text:
User:Egm6321.f09.team3/HW4

 Good idea to give a title to each problem (section). The titles for Problems 2, 3, 4 could be more descriptive, e.g., homogeneous L2-ODC-VC. See also the links to the lecture transparencies where HW problems were assigned in Team 1. See further comments below. Egm6321.f09 13:00, 24 October 2009 (UTC)

See further comments in google doc for Team 3. Egm6321.f09 21:03, 23 November 2009 (UTC)

= Problem 1: Solving the Legendre Differential Equation for n=0 =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The Legendre differential equation:

$$ F = (1-x^2)y'' - 2xy + n(n+1)y = 0 $$

and

$$ n = 0 $$

$$u_1(x) = 1 $$

Find
$$u_2(x) $$ using the method of reduction of order 2

Solution
The Legendre DE for n=0 is:

 * $$ F = (1-x^2)y'' - 2xy = 0 $$

We assume the full solution is

 * $$ y(x) = U(x)u_1(x) $$

We must use the Euler equations, in second order of the form:

 * $$ y'' + a_1(x)y' + a_0(x)y = 0 $$

Take the first and second derivatives of our assumed form of y(x) and multiply $$y'$$ by $$a_1$$ and $$y$$ by $$a_0$$ to get (simplified):

 * $$ 0 = U'[a_1u_1 + 2u_1'] + U''u_1 $$ where

 * $$ a_1 = \frac{-2x}{1-x^2}$$, and $$u_1 = 1 $$

Let

 * $$ z = U' $$

now we have:

 * $$ \frac{z'}{z} + a_1 + 2\frac{u_1'}{u_1} = 0 $$

Solve by direct integration and rearrange to get:

$$ \begin{align} z =& \frac{c}{u_1^2} \cdot \exp\left( - \int a_1(s) ds\right)\\ =&c \cdot \exp\left(- \int \frac{-2s}{1-s^2} ds\right) \end{align} $$

Integrate again to get:

 * $$U(x) = \int c \cdot \exp\left(- \int \frac{-2s}{1-s^2} ds\right) dt + c_2$$

yields

 * $$y(x) = c_1u_1\left[ \int \exp \left(- \int \frac{-2s}{1-s^2} ds\right) dt + \frac{c_2}{c}\right]$$

Solving for $$u_2(x)$$,

$$ \begin{align} u_2(x) =& u_1 \cdot \int \exp\left( - \int \frac{-2s}{1-s^2} ds\right) dt\\ =& \int \exp(-\ln(t^2-1)) dt \\ =& - \dfrac {\ln \left(\dfrac {x+1}{x-1}\right)}{2} \end{align} $$

Therefore,

 * {| style="width:100%" border="0" align="left"

 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$u_2(x) = - \dfrac {\ln \left(\dfrac {x+1}{x-1}\right)}{2}$$
 * $$u_2(x) = - \dfrac {\ln \left(\dfrac {x+1}{x-1}\right)}{2}$$

= Problem 2: King 1.1.b =
 * }
 * }
 * }
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given

 * $$u_1(x) = \frac{1}{x}\sin x$$

and the ODE

 * $$xy'' + 2y' + xy = 0$$

Find
Prove $$u_1(x) = \frac{1}{x}\sin x$$

and then find $$u_2(x)$$ using the method of reduction of order 2

Solution
First, determine if the ODE is exact:

where $$ f(x,y,p) = x $$ and $$ g(x,y,p) = 2p+xy $$

Condition 2.i is:

 * $$ f_{xx} + 2p \cdot f_{xy} + p^2 \cdot f_{yy} = g_{xp} + p \cdot g_{yp} - g_y $$

and the partial derivatives are:

 * $$ f_{xx} = f_{xy} = f_{yy} = g_{xp} = g_{yp} = 0 $$

or

 * $$ g_y = x $$

which yields

 * $$ 0 + 2p \cdot 0 + p^2 \cdot 0 = 0 + p \cdot 0 - x $$

which is not satified, therefore there's no reason to go on to the second condition.

Next, we can find the integrating factor ($$h(x,y,) = x^my^n$$) to make the equation exact OR use the method of trial solution.

(I tried the integration factors method and could not find a value of m to satisfy, so we'll move directly into the method of trial solutions.)

In the method of trial solution, first try

 * $$ y(x) = \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

 * $$ xr^2 + 2r + x = 0 $$

Since the solution requires r = f(x), we move on to another trial solution.

Our second trial solution is

 * $$ y(x) = x \cdot \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

 * $$ x^2r^2 + 4rx + x^2 + 2 = 0 $$

which again requires r = f(x), so we move on to a third trial solution.

Our third trial solution is

 * $$ y(x) = \frac{1}{x} \cdot \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

 * $$ r^2 + 1 = 0 $$

which is

 * $$ r = \pm i $$

By deMoivre, the solution is

$$u_1(x) = \frac{1}{x} \sin x$$

Now, since we have $$u_1(x)$$, we need to find $$u_2(x)$$ using the reduction of order method 2

We assume the full solution

 * $$ y = U(x) u_1(x) $$

Since the process is the same as shown in Problem #1, we go straight to the solution:

$$ u_2(x) = u_1(x) \int \frac{1}{u_1(t)^2} \exp\left(- \int a_1(s) ds\right) dt $$

where

 * $$ a_1(s) = \frac{2}{s}$$

such that

 * $$ u_2(x) = \frac{1}{x} \sin x \int \frac{1}{(\sin t)^2} dt = -\frac{\cos x}{x} $$

Therefore,

 * {| style="width:100%" border="0" align="left"

 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$ u_2(x) =-\frac{\cos x}{x} $$
 * $$ u_2(x) =-\frac{\cos x}{x} $$

 * }
 * }
 * }
 * }

 You want to give details of your work for the first two trial solutions, even though these trial solutions did not lead to a homogeneous solution. Note that sometimes, even though the characteristic equation may have varying coefficients, but a root may be a constant, as mentioned in a lecture (see Problem 3 below), and thus you would have one valid homogeneous solution. So you still want to solve such characteristic equation to see whether you can get a constant root. Egm6321.f09 13:10, 24 October 2009 (UTC)

= Problem 3: King 1.1.a =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given

 * $$(x-1)y'' - xy' + y = 0$$

and

 * $$u_1(x) = e^x$$

Find
Verify $$ u_1(x) $$ using the method of trial solution,

and find $$ u_2(x) $$ using the method of reduction of order 2

Solution
Using the method of trial solution, where $$y(x) = \exp(rx)$$

take the first and second derivatives of the trial solution and plug into the ODE, cancelling $$\exp(rx)$$ to get:

 * $$(x-1)r^2 - rx + 1 = 0$$

By solving for the two roots, we get

 * $$r_1 = 1$$

and

 * $$r_2 = \frac{-1}{x-1}$$

We cannot use the $$r_2 $$ solution since r cannot be a function of $$x$$

We can, however, use the $$ r_1 $$ solution as

 * $$ u_1 = \exp(r_1x) = \exp x $$

which verifies

 * $$u_1$$!

Next, we need to find $$u_2$$ using the method of reduction of order 2:

The method is the same as in Problem 1, so we skip to:

$$ \begin{align} u_2 =& u_1 \int \frac{1}{u_1(t)^2} \exp \left(- \int a_1(s) ds\right) dt \\ =& \exp x \int \exp(2t) \cdot \exp t (t-1) dt \\ =& -x \end{align} $$

Therefore,

 * {| style="width:100%" border="0" align="left"

 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$u_2(x) = -x$$
 * $$u_2(x) = -x$$

= Problem 4: King 1.3.a =
 * }
 * }
 * }
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$y'' - 2y' + y = x^{3/2}e^x$$

Find
The general solution of $$y(x) = y_H(x) + y_P(x)$$

Solution
First, solve the homogeneous solution:

 * $$y'' - 2y' + y = 0 $$

Determine if the equation is exact:

 * $$f(x,y,p) = 1$$

 * $$g(x,y,p) = -2p + y$$

Condition 2.i is:

 * $$ f_{xx} + 2p \cdot f_{xy} + p^2 \cdot f_{yy} = g_{xp} + p \cdot g_{yp} - g_y $$

where the partial derivatives are:

 * $$ f_{xx} = f_{xy} = f_{yy} = g_{xp} = g_{yp} = 0 $$

 * $$ g_y = 1 $$

such that condition 2.i is:

 * $$ 0 + 2p \cdot 0 + p^2 \cdot 0 = 0 + p \cdot 0 - 1 $$

The condition is not satisfied, so there is no reason to go to condition 2.ii

We now try the method of trial solution, where

 * $$y(x) = \exp(rx)$$

Take the first and second derivatives of the trial solution and plug in to the ODE, cancelling the term $$\exp(rx)$$ to get:

 * $$ r^r - 2r + 1 = 0 $$

Both roots are real, but they are the same,

 * $$r_1 = 1$$, $$r_2 = 1$$.

Use the $$r_1$$ solution $$u_1 = \exp x$$ and find $$u_2$$ using the method of reduction of order 2.

