ODEs: an IVP example
. . . last time: we found general solutions and solutions of initial
value problems for some ODEs, especially separable ones.
Example: An object, initially at rest, is released in a constant
gravitational field, and subject to a drag force proportional to its
speed. Write an ODE IVP to model this. Then solve it.
v (t) = velocity (downward) as a function of time:
m
•
R
dv
= mg − kv ,
dt
v (0) = 0
(look familiar? – same as DDT in fish problem!)
R
dv
k
=
dt =⇒ − m
k
k ln(g − m v ) = t + C (why no | · |?)
g− v
m
k
k
• v (0) = 0 =⇒ − m
k ln(g ) = C =⇒ ln(g − m v ) = − m t + ln(g )
• =⇒ g −
k
mv
k
= e − m t g =⇒ v (t) =
• as t → ∞, v (t) →
mg
k
mg
k (1
k
− e− m t )
(terminal velocity)
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ODEs: a BVP example
Example of a boundary value problem: the wavefunction ψ(x)
for a particle of mass m at energy E inside a (1D) box 0 < x < L
solves the (stationary)√ Schrödinger equation:
d 2ψ
= −k 2 ψ, k = 2mE
~ . If the walls are infinitely high,
dx 2
formulate a boundary value problem, them solve it.
d 2ψ
= −k 2 ψ,
ψ(0) = 0, ψ(L) = 0.
dx 2
• general solution (by method of ‘clever guessing’):
ψ(x) = A sin(kx) + B cos(kx)
• 0 = ψ(0) = A·0+B ·1 = B =⇒ B = 0 =⇒ ψ(x) = A sin(kx)
• 0 = ψ(L) = A sin(kL) =⇒ either A = 0 (ψ = 0) or sin(kL) = 0
• sin(kL) = 0 =⇒ kL = nπ for some n = 0, 1, 2, 3, . . .
2k 2
~2 π 2 2
=⇒ E = ~2m
= 2mL
2 n (quantized energy levels!)
• the wavefunction for the n-th energy level is
ψn (x) = An sin( nπ
L x)
RL
(we can choose An to normalize 0 |ψn (x)|2 dx = 1)
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ODEs: Population Growth
Let P(t) be the population of some species as a function of time,
with initial population P(0) = P0 .
Recall: if the growth rate is proportional to the population,
dP
kt
dt = kP, we get exponential growth: P(t) = P0 e .
There may be limits to growth (eg. scarce resources) so that when
P is small it grows exponentially, but as P increases, its growth
rate slows down, vanishing when P reaches a “carrying capacity”
value K . The simplest such model is the logistic equation:
dP
= kP(1 − P/K ),
dt
P(0) = P0 .
This ODE is autonomous: the independent variable t is absent.
Exercise (blackboard): for the logistic equation:
• find the constant solutions (equilibria or steady-states)
• sketch some solutions by noting which P give dP
dt > 0 and < 0
• predict the fate limt→∞ P(t), depending on the value P0
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ODEs: Population Growth
Exercise (blackboard): perform the same “qualitative analysis” for
the first-order, autonomous ODE problem
dP
= kP(1 − P/K )(1 − m/P), P(0) = P0 ,
0 < m < K.
dt
What new effect (beyond the logistic equation) does this model
incorporate?
Challenge: solve the logistic equation IVP
dP
= kP(1 − P/K ), P(0) = P0 .
dt
R
R
K
1
kt dt = kt + C , P(KK−P) = P1 + K −P
(partial fractions)
P(K −P) dP =
R
P K
=⇒ kt + C = P(1−P/K
dP
=
ln
|P|
−
ln
|K
−
P|
=
ln
K −P )
0 P P0 P0 = P(0) =⇒ C = ln K P−P
=⇒
ln
=
kt
+ln
K −P
K −P0 0
0 kt
0
=⇒ K P−P = K P−P
e kt
e =⇒ K P−P = K P−P
0
0
=⇒ P(t) =
KP0
P0 +(K −P0 )e −kt
.
4/1