Find and classify critical points of $x^4-y^4-4xy^2-2x^2$, second derivative test is inconclusive.

Question

As in the title I found critical points to be $(1,0)$-saddle point, $(-1,0)$- local minimum and $(0,0)$ is the one I have problem with. Second derivative test is inconclusive in this case. If it was a saddle point I should find some curve in the graph that has inflection point at $(0,0)$ to prove it. Unfortunately, I suspect it to be local maximum and in this case I have no clue how can I prove that it grows in neighborhood of $(0,0)$ in every direction.

Answer 1

Consider the following graphs: \begin{align} y = x & \implies x^4-y^4-4xy^2-2x^2= -4x^3-2x^2\\ &\implies x=0\text{ is a local maximum along the curve}\\ y^2 = (-2\pm \sqrt{2})x, x<0 & \implies x^4-y^4-4xy^2-2x^2= x^4\\ &\implies x=0\text{ is a local minimum along the curve} \end{align} Hence, $(0,0)$ is a saddle point.

Answer 2

Here, I am following this definition of saddle points.

For the given function, $$f(x,y) = x^4 - y^4 - 4xy^2 -2x^2$$ $\nabla f = \vec 0$ , so it is a stationary point. Now looking at the level surfaces when $f(x,y) = 0$ $$\implies x^4 - y^4 - 4xy^2 - 2xy^2 = 0$$ $$\implies y = \pm \sqrt{-2x \pm x \sqrt{x^2 + 2}}$$ Whose implication is that that along these curve, value of $f(x,y) = 0$ remains the same, so following the definition of extremes, it should be obvious to conclude they are not one of them, hence they must be saddle point. Here is the graph of the function in the vicinity of $(0,0,0)$

Answer 3

The second-partial derivative test may leave us unedified when it is applied to functions which have terms involving power-functions with exponents larger than $ \ 2 \ . $ For the function here, the Hessian determinant function is

$$ \mathbf{H_f} \ \ = \ \ 4·(3x^2-1) \ · \ -4·(y^2 + 2x) \ - \ [ -8y ]^2 \ \ = \ \ -16·(3x^2-1) ·(y^2 + 2x) \ - \ 64y ^2 \ \ , $$

which equals zero at the origin and is inconvenient to "sort out" in the neighborhood.

Our experience with the second derivative test in single-variable calculus is that it is needful in such situations to investigate the properties of the function, and possibly the first-derivative function, in order to understand better what the behavior is in the vicinity of a point where the first two derivatives are zero.

This first graph is not u_sre's, but a graph of first partial-derivatives. The curve in dark blue is $ \ f_x \ = 0 \ \rightarrow \ y^2 \ = \ x^3 - x \ , $ an elliptic curve. The curve in green is $ \ f_y \ = \ 0 \ \rightarrow \ y·(y^2+4x) \ , $ which is the union of the $ \ x-$ axis and the "leftward-opening" horizontal parabola $ \ y^2 \ = -4x \ $ . As expected, the intersections are the three points $ \ (-1,0) \ , \ (0,0) \ , \ $ and $ \ (1,0) \ . $ The meeting of the parabola and the elliptic curve at the origin, however, suggests that complications may be present there.

This second graph is u_sre's, shaded in green where our function is positive and in red where it is negative. (The global minimum and the saddle point mentioned by Kran are marked as well.) If a local extremum occurs at a point, it would be in a single region of either color (as we see with the saddle point $ \ (1,0) \ , $ the converse does not follow). The origin, however, sits at the juncture of alternating "positive" and "negative" regions, indicating that the behavior of the function surface is not simple there. This is a sign that the origin is a saddle point of the function.