text:
User:Egm6321.f10.team5/hw1

=Problem #1=

Given

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

$$\displaystyle f(Y^1 (t),t) $$ $$
 * $$\displaystyle
 * }
 * }

Find
Derive the first and second time derivatives for the given equation shown by Equation 1 and 2 respectively.

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

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\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right) $$ $$\displaystyle (2) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solve
'''Step 1. Verify (1)''' Since $$\displaystyle f(y^1(t),t)$$ is defined as $$f(Y^1(t),t) = \left. f(s,t) \right |_{S=Y^1(t)}$$, we know that $$\displaystyle s$$ is a function of $$\displaystyle t$$.

Therefore, to take a derivative of $$\displaystyle f$$ with respect to $$\displaystyle t$$, we need to use the chain rule for several variables.

 * $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \frac {\partial t} {\partial t}$$ where $$\frac {\partial t} {\partial t} = 1$$.

So we reduce the equation as follows:
 * $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t}$$

Note that
 * $$\frac {\partial s} {\partial t} = \frac {\partial Y^1(t)} {\partial t} = \dot {Y}^1$$.

Expressing the first-order derivative formally, we have
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f} {\partial s}(Y^1(t),t) \dot {Y}^1 + \frac {\partial f} {\partial t}(Y^1(t),t)$$
 * $$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f} {\partial s}(Y^1(t),t) \dot {Y}^1 + \frac {\partial f} {\partial t}(Y^1(t),t)$$

 * }
 * }
 * }

'''Step 2. Verify (2)'''

 * $$\frac {d^2 f} {dt^2} = \frac d {dt} \left ( \frac {df} {dt} \right ) = \frac d {dt} \left (\frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \right ) = \frac {\partial s} {\partial t} \frac d {dt} \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \frac d {dt} \frac {\partial s} {\partial t} + \frac d {dt} \frac {\partial f} {\partial t}$$

Since the chain rule needs to be applied again, we write the equation as follows:
 * $$\frac d {dt} \left ( \frac {df} {dt} \right ) = \frac {\partial s} {\partial t} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial s} {\partial t} + \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial t}$$
 * $$= \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} \frac {\partial^2 s} {\partial s \partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} + \frac {\partial^2 f} {\partial t^2}$$

Because $$\frac {\partial^2 s} {\partial s \partial t} = 0$$ and $$\frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} = 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t}$$, we can simplify the equation as follows:
 * $$\frac {\partial^2 f} {\partial t^2} = \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial^2 f} {\partial t^2}$$:

Expressing the second-order derivative formally, we have
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\frac {d^2 f(Y^1(t),t)} {d t^2} = \frac {\partial^2 f(Y^1(t),t)} {\partial s^2} (\dot {Y}^1)^2 + 2 \frac {\partial^2 f(Y^1(t),t)} {\partial t \partial s} \dot {Y}^1 + \frac {\partial f} {\partial s} \ddot {Y}^1 + \frac {\partial^2 f(Y^1(t),t)} {\partial t^2}$$
 * $$\frac {d^2 f(Y^1(t),t)} {d t^2} = \frac {\partial^2 f(Y^1(t),t)} {\partial s^2} (\dot {Y}^1)^2 + 2 \frac {\partial^2 f(Y^1(t),t)} {\partial t \partial s} \dot {Y}^1 + \frac {\partial f} {\partial s} \ddot {Y}^1 + \frac {\partial^2 f(Y^1(t),t)} {\partial t^2}$$

 * }
 * }
 * }

'''Step 3. Background''' We are refer to the Coliolis effect and revised.

Coliolis Theorem
 * $$ \frac{d\mathbf{r}}{dt} =\frac{d'\mathbf{r}}{dt} +\mathbf{\Omega} \times \mathbf{r} $$

 * $$ \frac{d^2\mathbf{r}}{dt^2}= \frac{d'^2\mathbf{r}}{dt^2} + 2\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt} + \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

Using Newton's laws
 * $$ \mathbf{F} = m\frac{d'^2\mathbf{r}}{dt^2} + 2m\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt} + m\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) + m\frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

 * $$ m\frac{d'^2\mathbf{r}}{dt^2} = \mathbf{F} - \underbrace{2m\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt}}_{Coliolis Force} - \underbrace{m\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})}_{Centrifugal Force} - m\frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

Reynolds transport theorem

We are refer to the Reynolds transport theorem ,

Suppose $$\Omega(t)$$ is a region in Euclidean space with boundary $$\partial \Omega (t)$$, and let $$\mathbf{n}(\mathbf{x},t)$$ be the outward unit normal to the boundary at time $$t$$. Let $$\mathbf{x}(t)$$ be the positions of points in the region, $$\mathbf{v}(\mathbf{x},t)$$ the velocity field in the region, and let $$\mathbf{f}(\mathbf{x},t)$$ be a vector field in the region

\cfrac{\mathrm{d}}{\mathrm{d}t}\left(\int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)} (\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dA} ~. $$

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!Proof Let $$\Omega_0$$ be reference configuration of the region $$\Omega(t)$$. Let the motion and the deformation gradient be given by

\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t)~; \qquad\implies\qquad \boldsymbol{F}(\mathbf{X},t) = \boldsymbol{\nabla}_{\circ} \boldsymbol{\varphi} ~. $$ Let $$J(\mathbf{X},t) = \det[\boldsymbol{F}(\mathbf{X},t)]$$. Then, integrals in the current and the reference configurations are related by

\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV} = \int_{\Omega_0} \mathbf{f}[\boldsymbol{\varphi}(\mathbf{X},t),t]~J(\mathbf{X},t)~\text{dV}_0 = \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0 ~. $$ The time derivative of an integral over a volume is defined as

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t}    \left(\int_{\Omega(t + \Delta t)} \mathbf{f}(\mathbf{x},t+\Delta t)~\text{dV} -            \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) ~. $$ Converting into integrals over the reference configuration, we get

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t}    \left(\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t)~\text{dV}_0 -            \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0\right) ~. $$ Since $$\Omega_0$$ is independent of time, we have

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left[\lim_{\Delta t \rightarrow 0} \cfrac{ \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t) - \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)}{\Delta t} \right]~\text{dV}_0 \\ & = \int_{\Omega_0} \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)]~\text{dV}_0 \\ & = \int_{\Omega_0} \left(         \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+          \hat{\mathbf{f}}(\mathbf{X},t)~\frac{\partial }{\partial t}[J(\mathbf{X},t)]\right) ~\text{dV}_0 \end{align} $$ Now, the time derivative of $$\det\boldsymbol{F}$$ is given by

\frac{\partial J(\mathbf{X},t)}{\partial t} = \frac{\partial }{\partial t}(\det\boldsymbol{F}) = (\det\boldsymbol{F})(\boldsymbol{\nabla} \cdot \mathbf{v}) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\boldsymbol{\varphi}(\mathbf{X},t),t) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t) ~. $$ Therefore,

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left(         \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+          \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right) ~\text{dV}_0 \\ & =     \int_{\Omega_0} \left(\frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]+         \hat{\mathbf{f}}(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~J(\mathbf{X},t) ~\text{dV}_0  \\ & =     \int_{\Omega(t)} \left(\dot{\mathbf{f}}(\mathbf{x},t)+         \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} \end{align} $$ where $$\dot{\mathbf{f}}$$ is the material time derivative of $$\mathbf{f}$$. Now, the material derivative is given by

\dot{\mathbf{f}}(\mathbf{x},t) = \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) ~. $$ Therefore,

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \int_{\Omega(t)} \left(        \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) +         \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} $$ or,

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(        \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} +         \mathbf{f}~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\text{dV} ~. $$ Using the identity

\boldsymbol{\nabla} \cdot (\mathbf{v}\otimes\mathbf{w}) = \mathbf{v}(\boldsymbol{\nabla} \cdot \mathbf{w}) + \boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{w} $$ we then have

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(\frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\right)~\text{dV} ~. $$ Using the divergence theorem and the identity $$(\mathbf{a}\otimes\mathbf{b})\cdot\mathbf{n} = (\mathbf{b}\cdot\mathbf{n})\mathbf{a}$$ we have

{ \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dV} = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dV} \qquad \square } $$
 * }

=Problem #2=

Problem Statement
Do dimensional analysis of all terms in the following equation, and describe the physical meaning.
 * $$c_0 (y^1,t) = - F^1 [1 - \bar {R} u^2_{,ss} (y^1,t)] - F^2 u^2_{,s} - \frac T R + M \lbrace [1 - \bar R u^2_{,ss}] [u^1_{,tt} - \bar R u^2_{,stt}] + u^2_{,s} u^2_{,tt} \rbrace$$

Solution
We first do the dimesional analysis of each variable and state its physical meaning.

