## A nonlinear BVP

Posted March 08, 2013 at 09:19 AM | categories: pde | tags:

Adapted from Example 8.7 in _Numerical Methods in Engineering with Python_ by Jaan Kiusalaas.

We want to solve \(y''(x) = -3 y(x) y'(x)\) with $y(0) = 0 and \(y(2) = 1\) using a finite difference method. We discretize the region and approximate the derivatives as:

\(y''(x) \approx \frac{y_{i-1} - 2 y_i + y_{i+1}}{h^2} \)

\(y'(x) \approx \frac{y_{i+1} - y_{i-1}}{2 h} \)

We define a function \(y''(x) = F(x, y, y')\). At each node in our discretized region, we will have an equation that looks like \(y''(x) - F(x, y, y') = 0\), which will be nonlinear in the unknown solution \(y\). The set of equations to solve is:

\begin{eqnarray} y_0 - \alpha &=& 0 \\ \frac{y_{i-1} - 2 y_i + y_{i+1}}{h^2} + (3 y_i) (\frac{y_{i+1} - y_{i-1}}{2 h}) &=& 0 \\ y_L - \beta &=&0 \end{eqnarray}Since we use a nonlinear solver, we will have to provide an initial guess to the solution. We will in this case assume a line. In other cases, a bad initial guess may lead to no solution.

import numpy as np from scipy.optimize import fsolve import matplotlib.pyplot as plt x1 = 0.0 x2 = 2.0 alpha = 0.0 beta = 1.0 N = 11 X = np.linspace(x1, x2, N) h = (x2 - x1) / (N - 1) def Ypp(x, y, yprime): '''define y'' = 3*y*y' ''' return -3.0 * y * yprime def residuals(y): '''When we have the right values of y, this function will be zero.''' res = np.zeros(y.shape) res[0] = y[0] - alpha for i in range(1, N - 1): x = X[i] YPP = (y[i - 1] - 2 * y[i] + y[i + 1]) / h**2 YP = (y[i + 1] - y[i - 1]) / (2 * h) res[i] = YPP - Ypp(x, y[i], YP) res[-1] = y[-1] - beta return res # we need an initial guess init = alpha + (beta - alpha) / (x2 - x1) * X Y = fsolve(residuals, init) plt.plot(X, Y) plt.savefig('images/bvp-nonlinear-1.png')

That code looks useful, so I put it in the pycse module in the function BVP_nl. Here is an example usage. We have to create two functions, one for the differential equation, and one for the initial guess.

from pycse import BVP_nl import matplotlib.pyplot as plt def Ypp(x, y, yprime): '''define y'' = 3*y*y' ''' return -3.0 * y * yprime def init(x): return alpha + (beta - alpha) / (x2 - x1) * x x1 = 0.0 x2 = 2.0 alpha = 0.0 beta = 1.0 N = 11 x, y = BVP_nl(Ypp, x1, x2, alpha, beta, init, N) plt.plot(x, y) plt.savefig('images/bvp-nonlinear-2.png')

The results are the same.

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