Difference between revisions of "Airy Bi"

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The Airy function $\mathrm{Ai}$ and "Bairy" function $\mathrm{Bi}$ are given by the formulas
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__NOTOC__
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$$\mathrm{Ai}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \cos \left( \dfrac{t^3}{3} + xt \right) dt$$
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and
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The Airy function $\mathrm{Bi}$ (sometimes called the "Bairy function") is a solution of the [[Airy differential equation]]
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$$\mathrm{Bi}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \left[ e^{-\frac{t^3}{3} + xt} + \sin \left( \dfrac{t^3}{3}+xt \right) \right] dt.$$
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$$y' '(z)-zy(z)=0,$$
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which is [[linearly independent]] from the [[Airy Ai]] function.
  
 
<div align="center">
 
<div align="center">
 
<gallery>
 
<gallery>
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File:Airyai.png|Airy $\mathrm{Ai}$ function.
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File:Airybiplot.png|Aairy $\mathrm{Bi}$ function.
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File:Airybi.png|Bairy $\mathrm{Bi}$ function.
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File:Complexairybiplot.png|[[Domain coloring]] of $\mathrm{Bi}$.
 
</gallery>
 
</gallery>
 
</div>
 
</div>
  
 
=Properties=
 
=Properties=
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<div class="toccolours mw-collapsible mw-collapsed">
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[[Relationship between Airy Bi and modified Bessel I]]<br />
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<strong>Theorem:</strong> The function $\mathrm{Ai}$ is a solution to the differential equation
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[[Relationship between Scorer Gi and Airy functions]]<br />
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$$y''(z) - zy(z) = 0.$$
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[[Relationship between Scorer Hi and Airy functions]]<br />
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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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$$y(z) = \displaystyle\int_{\gamma} f(t)e^{zt} dt,$$
 
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where $\gamma$ 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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$$y''(z)=\displaystyle\int_{\gamma} f(t)t^2 e^{zt} dt.$$
 
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Thus we plug this representation into the differential equation to get
 
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$$(*) \hspace{35pt} y''(z)-zy(z) = \displaystyle\int_{\gamma} (t^2-z)f(t)e^{zt} dt = 0.$$
 
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Now we integrate by parts to see
 
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$$\begin{array}{ll}
 
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\displaystyle\int_{\gamma} zf(t)e^{zt} dt &= \displaystyle\int_{\gamma} f(t) \dfrac{d}{dt} e^{zt} dt \\
 
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&= -f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} f'(t)e^{zt} dt.
 
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\end{array}$$
 
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We will pick the contour $\gamma$ to enforce $f(t)e^{zt} \Bigg |_{\gamma}=0$. 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 |_{\gamma} + \displaystyle\int_{\gamma} (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 $\gamma$. 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 whose solution [http://www.wolframalpha.com/input/?i=t^2f%28t%29-f%27%28t%29%3D0 is]
 
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$$f(t)=c e^{\frac{t^3}{3}}.$$
 
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So we have derived
 
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$$y(z)=\displaystyle\int_{\gamma} e^{zt + \frac{t^3}{3}} dt.$$
 
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To pick the contour $\gamma$ note that the integrand of $y$ is an [[entire function]] and hence if $\gamma$ 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
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=Videos=
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$$e^{\frac{t^3}{3}} = \exp\left( \frac{|t|^3 e^{3i\theta}}{3} \right).$$
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[https://www.youtube.com/watch?v=HlX62TkR6gc&noredirect=1 Leading Tsunami wave reaching the shore (27 November 2009)]<br />
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Notice that the inequality $\mathrm{Re} \hspace{2pt} e^{3i\theta} \leq 0$ forces $\cos(3\theta)\leq 0$ yielding three sectors defined by $\theta$:
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[https://www.youtube.com/watch?v=0jnXdXfIbKk&noredirect=1 Series solution of ode: Airy's equation (3 November 2010)]<br />
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$$-\dfrac{\pi}{2} \leq \theta \leq -\dfrac{\pi}{6},$$
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[https://www.youtube.com/watch?v=oYJq3mhg5yE&noredirect=1 Airy differential equation (26 November 2013)]<br />
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$$\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2},$$
 
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$$\dfrac{9\pi}{6} \leq \theta \leq \dfrac{11\pi}{6}.$$
 
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Notice that the first two of these sectors includes the entire $y$-axis in the complex plane. Hence we will take our contour $\gamma$ to be the $y$ axis directed from $-i\infty$ toward $+i\infty$. This yields finally
 
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$$y(z)=\displaystyle\int_{-i\infty}^{i\infty} e^{zt+\frac{t^3}{3}} dt.$$
 
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</div>
 
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</div>
 
  
 
=References=
 
=References=
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[http://www.ams.org/samplings/feature-column/fcarc-rainbows The mathematics of rainbows]<br />
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[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 />
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[http://www.amazon.com/Special-Functions-Introduction-Classical-Mathematical/dp/0471113131 Special Functions: An Introduction to the Classical Functions of Mathematical Physics]<br />
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[http://www.people.fas.harvard.edu/~sfinch/csolve/ai.pdf Airy function zeros]
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=See Also=
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[[Airy Ai]] <br />
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[[Scorer Gi]] <br />
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[[Scorer Hi]] <br />
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[[Category:SpecialFunction]]

Latest revision as of 16:07, 21 October 2017


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.

Properties

Relationship between Airy Bi and modified Bessel I
Relationship between Scorer Gi and Airy functions
Relationship between Scorer Hi and Airy functions

Videos

Leading Tsunami wave reaching the shore (27 November 2009)
Series solution of ode: Airy's equation (3 November 2010)
Airy differential equation (26 November 2013)

References

The mathematics of rainbows
Tables of Weyl Fractional Integrals for the Airy Function
Special Functions: An Introduction to the Classical Functions of Mathematical Physics
Airy function zeros

See Also

Airy Ai
Scorer Gi
Scorer Hi

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