Riemann-Hilbert approach for the nonlocal modified Korteweg-de Vries equation with a step-like oscillating background
Yan Rybalko
TL;DR
This work addresses the Cauchy problem for the nonlocal mKdV equation $u_t+6u(x,t)u(-x,-t)u_x+u_{xxx}=0$ with oscillating step-like boundaries, and develops a rigorous Riemann-Hilbert formulation via the inverse scattering transform. The authors carefully characterize the spectral data associated to step-like backgrounds, including zeros of $a_1$ and $a_2$ and their behavior near $k=\pm B$, and show that, for small perturbations of the pure step, the data are determined by a single function $b(k)$ with determinant constraints. They classify spectral scenarios into several cases and provide trace formulas that express $a_1$ and $a_2$ in terms of $b(k)$, enabling a complete basic RH problem from which the solution $u(x,t)$ can be reconstructed. In the reflectionless setting, they construct three new families of two-soliton solutions with explicit formulas and demonstrate blow-up along certain curves, accompanied by detailed large-time asymptotics featuring decaying, transition, and periodic regimes. The results extend the Riemann-Hilbert approach to a nonlocal, two-place integrable system with oscillating backgrounds, revealing rich soliton dynamics with PT-symmetric structure.
Abstract
This work focuses on the Cauchy problem for the nonlocal modified Korteweg-de Vries equation $$ u_t(x,t)+6u(x,t)u(-x,-t)u_x(x,t)+u_{xxx}(x,t)=0, $$ with the oscillating step-like boundary conditions: $u(x,t)\to 0$ as $x\to-\infty$ and $u(x,t)\backsimeq A\cos(2Bx+8B^3t)$ as $x\to\infty$, where $A,B>0$ are arbitrary constants. The main goal is to develop the Riemann-Hilbert formalism for this problem, paying a particular attention to the case of the ``pure oscillating step'' initial data, that is $u(x,0)=0$ for $x<0$ and $u(x,0)=A\cos(2Bx)$ for $x\geq0$. Also, we derive three new families of two-soliton solutions, which correspond to the values of $A$ and $B$ satisfying $B<\frac{A}{4}$, $B>\frac{A}{4}$, and $B=\frac{A}{4}$.
