# Deriving the Airy functions from first principles

I have just started reading about the Airy functions and am stuck on a particular step of their derivation. But first here is some background information to give this question some meaning, more information can be found from a previous question of mine:

The general solution to Bessel's differential equation: $$\fbox{y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0}\tag{1}$$ is $$\fbox{y=x^aZ_p\left(bx^c\right)}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are constants. $J_p$ is called the Bessel function of the first kind of order $p$ and $N_p$ is any combination of $J_p$ and $J_{−p}$: $$N_p(x)=\frac{\cos(\pi p)J_p(x)-J_{-p}(x)}{\sin(\pi p)}\tag{3}$$ So equation $(2)$ can be written as $$y=x^{a}\left[AJ_{p}\left(bx^c\right)+BN_{p}\left(bx^c\right)\right]\tag{4}$$ or $$y=x^{a}\left[AJ_{p}\left(bx^c\right)+B\left(\frac{\cos(\pi p)J_p\left(bx^c\right)-J_{-p}\left(bx^c\right)}{\sin(\pi p)}\right)\right]\tag{5}$$

The Airy differential equation is $$y^{\prime\prime}-xy=0\tag{6}$$ and has solution $$y=x^{1/2}Z_{1/3}\left(\frac23 ix^{3/2}\right)\tag{7}$$

By using

$$I_p(x)=i^{-p}J_p(ix)\tag{8}$$ $$K_p(x)=\frac{\pi}{2}i^{p+1}\left[J_p(ix)+i\left(\frac{\cos(\pi p)J_p(ix)-J_{-p}(ix)}{\sin(\pi p)}\right)\right]\tag{9}$$

My objective is to show that $(7)$ can be written in terms of $I_{1/3}$ and $K_{1/3}$ to obtain

$$Ai(x)=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{1/3}\left(\frac23x^{3/2}\right)\tag{10}$$ $$Bi(x)=\sqrt{\frac{x}{3}}\left[I_{-1/3}\left(\frac23x^{3/2}\right)+I_{1/3}\left(\frac23x^{3/2}\right)\right]\tag{11}$$

Starting from $(7)$: \begin{align}y&=x^{1/2}Z_{1/3}\left(\frac23 ix^{3/2}\right)\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23 ix^{3/2}\right)+BN_{1/3}\left(\frac23 ix^{3/2}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+B\left(\frac{\cos\left(\frac{\pi}{3}\right)J_{1/3}\left(\frac23ix^{3/2}\right)-J_{-1/3}\left(\frac23ix^{3/2}\right)}{\sin\left(\frac{\pi}{3}\right)}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+B\left(\frac{\frac12J_{1/3}\left(\frac23ix^{3/2}\right)-J_{-1/3}\left(\frac23ix^{3/2}\right)}{\left(\frac{\sqrt3}{2}\right)}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+\frac{BJ_{1/3}\left(\frac23ix^{3/2}\right)}{\sqrt3}-\frac{2BJ_{-1/3}\left(\frac23ix^{3/2}\right)}{\sqrt3}\right]\\&=x^{1/2}\left[Ai^{1/3}I_{1/3}\left(\frac23x^{3/2}\right)+\frac{i^{1/3}BI_{1/3}\left(\frac23x^{3/2}\right)}{\sqrt3}-\frac{2i^{-1/3}BI_{-1/3}\left(\frac23x^{3/2}\right)}{\sqrt3}\right]\tag{a}\end{align} where in $(\mathrm{a})$ I used $(8)$ rearranged as $J_p(ix)=i^{p}I_p(x)$

I don't understand how to proceed with this calculation as I am unsure how to use $(9)$; I also have no idea what $i(x)$ means. Is $i$ really a function of $x$?

Is there anyone that could provide some hints or advice on how I can continue this calculation to obtain $(10)$ and $(11)$?

Below are some images showing some of the relevant formulae to this question:   • It is faster to check that both sides of $(10)$ or $(11)$ fulfill the same differential equation with the same initial conditions. You just have to apply a change of variable in the Bessel differential equation. – Jack D'Aurizio Jun 14 '16 at 3:07

At first note, that $i=\sqrt{-1}$ is the imaginary unit in (9) and we see a plain multiplication with $i$.

