Solve these systems of equations

Question

Consider the two equations below:

$$ y_{1}=\left(1-\frac{a_{1}}{x}\right)e^{-\dfrac{\alpha\, a_{1}}{x}}\\ y_{2}=\left(1-\frac{a_{2}}{x}\right)e^{-\dfrac{\alpha\, a_{2}}{x}} $$

Given $y_{1}$, $y_{2}$, $a_{1}$ and $a_{2}$, is there an analytical way to determine $\alpha$ and $x$? I tried several approaches. Any suggestions or comments will be appreciated

Answer 1

I tried using the parameters you gave. I rewrote user44197 approach reducing to same denominator, removing the common denominator. This made the equation better condioned.

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.

Answer 2

Note that $$ \frac{y_1^{a_2}}{y_2^{a_1}} = \frac{\left(1-\frac{a_{1}}{x}\right)^{a_2}} {\left(1-\frac{a_{2}}{x}\right)^{a_1}}$$ 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.

Answer 3

I do not think that analytical methods would allow to solve this problem. But, considering your equations, there are a few things which can be done at least to reduce the problem to a single equation in "alpha" (user44197 already proposed a way to solve first for x).

Just multiply the lhs and rhs by Exp[alpha]. We then have

y1 Exp[alpha] = (1 - a1 / x) Exp[alpha (1 - a1 / x)]
y2 Exp[alpha] = (1 - a2 / x) Exp[alpha (1 - a2 / x)]

Now, introduce z1 = (1 - a1 / x) and z2 = (1 - a2 / x). So, the previous equations write

y1 Exp[alpha] = z1 Exp[alpha z1]
y2 Exp[alpha] = z2 Exp[alpha z2]

So, we have

z1 = W(t1) / alpha
z2 = W(t2) / alpha

W(t) representing Lambert function with arguments

t1 = alpha y1 Exp[alpha]
t2 = alpha y2 Exp[alpha]

Back to x, we then have

x = alpha a1 / (alpha - W(t1)) = alpha a2 / (alpha - W(t2))

So, we need to solve one equation for "alpha" and this equation can write

a1 (alpha - W(t2)) = a2 (alpha - W(t1))

When the solution of this equation in "alpha" has been solved, the value of "x" is directly obtained. It must be noticed that this equation has an obvious solution for alpha=0 which needs to be discarded.

For illustration purposes, I used a1=2, a2=3, y1=1/2, y2=-1/4. The solution corresponds to
alpha = -0.852141 and x = 2.73205