International Journal of Mathematics and Statistics Invention (IJMSI) E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759 www.ijmsi.org ǁ Volume 2 ǁ Issue 3 ǁ March 2014 ǁ PP 48-63

Analyses of a mixing problem and associated delay models C. Ukwu Department of Mathematics, University of Jos, P.M.B 2084 Jos, Nigeria.

ABSTRACT: In this article, we gave an exposition on a class of mixing problems as they relate to scalar delay differential equations. In the sequel we formulated and proved theorems on feasibility and forms of solutions for such problems, in furtherance of our quest to enhance the understanding and appreciation of delay differential equations and associated problems. We obtained our results using the method of steps and forward continuation recursive procedure.

KEYWORDS: Delay, Feasibility, Mixing, Problems, Solutions. I.

INTRODUCTION

Dilution models are well known in the literature on ordinary differential equations. However literature on the extension of these models to delay differential equations is quite sparse and the associated analyses not thorough, detailed or general for the most part. See Driver (1977) for an example. This article leverages on the model in Driver to conduct detailed analyses of ordinary and associated delay differential equations models of mixing problems, with accompanying theorems, and corollary, together with appropriate feasibility conditions on the solutions.

II.

PRELIMINARY DEFINITION

A linear delay differential system is a system of the form:

x (t )  A(t ) x(t )  B(t ) x(t  h)  g (t ), where A and B are n  n matrix-valued functions on

R

(1)

and h > 0 is some constant and g(t) is continuous.

Remarks Any other appropriate conditions that could be imposed on g, A and B to guarantee existence of solution will still do. The need for appropriate specification of initial data will be looked at very shortly. If g  t   0 t  IR , then (1) is called homogeneous. If A and B are time-independent, the system (1) is referred to as an autonomous delay system.

III.

AN ORDINARY DIFFERENTIAL EQUATION DILUTION MODEL (MIXING PROBLEM)

Consider the following problem: A G-gallon tank initially contains s0 pounds of salt dissolved in W0 gallons of water. Suppose that

b 1 gallons of brine containing s0 pounds of dissolved salt per gallon runs into the tank every minute and that the mixture (kept uniform by stirring) runs out of the tank at the rate of b2 gallons per minute. Note: We assume continual instantaneous, perfect mixing throughout the tank. The following questions are reasonable: a) 1) 2) 3)

Set up a differential equation for the amount of salt in the tank after t minutes. What is the condition for the tankto overflow? Not to overflow? how much salt will be in the tank at the instant it begins to overflow?

b) Will the tank ever be empty?

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Analyses of a mixing problem and associated… Solution Let S  t  be the amount of salt in the tank at time t. Then b 1 gallon of brine flow into the tank every minute and each gallon contains s pounds of salt. Thus

b1 s pounds of salt flow into the tank each minute.

Amount of salt flowing out of the tank every minute: at time t we have S  t  lbs of salt and

W0  (b1  b2 )t gallons of solution in the tank, since there is a net increase of (b1  b2 ) gallons of solution every minute. Therefore, the salt concentration in the solution at time t is

S (t ) W0  (b1  b2

lbs per gallon, and salt

leaves the tank at the rate

b2 S (t ) S (t )   W  (b  b )t lbs/gallon  b2 gallons/minute  W  (b  b )t lbs/ min .  0  1 2 0 1 2 Hence the net rate of change,

dS dt

of salt in the tank is given by

Net rate of change = salt inflow per minute - salt outflow per minute, expressed as:

dS

 b1 s0 

dt (2) 

dS dt

b2 S W0  (b1  b2 )t

b2 S

 b1 s0 , a first order differential equation

p (t ) 

(2)

W0  (b1  b2 )t

b2 W0  (b1  b2 )t

dS dt

 p(t ) S  q (t ), with

and q (t )  b1 s0

The integrating factor is

I (t )  e

p ( t ) dt

e

b2  ( 0 b1b2 )t

W

dt

 e =

b2 Ln W0 (b 1b0 )t b1 b2

W0  (b1  b2 )t

, b1  b2

b2 b1 b2

Now W0  (b1  b2 ) t  0 makes practical sense, as the amount of brine cannot be negative or 0 at any time

t.

