User:Egm6321.f10.team5.abc/HW1

Problem 1
Derive the following equations. $$\frac {d} {dt} f(Y^1(t),t) = \frac {\partial f} {\partial S} (Y^1(t),t) \dot {Y}^1 + \frac {\partial f} {\partial t} (Y^1(t),t)$$ $$\frac {d^2 f} {d t^2} = f_{,S}(Y^1,t) \ddot{Y}^1 + f_{,SS} (\dot {Y}^1)^2 + 2 f_{,St} \dot {Y}^1 + f_{,tt}$$

Solution 1
Since $$f(y^1(t),t)$$ is defined as $$f(Y^1(t),t) = \left. f(s,t) \right |_{S=Y^1(t)}$$, we know that $$s$$ is a fuction of $$t$$. Therefore, to take a derivative of $$f$$ with respect to $$t$$, we need to use the chain rule for several variables. $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \frac {\partial t} {\partial t}$$ where $$\frac {\partial t} {\partial t} = 1$$. So we reduce the equation as following. $$\frac {df} {dt} = \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t}$$ Note that $$\frac {\partial s} {\partial t} = \frac {\partial Y^1(t)} {\partial t} = \dot {Y}^1$$. Expressing the first-order derivative formally, we have $$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f(Y^1(t),t)} {\partial s} \dot {Y}^1 + \frac {\partial f(Y^1(t),t)} {\partial t}$$ Therefore P.1-3(3) is derived completely.

Now we take the second derivative of $$f$$. $$\frac {d^2 f} {dt^2} = \frac d {dt} \left ( \frac {df} {dt} \right ) = \frac d {dt} \left (\frac {\partial f} {\partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial t} \right ) = \frac {\partial s} {\partial t} \frac d {dt} \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \frac d {dt} \frac {\partial s} {\partial t} + \frac d {dt} \frac {\partial f} {\partial t}$$ Since the chain rule needs to be applied again, we write the equation as the following. $$\frac d {dt} \left ( \frac {df} {dt} \right ) = \frac {\partial s} {\partial t} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial s} + \frac {\partial f} {\partial s} \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial s} {\partial t} + \left ( \frac \partial {\partial s} \frac {\partial s} {\partial t} + \frac \partial {\partial t} \right ) \frac {\partial f} {\partial t}$$ $$= \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial s} {\partial t} \frac {\partial^2 s} {\partial s \partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} + \frac {\partial^2 f} {\partial t^2}$$ Because $$\frac {\partial^2 s} {\partial s \partial t} = 0$$ and $$\frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial s} {\partial t} \frac {\partial^2 f} {\partial s \partial t} = 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t}$$, we can simplify the equation as following. $$\frac {\partial^2 f} {\partial t^2} = \frac {\partial^2 f} {\partial s^2} \left ( \frac {\partial s} {\partial t} \right )^2 + 2 \frac {\partial^2 f} {\partial t \partial s} \frac {\partial s} {\partial t} + \frac {\partial f} {\partial s} \frac {\partial^2 s} {\partial t^2} + \frac {\partial^2 f} {\partial t^2}$$ Expressing the second-order derivative formally, we have $$\frac {\partial^2 f(Y^1(t),t)} {\partial t^2} = \frac {\partial^2 f(Y^1(t),t)} {\partial s^2} (\dot {Y}^1)^2 + 2 \frac {\partial^2 f(Y^1(t),t)} {\partial t \partial s} \dot {Y}^1 + \frac {\partial f} {\partial s} \ddot {Y}^1 + \frac {\partial^2 f(Y^1(t),t)} {\partial t^2}$$ Therefore P.1-3(4) is derived completely.

