User:Egm6321.f12.team4/Report1

Statement
(3) p.1a-4:



\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1+\frac{\partial f(Y^1(t),t)}{\partial t} $$

Show the second derivative to be



\frac{d^2f}{dt^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) $$

Solution
According to chain rule,


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

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

The first term in (1):



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



= f_{,SS}(Y^1(t),t)(\dot{Y}^1)^2+f_{,St}(Y^1(t),t)\dot{Y}^1 $$

In the second step of it,



f_{,tS}(Y^1(t),t)=f_{,St}(Y^1(t),t) $$

is used.

The second term in (1) is derived as follow:



\frac{\partial f(Y^1(t),t)}{\partial S}{\frac{d}{dt}}(\dot{Y}^1)=f_{,S}(Y^1(t),t)\ddot{Y}^1 $$

Finally, the third term in (1) can be written as:



{\frac{d}{dt}}\left(\frac{\partial f(Y^1(t),t)}{\partial t} \right )=\frac{\partial^2 f(Y^1(t),t)}{\partial S\partial t}\frac{\partial S}{\partial t}+\frac{\partial^2 f(Y^1(t),t)}{\partial t^2} $$



=f_{,St}(Y^1(t),t){\dot Y}^1+f_{,tt}(Y^1(t),t) $$

Then by summing up the above three derived terms, we have:



\frac{d^2f}{dt^2}=f_{,SS}(Y^1(t),t)(\dot{Y}^1)^2+f_{,St}(Y^1(t),t)\dot{Y}^1+f_{,S}(Y^1(t),t)\ddot{Y}^1+f_{,St}(Y^1(t),t){\dot Y}^1+f_{,tt}(Y^1(t),t) $$

Note that there is a repeated term:



f_{,St}(Y^1(t),t){\dot Y}^1 $$

So after reorganization, we can have:



\frac{d^2f}{dt^2}=f_{,S}(Y^1(t),t)\ddot{Y}^1+f_{,SS}(Y^1(t),t)(\dot{Y}^1)^2+2f_{,St}(Y^1(t),t){\dot Y}^1+f_{,tt}(Y^1(t),t), $$

which is the final result. Proved.

Author and References

 * Solved and Typed by -- Che Rui (CH)

Statement


f(S,t)|_{S=Y^1(t)}=f(Y^1(t),t) $$

Derivation of (3) P.1a-4:

\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1+\frac{\partial f(Y^1(t),t)}{\partial t}, $$

(1) P.1a-5:



\frac{d^2f}{dt^2}=f_{,S}(Y^1(t),t)\ddot{Y}^1+f_{,SS}(Y^1(t),t)(\dot{Y}^1)^2+2f_{,St}(Y^1(t),t){\dot Y}^1+f_{,tt}(Y^1(t),t) $$

And the similarity with the derivation of Coriolis force:

\mathbf{F}_c=-2m \mathbf{\omega}_e\times\mathbf{v}_r=-2m(\dot x' \dot {\mathbf{i}}'+\dot y' \dot {\mathbf{j}}'+\dot z' \dot {\mathbf{k}}') $$

Solution
About Coriolis Force Coriolis force is a concept in kinetics or theoretical mechanics. It is due to the Coriolis acceleration:



\mathbf{a}_c=2 \mathbf{\omega}_e\times\mathbf{v}_r=2(\dot x' \dot {\mathbf{i}}'+\dot y' \dot {\mathbf{j}}'+\dot z' \dot {\mathbf{k}}'), $$

which comes from the decomposition of absolute acceleration(in the absolute coordinate) in a moving relative coordinate. Here the vector $$\mathbf{\omega}_e$$ is the angular velocity of the rotation of the relative coordinate in the absolute coordinate. Vector $$\mathbf{v}_{r}=(\dot x',\dot y',\dot z')=\dot x'\mathbf{i}'+\dot y'\mathbf{j}'+\dot z'\mathbf{k}'$$ represents the velocity of a particle in the relative coordinate. So $$\dot x',\dot y',\dot z'$$ are the three components of the relative velocity $$\mathbf{v}_r$$ in the relative coordinate and $$\mathbf{i}',\mathbf{j}',\mathbf{k}'$$ are the corresponding coordinate axis of the relative coordinate respectively. $$\dot{\mathbf{i}'}$$ is the change of $$\mathbf{i}'$$ in unit time.