Case i: b1  b2  tank overflows at some time t if mixing is continual, thus b2

I  t   W0  (b1  b2 )t  b1 b2 The general solution given by, S (t ) 

1

 q(t ) I (t )dt  C  , where C is an arbitrary constant. I (t )  

Hence,

S (t ) 

1 b2

W0  (b1  b2 )t  b b 1

2

b2 dt   b1 b2 b s W  ( b  b ) t C   1 2  1 0 0  

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Analyses of a mixing problem and associated…

b1 s0

=

b2

W0  (b1  b2 )t b1b2 =

b1 s0

W

0

 (b1  b2 )t

b2 b1  b2

b2     W0  (b1  b2 )t b1 b2 dt  C   

  1 1   (b1  b2 ) .  b  2    1  b1  b2  

W0  ( b1  b2 ) t

   C   

b2 b1 b2

=

b1 s0

W0  (b1  b2 )t 

b2 b1 b2

b1 1  b b2  C  . W  ( b  b ) t   1  0 1 2  b1 

At the instant the tank overflows, W0  (b1  b2 )t

(3)

 G , so that t 

G  W0 b1  b2

 G  w0   . Now, C can be obtained by noting that S (0)  S 0 ,  b1  b2 

The amount of salt in the tank at that instant is S  yielding:

b1 s0 b2 b1 b2

W0

b b  1 b b2b  s0 1 b b b b 1 2  C   S0 ,  C  W0  W0  W0 b1s0 b1  b1  2

1

1

2

1

2

Therefore,

 G  Wo  b1 s0 S = b2  b1  b2  b1 b2 G

=

s0 b2

G b1 b2

b2 1  1 b b1b b1b2 S0 1 b1b2  1 2  W0  W0  G  b1 s0 b1  b1  b

b2 b1  b1 b1b2 b1b2  1 b1 b2 G   S0W0  W0 s0    

Case ii: b1  b2  tank never overflows. b1  b2  0 

 b2   1 In (3) the expression,   b1  b2  Now the equation

b2 b1  b2

1

b2  b   1 b1 2

is feasible provided

 0.

b2   1. b1  b2

b2   b1  b2 ,  b1  0  no brine solution runs into the

tank at any minute  mixing or dilution does not take place; so we must have (3) is feasible, provided b1

b2   1 , implying that b1  b2

 b2 and the expression for S  t  is preserved if b1  b2 .

(c) The tank will be empty at an instant t if b1

 b2 and W0  (b1  b2 )t  0, yielding t 

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W0 . b2  b1

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Analyses of a mixing problem and associated…

S  t  undefined; thus the condition. W0  (b1  b2 )t  0 is infeasible. This agrees with our intuition and the physics of the problem: as long as b1  0 the tank is never empty. However, this would render

The case b1

 b2 implies that the tank never overflows. b1  b2  b1  b2  0. (2) yields: dS b2 S   b1 s0 dt W0  I t   e

b2 t W0 b2 W0

. 

t b1s0

b2

b2 t W0

 b2 t b s  S (t )  e W0  1 0 W0 e  b2

 S (t )  e

IV.

 b2 t W0

b2 t W0

 S0 

W0 W0 t e  C  b2 

     b1s0 e S (t )  e dt  C   e    bs b S (0)  S0  1 0 W0  C  S 0  C  S 0  1 s0W0 b2 b2  b2 t W0

 b1 s0 W0  b2 

  Wb2 t   b1  S0  S0W0  e 0  1   b2   

(4)

REFINEMENT OF THE MODEL TO A DELAY MODEL

Practical reality dictates that mixing cannot occur instantaneously throughout the tank. Thus the concentration of the brine leaving the tank at time t will be equal to the average concentration at some earlier instant, t  h say, where h  0 . Setting S (t )  x(t ) the ordinary differential equation (2) modifies to:

x (t ) 

b2 x(t  h)  b1s0 W0  (b1  b2 )t

(5)

This is a delay differential type of linear nonhomogeneous type, in the form (1) with

n  1, A(t )  0, B (t )  b(t ) 

b2 W0  (b1  b2 )t

, and g (t )

 b1 s0 . Clearly A, B and g are continuous. For

simplicity assume the following: i. s0  0 . This implies that the inflow is fresh water ii. b1 Then, set C 

b2 W0

 b2 . This implies that the inflow rate of fresh water equals the outflow rate of brine. to obtain the autonomous homogeneous linear delay differential equation:

x(t )   c x(t  h).

