Resurgence Theory Series — Part 1: Historical Development

1. What Awaits You in This Series?

Our primary aim throughout this series is to provide the mathematical background necessary to understand resurgence theory. Over the course of the series, we will examine topics such as the classification of differential equations, perturbation theory, complex analysis, and asymptotic analysis. In this first lesson, we will look at what resurgence theory is and how it developed historically. After all, no theory has ever emerged in a vacuum.

Throughout the lesson, you will encounter concepts in quotation marks. We will develop these concepts mathematically in later installments of our course and provide physical examples where appropriate. Without further ado, let us begin.

2. What Is Resurgence Theory?

The first milestone in the development of resurgence theory was actually laid by George Biddell Airy in 1838. So, what exactly is resurgence theory?

The mathematical models we use in physics often lead us to differential equations or integral equations. The "series solutions" of these equations frequently yield divergent results. Should we then conclude that the differential equation produces no sensible result and discard it? Of course not.

Before the theory of series had been placed on firm mathematical foundations, one scientist sought meaningful results by manipulating divergent series: Leonhard Euler (1707–1783). Euler was among the first to recognize the expression $1+2+3+\cdots=-\frac{1}{12}$, which many people today associate with Srinivasa Ramanujan (1887–1920).

Let us bring the discussion back into focus and summarize what resurgence tells us about these divergent solutions. According to resurgence theory, a solution diverges because of a "singularity" or singularities in the plane. The claim of resurgence theory is that physical results can be recovered once the contributions arising from these singularities are included.

3. George Biddell Airy and the Rainbow

George B. Airy, who served as Astronomer Royal from 1835 to 1881, noticed in rainbows a phenomenon analogous to the fringes seen in images obtained through telescopes. He thought that if he could explain these small bands in the rainbow—the supernumerary rainbow—he might also be able to sharpen telescopic images.

Using Thomas Young's then relatively new wave theory of light, published in 1802, Airy succeeded in modeling the formation of a rainbow [1].

$$\operatorname{Ai}(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \exp\left\{i\left( \frac{t^3}{3} + xt \right)\right\} dt. \tag{1}$$

The differential equation that gives rise to this integral equation is known as the Airy differential equation:

$$\frac{d^2y}{dx^2} = xy. \tag{2}$$

There was, however, a problem. In the integral equation he had obtained, the integrand oscillated extremely rapidly for large values of $t$. Airy consulted Augustus De Morgan about how he might solve this equation. De Morgan (1806–1871) advised him to construct a solution using convergent series.

Yet, as Figure 2 shows, when he solved it using convergent series, the first fringe did not appear until the nineteenth term. Convergent series therefore did not allow him to study the supernumerary bands as he wished. Why, then, had De Morgan recommended a convergent series? Because Niels Henrik Abel (1802–1829), one of the leading mathematicians of his day, had called divergent series an invention of the devil and vehemently opposed their use [2].

The bold scientist who used divergent series in the 1850s to calculate the supernumerary bands with very high precision was George Stokes.

4. George Stokes and Divergent Series

When George Stokes (1819–1903) examined the equation using divergent series, he noticed that the series behaved convergently up to a certain term, after which its terms grew exponentially and the series diverged. By truncating it judiciously, he discarded everything after its last convergent term. In this way, with a partial sum as short as five terms, he obtained a result very close to the true value, as Figure 3 shows [3].

Stokes's solution described the supernumerary bands with remarkable precision for the technology of the time. Although, like any scientist who had achieved his aim, he might have been expected to be satisfied, Stokes was troubled by the graph blowing up at the origin. Moreover, he was puzzled that the solution of a single equation differed on the positive and negative axes. Stokes completed his treatment of the subject in three separate works (1857, 1889, 1902). He realized that the function had different representations in different regions because of rays in the plane known as "Stokes lines." He observed that solutions changed character as they crossed these lines. This change is called the "Stokes jump" or "Stokes discontinuity."

