8 Differential Equations

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8
Differential Equations
8.1
Use separation of variables to solve the following differential equations with given initial
conditions.
(a)
(b)
dy
dt
dy
dt
= −2ty, y(0) = 10
= y(1 − y), y(0) = 0.5, (Hint:
1
y(y−1)
=
1
y−1
− y1 ).
Solution
(a)
dy
y
2
2
= −2tdt ⇒ ln |y| = −t2 + C0 ⇒ y = Ce−t . y(0) = 10 = C ⇒ y(t) = 10e−t .
1
= −t + C0 ⇒
= −dt ⇒ ( y−1
− y1 )dy = − dt ⇒ ln y−1
y ⇒ C = −1. Thus, y − 1 = −e−t y, which gives y(t) = 1+e1 −t .
(b)
dy
y(y−1)
R
R
y−1
y
= Ce−t . y(0) = 0.5
8.2
Consider the differential equation and initial condition
dy
dt
= 1 − y,
y(0) = 0.5.
(a) Solve for y as a function of t using separation of variables.
(b) Use the initial condition to find the value of C.
Solution
(a) The initial condition y(0) = 0.5 guarantees that 1 − y > 0, and therefore we can separate
variables:
Z
Z
dy
= dt = t + C =⇒ − ln |1 − y| = t + C
1−y
To find C we set t = 0 and y = 0.5:
− ln(0.5) = C =⇒ C = ln 2.
Solving for y we get:
1
1
ln |1 − y| = −t − ln 2 =⇒ 1 − y = e−t e− ln 2 = e−t =⇒ y = 1 − e−t .
2
2
Note: We should verify our calculations by doing the check, that is we should check that the
function
1
y = 1 − e−t
2
satisfies the differential equation
dy
=1−y
dt
and the initial condition y(0) = 0.5.
(b) C = 0.5.
1
8.3
Consider the differential equation and initial condition
dy
= 1 + y 2,
dt
y(0) = 0.5.
Repeat the process described in problem 8.2 and find the solution, i.e. determine the function
y(t) that satisfies this differential equation.
Solution
Since 1 + y 2 6= 0 we can separate variables:
Z
dy
=
1 + y2
Z
dt = t + C,
that is
tan−1 y = t + C.
Putting t = 0 and y = 0.5 gives
C = tan−1
1
≈ 0.464
2
and then solving for y we get
y = tan t + tan
−1
1
= tan(t + 0.464).
2
Again we should do the check.
8.4
Consider the differential equation and initial condition
dy
= 1 − y2,
dt
y(0) = 0.5.
Solve using separation of variables (Hint: use partial fractions.)
Solution
Separating variables gives
Z
Using partial fractions we get
Z
1
dy
=
2
1−y
2
Z
dy
=
1 − y2
Z
!
dt = t + C.
1
1
1
+
dy = (− ln |1 − y| + ln |1 + y|)
1−y 1+y
2
2
and therefore
Putting t = 0 and y = 0.5 gives
1 1 + y = t + C.
ln 2 1 − y 1
ln 3.
2
C=
Solving for y we get
1+y
= e2t+ln 3 = 3e2t =⇒ 1 + y = 3e2t − 3e2t y
1−y
3e2t − 1
.
3e2t + 1
A point worth noting is that the initial condition implies that
y(1 + 3e2t ) = 3e2t − 1 =⇒ y =
1+y
> 0.
1−y
Again we should do the check.
8.5
Solve the following differential equation and initial condition, assuming that k > 0 is a
constant:
dy
= ky 2/3
dt
y(0) = y0
Solution
1
y 2/3
Z
dy = k dt
y −2/3 dy = kt + C
y 1/3
= kt + C
1/3
1
y 1/3 = kt + C 0
3
1
0
where we have renamed the constant (C = 3 C). We note that the constant can be obtained
from the initial conditions. That is, plugging t = 0 into the above, we get
1/3
y0
= C0
3
Continuing to solve for the function, we get
1
y = ( kt + C 0 )3
3
Thus
1
1/3
y(t) = ( kt + y0 )3 .
3
8.6
A certain cylindrical water tank has a hole in the bottom, out of which water flows. The
√
height of water in the tank, h(t), can be described by the differential equation dh
h
=
−k
dt
where k is a positive constant. If the height of the water is initially h0 , determine how much
time elapses before the tank is empty.
Solution
h(0) = h0
√
dh
= −k h
dt
Z h(T )
Z T
dh
√ = −k
dt
h0
0
h
√ h(T )
= −kT
2 h
h0
q
2( h(T ) −
q
h0 ) = −kT
q
−k
T + h0
2
q
k
h(T ) = ( h0 − T )2
2
the tank is empty when h(T ) = 0, i.e.
q
h(T ) =
q
k
h0 − T = 0
2
√
2 h0
T =
k
4
8.7
The position of a particle is described by
dx √
= 1 − x2
dt
0≤t≤
π
2
(a)Find the position as a function of time given that the partical starts at x = 0 initially.
(b)Where is the particle when t = π2 ?
(c)At which position(s) is the particle moving the fastest? The slowest?
