Finite Horizon Optimal Dividend and Reinsurance Problem Driven by a Jump-Diffusion Process with Controlled Jumps

Abstract

In this paper, we discuss an optimal dividend and reinsurance problem for an insurance company facing two types of risks: unstable income and potential loses. The arrival of all loses is characterized as a compound Poisson process. We assumes that every possible loss can be reinsured for a part of it. The reserve is a combination of a diffusion process and a controllable compound Poisson process. We investigate the optimal dividend and reinsurance strategy by analyzing the corresponding variational inequality on the value function. A significant difference from the existing literature is that the HJB equation in this variational inequality is a partial integro-differential equation with a functional optimization problem appearing in the integral operator. We not only prove the existence of a classical solution to the problem and the continuity, strict monotonicity, boundedness of the dividend free boundary, but also discuss the properties of the optimal reinsurance policy, including the continuity, monotonicity of the optimal part covered by reinsurance for each possible loss, and the smoothness of the reinsurance free boundary.

Appendices

Appendix A Proof of Theorem 3.2

In this section, we prove Theorem 3.2. Suppose V is the value function defined in (2.5) and v is the solution of the problem (3.1), denote ${\widehat{V}}(x,t)=v(x,T-t)$. We come to prove ${\widehat{V}}(x,t)= V (x,t)$.

For any $(x,t)\in {\mathbb R}^+\times [0,T]$, and any admissible dividend control process ${\mathbb L}^t=\{L_s\}_{s\geqslant t-}$ and retained loss control process ${\mathbb H}^t=\{H_s\}_{s\geqslant t-}$. Suppose $\{X_s\}_{s\geqslant t-}$ is the solution of (2.2) and $\tau =\inf \{s\geqslant t \mid X_s\le 0\}$. Denote $L_s^c$ as the continuous part of $L_s$ and denote $\Delta L_s=L_s-L_{s-}$ as the jump part of $L_s$. The general Itô’s formula (see [27] Theorem 33 on page 81) and taking expected value gives

$$\begin{aligned} {\widehat{V}}(x,t)&= {\mathbb E}_{t,x}\Bigg [e^{-c (\tau \wedge T-t)}{\widehat{V}}(X_{\tau \wedge T},\tau \wedge T)\Bigg ]\nonumber \\&\qquad -{\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant }\tau \wedge T}e^{-c (s-t)}\left( {\widehat{V}}(X_s,s)-{\widehat{V}}(X_{s-},s)\right) \Bigg ] \nonumber \\&\qquad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\Big ({\widehat{V}}_t+\frac{\sigma ^2}{2}{\widehat{V}}_{xx}+\Big (\gamma +\lambda (1+\delta ){\mathbb E}_{s,X_{s-}}[H_s(Z_1)]\Big ) \nonumber \\&\quad {\widehat{V}}_x -c {\widehat{V}}\Big )(X_{s-},s) \textrm{d}s\Bigg ]\nonumber \\&\qquad +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)} {\widehat{V}}_x (X_{s-},s)\textrm{d}L^c_s\Bigg ] \nonumber \\&\qquad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\sigma {\widehat{V}}_x (X_{s-},s) \textrm{d}W_s\Bigg ]. \end{aligned}$$

(A.1)

The boundedness of ${\widehat{V}}_x$ implies the last expectation above is zero. The jump part of $X_s$, i.e. $\Delta X_s=X_s-X_{s-}$, can be decomposed into $\Delta X_s=-H_s(Z_1) \Delta N_s-\Delta L_s,$ where $\Delta N_s=N_s-N_{s-}$, so $X_s+\Delta L_s=X_{s-}-H_s(Z_1) \Delta N_s,$ then the second expectation on the right hand side of (A.1) can be decomposed into

$$\begin{aligned}{} & {} {\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\left( {\widehat{V}}(X_s,s) -{\widehat{V}}(X_s+\Delta L_s,s)\right) \Bigg ]\\{} & {} \quad +{\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\left( {\widehat{V}}(X_{s-}-H_s(Z_1) \Delta N_s,s)-{\widehat{V}}(X_{s-},s)\right) \Bigg ], \end{aligned}$$

