ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A midpoint BVP residual can have two simple solution branches

Example

For $y^{'} = 0$ on $[0, 1]$ with scalar boundary residual $B (y (0), y (1)) = y (0)^{2} - 1$ , the exact BVP has two constant solutions $y \equiv 1$ and $y \equiv - 1$ . For every uniform mesh, the normalized implicit-midpoint collocation residual has exactly those two roots, and each root is simple.

Verification

Given: The scalar vector field $f (t, y) = 0$ , the endpoint map $B (y_{0}, y_{1}) = y_{0}^{2} - 1$ , and a uniform mesh with $N \geq 1$ .

[F1] Square nonlinear two-point boundary-value problems and endpoint residuals: BVP solutions satisfy the ODE and endpoint residual.

[F2] Normalized implicit-midpoint collocation residual for a two-point BVP: the interior normalized rows are $(Y_{i + 1} - Y_{i}) / h$ and the last row is $Y_{0}^{2} - 1$ .

[F3] Zeros of the normalized midpoint residual are precisely midpoint collocation solutions: zeros of the residual are exactly the stated midpoint equations and boundary equation.

[F4] Finite-budget implicit-midpoint collocation for a nonlinear BVP: the algorithm applies finite-budget Newton to this square residual but does not select a global branch.

1.1givenF1algebra

Since $y^{'} = 0$ , 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.

2.1F2F3step 1.1algebra

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].

3.1F2step 2.1algebra

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 $2 Y_{0} z_{0} = 0$ ; since $Y_{0} = \pm 1$ , this forces $c = 0$ . The square Jacobian has trivial kernel, so it is invertible and each root is simple.

4.1F4step 3.1∎

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.

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Sources

T. A. Driscoll and R. J. Braun, Fundamentals of Numerical Computation, Chapter 10 (standard reference, not scraped)