User:Egm6321.f09.Team1.andy/HW1

Maglev Train: Basic Equations of Motion
Deriving the basic equations of motion for a simple vehicle-structure interaction, based on a high speed Magnetic levitated train moving on a flexible guide way, is the motivation for the course EGM 6321- Principles of Engineering Analysis. Please consider watching a video that depicts the motion of a Maglev [Magnetic Levitated] train.

The figure presented in Prof. Vu Quoc's lecture slide [1-1] shows a 2D diagram of vehicle-structure interaction. The following description for that diagram would help in deriving the basic equations of motion.

Y^1(t) $$ - Nominal Motion [Basically gives us the position of the vehicle if the guide way is rigid]
 * $$\displaystyle F^1, F^2$$ - Force Co-ordinates.
 * $$\displaystyle
 * $$\displaystyle u^1(t) $$ - Axial deformation [displacement]
 * $$\displaystyle u^2(t) $$ - Transversal deformation [displacement]
 * $$\displaystyle s $$ is the co-ordinate and $$\displaystyle Y^1 $$ is the position.

The functional $$\displaystyle f $$ is a function of $$\displaystyle Y^1(t) $$ and $$\displaystyle t $$.

The first total time derivative of $$\displaystyle f(Y^1(t),t) $$ is given by


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\frac{d}{dt}f(Y^1(t,t)) = \frac{\partial}{\partial s}f(Y^1(t),t) * \dot Y^1(t) + \frac{\partial}{\partial t}f(Y^1(t),t) $$
 * $$\displaystyle


 * }
 * }

Writing without the arguments $$\displaystyle Y^1 $$ and $$\displaystyle t $$


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{df}{dt} = \frac{\partial f}{\partial s} * \dot Y^1 + \frac{\partial f}{\partial t} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (1)


 * }
 * }


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{df}{dt} = f,_s(Y^1(t),t) * \dot Y^1 + f,_t(Y^1(t),t) $$
 * $$\displaystyle


 * }
 * }

And without the arguments the first total time derivative is,


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{df}{dt} = f,_s * \dot Y^1 + f,_t $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (2)


 * }
 * }

 Note: Partial derivative of function $$\displaystyle f $$ with respect to $$\displaystyle s $$ with arguments $$\displaystyle Y^1(t) $$ and $$\displaystyle t $$ is given by $$\displaystyle f,_s(Y^1(t),t) $$ 

The second derivative can be obtained as follows:

Equation (1) is given by,
 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{df}{dt} = \frac{\partial f}{\partial s} * \dot Y^1 + \frac{\partial f}{\partial t} $$
 * $$\displaystyle


 * }
 * }

Taking the total time derivative on the Right Hand Side applying chain rule,


 * {| style="width:70%" border="0" align="center"



\frac {d}{dt} \left [\frac{\partial f}{\partial s} * \dot Y^1 \right]= \left [\frac{\partial^2 f}{\partial s^2} * \dot Y^1 + \frac{\partial^2 f}{\partial s \partial t} \right]* \dot Y^1 + \frac{\partial f}{\partial s}* \ddot Y^1 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (3)
 * }
 * }

and,


 * {| style="width:70%" border="0" align="center"



\frac {d}{dt} \left [\frac{\partial f}{\partial t} \right]= \frac{\partial^2 f}{\partial t \partial s}* \dot Y^1 + \frac{\partial^2 f}{\partial t^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (4)
 * }
 * }

Adding (3) and (4),


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{d^2f}{dt^2} = \left [\frac{\partial^2 f}{\partial s^2} * \dot Y^1 + \frac{\partial^2 f}{\partial s \partial t} \right]* \dot Y^1 + \frac{\partial f}{\partial s}* \ddot Y^1 + \frac{\partial^2 f}{\partial t \partial s}* \dot Y^1 + \frac{\partial^2 f}{\partial t^2} $$
 * $$\displaystyle


 * }
 * }


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{d^2f}{dt^2} = \frac{\partial^2 f}{\partial s^2} * (\dot Y^1)^2 + 2 \frac{\partial^2 f}{\partial s \partial t}* \dot Y^1 + \frac{\partial f}{\partial s}* \ddot Y^1 + \frac{\partial^2 f}{\partial t^2} $$
 * $$\displaystyle


 * }
 * }


 * {| style="width:70%" border="0" align="center"



\Rightarrow \frac{d^2f}{dt^2} = f,_{ss} * (\dot Y^1)^2 + 2 f,_{st}* \dot Y^1 + f,_s* \ddot Y^1 + f,_{tt} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (5)
 * }
 * }

Equation (5) describes the equation of motion which is a second order nonlinear partial differential equation.

