## Population Differential Equations and Laplace Transform

Malthus Model
$\displaystyle \frac{dN}{dt}=BN-DN=kN$

$N$: Total population

$B$: Birth-rate per capita

$D$: Death-rate per capita

$k=B-D$

Solution to D.E.:
$\displaystyle \boxed{N(t)=\widehat{N}e^{kt}},$

where $\widehat{N}=N(0)$.

Logistic Equation
\begin{aligned} D&=sN\\ \frac{dN}{dt}&=BN-sN^2\\ \widehat{N}&=N(0)\\ N_\infty&=B/s \end{aligned}

Logistic Case 1: Increasing population ($\widehat{N})
\begin{aligned} N(t)&=\frac{B}{s+(\frac{B}{\widehat{N}}-s)e^{-Bt}}\\ &=\frac{N_\infty}{1+(\frac{N_\infty}{\widehat{N}}-1)e^{-Bt}} \end{aligned}

The second expression can be derived from the first: divide by $s$ in both the numerator and denominator.

Logistic Case 2: Decreasing population ($\widehat{N}>N_\infty$)
\begin{aligned} N(t)&=\frac{B}{s-(s-\frac{B}{\widehat{N}})e^{-Bt}}\\ &=\frac{N_\infty}{1-(1-\frac{N_\infty}{\widehat{N}})e^{-Bt}} \end{aligned}

Logistic Case 3: Constant population ($\widehat{N}=N_\infty$)
$\displaystyle N(t)=N_\infty$

Harvesting
Basic Harvesting Model: $\displaystyle \boxed{\frac{dN}{dt}=(B-sN)N-E}.$

$E$: Harvest rate (Amount harvested per unit time)

Maximum harvest rate without causing extinction: $\boxed{\dfrac{B^2}{4s}}$.

$\displaystyle \boxed{\beta_1,\beta_2=\frac{B\mp\sqrt{B^2-4Es}}{2s}}.$

$\beta_1$: Unstable equilibrium population

$\beta_2$: Stable equilibrium population

Extinction Time: $\displaystyle \boxed{T=\int_{\widehat{N}}^0\frac{dN}{N(B-sN)-E}}.$

Laplace transform of $f$
$\displaystyle F(s)=L(f)=\int_0^\infty e^{-st}f(t)\,dt$

Tip: Use this equation when the questions contains the words “show from the definition”.

Inverse transform of $F(s)$
$\displaystyle f(t)=L^{-1}(F(s))$

Linearity
\begin{aligned} L(af(t)+bg(t))&=aL(f)+bL(g)\\ L^{-1}(aF(s)+bG(s))&=aL^{-1}(F)+bL^{-1}(g) \end{aligned}

## List of common Laplace Transforms

\begin{aligned} L(e^{at})&=\frac{1}{s-a}\\ L(1)&=\frac{1}{s}\\ L(\cos wt)&=\frac{s}{s^2+w^2}\\ L(\sin wt)&=\frac{w}{s^2+w^2}\\ L(t^n)&=\frac{n!}{s^{n+1}}\\ L(f')&=sL(f)-f(0)\\ L(f'')&=s^2L(f)-sf(0)-f'(0)\\ L(f^{(n)})&=s^nL(f)-s^{n-1}f(0)\\ &\quad -s^{n-2}f'(0)-\dots-f^{(n-1)}(0)\\ L\left(\int_0^t f(\tau)\,d\tau\right)&=\frac{1}{s}L(f) \end{aligned}

$s$-shifting
If $L(f)=F(s)$, $s>a$, then $\displaystyle \boxed{L(e^{ct}f(t))=F(s-c)},$
$s-c>a$.

Tip: Use this when doing Laplace Transform of a function with an exponential factor $e^{ct}$. Note that the reverse direction can sometimes be used as well: $\displaystyle L^{-1}[F(s-c)]=e^{ct}f(t).$

$t$-shifting
If $L(f(t))=F(s)$, then $\displaystyle \boxed{L(f(t-a)u(t-a))=e^{-as}F(s)}.$

Tip: Frequently, we use the reverse direction $\displaystyle L^{-1}[e^{-as}F(s)]=f(t-a)u(t-a).$

Delta function
$\delta(t)$: infinitely tall and narrow spike at $t=0$.

$\delta(t-a)$: infinitely tall and narrow spike at $t=a$.

$\boxed{L[\delta(t-a)]=e^{-as}}$

Two properties of delta function
\begin{aligned} \int_0^\infty\delta(t-a)\,dt&=1\\ \int_0^\infty \delta(t-a)g(t)\,dt&=g(a) \end{aligned}
for $a\geq 0$.

Tip: Use delta function when the keywords “suddenly”, “burst”, etc. appear.

Unit step function
$\displaystyle u(t-a)=\begin{cases} 0, &ta. \end{cases}$

For $0, $\displaystyle u(t-a)-u(t-b)=\begin{cases} 0, &tb. \end{cases}$

Tip: Use unit step function for questions that require a force to “switch on / switch off” at certain times.

$\displaystyle \boxed{L(u(t-a))=\frac{e^{-as}}{s}}$

## Linear System of Differential Equations, Solutions, Phase Portrait Sketching

Solutions of Homogeneous Linear System of DE
$\displaystyle \mathbf{y}'=\mathbf{A}\mathbf{y}$
$\displaystyle \mathbf{y}(t)=\mathbf{v}e^{rt}$
where $r$ and $\mathbf{v}$ are eigenvalue and eigenvector for $\mathbf{A}$ respectively.

Superposition Principle
If $\mathbf{x_1}(t)$ and $\mathbf{x_2}(t)$ are two solutions to a homogeneous SDE $\mathbf{y'}=\mathbf{Ay}$, then $\displaystyle \mathbf{y}=c_1\mathbf{x_1}(t)+c_2\mathbf{x_2}(t)$ is also a solution for any scalars $c_1$, $c_2$.

Euler’s formula
$\displaystyle e^{i\theta}=\cos\theta+i\sin\theta$

General Solutions (Complex Eigenvalues)

1) Let $r_1=a+bi$ be an eigenvalue corresponding to eigenvector $\mathbf{v_1}$. (The eigenvectors are complex conjugates: $\mathbf{v_1,v_2}=\mathbf{p}\pm \mathbf{q} i$.)
2) Construct
$\displaystyle \mathbf{x}_\text{Re}(t)=e^{at}(\mathbf{p}\cos bt-\mathbf{q}\sin bt)$
$\displaystyle \mathbf{x}_\text{Im}(t)=e^{at}(\mathbf{p}\sin bt+\mathbf{q}\cos bt)$
3) The general solution is $\displaystyle \mathbf{y}=c_1\mathbf{x}_\text{Re}(t)+c_2\mathbf{x}_\text{Im}(t).$

## How to Sketch Phase Portrait

Probably the best video on how to sketch Phase Portrait:

## US “Common Core” Math gets a bit too convoluted

How would you solve 427-316?

Most people educated in the past would simply use the usual subtraction working to arrive at the answer of 111.

But lately, in the USA, the Common Core Math approach gets very complicated and teaches an approach that even the father who has a “Bachelor of Science Degree in Electronics Engineering, which included extensive study in differential equations and other higher math applications” cannot explain it, nor get the answer correct.