User:Egm6341.s10.team3.Min/HW7

= (1) Linearization about $$ x_{max} $$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg38.djvu|p.38-4]]

Problem Statement

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$$
 * $$\displaystyle x=\hat{x}+ y
 * $$\displaystyle x=\hat{x}+ y
 * }
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$$
 * $$\displaystyle f(x)=rx \left(1- \frac{x}{x_{max}} \right)
 * $$\displaystyle f(x)=rx \left(1- \frac{x}{x_{max}} \right)
 * }
 * }

At
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$$ Linearization about equil. pt $$\hat {x} $$
 * $$\displaystyle \hat{x}=x_{max}
 * $$\displaystyle \hat{x}=x_{max}
 * }
 * }

Solution

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$$
 * $$\displaystyle At \hat{x}=x_{max}
 * $$\displaystyle At \hat{x}=x_{max}
 * }
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\begin{align} f(x)&=f(x_{max}+y)\\ &=r(x_{max}+y)(1- \frac{x_{max}+y}{x_{max}})\\ &=-ry \left(1+ \frac{y}{x_{max}} \right) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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$$
 * $$\displaystyle where yy<<x_{max} thus 1+ \frac{y}{x_{max}} \approx 1
 * $$\displaystyle where yy<<x_{max} thus 1+ \frac{y}{x_{max}} \approx 1
 * }
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$$
 * $$\displaystyle f(x)\approx -ry
 * $$\displaystyle f(x)\approx -ry
 * }
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$$
 * $$\displaystyle \frac{dx}{dt}= \frac{d(x_{max}+y)}{dt}= \frac{dy}{dt}
 * $$\displaystyle \frac{dx}{dt}= \frac{d(x_{max}+y)}{dt}= \frac{dy}{dt}
 * }
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$$ -- By Min Zhong 12:00, 22 Apr 2010 (UTC)
 * $$\displaystyle \Rightarrow \frac{dy}{dt}=-ry \Rightarrow y=y_0e^{-rt}
 * $$\displaystyle \Rightarrow \frac{dy}{dt}=-ry \Rightarrow y=y_0e^{-rt}
 * }
 * }

= (2) Verify and solve the differential equation = Ref: Lecture Notes [[media:Egm6341.s10.mtg39.djvu|p.39-1]]

Problem Statement
The exact expression of x(t) is :
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$$ i) Verify that:
 * $$\displaystyle x(t)= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * $$\displaystyle x(t)= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * }
 * }
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$$
 * $$\displaystyle \frac{dx}{dt}=rx \left( 1- \frac{x}{x_{max}}\right)
 * $$\displaystyle \frac{dx}{dt}=rx \left( 1- \frac{x}{x_{max}}\right)
 * }
 * }

ii) solve the above differential equation with hint on Lecture Notes [[media:Egm6341.s10.mtg40.djvu|p.40-1]]

Solution
i) Verify the differential equation
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\begin{align} \frac{dx(t)}{dt}&= \frac {x_0x_{max}e^{rt}r}{x_{max}+x_0(e^{rt}-1)}+x_0x_{max}e^{rt}(-1) \left( \frac{1}{x_{max}+x_0(e^{rt}-1)} \right) ^2x_0e^{rt}r \\ &= \frac{x_ox_{max}e^{rt}}{x_{max}+x_0 \left( e^{rt}-1\right)}r+ \frac{r}{x_{max}}  \left( \frac{x_0x_{max}e^{rt}}{x_{max}+x_0 \left( e^{rt}-1\right)} \right)^2\\ &=xr-\frac{r}{x_{max}}x^2\\ &=rx \left(1- \frac{x}{x_{max}} \right) \end{align} $$ End the verification.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

ii)Solve the differential equation Partial fraction:
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\begin{align} \frac{1}{x \left(1- \frac{x}{x_{max}} \right)}&= \frac{a}{x}+ \frac{b}{1- \frac{x}{x_{max}}}\\ &= \frac{a \left(1- \frac{x}{x_{max}}\right)+bx}{x \left(1- \frac{x}{x_{max}} \right)}\\ \end{align} $$ Let
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
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$$
 * $$\displaystyle a\left(1- \frac{x}{x_{max}}\right)+bx=1 \Rightarrow a-x \left(b- \frac{a}{x_{max}} \right)=1
 * $$\displaystyle a\left(1- \frac{x}{x_{max}}\right)+bx=1 \Rightarrow a-x \left(b- \frac{a}{x_{max}} \right)=1
 * }
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\Rightarrow \begin{cases} & a= 1 \\ & b=  \frac{1}{x_{max}} \end{cases} $$ Separate the variables of the differential equation:
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
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$$
 * $$\displaystyle \int_{x_0}^{x}  \frac{dx}{x \left(1- \frac{x}{x_{max}} \right)}= \int_{0}^{t}rdt
 * $$\displaystyle \int_{x_0}^{x}  \frac{dx}{x \left(1- \frac{x}{x_{max}} \right)}= \int_{0}^{t}rdt
 * }
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$$
 * $$\displaystyle  \Rightarrow \int_{x_0}^{x}  \frac{1}{x} +  \frac{ \frac{1}{x_{max}}}{1- \frac{x}{x_{max}}}= \int_{0}^{t}rdt
 * $$\displaystyle  \Rightarrow \int_{x_0}^{x}  \frac{1}{x} +  \frac{ \frac{1}{x_{max}}}{1- \frac{x}{x_{max}}}= \int_{0}^{t}rdt
 * }
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$$ :{| style="width:100%" border="0" align="left" $$ :{| style="width:100%" border="0" align="left" $$
 * $$\displaystyle  \Rightarrow  \left[ lnx-ln \left(1- \frac{x}{x_{max}} \right) \right]_{x_0}^x =  \left[rt \right]_{0}^t
 * $$\displaystyle  \Rightarrow  \left[ lnx-ln \left(1- \frac{x}{x_{max}} \right) \right]_{x_0}^x =  \left[rt \right]_{0}^t
 * }
 * }
 * $$\displaystyle  \Rightarrow  \left[ lnx-ln \left(1- \frac{x}{x_{max}} \right) \right]_{x_0}^x =  rt
 * $$\displaystyle  \Rightarrow  \left[ lnx-ln \left(1- \frac{x}{x_{max}} \right) \right]_{x_0}^x =  rt
 * }
 * }
 * $$\displaystyle  \Rightarrow ln \left( \frac{x}{x_0} \right) \left( \frac{1- \frac{x_0}{x_{max}}}{1- \frac{x}{x_{max}}} \right) =  rt
 * $$\displaystyle  \Rightarrow ln \left( \frac{x}{x_0} \right) \left( \frac{1- \frac{x_0}{x_{max}}}{1- \frac{x}{x_{max}}} \right) =  rt
 * }
 * }


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$$
 * $$\displaystyle  \Rightarrow x= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * $$\displaystyle  \Rightarrow x= \frac{x_0x_{max}e^{rt}}{x_{max}+x_0(e^{rt}-1)}
 * }
 * }

-- By Min Zhong 12:00, 22 Apr 2010 (UTC)

= (13) Change of variable  = Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-2]]

Problem Statement
By Changing the variable of integration,
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$$ show that:
 * $$\displaystyle  x=cos \theta
 * $$\displaystyle  x=cos \theta
 * }
 * }
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 * $$\displaystyle  I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi}f(cos  \theta)sin \theta d\theta
 * $$\displaystyle  I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi}f(cos  \theta)sin \theta d\theta

$$
 * }
 * }

Solution
Let
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$$ and
 * $$\displaystyle  x=cos \theta \Rightarrow dx=-sin \theta d \theta
 * $$\displaystyle  x=cos \theta \Rightarrow dx=-sin \theta d \theta
 * }
 * }
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\begin{cases} & x=-1;  \theta =\pi \\ & x=+1;\theta=0 \end{cases} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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 * $$\displaystyle  I= \int_{-1}^{+1}f(x)dx= \int_{\pi}^{0}f(cos\theta)(-1)sin \theta d\theta=\int_{0}^{\pi}f(cos\theta)sin \theta d\theta
 * $$\displaystyle  I= \int_{-1}^{+1}f(x)dx= \int_{\pi}^{0}f(cos\theta)(-1)sin \theta d\theta=\int_{0}^{\pi}f(cos\theta)sin \theta d\theta

$$ End the proof -- By Min Zhong 12:00, 22 Apr 2010 (UTC)
 * }
 * }

= (14)Find the item $$a_k$$ in cosine series of $$f(cos\theta)$$  = Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-3]]

Problem Statement
The known cosine series of f(x) is as follows:
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$$ Find out the expression for $$a_k$$
 * $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right)
 * $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right)
 * }
 * }
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$$
 * $$\displaystyle a_k= \frac{2}{\pi} \int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta
 * $$\displaystyle a_k= \frac{2}{\pi} \int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta
 * }
 * }

Solution
The proof starts by multiplying $$cos(k\theta)$$ on both sides of cosine series of $$ f(cos\theta)$$
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f(cos \theta) = \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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f(cos \theta)cos(k\theta) = \frac{a_0}{2}cos(k\theta)+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right)^2 $$ Do integration from $$ [0,\pi] $$ on both sides of the above equation:
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
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\int_{0}^{\pi}f(cos \theta)cos(k\theta) = \int_{0}^{\pi}\frac{a_0}{2}cos(k\theta)+  \int_{0}^{\pi}\sum_{k=1}^{\infty}a_kcos \left( k\theta\right)^2 $$ where
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
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\begin{align} \int_{0}^{\pi}\frac{a_0}{2}cos(k\theta)&= \frac{a_0}{2k} \left[sin(k\theta) \right]_0^\pi\\ &= \frac{a_0}{2k} \left( sin(k\pi)-sin(0)\right)\\ &=0\\ \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
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\begin{align} \int_{0}^{\pi}\sum_{k=1}^{\infty}a_kcos \left( k\theta\right)^2&= \sum_{k=1}^{\infty}a_k \int_{0}^{\pi}cos^2(k\theta)d\theta\\ &=\sum_{k=1}^{\infty}a_k\int_{0}^{\pi} \frac{cos(2\theta+1)}{2}\\ &= \frac{1}{2} \sum_{k=1}^{\infty}a_k \left[ \frac{1}{2}sin(2\theta)+\theta\right]_0^\pi\\ &= \frac{1}{2} \sum_{k=1}^{\infty}a_k \left(  \frac{1}{2} sin(2\pi)+\pi- \frac{1}{2}sin0-0\right)\\ &= \frac{\pi}{2} \sum_{k=1}^{\infty}a_k\\
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }
 * }

Thus
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\int_{0}^{\pi}f(cos \theta)cos(k\theta) = \frac{\pi}{2} \sum_{k=1}^{\infty}a_k $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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\Rightarrow a_k= \frac{2}{\pi} \int_{0}^{\pi}f(cos\theta)cos(k\theta)d\theta $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

-- By Min Zhong 12:00, 23 Apr 2010 (UTC)

= (15)Find the expression of integration of fourier transfer  = Ref: Lecture Notes [[media:Egm6341.s10.mtg42.djvu|p.42-3]]

Problem Statement
The following is known:
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I= \int_{-1}^{+1}f(x)dx= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The known cosine series of f(x) is as follows:
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$$ Find the expression of Integration as:
 * $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right)
 * $$\displaystyle  f(cos \theta)= \frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos \left( k\theta\right)
 * }
 * }
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$$
 * $$\displaystyle I=a_0+ \sum_{j=1}^{\infty} \frac{2a_{2j}}{1- \left( 2j\right)^2}
 * $$\displaystyle I=a_0+ \sum_{j=1}^{\infty} \frac{2a_{2j}}{1- \left( 2j\right)^2}
 * }
 * }

Solution

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\begin{align} I&= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta\\ &= \int_{0}^{\pi} \left[  \frac{a_0}{2}+ \sum_{j=1}^{\infty}a_{2j}cos(2j\theta)\right]sin\theta d\theta\\ &= \frac{a_0}{2}\int_{0}^{\pi}sin\theta d \theta +  \sum_{j=1}^{\infty} a_{2j}\int_{0}^{\pi}sin\theta cos(2j\theta) d \theta\\ &= - \frac{a_0}{2} \left[ cos\theta\right]_0^\pi+ \sum_{j=1}^{\infty}a_{2j} \int_{0}^{\pi} \frac{1}{2} \left [sin(1+2j) \theta +sin (1-2j)\theta \right] d\theta\\ &= a_0+ \sum_{j=1}^{\infty}a_{2j} \frac{1}{2} \left[\left( \frac{-1}{1+2j} cos(1+2j) \theta \right)_0^\pi+ \left(  \frac{-1}{1-2j}cos(1-2j) \theta\right) _0^\pi \right]\\ &=a_0+ \sum_{j=1}^{\infty}a_{2j} \frac{1}{2} \left[ \frac{2}{1+2j} + \frac{2}{1-2j}\right]\\ &=a_0+ \sum_{j=1}^{\infty}a_{2j}  \left( \frac{1}{2}\right) \frac{4}{1-(2j)^2}\\ &=a_0+ \sum_{j=1}^{\infty} \frac{2a_{2j}}{1-(2j)^2}\\ \end{align} $$ The end of proof
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

-- By Min Zhong 12:00, 23 Apr 2010 (UTC)