Integrating Factor Method
General form of a first order Ordinary Differential Equations(ODE) is given by,


 * {| style="width:70%" border="0" align="center"



F(x, y(x),\dot y(x)) = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (6)


 * }
 * }

where, $$\displaystyle \dot y(x) = \frac{dy}{dx} $$

Exactness Conditions
There are two conditions that verify the exactness of a first order nonlinear ODE.

Condition: 1
General form $$\displaystyle F(x, y(x),\dot y(x)) = 0 $$ as in Equation (6) satisfies the exactness condition (1) if $$\displaystyle \exists $$ a function $$\displaystyle \phi(x,y(x)) $$ such that,


 * {| style="width:70%" border="0" align="center"



F(x, y(x),\dot y(x)) = \frac{d\phi}{dx}. $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (7)


 * }
 * }

and Equation (6) has the form,


 * {| style=";width:70%" border="0" align="center"



M(x, y(x)) + N(x,y(x)) * \dot y(x) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (8)


 * }
 * }


 * {| style="width:85%" border="0"




 * valign="top" | Note: 

\begin{align}

\frac{d\phi}{dx} & = \frac{d}{dx} \phi(x,y(x)) \\ & = \underbrace{\frac{\partial \phi(x,y)}{\partial x}}_{M(x, y(x))} + \underbrace{\frac{\partial \phi(x,y)}{\partial y}}_{N(x, y(x))} \left(\frac{dy}{dx}\right)\\ \end{align} \,\!$$


 * }
 * }

Condition: 2
If function $$\displaystyle \phi(x,y(x)) $$ exists then,


 * {| style=";width:70%" border="0" align="center"



M(x, y) = \phi_x(x,y) $$ N(x, y) = \phi_y(x,y) $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

where,


 * {| style=";width:70%" border="0" align="center"



\phi_x(x,y) = \frac{\partial}{\partial x} \phi(x,y) $$ \phi_y(x,y) = \frac{\partial}{\partial y} \phi(x,y) $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

General form $$\displaystyle F(x, y(x),\dot y(x)) = 0 $$ as in Equation (6) satisfies the exactness condition (2) if the above condtion (1) is satisfied and $$\displaystyle \exists $$ a function $$\displaystyle \phi(x,y(x)) $$ such that,


 * {| style="width:70%" border="0" align="center"



M(x, y(x))_y = N(x,y(x))_x \Rightarrow M_y = N_x $$
 * $$\displaystyle



\Rightarrow \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$
 * $$\displaystyle

$$
 * $$\displaystyle \longrightarrow (9)


 * }
 * }

Forming Exact Nonlinear ODEs
$$\displaystyle 1.\ \phi = 6x^4 + 2y^{\frac{3}{2}}$$


 * {| style=";width:70%" border="0" align="center"



\Rightarrow M(x, y) = \phi_x(x,y) = 24x^3 $$ \Rightarrow N(x, y) = \phi_y(x,y) = 6y^{\frac{1}{2}} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

$$\displaystyle 2.\ \phi = 10x^2 + 34y \sqrt{x}$$


 * {| style=";width:70%" border="0" align="center"



\Rightarrow M(x, y) = \phi_x(x,y) = 20x + 34y \frac{1}{\sqrt{x}} $$ \Rightarrow N(x, y) = \phi_y(x,y) = 34 \sqrt{x} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

$$\displaystyle 3.\ \phi = 5xy + \sqrt{xy}$$


 * {| style=";width:70%" border="0" align="center"



\Rightarrow M(x, y) = \phi_x(x,y) = 5y + \frac{\sqrt{y}}{\sqrt{x}} $$ \Rightarrow N(x, y) = \phi_y(x,y) = 5x + \frac{\sqrt{x}}{\sqrt{y}} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

$$\displaystyle 4.\ \phi = 2x^3y^2 + 5\sqrt{x^3y}$$


 * {| style=";width:70%" border="0" align="center"



\Rightarrow M(x, y) = \phi_x(x,y) = 6x^2y^2 + 7.5\sqrt{xy} $$ \Rightarrow N(x, y) = \phi_y(x,y) = 4x^3y + 5\sqrt{\frac{x^3}{y}} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle


 * }
 * }

Euler's Integrating Factors
If, for a given 1st order nonlinear ODE of Equation (6), condition (1) of Exactness is satisfied but not the condition (2), then we use Euler's method to find an Integrating Factor $$\displaystyle h(x,y)$$ with a constraint that the condition (2) is satisfied after multiplying $$\displaystyle h(x,y)$$ to the 1st order nonlinear ODE.


