Laguerre polynomials and transitional asymptotics of the modified Korteweg–de Vries equation for step-like initial data

Abstract

We consider the compressive wave for the modified Korteweg–de Vries equation with background constants \(c>0\) for \(x\rightarrow -\infty \) and 0 for \(x\rightarrow +\infty \). We study the asymptotics of solutions in the transition zone \(4c^2t-\varepsilon t<x<4c^2t-\beta t^{\sigma }\ln t\) for \(\varepsilon >0,\)\(\sigma \in (0,1),\)\(\beta >0.\) In this region we have a bulk of nonvanishing oscillations, the number of which grows as \(\frac{\varepsilon t}{\ln t}.\) Also we show how to obtain Khruslov–Kotlyarov’s asymptotics in the domain \(4c^2t-\rho \ln t<x<4c^2t\) with the help of parametrices constructed out of Laguerre polynomials in the corresponding Riemann–Hilbert problem.

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Appendix: A Asymptotics of modified Laguerre polynomials

Appendix: A Asymptotics of modified Laguerre polynomials

In this section we prove Lemma 5.1 and hence Corollary 5.1. Large \(n\rightarrow \infty \) asymptotics of the RHP as in Corollary 5.1 with \(\Lambda = n\) was studied in [57], but that treatment is not suitable for our present purposes.

In order to study large \(n\rightarrow \infty \) asymptotics, first we introduce a new matrix-valued function

$$\begin{aligned} W(\zeta )=\mathrm {e}^{n\ell \sigma _3}Y(\zeta )\mathrm {e}^{n(g(\zeta )-\ell )\sigma _3}, \end{aligned}$$

where \(g(\zeta )\) is the function analytic in \({\mathbb {C}}{\setminus }(-\infty ,a]\) defined by:

$$\begin{aligned} g(\zeta )= & {} \frac{2}{\pi a}\int \limits _{0}^a\ln (\zeta -s)\sqrt{\frac{a-s}{s}}\mathrm {d}s=\frac{-2}{a}\sqrt{\zeta (\zeta -a)} - \ln a\nonumber \\&+\ln \left( 2\zeta -a+2\sqrt{\zeta (\zeta -a)}\right) +\frac{2\zeta }{a}+\ell , \end{aligned}$$

A simple computation shows that \(g(\zeta )\) has the asymptotic expansion

$$\begin{aligned} g(\zeta )=\ln \zeta -\frac{a}{4\zeta }-\frac{a^2}{16\zeta ^2}-\frac{5a^3}{192\zeta ^3}+\cdots =\ln \zeta -\sum \limits _{j=1}^{\infty }\frac{2\ (2n-1)!}{(n+1)\ n!^2}\frac{a^n}{4^n \zeta ^n}. \end{aligned}$$
(114)

The constant \(a, \ell \) are given by

$$\begin{aligned} a=\frac{4n}{\Lambda },\quad \ell :=\ln \frac{a}{4\mathrm {e}}. \end{aligned}$$

At \(\zeta =\infty \) we have \(W(\zeta )=\left( I+{\mathcal {O}}(\zeta ^{-1})\right) , \) and W satisfies the boundary value condition

$$\begin{aligned}&W_+=W_-\begin{pmatrix}\mathrm {e}^{n(g_+-g_-)}&{}0\\ \sqrt{\zeta }\ \mathrm {e}^{n(g_-+g_+-2l-\frac{\Lambda \zeta }{n})}&{}\mathrm {e}^{-n(g_+-g_-)}\end{pmatrix},\zeta \in (0,+\infty ),\\&\qquad \qquad W_+=W_-\mathrm {e}^{n(g_+-g_-)\sigma _3},\zeta \in (-\infty ,0). \end{aligned}$$

Further, we introduce the “effective” potential

$$\begin{aligned} \varphi (\zeta ):=-\frac{\Lambda }{2n}\zeta +g(\zeta )-\ell , \end{aligned}$$

so that the jump condition can be written in the following form:

$$\begin{aligned} W_+=W_-\begin{pmatrix}\mathrm {e}^{n(\varphi _+-\varphi _-)}&{}0\\ \sqrt{\zeta }\ \mathrm {e}^{n(\varphi _-+\varphi _+)}&{}\mathrm {e}^{-n(\varphi _+-\varphi _-)}\end{pmatrix}. \end{aligned}$$

It is convenient now to scale the interval (0, a) into (0, 1),  i.e.