Problem 2
Do dimensional analysis of all terms in the following equation, and describe the physical meaning. $$c_0 (y^1,t) = - F^1 [1 - \bar {R} u^2_{,ss} (y^1,t)] - F^2 u^2_{,s} - \frac T R + M \lbrace [1 - \bar R u^2_{,ss}] [u^1_{,tt} - \bar R u^2_{,stt}] + u^2_{,s} u^2_{,tt} \rbrace$$

Solution 2
We first do the dimesional analysis of each variable and state its physical meaning.

$$c_0 = [\frac {ML} {T^2}]$$ : horizontal force acting on wheel

$$F^1 = [\frac {ML} {T^2}]$$ : horizontal component of force

$$\bar R = [L]$$ : rdistance from the center of the guideway to the center of the wheel

$$u^2_{,ss} = [\frac L {L^2}] = [\frac 1 L]$$ : curvature

$$F^2 = [\frac {M L} {T^2}]$$ : vertical component of force

$$u^2_{,s} = [\frac L L] = [1]$$ : slope of transversal deformation(display) of the guideway

$$T = [\frac {M L^2} {T^2}]$$ : torque

$$R = [L]$$ : radius of the wheel

$$M = [M]$$ : mass of the wheel

$$u^1_{,tt} = [\frac L {T^2}]$$ : acceleration of axial deformation (display) of the guideway

$$u^2_{,tt} = [\frac L {T^2}]$$ : acceleration of transversal deformation (display) of the guideway

$$u^2_{,stt} = [\frac L {L T^2}] = [\frac 1 {T^2}]$$ : acceleration of slope of transversal deformation (display) of the guideway

\noindent Now we do the complete dimensional analysis for the equation. $$[\frac {ML} {T^2}] = - [\frac {M L} {T^2}] ([1] - [L \frac 1 L] ) - [\frac {M L} {T^2}] [1] - [\frac {M L^2} {T^2} L] + [M \lbrace ([1] - [L \frac 1 L])([\frac L {T^2}] - [L \frac 1 {T^2}]) + [1 \frac L {T^2}] \rbrace$$ $$= - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [M \lbrace ([\frac L {T^2}] - [\frac L {T^2}] - [\frac L {T^2}] + [\frac L {T^2}] + [\frac L {T^2}] \rbrace$$ $$=- [\frac {M L} {T^2}] - [\frac {M L} {T^2}] - [\frac {M L} {T^2}] + [\frac {ML} {T^2}] - [\frac {ML} {T^2}] - [\frac {ML} {T^2}] + [\frac {ML} {T^2}] + [\frac {ML} {T^2}]$$ Is it verified that each term in this equation has the dimension of $$[\frac {ML} {T^2}]$$ which is the unit of force.

Problem 3
Show $$c_3(Y^1,t) \ddot {Y}^1$$ is nonlinear with respect to $$Y^1$$ where $$c_3(Y^1,t) = M[1 - \bar R u^2_{,ss} (Y^1,t)]$$.

Solution 3
Nonlinearity is defined as $$F(\alpha u + \beta v) \ne \alpha F(u) + \beta F(v)$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R u^2_{,ss} (Y^1,t)] \ddot {Y}^1$$ Given that $$u^2_{,ss} (Y^1,t)$$ is curvature, we define it as $$X$$. $$c_3(Y^1,t) \ddot {Y}^1 = M[1 - \bar R X] \ddot {Y}^1$$ Define a differential operator as below. $$F(\cdot) = M (1 - \bar R X) \frac {\partial^2 (\cdot)} {\partial X^2}$$ This operator must satisfty the condition that $$F(\alpha u + \beta v) = \alpha F(u) + \beta F(v)$$ where $$\alpha$$ and $$\beta$$ are real numbers while $$u$$ and $$v$$ are functions of $$X$$. Now we test the condition of linearity. $$F(\alpha u + \beta v) = M (1 - \bar R X) \frac {\partial^2 (\alpha u + \beta v)} {\partial X^2}$$ $$\alpha F(u) + \beta F(v) = \alpha M (1 - \bar R X) \frac {\partial^2 u} {\partial X^2} + \beta M (1 - \bar R X) \frac {\partial^2 v} {\partial X^2}$$

Problem 4
Boundary Value Problem $$y(a)=\alpha$$, $$y(b)=\beta$$. Find $$c$$,$$d$$ in terms of $$\alpha$$, $$\beta$$.