moreover, the above second term

$$\begin{aligned}&{\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\Big ({\widehat{V}}(X_{s-}-H_s(Z_1) \Delta N_s,s)-{\widehat{V}}(X_{s-},s)\Big )\Bigg ]\\&\quad ={\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\Big ({\widehat{V}}(X_{s-}-H_s(Z_1),s)-{\widehat{V}}(X_{s-},s)\Big )\Delta N_s\Bigg ]\\&\quad ={\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T} e^{-c (s-t)}\Big ({\widehat{V}}(X_{s-}-H_s(Z_1) ,s)-{\widehat{V}}(X_{s-},s)\Big ) \lambda \textrm{d}s\Bigg ]\\&\quad ={\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T} e^{-c (s-t)}\Big ({\mathbb E}_{s,X_{s-}}[{\widehat{V}}(X_{s-}-H_s(Z_1) ,s)]-{\widehat{V}}(X_{s-},s)\Big ) \lambda \textrm{d}s\Bigg ], \end{aligned}$$

where the second equality is followed from the compensated Poisson process $\{N_s- \lambda s\}_{s\geqslant t-}$ is a martingale (see [32]), the last equality is followed from the double expectations formula. Therefore, (A.1) becomes

$$\begin{aligned} {\widehat{V}}(x,t)&={\mathbb E}_{t,x}\Bigg [e^{-c (\tau \wedge T-t)}{\widehat{V}}(X_{\tau \wedge T},\tau \wedge T)\Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\left( {\widehat{V}}(X_{s},s) -{\widehat{V}}(X_s+\Delta L_s,s)\right) \Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\Big ({\widehat{V}}_t+\frac{\sigma ^2}{2}{\widehat{V}}_{xx}+\gamma {\widehat{V}}_x -(c+\lambda ) {\widehat{V}}\Big )(X_{s-},s) \textrm{d}s\Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\lambda \textrm{d}s \int _0^\infty \Big ((1+\delta )H_s(z){\widehat{V}}_x(X_{s-},s)\nonumber \\&\quad + {\widehat{V}}(X_{s-}-H_s(z) ,s)\Big )\textrm{d}F(z)\Bigg ]\nonumber \\&\quad +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)} {\widehat{V}}_x (X_{s-},s)\textrm{d}L^c_s\Bigg ] \end{aligned}$$

(A.2)

The first expectation is non-negative, the variational inequality in (3.1) implies the third expectation together with the fourth expectation is non-positive. Combining with ${\widehat{V}}_x \geqslant 1$ and $\Delta L_s> 0$, we have

$$\begin{aligned} {\widehat{V}}(X_s+\Delta L_s,s)-{\widehat{V}}(X_{s},s)\geqslant \Delta L_s \end{aligned}$$

so

$$\begin{aligned} {\widehat{V}}(x,t)&\geqslant {\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau \wedge T}}e^{-c (s-t)}\Delta L_s\Bigg ] +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\textrm{d}L^c_s\Bigg ] \\&= {\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\textrm{d}L_s\Bigg ]. \end{aligned}$$

Since $({\mathbb H}^t,{\mathbb I}^t)$ is arbitrary, we obtain ${\widehat{V}}(x,t)\geqslant V(x,t)$.

We now prove the opposite inequality ${\widehat{V}}(x,t)\leqslant V(x,t)$ and $\{(L^*_s,H^*_s)\}_{s\geqslant t-}$, defined in Theorem 3.2, is an optimal strategy. Set $f(s)=d(T-s)$, then the dividend payout policy $\{L^*_s\}_{s\geqslant t-}$ is the local time of the corresponding reserve $X_s^*$ at the level f(s). By the Skorohod problem(see [33]). $\{(X^*_s,L^*_s)\}_{s\geqslant t-}$ satisfies

$$\begin{aligned}&\textrm{d}X^*_s= (\gamma +\lambda (1+\delta ){\mathbb E}_{X_{s-}}[H^*_s(Z_1)])\textrm{d}s+\sigma \textrm{d}W_s -H^*_s(Z_1)\textrm{d}N_s-\textrm{d}L^*_s,\nonumber \\&X^*_s\leqslant f(s),\quad a.s.\quad t\leqslant s\leqslant T, \end{aligned}$$

(A.3)

$$\begin{aligned}&\int _{ t }^{T\wedge \tau ^*}1_{\{X^*_s\ne f(s)\}}\textrm{d}L^*_s=0. \end{aligned}$$

(A.4)

where $\tau ^*=\inf \{s\geqslant t\mid X^*_s\leqslant 0\}$. Note that (A.2) is true when $X_s=X^*_s$, $H_s=H^*_s$ and $L_s=L^*_s$, i.e.

