User:EGM6321.f10.team6.cook/hw1

= Total Derivative =

Required
Derive the first and second derivative of $$f(s,t)_{s=Y^1(t)}$$ with respect to $$t$$ and improve on the existing solutions.

== Solution ==

Chain Rule
Since $$f = f_{(s,t)}$$ and $$s=s_{(Y^1(t), t)}$$ we must apply the chain rule. The reader will likely remember that when applying the chain rule to a function that the result is the sum of the chain rule when applied to each component that depends on a given variable. That is to say that since f not only depends on s and t, but s depends on Y1 and t that the chain rule becomes the additive sum of the derivative of all the terms that are dependent upon t.

The component of the partial derivative that is a result of the dependency upon t for both f and s is shown by 1 and the component that is a result of the dependency upon t for f is shown by 2.

First Derivative
Substituting for $$f$$ evaluated at $$s$$, $$f(s,t) = f(Y^1(t),t)$$, $$s=Y^1(t)$$ and using the notation $$\frac{\partial}{\partial t}Y^1(t)=\dot{Y}^1(t)$$.

Second Derivative
For the second derivative define

Then again by chain rule

substitute in for $$g$$

distributing the derivatives

Schwarz&rsquo;s Theorem
Assuming real and continuous functions so that the order of the partial derivatives may be changed (Schawrz&rsquo;s theorem )

Substituting for $$f$$ evaluated at $$s$$, $$f(s,t) = f(Y^1(t),t)$$, $$s=Y^1(t)$$ and using the notation $$\frac{\partial}{\partial t}Y^1(t)=\dot{Y}^1(t)$$, the final form is found.

Discussion
The general derivation of this problem is relatively easy, however, familiarity with the chain rule and how it applies to functions of functions is probably the location where most people will have difficulty. A relatively clear and concise explanation of the chain rule may be found at Paul’s Online Math Notes.

Developing the Reynolds' Transport Theorem
Developing the Reynolds' Transport Theorem is an application of the first derivative of $$\Phi (\text{x},t)$$, and it is presents as follows:

Suppose we have a spatial scalar field $$\Phi (\text{x},t)$$ describing some physical quantity (for example, mass, internal energy, entropy, heat, or entropy sources) of a particle in space per unit volume at time t.

Assume $$\Phi (\text{x},t)$$ to be smooth, so that it is continuously differentiable. Hence, the present status of a continuum body in some three-dimensional region Ω with volume $$\upsilon $$ at given time t may be characterized by the scalar-valued function

The aim is now to compute the material time derivative of the volume integral $$I(t)$$ Since the region of integration Ω depends on time $$t$$, integration and time differentiation do not commute. Therefore, as a first step $$I(t)$$ must be transformed to the reference configuration. By changing variables using the motion $$\text{x}=\chi (\text{X},t)$$ and the relation $$\text{d}\upsilon \text{= }J(\text{X},t)\text{d}V$$ we find the time rate of change of $$I(t)$$ to be

Since the region of integration is now time-independent, integration and differentiation commute. Hence, as a second step, from (1.10)we obtain, using the product rule of differentiation,

where $${\dot{\Phi }}$$ denotes the material time derivative of the spatial scalar field $$\Phi $$. In a last step we undo the change of variables and convert the volume integral back to the current configuration. By means of the material time derivative of the scalar J, $$\text{d}\upsilon \text{= }J(\text{X},t)\text{d}V$$ and motion $$\text{x}=\chi (\text{X},t)$$, we find finally that

In the following the arguments of the tensor quantities are dropped in order to simplify the notation. However, in cases where additional information is needed, they will be employed. Hence, relation (1.12) reads as

Where we have assumed smoothness of the spatial velocity field v.

Other forms of the time rate of change of the integral (1.9) result from (1.13) by means of the material time derivative of the spatial scalar field $$\Phi $$ can be developed, i.e. we can write the first derivative of the homework equation as

We can write the integral as

And finally, using the divergence theorem

The first term on the right-hand side of eq.(1.20) characterizes the rate of transport (or the outward normal flux) of $$\Phi \nu $$ across the surface $$\partial \Omega $$ out of region $$\Omega $$, which is assumed to be fixed. This contribution arises from the moving region. The second term denotes the local time rate of change of the spatial scalar field $$\Phi $$ within region $$\Omega $$. In (1.20) n denotes the outward unit normal field acting along $$\partial \Omega $$. Relation (1.20) is referred to as Reynolds’ transport theorem.

