Let's solve the given **Nonlinear Schrödinger Equation (NLS)** step by step. --- ### **Given:** \[ i u_t + u_{xx} + 2|u|^2 u = 0 \] We are to **derive the solitary wave solution** (also called the soliton solution). --- ## **Step 1: Assume a Solitary Wave Solution** Assume a solution of the form: \[ u(x, t) = f(\xi) e^{i(\theta)} \] where \(\xi = x - vt\), and \(f(\xi)\) is a real function, \(\theta = kx - \omega t\). --- ## **Step 2: Substitute and Simplify** Let \[ u(x, t) = f(\xi) e^{i(kx - \omega t)} \] So, - \(\frac{\partial u}{\partial t} = (f'(\xi) \frac{d\xi}{dt}) e^{i\theta} + f(\xi) e^{i\theta} (i(-\omega))\) - \(\frac{\partial u}{\partial t} = -v f'(\xi) e^{i\theta} - i\omega f(\xi) e^{i\theta}\) So, \[ u_t = -v f' e^{i\theta} - i\omega f e^{i\theta} \] \[ u_{xx} = (f'' - 2ik f' - k^2 f) e^{i\theta} \] \[ |u|^2 u = f^3 e^{i\theta} \] --- ## **Step 3: Plug into the NLS Equation** Plug all into the equation: \[ i u_t + u_{xx} + 2 |u|^2 u = 0 \] \[ i \left[-v f' e^{i\theta} - i\omega f e^{i\theta}\right] + [f'' - 2ik f' - k^2 f] e^{i\theta} + 2 f^3 e^{i\theta} = 0 \] \[ [-i v f' - \omega f + f'' - 2ik f' - k^2 f + 2 f^3] e^{i\theta} = 0 \] Since \(e^{i\theta} \neq 0\), we have: \[ f'' - (v + 2k)i f' + [-\omega - k^2] f + 2f^3 = 0 \] --- ## **Step 4: Separate Real and Imaginary Parts** - Real part: \(f'' + [-\omega - k^2] f + 2f^3 = 0\) - Imaginary part: \((v + 2k) f' = 0\) For nontrivial solution, \(f' \neq 0\), so \(v + 2k = 0 \implies v = -2k\). So, the real part becomes: \[ f'' + [-\omega - k^2] f + 2f^3 = 0 \] Let \(\lambda = -\omega - k^2\), so: \[ f'' + \lambda f + 2f^3 = 0 \] --- ## **Step 5: Soliton Solution (Sech Form)** We look for a localized solution. Assume: \[ f(\xi) = A \, \mathrm{sech}(B\xi) \] Compute derivatives: - \(f' = -A B \, \mathrm{sech}(B\xi) \tanh(B\xi)\) - \(f'' = -A B^2 \, \mathrm{sech}(B\xi) [1 - 2 \mathrm{sech}^2(B\xi)]\) Plug into ODE: \[ -A B^2 \, \mathrm{sech}(B\xi) [1 - 2 \mathrm{sech}^2(B\xi)] + \lambda A \, \mathrm{sech}(B\xi) + 2A^3 \, \mathrm{sech}^3(B\xi) = 0 \] Divide by \(\mathrm{sech}(B\xi)\), group like powers: \[ -A B^2 [1 - 2 \mathrm{sech}^2(B\xi)] + \lambda A + 2A^3 \mathrm{sech}^2(B\xi) = 0 \] \[ -AB^2 + 2AB^2 \mathrm{sech}^2(B\xi) + \lambda A + 2A^3 \mathrm{sech}^2(B\xi) = 0 \] Group \(\mathrm{sech}^0\) and \(\mathrm{sech}^2\) terms: \[ (-AB^2 + \lambda A) + (2AB^2 + 2A^3) \mathrm{sech}^2(B\xi) = 0 \] Set coefficients to zero: 1. \( -AB^2 + \lambda A = 0 \implies B^2 = \lambda \) 2. \( 2A B^2 + 2A^3 = 0 \implies B^2 = -A^2 \) So, - \(B^2 = \lambda\) - \(B^2 = -A^2 \implies \lambda = -A^2\) --- ## **Step 6: Final Soliton Solution** Therefore: \[ f(\xi) = A \, \mathrm{sech}(A \xi) \] where \(A^2 = -\lambda = \omega + k^2\). Recall: \(\xi = x - vt = x + 2k t\). Plug back into full solution: \[ u(x, t) = A \, \mathrm{sech}\left[A(x + 2k t)\right] e^{i(kx - \omega t)} \] --- ## **Step 7: Standard Form** Often, the **soliton solution** is written as: \[ u(x, t) = \eta \, \mathrm{sech} \left[\eta (x - 2\xi t - x_0)\right] e^{i (\xi x - (\xi^2 - \eta^2) t)} \] where \(\eta, \xi\) are real parameters. --- ## **Final Answer:** \[ \boxed{ u(x, t) = \eta \, \mathrm{sech} \left[\eta (x - 2\xi t)\right] e^{i \left(\xi x - (\xi^2 - \eta^2) t \right)} } \] where \(\eta\) and \(\xi\) are real constants that set the amplitude and velocity of the soliton. --- **This is the solitary wave solution (soliton) of the nonlinear Schrödinger equation.**