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| − | The Airy functions $\mathrm{Ai}$ and $\mathrm{Bi}$ are linearly independent solutions of the Airy differential equation | + | __NOTOC__ |
| − | $$y''(z)-zy(z)=0.$$ | + | |
| | + | The Airy function $\mathrm{Bi}$ (sometimes called the "Bairy function") is a solution of the [[Airy differential equation]] |
| | + | $$y' '(z)-zy(z)=0,$$ |
| | + | which is [[linearly independent]] from the [[Airy Ai]] function. |
| | | | |
| | <div align="center"> | | <div align="center"> |
| | <gallery> | | <gallery> |
| − | File:Airyai.png|Airy $\mathrm{Ai}$ function. | + | File:Airybiplot.png|Aairy $\mathrm{Bi}$ function. |
| − | File:Airybi.png|Bairy $\mathrm{Bi}$ function. | + | File:Complexairybiplot.png|[[Domain coloring]] of $\mathrm{Bi}$. |
| | </gallery> | | </gallery> |
| | </div> | | </div> |
| | | | |
| | =Properties= | | =Properties= |
| − | <div class="toccolours mw-collapsible mw-collapsed"> | + | [[Relationship between Airy Bi and modified Bessel I]]<br /> |
| − | <strong>Theorem:</strong> The complex function
| + | [[Relationship between Scorer Gi and Airy functions]]<br /> |
| − | $$\mathrm{Ai}(z)=\dfrac{1}{2 \pi i} \displaystyle\int_{-i\infty}^{i \infty} e^{-zt + \frac{t^3}{3}} dt$$
| + | [[Relationship between Scorer Hi and Airy functions]]<br /> |
| − | solves the Airy differential equation and moreover, if $z=x$ is a real number then $\mathrm{Ai}(x)$ is a real number and
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| − | $$\mathrm{Ai}(x)=\dfrac{1}{\pi} \displaystyle\int_{-\infty}^{\infty} \cos \left( zu + \dfrac{u^3}{3} \right) du.$$
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| − | <div class="mw-collapsible-content">
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| − | <strong>Proof:</strong> Suppose that $y$ has the form
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| − | $$y(z) = \displaystyle\int_{C} f(t)e^{-zt} dt,$$
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| − | where $C$ is an as-of-yet undefined contour in the complex plane. Assuming that we may differentiate under the integral it is clear that
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| − | $$y''(z)=\displaystyle\int_{C} f(t)t^2 e^{-zt} dt.$$
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| − | Thus we plug this representation into the differential equation to get
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| − | $$(*) \hspace{35pt} y''(z)-zy(z) = \displaystyle\int_{C} (t^2-z)f(t)e^{-zt} dt = 0.$$
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| − | Now we integrate by parts to see
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| − | $$\begin{array}{ll}
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| − | \displaystyle\int_{C} zf(t)e^{-zt} dt &= -\displaystyle\int_{C} f(t) \dfrac{d}{dt} e^{-zt} dt \\
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| − | &= -f(t)e^{zt} \Bigg |_{C} + \displaystyle\int_{C} f'(t)e^{-zt} dt.
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| − | \end{array}$$
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| − | We will pick the contour $C$ to enforce $f(t)e^{zt} \Bigg |_{C}=0$. We will do this by first determining the function $f$. Plugging this back into the formula $(*)$ yields
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| − | $$\begin{array}{ll}
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| − | 0 &= y''(z) - zy(z) \\
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| − | &= f(t)e^{zt} \Bigg |_{C} + \displaystyle\int_{C} (t^2f(t)-f'(t))e^{zt} dt.
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| − | \end{array}$$
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| − | We have the freedom to choose $f$ and $C$. We will choose $f$ so that
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| − | $$t^2f(t)-f'(t)=0.$$
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| − | This is a simple differential equation with [http://www.wolframalpha.com/input/?i=t^2f%28t%29-f%27%28t%29%3D0 a solution]
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| − | $$f(t)=\xi e^{\frac{t^3}{3}},$$
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| − | for some constant $\xi$ (later when we define $\mathrm{Ai}$, the convention is to choose $\xi=\dfrac{1}{2\pi i}$, but we will proceed the argument right now as if $\xi=1$). So we have derived
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| − | $$y(z)=\displaystyle\int_{C} e^{-zt + \frac{t^3}{3}} dt.$$
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| − | To pick the contour $C$ note that the integrand of $y$ is an [[entire function]] and hence if $C$ is a simple closed curve we would have $y(z)=0$ for all $z \in \mathbb{C}$.
