User:Egm6321.f11.team2/HW4

=R4.1 - A Physical Derivation of the Reynolds Transport Theorem=

Given
$$\displaystyle \frac{D}{Dt}\int_{\mathcal B_t} f(x,t)\, d\mathcal B_t = \int_{\mathcal B_t} \frac{\partial f}{\partial t}\, d\mathcal B_t + \int_{\partial \mathcal B_t} \mathbf n \cdot (f\mathbf u)\, d (\partial \mathcal B_t) $$

Find
A direct, physical derivation of the given equation, given by Malvern, 1969, p. 211.

Solution
Consider a region of space that moves with a given quantity of matter, such that it always contains exactly the same particles.

Let f(x, t) be any physical property of the material per unit mass.

As Malvern states, the rate of increase of the total amount of f within the region of space is equal to the rate of increase of the amount of f possessed by the material instantaneously inside the region minus the net rate of outward flux of f carried by mass transport through the control surface. .

Translated into mathematics, this states

$$\displaystyle \int_{\mathcal B_t} \frac{\partial f}{\partial t}\, d\mathcal B_t = \frac{D}{Dt} \int_{\mathcal B_t} f(x, t)\, d\mathcal B_t - \int_{\partial \mathcal B_t} \mathbf n \cdot (f \mathbf u)\, d(\partial \mathcal B_t) $$

which, upon a slight rearrangement of the terms, is the sought after equation!

=R*4.2 - Verify the exactness for L2-ODE-VC =

Given
$$\displaystyle \sqrt{x} \, y'' + 2xy' + 3y = 0 $$  (4.2.1)

Find
Verify the exactness of eq (4.2.1)

Solution
Let $$\displaystyle g(x,y,p) = 2xy' + 3y, \, \, f(x,y,p) = \sqrt{x} $$ $$\displaystyle p := y' $$

$$\displaystyle \underbrace {\sqrt{x}}_{f(x,y,p)}\, y'' + \underbrace {2xp + 3y}_{g(x,y,p)} = 0 $$ It shows that above equation satisfy the first exactness. In order to satisfy 2nd exacntess condition, $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$ (4.2.2)

$$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp}$$ (4.2.3) From the eq (4.2.1),

$$\displaystyle f_{xx} = -\frac {1}{4}x^{-\frac{3}{2}}, \, \, f_{xy} = 0, \, \, f_{yy} = 0 $$ $$\displaystyle g_{xp} = 2, \, \, g_{yp} = 0, \, \, g_{y} = 3, \, \, f_{xp} = 0, \, \, f_{yp} = 0, \, \, f_{y} = 0, \, \, g_{pp} = 0 $$ From the above values, eq (4.2.2) and eq (4.2.3) is that $$\displaystyle f_{xx} = g_{xp} - g_{y}$$  (4.2.4) $$\displaystyle f_{xx} = \frac {1}{4}x^{-\frac{3}{2}}, \, \, \, g_{xp} - g_{y} = -1 $$ $$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp} = 0 $$ (4.2.5)

This shows that although the eq (4.2.5) satisfies the 2nd exactness condition, eq (4.2.4) does not satisfy the 2nd exactness condition. Thus, the eq (4.2.1) is not exact.

=R*4.3 - Integrating Factor Methods - Power-Form=

Given
Giving the following N2-ODE $$\displaystyle \sqrt{x}y''+2xy'+3y=0$$ (4.3.1)

Find
find $$ \displaystyle m,n\in\mathbb{R} $$ such that 4.3.2 is exact: $$ \displaystyle (x^{m}y^{n})\left[\sqrt{x}y''+2xy'+3y\right]=0 $$ (4.3.2) show that the first integral is a L1-ODE-VC $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$ (4.3.3) with $$ \displaystyle p(x):=y'(x) $$ Solve 4.3.3 for $$ \displaystyle y $$.