$$\begin{aligned} W^{(1)}\left( \lambda \right) :=W\left( \zeta \right) ,\qquad \lambda :=\frac{\zeta }{a}=\frac{\zeta \ \Lambda }{4n}. \end{aligned}$$

Denote

$$\begin{aligned}&\psi \left( \lambda \right) := \varphi (\zeta )=-2\int \limits _{1}^{\lambda }\sqrt{\frac{s-1}{s}}\ \mathrm {d}s=-2\sqrt{\lambda (\lambda -1)}+\ln \left( -1+2\lambda +2\sqrt{\lambda (\lambda -1)}\right) \\&\quad =-2\lambda +\ln \lambda +\ln 4e-h(\lambda )=-2\lambda +\ln \lambda +\ln 4e+{\mathcal {O}}(\lambda ^{-1}), \end{aligned}$$

where h(.) is as in (92). The scaled effective potential \(\psi (\lambda ),\) which is analytic in \(\lambda \in {\mathbb {C}}{\setminus }(-\infty ,1],\) has the following properties on the interval \((-\infty ,1]\):

$$\begin{aligned} \psi _+-\psi _-=2\pi \mathrm {i},\quad \lambda \in (-\infty ,0),\qquad \qquad \psi _-+\psi _+ = 0,\ \ \ \lambda \in (0,1). \end{aligned}$$
(115)

Due to the above properties (115), function

$$\begin{aligned} W^{(2)}(\lambda ) = a^{\frac{\sigma _3}{4}}W^{(1)}(\lambda )a^{\frac{-\sigma _3}{4}} \end{aligned}$$

does not have jump along \(\lambda \in (-\infty ,0),\) and solves a RHP of the form

$$\begin{aligned}&W^{(2)}_+=W^{(2)}_-\begin{pmatrix}\mathrm {e}^{n(\psi _+-\psi _-)} &{} 0 \\ \sqrt{\lambda } &{} \mathrm {e}^{-n(\psi _+-\psi _-)}\end{pmatrix},\quad \lambda \in (0,1),\\&W^{(2)}_+=W^{(2)}_-\begin{pmatrix}1 &{} 0 \\ \sqrt{\lambda }\ \mathrm {e}^{2n\psi } &{} 1\end{pmatrix},\quad \lambda \in (1,\infty ), \\&W^{(2)}(\lambda )=I+\mathrm {O}\left( \frac{1}{\lambda }\right) . \end{aligned}$$

Tracking back the relation between \(W^{(2)}(\lambda )\) and \(W^{(1)}(\lambda ),\)\(W(\zeta )\), \(Y(\zeta ),\)\(L(\zeta ),\) we see that

$$\begin{aligned} W^{(2)}(\lambda ) = \left( \frac{\sqrt{2}\ n^{n+\frac{1}{4}}}{\mathrm {e}^n}\right) ^{\sigma _3}L(4n\lambda )(4n\lambda )^{n\sigma _3}\mathrm {e}^{-nh(\lambda )\sigma _3}\left( \frac{\mathrm {e}^n}{\sqrt{2}\ n^{n+\frac{1}{4}}}\right) ^{\sigma _3}, \end{aligned}$$

Important role now is played by the signature table of \({\text {Re}}\psi .\) It is shown in the Fig. 6.

Fig. 6
figure6

Distribution table of signs for \(\mathrm {Re}\ \psi (\lambda )\)

Now we introduce the \(\delta -\) transformation, which removes the term \(\sqrt{\lambda }\) from the jump matrix. It must solve the following scalar conjugation problem:

$$\begin{aligned} \delta _+\delta _-=\sqrt{\lambda },\ \lambda \in (0,1). \end{aligned}$$

The solution is given by the formula

$$\begin{aligned}&\delta (\lambda )=\exp \left\{ \frac{R(\lambda )}{2\pi \mathrm {i}}\int \limits _{0}^{1}\frac{\ln \sqrt{s}\ \mathrm {d}s}{(s-\zeta )\ R_+(s)}\right\} ,\quad R(\lambda )=\sqrt{\lambda (\lambda -1)},\\&\delta (\lambda )=\left( \frac{\lambda }{2\lambda -1+2\sqrt{\lambda (\lambda -1)}}\right) ^{1/4},\quad \delta (\infty ) = \frac{1}{\sqrt{2}}. \end{aligned}$$

With a series of transformations, we now show how to remove the oscillating jump from the interval (0, 1). First of them is

$$\begin{aligned} W^{(3)}(\lambda ):=\delta ^{\sigma _3}(\infty )W^{(2)}(\lambda )\delta ^{-\sigma _3}(\lambda ). \end{aligned}$$