Derivation of Coriolis Force Consider a particle A represented in a absolute three-dimensional coordinate ($$\mathbf{i},\mathbf{j},\mathbf{k}$$). There is also a moving relative coordinate ($$\mathbf{i}',\mathbf{j}',\mathbf{k}'$$), the origin of which is $$O'$$ and its position in the absolute coordinate is $$\mathbf{r}_{O'}$$:



\mathbf{r}_{O'}=x_{O'}\mathbf{i}+y_{O'}\mathbf{j}+z_{O'}\mathbf{k} $$

So the coordinate of A in the absolute coordinate　can be described by：



\mathbf{r}=x\mathbf{i}+ y\mathbf{j}+ z\mathbf{k}=\mathbf{r}_{O'}+x'\mathbf{i}'+ y'\mathbf{j}'+ z'\mathbf{k}' $$

We first consider the first representation of $$\mathbf{r}$$ above: the absolute velocity of the particle(in the absolute coordinate) is the its derivative with respect to time:



\mathbf{v}=\frac{d}{dt}\mathbf{r}=\frac{d}{dt}(x\mathbf{i}+ y\mathbf{j}+ z\mathbf{k}) $$

Since the absolute coordinate is not moving, the derivative operation on the axis vectors $$\mathbf{i},\mathbf{j},\mathbf{k}$$ will equal to zero, namely:



\frac{d}{dt} \mathbf{i}=\mathbf{0},\frac{d}{dt} \mathbf{j}=\mathbf{0},\frac{d}{dt} \mathbf{k}=\mathbf{0} $$

So the result is:



\mathbf{v}=\dot x\mathbf{i}+ \dot y\mathbf{j}+ \dot z\mathbf{k}, $$

Which is the description of the absolute velocity of particle A by using only the absolute coordinate. Next if we use the second description of $$\mathbf{r}$$ to take the derivation:



\mathbf{v}=\frac{d}{dt}\mathbf{r}=\frac{d}{dt}(\mathbf{r}_{O'}+x'\mathbf{i}'+ y'\mathbf{j}'+ z'\mathbf{k}') $$

Note that the relative coordinate is moving with respect to the absolute coordinate. So the derivation of the axis vector $$\mathbf{i}',\mathbf{j}',\mathbf{k}'$$ with respect to time is not zero:



\frac{d}{dt} \mathbf{i}'=\dot{\mathbf{i}}'\neq {\mathbf{0}},\frac{d}{dt} \mathbf{j}'=\dot{\mathbf{j}}'\neq {\mathbf{0}},\frac{d}{dt} \mathbf{k}'=\dot{\mathbf{k}}'\neq {\mathbf{0}} $$

So the derivation result is found to be:



\mathbf{v}=(\dot{\mathbf{r}}_{O'}+x'\dot{\mathbf{i}}'+y'\dot{\mathbf{j}}'+z'\dot{\mathbf{k}}')+(\dot x'\mathbf{i}'+\dot y'\mathbf{j}'+\dot z'\mathbf{k}'), $$

Where the first term represents the contribution of the moving (translation plus rotation) of the relative coordinate to the absolute velocity of the particle. Here $$\dot{\mathbf{r}}_{O'}=\dot x_{O'}\mathbf{i}+\dot y_{O'}\mathbf{j}+\dot z_{O'}\mathbf{k}$$ is the translational velocity of the relative coordinate, while $$x'\dot{\mathbf{i}}'+y'\dot{\mathbf{j}}'+z'\dot{\mathbf{k}}'$$ is due to the rotation of the relative axis with respect to the origin of the relative coordinate $$O'$$. The second term here is the relative velocity $$\mathbf{v}_{r}=\dot x'\mathbf{i}'+\dot y'\mathbf{j}'+\dot z'\mathbf{k}'$$, which has been introduced in the last section.

