Transform three inequalities to just two (what is the general method for this problem?)

Question

I am very anxious to know the answer to the following question, thus I might have missed threads that adress this topic. I suppose this is part of elementary calculus, nevertheless I found myself wodering if my parameter intervals where the solution is valid could be made tidier.

I have the inequalities

$$ \begin{align} \beta &> 0, & \alpha &\geq \frac{4\beta}{\Delta x^{2}},\quad & 1\geq \frac{\Delta t}{\Delta x^{2}}\left(\alpha-\frac{\beta}{\Delta x^{2}}\right). \end{align} $$

What ways are allowed to combine such to show all relationships? I would like to have intervals for each of the parameters $\alpha$ and $\beta$, I made an attempt and found

$$ \begin{align} \beta &> 0\quad\quad\text{and} & \alpha &\geq \frac{4\beta}{\Delta x^{2}}\iff \frac{\Delta x^{2}\alpha}{4} \geq\beta\implies & \frac{\Delta x^{2}\alpha}{4} \geq\beta>0,\\[3mm] \end{align} $$ $$ \begin{align} \alpha&>0\quad\quad\text{and}&1\geq \frac{\Delta t}{\Delta x^{2}}\left(\alpha-\frac{\beta}{\Delta x^{2}}\right)\iff\frac{\Delta x^{2}}{\Delta t}+\frac{\beta}{\Delta x^{2}}\geq\alpha\implies & \frac{\Delta x^{2}}{\Delta t}+\frac{\beta}{\Delta x^{2}}\geq\alpha>0. \end{align} $$

But I am not very happy with my inequalities, since each parameter depends on the other. Is there a better way so that to avoid this, and still have all relationships expressed in the two intervals?

Edit: I missed to mention that $\alpha > 0$.

Answer 1

I don't know if there is a "general method" for this. Especially since you just want it to look nicer. And you can't expect the parameters not to depend on each other; they will, unless the region is a box. But drawing the region is always good! In this case we get a triangle. I would probably express the region in either of two ways: $$ 0 < \beta, \quad\frac{4}{(\Delta x)^2}\beta \le \alpha \le \frac{1}{(\Delta x)^2}\beta + \frac {(\Delta x)^2}{\Delta t} $$ or $$ 0 < \beta, \quad (\Delta x)^2\alpha - \frac{(\Delta x)^4}{\Delta t} \le \beta \le \frac{(\Delta x)^2}{4}\alpha $$ Either of these should be equivalent to the original system.

EDIT: This is under the assumption $\Delta t> 0$. Otherwise I think there are no solutions.