User:Egm6321.f12.team7/report1

= Problem 1 =  We have solved this problem by ourselves.

Given

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$$\displaystyle\left.f(S,t)\right|_{S=Y^{1}(t)}=f(Y^{1}(t),t)$$  (1.1)
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Find

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$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=f_{,S}(Y^1,t)\ddot{Y}^1+f_{,SS}(Y^1,t)(\dot{Y}^1)^2+2f_{,St}(Y^1,t)\dot{Y^1}+f_{,tt}(Y^1,t)$$, where $$\displaystyle\ f_{,S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$ and  $$\displaystyle\ f_{,St}(Y^1,t):=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}$$  (1.2)
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Solution

 * First, we get the first derivative of $$\displaystyle f$$ with respect to time,
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$$\displaystyle\frac{\mathrm{d} f(Y^1(t),t)}{\mathrm{d} t}=\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$, with  $$\displaystyle\dot{Y}^1:=\frac{\mathrm{d} Y^1(t)}{\mathrm{d} t}$$  (1.3)
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 * Then we get,
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$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=\underbrace{\frac{\mathrm{d} }{\mathrm{d} t}[\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1]}_{\color{blue}{RHS1}}+\underbrace{\frac{\mathrm{d} }{\mathrm{d} t}[\frac{\partial f(Y^1(t),t)}{\partial t}]}_{\color{blue}{RHS2}}$$  (1.4)
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$$RHS1=\frac{\partial f(Y^1(t),t)}{\partial S}\ddot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial S^2}(\dot{Y}^1)^2+\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y}^1$$  (1.5)
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$$RHS2=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial t^2}$$  (1.6)
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 * Bring (1.5) and (1.6) into (1.4), we can get,
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$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=\frac{\partial f(Y^1(t),t)}{\partial S}\ddot{Y}^1+\frac{\partial^2 f(Y^1(t),t)}{\partial S^2}(\dot{Y}^1)^2+2\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\dot{Y^1}+\frac{\partial^2 f(Y^1(t),t)}{\partial t^2}$$  (1.7)
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 * i.e.
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$$\frac{\mathrm{d^2} f}{\mathrm{d} t^2}=f_{,S}(Y^1,t)\ddot{Y}^1+f_{,SS}(Y^1,t)(\dot{Y}^1)^2+2f_{,St}(Y^1,t)\dot{Y^1}+f_{,tt}(Y^1,t)$$, where $$\displaystyle\ f_{,S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$  and  $$\displaystyle\ f_{,St}(Y^1,t):=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}$$ (1.8)
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= Problem 2 =  We have solved this problem by ourselves.

Given
The Equation of Motion (EOM) for the meglev train is modeled by:

Find

 * Get the derivation of the 1st total time derivative.
 * Get the derivation of the 2nd total time derivative.
 * Show the similarity with the derivation of the Coriolis force.

* Get the derivation of the 1st total time derivative
Apply the chain rule to $$\displaystyle f(s,t) $$, we have, Given that for $$\displaystyle s $$ is evaluated at $$\displaystyle Y^1(t) $$ in the Eq. (2-1). Then the equation can be rewritten as, Define the notation as below, As the result, Eq. (2-1) can be simplified to,

* Get the derivation of the 2nd total time derivative
Take the 2nd derivative of the Eq. (2.2) with respect to $$\displaystyle t $$, and apply the chain rule yields. We have, Define the notations as below, Then the Eq. (2.6) can be rewritten as below,

* Show the similarity with the derivation of the Coriolis force
Define $$\displaystyle \mathbf{r} $$ as the position vector indicating the position of the origin of the reference frame, and define $$\displaystyle A $$ is a point fixed on an object rotating with an angular velocity of $$ \displaystyle {}^{N}\boldsymbol{\omega}^{A} $$ with respect to inertial reference frame $$\displaystyle N $$. And define the three mutual perpendicular unit vectors $$\displaystyle \mathbf{i}, \mathbf{j}, \mathbf{k} $$. Then, $$\displaystyle \mathbf{r} $$ can be defined as, The velocity of the object as viewed by an observer fixed to the inertial reference frame $$\displaystyle N $$ is defined as below, The acceleration of the object as viewed by an observer fixed to the inertial reference frame $$\displaystyle N $$ is defined as below, Substitute $$\displaystyle {}^{N} \mathbf{v} $$ from the Eq. (2.10) into the Eq. (2.11),and we have,

