asymptotic expansion from 3 leading terms

To complement Dr. MV's answer I'd like to prove that the differential equation in $f$,

$$ 0 = f''(x)-2xf'(x)-2(1+\beta)f(x), \tag{1} $$

has a solution with an asymptotic expansion of the form

$$ f(x)\sim \sum_{n=0}^\infty a_nx^{-(n+r)} \qquad \text{as } x \to \infty \tag{2} $$

for some $r \in \mathbb R$. I'll use the same method that is used to calculate the asymptotics of the Airy functions.

To begin, we will search for a solution of $(1)$ of the form

$$ f(x) = \int_\gamma g(z) e^{xz}\,dz, \tag{3} $$

where $\gamma$ is a (possibly infinite) contour in the complex plane which will be determined.

We calculate

$$ f'(x) = \int_\gamma z g(z) e^{xz}\,dz \qquad \text{and} \qquad f''(x) = \int_\gamma z^2 g(z) e^{xz}\,dz, $$

and integrating by parts yields

$$ xf'(x) = \int_\gamma [zg(z)]\cdot[xe^{xz}]\,dz = \Bigl[zg(z) e^{xz} \Bigr]_{z=a}^{z=b} - \int_\gamma [g(z) + zg'(z)] e^{xz}\,dz, $$

where $a$ and $b$ are the (possibly infinite) endpoints of $\gamma$. Substituting these into $(1)$ we get

$$ 0 = \int_\gamma \Bigl[ 2zg'(z) + (z^2-2\beta)g(z) \Bigr] e^{xz}\,dz - \Bigl[zg(z) e^{xz} \Bigr]_{z=a}^{z=b}. \tag{4} $$

The integral vanishes if

$$ 0 = 2zg'(z) + (z^2-2\beta)g(z), $$

and one solution to this equation is

$$ g(z) = z^\beta e^{-z^2/4}, \tag{5} $$

where the branch cut for $z^\beta$ is taken to be the negative real axis: $\{x : x<0\}$.

With this $g$ the boundary term in $(3)$, namely

$$ \Bigl[zg(z) e^{xz} \Bigr]_{z=a}^{z=b}, $$

vanishes if $\gamma$ begins and ends at $\infty$ in the quadrants defined by $\operatorname{Re} z^2 > 0$, as shown in the image below:

Taking into account the branch cut of $g$ along $\{x : x < 0\}$ we observe that there are exactly three possible contours (up to deformation and changes in orientation):

$$ \begin{align} &\gamma_1 : -\infty-i\epsilon \longrightarrow (0+) \longrightarrow -\infty + i\epsilon, \\ &\gamma_2 : -\infty+i\epsilon \longrightarrow \infty, \\ &\gamma_3 : -\infty-i\epsilon \longrightarrow \infty, \end{align} $$

where $(0+)$ means that the contour passes around the point $z=0$ with positive orientation. These three contours are shown in the figure below, and the branch cut of $g$ is shown as a black jagged line:

Since the differential equation $(1)$ is of second order we expect that the solutions $f$ given by one of these contours must be dependent on the other two, and indeed we have

$$ \int_{\gamma_3} = \int_{\gamma_1} + \int_{\gamma_2}. $$

So we can throw the contour $\gamma_3$ out, and we are finally left with two linearly independent solutions of $(1)$ which we call $f = F_1$ and $f = F_2$, given by

$$ \begin{align} F_1(x) &= \int_{\gamma_1} z^\beta \exp(xz - z^2/4)\,dz, \\ F_2(x) &= \int_{\gamma_2} z^\beta \exp(xz - z^2/4)\,dz. \end{align} \tag{6} $$

I'll omit the analysis of $F_2(x)$ as $x \to \infty$, but suffice it to say that a relatively straightforward application of the saddle point method yields the asymptotic expansion

$$ F_2(x) \sim (2x)^\beta \exp(x^2) \sum_{n=0}^{\infty} \binom{\beta}{2n} \Gamma\!\left(n + \frac{1}{2}\right) x^{-2n} \qquad \text{as } x \to \infty. $$ To leading order, $$ F_2(x) \sim \sqrt{\pi} (2x)^\beta e^{x^2} \qquad \text{as } x \to \infty. $$

Since $F_2(\infty) \neq 0$ this is not the solution we're looking for. Let's take a closer look at $F_1$.

