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A dissipative electromechanical toy model under slow-fast blocking

Example

Use the linear toy model with mechanical position $q$ , velocity $v$ , and electrical current $i$ : $q^{'} = v, v^{'} = - q - v + i, i^{'} = v - 8 i .$ The unit mass, unit spring and damping coefficients, unit inductance, resistance $8$ , force-current coupling, and back-emf coupling are stipulated model data. Its mechanical block is $x = (q, v)$ and its electrical block is $z = i$ . For the quadratic energy $E (q, v, i) = \frac{1}{2} (q^{2} + v^{2} + i^{2})$ , $E^{'} = - (v - i)^{2} - 7 i^{2} \leq 0.$ Thus the exact model dissipates this energy. Starting from $(q, v, i) = (0, 1, 0)$ , apply Frozen-slow explicit Euler with fast microsteps with $H = 1 / 10$ , $m = 4$ , and $h = 1 / 40$ . The macro method returns $(q^{+}, v^{+}, i_{4}) = (1 / 10, 9 / 10, 369 / 5000)$ .

Facts & Assumptions

Given: The stipulated electromechanical model, initial state, macro step $H = 1 / 10$ , and four fast microsteps of length $h = 1 / 40$ .

[F1]

The assigned algorithm updates the slow block once by explicit Euler and then updates the fast block four times while keeping the slow state fixed at its initial value (Frozen-slow explicit Euler with fast microsteps).

[F2]

If $E$ is continuous on $[t_{1}, t_{2}]$ and differentiable on $(t_{1}, t_{2})$ , then $E (t_{2}) - E (t_{1}) = E^{'} (c) (t_{2} - t_{1})$ for some $c \in (t_{1}, t_{2})$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[A1]

In this example $f (q, v, i) = (v, - q - v + i)$ and $g (q, v, i) = v - 8 i$ ; the mechanical and electrical labels refer to this stipulated block partition.

Proof

1.1A1F2algebra

Differentiating $E = \frac{1}{2} (q^{2} + v^{2} + i^{2})$ along [A1] gives $E^{'} = q v + v (- q - v + i) + i (v - 8 i) = - v^{2} + 2 v i - 8 i^{2} = - (v - i)^{2} - 7 i^{2} \leq 0$ . On any solution interval and any $t_{1} < t_{2}$ , [F2] gives $E (t_{2}) - E (t_{1}) = E^{'} (c) (t_{2} - t_{1}) \leq 0$ , so the energy is nonincreasing.

1.2F1A1algebra

By [F1] and [A1], the mechanical update uses the initial state $(q, v, i) = (0, 1, 0)$ : $q^{+} = 0 + (1 / 10) 1 = 1 / 10$ and $v^{+} = 1 + (1 / 10) (- 0 - 1 + 0) = 9 / 10$ .

1.3F1A1inductionalgebra

The fast recurrence uses frozen $v = 1$ , so $i_{j + 1} = i_{j} + (1 / 40) (1 - 8 i_{j})$ . Starting with $i_{0} = 0$ , successive substitution gives $i_{1} = 1 / 40$ , $i_{2} = 9 / 200$ , $i_{3} = 61 / 1000$ , and $i_{4} = 369 / 5000$ . Thus the stated macro output follows.

2.1F1step 1.2step 1.3algebra∎

By [F1], the method evaluates the two-component mechanical field once and the electrical field four times. Its fast loop reads the initial slow value $v = 1$ at each stage, so an implementation may transmit that frozen value once at the block boundary and cache it; no fast-to-slow refresh occurs inside this macro block. This count describes the procedure and does not claim a wall-clock saving.

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