Nonlinear Boundary-Value Solvers — Examples

Since $y^{\mathrm{\prime }}=0$, every BVP solution is constant; the boundary equation is $y(0{)}^2-1=0$, whose real roots are $1$ and $-1$. Both constant curves satisfy the equation and endpoint condition.

At a residual zero, the interior rows from [F2] force $Y_{i+1}=Y_i$ for every $i$, so all nodes equal some $c$; the endpoint row then gives $c^2-1=0$. Thus the only residual zeros are the two constant node vectors $Y_i=1$ and $Y_i=-1$, in accord with [F3].

At either root, a vector $z=(z_0,\dots ,z_N)$ in the kernel of the residual Jacobian satisfies $z_{i+1}-z_i=0$ in every interior row, hence all $z_i=c$. The boundary row is $2Y_0{z}_0=0$; since $Y_0=\pm 1$, this forces $c=0$. The square Jacobian has trivial kernel, so it is invertible and each root is simple.

Fact [F4] defines a finite Newton procedure but makes no branch-selection or global-convergence claim. The existence of both roots shows directly that the BVP and its collocation residual have no unique global solution branch.

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.

The exact ODE flow for $y^{\mathrm{\prime }}=0$ is $\varphi (t;s)=s$, so [F1] gives $R(s)=B(s,s)=s^2-1$, and [F2] certifies that the exact evaluator used by the algorithm has zero IVP error. Thus $R(1)=R(-1)=0$ and $R^{\mathrm{\prime }}(s)=2s$, making both roots simple.

At the initial guess $s_0=0$, the residual is $-1$, so a zero tolerance is not met; the Jacobian is the zero $1\times 1$ map because $R^{\mathrm{\prime }}(0)=0$. The linear correction therefore has no solution in the sense required by Newton and [F4] returns singular-jacobian before any update.

The roots at $\pm 1$ have nonzero derivative, so the simple-root condition can hold on a sufficiently small ball around either root, where the polynomial derivative is Lipschitz. Fact [F5] then gives a local Newton basin there, while step 2.1 exhibits a valid start outside both guaranteed basins for which the method stops.