Runge Kutta method to solve chapman kolmogorov differential equations

Question

fakhar jahan 2019-7-24

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回答： Le Duc Long 2020-6-29

I need to solve these 2 equations by runge kutta 4th order method. Could anyone help me out to imelemnt these equation without using ode45 function. Please assume any initial conditions of your choice.

Equation (3.7) states that the probability of being in state 0 with no one in the system a very short time from now is equal to (i) the probability that the system is in state 0 now and no customer arrives plus (ii) the probability that there is one customer in the system and that person finishes his or her service during the very short time. Equation (3.8) states that the probability that the system is in state i a very short time from now is equal to the sum of (i) the probability that the system is in state i-l now and there is one arrival during the very short time, (ii) the probability that the system is in state i right now and there are no arrivals or service completions during the short interval of time, and (iii) the probability that the system is in state i + 1 right now and there is one service completion during the short interval.

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fakhar jahan 2019-7-24

Hi.

I have solved these equations for only one value of i (i=1) and the code is given below. But I want to solve these equations for i=1,2,3,…,100. Besides, I am not sure that the runge kutta implementation which I have done through this code is correct for these equations. I will be very thankful for your help.

%Matlab code

clear;clc

L=10.653; % Lambda remains constant at each time step (Lambda (1)=Lambda (2)=L)

M=1.5; % Mu remains constant at each time step (Mu (1)=Mu(2)=M)

fR =@(t,P0,P1)-L*P0+M*P1; % equation 3.7

% equation 3.8

fF =@(t,P0,P1,P2)L*P0-[L+M]*P1 +M*P2; %I have placed P0, P1, P2 because i=1

t(1)=0; %initial condition

P0(1)=1; %initial condition

P1(1)=0; %initial condition

P2(1)=0; %initial condition

N=100;

h=0.01; %Time step for runge kutta

for j=1:N

%update time

t(j+1)= t(j)+h;

%update equations

k1R=fR(t(j) ,P0(j) ,P1(j) );

k1F=fF(t(j) ,P0(j) ,P1(j) ,P2(j) );

k2R =fR(t(j)+h/2,P0(j)+h/2*k1R,P1(j)+h/2*k1F);

k2F =fF(t(j)+h/2,P0(j)+h/2*k1R,P1(j)+h/2*k1F ,P2(j)+h/2*k1F);

k3R =fR(t(j)+h/2,P0(j)+h/2*k2R,P1(j)+h/2*k2F);

k3F =fF(t(j)+h/2,P0(j)+h/2*k2R,P1(j)+h/2*k2F, P2(j)+h/2*k2F);

k4R =fR(t(j)+h,P0(j)+h*k3R,P1(j)+h*k3F);

k4F =fF(t(j)+h,P0(j)+h*k3R,P1(j)+h*k3F ,P2(j)+h*k3F);

P0(j+1)=P0(j)+h/6*(k1R+2*k2R+2*k3R+k4R);

P1(j+1)=P1(j)+h/6*(k1F+2*k2F+2*k3F+k4F);

P2(j+1)=P2(j)+h/6*(k1F+2*k2F+2*k3F+k4F);

end

Torsten 2019-7-25

编辑：Torsten 2019-7-25

And what is P_(i+1) since you only have equations for P_0,...,P_i ?

Maybe P_(i+1) = 0 ?

fakhar jahan 2019-7-25

In my attempt, I have kept i=1, so (Pi+1=P2). Pi+1 is the probability for the next increment in "i" .In my attempt I have taken its initial value for P2=0. Although, I am not hundred precent sure that how it will be computed in these equations and htrough runge kutta method.

mu (M) and lambda (L) could be treated as constant for the ease of this computation. Please only take P indexed with respect to i.

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Answer 1

fakhar jahan 2019-7-24

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i=2 will change the P0 to P1, P1 to P2 and P2 to P3 in the equation 3.8. As per my understanding i-1 is dependent on previous iteration of "i".

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darova 2019-7-24

Is it possible to use loop?

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Answer 2

fakhar jahan 2019-7-24

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Loop for "i" values? I think yes.

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Answer 3

Le Duc Long 2020-6-29

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Hi Fakhar Jahan,

This is Kolmogorov differential equations in theory queue. I am also concerned about Your problem? Do you have the code to solve this system of equations? You can share me at the address: lelongbg@gmail.com.

Thank you so much!

Best Regards!