Assuming $x > 0$ we calculate

$$ \begin{align} F_1(x) &= \int_{\gamma_1} z^\beta \exp\!\left(xz - z^2/4\right) dz \\ &= e^{-i\pi \beta} \int_{-\infty}^0 (-z)^\beta \exp\!\left(xz - z^2/4\right) dz - e^{i\pi \beta} \int_{-\infty}^0 (-z)^\beta \exp\!\left(xz - z^2/4\right) dz \\ &= -2i\sin(\beta \pi) \int_0^\infty t^\beta \exp\!\left(-xt - t^2/4\right) dt \\ &= -2i\sin(\beta \pi) x^{\beta+1} \int_0^\infty \exp\!\left[-x^2 \left(u+u^2/4\right)\right] du \\ &= -i\sin(\beta \pi) (2x)^{\beta+1} \int_0^\infty \frac{\left(\sqrt{1+v} - 1\right)^\beta}{\sqrt{1+v}} \exp\!\left(-x^2v\right) dv, \end{align} $$

where we have made the successive substitutions $z = -t$, $t=xu$, and $u+u^2/4 = v$. The non-exponential factor in the integrand can be represented by a convergent power series,

$$ \frac{\left(\sqrt{1+v} - 1\right)^\beta}{\sqrt{1+v}} = v^\beta \sum_{n=0}^{\infty} a_n v^n, \qquad |v| < 1, $$

so an application of Watson's lemma tells us that $F_1(x)$ has an asymptotic expansion of the form

$$ F_1(x) \sim x^{-\beta+1} \sum_{n=1}^{\infty} b_n x^{-2n} \qquad \text{as } x \to \infty $$ for some coefficients $b_n$. To leading order, $$ F_1(x) \sim B x^{-\beta-1} \qquad \text{as } x \to \infty $$ for some constant $B$.

We have therefore shown that the differential equation $(1)$ has a solution with an asymptotic expansion of the form $(2)$, and in principle we could calculate the coefficients from this integral representation of the solution.

We start with the ODE

$$w''(x)+2xw'(x)-2\beta w(x)=0 \tag 1$$

and make the substitution $w(x)=e^{-x^2}f(x)$. Then, the ODE given in $(1)$ becomes

$$f''(x)-2xf'(x)-2(1+\beta)f(x)=0 \tag 2$$

Now, using the Method of Frobenius, we expand $f$ is the series

$$f(x)\sim \sum_{n=0}^\infty a_nx^{-(n+r)} \tag 3$$

where we assume that $a_0\ne 0$. Substituting $(3)$ into $(2)$ and equating coefficients on powers of $x$ to zero reveals that for $x^{-r}$ and $x^{-r-1}$

$$[2r-2(1+\beta)]a_0=0 \tag 4$$

$$[2(1+r)-2(1+\beta)]a_1=0 \tag 5$$

and that for $x^{-r-n}$, $n\ge 2$,

$$(n+r-2)(n+r-1)a_{n-2}+2[(n+r)-(1+\beta)]a_{n} \tag 6$$

Since $a_0\ne 0$, we find from $(4)$ that $r=1+\beta$. Using this value for $r$ in $(5)$ shows that $a_1=0$. And using this value of $r$ in $(6)$ yields the recurrence relationship

$$a_{n}=\frac{(n-1+\beta)(n+\beta)}{2n}a_{n-2} \tag 7$$

From $(7)$, we find that $a_2=\frac{(1+\beta)(2+\beta)}{4}$ and $a_3=0$. Therefore, we have

$$f(x)\sim a_0x^{-\beta}\left(x^{-1}+\frac{(1+\beta)(2+\beta)}{4}x^{-3}\right)$$

Finally, we have for

$$\bbox[5px,border:2px solid #C0A000]{w(x)\sim a_0x^{-\beta}\left(x^{-1}+\frac{(1+\beta)(2+\beta)}{4}x^{-3}\right)e^{-x^2}}$$

which is the third-order asymptotic expansion for $w$.