We start with the representation (5) and use (7) to obtain \begin{align*} y&=x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\\ &=x^{\frac{1}{2}}\left[AJ_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) +B\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) -\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right]\tag{f} \end{align*} With $p=\frac{1}{3}$ we can write $K_p$ as \begin{align*} K_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right)=\frac{\pi}{2}i^{\frac{4}{3}} \left[J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)+i\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) -\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right] \end{align*} Now we can write $Ai(x)$ using $K_{\frac{1}{3}}$ and see the structural connection with $(\mathrm{f})$.

\begin{align*} Ai(x)&=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right)\\ &=\frac{1}{2\sqrt{3}}i^{\frac{4}{3}}x^{\frac{1}{2}} \left[J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)+i\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) -\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right] \end{align*}

Comparison with $(\mathrm{f})$ results in \begin{align*} A=\frac{1}{2\sqrt{3}}i^{\frac{4}{3}}\qquad\qquad B=-\frac{1}{2\sqrt{3}}i^{\frac{1}{3}} \end{align*}

We can see the connection of $B_i(x)$ with (5) by substituting $I_p(x)$ with $J_p(x)$. We obtain \begin{align*} Bi(x)&=\frac{1}{\sqrt{3}}x^{\frac{1}{2}}\left[I_{-\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right) +I_\frac{1}{3}\left(\frac{2}{3}x^{\frac{3}{2}}\right)\right]\\ &=\frac{1}{\sqrt{3}}x^{\frac{1}{2}}\left[i^{\frac{1}{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right) +i^{-\frac{1}{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right] \tag{g}\end{align*} Comparison of coefficients of $J_{-\frac{1}{3}}$ gives \begin{align*} B=-\frac{1}{2}i^{\frac{1}{3}} \end{align*} Comparison of coefficients of $J_{\frac{1}{3}}$ gives \begin{align*} A=\frac{1}{\sqrt3}\left(i^{-1/3}-\frac{i^{1/3}}{2}\right) \end{align*}

Hint: Note that the Airy function is defined according to (10) in terms of $K_p$ which is defined according to (9). So, the right-hand side of the blue part is just the definition of the Airy function in terms of $J_p$ and $J_{-p}$. From this representation we can deduce $A$ and $B$ by a rather simple comparison with $(\mathrm{f})$.

• @BLAZE: Note that $Ai$ is the name of the Airy function. So, $Ai(x)$ is just the Airy function evaluated at $x$. – Markus Scheuer Jun 20 '16 at 17:47
• @BLAZE: I've added a hint which might clarify the status of the definition of the Airy function and the derivation of $A$ and $B$. – Markus Scheuer Jun 22 '16 at 6:25
• @BLAZE: Two pairs of $A$ and $B$, one for $Ai(x)$ the other one for $Bi(x)$. Note that $J_p$ and $J_{-p}$ are independent. So, you can find $B$ immediately by comparing coefficients of $J_{-p}$. Then you can derive $A$. Regards, – Markus Scheuer Jun 26 '16 at 16:44
• @BLAZE: Comparing coefficients of (f) and (g) addresses $Bi(x)$. Substituting $B=-\frac{1}{2}i^{\frac{1}{3}}$ gives $A-\frac{1}{2\sqrt{3}}i^{\frac{1}{3}}=\frac{1}{\sqrt{3}}i^{-\frac{1}{3}}$. It follows ... – Markus Scheuer Jun 26 '16 at 20:42
• @BLAZE: Thanks for the bounty and accepting my answer! ☺ Two different pairs of $A,B$ which are not multiples of each other give two independent, different solutions of Bessel's differential equation. So, $Ai$ and $Bi$ are related, but different solutions. Since the determination of both pairs seems to be quite clear to you, I think the answer is complete. Best regards , – Markus Scheuer Jun 28 '16 at 6:31