I am trying to solve the following problem:
Let $f: \mathbb R^2\rightarrow\mathbb R$ be a function defined by
$$
f(x,y) = x^{2n} + y^{2n} – nx^2 + 2nxy – ny^2,
$$
where $n$ is a natural number greater than 1.
Decide whether $f$ has a (global) minimum. Also find all the points at which $f$ attains its local maxima and local minima.
I calculated the partial derivatives $f_x = 2nx^{2n-1}-2nx+2ny$, $f_{xx} = 2n(2n-1)x^{2n-2}-2n$, $f_{xy} = 2n$, and so on. But I got an equation system $f_x = f_y = 0$ of degree three, which I had hard times solving it. I found out that $(x,y) = (0,0)$ is one of its solutions, but at that point the the determinant of the Hessian is zero, from which I could not conclude whether it was a local extremum.
I was not sure about how to prove that the minimum value of $f$ exists, either.
I would be most grateful if you could help me solve this problem.,
Best Answer
The point $(0,0)$ is a saddle point for $n>0$.
To see this, look at the function along the lines $y=x$ and $y=-x$: $$y=\ \ x\quad \ \to \quad f(t) = 2 x^{2 n}\quad\quad\quad\quad\quad\ \ \ $$ $$y=-x\quad \ \to \quad f(t) = (-x)^{2 n} - 4 n x^2 + x^{2 n}$$
Now take the second derivative on the line $y=-x$ to get $-8n<0$, and take the $2n$-th derivative on the line $y=x$ to get $2(2n!)>0$, thus showing this point is a saddle point and not a local maximum or minimum.
To find the minima, note that $f_x =0, f_y=0$ leads to: $$2 n x^{2 n-1} = -2 n y^{2 n-1} \quad \to \quad y = -x$$ Using the fact that $2n-1$ is odd. Now plug this into $f_x=0$ to get:
$$x\to \pm2^{-1/(2-2 n)}, \ y=-x$$ You can verify these are indeed minima.
In the graphs below:
$f(x,y)$ along the lines $y=x$ and $y=-x$ for $n=3$.