$$\begin{aligned} {\widehat{V}}(x,t)&= {\mathbb E}_{t,x}\Bigg [e^{-c (\tau \wedge T-t)}{\widehat{V}}(X^*_{\tau ^*\wedge T},\tau ^*\wedge T)\Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau ^*\wedge T}}e^{-c (s-t)}\Big ({\widehat{V}}(X^*_s,s) -{\widehat{V}}(X^*_s+\Delta L^*_s,s)\Big )\Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\Big ({\widehat{V}}_t+\frac{\sigma ^2}{2}{\widehat{V}}_{xx}+\gamma {\widehat{V}}_x -(c+\lambda ) {\widehat{V}}\Big )(X^*_{s-},s) \textrm{d}s\Bigg ]\nonumber \\&\quad -{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau \wedge T}e^{-c (s-t)}\lambda \textrm{d}s \int _0^\infty \Big ((1+\delta )H^*_s(z){\widehat{V}}_x(X_{s-},s)\nonumber \\&\quad + {\widehat{V}}(X^*_{s-}-H^*_s(z) ,s)\Big )\textrm{d}F(z)\Bigg ]\nonumber \\&\quad +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau ^*\wedge T}e^{-c (s-t)} {\widehat{V}}_x (X^*_{s-},s)\textrm{d}L^{*c}_s\Bigg ]. \end{aligned}$$

(A.5)

Due to $f(T)=d(0)=0$ and (A.3), we have $X^*_T= 0$, so $X^*_{\tau ^*\wedge T}=0$, so the first expectation above is zero. The variational inequality in (3.1) implies the third expectation together with the fourth expectation is zero. Owing to (A.4) and ${\widehat{V}}_x(f(s),s)=1$,

$$\begin{aligned} {\widehat{V}}_x(X^*_{s-},s)\textrm{d}L^{*c}_s={\widehat{V}}_x(f(s),s)\textrm{d}L^{*c}_s=\textrm{d}L^{*c}_s. \end{aligned}$$

Therefore (A.5) becomes

$$\begin{aligned} {\widehat{V}}(x,t)&= {\mathbb E}_{t,x}\Bigg [\sum _{t\leqslant s{\leqslant } {\tau ^*\wedge T}}e^{-c (s-t)}\Big ({\widehat{V}}(X^*_s+\Delta L^*_s,s)-{\widehat{V}}(X^*_s,s) \Big )\Bigg ]\nonumber \\&\quad +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau ^*\wedge T}e^{-c (s-t)}\textrm{d}L^{*c}_s\Bigg ]. \end{aligned}$$

(A.6)

Now, if $x\leqslant f(t)$, $L^*_s$ is continuous and $\Delta L^*_s=0$ for all $t\leqslant s\leqslant T$, so

$$\begin{aligned} {\widehat{V}}(x,t) ={\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau ^*\wedge T}e^{-c (s-t)}\textrm{d}L^{*}_s\Bigg ]; \end{aligned}$$

And if $x>f(t)$, then $L^*_s$ is only discontinuous at $s=t$ and $\Delta L^*_t=x-f(t).$ Noting that ${\widehat{V}}_x (y,t)=1$ for $y\in (f(t),x)$, thus (A.6) becomes

$$\begin{aligned} {\widehat{V}}(x,t)&={\mathbb E}_{t,x}[e^{-c (t-t)}(x-f(t))] +{\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau ^*\wedge T}e^{-c (s-t)}\textrm{d}L^{*c}_s\Bigg ]\\&={\mathbb E}_{t,x}\Bigg [\int _{ t }^{\tau ^*\wedge T}e^{-c (s-t)}\textrm{d}L^{*}_s\Bigg ]. \end{aligned}$$

In view of the definition of V(x, t), we obtain ${\widehat{V}}(x,t)\leqslant V(x,t)$. Therefore, ${\widehat{V}}(x,t)= V(x,t)$ and $\{(L^*_s,H^*_s)\}_{s\geqslant t-}$ is an optimal strategy.