Solution
(a)
Z
√
dx
= dt
1 − x2
Z
sin−1 x = t + C
x = sin(t + c)
at t = 0, x = 0, so C = 0 Therefore, x = sint
(b)at t =
π
2
x = sin π2 = 1
(c)
dx √
= 1 − x2
dt
Therefore v is greatest when x = 0 (then v = 1), and v is smallest when x = +1 (then
v = 0). (We do not consider x = −1 because it is not in the domain.)
v=
8.8
In an experiment involving yeast cells, it was determined that mortality of the cells increased
at a linear rate, i.e. that at time t after the beginning of the experiment the mortality rate
was
m(t) = m0 + rt.
The experiment was started with No cells at time t = 0. Let N (t) represent the population
size of the cells at time t.
(a) If no “birth” occurs, then
dN (t)/dt = −m(t)N (t).
Solve this differential equation by separation of variables, i.e find N (t) as a function of time.
5
(b) Suppose cells are “born” at a constant rate b, so that the differential equation is
dN (t)/dt = −m(t)N (t) + bN (t).
Determine how this affects the population size at time t.
Solution
(a) We have to solve the differential equation
dN (t)/dt = −(m0 + rt)N (t).
Separation of variables yields
Z
dN
=
N
Z
−(m0 + rt)dt,
hence
Taking exponentials, this yields
ln(N ) = −m0 t − rt2 /2 + c.
N (t) = Ce−m0 t−rt
2 /2
where the constant C must be determined from the initial condition N (0) = N0 , hence
C = N0 , i.e.
2
N (t) = N0 e−m0 t−rt /2 .
(b) The only difference to the previous question is that we add a constant b to the constant
−m0 in (a):
dN (t)
= −(m0 + rt + b)N (t)
dt
Z
Z
dN
=
−(m0 + b + rt)dt
N
rt2
ln(N ) = −m0 t − bt −
+C
2
2
N (t) = N0 e−(m0 +b)t−rt /2
8.9
Muscle cells are known to be powered by filaments of the protein actin which slide past one
another. The filaments are moved by cross-bridges of myosin, that act like little motors,
which attach, pull the filaments, and then detach. Let n be the fraction of myosin crossbridges that are attached at time t. A model for cross-bridge attachment is:
dn
= k1 (1 − n) − k2 n,
dt
n(0) = n0
6
where k1 > 0, k2 > 0 are constants.
(a) Solve this differential equation and determine the fraction of attached cross-bridges n(t)
as a function of time t.
(b) What value does n(t) approach after a long time?
(c) Suppose k1 = 1.0 and k2 = 0.2. Starting from n(0) = 0, how long does it take for 50% of
the cross-bridges to become attached ?
Solution
(a) We have to solve the differential equation
dn
= k1 (1 − n) − k2 n = k1 − k1 n − k2 n = k1 − n(k1 + k2 )
dt
Separation of variables yields
Z
hence
dn
=
k1 − n(k1 + k2 )
Z
dt
−1
ln |k1 − n(k1 + k2 )| = t + c
k1 + k 2
and thus
k1 − n(k1 + k2 ) = Ce−(k1 +k2 )t .
(1)
Thus, the general solution is
n(t) =
−Ce−(k1 +k2 )t + k1
.
k1 + k 2
The constant C must be determined by the initial condition n(0) = n0 using equation (1):
C = k1 − n0 (k1 + k2 ).
1
(b) As t → ∞, we get n(t) → k1k+k
, because the exponential term approaches 0.
2
(c) Plugging in k1 = 1.0 and k2 = 0.2, we have
n(t) =
−Ce−1.2t + 1
.
1.2
From (a) we find that C = 1 − 0(1.2) = 1. We have to solve for n(t) = 0.5, hence
0.5 = (−e−1.2t + 1)/1.2
which yields
e−1.2t = 0.4 =⇒ t = − ln(0.4)/1.2 ≈ 0.7636.
7
8.10
The velocity of an object falling under the effect of gravity with air resistance is given by:
dv
= f (v) = g − kv, v(0) = vo , where g > 0 is acceleration due to gravity and k > 0 is a
dt
frictional coefficient (both constant).
(a) Sketch the function f (v) as a function of v. Identify a value of v for which no change
occurs, i.e., for which f (v) = 0. This is called a steady state value of the velocity or a fixed
point. Interpret what this value represents.
(b) Explain what happens if the initial velocity is larger or smaller than this steady state
value. Will the velocity increase or decrease ? (Recall that the sign of the derivative dv/dt
tells us whether v(t) is an increasing or decreasing function of t.
(c) Use separation of variables to find the function v(t).
Solution
(a) See the sketch below. The steady state value is v = kg . At this value of v there is no
change in the velocity (that is dv
= 0) and so v(t) = kg for all time t that the differential
dt
equation remains valid.
dv/dt = g − kv
g/k
v
Stable steady state = g/k
Figure 1: Plot for problem 8.10
(b) If v(0) >
g
k
then
dv
<0
dt
for all time t that the differential equation remains valid, and so v(t) decreases to the terminal
velocity kg . On the other hand, if v(0) < kg , then
dv
>0
dt
8
for all time t that the differential equation remains valid, and so v(t) increases to the terminal
velocity kg .
