Controllability and qualitative properties of the solutions to SPDEs driven by boundary Lévy noise

Abstract

In the present paper we are interested in the qualitative properties of the Markovian semigroups ${{\mathcal P }}=({{\mathcal P }}_t)_{t\ge 0}$ associated to the solutions of certain stochastic partial differential equations (SPDEs) with boundary noise. We assume that these problems can be written as an abstract stochastic PDE on a Hilbert space $H$ taking the following form:

$$\begin{aligned} \left\{ \begin{array}{rcl} du(t,x)&{} = &{} A u(t,x)\, dt + B\;\sigma (u(t,x)) \, dL(t),\quad t>0;\\ u(0,x)&{} =&{}x\in H.\end{array}\right. \end{aligned}$$

(1)

Here $L$ is a real-valued Lévy process, $A:D(A)\subset H\rightarrow H$ is an infinitesimal generator of a strongly continuous semigroup, $\sigma :H\rightarrow {\mathbb {R}}$ is a Lipschitz continuous map bounded from below and above, and $B:{\mathbb {R}}\rightarrow H$ a possibly unbounded operator. As typical examples of such stochastic evolution equation we mainly consider and treat the damped wave equation and heat equation both driven by boundary Lévy noise. In this article, we first show that, if the system

$$\begin{aligned} \left\{ \begin{array}{rcl} du(t,x)&{} = &{}A u(t,x)\, dt + B\;v(t)\, dt,\quad t>0;\\ u(0,x)&{} =&{}x\in H\end{array}\right. \end{aligned}$$

(2)

is approximate controllable at time $T>0$ with control $v$, then, under some additional conditions on $B$, $A$ and $L$, the probability measure on $H$ induced by $u(t,x)$ at a given time $t>0$, $x\in H$, is positive on open subsets of $H$. Secondly, we investigate under which conditions on the Lévy process $L$ and on the operators $A$ and $B$ the solution to Eq. (1) is asymptotically strong Feller. It follows from our results that the wave equation with boundary Lévy noise has at most one invariant measure which is non-degenerate.

Appendices

Appendix A: Technical Preliminaries

Let $\lambda$ be the Lebesgue-measure on ${\mathbb {R}}$ and $c:{\mathbb {R}}\rightarrow {\mathbb {R}}$ given by (17). Let $r_0$ as in Hypothesis 1. As in page 8 let $R\ge r_0$ and $g_R$ be as in (22).

In this section we will show that one can find a transformation $\theta ^{(R)}:[0,T]\times {\mathbb {R}}\backslash \{0\}\rightarrow {\mathbb {R}}$ such that for a given mapping $V: [0,T]\rightarrow {\mathbb {R}}$ we have

$$\begin{aligned} -\left( {V(s)}+g_R\right) =\int _{{\mathbb {R}}\setminus B_{\mathbb {R}}(R)} \,[c(z)-c(\theta ^{(R)}(s,z))]\lambda (dz), \quad s\in [0,T]. \end{aligned}$$

(72)

For simplicity, we will first reformulate this problem in the following from. For any $R\ge 10$ and $R\ge r_0$ find a function $\vartheta ^R:{\mathbb {R}}\times {\mathbb {R}}\rightarrow {\mathbb {R}}$, such that

$$\begin{aligned} v=\int _{{\mathbb {R}}\setminus B_{\mathbb {R}}(R)} [c(z)-c(\vartheta ^R(v,z))] \lambda (dz),\quad \forall v\in {\mathbb {R}}. \end{aligned}$$

Setting

$$\begin{aligned} \theta ^{(R)}(s,z) =\vartheta ^R \left( -\left( {V(s)}+g_R\right) ,z\right) , \quad s\in [0,T]. \end{aligned}$$

It is straightforward to calculate that $\theta$ satisfies (72).

To find such a transformation, we perturb the jumps a little bit. By the symmetricity of the Lévy measure, it is then sufficient to consider only positive jumps and only positive values of $v$ in the first step. In the second step we construct a transformation which works for perturbation $v\in {{ {\mathbb {R}}}}$ and positive and negative jumps.

Now, let $\tilde{\delta }\in (0,\frac{2}{\alpha }(\alpha -1))$ and put

$$\begin{aligned} \gamma _1 = { 2\alpha -\alpha \tilde{\delta }-2\over \alpha (2-\alpha +\alpha \tilde{\delta })} \end{aligned}$$

and

$$\begin{aligned} \beta _1=-\frac{1}{2\alpha \gamma _1} \, \left( \alpha -2 \gamma _1 -\alpha \gamma _1 - \sqrt{ (\alpha +2\gamma _1+\alpha \gamma _1)^ 2 - 4\alpha \gamma _1(2-\alpha +2\gamma _1)}\right) . \end{aligned}$$

Let also $\beta _2>-1$ be an arbitrary number and $\gamma _2=2$. Note that $-\beta _1-1=-\tilde{\delta }$ and because $\alpha >1$ we also have $1-\gamma _1(\beta _1+1)\le 0$.

