Difference between revisions of "Airy Bi"
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| − | The Airy | + | The Airy function $\mathrm{Ai}$ and "Bairy" function $\mathrm{Bi}$ are given by the formulas |
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$$\mathrm{Ai}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \cos \left( \dfrac{t^3}{3} + xt \right) dt$$ | $$\mathrm{Ai}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \cos \left( \dfrac{t^3}{3} + xt \right) dt$$ | ||
and | and | ||
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[[File:Airybi.png|500px]] | [[File:Airybi.png|500px]] | ||
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| + | =Properties= | ||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Theorem:</strong> The function $\mathrm{Ai}$ is a solution to the differential equation | ||
| + | $$y''(z) - zy(z) = 0.$$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> Suppose that $y$ has the form | ||
| + | $$y(z) = \displaystyle\int_{\gamma} f(t)e^{zt} dt,$$ | ||
| + | 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 | ||
| + | $$y''(z)=\displaystyle\int_{\gamma} f(t)t^2 e^{zt} dt.$$ | ||
| + | Thus we plug this representation into the differential equation to get | ||
| + | $$(*) \hspace{35pt} y''(z)-zy(z) = \displaystyle\int_{\gamma} (t^2-z)f(t)e^{zt} dt = 0.$$ | ||
| + | Now we integrate by parts to see | ||
| + | $$\begin{array}{ll} | ||
| + | \displaystyle\int_{\gamma} zf(t)e^{zt} dt &= \displaystyle\int_{\gamma} f(t) \dfrac{d}{dt} e^{zt} dt \\ | ||
| + | &= -f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} f'(t)e^{zt} dt. | ||
| + | \end{array}$$ | ||
| + | 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 | ||
| + | $$\begin{array}{ll} | ||
| + | 0 &= y''(z) - zy(z) \\ | ||
| + | &= f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} (t^2f(t)-f'(t))e^{zt} dt. | ||
| + | \end{array}$$ | ||
| + | We have the freedom to choose $f$ and $\gamma$. We will choose $f$ so that | ||
| + | $$t^2f(t)-f'(t)=0.$$ | ||
| + | This is a simple differential equation whose solution [http://www.wolframalpha.com/input/?i=t^2f%28t%29-f%27%28t%29%3D0 is] | ||
| + | $$f(t)=c e^{-\frac{t^3}{3}}.$$ | ||
| + | So we have derived | ||
| + | $$y(z)=\displaystyle\int_{\gamma} e^{zt - \frac{t^3}{3}} dt.$$ | ||
| + | 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 | ||
| + | $$e^{-\frac{t^3}{3}} = \exp\left( \frac{-|t|^3 e^{3i\theta}}{3} \right).$$ | ||
| + | Notice that the inequality $\mathrm{Re} \hspace{2pt} e^{3i\theta} > 0$ forces $\cos(3\theta)>0$ yielding three sectors defined by $\theta$: | ||
| + | $$-\dfrac{\pi}{6} < \theta < \dfrac{\pi}{6},$$ | ||
| + | $$\dfrac{\pi}{2} < \theta < \dfrac{5\pi}{6},$$ | ||
| + | $$\dfrac{7\pi}{6} < \theta < \dfrac{3\pi}{2}.$$ | ||
| + | Picking the contour $\gamma$ so that it tends to complex infinity inside two of the sectors forces the condition $f(t)e^{zt} \Bigg |_{\gamma}=0$. █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | =References= | ||
Revision as of 08:15, 14 January 2015
The Airy function $\mathrm{Ai}$ and "Bairy" function $\mathrm{Bi}$ are given by the formulas $$\mathrm{Ai}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \cos \left( \dfrac{t^3}{3} + xt \right) dt$$ and $$\mathrm{Bi}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \left[ e^{-\frac{t^3}{3} + xt} + \sin \left( \dfrac{t^3}{3}+xt \right) \right] dt.$$
Properties
Theorem: The function $\mathrm{Ai}$ is a solution to the differential equation $$y(z) - zy(z) = 0.$$
Proof: Suppose that $y$ has the form $$y(z) = \displaystyle\int_{\gamma} f(t)e^{zt} dt,$$ 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 $$y(z)=\displaystyle\int_{\gamma} f(t)t^2 e^{zt} dt.$$ Thus we plug this representation into the differential equation to get $$(*) \hspace{35pt} y(z)-zy(z) = \displaystyle\int_{\gamma} (t^2-z)f(t)e^{zt} dt = 0.$$ Now we integrate by parts to see $$\begin{array}{ll} \displaystyle\int_{\gamma} zf(t)e^{zt} dt &= \displaystyle\int_{\gamma} f(t) \dfrac{d}{dt} e^{zt} dt \\ &= -f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} f'(t)e^{zt} dt. \end{array}$$ 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 $$\begin{array}{ll} 0 &= y(z) - zy(z) \\ &= f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} (t^2f(t)-f'(t))e^{zt} dt. \end{array}$$ We have the freedom to choose $f$ and $\gamma$. We will choose $f$ so that $$t^2f(t)-f'(t)=0.$$ This is a simple differential equation whose solution is $$f(t)=c e^{-\frac{t^3}{3}}.$$ So we have derived $$y(z)=\displaystyle\int_{\gamma} e^{zt - \frac{t^3}{3}} dt.$$ 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 $z \in \mathbb{C}$.
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 $$e^{-\frac{t^3}{3}} = \exp\left( \frac{-|t|^3 e^{3i\theta}}{3} \right).$$ Notice that the inequality $\mathrm{Re} \hspace{2pt} e^{3i\theta} > 0$ forces $\cos(3\theta)>0$ yielding three sectors defined by $\theta$: $$-\dfrac{\pi}{6} < \theta < \dfrac{\pi}{6},$$ $$\dfrac{\pi}{2} < \theta < \dfrac{5\pi}{6},$$ $$\dfrac{7\pi}{6} < \theta < \dfrac{3\pi}{2}.$$ Picking the contour $\gamma$ so that it tends to complex infinity inside two of the sectors forces the condition $f(t)e^{zt} \Bigg |_{\gamma}=0$. █
