The aim of this paper is to show the controlled SDEwMJ (6) is almost surely exponentially stable. Before proving this result, a number of lemmas should be proved.
Lemma 3.1
Let (A1) hold, and \(p\in (0,1)\). Then, for any \(t_{0}\geq 0\) and \(T\geq 0\),
$$\begin{aligned}& \sup_{t_{0}\leq t \leq t_{0}+T+\delta }E \bigl\vert x(t) \bigr\vert ^{p}\leq M_{1}^{ \frac{p}{2}} E \Vert \varphi \Vert ^{p}, \end{aligned}$$
(16)
$$\begin{aligned}& \sup_{t_{0}\leq t\leq t_{0}+T}E \Bigl(\sup_{0\leq \nu \leq \delta } \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{p} \Bigr)\leq M^{\frac{p}{2}}_{2}E \Vert \varphi \Vert ^{p}, \end{aligned}$$
(17)
and
$$\begin{aligned} E \Bigl(\sup_{t_{0}\leq t\leq t_{0}+T+\delta } \bigl\vert x(t) \bigr\vert ^{p} \Bigr)\leq M^{\frac{p}{2}}_{3}E \Vert \varphi \Vert ^{p}, \end{aligned}$$
(18)
where
$$\begin{aligned} &M_{1}\triangleq :M_{1}(p,\delta ,T)= (1+k\delta )e^{[2(L_{1}+ \lambda L_{3})+L_{2}^{2}+\lambda L^{2}_{3}+2K](T+\delta )}, \end{aligned}$$
(19)
$$\begin{aligned} &M_{2}\triangleq :M_{2}(p,\delta ,T)= 4 \bigl(L^{2}_{1}\delta +4L^{2}_{2}+2 \lambda ^{2}L^{2}_{3}\delta +8\lambda L^{2}_{3}+K^{2}\delta \bigr)M_{1} \delta , \end{aligned}$$
(20)
$$\begin{aligned} &M_{3}\triangleq :M_{3}(p,\delta ,T)= 5 \bigl[1+L^{2}_{1}(T+\delta )+K^{2}(T+ \delta )+4L^{2}_{2}+2\lambda ^{2} L^{2}_{3}(T+ \delta ) \\ &\hphantom{M_{3}\triangleq :} {}+8\lambda L^{2}_{3}\bigr])M_{1}(T+\delta ). \end{aligned}$$
(21)
Proof
(1) Applying the Itô’s formula to \(|x(t)|^{2}\) and taking the expectation, one obtains
$$\begin{aligned} E \bigl\vert x(t) \bigr\vert ^{2}= {}& E \bigl\vert x(t_{0}) \bigr\vert ^{2}+2E \int ^{t}_{t_{0}}x^{T}(s)f\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+2E \int ^{t}_{t_{0}}x^{T}(s)u\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \\ & {}+E \int ^{t}_{t_{0}} \bigl\vert g\bigl(x(s),s,r(s)\bigr) \bigr\vert ^{2}\,ds \\ & {}+\lambda E \int ^{t}_{t_{0}}\bigl[h^{2}\bigl(x(s),s,r(s) \bigr)+2x^{T}(s)h\bigl(x(s),s,r(s)\bigr)\bigr]\,ds. \end{aligned}$$
(22)
According to (9), one gains
$$\begin{aligned} E \bigl\vert x(t) \bigr\vert ^{2}\leq {}& E \bigl\vert x(t_{0}) \bigr\vert ^{2}+2L_{1}E \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds +2KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert \bigl\vert x(s-\delta ) \bigr\vert \,ds \\ & {}+L^{2}_{2}E \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds +\lambda L^{2}_{3} E \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ & {}+2\lambda L_{3}E \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds. \end{aligned}$$
(23)
By basic inequality, one can get
