Omega Model for a Jump-Diffusion Process with a Two-Step Premium Rate and a Threshold Dividend Strategy

Abstract

In this paper, a jump-diffusion Omega model with a two-step premium rate and a threshold dividend strategy is studied. For this model, the surplus process is a perturbation of a compound Poisson process by a Brownian motion. Firstly, using the strong Markov property, the integro-differential equations for the expected discounted dividend payments function, the Gerber-Shiu expected discounted penalty function and bankruptcy probability are derived. Secondly, for a constant bankruptcy rate function, the renewal equations satisfied by the expected discounted dividend payments function and the Gerber-Shiu expected discounted penalty function are obtained, respectively, and by iteration, their closed-form solutions are also given. Furthermore, the explicit solutions of the two kinds of functions are obtained when the individual claim size is subject to exponential distribution. Finally, a numerical example is presented to illustrate some properties of the Omega model.

Appendix

Theorem 8

The integro-differential equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} \frac{\sigma^{2}}{2} {v}_{1}^{\prime\prime}(u)+c_{1} v_{1}^{\prime}(u)-(\lambda+\delta+\omega) v_{1}(u)+\lambda {\int}_{0}^{+\infty} v_{1}(u-x) p(x) \mathrm{d}x=f(u), \\ v_{1}(-\infty)=0,\ \ \ \ v_{1}(0)=a_{1}(0), \end{array} \right. \end{array} $$

the general solution of v₁(u) is a renewal equation of the form

$$ \begin{array}{@{}rcl@{}} &v_{1}(u)={\varDelta}_{1}(u)+\widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)F_{1}(u,z)\mathrm{d}z, \end{array} $$

where $\widehat {\lambda }=\frac {2 \lambda }{\sigma ^{2}}$, and

$$ \begin{array}{@{}rcl@{}} &&F_{1}(u,z)=-{{\int}_{z}^{0}}K(u,y)p(y-z)\mathrm{d}y, \end{array} $$

(56)

$$ \begin{array}{@{}rcl@{}} &&{\varDelta}_{1}(u)=a_{1}(0)f_{0}(u)+\frac{2}{\sigma^{2}}{\int}_{-\infty}^{0} f(y)K(u,y)\mathrm{d}y, \end{array} $$

(57)

$$ \begin{array}{@{}rcl@{}} &&f_{0}(u)=e^{u}-\widehat{\lambda}{\int}_{-\infty}^{0}e^{z}F_{1}(u,z)\mathrm{d}z +\frac{2}{\sigma^{2}}{\int}_{-\infty}^{0}f_{1}(y)K(u,y)\mathrm{d}y, \end{array} $$

(58)

$$ \begin{array}{@{}rcl@{}} && f_{1}(u)=e^{u}(-\lambda{\int}_{0}^{+\infty}e^{-x}p(x)\mathrm{d}x-\frac{\sigma^{2}}{2} -c_{1}+\lambda+\delta+\omega), \end{array} $$

(59)

and K(u,y) is defined by Eqs. 60 and 61.

Proof

Let K(u,y) be the solution of the following equation

$$ \begin{array}{@{}rcl@{}} \left \{ \begin{array}{ll} \frac{\sigma^{2}}{2} {K}_{yy}^{\prime\prime}(u,y)-c_{1} {K}_{y}^{\prime}(u,y)-(\lambda+\delta+\omega) K(u,y)=0, \\ K(u,-\infty)=0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ K(u,0)=0,\\ K(u,u+)=K(u,u-),\ \ \ \ \ {K}_{y}^{\prime}(u,u+)-{K}_{y}^{\prime}(u,u-)=1. \end{array} \right. \end{array} $$

(60)

Then the solution is

$$ \begin{array}{@{}rcl@{}} {K(u,y)=}\left \{ \begin{array}{ll} C_{0}(u)e^{{\beta}_{1}y}, \ \ \ \ &(-\infty<y\leq u), \\ C_{1}(u)(e^{{\beta}_{1}y}-e^{{\beta}_{2}y}), \ \ \ \ &(u<y\leq 0), \end{array} \right. \end{array} $$

