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I have the following equations and I am trying to solve them while taking $\lambda$ $\neq$ $0$.

Eq 1: $4x_1^3$ - $2 \lambda x_1$ = $0$

Eq 2: $-4x_2^2$ - $2 \lambda x_2$ - $\lambda$ - $1$ = $0$

Eq 3: $\lambda ( x_1^2+ x_2^2 + x_2)$ = $0$

I have been trying to solve $\lambda$, $x_1$, and $x_2$ but going around in circles for this case. I found (0,0) and (0, -1) but nothing afterwards.

Kurapika
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  • As written, your second equation can be simplified to $\ -4x_2^2-2\lambda x_2-2=0\ .$ Is that what you really meant, or is there a typo in what you've written? – lonza leggiera Dec 19 '23 at 11:05
  • @Kurapika are you sure about equation 2? This seems to be a system arising from the application of the KKT conditions and the last $-1$ should probably be $-\lambda$. – PierreCarre Dec 19 '23 at 11:06
  • @lonzaleggiera I made a typo. I have edited the question to include the correct equations that have been giving trouble. – Kurapika Dec 19 '23 at 11:33

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If you are trying to find the minimum of $f(x_1,x_2)=x_1^2-2 x_2^2-x_2$, subject to $x_1^2+x_2^2+x_2 \leq 0$, the proper KKT system would be $$ \begin{cases} 2x_1 - 2 \lambda x_1 = 0\\ -4 x_2-1 -\lambda(2x_2+1)=0\\ \lambda (x_1^2+x_2^2+x_2) = 0\\ x_1^2+x_2^2+x_2 \leq 0\\ \lambda \leq 0 \end{cases}. $$ How did you get to the system you presented?

Regarding the amended system, you still have a typo in the second equation, it should be: $-4x_2 - 1 -2 \lambda x_2 -\lambda = 0$. Once you solve the system you'll find the following solutions: $(0,0)$, $(0, -\frac 14)$ and $(0,-1)$, all with $\lambda \leq 0$.

The solution $(0,-\frac 14)$ is in fact an interior saddle point and you end up with the two points you had already mentioned in your post. Why were you expecting more minima?

PierreCarre
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  • Thanks this makes sense and it brought to my attention that I actually made two massive typos. No wonder my question wasn't making sense. I have fixed it in the original question now : f(x1, x2)= x1^4-2x^2-x2. – Kurapika Dec 19 '23 at 11:45