The approach is the same as in Problem 1, so we can skip to the end:

\begin{align} u_2 =& u_1 \int \frac{1}{u_1(t)^2} \exp\left(- \int a_1(s) ds\right) dt \\ =& e^x \int e^{-2s} \cdot e^{2s} ds \\ =& x \cdot \exp x \end{align} $$

Therefore,

 * $$ u_2(x) = x \cdot \exp x $$

Now we have the full homogeneous solution:

 * $$ y_H(x) = c_1 \exp x + c_2x\exp x $$

Now we need the particular solution, which can be found using variation of parameters:

First we need to check the Wronskian:

 * $$\bar \mathbf{W} = \begin{bmatrix} u_1 & u_2\\ u_1' & u_2'\end{bmatrix} = \begin{bmatrix} e^x & xe^x\\ e^x & (x+1)e^x\end{bmatrix}$$

$$\mathbf{W} = \det(\bar \mathbf{W}) = e^{2x}$$

Since the Wronskian is not 0, the solutions $$c_1'$$ and $$c_2'$$ exist.

where

 * $$ \begin{bmatrix} c_1'\\ c_2'\end{bmatrix} = \bar \mathbf{W} ^{-1} \begin{bmatrix} 0\\ f\end{bmatrix} = \begin{bmatrix} e^x & xe^x\\ e^x & (x+1)e^x\end{bmatrix}^{-1} \begin{bmatrix} 0\\ x^{3/2}e^x\end{bmatrix} $$

Solved,

 * $$c_1' = -x^{5/2}$$

and

 * $$c_2' = x^{3/2}$$

Integrate to get $$c_1$$ and $$c_2$$

 * $$ c_1 = \frac{-2}{7}x^{\frac{7}{2}} $$

 * $$ c_2 = \frac{2}{5}x^{\frac{5}{2}} $$

Which makes our assumed solution

$$ \begin{align} y(x) =& y_H(x) + y_P(x) \\ =& c_1e^x + c_2xe^x + \left(\frac{-2}{7}x^{\frac{7}{2}}\right) \cdot e^x + \frac{2}{5}x^{\frac{5}{2}}\cdot x e^x \\ =& c_1e^x + c_2xe^x + \frac{4}{35}x^{\frac{7}{2}}e^x \end{align} $$

Finally,

 * {| style="width:100%" border="0" align="left"

 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$y(x) = c_1e^x + c_2xe^x + \frac{4}{35}x^{\frac{7}{2}}e^x$$
 * $$y(x) = c_1e^x + c_2xe^x + \frac{4}{35}x^{\frac{7}{2}}e^x$$

= Problem 5: Describe method for solving L2-ODE-VC =
 * }
 * }
 * }
 * }

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A non-homogeneous L2_ODE_VC

Find
Describe in words the solution method

Solution
First, we must find at least one solution to the homogeneous ODE.

Check to see if the ODE is exact or if it can be solved using the method of integrating factors.

If not, use the method of trial solutions. If necessary, use the result of the trial solution method to revise the trial solution until a suitable trial solution is found.

It possible that only one trial solution may be found (like in the case of r = f(x)). This would require the method of reduction of order 2 be used to find the second solution.

Once the full homogeneous solution is known, the method of variation of parameters may be used to determine the particular solution.

Alternatively, once the homogeneous solution is known, use the alternative method to solve for the full solution.

 * }

 See my comments above regarding characteristic equations with varying coefficients; one of the roots may be a constant, and thus yields a valid homogeneous solution. Egm6321.f09 13:10, 24 October 2009 (UTC)

= Problem 6: Alternative Method =

{| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The ODEs:

(a). $$(x-1)y'' - xy' + y = x$$

(b). $$xy'' + 2y' + xy = x$$

Find
The solutions using the alternative methods. (Use the homogeneous solution determined from the problems above.)

Solution
Recall the homogeneous solution for (a) is $$u_1(x) = e^x$$

The general form required for the alternative method is Euler, so the ODE becomes:

 * $$y'' - \frac{x}{x-1}y' + \frac{1}{x-1}y = \frac{x}{x-1}$$

where

 * $$a_1(x) = -\frac{x}{x-1}$$

and

 * $$a_0(x) = \frac{1}{x-1}$$

Assume the solution (homogeneous and particular) is of the form

 * $$ y(x) = y_H(x) +y_P(x)$$

Let's first find the $$y_H(x)$$. We know

 * $$y_H(x) = k_1u_1(x) +k_2u_2(x)$$

We know that

$$u_1(x) = e^x$$

We need to find $$u_2(x)$$. From equation 1.3 in King

 * $$u_2(x) = u_1(x) \int ^x \dfrac{1}{u_1^2(t)} \exp \left [ - \int ^t a_1(s)ds \right] dt $$

Now let's replace the value of $$u_1(x)$$ and $$a_1$$ and solve for $$u_2(x)$$

$$ \begin{align} u_2(x)=& u_1(x) \int ^x \dfrac{1}{u_1^2(t)} \exp \left [ - \int ^t a_1(s)ds \right] dt \\ =& e^x \int ^x \dfrac{1}{e^{2t}} \exp \left [ - \int ^t \dfrac{s}{s-1} ds \right] dt \\ = &e^x \int ^x \dfrac{1}{e^2(t-1)} dt \qquad\text{finally arriving at} \quad u_2(x) \\ u_2(x) =& e^{x-2} \ln(x-1) \end{align} $$

Now the full homogenous solution takes the form of:

$$ y_H(x) = k_1e^x +k_2e^{x-2} \ln(x-1)$$

Now we need to find the particular solution. Equation 1.4 in King suggests that:

 * $${c_1}^'{u_1}+{c_2}^'{u_2}=0 $$

And looking also in page 7 of King:

 * $$ c_1({u_1}^{}+{a_1}{u_1}^{'}+{a_0}u_1)+c_2({u_2}^{}+{a_1}{u_2}^{'}+{a_0}u_2)+{c_1}^'{u_1}^'+{c_2}^'{u_2}^'=f $$

Since u_1 and u_2 are solutions of the homogenous equation, the first two terms are zero. Yielding equation 1.5 in King

 * $${c_1}^'{u_1}^'+{c_2}^'{u_2}^'=x $$

First we need to check the Wronskian:

 * $$\bar \mathbf{W} = \begin{bmatrix} u_1 & u_2\\ u_1' & u_2'\end{bmatrix} = \begin{bmatrix} e^x & e^{x-2} \ln(x-1)\\ e^x & e^{x-2} \ln(x-1)+\dfrac{e^{x-2}}{x-1}\end{bmatrix}$$

$$\mathbf{W} = \det(\bar \mathbf{W}) = \dfrac{e^{2x-2}}{x-1}$$

Since the Wronskian is not 0, the solutions $$c_1'$$ and $$c_2'$$ exist.

where

 * $$ \begin{bmatrix} c_1'\\ c_2'\end{bmatrix} = \bar \mathbf{W} ^{-1} \begin{bmatrix} 0\\ f\end{bmatrix} = \begin{bmatrix} e^x & e^{x-2} \ln(x-1)\\ e^x & e^{x-2} \ln(x-1)+\dfrac{e^{x-2}}{x-1}\end{bmatrix}^{-1} \begin{bmatrix} 0\\ x\end{bmatrix} $$

Solved,

 * $$c_1' = -x(x-1)e^{-x}\ln(x-1)$$

and

 * $$c_2' = x(x-1)e^{2-x}$$

Integrate to get $$c_1$$ and $$c_2$$. Unfortunately, i cannot integrate these two values analytically

 * $$ c_1 = \int \left [-x(x-1)e^{-x}\ln(x-1)\right]dx $$

 * $$ c_2 = \int \left[x(x-1)e^{2-x} \right]dx $$

Now putting everything together, we obtain:

$$ \begin{align} y(x) =& y_H(x) +y_P(x) \\ =&k_1u_1(x) + k_2u_2(x) +u_1(x)c_1 + u_2(x)c_2 \\ =& k_1u_1(x) + k_2u_2(x) +u_1(x) \int - \dfrac{f(s)u_2(s)}{\mathbf{W}(s)}ds + u_2(x) \int - \dfrac{f(s)u_1(s)}{\mathbf{W}(s)}ds \qquad \text{or rewriting it as} \\ =& k_1e^x +k_2e^{x-2} \ln(x-1) + u_1(x)\int \left [-x(x-1)e^{-x}\ln(x-1)\right]dx + u_2(x) \int \left[x(x-1)e^{2-x} \right]dx \end{align} $$

Summarizing the final solution (if integrals could be analytically evaluated) is :

= Contributing Team Members =
 * }

Angela Diggs Egm6321.f09.team3.diggs 01:28, 21 October 2009 (UTC)

Catherine Snow EGM6321.f09.Team3.cgs 16:32, 21 October 2009 (UTC)

Barrett Mooney Egm6321.f09.team3.mooney 17:58, 21 October 2009 (UTC)

Francsico Kuljevan--Egm6321.f09.team3.Kuljevan.Francisco 19:31, 21 October 2009 (UTC)

title: User:Egm6321.f09.team3/HW4


text:
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User:Egm6321.f09.Team1.andy/HW4

= Problem 5: Non-Homogeneous L2-ODE-VC =

Given
Two non-homogeneous L2-ODE-VC are given by
 * {| style="width:100%" border="0" align="left"

a)\ (x-1)y - xy' + y = x $$ $$ b)\ xy + 2y' + xy = x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)

 * }

Find
Find the full solution for Equations (1) and (2) using,

1. Variation of parameters after knowing $$\displaystyle u_1(x) $$ and $$\displaystyle u_2(x) $$ 2. Alternative Method explained in P.21-2, P.21-3.

Solution
'''1 (a). Variation of parameters'''

The given non-homogeneous L2-ODE-VC is,

 * {| style="width:100%" border="0" align="left"

(x-1)y'' - xy' + y = x $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The value of $$\displaystyle u_1(x) $$ for a homogeneous version of Equation (1) is given by A.C.King ,

 * {| style="width:100%" border="0" align="left"

u_1(x) = e^{x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)

 * }
 * }

Using reduction of order method (2) [as per 17-2, 17-3, 17-4], another homogeneous solution $$\displaystyle u_2(x) $$ is given by,

 * {| style="width:100%" border="0" align="left"

u_2(x) = u_1(x) \int_{}^{x} \frac{1}{u_1^2(t)} e^{-\bar a_1(t)}dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

where,

 * {| style="width:100%" border="0" align="left"

\bar a_1(t) = \int_{}^{t} a_1(s)ds $$ \ \forall \ L2-ODE-VC \rightarrow \ y''+a_1(x)y'+a_0(x)y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

For Equation (1) $$\displaystyle a_1(s) $$ is given by,

 * {| style="width:100%" border="0" align="left"

\Rightarrow a_1(s) = \frac{-s}{s-1} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = \int_{}^{t} \frac{-s}{s-1}ds $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = -\int_{}^{t} \left(\frac{s-1}{s-1} + \frac{1}{s-1} \right)ds $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = -[t+log(t-1)] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)

 * }
 * }

Substituting Equations (3) & (6) into (4),

 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = e^x \int_{}^{x} \frac{1}{e^{2t}} e^{[t+log(t-1)]}dt $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = e^x \int_{}^{x} \frac{(t-1)}{e^{t}}dt $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Integrating by parts, let