Solution
The $$ 2^{nd} $$ exactness condition is , $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}$$  (4.3.4) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$  (4.3.5) From 4.3.2, we have $$ \displaystyle \underbrace{(x^{m+\frac{1}{2}}y^{n})}_{f(x,y,p)}y''+\underbrace{2(x^{m+1}y^{n})p+3x^{m}y^{n+1}}_{g(x,y,p)}=0 $$  (4.3.5) Then $$\displaystyle f_{x}=(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n} $$ $$\displaystyle f_{xx}=(m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n} $$ $$\displaystyle f_{xy}=n(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1} $$ $$\displaystyle f_{yy}=n(n-1)x^{m-\frac{1}{2}}y^{n-2} $$ $$\displaystyle f_{xp}=0,\qquad f_{yp}=0 $$ $$\displaystyle g_{x}=2(m+1)x^{m}y^{n}p+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{y}=2nx^{m+1}y^{n-1}p+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{xp}=2(m+1)x^{m}y^{n}+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{yp}=2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{pp}=0 $$

Substitute the above equation to 4.3.4 and 4.3.5, we have $$\displaystyle (m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n}+2pn(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1}+p^{2}n(n-1)x^{m-\frac{1}{2}}y^{n-2}=2(m+1)x^{m}y^{n} $$ $$\displaystyle +3mx^{m-1}y^{n+1}+p\left(2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n}\right)-2nx^{m+1}y^{n-1}p-3(n+1)x^{m}y^{n} $$  (4.3.6) $$\displaystyle 2nx^{m+\frac{1}{2}}y^{n-1}=0 $$  (4.3.7) From 4.3.7 we know $$\displaystyle n=0 $$ , then examine 4.3.6 we have $$\displaystyle m= \frac{1}{2} $$. Thus $$\displaystyle f(x,y,p)=\underbrace{x}_{\phi_{p}},\quad g(x,y,p)=\underbrace{3x^{\frac{1}{2}}y}_{\phi_{x}}+\underbrace{2x^{\frac{3}{2}}}_{\phi_{y}}p $$  (4.3.8) Integrating 4.3.8, we have $$\displaystyle \begin{array}{l} \phi=xp+a(x,y)\\ \phi=2x^{\frac{3}{2}}y+b(y,p)\\ \phi=2x^{\frac{3}{2}}y+c(x,p)\end{array} $$ Thus $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$  (4.3.9) From 4.3.9, we have $$\displaystyle \underbrace{1}_{a_{1}(x)}\cdot y'+\underbrace{(2x^{\frac{1}{2}}-\frac{1}{x})}_{a_{0}(x)}y=\underbrace{\frac{k}{x}}_{b(x)} $$ So, using Maple, we have $$\displaystyle y\left( x\right) =x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}\,k_2-\frac{8\,k\,\Pi\,x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}}{\sqrt{3}\,{36}^{\frac{1}{3}}\,\Gamma \left( \frac{2}{3}\right) }+\frac{4\,\Gamma \left( \frac{1}{3},-\frac{4\,{x}^{\frac{3}{2}}}{3}\right) \,k\,x\,{e}^{-\frac{4\,{x}^{\frac{3}{2}}}{3}}}{{36}^{\frac{1}{3}}}-k $$

=R*4.4 - Derivation of a class of exact L2-ODE-VC=

Given
The form of general N2-ODE is

The (4.4.1) is considered as follows:

A generated class of exact L2-ODE-VC is

Find
Show that (4.4.2) and (4.4.3) leads to (4.4.4).

Solution
From (4.4.1) and (4.4.2), we can obtain

Integrating the both sides of (4.4.5), the following is obtained.

Using (4.4.6),

whrere

$$ \displaystyle \begin{align} &p = y'(x) \\ &\phi_x = P'(x)p + \frac{\partial k(x,y)}{\partial x} \\ &\phi_y = \frac{\partial k(x,y)}{\partial y} \end{align} $$

Integrating the term of R(x)y in (4.4.7),

With (4.4.8), (4.4.6) can be rewritten as follows:

To find k(y), $$\displaystyle \phi_y $$ is found and compared to Q(x) because $$\displaystyle \phi_y = Q(x) $$ in (4.4.7).

