Nonlinear ODE System: Computing w' & Finding R

[zokomoko]

Given the ODE system:
v' = u(u²-1)
u' = v-u

Define w=u²+v². Compute w'.
Find the largest radius R for which u²+v²<R so that the if the solution curve (u,v) is inside that circle the solution tends to (0,0) as t--> +\infty


Any guidance would be appriciated !

 

[Dickfore]

First, use the chain rule of differentiation and the expressions for u&#039; and u&#039; to find w&#039;. Please show us the result of your work.

 

[zokomoko]

sorry I forgot to mention I only had difficulty with the second part of the question.

W'= 2uu'+2vv'=2u[v-u]+2v[u(u²-1)]=2vu-2u²+2vu³-2uv
W' = -2u²+2vu³

I think (but perhaps I'm mistaken) W is suppose to be a lyoponouv function and I'm suppose to find the radius R in which W is monotically decreasing thus proving that the origin is stable fixed point (so every solution tends to the origin) in the said circle.

so my problem is the second part, how to find the radius R in which W is monotically decreasing, if what I've written earlier is even correct..

thank you for your reply :-)

 

[elibj123]

What is the condition so W(t) would monotonically decrease?

 

[zokomoko]

W'=-2u²+2vu³<0

v>0, u<0 no problem
v<0, u>0 no problem

u,v both positive or both negitive are problematic because W' can be positive in those regions, no ?

 

[Dickfore]

W'=-2u²+2vu³<0

v>0, u<0 no problem
v<0, u>0 no problem

u,v both positive or both negitive are problematic because W' can be positive in those regions, no ?

This is not correct.

 

[zokomoko]

Could you please elaborate ?

 

[elibj123]

Simplify the inequality you got to get a simpler relation.

 

[Dickfore]

Could you please elaborate ?

You can factorize the inequality you got. Then use the following rule:

<br /> A B &lt; 0 \Leftrightarrow \left[\begin{array}{l}<br /> \left\{\begin{array}{l}<br /> A &gt; 0 \\<br /> <br /> B &lt; 0<br /> \end{array}\right. \\<br /> <br /> \left\{\begin{array}{l}<br /> A &lt; 0 \\<br /> <br /> B &gt; 0<br /> \end{array}\right.<br /> \end{array}\right.<br />