Appendix B Proof of Proposition 4.1

In this section, we prove Proposition  4.1. To prove the strict monotonicity, we first derive the equation on $u_t$ in $\mathcal{N}\mathcal{D}$. Differentiating the equation in (3.7) w.r.t. t we have

$$\begin{aligned}{} & {} \Big ({\partial }_t u_t-\mathcal{L}u _t+(\mathcal{Y}u){\partial }_x u _t+{\partial }_t(\mathcal{Y}u)u _x\Big )(x,t)\nonumber \\{} & {} \quad +\lambda \int _0^x 1_{\mathbb B}[(1+\delta )u _t(x,t)-u _t(x-z,t)]\textrm{d}F(z)=0\quad \hbox {in}\; \mathcal{N}\mathcal{D}, \end{aligned}$$

(B.1)

where ${\mathbb B}:=\{(1+\delta )u(x,t)-u(x-z,t)\geqslant 0\}$. Noting that

$$\begin{aligned} {\partial }_t (\mathcal{Y}u)= & {} \lambda (1+\delta )\int _0^\infty {\partial }_t Y^u(x,t,z) \textrm{d}F(z)\nonumber \\= & {} \lambda (1+\delta ) \int _0^\infty \frac{(1+\delta )u_t(x,t)-u_t(Y^u(x,t,z),t)}{u_x(Y^u(x,t,z),t)} 1_{{\mathbb D}}\textrm{d}F(z), \end{aligned}$$

(B.2)

where ${\mathbb D}:=\{u((x-z)^+,t)>(1+\delta )u(x,t)\}$, (B.1) becomes

$$\begin{aligned}{} & {} \Big ({\partial }_t u_t-\mathcal{L}u _t+(\mathcal{Y}u){\partial }_x u _t\Big )(x,t)\nonumber \\{} & {} \qquad +\left[ \lambda \int _0^\infty \frac{u _x(x,t)}{u_x(Y^u(x,t,z),t)}(1+\delta )^21_{{\mathbb D}}\textrm{d}F (z)\right] u_t(x,t)\nonumber \\{} & {} \qquad +\left[ \lambda \int _0^x (1+\delta )1_{\mathbb B}\textrm{d}F (z)\right] u _t(x,t)\nonumber \\{} & {} = \lambda \int _0^\infty \frac{(1+\delta )u _x(x,t)}{u_x(Y^u(x,t,z),t)}1_{{\mathbb D}}u_t(Y^u(x,t,z),t) \textrm{d}F (z)\nonumber \\{} & {} \qquad + \lambda \int _0^x (1+\delta )1_{\mathbb B}u _t(x-z,t) \textrm{d}F (z)\geqslant 0 \end{aligned}$$

(B.3)

in $\mathcal{N}\mathcal{D}$. Here, all integrations in [...] regarded as coefficients are bounded.

If $d(\cdot )$ is not strict monotonicity, there exists $0<t_1<t_2$ such that $d(t_1)=d(t_2)$, denote $x_0=d(t_1)=d(t_2)$ and $\Gamma =\{x_0\}\times [t_1,t_2]$, then on the line $\Gamma$, there exists

$$\begin{aligned} u=1,\quad u_t=0,\quad u_x=0,\quad u_{xt}=0. \end{aligned}$$

(B.4)

Recalling that $u_t\geqslant 0$ in $\mathcal{N}\mathcal{D}$ and $u_t=0$ on $\Gamma$, by the Hopf’s boundary point theorem, either $u_t\equiv 0$ in $\mathcal{N}\mathcal{D}$ or $u_{xt}<0$ in $\Gamma$, but the former leads to $u\equiv 1$ in $\mathcal{Q}_T$, the latter contradicts to (B.4).

Appendix C Proof of Lemma 5.1

In this section, we prove Lemma 5.1. In order to convert the mixed boundary condition to a second boundary condition, we first make a transformation $w=e^{\eta x}(u_1-u_2)$, then w satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} w_t-\mathcal{L}_3 w-g +{J}'(\cdot )w\leqslant 0\quad \hbox {in } {\mathcal{Q}_T^L},\\ w_x(0,t)\geqslant 0 \;(\textrm{or}\,w(0,t)\leqslant 0),\; w(L,t)\leqslant 0,\\ w(x,0)\leqslant 0. \end{array} \right. \end{aligned}$$

(C.1)

where

$$\begin{aligned} \mathcal{L}_3 w&:=\frac{1}{2}\sigma ^2 w_{xx}+\Big ( b -\sigma ^2\eta \Big )w_x+\Big ( d - b \eta +\frac{1}{2}\sigma ^2\eta ^2\Big )w,\\ g\;&:=e^{\eta x}\Big (G (u_1(\cdot ,t),x,t)-G (u_2(\cdot ,t),x,t)\Big ). \end{aligned}$$

