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\newcommand{\ds}{\displaystyle}
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\begin{document}


{\bf Question}

Use the WKB solution to estimate the eigenvalues $k$ of the
problem

$$y''+\df{k^2}{x^2}y=0,\ y(1)=0,\ y(e)=0,\ k \to +\infty$$

Show that the equation can be solved exactly to give

$$y=a\sqrt{x}\cos\left(\sqrt{k^2-\df{1}{4}}\log
x\right)+b\sqrt{x}\sin l\left(\sqrt{k^2-\df{1}{4}} \log x\right)$$

Hence find the exact eigenvalues and compare them with the WKB
approximations. Will higher order terms in the WKB expansion
improve the accuracy of that approximation?

\vspace{.25in}

{\bf Answer}

$y''+\df{k^2}{x^2}y=0,\ \undb{y(1)=0,\ y(e)=0}$

\hspace{1.1in} $\Rightarrow x=O(1),\ k\to\infty$ limit

Can solve by $y \sim e^{g_0(k)\psi_0(x)+\cdots},\ k\to\infty$
limit ansatz, or jump to equations (5.61) (to save space here!) in
section 5.5 of the notes.  Identify $h(x)=\df{1}{x^2}$ with
$h(x)&gt;0$ in $1 \leq x \leq e$.

Therefore from (5.61) we have

$y \sim \df{A}{(\frac{1}{x^2})^\frac{1}{4}}\cos\left\{k\ds\int^x
\sqrt{\df{1}{x^2}}\,
dx\right\}+\df{B}{(\frac{1}{x^2})^\frac{1}{4}}\sin\left\{k\ds\int^x
\sqrt{\df{1}{x^2}}\, dx\right\},$

$k\to\infty,\ x=O(1)$

so

$y \sim A\sqrt{x}\cos(k\ln x)+B\sqrt{x}\sin(k \ln x),\
k\to\infty,\ x=O(1)$

$A$ and $B$ from boundary conditions

$\Rightarrow \begin{array}{l}0 \sim A\sqrt{1}\cos(k\ln
1)+B\sqrt{1}\sin(k \ln 1)\\ 0 \sim A\cos 0+B\sin 0 \end{array}$

$\Rightarrow \un{A=0}$

and

$0 \sim A\sqrt{e}\cos(k \ln e)+B\sqrt{e}\sin(k \ln e)$

$0 \sim B \sqrt{e}\sin k$ since $A=0$. $\ln e=1$ by definition

so either $B=0$ BORING! ($y \sim 0$) or $\sin k=0 \Rightarrow
k=n\pi,\ \ n \in \bf{Z}$

Therefore $y \sim B\sqrt{x}\sin(n\pi \ln x)$ with eigenvalues
$\un{k_n \sim \pi}, n$ integer.

Given trial solution:

