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I have a surface which is defined by equation

$$V = \frac{\frac{C}{D}}{\sqrt{\left(1-\frac{a}{k b}\frac{1-a^{-2}}{b^{-2}-1}\right) \left(1-\frac{a}{k b}\frac{b^{-2}-a^{-2}}{b^{-2}-1}\right)}}$$

Where everything else is a constant except $a, b, V$. $a$ and $b$ have some conditions so that $a \neq 0, b \neq 0, a > 1, b < 1$ and all the values are real.

What I would like to do is to set $V$ to a constant and solve the equation so that $a$ is dependant on $b$. As far as I can see this is equivelant to looking for the boundary between the surface and a plane set by $V$. The surface plot looks like this Surface plot, b, a, V

And I am interested in finding the boundary of the surface and the plane (the surface is clipped by a plane that is not visible in the plot) when approaching from the left. I am not interested on the right edge, but I do not know how do I constrain it outside. The values in between the edges (in the "valley") are imaginary (not real like the plot might suggest).

I have tried solving the equation for $a$ or $b$ but it is either impossible or the solution is extremely long/complex (I've managed to get "something" out as a solution with Maxima but it was literally hundreds of lines long) or I'm missing some constraint that simplifies it. If the solution really is so long or complex then I am happy to find just an approximation for the boundary. Numerical solution with curve fitting is also a possibility I guess but I am not sure how do I get rid of the right side edge. How can I find the boundary?

Michael Hardy
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Dago
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1 Answers1

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HINT:

To handle the zeros and poles it is suggested to treat $(a,b)$ as variables.

$$ \frac1z= \frac {VD}{C};\quad(x,y)=(a,b) ;\\ \quad z^2 =Z_1Z_2=\left( 1- \frac{(x^2-1)y}{k x (y^2-1)} \right) \cdot\left( 1- \frac{ x^2 y^2}{k x^2 (1-y^2)} \right)$$

where $ y= \pm 1 ,x=\pm1$ can be looked to for zeros/ poles. Factored functions can be considered at first as:

$$Z_1= \left( 1- \frac{(x^2-1)y}{k x (y^2-1)} \right) ,\quad Z_2=\left( 1- \frac{ x^2 y^2}{k x^2 (1-y^2)} \right) $$

Narasimham
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