User:Egm6321.f10.team2.oztekin/HW1

=Problem 1: Deriving Function Relate to Maglev=

Problem Infrastructure
Any fuction like
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$$ F\left ( x,y(x) \right )=0 \displaystyle  $$ (Eq 1.1) is called a closed fuction.
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A total differential of F is as follows:
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$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy \displaystyle $$
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(Eq 1.2)
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From that we can write:
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$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \displaystyle $$
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(Eq 1.3)
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Given
From the maglev train which mentioned in meeting one

$$Y'\left ( t \right )$$:nominal position of the wheel

$$U'\left ( s,t \right )$$:axial deformation of the guide way

$$U^{2}\left ( s,t \right )$$:transverse deformation of the guide way

And we conclude that there is a impilicit and explicit dependence between U and Y(t)

Find

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$$\frac{\mathrm{d}^2 f(Y'(t),t) }{\mathrm{d} t}=\frac{\partial f(Y'(t),t)}{\partial s}\left ( \ddot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\left ( \dot{Y'} \right )^2+2\frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}\left ( \dot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial t^2} \displaystyle $$ (Eq 1.4)
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Solution

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$$\frac{\mathrm{d} f(Y'(t),t)}{\mathrm{d} t} =\frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t} \displaystyle $$ (Eq 1.5)
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$$\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t}\right )=\left (\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\dot{Y'}+\underbrace{ \frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}}  \right )\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial s}\ddot{Y'}+\underbrace{\frac{\partial^2 f(Y'(t),t)}{\partial t\partial s}}\dot{Y'}+\frac{\partial^2 f(Y'(t),t) }{\partial t^2}) \displaystyle $$
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(Eq 1.6)
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According to the Schwarz Theorem


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$$\frac{\partial^2 f}{\partial s\partial t}=\frac{\partial^2 f}{\partial t\partial s} \displaystyle $$ (Eq 1.7)
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$$\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial f(Y'(t),t)}{\partial s}\dot{Y'}+\frac{\partial f(Y'(t),t)}{\partial t}\right )=\frac{\partial f(Y'(t),t)}{\partial s}\left ( \ddot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial s^2}\left ( \dot{Y'} \right )^2+2\frac{\partial^2 f(Y'(t),t)}{\partial s\partial t}\left ( \dot{Y'} \right )+\frac{\partial^2 f(Y'(t),t)}{\partial t^2} \displaystyle $$ (Eq 1.8)
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Relation To Engineering Equations
If an object turning around an axis with an angular velocity of '$$\omega$$',its velocity according to a motionless observer can be written as;
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$$\vec{v}=\dot{\vec{r}}+\vec{\omega}*\vec{r} \displaystyle $$
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(Eq 1.9)
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And from this we can write acceleration as:
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$$\vec{a}=\ddot{\vec{r}}+2*\vec{\omega }*\dot{\vec{r}}+\vec{\omega }\left ( \vec{\omega *\vec{r}} \right ) \displaystyle$$
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(Eq 1.10)
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For an object moving under gravity force and another force(F), the equation of motion is:
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$$m\vec{a}=m\vec{g}+\vec{F}\displaystyle$$ (Eq 1.11)
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Substituting $$\vec{a}$$ into the (Eq 1.11) gives:
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$$m\ddot{\vec{r}}=m\vec{g}+\vec{F}-2*m*\vec{\omega }*\dot{\vec{r}}-m*\vec{\omega }\left ( \vec{\omega *\vec{r}} \right ) \displaystyle $$ (Eq 1.12)
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We can then show that $$m*\vec{\omega }\left ( \vec{\omega *\vec{r}} \right )$$ is 'centrifugal force'. And the term $$2*m*\vec{\omega }*\dot{\vec{r}}$$ is a 'Coriolis Force'.
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= Problem 2: Boundary Value Problem =

Given
From the note 5-3 and 5-4 we know,


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$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}(x)+{{y}_{P}}(x) \displaystyle$$ (Eq 2.1)
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$$y(a)=\alpha \displaystyle $$ (Eq 2.2)
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$$\displaystyle y(b)=\beta $$
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Find
$$\displaystyle c, d$$ in terms of $$\displaystyle \alpha, \beta. $$

