By the Lipschitz assumption(A2)we have
kU(·,0;q)u0−U(·,0; ˜q)u0kC(Iε,X)≲kq−q˜kC(Iε) ku0k+λ(Tε) . Thus for the initial Lipschitz estimate:
kSε(q)−Sε(˜q)kC(Iε)≲εTεkq−q˜kC(Iε)(1 +λ(Tε)).
With Tε sufficiently small, this implies that Sε is a contraction and hence Banach’s fixed-point theorem yields existence and uniqueness ofqε.
For the breakdown result, note that by g’s boundedness, limt↑Tεqε(t) ex-ists. If qε(Tε) ∈ Q we could extend qε beyond Tε locally, contradicting the
maximality of Tε. Henceqε(Tε)∈∂Q. □
3.4 Proof of Multiscale Convergence
We use the following classical result for averaging of an ordinary differential equations with periodic right-hand side, cf. [SVM07, Theorem 2.8.1]:
Theorem 3.4.1. LetQ⊂Rn be connected, bounded and open. Letf: R×Q→ Rn be 1-periodic, continuous in its first and uniformly Lipschitz continuous in its second argument. Then for anyx0∈Qthere is aT >0such that a solution x0: [0, ε−1T]→Qof
x′0(t) =εf(x0(t)) :=ε Z 1
0
f(s, x0(t))ds, x0(0) =x0
exists. The solution x0 is a first order approximation of the solution xεof x′ε(t) =εf(t, xε(t)), xε(0) =x0
in the sense that there is an ε∗>0 such that for any0 < ε < ε∗ the solution xεexists for all t∈[0, ε−1T] and there holds
max
t∈[0,ε−1T]|xε(t)−x0(t)|≲ε.
Proof. The existence of limit solutions on the O(ε−1)-timescale follows by rescaling to slow time, τ := εt, and using that x 7→ R1
0 f(s, x)ds is Lips-chitz continuous, since f is. The main ingredient for the averaging result is a comparison ofx0 andxεby keepingxεfixed on small intervals, see [Art07] for
details. □
Corollary 3.4.2. There exists T > 0 independent of ε such that the limit equation
q′0(t) =ε Z 1
0
g(q0(t), uπ(s;q0(t)))ds, q0(0) =q0 (3.2) has a unique solutionq0: [0, ε−1T]→Q. Furthermore, there is anε∗>0such that for all0< ε < ε∗ the solutionqε,π of
qε,π′ (t) =εg(qε,π(t), uπ(t;qε,π(t))), qε,π(0) =q0. (3.7) exists for t∈[0, ε−1T] with
max
t∈[0,ε−1T]|qε,π(t)−q0(t)|≲ε.
Proof. Define G(t, q) := g(q, uπ(t;q)). Then qε,π′ (t) = εG(t, qε,π(t)) and G is continuous and1-periodic intand Lipschitz continuous inq, the latter following from the Lipschitz continuity of g by assumption and ofuπ byLemma 3.3.2.
The time-average ofG is the right-hand side of the limit equation and hence
we can apply the averaging theorem. □
With this result, it remains to prove that qε is close to qε,π. For this, we first make the additional assumption that g is bounded by M > 0. By careful analysis of the constants involved, we will lift this restriction later on.
To this end, let the constant in ≲ be independent ofM. We will first prove that uε(t) := U(t,0;qε)u0 reaches anε-neighborhood of uπ(t;qε(t), ε) after a boundary layer of sizeO(|lnε|)and stays there, i.e.uεis slaved touπ. Lemma 3.4.3. Let K ∈N and (ak)Kk=0 be non-negative with ak ≤bak−1+c forb, c≥0andk= 1, . . . , K. Then ak≤bka0+cPk−1
j=0bj fork= 1, . . . , K.
Proof. Follows by induction. □
Lemma 3.4.4. There existsC >0 such that there holds
kU(t,0;qε)u0−uπ(t;qε(t))k≲e−αtku0−u0π(q0)k+εM for all t∈Iε with constants independent of|Iε|.
Proof. First, using the assumption of exponential stabilityA1we get kU(t,0;qε)u0−U(t,0;qε)u0π(q0)k≲e−αtku0−u0π(q0)k.
Hence it remains to prove that kU(t,0;qε)u0π(q0)−uπ(t;qε(t))k ≲ εM. For this, letL∈Nand the constant in≲in the following be independent ofL. Let Ik:= [tk, tk+1)fork= 0, . . . , K with0 =t0< t1< . . . < tK+1=Tε such that
|Ik|=Lfork= 0, . . . , K−1 and|IK| ≤L.
Let t ∈ Ik for k = 0, . . . , K and define qεk := qε(tk). Note that by the boundedness assumption ongand sincet−tk ≤L we have
|qε(t)−qεk|=|qε(t)−qε(tk)| ≤ε Z t
tk
|g(qε(s), uε(s))|ds≤εLM. (3.8)
We estimate
kU(t,0;qε)u0π(q0)−uπ(t;qε(t))k
≤ kU(t,0;qε)u0π(q0)−uπ(t;qkε)k+kuπ(t;qε(t))−uπ(t;qkε)k. The Lipschitz property ofuπ byLemma 3.3.2andineq. (3.8) yields
kuπ(t;qε(t))−uπ(t;qkε)k≲|qε(t)−qεk|≲εLM and hence
kU(t,0;qε)u0π(q0)−uπ(t;qε(t))k ≤ kU(t,0;qε)u0π(q0)−uπ(t;qεk)k+εLM.
