/sci/ - Science & Math

I am currently studying the equilibria of some non-linear ODE system :
[math]

\left\{
\begin{array}{ll}
V(t)= \theta_0 - \theta_1 N(t) \\
\dfrac{ \text{d}N(t)}{\text{dt}} =F(N,t,\theta_0,\theta_1,\theta_2) \\
\end{array}
\right.
[/math]
With [math] V [/math], [math] N [/math] the two unknowns and [math]\theta_0,\theta_1,\theta_2 [/math] some parameters. [math] F [/math] is non-linear but is [math] \mathcal{C^{\infty}}[/math].
To study the equilibria over [math] N [/math], I extracted several conditions (inequalities) on the parameters. For instance if [math] G_{0}(\theta_0) < G_{1}(\theta_0,\theta_2) [/math] then we have two equilibria , etc...
Since [math] V [/math] is just a linear translation of [math] N [/math] (as per the first line of the system), I know the equilibria states of [math] V [/math] and [math] N [/math] are equivalent.

My question is : does that equivalence also applies to the equalities I used to find the equilibria states of [math] N [/math]? More precisely, if I were to rewrite the second line with [math] V [/math] only, would I find the same inequalities to verify when working on [math] N [/math] in order to have the equilibrium states of [math] V [/math]? I am not sure if I am retarded or not.

I am trying to calculate the mean velocity of a fluid in a rectangular pipe.
Height is [math]h [/math], width is [math]w [/math], so a cross section of the pipe is [math]S=h×w [/math]. My velocity is an Hagen-Poiseuille thingy like this : [math] v(y) = -\frac{G_1}{2 \rho \nu}y^2 +\frac{G_1}{2 \rho \nu}hy [/math].
The maximal velocity is [math] v(\frac{h}{2}) = v_{max} = \frac{3G_1}{2 \rho \nu}h^2 [/math]
To compute the average value over the cross section [math]S [/math] , I did :
[math] \displaystyle v_{avg} = \frac{1}{S} \iint_{[0,w]×[0,h]} \left (-\frac{G_1}{2 \rho \nu}y^2 +\frac{G_1}{2 \rho \nu}hy \right ) \,dy \,dw' = \frac{G_1}{12 \rho \nu}h^2 = \frac{1}{18}v_{max} [/math]
Am I retarded? Why is the average speed is so low compared to the max?
When I did the same shit for a tube flow, I got [math] v_{avg} = \frac{1}{2}v_{max} [/math] which seems much more reasonable. I can't believe it changes that much if you add four corners in your pipe. Can anyone confirm if my approach is correct?