The Method of Steepest Descent and Asymptotic Forms of Airy Function

 
  If you have read some textbooks or have taken any lectures pertaining to the WKB approximation while studying quantum physics in undergraduate level, then you might, with high probability, be irritated from the situation that both textbooks and professors just apprise us the formulae of asymptotic forms of Airy function without any sophisticated explanations. (Of course I know it is inevitable bcz of deficient space of textbooks and insufficient time for explaning every underlying mathematics in lectures. Hahaha...). Nevertheless, if you are a student who majors in theoretical physics, then you are obliged to ask a question "WHY?" albeit nobody tells you any underlying reasons!!!

  Hence, in this post, we are going to derive an approximation method which is called The Method of Steepest Descent (or Saddle-Point Approximation), and will apply it to Airy function in purpose to obtain its asymptotic forms.

1. The Method of Steepest Descent (Saddle-Point Approximation)

" How do we obtain an approximation form of a real-valued function $f(t)$ defined on a real domain? "

Let us write the real-valued function $f(t)$ in a form of a contour integral along the contour $C$ \[f(t) = \int_{C} dz\, F(z,t)\] where $F$ is an analytic function and $\forall t \in \mathbb{R}$.

Since $F$ is analytic and an exponential function preserves holomorphism, it can be expressed in terms of an exponential function whose exponent is an analytic function. i.e. \[F(z,t) = e^{w(z,t)}\] where \[w(x,y) = u(x,y) + i \, v(x,y) \quad \forall x,y \in \mathbb{R} \;\;.\]

Lemma. Both $u$ and $v$ cannot have extremum if $w = u + i\,v$ is analytic.

proof. Assume that $u(x,y)$ has a local minimum(maximum) at $(x,y) = (x_0, y_0)$. Then the Hessian matrix of $u(x,y)$ at $(x_0, y_0)$ should be positively(negatively) definite, and this implies $\left.\frac{\partial^2 u}{\partial x^2}\right|_{x_0,y_0}$ and $\left.\frac{\partial^2 u}{\partial y^2}\right|_{x_0,y_0}$ have same signs. (If not, then the determinant of Hessian matrix becomes negative and this violates the condition of positively(negatively) definite Hessian matrix since determinant should be positive in both cases.) Since $w(x,y)$ is analytic, however, $u(x,y)$ must satisfy ${\nabla}^2 u(x,y) = 0$. This implies $\left.\frac{\partial^2 u}{\partial x^2}\right|_{x_0,y_0}$ and $\left.\frac{\partial^2 u}{\partial y^2}\right|_{x_0,y_0}$ cannot have same signs, and we get a contradiction. By applying the identical procedure to $v(x,y)$, we may show that there is NO extremum for $u$ and $v$ if $w = u + i v$ is analytic.$\quad \blacksquare$


Although $u(x,y)$ and $v(x,y)$ cannot have extremum, both can have a saddle point(say $z_0$), and this point has a crucial rule for developing this methodology.

Let us expand $w(z,t)$ near the saddle point $z=z_0$ up to second order of $z$. \[w(z,t) = w(z_0,t) + \frac{1}{2} \left.\frac{\partial^2 w}{\partial z^2}\right|_{z_0} (z-z_0)^2 + \cdots \] Note that $\left.\frac{dw}{dz}\right|_{z_0}$ is vanished since $z_0$ is a saddle point.

For convenience, let us introduce some abbreviations, and let’s denote the phases explicitly. \[w(z_0,t) = w_0 , \quad \left.\frac{\partial^2 w}{\partial z^2}\right|_{z_0} = w''_0 = \left|{w''_0}\right| e^{i\alpha} , \quad z-z_0 = re^{i\theta} \;\;.\]
Then we can rewrite the $w(z,t)$ as \[\begin{aligned} w(z,t) &= w_0 + \frac{1}{2}\left|w''_0\right| e^{i\alpha} (r e^{i\theta})^2 + \mathrm{O}(r^3) \\ &= w_0 + \frac{1}{2}\left|w''_0\right| r^2 e^{ i(\alpha + 2\theta) } + \mathrm{O}(r^3) \\ &= w_0 + \frac{1}{2} \left|w''_0\right| r^2 \left[ \cos(\alpha + 2\theta) + i \, \sin(\alpha + 2\theta) \right] + \mathrm{O}(r^3) \\ &\approx \left[ w_0 + \frac{1}{2} \left|w''_0\right| r^2 \cos(\alpha + 2\theta) \right] + i \, \left[ \frac{1}{2} \left|w''_0\right| r^2 \sin(\alpha + 2\theta) \right] \;\;. \end{aligned}\]
Remember that we have defined $w(z,t)$ by $w = u + i v$. Hence we obtain \[\begin{cases} \quad u(z,t) = w_0 + \frac{1}{2} \left|w''_0\right| r^2 \cos(\alpha + 2\theta) \\ {} \\ \quad v(z,t) = \frac{1}{2} \left|w''_0\right| r^2 \sin(\alpha + 2\theta) \end{cases}\]

Consider the situation in which $z$ departs from $z_0$ (i.e. as r increases from 0).

