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

= Problem #1 =

Given

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$$\displaystyle f(Y^1 (t),t) $$ $$
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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) $$
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\frac{d^2}{dt^2} f\left(Y^1(t),t\right) = f_{,s} \left(Y^1(t),t\right) \ddot Y^1 + f_{,ss} \left(Y^1(t),t\right) (\dot Y^1)^2 + 2 f_{,st} \left(Y^1(t),t\right) \dot Y^1 + f_{,tt} \left(Y^1(t),t\right) $$ $$\displaystyle (2) $$
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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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'''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 } $$
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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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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


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= 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


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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 $$\displaystyle \beta$$.


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


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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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= 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)


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Find

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


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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)


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


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$$\displaystyle {{L}_{2}}(y_{H}^{1})=(1-x^2)(0)-2x(1)+2x=0 $$ $$
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 * $$\displaystyle (9)


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''' 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)


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


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$$\displaystyle {{L}_{2}}(y_{H}^{2})=(1-x^2)(\frac{2}{(1-x^2)^2})-2x(\frac{1}{2} \log \left( \frac{1+x}{1-x} \right ) + \frac{x}{1-x^2})+2(\frac{x}{2} \log \left( \frac{1+x}{1-x} \right ) - 1) $$ $$\displaystyle = \frac{2}{1-x^2} - x \log \left( \frac{1+x}{1-x} \right ) - \frac{2x^2}{1-x^2} + x \log \left( \frac{1+x}{1-x} \right ) - 2 $$
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$$\displaystyle = \frac{2-2x^2}{1-x^2} - 2 = \frac{2(1-x^2)}{1-x^2} - 2 = 0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (13)


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According to (9) and (13), Therefore


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