$$\begin{aligned} & 2KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert \bigl\vert x(s-\delta ) \bigr\vert \,ds \\ & \quad \leq KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds+KE \int ^{t}_{t_{0}} \bigl\vert x(s- \delta ) \bigr\vert ^{2}\,ds \\ & \quad =KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds+KE \int ^{t-\delta }_{t_{0}-\delta } \bigl\vert x( \nu ) \bigr\vert ^{2}\,d\nu \\ & \quad =KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds+KE \int ^{t_{0}}_{t_{0}-\delta } \bigl\vert x( \nu ) \bigr\vert ^{2}\,d\nu +KE \int ^{t-\delta }_{t_{0}} \bigl\vert x(\nu ) \bigr\vert ^{2}\,d\nu \\ &\quad \leq 2KE \int ^{t}_{t_{0}} \bigl\vert x(s) \bigr\vert ^{2}\,ds+K\delta E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(24)
According to (6) and substituting (24) into (23), one can gain
$$\begin{aligned} & E \bigl\vert x(t) \bigr\vert ^{2} \\ & \quad \leq E \bigl\vert x(t_{0}) \bigr\vert ^{2}+ \bigl[2(L_{1}+\lambda L_{3})+L^{2}_{2}+ \lambda L^{2}_{3}+2K\bigr] \int ^{t}_{t_{0}}E \bigl\vert x(s) \bigr\vert ^{2}\,ds+K\delta E \Vert \varphi \Vert ^{2} \\ &\quad \leq (1+K\delta )E \Vert \varphi \Vert ^{2}+ \bigl[2(L_{1}+\lambda L_{3})+L^{2}_{2}+ \lambda L^{2}_{3}+2K\bigr] \int ^{t}_{t_{0}} \Bigl(\sup_{t_{0}\leq \nu \leq s}E \bigl\vert x(\nu ) \bigr\vert ^{2} \Bigr)\,ds. \end{aligned}$$
(25)
Therefore,
$$\begin{aligned} & \sup_{t_{0}\leq \nu \leq t}E \bigl\vert x(\nu ) \bigr\vert ^{2} \\ &\quad \leq (1+K\delta )E \Vert \varphi \Vert ^{2}+ \bigl[2(L_{1}+\lambda L_{3})+L^{2}_{2}+ \lambda L^{2}_{3}+2K\bigr] \\ & \qquad {}\times \int ^{t}_{t_{0}} \Bigl(\sup_{t_{0}\leq \nu \leq s}E \bigl\vert x( \nu ) \bigr\vert ^{2} \Bigr)\,ds. \end{aligned}$$
(26)
By the Gronwall inequality, one obtains
$$\begin{aligned} \sup_{t_{0} \leq \nu \leq t_{0}+T+\delta }E \bigl\vert x(\nu ) \bigr\vert ^{2} \leq{} & (1+K \delta )e^{[2(L_{1}+\lambda L_{3})+L^{2}_{2}+\lambda L^{2}_{3}+2K](T+ \delta )}E \Vert \varphi \Vert ^{2} \\ \triangleq :{} & M_{1}E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(27)
Applying the Hölder inequality, one has
$$\begin{aligned} \sup_{t_{0}\leq t \leq t_{0}+T+\delta }E \bigl\vert x(t) \bigr\vert ^{p} \leq M^{ \frac{p}{2}}_{1}E \Vert \varphi \Vert ^{p}. \end{aligned}$$
(28)
(2) For \(0\leq \nu \leq \delta \), it follows from (6) that one can get
$$\begin{aligned} x(t+\nu )-x(t)={}& \int ^{t+\nu }_{t}\bigl[f\bigl(x(s),s,r(s)\bigr)+u \bigl(x(s-\delta ),s,r(s)\bigr)\bigr]\,ds \\ & {}+ \int ^{t+\nu }_{t}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s) \\ & {}+ \int ^{t+\nu }_{t}h\bigl(x(s),s,r(s)\bigr)\,dN(s). \end{aligned}$$
(29)
Then
$$\begin{aligned} \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{2}\leq {}& 4 \biggl\vert \int ^{t+\nu }_{t}f\bigl(x(s),s,r(s)\bigr)\,ds \biggr\vert ^{2} \\ & {}+ 4 \biggl\vert \int ^{t+\nu }_{t}u\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \biggr\vert ^{2} \\ & {}+4 \biggl\vert \int ^{t+\nu }_{t}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s) \biggr\vert ^{2} \\ & {}+4 \biggl\vert \int ^{t+\nu }_{t}h\bigl(x(s),s,r(s)\bigr)\,dN(s) \biggr\vert ^{2} \\ \leq {}& 4\nu \int ^{t+\nu }_{t}f^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+ 4\nu \int ^{t+\nu }_{t}u^{2}\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \\ & {}+4 \biggl\vert \int ^{t+\nu }_{t}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s) \biggr\vert ^{2} \\ & {}+8\lambda ^{2}\nu \int ^{t+\nu }_{t}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+8 \biggl\vert \int ^{t+\nu }_{t}h\bigl(x(s),s,r(s)\bigr)\,d\tilde{N}(s) \biggr\vert ^{2}. \end{aligned}$$
