User:Egm6321.f12.team6/report1

=Report 1=

Problem Statement

 * Show That

\displaystyle\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) $$

Given


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

Nomenclature

 * Short Hand for Derivatives Follow the Examples Below

\displaystyle f_{,S}(Y^1(t),t):=\frac{\partial f(Y^1(t),t)}{\partial S} $$

\displaystyle f_{,St}(Y^1(t),t):=\frac{\partial ^2f(Y^1(t),t)}{\partial S \partial t} $$

\displaystyle \frac{d}{dt}Y^1(t)=\dot Y^1 $$

\displaystyle \frac{d}{dt}f(t)=f'(t) $$


 * Substitutions of the Following are Made

\displaystyle S=Y^1(t)=g(t) $$

Solution

 * Step 1 - we are given

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


 * Step 2 - substitute $$ \displaystyle g(t)=Y^1(t) $$

\displaystyle\frac{df}{dt}=\frac{\partial f(g(t),t)}{\partial g}\dot g+\frac{\partial f(g(t),t)}{\partial t} $$


 * Step 3 -

\displaystyle\frac{d^2f}{dt^2}=\frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial g}\dot g\right)+\frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial t}\right) $$


 * Step 4 - we analyze the problem by parts, and look at, $$ \frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial g}\dot g\right) $$ . We recognize that it's the product of two functions and
 * the product rule applies to its derivation. Furthermore, the first constituent in this product, $$ \displaystyle\frac{\partial f(g(t),t)}{\partial g}\ $$, is a composite function and the chain rule applies. Therefore,
 * $$ \frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial g}\dot g\right)= \left(\frac{\partial ^2f}{\partial g^2}\dot g+\frac{\partial ^2f}{\partial t\partial g}\frac{dt}{dt}\right)\dot g+\ddot g\frac{\partial f}{\partial g} $$


 * Step 5 - The additional part of the equation, $$ \frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial t}\right) $$, is recognized as a composite function and the chain rule applies.
 * Therefore,
 * $$ \frac{d}{dt}\left(\frac{\partial f(g(t),t)}{\partial t}\right)=\frac{\partial ^2f}{\partial t\partial g}\dot g+ \frac{\partial ^2f}{\partial t^2}\frac{dt}{dt}$$


 * Step 6 - Placing the pieces from the fourth and fifth steps back together gives

\displaystyle\frac{d^2f}{dt^2}=\left(\frac{\partial ^2f}{\partial g^2}\dot g+\frac{\partial ^2f}{\partial t\partial g}\frac{dt}{dt}\right)\dot g+\ddot g\frac{\partial f}{\partial g}+\frac{\partial ^2f}{\partial t\partial g}\dot g+ \frac{\partial ^2f}{\partial t^2}\frac{dt}{dt} $$


 * Step 7 - Changing the nomenclature as defined in the nomenclature section gives

\displaystyle\frac{d^2f}{dt^2}=\left(f_{,gg}\dot g+f_{,gt}\frac{dt}{dt}\right)\dot g+\ddot g f_{,g}+f_{,gt}\dot g+ f_{,tt}\frac{dt}{dt} $$


 * Step 8 - Distributing, combining like terms, and recognizing that, $$ \frac{dt}{dt}=1 $$, we get
 * $$ \displaystyle\frac{d^2f}{dt^2}=f_{,g} \ddot g+f_{,gg}(\dot g)^2+2f_{,gt}\dot g+f_{,tt} $$


 * Step 9 - Substitute, $$ \displaystyle Y^1(t)=g(t) $$, also recognize that, $$ \displaystyle S=Y^1(t) $$ , and by writing the function, $$ f $$ , in its formal form of,
 * $$ \displaystyle f=f(Y^1(t),t) $$, we arrive at our concluding equation,
 * {| style="width:100%" border="0"|-

$$ \displaystyle\frac{d^2f}{dt^2}=f_{,S}(Y^1(t),t)\ddot Y^1+f_{,SS}(Y^1(t),t)(\dot Y^1)^2+2f_{,St}(Y^1(t),t)\dot Y^1+f_{,tt}(Y^1(t),t) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * Derive
 * a) $$\displaystyle\;$$$$\displaystyle\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$
 * and
 * b) $$\displaystyle\;$$ $$\displaystyle\frac{d^2f}{dt^2}=f_{,S}(Y^1(t),t)\ddot Y^1+f_{,SS}(Y^1(t),t)(\dot Y^1)^2+2f_{,St}(Y^1(t),t)\dot Y^1+f_{,tt}(Y^1(t),t)$$,
 * and
 * c) $$\displaystyle\;$$ show the similarity with the derivation of the Coriolis force.

Given

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

Nomenclature

 * $$\displaystyle \dot{Y}^1:=\frac{dY^1(t)}{dt}$$
 * $$\displaystyle f,_{S}(Y^1,t):=\frac{\partial f(Y^1(t),t)}{\partial S}$$
 * $$\displaystyle f,_{S\,t}(Y^1(t),t):=\frac{\partial ^2f(Y^1(t),t)}{\partial S\partial t}$$

Part a)

 * By using Chain Rule,
 * $$\displaystyle \frac{d}{d t}f(S,t)=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}\cancelto{1}\frac{d t}{d t}=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}$$.
 * Also,
 * $$\displaystyle S=Y^{^{1}}(t)$$
 * is a given condition. This implies that
 * $$\displaystyle\frac{d S}{d t}=\frac{dY^1(t)}{dt}=:\dot{Y^1}$$.
 * Therefore, when the equation is rewritten in terms of $$\displaystyle f(Y^{^{1}}(t),t)$$, then the equation for $$\displaystyle\frac{d}{dt}f(Y^1(t),t)$$ becomes following:
 * {| style="width:100%" border="0"|-

$$\displaystyle\;$$ $$\displaystyle\frac{d}{dt}f(Y^1(t),t)=\frac{\partial f(Y^1(t),t)}{\partial S}\dot Y^1+\frac{\partial f(Y^1(t),t)}{\partial t}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Part b)

 * By using Part a) derivation
 * $$\displaystyle \frac{d }{d t}f(S,t)=\frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}$$,
 * The second derivate of $$\displaystyle f(Y^1(t),t)$$, can be rewritten in following:
 * $$\displaystyle \frac{d^2 f}{d t^2}=\frac{d}{d t} \left [ \frac{d}{d t} f(S,t) \right ]=\frac{d}{d t} \left [ \frac{\partial f}{\partial S}\frac{d S}{d t}+\frac{\partial f}{\partial t}\right ]$$
 * Then, the expression above can expanded by using Chain and Product Rule.
 * $$\displaystyle \frac{d^2 f}{d t^2} = \frac{\partial f}{\partial S}\frac{d}{d t}\left ( \frac{d S}{d t} \right )+\frac{d }{d t} \left ( \frac{\partial f}{\partial S} \right )\frac{d S}{d t}+\frac{d}{d t}\left ( \frac{\partial f}{\partial t} \right )$$
 * $$\displaystyle = \displaystyle \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \left ( \frac{\partial^2 f}{\partial S^2} \frac{d S}{d t} + \frac{\partial^2 f}{\partial S \, \partial t} \cancelto{1}\frac{d t}{d t} \right )\frac{d S}{d t} + \left ( \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2} \cancelto{1}\frac{d t}{d t} \right ) \cancelto{1}\frac{d t}{d t}$$
 * $$\displaystyle = \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \frac{\partial^2 f}{\partial S^2} \left ( \frac{d S}{d t} \right )^2 + \frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2}$$
 * Also, in this case,
 * $$\displaystyle\frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} = \frac{\partial^2 f}{\partial t \, \partial S} \frac{d S}{d t}$$
 * Adding both terms will give
 * $$ \frac{d^2 f}{d t^2} = \frac{\partial f}{\partial S}\frac{d^2 S}{d t^2} + \frac{\partial^2 f}{\partial S^2} \left ( \frac{d S}{d t} \right )^2 + 2\frac{\partial^2 f}{\partial S \, \partial t} \frac{d S}{d t} + \frac{\partial^2 f}{\partial t^2} $$.
 * By using $$\displaystyle f(Y^{^{1}}(t),t)$$, terms and different nonmenclature technique, then equation becomes following:
 * {| style="width:100%" border="0"|-

$$\displaystyle\;$$ $$\displaystyle\frac{d^2f}{dt^2} = f_{,S}(Y^1(t),t)\ddot Y^1 + f_{,SS}(Y^1(t),t)(\dot Y^1)^2 + 2f_{,St}(Y^1(t),t)\dot Y^1 + f_{,tt}(Y^1(t),t) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

====Part c) ====
 * Consider a two frames of reference centered at O, a fixed frame OXY and frame Oxy which rotates about the fixed axis OA; let $$\displaystyle \Omega$$ denote the angular velocity of the frame Oxy at a give instant.
 * $$ \displaystyle\boldsymbol{r}$$ can be expressed in vector form:
 * $$ \displaystyle\boldsymbol{r} = r_{x}\boldsymbol{i} + r_{y}\boldsymbol{j} $$
 * Differentiating with repect to t and considering the unit vectors as fixed, rate of change of $$\displaystyle\boldsymbol{r}$$ with respect to the rotating frame Oxy can be obtained.
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{Oxy} = \dot{r}_{x}\boldsymbol{i} + \dot{r}_{y}\boldsymbol{j} $$
 * To obtain the rate of change of $$\displaystyle\boldsymbol{r}$$ with respect to the fixed frame OXY, we must consider the unit vectors as variable when differentiating the above equation. Therefore the equation becomes
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \dot{r}_{x}\boldsymbol{i} + \dot{r}_{y}\boldsymbol{j} + r_{x}\frac{d\boldsymbol{i}}{dt} + r_{y}\frac{d\boldsymbol{j}}{dt} $$
 * Following equation can be simplified into
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \left (\boldsymbol{\dot{r}}\right)_{Oxy} + \boldsymbol{\Omega \times r} $$
 * or
 * $$ \displaystyle \left (\boldsymbol{\dot{r}}\right )_{OXY} = \boldsymbol{\Omega \times r} + \left (\boldsymbol{\dot{r}}\right)_{Oxy} $$
 * From above equation, rate of change with respect to OXY of the terms will give absolute acceleration.
 * $$ \displaystyle \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right )_{OXY} \right ] = \left (\boldsymbol{\ddot{r}}\right)_{OXY} = \frac{d}{dt} \left [ \boldsymbol{\Omega \times r}\right ] + \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right)_{Oxy}\right ] $$
 * $$ = \boldsymbol{\dot{\Omega} \times r} + \boldsymbol{\Omega \times \dot{r}} + \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right)_{Oxy}\right ] $$
 * Similary with previous part, we can investigate that
 * $$ \displaystyle \frac{d}{dt}\left [ \left (\boldsymbol{\dot{r}}\right )_{Oxy} \right ] = \left (\boldsymbol{\ddot{r}}\right)_{Oxy} + \boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} $$
 * Therefore, the equation for coriolis acceleration is
 * {| style="width:100%" border="0"|-

$$ \displaystyle \frac{d^2 \left (\boldsymbol{r}\right )_{OXY}}{dt^2} = \boldsymbol{\dot{\Omega} \times r} + \boldsymbol{\Omega \, \times} \left (\boldsymbol{\Omega \times r}\right ) + 2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} + \left (\boldsymbol{\ddot{r}}\right)_{Oxy} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }
 * The table below summarizes the similarity between the equation from Part b) and Part c)
 * {| class="prettytable"


 * colspan="3" | Comparison Table
 * Part b) Expression
 * Part c) Expression
 * 1st Term
 * $$f_{,S}(Y^1(t),t)\ddot Y^1 \,$$
 * $$\boldsymbol{\dot{\Omega} \times r} \,$$
 * 2nd Term
 * $$f_{,SS}(Y^1(t),t)(\dot Y^1)^2 \,$$
 * $$\boldsymbol{\Omega \, \times} \left (\boldsymbol{\Omega \times r}\right ) \,$$
 * 3rd Term
 * $$2f_{,St}(Y^1(t),t)\dot Y^1 \,$$
 * $$2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} \,$$
 * 4th Term
 * $$\rho (f_{,tt}(Y^1(t),t) \,$$
 * $$\left (\boldsymbol{\ddot{r}}\right)_{Oxy} \,$$
 * }
 * $$2\boldsymbol{\Omega \, \times} \left (\boldsymbol{\dot{r}}\right)_{Oxy} \,$$
 * 4th Term
 * $$\rho (f_{,tt}(Y^1(t),t) \,$$
 * $$\left (\boldsymbol{\ddot{r}}\right)_{Oxy} \,$$
 * }
 * }

Problem Statement

 * Analyze the dimension of all terms, and give the physical meaning for the equation:
 * $$ \displaystyle c_0(Y^1,t)= -F^1[1-\bar Ru^2_{,SS}(Y^1,t)]-F^2u^2_{,S}-\frac{T}{R}+M\left[(1-\bar Ru^2_{,SS})(u^1_{,tt}-\bar Ru^2_{,Stt})+u^2_{,S}u^2_{,tt}\right]

$$

Given

 * The dimensional analysis of, $$ \displaystyle c_0(Y^1,t) $$, requires no additional information than what is already given in the problem statement.

Nomenclature

 * Equations of the form, $$ [V]=D $$, are interpreted as, "The dimension of 'V' is 'D'." The following table identifies the dimensional symbols and their meaning.
 * {| class="wikitable"

! Unit Symbol !! Interpretation
 * 1 || Dimensionless
 * A || Acceleration
 * F || Force
 * L || Length
 * M || Mass
 * t || Time
 * }
 * L || Length
 * M || Mass
 * t || Time
 * }
 * t || Time
 * }

Solution

 * Step 1 - Perform a term by term identification of dimensions:
 * $$ [F^1]=F $$
 * $$ [1]=1 $$
 * $$ [\bar R]=L $$
 * $$ [u^2_{,SS}(Y^1,t)]=\left[\frac {\partial ^2u}{\partial S^2}\right]=\frac{L}{L^2}=L^{-1} $$
 * $$ [F^2]=F $$
 * $$ [u^2_{,S}]=\left[\frac {\partial u}{\partial S}\right]=\frac{L}{L}=1 $$
 * $$ [T]=FL $$
 * $$ [R]=L $$
 * $$ [M]=M $$
 * $$ [u^1_{,tt}]=\left[\frac {\partial ^2u}{\partial t^2}\right]=\frac{L}{t^2} $$
 * $$ [u^2_{,Stt}]=\left[\frac {\partial ^3u}{\partial S \partial t^2}\right]=\frac{L}{Lt^2}=t^{-2} $$


 * Step 2 - Substitute the identified dimensions into the equation:
 * $$ \displaystyle [c_0(Y^1,t)]= -F[1-LL^{-1}]-F(1)-\frac{FL}{L}+M\left[(1-LL^{-1})\left(\frac {L}{t^2}-Lt^{-2}\right)+1\left(\frac{L}{t^2}\right)\right]

$$


 * Step 3 - Simplify and recognize that $$ \frac {L}{t^2}=A $$:
 * $$ \displaystyle [c_0(Y^1,t)]= -F[1-1]-F(1)-F+M[(1-1)(A-A)+(1)(A)]

$$


 * Step 4 - Reduce with the following considerations necessary for dimensional analysis:


 * There is no regard for the quantity of a dimensioned parameter, $$ V $$. Therefore $$ V+V+V $$ isn't evaluated as, $$ 3V $$, rather it is just $$ V $$.
 * There is no regard for the quantity of dimensionless parameters, $$ 1 $$. Therefore $$ 1+1+1 $$ isn't evaluated as, $$ 3 $$, rather it is just $$ 1 $$.
 * A dimensional parameter multiplied by a dimensionless parameter is the same dimension as the dimensional parameter, so, $$ [V*1]=V $$.
 * Recognize that $$ F=MA $$
 * Therefore,
 * $$ [c_0(Y^1,t)]= -F-F-F+F $$


 * Step 5 - Conclusion of the dimensional analysis and each terms physical meaning,
 * {| style="width:100%" border="0"|-

$$\displaystyle\;$$ $$ [c_0(Y^1,t)]= F $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle -F^1[1-\bar Ru^2_{,SS}(Y^1,t)] $$ - The transverse force scaled by the guideway deformation in the transverse direction.
 * $$ F^2u^2_{,S} $$ - Axial force due to the slope of the guideway in the transverse direction.
 * $$ \frac{T}{R} $$ - Force generated at the wheel's outer edge.
 * $$ M\left[(1-\bar Ru^2_{,SS})(u^1_{,tt}-\bar Ru^2_{,Stt})+u^2_{,S}u^2_{,tt}\right] $$ - Inertial force from the mass of the wheel.
 * }

Problem Statement

 * Draw the polar coordinate lines, in a 2-D plane emanating from a point, not at the origin.