 * {| style=";width:70%" border="0" align="center"



\begin{align} (8) \longrightarrow & M(x, y(x)) + N(x,y(x)) * \dot y(x) = 0 \\ \Rightarrow & M dx + N dy = 0 \end{align} $$
 * $$\displaystyle


 * }
 * }

Including Euler's Integrating Factor $$\displaystyle h(x,y) $$, we get,


 * {| style=";width:70%" border="0" align="center"




 * $$\displaystyle

\Rightarrow \underbrace{(hM)}_\bar{M} dx + \underbrace{(hN)}_\bar{N} dy = 0

$$


 * }
 * }

We need to find the Euler Integrating Factor $$\displaystyle h(x,y) $$ such that,


 * {| style=";width:70%" border="0" align="center"



\Rightarrow \bar{M}_y = \bar{N}_x $$
 * $$\displaystyle



\Rightarrow (hM)_y = (hN)_x $$
 * $$\displaystyle
 * }
 * }

We know, $$\displaystyle \bar{M}_y \Rightarrow (hM)_y = h_y M + h M_y $$ and $$\displaystyle \bar{N}_x \Rightarrow (hN)_x = h_x N + h N_x $$


 * {| style=";width:70%" border="0" align="center"



\Rightarrow h_x N - h_y M + h (N_x - M_y) = 0 $$
 * $$\displaystyle

$$
 * $$\displaystyle \longrightarrow (10)


 * }
 * }

Solving Equation (10) for $$\displaystyle h(x,y) \Rightarrow$$ Not easy in general but Euler gives us 2 cases.

Case: 1
If $$\displaystyle h_y M = 0$$, then Equation (10) becomes,


 * {| style=";width:70%" border="0" align="center"



\Rightarrow h_x N + h (N_x - M_y) = 0 $$
 * $$\displaystyle





\Rightarrow \frac{h_x}{h} = -\frac{1}{N}(N_x - M_y) $$
 * $$\displaystyle





\Rightarrow \underbrace{\int\frac{h_x}{h}dx}_{log\ h} = -\int\underbrace{\frac{1}{N}(N_x - M_y)dx}_{f(x)} $$
 * $$\displaystyle





\Rightarrow h = \exp \left[-\int\frac{1}{N}(N_x - M_y)dx \right] $$
 * $$\displaystyle





\Rightarrow h(x) = \exp \left[-\int_{}^{x} f(s)ds\right] $$
 * $$\displaystyle




 * }


 * {| style="width:85%" border="0"




 * valign="top" | Note: 



h_y M = 0 \Rightarrow h_y = 0,\ since\ M\ \ne\ 0

\,\!$$



\Rightarrow h\ is\ only\ a\ function\ of\ x\,\!$$.




 * }

Case: 2
If $$\displaystyle h_x N = 0$$, then Equation (10) becomes,


 * {| style=";width:70%" border="0" align="center"



\Rightarrow - h_y M + h (N_x - M_y) = 0 $$
 * $$\displaystyle





\Rightarrow \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) = -g(y) $$
 * $$\displaystyle





\Rightarrow \underbrace{\int\frac{h_y}{h}dy}_{log\ h} = \int\frac{1}{M}(N_x - M_y) dy $$
 * $$\displaystyle





\Rightarrow h(y) = \exp \left[\int\frac{1}{M}(N_x - M_y)dy\right] $$
 * $$\displaystyle





\Rightarrow h(y) = \exp \left[-\int_{}^{y} g(s)ds\right] $$
 * $$\displaystyle


 * }
 * }

In Class Problem
To find the solution to the equation $$\displaystyle p' + p - x = 0$$ using Integrating Factors.

Let $$\displaystyle y = p $$ for convenience.