(c) Separating variables gives
Z
Z
dv
= dt = t + C :
g − kv
that is
1
− ln |g − kv| = t + C
k
ln |g − kv| = −kt − kC
g − kv = C 0 e−kt
Putting t = 0 and v(0) = v0 gives
C 0 = g − kv0
and then solving for v we get:
g − (g − kv0 )e−kt = kv
g
g
−
− v0 e−kt .
v(t) =
k
k
8.11
A model for the velocity of a sky diver (slightly different from the model we have already
studied) is
dv
= 9 − v2
dt
(a) What is this skydiver’s ”terminal velocity”; that is, near what velocity will the sky diver
eventually stabilize?
(b) Starting from rest, how long will it take to reach half of this velocity?
Solution
(a) Velocity will stop changing and reach a ”terminal” value when
dv
=0
dt
i.e.
0 = 9 − v2
v = 3.
We reject v = −3 because it is in the wrong direction.
9
(b)
Z
(time to reach velocity v =
3
)
2
1.5
0
dv
= 9 − v2
dt
dv
= dt
9 − v2
Z T
dv
=
dt = T
(3 − v)(3 + v)
0
Partial Fractions:
a
B
1
=
+
(3 + v)(3 − v)
(3 + v) 3 − v
1
T =
6
8.12
Z
1.5
0
1 = A(3 − v) + B(3 + v)
1
A=B=
6
1.5
1
1
1
1
4.5
+
dv = (ln |3 + v| − ln |3 − v|) = ln | |
(3 + v) (3 − v)
6
6
1.5
0
1
T = ln 3 = 0.183 sec
6
Newton’s Law of Cooling
Consider the differential equation for Newton’s Law of Cooling
dT
= −k(T − E)
dt
and the initial condition: T (0) = T0 . Solve this differential equation by the method of
separation of variables i.e. find T (t). Interpret your result.
Solution
dT
= −k(T − E), solve for T (t).
dt
Separation of variables:
dT
= −k dt
T −E
Integration:
dT
= − k dt
T −E
ln |T − E| = −kt + C0
Z
Z
10
If T > E, we get:
T − E = Ce−kt
T = E + Ce−kt
Using the initial condition, T (0) = T0 , we find that T0 = E + C ⇒ C = T0 − E. This gives
us Newton’s Law of Cooling, T (t) = E + (T0 − E)e−kt .
However, if T < E, we get:
−T + E = Ce−kt
T = E − Ce−kt
In this case, with the initial condition we get T0 = E − C, or C = E − T0 . Plugging this into
the equation gives:
T = E − (E − T0 )e−kt =⇒ T = E + (T0 − E)e−kt .
Again, we end up with the familiar equation for Newton’s Law of Cooling.
Interpret result: For t → ∞, T (t) → E + very small quantity ≈ E.
Thus, the coffee temperature approaches the value of the environment as time goes on. If
initially it is greater than the environment, i.e. if T0 > E, then it approaches equilibrium
from above. If T0 < E, then the temperature approaches equilibrium from below.
8.13
Let V (t) be the volume of a spherical cell that is expanding by absorbing water from its
surface area. Suppose that the rate of increase of volume is simply proportional to the
surface area of the sphere, and that, initially, the volume is V0 . Find a differential equation
that describes the way that V changes. Use the connection between surface area and volume
in a sphere to rewrite your differential equation in terms of the volume alone. You should
get an equation in the form
dV
= kV 2/3 ,
dt
where k is some constant. Solve the differential equation to show how the volume changes
as a function of the time. (Hint: recall that for a sphere, V = (4/3)πr 3 , S = 4πr 2 .)
Solution
The volume changes at a rate proportional to the surface area, i.e.
dV
= αS.
dt
V = (4/3)πr 3 ⇒ r = (3V /4π)1/3
S = 4πr 2 = 4π[(3V /4π)1/3 ]2 = βV 2/3
11
where β is a constant
β = 4π(3/4π)2/3
Thus
dV
= αβV 2/3 = kV 2/3 ,
dt
where k is just a constant. This equation is in the same form as the one solved in problem
8.5, and the method of solution will be similar
Thus
1
1/3
V (t) = ( kt + V0 )3 .
3
Qualitative Methods
8.14
In problems 8.2, 8.3, and 8.4 above, your method was to calculate a formula for the unknown
function y(t). Now you will use qualitative methods for these same problems instead: Sketch
a graph of the expression dy/dt versus y in each case. (E.g., in problem 8.2 you will be
sketching the straight line f (y) = 1 − y). In each case, determine the values of y at which
dy/dt = 0 (steady states, or critical points).
Solution
See the sketches in Figure 2.
8.15
Find the fixed points (steady states, critical points) of the following differential equations
and classify their stability.
dy
= 4 − y2
dt
(a)
dy
= y2 − 4
dt
(c)
dy
= y(y − 1)(y + 1)
dt
(b)
(d)
dy
= cos(2πy)
dt
Solution
(a) f (y) = y 2 − 4. f (y) = 0 gives us two steady states: y = −2 and y = 2. Since f 0 (y) = 2y,
f 0 (−2) = −4 < 0 ⇒ y = −2 is stable. f 0 (2) = 4 > 0 ⇒ y = 2 is unstable.
(b) f (y) = 4 − y 2 . f (y) = 0 gives us two steady states: y = −2 and y = 2. But f 0 (y) = −2y.
The stability is the reverse of (a). y = −2 is unstable, y = 2 is stable.
