Is this system of equations "solvable"?

Question

I've been trying to solve the following system of equations for $p_1>0$ and $p_2>0$, but have failed to do so (in a reasonable fashion). Can anybody confirm that this is indeed not possible or show me how?

$$ 0\overset{!}{=}-{K}_1p_{1}^2-2{K}_2 p_{2} p_{1}-\delta_{1} p_{1}+1\\ 0\overset{!}{=}-2{K}_1p_{1}p_{2}-{K}_2 p_{2}^2-\delta_{2} p_{2}+1 $$

where $K_i,\delta_i>0$ for $i\in\{1,2\}$.

The $+1$-term seems to ruin everything...

Thanks a lot in advance!

Answer 1

These equations represent two conics (hyperbolas if I am right), which don't seem to enjoy particular properties.

As the conics potentially have four intersection points, the problem is of the quartic type. Elimination of one of the unknowns should lead to a quartic polynomial, and there's no shortcut.

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From the first equation, draw

$$2{K}_2 p_{2} p_{1}=-{K}_1p_{1}^2-\delta_{1} p_{1}+1.$$

Then multiply the second by $4K_2p_1^2$ to get

$$0=-4{K}_1p_1^22K_2p_{1}p_{2}-4K_2^2p_1^2p_{2}^2-2\delta_{2}p_12K_2p_1 p_{2}+4K_2p_1^2\\ =-4{K}_1p_1^2(-{K}_1p_{1}^2-\delta_{1} p_{1}+1)-(-{K}_1p_{1}^2-\delta_{1} p_{1}+1)^2-2\delta_{2}p_1(-{K}_1p_{1}^2-\delta_{1} p_{1}+1)+4K_2p_1^2,$$

the announced quartic.

Answer 2

I would take the first equation and solving this for $p_2$: so $$p_2=-\frac{K_1p_1}{2K_2}-\frac{\delta_1}{2K_2}+\frac{1}{2K_2p_1}$$ and plug this in the second one. Should a post a result for you?