User:Egm6321.f10.team5/hw1

=Problem #1=

Given

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


 * {| 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 + 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)


 * }
 * }


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


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


 * }
 * }


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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)


 * }
 * }


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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.