(30)
For \(\int ^{t+\nu }_{t}g(x(s),s,r(s))\,d\omega (s)\) and \(\int ^{t+\nu }_{t}h(x(s),s,r(s))\,d\tilde{N}(s)\) are martingales, thus by the Doob martingale inequality, one gains
$$\begin{aligned} E \Bigl(\sup_{0\leq \nu \leq \delta } \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{2} \Bigr) \leq{} & 4\delta E \int ^{t+\delta }_{t}f^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+4\delta E \int ^{t+\delta }_{t}u^{2}\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \\ & {}+16E \int ^{t+\delta }_{t}g^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+8\lambda ^{2}\delta E \int ^{t+\delta }_{t}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+32\lambda E \int ^{t+\delta }_{t}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds. \end{aligned}$$
(31)
Applying (A1), one can further get
$$\begin{aligned} & E \Bigl(\sup_{0\leq \nu \leq \delta } \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{2} \Bigr) \\ & \quad \leq 4\delta L^{2}_{1}E \int ^{t+\delta }_{t} \bigl\vert x(s) \bigr\vert ^{2}\,ds+4\delta K^{2}E \int ^{t+\delta }_{t} \bigl\vert x(s-\delta ) \bigr\vert ^{2}\,ds \\ &\qquad {}+16L^{2}_{2}E \int ^{t+\delta }_{t} \bigl\vert x(s) \bigr\vert ^{2}\,ds+8\lambda ^{2} L^{2}_{3} \delta E \int ^{t+\delta }_{t} \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ & \qquad {}+32\lambda L^{2}_{3}E \int ^{t+\delta }_{t} \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ &\quad \triangleq : \bigl(4 L^{2}_{1}\delta +16 L^{2}_{2}+8\lambda ^{2} L^{2}_{3} \delta +32\lambda L^{2}_{3}\bigr)E \int ^{t+\delta }_{t} \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ & \qquad {}+4 K^{2}\delta E \int ^{t+\delta }_{t} \bigl\vert x(s-\delta ) \bigr\vert ^{2}\,ds. \end{aligned}$$
(32)
In fact, for \(t\in [t_{0}-\delta , t_{0}]\), \(E|x(t)|^{2}\leq E\|\varphi \|^{2}\leq M_{1}E\|\varphi \|^{2}\). Together with (27), one can get \(E|x(t)|^{2}\leq M_{1}E\|\varphi \|^{2}\) for \(t\in [t_{0}-\delta ,t_{0}+T+\delta ]\). Then
$$\begin{aligned} & \sup_{t_{0}\leq t\leq t_{0}+T} \Bigl(E \Bigl(\sup _{0\leq \nu \leq \delta } \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{2} \Bigr) \Bigr) \\ &\quad \leq \bigl(4 L^{2}_{1}\delta +16 L^{2}_{2}+8 \lambda ^{2} L^{2}_{3} \delta +32\lambda L^{2}_{3}\bigr)M_{1}\delta E \Vert \varphi \Vert ^{2}+4\delta K^{2} M_{1}\delta E \Vert \varphi \Vert ^{2} \\ &\quad = 4\bigl(L^{2}_{1}\delta +4 L^{2}_{2}+2 \lambda ^{2} L^{2}_{3}\delta +8 \lambda L^{2}_{3}+K^{2}\delta \bigr)M_{1}\delta E \Vert \varphi \Vert ^{2} \triangleq : M_{2}E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(33)
By the Hölder inequality, one has
$$\begin{aligned} \sup_{t_{0}\leq t\leq t_{0}+T} \Bigl(E \Bigl(\sup _{0\leq \nu \leq \delta } \bigl\vert x(t+\nu )-x(t) \bigr\vert ^{{p}} \Bigr) \Bigr) \leq M^{ \frac{p}{2}}_{2}E \Vert \varphi \Vert ^{p}. \end{aligned}$$
