The Method of Steepest Descent and Asymptotic Forms of Airy Function

 
  Airy function has a crucial rule in quantum physics because it is the wave function of a linear potential. In WKB-approximation, we consider a smooth potential function and approximate it as a linear potential on a specific spacial interval. And then we match wave functions outside and approximated wave functions inside (which are in forms of asymptotic forms of Airy function) on each boundary. By matching each wave function with each other's pair, we get additional information which alludes us quantization condition or et cetera.

  Although studying limiting behaviors of Airy function is important, however, detailed derivations are seldom discussed in textbooks. Hence, in this post, we first introduce the method of steepest descent, a powerful integral approximation technique, and apply it to derive the asymptotic forms of Airy function, particularly the first kind.

1. The Method of Steepest Descent

" How do we obtain an asymptotic form of a real-valued function $f:\mathbb{R}\to\mathbb{R}$? "

Write a real-valued function $f(t)$ be in a form of a contour integral, along the contour $C$, by \[ f(t) = \int_{C} dz\, F_t (z) \] where $F_t$ is an analytic function for each ${t} \in \mathbb{R}$.

Since $F_t$ is analytic and an exponential function preserves holomorphicity, $F_t$ can be rewritten by \[ F_t (z) = e^{w_t (z)} \] where \[w_t (x,y) = u_t (x,y) + i \, v_t (x,y) \quad \forall x,y \in \mathbb{R}\] is also analytic.


Observe that neither $u_t$ nor $v_t$ can have extremum!


Lemma. Neither $u_t$ nor $v_t$ can have extremum if $w_t = u_t + i\,v_t$ is analytic.
proof. Assume that $u_t (x,y)$ becomes locally minimum(maximum) at $(x_0, y_0)$. Then Hessian matrix of $u_t$ at $(x_0, y_0)$ should be positively(negatively) definite. This implies $\left.\frac{\partial^2 u_t}{\partial x^2}\right|_{x_0,y_0}$ and $\left.\frac{\partial^2 u_t}{\partial y^2}\right|_{x_0,y_0}$ have same signs (If not, the determinant of Hessian matrix becomes negative and this violates the condition of positively(negatively) definite.). Since $w_t (x,y)$ is analytic, however, $u_t (x,y)$ must satisfy $ {\nabla}^2 u_t (x,y) = 0 $. This implies $\left.\frac{\partial^2 u_t}{\partial x^2}\right|_{x_0,y_0}$ and $\left.\frac{\partial^2 u_t}{\partial y^2}\right|_{x_0,y_0}$ cannot have same signs and we get a contradiction!
Applying the same logic to $v_t (x,y)$, we can check $v_t$ has no extremum. $\blacksquare$


Although $u_t (x,y)$ and $v_t (x,y)$ cannot have any extremum, both can have a saddle point.


Let us expand $w_t (z)$ near the saddle point, say $z=z_0$, up to second order of $z$. \[w_t (z) = w_t (z_0) + \frac{1}{2} \left.\frac{\partial^2 w_t}{\partial z^2}\right|_{z_0} (z-z_0)^2 + \cdots \tag{1} \] Note that $ \left.\frac{dw}{dz}\right|_{z_0} $ vanished since $z_0$ is a saddle point. Moreover, the holomorphicity of $w_t(z)$ assures that no singular term exists.

