b’Subject: Mathematics’

b’Topic: Mathematical modeling’

b’Help me answer Questinyzon 8, its on the problem.pdf’

4. Compartmental Systems

SM-4312: Mathematical Modelling

Semester I, Session 2020-21

Assoc Prof Dr Sannay Mohamad

Faculty of Science, Universiti Brunei Darussalam, Gadong BE1410, Brunei Darussalam

Emails: sannay.mohamad@ubd.edu.bn, sannay.mohamad@gmail.com

Contents Open systems Closed systems

Contents

1 Open systems

Model formulation

Solution by the elimination method

Solution by the eigenvalue-eigenvector method

2 Closed systems

Model formulation

Solution method

3. Compartmental Systems Pg. 2–26

Contents Open systems Closed systems

Model formulation

Problem Consider an open system of three tanks

containing V1, V2, V3 litres of salt solutions.

Salt solution of cin g/l flows into Tank 1 at rin l/min,

while the well-stirred salt solution flows from Tank 1

into Tank 2 at r12 l/min, from Tank 2 into Tank 3 at

r23 l/min, and out of Tank 3 at rout l/min.

Determine xi = xi(t) the amount of salt (in grammes)

in Tank i (i = 1, 2, 3) at time t > 0, if the initial

amount of salt in the respective tanks are

x1(0) = x1 0, x2(0) = x2 0, x3(0) = x3 0 grammes.

rin cin

r12

r23

ro

ut

Tank 1

x1(t)

V1(t)

Tank 2

x2(t)

V2(t)

Tank 3

x3(t)

V3(t)

Fig. 4.1: Open system

3. Compartmental Systems Pg. 3–26

Contents Open systems Closed systems

Assumption The change ∆xi within time ∆t is

∆xi = (input amount per time)∆t − (output amount per time)∆t, i = 1, 2, 3. (4.1)

Formulation

∆xi

∆t

= (input amount per time) − (output amount per time), i = 1, 2, 3.

Applying the limit ∆t → 0,

dxi

dt

= (input amount per time) − (output amount per time), i = 1, 2, 3. (4.2)

Model is therefore given by

dx1

dt

= rincin − r12

x1

V1(t)

dx2

dt

= r12

x1

V1(t) − r23

x2

V2(t)

dx3

dt

= r23

x2

V2(t) − rout

x3

V3(t)

(4.3)

where

V1(t) = V1 + (rin − r12)t, V2(t) = V2 + (r12 − r23)t, V3(t) = V3 + (r23 − rout)t

3. Compartmental Systems Pg. 4–26

Contents Open systems Closed systems

This is an open system (non-conservative) because x #1 + x #2 + x #3 = rincin − rout Vx 33 (t) ! 0.

In matrix form,

!””””#

x #1

x #2

x #3

$%%%%&

=

!”””””#

−

r12

V1(t) 0 0

r12

V1(t) −

r23

V2(t) 0

0 r23

V2(t) −

ro

ut

V3(t)

$%%%%%&

!””””#

x1

x2

x3

$%%%%&

+

!””””#

rincin

0 0

$%%%%&

(4.4)

The linear system (4.4) is homogeneous if rincin = 0 i.e. x # = Ax, and nonhomogeneous if

otherwise i.e. x # = Ax + I.

The general solution of (4.4) is

x(t) = xc(t) + xp(t), (4.5)

where x(t) = [x1(t), x2(t), x3(t)]T, xc(t) denotes a complimentary function (i.e. solution of

x # = Ax), and xp(t) the particular integral of (4.4) (i.e. solution of x # = Ax + I).

3. Compartmental Systems Pg. 5–26

Contents Open systems Closed systems

Solution by the elimination method

Problem

rin cin

r12 r23

ro

Tank 1 ut

x1(t)

V1(t)

Tank 2

x2(t)

V2(t)

Tank 3

x3(t)

V3(t)

Fig. 4.2: Open compartmental system

Given

rin = r12 = r23 = rout = 10 l/min, cin = 10e−0.5t g/l

V1 = 200 l, V2 = 400 l, V3 = 300 l

x1(0) = 50 g, x2(0) = 0 g, x3(0) = 0 g

To determine x1(t), x2(t) and x3(t) for t > 0.

3. Compartmental Systems Pg. 6–26

Contents Open systems Closed systems

Model

dx1

dt

= 100e−0.5t − 1

20

x1

dx2

dt

=

1 20

x1 −

1 40

x2

dx3

dt

=

1 40

x2 −

1 30

x3

Solution (elimination method) The first equation is uncoupled and can be solved immediately. The

solution x1(t) is substituted into the second equation and now this equation can be solved

subsequently. Then x2(t) is substituted into the third equation and so this equation be solved

subsequently.