(6)

Above equation can be solved using the steps method. This consists in specifying appropriate initial conditions on prior intervals of length h and extending the solutions to the next intervals of length h. Since the tank contained S 0 lbs of salt thoroughly mixed in W0 lbs of brine prior to time t  t 0 , the commencement of the flow process, we can specify the initial conditions solution on the intervals

t

0

x(t )  S0 for t 0  h  t  t 0 and then obtain the

 ( k  1)h, t 0  kh for k  1, 2, successively.

Therefore given:

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Analyses of a mixing problem and associated…

(7) and x(t )  S 0 on t0  h, t0  , we wish

x(t )   c x(t  h)

t  t0 . Set S 0   0 . Note that t0  h  t  t0 on t0 , t0  h . Therefore, x  t   c0 on t0 , t0  h , with to obtain the solution for

initial condition:

x(t0 )   0  x  t    c 0t  k1 ; x  t0    0   c 0 t0  k1  k1  1  c t0   0

 x  t   0 1  c  t  t0  for t t0 , t0  h

On

t

(8)

x t   0, if 1 ch  0 or ch  1 0

 h, t 0  2h , t  h  t 0 , t 0  h . Hence,

x  t    c 0  c0  t  t0  h  on t 0  h, t 0  2h, leading to the solution: 2 c 2 0 x  t    c0t  t   t0  h    k2 , on [t0  h, t0  2h]. 2

(9)

By direct substitution and use of (8) we get:

x t0  h    c0 t0  h   k2  0 1  ch   c2 2   k2  0 1  ct0   x  t   1  c  t  t0   t   t0  h   0 2!  

x t   0, if 1 c t  t0   0

Noting that

(10)

(11)

 t  t 0    2h on t 0  h, t 0  2h , we infer that

1  1  if ch   2  2!  Next, consider the interval t0  2h, t0  3h  . Then t  h  t0  h, t0  2h  . Therefore:

1  c t  t0   0 if 1  2ch  0 or 2ch  1. Hence x  t   0, if ch  

x t    c 1  c  t   t0  h   

 on the open set  t0  2h, t0  3h  . Integrating over the interval

t

0

2 c2    0 , t  t  2 h   0  2!  

(12)

 2h, t 0  3h yields:

2  3 c2  t  (t0  h  x  t    c t  c   t   t0  2h     0  k3  2 3!     Direct substitution into (13) and use of (10) yields:   c2 2  c 2 h2   k  1 c t x  t0  2h    c t0  2h  h  0  k3  1  2ch  0 3 0 2 2    

13

0

 k3  1 c t0 0

2 3   c2 c3  x  t    1  c  t  t0   t   t0  h        t   t0  2h     0 2! 3!   Assertion 1: 1 x  t   0, on t0  2h, t0  3h  , if ch  3! Proof

t  t0  h  h, 2h

on

t

0

14  (15)

 2h, t 0  3h

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Analyses of a mixing problem and associated…

t  t 0  2h  0, h on [t 0  2h, t 0  3h]

t  t 0  2h, 3h on [t 0  2h, t 0  3h] From these facts, we deduce the following: c  t  t0    3ch;  c  t   t0  2h      ch,

16

3  ch  , c3  t   t0  2h    3! 3!

17 

2 c2 c2 h2 . t   t  h    2! 2!

18

3

Plug in (16), (17) and (18) into (14) to deduce that:

 c2h2 c 2h2  x  t   1  3ch    0 2! 3!   

 1  

1 1  2 64

 0 , if ch  Let

1 3!