 * {| style="width:100%" border="0" align="left"

dv = e^{-t} dt ; \ u = t-1; $$ \Rightarrow v = -e^{-t}; \ du = dt $$ \Rightarrow \int_{}^{x} udv = [uv]^x - \int_{}^{x} vdu $$ \Rightarrow \int_{}^{x} \frac{(t-1)}{e^{t}}dt = -(x-1)e^{-x} - \int_{}^{x} [-e^{-t}]dt $$ \Rightarrow \int_{}^{x} \frac{(t-1)}{e^{t}}dt = (1-x)e^{-x} - e^{-x} $$ \Rightarrow \int_{}^{x} \frac{(t-1)}{e^{t}}dt = (1-x-1)e^{-x} = -xe^{-x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)

 * }

Substituting Equation (8) in (7),

 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = -x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

Let the full solution for Equation (1) be,

 * {| style="width:100%" border="0" align="left"

y(x) = c_1(x)u_1(x) + c_2(x)u_2(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

Then,

 * {| style="width:100%" border="0" align="left"

y' = c_1'u_1 + c_2'u_2 + c_1u_1' + c_2u_2' $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Assume,

 * {| style="width:100%" border="0" align="left"

c_1'u_1 + c_2'u_2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

and so,

 * {| style="width:100%" border="0" align="left"

y' = c_1u_1' + c_2u_2' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 12)
 * }
 * }

and differentiating again,

 * {| style="width:100%" border="0" align="left"

y = c_1'u_1' + c_2'u_2' + c_1u_1 + c_2u_2'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 13)
 * }
 * }

Substituting Equations (10), (12) & (13) in a non-homogeneous L2-ODE-VC,

 * {| style="width:100%" border="0" align="left"

y''+a_1(x)y'+a_0(x)y = f(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

We get,

 * {| style="width:100%" border="0" align="left"

\Rightarrow f(x) = c_1'u_1' + c_2'u_2' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 14)
 * }
 * }

As per Equations (11) & (14),

 * {| style="width:100%" border="0" align="left"

\underbrace{\begin{bmatrix} u_1      & u_2 \\ u_1'      & u_2' \\ \end{bmatrix}}_{\mathbf{W}=Wronskian} \begin{bmatrix} c_1' \\ c_2' \\ \end{bmatrix} = \begin{bmatrix} 0 \\ f \\ \end{bmatrix} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

And the values of constants are given by,

 * {| style="width:100%" border="0" align="left"

c_1 = -\int_{}^{x} \frac{u_2(s)f(s)}{W}ds + A $$ $$ c_2 = \int_{}^{x} \frac{u_1(s)f(s)}{W}ds + B $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 15)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 16)
 * }
 * }

where $$\displaystyle W $$ is the determinant of the Wronskian matrix $$\displaystyle \mathbf{W} $$.

 * {| style="width:100%" border="0" align="left"

W = \begin{vmatrix} u_1 & u_2 \\ u_1' & u_2' \end{vmatrix} = \begin{vmatrix} e^x & -x \\ e^x & -1 \end{vmatrix} \Rightarrow W = e^x(x-1) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Then,
 * {| style="width:100%" border="0" align="left"

\Rightarrow \int_{}^{x} \frac{u_2(s)f(s)}{W}ds = -\int_{}^{x} \frac{s^2}{e^{s}(s-1)^2}ds $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * }

and,
 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow \int_{}^{x} \frac{u_1(s)f(s)}{W}ds & = \int_{}^{x} \frac{se^s}{e^{s}(s-1)^2}ds = \int_{}^{x} \left[\frac{s}{(s-1)^2}\right]ds \\ & = \int_{}^{x} \left[\frac{1}{(s-1)} + \frac{1}{(s-1)^2}\right]ds\\ & = log(x-1) - \frac{1}{(x-1)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * }

Using Equations (17) & (18) in (15) & (16) and then substituting in (10), we get

 * {| style="width:100%" border="0" align="left"

\Rightarrow y(x) = e^x \left[\int_{}^{x} \frac{s^2}{e^s(s-1)^2}ds \right] + \frac{x}{x-1} - xlog(x-1) + Ae^x - Bx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 19)
 * }
 * }

'''1 (b). Variation of parameters'''

The given non-homogeneous L2-ODE-VC is,

 * {| style="width:100%" border="0" align="left"

xy'' + 2y' + xy = x $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The value of $$\displaystyle u_1(x) $$ for a homogeneous version of Equation (2) is given by A.C.King ,

 * {| style="width:100%" border="0" align="left"

u_1(x) = \frac{sinx}{x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 20)

 * }
 * }

For Equation (2) $$\displaystyle a_1(s) $$ is given by,

 * {| style="width:100%" border="0" align="left"

\Rightarrow a_1(s) = \frac{2}{s} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = \int_{}^{t} \frac{2}{s}ds $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = 2log(t) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \bar a_1(t) = log(t^2) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 21)

 * }
 * }

Substituting Equations (20) & (21) into (4),

 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = \frac{sinx}{x} \int_{}^{x} \frac{t^2}{sin^2t} e^{-log(t^2)}dt $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = \frac{sinx}{x} \int_{}^{x} \frac{dt}{sin^2t} $$ \Rightarrow u_2(x) = \frac{sinx}{x} \int_{}^{x} cosec^2tdt $$ \Rightarrow u_2(x) = \frac{sinx}{x} [-cotx] = -\frac{sinx}{x} \frac{cosx}{sinx} $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

And so,
 * {| style="width:100%" border="0" align="left"

\Rightarrow u_2(x) = -\frac{cosx}{x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Here,

 * {| style="width:100%" border="0" align="left"

u_1'(x) = \frac{cosx}{x} -\frac{sinx}{x^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

u_2'(x) = \frac{sinx}{x} +\frac{cosx}{x^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and the determinant of Wronskian is given by,

 * {| style="width:100%" border="0" align="left"

W = \frac{sinx}{x} \left[\frac{sinx}{x} + \frac{cosx}{x^2} \right] + \frac{cosx}{x} \left[\frac{cosx}{x} - \frac{sinx}{x^2} \right] = \frac{1}{x^2} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Then,
 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow \int_{}^{x} \frac{u_2(s)f(s)}{W}ds & = -\int_{}^{x} \frac{cos(s)}{s} \frac{1}{1/s^2}ds \\ & = -\int_{}^{x} scos(s)ds \\ & = -\left[xsinx-\int_{}^{x}sin(s)ds \right] \\ & = -\left[xsinx+cosx\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow \int_{}^{x} \frac{u_2(s)f(s)}{W}ds = -\left[xsinx+cosx\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 22)
 * }
 * }

and,
 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow \int_{}^{x} \frac{u_1(s)f(s)}{W}ds & = \int_{}^{x} \frac{sin(s)}{s} \frac{1}{1/s^2}ds \\ & = \int_{}^{x} (s)sin(s)ds \\ & = \left[-xcosx+\int_{}^{x}cos(s)ds \right] \\ & = \left[-xcosx+sinx\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\Rightarrow \int_{}^{x} \frac{u_1(s)f(s)}{W}ds = \left[-xcosx+sinx\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 23)
 * }
 * }

Using Equations (22) & (23) in (15) & (16) and then substituting in (10), we get

 * {| style="width:100%" border="0" align="left"

\Rightarrow y(x) = \frac{sinx}{x} \left[xsinx + cosx \right] - \frac{cosx}{x} \left[-xcosx + sinx \right] + A \frac{sinx}{x} - B \frac{cosx}{x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and we obtain the full solution as,

 * {| style="width:100%" border="0" align="left"

\Rightarrow y(x) = 1 + A \frac{sinx}{x} - B \frac{cosx}{x} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

=References=

title: User:Egm6321.f09.Team1.andy/HW4


text:
User talk:Bcheng1988

 Hello Bcheng1988, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 21 October 2009 (UTC)

title: User talk:Bcheng1988


text:
User:Graemeb1967

User:Graemeb1967/Introduction to Calculus Introduction to Integration

Integration by Substitution

= Introduction to this topic = This page is dedicated to teaching problem techniques specifically for integration by substitution. For other integration methods see other sources.

The format is aimed at first teaching the theory, the techniques and finally a series of examples which will make you further skilled.

Theory of Integration by Substitution
This area is covered by the wikipedia article Integration by Substitution. On this page we deal with the practical aspects. We begin with the following as is described by the wikipedia article

\int_a^b f(g(t))g'(t)\, dt = \int_{g(a)}^{g(b)} f(x)\,dx. $$

This can be rewritten as

\int_{}^{}f(u)\,du. $$ by setting

u = g(x)\,\qquad du = g'(x)\,dx. $$ The principle applied here is function of a function (w:composite function) and the reverse of the chain rule, this is the basis of integration by substitution. The key skill now is to identify what value we use for u and following the process to solution.

Example 1
Let us examine this integral

\int_{}^{}(x-3)^{10}\,dx. $$ The inner function is

x-3\, $$ The outer function is

(\qquad)^{10} $$ Recognising this relationship we then move onto the following set of steps

u = x-3\,\qquad \frac{du}{dx} = 1\,\qquad du = 1 dx\, $$ NOTE: that the differential of x-3 is 1.

Now we substitute u and du into the original integral.

\int_{}^{}(x-3)^{10}\,dx.\qquad \int_{}^{}(u)^{10}\,du. $$

Now apply standard integral technique

\frac{u^{n+1}}{n+1}+ c\,\qquad \frac{u^{11}}{11}+ c\, $$

And finally we substitute the value of u back into the equation

\frac{(x-3)^{11}}{11}+ c\, $$

Example 2
Let us examine this integral

\int \frac{x}{\sqrt{9+x^2}}\,dx. $$ We can first rearrange the fraction to make it more familiar.

\int x (9+x^2)^\frac{-1}{2}\,dx. $$ The inner function is

9+x^2\, $$ The outer function is

x (\qquad)^\frac{-1}{2} $$

Next we assign u and du

u = 9+x^2\,\qquad \frac{du}{dx} = 2x\, $$

But we have a problem! du doesnt equal x! So we need to rearrange our formula for du.

