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Averaging estimate on [0,100] for the scalar equation x′=εcos⁡t

Example

Take n=1, P=2π, O=R, f(t,x)=cos⁡t (independent of x), x0=0, and the trapping data R=1, ρ=12, A=1. Then f is continuous, C1 in x with Dxf=0, and 2π-periodic in t; on R×B‾(0,1) one may take M=1 and L=0. The average and deviation of Periodic averaging data: the average field, the oscillatory part and the periodic deviation are

fˉ=0,g(t,x)=sin⁡t(0≤t≤2π, extended 2π-periodically).

The averaged system is y′=0, so y(t)≡0 solves it and stays in B‾(0,12). With B=2MP=4π, B1=2LP=0 and K=eLA[B+(LB+B1M)A]=4π, the smallness hypothesis reads ε⋅4π<12; for ε=1100 it holds because εK=π25<12. The averaging theorem then gives

∣x(t)−y(t)∣=∣x(t)∣≤π25(0≤t≤100).

The exact solution is x(t)=1100sin⁡t, whose modulus is at most 1100; the theorem's bound is valid but not sharp here.

Facts & Assumptions

Given: The data above and the scalar ODE x′=εcos⁡t, x(0)=0.

[L1]

For this f, fˉ(x)=12π∫02πcos⁡s ds=0 and g(t,x)=∫0tcos⁡s ds=sin⁡t on [0,2π], extended periodically; this is the construction of Periodic averaging data: the average field, the oscillatory part and the periodic deviation, evaluated with the fundamental theorem for continuous integrands (Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G, The derivatives of sine and cosine are cosine and minus sine); the endpoint values sin⁡0=sin⁡(2π)=0 and periodicity used here are supplied by [L4]. [given]

[L2]

The averaging theorem (Averaging for fast periodic oscillations: order-ε agreement on the O(1/ε) time scale): under its hypotheses with a solution y:[0,A/ε]→B‾(x0,R−ρ) of the averaged system and εK<ρ, the solution x of x′=εf(t,x), x(0)=x0 exists on [0,A/ε] and satisfies ∥x(t)−y(t)∥2≤εK there, with K=eLA[B+(LB+B1M)A], B=2MP, B1=2LP. [given]

[L3]
[L4]

Sine and cosine are 2π-periodic, and sin⁡(2π)=sin⁡0=0 by their zero set (The zero sets of sine and cosine and the least positive common period 2 pi). Their ranges are [−1,1] (Signs, monotonicity intervals, and ranges of sine and cosine); in particular ∣cos⁡t∣,∣sin⁡t∣≤1. Their differentiability in [L1] gives continuity.

Verification

technique · direct; evaluate the averaging data, check the hypotheses, and compare the bound with the exact solution
1.1givenL1L3L4algebra

By [L4], the continuous field f(t,x)=cos⁡t is 2π-periodic and bounded by one; its state derivative is zero, so M=1 and L=0 meet the hypotheses on O=R. By [L1], fˉ=0 and g(t,x)=sin⁡t, so the averaged system is y′=0; the constant curve y≡0 is its solution with y(0)=0 on [0,100], and ∥y(t)−0∥=0≤R−ρ=12. The constants are M=1, L=0, B=4π, B1=0, K=e0[4π+0]=4π, and with ε=1100, [L3] gives εK=π25<425<1/2=ρ.

2.1step 1.1L1L2L4algebra∎

By [L2] the solution x of x′=1100cos⁡t, x(0)=0 exists on [0,100] and satisfies ∣x(t)∣≤π25 there. Directly, the primitive sin⁡t gives x(t)=1100sin⁡t by [L1], so ∣x(t)∣≤1100 by [L4]; the stated bound holds and is not attained.

Depends on

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