$$\displaystyle c_0 = [\frac {ML} {T^2}]$$ : horizontal force acting on wheel

$$\displaystyle F^1 = [\frac {ML} {T^2}]$$ : horizontal component of force

$$\displaystyle \bar R = [L]$$ : rdistance from the center of the guideway to the center of the wheel

$$\displaystyle u^2_{,ss} = [\frac L {L^2}] = [\frac 1 L]$$ : curvature

$$\displaystyle F^2 = [\frac {M L} {T^2}]$$ : vertical component of force

$$\displaystyle u^2_{,s} = [\frac L L] = [1]$$ : slope of transversal deformation(display) of the guideway

$$\displaystyle T = [\frac {M L^2} {T^2}]$$ : torque

$$\displaystyle R = [L]$$ : radius of the wheel

$$\displaystyle M = [M]$$ : mass of the wheel

$$\displaystyle u^1_{,tt} = [\frac L {T^2}]$$ : acceleration of axial deformation (display) of the guideway

$$\displaystyle u^2_{,tt} = [\frac L {T^2}]$$ : acceleration of transversal deformation (display) of the guideway

$$\displaystyle u^2_{,stt} = [\frac L {L T^2}] = [\frac 1 {T^2}]$$ : acceleration of slope of transversal deformation (display) of the guideway

Now we do the complete dimensional analysis for the equation. $$[\frac {ML} {T^2}] = - [\frac {M L} {T^2}] ([1] - [L \frac 1 L] ) - [\frac {M L} {T^2}] [1] - [\frac {M L^2} {T^2} L] + [M \lbrace ([1] - [L \frac 1 L])([\frac L {T^2}] - [L \frac 1 {T^2}]) + [1 \frac L {T^2}] \rbrace$$ $$= - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [M \lbrace ([\frac L {T^2}] - [\frac L {T^2}] - [\frac L {T^2}] + [\frac L {T^2}] + [\frac L {T^2}] \rbrace$$ $$=- [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [\frac {ML} {T^2}] - [\frac {ML} {T^2}] - [\frac {ML} {T^2}] + [\frac {ML} {T^2}] + [\frac {ML} {T^2}]$$ Is it verified that each term in this equation has the dimension of $$[\frac {ML} {T^2}]$$ which is the dimension of force.

=Problem #3=

Given
The vehicle component equation of motion for a Maglev train is represented by a second-order nonlinear partial differential equation (Eq. 2.5e).

The coefficient for the second-order term is defined as follows (assuming $$ I_w = 0 $$):

$$ C_3(Y^1,t) := M \left [ 1 - \bar{R} u,_{SS}^2 (Y^1,t) \right ]^2 $$

Find
Show that $$ C_3 \! $$ is non-linear with respect to $$ Y_1 \! $$.

Solution
If the coefficient $$ C_3 \! $$ is linear, it must satisfy both properties of homogeneity and additivity.

Homogeneity $$ \Rightarrow C_3(\alpha Y^1,t) = \alpha C_3(Y^1,t) $$, where $$ \alpha \! $$ is an arbitrary constant.

$$ C_3(\alpha Y^1,t) = M \left [ 1 - \bar{R} u,_{SS}^2 (\alpha Y^1,t) \right ]^2 = M \left [ 1 - \alpha \bar{R} u,_{SS}^2 (Y^1,t) \right ]^2 \not= \alpha M \left [ 1 - \bar{R} u,_{SS}^2 (Y^1,t) \right ]^2 $$

$$ C_3(\alpha Y^1,t) \not= \alpha C_3(Y^1,t) $$

Since $$ C_3 \! $$ is not homogeneous, it is therefore nonlinear with respect to $$ Y_1 \! $$.

=Problem #4=

Given

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y(x)=cy^1_H(x)+dy^2_H(x)+y_p(x) $$ y(a)=\alpha $$ y(b)=\beta $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

Find
Solve the boundary value problem (BVP), by finding the $$c$$ and $$d$$ terms of $$\alpha$$ and $$\beta$$.

Solve
The function $$y(x)$$ is defined accordingly. $$y(x) = c y_h^1(x) + d y_h^2(x) + y_p(x)$$ Using the boundary value conditions, we get the following relations. $$y(a) = c y_h^1(a) + d y_h^2(a) + y_p(a) = \alpha$$ $$y(b) = c y_h^1(b) + d y_h^2(b) + y_p(b) = \beta$$ Now we need to express $$c$$ and $$d$$ in terms of $$\alpha$$ and $$\beta$$. $$c = \frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} \qquad d = \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)}$$ $$y_h^1(a) + \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)} y_h^2(a) + y_p(a) = c \left (y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)} \right ) + y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} + y_p(a) = \alpha$$ $$c = \frac {\alpha - y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} - y_p(a)} {y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)}} = \frac {\alpha {y_h^2(b)} - y_h^2(a) (\beta - y_p(b)) - y_p(a) {y_h^2(b)}} {y_h^1(a) {y_h^2(b)} - y_h^2(a) {y_h^1(b)}}$$ $$\frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} y_h^1(a) + d y_h^2(a) + y_p(a) = d \left (y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)\right ) + \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) + y_p(a) = \alpha$$ $$d = \frac {\alpha - \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) - y_p(a)} {y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)} = \frac {\alpha y_h^1(b) - (\beta - y_p(b)) y_h^1(a) - y_p(a) y_h^1(b)} {y_h^2(a) y_h^1(b) - y_h^2(b) y_h^1(a)}$$

=Problem #5=

Given

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L_2(y):=(1-x^2) {y}^{''}-2x{y}^{'}+n(n+1)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1)

 * }
 * }

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

n=1 => (1-x^2) {y}^{''}-2x{y}^{'}+2y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (2)

 * }
 * }

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y_H^1=x \equiv P_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (3)
 * }
 * }

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

y_H^2=\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 \equiv Q_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (4)

 * }
 * }

Find

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Verify : {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5)

 * }
 * }

Solve
''' Step1. Verify : $${{L}_{2}}(y_{H}^{1})=0$$'''

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P_1(x)=x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)

 * }
 * }

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{P_1}^{'}(x)=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (7)

 * }
 * }

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{P_1}^{''}(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8)

 * }
 * }

combined (6)~(8) using (2)

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$$\displaystyle {{L}_{2}}(y_{H}^{1})=(1-x^2)(0)-2x(1)+2x=0 $$ $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (9)

 * }
 * }

''' Step2. Verify : $${{L}_{2}}(y_{H}^{2})=0$$'''

 Logarithmic derivative and Quotient rule
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Q_1(x) = \frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (10)

 * }
 * }

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

Q_1'(x) = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{2} (\frac{\frac{(1)(1-x)-(1+x)(-1)}{(1-x)^2}}{\frac{1+x}{1-x}}) = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{1-x^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11)

 * }
 * }

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

Q_1''(x) = \frac{1}{2} (\frac{2}{1-x^2}) + \frac{(1)(1-x^2)-x(-2x)}{(1-x^2)^2} = \frac{2}{(1-x^2)^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (12)

 * }
 * }

combined (10)~(12) using (2)

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$$\displaystyle {{L}_{2}}(y_{H}^{2})=(1-x^2)(\frac{2}{(1-x^2)^2})-2x(\frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{1-x^2})+2(\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1) $$ $$\displaystyle = \frac{2}{1-x^2} - x \log \left( \frac{1+x}{1-x} \right ) - \frac{2x^2}{1-x^2} + x \log \left( \frac{1+x}{1-x} \right ) - 2 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle = \frac{2-2x^2}{1-x^2} - 2 = \frac{2(1-x^2)}{1-x^2} - 2 = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (13)

 * }
 * }

According to (9) and (13), therefore

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

 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * $$\displaystyle

{{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0 $$
 * }
 * }
 * }

=References=

=Contributions=

Raul Riveros Created team wiki site, coordinated content addition, and reviewed problem 4.

Sang Min Oh Solved problem 1 and 5 / Proof-read and revised problem 2 and 4.

Michael Faraone Provided Solution for Problem 4, Checked Problem 1.

Scott Elliot Co-authored problem 2.Egm6321.f10.team5.elliott 20:57, 15 September 2010 (UTC)

Rob Carroll provided secondary solutions for problems 4 and 5. Also reviewed problems 4 and 5.

Jo Solved and typed the derivation and verification parts of problem 1. Solved and typed problem 2 except for the physical meanings. Improved problem 4. Checked problem 5.

Michael Steinberg Solved Problem 3 and proof-read Problem 4.

title: User:Egm6321.f10.team5/hw1


text:
User:Egm6321.f2010.team5.riveros

Hi.

$$\Re$$

HW2 - Problem 1

HW3

 * HW3 Problem 9
 * HW3 Problem 6

HW4

 * HW4 Problem 2
 * HW4 Problem 1

HW5

 * HW5 Problem 4

HW6

 * HW6 Problem 6

HW7

 * HW7 Problem 10

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6321.f2010.team5.riveros


text:
User talk:Egm6321.f10.team5.oh/hw1

= Problem #1 =

Given

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

$$\displaystyle f(Y^1 (t),t) $$ $$
 * $$\displaystyle
 * }
 * }

Find
Derive the first and second time derivatives for the given equation shown by Equation 1 and 2 respectively.

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

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\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right) $$ $$\displaystyle (2) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solve
'''Step 1. Verify (1)''' Since $$\displaystyle f(y^1(t),t)$$ is defined as $$f(Y^1(t),t) = \left. f(s,t) \right |_{S=Y^1(t)}$$, we know that $$\displaystyle s$$ is a fuction of $$\displaystyle t$$.

Therefore, to take a derivative of $$\displaystyle f$$ with respect to $$\displaystyle t$$, we need to use the chain rule for several variables.

$$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \frac {\partial t} {\partial t}$$ where $$\frac {\partial t} {\partial t} = 1$$.

So we reduce the equation as following. $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t}$$

Note that $$\frac {\partial s} {\partial t} = \frac {\partial Y^1(t)} {\partial t} = \dot {Y}^1$$.

Expressing the first-order derivative formally, we have
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$$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f} {\partial s}(Y^1(t),t) \dot {Y}^1 + \frac {\partial f} {\partial t}(Y^1(t),t)$$
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'''Step 2. Verify (2)'''

$$\frac {d^2 f} {dt^2} = \frac d {dt} \left ( \frac {df} {dt} \right ) = \frac d {dt} \left (\frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \right ) = \frac {\partial s} {\partial t} \frac d {dt} \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \frac d {dt} \frac {\partial s} {\partial t} + \frac d {dt} \frac {\partial f} {\partial t}$$

Since the chain rule needs to be applied again, we write the equation as the following. $$\frac d {dt} \left ( \frac {df} {dt} \right ) = \frac {\partial s} {\partial t} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial s} {\partial t} + \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial t}$$ $$= \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} \frac {\partial^2 s} {\partial s \partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} + \frac {\partial^2 f} {\partial t^2}$$ Because $$\frac {\partial^2 s} {\partial s \partial t} = 0$$ and $$\frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} = 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t}$$, we can simplify the equation as following. $$\frac {\partial^2 f} {\partial t^2} = \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial^2 f} {\partial t^2}$$

Expressing the second-order derivative formally, we have
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$$\frac {d^2 f(Y^1(t),t)} {d t^2} = \frac {\partial^2 f(Y^1(t),t)} {\partial s^2} (\dot {Y}^1)^2 + 2 \frac {\partial^2 f(Y^1(t),t)} {\partial t \partial s} \dot {Y}^1 + \frac {\partial f} {\partial s} \ddot {Y}^1 + \frac {\partial^2 f(Y^1(t),t)} {\partial t^2}$$
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 * }
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'''Step 3. Background''' We are refered to Coliolis effect(korean Version) and revised.

Coliolis Theorem $$ \frac{d\mathbf{r}}{dt} =\frac{d'\mathbf{r}}{dt} +\mathbf{\Omega} \times \mathbf{r} $$

$$ \frac{d^2\mathbf{r}}{dt^2}= \frac{d'^2\mathbf{r}}{dt^2} + 2\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt} + \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

Using Newton's laws $$ \mathbf{F} = m\frac{d'^2\mathbf{r}}{dt^2} + 2m\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt} + m\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) + m\frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

$$ m\frac{d'^2\mathbf{r}}{dt^2} = \mathbf{F} - \underbrace{2m\mathbf{\Omega} \times \frac{d'\mathbf{r}}{dt}}_{Coliolis Force} - \underbrace{m\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})}_{Centrifugal Force} - m\frac{d\mathbf{\Omega}}{dt} \times \mathbf{r} $$

Reynolds transport theorem We are refered to Reynolds transport theorem Suppose $$\Omega(t)$$ is a region in Euclidean space with boundary $$\partial \Omega (t)$$, and let $$\mathbf{n}(\mathbf{x},t)$$ be the outward unit normal to the boundary at time $$t$$. Let $$\mathbf{x}(t)$$ be the positions of points in the region, $$\mathbf{v}(\mathbf{x},t)$$ the velocity field in the region, and let $$\mathbf{f}(\mathbf{x},t)$$ be a vector field in the region

\cfrac{\mathrm{d}}{\mathrm{d}t}\left(\int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)} (\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dA} ~. $$

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!Proof Let $$\Omega_0$$ be reference configuration of the region $$\Omega(t)$$. Let the motion and the deformation gradient be given by

\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}, t)~; \qquad\implies\qquad \boldsymbol{F}(\mathbf{X},t) = \boldsymbol{\nabla}_{\circ} \boldsymbol{\varphi} ~. $$ Let $$J(\mathbf{X},t) = \det[\boldsymbol{F}(\mathbf{X},t)]$$. Then, integrals in the current and the reference configurations are related by

\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV} = \int_{\Omega_0} \mathbf{f}[\boldsymbol{\varphi}(\mathbf{X},t),t]~J(\mathbf{X},t)~\text{dV}_0 = \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0 ~. $$ The time derivative of an integral over a volume is defined as

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t}    \left(\int_{\Omega(t + \Delta t)} \mathbf{f}(\mathbf{x},t+\Delta t)~\text{dV} -            \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) ~. $$ Converting into integrals over the reference configuration, we get

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \lim_{\Delta t \rightarrow 0} \cfrac{1}{\Delta t}    \left(\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t)~\text{dV}_0 -            \int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\text{dV}_0\right) ~. $$ Since $$\Omega_0$$ is independent of time, we have

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left[\lim_{\Delta t \rightarrow 0} \cfrac{ \hat{\mathbf{f}}(\mathbf{X},t+\Delta t)~J(\mathbf{X},t+\Delta t) - \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)}{\Delta t} \right]~\text{dV}_0 \\ & = \int_{\Omega_0} \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)]~\text{dV}_0 \\ & = \int_{\Omega_0} \left(         \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+          \hat{\mathbf{f}}(\mathbf{X},t)~\frac{\partial }{\partial t}[J(\mathbf{X},t)]\right) ~\text{dV}_0 \end{align} $$ Now, the time derivative of $$\det\boldsymbol{F}$$ is given by

\frac{\partial J(\mathbf{X},t)}{\partial t} = \frac{\partial }{\partial t}(\det\boldsymbol{F}) = (\det\boldsymbol{F})(\boldsymbol{\nabla} \cdot \mathbf{v}) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\boldsymbol{\varphi}(\mathbf{X},t),t) = J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t) ~. $$ Therefore,

\begin{align} \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) & = \int_{\Omega_0} \left(         \frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]~J(\mathbf{X},t)+          \hat{\mathbf{f}}(\mathbf{X},t)~J(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right) ~\text{dV}_0 \\ & =     \int_{\Omega_0} \left(\frac{\partial }{\partial t}[\hat{\mathbf{f}}(\mathbf{X},t)]+         \hat{\mathbf{f}}(\mathbf{X},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~J(\mathbf{X},t) ~\text{dV}_0  \\ & =     \int_{\Omega(t)} \left(\dot{\mathbf{f}}(\mathbf{x},t)+         \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} \end{align} $$ where $$\dot{\mathbf{f}}$$ is the material time derivative of $$\mathbf{f}$$. Now, the material derivative is given by

\dot{\mathbf{f}}(\mathbf{x},t) = \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) ~. $$ Therefore,

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)~\text{dV}\right) = \int_{\Omega(t)} \left(        \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t} + [\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t)]\cdot\mathbf{v}(\mathbf{x},t) +         \mathbf{f}(\mathbf{x},t)~\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)\right)~\text{dV} $$ or,

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(        \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} +         \mathbf{f}~\boldsymbol{\nabla} \cdot \mathbf{v}\right)~\text{dV} ~. $$ Using the identity

\boldsymbol{\nabla} \cdot (\mathbf{v}\otimes\mathbf{w}) = \mathbf{v}(\boldsymbol{\nabla} \cdot \mathbf{w}) + \boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{w} $$ we then have

\cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)} \left(\frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\right)~\text{dV} ~. $$ Using the divergence theorem and the identity $$(\mathbf{a}\otimes\mathbf{b})\cdot\mathbf{n} = (\mathbf{b}\cdot\mathbf{n})\mathbf{a}$$ we have

{ \cfrac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}~\text{dV}\right) = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dV} = \int_{\Omega(t)}\frac{\partial \mathbf{f}}{\partial t}~\text{dV} + \int_{\partial \Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}~\text{dV} \qquad \square } $$
 * }

Other References 1. Fictitious_force 2. Mechanics of planar particle motion 3. Continuum mechanics 4. Coriolis effects

= Problem #2 =

Given
$$\displaystyle c_0 (y^1,t) = - F^1 [1 - \bar {R} u^2_{,ss} (y^1,t)] - F^2 u^2_{,s} - \frac T R + M \lbrace [1 - \bar R u^2_{,ss}] [u^1_{,tt} - \bar R u^2_{,stt}] + u^2_{,s} u^2_{,tt} \rbrace$$

Find
Do dimensional analysis of all terms in the following equation, and describe the physical meaning.