Let us now go a little further into the mathematics and try to solve the Airy differential equation.

5. Solution of the Airy Differential Equation — Part 1

In this section, we will seek a solution to the Airy differential equation. At one stage, we will learn a technique from asymptotic analysis needed to complete the solution, and then return to it.

As we mentioned earlier, the Airy differential equation is

$$y'' = xy.$$

There are several ways to solve a differential equation of this kind. For instructional purposes, however, let us seek a solution using the Fourier transform.

For those who have not encountered it before, the Fourier transform is essentially a transformation that allows us to pass between position space and momentum space. Consequently, once the equation has been solved after this transformation, it must be converted back into position-space form. To do this, we must apply the inverse Fourier transform. For readers encountering these transformations for the first time, let us briefly state their definitions:

$$Y(k) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dx \ e^{-ikx} \ y(x), \tag{3}$$ $$y(x) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dk \ e^{+ikx} \ Y(k), \tag{4}$$

Having given the definitions, what we need to do is find the Fourier transforms of $\frac{d^2y}{dx^2}$ and $xy$. Let us see what transformation results when we differentiate the definition of the Fourier transform with respect to $k$.

The only part of the integrand that depends on $k$ is the exponential. Therefore, the derivative operator acts only on the exponential term.

$$\frac{dY(k)}{dk} = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dx \ e^{-ikx} \ [-ix y(x)], \tag{5}$$

The expression $xy(x)$ inside the integral is the transform we are looking for. Thus, we have found the result for the right-hand side of the equation. Let us now focus on the left-hand side. If we place the desired expression inside the integral and integrate by parts twice, we obtain

$$\frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dx \ e^{-ikx} \ y''(x) = -\frac{k^2}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dx \ e^{-ikx} \ y(x) = -k^2 Y(k), \tag{6}$$

So the second-order differential equation in our hands has been transformed into a simple first-order differential equation.

$$-k^2 Y = i \frac{dY}{dk}. \tag{7}$$

If we approach it like a physicist and move $Y$ and $dk$ to opposite sides—the method of separation of variables—we are left with a Calculus I problem. Solving it directly gives

$$ik^2 dk = \frac{dY}{Y} \quad \Rightarrow \quad Y(k) = A \exp\left\{i\frac{k^3}{3}\right\}. \tag{8}$$

Now let us apply the inverse Fourier transform to carry the solution of the equation back into position space.

$$y(x) = \frac{A}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} dk \ \exp\left\{+i\left(\frac{k^3}{3}+kx\right)\right\}. \tag{9}$$

Notice that (1) and (9) have the same integrand. Thus, starting from the differential equation, we have recovered the integral representation of the Airy function. Solving this integral equation requires knowledge of "asymptotic analysis" and "complex analysis." We will therefore pause this solution briefly and fill in the necessary background.

6. A Brief Look at Complex Analysis: Euler's Identity

Learning the ending while still at the beginning of a subject can sometimes dampen one's enthusiasm. In mathematics, however, it can be very useful because it gives us a goal. Let us therefore begin by stating Euler's identity:

$$e^{i\theta} = \cos{\theta} + i \sin{\theta}. \tag{10}$$

Students who have taken mathematical physics or complex analysis will know this identity and its proof. For readers who have not yet taken either course, let us briefly explain the basic concepts of complex analysis and prove the identity. Readers already familiar with these topics may proceed to §7 A Brief Look at Asymptotic Analysis.

Since the main purpose of our lesson is not to teach complex analysis, we will mention only the most fundamental concepts relevant to the calculations we will perform and then move on.

At some point in our education, all of us have encountered the imaginary number ($i = \sqrt{-1}$). Complex numbers are generally denoted by $z$, and the set of complex numbers by $\mathbb{C}$. The set of complex numbers and a complex number are defined as follows:

$$\begin{aligned} &\mathbb{C}=\{x+iy: \ \ x,y \in \mathbb{R}\}, \\ &z=x+iy \in \mathbb{C}. \end{aligned} \tag{11}$$

After just a few iterations of multiplication by $i$, it is easy to see that we return to the starting point after every four multiplications.