Now define a function $\vartheta ^{+,R}$ by

$$\begin{aligned}{}[0,\infty ) \ni K \mapsto \vartheta ^{+,R} (K) := \int _{{\mathbb {R}}^+} \left( c(z) - c(z+ \varsigma ^{(R)} (K,z))\right) \; dz\in {{ {\mathbb {R}}}}, \end{aligned}$$

(73)

where $\varsigma ^{(R)} :{\mathbb {R}}^ +\times {\mathbb {R}}^ + \rightarrow {\mathbb {R}}^ +$ is defined by

$$\begin{aligned} \varsigma ^{(R)} (K,z) := {\left\{ \begin{array}{ll}\begin{array}{ll} K z ^ {-\beta _1} &{} z \in \left( \frac{4}{3} R K^{\gamma _1},\frac{8}{3} K^ {\gamma _1}R\right) \text{ and } K\ge 1,\\ 0 &{} z \not \in \left( K^{\gamma _1}R, \frac{10}{3} K^{\gamma _1}R\right) \text{ and } K\ge 1, \\ C z ^ {-\beta _2} &{} z \in \left( \frac{4}{3} R , \frac{4}{3} R (1+ K^{\gamma _2} )\right) \text{ and } K< 1, \\ 0&{} z \not \in \left( R, \frac{5}{3} R(1+ K^{\gamma _2} )\right) \text{ and } K< 1, \\ \end{array}\\ \text{ differentiable } \text{ interpolated } \text{ elsewhere }., \end{array}\right. }\end{aligned}$$

(74)

Here, the constant $C>0$ has to be chosen in such a way that ${\mathbb {R}}_0^ +\ni K\mapsto \vartheta ^{+,R} (K)$ is a continuous function. From the definition of $\varsigma ^{(R)}$ we see in particular that for $K\ge 1$ and $z \in \left( K^{\gamma _1}R, \frac{4}{3} K^{\gamma _1}R)\cup (\frac{8}{3} K^{\gamma _1}R, \frac{10}{3} K^{\gamma _1}R\right)$, $\varsigma ^{(R)}_z(K,z)\le K z^{-\beta _1-1}$ and for $K< 1$ and $z \in \left( R, \frac{4}{3} R)\cup (\frac{4}{3} (1+K^{\gamma _2}) R,\frac{5}{3} (1+K^{\gamma _2}) R\right)$, $\varsigma ^{(R)}_z(K,z)\le C z^{-\beta _2-1}$, where $\varsigma ^{(R)}_z$ is the partial derivative of $\varsigma ^{(R)}$ with respect to $z$.

Lemma 6.1

Under the Hypothesis 1 the function $\vartheta ^{+,R} :{{ {\mathbb {R}}}}_0^+\rightarrow {{ {\mathbb {R}}}}_0^+$ is invertible.

Proof

We start by verifying the following properties

• (1) $\vartheta ^{+,R} (K)\in {\mathbb {R}}^+_0$;

• (2) the function ${\mathbb {R}}^+_0\ni K\mapsto \vartheta ^{+,R}(K) \in {\mathbb {R}}^+ _0$ is continuous.

• (3) the function ${\mathbb {R}}^+\ni K\mapsto \vartheta ^{+,R}(K) \in {\mathbb {R}}^+_0$ is injective.

• (4) the function ${\mathbb {R}}^+\ni K\mapsto \vartheta ^{+,R}(K) \in {\mathbb {R}}^+ _0$ is surjective.

It will follow from (2) to (4) that the function $\vartheta ^{+,R}$ is invertible.

Item (1) is clear by the definition of $c$. In order to show Items (2) and (3) we take into account that the function ${\mathbb {R}}^+_0 \ni K\mapsto \vartheta ^{+,R} (K) \in {\mathbb {R}}^ +_0$ is strictly increasing and continuous.

Since $\vartheta ^{+,R} (0)=0$ and $\vartheta ^{+,R}$ is continuous on ${\mathbb {R}}^+_0$, item (4) will follow if we can show that $\lim _{K\rightarrow \infty } \vartheta ^{+,R}(K) =\infty$. For this purpose let us first recall that for $z>0$ we have $c(z)=U^{-1}(z)$, where $U:{\mathbb {R}}^+\rightarrow {\mathbb {R}}_0^+$ denotes the tail integral given by

$$\begin{aligned} U(z)=\nu (z,\infty ),\quad z>0. \end{aligned}$$

From Hypothesis 1 it follows that there exist constants $C_1>0$ and $C_2>0$ such that

$$\begin{aligned} c(z) \ge C_1 z^ {-\frac{1}{\alpha }},\quad \forall \, z\ge C_2. \end{aligned}$$

(75)

In fact, since by definition $U(\cdot )=\nu (\cdot ,\infty )$, and by Hypothesis 1, there exists $K_2>0$ such that for all $z\in (0,r_0)$ we have

$$\begin{aligned} U(z) = \nu (z,r_0) + \nu (r_0,\infty ) \ge K_2\int _{z}^ {r_0} x^ {-1-\alpha }\, dy + \nu (r_0,\infty ) \\ \ge \alpha K_2 \left( z^{-\alpha } - {r_0^{-\alpha }} \right) + \nu (r_0,\infty ). \end{aligned}$$