12
dy/dt = 1− y
dy/dt = 1+y2
1
y
y
Stable steady state = 1
Stable steady state: NONE
dy/dt = 1−y2
1
−1
y
Stable steady state: ± 1
Figure 2: Plots for problem 8.14
(c) f (y) = y(y − 1)(y + 1). The steady states are: y = −1, 0, 1. f 0 (y) = (y − 1)(y + 1) +
y(y + 1) + y(y − 1). Thus, f 0 (−1) = 2, f 0 (0) = −1, f 0 (1) = 2. Therefore, y = −1, 1 are
unstable, y = 0 is stable.
(d) f (y) = cos(2πy). For 0 < y < 1, there are 2 steady states: y = 1/4, 3/4. Since
f 0 (y) = −2π sin(2πy), f 0 (1/4) = −2π and f 0 (3/4) = 2π. Thus, y = 1/4 is stable and
y = 3/4 is unstable.
8.16
For each part of problem 8.15, sketch the qualitative behavior, i.e. the flows along the y
axis.
Solution
See Figure 3.
13
dy/dt
dy/dt
−2
y
2
2
−2
y
dy/dt
1.25
−1
0
1
−1.25
y
−.75
−.25
.25
.75
y
Figure 3: Plots for problem 8.16
8.17
Determine the stability type of the equilibrium solution (fixed point) in the sky-diving equation
dv
= g − kv.
dt
Interpret your result in the context of the terminal velocity of the sky-diver.
Solution
The only steady state is v = kg and it is stable. This follows from the Linear Stability
Condition since f 0 (v) = −k < 0, where f (v) = g − kv. This means that any object in free
fall will eventually reach the terminal velocity v = kg .
8.18
Newton’s Law of Cooling states that the temperature of an object T (t) changes at a rate that
depends on the difference between that temperature and the temperature of the environment,
E: dT
= −k(T − E), T (0) = T0 . Assume that E, k are positive constants.
dt
= −k(T − E) as a function of T . Identify points on your
(a) Sketch the expression dT
dt
sketch that correspond to value(s) of the temperature that would not change (i.e. at which
dT /dt = 0).
14
(b) Use your sketch to identify ranges of the temperature T for which dT /dt is negative or
positive, i.e. for which the temperature is increasing or decreasing. Use arrows on the T axis
to indicate the “flow”.
(c) Interpret your sketch. Give a verbal description of what Newton’s Law of Cooling is
saying about the temperature T (t) of the body as the time t increases.
Solution
(a) See the sketch in Figure 4. The steady state temperature is just the ambient temperature,
that is, T = E.
dT/dt = −k (T−E)
E
T
Stable steady state = E
Figure 4: Plot for problem 8.18
(b) From the differential equation dT
= −k(T − E) it is clear that if T > E then dT
is
dt
dt
dT
negative and if T < E then dt is positive. See the sketch in Figure 4 for the “direction of
flow”.
(c) Newton’s law of cooling is saying that objects either warm up to the room temperature
(if T (0) < E) or cool down to it (if T (0) > E) , and this happens exponentially. In all cases
T (t) → E as t → ∞.
8.19
Find a differential equation that has exactly three steady states, at y = 0, 2, 4 and for which
only y = 2 is an unstable steady state. How would you change this to an equation for which
y = 2 is stable and the other two states are unstable?
Solution
= ry(y − 2)(y − 4) has exactly 3 steady states at
The differential equation f (x) = dy
dt
y = 0, 2, 4. Now we need to determine if the constant r should be positive or negative in
order that y = 2 be unstable. The derivative of this function is:
f 0(y) = r(y − 2)(y − 4) + ry(y − 4) + ry(y − 2).
15
Thus f 0(2) = ry(y − 4) = −4r. For y = 2 to be an unstable steady state, f 0(2) > 0 ⇒ −4r >
0 and r < 0. Therefore, the differential equation is dy
= −ry(y − 2)(y − 4) (r > 0 constant).
dt
If we change the sign of the right hand side of the equation, i.e. dy
= ry(y − 2)(y − 4), y = 2
dt
becomes stable while the other two become unstable.
8.20
Consider the differential equation
dy
= y 2 − r,
dt
where r is some positive constant. Find the fixed points of this equation and determine their
stability. What happens as r is decreased ? What happens when r = 0? The Linear Stability
Condition fails (WHY?). But would you describe the fixed point as stable or unstable in
that case?
Solution
√
dy
= y 2 − r are y = ± r. From
The fixed points (steady states) of the differential equation
dt
√
√
the
graph
we
see
that
−
r
is
stable;
whereas
r
is
unstable.
As r → 0+ the steady states
√
± r coalesce and both become 0. The linear stability condition fails because the derivative
of f (y) = y 2 at y = 0 is 0 i.e. the slope is neither positive nor negative. On the other hand,
y = 0 is an unstable steady state because the trajectories point away from it for y > 0.
8.21
You are a resource ecologist, hired to manage the population of fish in some lakes. Records
indicate that in some lakes, the population of fish stabilizes at a very low level, and that in
other lakes, the population stabilizes at a high level. The outcome seems to depend on the
starting density of fish in the lake. You are asked to model the density of the fish population
y(t) in a lake by a differential equation:
dy
= f (y)
dt
(a)You are asked to predict whether any major changes in the behavior of the fish population
would occur if the lake was continuously stocked (i.e. new fish were continuously brought
from a hatchery and added to the lake). How would the model change? Use a diagram
and explain clearly what you expect would happen, emphasizing any abrupt transitions
(bifurcations) that might occur.
