MA 2327
Assignment 1
Due 12 October 2015
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1. Solve the initial value problem x(t0 ) = x0 for the differential equation
x′ (t) = −p + qx(t) − rx(t)2
in the exceptional case q 2 = 4pr. What is the largest interval on which
wyour solution is defined? What is the largest interval on which it
is biologically meaningful, with the usual interpretation of x as the
population of a species? Don’t forget to check that the solution you
find really is a solution and that it’s the only solution.
Solution:
It’s easier to do the calculations if we write the equation as
x′ (t) = −r(x(t) − α)2
where
α=
q
.
2r
Formally,
dx
= −r(x − α)2 ,
dt
dx
= dt,
−r(x − α)2
1
1
1
= t − t0 ,
−
r x − α x0 − α
1
1
=
− rt0 + rt,
x−α
x0 − α
1
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x−α=
x=α+
1
x0 −α
x0 − α
1
,
=
1 + r(x0 − α)(t − t0 )
− rt0 + rt
x0 − α
x0 + rα(x0 − α)(t − t0 )
=
.
1 + r(x0 − α)(t − t0 )
1 + r(x0 − α)(t − t0 )
If x0 = α then
x(t) = α
is a valid solution, both mathematically and biologically, on the whole
real line. If x0 > α then the largest interval on which
x0 + rα(x0 − α)(t − t0 )
1 + r(x0 − α)(t − t0 )
x(t) =
is well defined is
t0 −
1
< t < ∞.
r(x0 − α)
The solution positive, and hence biologically meaningful on this whole
interval. If x0 < α then the largest interval on which x(t) is well defined
is
1
−∞ < t < t0 +
,
r(α − x0 )
but it is only positive on the smaller interval
−∞ < t < t0 +
x0
,
rα(α − x0 )
and it is only on this interval that the solution is biologically meaningful.
We still need to justify the formal calculation. We start by checking
that
x0 + rα(x0 − α)(t − t0 )
x(t) =
1 + r(x0 − α)(t − t0 )
really is a solution to the initial value problem. The initial condition
x(t0 ) = x0 is clearly satisfied. Differentiating,
x′ (t) =
−r(x0 − α)
.
(1 + r(x0 − α)(t − t0 ))2
Also
x(t) − α =
x0 − α
,
1 + r(x0 − α)(t − t0 )
so
x′ (t) = −r(x(t) − α)2 ,
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as expected.
To show that this is the only solution we start from an arbitrary solution
of the differential equation
x′ (t) = −r(x(t) − α)2 ,
and consider the quantity
I(t, x) =
1
− rt.
x(t) − α
Differentiating,
d
d
I(t, x(t)) =
dt
dt
!
1
x′ (t)
− rt = −
− r = 0,
x(t) − α
(x(t) − α)2
so I(t, x(t)) is independent of t. If x satisfies the initial conditions then
I(t, x(t)) = I(t0 , x(t0 )) = I(t0 , x0 ),
or
1
1
− rt =
− rt0 .
x(t) − α
x0 − α
From this it follows that
1
1 − r(x0 − α)(t − t0 )
1
=
− r(t − t0 ) =
x(t) − α
x0 − α
x0 − α
and then
x(t) − α =
and
x(t) =
x0 − α
1 − r(x0 − α)(t − t0 )
x0 + rα(x0 − α)(t − t0 )
.
1 + r(x0 − α)(t − t0 )
Strictly speaking there is a slight problem with this argument: It assumes x(t) 6= α. This is true for all t if x0 6= α, but that’s not immediately obvious. We’ll see how to deal with this problem later.
2. Show that if q 2 < 4pr then
x′ (t) = −p + qx(t) − rx(t)2
has no solutions which extend to t = ∞ and remain positive, assuming,
as usual, that p, q, r > 0. Biologically this means there are no solutions
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which do not lead sooner or later to extinction of the species.
Note: You can do this by solving the equation explicitly, but please
don’t. Instead show that
4pr − q 2
x (t) ≤ −
<0
4r
′
and then use that inequality. You will need the Mean Value Theorem,
or something equivalent.
Solution: By simple algebra
−p + qx − rx2 = −r(x − α)2 −
4pr − q 2
4r
where
q
.
2r
Because r > 0 and (x − α)2 ≥ 0 it follows that
α=
−p + qx − rx2 ≤ −
4pr − q 2
4r
and hence
4pr − q 2
4r
for all t. Assume x(t) > 0 for all t. Then
x′ (t) ≤ −
x (t1 ) > 0
where
t1 = t0 +
4rx0
.
(4pr − q 2 )
By the Mean Value Theorem there is some point t in the interval (t0 , t1 )
such that
x′ (t) =
x(t1 ) − x(t0 )
x0
4pr − q 2
>−
=−
,
t1 − t0
t1 − t0
4r
which contradicts the inequality derived earlier.