Thus,

$$\begin{aligned} {U(z) - \nu (r_0,\infty ) \over \alpha K_2} +r_0^ {-\alpha } \ge z ^ {-\alpha } \end{aligned}$$

which implies that

$$\begin{aligned} z^ \alpha \ge {\alpha K_2\over {U(z) - \nu (r_0,\infty ) } +\alpha K_2r_0^ {-\alpha }} . \end{aligned}$$

Hence,

$$\begin{aligned} z \ge \left( {\alpha K_2\over {U(z) - \nu (r_0,\infty ) } +\alpha K_2r_0^ {-\alpha }}\right) ^ \frac{1}{\alpha }. \end{aligned}$$

If $\alpha K_2r_0^ {-\alpha } \le \nu (r_0,\infty )$, then (75) follows. If $\alpha K_2r_0^ {-\alpha } > \nu (r_0,\infty )$, then one can show by using elementary calculations that for any $c>0$ and $K\in (0,1)$

$$\begin{aligned} {1\over (y+c)^ \frac{1}{\alpha }} \ge {K\over y^ \frac{1}{\alpha }} ,\quad \forall y\ge {c\over 1-K^ \alpha }. \end{aligned}$$

Thus (75) is also valid.

By similar arguments one can show that there exist constants $\tilde{C}_1>0$ and $\tilde{C}_2>0$ such that

$$\begin{aligned} c(z) \le \tilde{C}_1 z^ {-\frac{1}{\alpha }},\quad \forall \, z\ge \tilde{C}_2. \end{aligned}$$

(76)

In fact, again by Hypothesis 1, there exists a constant $\tilde{K}_2>0$ such that for all $\tilde{z}\in (0,r_0)$ we have

$$\begin{aligned} U(z)= & {} \nu (z,r_0) + \nu (r_0,\infty ) \le \tilde{K}_2\int _{r_0}^ z x^ {-1-\alpha }\, dy + \nu (\tilde{r}_0,\infty ) \\\le & {} \alpha \tilde{K}_2 \left( z^{-\alpha } - {\tilde{r}_0^{-\alpha }} \right) + \nu (\tilde{r}_0,\infty ). \end{aligned}$$

Thus, by the same calculations as before we can infer that

$$\begin{aligned} z \le \left( {\alpha \tilde{K}_2\over {U(z) - \nu (\tilde{r}_0,\infty ) } +\alpha \tilde{K}_2\tilde{r}_0^ {-\alpha }}\right) ^ \frac{1}{\alpha }. \end{aligned}$$

If $\alpha K_2r_0^ {-\alpha } \ge \nu (r_0,\infty )$, then the assertion follows. If $\alpha K_2r_0^ {-\alpha } < \nu (r_0,\infty )$, then again it follows by direct calculations give that for any $c>0$ and $K>1$ we have

$$\begin{aligned} {1\over (y+c)^ \frac{1}{\alpha }} \ge {K\over y^ \frac{1}{\alpha }} ,\quad \forall y\ge {cK^ \alpha \over K^ \alpha -1}. \end{aligned}$$

Hence, (76) follows.

Next, for $z>0$ the mapping $U(\cdot )=\nu (\cdot , \infty )$ is invertible and its inverse $U^{-1}(\cdot )$ coincides with $c(\cdot )$ on $(0,\infty )$. Therefore, there exists a constant $C>0$ such that $c$ is differentiable for all $z\ge C$ and

$$\begin{aligned} c^ \prime (z) = {1\over U'(y)\Big |_{y=U^ {-1}(z)}}, \end{aligned}$$

for any $z\ge C$. We now derive a lower estimate for $c^ \prime$. For this aim observe that Hypothesis 1 implies that there exist two constants $C_3=\frac{1}{K_1}>0$ and $C_4=K_2>0$ such that

$$\begin{aligned} {1\over U'(y)} \ge {C_3\over y^{-\alpha -1}}, \quad \text{ for } \text{ all } 0<y\le C_4. \end{aligned}$$

(77)

In fact, since $U(y)=\nu (y,\infty )$ we know that $U'(y)=k(y)$. By Hypothesis 1 we have

$$\begin{aligned} k(y)\le K_2 y^ {-\alpha -1} ,\quad \forall y\le r_0. \end{aligned}$$

Hence

$$\begin{aligned} {K_2 \over k(y)} \ge {1\over y^ {-\alpha -1}} ,\quad \forall y\le r_0, \end{aligned}$$

which implies (77). Now, let $z\ge U(r_0)$. Then $U^ {-1}(z)\le r_0$ and

$$\begin{aligned} c^ \prime (z) \ge {1\over K_2 y^ {-\alpha -1}\Big |_{y=U^ {-1}(z)}}. \end{aligned}$$

Estimate (75) gives that there exists two constants $C_5>0$ and $C_6>0$ such that

$$\begin{aligned} c^ \prime (z) \ge {C_5\over z^{1+\frac{1}{\alpha }}} ,\quad \text{ for } \text{ all } z\ge C_6. \end{aligned}$$

(78)