For convenience, let us denote \[w_t (z_0) = w_0 , \quad \left.\frac{\partial^2 w}{\partial z^2}\right|_{z_0} = w''_0 = \left|{w''_0}\right| e^{i\alpha} , \quad z-z_0 = re^{i\theta} \;\;.\]
Then we can rewrite the $w_t (z)$ as \[ \begin{aligned} w_t (z) &= w_0 + \frac{1}{2}\left|w''_0\right| e^{i\alpha} (r e^{i\theta})^2 + \mathrm{O}(r^3) \\ &= w_0 + \frac{1}{2}\left|w''_0\right| r^2 e^{ i(\alpha + 2\theta) } + \mathrm{O}(r^3) \\ &= w_0 + \frac{1}{2} \left|w''_0\right| r^2 \left[ \cos(\alpha + 2\theta) + i \, \sin(\alpha + 2\theta) \right] + \mathrm{O}(r^3) \\ &\approx \left[ w_0 + \frac{1}{2} \left|w''_0\right| r^2 \cos(\alpha + 2\theta) \right] + i \, \left[ \frac{1}{2} \left|w''_0\right| r^2 \sin(\alpha + 2\theta) \right] \;. \end{aligned}\]
Since we defined $w_t (z)$ by $ w_t = u_t + i v_t $, we obtain \[u_t (z) = w_0 + \frac{1}{2} \left|w''_0\right| r^2 \cos(\alpha + 2\theta)\] and \[v_t (z) = \frac{1}{2} \left|w''_0\right| r^2 \sin(\alpha + 2\theta) \;.\]

Consider a situation that $z$ leaves from $z_0$ (i.e. $r$ increases from 0) and observe the followings:

$\mathbf{\cdot}$ For $u_t (z)$

$\qquad$ i) $u_t$ increases most rapidly when \[ \cos(\alpha + 2\theta) = 1 \quad \Rightarrow \quad \alpha + 2\theta = 2n\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} +n\pi \; . \] $\qquad$ ii) $u_t$ decreases most rapidly when \[ \cos(\alpha + 2\theta) = -1 \quad \Rightarrow \quad \alpha + 2\theta = (2n+1)\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} + \left(n+\frac{1}{2}\right)\pi \; . \] $\qquad$ iii) $u_t$ doesn't change when \[ \cos(\alpha + 2\theta) = 0 \quad \Rightarrow \quad \alpha + 2\theta = \left(n+\frac{1}{2}\right)\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} + \left(\frac{n}{2}+\frac{1}{4}\right)\pi \; . \]

$\mathbf{\cdot}$ For $v_t (z)$

$\qquad$ i) $v_t$ increases most rapidly when \[ \sin(\alpha + 2\theta) = 1 \quad \Rightarrow \quad \alpha + 2\theta = \left(2n+\frac{1}{2}\right)\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} +\left(n+\frac{1}{4}\right)\pi \; . \] $\qquad$ ii) $v_t$ decreases most rapidly when \[ \sin(\alpha + 2\theta) = -1 \quad \Rightarrow \quad \alpha + 2\theta = \left(2n+\frac{3}{2}\right)\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} + \left(n+\frac{3}{4}\right)\pi \; . \] $\qquad$ iii) $v_t$ doesn't change when \[ \sin(\alpha + 2\theta) = 0 \quad \Rightarrow \quad \alpha + 2\theta = n\pi \quad \Rightarrow \quad \theta = -\frac{\alpha}{2} + \frac{n\pi}{2} \; . \]

Fig 1. Each arrow's head points toward each direction of steepest increase of $u_t$ and $v_t$.

Keep in mind that we haven't imposed any restriction to the integration contour $C$. This alludes us that we can set the $C$ however we want. Before proceeding, therefore, we should agonize for a while what contour will give us the most reasonable consequence.

• Why are we specifically working with the saddle point?

  • If $z_0$ is any point other than the saddle point, the series expansion of $w_t$ contains a non-zero linear term in $r$ (i.e. $z-z_0$) because $ \left.\frac{\partial u}{\partial z}\right|_{z_0} $ doesn't vanish this time. Then, it is almost impossible to find a contour that passes through $z_0$ making $\vert F \vert = e^u $ decreases and keeping the phase term of $F$ (i.e. $e^{iv}$) invariant, as $z$ departs from $z_0$. This is why we are dealing with the saddle point. Thus, it is natural to set $z_0$ to the saddle point.

• What is the most adequate integration contour?