(34)
(3) It follows from (6) that
$$\begin{aligned} x(t)={}& x(t_{0})+ \int ^{t}_{t_{0}}\bigl[f\bigl(x(s),s,r(s)\bigr)+u \bigl(x(s-\delta ),s,r(s)\bigr)\bigr]\,ds \\ & {}+ \int ^{t}_{t_{0}}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s)+ \int ^{t}_{t_{0}}h\bigl(x(s),s,r(s)\bigr)\,dN(s). \end{aligned}$$
(35)
Using the Hölder inequality and (A1), one gains
$$\begin{aligned} \bigl\vert x(t) \bigr\vert ^{2}\leq {}& 5 \bigl\vert x(t_{0}) \bigr\vert ^{2}+5 \biggl\vert \int ^{t}_{t_{0}}f\bigl(x(s),s,r(s)\bigr)\,ds \biggr\vert ^{2} \\ & {}+ 5 \biggl\vert \int ^{t}_{t_{0}}u\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \biggr\vert ^{2} \\ & {}+5 \biggl\vert \int ^{t}_{t_{0}}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s) \biggr\vert ^{2} \\ & {}+5 \biggl\vert \int ^{t}_{t_{0}}h\bigl(x(s),s,r(s)\bigr)\,dN(s) \biggr\vert ^{2} \\ \leq {}& 5 \bigl\vert x(t_{0}) \bigr\vert ^{2}+5(t-t_{0}) \int ^{t}_{t_{0}}f^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+ 5(t-t_{0}) \int ^{t}_{t_{0}}u^{2}\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \\ & {}+5 \biggl\vert \int ^{t}_{t_{0}}g\bigl(x(s),s,r(s)\bigr)\,d\omega (s) \biggr\vert ^{2} \\ & {}+10\lambda ^{2}(t-t_{0}) \int ^{t}_{t_{0}}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & {}+10 \biggl\vert \int ^{t}_{t_{0}}h\bigl(x(s),s,r(s)\bigr)\,d\tilde{N}(s) \biggr\vert ^{2}. \end{aligned}$$
(36)
By the Doob martingale inequality, one can gain
$$\begin{aligned} & E \Bigl(\sup_{t_{0}\leq t\leq t_{0}+T+\delta } \bigl\vert x(t) \bigr\vert ^{2} \Bigr) \\ & \quad \leq 5E \bigl\vert x(t_{0}) \bigr\vert ^{2}+5(T+ \delta )E \int ^{t_{0}+T+\delta }_{t_{0}}f^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ &\qquad {}+ 5(T+\delta )E \int ^{t_{0}+T+\delta }_{t_{0}}u^{2}\bigl(x(s-\delta ),s,r(s)\bigr)\,ds \\ & \qquad {}+20E \int ^{t_{0}+T+\delta }_{t_{0}}g^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & \qquad {}+10\lambda ^{2}(T+\delta )E \int ^{t_{0}+T+\delta }_{t_{0}}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & \qquad {}+40\lambda E \int ^{t_{0}+T+\delta }_{t_{0}}h^{2}\bigl(x(s),s,r(s) \bigr)\,ds \\ & \quad \leq 5E \Vert \varphi \Vert ^{2}+5L^{2}_{1}(T+ \delta ) \int ^{t_{0}+T+\delta }_{t_{0}}E \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ & \qquad {}+ 5K^{2}(T+\delta ) \int ^{t_{0}+T+\delta }_{t_{0}}E \bigl\vert x(s-\delta ) \bigr\vert ^{2}\,ds \\ & \qquad {}+20L^{2}_{2} \int ^{t_{0}+T+\delta }_{t_{0}}E \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ & \qquad {}+10\lambda ^{2}L^{2}_{3}(T+\delta )E \int ^{t_{0}+T+\delta }_{t_{0}}E \bigl\vert x(s) \bigr\vert ^{2}\,ds \\ &\qquad {}+40\lambda L^{2}_{3}E \int ^{t_{0}+T+\delta }_{t_{0}}E \bigl\vert x(s) \bigr\vert ^{2}\,ds. \end{aligned}$$
(37)
By (27), one obtains