Now, we can start with proving Item (4). For simplicity, we put $r_1=\frac{4}{3} R$ for the rest of the proof of (4). For any $K>1$ we have

$$\begin{aligned} \vartheta ^{+,R} (K)= & {} \int _0 ^ \infty \left[ c(z) - c(z+\varsigma ^{(R)} (K,z))\right] \; dz\\= & {} \int _{ r_1K^{\gamma _1}} ^ {K^{\gamma _1}2 r_1} \int _z ^ { z+\varsigma ^{(R)} (K,z)} \; \,c'(y) \; dy\; dz\\\ge & {} \int _{K^ {\gamma _1}r_1} ^ {K^ {\gamma _1}2 r_1} \varsigma ^{(R)} (K,z)\; \, c'(z+ \varsigma ^{(R)} (K,z)) \; dz . \end{aligned}$$

For $\tilde{\gamma _1}= \gamma _1(1+\beta _1)$ we can derive from the estimate (78) that

$$\begin{aligned} \ldots\ge & {} \delta _0 K\; \int _{ \frac{4}{3} K^ {\gamma _1}R} ^ {\frac{8}{3} K^ {\gamma _1}R} { z ^{-{\beta _1}}\over \left( z+{K\over z ^ {\beta _1}}\right) ^ {\frac{1}{\alpha }+1}}\; dz \ge \delta _0 K\; \int _{ r_1K^ {\gamma _1}} ^ {K^ {\gamma _1}2 r_1} { z ^{-{\beta _1}+{\beta _1} ( {\frac{1}{\alpha }+1}) }\over \left( z ^ {1+{\beta _1}}+K\right) ^ {\frac{1}{\alpha }+1}}\; dz \\= & {} \delta _0\, K \, K ^ {- \frac{1+\alpha }{\alpha }} \int _{ r_1K^ {\gamma _1}} ^ {K^{\gamma _1}2 r_1} { z ^{{\beta _1} \over \alpha } \over \left( {z ^ {1+{\beta _1}} \over K} +1 \right) ^ {\frac{1}{\alpha }+1}}\; dz\\= & {} K ^{-\frac{1}{\alpha }} \int _{r_1^ {1+{\beta _1}} K^{\tilde{\gamma _1}}} ^ { (2r_1)^{1+{\beta _1}} K^{\tilde{\gamma _1}}} { \left( K u \right) ^{{1\over 1+{\beta _1}}\, {\beta _1} \over \alpha } \over \left( {u } +1 \right) ^ {\frac{1}{\alpha }+1}}\; K ^{\frac{1}{{\beta _1}+1}} u ^{\frac{1}{{\beta _1}+1} -1} du \\= & {} K ^{ - (1+{\beta _1}) +{\beta _1} +\alpha \over \alpha (1+{\beta _1})} \int _{{r_1^ {1+{\beta _1}}K^{\tilde{\gamma _1}}}} ^{(2r_1)^{1+{\beta _1}}K^{\tilde{\gamma _1}}} { u ^{ {\beta _1} +\alpha - ({\beta _1}+1)\alpha \over ({\beta _1}+1)\alpha } \over \left( {u } +1 \right) ^ {\frac{1}{\alpha }+1}}\; du \\= & {} K ^{ \alpha - 1 \over \alpha (1+{\beta _1})} \int _{{r_1^ {1+{\beta _1}}K^{\tilde{\gamma _1}}}} ^ {(2r_1)^{1+{\beta _1}}K^{\tilde{\gamma _1}}} { u ^{ {\beta _1} (1-\alpha ) \over ({\beta _1}+1)\alpha } \over \left( {u } +1 \right) ^ {\frac{1}{\alpha }+1}}\; du. \end{aligned}$$

Since ${ {\beta _1} (1-\alpha ) \over ({\beta _1}+1)\alpha }<{\frac{1}{\alpha }+1}$, there exists a constant $C=C(\beta _1,\gamma _1,\alpha )>0$, such that

$$\begin{aligned} { u ^{ {\beta _1} (1-\alpha ) \over ({\beta _1}+1)\alpha } \over \left( {u } +1 \right) ^ {\frac{1}{\alpha }+1}}\;\ge C\, u^{-{1 + \alpha + 2 \alpha \beta _1\over \alpha (1+\beta _1)}}= C\, u^{-{1 + \alpha \over \alpha (1+\beta _1)}-1}, \quad u\ge 1. \end{aligned}$$

Integrating gives

$$\begin{aligned} \ldots\ge & {} C\, r_1^{2\beta _1-\alpha \beta _1 +1} K^{\Gamma }, \end{aligned}$$

(79)

where, for simplicity, we have put

$$\begin{aligned} \Gamma = {\alpha -1 -\gamma _1(1+\beta _1)(\alpha \beta _1+1) \over \alpha (1+\beta _1)} = {\alpha -\alpha \beta _1^2 \gamma _1-\alpha \beta _1\gamma _1-\beta _1\gamma _1 -\gamma _1-1 \over \alpha (1+\beta _1)}.\nonumber \\ \end{aligned}$$

(80)

It follows from (79) that

$$\begin{aligned} \lim _{K\rightarrow \infty } \int _{r_1} ^ \infty \left[ c(z) - c(z+\varsigma ^{(R)} (K,z))\right] \; dz =\infty , \end{aligned}$$

which concludes the proof of Item (4). From (1) to (4) it follows that the function $\vartheta ^{+,R} :{\mathbb {R}}^+\rightarrow {\mathbb {R}}^+$ defined by (73) is invertible. $\square$

We also state the following remark which is very crucial for our analysis.

