Fractional power equation. Solution domain.by Final ansatz Tags: domain, equation, fractional, power, solution 

#1
Mar1512, 05:51 AM

P: 6

Hi everyone,
I'm currently looking to solve an equation of the general form: [itex] \sqrt{x^2y^2}+\sqrt{\epsilon x^2y^2} = \beta[/itex]. I'm interested in solving this equation for [itex]x[/itex] assuming [itex]y>0[/itex], [itex]\epsilon>1[/itex] and [itex]\beta \in \mathbb{C}[/itex]. By squaring the equation twice I can find four potential solutions of the form: [tex]x = (1)^n \sqrt{ \frac{\beta^2}{(1\epsilon)^2}\Big[1+\epsilon+(1)^m \frac{2}{\beta}\sqrt{\epsilon y^2(2\epsilon)+\epsilon\beta^2y^2}\Big]} \ \ \ \ \ \ \ \mathrm{for}\ \{n,m\}\in \{1,2\}.[/tex] Now, I have tested these solutions numerically for parameters roughly in the range [itex]y\in ]1.67, 2[ [/itex] and with [itex] \epsilon \in ]1.01, 4[ [/itex]. Generally, I seem to be getting proper solutions if [itex]\beta[/itex] is "large"  but if I set e.g. [itex]y = 1.9[/itex], [itex]\epsilon = 2[/itex] and [itex] \beta = 0.02 + i 0.01[/itex] then the solutions are wrong. I'm consequently quite convinced that [itex] \sqrt{x^2y^2}+\sqrt{\epsilon x^2y^2} = \beta[/itex] only has solutions for certain parameters choices  what I want to find out is; can I analytically express when the equation has a solution? I.e. when is the solution domain of the equation empty? I'll look forward to reading your replies! 



#2
Mar1512, 06:29 AM

P: 4,570

Hey Final ansantz and welcome to the forums.
For your question I'm going to assume that B is approximately zero. This implies: √(x^2  y^2) + √(ex^2  y^2) = β ~ 0 which implies: √(x^2  y^2) ~ √(ex^2  y^2) which implies x^2  y^2 ~ ex^2  y^2 which implies x^2  ex^2 ~ 0 which implies x^2[1  e] ~ 0 which implies x^2 = 0 or e = 1 for this approximation. But if this relation holds (or is a good approximation which will be the case for B being really small), then if e is significantly different from 1, then you will have a problem and get 'nonsensical' results. If you wanted to do a better analysis of saying 'what' values should be used to get a solution, then instead of making B = 0, you would introduce some kind of epsilon term that corresponds to information about the norm of B. But yeah given the above, if your e value is near two, then I can see where your answers would give radically different answers that don't make much sense. 



#3
Mar1512, 08:18 AM

P: 6

Thanks for your reply and your kind welcome chiro.
I agree that your suggestion is a good way to qualitatively understand the origin of the problems  and it's clear that the issue becomes significantly more complicated when [itex]\epsilon \neq 1[/itex]. I would however still be very interested in any quantitative, analytical statements about the solution domain of the problem. 



#4
Mar1612, 12:44 AM

P: 4,570

Fractional power equation. Solution domain.What will happen is that because you have constraints on your [itex]\epsilon[/itex] and [itex]\beta[/itex] this means you will have constraints on y and subsequently x as well. If the solutions go outside of these constraints then you will get answers that make no sense since basically these answers assume values that lie outside of your constraint which means that you from an equality to a nonequality (I won't say inequality because that's not what it is: it's a nonequality). Probably the best way I think you should go about this is to analyze what the solution space is for the lower and upper bounds of your [itex]\beta[/itex] and [itex]\epsilon[/itex] intervals and then using these results get all of the intervals for x and y. So for example with epilson, first set it to zero and get properties of the rest, set it to infinity, get the properties for the rest then do the same for beta and take the intersection of all the results for x and y and that will tell you what y should be and as a result what x should also be since x is just in terms of y. 



#5
Mar1612, 01:14 AM

P: 5

Thanks for your reply and your kind welcome chiro.



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