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{\bf Question}

Show that the eigenvalue problem $$y''+\lambda y=0,\ y'(0)=0,\
y(1)+y'(1)=0$$

has eigenvalues $\lambda=\mu^2$ with $\mu$ any root of $\mu
\tan\mu=1$. By means of a suitable sketch, show that this equation
has a solution $\mu_n$ satisfying

$$n\pi<\mu_n<\left(n+\df{1}{2}\right)\pi,\ n=0,1,2,\cdots$$

Hence justify the approximation $\mu_n=n\pi+m_1$ where
$m_1=o(n\pi)$. Substitute this into $\mu\tan\mu=1$ and expand in
powers of $m_1$ showing that $m_1=\df{1}{n\pi}$.


\vspace{.25in}

{\bf Answer}

$y''+\lambda y=0,\ y'(0)=0\ \ (A),\ y(1)+y'(1)=0\ \ (B)$

Clearly $y=A\sin \sqrt{\lambda}x+B\cos\sqrt{\lambda}x\ \ A,\ B
const$

${}$

Boundary conditions:

$(A): 0=A\sqrt{\lambda}\cos 0-B\sqrt{\lambda}\sin 0$

$(B): 0=B\cos \sqrt{\lambda}-B\sqrt{\lambda}\sin\sqrt{\lambda}$

Assume $B \ne 0$ then must have

$$\cos \sqrt{\lambda}=\sqrt{\lambda} \sin \sqrt{\lambda}
\Rightarrow \sqrt{\lambda} \tan \sqrt{\lambda}=1$$

or if $\sqrt\lambda=\mu^2,\ \mu\tan\mu=1$

PICTURE \vspace{2in}

From diagram (for $\mu$ not so large and positive) we have

$$n\pi<\mu_n<\left(n+\df{1}{2}\right)\pi\ \ n\in {\bf{Z}}^+$$

Since root $\rightarrow 0$ in limit from above and $\tan\mu>0$ for
$n\pi<\mu<\left(n+\df{1}{2}\right)\pi\ \ n \in {\bf{Z}}^+$.

Clearly root is small and $\tan\mu=0$ when $\mu=n\pi$.

Therefore $\mu_n=n\pi+m_1$ where $m_1$ is small, say $o(n\pi)$

Substitute into $\mu\tan\mu=1$:

$$\begin{array}{l} (n\pi+m_1)\tan(n\pi+m_1)=1\\
(n\pi+m_1)\left[\df{\tan n\pi+\tan m_1}{1-\tan n\pi\tan
m_1}\right]=1\\ (n\pi+m_1)\tan m_1=1\\
\left(1+\df{m_1}{n\pi}\right)\tan m_1=\df{1}{n\pi} \end{array}$$

So to leading order in $n$

$$\tan m_1=\df{1}{n\pi}$$

As $n\to +\infty,\ m_1 \approx \tan m_1$ as $m_1$ is small
$\Rightarrow m_1 \approx \df{1}{n\pi}$

Therefore $\mu_n \sim
n\pi+\df{1}{n\pi}+o\left(\df{1}{n\pi}\right)$.



\end{document}