13.3 Asymptotic ODE behavior

In the case of polarized Gowdy, the solution asymptotically behaves like a spatially homogeneous solution to the same equation (at least in some respects). Is it possible to prove something similar in the general case? In the polarized case, the averaged equation is equivalent to the statement that an integral is conserved; there is a constant K such that
∫ tPtd 𝜃 = K. (58 ) S1
Furthermore, this equation can be interpreted as an ODE for ⟨P ⟩. Are there analogous conserved quantities in the general case?

13.3.1 Conserved quantities

As has been noted before, we have the conserved quantities A, B and C given by Equations(25View Equation) – (27View Equation). We shall also use the notation α = A ∕2π, β = B ∕2π, γ = C∕2 π and

∘ ----------- |α2 + 4βγ| δ = ------------. (59 ) 2
As was mentioned in Section 6.4, applying an isometry of hyperbolic space to a solution yields a new solution. It is of interest to know what the corresponding change in the conserved quantities A, B and C is. It turns out that the quantity A2 + 4BC remains unchanged; see the proof of [75Jump To The Next Citation Point, Lemma 8.2, p. 681]. Note that in the spatially homogeneous case,
2 2 2 2P 2 α + 4βγ = 4t (P t + e Q t). (60 )
In other words, A2 + 4BC ≥ 0 for spatially homogeneous solutions. However, in the inhomogeneous case, A2 + 4BC can take any value; see [75Jump To The Next Citation Point, (6.9) – (6.10), p. 676] as well as the adjacent text. Consequently, it seems unlikely that it should be possible to approximate solutions such that 2 A + 4BC is negative by spatially homogeneous solutions.

In practice, it is often convenient to apply an isometry to a solution so that the conserved quantities become as simple as possible. This is achieved in [75Jump To The Next Citation Point, Lemma 8.2, p. 681]:

Lemma 1 Consider a solution to Equations (11View Equation)–(12View Equation). If 2 A + 4BC > 0, there is an isometry such that if A1, B1 and C1 are the constants of the transformed solution, then √ ----------- A1 = − A2 + 4BC and B1 = C1 = 0. If A2 + 4BC = 0, there is an isometry such that the constants of the transformed solution are A1 = B1 = 0 and C1 = 4 π or C1 = 0.

Analyzing the asymptotic behavior of the transformed solution and then transforming back is often more convenient than analyzing the original solution.

13.3.2 Interpreting the conserved quantities as ODEs for the averages

Returning to the question of the asymptotics, we wish to interpret the conserved quantities as ODEs for ⟨P ⟩ and ⟨Q ⟩. Due to [75Jump To The Next Citation Point, Lemma 8.1, p. 680], we have the following result:

Lemma 2 Consider a solution to Equations (11View Equation)–(12View Equation). Then

∫ t⟨P ⟩ = β⟨Q ⟩ − α-+ 1-- te2P(Q − ⟨Q⟩)Q d𝜃, (61 ) t 2 2π S1 t 1 ∫ te⟨P⟩⟨Qt⟩ = βe− ⟨P ⟩ −---e⟨P⟩ (e2P− 2⟨P⟩ − 1)tQtd𝜃, (62 ) 2π S1 ∫ ∫ 2 t- -t- 2P 2 t⟨Qt⟩ = γ + α⟨Q ⟩ − β⟨Q ⟩ + π S1(⟨Q⟩ − Q )Ptd𝜃 + 2π S1 e Qt(Q − ⟨Q ⟩) d𝜃. (63 )

Naively, it would seem natural to consider the integral expressions to be error terms and to interpret what remains as ODEs for the averages. On the other hand, it would then seem that we have too many equations. In the end, the situation turns out to be somewhat more complicated; see below.