• $i^1=i$
• $i^2=-1$
• $i^3=-i$
• $i^4=1$

Multiplying a number by $i$ actually corresponds to a rotation of $90^\circ$ in the plane, as shown in Figure 4. For an instructive example, the graph takes $z=1+2i$. Showing that $iz=-2+i$ is perpendicular to $z=1+2i$ is left as an exercise for the reader.

With these basic properties in hand, let us quickly proceed to the proof of Euler's identity. We will use the Taylor/Maclaurin series learned in Calculus I. To recall them briefly:

$$\begin{aligned} & e^x = \sum_{n \geq 0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots, \\ & \cos{x} = \sum_{n\geq 0} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \cdots, \\ & \sin{x} = \sum_{n\geq 0} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \cdots. \end{aligned} \tag{12}$$

A little inspection of the equations above shows that $\cos{x} + \sin{x}$, apart from differences in sign ($\pm$), closely resembles $e^x$. Let us therefore expand $e^{i\theta}$, $\cos{\theta}$, and $\sin{\theta}$ in series and see whether the powers of $i$ account for those sign differences:

$$\begin{aligned} & e^{i\theta} = \sum_{n \geq 0} \frac{(i\theta)^n}{n!} = 1 + i\theta - \frac{\theta^2}{2!} - i \frac{\theta^3}{3!} + \frac{\theta^4}{4!}+ \cdots, \\ & \cos{\theta} = \sum_{n\geq 0} \frac{(-1)^n \theta^{2n}}{(2n)!} = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!}+\cdots, \\ & \sin{\theta} = \sum_{n\geq 0} \frac{(-1)^{n}\theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!} + \cdots. \end{aligned} \tag{13}$$

As is perfectly clear from the expressions above, examining the terms one by one yields Euler's identity. Thus,

$$e^{i\theta} = \cos{\theta} + i \sin{\theta}.$$

Before completing this section, let us add one final note. If we choose $\theta=\frac{\pi}{2}=90^\circ$, Euler's identity gives $e^{i\frac{\pi}{2}} = i$. We previously said that multiplying a number by $i$ corresponds to a rotation of $90^\circ$. It follows that multiplying a number by $e^{i\theta}$ corresponds to a rotation through $\theta$. As an example, Figure 5 shows the graph for $z=3+4i$ and $z\exp{\left[i\frac{37\pi}{180}\right]}$. Demonstrating that the angle between the two vectors in the graph is $37^\circ$ is again left to the reader.

7. A Brief Look at Asymptotic Analysis: The Method of Stationary Phase

Let us begin by briefly explaining what asymptotic analysis means. Roughly speaking, asymptotic analysis is a mathematical technique that simplifies the process of determining how a complicated integral equation behaves at very large or very small values. Therefore, when constructing an asymptotic solution, we must also specify the conditions under which that solution is being examined.

Suppose that $g(k)$ and $\phi(k)$ are "well-behaved" real-valued functions of a real variable for every $k \in [a,b]$, and that we have an integral equation of the following form:

$$I(t) = \int_{a}^{b} dk \ g(k) e^{it \phi(k)}. \tag{14}$$

As discussed in §6, Euler's identity allows us to write $e^{it \phi(k)} = \cos{[t \phi(k)]} + i \sin{[t \phi(k)]}$. Thus, the oscillation becomes increasingly rapid as $t$ grows. Consequently, when $t$ is large, intervals outside a neighborhood of the value at which the phase—the exponential expression—is stationary ($\phi'(k)=0$) make no contribution. We can see this more concretely in Figure 6 for different values of $t$, with $\phi(k)=k^2$, $a=0$, and $b=1$.

