Math 340-220 Projects, Spring 2011
Select one of the 6 projects below. Reports can be turned in by groups of at most two students.
Project Report due: Friday, May 6, in class.
Project 1: Modelling a Thermostat
A model for the heating of a building is provided by a generalization of Newton’s law of cooling:
dT
= k0 [A(t) − T ] + F (t).
dt
Here, A(t) is the outside temperature and F (t) is the heat supplied by the furnace. A simple model
for a thermostat controlled furnace is
F (t) = k1 [Tc − T (t)],
where Tc is a fixed reference temperature. The daily fluctuations of the outside temperature about
an average temperature A0 can be modelled by
A(t) = A0 − B cos(ωt),
where B is the amplitude of the fluctuation and ω = 2π/24 (time measured in hours). Thus the
temperature oscillates between A0 − B at t = 0 (midnight) and A0 + B at t = 12 (noon).
(a) Show that under the assumptions above the differential equation for the temperature T (t)
in the building takes the form
dT
+ kT = kTD − k0 B cos(ωt),
dt
(1)
where
1
(k0 A0 + k1 Tc ).
k
(b) Equation (1) is a linear, nonhomogeneous ODE for T . Apply the method of integrating factors
or the method of variation parameters to find the solution to this equation with general initial
condition T (0) = T0 . You should get the following result:
k = k0 + k1 ; TD =
T (t) = e−kt (T0 − TD ) + Tsp (t),
where
Tsp (t) = TD −
k0 B
[ω sin(ωt) + k cos(ωt)].
+ ω2
k2
Observe that the first term in T (t) is exponentially decaying. After a certain transient time this
term will be negligibly small. Then T (t) ≈ Tsp (t) which is called the steady periodic solution.
Tsp (t) consists of a constant part, TD , and an oscillating part with zero mean. Typically the goal
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will be to adjust TD to a prescribed value, e.g. 70o F , which can be achieved by choosing Tc , the
reference temperature of the thermostat, as
Tc = TD +
k0
(TD − A0 ).
k1
(b) Typical values of k0 and k1 are k0 between 1/4 and 1/2 and k1 somewhat smaller than 2. Set
k = 2, k0 = 0.4 and choose typical winter data: TD = 70, B = 15, A0 = 20 (in o F ). Recall
ω = 2π/24. Graph T (t) for a number of initial conditions T0 over a period of 2 days in the same
plot. You should see that after about 10 hours the transient part has disappeared.
(c) To compare the temperature fluctuations inside and outside, graph (in the same plot) the
oscillating parts
Tf o (t) = −B cos(ωt),
Tf i (t) = −
k0 B
[ω cos(ωt) + k sin(ωt)]
k2 + ω2
over a period of 2 days for the parameter values given in (b). You should see that (1) the amplitude
of Tf i (t) is smaller than the amplitude of Tf o (t) and (2) that Tf i (t) lags behind Tf o (t). To quantify
this observation, note that Tf i (t) can be written in the form
Tf i (t) = −aB cos(ω(t − tlag )),
which follows from a well known trig-identity. Compute the attenuation factor a and the time lag
tlag (you don’t need a computer for that).
Project 2: Modeling the US Population
The logistic equation,
dP
= kP (M − P ),
dt
models the growth or decay of of a population towards a carrier capacity M . The solution for the
initial condition P (0) = P0 is given by
P (t) =
M P0
.
P0 + (M − P0 )e−kM t
Table 1 shows the US population (in millions) between 1800 and 1900 and between 1910 and 1990
every 10 years. The left table shows in addition approximate increase rates Pi0 in these years. If the
logistic model were correct, P 0 (t)/P (t) = k(M − P (t)), and the given points (xi , yi ) = (Pi , Pi0 /Pi )
would be on a straight line.
(a) Find parameters a and b such that the line y = ax + b is the line that gives
P11 the best match
to the given points. This line is found by minimizing the least square error, i=1 (yi − axi − b)2 .
Show in general, given N points (xi , yi ), i = 1 . . . N , that the least square error minimization leads
to the following equations for a and b,
a=
W − XY
DY − XW
, b=
,
D − X2
D − X2
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Year
1800
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
i
1
2
3
4
5
6
7
8
9
10
11
ti
0
10
20
30
40
50
60
70
80
90
100
Population
5.308
7.240
9.638
12.861
17.064
23.192
31.443
38.559
50.189
62.980
76.212
Slope Pi0
0.166
0.217
0.281
0.371
0.517
0.719
0.768
0.937
1.221
1.301
1.462
Year
1910
1920
1930
1940
1950
1960
1970
1980
1990
Population
92.228
106.022
123.203
132.165
151.326
179.323
203.302
226.542
248.710
Table 1: Us-population in millions during 1800 and 1990.
where
X=
N
N
N
N
1 X
1 X
1 X
1 X 2
xi , Y =
yi , W =
xi yi , D =
xi .
N
N
N
N
i=1
i=1
i=1
i=1
Apply the formula for a and b to the given data (N = 11). In Matlab you can edit two rowarrays containing the above data,
À x=[5.308 7.24 9.638 . . . 76.212];
À pp=[0.166 0.217 0.281 . . . 1.462];
and then get the y-array using
À y=pp./x;
The commands
À X=mean(x);Y=mean(y);
À D=mean(x.∗ x);W=mean(x.∗ y);
generate the four relevant numbers needed for computing a and b. Create a plot showing the
data points and the best straight line.
(b) Identifying a = −k and b = kM gives values for the parameters in the logistic equation.
Use as initial value the population in 1800. Create a plot showing the solution curve and the actual
population data from the table. You should observe reasonable agreement until 1940, but strong
disagreement afterwards. Explain if you have any idea what the reason for this disagreement might
be. What’s about the carrier capacity?
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Project 3: Euler’s Method, Period Doubling, and Chaos
Consider the logistic equation with fixed parameters and initial condition,
dP
= 10P (1 − P ), P (0) = 0.1.
dt
Recall that the exact solution to this initial value problem approaches the limiting population
M = 1 when t → ∞.
(a) Show that Euler’s method with step size h to approximate the solution to the initial value
problem gives
Pn+1 = (1 + 10h)Pn − (10h)Pn2 , P0 = 0.1.
(2)
(b) For h = 0.18, 0.23, 0.25 and 0.3 plot 40 iterates Pn of (2). You should observe that Pn
(i) converges to 1 when h = 0.18,
(ii) after some iterates jumps between 1.8 and 0.69 when h = 0.23,
(iii) jumps between the 4 values 1.23, 0.54, 1.16 and 0.70 when h = 0.25, and
(iv) appears to be random (no discernible pattern) when h = 0.3.
The transitions from convergence to jumps between 2 values and jumps between 4 values are
called period doublings. The behavior observed in (iv) is called chaos. Obviously we can only trust
the results of the Euler computation as approximation of the logistic ODE when h = 0.18. The
other values of h are too large.
Project 4: Spruce Budworm
Perform tasks 1-6 of Project 3.5 in the text, p.132.
Project 5: Long-Term Behavior of Solutions for Dimension 1 and
2
Perform the tasks described in Project 3.5 in the text for Dimension D=1 and Dimension D=2 (p.
321). Numerical solutions can be generated using pplane6.
Project 6: Long-Term Behavior of Solutions for Dimension 3
Perform the tasks described in Project 3.5 in the text for Dimension D=3 (p. 321). Matlab code
for computing and plotting solutions will be provided if you choose this project.
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