MA1506
Tutorial 11
·
¸
1
3
1.
T r = −1, det = −5 ⇒ SADDLE
1 −2
·
·
·
·
·
2.
2 −2
4
0
¸
T r = 2, det = 8, Tr2 − 4 det = −28 < 0 ⇒ SPIRAL SOURCE
−2 −4
10
0
−5
−2
4
1
5 −4
2 −1
Solutions
¸
T r = −2, det = 40, Tr2 − 4 det = −36 < 0 ⇒ SPIRAL SINK
¸
T r = −4, det = 3, Tr2 − 4 det = 4 > 0 ⇒ NODAL SINK
¸
0 1
−10 0
T r = 4, det = 3, Tr2 − 4 det = 4 > 0 ⇒ NODAL SOURCE
¸
T r = 0, det = 10, Tr2 − 4 det = −40 < 0 ⇒ CENTRE
Let E(t) be the number of elves, D(t) the number of Dwarves. Let BE , DE be the
birth and death rates per capita for Elves, similarly BD , DD for Dwarves. We are
told that BE > BD and DE < DD . So BE − DE > BD − DD > 0 (more births
than deaths).
Now dE
dt is controlled by (BE − DE )E in the usual Malthus model, but here we are
told that dE
dt is reduced by the presence of Dwarves, so we propose
dE
= (BE − DE )E − PE D
dt
where PE is a constant that measures the prejudice of the elves. Similarly
dD
= (BD − DD )D − PD E
dt
Where PD represents the prejudice of the Dwarves. So


dE
dt
dD
dt

·
¸µ ¶
−PE
E
 = (BE − DE )
−PD
(BD − DD )
D
So in a concrete case, the constants in the matrix should
satisfy
·
¸ BE − DE >
5 −4
BD − DD > 0, and we are told that PE > PD . Indeed
satisfies all of
−1
2
these. T r = 7, det=6, Tr2 − 4 det = 49 − 24 = 25 > 0 so we have a nodal source.
Eigenvalues
(5 − λ)(2 − λ) − 4 = 0 ⇒ λ = 1 or 6.
1
µ ¶ µ
¶
1
1
,
.
Eigenvectors are
1
−1/4
So the phase plane diagram
(we only care about the
first quadrant, since
D ≥ 0, E ≥ 0) is
bisected by the line D = E. All points ABOVE that line will move along trajectories
that eventually hit the D axis (that is, E = 0). So if at any time D > E, then the
Elf population may increase for a while, but eventually it will reach a maximum
and then collapse to zero. Rivendell is completely taken over by Dwarves, EVEN
THOUGH BE > BD AND DE < DD . The prejudice of the Elves cancels out their
other advantages and causes them to lose the competition.
3.
Let S(t) be the number of Spartans, P (t) the number of Persians. We have
dS
= −P
dt
dP
= −11, 111, 111.1S
dt