$$\begin{aligned} & E \Bigl(\sup_{t_{0}\leq t\leq t_{0}+T+\delta } \bigl\vert x(t) \bigr\vert ^{2} \Bigr) \\ & \quad \leq 5\bigl[1+L^{2}_{1}(T+\delta )+K^{2}(T+ \delta )+4L^{2}_{2}+2\lambda ^{2}L^{2}_{3}(T+ \delta )+8\lambda L^{2}_{3}\bigr] \\ & \qquad {}\times (T+\delta )M_{1}E \Vert \varphi \Vert ^{2} \\ & \quad \triangleq :M_{3}E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(38)
Applying the Hölder inequality, one can compute
$$\begin{aligned} E \Bigl(\sup_{t_{0}\leq t\leq t_{0}+T+\delta } \bigl\vert x(t) \bigr\vert ^{2} \Bigr) \leq M^{\frac{p}{2}}_{3}E \Vert \varphi \Vert ^{p}. \end{aligned}$$
(39)
The proof is complete. □
Lemma 3.2
Let (A1) hold and \(p\in (0,1)\). Given \(t_{0}\geq \delta \) and \(T\geq 0\) arbitrarily. Denote \(y(t;x(t_{0}),r(t_{0}),t_{0})=y(t)\) for all \(t\geq t_{0}\). Then
$$\begin{aligned} E \bigl\vert x(t)-y(t) \bigr\vert ^{p}\leq M_{4}^{\frac{p}{2}} E\|\varphi |^{p},\quad \textit{for } \forall t \in [t_{0},t_{0}+T+\delta ], \end{aligned}$$
(40)
where
$$\begin{aligned} M_{4}\triangleq :M_{4}(p,\delta ,T)= M_{2}K(T+\delta )e^{[2L_{1}+L_{2}^{2}+ \lambda L^{2}_{3}+2\lambda L_{3}+3K](T+\delta )}. \end{aligned}$$
(41)
Proof
It follows from (6) and (10) that one gains
$$\begin{aligned} d\bigl[x(t)-y(t)\bigr]={}& \bigl[f\bigl(x(t),t,r(t)\bigr)-f\bigl(y(t),t,r(t)\bigr) \bigr]\,dt \\ & {}+\bigl[u\bigl(x(t-\delta ),t,r(t)\bigr)-u\bigl(y(t),t,r(t)\bigr)\bigr]\,dt \\ & {}+\bigl[g\bigl(x(t),t,r(t)\bigr)-g\bigl(y(t),t,r(t)\bigr)\bigr]\,d\omega (t) \\ & {}+\bigl[h\bigl(x(t),t,r(t)\bigr)-h\bigl(y(t),t,r(t)\bigr)\bigr]\,dN(t). \end{aligned}$$
(42)
Applying the Itô’s formula to \(|x(t)-y(t)|^{2}\), one gets
$$\begin{aligned} & E \bigl\vert x(t)-y(t) \bigr\vert ^{2} \\ &\quad =E \bigl\vert x(t_{0})-y(t_{0}) \bigr\vert ^{2} +2E \int ^{t}_{t_{0}}\bigl(x(s)-y(s)\bigr)^{T} \bigl(f\bigl(x(s),s,r(s)\bigr)-f\bigl(y(s),s,r(s)\bigr)\bigr)\,ds \\ &\qquad {}+ 2E \int ^{t}_{t_{0}}\bigl(x(s)-y(s)\bigr)^{T} \bigl(u\bigl(x(s-\delta ),s,r(s)\bigr)-u\bigl(y(s),s,r(s)\bigr)\bigr)\,ds \\ &\qquad {}+ E \int ^{t}_{t_{0}} \bigl\vert g\bigl(x(s),s,r(s) \bigr)-g\bigl(y(s),s,r(s)\bigr) \bigr\vert ^{2}\,ds \\ &\qquad {}+\lambda E \int ^{t}_{t_{0}} \bigl\vert h\bigl(x(s),s,r(s) \bigr)-h\bigl(y(s),s,r(s)\bigr) \bigr\vert ^{2}\,ds \\ &\qquad {}+2\lambda E \int ^{t}_{t_{0}}\bigl(x(s)-y(s)\bigr)^{T} \bigl(h\bigl(x(s),s,r(s)\bigr)-h\bigl(y(s),s,r(s)\bigr)\bigr)\,ds. \end{aligned}$$
(43)
Using (A1), one can further obtain
$$\begin{aligned} & E \bigl\vert x(t)-y(t) \bigr\vert ^{2} \\ &\quad \leq 2L_{1} \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds+2KE \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert \bigl\vert x(s- \delta )-y(s) \bigr\vert \,ds \\ &\qquad {}+ L^{2}_{2} \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds +\lambda L^{2}_{3} \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ &\qquad {}+2\lambda L_{3} \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ &\quad \triangleq : \bigl(2L_{1}+L^{2}_{2}+\lambda L^{2}_{3}+2\lambda L_{3}\bigr) \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ &\qquad {}+2KE \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert \bigl\vert x(s-\delta )-y(s) \bigr\vert \,ds. \end{aligned}$$