Therefore, for $\phi(k)$, if we Taylor-expand around a point satisfying $\phi'(k)=0$, namely $k_0$, we obtain

$$\phi(k) = \phi(k_0) + \frac{\phi''(k_0)}{2}(k-k_0)^2 + \mathcal{O}((k-k_0)^3), \tag{15}$$

If we seek a convergent solution using the first two terms, our equation takes the form

$$I(t) \approx e^{it\phi(k_0)} \int_{a}^{b} dk \ g(k) e^{it\frac{\phi''(k_0)}{2}(k-k_0)^2}. \tag{16}$$

The term $e^{it\phi(k_0)}$ can be taken outside because it is independent of the variable $k$. Moreover, our integral has now been reduced to a Gaussian-type integral. One point still requires attention, however: our limits are not yet $[0,\infty)$ or $(-\infty,+\infty)$. Nor have we yet used our assumption that $t$ is large.

Let us therefore apply the first idea from Calculus I that comes to mind to simplify the Gaussian-type integral: a change of variables. If we choose $u=(k-k_0)\sqrt{t}$, the Gaussian kernel is normalized. Using at this stage our assumption that $t$ is very large ($t \to \infty$), our integration interval becomes $(-\infty,+\infty)$. At the same time, since $g(k)$ contributes only near $k=k_0$, we may take $g(k) \approx g(k_0)$. We thereby obtain the asymptotic form of our original integral.

$$I(t) \sim \frac{e^{it \phi(k_0)} g(k_0)}{\sqrt{t}} \int_{-\infty}^{+\infty} du \ e^{i\frac{\phi''(k_0)}{2}u^2}, \quad t \to \infty. \tag{17}$$

At this point, we have a complete Gaussian integral and can readily obtain the following result:

$$I(t) \sim \sqrt{\frac{2\pi}{t |\phi''(k_0)|}} g(k_0) e^{i\left(t \phi(k_0) \pm \frac{\pi}{4}\right)}, \quad t \to \infty. \tag{18}$$

We can now turn to the Airy integral equation (1), the reason we introduced all these concepts in the first place.

8. Solution of the Airy Differential Equation — Part 2

Since we took a long detour, let us begin by recalling the Airy integral equation:

$$\operatorname{Ai}(x) = \frac{1}{\pi} \int_{0}^{\infty} \exp\left\{i\left(\frac{t^3}{3}+xt\right)\right\} dt.$$

Here, the independent variable we will take to be large is $x$, while $g(t)=1$ and our phase function is $\phi(t)=\frac{t^3}{3}+xt$. Let us begin the calculation:

$$\phi(t) = \frac{t^{3}}{3}+tx \quad \Rightarrow \quad \phi'(t_0)=t_0^2+x \quad \Rightarrow \quad t_0=\pm i\,x^{1/2}. \tag{19}$$

Now that we have found the stationary phase, we can substitute everything into the equation:

$$\operatorname{Ai}(x) \sim \frac{e^{\pm \frac{2}{3}x^{3/2}}}{\pi} \int\limits_{0}^{+\infty} e^{\pm x^{1/2}(t-t_0)^2}dt, \quad x \to \infty. \tag{20}$$

Evaluating the integral then gives the asymptotic solutions of the Airy function.

$$\operatorname{Ai}(x) \sim \frac{x^{-1/4}}{2\sqrt{\pi}} e^{\pm \frac{2}{3}x^{3/2}}, \quad x \to \infty. \tag{21}$$

To understand how meaningful the solution we obtained is, let us compare the graph of $\operatorname{Ai}(x)$ with the graph of our asymptotic solution in Figure 7.

A striking result emerges here. For the asymptotic solution, we had assumed that $x$ was very large. We even wrote $x \to \infty$ to indicate this largeness. The result shows that $x=5$ is a number sufficiently close to infinity. In other words, asymptotic solutions approximate the function far more powerfully than we might expect. Before wrapping up the discussion, let us examine in Figure 8 what happens when we take $x \to -x$ in our solution—that is, when we include the negative axis—and compare the asymptotic solution with the Airy function.

