User:Egm6321.f11.team4/HW1

= Problem 1.1 - First and Second Derivatives of the Equation of Maglev Trains =

Given
The equation of the meglev trains:$$\displaystyle f(Y^1 (t),t) $$

Find
Derive the first and second time derivatives for the given equation.

Partial Derivation of Multivariable Functions
Suppose $$\displaystyle u(x) $$ and $$\displaystyle v(x) $$ are functions of $$\displaystyle x $$. Then the first partial derivative with respect to $$\displaystyle x $$ of the multivariable function $$\displaystyle f(u,v)=f(u(x),v(x)) $$ is:

First Derivative
Now let $$\displaystyle s=Y^1(t) $$ for ease,so the first time derivative of the given function should be:

Use $$\displaystyle \dot Y^1(t) = \frac{dY^1}{dt} $$ for ease:

Second Derivative
Derive Eq. (1.3) with respect to $$\displaystyle t $$ again:

For ease, use $$\displaystyle f_{,x} := \frac{\partial f}{\partial x} $$ and $$\displaystyle f_{,xy} := \frac{\partial^2 f}{\partial x \partial y} $$, so (1.4) is actually:

Author
Contributed by Yu Chen

=Problem 1.2 - Derivation of the 1st and 2nd derivatives and its similarity with the Coriolis force= From [[media:Pea1.f11.mtg3.djvu|Mtg 3-2]]

Given
The equation of the meglev trains:

Find
Derive the first and the second derivatives of the Eq. (2.1) and show the similarity with the derivation of the Coriolis force.

Solution
To compute the first derivative of the function $$\displaystyle f(s,t) $$, the chain rule should be applied.

First Derivative
Substituting $$\displaystyle Y^1(t) $$ for $$\displaystyle s $$ in the Eq. (2-2).

Defining the notation as below.

Second Derivative
Taking a derivative of the Eq. (2.2) with respect to $$\displaystyle t $$.

Defining the notations below. Then, the Eq. (2.6) is redefined using the notations in the Eq. (2.7) as below. Substituting $$\displaystyle Y^1(t) $$ for $$\displaystyle s $$ in the Eq. (2.8).

Derivation for the Coriolis acceleration
Define $$\displaystyle \mathbf{r} $$ as the vector indicating the position of the origin of the reference frame $$\displaystyle A $$ on an object rotating with an angular velocity of $$ \displaystyle {}^{N}\boldsymbol{\omega}^{A} $$ with respect to inertial reference frame $$\displaystyle N $$ ( not accelerating observer ). Then, the velocity of the object as viewed by an observer fixed to the inertial reference frame $$\displaystyle N $$ is defined as below. Also, the acceleration of the object as viewed by an observer fixed to the inertial reference frame $$\displaystyle N $$ is defined as below. Substituting $$\displaystyle {}^{N} \mathbf{v} $$ from the Eq. (2.11) into the Eq. (2.12), Expanding the Eq. (2.13), In the Eq. (2.14), the term $$\displaystyle {}^{N}\boldsymbol{\alpha}^{A} $$ represents the angular acceleration of the object in the inertial reference frame. The term $$\displaystyle 2{}^{N}\boldsymbol{\omega}^{A} \times {}^{A}\mathbf{v} $$ in the last equation in the Eq. (2.14) is the so-called Coriolis acceleration.

Similarities
The Eq. (2-9) and the Eq. (2-14) are related as below.

Author
Contributed by Shin

=Problem 1.3 - Dimensional Analysis= From [[media:Pea1.f11.mtg3.djvu|Mtg 3-5]]

Given
Equation of motion (EOM) of wheel/magnet can be expressed as

thus, we have the $$\displaystyle {c_0(Y^1,t)}$$ term

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

First term
in which so

Second term
in which so

Third term
in which so

Fourth term
in which so

Physical meaning
In conclusion, the external forces (F1,F2,T) as well as the inertia term Ma contribute to each element from the physical aspect.

Author
Contributed by Kexin Ren

=Problem 1.4 - Polar coordinate lines= From [[media:Pea1.f11.mtg4.djvu|Mtg 4-5]]

Given
A point in a 2-D plane

Find
Draw the polar coordinate lines emanating from the point

Author
Contributed by Kexin Ren

=Problem 1.5 - Simplification of Separated Equations= From [[media:Pea1.f11.mtg4.djvu|Mtg 4-5]]

Given: Separated Equations in curvilinear coordinates
A point $$\displaystyle \xi$$ in 3-D can be written in curvilinear coordinates as $$\displaystyle \xi = (\xi_1,\xi_2,\xi_3)$$

Applying separation of variables to a function, $$\displaystyle X(\xi) $$, i.e. $$\displaystyle X(\xi) = X_1(\xi_1)X_2(\xi_2)X_3(\xi_3)$$, we get following three decoupled 2nd order partial differential equations in $$\displaystyle X_i(\xi_i) $$,

Above, $$\displaystyle \xi_i$$ is the independent variable and $$\displaystyle X_i$$ is the dependent variable, and functions, $$\displaystyle g_i$$ and $$\displaystyle f_i$$ are dependent on $$\displaystyle \xi_i$$

Find: Simplified form of separated equations
Show that by changing the symbols to,

$$\displaystyle \xi_i \rightarrow x$$

$$\displaystyle X_i(\xi_i) \rightarrow y(x)$$

$$\displaystyle g_i(\xi_i) \rightarrow g(x)$$

$$\displaystyle f_i(\xi_i) \rightarrow a_0(x)$$

We can find following simplified form of the separated equations,

This is a 2nd order linear ordinary differential equation in y, similar to the general equation given in Differential Equations, King 2003, p.3

