Example-1
Given the second-order differential equation:
\[ y'' = -y + 2 \frac{\left(y'\right)^2}{y} \]
Boundary Conditions
The equation has the following boundary conditions:
\[ y(-1) = \frac{1}{e} + \frac{1}{e}, \quad y(1) = \frac{1}{e} + \frac{1}{e} \]
Exact Solution
We want to find an approximate numerical solution for \( y(x) \) using the shooting method and compare it with the exact solution:
\[ y(x) = \frac {1}{e^x + e^{-x}} \]
The solution will be evaluated over the interval:
\[ x \in [-1, 1] \]
Convert to a System of First-Order Differential Equations
Let \( y = y(x) \) and \( z = y'(x) \). Then the second-order equation:
\[ y'' = -y + \frac{2(y')^2}{y} \]
can be expressed as two first-order equations:
\[ y' = z \]
\[ z' = -y + \frac{2z^2}{y} \]
This system can be represented as a function \( f(u, x) \), where \( u = [y, z] \).
Initial Guess for the Derivative at the Start Point
Start with two initial guesses for \( z(-1) \), denoted \( z_1 \) and \( z_2 \). For example:
\[ z_1 = 0.1, \quad z_2 = 0.25 \]
Iterative Shooting Method
For each guess \( z_i \), solve the system of equations from \( x = -1 \) to \( x = 1 \).
Let \( w_1 \) and \( w_2 \) represent the values of \( y(1) \) obtained with initial values \( z_1 \) and \( z_2 \), respectively.
Convergence Check and Update of \( z \) Values
Check if the error \( |y_1 - w_2| \) is within a tolerance, e.g., 0.001. If so, stop.
Otherwise, update the guess for \( z \) as follows:
\[ z_{\text{new}} = z_2 + (z_2 - z_1) \cdot \frac{y_1 - w_2}{w_2 - w_1} \]
Set \( z_1 = z_2 \) and \( z_2 = z_{\text{new}} \) and repeat the process.