User:Egm6321.f10.team03/Hwk1

=Homework One Summary= Problems assigned in the first two weeks of class. = Problem 1 - Maglev First and Second Derivative Derivations =

Reading convention
A function $$\,f^1$$ or $$\,f^1(.)$$ has to be read as: "$$f$$ superscript 1". A function $$\,(f^1)^2$$ or $$\,(f^1(.))^2$$has to be read as: "$$f$$ superscript 1, power 2".

Given
Based on the definitions and equations in Maglev Train EOM the axial deformation of the guidance for a High-Speed Train is represented by a function $$\,u^1(s,t)$$. The argument $$\,s$$ itself is also a function of the time variable $$\,t$$, so that for a general function $$\,f$$ the following is valid:
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$$\,{{\left. f(s,t) \right|}_{s={{y}^{1}}(t)}}=f({{y}^{1}}(t),t)$$
 * $$\,(Eq.1.1)$$
 * }

Find
The first and second derivation of $$\,f({{y}^{1}}(t),t)$$ has to be calculated with respect to $$\,t$$.

Solution
First derivation $$\frac{df}{dt}$$: Let's set
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$$\,z:=f({{y}^{1}}(t),t)$$ $$ and
 * $$\,(Eq. 1.2)
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$$\,s:={{y}^{1}}(t)$$ $$ thus we get for $$z$$ :
 * $$\,(Eq. 1.3)
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$$\,z=f(s,t)$$ $$ The absolute derivation of $$z$$ is:
 * $$\,(Eq. 1.4)
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$$\,dz={{f}_{s}}\ ds+{{f}_{t}}\ dt$$ $$ Following the suggestion in Papula, we devide $$(Eq.1.5)$$ by $$\,dt$$ and get:
 * $$\,(Eq. 1.5)
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 * }
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$$\frac{dz}{dt}={{f}_{s}}\ \frac{ds}{dt}+{{f}_{t}}$$ $$ substituting $$z$$ $$(Eq. 1.2)$$ gives:
 * $$\,(Eq. 1.6)
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 * }
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$$\frac{df(s,t)}{dt}=\frac{\partial f(s,t)}{\partial s}\frac{\partial s}{\partial t}+\frac{\partial f(s,t)}{\partial t}$$ $$ substituting $$s$$ this time, gives:
 * $$\,(Eq. 1.7)
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 * }
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$$\frac{df({{y}^{1}}(t),t)}{dt}=\frac{\partial f({{y}^{1}}(t),t)}{\partial s}\frac{\partial {{y}^{1}}(t)}{\partial t}+\frac{\partial f({{y}^{1}}(t),t)}{\partial t}$$ $$ So that we can finally say:
 * $$\,(Eq. 1.8)
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 * }
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$$\frac{df}{dt}={{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$
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 * $$\,(Eq. 1.9)
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 * }

Second derivation $$\frac{{{d}^{2}}f}{d{{t}^{2}}}$$: The procedure to calculate the second derivative of $$f$$ is the same used to calculate the first derivative. For a better overview let's assign $$(Eq. 1.9)$$ to a new variable and also parts of the equation to new variables like follows:
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$$\,q:= {{f}_{s}}\cdot \dot{y}^{1}+{{f}_{t}}$$ $$\,r:= {{f}_{s}}$$ $$\,m:= \dot{y}^{1}$$ $$\,h:= {{f}_{t}}$$ Equation $$(Eq. 1.9)$$ can then be written as:
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 * }
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$$\,q=r\cdot m+h$$ $$ Similarly to the calculation of the first derivative, we calculate the total derivative of $$q$$, paying attention to the fact that now we have to derive the prduct $$(r\cdot m)$$ with $$r$$ being a function of $$s$$ and $$t$$ :
 * $$\,(Eq. 1.10)
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 * }
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$$dq=({{r}_{s}}\cdot ds+{{r}_{s}}\cdot dt)\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ multiplying $$m$$ into the braket expression gives:
 * $$\,(Eq. 1.11)
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$$dq={{r}_{s}}\cdot ds\cdot m+{{r}_{s}}\cdot dt\cdot m+r\cdot {{m}_{t}}\cdot dt+{{h}_{t}}\cdot dt+{{h}_{s}}\cdot ds$$ $$ Dividing also here $$dq$$ by $$dt$$ and substituiting $$r, m, h$$ with their original terms we get:
 * $$\,(Eq. 1.12)
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 * }
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$$\frac{dq}{dt}=\frac{\partial {{f}_{s}}}{\partial s}\ \frac{ds}{dt}\ {{{\dot{y}}}^{1}}+\frac{\partial {{f}_{s}}}{\partial t}\ {{{\dot{y}}}^{1}}+{{f}_{s}}\frac{\partial {{{\dot{y}}}^{1}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial t}+\frac{\partial {{f}_{t}}}{\partial s}\ \frac{ds}{dt}$$ $$ That in fact after deriving becomes:
 * $$\,(Eq. 1.13)
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 * }
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$$\frac{dq}{dt}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}+{{f}_{ts}}\cdot {{{\dot{y}}}^{1}}$$ $$ Since the derivation sequence doesn't matter for partial derivations, we finally after ordering the terms in $$(Eq. 1.14) $$ get for the second derivative:
 * $$\,(Eq. 1.14)
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 * }
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$$\frac{{{d}^{2}}f}{d{{t}^{2}}}={{f}_{ss}}\cdot {{({{{\dot{y}}}^{1}})}^{2}}+2{{f}_{st}}\cdot {{{\dot{y}}}^{1}}+{{f}_{s}}\cdot {{{\ddot{y}}}^{1}}+{{f}_{tt}}$$ $$ Mike SigilloEgm6321.f10.team03 16:46, 14 September 2010 (UTC)
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 * $$\,(Eq. 1.15)
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Kurt SchulzeEgm6321.f10.team03 16:46, 14 September 2010 (UTC)