Solution
By substituting $$\displaystyle a $$ and $$\displaystyle b $$ into $$\displaystyle Eq 2.1 $$ and combining with $$\displaystyle Eq 2.2 $$ and $$\displaystyle Eq 2.3 $$, we derive,


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$$\displaystyle cy_{H}^{1}(a)+dy_{H}^{2}(a)+{{y}_{P}}(a)=\alpha, $$ (Eq 2.4)
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$$\displaystyle cy_{H}^{1}(b)+dy_{H}^{2}(b)+{{y}_{P}}(b)=\beta. $$ (Eq 2.5)
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By multipling $$\displaystyle y_{H}^{2}(b) $$ in both sides of $$\displaystyle Eq. 4 $$ and $$\displaystyle y_{H}^{2}(a) $$ of $$\displaystyle Eq 2.5 $$, we obtain,


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$$\displaystyle y_{H}^{1}(a)y_{H}^{2}(b)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(a)y_{H}^{2}(b)=\alpha y_{H}^{2}(b),$$ (Eq 2.6)
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$$\displaystyle y_{H}^{1}(b)y_{H}^{2}(a)*c+y_{H}^{2}(a)y_{H}^{2}(b)*d+{{y}_{P}}(b)y_{H}^{2}(a)=\beta y_{H}^{2}(a).$$ (Eq 2.7)
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Let $$\displaystyle Eq 2.6 $$ minus $$\displaystyle Eq 2.7 $$, then $$\displaystyle d $$ is cancelled, so we get,


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$$\displaystyle c=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{2}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{2}(a)}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)}. $$ (Eq 2.8)
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Note the structrue of $$\displaystyle Eq 2.4 $$ and $$\displaystyle Eq 2.5 $$, we find that $$\displaystyle c $$ is bound to $$\displaystyle y_{H}^{1} $$ while $$\displaystyle d $$ is bound to $$\displaystyle y_{H}^{2} $$, that is, when we switch the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$, $$\displaystyle d $$ will become the former $$\displaystyle c $$ thus the same result of $$\displaystyle Eq 2.8 $$. After that we switch back the superscript between $$\displaystyle y_{H}^{1} $$ and $$\displaystyle y_{H}^{2} $$ then we get the right answer of $$\displaystyle d $$. By doing this, simply exchange the superscript of right side of $$\displaystyle Eq 2.8 $$, we have,


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$$\displaystyle d=\frac{\left[ \alpha -{{y}_{P}}(a) \right]y_{H}^{1}(b)-\left[ \beta -{{y}_{P}}(b) \right]y_{H}^{1}(a)}{y_{H}^{2}(a)y_{H}^{1}(b)-y_{H}^{2}(b)y_{H}^{1}(a)}. $$ (Eq 2.9)
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= Problem 3: Verifying 2 homogeneous solutions =

Given

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$$\displaystyle {{L}_{2}}(y)=(1-{{x}^{2}})\ddot{y}-2x\dot{y}+2y$$ (Eq 3.1)
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where $$\displaystyle {{L}_{2}} $$ is an operator of functions.


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$$\displaystyle y_{H}^{1}(x)=x,$$ (Eq 3.2)
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$$\displaystyle y_{H}^{2}(x)=\frac{x}{2}\log (\frac{1+x}{1-x})-1,$$ (Eq 3.3)
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where $$\displaystyle y_{H}^{1}(x) $$ and $$\displaystyle y_{H}^{2}(x) $$ are homogeneous solutions of the equation $$\displaystyle {{L}_{2}}(y)=0$$.

Find
Verify:
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$$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$
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Solution
From $$\displaystyle Eq 3.2 $$, we have,


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$$\displaystyle \dot{y}_{H}^{1}(x)=1,$$ (Eq 3.4)
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$$\displaystyle \ddot{y}_{H}^{1}(x)=0.$$ (Eq 3.5)
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Then by sustituting $$\displaystyle \dot{y}_{H}^{1}(x)$$ and $$\displaystyle \ddot{y}_{H}^{1}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,


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$$\displaystyle \begin{align} {{L}_{2}}(y_{H}^{1})& =0-2x+2x \\ & =0. \end{align}$$ (Eq 3.6)
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From $$\displaystyle Eq 3.3 $$, we derive,