Remark 6.2

For any $R\ge r_0$ and $0<K\le 1$ we have

$$\begin{aligned} \int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz\ge C(R) \, K. \end{aligned}$$

(81)

Proof of Estimate (81)

As above we set $r_1=\frac{4}{3}R$. For any $0<K<1$ we have

$$\begin{aligned}&\int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz\\&\quad = \int _{r_1}^{r_1(1+K)^{\gamma _2}} \varsigma ^{(R)} (K,z) \frac{d}{dz}c(z+\varsigma ^{(R)}(K,z)) dz \; du, \end{aligned}$$

which implies the existence of a positive constant $C(r_1, \alpha , \gamma _2,\beta _2)$ such that

$$\begin{aligned} \int _0 ^ \infty \left[ c(z) \!-\! c(z\!+\! \varsigma ^{(R)} (K,z))\right] \; dz&\!=\! C(r_1, \alpha , \gamma _2,\beta _2) \int _{r_1}^{r_1(1\!+\!K^{\gamma _2})}\frac{z^{-\beta _2}}{(z\!+\!z^{-\beta _2})^{1+\frac{1}{\alpha }}}dz \\&\!=\! C(r_1, \alpha , \gamma _2,\beta _2) \int _{r_1}^{r_1(1+K^{\gamma _2})}\frac{z^{-\beta _2{\alpha -1\over \alpha }}}{(z^{1+\beta _2}\!+\!1)^{1+\frac{1}{\alpha }}}dz. \end{aligned}$$

By change of variables we get that

$$\begin{aligned}&\int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz \\&\quad \ge C(r_1, \alpha , \gamma _2,\beta _2) \int _{{r_1}^{1+\beta _2}}^{r_1^{1+\beta _2} (1+K^{\gamma _2})^{(1+\beta _2)}}\frac{u^{-\frac{\beta _2(\alpha -1)}{\alpha (\beta _2+1)}}}{(u+1)^{1+\frac{1}{\alpha }}} u^{\beta _2\over 1+\beta _2} \, du \\&\quad \ge C(r_1, \alpha , \gamma _2,\beta _2) \int _{{r_1}^{1+\beta _2}}^{r_1^{1+\beta _2} (1+K^{\gamma _2})^{(1+\beta _2)}}\frac{u^{\frac{\beta _2}{\alpha (\beta _2+1)}}}{(u+1)^{1+\frac{1}{\alpha }}} \,du. \end{aligned}$$

Since ${r_1}^{1+\beta _2}\le u\le r_1^{1+\beta _2}(1+K^{\gamma _2})^{(1+\beta _2)}$, we get

$$\begin{aligned}&\int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz \\&\quad \ge C(r_1, \alpha , \gamma _2,\beta _2) \frac{1}{(1+2r_1^{1+\beta _2})^{(1+\frac{1}{\alpha }) }} \int _{{r_1}^{1+\beta _2}}^{[r_1(1+K^{\gamma _2})]^{(1+\beta _2)}} u^{\frac{\beta _2}{\alpha (\beta _2+1)}} du. \end{aligned}$$

The Taylor expansion gives

$$\begin{aligned} \int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz&\ge C(r_1, \alpha , \gamma _2,\beta _2) \frac{ r_1^ {{\beta _2\over \alpha }}}{(1+(2r_1)^{\beta _2+1} )^{1+\frac{1}{\alpha }}} K^{{\gamma _2}\,}\\&\ge C(r_1, \alpha , \gamma _2,\beta _2) K^{{\gamma _2}\, } . \end{aligned}$$

By the choice of $\beta _2$ and $\gamma _2$, we have

$$\begin{aligned} \int _0 ^ \infty \left[ c(z) - c(z+ \varsigma ^{(R)} (K,z))\right] \; dz\ge C(r_1) \, K,\quad 0<K\le 1. \end{aligned}$$

(82)

This proves Lemma 6.1. $\square$

Let $\kappa ^{+,R}$ be the inverse of $\vartheta ^{+,R}$, i.e. $\kappa ^{+,R}(z)=(\vartheta ^{+,R})^ {-1}(z)$, $z\in {\mathbb {R}}_0^+$. Taking into account the negative jumps and negative values of $v$, we will define the following transformation.

Corollary 6.3

Assume that Hypothesis 1 holds and let $r_0$ be as in Hypothesis 1. Then, for any $R\ge r_0$ the transformation

$$\begin{aligned} \vartheta ^R:{\mathbb {R}}\times {{ {\mathbb {R}}}}\setminus \{0\}\rightarrow {{ {\mathbb {R}}}}\end{aligned}$$

defined by

$$\begin{aligned} \vartheta ^R(v,z):= {\left\{ \begin{array}{ll}z+\varsigma ^{(R)}(\kappa ^{+,R}(|v|),z)&{} \text{ if } v\ge 0\, \text{ and } z> 0, \\ -z-\varsigma ^{(R)}(\kappa ^{+,R}(|v|),z)&{} \text{ if } v< 0 \, \text{ and } z< 0, \\ 0 &{} \text{ elsewhere } \end{array}\right. }\end{aligned}$$

satisfies

$$\begin{aligned} \int _{R} ^\infty \left[ c(z)-c(\vartheta ^R(v,z)) \right] \, dz\,\!+\! \int _{-\infty }^ {-R} \left[ c(z)-c(\vartheta ^R(v,z)) \right] \, dz\, \!=\! v . \quad \text{ for } \text{ all } \quad v \in {\mathbb {R}}. \end{aligned}$$

