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Math 308 - Differential Equations Fall 2003
Homework 1 Solutions
- The difference equation is
xn+1 = pxn + q, n ≥ 1 , and x 0 given (1)
where |p| < 1.
(a) We must show that xn = pn
x 0 − (^1) −qp
- (^1) −qp solves the difference equation (1) for any value of x 0. We have
pxn + q = p
pn
x 0 − q 1 − p
q 1 − p
= pn+
x 0 − q 1 − p
pq 1 − p
= pn+
x 0 − q 1 − p
pq + (1 − p)q 1 − p = pn+
x 0 − q 1 − p
q 1 − p = xn+1, so the formula does give the solution. (b) Because |p| < 1, (^) nlim→∞ |p|n^ = 0. Thus (^) nlim→∞ xn = q 1 − p
- The amount of the drug in the bloodstream, a(t), is determined by da dt = −ka,
where k > 0 is the proportionality constant. (a) The general solution to the differential equation is a(t) = Ce−kt, and to satisfy a(0) = 80, we must have C = 80, so a(t) = 80e−kt. We are given a(24) = 12.50, so 12.50 = 80e−^24 k. Solving for k gives k = 0.07735. (b) At the moment after the second dose is given, the amount of the drug in the bloodstream is 92. 50 (12.50 remains from the first dose, and 80.00 is added by the second dose). After the second dose, the drug continues to obey the same differential equation, and therefore the amount of the drug at the end of the second 24 hour period is 92. 50 e−^24 k^ = 14.45. (c) We have
a 2 = (a 1 + 80)e−^24 k^ = 14. 45 , a 3 = (a 2 + 80)e−^24 k^ = 14. 76 , a 4 = (a 3 + 80)e−^24 k^ = 14. 81 , etc. and in general, we have
an+1 = (an + 80)e−^24 k^ = e−^24 kan + 80e−^24 k.
This is a difference equation of the type discussed in problem 1, with p = e−^24 k^ = 0.1563, and q = 12.50. Given the meaning of an, we see that a 0 = 0. Therefore the solution is
an = pn
−q 1 − p
1 − e−^24 k^
1 − e−^24 k
1 − e−^24 k
1 − e−^24 kn
= (14.82) (1 − (0.1563)n)
(d) As n → ∞, an → q 1 − p
= 12.^50
1 − e−^24 k^
- The differential equation for ω(t) is dω dt = − 2 ω + v(t). (2)
(a) When v(t) = v 1 , the differential equation is dω dt = − 2 ω + v 1.
This equation has a single asymptotically stable equilibrium solution ω(t) = v 1 /2. Thus, after a long time, the display of the anemometer will be W (t) = cω(t) = cv 1 /2. The actual wind speed is v 1 , so we must choose c = 2 in order to have W (t) = v(t). (b) Under these conditions, ω(t) is the solution to the initial value problem dω dt = − 2 ω + 3 (t > 0), and ω(0) = 0.
i. The solution to this initial value problem is
ω(t) = −
e−^2 t^ +
and the error is v(t) − W (t) = 3 − 2 ω(t) = 3e−^2 t. ii. We see that as t increases, the error decays to zero exponentially. Initially, the display of the wind speed will be inaccurate; it takes some time for the anemometer to “spin up” towards the steady state. iii. The steady state value of W (t) is 3, and 90% of this 2.7. Let T be the time at which W (t) reaches 2.7; that is, W (T ) = 2ω(T ) = − 3 e−^2 T^ + 3 = 2.7. Solving for T gives
T = ln(10)/2 = 1. 151.
Thus it will take 1.151 seconds for the display to be within 90% of the correct value. iv. The following plot shows W (t) and v(t). The point where W (t) = 2.7 is also indicated.