Then we take the second derivative of r to get the acceleration with respect to the relative coordinate:



\mathbf{a}=\frac{d^2}{dt^2}\mathbf{r}=\frac{d}{dt}(\mathbf{v}) $$

The process is similar but trivial:



\mathbf{a}=\frac{d}{dt}(\mathbf{v})=\frac{d}{dt}(\dot{\mathbf{r}}_{O'}+x'\dot{\mathbf{i}}'+y'\dot{\mathbf{j}}'+z'\dot{\mathbf{k}}'+\dot x'\mathbf{i}'+\dot y'\mathbf{j}'+\dot z'\mathbf{k}') $$

Again, note that the relative axis is moving, so:



\frac{d}{dt} \mathbf{\dot i}'=\ddot{\mathbf{i}}'\neq {\mathbf{0}},\frac{d}{dt} \mathbf{\dot j}'=\ddot{\mathbf{j}}'\neq {\mathbf{0}},\frac{d}{dt} \mathbf{\dot k}'=\ddot{\mathbf{k}}'\neq {\mathbf{0}} $$

The derivation process is:



\mathbf{a}=\ddot{\mathbf{r}}_{O'}+\left (\frac{d}{dt}(x')\dot{\mathbf{i}}'+x'\frac{d}{dt}(\dot{\mathbf{i}}')+\frac{d}{dt}(y')\dot{\mathbf{j}}'+y'\frac{d}{dt}(\dot{\mathbf{j}}')+\frac{d}{dt}(z')\dot{\mathbf{k}}'+z'\frac{d}{dt}(\dot{\mathbf{k}}') \right ) $$



+\left (\frac{d}{dt}(\dot x')\mathbf{i}'+\dot x'\frac{d}{dt}(\mathbf{i}')+\frac{d}{dt}(\dot y')\mathbf{j}'+\dot y'\frac{d}{dt}(\mathbf{j}')+\frac{d}{dt}(\dot z')\mathbf{k}'+\dot z'\frac{d}{dt}(\mathbf{k}') \right ) $$



=\ddot{\mathbf{r}}_{O'}+\left (\dot x'\dot{\mathbf{i}}'+x'\ddot{\mathbf{i}}'+\dot y'\dot{\mathbf{j}}'+y'\ddot{\mathbf{j}}'+\dot z'\dot{\mathbf{k}}'+z'\ddot{\mathbf{k}}' \right )+\left (\ddot x'\mathbf{i}'+\dot x'\dot {\mathbf{i}'}+\ddot y'\mathbf{j}'+\dot y'\dot {\mathbf{j}'}+\ddot z'\mathbf{k}'+\dot z'\dot {\mathbf{k}'}  \right ) $$

There appear the repeated terms in the last two terms:



\dot x'\dot{\mathbf{i}}',\dot y'\dot{\mathbf{j}}',\dot z'\dot{\mathbf{k}}' $$

So after combination and reorganization, we have:



\mathbf{a}=\left (\ddot{\mathbf{r}}_{O'}+x'\ddot{\mathbf{i}}'+y'\ddot{\mathbf{j}}'+z'\ddot{\mathbf{k}}' \right )+2\left (\dot x'\dot {\mathbf{i}'}+\dot y'\dot {\mathbf{j}'}+\dot z'\dot {\mathbf{k}'}  \right )+\left( \ddot x'\mathbf{i}'+ \ddot y'\mathbf{j}'+ \ddot z'\mathbf{k}'\right ) $$

Here, the first term is again due to the translation and rotation of the relative axis. $$\ddot{\mathbf{r}}_{O'}$$ is the transitional acceleration of the relative coordinate. $$(x'\ddot{\mathbf{i}}'+y'\ddot{\mathbf{j}}'+z'\ddot{\mathbf{k}}')$$ is the contribution due to the rotation of the relative axis in the absolute coordinate. The third term is the relative acceleration, which is the measured acceleration with respect to the relative coordinate.