Thus, in this lesson, we have solved together Airy's supernumerary-band problem, which is regarded as the starting point of resurgence theory. But is the Airy differential equation used only to explain rainbows and the formation of supernumerary bands? The answer is no!

The Airy differential equation also appears in contemporary physics research. Let us conclude our lesson with an example.

9. The Airy Differential Equation and Quantum Mechanics

All students who have taken modern physics or quantum mechanics know the Schrödinger equation. For readers who have not taken these courses or have not yet encountered the Schrödinger equation, let us write it down and briefly explain it:

$$\left[ -\frac{\hbar^{2}}{2m}\,\frac{d^{2}}{dx^{2}} + V(x) \right]\psi = E\psi. \tag{22}$$

The expression in square brackets is called the Hamiltonian and is denoted by $\mathcal{H}$. It should be noted that physics students generally first encounter the concept of the Hamiltonian in a course on theoretical mechanics, and the Hamiltonian is not always equal to the total energy. In quantum mechanics, the Hamiltonian is the generator of time translations and the observable representing the total energy of the system.

Consider the problem of a particle in a finite potential well (Figure 9). Think briefly about the following question, which you can answer using your basic knowledge of physics: Where does the particle spend most of its time in such a system? The answer is that it spends more time in regions where its kinetic energy is smallest.

When, then, does the kinetic energy attain its minimum value? Near the position where the potential energy reaches its maximum. Let us therefore, under the condition $V(x_0)=E$, Taylor-expand the potential energy around $x_0$:

$$V(x) \approx E+V'(x_0)(x-x_0), \quad V'(x_0)\neq 0. \tag{23}$$

Substituting the first term of the Taylor series into the equation gives $E\psi$. Since the same term appears on the right-hand side, the equation takes a simpler form:

$$\left[ -\frac{\hbar^{2}}{2m}\,\frac{d^{2}}{dx^{2}} + V'(x_0)(x-x_0) \right]\psi = 0. \tag{24}$$

If we normalize the leading coefficient of the differential equation to one, it becomes

$$\frac{d^{2}\psi}{dx^{2}} - \frac{2m}{\hbar^{2}}V'(x_0)(x-x_0)\psi(x) = 0. \tag{25}$$

To simplify the equation, let us make a scaling-type change of variables, $z := c(x-x_0)$. We will determine $c$ from the equation itself. Applying the transformation gives

$$\frac{d^{2}\psi}{dz^{2}} - \frac{2m}{\hbar^{2}c^{3}} V'(x_0)\,z\,\psi(z) = 0. \tag{26}$$

At this stage, it is clear that the equation simplifies if we choose

$$c^{3} = \frac{2mV'(x_0)}{\hbar^{2}}$$

With this choice, the Schrödinger equation becomes the following differential equation:

$$\frac{d^{2}\psi}{dz^{2}} = z\,\psi. \tag{27}$$

As is immediately apparent, (2) and (27) are the same differential equation. Our analysis of the turning point and minimum kinetic energy for a particle in a potential well is known as the JWKB¹ (Jeffrey–Wentzel–Kramers–Brillouin) approximation.

10. What Is Coming in the Next Installments?

In the next installment, we will examine concepts from complex analysis such as singularities, analytic continuation, Riemann surfaces, and the calculus of residues. Where appropriate, we will reinforce these concepts with applications in physics.

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References

[1] A. B. O'Donnell, The Work of G. G. Stokes in Evaluating the Airy Rainbow Integral and Its Ramifications Today, Ph.D. thesis, Dublin Institute of Technology (1990).

[2] G. H. Hardy, Divergent Series (Clarendon Press, Oxford, 1949).

[3] G. G. Stokes, On the Discontinuity of Arbitrary Constants Which Appear in Divergent Developments, Trans. Cambridge Philos. Soc. 10, 105–128 (1857).

Footnotes

1. Most textbooks do not mention Jeffrey, and the approximation is also known as WKB. Since Jeffrey carried out work on the subject earlier, the designation JWKB is more accurate. ↩