The jump matrix \(W^{(3)}_+=W^{(3)}_+J^{(3)}\) is

$$\begin{aligned} J^{(3)}=\begin{pmatrix}\frac{\delta _-}{\delta _+}\mathrm {e}^{n(\psi _+-\psi _-)}&{}0\\ \mathrm {e}^{n(\psi _-+\psi _+)}&{}\frac{\delta _+}{\delta _-}\mathrm {e}^{-n(\psi _+-\psi _-)}\end{pmatrix},\ \lambda \in (0,1), \qquad J^{(3)}=\begin{pmatrix}1&{}0\\ \frac{\sqrt{\lambda }}{\delta ^2}\ \mathrm {e}^{2n\psi }&{}1\end{pmatrix},\ \lambda \in (1,+\infty ). \end{aligned}$$

Now we need to factorize the jump matrix on the interval [0, 1] according to the signature table, namely

$$\begin{aligned} \begin{pmatrix}\frac{\delta _-}{\delta _+}\mathrm {e}^{n(\psi _+-\psi _-)}&{}0\\ \mathrm {e}^{n(\psi _-+\psi _+)}&{}\frac{\delta _+}{\delta _-}\mathrm {e}^{-n(\psi _+-\psi _-)}\end{pmatrix} = \begin{pmatrix} 1 &{} \frac{\delta _-^2\ \mathrm {e}^{-2n\psi _-}}{\sqrt{\lambda }} \\ 0 &{} 1 \end{pmatrix} \begin{pmatrix} 0 &{} -\mathrm {e}^{-n(\psi _-+\psi _+)} \\ \mathrm {e}^{n(\psi _-+\psi _+)} &{} 0 \end{pmatrix} \begin{pmatrix} 1 &{} \frac{\delta _+^2\ \mathrm {e}^{-2n\psi _+}}{\sqrt{\lambda }} \\ 0 &{} 1 \end{pmatrix}. \end{aligned}$$

The next step is to “open lenses”, we transform the RH problem, “moving” the left and the right factor in the above formula to the lower and upper half-planes, respectively. For a new RH problem \(W^{(4)}\) the jumps \(W^{(4)}_+=W^{(4)}_-J^{(4)}\) are

$$\begin{aligned} J^{(4)} = {\left\{ \begin{array}{ll}\begin{pmatrix} 1 &{} \frac{\delta ^2\ \mathrm {e}^{-2n\psi }}{\sqrt{\lambda }} \\ 0 &{} 1 \end{pmatrix},\quad \lambda \in l_u (\text {the upper lense}), \quad \begin{pmatrix} 1 &{} \frac{\delta ^2\ \mathrm {e}^{-2n\psi }}{\sqrt{\lambda }} \\ 0 &{} 1 \end{pmatrix},\quad \lambda \in l_d (\text {the lower lense}), \\ \begin{pmatrix} 0 &{} -1 \\ 1 &{} 0 \end{pmatrix},\quad \lambda \in (0,1), \quad \begin{pmatrix}1&{}0\\ \frac{\sqrt{\lambda }}{\delta ^2}\ \mathrm {e}^{2n\psi }&{}1\end{pmatrix},\quad \lambda \in (1,+\infty ). \end{array}\right. } \end{aligned}$$

The RHP for \(W^{(4)}\) gives us the first statement of Lemma 5.1 concerning function

$$\begin{aligned} {\mathcal {Q}}_n(\lambda ):=W^{(4)}(\lambda )\left( \delta (\lambda )\delta ^{-1}(\infty )\right) ^{\sigma _3}. \end{aligned}$$

Indeed, the fact that \({\mathcal {Q}}_n(\lambda )\) decomposees into an asymptotic series follows, for example, from an explicit representation (62) of function L(.). Then RHP for \(W^{(4)}\) gives us uniformity with respect to \(n\ge 1\) of all the ingredients in this asymptotic series. Further, since for large \(n\rightarrow \infty \) the jump matrix is concentrated mainly on the interval (0, 1),  the solution to the RH problem for \(W^{(4)}\) is close everywhere except for a neighborhood of the points 0,  1 to the solution

$$\begin{aligned}&M_{mod}=\begin{pmatrix}a(\lambda )&{}\mathrm {i}b(\lambda )\\ i b(\lambda )&{}a(\lambda )\end{pmatrix},\quad a(\lambda )=\frac{1}{2}\left( \gamma (\lambda )+\gamma ^{-1}(\lambda )\right) ,\ b(\lambda )\\&\quad =\frac{1}{2}\left( \gamma (\lambda )-\gamma ^{-1}(\lambda )\right) ,\ \gamma (\lambda )=\root 4 \of {\frac{\lambda -1}{\lambda }} \end{aligned}$$

to the model problem with jumps only on (0, 1). This gives us the second statement of Lemma 5.1 concerning function

$$\begin{aligned} {\mathcal {E}}_n(\lambda ) := W^{(4)}(\lambda )M_{mod}^{-1}(\lambda ). \end{aligned}$$

