# Solutions for Math Homework 1

1. 2.1.1 Fixed points are points where sinx = 0, that is, integer multiples of π.

2.1.2 The greatest velocity to the right occurs where sinx is the greatest. That is, where x is π/2
plus an integer multiple of 2π.

2.1.3 a) implies by the chain rule , so Did you
remember that cosxsinx = sin(2x)/2? If not, here is an easy way to memorize it: Among the
trigonometric formulas , the only ones that I memorize are sin(a+b) = sinacosb+cosasinb
and cos(a+b)=cosacosb−sinasinb. All others fol low from these very quickly. For example,
the first with a = b = x gives you what you need for this problem. b) The acceleration to the
right is greatest where sin(2x) is greatest, that is where x = π/4 plus an integer multiple of π.

2. 2.2.1. Sketch of the vector field:

x = −2 is a stable fixed point, and x = 2 is unstable, as is apparent from the vector field.
Here is what the solutions look like:

(Solutions with x > 2 blow up in finite time.)

Solution in closed form: If then x(t) = ±2 for all t, since 2 and −2 are fixed points.
Let’s assume that now. Then x(t) ≠ ±2 for all t (this is true by the uniqueness part of
the existence and uniqueness theorem). This will be used in the following manipulations.

for some constant C, which is found by plugging in t = 0. Using the notation we find

So

Exponentiate both sides of this equation:

So

For t = 0, the correct sign is +. Therefore the correct sign must be + for all t. (If the sign
changed from + to − all of the sudden, x would have to be discontinuous at that point!) So

Multiply this equation by x+2, then solve for x :

3. 2.2.8 It is best to draw a suitable function first:

After that, it’s almost as easy to write down a formula defining a function that looks like that ,
qualitatively:

f (x) = x(x+1)2(x−2)

(In fact, the above plot shows precisely that function .)

2.2.9 The function must vanish ( be zero ) at x = 0 and x = 1. It must be negative between 0 and
1, and positive everywhere else. Its minimum must occur near x = 1/2. It must increase for
x > 1, and decrease for x < 0. (Can you see why?) A function that meets all these criteria is

f (x) = x(x−1) .

4. 2.3.4 To motivate this problem, suppose that in general

describes population growth. We certainly want

f (0) = 0.

This just expresses the fact that it is possible that there is no population at all, and never will be.
Now what are the simplest choices of functions f = f (N) with f (0)= 0? The very simplest
one is linear :

f (N) = rN.

This gives rise to exponential growth (or, if r < 0, decay). The second-simplest is quadratic:

This gives rise to logistic growth , as we discussed in class; is the carrying capacity. (The
physically interesting case is r > 0. What happens if r < 0?) This problem is about the third-simplest
case, that of a cubic function f . Since f (0) is to be zero, we must be able to write f (N)
as N times a quadratic function . Strogatz writes it like this:

f (N) = (r−a(N−b)2) N (1)

Any quadratic function can be written in the form r −a(N −b)2. So our population growth
equation is then.

(2)

Let us first consider the case when N = 0 is the only fixed point, that is, when the equation

r−a(N −b)2 = 0 (3)

has no solution. One case in which Eq. (3) has no solution is a = 0, r ≠ 0. In that case, Eq. (2)
becomes

the exponential growth (or, if r < 0, decay) model. This is not of interest to us here, so let us
assume a ≠ 0. In that case, Eq. (3) has no solution if and only if

(4)

There are two possibilities : Either N = 0 is a stable fixed point, or an unstable one. Which of
these two cases we are in depends on the sign of f ′(0). In general,

f ′(N) = r−a(N −b)2−2a(N −b)N,

and therefore

f ′(0) = r−ab2.

So N = 0 is stable if r < ab2, and unstable if r > ab2. Remember that we assume (4), that is, we
assume that r and a are of opposite signs. Therefore if a > 0, then certainly r < ab2, since then
r is negative and ab2 is positive. On the other hand, if a < 0, then certainly r > ab2, since then
r is positive and ab2 is negative. This argument shows that in fact, if (3) has no solution, then
N = 0 is a stable fixed point if a > 0, and an unstable one if a < 0. We could have seen this by
a different argument as well: For large N,

(compare Eq. (1). Therefore, if a < 0, then f (N) → −∞ as N → −∞, and f (N) → +∞ as
N →+∞. If f crosses through N = 0 only once, it must cross from negative values (for N < 0)
to positive ones (for N > 0), and therefore N = 0 must be unstable. One sees by a precisely
analogous argument that N = 0 must be a stable fixed point if Eq. (3) has no solution, and
a > 0.

But if N = 0 is the only fixed point, and is stable, the problem is uninteresting: The population
simply goes extinct, no matter where we start. If N = 0 is the only fixed point, and it
is unstable, then in fact Eq. (2) is unphysical, since it predicts blowup of the population size in
finite time. (Do you see why?)

So we have now concluded that the case in which Eq. (3) has no solution is uninteresting.
We will now assume

(5)

so that Eq. (3) has two solutions. Those two solutions are

(6)

Either one of the two solutions of Eq. (3) could be zero, depending on the parameter values,
but we will assume that they are not zero, so that there are three distinct solutions of f(N) = 0,
rather than a solution of multiplicity two at N = 0.

There are three cases to distinguish now:

(7)
(8)

and

(9)

The negative fixed points are of no physical interest to us, so the number of physically interesting
fixed points if one in case (7), two in case (8), and three in case (9). Case (7) either leads to
guaranteed extinction of the population, if a > 0, or blowup in finite time, if a < 0; the former
is uninteresting, the latter unphysical.

So let us proceed to case (8). Here there are again two possibilities. If a > 0, the fixed
point 0 is unstable, and the fixed point is stable. Therefore the population will level off at
for any initial value N(0) > 0. This is very similar to logistic growth, and therefore not so
interesting to us here. The other possibility is a< 0, in which case the population will go extinct
if N(0) < , and blow up in finite time if N(0) > — not a physical case.

So let us proceed to case (9). Again, we think about the two possibilities a > 0 and a < 0.
If a < 0, then the population will lever off to if 0 < N(0) < , and blow up in finite time if
N(0)>—not a physical case. The only potentially interesting and physical case is therefore
that in which (9) holds and a > 0. In this case, the population will go extinct if 0 < N(0) < ,
and level off to if N(0) > . Thus is an extinction threshold: If N(0) is below it, the
population goes extinct, if N(0) is above it, the population levels off to the “carrying capacity”
.

I summarize: The only choice of a cubic f which is interesting, physically plausible, and
qualitatively different from exponential and logistic growth is that in which

a > 0, r > 0,

In this case, there are three fixed points, namely

and

is the extinction threshold, and N+ the carrying capacity.

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