Expand the Eq. (2.12) as below. {| style="width:100%" border="0" $$ \displaystyle \begin{align} {}^{N}\mathbf{a} &= \frac{{}^{A}d}{dt} {}^{A}\mathbf{v} + \frac{{}^{A}d}{dt} \left( {}^{N} \boldsymbol{\omega}^{A} \times \mathbf{r} \right)+{}^{N} \boldsymbol{\omega}^{A} \times {}^{A}\mathbf{v} +{}^{N}\boldsymbol{\omega}^{A} \times \left({}^{N} \boldsymbol{\omega}^{A} \times \mathbf{r} \right) \\ &= {}^{A}\mathbf{a} + \frac{{}^{A}d}{dt} {}^{N} \boldsymbol{\omega}^{A}  \times \mathbf{r} + {}^{N} \boldsymbol{\omega}^{A} \times \frac{{}^{A}d\mathbf{r}}{dt} + {}^{N} \boldsymbol{\omega}^{A} \times {}^{A}\mathbf{v} + {}^{N}\boldsymbol{\omega}^{A} \times \left({}^{N} \boldsymbol{\omega}^{A} \times \mathbf{r} \right) \\ &={}^{A}\mathbf{a}+{}^{N}\boldsymbol{\alpha}^{A} \times \mathbf{r} + 2{}^{N}\boldsymbol{\omega}^{A} \times {}^{A}\mathbf{v}+{}^{N}\boldsymbol{\omega}^{A} \times \left({}^{N} \boldsymbol{\omega}^{A} \times \mathbf{r} \right) \\ \end{align} $$ As the result, the Eq.(2.12) can be rewritten as, by the definition,the expression on RHS can also be written as, The Eq. (2-8) and the Eq. (2-13) are related as below,
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=Problem 3=  We have solved this problem by ourselves.

Given
The expression of $$c_{0} ( Y^{1},t)$$ is given by, $$\displaystyle c_{0} ( Y^{1},t)=-F^1(1-\bar{R}u^2_{,SS})-F^2u^2_{,S}-\frac{T}{R}+M[(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})+u^2_{,S}u^2_{tt}]$$


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(3.1)
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Find
Analyze the dimension of each term and provide the physical meaning.

*Dimensional analysis of each term
Left hand side:

$$\displaystyle [c_{0} ( Y^{1},t)]=F $$
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(3.2)
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Right hand side:


 * First term

$$\displaystyle [ F^{1}]=F$$
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(3.3)
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$$\displaystyle [ 1]=1$$
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(3.4)
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$$\displaystyle [ \bar{R}]=L$$
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(3.5)
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$$\displaystyle [ u^2]=L\Rightarrow [u^2_{,SS}]=\frac{L}{L^2}=L^{-1}$$
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(3.6)
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$$\displaystyle [\bar{R}u_{,SS}^{2}]=L\cdot L^{-1}=1$$
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(3.7)
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$$\displaystyle [F^1(1-\bar{R}u^2_{,SS})]=F\cdot 1=F $$
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(3.8)
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 * Second term

$$\displaystyle [F^2]=F$$
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(3.9)
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$$\displaystyle [u^2]=L\Rightarrow [u^2_{,S}]=\frac{L}{L}=1$$
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(3.10)
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$$\displaystyle [F^2u^2_{,S}]=F\cdot{1}=F$$
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(3.11)
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 * Third term

$$\displaystyle [T]=F\cdot{L}$$
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(3.12)
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$$\displaystyle [R]=L$$
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(3.13)
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$$\displaystyle [\frac{T}{R}]=\frac{F\cdot{L}}{L}=F$$
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(3.14)
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 * Fourth term

