I have been following the thread " Convert Airy's Equation $y''-xy=0$ to Bessel equation $$t^2u''+tu'+(t^2-c^2)u$$ " but I can't join the dots to a solve similar equation $y''+x^2y=0$ so as to obtain a solution of the form $$\sqrt x\left(AJ_\frac{1}{4}+BJ_{-\frac{1}{4}}\right)$$ I actually get an equation that looks this way $$t^2\frac{du}{dt}+t\frac{du}{dt}+(t^2+\frac{5}{64})u$$ The above equation can not yield the desired solution. Please help me to clearly see this. Thank you.
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Actually, I think $\mathrm{Ai}(x)$ et al is related to the Bessels of order 1/3. – Ron Gordon Jan 03 '13 at 15:37
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@rlgordonma, Does that mean the method described in the Airy's differential equation can not be used to modify other likely equations into the Bessel model? Is there some other procedure for the Bessels of order $\frac{1}{4}$? – user55063 Jan 03 '13 at 18:06
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Not sure what you mean. See this, however: http://en.wikipedia.org/wiki/Airy_function#Relation_to_other_special_functions – Ron Gordon Jan 03 '13 at 19:49
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Perform the change of variables $y=\sqrt{x}f(x)$. Then, using the product rule, $f$ satisfies: $$ \frac{x^2}{4}f''+\frac{x}{4}f'+(\frac{x^4}{4}-\frac{1}{16})f=0. $$ Then make the change of variable $t=x^2/2$ and compute: $$ \frac{df}{dx}=x\frac{df}{dt}, $$ and $$\frac{d^2f}{dx^2}=\frac{df}{dt}+x^2\frac{d^2f}{dt^2}. $$ Thus, the equation above becomes $$ \frac{x^4}{4}f''+\frac{x^2}{2}f'+(\frac{x^4}{4}-\frac{1}{16})f=0, $$ where the primes now denote derivatives w.r.t. the variable $t$. Written in the variable $t$, the equation above becomes $$ t^2f''+tf'+(t^2-\frac{1}{16})f=0. $$ The solution of which is $$ f=A J_{1/4}(t)+B J_{-1/4}(t).$$ Thus since $y=\sqrt{x}f(x^2/2)$, you get the desired solution.
Gateau au fromage
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Here are maple solutions to $y''(x)+x^2y(x)=0$ in terms of the Bessel functions
$$ y \left( x \right) ={c_1}\,\sqrt {x} {{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}+{c_2}\,\sqrt {x} \left( {{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}-\sqrt {2} {{J_{-1/4}}\left(\frac{{x}^{2}}{2}\right)} \right), $$
or you can have the form
$$ y \left( x \right) ={c_1}\,\sqrt {x} {{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}+{c_2}\,\sqrt {x}\, {{Y_{1/4}}\left(\frac{{x}^{2}}{2}\right)},$$
where $Y_{1/4}\left(\frac{{x}^{2}}{2}\right)$ is the Bessel function of the second kind.
Mhenni Benghorbal
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1Very true @Mhenni Benghorbal. I forgot to place the $\sqrt{x}$ and a parenthesis so that $y=\sqrt{x}(AJ_{\frac{1}{4}}+BJ_{-\frac{1}{4}})$. Now my question is: why cant I get my Bessel equivalent to be $t^2\frac{d^2u}{dt^2}+t\frac{du}{dt}+(t^2-\frac{1}{16})u$? – user55063 Jan 03 '13 at 21:38
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My question is; where does the Bessel order $\frac{1}{4}$ in the solution come from? Please help me since following the method outlined in link yields the Bessel equation I gave above whose order is clearly not $\frac{1}{4}$. – user55063 Jan 05 '13 at 04:55
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1@PeterTamaroff:Have you seen a massage directed to moderators? – Mhenni Benghorbal Mar 09 '13 at 16:15
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1Yes:
@moderators: Can you tell us what's happening here? – Mhenni Benghorbal Jan 3 at 16:22– Did Mar 10 '13 at 19:49 -
@Did: On this website, we are supposed to assist people and share knowledge. – Mhenni Benghorbal Mar 16 '13 at 12:49
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What are you complaining about? You asked a question (@PeterTamaroff:Have you seen a massage directed to moderators?), I provided the answer, life is good, no? – Did Mar 16 '13 at 15:20
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