Does $F1(x,y) = \frac{1}{4}x^4 + x^2 y + y^2$ has a minimum through positive definite of the second derivative matrix?

Question

I took this question from the book Introduction to Linear Algebra (Gilbert Strang). It is from section 6.5, exercise 29:

For $F1(x,y) = \frac{1}{4}x^4 + x^2 y + y^2$ and $F2(x,y) = x^3 + xy - x$ find the second derivative matrix H1 and H2:

Test for minimun: $H = \begin{bmatrix}\frac{\partial^2 F}{\partial x^2} & \frac{\partial^2 F}{\partial xy}\\\frac{\partial^2 F}{\partial yx} & \frac{\partial^2 F}{\partial y^2}\end{bmatrix}$ is positive definite.

H1 is positive definite so F1 is concave up (= convex). Find the minimum point of F1 and the saddle point of F2

For $F1$, I found:

$H1 = \begin{bmatrix}3x^2+2y & 2x\\ 2x & 2\end{bmatrix}$,

so when I plug the function where the first derivative is equal to $0$, i.e. $y = \frac{-x^2}{2}$:

$H1 = \begin{bmatrix}2x^2 & 2x\\ 2x & 2\end{bmatrix}$,

it turns out that this matrix is not positive definite for any $x$.

My question is: Does it mean that this function F1 does not have a minimum?

Answer 1

Note that\begin{align}F_1(x,y)&=\frac14x^4+x^2y+y^2\\&=\left(\frac12x^2+y\right)^2.\end{align}So, it is always non-negative and it is $0$ if and only if $y=-\frac{x^2}2$. Therefore, $F_1$ has a minimum: $0$.

Answer 2

The first derivate of $F_1$ is a vector, $(4x^3+2xy, \ x^2+2y)$, and vanishing means both components are vanishing. We get $-x^3=2xy=-4x^3$, so $x=0$, which implies $y=0$. So only the point $(0,0)$ is a possible extremum.

Computing the Hessian matrix in this point we get $$ H_1(0,0)= \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix} \ . $$ We have a semidefinite Hessian matrix. (Eigenvalues are $\ge 0$). All we can say so far (by using the usual path of computing $F_1'$ and $F_1''=H_1$ in analogy to the one dimensional case) is that $(0,0)$ is still a possible extremum. (Also in the one dimensional case we have to work using other ideas in case of a zero second derivative in the critical point.) So other ideas/techniques are needed to proceed. The question can be answered by the remark that having a definite Hessian in the critical point is "only" a sufficient criterium for an extremum, having a semidefinite one is a necessary criterium. Sometimes we cannot decide, we do not have one sufficient and necessary criterium in the same time, other methods or (a closer look) are needed.

(Of course it is an absolute minimum, since the given function is the expression $a^2+ab+b^2\ge 0$ computed in $(x^2,y)$, but this was not the question, if i was interpreting it right.)