$$\displaystyle [M]=M$$
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(3.15)
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$$\displaystyle [1-\bar{R}u^2_{,SS}]=1$$
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(3.16)
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$$\displaystyle [u^1_{,tt}]=\frac{L}{T^2}=LT^{-2}$$
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(3.17)
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$$\displaystyle [u^2_{,Stt}]=\frac{L}{L\cdot{T^2}}=T^{-2}\Rightarrow[\bar{R}u^2_{,Stt}]=LT^{-2}$$
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(3.18)
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$$\displaystyle [(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})]=1\cdot{L}\cdot{T^{-2}}=LT^{-2}$$
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(3.19)
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$$\displaystyle [u^2_{,S}]=\frac{L}{L}=1$$
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(3.20)
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$$\displaystyle [u^2_{,tt}]=\frac{L}{T^2}=LT^{-2}$$
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(3.21)
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$$\displaystyle [u^2_{,S}u^2_{,tt}]=1\cdot{L}\cdot{T^{-2}}=LT^{-2}$$
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(3.22)
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$$\displaystyle [M[(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})+u^2_{,S}u^2_{tt}]]=M\cdot{L}\cdot{T^{-2}}=F$$
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(3.23)
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*Physical meaning
$$ c_0$$ is the horizontal force which act on wheel/magnet of the train. Other terms are forces from different source.

=Problem 4= <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color:white; text-align: center;"> We have solved this problem by ourselves.

Given

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The polar coordinate $$\displaystyle (\xi_1,\xi_2)=(r,\theta)$$ <p style="text-align:right">
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Find

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Draw the polar coordinate lines in a 2-D plane emanating from a point, not at the origin. <p style="text-align:right">
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Solution


Suppose a non-origin point $$\displaystyle (C_1,C_2)$$.
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Draw the polar coordinate lines in XY coordinate system from the point $$\displaystyle (C_1,C_2)$$, then we could derive:
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$$\displaystyle \left\{\begin{matrix} x=r\cos\theta+C_1\\y=r\sin\theta+C_2 \end{matrix}\right.$$ <p style="text-align:right"> (4.1)
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$$\displaystyle \Rightarrow\left\{\begin{matrix} (x-C_1)^2+(x-C_2)^2=r^2\\(y-C_1)/(x-C_2)=\tan\theta \end{matrix}\right.$$ <p style="text-align:right"> (4.2)
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According to (4.2), as $$\displaystyle (r,\theta)$$ changes, two sets of lines could be draw as shown in the following figure(4-1):
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=Problem 5= <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color:white; text-align: center;"> We have solved this problem by ourselves.

Given
From the lecture note, we have


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$$\displaystyle \frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$ (5.1)
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Find
Show that equation(5.1) becomes


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$$\displaystyle y''+\underbrace{\frac{g'(x)}{g(x)} }_{a_1(x)}y'+a_0(x)y=0$$ (5.2)
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Solution
According to the Chain Rule, equation (5.1) yields


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$$\displaystyle \frac{1}{g_i(\xi_i)}\left[g'_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}+g_i(\xi_i)\frac{d^2X_i(\xi_i)}{d\xi_i^2}\right]+f_i(\xi_i)X_i(\xi_i)=0.$$ (5.3)
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After simplification, it becomes


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$$\displaystyle X''_i(\xi_i)+\frac{g'_i(\xi_i)}{g_i(\xi_i)}X'_i(\xi_i)+f_i(\xi_i)X_i(\xi_i)=0 $$ (5.4)
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Obviously, When $$\displaystyle y=X_i(\xi_i)$$, equation(5.4) and equation(5.2) are the same.

=Problem 6= <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color:white; text-align: center;"> We have solved this problem by ourselves.

Given
The Equation of Motion of Wheel/Magnet has four terms, which are given by,
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$$\displaystyle C_3(Y^1,t)\ddot Y^1+C_2(Y^1,t)(\dot Y^1)^2+C_1(Y^1,t)\dot Y^1+C_0(Y^1,t)=0$$ (6.1) The general expression of $$\displaystyle C_3(Y^1,t)\ddot Y^1$$ is given by,
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$$\displaystyle c_3(Y^1,t)\ddot Y^1=[1-\bar{R}u_{,ss}(Y^1,t)]\ddot Y^1$$ (6.2) From the lecture note[}, we can Consider $$\displaystyle F(\cdot)$$ as an operator, $$\displaystyle F(\cdot)$$ is linear if and only if,
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$$\displaystyle F(\alpha f+\beta g)=\alpha F(f)+\beta F(g)$$ (6.3) Where$$\displaystyle f$$ and $$\displaystyle g$$ are two possible arguments for $$\displaystyle F(\cdot)$$, and $$\displaystyle \forall \alpha, \beta \in R.$$
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The derivative operator is a linear operator, which is given by,
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$$\displaystyle F(\cdot)=\frac{(\cdot)}{dx}$$ (6.4)
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Find
$$\displaystyle C_3(Y^1,t)\ddot Y^1$$ is nonlinear with repect $$\displaystyle Y^1.$$