Among other things, Equations (61View Equation) – (63View Equation) imply the existence of a constant K such that

K ⟨Pt⟩2 + e2⟨P ⟩⟨Qt ⟩2 ≤ -2 . (64 ) t
Note that this result is similar to the one obtained in the polarized case. Moreover, just as in the polarized case, it is a rather surprising result. Due to Theorem 5 and 57View Equation, the best estimate that can be obtained for
∫ (P 2 + e2PQ2 )d𝜃 S1 t t

is that it decays as t−1. In other words, first taking the average and then taking the square leads to decay of the form t−2. First taking the square and then taking the average leads to decay of the form − 1 t. This behavior reflects the same sort of asymptotics as those characterized by Equation (52View Equation).

Naively estimating the integral on the right-hand side of Equation (61View Equation) leads to the conclusion that it is bounded but no more. Consequently, it seems unreasonable to think of this term as an error term. On the other hand, integrating with respect to time might lead to an improvement. In fact, since ⟨Q ⟩ t decays very quickly, see Equation (64View Equation), replacing Qt with ⟨Qt⟩ in Equation (61View Equation) leads to a term, which tends to zero. Consequently, we can replace Qt by Qt − ⟨Qt ⟩ in (61View Equation) with a small error. Integrating Equation (61View Equation) and using such ideas leads to [75Jump To The Next Citation Point, Lemma 8.9, p. 685]:

Lemma 3 Consider a solution to Equations (11View Equation)–(12View Equation). Then if t > t0 ≥ 1

∫ t[ ] ⟨Pt⟩ − β-⟨Q ⟩ + α-- ds = O (t−1∕2). (65 ) t0 s 2s 0

Note that in the case of B = 0, this result gives detailed information concerning the asymptotics of ⟨P⟩; see [75Jump To The Next Citation Point, Corollary 8.10, p. 685]:

Corollary 1 Consider a solution to Equations (11View Equation)-(12View Equation). If B = 0 there is a constant cP and a T > 0 such that

⟨P⟩ + α-ln t − c = O(t−1∕2) 2 P

for all t ≥ T.

Clearly, Lemma 3 yields important information concerning the asymptotics. Is it possible to apply similar ideas to Equations (62View Equation) and (63View Equation)? It turns out to be necessary to combine both equations in order to obtain a single equation for ⟨Q ⟩. The problem is the last term in Equation (62View Equation) and the second to last term in Equation (63View Equation). However, combining partial integrations, Taylor expansions in the last term in Equation (62View Equation) with various estimates, such as Equation (64View Equation), leads to [75Jump To The Next Citation Point, Lemma 8.8, p. 684]:

Lemma 4 Consider a solution to Equations (11View Equation)–(12View Equation) and let f ∈ C ∞ (ℝ+, ℝ) satisfy

−⟨P⟩ −⟨P⟩ K--- |e f| ≤ K and |e ft| ≤ t1∕2 (66 )
for t ≥ T. Then, if t ≥ t0 ≥ T,
∫ t [ ] f 2⟨Qt⟩ − γ-− α⟨Q ⟩ + β-⟨Q ⟩2 − β-e−2⟨P ⟩ ds = O (t−1∕2). (67 ) t0 s s s s 0

In the case of B = 0 it is convenient to apply this result with f = t−α∕2. This leads to [75Jump To The Next Citation Point, Proposition 8.11, p. 685]:

Proposition 5 Consider a solution to Equations (11View Equation)–(12View Equation). If B = 0, then there is a constant cQ and a T > 0 such that for t ≥ T,

| ( ) | |⟨P⟩ γ- | −1∕2 |e ⟨Q⟩ + α − cQ| ≤ Kt (68 )
if α ⁄= 0 and
| | | γ- | − 1∕2 |⟨Q ⟩ − 2 ln t − cQ | ≤ Kt (69 )
if α = 0.

In case of B = 0, we consequently have detailed information concerning the asymptotic behavior of ⟨Q ⟩ as well. Furthermore, as was observed earlier, given a solution with the property that A2 + 4BC ≥ 0, there is an isometry of hyperbolic space such that the transformed solution is such that the corresponding B equals zero. Thus, the only case that remains to be analyzed is A2 + 4BC < 0. This case is more complicated (but more interesting). We shall therefore omit a description of the analysis.


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