Derivation for the Coriolis acceleration
Considering an arbitrary vector $$\mathbf{r}$$​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​:

where x, y, and z are the cartesian components of the vector and i, j, and k are unit vectors along these axes. In the case at hand, the components x, y, and z of the vector r are not constant, and, of course, the unit vectors i, j, and k are not constant either, as they rotate with the same angular velocity ω as the moving frame. Hence, the time

derivative of the vector r can be written in the form

where the time derivative of the unit vectors i, j, and k can be obtained from

as the derivative of any vector embedded in a rotating reference frame. Hence, replacing r by i,j, and k, we obtain

and then,

where,

can be identified as the time rate of change of r regarding the reference frame xyz as inertial.

the position vector of point P has the form

so that, we can write the absolute velocity of P as follows :

is the velocity of P relative to the moving frame xyz, in which $${x}_{AP}, {y}_{AP}$$ and $$ {z}_{AP}$$ are the cartesian components of $$ \mathbf{r}_{AP} $$ and $$ \mathbf{\omega}\times\mathbf{r}_{AP} $$ is the velocity of P due entirely to the rotation of the frame xyz. Similarly, we write the absolute acceleration of P in the form

where $$ \mathbf{a}_{A}=\mathbf{\dot{v}_{A}} $$ is the acceleration of A relative to the inertial space,
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$$ \mathbf{a'}_{AP}=\ddot{x}_{AP}\mathbf{i}+\ddot{y}_{AP}\mathbf{j}+\ddot{z}_{AP}\mathbf{k} $$      (1.29) is the acceleration of P relative to the rotating frame xyz, $$ 2\mathbf{\omega}\times\mathbf{v'}_{AP} $$ is the so-called Coriolis acceleration, and $$ \mathbf{\alpha}\times\mathbf{r}_{AP} + \mathbf{\omega}\times(\mathbf{\omega}\times\mathbf{r}_{AP}) $$ is the acceleration of P due entirely to the rotation of the frame xyz, in which $$ \mathbf{\alpha}=\mathbf{\dot{\omega}} $$ is the angular acceleration of the frame.
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Similarities and differences
Let's compare the equations between the general derivative and Coriolis. As the derivative for position vector equals velocity and the derivative for the velocity equals acceleration physically, we can use the general equation when we'd like to find the velocity and the acceleration from the position vector.

As for the position vector, $$ \mathbf{r}_{AP}(\theta,t)$$, we know

Which means we can use derivative of the position vector $$\mathbf{r}_{AP}$$ as the same to the position vector itself. So, if (1.30) is true, we can use follow relation,

and also,

In summary, we can relate the Coriolis equation to the general derivative.

Given
The equation of motion of a train (either Maglev or wheel on rail) can be shown to have the following form:

The $$c_0$$ term is :

Required
Perform dimensional analysis of the $$c_0$$ term and provide physical meaning of each term in $$c_0$$

Solution
According to Newton's second Law, the whole $$c_0$$ term should be the force term in the equation of motion.

First Term
For the first term,

in which,

so,

and

Second Term
For the second term,

Third Term
For the third term,

Fourth Term
For the last term,

in which,

so,

and

We have shown that each term in $${{c}_{o}}$$ has a dimension of force, $$[F]$$.

For the physical meaning of each term, in general, they are contributions of external forces $$({{F}^{1}},{{F}^{2}},T)$$ and the inertia term $$Ma$$.

Given
The equation of motion of a wheel on flexible guide rail type problem that is consistent with what occurs when the wheel of a high speed train interacts with its guiderail is given as :

where the coefficients may be expressed as:

Required
Prove the following is non-linear:

== Solution  ==

In general a function may be considered linear if the following conditions are met :

Differential Operator
Consider for a moment the differential operator. The differential operator is a linear operator, and as such the above conditions will be met if the the differential operator is applied to an arbitrary function. For example:

First let us examine the first condition of linearity given in (3.7). According to (3.7) the result if we multiply the differential operator by some constant C should be exactly the same as if we multiply function by some constant C and then take the derivative.

Clearly the placement of the scalar multiple C has no effect on the result, and therefore the differential operator is linear. If we were to apply the second condition of linearity it would also hold.

Demonstrating Non-linearity
For this particular problem we will apply the first condition of linearity to prove that the function given in (6) is non-linear. By applying the first condition of linearity using an arbitrary constant α we obtain:

Since the differential operator is linear we can take the scalar multiply from within the derivative terms to obtain:

Doing some simple algebraic manipulation we obtain the following:

Clearly the left and right sides of (3.15) don't match (there is second power scalar on the right, but only a first power on the left), therefore (3.6) is non-linear.

Given
Let $$L_2(y)$$ be a linear second-order differential equation with variable coefficients

with the homogeneous solution

and boundary values $$y(a)=\alpha$$, and $$y(b)=\beta$$.

