If you remove the radicals (`sympy.solvers.solvers.unrad(eq1)`) and replace
`Q1` and `Q2` with `x` and `y` and `Q_dp_1` with a and `s1` with `b` you
will get an expression that is of degree 4 in every variable and can be
split into a term with `a` and `b` and a term with only `b` -- both with
`x` and `y`.
u1 = a*(a**3 - 4*a**2*y*(2 - b) - 2*a*y**2*(-3*b**2 + 8*b - 8) -
4*b**3*x**3 + 2*b**2*(-x**2*(-3*a + 2*y*(b - 2)) + 2*y**3*(b - 2)) -
4*b*x*(a**2 + a*y*(b - 2) + y**2*(-b**2 - 8*b + 8))) + \
b**2*(b*(x**2 + y**2) + 2*x*y*(b - 2))**2
Replace a,b with c,d (for `q_dp_2` and `s2`) to get `u2`. I can't imagine
that solving a pair of quartics is going to give a nice solution. But
solving this system with known values for `a` and `b` would be
straightforward with `nsolve`.
/c
On Thursday, August 4, 2022 at 11:54:24 PM UTC-5 klp...@gmail.com wrote:
> And eq1=Q_1*s_1 - Q_2*s_1 + 2*Q_2 - Q_dp_1 - 2*sqrt(Q_2*(Q_1*s_1 - Q_2*s_1
> + Q_2 + 2*sqrt(Q_1*Q_2*s_1*(1 - s_1)))) + 2*sqrt(Q_1*Q_2*s_1*(1 - s_1)) ...
> sorry, long day!
>
> On Thursday, August 4, 2022 at 9:39:53 PM UTC-7 Kevin Pauba wrote:
>
>> Sorry, Jeremy. Good suggestion!
>>
>> s1, s2, q_dp1, q_dp2 = sym.symbols('s_1, s_2, Q_dp_1, Q_dp_2')
>> eq1 = equ1.subs({ s: s1 }) - q_dp1
>> md( "$" + sym.latex(eq1) + " = 0$\n" )
>>
>> eq2 = equ1.subs({ s: s2 }) - q_dp2
>> md( "$" + sym.latex(eq2) + " = 0$\n" )
>>
>> soln = sym.solve([eq1, eq2], q1, q2)
>> print(f"soln = {soln}")
>>
>> I'll set simplify to False and see how it goes ...
>> On Thursday, August 4, 2022 at 8:18:00 PM UTC-7 Kevin Pauba wrote:
>>
>>> I've attached a portion of a jupyter notebook. I'm attempting to solve
>>> a simultaneous equation using sympy. The sym.solve() in the green input
>>> box doesn't return (well, I waited over night on my macbook pro). Might
>>> the solution be intractable? Is there another way to get a solution? Any
>>> help is greatly appreciated.
>>
>>
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