The solutions for t > 0 are given by

x1(t) = 2450

9

e−t/20 − 2000

9

e−t/2,

x2(t) = − 4900

9

e−t/20 + 9900

19

e−t/40 + 4000

171

e−t/2,

x3(t) = 2450

3

e−t/20 + 29700

19

e−t/40 − 16650

7

e−t/30 − 500

399

e−t/2.

3. Compartmental Systems Pg. 7–26

0

Contents Open systems Closed systems

0 100

50

100

150

200

200 300

x1(t)

x2(t)

x3(t)

Salts

t

Fig. 4.3 Variation of salt amount in each Tank for t > 0.

3. Compartmental Systems Pg. 8–26

Contents Open systems Closed systems

Solution by the eigenvalue-eigenvector method

Problem

rin cin

r12

r23

r32

ro

ut

Tank 1

x1(t)

V1(t)

Tank 2

x2(t)

V2(t)

Tank 3

x3(t)

V3(t)

Fig. 4.4 Open compartmental system

Given

rin = 5 l/h, r12 = 5 l/h, r23 = 7 l/h, r32 = 2 l/h, rout = 5 l/h, cin = 15 g/l

V1 = 20 l, V2 = 30 l, V3 = 10 l, x1(0) = x2(0) = x3(0) = 0 g

To determine x1(t), x2(t) and x3(t) for t > 0.

3. Compartmental Systems Pg. 9–26

Contents Open systems Closed systems

The model is

dx1

dt

= −

1 4

x1 + 75,

dx2

dt

=

1 4

x1 −

7

30

x2 +

1 5

x3,

dx3

dt

=

7

30

x2 −

7

30

x3.

Note The first equation is uncoupled and it can be solved. After substituting the solution

x1(t) into the second equation, this equation, however, cannot be solved subsequently for x2

since the variable x3 still exists. Therefore the elimination method cannot be applied here.

In matrix form,

x # =

!”””””#

−

14

0 0

14

−

7

30

15

0 7

30 −

7

10

$%%%%%&

x +

!”””””#

75

0 0

$%%%%%&

3. Compartmental Systems Pg. 10–26

Contents Open systems Closed systems

Eigenvalue & eigenvector equations

In solving the homogeneous part of (4.4) i.e.

x # = Ax, (4.6)

each solution is of the form x = veλt, where v is a nonzero constant vector and λ is a

constant.

Substitute x = veλt into (4.6),

λveλt = Aveλt ⇒ Av − λv = 0 ⇒ (A − λI)v = 0, (4.7)

where I is an identity matrix. Here, λ is called an eigenvalue of A and v the eigenvector of

A.

Require

det(A − λI) = 0; (4.8)

it is called the characteristic (i.e. eigenvalue) equation of A. It is used for finding λ.

For each λ, we then find v from the eigenvector equation (4.7).

Hence, x(t) = veλt of x # = Ax is found. This forms xc(t).

3. Compartmental Systems Pg. 11–26

Contents Open systems Closed systems

Finding x c(t)

To find λ, we solve det(A − λI) = 0 i.e.

”””’

−

14

− λ 0 0

14

−

7

30 − λ 1 5

0 7

30 −

7

10 − λ

”””’

= 0

∴ λ1 = −

1 4

, λ2 = −

7

15

−

√91

30

, λ3 = −

7

15

+

√91

30

.

To find v1 corresponding to λ1 = − 1 4, we solve (A − λ1I)v1 = 0 i.e.

!”””””#

−

14

− λ1 0 0

14

−

7

30 − λ1 1 5

0 7

30 −

7

10 − λ1

$%%%%%&

!””””#

a b c

$%%%%&

=

!””””#

0 0 0

$%%%%&

⇒ a = −13, b = 27, c = 14

∴ v1 =

!””””#

−13

27

14

$%%%%&

∴ x1(t) =

!””””#

−13

27

14

$%%%%&

e−t/4 (1st fundamental solution)

3. Compartmental Systems Pg. 12–26

Contents Open systems Closed systems

To find v2 corresponding to λ2 = − 1 75 − √ 39 01, we solve (A − λ2I)v2 = 0 i.e.