  

19 

proving assertion 1.

x  t   yk  t  on t 0   k  1h, t 0  kh . Then,

Theorem 1

x  t   yk  t  :

  1  

k

j j  1  c t   t0   j  1 h  

j 1

j!

on the interval

 0 

(20)

J k  t0   k 1 h, t0  kh  , with initial function x t   yk 1 t  , on the interval J k 1

for k = 1,2, …, where,

x t   y0  0 , on I0.

Moreover, yk

t   0 , wherever ch

x t   0, for t  t0 .

1 . Hence, k!

Proof Proof is by inductive reasoning on k. The result is definitely true for k = 1, 2 and 3, following the solutions obtained on J k , for k =1, 2 and 3. Assume that (20) is valid for 1 k  m for some integer m > 3. Then, t  h  J m for t  J m 1 and hence Now,

x t    c x t  h

on

t

0

x t  h  0 on I m1 if ch 

1 .  m  1!

 mh t0   m  1 h   x t    c ym t  h 

j  m  1 c t  h  (t0   j  1 h  c 1    j 1 j! 

j

j j   c t   t0  jh    m  1     c 1      0  0 j ! j 1   

Thus:

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j  m  1 c t   t0  jh    x t    c t     j  ! j 1 

Plug

j 1

    k , on J m 1 m 1  0 

 21

x  t0  mh  into (20) with k = m and into (21) and set the results equal to each other to obtain,

j  m  1 c t  mh   t  ( j  1)h   0 0  1    j 1 j! 

j

   0 

j 1  m  1 c t0  mh   t0  jh     c  t0  mh      j  1! j 1 

j 1

   K m1  0 

The last summation notation can be rewritten as:

j    1 j  c t 0  mh  t 0   j  1h   m

j2

m

T j2

j

, so that

Km1  1  ct 0  mh  cmh 0  T j  T j 0  1  c t 0 0 m

j 2

Thus:

K j  (1  ct0 ) 0 , j  1, 2,

(22)

Now plug (22) into (21) to obtain

j 1 j 1   m  1 c t   t0  jh       (1  ct ) x  t   ct  0 0   0 j  1 !   j 1   j 1 j 1   m  1 c t   t0  jh      1  c  t  t0      0 j  1!  j 1  

j  m 1  1 c t   t0   j  1 h     1  c  t  t0     j! j 2 

 m1 1 j c t  t  j  1 h      0     1    j 1 j! 

j

  ,  0 

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   j

 

0

(23)

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Analyses of a mixing problem and associated… m 1

since

 1  c t   t0   j  1 h  j

j!

j 1

j

evaluated at

j  1 yields c  t  t0 

Therefore, (20) is valid for all k = 1, 2,……, as set out to be proved.

yk  t   0 whenever ch 

Next, we need to prove that;

 c t   t0   j  1 h   x  t   1   j! 1  j k  j odd

   , on I

1 . k!

From (20)

j

 

0

k

 24 

Observe that on I k , t  (t0 +  j 1 h) t0   k 1 h  t0   j  1 h  , t0  kh  t0   j  1 h  , that is,

t   t0   j  1 h    k  j  h,  k  1  j  h  .

Therefore,

t  t0   j 1 h    k  1  j  h,  c t   t0   j  1 h     k  1  j  ch

Clearly, x  t   0 if:

j 1    k  1  j  ch   1   0  1   0 j  1 !  jk j odd  

 25

(25) would hold if:

 (k  1  j )ch    j  1 ! 1 j  k

j 1

 26 

1

j odd

Suppose that

1 on J k , then (26) would be valid if: k!

ch 

k 1 j 

1 1 k!

  j  1! 1 j  k j odd

 27 

(27) would in turn be valid if:

 1  j  k k !  j  1 ! k

1

 28 

 1,

j odd

that is, if:

1  k  1! 1  jk

1 1 j  1 !

 29  .

j odd

However,

 j 1!  j 2 , so that

1 1  2  j  1! j

.

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Analyses of a mixing problem and associated… Hence (29) would be true if:

1 1 1   k  1 ! 1  j  k j 2

(30)

j odd

(30) would be valid if:

1 k 1 11  k  1!  j 1

Obviously (32) is valid for

1 k!

k 1 1  k  1!

i.e. if

1 1  k  2 !

k  2.

