Math 5410 Exam 1 Solutions
1) The claim that the limit function y(t) is a solution of z 0 = f (t, z) is true.
Since yn is a solution of the differential equation, we have for any t0 and t,
yn (t) = yn (t0 ) +
Z
t
f (s, yn (s)) ds.
t0
Since yn converges uniformly to y, yn (t) → y(t) and yn (t0 ) → y(t0 ). Since f
is Hőlder continuous, we can conclude that f (s, yn (s)) converges uniformly
to f (s, y(s)) by the following argument. Let ≥ 0. We need to show that
there exists an N such that if n ≥ N, then |f (s, yn(s)) − f (s, y(s))| ≤ for all s. By the Hőlder condition, we know that |f (s, yn(s)) − f (s, y(s))| ≤
K|yn (s) − y(s)|α for all s. Since yn converges uniformly to y, there exists a
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N1 such that if n ≥ N1 , then |yn (s) − y(s)| ≤ (/K) α for all s. Combining
this inequality with the last one, we obtain that |f (s, yn (s)) − f (s, y(s))| ≤ for all s and all n ≥ N1 . Therefore,
Z
Z
t
t0
f (s, yn (s)) ds →
which implies that
y(t) = y(t0 ) +
Z
t
t0
f (s, y(s)) ds,
t
t0
f (s, y(s)) ds
so by differentiation, we obtain that y solves the differential equation.
2) It is convenient to divide the population into three groups: those who
are immune to the disease (including those naturally immune and those who
have recovered from it), those susceptible to the disease (those who are not
naturally immune and have not been infected), and those currently infected.
Let y(t) be the proportion of the population that is infected at time t, so
that y(t) is the number currently infected divided by the (constant) total
population of the community. Let z(t) be the proportion of the population
immune at t, and w(t) be the proportion susceptible to the disease at time
t. Since every member of the population belongs to one of these groups, we
have y(t) + z(t) + w(t) = 1 at all times t, and differentiating this equation,
we get y 0 (t) + z 0 (t) + w 0(t) = 0. Our initial conditions are: y(0) = y0 , where
y0 is the proportion of the population that is initally exposed to the disease,
0 < y0 < .85, because if y0 = 0 the disease will not spread and we know
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that 15% of the population is naturally immune, i.e. z(0) = .15, and then
we must have that w(0) = .85 − y0 . The rate of change of w is given by
the rate of new infections and reduces w, so this rate is negative. The rate
of new infections if proportional to the product of the proportion infected
and the proportion susceptible, so w 0 = −Ayw for some positive constant
A. The rate of change of y is equal to the rate of new infections minus the
rate of of recovery. We have previously seen that the rate of new infections is
Ayw. The disease lasts ten days, so the rate of recovery is equal to the rate of
infection from ten days earlier. Since no one gets better for the first ten days,
this is given by step(t − 10)w 0 (t − 10). Therefore, y 0 (t) = Ay(t)w(t)−step(t −
10)w 0(t − 10). Finally, using the fact that y 0(t) + z 0 (t) + w 0 (t) = 0, we see
that z 0 (t) =step(t − 10)w 0(t − 10). This gives us a system of three differential
equations that describe the spread of the disease and the development of
immmunity to it.
As for whether this model is reasonable, many answers were possible.
The rate of new infections is highest when both the number of infected and
susceptible is predicted to be high. This makes sense because if the number
of infected is low, there aren’t many people spreading the disease, and if
there is a small number of susceptibles there aren’t that many people who
can contract the disease. This model does assume a uniform level of contact
among people in the community: it takes no account for people who have
contact with very large or very small numbers of other people. Also, most
diseases do not last precisely ten days.
3) a) Let z = (y − g)−1. Then z 0 = (g 0 − y 0 )(y − g)−2 .
z 0 + (a + 2bg)z =
g0 − y0
a + 2bg
+
=
2
(y − g)
y−g
g 0 − y 0 + (a + 2bg)(y − g)
(y − g)2
(1)
Since y and g both solve the nonlinear equation, a calculation shows that
g 0 − y 0 = −[a(y − g) + b(y 2 − g 2)].
Therefore, using this and multiplying out, equation (1) becomes
−a(y − g) − b(y + g)(y − g) + a(y − g) + 2bgy − 2bg 2
=
(y − g)2
2
−b(y + g) + 2bg
bg − by
=
= −b,
y−g
y−g
which is what we wanted to show.
b) Suppose g is a particular solution of the nonlinear equation, and y is
any other solution. Then z = (y − g)−1 solves the linear equation, so z
is an element (say w) of the general solution (of the linear equation), and
rearranging, we see that y = g + 1/w, showing that every solution of the
nonlinear equation has to be of the given form. On the other hand, every
function of the form y = g + 1/w, where w solves the linear equation, must
solve the nonlinear equation (can be checked directly), thus showing that
y = g + 1/w, where w is the general solution of the linear equation, is the
general solution of the nonlinear equation. Recall that to show a given family
of functions is the general solution of a differential equation, you must show
that every function in the family solves the DE and that every solution of
the DE is a member of the family.
c) After some algebra on the original equation, we obtain a nonlinear equation
of the type considered in part a); the coefficients are a(t) = t−1 −2t4 , b(t) = t3
and F (t) = t5 . g(t) = t is one solution of the given equation. Using the idea
of part a), make the change of variables z = (y − t)−1 . After some more
algebra, we get the following linear equation in z:
1
z 0 + z = −t3 .
t
4
This equation has general solution −t5 + C/t, where C is an arbitrary constant. Therefore, using the result of part b), we get that the general solution
for the nonlinear equation is
y =t+
−t4
5
1
+
C
t
=t+
5t
.
5C − t5
Note: Other substitutions will also result in the correct solution. If you
tried a different approach, and were penalized, please see me, you may be
entitled to some addtional credit.
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4) a) x0 = I − k1 x, x(0) = 0, y 0 = k1 x − k2 y, y(0) = 0.
b) The equation for x is linear and the IVP has solution
x(t) =
I
(1 − e−k1 t ).
k1
Using this result in the (linear) equation for y, using an integrating factor
and the initial condition, we obtain that
y(t) =
I
I
Ik1
−
e−k1 t +
e−k2 t .
k2 k2 − k1
k2 (k2 − k1 )
c) By looking at the formulas for x and y in part b), we see that limt→∞ x(t) =
I/k1 and limt→∞ y(t) = I/k2 .
d), e), f) call for plotting solution curves for several values for k1 and k2 . If
you have questions about how to do this, I would be happy to show you.
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