<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

Find the eigenvalues and eigenfunctions for the differential
equation $$ y''+\lambda y=0 $$ with the following boudary
conditions
\begin{description}
\item[(a)]$y(0)=0, \qquad y'(1)=0$
\item[(b)]$y'(0)=0, \qquad y(1)=0$
\item[(c)]$y'(0)=0, \qquad y'(1)=0$
\item[(d)] $y'(0)=0, \qquad y'(1)+y(1)=0$
\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]

 $$ y''+\lambda y=0 \hspace{1in}(*) $$ There are three
different cases to consider; $\lambda&lt;0$, $\lambda=0$ and
$\lambda&gt;0$. We consider them each in turn.

(i) $\lambda&lt;0$. In this case we let $\lambda=-k^2$, (where $k \neq
0$) and equation (*) becomes

$\ds y''-k^2y=0, \quad \Rightarrow \quad y=A\cosh (kx)+B\sinh
(kx)$.

Using the boundary condition $y(0)=0, \Rightarrow A=0$, so

$y=B\sinh (kx)$.

Differentiating gives $y'=kB\cosh(kx)$ and hence $y'(1)=0$ gives

$kB\cosh k=0 \Rightarrow B=0$ (since $k \neq 0$).

So there is no non-trivial solution if $\lambda&lt;0$.

(ii) $\lambda=0$. In this case the equation becomes $y''=0,
\Rightarrow y=A+Bx$.

The boundary condition $y(0)=0 \Rightarrow A=0$, and the boundary
condition $y'(1)=0, \Rightarrow B=0$. So there is no non-trivial
solution for $\lambda=0$.

(iii) $\lambda&gt;0$. In this case we let $\lambda=k^2$ (where $k \neq
0$) and equation (*) becomes

$y''+k^2y=0, \quad \Rightarrow \quad y=A\cos(kx)+B\sin(kx)$.

The boundary condition $y(0)=0$ gives $A=0$, so that $y=B\sin(kx)$.

Hence $y'=Bk\cos(kx)$ and the boundary condition $y'(1)=0$ gives
$Bk\cos k=0$.

For non-trivial solutions we require $\cos k=0$ and hence

$\ds k={{(2n-1)\pi} \over 2}, \quad n=1,2,3,\ldots$.

The corresponding eigenvalues and eigenfunctions are

$\ds \lambda_n={{(2n-1)^2\pi^2} \over 4}, \quad
y_n=\sin\left[{{(2n-1)\pi x} \over 2}\right], \quad n=1,2,3,\ldots$.

\item[(b)]
As in part (a) there are no non-trivial solutions unless
$\lambda&gt;0$. We write $\lambda=k^2$ (with $k \neq 0$) and (*) becomes

$y''+k^2y=0$, with solution $y=A\cos(kx)+B\sin(kx)$, and derivative
$y'=-Ak\sin(kx)+Bk\cos(kx)$.

Using the boundary condition $y'(0)=0$ gives $B=0$ and hence
$y=A\cos(kx)$.

Using the boundary condition $y(1)=0$ gives $A\cos k=0$ so for
non-trivial solutions we require:

$\ds \cos k=0, \quad \Rightarrow \quad k={{(2n-1)\pi} \over 2},
\quad n=1,2,3,\ldots$

The corresponding eigenvalues and eigenfunctions are

$\ds \lambda_n={{(2n-1)^2\pi^2} \over 4}, \quad
y_n=\sin\left[{{(2n-1)\pi x} \over 2}\right], \quad n=1,2,3,\ldots$.

\item[(c)]
There are no non-trivial solutions when $\lambda&lt;0$, however in this
case there are non-trivial solution when $\lambda=0$ or $\lambda&gt;0$.

When $\lambda=0$ the equation becomes $y''=0$ with solution $y=A+Bx$.

Hence $y'=B$ and the boundary condition $y'(0)=0$ requires
$B=0$. However with this condition the other boundary condition
$y'(1)=0$ is automatically satisfied and hence $y=A$ satisfies the DE
and the boundary conditions.

Hence $\lambda_0=0$ is an eigenvalue with $y_0=1$ the
corresponding eigenfunction.

For $\lambda&gt;0$ the solution is as usual $y=A\cos(kx)+B\sin(kx)$, with
derivative $y'=-Ak\sin(kx)+Bk\cos(kx)$.

The boundary condition $y'(0)=0$ gives $B=0$ and hence

$y'=-Ak\sin(kx)$

The other boundary condition $y'(1)=0$ now gives $-Ak\sin k=0$ so for
non-trivial solutions we require:

$\sin k=0, \quad \Rightarrow \quad k=n\pi, n=1,2,3,\ldots.$

The corresponding eigenvalues and eigenfunctions are

$\ds \lambda_n=n^2\pi^2, \quad
y_n=\cos(n\pi) \quad n=1,2,3,\ldots$.

Note that if we allow $n=0$ this includes the case of the zero
eigenvalue.

\item[(d)]
As in part (a) there are no non-trivial solutions unless
$\lambda&gt;0$. We write $\lambda=k^2$ (with $k \neq 0$) and (*) becomes

$y''+k^2y=0$, with solution $y=A\cos(kx)+B\sin(kx)$, and derivative
$y'=-Ak\sin(kx)+Bk\cos(kx)$.

The boundary condition $y'(0)=0$ gives $B=0$ so $y=A\cos(kx)$
and $y'=-Ak\sin(kx)$.

Applying the second boundary condition $y(1)+y'(1)=0$ gives $\cos
k-k\sin k=0$ which implies $k=\cot k$.

By drawing the graphs of $y=\cot x$ and $y=x$ we see that $k=\cot k$
has an infinite number of positive roots; $k_1, k_2, k_3, \ldots$.

The corresponding eigenvalues and eigenfunctions are

$\ds \lambda_n=k_n^2, \quad
y_n=\cos(k_n\pi) \quad n=1,2,3,\ldots$.

Where $k_n$ is the $n$-th positive root of $x=\cot x$.

Note this eigenvalue problem arises in a problem in quantum mechanics.
\end{description}


\end{document}
</pre></body></html>