!”””””#

−

14

− λ2 0 0

14

−

7

30 − λ2 1 5

0 7

30 −

7

10 − λ2

$%%%%%&

!””””#

a b c

$%%%%&

=

!””””#

0 0 0

$%%%%&

⇒ a = 0, b = 6, c = −7 − √91

∴ v1 =

!””””#

0 6

−7 − √91

$%%%%&

∴ x2(t) =

!””””#

0 6

−7 − √91

$%%%%&

e!− 1 7 5 − √ 39 01 “t (2nd fundamental solution)

3. Compartmental Systems Pg. 13–26

Contents Open systems Closed systems

To find v3 corresponding to λ3 = − 1 75 + √ 39 01, we solve (A − λ3I)v3 = 0 i.e.

!”””””#

−

14

− λ3 0 0

14

−

7

30 − λ3 1 5

0 7

30 −

7

10 − λ3

$%%%%%&

!””””#

a b c

$%%%%&

=

!””””#

0 0 0

$%%%%&

⇒ a = 0, b = 6, c = −7 + √91

∴ v3 =

!””””#

0 6

−7 + √91

$%%%%&

∴ x3(t) =

!””””#

0 6

−7 + √91

$%%%%&

e!− 1 75 + √ 39 01 “t (3rd fundamental solution)

The complementary function is therefore

x

c(t) = c1x1(t) + c2x2(t) + c3x3(t)

= c1

!””””#

−13

27

14

$%%%%&

e−t/4 + c2

!””””#

0 6

−7 − √91

$%%%%&

e!− 1 75 − √ 39 01 “t + c3

!””””#

0 6

−7 + √91

$%%%%&

e!− 17 5 + √ 39 01 “t,

where c1, c2, c3 denote arbitrary constants.

3. Compartmental Systems Pg. 14–26

Contents Open systems Closed systems

Finding xp(t)

The nonhomogeneous term (i.e. input) of (4.4) is

!””””#

rincin

0 0

$%%%%&

=

!””””#

75

0 0

$%%%%&

.

By the method of undetermined coefficients, we assume

x

p(t) = w,

where w is a nonzero constant vector to be determined. Note that w is not an eigenvector

of A.

3. Compartmental Systems Pg. 15–26

Contents Open systems Closed systems

Substitute x

p(t) into system (4.4) i.e. x # p = Axp + I to get

0 = Aw +

!””””#

75

0 0

$%%%%&

⇒ Aw = −

!””””#

75

0 0

$%%%%&

!”””””#

−

14

0 0

14

−

7

30

15

0 7

30 −

7

10

$%%%%%&

!””””#

w1

w2

w3

$%%%%&

=

!””””#

−75

0 0

$%%%%&

∴ x

p(t) =

!””””#

300

450

150

$%%%%&

.

These two vector solutions, xc(t) and xp(t), form the general solution x(t) given by

!””””#

x1(t)

x2(t)

x3(t)

$%%%%&

= c1

!””””#

−13

27

14

$%%%%&

e−t/4 + c2

!””””#

0 6

−7 − √91

$%%%%&

e!− 17 5 − √ 39 01 “t + c3

!””””#

0 6

−7 + √91

$%%%%&

e!− 175 + √ 39 01 “t +

!””””#

300

450

150

$%%%%&

,

where c1, c2 and c3 are arbitrary constants.

3. Compartmental Systems Pg. 16–26

Contents Open systems Closed systems

Applying initial condition

To determine c1, c2, c3, we solve x(0) = x c(0) + x p(0)—namely,

!””””#

0 0 0

$%%%%&

= c1

!””””#

−13

27

14

$%%%%&

+ c2

!””””#

0 6

−7 − √91

$%%%%&

+ c3

!””””#

0 6

−7 + √91

$%%%%&

+

1 3

!””””#

300

450

150

$%%%%&

∴ c1 =

300

3

, c2 = 0.99, c3 = −179.84.

Therefore the amounts of salt in the tanks are described by

x1(t) = −300e−0.25t + 300,

x2(t) = 623.08e−0.25t + 5.94e−0.785t − 1079.04e−0.149t + 450,

x3(t) = 323.08e−0.25t − 16.37e−0.785t − 456.68e−0.149t + 150.

3. Compartmental Systems Pg. 17–26

Contents Open systems Closed systems

Interpretations

0 10 20 30 40

100

200

300

400

salts xi(t)

x1(t)

x2(t)

x3(t)

time, t

Fig. 4.5: Variation of x1(t), x2(t), x3(t) for t > 0.