(61)

where β₁ > 0 > β₂ are two roots of the equation $\frac {\sigma ^{2}}{2} {\beta }^{2}-c_{1} {\beta }-(\lambda +\delta +\omega )=0$, and

$$ \begin{array}{@{}rcl@{}} &C_{0}(u)=\frac{e^{{\beta}_{1}u}-e^{{\beta}_{2}u}}{({\beta}_{1}-{\beta}_{2}) e^{({\beta}_{1}+{\beta}_{2})u}},\ \ \ C_{1}(u)=\frac{1}{e^{{\beta}_{2}u}({\beta}_{1}-{\beta}_{2})}. \end{array} $$

Let

$$ W_{1}(u)=v_{1}(u)-a_{1}(0)e^{u}, $$

then W₁(u) satisfies the following equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} \frac{\sigma^{2}}{2} W_{1}^{\prime\prime}(u)+c_{1} {W}_{1}^{\prime}(u)-(\lambda+\delta+\omega) W_{1}(u)=f(u)-\lambda {\int}_{0}^{+\infty} W_{1}(u - x) p(x) \mathrm{d}x+a_{1}(0)f_{1}(u), \\ W_{1}(-\infty)=0,\ \ \ \ W_{1}(0)=0, \end{array} \right. \end{array} $$

where f₁(u) is defined as Eq. 59. Multiplying both sides of the above equation by K(u,y) and integrating from $-\infty$ to 0, integration by part and in view of Eq. 60, one can get

$$ \begin{array}{@{}rcl@{}} &W_{1}(u) = -\widehat{\lambda}{\int}_{-\infty}^{0}K(u,y){\int}_{0}^{+\infty}W_{1}(y - x)p(x) \mathrm{d}x\mathrm{d}y+\frac{2}{\sigma^{2}}{\int}_{-\infty}^{0}[f(y)+a_{1}(0) f_{1}(y)]K(u,y)\mathrm{d}y. \end{array} $$

Let $F_{1}(u,z)=-{{\int \limits }_{z}^{0}}K(u,y)p(y-z)\mathrm {d}y$, one have

$$ \begin{array}{@{}rcl@{}} &W_{1}(u)=\widehat{\lambda}{\int}_{-\infty}^{0}W_{1}(z) F_{1}(u,z)\mathrm{d}z+ \frac{2}{\sigma^{2}}{\int}_{-\infty}^{0}[f(y)+a_{1}(0) f_{1}(y)]K(u,y)\mathrm{d}y. \end{array} $$

Substituting W₁(u) = v₁(u) − a₁(0)e^u into the above equation, one can obtained

$$ \begin{array}{@{}rcl@{}} &v_{1}(u)={\varDelta}_{1}(u)+\widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)F_{1}(u,z)\mathrm{d}z,\ \ \ \ \ (-\infty<u\leq 0), \end{array} $$

where Δ₁(u) is defined in Eq. 57. □

Then, by iteration, one can get the closed-form solution of v₁(u) that

$$ \begin{array}{@{}rcl@{}} &v_{1}(u)={\varDelta}_{1}(u)+{\sum}_{k=1}^{\infty}\widehat{\lambda}^{k}{\int}_{-\infty}^{0}F_{1}^{(k)}(u,z){\varDelta}_{1}(z)\mathrm{d}z, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} {F}_{1}^{(1)}(u,z)&=&F_{1}(u,z),\\ {F}_{1}^{(k)}(u,z)&=&{\int}_{-\infty}^{0}F_{1}(u,s)F_{1}^{(k-1)}(s,z)\mathrm{d}s, \ \ \ \ (k=2,3,\cdot\cdot\cdot). \end{array} $$

(62)