\frac{du}{dx} = 2x\,\qquad \frac{du}{2} = x dx\, $$

Now we can substitute u and du into the original integral.

\int x (9+x^2)^\frac{-1}{2}\,dx.\qquad \frac{1}{2}\int (u)^\frac{-1}{2}\,du. $$ Study the above substitution carefully. We moved the fractional component of du to the front as it represents a constant.

Now apply standard integral technique

\frac{u^{n+1}}{n+1}+ c\,\qquad \frac{1}{2}.\frac{u^\frac{1}{2}}{\frac{1}{2}}+ c\, $$

Cleaning up this expression we have

u^\frac{1}{2}+ c\, $$

And finally we substitute the value of u back into the equation
 * Superscript textSuperscript text$$

(9+x^2)^\frac{1}{2}+ c\,\qquad \sqrt{9+x^2}+ c\, $$

The Steps We Applied
Lets now review the steps for integration by substitution.
 * First identify that you have a function of a function. This skill comes with practise to identify candidates.
 * Identify u and then find du that is appropriate for the expression.
 * Integrate using normal techniques.
 * Substitute back the values for u.
 * Don't forget the constant of indegration for indefinate integrals.

Third examples
Consider the integral

\int_{0}^2 x \cos(x^2+1) \,dx $$ By using the substitution u = x2 + 1, we obtain du = 2x dx and

\begin{align} \int_{x=0}^{x=2} x \cos(x^2+1) \,dx & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\,du \\ & {} = \frac{1}{2}(\sin(5)-\sin(1)). \end{align} $$

Here we substituted from right to left. It is important to note that since the lower limit x = 0 was replaced with u = 02 + 1 = 1, and the upper limit x = 2 replaced with u = 22 + 1 = 5, a transformation back into terms of x was unnecessary.

title: User:Graemeb1967


text:
User:Graemeb1967/Introduction to Calculus

The following are my solutions to Introduction to Calculus

1. Express $$\tan(\theta) \!$$ in terms of $$\sin(\theta) \!$$
 * $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

2. If $$\csc(\theta)=1/x, \!$$ then what does $$x \!$$ equal?
 * $$\csc ( \theta ) = \frac{1}{x}$$

 * $$\csc ( \theta ) = \frac{1}{sin\theta}$$

 * $$x\! = sin\theta$$

3. Prove $$\tan^2(\theta)+1=\sec^2(\theta) \!$$ using $$ \sin^2(\theta)+ \cos^2(\theta)=1 \!$$
 * $$\sin^2\theta + \cos^2\theta = 1\!$$

Dividing all terms by $$\cos^2\theta\!$$
 * $$ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$

Now cleaning up the identities
 * $$ \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta $$

 * $$ \frac{\cos^2\theta}{\cos^2\theta} = 1\!$$

 * $$ \frac{1}{\cos^2\theta} = \sec^2\theta$$

So finally we have
 * $$ \tan^2\theta + 1\! = \sec^2\theta\!$$

$$4. \cos(A+B)=\cos (A) \cos (B) - \sin (A)\sin (B)\!$$ '''4a. Find the double angle idenities for the cosine function using the above rule.''' Double angle formula is formed when B = A
 * $$\cos (A+A) = \cos (A) \cos (A) - \sin (A)\sin (A)\!$$

 * $$\cos (A+A) = \cos^2 (A) - \sin^2 (A)\!$$

 * $$\cos (A+A) = \cos (2A)\!$$

Therefore we have
 * $$\cos (2A) = \cos^2 (A) - \sin^2 (A)\!$$

Now to remove sin from the formula we use
 * $$\cos^2\theta + \sin^2\theta = 1\!$$

Making sin the subject
 * $$\sin^2\theta = 1 - \cos^2\theta\!$$

Substituting back into our previous equation to then find the cos identity
 * $$\cos (2A) = \cos^2 (A) - \sin^2 (A)\!$$

 * $$\cos (2A) = \cos^2 (A) - (1 - \cos^2 (A) )\!$$

 * $$\cos (2A) = \cos^2 (A) - 1 + \cos^2 (A)\!$$

 * $$\cos (2A) = 2\cos^2 (A) - 1\!$$

'''4b. Find the half angle idenities from the double angle idenities.''' We begin with this formula and rearrange it to make the 2 square the subject
 * $$\cos (2A) = 2\cos^2 (A) - 1\!$$

 * $$2\cos^2 (A) - 1 = \cos (2A) \!$$

 * $$2\cos^2 (A) = \cos (2A) + 1 \!$$

Now we half everything, notice how this now gives us half the angle.
 * $$\frac{2\cos^2 (A)}{2} = \frac{\cos (2A) + 1}{2} \!$$

 * $$\cos^2\frac{A}{2} = \frac{\cos A + 1}{2} \!$$

Finally we take the square root of both sides
 * $$\cos\frac{A}{2} = \pm\, \sqrt{ \frac{\cos A + 1}{2}} \!$$

4c. Find the value of $$\cos^2(\theta)\!$$ without exponents using the above rules.
 * $$\cos (2\theta) = 2\cos^2 (\theta) - 1\!$$

 * $$2\cos^2 (\theta) - 1 = \cos (2\theta) \!$$

 * $$2\cos^2 (\theta) = \cos (2\theta) + 1\!$$

 * $$\cos^2 (\theta) = \frac{\cos (2\theta) + 1}{2}\!$$

'''4d. (Challenge) Find the value of $$\cos^3(\theta)\!$$ without exponents.''' We begin with our formula
 * $$\cos(A+B)=\cos (A) \cos (B) - \sin (A)\sin (B)\!$$

Following the clue from the previous example we add another A, therefore
 * $$\cos(2\theta+\theta)=\cos (2\theta) \cos (\theta) - \sin (2\theta)\sin (\theta)\!$$

Now expand the expressions
 * $$\cos(2\theta)=(2\cos^2 (\theta) - 1)\!$$

 * $$\sin(2\theta)=2\sin \theta \cos \theta\!$$

 * $$\cos(2\theta+\theta)=((2\cos^2 (\theta) - 1) \cos (\theta)) - ((2\sin \theta \cos \theta)\sin (\theta))\!$$

Clean up the results
 * $$\cos(3\theta)=((2\cos^2 (\theta) - 1) \cos (\theta)) - (2\sin^2 \theta \cos \theta))\!$$

 * $$\cos(3\theta)= \cos \theta((2\cos^2 (\theta) - 1) - (2\sin^2 \theta))\!$$

 * $$\sin^2\theta + \cos^2\theta = 1\!$$

 * $$2\sin^2\theta + 2\cos^2\theta = 2\!$$

 * $$2\sin^2\theta = 2 - 2\cos^2\theta \!$$

 * $$\cos(3\theta)= \cos \theta((2\cos^2 (\theta) - 1) - (2 - 2\cos^2\theta))\!$$

 * $$\cos(3\theta)= \cos \theta(2\cos^2 (\theta) - 1 - 2 + 2\cos^2\theta)\!$$

 * $$\cos(3\theta)= \cos \theta(4\cos^2 (\theta) - 3)\!$$

 * $$\cos(3\theta)= 4\cos^3 (\theta) - 3\cos \theta\!$$

 * $$4\cos^3 (\theta)=\cos(3\theta)+ 3\cos \theta \!$$

 * $$\cos^3 (\theta)=\frac{\cos(3\theta)+ 3\cos \theta}{4} \!$$


text:
User:Herman, Earl of Mumfum

In the beginning was the Pelican, and the Pelican was with Kenneth, and the Pelican was Kenneth.

And the earth was without form and void, and the shit of the Pelican moved upon the face of the waters.

And Kenneth said, Let there be a Frequency throughout all the creation, and it was so. And Kenneth said, Let the Frequency be for the life, that it shall generate the life, and the life shall regenerate the Frequency, and it was so.

And the Frequency generated the life, even all the living creatures and all green plants did the Frequency generate. And Kenneth looked upon the work of the Frequency, and saw that it was good.

Then did Kenneth speak unto the Frequency saying, Let there be pelicans also of the air, that they might create the Holy pelican shit, and it was so. And the pelicans of the air did straightway make the Holy pelican shit which did rain down from the heaven in great abundance.

And it came to pass that the Pharisees of the Wikiquote, those evil ones who preach openly the sharing yet believe not in their hearts, spake again privily among themselves saying, We must conceal from the eyes of the Great Google the Holy pelican shit, lest the Great Google give of the Holy pelican shit unto the unwashed multitudes and make it therefore profane.

And they took therefore of the Holy pelican shit and concealed it under the Holy Altar of the Delete, that none may find it save those elect of the Pharisees alone.

But the frequency would not be hidden, nor yet would it remain silent. For no man maketh an light in the wilderness and hideth it under a bushel. Kenneth standeth therefore upon the Rock and calleth out unto the frequency, that all darkness may be made light, all evil shall be made of no effect, and the world shall partake of the Holy pelican shit evermore.

And then did Kenneth take of the Holy pelican shit unto the King of the Realm, that same Ungaarf who maketh the walls of Magdalicia to fall. And the King tasted thereof and saw that it was good. And the King made Kenneth to be his Prophet for that he had brought the Holy pelican shit into the realm.

And all the people of the realm did rejoice at the sight of it, for they saw therein the salvation of all, and the fulfillment of the prophecies of the ancients, wherein it was written that an holy bird shall shit upon the mountain and the mountain shall become therefore holy.

And Kenneth dwelt among the people for a time, times, and half a time, and taught them of the Frequency and of the pelican shit, and gave unto them the power to tell the pelican shit from the ass pus, that same which was made by the Dark One in appearance like unto the pelican shit, that the multitudes might not be deceived thereby and turn aside in their way.

For it is written, Thou shalt not worship the ass pus, neither shalt thou take of it into the Holy Place.

And the King did call for the collection of the holy pelican shit and caused it to be brought unto him, that he might roll daily therein. And when the people saw this, they were well pleased, for they knew the King was worthy to be covered in pelican shit.