Given

 * $$\displaystyle \left (\xi_1, \xi_2 \right ) = \left (R, \theta \right )  $$

Problem Statement

 * Show that
 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)]\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * is equivalent to
 * $$\displaystyle y''+\frac{g'(x)}{g(x)}\ y'+a_0(x)y=0$$

Given

 * $$\xi_i=x$$
 * $$X_i(\xi_i)=y(x)$$
 * $$g_i(\xi_i)=g(x)$$
 * $$f_i(\xi_i)=a_0(x)$$

Solution

 * We are given that
 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * Substitue $$\displaystyle \xi_i, X_i(\xi_i), g_i(\xi_i), f_i(\xi_i)$$
 * with $$\displaystyle x, y(x), g(x), a_0(x)$$ respectively
 * Then we have
 * $$\displaystyle\frac{1}{g(x)}\frac{d}{dx}\left[g(x)\frac{dy(x)}{dx}\right]+a_0(x)y(x)$$
 * $$=\displaystyle\frac{1}{g(x)}\left[\frac{dg(x)}{dx}\frac{dy}{dx}\ +g(x)\frac{d^2y}{dx^2}\right]+a_0(x)y $$
 * $$=\displaystyle\frac{g'(x)}{g(x)}\ y'+\cancelto{1}\frac{g(x)}{g(x)}\ y''+a_0(x)y$$
 * {| style="width:25%" border="0"|-

$$=\displaystyle y''+\frac{g'(x)}{g(x)}\ y'+a_0(x)y$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }


 * Therefore, we can conclude that


 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * is equivalent to


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

Problem Statement

 * Show $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear with respect to $$\displaystyle Y^1 $$.

Given

 * $$\displaystyle c_3(Y^1,t) \ddot Y^1 $$
 * and according to the PDF
 * $$\displaystyle c_3(Y^1,t)=M[1-\bar{R} u_{,ss}^2 (Y^1,t)] $$

Solution

 * Definition of Linearity : The equation must satisfy
 * $$\displaystyle F(\alpha x + \beta y) = \alpha F(x) +\beta F(y), \forall \alpha, \beta\in\mathbb R$$
 * Therefore, the equation is
 * $$\displaystyle F(\alpha x+\beta y)\neq\alpha F(x)+\beta F(y), \forall \alpha, \beta\in\mathbb R $$
 * and it is nonlinear
 * Suppose $$\displaystyle F(x)=M[1-\bar{R} u_{,ss}^2 (x,t)] \ddot x $$
 * Then
 * $$ \displaystyle \alpha F(x) = \alpha M[1-\bar{R} u_{,ss}^2 (x,t)] \ddot x $$
 * $$ \displaystyle \beta F(y) = \beta M[1-\bar{R} u_{,ss}^2 (y,t)] \ddot y $$
 * and
 * $$\displaystyle F(\alpha x + \beta y) = M[1-\bar{R} u_{,ss}^2 (\alpha x + \beta y,t)](\alpha \ddot x + \beta \ddot y)$$
 * If it is linear, then
 * $$\displaystyle M[1-\bar{R} u_{,ss}^2 (\alpha x + \beta y,t)] \ddot {(\alpha x + \beta y)} = \alpha M[1-\bar{R} u_{,ss}^2 (x,t)] \ddot x + \beta M[1-\bar{R} u_{,ss}^2 (y,t)] \ddot y$$
 * $$ \displaystyle M[1-\bar{R} u_{,ss}^2 (\alpha x + \beta y,t)] (\alpha \ddot x + \beta \ddot y)= \alpha M[1-\bar{R} u_{,ss}^2 (x,t)] \ddot x + \beta M[1-\bar{R} u_{,ss}^2 (y,t)] \ddot y$$
 * $$\displaystyle M\bar{R} u_{,ss}^2 (\alpha x + \beta y,t) (\alpha \ddot x + \beta \ddot y) = \alpha M\bar{R} u_{,ss}^2 (x,t) \ddot x + \beta M\bar{R} u_{,ss}^2 (y,t) \ddot y$$
 * $$\displaystyle \alpha \bar{R} u_{,ss}^2 (\alpha x + \beta y,t)\ddot x + \beta \bar{R} u_{,ss}^2 (\alpha x + \beta y,t)\ddot y = \alpha \bar{R} u_{,ss}^2 (x,t) \ddot x + \beta \bar{R} u_{,ss}^2 (y,t) \ddot y$$
 * ==>In linear situation
 * $$\bar{R}u^2_{,ss}(\alpha x +\beta y ,t )=\bar{R}u^2_{,ss}(x ,t)=\bar{R}u^2_{,ss}(y, t)$$
 * {| style="width:100%" border="0"|-
 * ==>In linear situation
 * $$\bar{R}u^2_{,ss}(\alpha x +\beta y ,t )=\bar{R}u^2_{,ss}(x ,t)=\bar{R}u^2_{,ss}(y, t)$$
 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |

But the $$\alpha$$ and $$\beta$$ are all real numbers; therefore, the equation can't be guaranteed to be true.

So $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear with respect to $$\displaystyle Y^1 $$.
 * }

Problem Statement

 * Show that
 * $$\displaystyle L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$
 * is Linear

Solution

 * To prove it's linear following equation should be established
 * $$\displaystyle L_2(\alpha u+\beta v)=\alpha L_2(u)+\beta L_2(v), \forall \alpha, \beta\in\mathbb R$$
 * Substitue $$\displaystyle (\cdot)$$ with $$\displaystyle (\alpha u+\beta v)$$
 * Then we have
 * $$\displaystyle L_2(\alpha u+\beta v)=\frac{d^2(\alpha u+\beta v)}{dx^2}+a_1(x)\frac{d(\alpha u+\beta v)}{dx}+a_0(x)(\alpha u+\beta v)$$
 * $$=\displaystyle \alpha \frac{d^2(u)}{dx^2}+\beta \frac{d^2(v)}{dx^2}+a_1(x)(\alpha \frac{du}{dx}+\beta \frac{dv}{dx})+a_0(x)(\alpha u+\beta v)$$
 * $$=\displaystyle \alpha [\frac{d^2(u)}{dx^2}+a_1(x) \frac{du}{dx}+a_0u(x)]+\beta [\frac{d^2(v)}{dx^2}+a_1(x) \frac{dv}{dx}+a_0v(x)]$$
 * Also we have
 * $$\displaystyle L_2(u)=\frac{d^2(u)}{dx^2}+a_1(x) \frac{du}{dx}+a_0u(x)$$
 * $$\displaystyle L_2(v)=\frac{d^2(v)}{dx^2}+a_1(x) \frac{dv}{dx}+a_0v(x)$$
 * {| style="width:100%" border="0"|-

Therefore, since $$\displaystyle L_2(\alpha u+\beta v)=\alpha L_2(u)+\beta L_2(v)$$, we can conclude that $$\displaystyle L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$ is linear.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * Problems
 * Solved and Typed By
 * Inspected By
 * Skimmed By
 * Problem 1.1
 * Brandon Taylor
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 1.2
 * Kyung-Hoon Bang
 * Brandon Taylor
 * All Team Members
 * Problem 1.3
 * Brandon Taylor
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 1.4
 * Jiajun Han
 * Brandon Taylor
 * All Team Members
 * Problem 1.5
 * Tianyu Gu
 * Che-Ming Lee
 * All Team Members
 * Problem 1.6
 * Che-Ming Lee
 * Mutian Hu
 * All Team Members
 * Problem 1.7
 * Mutian Hu
 * Tianyu Gu
 * All Team Members
 * }
 * Mutian Hu
 * All Team Members
 * Problem 1.7
 * Mutian Hu
 * Tianyu Gu
 * All Team Members
 * }
 * }
 * }