Frrom (4.4.10), we can find the following:

Therefore, k(y) is verified with constant k and (4.4.9) is resulted in

$$ \displaystyle \phi(x,y,p) = P(x)p + T(x)y + k $$

=R*4.5 - To show an L2-ODE-VC to be exact and solve for first integral and final solution=

Find
a) Is F exact?

b) Find $$\phi$$

c) Solve for $$y$$

Solution
Part a

In order for F to be exact, it should follow the two exactness condition. It is already in the particular form but it has to satisfy the two equations of second exactness condition. From equation 4.5.1 :

The partial derivatives are as follows


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

$$\displaystyle g_y=2x $$
 * }
 * }


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

$$\displaystyle g_{yp}=0 $$
 * }
 * }


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

$$\displaystyle g_x=2y+(2x-cox(x))p $$
 * }
 * }


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

$$\displaystyle g_{xp}=2x-cos(x) $$
 * }
 * }


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

$$\displaystyle f_x=-sin(x) $$
 * }
 * }


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

$$\displaystyle f_{xp}=0 $$
 * }
 * }


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

$$\displaystyle f_{xy}=0 $$
 * }
 * }


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

$$\displaystyle f_{xx}=-cos(x) $$
 * }
 * }


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

$$\displaystyle f_y=0 $$
 * }
 * }


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

$$\displaystyle g_{pp}=0 $$
 * }
 * }


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

$$\displaystyle f_{yp}=0 $$
 * }
 * }


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

$$\displaystyle f_{yy}=0 $$
 * }
 * }

If we take the above differentials and replace them in Equation 4.5.2 and 4.5.3. Both conditions are satisfied.

Part b

Equation 4.5.1 can be written as: $$\displaystyle {\phi}_x + {{\phi}_y}y' + {{\phi}_p}y'' $$ Where : $$\displaystyle {\phi}_x + {{\phi}_y}y' = 2xy+(x^2-sin x)p $$ And: From Equation 4.5.6 Taking partial derivatives w.r.t 'x' and 'y' for equation 4.5.7 we get: And Now: Putting the values of $$\displaystyle {\phi}_x, {\phi}_y $$ from Equation 4.5.8 and 4.5.9 in Equation 4.5.10 we get: From Equation 4.5.11 Partially differentiating Equation 4.5.13 Comparing Equation 4.5.14 and 4.5.12
 * {| style="width:100%" border="0" align="left"

$$\displaystyle k'(y)=0$$
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle k(y)=k$$ (constant). Hence Equation 4.5.13 becomes And the first integral i.e Equation 4.5.7 becomes:
 * }
 * }

Part c Hence Let $$\displaystyle k_2=0 $$ Equation 4.5.17 becomes Integrating both sides: We get:

=R*4.6 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 1=

Given
The second exactness condition for a Nn-ODE to be exact is
 * $$ \displaystyle \sum^n_{j=0} (-1)^j g^{(j)}_j = 0$$(4.6.1)

Where,
 * $$\displaystyle g^{(j)}_j = \frac{d^j}{dx^j} \frac{\partial G}{\partial y^{(j)}}$$


 * $$\displaystyle G(x,y',y'') = \frac{d}{dx}\phi (x,y,y') = 0$$

Find
Show that the first and second forms of the 2nd exactness condition are equivalent, where the first form is,
 * $$\displaystyle \phi_{xy} = \phi_{yx}, \ \ \phi_{yp} = \phi_{py}, \ \ \phi_{px} = \phi_{xp}$$

where,
 * $$\displaystyle p = y'$$

or equivalently,
 * $$\displaystyle f_{xx} + 2pf_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y$$(4.6.2)


 * $$\displaystyle f_{xp} + p f_{yp} + 2 f_y = g_{pp}$$(4.6.3)

And the second form is
 * $$\displaystyle g_o - \frac{d g_1}{dx} + \frac{d^2 g_2}{dx^2} = 0$$(4.6.4)

Solution
First we should derive the expressions for the terms of equation (4.6.4).
 * $$\displaystyle g_0 := \frac{\partial G}{\partial y} = \frac{\partial}{\partial y} (\frac{d \phi}{dx})$$


