$f(x,y) = x^3+y^2-3x$
Then $H(\vec{x}) = \begin{pmatrix} \frac{\partial^{2}f}{\partial x^2} & \frac{\partial^{2}f}{\partial x \partial y}\\ \frac{\partial^{2}f}{\partial y \partial x} & \frac{\partial^{2}f}{\partial y^{2}} \end{pmatrix} =\begin{pmatrix} 6x & 0\\ 0 & 2 \end{pmatrix}$
Let $\lambda_{1}, \lambda_{2}$ be the eigenvalues of $H(\vec{x})$
Let $\tau = tr(H)$
Let $ \triangle = $det$(H) $
So I found the stationary points:
$P_{1} = (-1, 0)$
$P_{2} = (1,0)$
At $P_{1}$, we have:
$H(P_{1}) = \begin{pmatrix} -6 & 0\\ 0 & 2 \end{pmatrix} $
We have $\lambda_{1}<0, \lambda_{2}>0, \tau < 0, \triangle <0 \implies$saddle point at $P_{1}$
At $P_{2}$, we have:
$H(P_{2}) = \begin{pmatrix} -6 & 0\\ 0& 2 \end{pmatrix}$
$\lambda_{1},\lambda_{2} > 0, \triangle >0, \tau > 0 \implies$ minimum at $P_{2}$
I do not know how to go about sketching the graph/how to visualise it. Could someone please provide an answer which includes some general statements about sketching contour plots. Thanks.
The level curve $f(x,y) = c$ is $y = \pm \sqrt{c + 3x-x^3}$.
Note the symmetry about the $x$ axis.
Of course the quantity inside the square root must be nonnegative for $y$ to be real.
If $c = f(P_1) = 2$, $2 + 3 x - x^3 = (2-x)(x+1)^2$, so this is $0$ at $x= -1$ and $x=2$, positive on both sides of $x=-1$, but negative for $x > 2$. There is a local maximum at $x=1$ in the upper half plane and local minimum at $x=1$ in the lower half plane. Moreover, as $x \to -\infty$, $\sqrt{c + 3 x - x^3} \sim (-x)^{3/2}$. Thus the level curve $f(x,y) = 2$ looks like this:
For $ c > 2$, $\sqrt{c+3x - x^3}$ will be positive for all $x$ up to some value $> 2$. The curve will be on the "outside" of the previous one. In the upper half plane the curve has a local minimum at $x=-1$, local maximum at $x=1$. Again there is symmetry about the $x$ axis.
For $-2 < c < 2$, where $-2 = f(P_2)$, the curve has two parts, one to the left of the $c=2$ curve in $x < -1$, the other a closed loop inside the loop of the $c=2$ curve. For $c < -2$, there is no closed loop.