To estimate the first term on the right we add and subtractU(t, tk;qε)u0π(qkε):
kU(t,0;qε)u0π(q0)−uπ(t;qkε)k ≤ kU(t,0;qε)u0π(q0)−U(t, tk;qε)u0π(qkε)k +kU(t, tk;qε)u0π(qkε)−uπ(t;qεk)k. (3.9) For the first term, we split the evolution on[0, t]into[0, tk]and the remainder [tk, t]. Using the assumption of exponential stability(A1)this results in
kU(t,0;qε)u0π(q0)−U(t, tk;qε)u0π(qεk)k
≲e−α(t−tk)kU(tk,0;qε)u0π(q0)−u0π(qεk)k. (3.10) For the second term the periodicity andtk∈Nimply
uπ(t;qkε) =U(t, tk;qεk)uπ(tk;qεk) =U(t, tk;qkε)u0π(qkε) and thus the Lipschitz assumption(A2)yields
kU(t, tk;qε)u0π(qεk)−uπ(t;qεk)k≲kqε−qkεkC(tk,t) ku0π(qkε)k+λ(L)
≲εL(1 +λ(L))Mkuπ(·;qkε)kCπ(X)
≲εLλ(L)M,
where we used Lemma 3.3.3 and assumed, without restriction of generality, that λ(L)≥1. Using the last estimate andineq. (3.10)inineq. (3.9) leads to
kU(t,0;qε)u0π(q0)−uπ(t;qε(t))k
≲e−α(t−tk)kU(tk,0;qε)u0π(q0)−u0π(qεk)k+εLλ(L)M (3.11) for anyt∈Ik. By passing to the limitt↑tk+1 fork < K this yields
kU(tk+1,0;qε)u0π(q0)−uπ(tk+1;qk+1ε )k
≲e−αLkU(tk,0;qε)u0π(q0)−u0π(qkε)k+εLλ(L)M.
This is an inequality of the form of Lemma 3.4.3. Together with kU(tk,0;qε)u0π(q0)−u0π(qεk)k= 0
fork= 0, an application of this lemma yields fork= 0, . . . , K that kU(tk,0;qε)u0π(q0)−uπ(tk;qεk)k≲εLλ(L)M
k−1
X
j=0
(Ce−αL)j
with a constantCindependent of LandM. We may chooseLindependent of M such thatCe−αL<1. This implies, with≲now depending onL, that
kU(tk,0;qε)u0π(q0)−uπ(tk;qεk)k≲εM
fork= 0, . . . , K. Using this estimate inineq. (3.11)yields the claimed bound
for allt∈Ik and concludes the proof. □
We now use the previous result to establish thatqε is close to qε,π, where qε,π was defined, repeated here for convenience, as solution of
qε,π′ (t) =εg(qε,π(t), uπ(t;qε,π(t))), qε,π(0) =q0. (3.7) Lemma 3.4.5. LetT >0 be fixed and|Iε| ≤ε−1T. Then we have
maxt∈Iε|qε(t)−qε,π(t)|≲εM, with constant depending onT.
Proof. Using the Lipschitz continuity ofg anduπ, we have fort∈Iε that
For the second term we use our estimate fromLemma 3.4.4to see that ε Lemma 3.3.3. By Lipschitz continuity ofuπ there holds for the third term:
ε and application of Gronwall’s inequality yields the claimed estimate
|qε(t)−qε,π(t)| ≤εMeCεt ≲εM,
Proof. By the previous result and Corollary 3.4.2 we see that the estimate is valid on a possibly ε-dependent interval Iε for ε small enough. It remains to prove that the interval of existence may be extended to [0, ε−1T]. Since the trajectory of the averaged equation Γ0 := q0([0, ε−1T]) is compact and contained inQ, we can findη >0such that aη-neighborhood ofΓ0is contained in Q. Choosing ε∗ small enough such that εM ≲ η for all 0 < ε < ε∗ the previous corollary implies thatqε cannot leave this neighborhood (and hence Q) for any such ε. ByLemma 3.3.4solutions only cease to exist if qε crosses the boundary ofΩ. Hence for all0< ε < ε∗ the solutionqεmust exist at least
untilε−1T. □
A simple consequence of our results is an estimate on the fast component, albeit limited by a boundary layer att= 0:
Lemma 3.4.7. For any δ∈(0,T)there exists ε∗ =ε∗(δ, M)>0, such that for all 0< ε < ε∗ there holds
max
t∈[ε−1δ,ε−1T]kU(t,0;qε)u0−uπ(t;q0(t))k≲εM.
Proof. ByLemma 3.4.4, we have fort∈[0, ε−1T]that
kU(t,0;qε)u0−uπ(t;qε(t))k≲e−αtku0−u0πk+εM.
