```python
import numpy as np

def solve_kdv_burgers():
    # Parameters
    L = 10.0
    T = 10.0
    nx = 200
    nt = 5000
    a = 1e-4
    b = 2e-4
    n = 20

    dx = L / nx
    dt = T / nt

    # Spatial grid
    x = np.linspace(0, L, nx, endpoint=False)

    # Initial condition
    u = 0.5 / n * np.log(1 + np.cosh(n)**2 / np.cosh(n * (x - 0.2 * L))**2)

    # Numerical solution using finite difference method
    for _ in range(nt):
        u_new = np.copy(u)
        for i in range(nx):
            # Periodic boundary conditions
            i_minus_1 = (i - 1) % nx
            i_plus_1 = (i + 1) % nx
            i_plus_2 = (i + 2) % nx
            i_minus_2 = (i - 2) % nx

            # Spatial derivatives (central difference)
            du_dx = (u[i_plus_1] - u[i_minus_1]) / (2 * dx)
            d2u_dx2 = (u[i_plus_1] - 2 * u[i] + u[i_minus_1]) / (dx**2)
            d3u_dx3 = (-u[i_minus_2] + 2 * u[i_minus_1] - 2 * u[i_plus_1] + u[i_plus_2]) / (2 * dx**3)

            # Update using forward Euler
            u_new[i] = u[i] + dt * (-u[i] * du_dx + a * d2u_dx2 + b * d3u_dx3)

        u = u_new

    # Save the final solution
    np.save('u.npy', u)

solve_kdv_burgers()
```