  • We should pick a contour which passes through the saddle point $z_0$ along the directions that $u$ decreases most rapidly as $z$ leaves from $z_0$. Observe that this contour is exactly the level line of $v_t$ (see the fig.1). Hence, $|F|=\exp(u_t)$ decreases most rapidly along this contour keeping the phase contribution of $F$, $\exp(i v_t)$, constant.

• Evaluation of the integral

  • From now onwards, we assume that ${\vert w''_0 \vert}$ (the measure of the rate of decrease in $|F|$ as we leave $z_0$) is large enough that the value of the integral has already been attained only from the vicinity of the saddle point. (i.e. The significant contribution to the integral comes from the tiny segment of the contour near the saddle point where $R\ll1$.)

Let us split the integration contour $C$ as below. \[ \begin{aligned} f(t) \; &=\; \int_{C} dz\,F_t (z) \; \\ &{} \\ &= \; \int_{\text{near the saddle point}} dz \, F_t (z) \;+\; \int_{\text{error contribution}} dz \, F_t (z) \\ &{} \\ &= \; \int_{\text{descend from } F_t (z_0)} dz \, F_t (z) \;+\; \int_{\text{ascend to } F_t (z_0)} dz \, F_t (z) \;+\; \int_{\text{error contribution}} dz \, F_t (z) \end{aligned} \]

Because we assumed "the significant contribution to the integral only comes from the immediate vicinity of the saddle point", we can approximate the integration contour by a straight line penetrating through the saddle point, without making significant error. Since $ z = z_0 + r e^{i\theta} $ gives $ dz = e^{i\theta} dr $ and the angle $\theta$ is invariant as $z$ varies along the contour $C$ according to our 'straight-line approximation', we get \[ f(t) \;=\; \int_{0}^{R} dr \, r^2 e^{i\theta} \exp{ \left( w_0 + \frac{1}{2} {\vert w''_0 \vert} e^{i(\alpha+2\theta)} \right)} \;+\; \int_{R}^{0} dr \, r^2 e^{i\bar{\theta}} \exp{ \left( w_0 + \frac{1}{2} {\vert w''_0 \vert} e^{i(\alpha+2\bar{\theta})} \right)} \;+\; \text{error term} \] where $ \bar{\theta} = \theta - \pi$.

Note that we set the integration contour to satisfy $ e^{i(\alpha + 2\theta)} = e^{i(2n+1)\pi} = -1 $ right near the saddle point. Thus, \[ \begin{aligned} f(t) \;&=\; e^{w_0 + i\theta} \int_{0}^{R} dr \, r^2 e^{ - \frac{1}{2} {\vert w''_0 \vert}^2 } \;+\; e^{w_0 + i(\theta-\pi)} \int_{R}^{0} dr \, r^2 e^{ - \frac{1}{2} {\vert w''_0 \vert} } \;+\; \text{error term} \\ &=\; 2 \, e^{w_0 + i\theta} \int_{0}^{R} dr\, e^{-\frac{1}{2} {\vert w''_0 \vert} r^2 } \;+\; \text{error term} \;\;. \end{aligned} \]

Furthermore, since we assumed ${\vert w''_0 \vert}$ is large enough that only the neighborhood of the saddle point matters, we can set $ R \to \infty$ without making a significant error.

