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8 Solving Ordinary Differential Equations
where k is some arbitrary integer. A computer program can easily compute N((k +
1)Δt) for us with the aid of a little loop.
The initial condition
Observe that the computational formula cannot be started unless we have an
initial condition!
The solution of N = rN is N = Ce rt for any constant C, and the initial
condition is needed to fix C so the solution becomes unique. However, from
a mathematical point of view, knowing N(t) at any point t is sufficient as
initial condition. Numerically, we more literally need an initial condition: we
need to know a starting value at the left end of the interval in order to get the
computational formula going.
In fact, we do not really need a computer in this particular case, since we see
a repetitive pattern when doing hand calculations. This leads us to a mathematical
formula for N((k + 1)Δt):
N((k + 1)Δt) = N(kΔt) + Δt rN(kΔt) = N(kΔt)(1 + Δt r)
= N((k − 1)Δt)(1 + Δt r)
2
. . .
= N 0 (1 + Δt r)
k+1 .
Rather than using (8.2) as a computational model directly, there is a strong
tradition for deriving a differential equation from this difference equation. The idea
is to consider a very small time interval Δt and look at the instantaneous growth
as this time interval is shrunk to an infinitesimally small size. In mathematical
terms, it means that we let Δt → 0. As (8.2) stands, letting Δt → 0 will
just produce an equation 0 = 0, so we have to divide by Δt and then take the
limit:
lim
Δt →0
N(t + Δt) − N(t)
Δt
= rN(t) .
The term on the left-hand side is actually the definition of the derivative N (t), so
we have
N
(t) = rN(t),
which is the corresponding differential equation.
There is nothing in our derivation that forces the parameter r to be constant—it
can change with time due to, e.g., seasonal changes or more permanent environmental changes.
8 Solving Ordinary Differential Equations
where k is some arbitrary integer. A computer program can easily compute N((k +
1)Δt) for us with the aid of a little loop.
The initial condition
Observe that the computational formula cannot be started unless we have an
initial condition!
The solution of N = rN is N = Ce rt for any constant C, and the initial
condition is needed to fix C so the solution becomes unique. However, from
a mathematical point of view, knowing N(t) at any point t is sufficient as
initial condition. Numerically, we more literally need an initial condition: we
need to know a starting value at the left end of the interval in order to get the
computational formula going.
In fact, we do not really need a computer in this particular case, since we see
a repetitive pattern when doing hand calculations. This leads us to a mathematical
formula for N((k + 1)Δt):
N((k + 1)Δt) = N(kΔt) + Δt rN(kΔt) = N(kΔt)(1 + Δt r)
= N((k − 1)Δt)(1 + Δt r)
2
. . .
= N 0 (1 + Δt r)
k+1 .
Rather than using (8.2) as a computational model directly, there is a strong
tradition for deriving a differential equation from this difference equation. The idea
is to consider a very small time interval Δt and look at the instantaneous growth
as this time interval is shrunk to an infinitesimally small size. In mathematical
terms, it means that we let Δt → 0. As (8.2) stands, letting Δt → 0 will
just produce an equation 0 = 0, so we have to divide by Δt and then take the
limit:
lim
Δt →0
N(t + Δt) − N(t)
Δt
= rN(t) .
The term on the left-hand side is actually the definition of the derivative N (t), so
we have
N
(t) = rN(t),
which is the corresponding differential equation.
There is nothing in our derivation that forces the parameter r to be constant—it
can change with time due to, e.g., seasonal changes or more permanent environmental changes.