Multiplying the equation in (C.1) by $w^+$ and integrating in ${\mathcal{Q}_\tau ^L}:=[0,L]\times [0,\tau ]$, $\tau \in [0,T]$, we obtain

$$\begin{aligned}{} & {} \frac{1}{2}\int _0^L (w^+)^2(x,\tau ) \textrm{d}x-\frac{1}{2}\int _0^L (w^+)^2(x,0) \textrm{d}x\nonumber \\{} & {} \quad +\frac{\sigma ^2}{2}\Bigg (\int _{\mathcal{Q}_\tau ^L} ({\partial }_x w^+)^2 \textrm{d}x \textrm{d}t-\int _0^\tau (w^+{\partial }_x w)(L,t)\textrm{d}t+\int _0^\tau (w^+{\partial }_x w)(0,t)\textrm{d}t\Bigg )\nonumber \\{} & {} \quad -\int _{\mathcal{Q}_\tau ^L}\Big ( b -\sigma ^2\eta \Big ) w^+ {\partial }_x w^+\textrm{d}x \textrm{d}t-\int _{\mathcal{Q}_\tau ^L} \Big ( d - b \eta +\frac{1}{2}\sigma ^2\eta ^2\Big ) (w^+)^2 \textrm{d}x \textrm{d}t\nonumber \\{} & {} \quad +\int _{\mathcal{Q}_\tau ^L} {J}'(\cdot )(w^+)^2 \textrm{d}x \textrm{d}t= \int _{\mathcal{Q}_\tau ^L}g w^+ \textrm{d}x \textrm{d}t. \end{aligned}$$

(C.2)

The initial condition in (C.1) yields $\frac{1}{2}\int _0^L (w^+)^2(x,0) \textrm{d}x=0$. The right boundary condition in (C.1) yields $\int _0^\tau (w^+{\partial }_x w)(L,t)\textrm{d}t=0$ and the left boundary condition yields $\int _0^\tau (w^+{\partial }_x w)(0,t)\textrm{d}t$ $\geqslant 0$. Moreover, by the assumption of G and the embedding theorem, we have

$$\begin{aligned} g(x,t)\leqslant & {} C e^{\eta L} \Big ( | w^+(\cdot ,t) | _{\textrm{L}^\infty [0,L]}+|w(x,t)|\Big )\\\leqslant & {} \epsilon | {\partial }_x w^+(\cdot ,t) | _{\textrm{L}_2 [0,L]}+C_\epsilon | w^+(\cdot ,t) | _{\textrm{L}_2 [0,L]}+C e^{\eta L} |w(x,t)| \end{aligned}$$

for any $\epsilon >0$ and $C_\epsilon$ only depends on $\epsilon$. By using the Cauchy’s inequality, the right hand side of (C.2)

$$\begin{aligned} \int _{\mathcal{Q}_\tau ^L}g w^+ \textrm{d}x \textrm{d}t\leqslant \frac{\epsilon L}{2} \int _{\mathcal{Q}_\tau ^L} ({\partial }_x w^+)^2\textrm{d}x \textrm{d}t + \frac{C_\epsilon L+1+C e^{\eta L}}{2} \int _{\mathcal{Q}_\tau ^L}(w^+)^2\textrm{d}x \textrm{d}t. \end{aligned}$$

Choosing $\epsilon L=\sigma ^2/4$ and using the Cauchy’s inequality again, there exists a constant $C_1>0$ such that

$$\begin{aligned} \int _0^L (w^+)^2(x,\tau ) \textrm{d}x\leqslant C_1 \int _{\mathcal{Q}_\tau ^L} (w^+)^2 \textrm{d}x \textrm{d}t. \end{aligned}$$

Applying the Gronwall’s inequality we deduce that $w^+=0,$ which implies $u_1\leqslant u_2.$

Appendix D Hölder Space and $W^{l,l/2}_p(Q_T)$ Space

In this section, we give the definitions of Hölder space and $W^{l,l/2}_p(Q_T)$ space with parabolic distance which we used in this paper.

1.1 D.1 Parabolic Distance

Recall that $\mathbb {R}^n$ is the n-dimensional Euclidean space, the point on which is denoted by $x=(x_1,x_2,\cdots ,x_n)$. By introducing a time variable t, denote the point on the constructed $n+1$-dimensional space $\mathbb {R}^{n+1}$ as $X=(x,t_X)$. We now introduce the distance in $\mathbb {R}^{n+1}$. $\delta (X,Y)$ is called the parabolic distance if

$$\begin{aligned} \delta (X,Y)=\textrm{max}\{|x-y|,|t_X-t_Y|^{\frac{1}{2}}\}. \end{aligned}$$

Let $\Omega$ be a bounded region in $\mathbb {R}^n$ and $Q_T=\Omega \times (0,T]$ is a cylinder in $\mathbb {R}^{n+1}$. Now we introduce some spaces.