Finally, I am pleased to introduce the second term, which is the Coriolis acceleration as described at the beginning. Its appearance is due to the fact that in the second derivative of the position of particle A, each of the coordinate value terms $$x',y',z'$$ and the coordinate axis $$\mathbf{i}',\mathbf{j}',\mathbf{k}'$$ is taken derivative of once. So it is the coupling term of the relative velocity of the particle and the rotation velocity of the coordinate axis.

Derivation of (3) p.1a-4 and (1) p.1a-5 Since (1) p.1a-5 has been derived in R1.1, here we show the derivation of (3) p.1a-4:



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



=\frac{\partial f(Y^1(t),t)}{\partial S}\dot{Y}^1+\frac{\partial f(Y^1(t),t)}{\partial t} $$

Similarity with the derivation of the Coriolis force We can recall from answer to R1.1 that the term $$2f_{,St}(Y^1(t),t)\dot{Y}^1$$ is similar to the Coriolis acceleration term $$\mathbf{F}_c=-2m(\dot x' \dot {\mathbf{i}}'+\dot y' \dot {\mathbf{j}}'+\dot z' \dot {\mathbf{k}}')$$, because they are both the repeated terms and are derived from the second derivation. $$2f_{,St}(Y^1(t),t)\dot{Y}^1$$ is derived from taking partial derivative of $$f(Y^1(t),t)$$ with respect to each of t and S once. The Coriolis acceleration, as stated above, is due to each of the coordinate value terms $$x',y',z'$$ and the coordinate axis $$\mathbf{i}',\mathbf{j}',\mathbf{k}'$$ being taken derivative of with respect to time t once. So they are both the coupling terms, which comes from the derivation with respect to two different compositions. In addition, the other terms in the two expressions are also similar in this way.

Author and References

 * Solved and Typed by -- Che Rui (CH)

Statement
Equation of motion (EOM) of wheel/magnet is expressed as

$$C_{3}\left ( Y^{1},t \right )\ddot{Y}+C_{2}\left ( Y^{1},t \right )(\dot{Y}^{1})^{2}+C_{1}\left ( Y^{1},t \right )\dot{Y}^{1}+C_{0}\left ( Y^{1},t \right )=0$$                             (1.3.1)

Where

$$C_{0}\left ( Y^{1},t \right )=-F^{1}[1-\overline{R}U_{SS}^{2}\left ( Y^{1},t \right )]-F^{2}U_{S}^{2}-\frac{T}{R}+M[(1-\overline{R}U_{SS}^{2})(U_{tt}^{1}-\overline{R}U_{Stt}^{2})+U_{S}^{2}U_{tt}^{2}]$$          (1.3.2)

Analyze the dimension of all terms in the equation, and provide the physical meaning.

Solution
 Dimensional analysis: 


 * M: mass, [kg]


 * L: length, [m]


 * T: time, [s]

 First term 


 * $$F^{1}[1-\overline{R}U_{SS}^{2}\left ( Y^{1},t \right )]$$

Where


 * $$[F]=MLT^{-2}$$


 * $$[1]=1$$


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


 * $$[U_{SS}^{2}\left ( Y^{1},t \right )]=L^{-1}$$


 * $$[\overline{R} U_{SS}^{2}\left ( Y^{1},t \right )]=1$$

We have

$$[F^{1}[1-\overline{R}U_{SS}^{2}\left ( Y^{1},t \right )]]= MLT^{-2}$$                       (1.3.3)

 Second term 


 * $$F^{2}U_{S}^{2}$$

Where


 * $$[F^{2}]=MLT^{-2}$$


 * $$[U_{S}^{2}]=1$$

We have

$$[F^{2}U_{S}^{2}]= MLT^{-2}$$                                        (1.3.4)