Solution 4
The function $$y(x)$$ is defined accordingly. $$y(x) = c y_h^1(x) + d y_h^2(x) + y_p(x)$$ Using the boundary value conditions, we get the following relations. $$y(a) = c y_h^1(a) + d y_h^2(a) + y_p(a) = \alpha$$ $$y(b) = c y_h^1(b) + d y_h^2(b) + y_p(b) = \beta$$ Now we need to express $$c$$ and $$d$$ in terms of $$\alpha$$ and $$\beta$$. $$c = \frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} \qquad d = \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)}$$ $$y_h^1(a) + \frac {\beta - c y_h^1(b) - y_p(b)} {y_h^2(b)} y_h^2(a) + y_p(a) = c \left (y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)} \right ) + y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} + y_p(a) = \alpha$$ $$c = \frac {\alpha - y_h^2(a) \frac {\beta - y_p(b)} {y_h^2(b)} - y_p(a)} {y_h^1(a) - y_h^2(a) \frac {y_h^1(b)} {y_h^2(b)}} = \frac {\alpha {y_h^2(b)} - y_h^2(a) (\beta - y_p(b)) - y_p(a) {y_h^2(b)}} {y_h^1(a) {y_h^2(b)} - y_h^2(a) {y_h^1(b)}}$$ $$\frac {\beta - d y_h^2(b) - y_p(b)} {y_h^1(b)} y_h^1(a) + d y_h^2(a) + y_p(a) = d \left (y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)\right ) + \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) + y_p(a) = \alpha$$ $$d = \frac {\alpha - \frac {\beta - y_p(b)} {y_h^1(b)} y_h^1(a) - y_p(a)} {y_h^2(a) - \frac {y_h^2(b)} {y_h^1(b)} y_h^1(a)} = \frac {\alpha y_h^1(b) - (\beta - y_p(b)) y_h^1(a) - y_p(a) y_h^1(b)} {y_h^2(a) y_h^1(b) - y_h^2(b) y_h^1(a)}$$

Problem 5
Verify that $$L_2(y_H^1) = L_2(y_H^2) = 0$$.

Solution 5
The fuction $$L_2(y)$$ is defined as $$L_2(y)=(1-x^2)y''-2xy'+2y$$ Then we easily verified that $$L_2(y_H^1(x))$$ is 0. $$L_2(y_H^1(x)) = L_2(x) = (1-x^2) 0 - 2 x 1 + 2 x = 0$$ Meanwhile, checking $$L_2(y_H^2(x))$$ is cumbersome. $$L_2(y_H^2(x)) = L_2(\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)$$ We calculate $$y_H^2(x)'$$ first. $$y_H^2(x)' = (\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)' = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x 2 \frac {1-x} {1+x} \left( \frac 1 {1-x} + \frac {1+x} {(1-x)^2} \right)$$ $$ = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x 2 \frac {1-x} {1+x} \frac {1-x+1+x} {(1-x)^2} = \frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2}$$ Now we calculate $$y_H^2(x)''$$ using $$y_H^2(x)'$$. $$y_H^2(x)'' = (\frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2})' = \frac 1 2 \frac {1-x} {1+x} \left( \frac 1 {1-x} + \frac {1+x} {(1-x)^2} \right) + \left( \frac 1 {1-x^2} + \frac {2 x^2} {(1-x^2)^2} \right)$$ $$=\frac 1 2 \frac {1-x} {1+x} \frac {1-x+1+x} {(1-x)^2} + \frac {1-x^2+2x^2} {(1-x^2)^2} = \frac 1 {1-x^2} + \frac {1+x^2} {(1-x^2)^2} = \frac 2 {(1-x^2)^2}$$ Substituing the derivatives into the equation of $$L_2(y)$$ $$L_2(y_H^2(x))=(1-x^2) \frac 2 {(1-x^2)^2} - 2 x (\frac 1 2 \log \left(\frac {1+x} {1-x} \right) + \frac x {1-x^2}) + 2 (\frac x 2 \log \left(\frac {1+x} {1-x} \right)- 1)$$ $$= \frac 2 {1-x^2} - \frac {2 x^2} {1-x^2} - 2 = \frac {2(1-x^2)} {1-x^2} - 2 = 2 - 2 =0$$ It is verified that $$L_2(y_H^2(x))$$ is 0.

Therefore we verified that $$L_2(y_H^1(x)) = L_2(y_H^2(x)) = 0$$.