\magnification=1440 \input amstex \documentstyle{amsppt} \TagsOnRight \tolerance=900 \tenpoint \hsize=10cm \vsize=17cm \voffset=-10mm \hoffset=5mm \def\Re{\operatorname{Re}} \def\conv{\operatorname{conv}} \def\bPhi{{\bold \Phi}} \def\bq{{\bold q}} \def\bp{{\bold p}} \def\bI{{\bold I}} \def\bC{{\bold C}} \def\b(f){{\bold f}} \document %\pageno=-1 \noindent \centerline{\bf NUMERICAL METHOD FOR INTEGRATION OF} \centerline{\bf EQUATION OF CONVOLUTION UNDER} \centerline{\bf THE INTEGRAL RESTRICTIONS ON THE KERNEL} \bigskip \centerline{ A.P. Zhabko,\ \ A.V. Ekimov,\ \ N.V. Smirnov} \bigskip \centerline{\it Faculty of Applied Mathematics, St.-Petersburg State University} \centerline{\it Bibliotechnaja Pl.2, Petrodvorets, St.-Petersburg, 198904, Russia} \centerline{\it e-mail: Alexandr.Ekimov\@pobox.spbu.ru} \bigskip Let us consider the integral equation of a convolution [1] $$\bq(t)=\bPhi(t)\bp^{(0)}+\int\limits_0^t\bPhi(t-x)\bq(x)dx. \eqno(1)$$ It is a model of stochastic process in the case of a finite number of a states. In (1) $\bPhi(t)=\{\varphi_{ij}(t)\}_{i,j=\overline{1,n}}$ is the $(n\times n)$-matrix of the transition probability densities with the elements which are continuous functions for $t\in [0,T]$ and $\varphi_{ij}(t)\geq 0$, $i,j=\overline{1,n},\$ $t\in [0,T]$; the columns of $\bPhi(t)$ satisfy the condition of the normalization $$\sum_{i=1}^n\int\limits_0^{\infty}\varphi_{ij}(t)dt=1, \quad j=\overline{1,n};\eqno(2)$$ $\bp^{(0)}$ is a n-dimensional vector of the initial probabilities and $$\sum_{i=1}^np_i^{(0)}=1, \quad p_i^{(0)}\geq 0.$$ Note that (1) can be written in the form $$\bq(t)=\bPhi(t)\bp_0+{1\over (1+\varepsilon)}{1\over (1-\delta)} \int\limits_0^t\bPhi(t-x)\bq(x)dx, \eqno(3)$$ where $\delta={\varepsilon\over (1+\varepsilon)}$, for every $\varepsilon >0$ and $\delta\in (0,1)$. It is obvious that the series $${1\over (1-\delta)}=1+\delta+\delta^2+\ldots+\delta^n+\ldots= \sum_{k=0}^{\infty}\delta^k \eqno(4)$$ is convergent. We'll find the solution of (3) in the form of a series $$\bq(t)=\sum_{k=0}^{\infty}\bq^{(k)}(t)\delta^k. \eqno(5)$$ Substituting series (5) into eqn. (3) we obtain $$\gathered \sum_{k=0}^{\infty}\bq^{(k)}(t)\delta^k= \bPhi(t)\bp^{(0)}+{1\over (1+\varepsilon)}\sum_{k=0}^{\infty}\delta^k \times\\ \times\int\limits_0^t\bPhi(t-x)\sum_{k=0}^{\infty}\bq^{(k)}(x)\delta^kdx.\\ \endgathered\tag6$$ If we equate the coefficients in the left hand side of eqn. (6) to the coefficients in the right hand side of eqn. (6) for the same powers of $\delta$, then we get $$\bq^{(0)}(t)= \bPhi(t)\bp^{(0)}+{1\over (1+\varepsilon)} \int\limits_0^t\bPhi(t-x)\bq^{(0)}(x)dx,$$ $$\bq^{(1)}(t)={1\over (1+\varepsilon)} \int\limits_0^t\bPhi(t-x)\bq^{(1)}(x)dx+$$ $$+{1\over (1+\varepsilon)} \int\limits_0^t\bPhi(t-x)\bq^{(0)}(x)dx$$ and for $n\geq 1$ $$\gathered \bq^{(n)}(t)={1\over (1+\varepsilon)} \int\limits_0^t\bPhi(t-x)\bq^{(n)}(x)dx+\\ +{1\over (1+\varepsilon)} \sum_{k=0}^{n-1}\int\limits_0^t\bPhi(t-x)\bq^{(k)}(x)dx.\\ \endgathered\tag7$$ For all $n$ eqn. (7) is the integral equation of a convolution with the initial conditions which are $$\bq^{(0)}(0)=\bPhi(0)\bp^{(0)}, \quad \bq^{(k)}(0)=0, \quad k\geq 1.$$ \bigskip {\bf Theorem 1.} For every $T>0$ series (5) is uniform convergent in $t\in [0,T]$. \bigskip {\bf Proof.} For the proof of Theorem 1 it is sufficient to show that $\{\bq^{(n)}(t)\}$ is uniform bounded for $t\in [0,T]$. Let us use the induction on $n$. For $n=0$ $$\bq^{(0)}(t)= \bPhi(t)\bp^{(0)}+{1\over (1+\varepsilon)} \int\limits_0^t\bPhi(t-x)\bq^{(0)}(x)dx$$ and $$\|\bq^{(0)}(t)\|\leq \|\bPhi(t)\|\|\bp^{(0)}\|+{1\over (1+\varepsilon)} \int\limits_0^t\|\bPhi(t-x)\|\|\bq^{(0)}(x)\|dx.