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

problem 2: Solution of the Logistic equation

Solution

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

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


Separating variables and integrating we have

     ${\displaystyle \int _{x_{0}}^{x}{\frac {x_{max}}{x(x_{max}-x)}}dx=\int _{t_{0}=0}^{t}rdt}$


which can be written as

     ${\displaystyle \int _{x_{0}}^{x}{\frac {1}{x}}dx+\int _{x_{0}}^{x}{\frac {1}{(x_{max}-x)}}dx=\int _{t_{0}=0}^{t}rdt}$


We get

     ${\displaystyle log_{e}({\frac {x}{x_{0}}})-log_{e}({\frac {x_{max}-x}{x_{max}-x_{0}}})=r(t-0)}$


     ${\displaystyle {\frac {x(x_{max}-x_{0})}{x_{0}(x_{max}-x)}}=e^{rt}}$


Rearranging we have

     ${\displaystyle x={\frac {x_{max}x_{0}e^{rt}}{x_{max}+x_{0}(e^{rt}-1)}}}$


problem 14: Constants of the Cosine Series

Statement

We have the cosine series expressed as ${\displaystyle f(cos\theta )={\frac {a_{0}}{2}}+\sum _{k=1}^{\infty }a_{k}cos(k\theta )}$, we need to express teh constants ${\displaystyle a_{k}}$ as

${\displaystyle a_{k}={\frac {2}{\pi }}\int _{0}^{\pi }f(cos\theta )cos(k\theta )d\theta }$

Solution

We have the given expression for the cosine series as

${\displaystyle f(cos\theta )={\frac {a_{0}}{2}}+\sum _{k=1}^{\infty }a_{k}cos(k\theta )}$

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

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

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

${\displaystyle \int _{0}^{2\pi }f(cos\theta )(cos(k\theta ))d\theta ={\frac {a_{0}}{2}}\int _{0}^{2\pi }cos(m\theta )d\theta +a_{k}\int _{0}^{2\pi }cos^{2}(k\theta )d\theta }$

wkt,

${\displaystyle \int _{0}^{2\pi }cos(m\theta )d\theta =0}$ and

${\displaystyle \int _{0}^{2\pi }cos^{2}(k\theta )d\theta ={\frac {2\pi -0}{2}}=\pi }$

Thus we have by substituting and rearranging terms,

${\displaystyle a_{k}={\frac {2}{\pi }}\int _{0}^{\pi }f(cos\theta )cos(k\theta )d\theta }$

Author

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