I was tasked with solving (obtain explicit solutions without utilizing complex coordinates) this system of non-linear equations as part of an exam,

\begin{cases} 2z e^{xy} +2 \lambda z - 2 \lambda = 0 \\ yz^2 e^{xy} -2x \lambda = 0\\ xz^2 e^{xy} - 2y \lambda = 0 \end{cases}

unfortunately I failed this task.

I'm hoping that if I post my work here someone could tell me why I am approaching the problem incorrectly.

When I have a system of linear equations I can use notions from linear algebra to easily find solutions, but with non-linear equations I don't know when I have obtained all the solutions to the system nor am I sure how to proceed.

I proceeded as follows:

adding the 2 equation to the last one I obtain $(y +x) (z^2 e^{xy} - 2 \lambda) = 0$ so either 1) $x = - y$ or 2) $z^2 e^{xy} = 2 \lambda$

In the case where 1) is true I obtain substituting in the second equation that $x(- z^2e^{-x^2} -2 \lambda) = 0$, so either

1.a) $x = 0$ or 1.b) $z^2e^{-x^2} = -2 \lambda$.

From 1.a) I get the "solution" $(0,0, \lambda \ (1 + 2\lambda), \lambda)$ except this is not an explicit solution so it's useless for the optimization problem this system is connected to.

From 1.b) I substitute in the first equation and get $z (2 z^2e^{-x^2}+ ze^{-x^2}+ 2e^{-x^2}) = 0 $ that gives me either

1.b. 1) $z = 0$ or 1.b.2) $z = \frac{1+ \sqrt{17}}{4}$, $z = \frac{1- \sqrt{17}}{4}$

From 1.b. 1) $z = 0$ I obtain the solution $(0,0,0,0)$.

From 1.b.2) I am unsure how to proceed. And this is only the first case.

I feel like I am approaching this with the wrong method, anyone care to solve the system and teach me a better way to approach systems of non-linear equations?


1 Answer 1


You got from the last two equations $$ (y+x)(z^2e^{xy}−2λ)=0 $$ In the same way using the difference you get $$ (y-x)(z^2e^{xy}+2λ)=0 $$ which gives the additional information that your cases should read, if not $x=y=0$

1) $y=-x$ and from the second product $2λ=-z^2e^{-x^2}$ and

2) $x=y$ and $2λ=z^2e^{x^2}$

Thus the first equation amounts in case 1) to $$ 2ze^{-x^2}+(-z^2e^{-x^2})(z-1)=0 $$ so that either $z=0$ or $2-z(z-1)=0$ with solutions $z=2$ and $z=-1$.

Similarly in case 2)


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