(b)Repeat part (b) but assume instead that the lake is used for fishing, and that fish are
continuously harvested at some constant rate (and no new fish are added). What happens
in this case?
16
Solution
(a)The equation would then be dy
= f (y) + s where s is a positive constant ”stocking rate”.
dt
Expect the transition shown in figure 5 .
increase s
Figure 5:
The fish population should always stabilize at the high level beyond some thus hold value
of the stocking rates.
(b) The model would change to dy
= f (y) − h where h is constant harvesting rate. We
dt
expect the transition shown in figure 6 .
increase in h
Figure 6:
The fish population may be driven to extinction.
8.22
= sin(y). By plotting dy/dt versus y, identify
(a) Consider the differential equation dy
dt
values of y which would not change (steady states), and regions for which y is increasing or
17
dy/dt = sin(y)
y
Figure 7: Plot for problem 8.22 (a)
decreasing. Use this diagram to summarize all the possible outcomes (behaviors of y(t)) for
various initial values of y.
(b) Now consider a related equation, dy
= sin(y)+A where A is some positive constant. How
dt
does the behavior that you described in (a) change as the constant A is increased gradually?
For what value(s) of A does the behavior change dramatically ? Such changes are called
bifurcations.
Solution
(a) The steady states are those values of y such that sin y = 0, thus y = nπ, where n is
any integer. See the plot in Figure 7. The regions where y(t) is increasing are given by
2nπ ≤ y ≤ (2n + 1)π, (i.e between an even and an odd multiple of π) and the regions where
y(t) is decreasing are given by (2n−1)π ≤ y ≤ 2nπ, where n is any integer (i.e between an odd
and an even multiple of π). If y(0) = nπ then y(t) = nπ for all t, if 2nπ < y(0) ≤ (2n + 1)π
then y(t) → (2n + 1)π as t → ∞, and if (2n − 1)π ≤ y(0) < 2nπ then y(t) → (2n − 1)π.
dt
= sin y + A are those values of y, if any, such that
dt
sin y + A = 0. If |A| < 1 then there are infinitely many such y values and the behavior of
the solutions is as in part (a). See Figure 8. If A > 1 (resp. A < 1) the behavior is quite
different, in that all solutions tend to +∞ (resp. −∞). If A = 1 (resp. A = −1) then steady
states occur at intervals of 2π, and all solutions tend to increase (resp. decrease) to the next
steady state. Thus, bifurcation occurs for A = ±1. (But, note that the question only asked
for positive A values.)
(b) The steady state solutions of
18
dy/dt = sin(y) + A
y
typical picutre if −1 < A < 1
dy/dt = sin(y) + A
y
A=1
A = −1 similar
dy/dt = sin(y) + A
A > 1 (A < −1 similar)
Stable steady state: NONE
Figure 8: Plots for problem 8.22 (b)
19
y
Orthogonal Trajectories
8.23
Find the curve passing through the point (1,2) and orthogonal to the family of curves x2 y = k.
Solution
We find the slope of the curve x2 y = k at the point (x,y) by implicit differentiation: 2xy +
dy
dy
= 0 or dx
= − 2y
. We are interested in moving in a perpendicular direction; that is, in
x2 dx
x
the direction such that the slope is the negative reciprocal of − 2y
, i.e.:
x
x
dy
=
dx
2y
This is a separable equation and we solve it in the usual way:
Z
2y
dy
=
dx
Z
2ydy =
Z
xdx
This means that
x2
+ C.
2
The constant C can be determined by requiring that this curve pass through (1,2). This
means that the curve is
x2 7
2
+ .
y =
2
2
y2 =
8.24
Sketch the family of curves
x2
+ y 2 = k2
9
for k constant. Find the orthogonal trajectories by setting up and solving the appropriate
differential equation. Sketch these on the same picture.
Solution
2
See Figure 9 for the sketch. The tangent line at a point (x, y), x 6= 0, on the curve x9 + y 2 =
x
k 2 has slope − 9y
, and therefore the differential equation for the orthogonal trajectories is
R dy
R dx
9y
dy
=
.
Assume
y
=
6
0.
Then,
separating
variables
leads
to
=
9
, that is y = Kx9 ,
dx
x
y
x
where K is a constant. Thus, the orthogonal trajectories are x = 0 and the power curves
y = Kx9 , where K is a constant. Note: K = 0 gives y = 0.
20
k
(1/9) x^2 + y^2 = k^2
−3k
3k
−k
Figure 9: Plot for problem 8.24
8.25
Find the family of curves that are orthogonal to the parabola
y = Cx2
Important remark about a possible pitfall: C is a constant only for a given parabola, not for
the whole family. For this reason, we need to eliminate it from the general expression for the
slope of the parabola (dy/dx = 2Cx). This can be done by plugging in C = y/x2 in place
of the constant C, (i.e getting dy/dx = 2y/x, a relationship that holds for all the parabola).
Proceed from this point to find the orthogonal curves.