Theorem 9

The integro-differential equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} \frac{\sigma^{2}}{2} {v}_{2}^{\prime\prime}(u) + c_{2} {v}_{2}^{\prime}(u) - (\lambda + \delta) v_{2}(u)+\lambda {{\int}_{0}^{u}} v_{2}(u - x) p(x) \mathrm{d}x+\lambda {\int}_{u}^{+\infty} v_{1}(u - x) p(x) \mathrm{d}x=g(u), \\ v_{2}(0)=a_{2}(0),\ \ \ \ v_{2}(b)=a_{2}(b), \end{array} \right. \end{array} $$

the general solution of v₂(u) is a renewal equation of the form

$$ \begin{array}{@{}rcl@{}} &v_{2}(u)={\varDelta}_{2}(u)+\widehat{\lambda} {{\int}_{0}^{b}}v_{2}(z)G_{1}(u,z)\mathrm{d}z+\widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)G_{2}(u,z)\mathrm{d}z, \end{array} $$

where $\widehat {\lambda }=\frac {2 \lambda }{\sigma ^{2}}$,

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!G_{1}(u,z)&=&-{{\int}_{z}^{b}}\widehat{K}(u,y)p(y-z)\mathrm{d}y, \end{array} $$

(63)

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!G_{2}(u,z)&=&-{{\int}_{0}^{b}}\widehat{K}(u,y)p(y-z)\mathrm{d}y, \end{array} $$

(64)

$$ \begin{array}{@{}rcl@{}} \theta_{1}(u)&=&\frac{e^{-b}-e^{-u}}{e^{-b}-1}, \end{array} $$

(65)

$$ \begin{array}{@{}rcl@{}} \theta_{2}(u)&=&\frac{e^{b}-e^{b-u}}{e^{b}-1}, \end{array} $$

(66)

$$ \begin{array}{@{}rcl@{}} {\varDelta}_{2}(u)&=&a_{2}(0)g_{1}(u)+a_{2}(b)g_{2}(u)+\frac{2}{\sigma^{2}}{{\int}_{0}^{b}} g(y)\widehat{K}(u,y)\mathrm{d}y, \end{array} $$

(67)

$$ \begin{array}{@{}rcl@{}} g_{01}(u)&=&-{\lambda{\int}_{0}^{u}}\theta_{1}(u-x)p(x)\mathrm{d}x +\left( c_{2}+\lambda+\delta - \frac{\sigma^{2}}{2}\right)\theta_{1}(u) +\left( \frac{\sigma^{2}}{2} - c_{2}\right) \frac{e^{-b}}{e^{-b}-1},\ \ \ \end{array} $$

(68)

$$ \begin{array}{@{}rcl@{}} g_{02}(u)&=&-{\lambda{\int}_{0}^{u}}\theta_{2}(u-x)p(x)\mathrm{d}x +\left( c_{2}+\lambda+\delta-\frac{\sigma^{2}}{2}\right)\theta_{2}(u) +\left( \frac{\sigma^{2}}{2}-c_{2}\right)\frac{e^{b}}{e^{b}-1},\ \ \ \ \end{array} $$

(69)

$$ \begin{array}{@{}rcl@{}} g_{1}(u)&=&-\widehat{\lambda}{{\int}_{0}^{b}}\theta_{1}(z)G_{1}(u,z)\mathrm{d}z+ \frac{2}{\sigma^{2}}{{\int}_{0}^{b}}g_{01}(y)\widehat{K}(u,y)\mathrm{d}y +\theta_{1}(u), \end{array} $$

(70)

$$ \begin{array}{@{}rcl@{}} g_{2}(u)&=&-\widehat{\lambda}{{\int}_{0}^{b}}\theta_{2}(z)G_{1}(u,z)\mathrm{d}z+ \frac{2}{\sigma^{2}}{{\int}_{0}^{b}}g_{02}(y)\widehat{K}(u,y)\mathrm{d}y +\theta_{2}(u), \end{array} $$

(71)

and $\widehat {K}(u,y)$ is defined by Eqs. 72 and 73.