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$$\displaystyle \begin{align} \dot{y}_{H}^{2}(x)& ={{\left[ \frac{x}{2}\log \left( \frac{2}{1-x}-1 \right)-1 \right]}^{\prime }} \\ & ={{\left( \frac{x}{2} \right)}^{\prime }}*\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{2}*{{\left[ \log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }} \\ & =\frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}} \end{align}$$ (Eq 3.7)
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$$\displaystyle \begin{align} \ddot{y}_{H}^{2}(x) & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right)+\frac{x}{1-{{x}^{2}}} \right]}^{\prime }} \\ & ={{\left[ \frac{1}{2}\log \left( \frac{2}{1-x}-1 \right) \right]}^{\prime }}+{{\left( \frac{x}{1-{{x}^{2}}} \right)}^{\prime }} \\ & =\frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}} \end{align}$$ (Eq 3.8)
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Then by substituting $$\displaystyle \dot{y}_{H}^{2}(x)$$ and $$\displaystyle \ddot{y}_{H}^{2}(x)$$ into $$\displaystyle {{L}_{2}}(y)$$ we have,


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$$\displaystyle \begin{align} {{L}_{2}}(y_{H}^{2}) & =(1-{{x}^{2}})\left( \frac{1}{1-{{x}^{2}}}+\frac{1+{{x}^{2}}} \right)-2x\left( \frac{1}{2}\log M+\frac{x}{1-{{x}^{2}}} \right)+2\left( \frac{x}{2}\log M-1 \right) \\ & =1+\frac{1+{{x}^{2}}}{1-{{x}^{2}}}-x\log M-\frac{2{{x}^{2}}}{1-{{x}^{2}}}+x\log M-2 \\ & =0, \end{align}$$ (Eq 3.9)
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where $$\displaystyle M=\frac{2}{1-x}-1$$

By $$\displaystyle Eq 3.6$$ and $$\displaystyle Eq 3.9$$, we have proved $$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0$$.

=Problem 4: Perform Dimensional analysis on Governing Equation of Force acting along Maglev Deformed Guideway=

Given
Governing Equation of force acting along a Maglev deformed guideway
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$$C_0\left ( Y^1,t \right)=-F^1\left [ 1-\overline{R}{u^2_{,ss}}\left ( Y^1,t \right) \right] -F^2 {u^2_{,s}}\left ( Y^1,t \right) - \frac{T}{R} +M\left [ \left [ 1-\overline{R}{u^2_{,ss}}\left ( Y^1,t \right) \right ] \left [ {u^1_{,tt}}\left ( Y^1,t \right)-\overline{R}{u^2_{,stt}}\left ( Y^1,t \right) \right ]+{u^2_{,s}}\left ( Y^1,t \right){u^2_{,tt}}\left ( Y^1,t \right) \right ]
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$$     (Eq 4.1)
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Note: Argument $$\left ( Y^1, t \right)$$ will be omitted for clarity in subsequent steps.

The following are the known variables of (Eq 4.1):

$$F^{1}$$:vertical force

$$F^{2}$$:horizontal force

$$F^{1}$$:vertical force

$$\overline{R}$$:distance from deformed guide way centerline to wheel centerline

$$ u^2_{,s}$$:slope of deform guide way

$$T$$:Torque exerted on wheel

$$R$$:radius of wheel

$$M$$:mass of wheel

The following are the derived variables of (Eq 4.1):

$$ u^2_{,ss}$$

$$ u^1_{,tt}$$

$$ u^2_{,stt}$$

$$ u^2_{,tt}$$

Find
Verify validity of (Eq 4.1) via “dimensional analysis”

Solution
Dimensional analysis of each variable must be completed for each variable in the equation prior to determining the overall dimension of the equation.


