User:Egm6341.s10.team3.Min/HW6

2 Show $$ dp_{\overline{y}}=mVdr$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg33.djvu|p.33-2]]

Problem Statement
At t+dt, consider V+dV Show that $$ dp_{\overline{y}}=mVdr$$

Solution


At t+dt, V+dV,see the above figure
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$$ By negnecting higher oder terms, mdVdr, We have: At t+dt, V+dV,see the above figure
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * }
 * }
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$$ End the proof
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * }
 * }

-- By Min Zhong 12:00, 7 Arip 2010 (UTC)

7 Show that $$ Z(\varsigma= \frac{1}{2})= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$ Z(\varsigma=2)= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$

In Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2]](2), We have:
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$$ $$
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle (Eq. 7-1)
 * }
 * }

In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.35-4]](1), we have:


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\begin{Bmatrix} C_0 \\ C_1 \\ C_2 \\ C_3 \\ \end{Bmatrix}= \begin{bmatrix} 1 & 0 & 0&0 \\ 0 & 1 & 0&0 \\  -3 & -2 & 3&-1 \\  2 & 1 & -2&1\\ \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^\prime \\ Z_{i+1} \\ Z_{i+1}^\prime \\ \end{Bmatrix}
 * $$\displaystyle
 * $$\displaystyle

$$ $$ In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.36-1]](1), we have:
 * $$\displaystyle (Eq. 7-2)
 * }
 * }
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$$ $$
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle (Eq. 7-3)
 * }
 * }

Solution
We start the proof from Equation 1.
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\begin{align} Z(\varsigma= \frac{1}{2})& = \sum_{n=0}^{3}C_i{ \frac{1}{2}}^i\\ &=C_0+ \frac{1}{2}C_1+ \frac{1}{4}C_2+ \frac{1}{8}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-4)
 * }
 * }

Next, we will find the expression of Ci by Zi and fi From the matrix Eq. 7-2, we have:
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$$ $$
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle (Eq. 7-5)
 * }
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$$ $$
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle (Eq. 7-6)
 * }
 * }


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$$ $$
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-7)
 * }
 * }


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$$ $$
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-8)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.7-4,We can get:
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\begin{align} Z(\varsigma= \frac{1}{2})& =Z_i+ \frac{1}{2}Z_i^\prime + \frac{1}{4}(-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime)+ \frac{1}{8}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime)\\ &=\frac{1}{2}Z_i+\frac{1}{2}Z_{i+1}+\frac{1}{8}Z_i^\prime-\frac{1}{8}Z_{i+1}^\prime\\ &=\frac{1}{2}(Z_i+Z_{i+1})+\frac{1}{8}(Z_i^\prime-Z_{i+1}^\prime)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-9)
 * }
 * }

Plug Eq 3 into above equation, we have


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$$\displaystyle Z(\varsigma=\frac{1}{2})=\frac{1}{2}(Z_i+Z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
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 * }
 * }

-- By Min Zhong 12:00, 7 Apri 2010 (UTC)

8 Show $$ Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime)$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime) From Lecture Notes [[media:Egm6341.s10.mtg56.djvu|p.35-1]](2)


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$$ $$
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle (Eq. 8-1)
 * }
 * }

Solution
We start the proof from Eq.8-1,


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$$ $$
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle (Eq. 8-2)
 * }
 * }


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\begin{align} Z(\varsigma={\frac{1}{2}}^\prime)&=C_1+ 2C_2\frac{1}{2}+ 3C_3{\frac{1}{2}}^2\\ &=C_1+ C_2 +\frac{3}{4}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-3)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.8-3, we get:


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\begin{align} Z(\varsigma=\frac{1}{2})^\prime & = Z_i^\prime-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime+\frac{1}{2}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime) \\ &=-\frac{3}{2}Z_i+\frac{3}{2}Z_{i+1}-\frac{1}{4}Z_i^\prime-\frac{1}{4}Z_{i+1}^\prime\\ &=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime+Z_{i+1}^\prime)\\ \end{align}
 * $$\displaystyle
 * $$\displaystyle

$$ -- By Min Zhong 12:00, 7 Apri 2010 (UTC)
 * }
 * }

9 Show $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$
Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]]

Problem Statement
Proof that $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]](1)


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$$ $$
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle (Eq. 9-1)
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]


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\dot Z_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
When $$\triangle=0$$, from Eq.9-1, we have :
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$$ $$
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle (Eq. 9-3)
 * }
 * }

Plug Eq. 9-3 into Eq, 9-2, we can get:
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f_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9-4)
 * }
 * }

Rearrange Eq. 9-4 we have:


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\begin{align} Z_{i+1}&=Z_i+ \frac{h}{6}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ &=Z_i+ \frac{h/2}{3}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 9-2)
 * }
 * }