MAT244-2018S > Final Exam

FE-P6

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Tim Mengzhe Geng:

--- Quote from: Syed_Hasnain on April 12, 2018, 01:25:37 AM ---there is a small mistake..... in step 5 you have mentioned that h(y) = 0
it is notzero, it is a constant

--- End quote ---
Please note that at I state, I just choose $h(y)=0$ for simplification.

Tim Mengzhe Geng:
For Part(a), Note that for stationary points, we should have
\begin{equation}
    x^2+y^2-1=0
\end{equation}
And at the same time
\begin{equation}
    -2xy=0
\end{equation}
Therefore there're totally four stationary points. They are
\begin{equation}
    (x,y)=(1,0), (-1,0), (0,1) or (0,-1).
\end{equation}
\begin{equation}
   J={
\left[\begin{array}{ccc}
2x & 2y \\
-2y & -2x
\end{array}
\right ]},
\end{equation}
At point (1,0),
\begin{equation}
   J[1,0]={
\left[\begin{array}{ccc}
2 & 0 \\
0 & -2
\end{array}
\right ]},
\end{equation}
At point (-1,0),
\begin{equation}
   J[-1,0]={
\left[\begin{array}{ccc}
-2 & 0 \\
0 & 2
\end{array}
\right ]},
\end{equation}
At point (0,1),
\begin{equation}
   J[0,1]={
\left[\begin{array}{ccc}
0 & 2 \\
-2 & 0
\end{array}
\right ]},
\end{equation}
At point (0,-1),
\begin{equation}
   J[0,-1]={
\left[\begin{array}{ccc}
0 & -2 \\
2 & 0
\end{array}
\right ]},
\end{equation}

Nikola Elez:
Ah I seem to have missed two points, thanks for adding the full solution Tim!

Victor Ivrii:
Observe that Hessian of $H(x,y)$ is
$$
\begin{pmatrix}
2x &2y\\
2y &2x
\end{pmatrix};
$$
compare with the  Jacobi matrix (Jacobian is its determinant). In this particular case (of exact system) sometimes it is called skew-Hessian.

I attach the Contour plot of $H(x,y)$; note that $(-1,0)$ is the local maximum and $(1,0)$ is the local minimum, while $(0,\pm 1)$ are two saddle points

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