Combine this with the fact that x whenever ch 

i.e. if

t   0 for t  I , if ch  1 to deduce that

(31)

 32  yk t   0 on I k

, k  1, 2,....

The validity of (32) implies that

lim k 

1 1    1, lim yk  t   0 or x  t   0  t  t0 . Note:  0   1 . k     k  2 ! 

This completes the proof of the theorem.

V.

ASSOCIATED NONHOMOGENOUS MODELS

Suppose that the inflow is not fresh water, that is s0 equation:

 0 . Then the nonhomogeneous differential difference

x t   c x t  h   b1s0 ,

33

x t   0 , t0  h  t  t0

34

with initial data:

referred to as the initial function, can be solved by a transformation of variables. In the sequel,

b1 b1s0 s on t0  h t0  , x  t   y t  and . Then, y t   0  c 0 c b x t  h  y t  h  1 s0 c b1 b1    y  t    c  y t  h  s0   b1 s0  y  t    c y t  h with y t   0  s0 c  c  on t0  h, t0  let y  t  

5.1

x t  

Corollary 1

i. The transformation

b ~ y  t   x t   1 s0 converts the linear nonhomogeneous delay differential equation c

(33) with initial function specification (34) to the following linear homogeneous delay equation with corresponding initial function:

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y t   c y t  h  b ~ y  t   0  1 s0 on c

t0  h

ii. If

35

t0 

x t  is the solution of (35) on J k then:

 36

x  t   yk  t 

j  k  1 c t   t0   j  1 h   1    j 1 j! 

x  t     0 

Moreover, yk 1 iii.

j

 

1 0

(37)

c

x t   y k 1  t  on J k 1 , k  1, 2,...

on J k , with initial function where

    b s

b1 s0 , on J 0 . c

t   0 whenever ch

c0  b1 s0  0 .

1 . Hence, x  t   0, for t  t0 . k!

More generally, given the constant initial function problem:

x  t   a x  t   b x  t  h   c

 38  39 

x  t    0 , t0  h  t  t 0 ,

where

y  t   x t   d a, b, c, are given constants, the change of variables ~  y  t   x (t ) and y  t   a  y  t   d   b  y  t  h   d   c  y  t   a y  t   b y  t  h   c   a  b  d

Setting

c   a  b d  0  d 

c c y  t   0  . Also ~ , leading to the following proposition. ab ab

5.2 Proposition 1 The initial function problem:

x  t   ax  t   bx  t  h   c x  t    0 , t0  h  t  t0 where a , b and c are given constants such that linear homogeneous type:

 40   41

a  b  0 , is equivalent to the initial function problem of the

y  t   ay  t   by  t  h  y  t   0 , t0  h  t  t0

 42   43

where:

0  0 

c . ab

 44 www.ijmsi.org

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Furthermore

y  t  by the equation: x t  is related to ~ x  t   y  t  

c ab

 45

Remark 1 Instead of solving (40) and (41), solve the easier equivalent problem (42) through (45). If b = 0, (40) and (41) degenerate to the IVP

x  a x  c

x  t0   0 , c  a  t t  c  x  t   0   e 0  . a a 

with the unique solution

Equivalently, the solution of (4) through (45) is given by:

x  t   y  t    x  t   o e

c a

a  t  t0 

c  c  a t t c  0   e  0   , a  a a

 46 

as desired. Remark 2

y  t   x t   d  t  leads c(t ), as ~ to c t   ad  t   bd  t  h  0 , and it is impossible to determine d (t ). The transformation (45) is doomed if c in (40) is replaced by nonconstant

Linear translation of the initial interval and representation of the unique solution. Consider the constant initial function problem:

x  t   ax  t   bx  t  h  , t  t0`

 47  .  48

x  t    0 , t0  h  t  t0 Define:

z t   x t  t0  .

 49

Then

z  t0   x  t0  t0   x  0  and z  t0  h   x  t0  h  t0   x   h  .

