Question 5.56: Use the shooting method to find the solution of the boundary......
Use the shooting method to find the solution of the boundary value problem
\begin{aligned} y^{\prime \prime} & =6 y^{2}, \\ y(0) & =1, y(0.5)=4 / 9 . \end{aligned}
Assume the initial approximations
y^{\prime}(0)=\alpha_{0}=-1.8, y^{\prime}(0)=\alpha_{1}=-1.9 \text {,}
and find the solution of the initial value problem using the fourth order Runge-Kutta method with h=0.1. Improve the value of y^{\prime}(0) using the secant method once. Compare with the exact solution y(x)=1 /(1+x)^{2}.
We use the fourth order Runge-Kutta method to solve the initial value problems :
\begin{aligned} & I: \quad y^{\prime \prime}=6 y^{2} \text {, } \\ & y(0)=1, y^{\prime}(0)=-1.8, \\ & \text { II : } \quad y^{\prime \prime}=6 y^{2} \text {, } \\ & y(0)=1, y^{\prime}(0)=-1.9 \text {, } \end{aligned}
and obtain the solution values at x=0.5. We then have
g\left(\alpha_{0}\right)=y\left(\alpha_{0} ; b\right)-\frac{4}{9}, g\left(\alpha_{1}\right)=y\left(\alpha_{1} ; b\right)-\frac{4}{9}.
The secant method gives
\alpha_{n+1}=\alpha_{n}-\left[\frac{\alpha_{n}-\alpha_{n-1}}{g\left(\alpha_{n}\right)-g\left(\alpha_{n-1}\right)}\right] g\left(\alpha_{n}\right), n=1,2, \ldots
The solution values are given by
| x | y(0)=1 \alpha_0= – 1.8 |
y(0)=1 \alpha_0= – 1.9 |
y(0)=1 \alpha_{sc}=1.998853 |
y(0)=1 y^{\prime}(0)=-2 |
| 0.1 | 0.8468373 | 0.8366544 | 0.8265893 | 0.8264724 |
| 0.2 | 0.7372285 | 0.7158495 | 0.6947327 | 0.6944878 |
| 0.3 | 0.6605514 | 0.6261161 | 0.5921678 | 0.5917743 |
| 0.4 | 0.6102643 | 0.5601089 | 0.5108485 | 0.5102787 |
| 0.5 | 0.5824725 | 0.5130607 | 0.4453193 | 0.4445383 |