CHAPTER 7
SYSTEMS OF FIRST-ORDER ODEs
7.1. ROMEO AND JULIET
We all know that many (most) relationships
have their ups and downs. Let’s try to model
this fact.
Romeo loves Juliet, but Juliet believes in a
more subtle approach and finds Romeo’s excessive enthusiasm rather repulsive - the more he
loves her, the less she likes him. On the other
hand, when he loses interest, she fears losing
him and begins to see his good side. Romeo
is more straightforward: his love for Juliet increases when she is warm to him, and decreases
1
when she is cold. Let R(t) represent Romeo’s
feelings and J(t) Juliet’s. We can model the
lovers’ feelings by:
dR
dt = aJ
dJ
dt = −bR
R(0) = α
J(0) = β
(1)
where a, b are positive constants and α and
β represent their feelings when they first meet.
This is a system of simultaneous firstorder ODEs. In this case, the equations are
linear, so it is easy to solve them. Put, as a
trial,
R = Aeλt
J = Beλt
where λ could turn out to be complex - if so, we
will as usual interpret the exponential to mean
2
that we are really dealing with sine and cosine
functions. [The final solutions for R and J must
be real — the feelings are real, not complex!]
So Aλeλt = aBeλt and Bλeλt = −bAeλt.
So we get Aλ = aB, Bλ = −bA so ignoring
special cases we get λ2 = −ab < 0.
Since eiθ = cos θ+i sin θ, this is telling us that
√
the solution is some combination of cos( abt)
√
and sin( abt) i.e.:
√
√
R(t) = C cos( abt) + D sin( abt)
√
√
J(t) = E cos( abt) + F sin( abt)
3
So
√
R(0) = C
Ṙ(0) =
abD
√
˙
J(0) = abF
J(0) = E
But from equations (1), we see that
R(0) = α
J(0) = β
Ṙ(0) = aJ(0) = aβ
˙ = −bR(0) = −bα
J(0)
so we have:
r
C = R(0) = α
E = J(0) = β
Ṙ(0)
a
D= √ =β
b
ab
r
˙
J(0)
b
F = √ = −α
a
ab
Hence
√
R(t) = α cos( abt) + β
√
a
sin( abt)
rb
√
√
b
sin( abt)
J(t) = β cos( abt) − α
a
4
r
Suppose (for example) that β = 0 and α > 0.
Then
√
R(t) = α cos(
r abt)
√
b
J(t) = −α
sin( abt)
a
Then the graphs of R(t) and J(t) are:
0,8
0,4
-2,5
0
2,5
5
7,5
10
12,5
15
-0,4
-0,8
-1,2
Actually it is more useful to eliminate t and
get a direct relation between R and J. A bit of
-1,6
algebra will convince you that:
R2
J2
+ 2 =1
2
Rmax
Jmax
5
which can be sketched in the R − J plane: it is
an ellipse.
This is just for these particular initial conditions. Different initial conditions will result in
a smaller or a larger ellipse. The full set of ALL
possible love-affairs is represented by an infinite
set of concentric ellipses: So this diagram tells
6
us everything there is to know about love.
Note that at a particular time t, R(t) and J(t)
have definite values, giving a point (R(t), J(t))
in these pictures. The arrows indicate the direction of motion of such a point as time goes
by. You can check this against the graphs on
page 5 - notice that J becomes negative immediately after t = 0 if the initial coordinates are
7
(R, J) = (α, 0). A picture like this, where we
have two functions R(t) and J(t) but where we
eliminate t (and regard it as a parameter)
is called a PHASE PLANE DIAGRAM.
It tells us many things that are not so obvious from the diagram on page 5. For example, suppose we ask: can Romeo and Juliet
ever have a steady relationship with R= constant and J=constant? The answer is clear
from the picture on page 7: this is possible only
if R = J = 0 (at the centre). We say that
this is a point of equilibrium, and clearly
it is the only one. Furthermore, it is obvious
from the diagram that this equilibrium is sta8
ble.
