Semi-Markov Models

Appendix C

The Solution of the System of Integral Equation (3.6)

Let us prove the formula (3.9) to determine the solution of (3.6).

Introduce the operator: $\left(A_r\varphi \right)(x_1,x_2)=\underset{0}{\overset{\infty }{\int }}\varphi (x_1+t,x_2+t)r(t)dt$, then the second equation of the system (3.6) can be rewritten as follows:

${\varphi }_1={\rho }_0{A}_r\left[f_1(x_1)f_2(x_2)\right]+A_r{\varphi }_1+A_r\left[f_1(x_1){\varphi }_4(x_2)\right]+A_r\left[{\varphi }_5(x_1)f_2(x_2)\right],$

then,

$\begin{array}{c}\hfill \left(I-A_r\right){\varphi }_1={\rho }_0{A}_r\left[f_1(x_1)f_2(x_2)\right]+A_r\left[f_1(x_1){\varphi }_4(x_2)\right]+A_r\left[{\varphi }_5(x_1)f_2(x_2)\right],\hfill \\ \hfill {\varphi }_1={\rho }_0{\left(I-A_r\right)}^{-1}{A}_r\left[f_1(x_1)f_2(x_2)\right]+{\left(I-A_r\right)}^{-1}{A}_r\left[f_1(x_1){\varphi }_4(x_2)\right]+\hfill \\ \hfill +{\left(I-A_r\right)}^{-1}{A}_r\left[{\varphi }_5(x_1)f_2(x_2)\right],\hfill \\ \hfill {(I-A_r)}^{-1}=I+\sum _{n=1}^{\infty }{A}_r^n,\left(A_r^n\varphi \right)(x_1,x_2)=\underset{0}{\overset{\infty }{\int }}\varphi (x_1+t,x_2+t)r^{*(n)}(t)dt,\hfill \\ \hfill \left(\sum _{n=1}^{\infty }{A}_r^n\right)\varphi =\underset{0}{\overset{\infty }{\int }}\varphi (x_1+t,x_2+t)h_r(t)dt\hfill \end{array}$

where