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Exponential growth of the Nyquist mode above the CFL bound

Physical scope

An exact mathematical failure mode of the scalar benchmark above its stated time-step bound; no Einstein solver or physical evolution is asserted.

Statement

Let $N \geq 4$ be even, let $h = 2 π / N$ , and choose the Courant--Friedrichs--Lewy (CFL) step ratio $λ : = k / h > 1$ . Run the periodic wave algorithm with $u_{j}^{0} = (- 1)^{j}$ and $v_{j}^{0} = 0$ , with any integer $n_{\max} \geq 1$ . For every integer $0 \leq n \leq n_{\max}$ the output has the form $u_{j}^{n} = a_{n} (- 1)^{j}$ , where, with $c : = 1 - 2 λ^{2} < - 1$ ,

a_{n} = \frac{z_{g r o w}^{n} + z_{d e c a y}^{n}}{2}, z_{g r o w} : = c - \sqrt{c^{2} - 1} < - 1, z_{d e c a y} : = c + \sqrt{c^{2} - 1} \in (- 1, 0) .

Consequently $∣ a_{n} ∣ \geq ∣ z_{g r o w} ∣^{n} / 2$ , so allowing more time steps causes exponential growth of this alternating mode. The strict hypothesis $0 < k / h < 1$ in the preceding mode theorem excludes this example; no endpoint claim at $k / h = 1$ is made here.

Statement references

A finite periodic scalar-wave benchmark

Termination and operation count of the periodic wave benchmark

Fourier-mode stability of the periodic leapfrog benchmark

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$

Absolute value in an ordered field

Basic properties of the absolute value

Facts & Assumptions

Given: An even integer $N \geq 4$ , the benchmark step ratio $λ > 1$ , and the real initial arrays $u_{j}^{0} = (- 1)^{j}$ , $v_{j}^{0} = 0$ .

[F1]

A finite periodic scalar-wave benchmark: the initialization and periodic update are $u^{1} = u^{0} + k v^{0} + (k^{2} / 2) D_{h} u^{0}$ and $u^{n + 1} = 2 u^{n} - u^{n - 1} + k^{2} D_{h} u^{n}$ .

[F2]

Termination and operation count of the periodic wave benchmark: for every finite $n_{\max}$ the algorithm outputs every level through $u^{n_{\max}}$ .

[F3]

Fourier-mode stability of the periodic leapfrog benchmark: the preceding unit-circle result assumes $0 < k / h < 1$ and makes no claim for $k / h > 1$ .

[F4]

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ : a positive real number has a unique positive square root.

[F5]

Absolute value in an ordered field: for real $x$ , $∣ x ∣ = x$ if $x \geq 0$ and $∣ x ∣ = - x$ if $x < 0$ .

[F6]

Basic properties of the absolute value: $∣ x ∣ \geq 0$ , $∣ x ∣ = 0$ exactly when $x = 0$ , and $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ for real $x, y$ .

Verification

Given: The even periodic grid, the stated time-step ratio, and the alternating initial array.

1.1givenF1F3algebra

Because $N$ is even, both periodic neighbors of $(- 1)^{j}$ equal $- (- 1)^{j}$ . Hence $(D_{h} u^{0})_{j} = - 4 h^{- 2} (- 1)^{j}$ , and the startup in [F1] gives $a_{0} = 1$ and $a_{1} = 1 - 2 (k / h)^{2} = c$ . This mode has index $ℓ = N / 2$ ; it is outside the strict range of [F3].

2.1givenF1F4step 1.1algebra

Substitution of $u_{j}^{n} = a_{n} (- 1)^{j}$ into the update [F1] gives $a_{n + 1} - 2 c a_{n} + a_{n - 1} = 0$ . Since $c < - 1$ , [F4] supplies $r = \sqrt{c^{2} - 1} > 0$ . The two roots $z_{g r o w} = c - r$ and $z_{d e c a y} = c + r$ satisfy $z_{g r o w} z_{d e c a y} = 1$ ; also $z_{g r o w} < - 1$ and $z_{d e c a y} = 1 / z_{g r o w} \in (- 1, 0)$ .

3.1step 1.1step 2.1F1F2F5F6algebra∎

The sequence $(z_{g r o w}^{n} + z_{d e c a y}^{n}) / 2$ satisfies the recurrence because each root satisfies $z^{2} - 2 c z + 1 = 0$ , and its values at $n = 0, 1$ are $1, c$ . Induction identifies it with $a_{n}$ . For even $n$ the two powers are nonnegative, and for odd $n$ they are both negative; since $1 / 2 > 0$ , their half-sum has the same sign. The piecewise definition [F5] therefore gives $∣ a_{n} ∣ = (∣ z_{g r o w}^{n} ∣ + ∣ z_{d e c a y}^{n} ∣) / 2$ , while induction using multiplicativity [F6] gives $∣ z_{g r o w}^{n} ∣ = ∣ z_{g r o w} ∣^{n}$ and $∣ z_{d e c a y}^{n} ∣ = ∣ z_{d e c a y} ∣^{n}$ . Since $z_{g r o w} < - 1$ , [F5] gives $∣ z_{g r o w} ∣ = - z_{g r o w} > 1$ , while $∣ z_{d e c a y} ∣^{n} \geq 0$ by [F6]; hence $∣ a_{n} ∣ \geq ∣ z_{g r o w} ∣^{n} / 2$ .

Diagram

The following $N = 8$ picture illustrates the periodic alternating data; the last edge identifies the two ends of the grid. The proof above uses the recurrence and applies to every even $N \geq 4$ ; the drawing is not a proof.

Exponential growth of the Nyquist mode above the CFL bound

Statement

Statement references

Facts & Assumptions

Verification

Diagram

Depends on

Dependency tree · 2 levels

Sources