Difference between revisions of "Airy Ai"

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{{:Relationship between Scorer Gi and Airy functions}}

Revision as of 17:37, 14 May 2016

The Airy function $\mathrm{Ai}$ is a solution of the Airy differential equation linearly independent from the Airy Bi function.

Properties

Theorem

The Airy Ai function $$\mathrm{Ai}(z)=\dfrac{1}{2 \pi i} \displaystyle\int_{-i\infty}^{i \infty} e^{-zt + \frac{t^3}{3}} \mathrm{d}t$$ solves the Airy differential equation and moreover, if $z=x$ is a real number then $\mathrm{Ai}(x)$ is a real number and $$\mathrm{Ai}(x)=\dfrac{1}{\pi} \displaystyle\int_{0}^{\infty} \cos \left( xu + \dfrac{u^3}{3} \right) \mathrm{d}u.$$

Proof

Suppose that $y$ has the form $$y(z) = \displaystyle\int_{C} f(t)e^{-zt} dt,$$ 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 $$y^{\prime\prime}(z)=\displaystyle\int_{C} f(t)t^2 e^{-zt} dt.$$ Thus we plug this representation into the differential equation to get $$(*) \hspace{35pt} y^{\prime\prime}(z)-zy(z) = \displaystyle\int_{C} (t^2-z)f(t)e^{-zt} dt = 0.$$ Now we integrate by parts to see $$\begin{array}{ll} \displaystyle\int_{C} zf(t)e^{-zt} dt &= -\displaystyle\int_{C} f(t) \dfrac{d}{dt} e^{-zt} dt \\ &= -f(t)e^{zt} \Bigg |_{C} + \displaystyle\int_{C} f^{\prime}(t)e^{-zt} dt. \end{array}$$ 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 $$\begin{array}{ll} 0 &= y^{\prime\prime}(z) - zy(z) \\ &= f(t)e^{zt} \Bigg |_{C} + \displaystyle\int_{C} (t^2f(t)-f^{\prime}(t))e^{zt} dt. \end{array}$$ We have the freedom to choose $f$ and $C$. We will choose $f$ so that $$t^2f(t)-f^{\prime}(t)=0.$$ This is a simple differential equation with general solution $$f(t)=\xi e^{\frac{t^3}{3}},$$ 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 $$y(z)=\displaystyle\int_{C} e^{-zt + \frac{t^3}{3}} dt.$$ 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 $z \in \mathbb{C}$.

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 $\mathrm{Re} \hspace{2pt} e^{3i\theta} \leq 0$ forces $\cos(3\theta)\leq 0$ yielding three sectors defined by $\theta$:

Airysectors.png

$$-\dfrac{\pi}{2} \leq \theta \leq -\dfrac{\pi}{6},$$ $$\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2},$$ $$\dfrac{5\pi}{6} \leq \theta \leq \dfrac{7\pi}{6}.$$

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: $$y_i(z) = \displaystyle\int_{C_i} e^{-zt + \frac{t^3}{3}} dt;i=1,2,3$$ Since these functions satisfy a second order differential equation, it is impossible for them to be linearly independent. Now notice that we can compute $$\displaystyle\int_{C_1\cup C_2 \cup C_3} e^{-zt + \frac{t^3}{3}} dt = 0.$$ Therefore $$y_1(z)+y_2(z)+y_3(z)=0.$$

By convention we define $$\mathrm{Ai}(z) = \dfrac{1}{2\pi i} \displaystyle\int_{C_1} e^{-zt + \frac{t^3}{3}} dt,$$ 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 $t=-ui$), $$\begin{array}{ll} \mathrm{Ai}(z) &= \dfrac{1}{2 \pi i} \displaystyle\int_{-i\infty}^{i \infty} e^{-zt + \frac{t^3}{3}} dt \\ &= \dfrac{1}{2\pi i} \displaystyle\int_{\infty}^{-\infty} (-i) e^{zui+\frac{(-ui)^3}{3})} du \\ &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} e^{i(zu + \frac{u^3}{3})} du \\ &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \cos\left( zu + \dfrac{u^3}{3} \right) + i \sin \left( zu + \dfrac{u^3}{3} \right) du. \end{array}$$ Now if $z=x$ is a real number, then notice that $$\displaystyle\int_{-\infty}^{\infty} \sin \left( xu + \dfrac{u^3}{3} \right) du = \displaystyle\lim_{b \rightarrow \infty} \int_{-b}^b \sin \left( xu + \dfrac{u^3}{3} \right) du = 0,$$ because $xu+\dfrac{u^3}{3}$ 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: $$\begin{array}{ll} \mathrm{Ai}(x) &= \dfrac{1}{2\pi} \displaystyle\int_{-\infty}^{\infty} \cos \left( xu + \dfrac{u^3}{3} \right) du \\ &= \dfrac{1}{\pi} \displaystyle\int_{0}^{\infty} \cos \left( xu + \dfrac{u^3}{3} \right) du, \end{array}$$ using the fact that the cosine function is even.█

Theorem

The following formula holds: $$\mathrm{Ai}(0)=\dfrac{1}{3^{\frac{2}{3}}\Gamma\left(\frac{2}{3} \right)}=0.3550280538\ldots,$$ where $\mathrm{Ai}$ denotes the Airy Ai function and $\Gamma$ denotes the gamma function.

Proof

References

Theorem

The following formula holds: $$\mathrm{Ai}(z)=\dfrac{1}{\pi} \sqrt{\dfrac{z}{3}} \mathrm{K}_{\frac{1}{3}} \left( \dfrac{2}{3} x^{\frac{3}{2}} \right),$$ where $\mathrm{Ai}$ is the Airy Ai function and $K_{\nu}$ denotes the modified Bessel $K$.

Proof

References

Theorem

The following formula holds: $$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t)\mathrm{d}t + \mathrm{Ai}(x)\displaystyle\int_0^x \mathrm{Bi}(t) \mathrm{d}t,$$ where $\mathrm{Gi}$ denotes the Scorer Gi function, $\mathrm{Ai}$ denotes the Airy Ai function, and $\mathrm{Bi}$ denotes the Airy Bi function.

Proof

References

Theorem

The following formula holds: $$\mathrm{Hi}(x)=\mathrm{Bi}(x)\displaystyle\int_{-\infty}^x \mathrm{Ai}(t) \mathrm{d}t - \mathrm{Ai}(x)\displaystyle\int_{-\infty}^x \mathrm{Bi}(t)\mathrm{d}t,$$ where $\mathrm{Hi}$ denotes the Scorer Hi function, $\mathrm{Ai}$ denotes the Airy Ai function, and $\mathrm{Bi}$ denotes the Airy Bi function.

Proof

References

Videos

Airy differential equation
Series solution of ode: Airy's equation
Leading Tsunami wave reaching the shore

See Also

Airy Bi
Scorer Gi
Scorer Hi

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