Warning: jsMath requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

2005 I 1

From QED

< PlasmaWiki | Generals(Link to this page as [[PlasmaWiki/Generals/2005 I 1]])

Jump to: navigation, search

We consider the differential equation:

$\frac{d^{2}y}{dx^{2}}+\frac{1}{x^{2}}\frac{dy}{dx}+\frac{1}{x^{2}}y=0$

For $x \rightarrow 0$ , the equation is singular, so we guess $y ˜ e S$ :

$\left(S^{\prime\prime}+\left(S^{\prime}\right)^{2}\right)+\frac{S^{\prime}}{x^{2}}+\frac{1}{x^{2}}=0$

Balance between $S^{\prime}/x^{2}$ and $(S^{\prime})^{2}$ gives $S^{\prime}=-x^{-2}$ , which is dominant. There is an alternate balance between $S^{\prime}=-1$ . Finding the next term $g$ for the first case:

$\left(\frac{2}{x^{3}}+g^{\prime\prime}-\frac{2g^{\prime}}{x^{2}}+\left(g^{\prime}\right)^{2}\right)+\frac{g^{\prime}}{x^{2}}+\frac{1}{x^{2}}=0$

So balance between $\frac{2}{x^{3}}$ , $-\frac{2g^{\prime}}{x^{2}}$ , and $\frac{g^{\prime}}{x^{2}}$ gives $g^{\prime}=2x^{-1}$ . The other term is already less singular that $ln x$ . The leading asymptotic forms are then:

$y\sim x^{2}e^{1/x},\quad e^{-x}$

For $x\rightarrow\infty$ , the equation is again singular, so, guessing $y ˜ e S$ :

$\left(S^{\prime\prime}+\left(S^{\prime}\right)^{2}\right)+\frac{S^{\prime}}{x^{2}}+\frac{1}{x^{2}}=0$

This time there is balance between $S^{\prime\prime}+\left(S^{\prime}\right)^{2}+x^{-2}$ , giving $S\sim\frac{1}{2}\left(1\pm i\sqrt{3}\right)\ln x$ . So:

$y\sim x^{\left(1\pm i\sqrt{3}\right)/2}$

Or:

$y\sim x^{1/2}\sin\left(\sqrt{3}\ln x/2\right);\quad x^{1/2}\cos\left(\sqrt{3}\ln x/2\right)$

Warning- The below contains a sign error!

For $x\rightarrow\infty$ the equation is not singluar, so, writing

y = \sum a n x - n - α n

$\sum_{n}a_{n}\left(n+\alpha\right)\left(n+\alpha+1\right)x^{-n-2-\alpha}-\sum_{n}a_{n}\left(n+\alpha\right)x^{-n-3-\alpha}+\sum_{n}a_{n}x^{-n-2-\alpha}=0$

Shifting: