User:Egm6321.f12.team1.cla/report1

=Problem 1: Second total time derivative=

Given:

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


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


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

Find:

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

Solution:

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

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


 * $$\frac{d^2f}{dt^2}=\frac{d}{dt}(\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial t})$$


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


 * $$=\frac{\partial f(S,t)}{\partial S}(\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial t})\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2S}{\partial t^2}+\frac{\partial f(S,t)}{\partial t}(\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t})+\frac{\partial f(S,t)}{\partial t}$$


 * $$=\frac{\partial^2f(S,t)}{\partial S^2}(\frac{\partial S}{\partial t})^2+\frac{\partial^2f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2 S}{\partial t^2}+\frac{\partial^2 f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t}+\frac{\partial^2 f(S,t)}{\partial t^2}$$


 * $$=\frac{\partial^2 f(S,t)}{\partial S^2}(\frac{\partial S}{\partial t})^2+2(\frac{\partial^2f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t})+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2 S}{\partial t^2}+\frac{\partial^2 f(S,t)}{\partial t^2}$$


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

=Problem 2: Derive the first and second total time derivative to show similarity to derivation of Coriolis Force =

Given:

 * $$ f(S,t)=f(Y^1(t),t)$$

Show that:

 * $$\frac{df}{dt}$$, $$\frac{d^2f}{dt^2}$$ have a similar derivation to Coriolis Force

Solution:
First total time derivative,


 * $$ f(S,t)=f(Y^1(t),t)$$


 * $$\frac{d}{dt}f(S,t):=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial S}\frac{dS}{dt}$$


 * $$\frac{d}{dt}f(S,t):=\frac{\partial f(S,t)}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{dS}{dt}$$

Then using the first total time derivative find the second total time derivative,


 * $$\frac{d^2}{dt^2}f(S,t):=\frac{d}{dt}(\frac{\partial f(S,t)}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{dS}{dt})$$


 * $$\frac{d^2}{dt^2}f(S,t):=\frac{d}{dt}\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{d}{dt}\frac{\partial S}{\partial t}+\frac{d}{dt}\frac{\partial f(S,t)}{\partial t}$$


 * $$\frac{d^2}{dt^2}f(S,t):=\frac{\partial f(S,t)}{\partial S}(\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial t})\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2S}{\partial t^2}+\frac{\partial f(S,t)}{\partial t}(\frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t})+\frac{\partial^2 f(S,t)}{\partial t^2}$$


 * $$\frac{d^2}{dt^2}f(S,t):=\frac{\partial^2f(S,t)}{\partial S^2}(\frac{\partial S}{\partial t})^2+\frac{\partial^2f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t}+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2 S}{\partial t^2}+\frac{\partial^2 f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t}+\frac{\partial^2 f(S,t)}{\partial t^2}$$


 * $$\frac{d^2}{dt^2}f(S,t):=\frac{\partial^2 f(S,t)}{\partial S^2}(\frac{\partial S}{\partial t})^2+2(\frac{\partial^2f(S,t)}{\partial S \partial t}\frac{\partial S}{\partial t})+\frac{\partial f(S,t)}{\partial S}\frac{\partial^2 S}{\partial t^2}+\frac{\partial^2 f(S,t)}{\partial t^2}$$

The acceleration of a particle in a rotating frame causes the Coriolis Force so first we define the position of the particle,


 * $$ Position of Particle=\bar{R}$$

By taking the derivative of the position we obtain the velocity of the particle to be,


 * $$\frac{d}{dt}\bar{R}=\bar{v}$$

Therefore the particle's velocity is the sum of the velocity with respect to the rotation of the frame and with respect to the frame,


 * $$\bar{v}_{tot}= \frac{d}{dt}\bar{R}_{frame}+\frac{d}{dt}(\bar{T}x\bar{R})$$

With T being the rotation of the frame Take the second derivative to obtain the acceleration of the particle,
 * $$\bar{a}_{tot}=\frac{d}{dt}\bar{v}_{tot}=\frac{d^2}{dt^2}\bar{R}+(\bar{T}x\frac{d}{dt}\bar{R})+\bar{T}x(\bar{T}x\bar{R})+(\bar{T}x\frac{d}{dt}\bar{R})+\frac{d}{dt}\bar{T}x\bar{R}$$


 * $$\bar{a}_{tot}=\frac{d^2}{dt^2}\bar{R}+2(\bar{T}x\frac{d}{dt}\bar{R})+\bar{T}x(\bar{T}x\bar{R})+\frac{d}{dt}\bar{T}x\bar{R}$$

I referred to this page for the derivation of coriolis acceleration

=Problem 3: Dimensional analysis of all the terms in given Equation =

Given:

 * $$ c_0(Y',t)= - F^1 [1-\bar{R}u^2_{,SS}(Y',t)] - F^2 u^2_{,S} -\frac{T}{R} +M[1-\bar{R}u^2_{,SS}(u^1_{,Stt})+u^2_{S}u^2_{,tt}] $$

Equation from

Solution
{| style="width:100%" border="0" $$\begin{align} \underbrace{c_0(Y',t)}_{Resultant Force}= - F^1 [1-\bar{R}u^2_{,SS}(Y',t)] - \underbrace{F^2 u^2_{,S}}_{Force} -\underbrace{\frac{T}{R}}_{Force} +M[1-\underbrace{\bar{R}u^2_{,SS}(u^1_{,Stt})+u^2_{S}u^2_{,tt}}_{Accleration}] \end{align}$$
 * style="width:95%" |
 * style="width:95%" |

Analyzing term by term:
 * $$  \left[c_0(Y',t)\right] =  F  $$


 * $$  \left[F^1\right] = F    $$


 * $$  \left[\bar{R}\right] = L  $$


 * $$  \left[F^2\right] = F   $$


 * $$  \left[u^2\right] = L   $$


 * $$  \left[u^2_{,S}\right] = \frac{L}{L} = 1   $$


 * $$  \left[u^2_{,SS}\right] = \frac{L}{L*L} = L^{-1}  $$


 * $$  \left[u^2_{,Stt}\right] = \frac{L}{L*T*T} = T^{-2}  $$


 * $$  \left[u^2_{,tt}\right] = \frac{L}{T*T} = \frac{L}{T^2}   $$

Using the above dimensional relations

F = F[1- L L^{-1}] - F - \frac{F L}{L} + M [(1 - L L^{-1}) (\frac{L}{T^2})- L \frac{L}{T^2} - \frac{L}{T^2}] $$

Simplifying
 * $$  F = F + M \frac{L}{T^2}   $$
 * $$ M \frac{L}{T^2} = F  $$

Hence we see the two sides are dimensionally equal.

=Problem 4: Separation of Variables=

Draw:

 * 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:
Coordinate lines are lines that represent the direction of change of a particular coordinate, but orthogonal to other coordinates, i.e. other coordinates, than the one which is represented by the coordinate lines are constant at those coordinate lines.


 * The relation between Cartesian and polar coordinates are given by


 * $$x=r \cos \theta$$
 * and
 * $$y=r \sin \theta$$

According to the definition, the coordinate lines can be generated by eliminating corresponding coordinates from the equation.

At a point $$(r,\theta)$$

The coordinate lines corresponding to $$(\xi _{1})$$ or $$r$$ can be generated by eliminating $$r$$ from the equations. So in Cartesian coordinate, corresponding equation will be


 * $$\frac {y} {x} = tan \theta$$

Similarly the coordinate lines corresponding to $$(\xi _{2})$$ or $$\theta$$ is given by


 * $$ x^2+y^2 = r^2$$

=Problem 5: Simplify Equation =

Given:

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

Show that:
The equation becomes $$ y''+\frac{g'(x)}{g(x)}y'+a_0(x)y(x)=0 $$ by changing
 * $$\xi_i \qquad \rightarrow x$$
 * $$X_i(\xi_i)\,\rightarrow y(x)$$
 * $$g_i(\xi_i)\; \rightarrow g(x)$$
 * $$f_i(\xi_i)\; \rightarrow a_0(x)$$

Solution:

 * Replace the $$\xi_i$$ with $$x$$
 * Replace the $$X_i(\xi_i)$$ with $$g(x)$$
 * Repalce the $$g_i(\xi_i)$$ with $$g(x)$$
 * Replace the $$f_i(\xi_i)$$ with $$a_0(x)$$


 * then equation $$\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$ becomes


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


 * $$\frac{1}{g(x)}\left[g'(x)y'+g(x)y'')\right]+a_0(x)y(x)=0$$


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

=Problem 6: Proof of Non-linearity =

Given:

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

Proof That:

 * $$c_3 (Y^1, t) \ddot Y^1 $$ is non-linear with respect to $$Y^1$$

Solution:

 * A function $$F(x)$$ is non-linear with respect to $$x$$ if $$\exists \alpha, \beta \in \mathbb R $$ such that $$F(\alpha x + \beta y) \neq \alpha F(x)+\beta F(y)$$


 * Let us assume that $$c_3 (Y^1, t) \ddot Y^1 $$ is linear with respect to $$Y^1$$


 * $$\Rightarrow c_3(\alpha Y^1_1 + \beta Y^1_2, t)

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha c_3(Y^1_1)\ddot Y^1_1+\beta c_3(Y^1_2)\ddot Y^1_2,\qquad \forall \alpha, \beta \in \mathbb R $$
 * $$\Rightarrow M[1-\bar R u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)]

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha M[1-\bar R u^2_{,ss}(Y^1_1,t)]\ddot Y^1_1+ \beta M[1-\bar R u^2_{,ss}(Y^1_2,t)]\ddot Y^1_2$$
 * Subtracting $$ M(\alpha \ddot Y^1_1+\beta \ddot Y^1_2)$$ from both sides


 * &#160;&#160;&#160;&#160;&#160;&#160;$$ M \bar R u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)

(\alpha \ddot Y^1_1+\beta \ddot Y^1_2) = \alpha M\bar R u^2_{,ss}(Y^1_1,t) \ddot Y^1_1+ \beta M \bar R u^2_{,ss}(Y^1_2,t) \ddot Y^1_2$$
 * $$\Rightarrow \alpha u^2_{,ss}(\alpha Y^1_1 + \beta Y^1_2, t)\ddot Y^1_1+\beta u^2_{,ss}

(\alpha Y^1_1 + \beta Y^1_2, t)\ddot Y^1_1 = \alpha u^2_{,ss}(Y^1_1,t) \ddot Y^1_1+ \beta u^2_{,ss}(Y^1_2,t) \ddot Y^1_2,\qquad \forall \alpha, \beta \in \mathbb R $$
 * Which is clearly impossible whether $$u^2(Y^1_2,t)$$ is linear or non-linear with respect to $$Y^1_2$$


 * This implies the the only assumption we made was incorrect.


 * So $$c_3 (Y^1, t) \ddot Y^1 $$ is non-linear with respect to $$Y^1$$

Pavel Bhowmik (talk) 01:35, 16 October 2012 (UTC)

=Problem 7: Proof of Linearity =

Given:

 * $$L_2(y) = y''+a_1(x)y'+a_0(x)y$$

Proof That:

 * $$L_2(.)$$ is linear.

Solution:

 * A function $$F(.)$$ is linear if $$\forall \alpha, \beta \in \mathbb R, \quad F(\alpha y_1 + \beta y_2) = \alpha F(y_1)+\beta F(y_2)$$


 * $$L_2( y_1 ) = y_1+a_1(x)y'_1+a_0(x)y_1$$ and $$L_2( y_2 ) = y_2+a_1(x)y'_2+a_0(x)y_2$$


 * $$L_2( \alpha y_1 + \beta y_2) = \frac {d^2} {dx^2} ( \alpha y_1 + \beta y_2) +

a_1(x) \frac {d} {dx} ( \alpha y_1 + \beta y_2) +a_0(x)( \alpha y_1 + \beta y_2) $$
 * $$\;=\frac {d^2} {dx^2} ( \alpha y_1) +\frac {d^2} {dx^2} ( \beta y_2)+

a_1(x) \frac {d} {dx} ( \alpha y_1 )+ a_1(x) \frac {d} {dx} ( \beta y_2)+ a_0(x)( \alpha y_1 ) + a_0(x)( \beta y_2) $$
 * $$\;= \alpha y_1 + \beta y_2+ a_1(x) \alpha y'_1 + a_1(x) \beta y'_2+ a_0(x)\alpha y_1+a_0(x) \beta y_2 $$


 * $$\;= \alpha (y_1 + a_1(x) y'_1 + a_0(x) y_1)+\beta (y_2 + a_1(x) y'_2+ a_0(x) y_2) $$


 * $$\Rightarrow L_2( \alpha y_1 + \beta y_2) = \alpha L_2( y_1 ) + \beta L_2 (y_2) $$


 * So $$ L_2(.) $$ is a linear operator.

Pavel Bhowmik (talk) 01:35, 16 October 2012 (UTC)

=Contribution Table=

= References =