Contrast this with a phase diagram
like the one in the next picture: In this picture
(R, J) = (0, 0) is a point of equilibrium, but a
slight push along the positive J axis will cause
the point to be swept off to infinity in the second
quadrant, with J → +∞, R → −∞. Clearly
9
unstable equilibrium! So stability can be determined just by a glance at the phase diagram.
VERY OFTEN, WE DON’T CARE ABOUT
THE DETAILS OF THE SOLUTIONS — THE
PHASE PLANE DIAGRAM ALREADY TELLS
US ALL WE NEED TO KNOW FOR MANY
APPLIED PROBLEMS!
So in this part of the course, we are more interested in the shape of the phase plane diagram
than in the explicit solution.
7.2. HOW TO SOLVE SYSTEMS OF
SIMULTANEOUS ODEs
10
Let’s consider the general system:
d x
x , with a, b, c and d
a b
=
constants
y
c d
dt y
(i.e.
dx
dt
= ax + by ,
dy
dt
= cx + dy)
Trial solution:
x
x0
~u =
= ert~u0 , ~u0 =
= constant
y
y
0 a b
So we get (defining B =
)
c d
rert~u0 = Bert~u0
or
B~u0 = r~u0
so the possibilities for r are given by the eigenvalues of B. We have
(B − rI2)~u0 = ~0
11
so non-trivial solutions (i.e. ~uo 6= ~0) exist only
if det(B − rI2) = 0 i.e. (a − r)(d − r) − bc = 0
i
p
1h
⇒ r = a + d ± (a + d)2 − 4(ad − bc)
2
i
p
1h
⇒ r = T r[B] ± (T r[B])2 − 4(Det[B]) .
2
Except in the very special case where (T rB)2 is
exactly equal to 4 det B, there are two solutions
r+, r− and so
~u(t) = c+er+t~u+ + c−er−t~u−
where c+ and c− are constants and ~u+ and ~u−
are eigenvectors corresponding to r+, r−. Naturally r+ and r− could be complex so we might
have to interpret the exponentials as sin and
cos.
12
Example
Solve
dx
dt = −4x + 3y
dy
dt = −2x + y
−4 3
→ T rB = −3 , det B = 2
−2 1
√
1
r = 2 [−3 ± 9 − 8] = −1 or −2.
Here B =
First eigenvector satisfies:
−4 3
x0
x
= − 0 → −4x0+3y0 = −x0
−2 1
y0
y0
x0
1
will work. (Note: if B~u0 =
So
=
1
y0
r~u0, then B(c~u0) = r(c~u0) for any number c,
so clearly ~u0 cannot be found uniquely.
2
13.59
So if you prefer
or
, go ahead - it
2
13.59
won’t matter!)
13
Second eigenvector satisfies
x0
x
−4 3
= −2 0 → −4x0+3y0 = −2x0
−2 1
y0
y0
x0
3
i.e. −2x0 + 3y0 = 0 so take
=
2 y0
x
1
= c+e−t
+
General solution is now:
y
1
3
where c+ and c− are arbitrary conc−e−2t
2
stants. i.e.
x = c+e−t + 3c−e−2t
y = c+e−t + 2c−e−2t
and we can work out c+ and c− given x(0) and
y(0).
Example
Solve
dx
dt
dy
dt
= 4x − 5y
= 2x − 2y
14
4 −5
→ T rB = 2 , det B = 2
2 −2
√
1
r = [2 ± 4 − 8] = 1 ± i Eigenvectors are
2
5
5
and
. So,
3−i
3+i
5
5
(1+i)t
(1−i)t
~u(t) = c+e
+ c− e
3−i
3+i
Here B =
Of course, we only want the real part of this. As
usual, the presence of all these complex numbers
is just telling us that there must be sines and
cosines [as well as exponentials of course] in the
solution. We don’t really care too much about
the details of the solution, but anyway here it
15
is:
x(t) = 10(α cos t − β sin t)et
y(t) = 6(α cos t − β sin t)et + 2 (β cos t + α sin t) et
= 2(3α + β)et cos t + 2(α − 3β)et sin t
What to notice about these examples First, the solutions are always combinations of et, e−t, sin and cos. If it’s e−t or sin or
cos, the solution is bounded as t → ∞ and we
associate that with stable behaviour, i.e.
if a system is in a stable state then [by definition] if you push it slightly away from that
state, it will not stray very far from the original
16
state. On the other hand, et soon becomes large
even if it is small at first, so we associate et (or
any other positive exponent) with unstable
behaviour.
Now in the first example, we got e−t and
e−2t because the eigenvalues of B were −1 and
−2, i.e. both real and both negative.
That solution is stable because both e−t → 0
and e−2t → 0 as t → ∞. If the eigenvalues had
been +1 and +2 then of course we would have
an unstable situation, and similarly if they had
been +1 and −2 or −1 and +2.
17
Now we know that
i
p
1h
r = T rB ± (T rB)2 − 4 det B
2
so the condition to have real solutions is (T rB)2 >
4 det B. To have both solutions negative, we
√
obviously need T rB < 0, otherwise T rB + ...
√
will be positive. If T rB < 0 then T rB − ...
√
is OK, but what about T rB + ... ?
If T rB +
then
√
√
... < 0
... < −T rB(which is positive)
square: (T rB)2 − 4 det B < (T rB)2
⇒ det B > 0
Conclusion: If both eigenvalues are real,
18
the system is stable when:
T rB < 0
(2)
det B > 0
(3)
Now we turn to the other example. We saw
that you get sin and cos only when the eigenvalues are complex i.e. (T rB)2 < 4 det B. In
our example, the eigenvalues were 1 ± i so we
obtained:
e(1±i)t = et(cos t ± i sin t)
In general, if the eigenvalues are Φ±iΨ, we get:
e(Φ±iΨ)t = eΦt(cos(Ψt) ± i sin(Ψt))
so we need Φ ≤ 0 for stability, i.e. if the
eigenvalues are complex the condi19
tion for stability is T rB ≤ 0. You might
find it helpful to remember this diagram of the
”det B vs Tr B” plane. Only one quadrant corresponds to stable systems.
SUMMARY
The stability of the linear system of ODEs
with constant coefficient can be decided by ex20
amining the trace and determinant of the coefficient matrix B.
7.3. PHASE PLANE: REAL EIGENVALUES Our objective is to be able to say
something about the shape of the paths in the
phase plane, just by studying the matrix B.
Let’s talk about the case where the eigenvalues of B are REAL, because that case is easy to
picture.
We have
d~u
dt
= B~u and we set ~u = ert~u0. If
~u0 is an eigenvector of B and r is the corresponding eigenvalue, then ert~u0 is of course a
solution. But what does this mean geometrically? When you multiply a vector by a scalar,
21
you are stretching or shrinking that vector
So ert~u0 represents a vector that always
points in the same direction but which
grows if r > 0, shrinks if r < 0.
So the eigenvalues and eigenvectors give us
two solutions which are straight lines in the
phase plane, with one of them shooting out
to infinity [the one with r > 0] and the other
shrinking into the origin [the one with r < 0].
0 1
So for the case B =
, the eigenvectors
1 0
are shown here:
22
[ remember that if ~u0 is an eigenvector, so is
−~u0] so we can get two solutions just by letting
1
−1
−1
and
grow and by letting
1 −1
1
1
and
shrink. The solutions must look
−1
like the ones shown.
So you can see what will happen in other cases
23
like this: suppose we have a different system but
with eigenvalues which are both real but one
is positive and the other negative.
Maybe the eigenvectors look like this:
with the eigenvalues as shown, and so we get:
24
The point is this: All of the solutions with both eigenvalues real but
of opposite signs have phase plane diagrams of the same general shape!
This shape is called a saddle.
Let’s look at another family: Both eigenval25
ues real, both of the same sign.
and it’s not too hard to guess what the solutions will look like: for example, suppose that
we consider
dx
dt = 2x
dy
dt = y
B=
2 0
1
eigenvalues 2, 1 eigenvectors:
0 1
0
26
0
1
x = x0e2t
y = y 0 et
Actually x = ky 2 so the solution curves (apart
from the straight lines) are parabolas.
Of course for the system
dx
dt = 22x
dy
dt = y
the details will be different, but the general
shape is the same. We call a phase diagram
27
with this kind of shape a nodal source.
dx
= −2x
In the case of dtdy
the shape will
=
−y
dt
be identical but the arrows all point in - this
is called a nodal sink. In all these cases,
x =constant= y = 0 is an equilibrium point,
but obviously nodal sources are unstable, nodal
28
sinks are stable. And so on: All phase
diagrams can be classified into families: Saddles, nodal sources, and nodal sinks
are the ONLY possibilities when the eigenvalues
are REAL. When the eigenvalues and eigenvectors are complex it is not so easy to picture
things, so I will just tell you the answer!
7.4. PHASE PLANE: COMPLETE CLASSIFICATION It turns out that all phase di x
u
=
B~
u
,
~
u
=
, belong to one
agrams for d~
dt
y
of the following families.
(Note: both of the above can be either clockwise or anticlockwise)
29
Figure 1: Spiral Sink
Figure 2: Spiral Source
Figure 3: Saddle Point
30
Figure 4: Nodal Sink
Figure 5: Nodal Source
31
Figure 6: Centre
In words:
1. A spiral sink has all trajectories (i.e. solution curves) moving on a spiral towards
a certain equilibrium point.
2. A spiral source has all trajectories moving
on a spiral away from a certain equilibrium point.
3. A saddle has some trajectories moving toward, and some away from an equilibrium
32
point
4. A nodal sink has all trajectories moving not
on a spiral, towards a certain equilibrium
point
5. A nodal source has all trajectories moving
not on a spiral, away from a certain
equilibrium point
6. A centre has the trajectories orbiting around
an equilibrium point
To understand what is happening in terms of
eigenvalues, recall
i
p
1h
r = T rB ± (T rB)2 − 4 det B
2
We know that if they are real, then:
33
• Both positive −→ Nodal source
• Both negative −→ Nodal sink
• One positive, one negative −→ saddle
By inspection we see:
• T rB > 0, det B > 0, real ⇒ Nodal Source
• T rB < 0, det B > 0, real ⇒ Nodal Sink
• T rB =anything, det B < 0 ⇒ Saddle
What if the eigenvalues are complex? Then
i
p
1
2
2 [T rB ± (T rB) − 4 det B
exponential ↑
sin, cos ↑
Clearly T rB controls the exponential part and
√
± ... controls sin, cos part.
34
Now a spiral, by definition, is a curve which
is (a) steadily moving away from (or towards)
the origin and (b) moves around the origin
infinitely many times.
So we get spirals when the eigenvalues are complex and when T rB 6= 0
35
(Source if T rB > 0, sink if T rB < 0). What
if T rB = 0? then if r is complex we still have
sin and cos, but we have no exponential, e.g.
x = cos(t)
circle.
y = sin(t)
In general, we get a centre when the
eigenvalues are pure imaginary i.e.
T rB = 0
SUMMARY
dx
dt = ax + by i.e. d~u = B~
u, if you
Given dy
dt
=
cx
+
dy
dt
want to sketch the phase plane, you can do it
in either of two ways:
Method 1
36
Find eigenvalues of B, given by
i
p
1h
r = T rB ± (T rB)2 − 4 det B,
2
and then classify as follows:
Method 2
Look at the entries in the matrix B, compute the TRACE and the DETERMINANT,
and classify as follows:
37
Romeo + Juliet
0 a
, a, b > 0, T rB = 0,
B =
−b 0
Example
det B = ab > 0
(T rB)2 − 4 det B < 0 → complex
38
√
0 − 4ab =
√
abi, Real part = 0
⇒ centre
Clearly, from the equations themselves, J is
decreasing anywhere along the positive R axis,
so the arrows have to be pointing DOWN there
[because down is the direction of decreasing J],
and this tells us that the direction of the arrows
must be clockwise, as in the diagram.
39
(page 13)
−4 3
→ T rB = −3 , det B = 2
B=
−2 1
Example
(T rB)2 − 4 det B = 1 > 0 → real
T rB < 0 , det B > 0 → nodal sink
The solution is actually
x
1
3
= c+e−t
+ c−e−2t
y
1
2
and the phase diagram is shown in the next
diagram.
Don’t worry too much about the details of
this picture: the main thing is for you to understand that a nodal sink has this kind of shape,
with everything being sucked into the origin
along a sort of S-shaped pattern.
40
(page 14)
4 −5
T rB = 2
B=
→
→ spiral source
2 −2
det B = 2
Example
The phase diagram is shown in the next diagram.
How did I know that the spirals move in the
41
anticlockwise direction? Easy: just check what
happens along the positive x axis. On that axis,
y is zero so dy/dt = 2x - 2y = 2x which is
of course positive, so y must be INCREASING
there, which means that the direction is upward,
that is, anticlockwise!
42
7.5. WARFARE
A long and bitter battle is being fought on
the slopes of Mount Doom between 15,000 Men
of Gondor and 11,000 Orcs of Mordor. The
Men die at a rate proportional to the number
of Orcs, and also from a dread disease spread
among them by the servants of Sauron, while
the Orcs only die at a rate proportional to the
number of Men – Orcs never get sick.
Let G(t) denote the number of Gondorians
and M(t) denote the number of Mordor citizens
in the battle. Then the above information tells
us that we have a pair of differential equations
which might have this form:
43
d G
dt M
=
−1 −0.75
−1 0
G
M
,
where the −1 in the top left corner is the death
rate per capita of the Gondorians due to disease,
the −0.75 describes the rate at which Mordorians kill Gondorians, and the −1 in the second
row describes the rate at which Gondorians kill
Mordorians. [So the Gondorians are better soldiers.]
ODEs have actually been used to study real
battles in this way! Of course, AS USUAL, you
would want to make the model a lot more complicated than this model if you are really serious.
You can easily show that this is a saddle. The
44
1
1
and
, but we
−2
2/3
are only interested in the first quadrant, since
two eigenvectors are
G and M cannot be negative! So we focus on
the second vector. If we take the G axis to be
horizontal and the M axis to be vertical, you can
show that this vector points along the line M
= (2/3)G. The points in the phase plane move
towards the origin along this line, because this
vector has a negative eigenvalue.
If you draw the phase plane diagram for this
saddle, you will soon see that any initial point
lying BELOW this line will lead to victory for
Gondor. Any initial value ABOVE it results in
45
victory for Mordor. [Any initial point ON this
line results in a situation where the last Gondorian falls dead while killing the last Mordorian!]
Alas, you should be able to see that MORDOR
WINS this particular battle! That wasn’t obvious, since the Gondorians heavily outnumbered
the Mordorians, and furthermore they are better fighters. Sad.
7.6. INHOMOGENEOUS SYSTEMS
We now know how to analyse systems of LINEAR first-order differential equations of the form
d x
a b
x , with a, b, c and d
=
constants
c d
y
dt y
(i.e.
dx
dt
= ax + by ,
dy
dt
46
= cx + dy.)
Let’s write this as
d~r
= B~r,
dt
where ~r is the position vector in the phase plane,
and B is the matrix of coefficients as usual. Now
clearly this is not the most general linear system; more generally we could let B become a
function of time [complicated!] or we could consider the INHOMOGENEOUS system
d~r
= B~r + F~ (t),
dt
where F~ (t) is some given vector function of
time. This just means that we are considering a pair of linear equations of the form
ax + by + f (t) ,
dy
dt
dx
dt
=
= cx + dy + g(t), where
47
f (t) and g(t) are given functions of time. This
is of course interesting if we wish to apply an
external “force” to the system — for example,
if we are modelling warfare, and the two armies
are steadily being reinforced at given rates.
Such problems can actually be quite tricky,
but there is a very nice way of solving them,
as follows. If we didn’t know that our equation
is an equation involving vectors and matrices,
then we would immediately know what to do:
we would try to find an INTEGRATING FACTOR, because our equation is linear and firstorder. If B is constant, then that integrating
48
factor is just e−Bt, and we would get
d −Bt
[e ~r] = e−BtF~ (t),
dt
by the usual method of the integrating factor.
Then the solution would be
Z t
~r = eBt~c + eBt
e−BxF~ (x)dx,
0
where ~c is a constant vector and where x is a
dummy variable of integration. Actually ~c is
just the value of ~r at time t = 0.
We know that all this works very well for
NUMBERS. But it doesn’t seem to make sense
for MATRICES! Or does it? Let’s DEFINE [I
= identity matrix] the EXPONENTIAL OF A
49
MATRIX by
e
−Bt
1
1
2
= I − Bt + [Bt] − [Bt]3 + ....
2
6
Of course, you should also DEFINE what it
means for such a series to converge, etc etc etc.
Let’s not worry about that [it does all work out
nicely in fact] and just accept this definition.
Then, with this definition, you can easily verify
that
eBte−Bt = I
and that
d −Bt
e
= −Be−Bt = −e−BtB.
dt
But in fact these are the only properties we need
for the above argument to go through! [Check
50
this, step by step! Of course, in dealing with
matrices, you have to be careful about the ordering of the symbols.] It follows that the solution we found above is correct, provided that we
interpret the matrix exponentials in this way.
There’s just one catch.....given B, how do you
actually work out eBt?? Do we really have to
write out the infinite sum and somehow work
that out? The cure is worse than the disease!
Luckily, the answer is no. Suppose that, like
most matrices, B is DIAGONALIZABLE. That
is,
B = P DP −1,
where P is a matrix that you get by comput51
ing the eigenvectors of B and D is the diagonal
matrix that you get from the eigenvalues of B.
Then clearly B n = P DnP −1 for any integer n,
and using this you can easily show that
eBt = P eDtP −1.
The point, of course, is that eDt is easy to compute: if
D=
λ1 0
,
0 λ2
then
Dn =
λn1
0
,
0 λn2
and so
eDt =
λ1 t
e
0
0 eλ2t
.
So now we have a concrete expression for the
52
exponentials of Bt and − Bt. Substituting into
our solution, we therefore get:
Z t
~r = P eDtP −1~c + P eDt
e−DxP −1F~ (x)dx,
0
where we now know how to compute everything
on the right hand side! [Remember that ~c is just
~r(t = 0).]
Notice that this also works in the case we have
already considered in earlier sections [that is,
the case where F~ = ~0]. So you can use it to
solve, for example, the problem of the war between Gondor and Mordor; we already know the
eigenvectors and the eigenvalues in that case.
Now suppose for example that both the Gondorians and the Mordorians are reinforced at
53
constant rates: the Gondorians send g new troops
per day to the front, and the Mordorians send
m new troops per day, where g and m are constants. Then the equations become
d G
−1 −0.75
G
g
=
+
.
M
−1
0
M
m
dt
Since F~ (x) here is a constant, the integral can
be done easily:
Z t
e−Dxdx = −D−1[e−Dt − I].
0
So the solution is
G
g
= P eDtP −1~c − P D−1[I−eDt]P −1
.
M
m
Since we know everything on the right hand
side, the problem is solved.