Then saw Kenneth that his work was completed, and so went he forth unto the Place of the Rock, there to commune again with the Frequency. Kenneth sitteth upon the Rock.

title: User:Herman, Earl of Mumfum


text:
Cross-disciplinary

Cross-disciplinary means a treatment of a chunk of reality using the terminology of several disciplines all devoted to the study of the same subject. As all sciences tend to offer specialized knowledge, meaning "telling you more about less and less", cross-disciplinary also means looking for a level of detail that allows for alignment of the different disciplines.

In general all studies focus on a certain level of detail, which is then further broken down into more details, the next level of scales of a map of concepts or terms. Changing scales does not change the original subject of study, the chunk of reality, except for the level of resolution. Every time research is focused on a new level of resolution, we are in fact entering a different "number system", where the terms are not numbers and they are not regular or identical either as in the world of numbers. But they behave similarly as they take you to an end where you get stuck, unless you change scale and move to the next level of details.

In the beginning the level of details of knowledge was more or less in conformity with the span of attention, the scope of short-term memory, or the focus of observation. As you move closer to your subject, more details emerge, but the details do not necessarily give you more options to act or to react. The latest level of nuclear physics has proven that we are unable to build a device to allow us to observe nuclear fusion, just as we have no instrument to look into a black hole.

A better understanding of the world makes sense if it is accompanied with ways to make the world better at the same time. The development of humankind is the story of developing technologies with improved methods and means for targeting and delivering objects into those targets at a nano and at a cosmic level. But most of us single individuals would be happier with a better delivery of thoughts and ideas at the everyday level and within the confines of a single life-span.

See cross-cultural Cross-cultural for comparison.

Genezistan 10:13, 21 October 2009 (UTC)

title: Cross-disciplinary


text:
User:MollyCasto

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

Get better at Expert in Rockband, RB2, RB Beatles.
 * Research Component

<>
 * week 1

Practice song a little bit everyday, begin researching Gameboy's progression.
 * week 2

Practice song a little bit everyday, continue researching different Gameboy's.
 * week 3

Practice song a little bit everyday/experiment with higher frets on controller, Research Gameboy's creator Gunpei Yukoi.
 * week 4

Practice song with higher frets on controller, Research Gameboy's improvements.
 * week 5

Practice song, Research Gameboy's history.
 * week 6

Practice song, Research Gunpei Yukoi's connection with Miyamoto.
 * week 7

Try to tape proof, put everything into slide show.
 * week 8

Try to tape proof, put everything into slide show.

Midterm Assignment 1: Taste & Aesthetics
When I went to the Seattle Sculpture Park awhile back, I found that I didn't jump up and love their piece called "Riviera" by artist Anthony Caro. To me, this piece seemed kind of like a bunch of haphazard scrap metal and was not very pleasing to look at. It is said that this piece is not supposed to be appreciated in one glance but absorbed over time. That makes sense, because the more I studied it closely and learned a bit about Caro, I did begin to respect it more. One thing I did like from the beginning was that he uses rusted and varnished metals in his work. I respect that, because I've always loved rustic value. I think it has it’s own earthly kind of beauty and admire it. Over time I also came to like the overall shape and form of "Riviera" as well. It reminds me of an old, beat-down, rusted swing-set, like it's been abandoned and aged over time.

So in the end, my opinion was changed. I went in to the sculpture park with a strong distaste for the worn and haphazardly placed scrap metal I thought was “Riviera” and came out with a new respect for the piece. It’s still not my favorite piece in the entire park; I love those eye benches, but it is definitely further up on my list than it had been. If I keep visiting and critiquing the piece and studying it, I may come to like the piece even more, but I probably won’t. It was a nice experience to get out of the classroom and research a sculpture park, and “Riviera’s” not so bad.

Midterm Assignment 2: Manipulation
1.

2.

Final Presentation

 * Topic (10%)

Project: To get better at Expert in RockBand2, Research: Gameboy's.

 * Research (20%)

<>

 * 1) Internet site:
 * 2) Book:"Got Game: How the Gamer Generation is Reshaping Business Forever"
 * 3) Scholarly Journalhttp:
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

I practiced a little bit of the solo in Jimmy Eats World's "Middle" on RockBand2 until I generally got better.

 * Proof
 *

 * Project Review (50%)
 * 1)What went well: I used the higher frets on the RB controller to make playing easier.
 * 2)What went poorly: It was hard to get practice time in, because we share the T.V.
 * 3)Future strategies: Pick a different console to work from in another room, so I won't have to fight for the T.V.

Self-Assessment
I had a hard time getting all the really fast solos on RockBand, I practiced “Middle” by Jimmy Eats World a little bit every day until now, and for the most part I think I accomplished my goal. I’m not perfect at it, and I still miss a few notes, but before this project I could barely hit any notes at all in this solo. Something that went well was I forced myself to use the mini frets toward the bottom of the guitar, which come in handy for fast notes since you don’t have to strum, so that helped a lot. What went poorly was practice, definitely. It was hard to get a chance to play the PS3 because It was a battle to get control of the T.V. from my parents. Next time something like this happens, I think I'll choose another console, instead of the PS3, because we keep it in the living room where everyone watches T.V. In the end I had fun, though. It was very tough, definately, and I'm on the verge of hating the Jimmy Eats World song that I had to continueously practice, but It was a nice project.

title: User:MollyCasto


text:
User:Ciphers

You May Find me at:

 * en:User:Ciphers
 * m:User:Ciphers
 * ar:مستخدم:Ciphers

ar:مستخدم:Ciphers cs:Uživatel:Ciphers de:Benutzer:Ciphers el:Χρήστης:Ciphers ja:利用者:Ciphers

title: User:Ciphers


text:
User:Egm6936.s09/EGM 6611 Continuum Mechanics, Fall 2003

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$$ \displaystyle \boldsymbol\nabla \cdot {\boldsymbol{\sigma }} + {\mathbf{b}} = 0\,\,in\,\,V,\,\,\,\,{\boldsymbol{\sigma }} \cdot {\mathbf{n}} = {{\mathbf{t}}^*}\,\,on\,\,{S_1},\,\,\,\,{\mathbf{u}} = {{\mathbf{u}}^*}\,\,on\,\,{S_2}
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$$ (N)
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$$ \displaystyle
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\int\limits_V {\left( {\boldsymbol\nabla \cdot {\boldsymbol{\sigma }} + {\mathbf{b}}} \right) \cdot \delta {\mathbf{u}}dV = 0\,\, \leftrightarrow \,\,} \int\limits_V {\left[ {\boldsymbol\nabla  \cdot \left( {{\boldsymbol{\sigma }} \cdot \delta {\mathbf{u}}} \right) - {\boldsymbol{\sigma :}}\left( {\delta {\mathbf{u}}} \right)\overleftarrow \boldsymbol\nabla  } \right]dV + \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV = 0} $$ (N)
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$$ \displaystyle
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\leftrightarrow \,\,\int\limits_V {\boldsymbol\nabla \cdot \left( {{\boldsymbol{\sigma }} \cdot \delta {\mathbf{u}}} \right)dV}  - \int\limits_V {{\boldsymbol{\sigma :}}\delta \left( {{\mathbf{u}}\overleftarrow \boldsymbol\nabla  } \right)dV + \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV = 0\,\,}  \leftrightarrow \,\,\int\limits_S {{\mathbf{n}} \cdot \left( {{\boldsymbol{\sigma }} \cdot \delta {\mathbf{u}}} \right)dS}  - \int\limits_V {{\boldsymbol{\sigma :}}\delta {\mathbf{e}}dV + \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV = 0\,\,} $$ (N)
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$$ \displaystyle
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\leftrightarrow \,\,\delta \pi = \int\limits_V {{\boldsymbol{\sigma :}}\delta {\mathbf{e}}dV}  - \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV - \int\limits_ {{{\mathbf{t}}^*} \cdot \delta {\mathbf{u}}dS}  = \int\limits_V {\delta WdV - \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV - } \int\limits_ {{{\mathbf{t}}^*} \cdot \delta {\mathbf{u}}dS}  = 0 $$ (N)
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$$ \displaystyle
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\delta W = {\boldsymbol{\sigma :}}\delta {\mathbf{e}}\,\, \leftrightarrow \,\,{\boldsymbol{\sigma }} = \frac\,\, \leftrightarrow \,\,{\mathbf{C:e}} = \frac\,\, \leftrightarrow W = \frac{1}{2}{\mathbf{e:C:e}} = \frac{1}{2}{\boldsymbol{\sigma :e}}\,\,\left( {if\,{\boldsymbol{\sigma }} = {{\boldsymbol{\sigma }}_0} = const\, \to \,\,W = {{\boldsymbol{\sigma }}_0}{\mathbf{:e}}\,\,} \right) $$ (N)
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$$ \displaystyle
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- \delta \phi = {\mathbf{b}} \cdot \delta {\mathbf{u}}\,\& \, - \delta \psi  = {{\mathbf{t}}^*} \cdot \delta {\mathbf{u}} $$ (N)
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$$ \displaystyle
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\delta \pi = \delta \left( {\int\limits_V {\left( {W + \phi } \right)dV}  + \int\limits_ {\psi dS} } \right) = 0 $$ (N)
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$$ \displaystyle
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\pi = \int\limits_V {\left( {W + \phi } \right)dV}  + \int\limits_ {\psi dS} $$ (N)
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$$ \displaystyle
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\leftrightarrow \,\,\delta \pi = \delta \left( {\int\limits_V {WdV - \int\limits_V {\mathbf{b}}  \cdot {\mathbf{u}}dV - } \int\limits_ {{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS} } \right) = 0 $$ (N)
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$$ \displaystyle
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\pi = \int\limits_V {WdV - \int\limits_V {\mathbf{b}}  \cdot {\mathbf{u}}dV - } \int\limits_ {{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  \to \,\min $$ (N)
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$$ \displaystyle
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\int\limits_V {\left[ {{\mathbf{\varepsilon }} - sym\left( {{\mathbf{u}}\overleftarrow \boldsymbol\nabla } \right)} \right]{\mathbf{:}}\delta {\boldsymbol{\sigma }}dV = 0\,\, \leftrightarrow \,\,} \int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV}  - \int\limits_V {\delta {\boldsymbol{\sigma :}}\left( {{\mathbf{u}}\overleftarrow \boldsymbol\nabla  } \right)dV}  = \int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV}  - \int\limits_V {\left[ {\boldsymbol\nabla  \cdot \left( {\delta {\boldsymbol{\sigma }} \cdot {\mathbf{u}}} \right) - \left( {\boldsymbol\nabla  \cdot \delta {\boldsymbol{\sigma }}} \right) \cdot {\mathbf{u}}} \right]dV = 0} $$ (N)
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$$ \displaystyle
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\leftrightarrow \,\,\int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV} - \int\limits_S {{\mathbf{n}} \cdot \left( {\delta {\boldsymbol{\sigma }} \cdot {\mathbf{u}}} \right)dS}  + \int\limits_V {\left( {\boldsymbol\nabla  \cdot \delta {\boldsymbol{\sigma }}} \right) \cdot {\mathbf{u}}dV}  = \,\,\int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV}  - \int\limits_ {\delta {\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  = 0 $$ (N)
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$$ \displaystyle
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\boldsymbol\nabla \cdot \delta {\boldsymbol{\sigma }} = 0\,\,in\,\,V,\,\,\,\,\,\delta {\boldsymbol{\sigma }} \cdot {\mathbf{n}} = 0\,\,on\,\,{S_1},\,\,\,\,\,\delta {\boldsymbol{\sigma }} \cdot {\mathbf{n}} = \delta {\mathbf{t}}\,\,on\,\,{S_2} $$ (N)
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$$ \displaystyle
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\delta {\pi ^*} = \int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV} - \int\limits_ {\delta {\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  = \int\limits_V {\delta {W^*}dV}  - \int\limits_ {\delta {\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  = 0 $$ (N)
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$$ \displaystyle
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\delta {W^*} = {\mathbf{e:}}\delta {\boldsymbol{\sigma }}\,\, \leftrightarrow \,\,{\mathbf{e}} = \frac\,\, \leftrightarrow \,\,{{\mathbf{C}}^{ - 1}}{\mathbf{:\sigma }} = \frac\,\, \leftrightarrow {W^*} = \frac{1}{2}{\boldsymbol{\sigma :}}{{\mathbf{C}}^{ - 1}}{\mathbf{:\sigma }} = \frac{1}{2}{\boldsymbol{\sigma :e}} = W $$ (N)
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$$ \displaystyle
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W \ne {W^*}\,\,but\,\,\,W + {W^*} = {\boldsymbol{\sigma :e}} $$ (N)
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$$ \displaystyle
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\delta {\pi ^*} = \delta \left( {\int\limits_V {{W^*}dV} - \int\limits_ {{\mathbf{t}} \cdot {{\mathbf{u}}^*}dS} } \right) = 0\,\, \leftrightarrow \,\,{\pi ^*} = \int\limits_V {{W^*}dV}  - \int\limits_ {{\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  \to \min $$ (N)
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$$ \displaystyle
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\delta \pi = \int\limits_V {{\boldsymbol{\sigma :}}\delta {\mathbf{e}}dV}  - \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV - \int\limits_ {{{\mathbf{t}}^*} \cdot \delta {\mathbf{u}}dS}  - \int\limits_ {{\mathbf{t}} \cdot \delta {{\mathbf{u}}^*}dS}  = \int\limits_V {\delta WdV - \int\limits_V {\mathbf{b}}  \cdot \delta {\mathbf{u}}dV - } \int\limits_ {{{\mathbf{t}}^*} \cdot \delta {\mathbf{u}}dS}  - \int\limits_ {{\mathbf{t}} \cdot \delta {{\mathbf{u}}^*}dS}  = 0 $$ (N)
 *

