A Note on Parabolic Homogenization with a Mismatch between the Spatial Scales

Abstract

We consider the homogenization of the linear parabolic problem ( / 2 ) ( , ) − ∇ ⋅ ( ( / 1 , / 2 1 )∇ ( , )) = ( , ) which exhibits a mismatch between the spatial scales in the sense that the coefficient ( / 1 , / 2 1 ) of the elliptic part has one frequency of fast spatial oscillations, whereas the coefficient ( / 2 ) of the time derivative contains a faster spatial scale. It is shown that the faster spatial microscale does not give rise to any corrector term and that there is only one local problem needed to characterize the homogenized problem. Hence, the problem is not of a reiterated type even though two rapid scales of spatial oscillation appear.

Homogenization of a linear parabolic problem with a certain type of matching between the microscopic scales

“…It was established already in [4] that parabolic problems normally have this property when the temporal microscopic scale is the square of the spatial scale, see also e.g. [10], [15], [8], [21] or [9]. But, in our case we have resonance even though the spatial and the temporal scale do not relate to each other in that way.…”

“…There are a number of other articles treating problems related to (1.1) in the sense that the coefficient in front of the time derivative depends on the parameter ε, see e.g. [17], [3], [6], [8], [21], and [5]. A significant difference is that in those articles the coefficient oscillates, while in our case it vanishes, as ε tends to zero.…”

Homogenization of the Heat Equation with a Vanishing Volumetric Heat Capacity

“…Several other studies of parabolic equations, both equations where the coefficient in front of the time derivative is identical to one and equations with oscillating coefficients, show resonance for the same type of matching, see e.g. [10], [15], [3], [7], [8], [19], [9] and [5]. As we will see in the homogenization result, equation (1) will have resonance if r = q + 2, i.e.…”

“…for v 1 ∈ D(Ω), c 1 ∈ D(0, T ) and c 2 ∈ C ∞ ♯ (S) and r > 0. By using the weak form (8) with the test function…”

“…To obtain the homogenized problem we choose, in the weak form(8), the test function v (x) c (t) = v 1 (x) c 1 (t) , where v 1 ∈ H 1 0 (Ω) and c 1 ∈ D(0, T ), giving ΩT −ε q u ε (x, t) v 1 (x) ∂ t c 1 (t) + a x ε , t ε r ∇u ε (x, t) · ∇v 1 (x) c 1 (t) dxdt = ΩT f (x, t) v 1 (x) c 1 (t) dxdt.and applying the Variational lemma we get S −u 1 (x, t, y, s) ∂ s c 2 (s) ds = 0 a.e. in Ω T × Y , which implies that u 1 is independent of s. Now, to find the local problem, we choose the test function v (x) c (t) = εv 1 (x) v 2 x ε c 1 (t) ,…”