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$$[F^{1}]\longrightarrow$$ $$F$$ (Eq 4.2)
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$$[F^{2}]\longrightarrow$$ $$F$$ (Eq 4.3)
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$$[\overline{R}]\longrightarrow$$ $$L$$ (Eq 4.4)
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$$ [u^2_{,s}]=\frac{\partial{u}^2}{\partial s}\longrightarrow $$ $$\frac{L}{L}$$ $$ \longrightarrow [1] $$ (Eq 4.5)
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$$[T] \longrightarrow$$ $$F\cdot L$$ (Eq 4.6)
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|| <p style="text-align:right"> (Eq 4.7)
 * $$[R]\longrightarrow$$ $$L$$
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(Eq 4.8)
 * $$[M]\longrightarrow $$ $$L$$
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(Eq 4.9)
 * $$ [u^2_{,ss}]=\frac{\partial^2{u}^2}{\partial s^2}\longrightarrow $$$$\frac{L}{{L}^2}$$ $$ \longrightarrow \frac{1}{L}$$
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(Eq 4.10)
 * $$ [u^1_{,tt}]=\frac{\partial^2{u}^1}{\partial s^2}\longrightarrow $$$$\frac{L}{T}^2$$
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(Eq 4.11)
 * $$ [u^2_{,stt}]=\frac{\partial {u}^2}{\partial s\partial^2 {t}}\longrightarrow $$$$\frac{L}{L\cdot{T}^2}$$ $$ \longrightarrow \frac{1}{T}^2$$
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(Eq 4.12)
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The dimensional analysis of the entire equation can now be applied to (Eq 4.1) by substitution of (Eq 4.2) to (Eq 4.12).


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$$ [C_0\left ( Y^1,t \right)]=F\left [ 1-L \frac{1}{L}\right] -F \cdot\frac{L}{L} -\frac{F \cdot L}{L} +M\left [ \left [ 1-L \cdot \frac{1}{L} \right] \left [ \frac{L}{T^2} - \frac{L}{T^2} \right] +1 \cdot \frac{L}{T^2} \right]
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$$     (Eq 4.13)
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In conclusion, it can be deduced by simplification of fractions that the components of equation (Eq 4.13) will reduce down to force ($$F$$) dimensions and mass x acceleration ($$M \cdot \tfrac{L}{T^2}$$), which further reduces to force ($$F$$) dimensions.

=Problem 5:Proof of Nonlinearity=

Given
Show why $${{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}$$ is nonlinear with respect to Y1

Solution
A nonlinear ordinary differential equation (ODE) is as an ODE that is not linear. There are few properties of ODEs that distinguish them as linear. First, the dependent variable and all of the derivatives of the dependent variable are of a degree one. Second, nonlinear functions of the dependent variable or its derivatives aren’t allowed.

In general, linear ODEs must satisfy the following statement:
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$$F\left( \alpha u+\beta v \right)=\alpha F\left( u \right)+\beta F\left( v \right)\displaystyle$$ (5.1)
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Evaluating the left hand side (LHS) of the equation for $$F\left( {{Y}^{1}},t \right)={{c}_{3}}\left( {{Y}^{1}},t \right){{\ddot{Y}}^{1}}\left( t \right)$$ yields:


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$$\begin{align} & F\left( \alpha u+\beta v \right)={{c}_{3}}\left( \alpha u+\beta v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                  }=M\left[ 1-\overline{R}u_{ss}^{2}\left( \alpha u+\beta v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}\displaystyle$$ (5.2)
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An evaluation of the right hand side (RHS) of the equation yields:


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$$\begin{align} & \alpha F\left( u \right)+\beta F\left( v \right)=\alpha {{c}_{3}}\left( u,t \right){{{\ddot{Y}}}^{1}}\left( t \right)+\beta {{c}_{3}}\left( v,t \right){{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=\alpha M\left[ 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right)+\beta M\left[ 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ & \text{                        }=M\left[ \alpha \left( 1-\overline{R}u_{ss}^{2}\left( u,t \right) \right)+\beta \left( 1-\overline{R}u_{ss}^{2}\left( v,t \right) \right) \right]{{{\ddot{Y}}}^{1}}\left( t \right) \\ \end{align}\displaystyle$$ (5.3)
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It is evident that the RHS does not equal the LHS and the expression is nonlinear because


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$$F\left( \alpha u+\beta v \right)\ne \alpha F\left( u \right)+\beta F\left( v \right)\displaystyle$$ (5.4)
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= References =

=Contrubituons Of Team Members=
 * Problem 1              $$\Rightarrow$$ Oztekin,Dube
 * Problem 2/Problem 3    $$\Rightarrow$$ Zou,Abudaram,Sahin (See ZOU'S PAGE)
 * Problem 4              $$\Rightarrow$$ Rasul (See RASUL'S PAGE)
 * Problem 5              $$\Rightarrow$$ Patterson (See PATTERSON'S PAGE)
 * Final Page Designer    $$\Rightarrow$$ Oztekin