User:Egm6321.f11.team1.colsonfe/HW4

= R4.1 Direct Derivation of Alternate Version of RTT =

Given
Another version of the Reynolds Transport Theorem:


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

$$
 * $$\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)$$
 * $$\displaystyle (Equation\;4.1.1)
 * $$\displaystyle (Equation\;4.1.1)
 * }
 * }

== Find ==

Provide a different and direct derivation of Equation 4.1.1.

Solution
Solution is adapted from Malvern. If $$\displaystyle f(x,t)$$ denotes any property of the material volume $$\displaystyle\mathcal B_t$$, for a spatial volume bounded by a control surface $$\displaystyle\partial\mathcal B_t$$, the following is true:

$$ \begin{bmatrix} \rm Total\,rate\,of\,change\,of\\ \rm{an\,arbitrary\,function} \\ f(x,t)\,\rm in\,region\,\mathcal B_t \end{bmatrix}=\begin{bmatrix} \rm Time\,rate\,of\,change\\ \rm of\,\mathit f(x,t)\,\rm within\,\mathcal B_t\\ \end{bmatrix}-\begin{bmatrix} \rm Flux\,of\,\mathit{f(x,t)}\\ \rm thru\,surface\,\partial\mathcal B_t \end{bmatrix} $$

Now expressing mathematically what is stated above:

= R*4.7 2nd Exactness Condition For n=2 Case of Nn-ODE (Method 2) =

Given
Equivalent form of 2nd Exactness Condition for N2-ODEs :


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

$$ Coefficient equalities, located in previous reference above.
 * $$g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=0$$
 * $$\displaystyle (Equation\;4.7.1)
 * $$\displaystyle (Equation\;4.7.1)
 * }
 * }


 * $$\displaystyle g_0=f_yq+g_y$$


 * $$\displaystyle g_1=f_pq+g_p$$


 * $$\displaystyle g_2=f$$

Note also that $$\displaystyle p(x):=y'(x)$$ and $$\displaystyle q(x):=y''(x).$$

== Find ==

Show the following equality is true:


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

Solution
Applying the chain rule to $$\frac{dg_1}{dx}$$ yields the following:

$$\frac{dg_1}{dx}=\frac{\partial g_1}{\partial x}+\frac{\partial g_1}{\partial y}\frac{dy}{dx}+\frac{\partial g_1}{\partial y^{(1)}}\frac{dy^{(1)}}{dx}$$

We know what $$\displaystyle g_1$$ is defined to be, we can substitute in $$\displaystyle p(x):=y'(x)$$ and $$\displaystyle q(x):=y''(x)$$ and carry out the derivatives:

$$\frac{dg_1}{dx}=\frac{\partial g_1}{\partial x}+\frac{\partial g_1}{\partial y}p+\frac{\partial g_1}{\partial p}q$$

$$\frac{dg_1}{dx}=\left(f_{xp}q+g_{xp}\right)+\left(f_{yp}q+g_{yp}\right)p+\left(0+g_{pp}\right)q$$

We can apply the same approach to $$\frac{d^2g_2}{dx^2}$$ :

$$\frac{d^2g_2}{dx^2}=\frac{\partial^2 g_2}{\partial x^2}+\frac{\partial^2 g_2}{\partial y\partial x}\frac{dy}{dx}+\frac{\partial^2 g_2}{\partial x\partial y}\frac{dy}{dx}+\frac{\partial^2 g_2}{\partial x\partial y^{(1)}}\frac{dy^{(1)}}{dx}+\frac{\partial^2 g_2}{\partial y^{(1)}\partial x}\frac{dy^{(1)}}{dx}+\frac{\partial^2 g_2}{\partial^2 y}\frac{dy^2}{dx^2}$$

$$\frac{d^2g_2}{dx^2}=f_{xx}+f_{yx}p+f_{xy}p+f_{xp}q+f_{px}q+f_{yy}p^2$$

Adding $$\displaystyle g_0 $$, $$\displaystyle \frac{dg_1}{dx}$$ , and $$\displaystyle \frac{d^2g_2}{dx^2}$$ according to Equation 4.7.1 yields our final solution:

$$ g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=\left(f_yq+g_y\right)-\left(f_{xp}q+g_{xp}+f_{yp}pq+g_{yp}p+g_{pp}q\right)+\left(f_{xx}+f_{yy}p^2+2f_{xy}p+2f_{xp}q\right)=0$$

= References =