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Let be the Sturm Liouville problem $ y(0)=0=y(\infty)=0 $

$$ y''(x)+e^{ax}y(x)=k^{2} _{n}y(x) $$

how could i solve it ?? i have tried with the cahnge of variable $ t= \frac{ln(u)}{a} $ i get

$$ 2a^{2}u^{2}f''(u)+uaf'(u)+uf(u)=k^{2}_{n}f(u) $$

I do not how to follow, any advice thanks ??

I only know the WKB solution valid for big $ x\to \infty $ of course

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This is just the issue that you don't familiarize the method of solving this kind of ODE. In fact this belongs to an ODE of the form eqworld.ipmnet.ru/en/solutions/ode/ode0232.pdf. –  doraemonpaul Nov 22 '12 at 22:20
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up vote 1 down vote accepted

In fact this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0232.pdf.

Let $u=e^{\frac{ax}{2}}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{ae^{\frac{ax}{2}}}{2}\dfrac{dy}{du}=\dfrac{au}{2}\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{au}{2}\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(\dfrac{au}{2}\dfrac{dy}{du}\right)\dfrac{du}{dx}=\biggl(\dfrac{au}{2}\dfrac{d^2y}{du^2}+\dfrac{a}{2}\dfrac{dy}{du}\biggr)\dfrac{au}{2}=\dfrac{a^2u^2}{4}\dfrac{d^2y}{du^2}+\dfrac{a^2u}{4}\dfrac{dy}{du}$

$\therefore\dfrac{a^2u^2}{4}\dfrac{d^2y}{du^2}+\dfrac{a^2u}{4}\dfrac{dy}{du}+u^2y=k_n^2y$

$\dfrac{a^2u^2}{4}\dfrac{d^2y}{du^2}+\dfrac{a^2u}{4}\dfrac{dy}{du}+(u^2-k_n^2)y=0$

Let $v=mu$ ,

Then $\dfrac{dy}{du}=\dfrac{dy}{dv}\dfrac{dv}{du}=m\dfrac{dy}{dv}$

$\dfrac{d^2y}{du^2}=\dfrac{d}{du}\left(m\dfrac{dy}{dv}\right)=\dfrac{d}{dv}\left(m\dfrac{dy}{dv}\right)\dfrac{dv}{du}=m\dfrac{d^2y}{dv^2}m=m^2\dfrac{d^2y}{dv^2}$

$\therefore\dfrac{a^2v^2}{4m^2}m^2\dfrac{d^2y}{dv^2}+\dfrac{a^2v}{4m}m\dfrac{dy}{dv}+\biggl(\dfrac{v^2}{m^2}-k_n^2\biggr)y=0$

$\dfrac{a^2v^2}{4}\dfrac{d^2y}{dv^2}+\dfrac{a^2v}{4}\dfrac{dy}{dv}+\biggl(\dfrac{v^2}{m^2}-k_n^2\biggr)y=0$

$v^2\dfrac{d^2y}{dv^2}+v\dfrac{dy}{dv}+\biggl(\dfrac{4v^2}{a^2m^2}-k_n^2\biggr)y=0$

Choose $m=\dfrac{2}{a}$ , the ODE becomes

$v^2\dfrac{d^2y}{dv^2}+v\dfrac{dy}{dv}+(v^2-k_n^2)y=0$

$y=\begin{cases}C_1J_{k_n}(v)+C_2Y_{k_n}(v)&\text{when}~k_n~\text{is an integer}\\C_1J_{k_n}(v)+C_2J_{-k_n}(v)&\text{when}~k_n~\text{is not an integer}\end{cases}$

$y=\begin{cases}C_1J_{k_n}\left(\dfrac{2u}{a}\right)+C_2Y_{k_n}\left(\dfrac{2u}{a}\right)&\text{when}~k_n~\text{is an integer}\\C_1J_{k_n}\left(\dfrac{2u}{a}\right)+C_2J_{-k_n}\left(\dfrac{2u}{a}\right)&\text{when}~k_n~\text{is not an integer}\end{cases}$

$y=\begin{cases}C_1J_{k_n}\biggl(\dfrac{2e^{\frac{ax}{2}}}{a}\biggr)+C_2Y_{k_n}\biggl(\dfrac{2e^{\frac{ax}{2}}}{a}\biggr)&\text{when}~k_n~\text{is an integer}\\C_1J_{k_n}\biggl(\dfrac{2e^{\frac{ax}{2}}}{a}\biggr)+C_2J_{-k_n}\biggl(\dfrac{2e^{\frac{ax}{2}}}{a}\biggr)&\text{when}~k_n~\text{is not an integer}\end{cases}$

The only problem is how to find $C_1$ and $C_2$ when $y(0)=y(\infty)=0$ .

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thanks for your answer.. :D wht would happen if i change $ exp(ax) $ by $ -exp(ax) $ ?? –  Jose Garcia Dec 17 '12 at 12:00
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