How to prove that $\vec{x}$ is a local minimum using level curves (if the hessian is inconclusive)?

Question

Suppose we have $f(x,y) = x^4+y^4+2x^3$, and trying to find the extreme values of the function. We have \begin{align*} & D_1f(x,y) = 4x^3+6x^2, \\ & D_2f(x,y) = 4y^3. \end{align*} and solving the following system \begin{align} 4x^3+6x^2 = 0, \\ 4y^3 = 0. \end{align} (1) gives $x^2(4x+6) = 0$, so $x = 0$ or $x = -\frac{3}{2}$. (2) gives $y = 0$. So we find two stationary points: $(0,0)$ en $(-\frac{3}{2},0)$.

For the Hessian we have \begin{align*} & D^2_1f(x,y) = 12x(x+1), \\ & D_{12}f(x,y) = 0, \\ & D^2_2(x,y) = 12y^2. \end{align*}

We note that $(0,0)$ is a saddle point. We have $f(0,0) = 0$. For $\epsilon \in (0,1)$ it holds that $f(-\epsilon,0) = (-\epsilon)^4 + 2(-\epsilon)^3 = \epsilon^4 -2\epsilon^3 < 0$, but on the other hand that $f(0,\epsilon) = \epsilon^4 > 0$. So $(0,0)$ is a saddle point.

Now I am supposed to prove that $\vec{x}_1 = (-\frac{3}{2},0)$ is a local minimum using level curves, i.e. $f(x,y) = c$. Note that $f(-\frac{3}{2},0) = -\frac{27}{16}$. The hessian is given by $$ H_f(-3/2,0) = \begin{bmatrix} 9 & 0 \\ 0 & 0 \end{bmatrix} $$ which is inconclusive, since it has eigenvalues $0$ and $9$. However, it follows that $\vec{x}_1$ cannot be a maximum since it is positive semidefinite.

For a level curve we have $x^4+y^4+2x^3 = c$, whence $y = \sqrt[4]{c-x^4-2x^3}$. So the level curve of level $c$ is parametrised by the curve $g(t) = (t, \sqrt[4]{c-t^4-2t^3})$, with $c \geq t^4-2t^3$.

How do I proceed from here to prove that $\vec{x}$ is a local minimum?

Answer 1

Taking

$$ f(x,y) = x^4+y^4+2x^3 $$

and making a coordinates change as

$$ x^2 = u\\ y^2 = v $$

the new function reads

$$ g(u,v) = u^2+v^2+2u\sqrt u $$

and the Hessian is

$$ H_g = \left( \begin{array}{cc} 2+\frac{3}{2 \sqrt{u}} & 0 \\ 0 & 2 \\ \end{array} \right) $$

We can observe that at $x= \epsilon$ we have

$$ H_f = \left( \begin{array}{cc} 2+\frac{3}{2 \epsilon^2} & 0 \\ 0 & 2 \\ \end{array} \right) $$

which is definite positive

Answer 2

To make things clearer, I recommend substituting $x=-\frac32+u$ for starters, and let's shift the function up by $27/16$ so that we're looking at a critical value of $0$ and the origin. Then we're considering the function $$g(u,y) = \frac92 u^2 - 4u^3 + u^4 + y^4.$$ From this, it is clear that $u=0, y=0$ is a local minimum, as $y^4\ge 0$ and $\frac92 u^2-4u^3+u^4\ge 0$ for $|u|$ near $0$. The same logic will tell you that there are no level curves with $c<-27/16$ near the point $u=0, y=0$.

Comment: If the point were a saddle point, you would expect to see a level curve through the origin that looks (locally) like a figure 8. On the other hand, it's clear that $y^4+\frac92 u^2 = c$ is empty for $c<0$ and a somewhat distorted ellipse for $c>0$.