Also:

t  t 0  h, t 0  t  t 0   h, 0 and

x  t   z  t0  t  z  t   x  t  t0   z  t   a z  t   b z  t  h  , t  0 z  t   0 ,  h  t  0

 50   51

Therefore, the constant initial function problem (47), (48) is equivalent to the constant initial function problem (50), (51) where

x (t )

is related to z  t  through the equation:

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Analyses of a mixing problem and associated…

x t   z t  t0  .

52

Consequently, without any loss in generality, given (47), (48) we can solve the equivalent problem:

x  t   a x  t   b x  t  h  , t  0

 53  54  Denote:  55

x  t    0 ,  h  t  t0 x    x    h  J k   k  1 h, kh  , k  0,1, 2, J k0    k  1 h, kh  , k  1, 2,.. Then, t  J k  t  h  J k 1 ; consequently x 0

0

Using (46) with c replaced by

56  .  57 

t   a x t   b0 , on J10.

b  b  b 0 we obtain x  t     0  0  e a t  0 on J1 . a  a 

Thus:

 b  b  x  t     1   eat  0  c1  c11 ea t  0 , on J1  a  a 

58

b  b  x 0  0 and x  h     0   0  e a h   0 . a  a  0 Consider t  J 2 . Then on J 2 , Clearly,

b  a   x t   a x t       1   e a  t  h  0 a  b   Denote the integrating factor by I

t  . Then, I  t   ea t

Hence:

 b    b x  t   ea t     e  a t dt   1   e  a h dt  0  C   a   a   b    b  ea t   2 e a t  1   e a h t  0  C   a  a   b  b x  h    1   e a h    0 a  a 

 59 

(using (58))

  b   b   ea h  1   h e a h  2 e a h  0  C  (using (59)) a    a  

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 b b   b  C    e a h  1   1  h e a h   2 e a h  0 a  a  a  b b b   b  b  x  t   eat  2 e a t  1   e a ht  e a h  2 e a h  1   1  h e a h   0 a a  a  a a    b  b b  b b  b   eat  1   t   2  h 1    e a h  1   2 e at  0 a a  a    a  a a 

 b  b b b  b   b  1     2  h 1    e a t  1  e aht e a t 0 2 a  a a a  a   a  at  c2   c21  c22t  e  0 on J 2 

 60 

where:

c2 

b b b b  b   b , c21  1     2  h 1    e  a h , c22  1   e  ah 2 a a a a  a   a

(61)

.

Notice that:

c2  

c1  1   1       c1 , c21  1   e ah  c1  1  hea h  c11 , c22  e a h c11. (62) a  a  a  

We propose the following result. 5.3 Theorem 2:

  k   x  t   ck    ck j t j 1  e a t   0  j 1    on J k for appropriately determinable constants ck , ck j , j  1, 2,..., k , where: b b b k 1 1 c1   , c11  1  ; ck  (1)k k , ckk  b k  2 e  ( k 1) ah c11 ; k  2,3, , a a a ( k  1)! and for j  k , ck j depends on

 63

(64)

h , but has no general mathematical representation.

Proof The proof is by inductive reasoning. From (58), we see that the theorem is valid on J1 , with

b b c1   , c11  1  . From (60), (61) and (62), it is clear that the theorem is true on J 2 , with: a a

c2 

b b b b 1  b   b , c21  1     2  h 1    e a h , c22  1   e ah  b22e (21)ah c11 2 a a a a (2  1)!  a   a

Consider J 3 . On t  J 3  t  h  J 2 . By (63), we have: 0

(65)

0

x t  h  c2  c21  c22  t  h  e a  t  h  0  x  t   ax t   b c2  c21  c22  t  h e a t h  0 www.ijmsi.org

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Integrating factor, I  t   e at a t h a t h  x  t   e at   b c2 e  at dt   b c21e   e  a t d t   b c22  t  h  e  e  at dt  C   0

 bc   t2   ea t   2 e a t  b c21 e  a ht  b c22   ht  e  a h  C   0 2   a 

Now plug

 66 

x 2h into (63) and (66) to obtain C as follows:  bc  x  2h   e 2 ah   2 e 2 ah  bc21 e  ah 2h  bc22  2h 2  2h 2  e  ah  C   0  a   bc     2  bc21 h e ah  C e 2 ah   0  a  2 ah  c2   c21  2h c22  e  0  bc   C  e 2 ah  2  bc21 h e ah  c2   c21  2h c22  e 2 ah   a 

