User:Egm6321.f10.team6.lee/HW1

= Problem 1 = Derive equation (1),(2) - improvement over existing solution on course wiki : better explanations : theoretical background related to the problem

Equation (1):
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$$ \frac{\mathrm{d}}{\mathrm{d} t}f\left ( Y^1\left (t \right ),t \right )= \frac{\partial f\left(Y^1\left(t\right),t\right)}{\partial s}\dot{Y}^1+ \frac{\partial f\left(Y^1\left(t\right),t\right)}{\partial t} $$ (1)
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Equation (2):
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$$ \frac{\mathrm{d}^2f }{\mathrm{d} t^2} = {f}_{s}(Y^1, t)\ddot{Y}^1+{f}_{ss}(\dot{Y}^1)^2+2{f}_{st}\dot{Y}^1+{f}_{tt} $$     (2)
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Derivation for the Coriolis acceleration
Considering an arbitrary vector r​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​
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$$ \mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k} $$ 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
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$$ \mathbf{\dot{r}}=\dot{x}\mathbf{i}+x\mathbf{\dot{i}}+\dot{y}\mathbf{j}+y\mathbf{\dot{j}}+\dot{z}\mathbf{k}+z\mathbf{\dot{k}} $$ where the time derivative of the unit vectors i, j, and k can be obtained from
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$$ \mathbf{\dot{r}}= \mathbf{\omega} \times \mathbf{r}$$ as the derivative of any vector embedded in a rotating reference frame. Hence, replacing r by i,j, and k, we obtain
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$$ \mathbf{\dot{i}}=\mathbf{\omega}\times\mathbf{i}, \mathbf{\dot{j}}=\mathbf{\omega}\times\mathbf{j}, \mathbf{\dot{k}}=\mathbf{\omega}\times\mathbf{k} $$ and then,
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$$ \mathbf{\dot{r}}=\dot{x}\mathbf{i}+\dot{y}\mathbf{j}+\dot{z}\mathbf{k}+\mathbf{\omega}\times(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})=\dot{r}'+\mathbf{\omega}\times\mathbf{r} $$ where,
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$$ \mathbf{\dot{r}'}=\dot{x}\mathbf{i}+\dot{y}\mathbf{j}+\dot{z}\mathbf{k} $$ can be identified as the time rate of change of r regarding the reference frame xyz as inertial.
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the position vector of point P has the form


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$$ \mathbf{R}=\mathbf{{r}_{A}}+\mathbf{{r}_{AP}} $$ 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
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$$ \mathbf{a}=\mathbf{\dot{v}}=\mathbf{\dot{a}_{A}}+\frac{d}{dt}(\mathbf{v'}_{AP})+\mathbf{\dot{\omega}}\times\mathbf{r}_{AP}+\mathbf{\omega}\times\mathbf{\dot{r}_{AP}} $$
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$$ =\mathbf{a}_{A}+\mathbf{a'}_{AP}+\mathbf{\omega}\times\mathbf{v'}_{AP}+\mathbf{\alpha}\times\mathbf{r}_{AP}+\mathbf{\omega}\times(\mathbf{v'}_{AP}+\mathbf{\omega}\times\mathbf{r}_{AP}) $$
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where $$ \mathbf{a}_{A}=\mathbf{\dot{v}_{A}} $$ is the acceleration of A relative to the inertial space,
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$$ \mathbf{a'}_{AP}=\ddot{x}_{AP}\mathbf{i}+\ddot{y}_{AP}\mathbf{j}+\ddot{z}_{AP}\mathbf{k} $$ 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 derivation and Coriolis. As the derivation for position vector equals velocity and the derivation 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 that $$ d\mathbf{r}_{AP}=d{\theta}\times\mathbf{r}_{AP}$$ $$ \frac{d\mathbf{r}_{AP}}{d{\theta}}=\mathbf{r}_{AP}$$

It means we can use derivation of the position vector $$\mathbf{r}_{AP}$$ as the same to the position vector itself. $$ \mathbf{r}_{AP,\theta\theta}=\mathbf{r}_{AP,\theta}=\mathbf{r}_{AP} $$

and also, $$ \mathbf{r}_{AP,\theta t}=\mathbf{r}_{AP,t}=\frac{\partial\mathbf{r}_{AP}}{\partial t}=\mathbf{v'}_{AP} $$ $$ \mathbf{r}_{AP,tt}=\frac{\partial^2\mathbf{r}_{AP}}{\partial t^2}=\mathbf{a'}_{AP} $$

Im summary, we can relate the Coriolis equation to the general derivation.

Given
yH1 and yH2 are homogeneous solution for L2_ODE_VC


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$$ y=c{y}_{H}^1+d{y}_{H}^2+{y}_{P} $$
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boundary values are,
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$$ y \left(a \right)=\alpha, y \left(b \right)=\beta $$
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Find
Find c, d in terms of $$ \alpha, \beta$$

Solve
Boundary conditions give,


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$$ y(a)=c{y}_{H}^1(a)+d{y}_{H}^2(a)+ {y}_{P}(a)=\alpha $$     (4-1)
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$$ y(b)=c{y}_{H}^1(b)+d{y}_{H}^2(b)+{y}_{P}(b)=\beta $$     (4-2)
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 * for c,
 * eliminate d by multiplying $${y}_{H}^2(b)$$ to (4-1), $${y}_{H}^2(a)$$ to (4-2)


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$$ c\cdot{y}_{H}^1(a){y}_{H}^2(b)+d\cdot{y}_{H}^2(a){y}_{H}^2(b)+{y}_{P}(a){y}_{H}^2(b)=\alpha \cdot{y}_{H}^2(b)$$ (4-3)
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$$ c\cdot{y}_{H}^1(b){y}_{H}^2(a)+d\cdot{y}_{H}^2(b){y}_{H}^2(a)+{y}_{P}(b){y}_{H}^2(a)=\beta \cdot{y}_{H}^2(a)$$ (4-4)
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by (4-3)-(4-4)


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$$ c= \frac{\alpha\cdot{y}_{H}^2(b)-\beta\cdot{y}_{H}^2(a)+{y}_{P}(b){y}_{H}^2(a)-{y}_{P}(a){y}_{H}^2(b)}{{y}_{H}^1(a){y}_{H}^2(b)-{y}_{H}^2(a){y}_{H}^1(b)} $$     (4-5)
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 * for d,
 * eliminate c by multiplying $${y}_{H}^1(b)$$ to (4-1), $${y}_{H}^1(a)$$ to (4-2)


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$$ c\cdot{y}_{H}^1(a){y}_{H}^1(b)+d\cdot{y}_{H}^2(a){y}_{H}^1(b)+{y}_{P}(a){y}_{H}^1(b)=\alpha \cdot{y}_{H}^1(b)$$ (4-6)
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$$ c\cdot{y}_{H}^1(b){y}_{H}^1(a)+d\cdot{y}_{H}^2(b){y}_{H}^1(a)+{y}_{P}(b){y}_{H}^1(a)=\beta \cdot{y}_{H}^1(a)$$ (4-7)
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by (4-6)-(4-7)


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$$ d= \frac{\alpha\cdot{y}_{H}^1(b)-\beta\cdot{y}_{H}^1(a)+{y}_{P}(b){y}_{H}^1(a)-{y}_{P}(a){y}_{H}^1(b)}{{y}_{H}^1(b){y}_{H}^2(a)-{y}_{H}^1(a){y}_{H}^2(b)} $$     (4-8)
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= Problem 5 =

Given

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$$ (1-x^2){y}''-2x{y}'+2y=0 $$     (5-1)
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Solution

 * for the 1st solution


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$$ {y}_{H}^1=x, \dot{y}_{H}^1=1, \ddot{y}_{H}^1=0 $$     (5-2)
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Let's put these values to equation (5-1). Then,


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$$ (1-x^2)\cdot0-2x\cdot1+2x=0 $$     (5-3)
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 * for the 2nd solution


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$$ {y}_{H}^2=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1 $$     (5-4)
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$$ \dot{y}_{H}^2=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{2}\cdot\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2} $$     (5-5)
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$$ \ddot{y}_{H}^2=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)+\frac{\left(1-x^2)+x(2x\right)}{(1-x^2)^2}=\frac{2}{(1-x^2)^2} $$     (5-6)
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Let's put these values to equation (5-1). Then,


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$$ (1-x^2)\cdot\frac{2}{(1-x^2)^2}-2x\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]+2\left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]=0$$ (5-7)
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