THE UNIVERSITY OF SYDNEY
Nonlinear Differential Equations & Applications
Semester 1
Tutorial Week 3
2016
SELECTED SOLUTIONS
1. (i)
(ii )
If N = 0 or N = 50 the solution is constant. If 0 < N < 50 the solution is increasing.
Otherwise the solution is decreasing.
If N = 2 or N = 8 the solution is constant. If 2 < N < 8 the solution is increasing.
Otherwise the solution is decreasing.
(iii ) If N = 0 or N = β/α the solution is constant. If 0 < N < β/α the solution is
decreasing. Otherwise the solution is increasing.
(iv ) If N = 0, N = M or N = K the solution is constant. If N > K the solution is
decreasing. If M < N < K the solution is increasing. If 0 < N < M the solution is
decreasing. If N < 0 the solution is increasing.
(v )
Thre are two constant solutions somewhere between 0 and 10. Specifically at
√
√
10 ± 100 − 4h
N=
= 5 ± 25 − h
2
√
√
The solition is increasing if 5 − 25 − h < N < 5 + 25 − h and otherwise it is
decreasing.
2. Phase Lines for the previous question
(i)
50
0
(ii)
2
(iii)
0
(iv)
(v)
0
M
5−√(25−h)
8
β/α
K
5+√(25−h)
N
N
N
N
N
3. Solution curves for the previous question
N
(i)
N
(ii)
50
8
t
0
2
t
N
N
(iii)
(iv)
K
β/α
M
0
0
(v)
t
N
5+√(25−h)
5−√(25−h)
t
MORE REVISION OF CURVE SKETCHING
4. (i)
(ii )
Passes through origin, peaks at (1, e−1 ) and then decays to zero,
Has a minimum at the origin (locally parabolic), peaks at (2, 4e−2 ) and then decays
to zero,
(iii ) It’s a hyperbola with asymptotes at x = 1 and y = 1 and is equivalent to
(x − 1)(y − 1) = 1,
(iv ) Vertical asymptote at x = 0, diagonal asymptote at y = x and a minimum at (1, 1),
(v )
Looks like famous Van der Waals potential,
(vi ) Ellipse,
(vii ) Hyperbola,
(viii ) Two diagonal lines that cross at the origin.
5. A sketch of y = tan(x) has vertical asymptotes at all the odd multiples of π/2. If you
superimpose a sketch of y = x then intersections occur just to the left of almost all of these
asymptotes; but since line y = x is tangent to the first branch of y = tan(x) there is no
other intersection point on this first branch. There is an exact solution at x = 0 but zero
is NOT a positive number. Thus the two smallest positive intersections occur just to the
5π
3π
left of
and
2
2
INTERPRETING & DESCRIBING SOLUTIONS
6. (i)
Linear, first-order, autonomous, separable x(t) = Aekt . Curve starts at A, if k > 0
exponentially growing, if k < 0 exponentially decaying
2
Linear, first-order, non-autonomous, separable x(t) = Ae−t . Curve starts at A, decays towards 0; has the same shape as the tail of the Gaussian or normal distribution
curve.
1
(iii ) Nonlinear, first-order, autonomous, separable x(t) =
, asymptotically tend1 + Ae−t
ing towards 1
1
, If A > 0 diverges when
(iv ) Nonlinear, first-order, autonomous, separable x(t) =
A−t
t reaches A, otherwise tends to zero from below.
(ii )
(v )
Linear, second-order, autonomous x(t) = Ae−2t + Be−3t , Decays as sum of two
decaying exponentials
(vi ) Linear, second-order, autonomous x(t) = Ae2t + Be3t , Grows as sum of two growing
exponentials
(vii ) Linear, second-order, autonomous x(t) = Ae−2t + Be+3t , The growing exponential
usually wins out over the decaying exponential unless B = 0, in which case it decays
asymptotically.
(viii) Linear, second-order, autonomous x(t) = A cos(3t) + B sin(3t), Periodic
(ix ) Linear, second-order, autonomous x(t) = e2t [A cos(3t) + B sin(3t)], Exponentially
growing oscillation
Linear, second-order, autonomous x(t) = e−2t [A cos(3t) + B sin(3t)], Exponentially
decaying oscillation
where A and B are arbitrary constants. There are of course many other equivalent formulas
for the general solutions.
(x )
7. (i )
(ii )
If k is positive then all non-zero solutions will tend to ±∞. If k is negative then all
non-zero solutions will tend to zero. If k = 0 are solutions are constant.
All non-zero solutions tend to zero.
(iii ) If A = 0 the solution is constant. If A > 0 the solution is bounded and tends to
1 from below. If −1 < A < 0 the solution blows up (diverges to infinity) when
t = ln(−A). If A = −1 the solution is undefined at t = 0. If A < −1 it blows up at
t = ln(−A) but diverges towards negative infinity.
(iv ) If A ≤ 0 the solution decays to zero. If A > 0 the solution blows up at t = A.
(v )
All non-zero solutions decay to zero.
(vi ) All non-zero solutions will tend to ±∞.
(vii ) Almost all non-zero solutions will tend to ±∞ unless B = 0 in which case the
solutions tend to zero.
(viii ) All non-zero solutions oscillate and so remain bounded.
(ix ) All non-zero solutions will oscillate and tend to ±∞.
(x )
All non-zero solutions decay to zero.
8. In part (vii) the reason for the almost always is that if the arbitrary constant B = 0 then
the exponentially growing term is absent and this entire subset of solutions will tend to
zero. This specific question is here to prepare students to think about saddle points in a few
weeks time which have both types of behaviour (stable and unstable manifolds) depending
upon from which direction you approach the saddle.
A NON-AUTONOMOUS MODEL
9. The RHS is clearly zero if p = 0 (no-one is infected) or p = 1 (everyone is infected) so both
of these values are possible equilibrium solutions and thus possible asymptotic limits. For
finite values of t these are the only possible equilibria. Unfortunately as t → ∞ the RHS
tends to zero for any value of p; so does this mean that any value of p can be the asymptotic
limit of the epidemic? Since the ODE is non-autonomous it makes understanding all the
possible behaviours very difficult.
The exact solution can be found by separation of variables and partial fractions:
Z
Z
Z
dp
1
1
= ( +
)dp =
r0 e−αt dt
p(1 − p)
p 1−p
r0
ln(p) − ln(1 − p) = − e−αt + C
α
p
r0 −αt
ln(
) = − e
+C
1−p
α
The asymptotic behaviour can be found by letting t → ∞ and doing a little bit of rearranging. Thus
p
ln(
) → C
1−p
eC
p →
1 + eC
Thus, the asymptotic value of p can be any number between 0 and 1 and it depends on
the arbitrary constant, which itself depends on the initial condition. This is why it is so
hard to make long-term predictions by just looking at the ODE.
For this model the final state of the epidemic depends critically on the initial state and the
relationship cannot be determined without knowing the exact formula for the solution.
Although the mathematical skills required to work through this problem are at a first year
level, the effort required demonstrates why autonomous equations are so much nicer to work
with.