Proof

Let $\widehat {K}(u,y)$ be the solution of the following equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} \frac{\sigma^{2}}{2} \widehat{K}_{yy}^{\prime\prime}(u,y)-c_{2} \widehat{K}_{y}^{\prime}(u,y)-(\lambda+\delta) \widehat{K}(u,y)=0, \\ \widehat{K}(u,0)=0,\ \ \ \ \widehat{K}(u,b)=0,\\ \widehat{K}(u,u+)=\widehat{K}(u,u-),\ \ \ \ \ \widehat{K}_{y}^{\prime}(u,u+)-\widehat{K}_{y}^{\prime}(u,u-)=1. \end{array} \right. \end{array} $$

(72)

Then the solution is

$$ \begin{array}{@{}rcl@{}} {\widehat{K}(u,y)=}\left \{ \begin{array}{ll} C_{2}(u)(e^{\xi_{1}y}-e^{\xi_{2}y}), \ \ \ \ &(0<y\leq u), \\ C_{3}(u)e^{\xi_{1}y}+C_{4}(u)e^{\xi_{2}y}, \ \ \ \ &(u<y\leq b), \end{array} \right. \end{array} $$

(73)

where ξ₁ > 0 > ξ₂ are two roots of the equation $\frac {\sigma ^{2}}{2} \xi ^{2}-c_{2} \xi -(\lambda +\delta )=0$, and

$$ \begin{array}{@{}rcl@{}} C_{2}(u)&=&\frac{e^{\xi_{1}u}-e^{(\xi_{1}-\xi_{2})b+\xi_{2}u}}{(\xi_{1}-\xi_{2}) e^{(\xi_{1}+\xi_{2})u}\left[e^{(\xi_{1}-\xi_{2})b}-1\right]},\ \ C_{3}(u)=\frac{e^{\xi_{1}u}-e^{\xi_{2}u}}{(\xi_{1}-\xi_{2})e^{(\xi_{1}+\xi_{2})u} \left[e^{(\xi_{1}-\xi_{2})b}-1\right]},\\ C_{4}(u)&=&-C_{3}(u)e^{(\xi_{1}-\xi_{2})b}. \end{array} $$

Let W₂(u) = v₂(u) − a₂(0)𝜃₁(u) − a₂(b)𝜃₂(u), where 𝜃₁(u) and 𝜃₂(u) are defined in Eqs. 65 and 66. Then W₂(u) satisfies the following equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} &\frac{\sigma^{2}}{2} {W}_{2}^{\prime\prime}(u) + c_{2} {W}_{2}^{\prime}(u) - (\lambda+\delta) W_{2}(u) = g(u) - \lambda {{\int}_{0}^{u}} W_{2}(u - x) p(x) \mathrm{d}x - \lambda{\int}_{u}^{+\infty}v_{1}(u - x)p(x)\mathrm{d}x \\&\ \ +a_{2}(0)g_{01}(u)+a_{2}(b) g_{02}(u), \\ &W_{2}(0)=0,\ \ \ \ W_{2}(b)=0, \end{array} \right. \end{array} $$

where g₀₁(u) and g₀₂(u) are defined as Eqs. 68 and 69. Multiplying both sides of the above equation by $\widehat {K}(u,y)$ and integrating from 0 to b, integration by part and in view of Eq. 72, one can get

$$ \begin{array}{@{}rcl@{}} &W_{2}(u)=-\widehat{\lambda}{{\int}_{0}^{b}}\widehat{K}(u,y){{\int}_{0}^{y}}W_{2}(y-x)p(x) \mathrm{d}x\mathrm{d}y-\widehat{\lambda}{{\int}_{0}^{b}}\widehat{K}(u,y){\int}_{y}^{+\infty} v_{1}(y-x)p(x) \mathrm{d}x\mathrm{d}y \\& +\frac{2}{\sigma^{2}}{{\int}_{0}^{b}}[g(y)+a_{2}(0) g_{01}(y)+a_{2}(b)g_{02}(y)]K(u,y)\mathrm{d}y. \end{array} $$

