WKB in dimension 1. 2


### Global theory

#### Global construction

We considered equation $$-ih^{-1}u_t + H(x,hD_x,h) u=0. \label{eq-6.2.1}$$ We constructed solution $$u_h(x,t)= e^{ih^{-1}S(x,t)} A(x,t,h) \label{eq-6.2.2}$$ with $$A(x,t,h)\sim \sum_{n\ge 0} A_n (x,t)h^n \label{eq-6.2.3}$$ before it hits points where $S(x,t)$ is no more a smooth function of $x$. However smooth Lagrangian manifold $\Lambda_t$ is constructed globally and $S(z,t)$ is a smooth function of $z\in \Lambda_t$.

Near points where $\pi_x:\Lambda_t\ni (x,p)\to x$ is no more local diffeomorphism we make $h$-Fourier transform ( $$v_h(p):= (F u)(p)= (2\pi h)^{-\frac{1}{2}} \int e^{-ih^{-1} p\cdot x } u_h(x)\,dx \label{eq-6.2.4}$$ and then according to Theorem 6.1.1 $$(Fu)(p,t) = e^{-ih^{-1}\tilde{S}(p,t)} \tilde{A}(p,h) \label{eq-6.2.5}$$ where $\tilde{S}(p)$ is a Legendre transform of $S(x)$: $$\tilde{S}(p)= p\cdot x(p) - S(x(p)) \label{eq-6.2.6}$$ where $x(p)$ is defined from $S_x (x)=p$, and $\tilde{A}(p,h)\sim \sum_n \tilde{A}_n(p) h^n$ with $$\tilde{A}_0(p)= \frac{1}{\sqrt{|S_{xx}|}} e^{-\frac{i\pi}{4}\sgn(S_{xx})} A(x(p)) \label{eq-6.2.7}$$ where $\sgn(S_{xx})$ is a sign of $S_{xx}$.

Further, it solves (\ref{6.2.1}) in $p$-representation $$-ih^{-1}v_t + H(-hD_p,p,h) v=0 \label{eq-6.2.8}$$ and therefore $\tilde{S}$ satisfies corresponding Hamilton-Jacobi equation $$-\tilde{S}_t + H_0(\tilde{S}_p,p)=0 \label{eq-6.2.9}$$ and $\tilde{A}_n$ satisfy corresponding transport equations and all those equations work as long as projection $\pi_p:\Lambda_t\ni (x,p)\to p$ is remains local diffeomorphism.

Furthermore, after $\pi_x:\Lambda_t\ni (x,p)\to x$ is again local diffeomorphism we can make inverse transform $$u_h (x,t) = F^{-1}v = (2\pi h)^{-\frac{1}{2}} \ \int e^{ih^{-1} p\cdot x } v_h(p,t)\,dp \label{eq-6.2.10}$$ and again get solution (\ref{eq-6.2.2}) with $S(z,t)$ already constructed globally and $\tilde{S}(z,t)$ too.

Then we can continue until $\pi_x$ is no more local diffeomorphism and so on.

#### Global construction: amplitude

What about amplitudes? We are looking mainly for a leading term $A_0(x,t)$. First, it has singularities as $\pi_x$ is no longer a local diffeomorphism. These singularities of $A_0(x,t)$ and of $\tilde{A}_0(p,t)$ could be "tamed" if we consider $$a_0(z,t)= A_0(z,t)|\frac{dx}{dz}|^{\frac{1}{2}} \label{eq-6.2.11}$$ and $$\tilde{a_0}(z,t)= \tilde{A}_0(z,t)|\frac{dx}{dz}|^{\frac{1}{2}} \label{eq-6.2.12}$$ where $dz$ is a measure on $\Lambda_t$ which is invariant with respect to Hamiltonian flow (so we take original $dz=dx$ as $t=0$ and push it forward).

Definition 1. We say that $a_0(z,.)$ is half-density because $|a_0(z,.)|^2$ is a density i.e. $|a_0(z,.)|^2\,dz$ does not change as we change $dz$.

Further, since $S_{xx}=\frac{dp}{dx}$ and $\tilde{S}_{pp}=\frac{dx}{dp}$ at points where both $\pi_x$ and $\pi_p$ are local diffeomorphisms, $a_0(x,t)$ and $\tilde{a}_0(p(x),t)$ almost coincide. What is the difference? Factor $$e^{-\frac{i\pi}{4}\sgn(\frac{dp}{dx})} \label{eq-6.2.13}$$ means that

1. $a_0$ acquires factor $1=e^{-\frac{i}{2}\eta(z^*)}$ with $\eta(z^*)=0$ if before point $z^*$ where $\frac{dx}{dp}=0$ and after it $\frac{dx}{dp}$ has the same sign.
2. $a_0$ acquires factor $i=e^{-\frac{i}{2}\eta(z^*)}$ with $\eta(z^*)=-1$ if before point point $z^*$ where $\frac{dx}{dp}=0$ $\frac{dx}{dp}<0$ and after it $\frac{dx}{dp}>0$.
3. $a_0$ acquires factor $i=e^{-\frac{i}{2}\eta(z^*)}$ with $\eta(z^*)=1$ if before point point $z^*$ where $\frac{dx}{dp}=0$ $\frac{dx}{dp}>0$ and after it $\frac{dx}{dp}<0$.