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| | | | |
| − | The variable of the integral defining $y$ is $t$ and for $t \in \mathbb{C}$ with $|t|$ very large, the cubic term in the exponent dominates. Hence consider polar form $t=|t|e^{i\theta}$ and compute
| + | =Videos= |
| − | $$e^{\frac{t^3}{3}} = \exp\left( \frac{|t|^3 e^{3i\theta}}{3} \right).$$
| + | [https://www.youtube.com/watch?v=HlX62TkR6gc&noredirect=1 Leading Tsunami wave reaching the shore (27 November 2009)]<br /> |
| − | Notice that the inequality $\mathrm{Re} \hspace{2pt} e^{3i\theta} \leq 0$ forces $\cos(3\theta)\leq 0$ [http://www.wolframalpha.com/input/?i=cos%283*theta%29%3C0 yielding] three sectors defined by $\theta$: [[File:Airysectors.png|200px]]
| + | [https://www.youtube.com/watch?v=0jnXdXfIbKk&noredirect=1 Series solution of ode: Airy's equation (3 November 2010)]<br /> |
| − | $$-\dfrac{\pi}{2} \leq \theta \leq -\dfrac{\pi}{6},$$
| + | [https://www.youtube.com/watch?v=oYJq3mhg5yE&noredirect=1 Airy differential equation (26 November 2013)]<br /> |
| − | $$\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2},$$
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| − | $$\dfrac{9\pi}{6} \leq \theta \leq \dfrac{11\pi}{6}.$$
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| | | | |
| − | We will consider three contours $C_1,C_2,C_3$, where each contour $C_i$ has endpoints at complex $\infty$ in different sectors. Call the left sector $\gamma$, the upper-right sector $\beta$ and the lower-right sector $\alpha$. Let $C_1$ be oriented from sector $\alpha$ to sector $\beta$ (this sort of curve is labelled as "$C$" in the image above), $C_2$ from sector $\beta$ to sector $\gamma$, and $C_3$ from sector $\gamma$ to sector $\alpha$. By our analysis we have derived three solutions to Airy's equation:
| + | =References= |
| − | $$y_i(z) = \displaystyle\int_{C_i} e^{-zt + \frac{t^3}{3}} dt;i=1,2,3$$
| + | [http://www.ams.org/samplings/feature-column/fcarc-rainbows The mathematics of rainbows]<br /> |
| − | Since these functions satisfy a second order differential equation, it is impossible for them to be [[linearly independent]]. Now notice that we can compute
| + | [http://www.ams.org/journals/mcom/1979-33-145/S0025-5718-1979-0514831-8/S0025-5718-1979-0514831-8.pdf Tables of Weyl Fractional Integrals for the Airy Function]<br /> |
| − | $$\displaystyle\int_{C_1\cup C_2 \cup C_3} e^{-zt + \frac{t^3}{3}} dt = 0.$$
| + | [http://www.amazon.com/Special-Functions-Introduction-Classical-Mathematical/dp/0471113131 Special Functions: An Introduction to the Classical Functions of Mathematical Physics]<br /> |
| − | Therefore
| + | [http://www.people.fas.harvard.edu/~sfinch/csolve/ai.pdf Airy function zeros] |
| − | $$y_1(z)+y_2(z)+y_3(z)=0.$$
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| − | | |
| − | By convention we define
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| − | $$\mathrm{Ai}(z) = \dfrac{1}{2\pi i} \displaystyle\int_{C_1} e^{-zt + \frac{t^3}{3}} dt,$$
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| − | where we take $C_1$ to be, specifically, the contour from $-i\infty$ to $i\infty$ along the $y$-axis in the complex plane. Hence we may compute by the substitution $u=it$ (hence $t=-ui$),
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| − | $$\begin{array}{ll}
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| − | \mathrm{Ai}(z) &= \dfrac{1}{2 \pi i} \displaystyle\int_{-i\infty}^{i \infty} e^{-zt + \frac{t^3}{3}} dt \\
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| − | &= \dfrac{1}{2\pi i} \displaystyle\int_{\infty}^{-\infty} (-i) e^{zui+\frac{(-ui)^3}{3})} du \\
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| − | &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} e^{i(zu + \frac{u^3}{3})} du \\
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| − | &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \cos\left( zu + \dfrac{u^3}{3} \right) + i \sin \left( zu + \dfrac{u^3}{3} \right) du.
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| − | \end{array}$$
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| − | Now if $z=x$ is a real number, then notice that
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| − | $$\displaystyle\int_{-\infty}^{\infty} \sin \left( zu + \dfrac{u^3}{3} \right) du = \displaystyle\lim_{b \rightarrow \infty} \int_{-b}^b \sin \left( zu + \dfrac{u^3}{3} \right) du = 0,$$
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| − | because $zu+\dfrac{u^3}{3}$ is an odd function of $u$. Hence we see that $\mathrm{Ai}$ is real-valued at real-valued inputs. This insight yields the real-valued formula for $\mathrm{Ai}$ for $z=x$ a real number:
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| − | $$\begin{array}{ll}
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| − | \mathrm{Ai}(x) &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \cos \left( zu + \dfrac{u^3}{3} \right) du \\
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| − | &= \dfrac{1}{\pi} \displaystyle\int_{-\infty}^{\infty} \cos \left( zu + \dfrac{u^3}{3} \right) du.
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| − | \end{array}$$
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| | | | |
| − | </div> | + | =See Also= |
| − | </div> | + | [[Airy Ai]] <br /> |
| | + | [[Scorer Gi]] <br /> |
| | + | [[Scorer Hi]] <br /> |
| | | | |
| − | =References=
| + | [[Category:SpecialFunction]] |
| − | [http://www.ams.org/journals/mcom/1979-33-145/S0025-5718-1979-0514831-8/S0025-5718-1979-0514831-8.pdf Tables of Weyl Fractional Integrals for the Airy Function]<br /> | |
| − | [http://www.amazon.com/Special-Functions-Introduction-Classical-Mathematical/dp/0471113131 Special Functions: An Introduction to the Classical Functions of Mathematical Physics]
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