Denote $G_{1}(u,z)=-{{\int \limits }_{z}^{b}}\widehat {K}(u,y)p(y-z)\mathrm {d}y$ and $G_{2}(u,z)=-{{\int \limits }_{0}^{b}}\widehat {K}(u,y)p(y-z)\mathrm {d}y$, one have

$$ \begin{array}{@{}rcl@{}} W_{2}(u)&=&\widehat{\lambda}{{\int}_{0}^{b}}W_{2}(z) G_{1}(u,z)\mathrm{d}z+ \widehat{\lambda}{\int}_{-\infty}^{0}v_{1}(z) G_{2}(u,z)\mathrm{d}z+ \frac{2}{\sigma^{2}}{{\int}_{0}^{b}}[g(y)+a_{2}(0) g_{01}(y)\\&&+a_{2}(b) g_{02}(y)]K(u,y)\mathrm{d}y. \end{array} $$

Substituting W₂(u) = v₂(u) − a₂(0)𝜃₁(u) − a₂(b)𝜃₂(u) into the above equation, one can obtained

$$ \begin{array}{@{}rcl@{}} &v_{2}(u)={\varDelta}_{2}(u)+\widehat{\lambda} {{\int}_{0}^{b}}v_{2}(z)G_{1}(u,z)\mathrm{d}z+\widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)G_{2}(u,z)\mathrm{d}z, (0<u\leq b), \end{array} $$

where Δ₂(u) is defined in Eq. 67. □

Then, by iteration, one can get the closed-form solution of v₂(u) that

$$ \begin{array}{@{}rcl@{}} &v_{2}(u)=\widehat{{\varDelta}}_{2}(u)+{\sum}_{k=1}^{\infty}\widehat{\lambda}^{k}{{\int}_{0}^{b}}G_{1}^{(k)}(u,z)\widehat{{\varDelta}}_{2}(z)\mathrm{d}z, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} \widehat{{\varDelta}}_{2}(u)&=&{\varDelta}_{2}(u)+\widehat{\lambda}{\int}_{-\infty}^{0}v_{1}(z) G_{2}(u,z)\mathrm{d}z,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ {G}_{1}^{(1)}(u,z)&=&G_{1}(u,z),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ {G}_{1}^{(k)}(u,z)&=&{{\int}_{0}^{b}}G_{1}(u,s)G_{1}^{(k-1)}(s,z)\mathrm{d}s, \ \ \ \ \ (k=2,3,\cdot\cdot\cdot). \end{array} $$

Theorem 10

The integro-differential equation

$$ \begin{array}{@{}rcl@{}} \left \{\begin{array}{ll} &\frac{\sigma^{2}}{2} {v}_{3}^{\prime\prime}(u)+(c_{2}-\alpha) v_{3}^{\prime}(u)-(\lambda+\delta) v_{3}(u)+\lambda {\int}_{0}^{u-b} v_{3}(u-x) p(x) \mathrm{d}x +\lambda {\int}_{u-b}^{u} v_{2}(u-x) p(x) \mathrm{d}x\\ &\ \ +\lambda{\int}_{u}^{+\infty}v_{1}(u-x)p(x)\mathrm{d}x=h(u), \\ &v_{3}(b)=a_{3}(b),\ \ \ \ v_{3}(+\infty)=0, \end{array} \right. \end{array} $$

the general solution of v₃(u) is a renewal equation of the form

$$ \begin{array}{@{}rcl@{}} &v_{3}(u) = {\varDelta}_{3}(u) + \widehat{\lambda} {\int}_{b}^{+\infty}v_{3}(z)H_{1}(u,z)\mathrm{d}z + \widehat{\lambda} {{\int}_{0}^{b}}v_{2}(z)H_{2}(u,z)\mathrm{d}z + \widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)H_{2}(u,z)\mathrm{d}z, \end{array} $$

where $\widehat {\lambda }=\frac {2 \lambda }{\sigma ^{2}}$,

$$ \begin{array}{@{}rcl@{}} H_{1}(u,z)&=&-{\int}_{z}^{+\infty}\widetilde{K}(u,y)p(y-z)\mathrm{d}y, \end{array} $$

(74)

$$ \begin{array}{@{}rcl@{}} H_{2}(u,z)&=&-{\int}_{b}^{+\infty}\widetilde{K}(u,y)p(y-z)\mathrm{d}y, \end{array} $$