$\begin{array}{rcl}y &amp; = &amp;
a\sqrt{x}\cos\left(\sqrt{k^2-\df{1}{4}}\log
x\right)+b\sqrt{x}\sin\left(\sqrt{k^2-\df{1}{4}}\log x\right)\\ y'
&amp; = &amp; \df{1}{2}ax^{-\frac{1}{2}}\cos\left(\sqrt{k^2-\df{1}{4}}\log
x\right)+bx^{-\frac{1}{2}}\sin\left(\sqrt{k^2-\df{1}{4}}\log
x\right)\\ &amp; &amp;
-a\df{\sqrt{x}\sqrt{k^2-\frac{1}{4}}}{x}\sin\left(\sqrt{k^2-\df{1}{4}}\log
x\right)\\ &amp; &amp;
+\df{b\sqrt{x}\sqrt{k^2-\frac{1}{4}}}{x}\cos\left(\sqrt{k^2-\df{1}{4}}\log
x\right)\\ &amp; = &amp;
\left(\df{a}{2}+b\sqrt{k^2-\df{1}{4}}\right)\df{\cos\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{\sqrt{x}}\\ &amp; &amp;
+\left(\df{b}{2}-a\sqrt{k^2\-df{1}{4}}\right)
\df{\sin\left(\sqrt{k^2-\frac{1}{4}}\log x\right)}{\sqrt{x}}\\ y''
&amp; = &amp;
-\df{(\frac{a}{2}+b\sqrt{k^2-\frac{1}{4}})\sqrt{k^2-\frac{1}{4}}
\sin\left(\sqrt{k^2-\frac{1}{4}}\log x\right)}{x^{\frac{3}{2}}}\\
&amp; &amp;  +\df{(\frac{b}{2}-a\sqrt{k^2-\frac{1}{4}})
\sqrt{k^2-\frac{1}{4}}\sin\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}\\ &amp; &amp;
-\df{\frac{1}{2}(\frac{a}{2}+b\sqrt{k^2-\frac{1}{4}})\cos\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}\\ &amp; &amp;
-\df{\frac{1}{2}(\frac{a}{2}+b\sqrt{k^2-\frac{1}{4}})\sin\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}\\ &amp; = &amp;
\left(-b\left[k^2-\df{1}{4}\right]-\df{b}{4}\right)\df{\sin\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}\\ &amp; &amp;
+\left(-a\left[k^2-\df{1}{4}\right]-\df{a}{4}\right)\df{\cos\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}\\ &amp; = &amp;
-bk^2\df{\sin\left(-b\left[k^2-\df{1}{4}\right]-\df{b}{4}\right)\df{\sin\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}}{x^{\frac{3}{2}}}\\ &amp; &amp;
-ak^2\df{\cos\left(-b\left[k^2-\df{1}{4}\right]-\df{b}{4}\right)\df{\sin\left(\sqrt{k^2-\frac{1}{4}}\log
x\right)}{x^{\frac{3}{2}}}}{x^{\frac{3}{2}}}\\ &amp; = &amp; \df{-k^2}{x}y
\surd\surd\end{array}$

so it is an exact solution of equation.

Moreover, it satisfies the boundary conditions provided.

$\begin{array}{rcl} 0 &amp; = &amp;
a\sqrt{1}\cos\left(\sqrt{k^2-\df{1}{4}}\log
1\right)+b\sqrt{1}\sin\left(\sqrt{k^2-\df{1}{4}}\log 1\right)\\
\Rightarrow 0 &amp; = &amp; a\\ \rm{and}\ 0 &amp; = &amp;
b\sqrt{e}\sin\left(\sqrt{k^2-\df{1}{4}}\log e\right)\\ \Rightarrow
0 &amp; = &amp; b\sin\left(\sqrt{k^2-\df{1}{4}}\right) \end{array}$

Therefore either $b=0$ BORING! $y=0$ for all $x$

or $\sqrt{k^2-\df{1}{4}}=n\pi \Rightarrow \begin{array}{rcl}
k^2=n^2\pi^2+\df{1}{4}\\ k &amp; = &amp; \pm \sqrt{n^2\pi^2+\df{1}{4}}\\ &amp;
= &amp; \pm \un{n\pi\sqrt{1+\df{1}{4n\pi}}} \end{array}$

Therefore boundary data $\Rightarrow$ eigenvalues are

$$k_n=\pm
n\pi\left(1+\df{1}{8n\pi}+O\left(\df{1}{n^2}\right)\right)\
\rm{as}\ n \to \infty$$

This agrees for $n\to\pm \infty$ with the WKB result $k_n \sim
n\pi$ as above.

Next WKB term (long calculation) gives evalues from

$$y \sim a\sqrt{x}\cos\left(\left(k-\df{1}{8k}\right)\log
x\right)+b\sqrt{x}\sin \left(\left(k-\df{1}{8k}\right)\log
x\right)$$

which when substituted into boundary conditions

$\begin{array}{rcl} \Rightarrow k-\df{1}{8k}=n\pi &amp; \Rightarrow &amp;
8k^2-8kn\pi-1=0\\ &amp; \Rightarrow &amp;
k=\df{8n\pi\pm\sqrt{64n^2\pi^2+32}}{16}\\ &amp; \Rightarrow &amp; k \sim
n\pi+\df{1}{8n\pi}+o\left(\df{1}{n^2}\right),\ n
\to\infty\end{array}$

which agrees with exact?



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