How Does the Wronskian Relate to Airy Functions Ai(x) and Bi(x)?

[Alone]

I am trying to show that that the Airy functions defined below satisfy: $W[Ai(x),Bi(x)]=1/\pi$.

$$Ai(x)=\frac{1}{\pi} \int_0^\infty \cos(t^3/3+xt)dt$$

$$Bi(x)=\frac{1}{\pi}\int_0^\infty \bigg[ \exp(-t^3/3+xt)+\sin(t^3/3+xt)\bigg]dt $$

I tried to compute it directly but I got stuck, here's the last term I got:

$$Ai(x)Bi'(x)-Ai'(x)Bi(x) = \frac{1}{\pi^2}\bigg[ \int_0^\infty \cos(t^3/3+xt)dt \int_0^\infty \bigg( s\exp(-s^3/3+xs)+s\cos(s^3/3+xs)\bigg) ds + \int_0^\infty \sin(t^3/3+xt)tdt\int_0^\infty \bigg(\exp(-s^3/3+xs)+\sin(s^3/3+xs)\bigg)ds \bigg]$$

I don't see how to proceed from here, I guess I need complex integration contour but how exactly?

Thanks.
I want also to show that $Bi(x),Bi'(x)>0 \forall x>0$, and to conclude the asymptotic identities:
$$Bi(x) \sim \pi^{-1/2}x^{-1/4}\exp(2/3 x^{3/2})$$

$$Bi'(x)\sim \pi^{-1/2}x^{1/4}\exp(2/3 x^{3/2})$$

 

[Alone]

I've asked my question also in MSE, it seems the Wronskian question is answered, I still will appreciate if someone were to show me how to show the asymptotic identities.

 

[GJA]

Hi Alan,

One idea that would get you there is to use the fact that $Ai(x)$ and $Bi(x)$ are solutions to $y''-xy=0.$ Compute $W'[Ai(x),Bi(x)]$ and use $y''-xy=0$ to obtain that $W[Ai(x),Bi(x)]$ is a constant. To show that $W[Ai(x),Bi(x)=1/\pi,$ use the values of $Ai(x), Ai'(x), Bi(x),$ and $Bi'(x)$ at zero (see https://en.wikipedia.org/wiki/Airy_function) and the duplication formula for the gamma function (see https://en.wikipedia.org/wiki/Gamma_function)

 

[Opalg]

I still will appreciate if someone were to show me how to show the asymptotic identities.

There is a detailed discussion of how to find the asymptotic expansion of $\operatorname{Bi}(x)$ http://math.arizona.edu/~meissen/docs/asymptotics.pdf (but be warned that it takes 16 pages). Presumably you get the formula for $\operatorname{Bi}'(x)$ by differentiating the one for $\operatorname{Bi}(x)$.

 

[Alone]

There is a detailed discussion of how to find the asymptotic expansion of $\operatorname{Bi}(x)$ http://math.arizona.edu/~meissen/docs/asymptotics.pdf (but be warned that it takes 16 pages). Presumably you get the formula for $\operatorname{Bi}'(x)$ by differentiating the one for $\operatorname{Bi}(x)$.

I am used of reading quite a lot, that's maths and physics for you... :-)

 

[Alone]

Hi Alan,

One idea that would get you there is to use the fact that $Ai(x)$ and $Bi(x)$ are solutions to $y''-xy=0.$ Compute $W'[Ai(x),Bi(x)]$ and use $y''-xy=0$ to obtain that $W[Ai(x),Bi(x)]$ is a constant. To show that $W[Ai(x),Bi(x)=1/\pi,$ use the values of $Ai(x), Ai'(x), Bi(x),$ and $Bi'(x)$ at zero (see https://en.wikipedia.org/wiki/Airy_function) and the duplication formula for the gamma function (see https://en.wikipedia.org/wiki/Gamma_function)

I think Euler's reflection formula for Gamma function solves this immediately.