(75)

$$ \begin{array}{@{}rcl@{}} {\varDelta}_{3}(u)&=&a_{3}(b)h_{0}(u)+\frac{2}{\sigma^{2}}{\int}_{b}^{+\infty} h(y)\widetilde{K}(u,y)\mathrm{d}y, \end{array} $$

(76)

$$ \begin{array}{@{}rcl@{}} \ \ \ \ \ h_{0}(u)&=&e^{b-u}+\frac{2}{\sigma^{2}}{\int}_{b}^{+\infty}h_{1}(y)\widetilde{K}(u,y)\mathrm{d}y- \widehat{\lambda}{\int}_{b}^{+\infty}e^{b-z}H_{1}(u,z)\mathrm{d}z, \end{array} $$

(77)

$$ \begin{array}{@{}rcl@{}} \ \ \ \ h_{1}(u)&=&e^{b-u}\left[-\lambda{\int}_{0}^{u-b}e^{x}p(x)\mathrm{d}x-\frac{\sigma^{2}}{2}+c_{2} -\alpha+\lambda+\delta\right], \end{array} $$

(78)

and $\widetilde {K}(u,y)$ is defined by Eqs. 79 and 80.

Proof

Let $\widetilde {K}(u,y)$ be the solution of the following equation

$$ \begin{array}{@{}rcl@{}} \left \{ \begin{array}{ll} \frac{\sigma^{2}}{2} \widetilde{K}_{yy}^{\prime\prime}(u,y)-(c_{2}-\alpha) \widetilde{K}_{y}^{\prime}(u,y)-(\lambda+\delta) \widetilde{K}(u,y)=0, \\ \widetilde{K}(u,b)=0,\ \ \ \ \widetilde{K}(u,+\infty)=0,\\ \widetilde{K}(u,u+)=\widetilde{K}(u,u-),\ \ \ \ \ \widetilde{K}_{y}^{\prime}(u,u+)-\widetilde{K}_{y}^{\prime}(u,u-)=1. \end{array} \right. \end{array} $$

(79)

Then the solution is

$$ \begin{array}{@{}rcl@{}} {\widetilde{K}(u,y)=}\left \{ \begin{array}{ll} C_{5}(u)e^{\eta_{1}y}+C_{6}(u)e^{\eta_{2}y}, \ \ \ \ &(b<y\leq u), \\ C_{7}(u)e^{\eta_{2}y}, \ \ \ \ &(u<y< +\infty), \end{array} \right. \end{array} $$

(80)

where η₁ > 0 > η₂ are two roots of the equation $\frac {\sigma ^{2}}{2} \eta ^{2}-(c_{2}-\alpha ) \eta -(\lambda +\delta )=0$, and

$$ \begin{array}{@{}rcl@{}} &C_{5}(u)=\frac{1}{(\eta_{2}-\eta_{1}) e^{\eta_{1}u}},\ C_{6}(u)=-C_{5}(u)e^{(\eta_{1}-\eta_{2})b},\ C_{7}(u)=\frac{e^{\eta_{1}u}-e^{(\eta_{1}-\eta_{2})b+\eta_{2}u}}{(\eta_{2}-\eta_{1}) e^{(\eta_{1}+\eta_{2})u}}. \end{array} $$

Let W₃(u) = v₃(u) − a₃(b)e^b−u. Then W₃(u) satisfies the following equation

$$ \begin{array}{@{}rcl@{}} \left \{\!\!\begin{array}{ll} &\frac{\sigma^{2}}{2} {W}_{3}^{\prime\prime}(u) + (c_{2} - \alpha) {W}_{3}^{\prime}(u) - (\lambda + \delta) W_{3}(u) = h(u) - \lambda {\int}_{0}^{u-b} W_{3}(u - x)p(x) \mathrm{d}x - \lambda{\int}_{u-b}^{u}v_{2}(u - x)p(x)\mathrm{d}x\\&\ \ \ -\lambda{\int}_{u}^{+\infty} v_{1}(u-x)p(x)\mathrm{d}x+a_{3}(b)h_{1}(u), \\ &W_{3}(b)=0,\ \ \ \ W_{3}(+\infty)=0, \end{array} \right. \end{array} $$

