# Discussion on Mathematical modeling

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

VRL 15 (Open Systems)

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### Free-revision policy

The 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.

The 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.

### Fair-cooperation guarantee

By 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.

## Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors: