University of Florida/Egm6341/s10.team2.niki/HW7

problem 2: Solution of the Logistic equation

---

Statement

P.39-1

Solution

We have Verhulst model or the Logistic equation as P.38-3

     $\dot{x}=\frac{dx}{dt}=rx(1-\frac{x}{x_{max}})$

Separating variables and integrating we have

     $\underset{x_0}{\overset{x}{\int }}\frac{x_{max}}{x(x_{max}-x)}dx=\underset{t_0=0}{\overset{t}{\int }}rdt$

which can be written as

     $\underset{x_0}{\overset{x}{\int }}\frac{1}{x}dx+\underset{x_0}{\overset{x}{\int }}\frac{1}{(x_{max}-x)}dx=\underset{t_0=0}{\overset{t}{\int }}rdt$

We get

     $lo{g}_e(\frac{x}{x_0})-lo{g}_e(\frac{x_{max}-x}{x_{max}-x_0})=r(t-0)$

     $\frac{x(x_{max}-x_0)}{x_0(x_{max}-x)}=e^{rt}$

Rearranging we have

     $x=\frac{x_{max}{x}_0{e}^{rt}}{x_{max}+x_0(e^{rt}-1)}$

---

---

---

problem 14: Constants of the Cosine Series

---

Statement

We have the cosine series expressed as $f(cos\theta )=\frac{a_0}{2}+\sum _{k=1}^{\mathrm{\infty }}{a}_{k}cos(k\theta )$, we need to express teh constants $a_k$ as

$a_k=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(cos\theta )cos(k\theta )d\theta$

Solution

We have the given expression for the cosine series as

$f(cos\theta )=\frac{a_0}{2}+\sum _{k=1}^{\mathrm{\infty }}{a}_{k}cos(k\theta )$

multiplying both sides by $cos(m\theta )$ where k is not the same as m and integrating we get,

$\underset{0}{\overset{2\pi }{\int }}f(cos\theta )(cos(m\theta ))=\underset{0}{\overset{2\pi }{\int }}\frac{a_{0}cos(m\theta )}{2}+\underset{0}{\overset{2\pi }{\int }}\sum _{k=1}^{\mathrm{\infty }}{a}_{k}cos(k\theta )cos(m\theta )$

Using the property of orthogonality we know that $\underset{0}{\overset{2\pi }{\int }}{a}_{k}cos(k\theta )cos(m\theta )$ exists only when k = m i.e

$\underset{0}{\overset{2\pi }{\int }}f(cos\theta )(cos(k\theta ))d\theta =\frac{a_0}{2}\underset{0}{\overset{2\pi }{\int }}cos(m\theta )d\theta +a_k\underset{0}{\overset{2\pi }{\int }}co{s}^2(k\theta )d\theta$

wkt,

$\underset{0}{\overset{2\pi }{\int }}cos(m\theta )d\theta =0$ and

$\underset{0}{\overset{2\pi }{\int }}co{s}^2(k\theta )d\theta =\frac{2\pi -0}{2}=\pi$

Thus we have by substituting and rearranging terms,

$a_k=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(cos\theta )cos(k\theta )d\theta$

Author

--Egm6341.s10.team2.niki 14:43, 23 April 2010 (UTC)

---