Single shooting amplifies an initial-slope error on a decaying-mode BVP

The substitution $w=u+1$ gives $w^{\mathrm{\prime }\mathrm{\prime }}={\lambda }^{2}w$ with $w(0)=0$ and $w(1)=1$. Solving this constant-coefficient equation gives $w(t)=\sinh (\lambda t)\mathrm{/}\sinh \lambda$, so $u(t)=w(t)-1\in [-1,0]$ because $\sinh (\lambda t)$ increases from $0$ to $\sinh \lambda$. Its initial slope is $s^{\ast }=w^{\mathrm{\prime }}(0)=\lambda \mathrm{/}\sinh \lambda$.

If the initial slope is $s^{\ast }+\delta$ while $u(0)=-1$ remains fixed, linearity of the equation for $w$ gives $w_{\delta }(t)=(s^{\ast }+\delta )\sinh (\lambda t)\mathrm{/}\lambda$. Subtracting the exact shot yields $w_{\delta }(1)-w(1)=\delta \sinh \lambda \mathrm{/}\lambda$.

Since $\sinh \lambda -\lambda$ has value zero at $0$ and derivative $\cosh \lambda -1>0$ for $\lambda >0$, we have $\sinh \lambda \mathrm{/}\lambda >1$. Also the derivative of $q(\lambda )=\sinh \lambda \mathrm{/}\lambda$ has numerator $g(\lambda )=\lambda \cosh \lambda -\sinh \lambda$, where $g(0)=0$ and $g\mathrm{\prime }(\lambda )=\lambda \sinh \lambda >0$; hence $q$ is increasing. For $\lambda \ge 2$, $\sinh \lambda \ge e^{\lambda -1}\mathrm{/}2$, so $q(\lambda )\ge e^{\lambda -1}\mathrm{/}(2\lambda )\to \mathrm{\infty }$. The perturbed terminal boundary residual vector changes by $(0,\delta \sinh \lambda \mathrm{/}\lambda )$, while [F2]–[F3] identify this as the endpoint residual evaluated by exact single shooting.