Now, plug this value of

 c  c 3

31

C into (66) and set the resulting expression equal to

 c32 t  c33 t 2  e a t 0 to get:

 b c2  at   t2  e  b c21e  a ht  b c22   h t  e  ah   a 2 at     0 x t   e  b c 2 ah   b c21 h e a h  c2 e 2 a h  c21  2hc22   2 e  a   c3   c31  c32t  c33 t 2    0

 c3  

b c2 b b   2 a aa

(67)

b2     a3 

c b c   c      1   c1 c2  c1   1    1 a a  a  a

2

Clearly (67)

c3   1

3

b31 1 1 1 , c33  b c22 e  ah  b e 2 ah c11  b3 2 e  31ah 3 a 2 2 (3  1)!

 1 c32  b c21e  ah  b h c22e  a h  1   be 2 ah c1  b 1  h e  a h e  ah c11  b he 2 ah c11  a  1    b 1   c1  e  ah c11  e 2 ah  a   bc c31  2 e 2 ah  b c21 he ah  c2e 2 ah  c21  2 h c22 a

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 68

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Analyses of a mixing problem and associated…

 1  c b  bc      1 e2 ah  1  b h e ah  1   e  a h c1  1  h e  ah  c11   1 e 2ah  2h 1   e  ah a a  a a    a 1 1  e2 ah c12  2 hc11 e  ah  e 2ah c1  1  bh e ah  1  1a  e  ah c1  1  h e  ah  c11  a a 1  1 1 1    e2 ah c12  1   e  ah  b h 1    e 2 ah  c1 a a a a      2 h e  ah  1  h e  ah   b h e ah  b h 2  c11

 69 

Therefore the theorem is also valid on J 3 . From the results already obtained for c21, c31 and c32, it is clear that no definite pattern can be postulated for ck j ,

j 1, 2,...., k 1 , even for the simplest initial

function problem. Now, we proceed to complete the proof of theorem. Assume that the theorem is valid on J k , k  4. t  J k 1  t  h  J k  the theorem is valid with t 0

0

replaced by t – h for t  J k 1 . Hence: 0

   k j 1  x  t   a x  t   b ck   ck j  t  h   e a t  h    0  j 1   

 70 

I  t   e a t   k  j 1  x  t   e at   b ck e  at dt    ck j  t  h  e  ah dt  C    0  j 1    j  1  k  t  h   ah   at  e   b ck e   ck j e  C  0  j 1  j  a   at

.

(71)

The constant term is: k 1 b k b  b  ck       1  0 using  (71)  a ak  a

  1 The term in

at k

e t

is

k 1

 k 1 1

k

b k 1 b k 1 b  0   1  0  ck 1 0  ck 1   1 k 1 k 1 a a a k 1

at

e e

 ah

.

 k 1 1

(72)

k tk b  k 1 ah at t ckk  e 0 e ah bk 2 e   c11 , k k  k  1!

by the induction hypothesis. Set

d k 1, k 1  e at

1  ah b 1 b e b k  2 e  k 1ah c11  e at e  ah b k  2 e  k 1ah c11. Then k k  1 ! k k  1 !    

bk 1  k ah b k 12  k 11 ah dk 1,k 1  e c11  e   c11  ck 1, k 1 , as desired. k!  k  1 1! Note that

j  j (t  h) j     t j r (h)r . r 0  r 

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Analyses of a mixing problem and associatedâ&#x20AC;Ś VI.

CONCLUSION

This article gave an exposition on how a delay could be incorporated into an ordinary differential equations dilution model to yield a delay differential equations dilution model. It went on to formulate and prove appropriate theorems on solutions and feasibility of such models. It also showed how a nonhomogeneous model with constant initial function could be converted to a homogeneous model. Some of the results relied on the use of integrating factors, change of variables technique and the principle of mathematical induction.

REFERENCE [1]

Driver, R. D. (1977). Ordinary and Delay Differential Equations. Applied Mathematical Sciences 20, Springer-Verlag, New York.

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