1550078481-Ordinary_Differential_Equations__Roberts_

398 Ordinary Differential Equations

the following assumptions:

The rate of change of the amount of substance in any container at time

t is equal to the sum over all inputs to the container of the concentration

of each input times the rate of input minus the concentration of the

substance in the particular container times the sum of the rates of flow

of outputs from the container.

That is , the rate of change of the amount of substance in container k at time

t is

where qk is the amount of substance in container k at time t, ci(t) is the

concentration of the substance in input i and ri(t) is the corresponding rate

of input, ck(t) is the concentration of the substance in container k at time t ,

and r 0 (t) is the rate of output to output o. Here the index of summation i is

over all inputs and the index of summation o is over all outputs.

As an example, suppose at time t = 0 an amount Ai of a particular sub-

stance is present in a solution that fills a container of volume Vi and an amount

A2 of the same substance is present in a solution that fills a second container

of volume V2. Assume the two containers are connected by tubes of negligible

length as shown in Figure 9.8. Further assume at time t = 0, (i) the solution

in the first container which is kept at uniform concentration ci ( t) is pumped

into the second container at the constant rate r and (ii) the solution in the

second container which is kept at uniform concentration c 2 (t) is pumped back

into the first container at the rater. The problem is to determine the amount

of substance qi ( t) in the first container and the amount of substance q 2 ( t) in

the second container as a function of time.

I r I

-+

V1

D

V2

~

I r I

Figure 9.8 Mixture Problem for Two Interconnected Tanks

Since the volume of both containers is constant, the concentration of the
substance in the first container is the amount of substance in the container

divided by the volume of the container- that is, ci(t) = qi(t)/Vi- and the

concentration of the substance in the second container is c 2 ( t) = q 2 ( t) /V 2.

Equating the rate of change 1of the amount of substance in each container to

the concentration of the substance in the other container times the rate at