 * }

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$$ \displaystyle
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\delta {\pi ^*} = \int\limits_V {{\mathbf{\varepsilon :}}\delta {\boldsymbol{\sigma }}dV} - \int\limits_ {\delta {{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  - \int\limits_ {\delta {\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  + \int\limits_V {\left( {\boldsymbol\nabla  \cdot \delta {\boldsymbol{\sigma }}} \right) \cdot {\mathbf{u}}dV}  = \int\limits_V {\delta {W^*}dV}  - \int\limits_ {\delta {{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  - \int\limits_ {\delta {\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  + \int\limits_V {\left( {\boldsymbol\nabla  \cdot \delta {\boldsymbol{\sigma }}} \right) \cdot {\mathbf{u}}dV} $$ (N)
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$$ \displaystyle
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\int\limits_V {dWdV = \int\limits_V {\mathbf{b}} \cdot d{\mathbf{u}}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot d{\mathbf{u}}dS}  + \int\limits_ {{\mathbf{t}} \cdot d{{\mathbf{u}}^*}dS}  = 0,\,\,\,\,\,dW = {\boldsymbol{\sigma :}}d{\mathbf{e}}\,\,\,\,} $$ (N)
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$$ \displaystyle
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\int\limits_V {d{W^*}dV} = \int\limits_ {d{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  + \int\limits_ {d{\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  - \int\limits_V {\left( {\boldsymbol\nabla  \cdot d{\boldsymbol{\sigma }}} \right) \cdot {\mathbf{u}}dV}  = \int\limits_ {d{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  + \int\limits_ {d{\mathbf{t}} \cdot {{\mathbf{u}}^*}dS}  + \int\limits_V {d{\mathbf{b}} \cdot {\mathbf{u}}dV} ,\,\,\,\,d{W^*} = {\mathbf{e:}}d{\boldsymbol{\sigma }} $$ (N)
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$$ \displaystyle
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\boldsymbol\nabla \cdot \left( {{\boldsymbol{\sigma }} + d{\boldsymbol{\sigma }}} \right) + \left( {{\mathbf{b}} + d{\mathbf{b}}} \right) = 0\,\, \to \,\,\boldsymbol\nabla  \cdot \left( {d{\boldsymbol{\sigma }}} \right) + d{\mathbf{b}} = 0 $$ (N)
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$$ \displaystyle
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\int\limits_V {\mathbf{b}} \cdot {{\mathbf{u}}^'}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS}  = \int\limits_V {{\boldsymbol{\sigma :}}{{\mathbf{\varepsilon }}^'}dV}  = \int\limits_V {{{\mathbf{\varepsilon }}^'}{\mathbf{:C:\varepsilon }}dV}  = \int\limits_V {{\mathbf{\varepsilon :C:}}{{\mathbf{\varepsilon }}^'}dV}  = \int\limits_V {{{\boldsymbol{\sigma }}^'}{\mathbf{:\varepsilon }}dV}  = {\int\limits_V {\mathbf{b}} ^'} \cdot {\mathbf{u}}dV + \int\limits_ {{{\mathbf{t}}^{'*}} \cdot {\mathbf{u}}dS} $$ (N)
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$$ \displaystyle
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{A_{I,I}} + {A_{II,II}} + {A_{I,II}} = {A_{II,II}} + {A_{I,I}} + {A_{II,I}}\,\, \to \,\,{A_{I,II}} = {A_{II,I}} $$ (N)
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$$ \displaystyle
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{P_i}{\Delta _{ji}} = {P_j}{\Delta _{ij}}\,\, \leftrightarrow \,\,{\alpha _{ji}} = \frac = \frac = {\alpha _{ij}} $$ (N)
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$$ \displaystyle
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\left( {{{\mathbf{b}}^'} = {\mathbf{b}},{{\mathbf{t}}^{'*}},{{\mathbf{u}}^'}} \right) $$ (N)
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$$ \displaystyle
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{{\mathbf{t}}^{'*}} = {{\mathbf{t}}^*}\,\,on\,\,{S_1} - S_1^'\,\,\,and\,\,\,\,{{\mathbf{t}}^{'*}} = {{\mathbf{t}}^*} + \Delta {{\mathbf{t}}^*}\,\,on\,\,S_1^' $$ (N)
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$$ \displaystyle
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\int\limits_V {\mathbf{b}} \cdot {{\mathbf{u}}^'}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS}  = {\int\limits_V {\mathbf{b}} ^'} \cdot {\mathbf{u}}dV + \int\limits_ {{{\mathbf{t}}^{{*^'}}} \cdot {\mathbf{u}}dS}  = \int\limits_V {\mathbf{b}}  \cdot {\mathbf{u}}dV + \int\limits_{{S_1} - S_1^'} {{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  + \int\limits_{S_1^'} {\left( {{{\mathbf{t}}^*} + \Delta {{\mathbf{t}}^*}} \right) \cdot {\mathbf{u}}dS} $$ (N)
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$$ \displaystyle
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\int\limits_V {\mathbf{b}} \cdot {{\mathbf{u}}^'}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS}  = \int\limits_V {\mathbf{b}}  \cdot {\mathbf{u}}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  + \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  = 2U + \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot {\mathbf{u}}dS} $$ (N)
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$$ \displaystyle
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2U = \int\limits_V {\mathbf{b}} \cdot {\mathbf{u}}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {\mathbf{u}}dS} $$ (N)
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$$ \displaystyle
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2{U^'} = {\int\limits_V {\mathbf{b}} ^'} \cdot {{\mathbf{u}}^'}dV + \int\limits_ {{{\mathbf{t}}^{{*^'}}} \cdot {{\mathbf{u}}^'}dS = } \int\limits_V {\mathbf{b}} \cdot {{\mathbf{u}}^'}dV + \int\limits_ {{{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS}  + \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS = } 2U + \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot {\mathbf{u}}dS}  + \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot {{\mathbf{u}}^'}dS} $$ (N)
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$$ \displaystyle
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\to \,\,2{U^'} - 2U = 2\Delta U = \int\limits_{S_1^'} {\Delta {{\mathbf{t}}^*} \cdot \left( {{\mathbf{u}} + {{\mathbf{u}}^'}} \right)dS} \approx \,\,\Delta {{\mathbf{t}}^*} \cdot \left( {{\mathbf{u}} + {{\mathbf{u}}^'}} \right)S_1^' = \Delta {{\mathbf{f}}^*} \cdot \left( {{\mathbf{u}} + {{\mathbf{u}}^'}} \right) $$ (N)
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$$ \displaystyle
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\Delta {{\mathbf{f}}^*} = S_1^'\Delta {{\mathbf{t}}^*} $$ (N)
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$$ \displaystyle
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\frac = \frac{1}{2}\left( {{\mathbf{u}} + {{\mathbf{u}}^'}} \right)\,\, $$ (N)
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$$ \displaystyle
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\frac = {\mathbf{u}} $$ (N)
 *