= Problem 2 - Dimensional Analysis=

Given
From the Maglev Train EOM we are given the horizontal force on the wheel/magnet $$\,{{c}_{0}}({{y}^{1}},t)$$ :
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$$\,{{c}_{0}}({{y}^{1}},t)=-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)]-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R}+M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}]$$ $$ with :
 * $$\,(Eq. 2.1)
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$$\,[{{F}^{1}}]=F(force)$$: Force acting in the lateral direction of the wheel $$\,[{{F}^{2}}]=F(force)$$: Force acting in the vertical direction of the wheel $$\,[T]=T(torque)$$: Torque at the edge of the wheel $$\,[M]=M(mass)$$: Mass of the wheel $$\,[\bar{R}]=L(length)$$: Distance between the rail-midpoint and the wheel-midpoint $$\,[R]=L(length)$$: Radius of the wheel $$\,[t]=t(time)$$ $$\,[{{u}^{2}}]=L(length)$$: Transverse deformation of the guideway

$$\,\left[ u_{s}^{2} \right]=\left[ \frac{\partial {{u}^{2}}}{\partial s} \right]=\frac{L}{L}= \left( \frac{length}{length} \right)$$: Slope of the deformation of the guideway in the transverse direction

$$\left[ u_{tt}^{1} \right]=\frac{d}{dt}\left[ \frac{\partial {{u}^{1}}}{\partial t} \right]=\left[ \frac{\partial {{t}^{2}}} \right]=\frac=\left( \frac{length}{{time} \cdot {time}} \right)$$ : Acceleration of the axial deformation of the guideway

$$\left[ u_{tt}^{2} \right]=\frac{d}{dt}\left[ \frac{\partial {{u}^{2}}}{\partial t} \right]=\left[ \frac{\partial {{t}^{2}}} \right]=\frac=\left( \frac{length}{{time} \cdot {time}} \right)$$ : Acceleration of the transverse deformation of the guideway $$\left[ u_{ss}^{2} \right]=\frac{d}{ds}\left[ \frac{\partial {{u}^{2}}}{\partial s} \right]=\left[ \frac{\partial {{s}^{2}}} \right]=\frac{L}=\frac{1}{L}\ \ \left( \frac{1}{length} \right)$$ : Curvature of the transverse deformation of the flexible guideway $$\left[ u_{stt}^{2} \right]=\frac=\left( \frac{length}{{length}\cdot{time}\cdot{time}} \right) $$: Acceleration of the slope of the deformation of the guideway in the transverse direction $$\,[1]=1$$
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Find
Dimensional Analysis of the terms in $$(Eq. 2.1)$$ and physical meaning.