Proof

Taken into account the symmetricity of the Lévy measure and the definition of $\kappa ^{+,R}$, the proof follows by Lemma 6.1 and direct calculations. $\square$

Corollary 6.4

Let $R$ as in Corollary 6.3 and define the function $\rho ^{(R)}:{\mathbb {R}}\times {\mathbb {R}}\setminus \{0\}\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \rho ^{(R)}(v,z):= {\left\{ \begin{array}{ll}\varsigma ^{(R)}(\kappa ^{+,R}(|v|),z)&{} \text{ if } v\ge 0\, \text{ and } z> 0,\\ -\varsigma ^{(R)}(\kappa ^{+,R}(|v|),z)&{} \text{ if } v< 0 \, \text{ and } z< 0, \\ 0 &{} \text{ elsewhere. } \end{array}\right. }\end{aligned}$$

(83)

Let us denote the derivative in the direction of the second variable by $\rho ^{(R)}_z$. Then

• (1) there exists a constant $C(R)>0$ such that

  $$\begin{aligned} \int _{{\mathbb {R}}\setminus \{0\}}|\rho ^{(R)}_z(x,z)|\, dz \le C(R)\,| x |^{2}, \quad \forall x\in {\mathbb {R}}; \end{aligned}$$

• (2) there exists $\tilde{R}>0$ such that for all $R>\tilde{R}$,

  $$\begin{aligned} |\rho ^{(R)}_z(x,z)|\le \frac{1}{2}, \forall \, x\in {\mathbb {R}},\, \forall z\in {\mathbb {R}}\setminus \{0\}. \end{aligned}$$

Proof

For sake of simplicity, we set $r_1=\frac{4}{3} R$ throughout this proof.

By the symmetricity assumption on the Lévy measure, it is enough to prove Item (1) for $v>0$. Let $v_0:= \inf \{v\ge 0: \kappa ^{+,R}(v)\ge 1\}$. First, let us assume, that $v\ge v_0$. Then $\kappa ^+(v) \ge 1$ and, by (79), $\kappa ^{+,R}(v)\sim v ^\frac{1}{\Gamma }$, where $\Gamma$ is defined in (80).⁸ By the definition of $\varsigma ^{(R)}$ we have that

$$\begin{aligned} \int _0 ^\infty \varsigma ^{(R)}_z(K,z)\,dz \sim K \int _{ r_1K ^{\gamma _1}} ^{ 2r_1 K ^{\gamma _1}} z ^{-\beta _1-1}\, dz \sim C(r_1)\, K ^{1-\beta _1\gamma _1}, \end{aligned}$$

for any $K>1$. Setting $K=v ^\frac{1}{\Gamma }$, we get by the choice of $\beta _1$ and $\gamma _1$ that

$$\begin{aligned} \int _0 ^\infty \varsigma ^{(R)}_z(K,z)\,dz\sim C(r_1)\, v ^{1-\beta _1\gamma _1\over \Gamma } \sim C(r_1)v ^2,\quad v\ge v_0. \end{aligned}$$

In case $0\le v\le v_0$, $\kappa ^+(v) < 1$. Thus, putting $K=\kappa ^ +(v)$ we have

$$\begin{aligned}&\int _0 ^\infty \varsigma ^{(R)}_z(K,z)\,dz \sim \int _{ r_1} ^{ r_1(1+ K ^{\gamma _2})} z ^{-\beta _2-1}\, dz \sim C(r_1,\beta _2)\\&\quad \times \left( (1+ K ^{\gamma _2} )^ {-\beta _2}-1\right) \sim C(r_1)\, K ^{\gamma _2}. \end{aligned}$$

Since by (81) we have $K\sim v$ and thanks to the choice of $\gamma _2$ we obtain

$$\begin{aligned} \int _0 ^\infty \varsigma ^{(R)}_z(K,z)\,dz\sim C(r_1)\, v ^{\gamma _2}\sim C(r_1,\beta _2)v ^2,\quad v\ge v_0. \end{aligned}$$

From this we conclude the proof of Item (1).