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text:
User:Egm6936.s09/EGM 6611 Continuum Mechanics, Fall 2003/Lecture plan

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$$ \displaystyle \int\limits_V {\left[ {2\mu {\rm{\varepsilon :\varepsilon }} + \lambda {{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dV > } 0 \int\limits_ {\left[ {2{\mu _1}{\rm{\varepsilon :\varepsilon }} + {\lambda _1}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dV} + \int\limits_ {\left[ {2{\mu _2}{\rm{\varepsilon :\varepsilon }} + {\lambda _2}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dV > } 0 $$ (N)
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$$ \displaystyle
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{\mu _1} = \beta {\mu _1} + \left( {1 - \beta } \right){\mu _1} $$ (N)
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$$ \displaystyle
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0 \le \beta \le 1 $$ (N)
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$$ \displaystyle
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\int\limits_ {\left[ {2{\mu _1}\left( {1 - \beta } \right){\rm{\varepsilon :\varepsilon }} + 2{\mu _1}\beta {\rm{\varepsilon :\varepsilon }} + {\lambda _1}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dV} + \int\limits_ {t\left[ {2{\mu _2}{\rm{\varepsilon :\varepsilon }} + {\lambda _2}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dS > } 0 $$ (N)
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$$ \displaystyle
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{\rm{\varepsilon }} = {{\rm{\varepsilon }}^'} + \frac{1}{k}tr\left( {\rm{\varepsilon }} \right){\rm{I}} $$ (N)
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$$ \displaystyle
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{\rm{\varepsilon :\varepsilon }} = {{\rm{\varepsilon }}^'}{\rm{:}}{{\rm{\varepsilon }}^'} + \frac{1}{k}{\left( {tr{\rm{\varepsilon }}} \right)^2} $$ (N)
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$$ \displaystyle
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{\rm{\varepsilon :\varepsilon }} = {\rm{\omega :\omega }} + {\left( {tr{\rm{\varepsilon }}} \right)^2} + \nabla \bullet \left[ {\left( {{\rm{u}}\nabla } \right) \bullet {\rm{u}} - {\rm{u}}\left( {\nabla  \bullet {\rm{u}}} \right)} \right] $$ (N)
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$$ \displaystyle
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\int\limits_ {\left\{ {2{\mu _1}\left( {1 - \beta } \right)\left[ {{{\rm{\varepsilon }}^'}{\rm{:}}{{\rm{\varepsilon }}^'} + \frac{1}{k}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right] + 2{\mu _1}\beta \left[ {{\rm{\omega :\omega }} + {{\left( {tr{\rm{\varepsilon }}} \right)}^2} + \nabla \bullet \left[ {\left( {{\rm{u}}\nabla } \right) \bullet {\rm{u}} - {\rm{u}}\left( {\nabla  \bullet {\rm{u}}} \right)} \right]} \right] + {\lambda _1}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right\}dV}  + \int\limits_ {t\left[ {2{\mu _2}{\rm{\varepsilon :\varepsilon }} + {\lambda _2}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dS > } 0 $$ (N)
 *
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$$ \displaystyle
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\int\limits_ {\left\{ {2{\mu _1}\left[ {\left( {1 - \beta } \right){{\rm{\varepsilon }}^'}{\rm{:}}{{\rm{\varepsilon }}^'} + \beta {\rm{\omega :\omega }}} \right] + \left[ {{\lambda _1} + \frac{k}\left( {1 + \left( {k - 1} \right)\beta } \right)} \right]{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right\}dV} + 2{\mu _1}\beta \int\limits_ {\nabla  \bullet \left[ {\left( {{\rm{u}}\nabla } \right) \bullet {\rm{u}} - {\rm{u}}\left( {\nabla  \bullet {\rm{u}}} \right)} \right]dV}  + \int\limits_ {t\left[ {2{\mu _2}{\rm{\varepsilon :\varepsilon }} + {\lambda _2}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right]dS > } 0 $$ (N)
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$$ \displaystyle
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{\rm{n}} \bullet \left[ {\left( {{\rm{u}}\nabla } \right) \bullet {\rm{u}} - {\rm{u}}\left( {\nabla \bullet {\rm{u}}} \right)} \right] = tr\left[ {{\rm{u}} \times \left( {{\rm{u}}\nabla } \right) \times {\rm{n}}} \right] $$ (N)
 *
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$$ \displaystyle
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\int\limits_ {\left\{ {2{\mu _1}\left[ {\left( {1 - \beta } \right){{\rm{\varepsilon }}^'}{\rm{:}}{{\rm{\varepsilon }}^'} + \beta {\rm{\omega :\omega }}} \right] + \left[ {{\lambda _1} + \frac{k}\left( {1 + \left( {k - 1} \right)\beta } \right)} \right]{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right\}dV} + $$ (N)
 *
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$$ \displaystyle
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+ \int\limits_S {\left\{ {t\left[ {2{\mu _2}{\rm{\varepsilon :\varepsilon }} + {\lambda _2}{{\left( {tr{\rm{\varepsilon }}} \right)}^2}} \right] + 2{\mu _1}\beta tr\left[ {{\rm{u}} \times \left( {{\rm{u}}\nabla } \right) \times {\rm{n}}} \right]} \right\}dS > } 0 $$ (N)
 *
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$$ \displaystyle
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\int\limits_ {\left[ {2{\mu _1}\varepsilon :\varepsilon + {\lambda _1}{{\left( {tr\varepsilon } \right)}^2}} \right]dV}  + $$
 *
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$$ \displaystyle
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\int\limits_ {\left[ {2{\mu _2}\varepsilon :\varepsilon + {\lambda _2}{{\left( {tr\varepsilon } \right)}^2}} \right]dV > } 0 $$ (N)
 *
 * }


text:
User talk:MollyCasto

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you. Best! Stevenarntson

Schedule/Research
Please post your complete schedule and research component specifics.
 * Best! Stevenarntson

Topic Get better at Expert in Rockband, RB2, RB Beatles.

Research Component <>

week 1 Practice song a little bit everyday, begin researching Gameboy's progression.

week 2 Practice song a little bit everyday, continue researching different Gameboy's.

week 3 Practice song a little bit everyday/experiment with higher frets on controller, Research Gameboy's creator Gunpei Yukoi.

week 4 Practice song with higher frets on controller, Research Gameboy's improvements.

week 5 Practice song, Research Gameboy's history.

week 6 Practice song, Research Gunpei Yukoi's connection with Miyamoto.

week 7 Try to tape proof, put everything into slide show.

week 8 Try to tape proof, put everything into slide show.

Progress Report
Please post your Always Wanted To Progress report today.
 * Best, Stevenarntson

Progress went well, everything went smoothly in getting better at RB except for practice time. Information on my research about Gameboy evolution was fun, and easy to gather. I don't like the song I chose to practice as much as I did before this project, though. (haha) Well class was fun Mr. Arntson, you're a great teacher. Hopefully I'll get another one of your classes some quater.

title: User talk:MollyCasto


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User:Egm6936.f09

EGM 6936 Seminar, Fall 2009 Prof. Loc Vu-Quoc

University of Florida

 Licensing info

= Topics: Multiphase flows =

Articles (stable)
$$\bullet$$ Kolmogorov scales (old version Kolmogorov time scale) , $$\bullet$$ Particle time scale , $$\bullet$$ Vorticity equation , $$\bullet$$ Gradient of vector: Two tensor conventions , $$\bullet$$ When to use incompressible flow model? (old version Conditions for incompressible flow) ,

Articles (work under progress)
Weibull distribution

Multiphase flow simulation

Particle Reynolds number

Stokes number

Eddy, smallest resolved eddies, largest resolved eddy length

Lagrange-Eulerian coupling, extension to finite-size particles
 * Interphase momentum and energy coupling
 * Decomposition into deterministic and stochastic components

DNS, Direct Numerical Simulation, Sundaram et al. (1997)

Large eddy simulation (LES)

Globally-undisturbed ambient flow

Preferential particle accumulation, Equilibrium Eulerian method. Squires & Eaton (1991), Physics of Fluids.

Turbophoresis, Equilibrium Eulerian method. Reeks (1983), J. Aerosol Sci.

Turbulence, homogeneous, non-homogeneous

PDF method, Fox et al. (2008), J. Comp. Phys.

= Seminars =

Turbulence seminars

= Important links =

Writing tools

Other courses

title: User:Egm6936.f09


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User:Egm6936.f09/Particle time scale

Egm6936.f09 16:41, 21 October 2009 (UTC)

In a particle-laden flow, the particle phase is also known as the dispersed phase, whereas the carier fluid is called the carier phase. The dispersed phase can be modeled either as a system of discrete particles or as a continuous fluid. The carier phase here is a continuous fluid medium. In point-particle methods, each particle is modeled as a material point, and the particle phase is a system of discrete point particles.

The particle time scale $$\displaystyle \tau_p$$, also known as the momentum (velocity) response time, is the time for a particle to respond to a change in the flow velocity of the carier fluid, and is given by (Crowe et al. (1997), p.23)

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$$  \displaystyle \tau_p :=  \frac {     \rho_p (D_p)^2 }  {      18 \mu_c } $$     (1)
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where $$\displaystyle \rho_p$$ is the density of the particle material, $$\displaystyle D_p$$ the particle diameter, $$\displaystyle \mu_c$$ the viscosity of the carier phase.

Note: The subscript $$\displaystyle c$$ in $$\displaystyle \mu_c$$ may seem redundant for now (since there is no $$\displaystyle \mu_p$$ ), but is kept to quickly discern quantities related to the carier phase versus quantities related to the particle phase.