However, rigorous proof of the fact that ingredients in asymptotic series for \({\mathcal {E}}_n\) tend to 0 as \(n\rightarrow \infty \) involves construction of local parametrices around the points 0, 1,  since jump matrix for \(W^{(4)}\) is not uniformly close to I in vicinities of these points. The details can be found for example in [57].

A.1 Airy Parametrix

For the benefit of the reader we recall the form of the so–called “Airy” parametrix, which is used to provide local solutions of a large class of Riemann Hilbert problems [23]. This is defined by

$$\begin{aligned} \Psi _{Ai}(\zeta ) = \left\{ \begin{array}{l} \begin{pmatrix}v_1&{}v_0\\ v'_1&{}v'_0\end{pmatrix},\arg \zeta \in (0,2\pi /3), \quad \begin{pmatrix}v_{-1}&{}v_0\\ v'_{-1}&{}v'_0\end{pmatrix},\arg \zeta \in (0,-2\pi /3), \\ \begin{pmatrix}v_1&{}-\mathrm {i}v_{-1}\\ v'_1&{}-\mathrm {i}v'_{-1}\end{pmatrix},\arg \zeta \in (2\pi /3,\pi ), \quad \begin{pmatrix}v_{-1}&{}\mathrm {i}v_{1}\\ v'_{-1}&{}\mathrm {i}v'_{1}\end{pmatrix},\arg \zeta \in (-\pi ,-2\pi /3), \end{array}\right\} \times \mathrm {e}^{\pi \mathrm {i}\sigma _3/4}. \end{aligned}$$
(116)

where we have used the notation

$$\begin{aligned} v_0(\zeta )=\sqrt{2\pi }Ai(\zeta ),\quad v_1(\zeta )=\sqrt{2\pi }\mathrm {e}^{-\pi \mathrm {i}/6}Ai(\zeta \mathrm {e}^{-2\pi \mathrm {i}/3}),\quad v_{-1}(\zeta )=\sqrt{2\pi }\mathrm {e}^{\pi \mathrm {i}/6}Ai(\zeta \mathrm {e}^{2\pi \mathrm {i}/3}). \end{aligned}$$
(117)

We have

$$\begin{aligned} v_0-\mathrm {i}v_1+\mathrm {i}v_{-1}\equiv 0. \end{aligned}$$

This function \(\Psi _{Ai}(\zeta )\) has the following jumps \(\Psi _{Ai,+}=\Psi _{Ai,-}J_{\Psi _{Ai}}\):

$$\begin{aligned} J_{\Psi _{Ai}}=\begin{pmatrix}1&{}0\\ 1&{}1\end{pmatrix},\arg \zeta =0,\ \begin{pmatrix}1&{}1\\ 0&{}1\end{pmatrix},\arg \zeta =\frac{2\pi }{3}, \ \begin{pmatrix}1&{}1\\ 0&{}1\end{pmatrix},\arg \zeta =\frac{-2\pi }{3},\ \begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix},\arg \zeta =\pi . \end{aligned}$$

and the following behaviour at infinity:

$$\begin{aligned} \Psi _{Ai}(\zeta )=\zeta ^{-\sigma _3/4}\frac{1}{\sqrt{2}}\begin{pmatrix}1&{}1\\ 1&{}-1\end{pmatrix}\left( I+ \begin{pmatrix} \frac{-1}{48}&{}\frac{-1}{8}\\ \frac{1}{8} &{} \frac{1}{48} \end{pmatrix}\zeta ^{-3/2}+{\mathcal {O}}(\zeta ^{-3})\right) \mathrm {e}^{\frac{2}{3}\zeta ^{3/2}\sigma _3}\mathrm {e}^{\pi \mathrm {i}\sigma _3/4}. \end{aligned}$$

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Bertola, M., Minakov, A. Laguerre polynomials and transitional asymptotics of the modified Korteweg–de Vries equation for step-like initial data. Anal.Math.Phys. 9, 1761–1818 (2019). https://doi.org/10.1007/s13324-018-0273-1

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