Rewriting in the form of $$\displaystyle M + N\dot y = 0$$,


 * {| style=";width:70%" border="0" align="center"



\underbrace{y - x}_M + \underbrace{1}_N y' = 0 $$
 * $$\displaystyle




 * }

$$\displaystyle \Rightarrow M_y = 1 $$ and $$\displaystyle N_x = 0 $$ and so, $$\displaystyle N_x \ne M_y $$

Using Euler's Integrating Factor $$\displaystyle h(x,y) $$,


 * {| style=";width:70%" border="0" align="center"



h(y - x)+ h y' = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (11)


 * }

Using Case: 1,


 * {| style=";width:70%" border="0" align="center"



\Rightarrow h(x) = \exp \left[-\int\frac{1}{N}(N_x - M_y)dx \right] $$
 * $$\displaystyle





\Rightarrow h(x) = \exp \left[-\int(0 - (1)dx \right] $$
 * $$\displaystyle





\Rightarrow h(x) = e^x $$
 * $$\displaystyle




 * }

which satisfies the fact $$\displaystyle (hM)_y = (hN)_x$$.

Substituting $$\displaystyle h(x) = e^x$$ in Equation (11),


 * {| style=";width:70%" border="0" align="center"



\Rightarrow e^x(y - x)+ e^x y'  = 0 $$
 * $$\displaystyle





\Rightarrow (e^x y)'  = e^x x $$
 * $$\displaystyle


 * }
 * }

Integrating on both sides,


 * {| style=";width:70%" border="0" align="center"



\Rightarrow e^x y  = \int_{}^{x}e^x xdx + A $$
 * $$\displaystyle





\Rightarrow e^x y  = xe^x - \int_{}^x e^x dx + A $$
 * $$\displaystyle





\Rightarrow e^x y  = xe^x - e^x + A $$
 * $$\displaystyle





\Rightarrow y  = x - 1 + Ae^{-x} $$
 * $$\displaystyle




 * }

Nonlinear ODEs
In order to verify whether a given ODE is Nonlinear or not, we use the conditions as per Linear Operators from Wikipedia.

Linearity of any equation is verified using the Linear operator $$\displaystyle L(.)$$ with the condition that,

$$\displaystyle \forall u, v\ functions\ of\ x\ and\ \forall \alpha,\beta \ scalars\ \in \mathbb{R} $$


 * {| style=";width:70%" border="0" align="center"



L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (12)
 * }
 * }

Verification of first order ODEs for non linearity
$$\displaystyle 1.\ 2x^2 + \sqrt y + x^5y^3y' = 0$$

Using the Linear Operator $$\displaystyle D(.)$$ the left hand side of Equation (12) for the given ODE,


 * {| style=";width:70%" border="0" align="center"



D(\alpha u + \beta v) = 2x^2 + \sqrt{(\alpha u + \beta v)} + x^5(\alpha u + \beta v)^3(\alpha u + \beta v)' $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (13)
 * }
 * }

The Right hand side of Equation (12) for the given ODE,


 * {| style=";width:70%" border="0" align="center"



\alpha D(u) + \beta D(v) =  \alpha \left [2x^2 + \sqrt{u} + x^5(u)^3(u)' \right ] +  \beta \left [2x^2 + \sqrt{v} + x^5(v)^3(v)' \right ] $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (14)
 * }
 * }

Since Equations (13) and (14) are not equal, the given equation is a first order non linear ordinary differential equation.

$$\displaystyle 2.\ x^2y^5 + 6(y')^2 = 0$$

Using the Linear Operator $$\displaystyle D(.)$$ the left hand side of Equation (12) for the given ODE,


 * {| style=";width:70%" border="0" align="center"



D(\alpha u + \beta v) = x^2 + (\alpha u + \beta v)^5 + 6[(\alpha u + \beta v)']^2 $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (15)
 * }
 * }

The Right hand side of Equation (12) for the given ODE,


 * {| style=";width:70%" border="0" align="center"



\alpha D(u) + \beta D(v) =  \alpha \left [x^2u^5 + 6u'^2 \right ] +  \beta \left [x^2v^5 + 6v'^2 \right ] $$ $$
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (16)
 * }
 * }

Since Equations (15) and (16) are not equal, the given equation is a first order non linear ordinary differential equation.