Required
Find $$c$$,$$d$$ in terms of $$\alpha$$, $$\beta$$.

Solution
Boundary conditions give,


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$$ y(a)=c{y}_{H}^1(a)+d{y}_{H}^2(a)+ {y}_{P}(a)=\alpha $$     (4.3)
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$$ y(b)=c{y}_{H}^1(b)+d{y}_{H}^2(b)+{y}_{P}(b)=\beta $$     (4.4)
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Coefficient C
for c, eliminate d by multiplying $${y}_{H}^2(b)$$ to (4.3), $${y}_{H}^2(a)$$ to (4.4)


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$$ c\cdot{y}_{H}^1(a){y}_{H}^2(b)+d\cdot{y}_{H}^2(a){y}_{H}^2(b)+{y}_{P}(a){y}_{H}^2(b)=\alpha \cdot{y}_{H}^2(b)$$ (4.5)
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$$ c\cdot{y}_{H}^1(b){y}_{H}^2(a)+d\cdot{y}_{H}^2(b){y}_{H}^2(a)+{y}_{P}(b){y}_{H}^2(a)=\beta \cdot{y}_{H}^2(a)$$ (4.6)
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by (4.5)-(4.6)


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$$ c= \frac{\alpha\cdot{y}_{H}^2(b)-\beta\cdot{y}_{H}^2(a)+{y}_{P}(b){y}_{H}^2(a)-{y}_{P}(a){y}_{H}^2(b)}{{y}_{H}^1(a){y}_{H}^2(b)-{y}_{H}^2(a){y}_{H}^1(b)} $$     (4.7)
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Coefficient D
for d, eliminate c by multiplying $${y}_{H}^1(b)$$ to (4.3), $${y}_{H}^1(a)$$ to (4.4)


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$$ c\cdot{y}_{H}^1(a){y}_{H}^1(b)+d\cdot{y}_{H}^2(a){y}_{H}^1(b)+{y}_{P}(a){y}_{H}^1(b)=\alpha \cdot{y}_{H}^1(b)$$ (4.8)
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$$ c\cdot{y}_{H}^1(b){y}_{H}^1(a)+d\cdot{y}_{H}^2(b){y}_{H}^1(a)+{y}_{P}(b){y}_{H}^1(a)=\beta \cdot{y}_{H}^1(a)$$ (4.9)
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by (4.8)-(4.9)


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$$ d= \frac{\alpha\cdot{y}_{H}^1(b)-\beta\cdot{y}_{H}^1(a)+{y}_{P}(b){y}_{H}^1(a)-{y}_{P}(a){y}_{H}^1(b)}{{y}_{H}^1(b){y}_{H}^2(a)-{y}_{H}^1(a){y}_{H}^2(b)} $$     (4.10)
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Given
Given, Legendre equation with $$n=1$$:
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$$\left(1-{{x}^{2}}\right){{y}^{''}}-2x{{y}^{'}}+2y=0$$, (5.1)
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and its two homogeneous solutions


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$$ \begin{align} {{y}_{h1}}=& x\\ {{y}_{h2}}=& \frac{x}{2}\cdot \log (\frac{1+x}{1-x})-1 \end{align} $$     (5.2)
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Required
Verify


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$${{L}_{1}}\left({{y}_{H1}}\right)={{L}_{2}}\left({{y}_{H2}}\right)=0$$ (5.3)
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== Solution  ==

First Equation
Legendre equation with $$n=1$$:
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$$\displaystyle (1-{{x}^{2}}){{y}^{''}}-2x{{y}^{'}}+2y=0$$ (5.4)
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1 st solution is given as


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$$  \displaystyle {{y}_{H1}}=x $$  (5.5)
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then
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$$ {y}_{H1}=x, {y'}_{H1}=1, {y''}_{H1}=0 \left. \right. $$  (5.6)
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Substituting the values into equation (5.1),


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$$ (1-x^2)\cdot0-2x\cdot1+2x=0 $$  (5.7)
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Second Equation
2 nd solution is given as
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$$  \displaystyle {y}_{H2}=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1 $$     (5.8)
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$$ \dot{y}_{H2}=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{2}\cdot\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2} $$     (5.9)
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$$ \ddot{y}_{H2}=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)+\frac{\left(1-x^2)+x(2x\right)}{(1-x^2)^2}=\frac{2}{(1-x^2)^2} $$     (5.10)
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Substituting these values into equation (5.1).


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$$ (1-x^2)\cdot\frac{2}{(1-x^2)^2}-2x\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]+2\left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]=0$$ (5.11)
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