where h₁(u) is defined as Eq. 78. Multiplying both sides of the above equation by $\widetilde {K}(u,y)$ and integrating from b to $+\infty$, integration by part and in view of Eq. 79, one can get

$$ \begin{array}{@{}rcl@{}} &W_{3}(u) = -\widehat{\lambda}{\int}_{b}^{+\infty}\widetilde{K}(u,y){\int}_{0}^{y-b} W_{3}(y - x)p(x)\mathrm{d}x\mathrm{d}y - \widehat{\lambda}{\int}_{b}^{+\infty}\widetilde{K}(u,y) {\int}_{y-b}^{y}v_{2}(y - x)p(x)\mathrm{d}x\mathrm{d}y \\&\ \ \ \ \ \ \!-\widehat{\lambda}{\int}_{b}^{+\infty} \widetilde{K}(u,y){\int}_{y}^{+\infty}v_{1}(y-x)p(x)\mathrm{d}x\mathrm{d}y +\frac{2}{\sigma^{2}}{\int}_{b}^{+\infty}[h(y)+a_{3}(b) h_{1}(y)]K(u,y)\mathrm{d}y. \end{array} $$

Denote $H_{1}(u,z)=-{\int \limits }_{z}^{+\infty }\widetilde {K}(u,y)p(y-z)\mathrm {d}y$ and $H_{2}(u,z)=-{\int \limits }_{b}^{+\infty }\widetilde {K}(u,y)p(y-z)\mathrm {d}y$, one have

$$ \begin{array}{@{}rcl@{}} &W_{3}(u)=\widehat{\lambda}{\int}_{b}^{+\infty}W_{3}(z) H_{1}(u,z)\mathrm{d}z+ \widehat{\lambda}{{\int}_{0}^{b}}v_{2}(z) H_{2}(u,z)\mathrm{d}z+ \widehat{\lambda}{\int}_{-\infty}^{0}v_{1}(z) H_{2}(u,z)\mathrm{d}z \\& + \frac{2}{\sigma^{2}}{\int}_{b}^{+\infty}[h(y)+a_{3}(b) h_{1}(y)]K(u,y)\mathrm{d}y.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} $$

Substituting W₃(u) = v₃(u) − a₃(b)e^b−u into the above equation, one can obtained

$$ \begin{array}{@{}rcl@{}} v_{3}(u)&=&{\varDelta}_{3}(u)+\widehat{\lambda} {\int}_{b}^{+\infty}v_{3}(z)H_{1}(u,z)\mathrm{d}z+\widehat{\lambda} {{\int}_{0}^{b}}v_{2}(z)H_{2}(u,z)\mathrm{d}z\\&&+\widehat{\lambda} {\int}_{-\infty}^{0}v_{1}(z)H_{2}(u,z)\mathrm{d}z, \ \ (b<u<+\infty), \end{array} $$

where Δ₃(u) is defined in Eq. 76. □

Then, by iteration, one can get the closed-form solution of v₃(u) that

$$ \begin{array}{@{}rcl@{}} &v_{3}(u)=\widetilde{{\varDelta}}_{3}(u)+{\sum}_{k=1}^{\infty}\widehat{\lambda}^{k}{\int}_{b}^{+\infty}{H}_{1}^{(k)}(u,z)\widetilde{{\varDelta}}_{3}(z)\mathrm{d}z, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} &&\widetilde{{\varDelta}}_{3}(u)={\varDelta}_{3}(u)+\widehat{\lambda}{{\int}_{0}^{b}}v_{2}(z) H_{2}(u,z)\mathrm{d}z+\widehat{\lambda}{\int}_{-\infty}^{0}v_{1}(z) H_{2}(u,z)\mathrm{d}z,\\ &&{H}_{1}^{(1)}(u,z)=H_{1}(u,z), \\ &&{H}_{1}^{(k)}(u,z)={\int}_{b}^{+\infty}H_{1}(u,s){H}_{1}^{(k-1)}(s,z)\mathrm{d}s, \ \ \ \ \ (k=2,3,\cdot\cdot\cdot). \end{array} $$