Weakly Nonlinear Wave Packets and the Nonlinear Schrödinger Equation

Dias, F., & Bridges, T. (2005). Weakly nonlinear wave packets and the nonlinear Schrödinger equation. In R. Grimshaw (Ed.), Nonlinear waves in fluids: Recent advances and modern applications (pp. 29–67). Springer. https://doi.org/10.1007/3-211-38025-6_2

The method of multiple scales

A derivation of the NLS (nonlinear Schrödinger) equation from the KdV (Korteweg-de Vries) equation based on the method of multiple scales.

• The KdV equation

  $$\mathcal{L}(u)+\mathcal{N}(u,u)=u_t+u_{xxx}+6uu_x=0,\quad(x\in\mathbb{R},t\ge0,u(x,t)\in\mathbb{R}).$$

  Introducting a small parameter $\epsilon$, and the slow scales $X=\epsilon x,T=\epsilon t,\tau=\epsilon^2 t$

  $$u=\epsilon(u_0+\epsilon u_1+\epsilon^2 u_2+\cdots)=\epsilon u_0+\epsilon^2 u_1+\epsilon^3 u_2+\cdots,$$

  $$\begin{aligned} \mathcal{L}&=\partial_t+\partial_{xxx} \\ &=\left(\frac{\partial}{\partial t}+\epsilon\frac{\partial}{\partial T}+\epsilon^2\frac{\partial}{\partial \tau}\right)+\left(\frac{\partial}{\partial x}+\epsilon\frac{\partial}{\partial X}\right)^3 \\ &=\left(\frac{\partial}{\partial t}+\frac{\partial^3}{\partial x^3}\right)+\epsilon\left(\frac{\partial}{\partial T}+3\frac{\partial^3}{\partial x^2\partial X}\right)+\epsilon^2\left(\frac{\partial}{\partial\tau}+3\frac{\partial^3}{\partial x\partial X^2}\right)+\cdots \\ &\equiv\mathcal{L}^{(0)}+\epsilon\mathcal{L}^{(1)}+\epsilon^2\mathcal{L}^{(2)}+\cdots, \end{aligned}$$

  $$\begin{array}{ll} \mathcal{O}(\epsilon) & \mathcal{L}^{(0)}u_0=0, \\ \mathcal{O}(\epsilon^2) & \mathcal{L}^{(0)}u_1=-\mathcal{L}^{(1)}u_0-3\partial_x(u_0^2), \\ \mathcal{O}(\epsilon^3) & \mathcal{L}^{(0)}u_2=-\mathcal{L}^{(2)}u_0-\mathcal{L}^{(1)}u_1-6\partial_x(u_0u_1)-3\partial_X(u_0^2). \end{array}$$

• At $\mathcal{O}(\epsilon)$: $\mathcal{L}^{(0)}u_0=0$
  the eigen mode

  $$u_0(x,t)=\psi_0(X,T)\exp[i(kx-\omega t)]+\text{c.c.}$$

  with the dispersion relation (decided by the linear operator $\mathcal{L}^{(0)}$)

  $$\omega=-k^3.$$

  the phase velocity $c=\omega/k$
  the group velocity $c_g=\mathrm{d}\omega/\mathrm{d}k=-3k^2$

• At $\mathcal{O}(\epsilon^2)$: $\mathcal{L}^{(0)}u_1=-\mathcal{L}^{(1)}u_0-3\partial_x(u_0^2)$
  the solvability condition ( remove the eigen mode term in RHS of the equation )

  $$\mathcal{L}^{(1)}u_0=0\Rightarrow(\partial_T+c_g\partial_X)u_0=0$$

  on the time scale $T$, the wave packet is just transported at the group velocity and thus depends only on the variable $\xi=X-c_gT$.
  

  ▶

  python script

  choose the function

  $$\begin{aligned} f(x,t)&=\psi(X,T)\cos(kx-\omega t) \\ &=\psi(X-c_g T)\cos(kx-\omega t) \\ &=\sin[\epsilon(x-c_g t)]\cos(kx-\omega t) \\ &=\sin[\epsilon(x+3k^2 t)]\cos(kx+k^3 t) \end{aligned}$$

  set $\epsilon=0.1, k=1$ so we have $\omega=-k^3=-1$
  phase velocity $c=\omega/k=-1$
  group veloctiy $c_g=\mathrm{d}\omega/\mathrm{d}k=-3k^2=-3$

  import numpy as np
  import matplotlib.pyplot as plt
  from matplotlib.animation import FuncAnimation
  
  # Parameter settings
  epsilon = 0.1
  k = 1.0
  
  x = np.linspace(-20, 100, 1000)
  t_frames = np.linspace(0, 100, 300)
  
  # Create the plot window
  fig, ax = plt.subplots(figsize=(10, 6))
  ax.set_xlim(-20, 100)
  ax.set_ylim(-1.1, 1.1)
  ax.set_xlabel('x')
  ax.set_ylabel('f(x,t)')
  ax.grid(True)
  