(44)
Noting that
$$\begin{aligned} & E \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert \bigl\vert x(s-\delta )-y(s) \bigr\vert \,ds \\ &\quad =E \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert \bigl\vert x(s-\delta )-x(s)+x(s)-y(s) \bigr\vert \,ds \\ & \quad \leq E \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert \bigl\vert x(s-\delta )-x(s) \bigr\vert \,ds +E \int ^{t}_{t_{0}} \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ & \quad \leq \frac{3}{2} \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds +\frac{1}{2} \int ^{t}_{t_{0}}E \bigl\vert x(s-\delta )-x(s) \bigr\vert ^{2}\,ds. \end{aligned}$$
(45)
Substituting (45) into (44), one has
$$\begin{aligned} E \bigl\vert x(t)-y(t) \bigr\vert ^{2} \leq {}& \bigl(2L_{1}+L^{2}_{2}+\lambda L^{2}_{3}+2 \lambda L_{3}+3K\bigr) \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ & {}+K \int ^{t}_{t_{0}}E \bigl\vert x(s-\delta )-x(s) \bigr\vert ^{2}\,ds. \end{aligned}$$
(46)
By (33), one can get for \(t\in [t_{0}, t_{0}+T+\delta ]\),
$$\begin{aligned} E \bigl\vert x(t)-y(t) \bigr\vert ^{2} \leq {}& \bigl(2L_{1}+L^{2}_{2}+\lambda L^{2}_{3}+2 \lambda L_{3}+3K\bigr) \int ^{t}_{t_{0}}E \bigl\vert x(s)-y(s) \bigr\vert ^{2}\,ds \\ & {}+ M_{2}K(T+\delta )E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(47)
According to the Gronwall inequality, one gets
$$\begin{aligned} E \bigl\vert x(t)-y(t) \bigr\vert ^{2} \leq M_{4}E \Vert \varphi \Vert ^{2}. \end{aligned}$$
(48)
Further, using the Hölder inequality, it follows that
$$ E \bigl\vert x(t)-y(t) \bigr\vert ^{p} \leq M^{\frac{p}{2}}_{4}E \Vert \varphi \Vert ^{p}. $$
(49)
The proof is complete. □
Theorem 3.3
Let (A1) and (A2) hold. There is a positive number δ̄ such that the solution of the controlled SDEwMJ (6) is almost surely exponentially stable provided \(\delta < \bar{\delta }\), that is,
$$\begin{aligned} \limsup_{t\rightarrow \infty } \frac{\log ( \vert x(t;x_{t_{0}},r(t_{0}),t_{0}) \vert )}{t}< 0 \quad \textit{a.s.} \end{aligned}$$
(50)
In practice, one can choose a constant \(\varepsilon \in (0,1)\) and set \(T=\frac{1}{\gamma }\log (\frac{4^{p}M}{\varepsilon })\). Let δ̄ is the unique root to the following equation
$$\begin{aligned} 2^{p}\bigl(M^{\frac{p}{2}}_{2}+2^{p}M^{\frac{p}{2}}_{4} \bigr)=1-\varepsilon , \end{aligned}$$
(51)
where \(M_{2}\) and \(M_{4}\) have been defined in (20) and (41), respectively.
Remark 2
Note that the left side of (51) is an increasing continuous function of δ. When \(\delta =0\), one can obtain \(M_{2}=M_{4}=0\). Therefore, the left side of (51) is 0 when \(\delta =0\). Moreover, the left side of (51) tends to +∞ when \(\delta \rightarrow +\infty \). Then, one can assert that the equation (51) admits a unique positive root δ̄. The root δ̄ can be obtained numerically, but its explicit form can not be expressed.