Solution
dy/dx = 2Cx = 2y/x. Thus, Rthe slope of Rthe orthogonal trajectories is dy/dx = −x/(2y).
Solve this separable equation: 2y dy = − x dx we obtain: y 2 = −x2 /2+C, i.e. x2 /2+y 2 =
C. As the constant, C, takes on values, this generates a family of ellipses.
8.26
Sketch the family of curves
xy = k
21
for k constant. Find the orthogonal trajectories by setting up and solving the appropriate
differential equation. Sketch these on the same picture.
Solution
xy = C
dy
dy
: x dx
+y =0⇒
Find slope dx
Orthogonal trajectories:
dy
dx
= − xy .
dy
x
=
⇒ y dy = x dx
dx
y
⇒
Z
y dy =
Z
x dx
y2
x2
=
+ const
2
2
⇒ x2 − y 2 = K
⇒
Applications to Biology
8.27
According to Klaassen and Lindstrom (1996) J theor Biol 183:29-34, the “fuel load” (nectar)
carried by a hummingbird, F (t) depends on the rate of intake (from flowers) and the rate of
consumption due to metabolism. They assume that intake takes place at a constant rate α.
They also assume that consumption increases when the bird is heavier (carrying more fuel).
Suppose that fuel is consumed at a rate proportional to the amount of fuel being carried
(with proportionality constant β).
(a) What features are being neglected or simplified in this model?
(b) Write down the differential equation model for F (t).
(c) Find F (t) as a function of time t.
(d) Klaassen and Lindstrom determined that for a hummingbird, α = 0.48 gm fuel /day and
β = 0.09 /day. Determine the steady state level of fuel carried by the bird.
Solution
(a) This model is simplified because it assumes that the rate of intake and the rate of
consumption are constant. It also does not take into account the fact that energy must be
expended at each flower for the hummingbird to get any fuel.
(b)
dF/dt = α − βF
22
dF
= dt, and hence −1
ln |α − βF | = t + C, which, as
(c) Separation of variables yields α−βF
β
−βt
−βt
in (9b), yields α − βF (t) = ce , hence F (t) = (α − ce )/β.
R
R
(d) The equilibrium level is reached when dF/dt = 0. This happens when α − βF = 0, i.e.
when F = α/β = 0.48/0.09 ≈ 5.33. This is the steady state level of fuel carried (i.e., when
the hummingbird carries that amount of fuel, then intake and consumption balance each
other).
8.28
A biochemical reaction S → P is catalysed by an enzyme E. The speed of the reaction
depends on the concentration of the substrate S. It is found that the substrate concentration,
x changes at the rate
dx
Kx
= rx −
dt
k+x
where r, K, k are positive constants. (The second term is often called “Michaelis Menten
kinetics” in chemistry.) How many steady state concentration values are there? What are
these values? Explain the behaviour of the solutions x(t) for very large and for very small
values of r. You may use sketches of the flow along the x axis.
Solution
Kx
To find the steady states we set dx
= 0 ⇒ rx − k+x
= 0 ⇒ x(rk+rx−K)
= 0. There are two
dt
k+x
K
K
s.s.: x = 0, r − k. For very small r, r − k is likely the positive s.s. that is unstable (i.e.
k < Kr ), x(t) decreases and approaches x = 0 as t increases if x(0) < Kr − k. For very large
r, Kr − k is likely the negative s.s. (i.e. k > Kr ) which has no meaning for the biochemical
reaction. In this case, x = 0 can become unstable. In case of a large r and very small k,
K
− k can remain positive and unstable. See Figure 10.
r
8.29
Harvesting 1
Consider the Logistic equation with harvesting,
dN
N
= rN 1 −
dt
K
−H
(a) Explain the distinction between this term and another possible term that might have
been used, −µy to represent a mortality or removal rate.
(b) Show that by defining a new variable, y = N/K, the equation in part can get rewritten
in the form:
dy
= ry(1 − y) − h
dt
23
Small r, K/r−k positive and unstable
Large r, K/r−k negative
dx/dt
dx/dt
0
K/r−k
x
K/r−k
0
x
Large r, very small k, K/r−k positive and unstable
dx/dt
0
K/r−k
x
Figure 10: Plots for problem 8.28
What is the new constant h?
(c) Find the steady states of this equation (all values of y for which dy/dt = 0) and describe
the stability properties of each one. There may be several cases to consider, and you are
asked to list the possible outcomes and how they depend on parameters.
(d) Use a sketch of dy/dt versus y to explain what happens to the population for very low,
intermediate, and high starting population sizes.
(e) What harvesting level is reasonable in this model? What happens if the harvesting is
increased to “unreasonable” levels? At what value of h does this transition take place?
Solution
Logistic equation with harvesting:
dN
N
= rN 1 −
dt
K
−H
where K is the carrying capacity and N is the population size.
24
(a) The harvesting term H is a constant removal rate, i.e. it does not depend on the population size. A term like −µy, where y = N/K, does depend on the population (linearly).
N
N
H
N
= K1 dN
= rK
= ry(1 − y) − h, where h = K
.
then dy
1− K
−H
(b) If y =
dt
dt
K
K
(c) The steady states are those values of y such that ry(1 − y) − h = 0. We rearrange
the equation such that
we
quadratic formula: ry 2 − ry + h = 0, and get y =
q can use the


√
1 ± 1 − 4h/r
r ± r 2 − 4rh
. Three cases now arise: 4h/r > 1, 4h/r = 1 or
= 
2r
2
dy
is always negative.