  # Initialize three curves: f(x,t) and the two envelope lines
  line_f, = ax.plot([], [], 'b-', lw=2, label='f(x,t)')
  line_env_pos, = ax.plot([], [], 'r--', lw=1.5, label='envelope')
  line_env_neg, = ax.plot([], [], 'r--', lw=1.5)
  
  # Create legend once and fix its position
  # bbox_to_anchor=(1, 1) places the legend at the upper-right corner of the axes,
  # with a small offset to avoid overlapping with the title.
  ax.legend(loc='upper right', bbox_to_anchor=(1.0, 1.0), frameon=True)
  
  # Initialization function (clear data)
  def init():
      line_f.set_data([], [])
      line_env_pos.set_data([], [])
      line_env_neg.set_data([], [])
      return line_f, line_env_pos, line_env_neg
  
  # Update function for each frame
  def update(frame):
      t = t_frames[frame]
      # Compute the waveform
      y_f = np.sin(epsilon * (x + 3 * k**2 * t)) * np.cos(k * x + k**3 * t)
      # Envelope (amplitude)
      env = np.sin(epsilon * (x + 3 * k**2 * t))
      y_env_pos =  env
      y_env_neg = -env
  
      line_f.set_data(x, y_f)
      line_env_pos.set_data(x, y_env_pos)
      line_env_neg.set_data(x, y_env_neg)
  
      ax.set_title(f't = {t:.2f}')
      return line_f, line_env_pos, line_env_neg
  
  # Create the animation (interval 50ms, i.e., 20 frames per second)
  # If legend still jitters, try setting blit=False (though slower)
  ani = FuncAnimation(fig, update, frames=len(t_frames), init_func=init,
                      blit=True, interval=50)
  
  # Save the animation (requires ffmpeg)
  ani.save('wave_animation.gif', writer='ffmpeg', fps=20)
  
  # plt.show()

  Then solve for $u_1$:

  $$\mathcal{L}^{(0)}u_1=-3\partial_x(\phi_0^2e^{2i(kx-\omega t)}+\text{c.c.})\tag{2.10}$$

  the general solution is

  $$u_1=A_2(X,T)e^{2i(kx-\omega t)}+\text{c.c.}+\psi_1 e^{i(kx-\omega t)}+\text{c.c.}+A_0(X,T)$$

  the term $\psi_1 e^{i(kx-\omega t)}$ can be incorporated into the term $\psi_0 e^{i(kx-\omega t)}$ by defining $\psi=\psi_0+\epsilon\psi_1$.

  $$\left\{ \begin{aligned} u_0&=\psi e^{i(kx-\omega t)}+\text{c.c.}, \\ u_1&=A_2(X,T)e^{2i(kx-\omega t)}+\text{c.c.}+A_0(X,T), \end{aligned} \right.$$

  $$u_1\rightarrow\mathcal{O}(\epsilon^2)\text{(2.10)}\Rightarrow A_2=\frac{-3(2ik)\psi^2}{-2i\omega+(2ik)^3}=\frac{\psi^2}{k^2}.$$

• At $\mathcal{O}(\epsilon^3)$: $\mathcal{L}^{(0)}u_2=-\mathcal{L}^{(2)}u_0-\mathcal{L}^{(1)}u_1-6\partial_x(u_0u_1)-3\partial_X(u_0^2)$
  the solvability condition

  $$-\partial_\tau\psi-3ik\partial_{XX}\psi-6ik|\psi|^2\psi/k-6ikA_0\psi=0\tag{2.13}$$

  Equating the coefficients of the oscillation free (time average) terms in (2.13)

  $$-\partial_T A_0-6\partial_X|\psi|^2=0\Rightarrow A_0=-\frac{2|\psi|^2}{k^2}$$

  noting that $\mathrm{d}^2\omega/\mathrm{d}k^2=-6k$, (2.13) leads to the cubic NLS equation

  $$i\frac{\partial\psi}{\partial\tau}+\frac{1}{2}\frac{\mathrm{d}^2\omega}{\mathrm{d}k^2}\frac{\partial^2\psi}{\partial\xi^2}+\frac{6}{k}|\psi|^2\psi=0.$$

The water-wave problem

• The governing equations:
  for incompressible, inviscid and irrotational fluid
  the conservation of mass $\Delta \phi=0$ ( $\boldsymbol{u}=\nabla\phi(x,y,z,t)$ )
  the Bernoulli’s equation

  $$\frac{\partial\phi}{\partial t}+\frac{1}{2}|\nabla\phi|^2+gz+\frac{p-p_0}{\rho}=0.$$

• The flow domain:
  the free surface $F(x,y,z,t)\equiv\eta(x,y,t)-z=0$
  the solid bottom $z=-h$
• Boundary conditions:
  the kinematic condition on the free surface: $DF/Dt=0\Rightarrow$ $\eta_t+\phi_x\eta_x+\phi_y\eta_y-\phi_z=0$ (the free surface is a streamline)
  the dynamic condition on the free surface:

  $$\phi_t+\frac{1}{2}\left(\phi_x^2+\phi_y^2+\phi_z^2\right)+g\eta-\frac{\sigma}{\rho}C=0$$

  where $C$ is the curvature of the free surface.
  the boundary condition at the bottom: $\phi_z(x,y,-h,t)=0$