Proof
Step 1: Denote \(x(t;x_{t_{0}},r(t_{0}),t_{0})=x(t)\) and \(r(t;r(t_{0}),t_{0})=r(t)\). Fix \(\delta \in (0, \bar{\delta })\). Let us consider \(x(t)\) on \([\delta , 2\delta +T]\), which can be regarded as the solution of (6) with the initial value \(x_{\delta }\) and \(r(\delta )\) at \(t=\delta \). Also consider the solution \(y(t; x(\delta ),r(\delta ),\delta )\) of (10) on \(t\in [\delta , \delta +T]\) with the initial value \(x(\delta )\) and \(r(\delta )\) at \(t=\delta \). Denote \(y(\delta +T;x(\delta ), r(\delta ),\delta )=y(\delta +T)\). By (A2), one can get
$$\begin{aligned} E \bigl\vert y(\delta +T) \bigr\vert ^{p} \leq Me^{-\gamma T}E \bigl\vert x(\delta ) \bigr\vert ^{p}. \end{aligned}$$
(52)
For
$$\begin{aligned} E \bigl\vert x(\delta +T) \bigr\vert ^{p} \leq 2^{p}E \bigl\vert x(\delta +T)-y(\delta +T) \bigr\vert ^{p}+2^{p}E \bigl\vert y( \delta +T) \bigr\vert ^{p}, \end{aligned}$$
(53)
then it follows from Lemma 3.2 and (52) that one gets
$$\begin{aligned} E \bigl\vert x(\delta +T) \bigr\vert ^{p} \leq {}& 2^{p} M^{\frac{p}{2}}_{4}E \Vert x_{\delta } \Vert ^{p}+2^{p}Me^{- \gamma T}E \bigl\vert x(\delta ) \bigr\vert ^{p} \\ \leq {}& 2^{p} \bigl(Me^{-\gamma T}+M^{\frac{p}{2}}_{4} \bigr)E \Vert x_{\delta } \Vert ^{p}. \end{aligned}$$
(54)
Applying (17) and (54), one can further get
$$\begin{aligned} E \Vert x_{2\delta +T} \Vert ^{p} & =E \Bigl(\sup _{-\delta \leq \theta \leq 0} \bigl\vert x(2 \delta +T+\theta ) \bigr\vert ^{p} \Bigr) =E \Bigl(\sup_{0\leq v\leq \delta } \bigl\vert x( \delta +T+v) \bigr\vert ^{p} \Bigr) \\ & =E \Bigl(\sup_{0\leq v\leq \delta } \bigl\vert x(\delta +T+v)-x(\delta +T)+x( \delta +T) \bigr\vert ^{p} \Bigr) \\ & \leq 2^{p}E \Bigl(\sup_{0\leq v\leq \delta } \bigl\vert x(\delta +T+v)-x( \delta +T) \bigr\vert ^{p} \Bigr) +2^{p}E \bigl\vert x(\delta +T) \bigr\vert ^{p} \\ & \leq 2^{p} M^{\frac{p}{2}}_{2}E \Vert x_{\delta } \Vert ^{p}+4^{p}\bigl(Me^{- \gamma T}+M^{\frac{p}{2}}_{4} \bigr)E \Vert x_{\delta } \Vert ^{p}. \end{aligned}$$
(55)
Choose \(T=\frac{1}{\gamma }\log (\frac{4^{p}M}{\varepsilon })\). Therefore,
$$\begin{aligned} E \Vert x_{2\delta +T} \Vert ^{p} \leq \bigl[ \varepsilon +2^{p}\bigl(M^{\frac{p}{2}}_{2}+2^{p}M^{ \frac{p}{2}}_{4} \bigr)\bigr]E \Vert x_{\delta } \Vert ^{p}. \end{aligned}$$
(56)
For \(\delta <\bar{\delta }\), then it follows from (51) that \(\varepsilon +2^{p}(M^{\frac{p}{2}}_{2}+2^{p}M^{\frac{p}{2}}_{4})<1\), one can assert that there exists a \(\tilde{\lambda }>0\) such that \(\varepsilon +2^{p}(M^{\frac{p}{2}}_{2}+2^{p}M^{\frac{p}{2}}_{4})=e^{- \tilde{\lambda }(\delta +T)}\). Then
$$\begin{aligned} E \Vert x_{2\delta +T} \Vert ^{p} \leq e^{-\tilde{\lambda }(\delta +T)}E \Vert x_{ \delta } \Vert ^{p}. \end{aligned}$$