Using the Lipschitz continuity of uπ and the estimate for qε−q0 from the Corollary 3.4.6, we hence get
kU(t,0;qε)u0−uπ(t;q0(t))k
≤ kU(t,0;qε)u0−uπ(t;qε(t))k+kuπ(t;qε(t))−uπ(t;q0(t))k
≲e−αtku0−u0πk+εM +|qε(t)−q0(t)|
≲e−αtku0−u0πk+εM.
Fort≥t∗(ε) =−α−1ln(ε(1+M))we have e−αt≤ε(1+M). Sinceε−1δ > t∗(ε) for all 0 < ε < ε∗ if ε∗ =ε∗(δ, M) is sufficiently small, the claimed estimate
follows. □
Proof for unbounded g
We will now prove that one can drop the assumption ofg being bounded. We reduce this to the previous case by replacing g with a bounded function gR outside a ball of radius R in X. Careful analysis of the preceding estimates then shows that forRsufficiently large andεsufficiently small this change has no effect on the solutions.
Lemma 3.4.8. ForR >0 definePR:X →X byPR(u) :=uforkuk ≤Rand PR(u) :=R∥uu∥ for kuk> R. Then PR is Lipschitz continuous with constant 1.
Proof. Letu1, u2∈Xand assume without restriction of generality thatku1k ≤ ku2k. If ku2k ≤ R we have kPR(u1)−PR(u2)k = ku1−u2k. For the case ku2k> Randku1k< Rletv∈X denote the orthogonal projection ofu1onto Ru2. Then it is easy to see that kv−PR(u2)k ≤ kv−u2kand hence
kPR(u1)−PR(u2)k2=ku1−PR(u2)k2≤ ku1−vk2+kv−u2k2=ku1−u2k2.
If ku1k > R first assume that ku1k =ku2k, i.e. there existsα > Rˆ such that u1= ˆα∥uu1
1∥ andu2= ˆα∥uu2
2∥. Sinceα7→αk∥uu11∥−∥uu22∥kis increasing the claim follows since α > R. In the last case,ˆ R <ku1k <ku2k, we use the already proven cases together withPR◦P∥u1∥=PR to see that
ku1−u2k ≥ ku1−P∥u1∥(u2)k ≥ kPR(u1)−[PR◦P∥u1∥](u2)k
≥ kPR(u1)−PR(u2)k. □
Lemma 3.4.9. For any R > 0 there exists a bounded, Lipschitz continuous function gR:Q×X → Rn such that gR(q, u) = g(q, u) for all q ∈ Q and kuk ≤Rwith the same Lipschitz constant as g.
Proof. DefinegR(q, u) :=g(q, PR(u)). ThengR(u) =g(u)forkuk ≤Rand the Lipschitz continuity follows from the previous lemma. Since
|gR(q, u)|=|g(q−q0+q0, PR(u−0))|≲|q−q0|+kPR(u)k+|g(q0,0)|≲1 +R, where we used thatQis bounded,gR is bounded. □
Lemma 3.4.10. ForR >0large enough the limit solution q0 solves q′0(t) =ε
Z 1 0
gR(q0(t), uπ(s;q0(t)))ds, q0(0) =q0 for all t∈[0, ε−1T].
Proof. The set Γ :={uπ(s;q0(t)) | s∈ [0,1], t∈ [0, ε−1T]} ⊂X is bounded.
Hence we can choose R >0 large enough such thatkuk ≤Rfor allu∈Γ and henceg(q0(t), uπ(s;q0(t))) =gR(q0(t), uπ(s;q0(t))). □
Now letqε,R denote the solution with right-hand sidegRwhereR >0is at least as large as required by the previous lemma. For the bound M >0 ofg we have M ≲1 +R ≲R for R sufficiently large. Then by the results of the previous section there is anε∗(R)>0such that for all0< ε < ε∗ the solution qε,R exists for[0, ε−1T]and there holds
kqε,R(t)−q0(t)k≲εR,
kU(t,0;qε,R)u0−uπ(t;qε,R(t))k≲e−αtku0−u0πk+εR
for allt∈[0, ε−1T], where we implicitly used thatq0is also the solution of the limit equation forgR. WithR-independent constantsC >0 we get
kU(t,0;qε,R)u0k ≤C ku0−u0πk+kuπ(t;qε,R(t))k+εR
≤C ku0−u0πk+kuπ(t;q0(t))k+|qε,R(t)−q0(t)|+εR
≤C ku0−u0πk+kuπ(t;q0(t))k+εR . Fixing R > 0 such that C ku0−u0πk+kuπ(t;q0(t))k
≤ R2 and making the corresponding ε∗(R) > 0 small enough such that 12 +Cε ≤ 1 holds for all 0< ε < ε∗ we getkU(t,0;qε,R)u0k ≤Rfor allt∈[0, ε−1T]and hence
qε,R(t)′ =εgR(qε,R(t), U(t,0;qε,R)u0) =εg(qε,R(t), U(t,0;qε,R)u0).
But this implies thatqε,Rfor suchRand0< ε < ε∗solves the originaleq.(3.1), henceqε,R =qεand of course the estimates from the previous section still hold.