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A one-sided Neumann closure gives only first-order error

Example

Let $h = 1 / N$ for an integer $N \geq 2$ , $x_{i} = i h$ , $δ_{h}^{2} V_{i} = (V_{i + 1} - 2 V_{i} + V_{i - 1}) / h^{2}$ , and $D_{h}^{+} V_{0} = (V_{1} - V_{0}) / h$ . The exact solution $u (x) = x^{2}$ solves $- u^{''} = - 2$ on $(0, 1)$ with $u^{'} (0) = 0$ and $u (1) = 1$ . Discretize the interior equation by $- δ_{h}^{2} V_{i} = - 2$ for $1 \leq i < N$ , impose the one-sided Neumann row $D_{h}^{+} V_{0} = 0$ , and impose $V_{N} = 1$ . The unique grid solution is

V_{i} = x_{i}^{2} + h (1 - x_{i}),

so $\max_{0 \leq i \leq N} ∣ V_{i} - u (x_{i}) ∣ = h$ . The boundary residual of the exact samples is $D_{h}^{+} u (0) - u^{'} (0) = h$ , and the global error is only first order.

Facts & Assumptions

Given: An integer $N \geq 2$ , $h = 1 / N$ , $x_{i} = i h$ , the displayed centered second difference and one-sided forward Neumann row, and the stated mixed discrete equations.

[A1]

The grid equations are $- δ_{h}^{2} V_{i} = - 2$ for $1 \leq i < N$ , $D_{h}^{+} V_{0} = 0$ , and $V_{N} = 1$ .

Verification

1.1givenA1algebra

For the exact samples $u_{i} = x_{i}^{2}$ , $D_{h}^{+} u_{0} = (h^{2} - 0) / h = h$ , whereas $u^{'} (0) = 0$ . Thus the imposed Neumann row has residual exactly $h$ .

1.2A1algebra

Define $V_{i} = x_{i}^{2} + h (1 - x_{i})$ . Then $V_{0} = h$ , $V_{1} = h$ , so $D_{h}^{+} V_{0} = 0$ ; also $V_{N} = 1$ . Since the second difference of the quadratic part is $2$ and the second difference of the affine part is $0$ , $- δ_{h}^{2} V_{i} = - 2$ at every interior node. Hence $V$ satisfies [A1].

2.1A1step 1.2algebra

If two grid solutions existed, their difference $d$ would satisfy $δ_{h}^{2} d_{i} = 0$ for $1 \leq i < N$ , $D_{h}^{+} d_{0} = 0$ , and $d_{N} = 0$ . The recurrence says $d_{i} - d_{i - 1}$ is constant in $i$ ; the forward boundary row makes that constant $0$ , and $d_{N} = 0$ then makes every $d_{i} = 0$ . Therefore the solution in step 1.2 is unique.

3.1step 1.1step 1.2algebra∎

The error is $V_{i} - u_{i} = h (1 - x_{i})$ . Because $0 \leq x_{i} \leq 1$ , its maximum absolute value is $h$ , attained at $i = 0$ . The boundary residual and global error are both first order in $h$ .

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Benjamin Seibold, MIT 18.336 Lecture 5: Finite Difference Approximation (standard reference, not scraped)