so
dS
dt
dP
dt

·
0
−1
=
−11, 111, 111.1 0
¸·
S
P
¸
Eigenvalues λ2 − 11, 111, 111.1 = 0 ⇒ λ = ±3333.3333
·
Eigenvectors
±3333.3333
−11, 111, 111.1
−1
±3333.3333
¸· ¸
1
=0
α
⇒ α = ±3333.3333.
Since we are interested only in the first quadrant, take +. We have a SADDLE.
2
The trajectory that passes
through the origin is a straight
line
to the eigenvector
µ parallel ¶
1
. That straight
3333.3333
line has equation P = 3333.3333S.
We want to start at the point on this straight line that has P = 1000000. So the
1000000
initial value of S has to be 3333.3333
= 300.
So Leonidas should take 299 soldiers, assuming that he intends to fight (and die)
himself.
4. (1) Use the rule:
Rate of change of the amount of U F6 = (concentration in) (flow rate in) - (concentration out) (flow rate out), where “concentration” is the mass of uranium hexafluoride
per unit volume of water − you can understand this equation by looking at the
units:
gallons
lbs
lbs
×
=
sec
gallons
sec
Let xA be the mass of U F6 in the first tank, so the concentration in that tank is
xA /100. Similarly let xB be the mass of U F6 in the second tank. Then from the
given data we have
2
6
ẋA =
xB −
xA
100
100
and
ẋB =
6
6
xA −
xB .
100
100
Notice that the 4 gallons/min of pure water flowing into the first tank does not
appear here − it contains no U F6 ; it is just there so that the amounts of water in
the tanks remain constant. The initial conditions are xA (0) = 25, xB (0) = 0.
In matrix form, we have
·
ẋA
ẋB
¸
¸·
¸
·
1 −6
2
xA
=
.
6 −6
xB
100
√
−6 + 2 3
The eigenvalues of the coefficient matrix can be found as λ1 =
and
100
√
·
¸
·
¸
−6 − 2 3
1
1
√
λ2 =
. The corresponding eigenvectors are √
and
.
3
− 3
100
Hence
·
xA
xB
¸
·
= c1 e
λ1 t
¸
·
¸
1
1
λ
t
2
√ + c2 e
√ .
3
− 3
Substituting the initial values, we have c1 = 12.5, c2 = 12.5, so we get xA (t) =
√
√
12.5(eλ1 t + eλ2 t ), xB (t) = 12.5( 3eλ1 t − 3eλ2 t ).
3
b) In the java applet, x corresponds to xA and y to xB . You have to click on a point
on the x axis because we are told that xB [that is, y] is zero initially. You should
see the red curve going steadily down, but the green curve goes up at first and then
down.
√
25
c) Yes. The two curves intersect when t0 = √
ln(2 + 3) ≈ 19, (which is in fact
3
the maximum point of xB ). Before t0 , the uranium hexafluoride in tank A is more
than that in tank B; after that, it’s always less.
d) Here Trace = −12/100 and det = 24/100, so Trace2 −4 det = (144-96)/10000 which
is positive; hence we have a nodal sink. That is also clear from the java applet.
5.
dL
dt
= BL L − DL L
(BL =birth rate per capita, DL =death rate per capita.)
dL
dt
But BL = uZ so
Similarly
= uZL − DL L.
1
(Units of u = year
.)
dZ
= BZ Z − DZ Z
dt
= BZ Z − sLZ
At equilibrium,
dL
dt
=
dZ
dt
= 0, so
DL
u
BZ
BZ − sL = 0 ⇒ L =
s
uZ − DL = 0 ⇒ Z =
⇒L=
0.05
0.004
= 12.5, Z =
0.20
0.0008
= 250.
Substitute L = 12.5 + x, Z = 250 + y, get
dx
dL
=
= 0.0008(250 + y)(12.5 + x) − 0.2(12.5 + x)
dt
dt
= 0.01y + 0.0008xy
≈ 0.01y
Hence
dZ
dy
=
= 0.05(250 + y) − 0.004(12.5 + x)(250 + y)
dt
dt
= −x − 0.004xy ≈ −x
 dx 
·
¸µ ¶
dt
0
0.01
x
 ≈
−1
0
y
dy
dt
T r = 0, det = 0.01
4
T r2 − 4 det < 0 ⇒ centre.
So near equilibrium, the Lion and Zebra populations should fluctuate up and down
but never vary much from their equilibrium values. The equilibrium is stable.
5
Question 6
From the last section of Chapter 7, we know that there is a line in the M - G plane
which separates victory from defeat. The discussion there shows that if the orcs get no
reinforcements, then this line [originally M = (2/3)G] gets moved vertically. We need to
move this line up so that it passes through the point (15, 11), since that corresponds to
the marginal situation in which the Gondor army is destroyed; any reinforcement rate
above this will result in victory for Gondor, since the initial point will then lie below
this line. [We measure numbers of soldiers in thousands; recall that initially there are 15
thousand men and 11 thousand orcs.] So we are interested in a line of the form
M − A = (2/3)G,
where A is a constant to be determined by the fact that the line has to pass through
(15,11). Clearly A = 1, so [again from the last part of Chapter 7] we have
− B −1 F~ =
0
1
!
,
where F~ is the vector which gives the number of reinforcements. Clearly
F~ = − B
0
1
!
= −
−1 −3/4
−1
0
!
0
1
!
=
3/4
0
!
,
so the minimum number of reinforcements needed is 751 per day [since in fact 750 would
lead to defeat for Gondor!].