,y^3)$ for some $x,y$. Substituting into eqs. 1) and 2) these expressions of $\alpha,\beta,\gamma,\delta$ in terms of $x,y$, we reduce your system of four eqs. to that of the following two: $P_1=0=P_2$, where $P_1,P_2$ are certain polynomials in $x,y$.

The resultant $\text{Res}_y(P_1,P_2)$ vanishes iff, for a given value of $x$, the polynomials $P_1,P_2$ have a common root $y$. We find that $\text{Res}_y(P_1,P_2)=0=\text{Res}_x(P_1,P_2)$ identically for all $x$ and $y$. So, $P_1$ and $P_2$ need some preliminary cleaning. Indeed, we see that \begin{equation} Q_1:=\frac{P_1}{y-x},\quad Q_2:=\frac{P_2}{2(y-x)} \end{equation} are polynomials in $x,y$. Thus, \begin{equation} \text{$(x,x)$ is a solution to our system for any complex $x$.} \tag{*} \end{equation}

The resultants $R_y:=\text{Res}_y(Q_1,Q_2)$ and $R_x:=\text{Res}_x(Q_1,Q_2)$ are polynomials in $x$ and in $y$, respectively, each having $16$ distinct complex roots, say $x_1,\dots,x_{16}$ for $R_y$ and $y_1,\dots,y_{16}$ for $R_x$. Details of all calculations here, as well as the particular enumeration of $x_1,\dots,x_{16}$ and $y_1,\dots,y_{16}$, can be seen in the Mathematica notebook or its pdf image.

So, for each $i\in[16]:=\{1,\dots,16\}$ there is at least one $j\in[16]$ such that the pair $(x_i,y_j)$ is a solution to the system $Q_1=0=Q_2$. Moreover, it is easy to see that for each $i\in[16]\setminus\{3\}$, there is at most one (and therefore the only one) $j=j_i\in[16]$ such that the pair $(x_i,y_{j_i})$ is a solution to the system $Q_1=0=Q_2$. The corresponding pairs $(i,j_i)$ are $(1, 3), (2, 4), (4, 2), (5, 6), (6, 5), (7, 11), (8, 12), (9, 10), (10, 9), (11, 7)$, $(12, 8), (13, 14), (14, 13), (15, 16), (16, 15)$. For the exceptional value $i=3$, we have at most two values of $j\in[16]$ (namely, $j=1$ and $j=3$) such that the pairs $(x_3,y_j)$ may be solutions to the system $Q_1=0=Q_2$; in fact we see that these pairs $(x_3,y_j)$, equal $(0,\sqrt3)$ and $(0,0)$, are indeed solutions to the system $Q_1=0=Q_2$; the pair $(0,0)$ has already been accounted for by (*).

Thus, we have described all the $16$ pairs $(x,y)$ that are (in addition to the "trivial" pairs given by (*)) solutions to the system $Q_1=0=Q_2$. All these solutions $(x,y)$ are now straightforward to transcribe into a complete set of solutions $(\alpha,\beta,\gamma,\delta)$ of the original system.

  • $\begingroup$ You mean that the solution is (0,0,3, 3 (3)^(1/2))......I don't know hwo I can thank you....Thanks a lot sir for your help $\endgroup$ – Sara yaqob Dec 25 '18 at 20:46
  • 1
    $\begingroup$ Solving numerically the original system, both Maple and Mathematica produce 13 different real solutions. Taking into account multiplicity, Mathematica counts 18 solutions (in particular, zero solution is of multiplicity 4 and the solution $(2.45653, -3.85019, 0., 0.)$ is of multiplicity 2 and the solution $(3., 5.19615, 0., 0.)$ is of multiplicity 2). $\endgroup$ – user64494 Dec 25 '18 at 21:48
  • $\begingroup$ @user64494 : How can solving numerically ensure that, among other things, you got all the solutions? $\endgroup$ – Iosif Pinelis Dec 26 '18 at 0:23
  • $\begingroup$ Here dropbox.com/s/6exiv40yikocwg5/sys.pdf?dl=0 is the result obtained with Mathematica and here dropbox.com/s/aqdh72m3cg8lwtg/sys%20in%20maple.pdf?dl=0 is the result in Maple. The DirectSearch is one of the best numerical solvers in the world. $\endgroup$ – user64494 Dec 26 '18 at 5:13
  • $\begingroup$ Solving numerically the system from the edited question dropbox.com/s/mkvvskm0i0m5mtx/screen26.12.2018.docx?dl=0, I obtain that the solution set is infinite (see dropbox.com/s/ul0u4wxzxuyv2jx/modifiedsys.pdf?dl=0 ). $\endgroup$ – user64494 Dec 26 '18 at 10:55

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