Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this exercise :

We consider the système :

$x_1'=x_2 , x_2'=-h_1(x_1)-x_2-h_2(x_3), x_3'=x_2-x_3$ ou $h_1$ et $h_2$ are locally lipschtizen , $h_i(0)=0$ and $yh_i(y)>0$ for all $y\neq0$ (i=1,2).

(a) Show that the origin is the unique equilibrium point of the system .

(b) Show that the functional $$V(x)=\int_0^{x_1} h_1(y) dy +\frac{x_2^2}{2}+\int_0^{x_3}h_2(y) dy$$ is positive definite for all $x=(x_1,x_2,x_3)\in \mathbb{R}^3$

(c) Show that the origin is asymptotically stable

For (a) and (b) that's ok , but to answer (c) i must calculate $V'$.

And i have a probleme with $V'$.

Can someone help me to find $V'$?

Please help me

Thank you .

share|improve this question
    
    
please who is $f_x (x,t)$ ? –  Vrouvrou Jul 11 '13 at 17:46

1 Answer 1

First look at: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign if you are not sure of how to differentiate V(x) w.r.t $x_1$ or $x_2$ etc.

Then,

$V'(x) = \nabla V(x).f(x)$, where $f(x)$ is the r.h.s of your system. i.e. $f(x)=[x_1' \: x_2' \: x_3']^T$

And $\nabla V(x)= [\frac{\partial(V)}{\partial x_1} \: \frac{\partial(V)}{\partial x_2} \: \frac{\partial(V)}{\partial x_3}]^T $

share|improve this answer
    
so $V'= \frac{\partial V}{\partial {x_1}}x_1'+\frac{\partial V}{\partial {x_2}}x_2'+\frac{\partial V}{\partial {x_3}}x_3'$ right ? ,we dont derive over y ? –  Vrouvrou Jul 11 '13 at 16:26
    
y is the variable that gets integrated (i.e. it is a dummy variable), so yes, you don't differentiate w.r.t y –  nonlinearism Jul 11 '13 at 18:33
    
$V'=- h_2(x_3)x_3-x_2^2 $? –  Vrouvrou Jul 11 '13 at 20:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.