Non-linear programming problem from a trapping region in a nonlinear ODE system?

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I have the next function... wich describes a trapping region from a glycolysys model of Sel'kov... I need to minimize

$z=h(u)=u^3(\frac{1}{A}-\frac{B}{A^2})+u^2(\frac{\epsilon B}{A^2}-1-\frac{\epsilon}{A}+\frac{B}{A})-\frac{u}{A}+\frac{b}{B}$

subject to

$A>0$

$B>0$

$\epsilon>0, \epsilon<<1$

$b>0$

For what value of $b$ can we find a minimum?. So this value of $b$ we'll call it $b^*$. For a general way to see this problem we can say that we need to prove (for a certain value of $b$)

$h(u)\leq 0$

Wich method should I use or what theory. I'm stuck in this part of this problem because I need to find $b^*$ to continue solving my nonlinear ODE system.

I'm a ODE student so I'm notvery familiar with non-linear programming problems.

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I assume here that $A$, $B$, $b$, and $\varepsilon$ are constants. If not, then what follows won't be of much use. I would start by writing $z$ as

$$ z = \frac{1}{A}(c_1 u^3 + c_2 u^2 - u) + \frac{b}{B}, $$

where

$$ c_1 = 1 - \frac{B}{A}, \quad c_2 = \varepsilon\left(\frac{B}{A}-1\right) + B - A $$

Now consider cases. If $B = A$, then $c_1 = c_2 = 0$ and $z = -u/A + b/B$, which is easy to work with. If not, then we can use calculus to try and find a local minimizer. Differentiating $z$ with respect to $u$ we get

$$ z' = 0 \iff 3c_1 u^2 + 2c_2 u - 1 = 0 $$

Applying the quadratic equation gives the roots of this equation:

$$ u_{1,2} = \frac{-c_2 \pm \sqrt{c_2^2 + 3c_1}}{3c_1} $$

Here again there are a few cases to consider. If $3c_1 \geq -c_2^2$, then there exist critical points (one or two). If not, then there do not exist any critical points. If $u$ is constrained to a closed and bounded interval $[x_1,x_2]$, then you need to test $z$ at $x_1$, $x_2$, and any critical point that gives you a local minimizer, should such a point exist. Otherwise, if $u$ belongs to an unbounded interval then $z$ may become arbitrarily small or large depending on the coefficients $c_1$ and $c_2$.

I hope this helps to get you started.