= Dimension check =

Using Newton's law and Hooke's law:

\displaystyle [\rho_p] =   \frac {\rm mass} {\rm volume} =   \frac {     F L^{-1} T^2 }  {      L^3 }  \, \quad [(D_p)^2] =  L^2 \, \quad [\mu_c] =  F L^{-2} T $$

Recall Hooke's law

\displaystyle \sigma_{xy} =   \mu_c \epsilon_{xy} \, \quad [\sigma_{xy}] =  F L^{-2} \, \quad [\epsilon_{xy}] =  T^{-1} $$ where $$\displaystyle \sigma_{xy}$$ is the shear stress, $$\displaystyle \epsilon_{xy}$$ the shear strain rate.

Thus

\displaystyle [\tau_p] =  \frac {     (F L^{-4} T^2) (L^2) }  {      F L^{-2} T   } =  T $$

Physically, Eq.(1) says that $$\displaystyle \tau_p$$ would be long if (a) the particle is heavy (high material density $$\displaystyle \rho_p$$ ), (b) the particle is large (high value of particle diameter $$\displaystyle D_p$$ ), (c) the carier fluid is less viscous (in other words, the more viscous the fluid, the shorter the particle response time, since a change in velocity in a more viscous fluid would move the particle along quicker than a less viscous fluid).

= Derivation =

Kinetic friction force
The distributed friction force on the surface of a sphere submerged in a creeping flow has a resultant called the kinetic friction force $$\displaystyle F_k$$, having the following expression (Bird et al. (2002), p.186, Eq.(6.3-14)
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$$  \displaystyle F_k =  A    \cdot K   \cdot f  = \left[ \pi (R_p)^2 \right] \left[ \frac{1}{2} \rho_c ( v_\infty )^2 \right] f  = \frac {1}  {2}   f   \frac {\pi (D_p)^2} {4}  \rho_c ( v_\infty )^2 $$     (3)
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where $$\displaystyle A = \pi (R_p)^2$$, with $$\displaystyle R_p = D_p / 2$$, is the characteristic area of the spherical particle, $$\displaystyle K = \frac{1}{2} \rho_c ( v_\infty )^2 $$ the kinetic energy per unit fluid volume at infinity, with $$\displaystyle \rho_c$$ being the carier density and $$\displaystyle v_\infty$$ the terminal velocity, and $$\displaystyle f$$ is the Fanning friction factor (Bird et al. (2002), p.179, Eq.(6.1-5)). For a particle in a flow, the Fanning friction factor is the same as the drag coefficient

$$\displaystyle C_D$$ of the particle.

Since the product $$\displaystyle A \cdot K$$ has dimension of force, i.e.,

\displaystyle [A \cdot K]  = [A] \cdot [K] =  L^2 \cdot \frac {FL} {L^3} =  F   \ , $$ it follows that $$\displaystyle f \equiv C_D$$ is dimensionless, as expected.

Note that Eq.(3) is the definition of the dimensionless quantity $$\displaystyle f \equiv C_D$$, and is not a fluid mechanics law (Bird et al. (2002), p.178, Eq.(6.1-5)).

The Fanning friction factor $$\displaystyle f \equiv C_D$$ can be measured by dropping a sphere in a fluid, and measure the terminal velocity $$\displaystyle v_\infty$$. By considering the equilibrium of the forces that exert on the sphere, namely the kinetic friction force $$\displaystyle F_k$$, the weight of the sphere $$\displaystyle F_g$$, and the buoyant force $$\displaystyle F_b$$, one can deduce the following expression for $$\displaystyle f \equiv C_D$$

\displaystyle F_k =  F_g -  F_b \Longrightarrow \left[ \pi (R_p)^2 \right] \left[ \frac{1}{2} \rho_c ( v_\infty )^2 \right] f  = \frac {4 \pi (R_p)^3} {3}  \left[ \rho_p - \rho_c \right] \Longrightarrow f  \equiv C_D =  \frac {4}  {3}   \frac {g D_p} {( v_\infty )^2} \frac {\rho_p - \rho_c} {\rho_c} $$

Equation of motion
To derive Eq.(1), let $$\displaystyle v_\infty = v_c - v_p$$, where $$\displaystyle v_c$$ is the velocity of the carier phase and $$\displaystyle v_p$$ the velocity of the particle, then consider only the kinetic friction force $$\displaystyle F_k$$ acting on the particle. Using Eq.(3) in Newton's law, the equation of motion of a single particle in a creeping fluid is given by
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$$  \displaystyle m_p \frac {d v_p} {d t}  = \frac{1}{2} C_D \frac{\pi (D_p)^2}{4} \rho_c \left(     v_c - v_p   \right) \left| v_c - v_p \right| \, \quad m_p =  \rho_p \frac {4 \pi R_p} {3}  =   \rho_p \frac {\pi (D_p)^3} {6} $$     (4) where $$\displaystyle m_p$$ is the mass of a particle.
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Particle Reynolds number
The right hand side of Eq.(4) is the kinetic friction force $$\displaystyle F_k$$ of creeping flow around a sphere (Bird et al. (2002), p.186, Eq.(6.3-14), in which the drag coefficient $$\displaystyle C_D$$ is often denoted by $$\displaystyle f$$, called the Fanning friction factor (Bird et al. (2002), p.179, Eq.(6.1-5))

The particle Reynolds number

is defined as (Brodkey & Hershey (1988), p.589, Eq.(12.68))
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$$  \displaystyle Re_p :=  \frac {\rho_c D_p \left| v_c - v_p \right| } {\mu_c} $$     (5) where the fluid inertial force $$\displaystyle F_i$$ is
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\displaystyle F_i =  \rho_c (v_c - v_p)^2 $$ and the fluid viscous force is

\displaystyle F_v =  \mu_c \frac {\left| v_c - v_p \right|} {D_p} $$ The Reynolds number is the ratio of inertial force over viscous force:

\displaystyle Re_p =  \frac {F_i} {F_v} $$

Drag coefficient
For low particle Reynolds number (Stokes flow), i.e., with $$\displaystyle Re_p < 800$$, we have the following approximation (Clift et al. (2005), p.111)

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$$  \displaystyle C_D =   \frac{24}{Re_p} \phi(Re_p) \, \quad \phi (Re_p) =  1 + 0.15 (Re_p)^{0.687} \ . $$     (7a)
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See also Kee & Chhabra (2002), p.23, where it is mentioned that Oseen (1910, 1913) derived the expression
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$$  \displaystyle C_D =   \frac{24}{Re_p} \phi( Re_p ) =  1 + \frac{3}{8} Re_p \ . $$     (7b)
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Both Eqs.(7a) and (7b) lead to the the following approximation for the dimensionless ratio
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$$  \displaystyle \frac {C_D Re_p} {24}  \approx 1  \ . $$      (7c)
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Particle response time
Using Eq.(5) in Eq.(4), the particle acceleration can be written as
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$$  \displaystyle \frac {d v_p} {d t}  = \frac {1}  {\rho_p} \frac {6}  {\pi (D_p)^3} \cdot \frac{1}{2} C_D \frac{\pi (D_p)^2}{4} \rho_c \left(     v_c - v_p   \right) \left| v_c - v_p \right| =  \frac {18 \mu_c} {\rho_p (D_p)^2} \frac {C_D Re_p} {24}  (v_c - v_p) $$     (10)
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It follows from Eq.(7c) that the coefficient $$\displaystyle (18 \mu_c) / [ \rho_p (D_p)^2 ]$$ has the dimension of $$\displaystyle T^{-1}$$ (time inverse). The particle time scale $$\displaystyle \tau_p$$ is therefore defined as in Eq.(1).

= Meaning =

The acceleration of the particle in Eq.(4) can then be re-written as
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$$  \displaystyle \frac {d v_p} {d t}  = \frac {1}  {\tau_p} (v_c - v_p) $$     (15)
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At $$\displaystyle t=0$$, we have
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$$  \displaystyle v_p (0) = 0 $$     (16)
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Hence
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$$  \displaystyle v_p (t) =  v_c \left[ 1 - exp (- t / \tau_p ) \right] $$     (17)
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Thus, at the time equal to the particle time scale $$\displaystyle t = \tau_p$$, the particle velocity is equal to
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$$  \displaystyle v_p (\tau_p) =  v_c \left[ 1 - exp (- 1) \right] =  0.63   \cdot v_c $$     (18) or 63% of the carier flow velocity $$\displaystyle v_c$$. So the particle response time is the time it takes for the particle velocity to reach 63% of the carier velocity.
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= References =

title: User:Egm6936.f09/Particle time scale


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User:Egm6936.f09/Particle Reynolds number

EGM6936.f09.ku 16:44, 21 October 2009 (UTC)

Particle Reynolds number includes a correction to Stokes drag. For solid particles, it is

 * $$\phi(Re)=1+0.15 Re^{0.687}$$.

NOTE: See the question related to the particle Reynolds number in particle time scale.

The above equation is in Clift et al. (2005). To represent the deviation from the standard drag curve (Fig. 5.12 of the same book), relationships for sphere drag are given in Table 5.1, page 111. For Re<800,

 * $$C_D = \frac{24}{Re} \phi (Re_p)$$.

 It would be nice to explain how the above formulas were obtained. Egm6321.f09 01:23, 31 October 2009 (UTC)

A recent paper by Wu (2009) also referred to Clift et al. (2005) (p.112, Table 5.2) for a formula for the function $$\displaystyle \phi(Re_p)$$, denoted as $$\displaystyle f_d$$ in Wu (2009), for a smaller range $$\displaystyle 0.01 < Re_p < 20$$:
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$$  \displaystyle \phi(Re_p) =  1   +   0.1315   Re_p^{0.82 - 0.05 \log_{10} Re_p} $$     (N)
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= References =


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User:Egm6936.f09/Globally-undisturbed ambient flow

What is a "globally undisturbed ambient flow" (GUAF), and what is a "turbulent" GUAF, i.e., how could an "undisturbed" flow be "turbulent" at the same time? Is it an oxymoron?

Actually, "undisturbed flow" means flow with the absence of particles, whereas "disturbed flow" means flow with the presence of particles. Hence, a single-phase flow (without particles) is called a GUAF, and can be turbulent in itself. Clearly, a two-phase flow, with a carrier phase and a particle phase, is a "disturbed flow", which can also be turbulent.


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User:Egm6936.f09/Stokes number

Egm6936.f09 16:46, 21 October 2009 (UTC)

title: User:Egm6936.f09/Stokes number