User:Egm6321.f10.team4.Auerbach\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


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

Find
Derive the below equations and describe the physical meaning,

Axial deformation under moving wheel/guideway

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{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

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{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 $$\displaystyle s $$,


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

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


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$$\displaystyle \frac{d}{dt} f \left(Y^1(t),t\right) = \frac{\partial f}{\partial s} \left(Y^1(t),t\right) \dot Y^1(t) + \frac{\partial f}{\partial t} \left(Y^1(t),t\right) $$ $$
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 * $$\displaystyle (Eq. 6)
 * }
 * }

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


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


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$$\displaystyle = \frac{\partial^2 f}{\partial s^2} \left(\frac{\partial s}{\partial t}\right)^2 + \frac{\partial^2 f}{\partial s \partial t} \frac{\partial s}{\partial t} \underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial f}{\partial s} \underbrace{\frac{\partial^2 s}{\partial s \partial t}}_{=0}\frac{\partial s}{\partial t} + \frac{\partial f}{\partial s} \frac{\partial^2 s}{\partial t^2}\underbrace{\frac{\partial t}{\partial t}}_{=1} + \frac{\partial^2 f}{\partial s\partial t}\frac{\partial s}{\partial t} + \frac{\partial^2 f}{\partial t^2} \underbrace{\frac{\partial t}{\partial t}}_{=1} $$ $$
 * $$\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,
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$$\displaystyle \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \underbrace{\left(\frac{\partial s}{\partial t}\right)^2}_{\dot {Y^1}^{2}} + 2f_{,st} \underbrace{\frac{\partial s}{\partial t}}_{\dot Y^1} + f_{,s}\underbrace{\frac{\partial^2 s}{\partial t^2}}_{\ddot Y^1} + f_{,tt} $$ $$
 * $$\displaystyle (Eq. 9)
 * }
 * }

Finally, we can complete the desired solution by substituting equation 4 into equation 9.
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$$ \frac{d^2}{dt^2} f\left(Y^1(t),t\right) =f_{,ss} \left(\dot Y^1\right)^2 + 2f_{,st} \dot Y^1 + f_{,s}\ddot Y^1 + f_{,tt} $$
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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


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

Collecting terms,


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

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} $$&ensp;$$ \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

= Problem 2 - Dimensional Analysis=

From lecture page 2-3

Given

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

= Problem 3 - Proof of Nonlinearity =

From lecture page 4-3

Given

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

where


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

Find
Show that the second order differential equation given bellow is non-linear with respect to $$\displaystyle  y^1$$.

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

Property 1:
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$$\displaystyle H(\alpha )=\alpha H $$ $$
 * $$\displaystyle (Eq. 12)
 * }
 * }

Property 2:
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$$\displaystyle H(\alpha+\beta)=H(\alpha)+H(\beta) $$ $$
 * $$\displaystyle (Eq. 13)
 * }
 * }

Based on the first property (equation 12), Equation 10 will take the following form


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

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


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

= Problem 4 - Boundary Value Problem =

From lecture page 5-4

Given

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$$\displaystyle y(x)=cy_{H}^{1} (x)+dy_{H}^{2} (x)+y_{P} (x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 16)
 * }
 * }

with the following boundary conditions (B.C.s):


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$$\displaystyle y(a)=\alpha $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 17)
 * }
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$$\displaystyle y(b)=\beta $$ $$
 * <p style="text-align:right;">$$\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
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$$\displaystyle y(a)= \alpha = c y_{H}^{1}(a) + d y_{H}^{2}(a) + y_{P}(a) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 19)
 * }
 * }

Substituting equation 18 into equation 16 yields


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$$\displaystyle y(b)= \beta = c y_{H}^{1}(b) + d y_{H}^{2}(b) + y_{P}(b) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 20)
 * }
 * }

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


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$$\displaystyle \alpha - y_{P}(a) = c y_{H}^{1}(a) + d y_{H}^{2}(a) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 21)
 * }
 * }


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$$\displaystyle \beta - y_{P}(b) = c y_{H}^{1}(b) + d y_{H}^{2}(b) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 22)
 * }
 * }

To solve for $$\displaystyle c, d$$, equations 21 & 22 will be expressed in matrix form

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

Multiplying both sides by the inverse of the 2 x2 matrix 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


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

= Problem 5 - Legendre Polynomials =

From lecture page 6-1

Given
The Legendre function
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$$\displaystyle L_2 (y) := (1-x^2)y''-2xy'+n(n+1)y=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 23)
 * }
 * }

Find
For

$$\displaystyle n=1$$

verify that


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$$\displaystyle L_2 (y^1_H)=L_2 (y^2_H)=0 $$ $$ where
 * <p style="text-align:right;">$$\displaystyle (Eq. 24)
 * }
 * }
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$$\displaystyle y^1_H (x)=x\equiv P_1(x) $$ $$ and
 * <p style="text-align:right;">$$\displaystyle (Eq. 25)
 * }
 * }
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$$\displaystyle y^2_H (x)=\frac{x}{2}log\left ( \frac{1+x}{1-x}-1 \right ) \equiv Q_1(x) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 26)
 * }
 * }

Solution
Differntiating the equation 25:

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

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

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

These can then be substituted into the given differential equation 23 to find

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

Differntiating the equation 26:

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

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

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

Again, substituted into the given differential equation 23 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,
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$$\displaystyle L_{2} \left( {y}^{1}_{H}\right)=L_{2} \left( {y}^{2}_{H}\right)=0 $$
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 * }
 * }