As in the proof of (1) it is enough to prove that Item (2) is valid for any $v>0$. The general case follows from the symmetricity in Hypothesis 1. Let us assume first $v\ge v_0$, where again $v_0:= \inf \{v\ge 0: \kappa ^{+,R}(v)\ge 1\}$. Then $\varsigma ^{(R)}(z)=0$ for $z\le \frac{4}{3} K R$, where $K=\kappa ^{+,R}(v)$. In particular, by the definition of $v_0$, $\varsigma ^{(R)}(z)=0$ for $z\le \frac{4}{3} R$. We know that $\varsigma ^{(R)}_z(K,z)\le K z^{-\beta _1-1}$ for $K\ge 1$ and $z \in \left( K^{\gamma _1}R, \frac{4}{3} K^{\gamma _1}R)\cup (\frac{8}{3} K^{\gamma _1}R, \frac{10}{3} K^{\gamma _1}R\right)$, thus $\varsigma ^{(R)}_z(v,z)\le \frac{1}{2}$ for any $R> 2^\frac{1}{\tilde{\delta }}$. Next, let us assume that $0\le v\le v_0$, i.e. $K=\kappa ^{+,R}(v) \le 1$. Then $\varsigma ^{(R)}_z(K, z)\le C z^ {-\beta _2-1}$ for $z \in \left( R, \frac{4}{3} R)\cup (\frac{4}{3} (1+K^{\gamma _2}) R,\frac{5}{3} (1+K^{\gamma _2}) R\right)$. Therefore, $\varsigma ^{(R)}_z(z,K)\le \frac{1}{2}$ for any $R> [2C]^{\frac{1}{1+\beta _2}}.$ Choosing $\tilde{R} =\max \left( 2^\frac{1}{\tilde{\delta }}, [2C]^\frac{1}{\beta _2+1}\right) ,$ we easily see that for $R>\tilde{R}\vee r_0$ we have $\varsigma ^{(R)}_z(v,z)\le \frac{1}{2}$ for all $K$ and $z$. $\square$

Appendix B: Change of measure formula

Let $\mu$ be a Poisson random measure over $\bar{{\mathfrak {A}}}=(\bar{\Omega },\bar{{\mathbb {P}}},(\bar{{{\mathcal F }}}_t)_{t\ge 0},\bar{{{\mathcal F }}})$ with compensator $\gamma$ defined by $\gamma (U\times I)=\lambda (U) \lambda (I)$ for any $U\times I\in \mathcal {B}(\mathbb {R})\times \mathcal {B}([0,\infty ))$. Let $c:{{ {\mathbb {R}}}}\rightarrow {{ {\mathbb {R}}}}$ be the transformation defined by (49).

Let $g:\bar{\Omega }\times [0,\infty )\times {\mathbb {R}}\rightarrow {\mathbb {R}}$ be a predictable process with $g\in L^ 2([0,\infty )\times {\mathbb {R}};{\mathbb {R}})$ and $\psi$ be a mapping defined by

$$\begin{aligned} \psi :[0,\infty ) \times {{ {\mathbb {R}}}}\ni (t,z) \mapsto z+g(t,z) \, \in {{ {\mathbb {R}}}}. \end{aligned}$$

(84)

Combining Corollary 6.3 and Example 1.9 of [18] one can verify the following Lemma.

Lemma 6.5

There exists a probability measure ${\mathbb {Q}}^ \psi$ on ${\mathfrak {A}}$ such that the Poisson random measure $\mu _\psi$ defined by

$$\begin{aligned} {{\mathcal B }}({{ {\mathbb {R}}}})\times {{\mathcal B }}([0,\infty ))\ni A\times I \mapsto \int _I\int _{{ {\mathbb {R}}}}1_A(\psi (s,z ))\mu (dz,ds) \end{aligned}$$

(85)

has compensator $\gamma$. For $t\ge 0$ let ${\mathbb {Q}}^\psi _t$, respectively, $\bar{{\mathbb {P}}}_t$, be the projection of ${\mathbb {Q}}^\psi$(resp. $\bar{{\mathbb {P}}}$) onto $\bar{{{\mathcal F }}}_t$. Then the density process given by

$$\begin{aligned}{}[0,\infty ) \ni t\mapsto {{\mathcal G }}(t) := {d{\mathbb {Q}}^\psi _t\over d\bar{{\mathbb {P}}}_t},\quad t> 0, \end{aligned}$$

satisfies

$$\begin{aligned} \left\{ \begin{array}{rcl} d{{\mathcal G }}(t) &{}=&{} {{\mathcal G }}(t-) \int _{{ {\mathbb {R}}}}(1- \psi _z(z )) \,(\mu -\gamma )(dz,dt), \\ &{}=&{} {{\mathcal G }}(t-) \int _{{ {\mathbb {R}}}}g_z(t,z) \,(\mu -\gamma )(dz,dt), \\ {{\mathcal G }}(0) &{}=&{} 1. \end{array} \right. \end{aligned}$$

(86)

Remark 6.6

Let $v:\bar{\Omega }\times {\mathbb {R}}\rightarrow {\mathbb {R}}$ be a predictable process and let $\theta ^{(R)}(t,z):= \psi (t,z)+ \rho ^{(R)}(v(t),z)=z+\rho ^{(R)}(v(t),z)$, where $\rho ^{(R)}$ is defined in Corollary 6.4. Then, under ${\mathbb {Q}}^ \psi$ the Poisson random measure $\mu _\psi =\mu _\theta$ defined by (85) has compensator $\lambda \times \lambda$.