(57)
Step 2: Consider the solution \(x(t)\) on \(t\in [2\delta +T, \delta +2(\delta +T)]\), which can be regarded as the solution of (6) with the initial data \(x_{2\delta +T}\) and \(r(2\delta +T)\) at \(t=2\delta +T\). Similar to proof of (57), one can prove
$$\begin{aligned} E \Vert x_{\delta +2(\delta +T)} \Vert ^{p}\leq e^{-\tilde{\lambda }(\delta +T)}E \Vert x_{2\delta +T} \Vert ^{p} \leq e^{-2\tilde{\lambda }(\delta +T)}E \Vert x_{ \delta } \Vert ^{p}. \end{aligned}$$
(58)
By induction, one gains
$$\begin{aligned} E \Vert x_{\delta +k(\delta +T)} \Vert ^{p}\leq e^{-\tilde{\lambda }(\delta +T)}E \Vert x_{\delta +(k-1)(\delta +T)} \Vert ^{p} \leq \cdots \leq e^{-k \tilde{\lambda }(\delta +T)}E \Vert x_{\delta } \Vert ^{p}, \end{aligned}$$
(59)
for all \(k=1,2,\ldots \) . Using (18) and (59), one gets
$$\begin{aligned} E \Bigl(\sup_{\delta +k(\delta +T)\leq t\leq \delta +(k+1)(T+\delta )} \bigl\vert x(t) \bigr\vert ^{p} \Bigr) & \leq M^{\frac{p}{2}}_{3} E \Vert x_{\delta +k(\delta +T)} \Vert ^{p} \\ & \leq M^{\frac{p}{2}}_{3} e^{-k\tilde{\lambda }(\delta +T)}E \Vert x_{ \delta } \Vert ^{p}, \end{aligned}$$
(60)
for all \(k=0,1,2,\ldots \) . Hence, for \(t\in [\delta +k(\delta +T), \delta +(k+1)(\delta +T)]\), applying the Chebyshev inequality, one gets
$$\begin{aligned} & E \Bigl(\sup_{\delta +k(\delta +T)\leq t\leq \delta +(k+1)(T+ \delta )} \bigl\vert x(t) \bigr\vert ^{p}\geq e^{-\frac{1}{2}k\tilde{\lambda }(\delta +T)} \Bigr) \\ & \quad \leq M^{\frac{p}{2}}_{3} e^{-\frac{1}{2}k\tilde{\lambda }(\delta +T)}E \Vert x_{\delta } \Vert ^{p}. \end{aligned}$$
(61)
By the Borel–Cantelli lemma, one obtains for almost all \(\omega \in \Omega \) that there exists a positive integer \(k_{0}\) satisfying
$$\begin{aligned} \sup_{\delta +k(\delta +T)\leq t\leq \delta +(k+1)(T+\delta )} \bigl\vert x(t) \bigr\vert ^{p} < e^{-\frac{1}{2}k\tilde{\lambda }(\delta +T)}, \forall k\geq k_{0}, \end{aligned}$$
which implies
$$\begin{aligned} \limsup_{t\rightarrow \infty }\frac{1}{t}\log \bigl( \bigl\vert x(t) \bigr\vert \bigr)\leq - \frac{\tilde{\lambda }}{2p} a.s. \end{aligned}$$
The proof is complete. □
Remark 3
It follows from Theorem 3.3 that an unstable SDEwMJ can be stabilized by a delay feedback control. It is well known that SDEs, SDEwMSs, and SDEs with Poisson jumps are the special cases of SDEwMJs. Though [1, 16, 17] well studied the stabilization problem of SDEwMSs by state feedback controls, authors in these papers neither designed the feedback controls with delay nor took Poisson jumps into account. Authors in [18–20] designed the feedback controls depending on a past state, but they did not consider the Poisson jumps. Therefore, the results in this paper cover part of the results in [1, 16–20].