Actually in this particular case [constant F~ ]
54
there is a short-cut, as follows. The equation
can be written as
d~r
− B~r = F~ .
dt
We can think of this as an inhomogeneous linear differential equation. We already know how
to solve the homogeneous equation; the general
solution is
~r = P eDtP −1~c
for some constant vector ~c. To get a particular
solution of the inhomogeneous equation, we try
~ a constant vector since
to guess: let’s try ~r = S,
the right side of the equation is constant. Then
we get
~ = F~
− BS
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so, if [as is usually the case] the matrix B has
non-zero determinant, we can solve to get
~ = − B −1F~ .
S
The general solution of the inhomogeneous equation is the sum of the old solution with − B −1F~ .
But this too is just a constant vector. So to get
the solution here, we just have to take the old solution and push the entire phase diagram along
the vector − B −1F~ . In particular, the equilibrium point, which was originally at the origin
of coordinates, now moves to the point with
position vector − B −1F~ . Everything else [the
shape of the phase diagram etc] remains exactly
the same as before. This allows us to draw the
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phase diagram without doing any complicated
integrals or matrix multiplication.
Suppose for example that the Mordorians and
the Gondorians both begin to reinforce their
armies by 1000 men/orcs per day. Then [using units of 1000] we have
0
1
1
1
−1 ~
−B F =
=
.
4/3 −4/3
1
0
So the effect of these reinforcements is just to
shift the equilibrium point along the G axis to
the right. Instead of M = (2/3)G, we now have
to consider a line M = (2/3)[G − 1]. Any initial point above this line leads to victory for the
orcs, whereas any initial point on or below it
leads to victory for Gondor. [You can use the
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“Romeo and Juliet” java applet to check this.
You can use the column with “x+” and “y+” to
shift the origin; you have to use negative numbers to shift the origin to the right or upwards.]
That does not help the Gondorians, since the
initial point is still on the wrong side for them.
In fact, it just makes things worse! This is a
bit surprising, since after all the number of reinforcements is the same on both sides.
Suppose instead that the Gondorians get 1000
new men per day, but the Orcs have been cut
off and get no reinforcements at all. Then we
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have
−B −1F~ =
0
1
1
0
=
,
4/3 −4/3
0
4/3
so this time the equilibrium point moves vertically upwards by 4/3 units [of a thousand].
Now the crucial line between victory and defeat is M − 4/3 = (2/3)G; any initial point on
or above this leads to victory for the orcs, but
any initial point below it leads to victory for
Gondor. This does help the Gondorians, and
in fact they will now win, because the initial
point now lies below the crucial line, because
11 − (4/3) < (2/3) × 15. Note that it is important in all these discussions that the eigenvectors associated with the negative eigenvalue
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point into the first and third quadrants, and
that the eigenvectors associated with the positive eigenvalue point into the second and fourth
quadrants. This is always true for any matrix
of the form
−a −b
,
−c 0
where a, b, and c are all positive, as you can
easily prove. [Hint: multiply this matrix into a
1
. If this is an eigenvecvector with entries
α
tor with eigenvalue λ, then the second line of
the eigenvalue equation reads
−c = λα,
so α has to be positive when λ is negative and
vice versa.] This is important because, with a
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saddle phase plane diagram, no curve can cross
the straight lines defined by the eigenvectors:
once a point is in one of the four regions defined
by these straight lines, it has to stay in that
region. Using this you can compute a lower
bound on how many survivors there will be at
the end of the battle.
REMARK: with certain choices [eg if the orcs
are reinforced but the men are not] then the
equilibrium point moves into the fourth quadrant. [In this problem, with these particular
values for the matrix B, it cannot move into the
second or third because with any positive rein
0
1
forcement vector, multiplying by
4/3 −4/3
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gives a vector with a positive i-component.] In
that case the boundary of the orc-victory zone
expands, BUT some trajectories on the orc side
of the line will hit the G axis before they have
a chance to hit the M axis. In this case, the
eigenline is NOT the boundary between the orcvictory and the man-victory zones, as it always
is if the equilibrium point lies in the first quadrant.
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