4.3.1 Shooting Method

- Transform the given boundary value problem into an initial value problem with estimated parameters

- Adjust the parameters iteratively to reproduce the given boundary values

First trial shot:

Augment the $n_{1}$ boundary values given at $x=x_{1}$ by $n_{2} \equiv N-n_{1}$ estimated parameters

$$\mbox{$\bf a$}^{(1)} \equiv \{ a_{k}^{(1)}; \; k=1,\dots n_{2} \}^{T}$$

to obtain an IVP. Integrate numerically up $x=x_{2}$. The newly calculated values of $b_{k}$ at $x=x_{2}$,

$$\mbox{$\bf b$}^{(1)} \equiv \{ b_{k}^{(1)}; \; k=1,\dots n_{2} \}^{T}$$

will in general deviate from the given boundary values $\mbox{$\bf b$} \equiv \{ b_{k}; \dots \}^{T}$. The difference vector $\mbox{$\bf e$}^{(1)} \equiv \mbox{$\bf b$}^{(1)}-\mbox{$\bf b$}$ is stored for further use.

Second trial shot:

Change the estimated initial values $a_{k}$ by some small amount, $\mbox{$\bf a$}^{(2)} \equiv \mbox{$\bf a$}^{(1)}+\delta \mbox{$\bf a$}$, and once more integrate up to $x=x_{2}$. The values $b_{k}^{(2)}$ thus obtained are again different from the required values $b_{k}$: $\mbox{$\bf e$}^{(2)} \equiv \mbox{$\bf b$}^{(2)}-\mbox{$\bf b$}$.

Quasi-linearization:

Assuming that the deviations $\mbox{$\bf e$}^{(1)}$ and $\mbox{$\bf e$}^{(2)}$ depend linearly on the estimated initial values $\mbox{$\bf a$}^{(1)}$ and $\mbox{$\bf a$}^{(2)}$, compute that vector $\mbox{$\bf a$}^{(3)}$ which would make the deviations disappear:

$$\mbox{$\bf a$}^{(3)} = \mbox{$\bf a$}^{(1)}-\mbox{${\bf A}$}^{-1} \cdot \mbox{$\bf e$}^{... ...ith} \;\; A_{ij} \equiv \frac{b_{i}^{(2)}-b_{i}^{(1)}}{a_{j}^{(2)}-a_{j}^{(1)}}$$

Iterate the procedure up to some desired accuracy.

EXAMPLE:

$$\frac{d^{2}y}{dx^{2}} = -\frac{1}{(1+y)^{2}} \;\;\;\;{\rm with} \;\;\;y(0)=y(1)=0$$

* First trial shot: Choose $a^{(1)} \equiv y'(0)=1.0$. Applying 4th order RK with $\Delta x = 0.1$ we find $b^{(1)} \equiv y_{calc}(1)=0.674$. Thus $e^{(1)}\equiv b^{(1)}-y(1) =0.674$.

* Second trial shot: With $a^{(2)}=1.1$ we find $b^{(2)}=0.787$, i.e. $e^{(2)}=0.787$.

* Quasi-linearization: From

$$a^{(3)} = a^{(1)}-\frac{a^{(2)}-a^{(1)}}{b^{(2)}-b^{(1)}} \, e^{(1)}$$

we find $a^{(3)}=0.405 \, (\equiv y'(0))$.

Iteration: The next few iterations yield the following values for $a\, (\equiv y'(0))$ and $b\,(\equiv y(1))$:

$n$	$a^{(n)}$	$b^{(n)}$
3	0.405	- 0.041
4	0.440	0.003
5	0.437	0.000