A Strong Averaging Principle for Lévy Diffusions in Foliated Spaces with Unbounded Leaves

Abstract

This article extends a strong averaging principle for Lévy diffusions which live on the leaves of a foliated manifold subject to small transversal Lévy type perturbation to the case of non-compact leaves. The main result states that the existence of p-th moments of the foliated Lévy diffusion for $p\geqslant 2$ and an ergodic convergence of its coefficients in L ^p implies the strong L ^p convergence of the fast perturbed motion on the time scale t∕ε to the system driven by the averaged coefficients. In order to compensate the non-compactness of the leaves we use an estimate of the dynamical system for each of the increments of the canonical Marcus equation derived in da Costa and Högele (Potential Anal 47(3):277–311, 2017), the boundedness of the coefficients in L ^p and a nonlinear Gronwall-Bihari type estimate. The price for the non-compactness are slower rates of convergence, given as p-dependent powers of ε strictly smaller than 1∕4.

Appendix

Proposition 16.13 (Pachpatte [23])

Let u, f, g and h be nonnegative continuous functions defined on $\ensuremath {\mathbb {R}}^+$ . Let v be a continuous non-decreasing subadditive and submultiplicative function defined on $\ensuremath {\mathbb {R}}^+$ and v(u) > 0 on (0, ∞). Let e, ϕ be continuous and nondecreasing functions defined on $\ensuremath {\mathbb {R}}^+$ with p being strictly positive and ϕ(0) = 0. If

$$\displaystyle \begin{aligned} u(t) \leqslant e(t) + g(t) \int_0^t f(s) u(s) ds + \phi\Big(\int_0^t h(s) v(u(s)) ds\Big) \end{aligned} $$

for all $t\geqslant 0$ , then for any $0 \leqslant t \leqslant t_2$

$$\displaystyle \begin{aligned} u(t) \leqslant a(t) \Big[e(t) + \phi\Big(F^{-1}\big(F(A(t)) + \int_0^t h(s) v(a(s)) ds \big)\Big)\Big], \end{aligned} $$

where

$$\displaystyle \begin{aligned} a(t) &:= 1+ g(t) \int_0^t f(s) \exp\Big(\int_s^t g(\ensuremath{\sigma}) f(\ensuremath{\sigma}) d\ensuremath{\sigma}\Big)ds,\\ A(t) &:= \int_0^t h(s) v(a(s) e(s)) ds,\\ F(t) &:= \int_{0}^t \frac{ds}{v(\phi(s))}, \end{aligned} $$

F ⁻¹ is the inverse of F and $t_2\in \ensuremath {\mathbb {R}}^+$ such that

$$\displaystyle \begin{aligned} F(A(t)) + \int_0^t h(s) v(a(t)) ds \in \operatorname{\mathrm{dom}}(F^{-1}) \qquad \mathit{\mbox{ for all }}0\leqslant t\leqslant t_2. \end{aligned}$$

In the following special case of coefficients it is possible to drop the continuity assumption on u.

Corollary 16.14

Let Ψ a non-negative, measurable, increasing function and h be nonnegative, continuous, increasing function on the interval [0, T] satisfying for $p\geqslant 2$ , ε > 0, c > 0 and any t ∈ [0, T] the inequality

$$\displaystyle \begin{aligned} \varPsi(t) \leqslant \varepsilon c t^p + \varepsilon c \Big(\int_0^t \varPsi(s) + \varPsi(s)^{\frac{p-1}{p}} ds\Big), \qquad t\in [0, T]. \end{aligned} $$

(16.49)

Then there is a constant k > 0 such that for any ε ₀ ∈ (0, 1] such that ε ₀T < k we have for all t ∈ [0, T] and ε ∈ (0, ε ₀]

$$\displaystyle \begin{aligned} \varPsi(t) \leqslant C \Big(\varepsilon t^p+t^p (\varepsilon t)^{\frac{p-1}{p}}\Big). \end{aligned} $$

Proof

For e(t) = cεt ^p, g ≡ 1, f, h ≡ εc, ϕ(t) = t, $w(t) = t^{\frac {p-1}{p}}$ we calculate the coefficients of Proposition 16.13

$$\displaystyle \begin{aligned} a(t) &:= 1+ \varepsilon c \int_0^t \exp(\varepsilon c (t-s)) ds = \exp(\varepsilon c t) \end{aligned} $$

and in the limit of εt being small (εt ≪ 1) we have

$$\displaystyle \begin{aligned} \varepsilon \int_0^t a(s)^{\frac{p-1}{p}} ds &= \varepsilon t \Big(\frac{\exp(c \frac{p-1}{p} \varepsilon t) -1}{c \frac{p-1}{p} \varepsilon t}\Big)\leqslant_{\varepsilon t\ll 1} 2 \varepsilon t. \end{aligned} $$

Applying the change of parameter r = εs it follows that

$$\displaystyle \begin{aligned} A(t) & := \int_0^t \exp(\varepsilon c \frac{p-1}{p} s) (e(s))^{\frac{p-1}{p}} ds = \int_0^t \exp( c \frac{p-1}{p} \varepsilon s ) (c \varepsilon s^p)^{\frac{p-1}{p}} ds \\ &= \varepsilon^{\frac{p-1}{p}}\int_0^{\varepsilon t} \exp( c \frac{p-1}{p} r ) c^{\frac{p-1}{p}}\left(\frac{r}{\varepsilon}\right)^{p-1} \frac{dr}{\varepsilon} \leqslant t \frac{1}{\varepsilon^p t}\int_0^{\varepsilon t} \exp( c \frac{p-1}{p} r ) (c r)^{\frac{p-1}{p}} dr \\ &\leqslant_{\varepsilon t \ll 1} 2 t \exp( c \frac{p-1}{p} \varepsilon t ) (c \varepsilon t)^{\frac{p-1}{p}} \leqslant C_1 t \exp( c \frac{p-1}{p} \varepsilon t ) (\varepsilon t)^{\frac{p-1}{p}}. \end{aligned} $$

Finally, we obtain

$$\displaystyle \begin{aligned} F(t) := \int_{0}^t s^{-\frac{p-1}{p}} ds = p r^{\frac{1}{p}} \qquad \mbox{ and }\quad F^{-1}(t) := \frac{t^p}{p^p}. \end{aligned}$$

In the sequel we follow the proof of Theorem 2.4.2 in Pachpatte [23] and define the continuous, positive, non-decreasing function

$$\displaystyle \begin{aligned} n(t) := e(t) + \phi\Big(\int_0^t h(s) w(u(s)) ds\Big) = e(t) + \varepsilon c \int_0^t h(s) u(s)^{\frac{p-1}{p}} ds, \qquad t\geqslant 0, \end{aligned}$$

such that inequality (16.49) can be restated as

$$\displaystyle \begin{aligned} u(t) \leqslant n(t) + g(t) \int_0^t f(s) u(s) ds = e(t) + \varepsilon c \int_0^t u(s) ds . \end{aligned} $$

It is well-known, see for instance [1], that this integral estimate implies the following Gronwall-Bellmann inequality also in the case of u being merely positive measurable. The main reason is that the integral is absolutely continuous with a bounded density. This result yields

$$\displaystyle \begin{aligned} u(t) \leqslant a(t) n(t), \qquad t\geqslant 0. \end{aligned}$$

The remainder of the proof of Theorem 2.4.2 in [23] does use the continuity of u and remains intact.