grenivkah3z

2022-07-15

Consider the two equations below:

${y}_{1}=(1-\frac{{a}_{1}}{x}){e}^{-{\displaystyle \frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{1}}{x}}}\phantom{\rule{0ex}{0ex}}{y}_{2}=(1-\frac{{a}_{2}}{x}){e}^{-{\displaystyle \frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{2}}{x}}}$

Given ${y}_{1}$, ${y}_{2}$, ${a}_{1}$ and ${a}_{2}$, is there an analytical way to determine $\alpha $ and $x$?

${y}_{1}=(1-\frac{{a}_{1}}{x}){e}^{-{\displaystyle \frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{1}}{x}}}\phantom{\rule{0ex}{0ex}}{y}_{2}=(1-\frac{{a}_{2}}{x}){e}^{-{\displaystyle \frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{2}}{x}}}$

Given ${y}_{1}$, ${y}_{2}$, ${a}_{1}$ and ${a}_{2}$, is there an analytical way to determine $\alpha $ and $x$?

sniokd

Beginner2022-07-16Added 22 answers

A plot of the function reveals a solution close to x=0.01 and the solution, obtained using Newton, is x=0.0133409. From here, alpha=1.584997.

I also used the solution based on the Lambert function. Here again, the plot of the fuction reveal a root close to alpha=1.5. Starting from here, Newton iterations lead again to alpha=1.584997. Frm here, x=0.0133409.

Then, again, both approaches work and lead to the same results.

Please post where are your problems. As told earlier, the problem is a resonable starting guess for the solution; this can be easily obtained looking at the plot of the function.

I also used the solution based on the Lambert function. Here again, the plot of the fuction reveal a root close to alpha=1.5. Starting from here, Newton iterations lead again to alpha=1.584997. Frm here, x=0.0133409.

Then, again, both approaches work and lead to the same results.

Please post where are your problems. As told earlier, the problem is a resonable starting guess for the solution; this can be easily obtained looking at the plot of the function.

letumsnemesislh

Beginner2022-07-17Added 6 answers

Note that

$\frac{{y}_{1}^{{a}_{2}}}{{y}_{2}^{{a}_{1}}}=\frac{{(1-\frac{{a}_{1}}{x})}^{{a}_{2}}}{{(1-\frac{{a}_{2}}{x})}^{{a}_{1}}}$

This eliminates $\alpha $.

Not sure how you can solve this analytically. But once you get $x$, you can solve for $\alpha $ by taking logarithm of either equation.

$\frac{{y}_{1}^{{a}_{2}}}{{y}_{2}^{{a}_{1}}}=\frac{{(1-\frac{{a}_{1}}{x})}^{{a}_{2}}}{{(1-\frac{{a}_{2}}{x})}^{{a}_{1}}}$

This eliminates $\alpha $.

Not sure how you can solve this analytically. But once you get $x$, you can solve for $\alpha $ by taking logarithm of either equation.

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