Solve
''' Step1. Dimesional analysis of each variable'''
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$$c_0 = [\frac {ML} {T^2}], ~ F^1 = [\frac {ML} {T^2}], ~ 1 = [1], ~ \bar R = [L], ~ u^2_{,ss} = [\frac L {L^2}] = [\frac 1 L], ~ F^2 = [\frac {M L} {T^2}]$$ $$u^2_{,s} = [\frac L L] = [1], ~ T = [\frac {M L^2} {T^2}], ~R = [L], ~ M = [M], ~ u^1_{,tt} = [\frac L {T^2}], ~ u^2_{,stt} = [\frac L {L T^2}] = [\frac 1 {T^2}]$$
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 * }
 * }
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Now we do the complete dimensional analysis for the equation. $$[\frac {ML} {T^2}] = - [\frac {M L} {T^2}] ([1] - [L \frac 1 L] ) - [\frac {M L} {T^2}] [1] - [\frac {M L^2} {T^2} L] + [M \lbrace ([1] - [L \frac 1 L])([\frac L {T^2}] - [L \frac 1 {T^2}]) + [1 \frac L {T^2}] \rbrace$$ It is verified that each term in this equation has the dimension of $$[\frac {ML} {T^2}]$$.

'''Step2. Physical Meaning'''

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$$\displaystyle c_0$$ = > horizontal force acting on wheel

$$\displaystyle u^1$$ = > axial deformation(display) of the guideway

$$\displaystyle u^2$$ = > transversal deformation(display) of the guideway

$$\displaystyle F^1$$ = > horizontal component of force

$$\displaystyle F^2$$ = > vertical component of force

$$\displaystyle M$$ = > mass of the wheel

$$\displaystyle R$$ = > radius of the wheel

$$\displaystyle \overline{R}$$ = > distance from the center of the guideway to the center of the wheel

$$\displaystyle T$$ = > torque

$$\displaystyle u^2_{,s}$$ = > slope of transversal deformation(display) of the guideway

$$\displaystyle u^2_{,ss}$$ = > curvature

$$\displaystyle u^1_{,tt}$$ = > acceleration of axial deformation(display) of the guideway

$$\displaystyle u^2_{,tt}$$ = > acceleration of transversal deformation(display) of the guideway

$$\displaystyle u^2_{,stt}$$ = > acceleration of slope of transversal deformation(display) of the guideway

 * }
 * }
 * }

= Problem #3 =

Given
$$c_3(Y^1,t) = M[1 - \bar R u^2_{,ss} (Y^1,t)]$$

Find
$$c_3(Y^1,t) \ddot {Y}^1$$ is nonlinear with respect to $$Y^1$$

Solve
Nonlinearity is defined as $$F(\alpha u + \beta v) \ne \alpha F(u) + \beta F(v)$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R u^2_{,ss} (Y^1,t)] \ddot {Y}^1$$ Given that $$u^2_{,ss} (Y^1,t)$$ is curvature, we define it as $$X$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R X] \ddot {Y}^1$$ Define a differential operator as below. $$F(\cdot) = M (1 - \bar R X) \frac {\partial^2 (\cdot)} {\partial X^2}$$ This operator must satisfty the condition that $$F(\alpha u + \beta v) = \alpha F(u) + \beta F(v)$$ where $$\alpha$$ and $$\beta$$ are real numbers while $$u$$ and $$v$$ are functions of $$X$$. Now we test the condition of linearity. $$F(\alpha u + \beta v) = M (1 - \bar R X) \frac {\partial^2 (\alpha u + \beta v)} {\partial X^2}$$ $$\alpha F(u) + \beta F(v) = \alpha M (1 - \bar R X) \frac {\partial^2 u} {\partial X^2} + \beta M (1 - \bar R X) \frac {\partial^2 v} {\partial X^2}$$

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$$\displaystyle F(\alpha u + \beta v) \ne \alpha F(u) + \beta F(v)$$
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Therefore, $$\displaystyle c_3(Y^1,t) \ddot {Y}^1$$ is nonlinear with respect to $$\displaystyle Y^1$$
 * }
 * }
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=Problem #4=

Given
boundary value problem (BVP)
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y(x)=cy^1_H(x)+dy^2_H(x)+y_p(x) $$ y(a)=\alpha $$ y(b)=\beta $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

Find

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 * Find $$\displaystyle c$$ and $$\displaystyle d$$ in terms of $$\displaystyle \alpha$$ and $$\displaystyle \beta$$.
 * Find $$\displaystyle c$$ and $$\displaystyle d$$ in terms of $$\displaystyle \alpha$$ and $$\displaystyle \beta$$.

 * }
 * }

Solve
$$\displaystyle c = \frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} \qquad d = \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)}$$

$$\displaystyle c y_h^1(a) + \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)} y_h^2(a) + y_p(a) = c \left (y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)} \right ) + y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} + y_p(a) = \alpha$$

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$$c = \frac {\alpha - y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} - y_p(a)} {y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)}} = \frac {\alpha {y_h^2(b)} - y_h^2(a) (\beta - y_p(b)) - y_p(a) {y_h^2(b)}} {y_h^1(a) {y_h^2(b)} - y_h^2(a) {y_h^1(b)}}$$
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

$$\frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} y_h^1(a) + d y_h^2(a) + y_p(a) = d \left (y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)\right ) + \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) + y_p(a) = \alpha$$

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$$d = \frac {\alpha - \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) - y_p(a)} {y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)} = \frac {\alpha y_h^1(b) - (\beta - y_p(b)) y_h^1(a) - y_p(a) y_h^1(b)} {y_h^2(a) y_h^1(b) - y_h^2(b) y_h^1(a)}$$
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 * }
 * }
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= Problem #5 =

Given

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L_2(y):=(1-x^2) {y}^{''}-2x{y}^{'}+n(n+1)y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (1)

 * }
 * }

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n=1 => (1-x^2) {y}^{''}-2x{y}^{'}+2y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (2)

 * }
 * }

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y_H^1=x \equiv P_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (3)
 * }
 * }

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y_H^2=\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 \equiv Q_1(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (4)

 * }
 * }

Find

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Verify : {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5)

 * }
 * }

Solve
''' Step1. Verify : $${{L}_{2}}(y_{H}^{1})=0$$'''

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

P_1(x)=x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)

 * }
 * }

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{P_1}^{'}(x)=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (7)

 * }
 * }

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{P_1}^{''}(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (8)

 * }
 * }

combined (6)~(8) using (2)

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$$\displaystyle {{L}_{2}}(y_{H}^{1})=(1-x^2)(0)-2x(1)+2x=0 $$ $$
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (9)

 * }
 * }

''' Step2. Verify : $${{L}_{2}}(y_{H}^{2})=0$$'''

 Logarithmic derivative and Quotient rule
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Q_1(x) = \frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (10)

 * }
 * }

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

Q_1'(x) = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{2} (\frac{\frac{(1)(1-x)-(1+x)(-1)}{(1-x)^2}}{\frac{1+x}{1-x}}) = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{1-x^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (11)

 * }
 * }

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

Q_1''(x) = \frac{1}{2} (\frac{2}{1-x^2}) + \frac{(1)(1-x^2)-x(-2x)}{(1-x^2)^2} = \frac{2}{(1-x^2)^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (12)

 * }
 * }

combined (10)~(12) using (2)

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

$$\displaystyle {{L}_{2}}(y_{H}^{2})=(1-x^2)(\frac{2}{(1-x^2)^2})-2x(\frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{1-x^2})+2(\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1) $$ $$\displaystyle = \frac{2}{1-x^2} - x \log \left( \frac{1+x}{1-x} \right ) - \frac{2x^2}{1-x^2} + x \log \left( \frac{1+x}{1-x} \right ) - 2 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle = \frac{2-2x^2}{1-x^2} - 2 = \frac{2(1-x^2)}{1-x^2} - 2 = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (13)

 * }
 * }

According to (9) and (13), Therefore

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

$$\displaystyle {{L}_{2}}(y_{H}^{1})={{L}_{2}}(y_{H}^{2})=0 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

title: User talk:Egm6321.f10.team5.oh/hw1


text:
User:Egm6321.f10.team4/HW1

= Problem 1 - Derivation of Equations of Motion for High-Speed Maglev Trains =

From lecture page 1-3 (meeting 1 page 3)

Given
 The nominal motion of the Maglev train can be modeled by the following equation:

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

$$\displaystyle f(Y^1 (t),t) $$ $$
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Derive the below equations and describe the physical meaning.

Axial deformation under moving wheel/guideway

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

{d\over dt}f(Y^1(t),t)=\frac{\partial f}{\partial s}(Y^1(t),t)\dot{Y^1}+\frac{\partial f}{\partial t}(Y^1(t),t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

and

Transverse deformation of wheel/guideway

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

{d^2\over dt^2}f(Y^1(t),t)=\frac{\partial f}{\partial s}(Y^1(t),t)\ddot{Y^1}+\frac{\partial^2 f}{\partial s^2}(Y^1(t),t)(\dot{Y^1})^2+2\frac{\partial^2 f}{\partial st}\dot{Y^1}+\frac{\partial^2 f}{\partial t^2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Axial deformation under moving wheel/guideway
First, equating the function in Equation (1) to a dummy variable $$\displaystyle s $$,

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

$$\displaystyle Y^1(t)=s $$ $$
 * $$\displaystyle (Eq. 4)
 * }
 * }

substituting Equation (4) into the left side of Equation (1) and taking the time derivative,

\frac{d}{dt} f \left(Y^1(t),t\right) = \frac{d}{dt} f \left(s,t\right) $$

By the chain rule, Equation (5) now becomes
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{d}{dt} f \left(s,t\right) = \frac{\partial f}{\partial s} \underbrace {\frac{\partial s}{\partial t}}_{=\dot Y^1(t)}+ \frac{\partial f}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\displaystyle (Eq. 5)
 * }
 * }

Equation (5) can be rewritten by reverting function $$\displaystyle f $$ back to its original form:

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

$$\displaystyle \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 6)
 * }
 * }

Transverse deformation of wheel/guideway
Equation (3) can be rewritten as

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

$$\displaystyle \frac{d^2}{dt^2} f\left(Y^1(t),t\right) = \frac{d}{dt}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) $$ $$
 * $$\displaystyle (Eq. 7)
 * }
 * }

Substituting Equations (4) & (6) into Equation (7) yields,

= \frac{\partial}{\partial s}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) + \frac{\partial}{\partial t}\left( \frac{d}{dt} f\left(Y^1(t),t\right)\right) = \frac{\partial}{\partial s}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t}+ \frac{\partial f}{\partial t} \right) + \frac{\partial}{\partial t}\left( \frac{\partial f}{\partial s} \frac{\partial s}{\partial t} +  \frac{\partial f}{\partial t} \right) $$

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

$$\displaystyle = \frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\displaystyle (Eq. 8)
 * }
 * }

Collecting terms in the Equation (8)

=\frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + 2\frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2} +  \frac{\partial^2 f}{\partial t^2} $$

By transforming the Leivniz expression to Lagrange expression such as X,y instead of dx/dy,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \underbrace{\left(\frac{\partial s}{\partial t}\right)^2}_{\dot {Y^1}^{2}} + 2f_{,st} \underbrace{\frac{\partial s}{\partial t}}_{\dot Y^1} + f_{,s}\underbrace{\frac{\partial^2 s}{\partial t^2}}_{\ddot Y^1} + f_{,tt} $$ $$
 * $$\displaystyle (Eq. 9)
 * }
 * }

Finally, we can complete the desired solution by substituting Equation (4) into Equation (9).
 * {| style="width:60%" border="0"

$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Similaritry to/Difference from Derivation of Coriolis Force
in Rotating coordinate systems(i.e. the earth) To compare second order differential equation to derivation of Coriolis effect, the idea of fictitious forces is used. If the rotation of frame B can be described as a vector Ω pointed along the axis of rotation

 * $$ |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t) \, $$

then the time derivative of the three unit vectors consisting the vector Ω, defined as an observed frame with respect to frame A is

 * $$ \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t) \, $$

and

 * $$\frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \underbrace{\left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)}_{\frac {d \mathbf{u}_j (t)}{dt}}, $$

as is verified using the properties of the vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω ( t ). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting aAB = 0 to remove any translational acceleration, and focusing on only rotational properties

 * $$ \frac {d^2 \mathbf{x}_{A}}{dt^2}=\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} $$ $$+ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\ ,$$
 * $$\mathbf{a}_A=\mathbf{a}_B +\ 2\ \sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t)\ $$ $$+\   \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j  \  + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times  \mathbf{u}_j (t) \right)\ $$
 * $$=\mathbf{a}_B $$ $$+ 2\ \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) \  $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j  \   $$ $$+\ \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \sum_{j=1}^3 x_j  \mathbf{u}_j (t) \right)\ .$$

Collecting terms,

 * $$\mathbf{a}_A=\mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ $$ $$+ \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_B  \  + \boldsymbol{\Omega} \times \left(  \boldsymbol{\Omega} \times \mathbf{x}_B \right)\ .$$

Technically, the aceleration aA in the inertial frame A is not exact with the acceleration aB seen by observers in the rotational frame B but has several additional geometric acceleration terms associated with the rotation of B. Rearranging the above equation by the acceleration B

\mathbf{a}_{B} = \mathbf{a}_A - 2 \boldsymbol\Omega \times \mathbf{v}_{B} - \boldsymbol\Omega \times (\boldsymbol\Omega \times  \mathbf{x}_B )  - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_B \. $$

The force upon the object in the B coordinate system with respect to observers in the rotating frame A is FB = m aB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force Ffict is present, so the end result is FB = FA + Ffict. Thus, the fictitious force used by observers in B to get the correct behavior of the object from Newton's laws equals:

\mathbf{F}_{\mathrm{fict}} = \underbrace{- 2 m \boldsymbol\Omega \times \mathbf{v}_{B}}_{Coriolis force} \underbrace{- m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_B )}_{centrifugal force} $$ $$ \underbrace{\ - m \frac{d \boldsymbol\Omega }{dt} \times \mathbf{x}_B}_{Euler force=0}  \. $$

When the rate of rotation doesn't change, as is typically the case for a planet, the Euler force is zero.

Reference http://en.wikipedia.org/w/index.php?title=Fictitious_force&action=submit

Egm6321.f10.team4.Yoon 00:00, 12 September 2010 (UTC) - Primary Author & Editor

= Problem 2 - Dimensional Analysis=

From lecture page 2-3

Given

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

$$\displaystyle c_0(y^1,t)=-F^1\left[1-\overline{R}\frac{\partial^2}{\partial s^2}u^2 (y^1,t)\right]-F^2\frac{\partial}{\partial s}u^2(y^1,t)-\frac{T}{R}+M \left [ \left(1-\overline{R}\frac{\partial^2 u^2}{\partial s^2}\right)\left(\frac{\partial^2 u^1}{\partial t^2}-\overline{R} \frac{\partial^3 u^2}{\partial st^2} \right)+ \frac{\partial u^2}{\partial s}\frac{\partial^2 u^2}{\partial t^2}\right ] $$ $$
 * $$\displaystyle (Eq. 10)
 * }
 * }

where $$\displaystyle c_0$$ is a force and $$\displaystyle M$$ is the mass of the wheel/magnet

Find
Perform dimensional analysis of all terms of Equation (10).

Solution
If $$\displaystyle c_0(y^1,t)$$ is a force, then each term on the other side of the equation must also resolve into units of force. Each term will here be examined separately. All quantities must therefore be defined before dimensions can be determined.

$$\displaystyle u^1(y^1,t)$$ is the axial deformation of the guideway

$$\displaystyle u^2(y^1,t)$$ is the transverse deformation of the guideway

$$\displaystyle F^1$$ is the horizontal component of force

$$\displaystyle F^2$$ is the vertical component of force

$$\displaystyle M$$ is the mass of the wheel

$$\displaystyle R$$ is the radius of the wheel

$$\displaystyle \overline{R}$$ is a constant distance form the center of the guideway to the center of the wheel

$$\displaystyle T$$ is torque

To carry out our dimensional analysis, the techniques and conventions as outlined in Advanced Mathematics for Engineers will be utilized here. Each term will be broken down into fundamental units of mass (M), length (L), and time (T). The unit (k) will be implemented as a placeholder for dimensionless quantities. The dimensions for our defined quantities are as follows:

$$\displaystyle [u^1(y^1,t)] = L$$

$$\displaystyle [u^2(y^1,t)] = L$$

$$\displaystyle [F^1] = MLT^{-2}$$

$$\displaystyle [F^2] = MLT^{-2}$$

$$\displaystyle [M] = M$$

$$\displaystyle [R] = L$$

$$\displaystyle [\overline{R}] = L$$

$$\displaystyle [T] = ML^2T^{-2}$$

Dimensional analysis of the first term of Equation (10) yields:

$$\displaystyle [F^1[1-\overline{R}\frac{\partial^2}{\partial s^2}u^2 (y^1,t)]]=MLT^{-2}(k - L(L^{-1}))$$

$$\displaystyle =MLT^{-2}(k - k)$$

Since the parenthetical terms ($$\displaystyle k$$) are dimensionless, we are left with only units of force, $$\displaystyle MLT^{-2}$$.

Analysis of the second term of Equation (10) yields:

$$\displaystyle [F^2\frac{\partial}{\partial s}u^2(y^1,t)]=MLT^{-2} (L(L^{-1}))$$

Again, we are left with only units of force, $$\displaystyle MLT^{-2}$$.

Analysis of the third term of Equation (10) yields:

$$\displaystyle [\frac{T}{R}]=ML^2T^{-2} (L^{-1})$$

which also reduces to $$\displaystyle MLT^{-2}$$.

The last term of Equation (10) yields:

$$\displaystyle [M \left \{ \left[ (1-\overline{R}\frac{\partial^2 u^2}{\partial s^2}\right]\left[\frac{\partial^2 u^1}{\partial t^2}-\overline{R} \frac{\partial^3 u^2}{\partial st^2} \right ]+ \frac{\partial u^2}{\partial s}\frac{\partial^2 u^2}{\partial t^2}\right \}]=M((k - L(L^{-1}))(LT^{-2} + kLT^{-2}))$$

reducing to

$$\displaystyle =M (kLT^{-2} + k^{2}LT^{-2})$$

further simplifying to

$$\displaystyle =MLT^{-2}$$, or units of force.

Therefore, each of the terms within our original expression do reduce to units of force. It can then be concluded that $$\displaystyle c_0(y^1,t)$$ must represent a force in our analysis of the Maglev train problem.

68.17.105.125 01:07, 14 September 2010 (UTC)Primary Author: Joseph Osentowski

Egm6321.f10.team4.Auerbach 17:37, 15 September 2010 (UTC) Edited

= Problem 3 - Proof of Nonlinearity =

From lecture page 4-3

Given

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

$$\displaystyle {{c}_{3}}({{y}^{1}},t)=M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right] $$
 * }
 * }

Find
Show that the second order differential equation

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

$$\displaystyle {{c}_{3}}({{y}^{1}},t)\frac{\partial {{t}^{2}}} $$ $$
 * $$ \displaystyle (Eq. 11)
 * }
 * }

is nonlinear with respect to $$\displaystyle  y^1$$.

Solution
For an equation to be linear, it has to satisfy both of the following properties:

Property 1:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha )=\alpha H $$ $$
 * $$\displaystyle (Eq. 12)
 * }
 * }

Property 2:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha+\beta)=H(\alpha)+H(\beta) $$
 * $$\displaystyle (Eq. 13)

$$
 * }
 * }

Based on the first property (Equation (12)), Equation (11) will take the following form

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

$$\displaystyle M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u(\alpha {{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}(\alpha {{y}^{1}})}{\partial {{t}^{2}}}=\alpha M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}} $$ $$
 * $$\displaystyle (Eq. 14)
 * }
 * }

By definition, α is an arbitrary real number which would allow for the following expansion of Equation (14).

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

$$\displaystyle \alpha M\left[ 1-\alpha \bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}}=\alpha M\left[ 1-\bar{R}\frac{{{\partial }^{2}}u({{y}^{1}},t)}{\partial {{s}^{2}}} \right]\bullet \frac{{{\partial }^{2}}({{y}^{1}})}{\partial {{t}^{2}}} $$ $$
 * $$\displaystyle (Eq. 15)
 * }
 * }

It is now clear that the left and right hand sides of the Equation (15) are not equal and that Equation (10) is nonlinear with respect to $$\displaystyle y^1 $$.

128.227.164.111 15:50, 15 September 2010 (UTC)Primary Author: Naiara Petralanda

= Problem 4 - Boundary Value Problem =

From lecture page 5-4

Given

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

$$\displaystyle y(x)=cy_{H}^{1} (x)+dy_{H}^{2} (x)+y_{P} (x) $$ $$
 *
$$\displaystyle (Eq. 16)
 * }
 * }

with the following boundary conditions:

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

$$\displaystyle y(a)=\alpha $$ $$
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(b)=\beta $$ $$
 *
$$\displaystyle (Eq. 18)
 * }
 * }

Find
Find $$\displaystyle c, d$$ in terms of $$\displaystyle \alpha, \beta, y_{H}, y_{P}$$.

Solution
Substituting Equation (17) into Equation (16) yields
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y(a)= \alpha = c y_{H}^{1}(a) + d y_{H}^{2}(a) + y_{P}(a) $$ $$
 *
$$\displaystyle (Eq. 19)
 * }
 * }

Substituting Equation (18) into Equation (16) yields

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

$$\displaystyle y(b)= \beta = c y_{H}^{1}(b) + d y_{H}^{2}(b) + y_{P}(b) $$ $$
 *
$$\displaystyle (Eq. 20)
 * }
 * }

Equations (19) & (20) can be reduced further by subtracting the $$\displaystyle y_{P}$$ term from both sides

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

$$\displaystyle \alpha - y_{P}(a) = c y_{H}^{1}(a) + d y_{H}^{2}(a) $$ $$
 *
$$\displaystyle (Eq. 21)
 * }
 * }

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

$$\displaystyle \beta - y_{P}(b) = c y_{H}^{1}(b) + d y_{H}^{2}(b) $$ $$
 *
$$\displaystyle (Eq. 22)
 * }
 * }

To solve for $$\displaystyle c, d$$, Equations (21) & (22) will be expressed in matrix form:

$$ \begin{bmatrix} {\alpha - y_{P}(a)} \\ {\beta - y_{P}(b)} \\ \end{bmatrix} = \begin{bmatrix} {y_{H}^{1}(a)} & {y_{H}^{2}(a)}\\ {y_{H}^{1}(b)} & {y_{H}^{2}(b)} \\ \end{bmatrix} \begin{bmatrix} {c} \\ {d} \\ \end{bmatrix} $$
$$\displaystyle (Eq. 23) $$

 * Note: For a 2x2 matrix:
 * $$A=\begin{bmatrix}p & q\\r & s\end{bmatrix}$$,
 * The matrix inverse is given by:
 * $$ A^{-1}=\frac{\begin{bmatrix}s & -q\\-r & p\end{bmatrix}}{\left | A \right |} $$,
 * where $$ \left | A \right | = ps-qr $$ is the determinant of matrix A. /// End Note

Multiplying both sides of Equation (23) by the inverse of the 2x2 matrix on the right-hand side yields

$$ \begin{bmatrix} {c} \\ {d} \\ \end{bmatrix}

= \frac{1}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)} \begin{bmatrix} {y_{H}^{2}(b)} & {-y_{H}^{2}(a)}\\ {-y_{H}^{1}(b)} & {y_{H}^{1}(a)} \\ \end{bmatrix} \begin{bmatrix} {\alpha - y_{P}(a)} \\ {\beta - y_{P}(b)} \\ \end{bmatrix} $$

Multiplying the matrices

$$ \begin{bmatrix} {c} \\ {d} \\ \end{bmatrix}

= \frac{1}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)} \begin{bmatrix} {y_{H}^{2}(b)(\alpha - y_{P}(a))-y_{H}^{2}(a)(\beta - y_{P}(b))}\\ {-y_{H}^{1}(b)(\alpha - y_{P}(a))+y_{H}^{1}(a)(\beta - y_{P}(b))}\\ \end{bmatrix} $$

or

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

$$\displaystyle c=\frac{y_{H}^{2}(b)(\alpha - y_{P}(a))-y_{H}^{2}(a)(\beta - y_{P}(b))}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)} $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle d=\frac{-y_{H}^{1}(b)(\alpha - y_{P}(a))+y_{H}^{1}(a)(\beta - y_{P}(b))}{y_{H}^{1}(a)y_{H}^{2}(b)-y_{H}^{1}(b)y_{H}^{2}(a)} $$
 * }
 * }

Egm6321.f10.team4.ejm 20:29, 15 September 2010 (UTC) contributing author

= Problem 5 - Legendre Polynomials =

From lecture page 6-1

Given
The Legendre function
 * {| style="width:100%" border="0" align="left"

$$\displaystyle L_2 (y) := (1-x^2)y''-2xy'+n(n+1)y=0 $$ $$
 *
$$\displaystyle (Eq. 24)
 * }
 * }

Find
For

$$\displaystyle n=1$$

verify that

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

$$\displaystyle L_2 (y^1_H)=L_2 (y^2_H)=0 $$ $$ where
 *
$$\displaystyle (Eq. 25)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y^1_H (x)=x\equiv P_1(x) $$ $$ and
 *
$$\displaystyle (Eq. 26)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y^2_H (x)=\frac{x}{2}log\left ( \frac{1+x}{1-x} \right )-1 \equiv Q_1(x) $$ $$
 *
$$\displaystyle (Eq. 27)
 * }
 * }

Solution
Differentiating Equation (26):

$$\displaystyle {y}^{1}_{H}\left(x \right)=x$$

$$\displaystyle \frac{dy}{dx}=1,$$

$$\displaystyle \frac{d^2y}{dx^2}=0$$

These can then be substituted into the given differential equation (24) to find

$$ L_{2} \left( {y}^{1}_{H}\right)=\left(1-x^2 \right)\cdot \left( 0 \right)-2x\left( 1 \right)+2x=0$$

Differentiating Equation (27):

$$\displaystyle{y}^{2}_{H}\left(x \right)= \frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1$$

$$\displaystyle\frac{dy}{dx}= \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{1}{1+x}+ \frac{1}{1-x}\right)=\frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left(  \frac{x}{1-x^2}\right),  $$

$$\displaystyle\frac{d^2y}{dx^2}= \frac{1}{2} \left( \frac{1}{1+x}+ \frac{1}{1-x} \right)+ \frac{1\cdot \left( 1-x^2\right)-x\cdot \left( -2x\right)}{\left( 1-x^2 \right)^2}=\frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2}$$

Again, substituted into the given differential equation (24) to find

$$\displaystyle L_{2} \left( {y}^{2}_{H}\right)=\left(1-x^2 \right)\cdot \left( \frac{1}{ \left(1-{x}^{2} \right)}+ \frac{1+x^2}{\left( 1-x^2 \right)^2} \right)-2x\left( \frac{1}{2}log \left( \frac{1+x}{1-x} \right)+ \left( \frac{x}{1-x^2}\right) \right)+2\left(\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1\right)=1+\frac{1-x^2}{1-x^2}-2=0$$

Thus,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0 $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * }
 * }

Egm6321.f10.team4.Yoon 00:00, 12 September 2010 (UTC) - Primary Author & Editor

= Contributions=

Egm6321.f10.team4.ejm 14:58, 15 September 2010 (UTC) Co-author of Problem 4; formatted Problems 1-5; edited Problems 1-5

Egm6321.f10.team4.Corella.RL 16:27, 15 September 2010 (UTC) Co-Author of Problem 4; Edited Problems 2-4

Egm6321.f10.team4.Auerbach 17:37, 15 September 2010 (UTC) Edited Problems 1, 2, 4, 5

Egm6321.f10.team4.petralanda.n 14:58, 15 September 2010 (UTC) Author of Problem 1-3

Egm6321.f10.team4.osentowski 17:34, 15 September 2010 (UTC)Author of Problem 2

Egm6321.f10.team4.Yoon 00:00, 12 September 2010 (UTC) - Primary Author & Editor of Prob. 1&5; Edited HW format.

 * Note: This HW report is a collaboration, thus several team members have contributed to several problems. The final product is a result of a group effort.

title: User:Egm6321.f10.team4/HW1


text: File:Hungerappetite.jpg

title: File:Hungerappetite.jpg


text:
User:Slowertech

Currently an electrical engineering student at Cleveland State University.

Also studying German and Chinese as minors.

I am a huge proponent of open source work and non profits that promote open source.

--Slowertech 14:10, 13 September 2010 (UTC)--Slowertech 14:16, 13 September 2010 (UTC)

title: User:Slowertech


text:
User:Egm6321.f10.team5.oh

Team#5 Page

Homework#1

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Homework#5

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title: User:Egm6321.f10.team5.oh


text:
WikipediaOS/Release early, release often

Hypothesis
''Quick processes of development guarantee the possibility of growing articles and programs with the help of many participants. They have this way the opportunity to watch and/or test improvements or mistakes in non definitive versions, then solve them directly or by notifying other active members of the community''.

Votes/comments

 * Alpha, Beta and other software release life cycle indicators could be understand equivalent to stages of articles ("to be expanded", "needs citation", etc) --Esenabre 15:15, 13 September 2010 (UTC)
 * Agree with the wiki part, also that software should be released early to minimize not detecting important problems (verification vs. validation). One problem with too many releases can be: losing trust, because customers thinking why do they do this often, so buggy? Of course it can show an active development, but well... In the end it depends how one communicates this to the outside (e.g. software development process being used, marketing, ...) Erkan Yilmaz uses the Chat 18:06, 19 January 2011 (UTC)


text: File:Amygdalaappetite.jpg

title: File:Amygdalaappetite.jpg


text:
Ideas in Geometry/Probability Paradoxes

Topic:Ideas_in_Geometry Sorry I couldn't figure out how to upload my file so I added a page where you can see my work. Click on the link next to my file and it will take you straight to my work. cgonzal4

title: Ideas in Geometry/Probability Paradoxes


text:
User talk:Egm6321.f10.team6.Kim.MK/hw1

Homework 1 =Problem 3 - Proof of Non-Linearity in EOM of Train Wheel on Flexible Guide Rail=

Problem Statement
The equation of motion of a wheel on flexible guide rail type problem that is consistent with what occurs when the wheel of a high speed train interacts with its guiderail is given as :

$$ {C}_{3} \left(Y^1,t \right) \overset{..}{Y^1} + {C}_{2} \left(Y^1,t \right) (\overset{.}{Y^1})^2 + {C}_{1} \left(Y^1,t \right)\overset{.}{Y^1} + {C}_{0} (Y^1, t) = 0 $$(1)

where the coefficients may be expressed as:

$$\begin{align} & {C}_{0}(Y^1, t) : = - F^1[1- \bar{R} {u^2_{,SS}}(Y^1, t)] - F^2{u^2_{,S}}(Y^1, t) - \frac{T}{R} \\ & {}+ M [[1- \bar{R}{u^2_{,SS}}(Y^1, t)[u_{,tt}^{1}(Y^1,t) -\bar{R}{u_{,Stt}^{2}}(Y^1, t)] \\ & {}+ {u_{,s}^{2}(Y^1, t) u_{,tt}^{2}(Y^1,t)}] \end{align}$$ (2)

$$ C_{1}(Y^1, t) := 2M[[1 -\bar{R}u_{,SS}^{2}(Y^1,t)][u_{,St}^1(Y^1,t)-\bar{R}u_{,SSt}^{2}(Y^1,t)]+u_{,S}^2(Y^1,t)u^2_{,St}(Y^1,t)] $$(3)

$$C_{2}(Y^1, t) := M\left[[1-\bar{R}u_{,SS}^{2}(Y^1,t)][u_{,SS}^1(Y^1,t)-\bar{R}u_{,SSS}^{2}(Y^1,t)]+u_{,S}^2(Y^1,t)u^2_{,SS}(Y^1,t)]\right]$$ (4)

$$ C_{3}(Y^1,t):= M[1-\bar{R}u_{,SS}^{2}(Y^1,t)]^2 + \frac{I_w}{R^2} $$ (5)

Required
Prove the following is non-linear:

$${C}_{3} \left(Y^1,t \right) \overset{..}{Y^1}$$ (6)

Solution
In general a function may be considered linear if the following conditions are met :

 * {| style="width:60%" border="0"

$$f\left( \alpha x \right)=\alpha f\left( x \right)$$ (7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

 * {| style="width:60%" border="0"

$$f\left( {{x}_{1}}\pm {{x}_{2}} \right)=f\left( {{x}_{1}} \right)\pm f\left( {{x}_{2}} \right)$$ (8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

 * {| style="width:60%" border="0"

$${F\left({\alpha}u + {\beta}v\right) = {\alpha} F(u) + {\beta} F(v)}$$  (9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

 * {| style="width:60%" border="0"

$$ F\left( \alpha {Y^1} \right) = \alpha {F ( {Y^1})} $$ (10)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Consider for a moment the differential operator. The differential operator is a linear operator, and as such the above conditions will be met if the the differential operator is applied to an arbitrary function. For example:

First let us examine the first condition of linearity given in eq. 7. According to eq. 7 the result if we multiply the differential operator by some constant C should be exactly the same as if we multiply function by some constant C and then take the derivative.

$$\begin{align} & C\frac{d{f(y)}}{dy}=\frac{d{f(C\cdot y)}}{dy} \\ & C\frac{d\left( 5y \right)}{dy}=C\left( 5 \right)=5C \\ & \frac{d\left( 5Cy \right)}{dy}=5C \\ \end{align}$$

Clearly the placement of the scalar multiple C has no effect on the result, and therefore the differential operator is linear. If we were to apply the second condition of linearity it would also hold.

