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How do I show that the function $$ y(x) = \int_0^\infty \cos(t^3/3 + xt)~dt\qquad -\infty \lt x \lt \infty,$$ satisfies the differential equation $y''=xy$?

I can't simply differentiate under integral since the integrand is oscillatory and does not decay as $t$ becomes large.

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Ah, Airy's differential equation. Have you tried maybe a preliminary substitution, or integration by parts on your integral? Alternatively: you've tagged complex-analysis, so you could maybe try using a different integration path... –  J. M. Apr 17 '12 at 0:45
    
Does this integral even converge? I'm not answering OP's question at all by asking this, but I'm always too much suspicious when it comes to differential equation... and this has me worried. –  Patrick Da Silva Apr 20 '12 at 4:22

1 Answer 1

First we should note that the integral form solution of Airy's differential equation $y''-xy=0$ can be found by assuming a suitable integral kernel which is so-called the “integral kernel method”. It is a mutation of using integral transform. But it is better than using integral transform, since inverse transform should not be needed.

Let $y=\int_Ce^{xs}K(s)~ds$,

Then $(\int_Ce^{xs}K(s)~ds)''-x\int_Ce^{xs}K(s)~ds=0$

$\int_Cs^2e^{xs}K(s)~ds-\int_Ce^{xs}K(s)~d(xs)=0$

$\int_Cs^2e^{xs}K(s)~ds-\int_CK(s)~d(e^{xs})=0$

$\int_Cs^2e^{xs}K(s)~ds-[e^{xs}K(s)]_C+\int_Ce^{xs}~d(K(s))=0$

$\int_Cs^2e^{xs}K(s)~ds-[e^{xs}K(s)]_C+\int_Ce^{xs}K'(s)~ds=0$

$-[e^{xs}K(s)]_C+\int_C(K'(s)+s^2K(s))e^{xs}~ds=0$

$\therefore K'(s)+s^2K(s)=0$

$K'(s)=-s^2K(s)$

$\frac{K'(s)}{K(s)}=-s^2$

$\int\frac{K'(s)}{K(s)}~ds=\int-s^2~ds$

$\ln K(s)=-\frac{s^3}{3}+c_1$

$K(s)=ce^{-\frac{s^3}{3}}$

$\therefore y=\int_Cce^{-\frac{s^3}{3}+xs}~ds$

But since the above procedure in fact suitable for any complex number $s$,

$\therefore y_n=\int_{a_n}^{b_n}c_ne^{-\frac{((p_n+q_ni)t)^3}{3}+x(p_n+q_ni)t}~d((p_n+q_ni)t)$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{-\frac{(p_n^3+3p_n^2q_ni-3p_nq_n^2-q_n^3i)t^3}{3}+(p_n+q_ni)xt}~dt$

$=(p_n+q_ni)c_n\int_{a_n}^{b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}~dt$

For some $x$-independent real number choices of $a_n$ , $b_n$ , $p_n$ and $q_n$ such that:

$\displaystyle\lim_{t\to a_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}=\lim_{t\to b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}$

$\int_{a_n}^{b_n}e^{\frac{(3q_n^2-p_n^2)p_nt^3}{3}+p_nxt}e^{\left(\frac{(q_n^2-3p_n^2)q_nt^3}{3}+q_nxt\right)i}~dt$ converges

For $n=1$, the best choice is $a_1=-\infty$ , $b_1=\infty$ , $p_1=0$ , $q_1=1$

$\therefore y_1=ic_1\int_{-\infty}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt$

$=ic_1\left(\int_{-\infty}^{0}e^{\left(\frac{t^3}{3}+xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$

$=ic_1\left(\int_{\infty}^{0}e^{\left(\frac{(-t)^3}{3}+x(-t)\right)i}~d(-t)+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$

$=ic_1\left(\int_{0}^{\infty}e^{\left(-\frac{t^3}{3}-xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$

$=ic_1\left(\int_{0}^{\infty}e^{-\left(\frac{t^3}{3}+xt\right)i}~dt+\int_{0}^{\infty}e^{\left(\frac{t^3}{3}+xt\right)i}~dt\right)$

$=C_1\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

“integral kernel method” not only suitable for Airy's differential equation, it is in fact at least suitable for many high order linear differential equation with linear coefficients such as $y''+y'+xy=0$ (http://tw.group.knowledge.yahoo.com/ignored-knowledgeunion/article/view?aid=4) , $y''+2y'-2xy=0$ (http://tw.knowledge.yahoo.com/question/article?qid=1711011401283) , $ y''+xy'+xy=0$ (http://tw.knowledge.yahoo.com/question/question?qid=1009100207053) , even for $ y'''-xy''-4y'+3y=0$ (http://tw.knowledge.yahoo.com/question/question?qid=1011072501482) and $y^{(n)}=axy+b$ (http://eqworld.ipmnet.ru/en/solutions/ode/ode0411.pdf) , etc.

For proving $\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$ satisfies the differential equation $y''-xy=0$ , we should highly aware the convergences of the improper integrals of the form $\int_{c}^{\infty}x^n\sin(P(x))~dx$ and $\int_{c}^{\infty}x^n\cos(P(x))~dx$, where $c$ and $n$ are real numbers, $P(x)$ is a $m$-degree polynomial ($m\geq 2$) with real number coefficients. Since their convergences are fairly complicated, http://blog.yimg.com/2/vyvWdRF7s5_bqHpIzG6otLmp3eg5MBRGEnBi9N0YFnhA34nFqb3srA--/39/o/4oAekeQCUWteJ6Ch0NcAtg.jpg and http://blog.yimg.com/2/vyvWdRF7s5_bqHpIzG6otLmp3eg5MBRGEnBi9N0YFnhA34nFqb3srA--/40/o/52OMY4g6GgniM2CkxG9f1g.jpg have already discussed them in details.

$\left(\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt\right)''-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(-\int_{0}^{\infty}t\sin\left(\frac{t^3}{3}+xt\right)~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(-\int_{0}^{\infty}\frac{t^2\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(-\int_{0}^{\infty}\frac{(t^2+x)\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(-\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~d\left(\frac{t^3}{3}+xt\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(\int_{0}^{\infty}\frac{1}{t}~d\left(cos\left(\frac{t^3}{3}+xt\right)\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(\left[\frac{\cos\left(\frac{t^3}{3}+xt\right)}{t}\right]_{0}^{\infty}-\int_{0}^{\infty}cos\left(\frac{t^3}{3}+xt\right)~d\left(\frac{1}{t}\right)+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=\left(\int_{0}^{\infty}\frac{cos\left(\frac{t^3}{3}+xt\right)}{t^2}~dt+x\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt\right)'-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=-\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+\int_{0}^{\infty}\frac{\sin\left(\frac{t^3}{3}+xt\right)}{t}~dt+x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt-x\int_{0}^{\infty}\cos\left(\frac{t^3}{3}+xt\right)~dt$

$=0$

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Is it possible to find a general kernel for $y''-f(x).y=0$? Thanks a lot . My question is math.stackexchange.com/questions/99850/…. –  Mathlover May 20 '12 at 10:38
    
@Mathlover: This is a good mathematical topic. EqWorld has an article eqworld.ipmnet.ru/en/methods/methods-ode/… claims that $f''(x)+a_0f(x)=0$ have methods to solve generally, in math.stackexchange.com/questions/34228/… someone also claims that the similar method. But I doubt about their reliability. Since all special functions which are defined by 2nd order linear ODE (e.g. Mathieu function, Heun function, etc) still here and not shut down. –  doraemonpaul May 20 '12 at 22:04

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