MAT244--2018F > Final Exam

FE-P6

<< < (5/5)

Qinger Zhang:

--- Quote from: Victor Ivrii on December 16, 2018, 10:30:25 PM ---Question: For $H(x,y)$ what are points $(\pm 4,0)$ and $(0,0)$?

--- End quote ---
From the grape, I don't think any of them is max or min.

Victor Ivrii:
(a)  Since $x^2+y^2+4>0$  all stationary points are at $y=0$, and either $x=0$ or $x^2+y^2-16=0$; thus we have
$A_1=(0,0)$ and $A_{2,3}=(\pm 4,0)$.

(b) Let $f=2y(x^2+y^2+4)$, $g=-2x(x^2+y^2-16)$.

At $A_1$ we have $f_x=0$, $f_y=8$, $g_x=32$, $g_y=0$ and linearization is
$$\begin{pmatrix}X\\ Y\end{pmatrix}'=\begin{pmatrix} 0 &8\\32 &0\end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$$
 with eigenvalues $\pm 16$ and eigenvectors $\begin{pmatrix}1 \\\pm2\end{pmatrix}$ so  it is a saddle:

At points $A_{2,3}$ we have $f_x=0$, $f_y=16$, $g_x=-16$, $g_y=0$ with eigenvalues $16i$ and $-16i$; in virtue of (c) it is a center, and since the bottom left element is negative it is clockwise.


Rewriting the system as
$$
2x(x^2+y^2-16)\,dx + 2y(x^2+y^2+4)\,dy=0
$$
one can check that the form on the left is exact; $H_x= 2(x^3+xy^2-16x)\implies H(x,y)=\frac{1}{2}(x^4 +2x^2y^2 -32 x^2)+h(y)$ and plugging to $H_y=2x^2y +h'(y)= 2yx^2+2y^3 +8y\implies h'(y)=2y^3 +8y\implies h(y)=\frac{1}{2}y^4+4y^2$ (we take a constant equal $0$). Then
$$H(x,y)=\frac{1}{2}\bigl(x^4 +2x^2y^2 +y^4-32 x^2 +8y^2\bigr).$$

Points $(\pm 4,0)$ are minima of $H(x,y)$ and $(0,0)$ is the saddle point. So called symmetric two-well 2D potential.
See 3D plot generated by WolframAlpha

Navigation

[0] Message Index

[*] Previous page