I began to solve for $r_2$ from the following inequality, but it seems unlikely I would be able to obtain its solution because of its high nonlinearity (due to presence of $r_2^2$ and $lnr_2$ simultaneously). I would like the solution for $r_2$ from the following inequality.

$$x\leq y$$


\begin{aligned} x=\left[-(R^4-r_2^4) + 4\left\{\frac{r_2^2(1-\alpha)+R^2\alpha-r_1^2}{(1-\alpha)ln(r_2)+\alpha ln(R)-ln(r_1)}\right\}\left\{\frac{R^2 ln(R)-r_2^2 ln(r_2)}{2} - \frac{(R^2 - r_2^2)}{4}\right\} + 2\left\{\frac{(1-\alpha)(R^2 ln(r_2)-r_2^2 ln(R))-R^2 ln(r_1)+r_1^2 ln(R)}{(1-\alpha)ln(r_2)+\alpha ln(R)-ln(r_1)}\right\}(R^2 - r_2^2)\right] \end{aligned}


\begin{aligned} y=\left[-(R^4-r_1^4) + 4\left\{\frac{R^2-r_1^2}{ln(R)-ln(r_1)}\right\}\left\{\frac{R^2 ln(R)-r_1^2 ln(r_1)}{2} - \frac{(R^2 - r_1^2)}{4}\right\} + 2\left\{\frac{r_1^2 ln(R)-R^2 ln(r_1)}{ln(R)-ln(r_1)}\right\}(R^2 - r_1^2)\right] \end{aligned}

  • $\begingroup$ Is $y$ the same expression as $x$ only taken at $\alpha = 1$? $\endgroup$ – dxiv Oct 4 '16 at 20:15
  • $\begingroup$ Hello. No, $x$ and $y$ differ in terms of variables $\alpha$ and $r_1$, $r_2$... $\endgroup$ – user754 Oct 4 '16 at 20:17

I'd approach by looking at the 'special' values, as per dxiv's suggestion:

  • At $r_2=R$, $x$ vanishes.
  • At $\alpha=1$, $r_2=r_1$ is the break where the inequality doesn't/holds.
  • At $r_2\rightarrow\infty$, the start of the domain, again $x$ vanishes.
  • $x\rightarrow\infty$ as $r_2\rightarrow\infty$

Beyond these simple cases you can look at, I don't think you can solve this analytically, but note that since $y$ is independent of $r_2$, this is essentially an equality test. I suggest deciding on domains for $r_1$ and $R$ if possible, and numerically solving.

  • $\begingroup$ You need to read about the xyproblem in stack exchange: mywiki.wooledge.org/XyProblem. You did this twice, with a cross over to overflow. The Navier-Stocks formula is extremely complicated and almost never has an analytical solution. I'll try to think about that question and answer there! $\endgroup$ – kabanus Oct 22 '16 at 7:32

Firstly, I don't think you want to solve an inequality for a single value for a single variable (unless you mean that other variables are held constant?)

If you substitute a value of $\alpha = 1$ into x, you receive something similar to y (well, with $r_2$ substituted for $r_1$ in the difference of squares). So, my suggestion is to take the derivative of x with respect to $\alpha$, and examine for which regions of alpha.

Are you absolutely sure about the steps you've taken to get this far? This seems exceedingly difficult for precalc/algebra. Perhaps there is a step of simplification that has been missed here.

  • $\begingroup$ I didn't have enough reputation to do that $\endgroup$ – Owen Hempel Oct 22 '16 at 20:06
  • $\begingroup$ Well, I disagree. Read the first paragraph $\endgroup$ – Owen Hempel Oct 22 '16 at 20:26

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