 Third term 


 * $$\frac{T}{R}$$

Where


 * $$[T]=MLT^{-1}$$


 * $$[R]= L$$

We have

$$[\frac{T}{R}]= MLT^{-2}$$                                           (1.3.5)

 Fourth term 


 * $$M[(1-\overline{R}U_{SS}^{2})(U_{tt}^{1}-\overline{R}U_{Stt}^{2})+U_{S}^{2}U_{tt}^{2}]$$

Where


 * $$[M]=M$$


 * $$[1]=1$$


 * $$[{R}U_{SS}^{2}]=1$$


 * $$[\overline{R}U_{Stt}^{2}]=T^{-2}$$


 * $$U_{S}^{2}U_{tt}^{2}=L T^{-2}$$

We have

$$M[(1-\overline{R}U_{SS}^{2})(U_{tt}^{1}-\overline{R}U_{Stt}^{2})+U_{S}^{2}U_{tt}^{2}]= MLT^{-2}$$       (1.3.6)

 Physical meaning 

We know that, the dimension of the force is $$MLT^{-2}$$, so the equations from (1.3.3),(1.3.4),(1.3.5),(1.3.6) are showing force from different part of the physical aspect.

In conclusion, the external forces F1, F2 and moment T and mass M contribute to each element from the physical aspect.

Author and References

 * Solved and Typed by -- Jinchao Lu (JL)

Statement
Draw the polar coordinate lines $$(\xi_1,\xi_2)=(r,\theta)$$ in a 2-D plane emanating from a point not at the origin.

Solution


The polar coordinates may be represented by concentric circles around the point, which is not the origin (0,0)

Author and References

 * Solved and Typed by -- Jason Alphonso (JA)

Statement
Show that, $$ \frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}] + f_i(\xi_i)X_i(\xi_i) = 0 $$

becomes $$ y'' + \frac{g'(x)}{g(x)}y' + a_0(x)y = 0 $$

Solution
Change the symbols to simplify ,

$$ \xi_i \rightarrow x $$ ; $$ X_i(\xi_i)\rightarrow y(x) $$ ; $$ g_i(\xi_i)\rightarrow g(x) $$ ; $$ f_i(\xi_i)\rightarrow a_0(x) $$.

Hence, the equation becomes,

$$\frac{1}{g(x)} \frac{d}{dx}[g(x)\frac{dy(x)}{dx}] +a_0(x)y(x) = 0 $$

Using chain rule to differentiate the first term on the left hand side,

$$ \frac{1}{g(x)}[g(x)\frac{d^2y(x)}{dx^2}+g'(x)\frac{dy(x)}{dx}]+a_0(x)y(x)=0 $$

Multiplying the terms in square bracket by $$ \frac{1}{g(x)} $$ we get,

$$ \frac{g(x)}{g(x)} \frac{d^2y(x)}{dx^2}+\frac{g'(x)}{g(x)}\frac{dy(x)}{dx}+a_0(x)y(x)=0 $$

Using $$ \frac{g(x)}{g(x)}=1; \frac{d^2y(x)}{dx^2}=y''; \frac{dy(x)}{dx}=y'; y(x)=y $$

It follows that ,

$$ y''+\frac{g'(x)}{g(x)}y'+a_0(x)y=0 $$

Author and References

 * Solved and Typed by -- Pushkar Mishra (PM)

Statement
Show $$ c_3(Y^1,t)\ddot{Y^1} $$ is nonlinear, whether $$ u^2(Y^1,t) $$ is linear or nonlinear with respect to $$ Y^1 $$.

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

Solution
Definition of Linearity (Class note: sec4-6) $$ F(\alpha x + \beta y)=\alpha F(x)+\beta F(y) \, \forall \alpha, \beta \in \mathbb R $$

Let, $$ \alpha, \beta \in \mathbb R $$ and $$ Y^1, Y^2 $$ functions of t.

