Consider the following Sturm–Liouville problem $$u''+\lambda u=0, \ 0<x<1$$ $$u(0)-u'(0)=0, \ u(1)+u'(1)=0.$$ Obtain an asymptotic estimate for large eigenvalues.
I solved the problem and got the transcendental equation $\tan\sqrt{\lambda}=\frac{2\sqrt{\lambda}}{\lambda-1}.$
But how do I obtain an asymptotic estimate? I know I can use Herman Weyl formula $\int^b_a\sqrt{\frac{r(x)}{p(x)}}dx,$ but I don't know how to apply it here.
Let $\lambda = x^2$ so $\tan(x) = 2x/(x^2-1)$. It's easy to see that there is exactly one solution $x_n$ for $x$ in each interval $((n-1/2)\pi, (n+1/2)\pi)$, and (as has been remarked) $\tan x_n \to 0$ so $x_n - n \pi \to 0$. For a more detailed approximation, note that if $x_n = n \pi + a_1/n + a_2/n^2 + O(1/n^3)$ we have $$ \eqalign{\tan x_n &= \dfrac{\tan(a_1/n) + \tan(a_2/n^2) + O(1/n^3)}{1 - \tan(a_1/n) \tan(a_2/n^2) + O(1/n^3)} = \dfrac{a_1}{n} + \dfrac{a_2}{n^2} + O(1/n^3)\cr \dfrac{2 x_n}{x_n^2 - 1} &= \dfrac{2}{\pi n} + O(1/n^3)\cr}$$ so that $a_1 = 2/\pi$ and $a_2 = 0$. Thus $$ x_n = n \pi + \dfrac{2}{n\pi} + O(1/n^3)$$ and so $$ \lambda_n = n^2 \pi^2 + 4 + O(1/n^2)$$