Solution
Suppose we have,
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$$\displaystyle Y^1:[t_0,+\infty) \rightarrow R$$
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$$\displaystyle \forall u, v:[t_0,+\infty) \rightarrow R$$

$$\displaystyle \forall \alpha, \beta\in R$$ (6.5) For the right-hand side, we have,
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$$\displaystyle RHS:=\alpha C_3(u,t)\frac{\alpha d^2u}{dt^2}+\beta C_3(v,t)\frac{\beta d^2v}{dt^2}$$ (6.6) The left-hand side is given by,
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$$\displaystyle LHS:=C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}$$ (6.7) Since the derivative operator is linear operator, we can get,
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$$\displaystyle \frac{d^2 (\alpha u+\beta v)}{dt^2} = \frac{\alpha d^2u}{dt^2} +\frac{\beta d^2v}{dt^2}$$ (6.8) So (6.7) turns into,
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$$\displaystyle C_3(\alpha u+\beta v,t)\frac{d^2 (\alpha u+\beta v)}{dt^2}=C_3(\alpha u+\beta v,t)\frac{\alpha d^2u}{dt^2} +C_3(\alpha u+\beta v,t)\frac{\beta d^2v}{dt^2}$$ (6.9) If $$\displaystyle C_3(Y^1,t)$$ is nonlinear with respect to $$\displaystyle Y^1$$, by comparision between (6.5) and (6.8) we can find that,
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$$\displaystyle LHS\neq RHS $$ (6.10) Even if $$\displaystyle C_3(Y^1,t)$$ is linear with respect to $$\displaystyle Y^1$$, on the left-hsnd side we have,
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$$\displaystyle LHS:=[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\alpha d^2u}{dt^2}+[\alpha C_3(u,t)+\beta C_3(v,t)]\frac{\beta d^2v}{dt^2}$$ (6.11) Obviously,
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$$\displaystyle LHS\neq RHS $$ (6.12)
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=Problem 7= <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color:white; text-align: center;"> We have solved this problem by ourselves.

Given
$$\displaystyle L_2(\cdot ) = \frac{d^2 (\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$

Find
Show that $$\displaystyle L_2(\cdot) $$ is linear.

Solution
If $$\displaystyle L_2(\cdot )$$ is linear, then the following must be true: $$\displaystyle L_2(\alpha u +\beta v) = \alpha L_2(u) + \beta L_2(v) $$ $$\displaystyle \forall \alpha, \beta \in \mathbb{R} $$ $$\displaystyle L_2(\alpha u + \beta v) = \frac{d^2 (\alpha u + \beta v)}{dx^2}+a_1(x)\frac{d(\alpha u + \beta v)}{dx}+a_0(x)(\alpha u + \beta v) $$ $$\displaystyle L_2(\alpha u + \beta v)= \alpha u + \beta v +a_1(x)[\alpha u' + \beta v'] + a_0(x)[\alpha u + \beta v] $$ $$\displaystyle L_2(\alpha u + \beta v) = [\alpha u +a_1(x)\alpha u' + a_0(x)\alpha u] + [\beta v +a_1(x)\beta v' + a_0(x)\beta v] $$ $$\displaystyle L_2(\alpha u + \beta v) = \alpha[u +a_1(x)u' + a_0(x)u] + \beta[v +a_1(x)v' + a_0(x)v] $$ $$\displaystyle L_2(\alpha u + \beta v) = \alpha L_2(u) + \beta L_2(v) $$ Thus $$\displaystyle L_2(\cdot) $$ is a linear operator.

= Contribution =