Solution
Analysis of the first term $$\,(-{{F}^{1}}[1-\bar{R}\cdot u_{ss}^{2}({{y}^{1}},t)])$$ :
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$$\,F[1-L\cdot {{L}^{-1}}]=F[1-1]\to$$ Dimension results to be a Force Analysis of the second term $$\,(-{{F}^{2}}\cdot u_{s}^{2}-\frac{T}{R})$$ :
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$$F\cdot \frac{L}{L}-\frac{T}{L}=F-F\to $$ Dimension results to be a Force Analysis of third term $$\,(M[[1-\bar{R}\cdot u_{ss}^{2}][u_{tt}^{1}-\bar{R}\cdot u_{stt}^{2}]+u_{s}^{2}\cdot u_{tt}^{2}])$$ :
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$$\,M\left[ \left[ 1-L\cdot {{L}^{-1}} \right]\left[ \frac{L}-L\cdot \frac{L}{L\cdot {{t}^{2}}} \right]+\frac{L}{L}\cdot \frac{L} \right]=M\cdot \frac{L}+M\cdot \frac{L}=F+F\to $$ Dimension results to be a Force Mike SigilloEgm6321.f10.team03 16:46, 14 September 2010 (UTC)
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Kurt SchulzeEgm6321.f10.team03 16:46, 14 September 2010 (UTC)

Jimmy RoarkEgm6321.f10.team3.roark 18:20, 14 September 2010 (UTC)

= Problem 3 - Linearity Test of Maglev Train EOM Coefficient =

Given
$$\displaystyle C_3$$ from the Maglev Train EOM is defined as:


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C_3(Y^1,t) := M\left [\left( 1-\bar Ru^2,_{ss}\left(Y^1,t\right)\right)\right]^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.3.1)
 * }
 * }

Find
Using the definition of linearity of a function, show that $$\displaystyle C_3\ddot{Y^1}$$ is nonlinear, with respect to $$\displaystyle Y^1$$.

Solution
A linear function is a function which obeys two laws, shown below. $$\displaystyle H$$ is an arbitrary function while $$\displaystyle c$$ is an arbitrary constant.


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H(cx) = cH(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.3.2)
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 * }


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H(x+y) = H(x) + H(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.3.3)
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 * }

In order for $$\displaystyle C_3(Y^1,t)\ddot{Y^1}$$ to be linear, the following equation must be true.


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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = cC_3(x,t)\ddot{Y^1}$$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.3.4)
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 * }

Applying the cumulative property to (Eq.3.1) gives us


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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = M\left [\left( 1-\bar Ru^2,_{ss}\left(cx,t\right)\right)\right]^2\ddot{Y^1}.$$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.5)
 * <p style="text-align:right;">$$\displaystyle (Eq.3.5)
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 * }

Let us assume the function $$\displaystyle u^2,_{ss}$$ is linear, (Eq.3.5) becomes


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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} = M\left [\left( 1-c\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2\ddot{Y^1}.$$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.6)
 * <p style="text-align:right;">$$\displaystyle (Eq.3.6)
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 * }

Now we evaluate the right half of (Eq.3.4)


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$$
 * $$\displaystyle cC_3(x,t)\ddot{Y^1} = cM\left [\left( 1-\bar Ru^2,_{ss}\left(x,t\right)\right)\right]^2\ddot{Y^1}.$$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.7)
 * <p style="text-align:right;">$$\displaystyle (Eq.3.7)
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 * }

Since


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$$
 * $$\displaystyle C_3(cx,t)\ddot{Y^1} \not= cC_3(x,t)\ddot{Y^1},$$
 * <p style="text-align:right;">$$\displaystyle (Eq.3.8)
 * <p style="text-align:right;">$$\displaystyle (Eq.3.8)
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 * }

$$\displaystyle C_3(Y^1,t)\ddot{Y^1}$$ is nonlinear because it did not meet the linearity test of a function.

Adam Franklin 18:56, 12 September 2010 (UTC)

= Problem 4 - Coefficients of Linear Combination =

Given
Given a linear, second order ordinary differential equation with varying coefficients (L2-ODE-VC) and with boundary conditions:


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$$ y \left( a\right) =  \alpha  ;  y \left( b\right)  =   \beta $$
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 * }
 * }

From eqn. [4], pg [[media:2010_09_02_13_55_50.djvu|5-3]] of the notes, the ODE has a solution of the form:


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$$ y \left( x\right) =  c {y}^{1}_{H}\left( x\right) + d {y}^{2}_{H} \left( x\right) + {y}_{P} \left( x\right) $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4, Pg. 5-3)
 * }

Find
Find $$c,d$$ in terms of $$\alpha$$ and $$\beta$$

Solution
Boundary conditions
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$$ y \left( a\right) =  \alpha  ;  y \left( b\right)  =   \beta $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4.1)
 * }

Now


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$$ \alpha =  c {y}^{1}_{H}\left( a\right) + d {y}^{2}_{H} \left( a\right) + {y}_{P} \left( a\right) $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4.2)
 * }


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$$ \beta =  c {y}^{1}_{H}\left( b\right) + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4.3)
 * }

Solve for $$ c $$ in $$\displaystyle (Eq.4.2)$$


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$$ \Rightarrow c = \frac{\alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq.4.4)
 * }

Plug $$\displaystyle (Eq.4.4)$$ into $$\displaystyle (Eq.4.3)$$ to eliminate $$ c $$


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$$ \beta =  \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ \alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right) \right] + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$
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 * }
 * }

For ease of notation, Let $$ \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} = \xi $$


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$$ \Rightarrow \beta = \xi\left[ \alpha - {y}_{P} \left( a\right) - d {y}^{2}_{H} \left( a\right) \right] + d {y}^{2}_{H} \left( b\right) + {y}_{P} \left( b\right) $$ $$
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 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (Eq.4.5)
 * }

Solve $$\displaystyle (Eq.4.5)$$ for $$ d $$


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$$ d = \frac{ \beta - {y}_{P} \left( b\right) + \xi \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \xi {y}^{2}_{H} \left( a\right)} $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4.6)
 * }

Subsitute $$\displaystyle (Eq.4.6)$$ back into $$\displaystyle (Eq.4.4)$$ to obtain $$ c $$


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$$ c = \frac{\alpha - {y}_{P} \left( a\right) - \left[\frac{ \beta - {y}_{P} \left( b\right) + \xi \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) -  \xi {y}^{2}_{H} \left( a\right)} \right]{y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq.4.7)
 * }

$$ c $$ and $$ d $$ can be re-writen by substituting ​$$ \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} = \xi $$


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$$ c = \frac{\alpha - {y}_{P} \left( a\right) - \left[\frac{ \beta - {y}_{P} \left( b\right) + \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} {y}^{2}_{H} \left( a\right)} \right]{y}^{2}_{H} \left( a\right)}{{y}^{1}_{H}\left( a\right)} $$ $$
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 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq.4.8)
 * }


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$$ d = \frac{ \beta - {y}_{P} \left( b\right) + \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)} \left[ {y}_{P} \left( a\right) - \alpha\right]}{{y}^{2}_{H} \left( b\right) - \frac{{y}^{1}_{H}\left( b\right)}{{y}^{1}_{H}\left( a\right)}{y}^{2}_{H} \left( a\right)} $$ $$ Egm6321.f10.team3.roark 11:05, 14 September 2010 (UTC)
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq.4.9)
 * }

Egm6321.f10.team3.cook 11:12, 14 September 2010 (UTC)

Egm6321.f10.team3.Hong SJ 11:12, 14 September 2010 (UTC)

= Problem 5 - Verification of Homogenous Solutions of Legendre Differential Equation for n = 1=

Given
Legendre differential equation:
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{{L}_{2}}\left( y \right) = (1-x^2)y'' - 2xy' +n(n+1)y = 0 $$ $$ Homogenous solutions of the above equation for $$\displaystyle  n = 1 $$ [[media:2010_09_02_14_58_46.djvu|(Mtg 6-page1)]], are given by:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq.5.1)
 * }
 * }
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$$ $$
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * $$y_{H}^{1}\left( x \right)=x\equiv P_1(x)
 * <p style="text-align:right;">$$\displaystyle (Eq.5.2)
 * }
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 * {| style="width:100%" border="0" align="left"

$$ $$ Where $$y_{H}^{1}\left( x \right)$$ is the first homogenous solution and $$y_{H}^{2}\left( x \right)$$ is the second homogenous solution
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * $$y_{H}^{2}\left( x \right)=\frac{x}{2}.\log \left( \frac{1+x}{1-x} \right)-1\equiv Q_1(x)
 * <p style="text-align:right;">$$\displaystyle (Eq.5.3)
 * }
 * }

Find
We need to prove that:
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 * $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$
 * }
 * }
 * }

Solution
Let us consider first $${{L}_{2}}\left( y_{H}^{1} \right)$$:


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$$ Substitute Eq.2 for $${{L}_{2}}\left( y_{H}^{1} \right)$$ in Eq.5.4 =>
 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{1} \right)}^{''}}-2x{{\left( y_{H}^{1} \right)}^{'}}+2\left( y_{H}^{1} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq.5.4)
 * <p style="text-align:right;">$$\displaystyle (Eq.5.4)
 * }
 * }


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 * $${{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right){{\left( x \right)}^{''}}-2x{{\left( x \right)}^{'}}+2\left( x \right)$$
 * }
 * }
 * }