For this particular problem we will apply the first condition of linearity to prove that the function given in eq. 6 is non-linear. By applying the first condition of linearity using an arbitrary constant α we obtain:

$$\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {({{\alpha}{Y}^{1}},t)}} \right]}^{2}}+\frac \right)\ddot{Y}^{1} {\left( {{\alpha}{Y}^{1}},t \right)}\overset{?}{\mathop{=}}\,{\alpha}\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)}$$ (12)

Since the differential operator is linear we can take the scalar multiply from within the derivative terms to obtain:

$$\left( M{{\left[ 1-{\alpha}\bar{R}u{{_{,ss}^{2}} {({{Y}^{1}},t)}} \right]}^{2}}+\frac \right){\alpha}\ddot{Y}^{1} {\left( {{Y}^{1}},t \right)}\overset{?}{\mathop{=}}\,{\alpha}\left( M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)}$$ (13)

Doing some simple algebraic manipulation we obtain the following:

 * {| style="width:60%" border="0"

$$\left( {\alpha}M{{\left[ 1-{\alpha}\bar{R}u{{_{,ss}^{2}} {({{Y}^{1}},t)}} \right]}^{2}}+{\alpha}\frac \right){\alpha}\ddot{Y}^{1} {\left( {{Y}^{1}},t \right)}{\neq}{\alpha}\left({\alpha} M{{\left[ 1-\bar{R}u{{_{,ss}^{2}} {\left( {{Y}^{1}},t \right)}} \right]}^{2}}+{\alpha}\frac \right) \ddot{Y}^{1}{\left( {{Y}^{1}},t \right)}$$ (14)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

Clearly the left and right sides of eq. 12 don't match (there is second power scalar on the right, but only a first power on the left), therefore eq. 6 is non-linear.

title: User talk:Egm6321.f10.team6.Kim.MK/hw1


text:
User:EGM6321.f10.team6.cook/hw1/chris

= Problem 1 - Derivation of General Equations of Motion for a High Speed Train=

Problem Statement
The general setting for the motion of a high speed train, whether it be a Maglev or wheel on rail, can be given as some function (f) which is a function of a general function of space (s) and time (t).

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

f(s,t) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)

 * }
 * }

We can also say that the function of space (s) is really a function of time, which we shall call Y1(t)

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

s(Y_{1_{(t)}}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)

 * }
 * }

So in general we have the following function:

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

f(s_{{(Y_{1_{(t)}}}),t}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)

 * }
 * }

Based upon the general function above the first and second total derivatives with respect to time may be found. These derivatives are given as:

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

\frac{d}{dt} f \left(Y_{1_{(t)}},t\right) = \frac{\partial f}{\partial s} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + \frac{\partial f}{\partial t} \left(Y_{1_{(t)}},t\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)

 * }
 * }

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

\frac{d^2}{dt^2} f\left(Y_{1_{(t)}},t\right) = f_{s} \left(Y_{1_{(t)}},t\right) \ddot Y_{1_{(t)}} + f_{ss} \left(Y_{1_{(t)}},t\right) (\dot Y_{1_{(t)}})^2 + 2 f_{st} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + f_{tt} \left(Y_{1_{(t)}},t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 5)

 * }
 * }

Required
Derive equations 4 and 5 beginning with the function given in equation 1.

First Derivative WRT Time
We will begin with the first time derivative shown in equation 4.

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

\frac{d}{dt} f \left(s,t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 6)

 * }
 * }

Because f = f(s,t) and s=s(Y1 (t), t) we must apply the chain rule. The reader will likely remember that when applying the chain rule to a function that the result is the sum of the chain rule when applied to each component that depends on a given variable. That is to say that since f not only depends on s and t, but s depends on Y1 and t that the chain rule becomes the additive sum of the derivative of all the terms that are dependent upon t.

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

\frac{d}{dt} f \left(s,t\right) = \underbrace {\frac{\partial f}{\partial s} \frac{\partial s}{\partial t}}_{1} + \underbrace {\frac{\partial f}{\partial t} \frac{\partial t}{\partial t}}_{2}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 7)

 * }
 * }

The component of the partial derivative that is a result of the dependency upon t for both f and s is shown by 1 and the component that is a result of the dependency upon t for f is shown by 2.

Since the partial derivative of t with respect to itself is 1 equation 7 may be rewritten as follows:

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

\frac{d}{dt} f \left(s,t\right) =\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)

 * }
 * }

Additionally, since we have said that s is really just Y1 we can re-write equation 8 in the following form, which matches the form given in the problem statement as equation 4:

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

\frac{d}{dt} f \left(Y_{1_{(t)}},t\right) = \frac{\partial f}{\partial s} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + \frac{\partial f}{\partial t} \left(Y_{1_{(t)}},t\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)

 * }
 * }

Second Derivative WRT Time
We will now use the first derivative with respect to time to find the second derivative with respect to time. First let us define a new function F to be the first time derivative of f(s,t).

F = \frac{df(s,t)}{dt} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)

 * }
 * }

Using this new function F we will again apply the chain rule:

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

\frac{d}{dt} F\left(s,t\right) =\frac{\partial F}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial F}{\partial t}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 11)

 * }
 * }

Now we will substitute the first time derivative of f into equation 11:

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

\frac{d^2}{dt^2} f\left(s,t\right) =\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t} +  \frac{\partial }{\partial t} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} +  \frac{\partial f}{\partial t}\right)
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 12)

 * }
 * }

To make the derivative simpler, we will break it apart into two components and then recombine. First we have:

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

\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 13)

 * }
 * }

Recall that the product rule has the following form:

d\left(f g\right) = gdf +fdg
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 14)

 * }
 * }

By applying the product rule to equation 13 we obtain the following:

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

\frac{\partial }{\partial s} \left(\frac{\partial f}{\partial s} \frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right) \frac{\partial s}{\partial t} =  \left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}
 * $$\displaystyle

$$ || $$\displaystyle (Eq. 15) $$

 * }
 * }

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

\left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}=\frac{\partial^2f}{\partial s^2}\left(\frac{\partial s}{\partial t}\right)^2+\left(\frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s}\right) \frac{\partial s}{\partial t}+\frac{\partial^2 f}{\partial s^2}\frac{\partial s}{\partial t}
 * $$\displaystyle

$$ $$
 * $$\displaystyle (Eq. 16)

 * }
 * }

Since the partial derivative of s with respect to itself the above becomes zero we are left with:

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

\left(\frac{\partial^2f}{\partial s^2} \frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial s \partial t} \frac{\partial f}{\partial s} + \frac{\partial ^2f}{\partial s^2}\right) \frac{\partial s}{\partial t}=\left(\frac{\partial^2 f}{\partial s^2}\right)\left(\frac{\partial s}{\partial t}\right)^2+\left(\frac{\partial^2f}{\partial s \partial t}\right)\left(\frac{\partial s}{\partial t}\right)
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 17)

 * }
 * }

Now we will repeat the above procedure for the second piece:

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

\frac{\partial }{\partial t} \left(\frac{\partial f}{\partial s}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial t}\right)=\frac{\partial^2f}{\partial s \partial t}\frac{\partial s}{\partial t} + \frac{\partial^2s}{\partial t^2}\frac{\partial f}{\partial s} + \frac{\partial^2f}{\partial t^s}
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 18)

 * }
 * }

Now combining equations 17 and 18 we obtain the following:

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

\frac{d^2f}{dt^2}=\frac{\partial^2f}{\partial s^2}\left(\frac{\partial s}{\partial t}\right)^2+2\frac{\partial^2f}{\partial s \partial t}\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s} \frac{\partial^2s}{\partial t^2}+\frac{\partial^2f}{\partial t^2}
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 19)

 * }
 * }

Now substituting Y1 into equation 19 we obtain the following form, which is exactly the same as equation 5:

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

\frac{d^2}{dt^2} f\left(Y_{1_{(t)}},t\right) = f_{s} \left(Y_{1_{(t)}},t\right) \ddot Y_{1_{(t)}} + f_{ss} \left(Y_{1_{(t)}},t\right) (\dot Y_{1_{(t)}})^2 + 2 f_{st} \left(Y_{1_{(t)}},t\right) \dot Y_{1_{(t)}} + f_{tt} \left(Y_{1_{(t)}},t\right)
 * $$\displaystyle
 * $$\displaystyle

$$ $$
 *
$$\displaystyle (Eq. 20)

 * }
 * }

Discussion
The general derivation of this problem is relatively easy, however, familiarity with the chain rule and how it applies to functions of functions is probably the location where most people will have difficulty. A relatively clear and concise explanation of the chain rule may be found at Paul’s Online Math Notes.

title: User:EGM6321.f10.team6.cook/hw1/chris


text:
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User:Egm6321.f10.team4.Yoon

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title: User:Egm6321.f10.team4.Yoon