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 $\alpha > 0$.