4h/r < 1. In the first case there are no steady state solutions and
dt
(The level of harvesting is so great the population dies out.) In the second case there is only
one steady state solution, namely y = 1/2, q
and it is unstable. For the q
last case there are




1 − 1 − 4h/r
1 + 1 − 4h/r
 and y = 
. The
2 steady state solutions, namely y = 
2
2
smaller one is unstable and the larger is stable.
(d) y = p+ is stable while y = p− is unstable. For y(0) < p− , the population will become
extinct. For any y(0) > p− , the population will settle to the steady state size y = p+ . See
Figure 11.
dy/dt
p−
p+
y
Figure 11: Plot for problem 8.29
(e) A reasonable harvesting level is one that satisfies 4h
< 1, or h < r4 . If harvesting increases
r
r
to unreasonable levels (h > 4 ) then the population dies off. The transition takes place at
h = 4r . This corresponds to H = rK
.
4
8.30
Harvesting 2
(a) The Logistic Equation with harvesting discussed in a problem above has some unrealistic
behaviour. Explain why its behaviour for y = 0 is not biologically realistic.
(b) As a “fix” for this difficulty, we might consider a “corrected” equation of the form
ay
dy
= ry(1 − y) − h
dt
a+y
25
ay
. Explain what feature of this new equation makes it
where now the harvesting rate is h a+y
more convincing. Explain this new dependence of harvesting on the population level. (i.e.
what is being assumed about harvesting when the population is very low? when it is very
high?) It will be helpful to sketch how the harvesting depends on the population level.
(c) Determine the steady states of this new model equation by finding values of y for which
dy/dt = 0. Describe the stability properties of each one. There may be several cases to
consider, and you are asked to list the possible outcomes and how they depend on the
parameters.
(d) It is slightly awkward to try to graph dy/dt versus y since the expression on the RHS
(right hand side) of the equation is now a bit nasty. However, it is easy to plot each of the
terms in the equation, i.e. f1 (y) = ry(1 − y), f2 (y) = hay/(a + y) on the same graph and
ask where one term is larger than the other. (This would depend on the values of r and
h. You might want to pick values for these for your plot. Then ask yourself how your plot
changes if these values were larger or smaller.) By so doing, you can identify both steady
states (intersections) and regions where dy/dt is positive or negative. Use this idea to find a
qualitative description of the behaviour of y in this modified model.
Solution
Fixing the harvesting term:
dy
ay
= ry(1 − y) − h
dt
a+y
(a) In problem 8.29, the harvesting term was H, a constant. For y = 0 in this case, dy
=
dt
H
−h(= − K ).This says that the population will continue to decrease even when it is extinct.
ay
(b) The new harvesting term is h a+y
. It depends on the population size and unlike in
dy
problem 8.29, when y = 0, dt = 0 with this new harvesting term.
Sketch of harvesting versus y (in Figure 12):
ay
Low population: y small ⇒ h a+y
≈ hy (almost linear).
ay
High population: y → ∞ ⇒ h a+y ≈ ha (approaches some limiting value). Even if there is
a higher and higher population, you can only harvest it so fast!
(c)
dy
dt
= 0 ⇒ ry(1 − y) =
hay
a+y
⇒ y = 0, and r(1 − y) =
ha
.
a+y
⇒ r(1 − y)(a + y) = ha ⇒ ry 2 + (ra − r)y + ha − ra = 0.
y=
r − ra ±
q
(ra − r)2 − 4r(ha − ra)
2r
There are three steady-state values, at
y = 0,
r − ra −
q
(ra − r)2 − 4r(ha − ra) r − ra +
,
2r
26
q
(ra − r)2 − 4r(ha − ra)
2r
.
ay
h a+y
y=ha
y
Figure 12: Plot of modified harvesting term for problem 8.30
ay
h a+y
P
Q
y
ry(1−y)
Figure 13: Plot for problem 8.30 (d)
2
There are now three cases: (1)
q (ra−r) −4r(ha−ra) > 0. In this case,
q there are 3 steady state
2
r − ra − (ra − r) − 4r(ha − ra) r − ra + (ra − r)2 − 4r(ha − ra)
values at y = 0,
,
.
2r
2r
q
r − ra + (ra − r)2 − 4r(ha − ra)
The middle one is unstable, while y = 0,
are both stable.
2r
(2) (ra − r)2 − 4r(ha − ra) = 0. In this case there are 2 steady states at y = 0 (unstable)
1−a
r − ra
=
(stable). (3) (ra − r)2 − 4r(ha − ra) < 0. In this case there is only
and y =
2r
2
one steady state at y = 0, which is stable.
(d) See Figure 13.
ay
Intersections: ry(1 − y) = h a+y
⇒ y = 0, and y = P, y = Q.