ii. The error is proportional to m, so the faster the wind speed increases, the greater the error will be. As t increases, the exponential term in the error will approach zero, but the term m/2 remains. Thus, in a steadily increasing wind, the error does not go to zero, and instead it approaches the steady-state value of m/2. Roughly speaking, the anemometer can not keep up with the changing wind speed. iii. Actual and Displayed Wind Speed (Problem 3(c)iv)
v(t) W(t)
v(t)-W(t)
0
1
2
Wind Speed
1 2 3 4 5 t (d) In this case, the differential equation for ω(t) is
dω dt = − 2 ω + 1 − cos(kt),
i. We now have a linear equation with p(t) = 2 and g(t) = 1 − cos(kt). The integrating factor is still μ(t) = e^2 t, and
ω(t) =
μ(t)
μ(t)g(t) dt + C
= e−^2 t
e^2 t(1 − cos(kt) dt + C
= e−^2 t
e^2 t 2
2 cos(kt) + k sin(kt) 4 + k^2 e^2 t^ + C
=^1
− 2 cos(kt) +^ k^ sin(kt) 4 + k^2
The error is v(t) − W (t) = 1 − cos(kt) − 2 ω(t)
= − cos(kt) + 2 2 cos(kt) + k sin(kt) 4 + k^2 − 2 Ce−^2 t
=^2 k^ sin(kt)^ −^ k
(^2) cos(kt) 4 + k^2 − 2 Ce−^2 t
ii. As t increases, the exponential term in the error will go to zero, but the term involving sin(kt) and cos(kt) will not. The error will continue to oscillate, with an amplitude that depends on k. If k is “large” (e.g. much larger than 2), then the dominant term in the error is −k (^2) cos(kt) 4+k^2 ≈ −^ cos(kt), so the long term error is an oscillation with an amplitude of
approximately 1. If k is “small” (much less than 2), then the dominant term in the error is 2 k sin(kt) 4+k^2 ≈^
k sin(kt) 2 , so the long term error is an oscillation with an amplitude of approximately k/2. iii. When k = 1/3 we have
v(t) = 1 − cos(t/3) ω(t) =
cos(t/3) −
sin(t/3) + Ce−^2 t
|v(t) − W (t)| =
cos(t/3) +
sin(t/3) − 2 Ce−^2 t
It wasn’t stated in the problem, but in this part and the next, I will take the initial condition to be ω(0) = 0. To satisfy this initial condition, choose C = − 1 /74. The following plot shows the graphs. Note that the amplitude of the oscillation of the error is 0.16. Actual and Displayed Wind Speed (Problem 3(d)iii)
v(t)
W(t)
|v(t)-W(t)|
0
1
2
Wind Speed
5 10 15 20 25 30 t iv. When k = 2 we have v(t) = 1 − cos(2t) ω(t) =^1 2
cos(2t) − 1 4 sin(2t) + Ce−^2 t
|v(t) − W (t)| =
cos(2t) +
sin(2t) − 2 Ce−^2 ∗t
I’ll choose C = − 1 /4 so that ω(0) = 0. The following plot shows the graphs. Note that the amplitude of the oscillation of the error is 0.70, over four times the amplitude of the case where k = 1/3.
Changing λ or s will change the scale (and the numerical value of the non-zero equilibrium), but the graph of dk/dt versus k will always have the shape as the graph shown above. We see that the equilibrium k = 0 is unstable, and if k is small, k will increase rapidly. The graph of k(t) will have an inflection point when k reaches
( (^) s 3 λ
(where dk/dt has its maximum). k will then converge asymptotically to the non-zero equilibrium. The equilibrium k = (s/λ)^3 /^2 is asymptotically stable. In fact, all solutions with k(0) > 0 will converge asymptotically to this equilibrium. The following shows a direction field and some solutions when s = 1 and λ = 1. Direction Field with Some Solutions
0
1
k(t)
2 4 6 8 10 t Since k(t) = K(t)/L(t), and L(t) = L 0 eλt, if k(t) converges to an asymptotically stable equilibrium k 1 , then K(t) must behave asymptotically like k 1 L(t). This means that, in the long term, K(t) must grow exponentially, with the same exponent as L(t). This model predicts that in the long term, production will grow exponentially along with the labor. If, for example, production is too low, it will rapidly increase until it becomes proportional to the labor, and then it will settle into a long term behavior in which production remains proportional to the labor. (c) As shown above, there will always be two equilibria when s > 0 and λ > 0, and they will always have the stability properties described above. Changing s or λ changes the value of the non- zero equilibrium point, and also changes the rates at which solutions will grow or decay, but the qualitative picture does not change when s or λ are changed.
(d) A more general version of the Cobb-Douglas function that has constant returns to scale is f (L, K) = LαK^1 −α, where 0 < α < 1. This production function would result in the same qualitative analysis as the one given above. It results in g(k) = f (k, 1) = kα, and the differential equation for k is dk dt = −λk + skα. This equation has an unstable equilibrium at k = 0, and an asymptotically stable equilibrium at k =
( (^) s λ
) (^1) −^1 α .
- (a) Let v(t) be the volume of the balloon (in cm^3 ) at time t (in seconds). Then the differential equation for v is simply dv dt = 12. (b) Let r(t) be the radius (in cm) of the balloon at time t. Since v = 4 πr 3 3 , we have
dv dt = 4πr
2 dr dt , and from (a) we know dvdt = 12. We have dr dt
4 πr^2
dv dt
4 πr^2
πr^2
so the differential equation for r is dr dt
πr^2
(c) The differential equation in (a) looks a little easier than (b). We find
v(t) = 12t + C 1
where C 1 is an arbitrary constant (which, in this case, gives the initial volume of the balloon). Then, since r =
( (^3) v 4 π
, we have
r(t) =
36 t + 3C 1 4 π
9 t π
+ C 2
where C 2 = 34 Cπ^1.
