User:Egm6321.f11.team3.chang

Given
From lecture note 2-4, first total time derivative is given by,
 * $$\displaystyle

\frac{\mathrm{d} }{\mathrm{d} t}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} $$ with
 * $$\displaystyle

\dot{Y^{1}}:=\frac{\mathrm{d} Y^{1}(t)}{\mathrm{d} t} $$

Find
Show that
 * $$\displaystyle

\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) $$ with
 * $$\displaystyle

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

f_{,St}(Y^{1},t):=\frac{\partial^{2} f(Y^{1},t)}{\partial S \partial t} $$ Similarly for the other partial derivatives

Solution
Let
 * $$\displaystyle

\frac{\mathrm d^{2}f}{\mathrm dt^{2}}=\frac{\mathrm d}{\mathrm dt}(\frac{\mathrm d}{\mathrm dt}f(Y^{1}(t),t))=\frac{\mathrm d}{\mathrm dt}(\frac{\partial f(Y^{1}(t),t)}{\partial s}\dot{Y}^{1}+\frac{\partial f(Y^{1}(t),t)}{\partial t}) $$
 * $$\displaystyle

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

+\frac{\partial}{\partial t}(\frac{\partial f(Y^{1}(t),t)}{\partial S}\frac{\mathrm dY^{1}(t)}{\mathrm dt}+\frac{\partial f(Y^{1}(t),t)}{\partial t}) $$
 * $$\displaystyle

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

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

=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) $$

Given
The equation of motion (EOM) for the maglev train is modeled by:
 * $$\displaystyle

\displaystyle f(S,t)| _{S=Y^1(t)}=f(Y^1(t),t) $$ From lecture notes on page 2-4, the 1st total time derivative is given by: (Eq.2A)
 * $$\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} $$ and the 2nd total time derivative, on page 2-5, is given by: (Eq.2B)
 * $$\displaystyle

\displaystyle \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) $$

Problem
--Derive the 1st total time derivative. --Derive the 2nd total time derivative. --Show the similarity with the derivation of the Coriolis force.

Solution, First Total Time Derivative
Applying the chain rule to
 * $$\displaystyle

f(S,t) $$

Yields:


 * $$\displaystyle

\frac{df(S,t)}{dt}=\frac{\partial f}{\partial S}\frac{\partial S}{\partial t}+\frac{\partial f}{\partial t}\frac{\partial t}{\partial t} = \frac{\partial f(S,t)}{\partial S}\frac{\partial S}{\partial t} + \frac{\partial f(S,t)}{\partial t}\frac{\partial t}{\partial t} $$ Since:


 * $$\displaystyle

f(S,t)| _{S=Y^1(t)}=f(Y^1(t),t) $$ Then the equation can be rewritten as:
 * $$\displaystyle

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

And simplified to return to equation 2A:
 * $$\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} $$

Given
ξ = (ξ1,ξ2,ξ3) X(ξ) = X1(ξ1)X2(ξ2)X3(ξ3)

Find
Show that:
 * $$\displaystyle

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

y''+\underbrace{\frac{g'(x)}{g(x)}}_{a_a(x)}y'+a_0(x)\,y=0 $$

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

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
Let u(x) and v(x) be any 2 functions of x Let :$$\displaystyle \alpha \, and \, \beta $$ be and 2 real numbers We write :$$\displaystyle \forall u,v:\mathbb{R}\rightarrow \mathbb{R} \, x \mapsto u(x)  \,\,  x \mapsto v(x)  \,\, \forall \alpha ,\beta \in \mathbb{R} $$
 * $$\displaystyle

\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 \ddot{u}+\beta \ddot{v}+a_{1}(x)(\alpha \dot{u}+\beta \dot{v})+a_{0}(x)(\alpha u+\beta v)$$
 * $$\displaystyle

\frac{d^{2}(\alpha u)}{dx^{2}}+a_{1}(x)\frac{d(\alpha u)}{dx}+a_{0}(x)(\alpha u)$$
 * $$\displaystyle

= \alpha \ddot{u}++a_{1}(x)\alpha \dot{u}++a_{0}(x)\alpha u $$
 * $$\displaystyle

\frac{d^{2}(\beta v)}{dx^{2}}+a_{1}(x)\frac{d(\beta v)}{dx}+a_{0}(x)(\beta v)$$
 * $$\displaystyle

= \beta \ddot{v}+a_{1}(x)(\beta \dot{v})+a_{0}(x)(\beta v)$$
 * $$\displaystyle

\Rightarrow L_{2}(\alpha u+\beta v)= L_{2}(\alpha u)+L_{2}(\beta v) $$ So it is linear.

Given

 * $$\displaystyle

y(x)=y^{_{H}}(x)+y^{_{P}}(x)=c_{1}y_{H}^{1}(x)+c_{2}y_{H}^{2}(x)+y_{P}(x) $$

Find
Find the integration constants in terms of the initial conditions :$$\displaystyle y(a)=\alpha \,\, \, y{}'(a)=\beta $$

Solution

 * $$\displaystyle y(a)=c_{1}y_{H}^{1}(a)+c_{2}y_{H}^{2}(a)+y_{P}(a)=\alpha $$


 * $$\displaystyle y{}'(a)=c_{1}\dot{y}_{H}^{1}(a)+c_{2}\dot{y}_{H}^{2}(a)+\dot{y}_{P}(a)=\beta $$


 * $$\displaystyle \Rightarrow c_{1}= \frac{\alpha \dot{y}_{H}^{2}(a)-\beta y_{H}^{2}(a)-y_{P}(a)\dot{y}_{H}^{2}(a)+\dot{y}_{P}y_{H}^{2}(a)}{y_{H}^{1}(a)\dot{y}_{H}^{2}(a)-\dot{y}_{H}^{1}(a)y_{H}^{2}(a)}   $$


 * $$\displaystyle c_{2}= \frac{\alpha \dot{y}_{H}^{1}(a)-\beta y_{H}^{1}(a)-y_{P}(a)\dot{y}_{H}^{1}(a)+\dot{y}_{P}y_{H}^{1}(a)}{y_{H}^{2}(a)\dot{y}_{H}^{1}(a)-\dot{y}_{H}^{2}(a)y_{H}^{1}(a)}      $$

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