To summarize, solve for $\eta(x,y,t)$ and $\phi(x,y,z,t)$

$$\begin{aligned} \Delta\phi\equiv\phi_{xx}+\phi_{yy}+\phi_{zz}&=0,\text{ for }(x,y,z)\in\mathbb{R}\times\mathbb{R}\times[-h,\eta], \\ \eta_t+\phi_x\eta_x+\phi_y\eta_y-\phi_z&=0,\text{ on }z=\eta(x,y,t), \\ \phi_t+\frac{1}{2}\left(\phi_x^2+\phi_y^2+\phi_z^2\right)+g\eta-\frac{\sigma}{\rho}C&=0,\text{ on }z=\eta(x,y,t), \\ \phi_z&=0,\text{ on }z=-h \end{aligned}$$

Linearization of the 2D problem

Looks for solutions which are periodic in $x$ with wave number $k$ and in $t$ with frequency $\omega$.
Introduce dimensionless variables:

$$(x^*,z^*)=(kx,kz),\quad\eta^*=\frac{\eta}{a},\quad t^*=\omega t,\quad\phi^*=\frac{k}{\omega a}\phi,$$

where $a$ denotes the amplitude of the wave.

Dropping the stars, the surface wave problem linearized around the equilibrium state $\eta=0,\boldsymbol{u}=\boldsymbol{0}$ reads:

$$\begin{aligned} \phi_{xx}+\phi_{zz}&=0,\text{ for }(x,z)\in\mathbb{R}\times[-kh,0], \\ \eta_t-\phi_z&=0,\text{ on }z=0, \\ \phi_t+\left(\frac{gk}{\omega^2}\right)\eta-\left(\frac{\sigma k^3}{\rho\omega^2}\right)\eta_{xx}&=0,\text{ on }z=0, \\ \phi_z&=0,\text{ on }z=-kh \end{aligned}$$

assume $\eta(x,t)=\cos(x-t+\Theta)$, we can obtain

$$\phi(x-t,z)=\frac{1}{\tanh(kh)}\sin(x-t+\Theta)\frac{\cosh(z+kh)}{\cos(kh)}$$

and the dynamic condition yields the dispersion relation for linearized 2D capillary-gravity periodic waves in water of finite depth

$$\omega^2-gk\tanh(kh)\left(1+\frac{\sigma k^2}{\rho g}\right)=0.$$

• Interpretation of the group velocity:
  1. the travel velocity of the envelope;
  2. the speed of propagation of energy.

For 3D waves

• The extension of the dispersion relation to three-dimensional waves is

  $$\omega^2-g|\boldsymbol{k}|\tanh(|\boldsymbol{k}h|)\left(1+\frac{\sigma|\boldsymbol{k}|^2}{\rho g}\right)=0$$

  where $|\boldsymbol{k}|=\sqrt{k^2+l^2}$.

• The common ansatz used in the multiple scales:

  $$\begin{aligned} \eta &=\sum_{n=1}^3 \epsilon^n\eta_n(x_0,x_1,x_2;y_1,y_2;t_0,t_1,t_2)+\mathcal{O}(\epsilon^4), \\ \phi &=\sum_{n=1}^3 \epsilon^n\phi_n(x_0,x_1,x_2;y_1,y_2;z;t_0,t_1,t_2)+\mathcal{O}(\epsilon^4) \end{aligned}$$

  where

  $$\begin{array}{lll} x_0=x, & x_1=\epsilon x, & x_2=\epsilon^2 x, \\ & y_1=\epsilon y, & y_2=\epsilon^2 y, \\ t_0=t, & t_1=\epsilon t, & t_2=\epsilon^2 t. \end{array}$$

  The order one component of $\eta$ is the linearized monochromatic wave

  $$\eta_1=Ae^{i(kx-\omega t)}+\text{c.c.}$$

  Applying the method of multiple scales, the solvability condition at $\mathcal{O}(\epsilon^3)$ gives the evolution of the complex amplitude $A$ of the wave:

  $$\begin{aligned} 2i\frac{\partial A}{\partial t_2}+p\frac{\partial^2 A}{\partial \xi^2}+q\frac{\partial^2 A}{\partial y_1^2}+\gamma A|A|^2={}&-i\epsilon(sA_{\xi y_1y_1}+rA_{\xi\xi\xi}+uA^2A^*_\xi-v|A|^2A_\xi) \\ &+\epsilon A\left.\bar{\phi}_\xi\right|_{z_1=0}. \end{aligned}$$

The NLS equation:

$$iu_t+\alpha u_{xx}+\gamma|u|^2u=0,\quad u\in\mathbb{C}.$$