$\mathbf{\cdot}$ For $u(z,t)$

$\qquad$ i) $u$ will increase 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) $u$ will decrease 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) $u$ will be invariant 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 $v(z,t)$

$\qquad$ i) $v$ will increase 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) $v$ will decrease 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) $v$ will be invariant 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. The arrows indicate the directions in which u and v increase.

Let us discuss about some crucial notions on performing the integral.

• How do we choose an optimum integration contour?

  • We need to choose a contour that passes through the saddle point $z_0$ in the directions of maximum rate of decrease in $u$ with distance from $z_0$, and therefore also in $\vert F \vert$ (since $\vert F \vert = e^{u}$). These directions are level lines of $v(x,y)$, so the phase factor of $F$ (i.e. $e^{iv}$) will not change as we leave tha saddle point. If we chose $z_0$ to be a point other than a saddle point, then the expansion of $w$ (eqn. (1)) would have contained a non-zero linear term in $r$ (since $\left.\frac{\partial u}{\partial z}\right|_{z_0}$ does not vanish.), and it might be impossible to construct a curve which passes through $z_0$ that makes $\vert F \vert = e^u$ to decrease and keeps the phase of F (i.e. $e^{iv}$) constant, as $z$ departs from $z_0$.

• Evaluation of the integral

  • Let us assume that ${\vert w''_0 \vert}$, the measure of the rate of decrease in $\vert F \vert$ as we leave $z_0$, is large enough that the bulk of the value of the integral has already been attained for small $R$. (i.e. The significant contribution to the integral only comes from the immediate vicinity of the saddle point $z_0$.)

Split the integral as below. \[\begin{aligned} f(t) \;=\; \int_{C} dz\,F(z,t) \; &= \; \int_{\text{near the saddle point}} dz \, F(z,t) \;+\; \int_{\text{error contribution}} dz \, F(z,t) \\ &{} \\ &= \; \int_{\text{descend from } F(z_0)} dz \, F(z,t) \;+\; \int_{\text{ascend to } F(z_0)} dz \, F(z,t) \;+\; \int_{\text{error contribution}} dz \, F(z,t) \end{aligned}\]

Near the saddle point, $z = z_0 + r e^{i\theta}$ gives $dz = e^{i\theta} dr$ since we assumed that "the significant contribution to the integral only comes from the immediate vicinity of the saddle point" and this allows us to approximate the integration contour be a straight line right near the saddle point without making significant error. Hence, \[f(t) \;=\; \int_{0}^{R} dr \, r^2 e^{i\theta} \exp{ \left( w_0 + \frac{1}{2} {\vert w''_0 \vert} e^{i(\alpha+2\theta)} \right)} \;+\; \int_{R}^{0} dr \, r^2 e^{i\bar{\theta}} \exp{ \left( w_0 + \frac{1}{2} {\vert w''_0 \vert} e^{i(\alpha+2\bar{\theta})} \right)} \;+\; \text{error term}\] where $\bar{\theta} = \theta - \pi$.

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

Note that our assumption enables us to replace $R \to \infty$ without making any significant error.

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

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

2. The Asymptotic Form of Airy Function

To apply the saddle point approximation, first we need to express Airy function in a form of complex contour integral.

Consider a differential equation whose solutions are Airy functions. \[\begin{aligned} \frac{d^2 f}{d t^2} = t\, f(t) \tag{1} \end{aligned}\]

Let us write the solution $f(t)$ in a form of \[f(t) = \int_{C} dz \, h(z)\,e^{tz} \quad \forall z \in \mathbb{C} \;. \tag{2}\]
By plugging (2) into (1), L.H.S. 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, R.H.S. becomes \[t\,f(t) = \int_{C} dz\, h(z)\, t\, e^{tz} = \left[\, h(z)\,e^{tz} \,\right]_{\text{ends of } C} - \int_{C} dz\,e^{tz}\frac{dh}{dz} \;.\]

In order to make sure the equality in eqn(1), we must set \[\left[\, h(z)\,e^{tz} \,\right]_{\text{ends of } C} = 0 \;.\]
Accepting the constraint above, we can express $h(z)$ as \[z^2 h(z) = -\frac{dh}{dz} \quad \Rightarrow \quad h(z) = c \cdot e^{-\frac{1}{3}z^3} \; .\]
Since we’ve set the square bracket term to 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 \[\frac{2n\pi}{3} - \frac{\pi}{6} \, < \, \arg{(z_{\text{end}})} \,<\, \frac{2n\pi}{3} + \frac{\pi}{6} \tag{3}\]
and according to this constraint, the complex plane is divided as below.

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

Due to the condition (3), we must set our integration curve $C$ to start from an infinity in one gray sector, and terminate at infinity in the other gray sector. Note also that since there are three different sectors that satisfy the condition (3), we can make three different contours, say $C_1, C_2, C_3$, and each choice will give corresponding solutions $f_1, f_2, f_3$.

Consequently, the general solutions(in the contour integral form) to the differential equation (1) can be written as \[\begin{aligned} & \mathrm{Ai}\,(t) = \frac{1}{2\pi i} \int_{C_1} dz\,e^{zt-\frac{z^3}{3}} \\ &{} \\ & \mathrm{Bi}\,(t) = \frac{1}{2\pi} \int_{C_2 - C_3} dz\, e^{zt-\frac{z^3}{3}} \end{aligned}\]