2
Ay + By + C = 0 ⇒ y =
r − ra ±
q
(ra − r)2 − 4r(ha − ra)
2r
27
(= P, Q) .
dy
dt
P
Q
y
Figure 14: Plot for problem 8.30 (d): Case (i)
ay
h a+y
ry(1−y)
P=Q
1
dy
dt
y
P=Q
y
Figure 15: Plot for problem 8.30 (d): Case (ii)
Cases: for various combinations of r, a, and h, we may have
(i) both intersections at y = P , y = Q (distinct roots)
(ii) one intersection at y = P = Q
(iii) neither intersection (no real roots)
Note: y = 0 is always an intersection point (i.e. extinction is always a steady state here).
Case (i): See Figure 14. It occurs when (ra − r)2 − 4r(ha − ra) > 0 [B 2 − 4AC > 0].
dy
ay
= ry(1 − y) − h
dt
a+y
If y0 < P , then population goes extinct.
If y0 > P , then population approaches equilibrium size at y = Q.
Case (ii): See Figure 15. It occurs when (ra − r)2 − 4r(ha − ra) = 0 [B 2 − 4AC = 0].
Population approaches equilibrium size at y = P (= Q).
28
dy
dt
ay
h a+y
ry(1−y)
y
1
y
Figure 16: Plot for problem 8.30 (d): Case (iii)
Case (iii): See Figure 16. It occurs when (ra − r)2 − 4r(ha − ra) < 0 [B 2 − 4AC < 0].
Population goes extinct!
8.31
Mouse Population Type 1
A population of mice reproduces at a per capita rate r and is preyed upon by a cat. The
predation rate is proportional to the density of the mice. Explain the following simple model
for the population of mice:
dy
= ry − py
dt
Determine what happens to the mice. Show that the outcome depends on the relative values
of the parameters r and p.
Solution
This model assumes that the population size of the mice increases at a rate r, and decreases
only due to the predation rate p of the cat. Both rates are assumed proportional to the
vs. y, we see that there are 3 cases: (i) if
current population size. From the graphs of dy
dt
r > p, the population size will increase exponentially (ii) if r = p, the population size will
remain constant and (iii) if r < p, the population size will decrease to extinction.
8.32
Mouse population Type 2
Consider a mouse and cat model in which the cat (predator) has a so-called “Type II”
response, i.e. in which the differential equation is
y
dy
= ry − Pmax
.
dt
a+y
29
Explain the differences in this version of the model. What is being assumed about the
predator? One steady state of this equation is y=0. Find the other steady state. (Does the
other steady state always exist? What has to be true for this steady state to be biologically
meaningful?) Use the Linear Stability Condition to show that this second steady state is
unstable whenever it exists. Interpret your findings.
Solution
y
The predation term Pmax a+y
in this problem is similar to the modified harvesting term in
problem 8.30 (see Figure 12). When the population of mice is small, the cat’s predation rate
increases nearly linearly with population density (i.e. it is good at catching a small number
of mice). However, as the mice population grows and approaches infinity, the predation rate
levels off - the cat cannot keep catching more and more mice; there is a limit to its predation.
The second steady state can be found by setting f (y) = dy
= 0:
dt
ry(a + y) − Pmax y
= 0.
a+y
That is, ar + ry − Pmax = 0 and y =
meaningful, Pmax > ar.
To find stability we compute f 0(y):
f 0(y) = r − Pmax
a+y−y
(a + y)2
!
Pmax − ar
. For this steady state to be biologically
r
=r−
aPmax
(a + y)2
Linear Stability Condition:
f0
Pmax − ar
r
= r − Pmax
ar 2
ar
ar 2
=r 1−
=
r
−
2
(ar + Pmax − ar)
Pmax
Pmax
Since Pmax > ar for this steady state to exist, 1 −
ar
> 0. Therefore f 0 Pmaxr −ar is
Pmax
always positive, and this steady state is unstable.
This means that whenever the mice population size is less than Pmaxr −ar , the population will
decrease to extinction, while if it is larger than Pmaxr −ar , the population will continue growing
in size.
If the second steady state does not exist (i.e. if Pmax < ar) then y = 0 is the only steady
state, and it is unstable. The mice population will continue growing, no matter what its
original size.
8.33
Mouse population Type 3
Consider a population of rodents (density y(t)) that grow at an exponential rate (dy/dt = ry),
and suppose that a cat is brought in to control the population. The cat is a predator with
a “Holling Type III” response, i.e. it is not very good at catching a small number of mice,
30
it gets better at finding prey when there are quite a few of them, but if there are too many,
it can’t keep catching more and more of them. A model for the rodent population with
predation of this type might be:
y2
dy
= ry − P
dt
1 + y2
Explain this model. (You may want to use graphical arguments). Use qualitative methods to
determine the predictions for the outcome of predation on the prey. Pay particular attention
to the way that these predictions depend on the parameter r.
Solution
P
There are 3 steady states: y = 0, 2r
±
q
P 2
( 2r
) − 1 if
P
2r
> 1. In this case only the middle one:
P
−
− 1 is stable and is less than 1. When 2r
= 1, y = 1 is the steady state
y =
P
< 1,
resulting from the collapsing of the two nonzero steady states. It is unstable. When 2r
y = 0 becomes the only steady state which is unstable. y goes to infinity as time evolves.
See Figure 17.
P
2r
P 2
( 2r
)
q
P/2r>1
P/2r=1
dy/dt
dy/dt
y
0
y
P/2r<1
dy/dt
y
Figure 17: Plots for problem 8.33
31
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