Solving a system of two coupled quadratic equations

Questions
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1

I am currently researching dynamics of atomic clusters. Part of a problem depends on determining analytic behavior of solutions to the following non-linear system of two equations

sol = Solve[{
    b1 y^2 + b2 y + b3 z^2 + b4 z + b5 y z == A1,
    c1 z^2 + c2 z + c3 y^2 + c4 y + c5 y z == A2},
   {y, z}, Reals];

I expect solutions to be very large, but still derivable for analysis. Yet Mathematica does not give any result after quite long wait. I am looking for any suggestions on dealing with this type of systems. All coefficients are real numbers.

2

I’d instead do

sol = Solve[{b1 y^2 + b2 y + b3 z^2 + b4 z + b5 y z == A1, 
             c1 z^2 + c2 z + c3 y^2 + c4 y + c5 y z == A2}, 
            {y, z}, Cubics -> False, Quartics -> False];

Then for specific values of parameters you can weed out solutions that are not real valued. What you have now will, I believe, attempt to describe the parameter space. That can be both computationally quite hard and also give a solution that is too huge to be of much use.

3

Playing around a bit, it’s clear that finding some solutions symbolically is easier if you accept some (hopefully satisfiable) constraints on the values of b1, b2.. and c1, c2… In effect, if we allow there to be only two unknowns parameters for “b” and “c” we can simplify things significantly.

Consider replacing the left hand side with b1 (y + z)^2 + (-b2) (y - 1) (z - 1) + b2. Expanding it, we that it is a restriction on the original system:

Expand[b1 (y + z)^2 + (-b2) (y - 1) (z - 1) + b2]
b2 y + b1 y^2 + b2 z + 2 b1 y z - b2 y z + b1 z^2

So our equations are:

b1 (y + z)^2 - b2 (y - 1) (z - 1) == a1-b2
c1 (y + z)^2 - c2 (y - 1) (z - 1) == a2-c2

We have a linear system with the unknown values of (y + z)^2 and (y - 1) (z - 1)

{res1, res2} = LinearSolve[{{b1, -b2}, {c1, -c2}}, {a1 - b2, a2 - c2}]

We know have a simpler system to solve:

Solve[{(y + z)^2 == res1, (y - 1) (z - 1) == res2}, {y, z}, Reals]

The results are enormous, heavily conditional, incomplete, and probably of no practical use if I did everything right. But maybe this gives you an idea of what some of the solutions you are asking for look like. It’s also possible to change up the constraints and simplify the problem similarly.

4

Dear All,
Did you get the solution?..because I have similar problem…
and second problem If I want to vary one of the constants with for loop in your case (for instanse b1) …
Is it possible
thanks in advance!