\[ f(t) \;\approx \; 2\, e^{w_0 + i\theta} \int_{0}^{\infty} dr \, e^{-\frac{1}{2} {\vert w''_0 \vert} r^2} \;=\; 2\,e^{w_0 + i\theta} \frac{1}{2} \sqrt{ \frac{2\pi}{\vert w''_0 \vert} } \;=\; e^{w_t (z_0)} e^{i\theta} \sqrt{ \frac{2\pi}{\vert{ w''_t (z_0) \vert} } } \]
Therefore, for the region $\vert t \vert \gg 1$, we can approximate $f(t)$ by \[ \therefore \qquad f(t) \;\overset{|t| \gg 1} {\approx} \; F_t (z_0) \, e^{i\theta} \sqrt{ \frac{2\pi}{\vert w''_t (z_0) \vert} } \] where $z_0$ is a saddle point of $w_t (z)$ and \[ \theta = -\frac{1}{2} \mathrm{arg}{\left[w''_t (z_0)\right]} + { \left( { \frac{\pi}{2} \text{ or } \frac{3\pi}{2} } \right)\,. \; } \] The choice of $\theta$ (which affects only the sign of the final result) depends only to the sense in which the contour $C$ passes through the saddle point $z_0$. $\; \blacksquare$

Let's derive asymptotic forms of Airy function applying this approximation method!

2. The Asymptotic Forms of Airy Function

To apply the method of steepest descent, we need to express Airy function in a form of complex contour integral. \[ \begin{aligned} \frac{d^2 f}{d t^2} = t\, f(t) \tag{1} \end{aligned} \]

Take the solution $f(t)$ by \[ f(t) = \int_{C} dz \, h(z)\,e^{tz} \quad \forall z \in \mathbb{C} \;. \tag{2} \]
Plugging (2) into (1), the L.H.S. of (1) becomes \[ \frac{d^2 f}{d t^2} = \int_{C} dz \, h(z)\frac{d^2}{d t^2}e^{tz} = \int_{C} dz\, z^2 h(z) e^{tz} \;, \]
and by integrating by parts, the R.H.S. of (1) becomes \[ t\,f(t) = \int_{C} dz\, h(z)\, t\, e^{tz} = \left[\, h(z)\,e^{tz} \,\right]_{\text{ends of } C} - \int_{C} dz\,e^{tz}\frac{dh}{dz} \;. \]

To make sure the equality of eqn(1) holds, we must set \[ \left[\, h(z)\,e^{tz} \,\right]_{\text{ends of } C} = 0 \;. \]
Accepting the constraint above, $ h(z) $ is now given by \[ z^2 h(z) = -\frac{dh}{dz} \quad \Rightarrow \quad h(z) = c \cdot e^{-\frac{1}{3}z^3} \] where $c$ is just a constant.

Since the square bracket term should be annihilated, each end point of the integration contour $C$ must satisfy \[ \exp{\left[ -\frac{1}{3} {z^3_\text{end}} + t\, z_{\text{end}} \right]} = 0 \;. \]
This condition implies \[ \mathrm{Re}{(z^3_{\text{end}})} > 0 \qquad \text{and} \qquad \left| \, z_{\text{end}} \, \right| \to \infty \] because $z^3$ diverges much faster than $z$, as $|z|$ increases.

From $\mathrm{Re}{(z^3_{\text{end}})}>0$, we obtain obtain \[ \frac{2n\pi}{3} - \frac{\pi}{6} \, < \, \arg{(z_{\text{end}})} \,<\, \frac{2n\pi}{3} + \frac{\pi}{6} \; . \tag{3} \]
Accordingly, the complex plane is divided as Fig(2).

Fig 2. All the ends of the integration contours should be lied inside the shaded areas.

By the constraint (3), the integration contour $C$ should start from an infinity in one gray sector and terminate at infinity in the other gray sector. Note that, since there are three different shaded sectors, we can construct three different contours, say $C_1, C_2, C_3$, and each choice yields corresponding solutions $f_1, f_2, f_3$.

Consequently, the general solutions to the differential equation (1) are given by \[ \begin{aligned} & \mathrm{Ai}\,(t) = \frac{1}{2\pi i} \int_{C_1} dz\,e^{zt-\frac{z^3}{3}} \\ &{} \\ & \mathrm{Bi}\,(t) = \frac{1}{2\pi} \int_{C_2 - C_3} dz\, e^{zt-\frac{z^3}{3}} \end{aligned} \]