1.8: Solution to the dynamic equation Pₜ₊₁-Pₜ=rPₜ + b
- Page ID
- 38397
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The general form of the dynamic equation of the previous section is,
\[P_{t+1}-P_{t}=r P_{t}+b, \quad t=0,1, \cdots, \quad P_{0} \quad \text { a known value } \label{1.24}\]
The constant b represents influence from outside the system, such as flow or immigration into (b > 0) or flow or emigration out of (b < 0) the system.
Suppose \(r \neq 0\) and \(b \neq 0\). We convert
\[P_{t+1}-P_{t}=r P_{t}+b\]
to iteration form
\[P_{t+1}=R P_{t}+b, \quad \text { where } \quad R=r+1 \label{1.25}\]
Next we look for an equilibrium value E satisfying
\[E=R E+b \label{1.26}\]
(Think, if \(P_{t} = E\) then \(P_{t+1} = E\).) We can solve for E
\[E=\frac{b}{1-R}=-\frac{b}{r} \label{1.27}\]
Now subtract Equation \ref{1.26} from Equation \ref{1.25}.
\[\begin{aligned}
P_{t+1} &=R P_{t}+b \\
E &=R E+b
\end{aligned}\]
\[P_{t+1}-E=R\left(P_{t}-E\right) \label{1.28}\]
Let \(D_{t} = P_{t} − E\) and write
\[D_{t+1}=R D_{t}\]
This is iteration Equation 1.2.4 with DT instead of \(P_t\) and the solution is
\[D_{t}=R^{t} D_{0}\]
Remembering that \(D_{t} = P_{t} − E\) and \(E = \frac{b}{(1 − R)} = −\frac{b}{r}\) and \(R = 1 + r\) we write
\[P_{t}-E=R^{t}\left(P_{0}-E\right), \quad P_{t}=-\frac{b}{r}+\left(P_{0}+\frac{b}{r}\right)(1+r)^{t} \label{1.29}\]
You are asked to show in Exercise 1.8.2 that the solution to
\[P_{t+1}-P_{t}=b \quad \text { is } \quad P_{t}=P_{0}+t b \label{1.30}\]
Exercises for Section 1.8, Solution to the dynamic equation, \(P_{t+1}-P_{t}=r P_{t}+b\)
Exercise 1.8.1 Write a solution equation for the following initial conditions and difference equations or iteration equations. In each case, compute \(B_{100}\).
- \(B_{0}=100 \quad B_{t+1}-B_{t}=0.2 B_{t}+5\)
- \(B_{0}=138 \quad B_{t+1}-B_{t}=0.05 B_{t}+10\)
- \(B_{0}=138 \quad B_{t+1}-B_{t}=0.5 B_{t}-10\)
- \(B_{0}=100 \quad B_{t+1}-B_{t}=10\)
- \(B_{0}=100 \quad B_{t+1}=1.2 B_{t}-5\)
- \(B_{0}=100 \quad B_{t+1}-B_{t}=-0.1 B_{t}+10\)
- \(B_{0}=100 \quad B_{t+1}=0.9 B_{t}-10\)
- \(B_{0}=100 \quad B_{t+1}=-0.8 B_{t}+20\)
Exercise 1.8.2 Equation \ref{1.30}
\[P_{t+1}-P_{t}=b\]
represents a large number of equations
\[\begin{aligned}
P_{1}-P_{0} &=b \\
P_{2}-P_{1} &=b \\
P_{3}-P_{2} &=b \\
\vdots \\
P_{n-1}-P_{n-2} &=b \\
P_{n}-P_{n-1} &=b
\end{aligned}\]
Add these equations to obtain
\[P_{n}=P_{0}+n b\]
Substitute t for n to obtain
\[P_{t}=P_{0}+t b\]
Exercise 1.8.3 Suppose a quail population would grow at 20% per year without hunting pressure, and 1000 birds per year are harvested. Describe the progress of the population over 5 years if initially there are
- 5000 birds
- 6000 birds
- 4000 birds