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$$ {{L}_{2}}\left( y_{H}^{1} \right)=\left( 1-{{x}^{2}} \right)\left( 0 \right)-2x\left( 1 \right)+2\left( x \right)=0$$ $$
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 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq.5.5)
 * }
 * }

Now consider $${{L}_{2}}\left( y_{H}^{2} \right)$$:


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$$ Determine first derivative of $$y_{H}^{2}$$ (Eq.5.2) using Product rule, Logarithmic derivative rule ($$ (\ln f)'= \frac{f'}{f} \quad$$) and chain rule of differentiation:
 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right){{\left( y_{H}^{2} \right)}^{''}}-2x{{\left( y_{H}^{2} \right)}^{'}}+2\left( y_{H}^{2} \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq.5.6)
 * <p style="text-align:right;">$$\displaystyle (Eq.5.6)
 * }
 * }


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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}.\frac{d}{dx}\left( \log \left( \frac{1+x}{1-x} \right) \right)+\frac{d}{dx}\left( \frac{x}{2} \right).\log \left( \frac{1+x}{1-x} \right)-\frac{d}{dx}\left( 1 \right)$$
 * }
 * }
 * }


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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1}{\left( \frac{1+x}{1-x} \right)}.\frac{d}{dx}\left( \frac{1+x}{1-x} \right) \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)+0$$
 * }
 * }
 * }


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 * $${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$
 * }
 * }
 * }


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$${{\left( y_{H}^{2} \right)}^{'}}=\frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$ $$
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 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq.5.7)
 * }
 * }

Second derivative of $$y_{H}^{2}$$ (Eq.5.2) is:


 * {| style="width:100%" border="0"


 * $${{\left( y_{H}^{2} \right)}^{''}}={{\left( {{\left( y_{H}^{2} \right)}^{'}} \right)}^{'}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{d}{dx}\left( \frac{x}{1-{{x}^{2}}} \right)+\frac{d}{dx}\left( \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{\left( y_{H}^{2} \right)}^{''}}=\left( \frac{\left( 1-{{x}^{2}} \right)-x.\left( -2x \right)} \right)+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{\left( 1-x \right)+\left( 1+x \right)} \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}{2}\left( \frac{1-x}{1+x} \right)\left( \frac{2} \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{\left( y_{H}^{2} \right)}^{''}}=\frac{1+{{x}^{2}}}+\frac{1}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$${{\left( y_{H}^{2} \right)}^{''}}=\frac{2}$$ $$ Substituting Eqs.5.3,7,8 into Eq.5.5,6 yields,
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * style="width:20%; padding:1px; border:2px solid #8888aa" |
 * <p style="text-align:right;">$$\displaystyle (Eq.5.8)
 * }
 * }
 * {| style="width:100%" border="0"


 * $${{L}_{2}}\left( y_{H}^{2} \right)=\left( 1-{{x}^{2}} \right)\left( \frac{2} \right)-2x\left( \frac{x}{1-{{x}^{2}}}+\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right)+2\left( \frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 \right)$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2}{\left( 1-{{x}^{2}} \right)}-\frac{2{{x}^{2}}}{1-{{x}^{2}}}-x\log \left( \frac{1+x}{1-x} \right)+x\log \left( \frac{1+x}{1-x} \right)-2$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$${{L}_{2}}\left( y_{H}^{2} \right)=\frac{2-2{{x}^{2}}-2\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)}=0$$ $$
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * style="width:20%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq.5.9)
 * }
 * }

Eqs.5 and 9 verify that :{| style="width:10%" border="0" $${{L}_{2}}\left( y_{H}^{1} \right)={{L}_{2}}\left( y_{H}^{2} \right)=0$$ Egm6321.f10.team3.Sudheesh 18:11, 14 September 2010 (UTC)
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * }
 * }

Egm6321.f10.team3.Hong SJ 11:12, 14 September 2010 (UTC)

=Contributing Team Members=
 * Adam Franklin 18:56, 12 September 2010 (UTC) Author Problem 3
 * Sudheesh 18:11, 14 September 2010 (UTC) Author Problem 5
 * Jimmy Roark 14:24, 15 September 2010 (UTC) Co-Author Problem 4; Proofread and edited Problem 2
 * Michele Sigilló 22:07, 12 September 2010 (UTC) Author Problem 1 and 2
 * Kurt Schulze 01:10, 13 September 2010 (UTC) Proofread Problem Problem 2
 * Chris Cook 11:06, 14 September 2010 (UTC) Co-Author Problem 4
 * Hong SJ 11:12, 14 September 2010 (UTC) Proofread Problem 4 and 5

=Notes and references=