 * $$\displaystyle \frac{d}{dx} g_1 = \frac{d}{dx} \frac{\partial G}{\partial y'} = \frac{d}{dx} [ \phi_{xy'} + \phi_{yy'} y' + \phi_y + \phi_{y'y'} y'' ]$$


 * $$\displaystyle \frac{d^2}{dx^2} g_2 = \frac{d}{dx} \frac{d}{dx} \frac{\partial G}{\partial y} = \frac{d}{dx} [ \phi_{y'x} + \phi_{y'y} y' + \phi_{y'y'} y]$$

Replacing the terms in equation (4.6.4) with their derived expressions we obtain
 * $$\frac{\partial}{\partial y} (\frac{d \phi}{dx}) - \frac{d}{dx} [ \phi_{xy'} + \phi_{yy'} y' + \phi_y + \phi_{y'y'} y ] + \frac{d}{dx} [ \phi_{y'x} + \phi_{y'y} y' + \phi_{y'y'} y] = 0$$

It can be seen that by separating out like terms we can form the first form of the second exactness condition. $$\displaystyle \frac{\partial}{\partial y} (\frac{d \phi}{dx}) = \frac{d}{dx} \phi_y$$ $$\displaystyle \frac{d}{dx} \phi_{xy'} = \frac{d}{dx} \phi_{y'x}$$ $$\displaystyle \frac{d}{dx} \phi_{yy'} y' = \frac{d}{dx} \phi_{y'y} y'$$ By simplifying the equations and replacing $$\displaystyle y'$$ with $$\displaystyle p$$ we obtain $$\displaystyle \phi_{yx} = \phi_{xy}$$ $$\displaystyle \phi_{xp} = \phi_{px}$$ $$\displaystyle \phi_{yp} = \phi_{py}$$ Which is the first form of the 2nd exactness condition.

=R*4.7 - Equivalence of Two Forms of the Second Exactness Conditions for N2-ODEs, Method 2=

Given
The second exactness condition for N2-ODEs is given by

$$g_0 - \frac{dg_1}{dx} + \frac{d^2g_2}{dx^2} = 0$$

where

$$g_j := \frac{\partial G}{\partial y^{(j)}}, \quad y^{(i)} := \frac{\partial^i y}{\partial x^i}, \quad\text{and}\quad G := g + fy''$$

Find
Show that the form above is equivalent to the form of the second exactness condition for N2-ODEs given by

$$\displaystyle \begin{align} f_{xx} + 2pf_{xy} + p^2f_{yy} &= g_{xp} + pg_{yp} - g_y\\ f_{xp} + pf_{yp} + 2f_y &= g_{pp} \end{align} $$

Solution
Let $$p := y'$$ and $$q:= y''$$.

Then,

Plugging 4.7.1, 4.7.2, and 4.7.3 into the given form of the second exactness condition yields

$$\displaystyle ( f_{xx} + 2pf_{xy} + p^2f_{yy} - g_{xp} - pg_{yp} + g_y ) + ( f_{xp} + pf_{yp} + 2f_y - g_{pp} ) q = 0$$

By equating in turn the coefficients of q0 and q1 to zero, the desired relations are obtained.

=References=

=Contributing Members=

Egm6321.f11.team2.epps I solved R4.1 and R*4.7 and proofread all the other problems. 00:41, 19 October 2011 (UTC)

Egm6321.f11.team2.johri.y 00:50, 19 October 2011 (UTC) I solved R*4.5 and proofread all other problems.

Egm6321.f11.team2.wise 01:37, 19 October 2011 (UTC) I solved R*4.6 and proofread problems R4.4 and R4.5.

Egm6321.f11.team2.kim 14:23, 19 October 2011 (UTC) I solved R*4.2 and proofread problems R*4.3, R*4.6 and R*4.7

Egm6321.f11.team2.Xia 23:50, 19 October 2011 (UTC) I solved R*4.3 and proofread R*4.1,R*4.2 and R*4.7.

Egm6321.f11.team2.rho15:53, 19 October 2011 (UTC) I solved R*4.4 and proofread R*4.1, R*4.3, and R*4.5.