Therefore we arrive to

Definition 2. Consider a path $\gamma$ in which there are several points $z^*_k$, $k=1,\ldots, N$ in which $\frac{dx}{dy}=0$ and $\frac{dx}{dy}\ne 0$ in all other points. Then $\iota_M(\gamma)$ is called Maslov index of $\gamma$.

Remark 1.

1. We can define Maslov index without $t$, just on a single manifld $\Lambda$.
2. We are interested in Maslov index modulo $4$ since $i=e^{-\frac{i}{2}n}=1$ as $n\equiv 0 \mod 4$.
3. We are especially interested in Maslov index of the closed path $\gamma$. In this case Maslov index does not depend on the choice of the start point (which is also end point) of $\gamma$ but depends on orientation. However Maslov index $\mod 4$ does not depend on orientation.
4. For closed path Maslov index $\mod 4$ does not change if we permute $x$ and $p$.

Example 1. Let $\Lambda= \{(x,p): x^2+p^2=1\}$ and $\gamma$ is a single path around. Maslov index of $\gamma$ is $2\mod 4$.

#### Simple caustic points

Consider caustic point $\bar{z}$ $$\tilde{S}_{pp}(\bar{p},\bar{t})=0 \label{eq-6.2.14}$$ and assume that it is simple i.e. $$\tilde{S}_{ppp}(\bar{p},\bar{t})\ne 0. \label{eq-6.2.15}$$ We are interested in asymptotics of $$u_h(x,t)= (2\pi h)^{-\frac{1}{2}} \int e^{ih^{-1}(px -\tilde{S}(p,t)}\tilde{A}(p, t)\,dp \label{eq-6.2.16}$$ near such point. One can prove that under assumptions (\ref{eq-6.2.14}) and (\ref{eq-6.2.15}) one can make a change of variable $p$ such that \begin{equation*} u_h(x,t)= (2\pi h)^{-\frac{1}{2}} \int e^{ih^{-1}(\frac{1}{3}\pm{q^3}-\alpha (x,t)q +\beta (x,t))}B(q, x,t)\,dq \end{equation*} which after scaling becomes \begin{multline} u_h(x,t)= (2\pi )^{-\frac{1}{2}}h^{-\frac{1}{6}} e^{ih^{-1}\beta (x,t)} \int e^{i(\frac{1}{3}q^3-h^{-\frac{2}{3}}\alpha (x,t)q)} B(qh^{\frac{1}{3}}, x,t)\,dq\sim\\ e^{ih^{-1}\beta (x,t)} \operatorname{Ai} (-h^{-\frac{2}{3}}\alpha(x,t)) c(x,t) \label{eq-6.2.17} \end{multline} with $\alpha_x \ne 0$. Here $\operatorname{Ai}$ is Airy function and as $|\alpha (x,t)|\gg h^{\frac{2}{3}}$, $\alpha <0$ we have $u_h \sim 0$ and $|\alpha (x,t)|\gg h^{\frac{2}{3}}$, $\alpha >0$ we have a corresponding asymptotic formula via exponents $e^{ih^{-1}S^\pm (x,t)}b_\pm (x,t,h)$ with $S^\pm (x,t)=\beta (x,t)\pm \frac{2}{3}\alpha(x,t)^{\frac{3}{2}}$.

Example 2. Consider stationary solutions of the Schrödinger equation. Then $$\frac{1}{2}S_x^2 = E-V(x),\qquad V(x_E)=E,\ \ V'(x_E)> 0. \label{eq-6.2.18}$$ Then $(\beta \pm \frac{2}{3}\alpha^{\frac{3}{2}})_x)^2=2(E-V(x))\iff \beta_x^2 \pm (\frac{2}{3}\alpha^{\frac{3}{2}})_x)^2= 2(E-V(x))$ and then either $\beta_x =0$ or $\alpha_x=0$. The former is impossible, the latter means $\frac{2}{3}\alpha^{\frac{3}{2}}=-\int_x^{x_E} 2(E-V(x))^{\frac{1}{2}}\,dx$ and $$\alpha = \bigl(\int_x^{x_E} 3(E-V(x))^{\frac{1}{2}}\,dx\bigr)^{\frac{2}{3}},\qquad \beta=0. \label{eq-6.2.19}$$