3. Compartmental Systems Pg. 18–26

Contents Open systems Closed systems

Model formulation

Figure 3 describes the flow of brine solution in the direction

Tank 1 → Tank 2 → Tank 3 → Tank 1

with an equal flow rate r l/min.

Let x1 = x1(t), x2 = x2(t), and x3 = x3(t) denote

the amount of salt in grammes in respective

tanks at any time t > 0.

Let Vi(t) = Vi denote the volume of solution in

litres at time t in Tank i = 1, 2, 3.

Then, x1, x2, x3 are governed by the model

x #1 = −r(x1/V1) + r(x3/V3)

x #2 = r(x1/V1) − r(x2/V2)

x #3 = r(x2/V2) − r(x3/V3)

(4.9)

r

r r

Tank 1

x1(t)

V1

Tank 2

x2(t)

V2

Tank 3

x3(t)

V3

3. Compartmental Systems Fig. 4.6: Closed system Pg. 19–26

Contents Open systems Closed systems

Notice that x #1 + x #2 + x #3 = 0 for all t. So

x1(t) + x2(t) + x3(t) = x1(0) + x2(0) + x3(0) for all t. (4.10)

The model represents a closed system.

Determine the amounts x1(t), x2(t), x3(t) of salt in the three brine tanks at time t if V1 = 50,

V2 = 25, V3 = 50 litres, r = 10 l/min, x1(0) = 50, and x2(0) = x3(0) = 0 grammes.

3. Compartmental Systems Pg. 20–26

Contents Open systems Closed systems

Solution method

The system is

!””””#

x #1

x #2

x #3

$%%%%&

=

!””””#

−0.2 0 0.2

0.2 −0.4 0

0 0.4 −0.2

$%%%%&

!””””#

x1

x2

x3

$%%%%&

,

!””””#

x1(0)

x2(0)

x3(0)

$%%%%&

=

!””””#

50

0 0

$%%%%&

.

Characteristic equation is

”””

−0.2 − λ 0 0.2

0.2 −0.4 − λ 0

0 0.4 −0.2 − λ

”””

= 0

⇒ (−0.2 − λ)(−0.4 − λ)(−0.2 − λ) + (0.2)(0.2)(0.4) = 0

⇒ −λ3 − 0.8λ2 − 0.2λ = 0

⇒ λ[(λ + 0.4)2 + (0.2)2] = 0

∴ λ1 = 0, λ2 = −0.4 + 0.2i, λ3 = −0.4 − 0.2i

3. Compartmental Systems Pg. 21–26

Contents Open systems Closed systems

Solving (A − λ1I)v1 = 0 for v1:

(A − 0 · I)v1 =

!””””#

−0.2 0 0.2

0.2 −0.4 0

0 0.4 −0.2

$%%%%&

!””””#

a b c

$%%%%&

=

!””””#

0 0 0

$%%%%&

⇒ a = c, a = 2b ⇒ v1 = [2, 1, 2]T.

Therefore, the first solution is

x1(t) = v1eλ1t =

!””””#

2 1 2

$%%%%&

.

3. Compartmental Systems Pg. 22–26

Contents Open systems Closed systems

To find the real-valued solutions x2(t) and x3(t) corresponding to λ2 and λ3, we simply

work with one of the eigenvalues (since λ3 = λ2).

Let λ = λ3 = −0.4 − 0.2i. We solve (A − λI)v = 0 for v:

(A − (−0.4 − 0.2i)I)v =

!””””#

0.2 + 0.2i 0 0.2

0.2 0.2i 0

0 0.4 0.2 + 0.2i

$%%%%&

!””””#

a b c

$%%%%&

=

!””””#

0 0 0

$%%%%&

0.2a + 0.2ib = 0 ⇒ a = 1, b = i,

(0.2 + 0.2i)a + 0.2c = 0 ⇒ c = −1 − i,

check: 0.4b + (0.2 + 0.2i)c = 0.4i + (0.2 + 0.2i)(−1 − i)

= 0.4i − 0.2 − 0.2i − 0.2i + 0.2 = 0

∴ v =

!””””#

1 i

−1 − i

$%%%%&

.

3. Compartmental Systems Pg. 23–26

Contents Open systems Closed systems

The corresponding complex-valued solution x(t) = veλt is

x(t) =

!””””#

1 i

−1 − i

$%%%%&

e(−0.4−0.2i)t

=

!””””#

1 i

−1 − i

$%%%%&

e−0.4t(cos 0.2t − i sin 0.2t)

= e−0.4t

!””””#

cos 0.2t − i sin 0.2t

sin 0.2t + i cos 0.2t

− cos 0.2t − sin 0.2t − i cos 0.2t + i sin 0.2t

$%%%%&

.