$$ Under condition (2) $$\|\bPhi(t)\|, \quad \int\limits_0^t\|\bPhi(t)\|dt$$ are bounded for $t\in[0,+\infty)$. Let $\|\bPhi(t)\|\leq M, \|\bp_0\|\leq \mu$. Then $$\|\bq^{(0)}(t)\|\leq\lambda+{1\over (1+\varepsilon)} \int\limits_0^t\|\bPhi(t-x)\|\|\bq^{(0)}(x)\|dx,$$ where $\lambda=M\mu$. By Lemma Gronwall -- Bellman [2] $$\|\bq^{(0)}(t)\|\leq\lambda\ exp\biggl( \int\limits_0^t\|\bPhi(t-x)\|dx\bigg).$$ Under condition (2) $$\int\limits_0^t\|\bPhi(t-x)\|dx\leq 1.$$ Then $$\|\bq^{(0)}(t)\|\leq \lambda\ e^{1\over (1+\varepsilon)}. \eqno(8)$$ Let us prove that $$\|\bq^{(n)}(t)\|\leq {1+\varepsilon\over \varepsilon}\ \lambda\ e^{1\over (1+\varepsilon)} \eqno(9)$$ for every $n\geq 1$. From (8) it follows that inequality (9) is true for $n=0$. For $\bq^{(1)}(t)$ we have $$\|\bq^{(1)}(t)\|\leq {1\over (1+\varepsilon)}\ \lambda\ e^{1\over (1+\varepsilon)}+ {1\over (1+\varepsilon)}\int\limits_0^t\|\bPhi(t-x)\|\|\bq^{(1)}(x)\|dx.$$ By Lemma Gronwall -- Bellman $$\|\bq^{(1)}(t)\|\leq {1\over (1+\varepsilon)}\ \lambda\ e^{({1\over (1+\varepsilon)})^2} \leq {1\over (1+\varepsilon)}\ \lambda\ e^{1\over (1+\varepsilon)}.$$ It is obvious that $$\|\bq^{(n)}(t)\|\leq \biggl(\sum_{k=1}^n{1\over (1+\varepsilon)^k}\biggr) \lambda\ e^{1\over (1+\varepsilon)}, \quad n\geq 1.$$ Since $${1+\varepsilon\over \varepsilon}= \sum_{k=0}^{\infty}{1\over (1+\varepsilon)^k},$$ then $$\sum_{k=1}^n{1\over (1+\varepsilon)^k}<{1+\varepsilon\over \varepsilon}$$ for every $n\geq 1$. Hence, inequality (9) is true. Let us construct the majorant series for every component of vector $\bq(t)$ $$\sum_{k=0}^{\infty}\|\bq^{(k)}(t)\|\delta^k \leq {1\over (1+\varepsilon)}\ \lambda\ e^{1\over (1+\varepsilon)} \sum_{k=0}^{\infty}\delta^k.$$ Hence, for $\delta\in (0,1)$ and every $\varepsilon>0, T>0$ series (5) is absolutely uniform convergent. Q.E.D. {\bf Remark 1.} The degree of convergence of series (5) depend on the choice of $\varepsilon$. Thus the algorithm of the finding of the unique solution of equation (1) in the form of series (5) is suggested. For every step of this algorithm we consider an integral equation of a convolution $$\bq(t)=\b(f)(t)+\int\limits_0^t\bPhi(t-x)\bq(x)dx. \eqno(10)$$ Construct a numerical method for eqn. (10), $t\in [0,T]$. Consider the points $$t_0=0, \quad t_k=t_0+kh, \quad k=\overline{1,m}, \quad t_m=T,$$ where $h$ is a step of integration and substitute $t=t_k$ into eqn.(10) $$\bq(t_k)=\b(f)(t_k)+\int\limits_0^{t_k}\bPhi(t_k-x)\bq(x)dx. \eqno(11)$$ For calculation of integral $$J(t_k)=\int\limits_0^{t_k}\bPhi(t_k-x)\bq(x)dx$$ we'll use the trapezoid formula $$J(t_k)\approx {h\over 2} \bPhi(t_k)\bq(0)+h\sum_{i=1}^{k-1}\bPhi(t_k-i)\bq(t_i)+ {h\over 2}\bPhi(0)\bq(t_k). \eqno(12)$$ Denote by $\b(f)_i=\b(f)(t_i)$, $\bq_i\approx\bq(t_i)$, $\bPhi_i=\bPhi(t_i)$. Substitute (12) into eqn. (11), then we get $$\Bigl(\bI-{h\over 2}\bPhi_0\Bigr)\bq_k= \b(f)_k+{h\over 2}\bPhi_k\bq_0+ h\sum_{i=1}^{k-1}\bPhi_{k-i}\bq_i,\quad k=\overline{1,m}.$$ As a result the difference approximation for $\bq_k\approx\bq(t_k)$ takes the form $$\gathered \bq_k=\Bigl(\bI-{h\over 2}\bPhi_0\Bigr)^{-1}\Bigl( \b(f)_k+{h\over 2}\bPhi_k\bq_0+\\ +h\sum_{i=1}^{k-1}\bPhi_{k-i}\bq_i\Bigr), \quad k=\overline{1,m}.\\ \endgathered\tag13$$ {\bf Remark 2.} Let $\bC(h)=\bI-{h\over 2}\bPhi_0$. If $h$ is sufficient small, then $\bC(h)$ is nonsingular matrix and $$\lim_{h\to 0}\bC(h)=\bI, \quad \lim_{h\to 0}\|\bq(t_k)-\bq_k\|=0.$$ \bigskip \centerline{\bf REFERENCES} \bigskip \item {1.} R. Bellman, K.L. Cooke. Differential--Difference Equations. Academic Press, New York -- London, 1963. \item {2.} R. Bellman. Stability Theory of Differential Equations. Academic Press, New York -- Toronto -- London, 1953. \end 