### Author Topic: Final Review  (Read 6447 times)

#### Sabrina (Man) Luo

• Jr. Member
• Posts: 9
• Karma: 4
##### Final Review
« on: April 16, 2013, 02:29:29 AM »
As you said that if it is node, then the system is not integrable. However, what will be the way to know more about globally phase portrait?

Another one:We studied energy equation for saddle and centre in non-linear system, when it is integrable, so can we think that if it is integrable, then by looking for energy equation, the centre won't become a spiral?

#### Victor Ivrii

• Elder Member
• Posts: 2602
• Karma: 0
##### Re: Final Review
« Reply #1 on: April 16, 2013, 02:42:00 AM »
As you said that if it is node, then the system is not integrable. However, what will be the way to know more about globally phase portrait?

It is hard but possible in some cases

Quote
Another one:We studied energy equation for saddle and centre in non-linear system, when it is integrable, so can we think that if it is integrable, then by looking for energy equation, the centre won't become a spiral?

Integrable systems have neither nodes nor spirals

#### Victor Ivrii

• Elder Member
• Posts: 2602
• Karma: 0
##### Re: Final Review
« Reply #2 on: April 16, 2013, 04:34:45 PM »
I was asked today (by a student by an email) the following:

Quote
A. Can we say under proper node or improper node case in linear system that hardly (not easy to find) to have an integrable equation. It just like a spiral case?

B. If it is not like spiral case, can we say that when it is integrable, then proper node or improper node for linear system will stay the same for non-linear system? and when it is not integrable, it will become either node, or spiral for non-linear system?

Which one is ok, A or B?

If A is ok, then by which method (but not the integrable) we can tell anything globally(non-linear)?

If B is ok, then by which method we can tell it is a node, or spiral globally?

My answers (possibly useful others as well) were as follows:

1. Regarding methods to find the first integral see 2.6 , "not easy to find" is not a sufficient answer, but if you can repeat (shortly, but with a bit more details) an explanation of the Remark below it would definitely show you know:

a). If there is a 1st integral H(x,y) then the critical points of the system must be the critical points of H(x,y), i.e. gradient of H(x,y) at the points must vanish, and

b). If the Hessian of H at a critical point has a non zero determinant (we say that the critical point is non degenerate) then the phase portrait, i.e. the picture of the orbits of the system (which are level sets H(x,y)=constants), must be either like a center or like a saddle depending on the determinant being positive or, respectively, negative and consequently it can not be like either proper or improper node (answering your question A. to an extend we studied).

2. Regarding nonlinear systems phase portrait near its critical point versus the phase portrait of linearization of the system at the critical point see the table in 9.3 (beyond that it is clear that phase portrait for the original system may remain the same, e.g. linearization may coincide with the original system, but also the a). and b). of the Remark in 1. above apply ...).

In an email I can not write more (you should have made use of two fairly long "Question and Answers" sessions last Friday and Monday ...).

Anyway, good luck tomorrow.

Pierre Milman -- course coordinator.

#### Victor Ivrii

• Elder Member
• Posts: 2602
• Karma: 0
##### Re: Final Review
« Reply #3 on: April 16, 2013, 04:50:21 PM »
If system

\left\{\begin{aligned}
&\frac{dx}{dt}= f(x,y),\\
&\frac{dy}{dt}= g(x,y)
\end{aligned}\right.
\label{eq-1}

is integrable with integral $H(x,y)=C$ then

1. critical points of $H(x,y)$ (points where $H_x=H_y=0$) coincide with equilibrium points of the system (points where $f=g=0$).

2. Matrix of the linearized system at such point is

\begin{pmatrix}
&f_x & f_y\\
&g_x & g_y
\end{pmatrix}=
c\begin{pmatrix}
&H_{xy} & H_{yy}\\
&-H_{xx} &-H_{yx}
\end{pmatrix}
\label{eq-2}

and therefore eigenvalues of it  $\pm c r$ where $r=\sqrt{-\Delta}$ with $\Delta=-H_{xy}^2 + H_{xx}H_{yy}$ and therefore they are  purely imaginary opposite if $\Delta>0$ (in which case $H$ has either maximum or minimum and system has a center) and they are real and opposite and if $\Delta<0$ (in which case $H$ has a saddle and system has a saddle).

Integrable systems cannot have nodes or spiral points!!!