MHB F doesn't have in (0,0) a local minimum

[mathmari]

Hey! :giggle:

We have the function $\displaystyle{f(x,y)=y^2-3x^2y+2x^4}$ and the function $\displaystyle{g_v(t)=f(tv_1, tv_2)=t^2v_2^2-3t^3v_1^2v_2+2t^4v_1^4}$.

I have shown that $g$ has a local minimumat $t=0$

I want to show that $f$ has not a local minimum in $(0,0)$.

The gradient is \begin{equation*}\nabla f=\begin{pmatrix}-6xy+8x^3 \\ 2y-3x^2\end{pmatrix}\end{equation*} Acritical point is $(0,0)$, since $\displaystyle{\nabla f\begin{pmatrix}0 \\ 0\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}}$.
The Hessian matrix is \begin{equation*}H_f(x,y)=\begin{pmatrix}f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\end{pmatrix}=\begin{pmatrix}-6y+24x^2 & -6x\\ -6x & 2\end{pmatrix}\end{equation*}
The matrixin in $(0,0)$ is
\begin{equation*}H_f(0,0)=\begin{pmatrix}0 & 0\\ 0 & 2\end{pmatrix}\end{equation*}
The eigenvalues are, $\lambda_1=0$, $\lambda_2=2>0$. These are non-negativ, so the Hessian matrix is positiv semidefinite.
That means that $f$ has in$(0,0)$ either a minimum or a saddle point.

Is that correct so far? Now we have to show that in $(0,0)$ has a sddle point and not a minimum, right? But how? :unsure:

 

[I like Serena]

Hey mathmari!

In this case it does not work to look at what $f$ does at infinity, since there is more than 1 critical point.
Can we find all critical points and find the function values at those critical points?

 

[mathmari]

In this case it does not work to look at what $f$ does at infinity, since there is more than 1 critical point.
Can we find all critical points and find the function values at those critical points?

How does it help to check what kind of point we have at $(0,0)$ when we know all critical points? I got stuck right now. :unsure:

 

[mathmari]

Can we do also the following?

We consider the curves \begin{align*}&g(y)=f(0,y)=y^2\\ &h(x)=f(x,\frac{3}{2}x^2)= -\frac{x^4}{4}\end{align*}
Along the curve $g$ we go falling to a minimum at the origin but along the curve $h$ we go rising to a maximum at the origin.

This means that $(0,0)$ is a sattle point.

:unsure:

 

[I like Serena]

Yep. That works. (Nod)
We've found a direction up and a direction down.

Turns out there are 2 critical points. The 2nd one has a negative value, which implies that there is a direction down.