The real and imaginary parts of x(t) provide the real-valued solutions

x2(t) = e−0.4t

!””””#

cos 0.2t

sin 0.2t

− cos 0.2t − sin 0.2t

$%%%%&

, x3(t) = e−0.4t

!””””#

− sin 0.2t

cos 0.2t

− cos 0.2t + sin 0.2t

$%%%%&

.

Exercise: Find x2(t) and x3(t) by using λ = λ2 = −0.4 + 0.2i.

3. Compartmental Systems Pg. 24–26

Contents Open systems Closed systems

The general solution is

x(t) = c1x1(t) + c2x2(t) + c3x3(t)

or equivalently in scalar forms

x1(t) = 2c1 + e−0.4t(c2 cos 0.2t − c3 sin 0.2t),

x2(t) = c1 + e−0.4t(c2 sin 0.2t + c3 cos 0.2t),

x3(t) = 2c1 + e−0.4t([−c2 − c3] cos 0.2t + [−c2 + c3] sin 0.2t).

Observe that

x1(t) + x2(t) + x3(t) = 5c1 for all t.

Applying x1(0) = 50, x2(0) = x3(0) = 0, we obtain 5c1 = 50 ⇒ c1 = 10.

Thus, x1(t) → 20 g, x2(t) → 10 g, and x3(t) → 20 g as t → ∞.

3. Compartmental Systems Pg. 25–26

Contents Open systems Closed systems

0 10 20 30 40 50

0

10

20

30

40

50

Amount of salts xi(t)

x1(t)

x3(t) x2(t)

time, t

Fig. 4.7: Variation of amount of salts x1(t), x2(t), x3(t) at time t.

3. Compartmental Systems Pg. 26–26

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SM-4312: Mathematical Modelling, Semester I, Session 2020-2021 Page 5

(ii) Show that the amounts of pollutants in the lakes at time t > 0 are

x1(t) = r21

r12 + r21

(x1(0) + x2(0)) + r12x1(0) − r21x2(0)

r12 + r21

e−(r12+r21)t,

x2(t) = r12

r12 + r21

(x1(0) + x2(0)) − r12x1(0) − r21x2(0)

r12 + r21

e−(r12+r21)t.

(iii) Make a suggestion with appropriate mathematical analysis in order to achieve the eventual ratio

x1/x2 of the pollutants greater than one, that is, x1/x2 > 1.

Question 8

Tank 1

x1(t) x2(t)

Tank 2

r12 = r

r21 = r

In the figure, both Tank 1 and Tank 2 contain V1 l and V2 l

of the same chemical solution, respectively. The solution

flows from Tank 1 to Tank 2 and back to Tank 1 with flow

rates r12 = r l/min and r21 = r l/min, as indicated. The

solution in each tank is mixed uniformly and the amounts

of chemical in the tanks are denoted by x1 = x1(t) g and

x2 = x2(t) g at time t.

(i) Use the small change principle over a time period [t, t + ∆t], to formulate and derive the

compartmental system governing the chemical concentrations

Q1 = Q1(t) g/l and Q2 = Q2(t) g/l

for time t > 0 with Q1(0) and Q2(0) being the initial concentrations in the respective tanks.

(ii) Determine the chemical concentrations Q1(t) and Q2(t) at any time t > 0. Hence, show that the

limiting concentrations Q1(∞) and Q2(∞) satisfy

Q1(∞) = Q2(∞) = Q1(0) α1 + Q2(0) α2

α1 + α2

g/l,

where α1 = r/V1 and α2 = r/V2.

Question 9

Consider a competition model of two species X and Y whose population sizes x = x(t) and y = y(t)

are governed by

Species X :

dx

dt

= r1x − αy

Species Y :

dy

dt

= r2y − βx,

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Read moreThe Free Revision policy is a courtesy service that the Company provides to help ensure Customer’s total satisfaction with the completed Order. To receive free revision the Company requires that the Customer provide the request within fourteen (14) days from the first completion date and within a period of thirty (30) days for dissertations.

Read moreThe Company is committed to protect the privacy of the Customer and it will never resell or share any of Customer’s personal information, including credit card data, with any third party. All the online transactions are processed through the secure and reliable online payment systems.

Read moreBy placing an order with us, you agree to the service we provide. We will endear to do all that it takes to deliver a comprehensive paper as per your requirements. We also count on your cooperation to ensure that we deliver on this mandate.

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