1.2 D.2 Hölder Space

For $0<\alpha \leqslant 1$, let $C^\alpha (\overline{Q_T};\delta )$ denote the normed linear space composed by all functions u, where

$$\begin{aligned}{}[u]_{\alpha ,Q_T}:=\sup _{\begin{array}{c} X,Y\in Q_T\\ X\ne Y \end{array}}\frac{|u(X)-u(Y)|}{\delta (X,Y)^\alpha }<\infty . \end{aligned}$$

The norm is defined by

$$\begin{aligned} |u|_{\alpha ,Q_T}=\sup _{Q_T}|u|+[u]_{\alpha ,Q_T}. \end{aligned}$$

In order to describe spaces of higher orders, we introduce the definition of semi-norm as follows:

$$\begin{aligned} \begin{aligned} |u|_{0,Q_T}&=\sup _{Q_T}|u|, \\ [u]_{\alpha ,Q_T}&=\sup _{\begin{array}{c} X,Y\in Q_T\\ X\ne Y \end{array}}\frac{|u(X)-u(Y)|}{\delta (X,Y)^\alpha }, \quad 0<\alpha<1,\\ [u]_{\alpha ,Q_T}^{t}&=\sup _{\begin{array}{c} x\in \Omega .t\ne r\\ t,r\in [o,T] \end{array}}\frac{|u(x,t)-u(x,\tau )|}{|t-\tau |^\alpha },\quad 0<\alpha <1. \\ \end{aligned} \end{aligned}$$

The semi-norm of higher derivatives is introduced as

$$\begin{aligned}{}[u]_{k+\alpha ,Q_T}=\left\{ \begin{array}{ll} \sum \limits _{r+2s=k}[D_t^sD_x^ru]_{\alpha ,Q_T}, \quad k\mathrm {\; is\; even}, \\ \sum \limits _{r+2s=k}[D_t^sD_x^ru]_{\alpha ,Q_T}+\sum \limits _{r+2s=k-1}[D_t^sD_x^ru]_{\frac{1+\alpha }{2},Q_T}^t,\quad k\mathrm {\; is\; odd}. \end{array} \right. \end{aligned}$$

The linear space composed by functions u in ${C}(\overline{Q}_T)$ where

$$\begin{aligned} |u|_{k+\alpha ,Q_T}:=\sum \limits _{0\leqslant r+2s\leqslant k}[D_t^sD_x^ru]_{0,Q_T}+[u]_{k+\alpha ,Q_T}<\infty \end{aligned}$$

is denoted by $C^{k+\alpha ,\frac{k+\alpha }{2}}(\overline{Q}_T)$ or $C^{k+\alpha }(\overline{Q}_T;\delta )$. After introducing the norm $|u|_{k+\alpha ,Q_T}$, it will turn into a Banach space.

1.3 D.3 $W^{l,l/2}_p(Q_T)$ Space

For a positive integer l, $1\leqslant p<\infty$, denote

$$\begin{aligned} | u | _{W_p^{l,\frac{l}{2}}(Q_T)}=\left\{ \begin{array}{ll} \bigg \{\sum \limits _{0\leqslant r+2s\leqslant l} | D_t^sD_x^ru | ^p_{L^p(Q_T)}\bigg \}^{\frac{1}{p}}, \quad l\mathrm {\; is\; even}, \\ \bigg \{\sum \limits _{0\leqslant r+2s\leqslant l} | D_t^sD_x^ru | ^p_{L^p(Q_T)}+\sum \limits _{0\leqslant r+2s\leqslant l-1}[D_t^sD_x^ru]^p_{L^{\frac{1}{2}}_{p,t}(Q_T)}\bigg \}^{\frac{1}{p}},\quad l\mathrm {\; is\; odd}. \end{array} \right. \end{aligned}$$

where,

$$\begin{aligned}{}[u]_{L^{\frac{1}{2}}_{p,t}(Q_T)}=\Bigg \{\int _\Omega \textrm{d}x \int _0^T \textrm{d}t \int _0^T\frac{|u(x,t)-u(x,\tau )|^p}{|t-\tau |^{1+p/2}} \textrm{d}\tau \Bigg \}^{1/p}. \end{aligned}$$

Let $W_p^{l,\frac{l}{2}}(Q_T)$ denote the normed linear space composed by functions u where

$$\begin{aligned} | u | _{W_p^{l,\frac{l}{2}}(Q_T)}<\infty . \end{aligned}$$