Remark 6.7

In case there exists a $\delta >0$ such that , $g_z(t,z )\ge \delta -1$ $\forall \, z\in {\mathbb {R}}$, ${{\mathcal G }}$ is invertible and the inverse, ${\mathcal {H}}={{\mathcal G }}^ {-1}$ solves the following stochastic differential equation

Proof

The proof is done via the Laplace transform. Let $\xi =\{\xi (t):0\le t<\infty \}$ be given by

$$\begin{aligned} \left\{ \begin{array}{rcl} d\xi (t)&{}=&{} \int _{{ {\mathbb {R}}}}c(z) (\mu _\psi -\gamma ) (dz,dt), \\ \xi (0)&{}=&{} 0. \end{array} \right. \end{aligned}$$

Then under ${{\mathbb {Q}}^ \psi }$ the Laplace transform is given by

$$\begin{aligned} \mathbb {E}^{{\mathbb {Q}}^ \psi } e ^{-\lambda \xi (t)} = e ^{ \int _{{ {\mathbb {R}}}}\left[ e ^{-\lambda c(z)}-1+\lambda c(z)\, \right] \lambda (dz)} \end{aligned}$$

Rewriting $\xi$ gives

$$\begin{aligned} \left\{ \begin{array}{rcl} d\xi (t) &{}=&{} \int _{{ {\mathbb {R}}}}c(\psi (t,z)) (\mu -\gamma ) (dz,dt) +\int _{{ {\mathbb {R}}}}\left[ c(\psi (s,z)) -c(z)\right] \gamma (dz, dt), \\ \xi (0)&{}=&{} 0. \end{array}\right. \end{aligned}$$

Let $M _\lambda =\{ M_\lambda (t):0\le t<\infty \}$ be given by $M_\lambda (t) = e ^{-\lambda \xi (t)}$, $0\le t<\infty$. Now, we will show that $\mathbb {E}^{\bar{{\mathbb {P}}}} M _\lambda (t){{\mathcal G }}(t) = \mathbb {E}^{{\mathbb {Q}}^ \psi } e ^{-\lambda \xi (t)}$. First $M _\lambda (t)$ solves

$$\begin{aligned} dM _\lambda (t)= & {} -\lambda \, \int _{{ {\mathbb {R}}}}M _\lambda (t-) \left[ c(\psi (t,z)) -c(z)\right] \gamma (dz, dt)\\&+ \int _{{ {\mathbb {R}}}}M _\lambda (t-)\left[ e^{-\lambda c(\psi (t,z))}-1\right] (\mu -\gamma ) (dz,dt) \\&+ \int _{{ {\mathbb {R}}}}M _\lambda (t-) \left[ e^{-\lambda c(\psi (t,z))}- 1+\lambda c(\psi (t,z))\right] \gamma (dz, dt), \\ M_\lambda (0)= & {} 1. \end{aligned}$$

Therefore, ${\mathcal {Z}}_\lambda (t) = M _\lambda (t) \, {{\mathcal G }}(t)$ is given by

$$\begin{aligned} \mathbb {E}^{\bar{\mathbb {P}}} {\mathcal {Z}} _\lambda (t)= & {} -\lambda \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-) \left[ c(\psi (s,z)) - c(z)\right] \lambda (dz)\, ds \\&+\, \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-) \left[ e^{-\lambda c(\psi (s,z))}- 1+\lambda c(\psi (s,z))\right] \lambda (dz)\, ds \\&+\, \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t {\mathcal {Z}} _\lambda (s-) \int _{{ {\mathbb {R}}}}\left[ e^{-\lambda c(\psi (s,z))}- 1\right] \left[ \psi _z(s,z) -1\right] \lambda (dz)\, ds \\= & {} \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-)\Big [ \lambda c(z) - \lambda c(\psi (s,z)) + e^{-\lambda c(\psi (s,z))}- 1 \\&+\,\lambda c(\psi (s,z))+ e^{-\lambda c(\psi (s,z))}\psi _z(s,z) - \psi _z(s,z) -e^{-\lambda c(\psi (s,z))}\\&+\,1\Big ] \gamma (dz,ds)\\= & {} \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-)\Big [ e^{-\lambda c(\psi (s,z))}\psi _z(s,z) - \psi _z(s,z) +\lambda c(z) \Big ] \gamma (dz,ds).\\= & {} \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-)\Big [ e^{-\lambda c(\psi (s,z))}-1\Big ]\psi _z(s,z) \lambda (dz) ds\\&+\, \lambda \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-) c(z) \gamma (dz,ds). \end{aligned}$$

Substitution gives

$$\begin{aligned} \mathbb {E}^{\bar{\mathbb {P}}} {\mathcal {Z}} _\lambda (t)= & {} \mathbb {E}^{\bar{\mathbb {P}}} \int _0^t \int _{{ {\mathbb {R}}}}{\mathcal {Z}} _\lambda (s-)\Big [ e^{-\lambda c(z)} - 1 +\lambda c(z) \Big ] \gamma (dz,ds). \end{aligned}$$

Since

$$\begin{aligned} \mathbb {E}^{{\mathbb {Q}}_\psi } \left[ e^{-\lambda {\xi (t) }}\right]&= \mathbb {E}^{\bar{\mathbb {P}}} \left[ {{\mathcal G }}(t)\, e^{-\lambda {\xi (t)}}\right] = \mathbb {E}^{\bar{\mathbb {P}}} \left[ {\mathcal {Z}} _\lambda (t) \right] \\&= \exp \left( \int _0^t \int _{{ {\mathbb {R}}}}\left[ e^{-\lambda c(z)} - 1 +\lambda c(z) \right] \gamma (dz,dt)\right) , \end{aligned}$$

from which the Proposition follows. $\square$