Solution:Simplification Process
By substituting the symbols the above equation becomes, Taking the derivative $$\frac{d}{dx}$$, appearing in the first term on LHS into the square brackets (in the above equation), and performing the differentiation we get,

Denoting the first order derivative as ' on the variables, and simplifying equations we get,

Now the terms can be rearranged to get the final desired equation,

Author
Contributed by Ankush Bhatia

=Problem 1.6 - Proof of Nonlinearity of Certain Function= From [[media:Pea1.f11.mtg5.djvu|Mtg 5-5]]

Given
The first term in EOM:

where

Find
Show function (6.1) is nonlinear with respect to $$\displaystyle Y^1 $$.

Definition of Nonlinearity
If a function $$\displaystyle F(x) $$ does NOT satisfy:

then it is nonlinear.

Proof of Nonlinerarity
Let $$\displaystyle F(x)=M[1-\bar{R} u_{,ss}^2 (x,t)] \ddot x $$, then:

If $$\displaystyle F(\alpha x + \beta y) = \alpha F(x) +\beta F(y) $$ is true, then:

is true for any real number $$\displaystyle \alpha, \beta $$ and any $$\displaystyle x, y $$ in the domain. That means:

is true for any real number $$\displaystyle \alpha, \beta $$ and any $$\displaystyle x, y $$ in the domain, which is absolutely not true. So equation (6.3) is not true for $$\displaystyle c_3(Y^1,t) \ddot Y^1 $$, and $$\displaystyle c_3(Y^1,t) \ddot Y^1 $$ is nonlinear.

Author
Contributed by Yu Chen

='''Problem 1.7 - Show that L2 is linear. '''= From [[media:Nm1.s11.mtg23.djvu|Mtg 6-3]]

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

Find
$$L_2(\cdot)$$ is linear.

Solution
If $$y_1, y_2$$ are the solution of $$(\cdot)$$ then, let $$(\cdot) = c_1y_1+c_2y_2$$

As a result, $$L_2(\cdot)$$ is linear.

Author
Contributed by Chung

=Problem 1.8 - Find the Integration Constants in Terms of the Initial Conditions=

From [[media:Pea1.f11.mtg6.djvu|Mtg 6-5]]

Given
The general structure of Linear 2nd-Order ODEs with Varying Coefficients (L2-ODE-VC) is

$$\displaystyle P(x)y''+Q(x)y'+R(x)y=F(x). $$

The general solution can be written as

$$\displaystyle y(x)=C_1y_{H1}(x)+C_2y_{H2}(x)+y_P(x). $$

Find
Find the integration constants, $$\displaystyle C_1$$ and $$\displaystyle C_2$$, in terms of the following initial conditions:

$$\displaystyle y(a)=\alpha$$ and $$\displaystyle y'(b)=\beta$$.

Solution
First, we must differentiate the solution. This yields

$$\displaystyle y'(x)=C_1y'_{H1}(x)+C_2y'_{H2}(x)+y'_P(x)$$.

Next, we set up the two equations in order to solve for the two constants.

$$\displaystyle y(a)=\alpha=C_1y_{H1}(a)+C_2y_{H2}(a)+y_P(a)$$ (Eq. 1)

and

$$\displaystyle y'(a)=\beta=C_1y'_{H1}(x)+C_2y'_{H2}(a)+y'_P(a)$$. (Eq. 2)

Then, multiply Eq. 1 by $$\displaystyle y'_{H2}(a)$$ and Eq. 2 by $$\displaystyle y_{H2}(a)$$ which yields

$$\displaystyle y'_{H2}(a)\alpha=y'_{H2}(a)C_1y_{H1}(a)+y'_{H2}(a)C_2y_{H2}(a)+y'_{H2}(a)y_P(a)$$

and

$$\displaystyle y_{H2}(a)\beta=y_{H2}(a)C_1y'_{H1}(x)+y_{H2}(a)C_2y'_{H2}(a)+y_{H2}(a)y'_P(a)$$.

Solving the two equations for $$\displaystyle C_1$$ yields

$$\displaystyle C_1=\frac{y'_{H2}(a)\alpha-y_{H2}(a)\beta+y_{H2}(a)y'_P(a)-y'_{H2}(a)y_P(a)}{y'_{H2}(a)y_{H1}(a)-y_{H2}(a)y'_{H1}(a)}$$.

Similarly, multiply Eq. 1 by $$\displaystyle y'_{H1}(a)$$ and Eq. 2 by $$\displaystyle y_{H1}(a)$$ which yields

$$\displaystyle y'_{H1}(a)\alpha=y'_{H1}(a)C_1y_{H1}(a)+y'_{H1}(a)C_2y_{H2}(a)+y'_{H1}(a)y_P(a)$$

and

$$\displaystyle y_{H1}(a)\beta=y_{H1}(a)C_1y'_{H1}(x)+y_{H1}(a)C_2y'_{H2}(a)+y_{H1}(a)y'_P(a)$$.

Solving the two equations for $$\displaystyle C_2$$ yields

$$\displaystyle C_2=\frac{y'_{H1}(a)\alpha-y_{H1}(a)\beta+y_{H1}(a)y'_P(a)-y'_{H1}(a)y_P(a)}{y'_{H1}(a)y_{H2}(a)-y_{H1}(a)y'_{H2}(a)}$$.

Author
Contributed by Allen

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