To check linearity, let's substitute $$ Y^1(t) $$ with $$ \alpha Y^1(t) + \beta Y^2(t) $$ $$ c_3(\alpha Y^1(t) + \beta Y^2(t),t)( \ddot{\alpha Y^1(t) + \beta Y^2(t)} ) = M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] (\ddot{\alpha Y^1(t) + \beta Y^2(t)}) $$ $$ = \alpha M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] \ddot{Y^1(t)} +  \beta M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t)] \ddot{Y^2(t)} $$

Consequently, replacing $$ Y^1(t) $$ with $$ \alpha Y^1(t)$$ and replacing $$ Y^1(t) $$ with $$ \beta Y^2(t)$$ we have following equations: $$ c_3(\alpha Y^1(t),t)( \ddot{\alpha Y^1(t)} ) = M[1-\bar{R}u^2_{,SS}(\alpha Y^1(t),t)] (\ddot{\alpha Y^1(t)}) $$ $$ c_3(\beta Y^2(t),t)( \ddot{\beta Y^2(t)} ) = M[1-\bar{R}u^2_{,SS}(\beta Y^2(t),t)] (\ddot{\beta Y^2(t)}) $$

Looking at the above three equations, in order for the main function to be linear, $$ u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t) = u^2_{,SS}(\alpha Y^1(t),t) $$       (1) and, $$ u^2_{,SS}(\alpha Y^1(t) + \beta Y^2(t),t) = u^2_{,SS}(\beta Y^2(t),t) $$       (2)

Regardless of the linearity of $$ u^2(Y^1,t) $$ for the linearity of the equation $$ c_3(Y^1,t)\ddot{Y^1} $$, above equations must be satisfied. For equation (1) to be satisfied, $$ \beta $$ must be zero. Then the equation (2) cannot be satisfied unless $$ \alpha $$ is also a zero. The equations are not satisfied for any real values of $$ \alpha, \beta $$.

Therefore the equation $$ c_3(Y^1,t)\ddot{Y^1} $$ is nonlinear, regardless of the linearity of $$ u^2(Y^1,t) $$.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong

Statement
Show that $$ L_2(\cdot) $$ is linear

where

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

Solution
$$ L_2(\cdot)$$ is Linear if and only if (iff)

$$ L_2(\alpha u+\beta v)= \alpha L_2(u)+ \beta L_2(v)\, \forall\alpha,\beta \in R $$

where u and v are functions.

LHS

= $$ L_{2}\left ( \alpha u+\beta v \right )$$

= $$\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left ( \alpha u+\beta v \right )+ a_1(x)\frac{\mathrm{d} }{\mathrm{d} x}\left ( \alpha u+\beta v \right )+a_0(x)\left ( \alpha u+\beta v \right )$$

= $$ \alpha \frac{\mathrm{d^2}u }{\mathrm{d} x^2} + \beta \frac{\mathrm{d^2}v }{\mathrm{d} x^2}+ a_1(x)\alpha\frac{\mathrm{d} u}{\mathrm{d} x}+a_1(x)\beta\frac{\mathrm{d} v}{\mathrm{d} x} + a_0(x)\alpha u +a_0(x)\beta v $$

Separating out the $$ \alpha $$ and $$ \beta $$ terms,

= $$\alpha \left [ \frac{\mathrm{d^{2}u} }{\mathrm{d} x^{2}}+a_{1}(x)\frac{\mathrm{d}u }{\mathrm{d} x}+a_{0}(x)u \right ]+ \beta \left [ \frac{\mathrm{d^{2}v} }{\mathrm{d} x^{2}}+a_{1}(x)\frac{\mathrm{d}v }{\mathrm{d} x}+a_{0}(x)v \right ] $$

= $$\alpha L_2\left ( u \right )+\beta L_2\left ( v \right )$$

=RHS, hence linearity is proved.

Author and References

 * Solved and Typed by -- Jason Alphonso (JA)

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.