Past exam of the numerical solution of differential equations course of the University of Oxford / 2011 / B / 4 / a / Solution

For the second-order boundary value problem, introduce an unknown initial slope $s$ and solve the initial value problem $u^{\prime }=v$, $v^{\prime }=g(x,u,v)$ with $u(a)=\alpha$, $v(a)=s$. Its endpoint mismatch is $F(s)=u(b;s)-\beta$. The shooting method seeks a root $F(s)=0$; in general the dependence of the flow on $s$ is nonlinear. A sign-changing bracket for a continuous $F$ allows the bisection method, provided the intervening initial value solutions exist up to $b$.

For Newton's method, solve the parameter sensitivity equation alongside the trajectory. With $z={\partial }_{s}u$ and $w={\partial }_{s}v$,Then $F^{\prime }(s)=z(b)$ and the update is $\boxed{s_{\mathrm{new}}=s-F(s)/z(b)}$ whenever $z(b)\ne 0$.

For the third-order problem, shooting from $a$ leaves both $u^{\prime }(a)$ and $u^{\prime \prime }(a)$ unknown, requiring two shooting parameters to satisfy two right-endpoint conditions. Shooting backwards from $b$ already knows $u(b)=\beta$ and $u^{\prime }(b)=\gamma$; only $u^{\prime \prime }(b)$ is unknown, with the single residual $u(a)-\alpha$. Backward shooting reduces the nonlinear solve from two scalar unknowns to one. This parameter-count advantage does not by itself guarantee numerical stability of backward integration.