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Given equation

$$\begin{array}{cc} \omega(k)= k^3+8\Lambda \sin(k) & \textrm{ with } k>0, \Lambda>0 \end{array}$$

Clearly $\omega(k)$ has a minimum for $k \approx 4$ which i will call $k_{min}$. For greater $\Lambda$ the minimum appears at a smaller $\omega$ value.

Is there a way to determine analytically an explicit form of $\Lambda_c$ for which

$\frac{d\omega}{dk}(k_{min})=0 $ and $\omega (k_{min})=0$ ?

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2 Answers 2

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There is not much hope for a closed form solution.

Consider the two equations

$$3k^2+8\Lambda\cos k =0 \\ k^3+8\Lambda\sin k=0.$$

Dividing them (and excluding $k=0$) we obtain $$\tan k = \frac{\sin k}{\cos k} = \frac{-k^3}{-3k^2}\qquad\Rightarrow\qquad \tan k=\frac{k}{3}\qquad\text{(1)}$$

Adding them we have $$(8\Lambda)^2(\sin^2 k +\cos^2 k) = (-3k^2)^2 + (-k^3)^2$$ and as $\sin^2 k+\cos^2 k =1$ we obtain $$\Lambda = \frac{k^2\sqrt{k^2+1}}{8}.\qquad\text{(2)}$$

To answer your question you need to solve (1) for $k$ and then substitute it into (2). However, because there is no known analytical solution for (1), there is also none for your problem (otherwise putting this hypothetical solution into (2) and inversion would lead to an analytical solution for (1)).

The best you can do is solve (1) numerically or approximately based on additional assumptions, e.g. a perturbative expansion around $k\approx 4$.

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  • $\begingroup$ Interesting--but somehow I find this familiar... $\endgroup$
    – Did
    Apr 16, 2014 at 14:37
  • $\begingroup$ @Did interesting, how do you find this familiar? $\endgroup$
    – flonk
    Apr 16, 2014 at 18:01
  • $\begingroup$ Seriously? Look on the page, I am sure you will find out... $\endgroup$
    – Did
    Apr 16, 2014 at 18:03
  • $\begingroup$ @Did Sorry for me being slow, noticing irony in written language can be hard sometimes. If that was your point, yes I now see that my solution is very similar to yours, thanks. Should I delete it? $\endgroup$
    – flonk
    Apr 16, 2014 at 18:12
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Assume that $\omega(k)=\omega'(k)=0$ then $a\sin k+k^3=a\cos k+3k^2=0$ with $a=8\Lambda$ hence $a^2=(a\cos k)^2+(a\sin k)^2=(3k^2)^2+(k^3)^2$. Solving the equation $x^3+9x^2=a^2$ for $x\gt0$ yields some explicit unique solution $x(a)$ (for example, using Cardano's formula for cubics) hence $a$ must solve $a\cos\sqrt{x(a)}+3x(a)=0$. Not sure one can make this more explicit...

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  • $\begingroup$ Thank you, but i was looking for a possible way to get an explicit solution, and i just noticed that i forgot to specify it in the question. $\endgroup$
    – PhilipV
    Apr 16, 2014 at 14:15

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