User:Egm6321.f11.team3/Hwk4

Find
Provide a different and direct derivation of Reynolds Transport Theorem (4.1.1) by referring to pg.211 of Introduction to the Mechanics of a Continuous Medium by Malvern

Solution
Solved on our own

Let $$\displaystyle f(x,t)$$ denote any arbitrary function, $$\displaystyle\mathcal B_t$$, the material volume and $$\displaystyle\mathcal B_t$$ the control surface.

Then, the rate of change of an arbitrary function $$\displaystyle f(x,t)$$ in the material volume is the sum of the rate of change of the arbitrary function $$\displaystyle f(x,t)$$ inside the control surface $$\displaystyle\mathcal B_t$$ and the net rate of flux of the arbitrary function $$\displaystyle f(x,t)$$ through the control surface..

This can be expressed mathematically as,

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

Find
Verify the exactness of (4.2.1)

Solution
We solved this problem on our own.

The 1st exactness condition for N2-ODEs is that it has the particular form In this case, we can let Thus, (4,2,1) satisfies the 1st exactness condition for N2-ODEs. The 2nd exactness condition for N2-ODEs is Thus, Replace (4.2.7), (4.2.8) and (4.2.9) into (4.2.5) and (4.2.6). We can get Thus (4.2.1) is not exact

Find
Part 1. Find m,n$$\in \mathbb{R} $$such that 4.3.1 is exact. Part 2. Show that the first integral is the L1-ODE-VC 4.3.2 with 4.3.3.

Part 3. Solve 4.3.3 for y(x).

Solution
We solved this problem on our own.

PART 1.

The first condition for exactness of a N2-ODE is, $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ The second condition for exactness of a N2-ODE is, To meet the first exactness condition, the two terms must be $$\displaystyle g(x,y,p)= 2xy'+3y$$ $$\displaystyle f(x,y,p)= \sqrt x$$ And now considering distributing the integrating factor and applying 4.3.3, $$\displaystyle \bar{g}(x,y,p)= 2x^{m+1}y^np+3x^my^{n+1}$$ $$\displaystyle \bar{f}(x,y,p)=x^{1/2+m}y^n$$ And therefore, the terms are as follows, $$\displaystyle f_y=nx^{m+1/2}y^{n-1} $$ $$\displaystyle f_{xx}=(m+1/2)(m-1/2)x^{m-3/2}y^n $$ $$\displaystyle f_{xy}=n(m+1/2)x^{m-1/2}y^{n-1} $$ $$\displaystyle f_{yy}=n(n-1)x^{m+1/2}y^{n-2} $$ $$\displaystyle f_{xp}=0 $$ $$\displaystyle f_{yp}=0 $$ $$\displaystyle g_y=2nx^{m+1}y^{n-1}p+3(n+1)x^{m}y^n $$ $$\displaystyle g_{yp}=2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^n $$ $$\displaystyle g_{xp}=2(m+1)x^m y^n+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{pp}=0 $$ The terms are then put into equation 4.3.4 and 4.3.5, $$\displaystyle (m+1/2)(m-1/2)x^{m-3/2}y^n+2p(n(m+1/2)x^{m-1/2}y^{n-1})+p^2(n(n-1)x^{m+1/2}y^{n-2})$$
 * $$\displaystyle =2(m+1)x^m y^n+3mx^{m-1}y^{n+1}+p(2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^n)-2nx^{m+1}y^{n-1}p-3(n+1)x^{m}y^n $$

$$\displaystyle 2(nx^{m+1/2}y^{n-1})=0 $$ From this we can see that n=0 for 4.3.1 to be exact. This simplifies the equation to $$\displaystyle (m+1/2)(m-1/2)x^{m-3/2} = 2(m+1)x^m+3mx^{m-1}y+3px^{m}-3x^m $$ $$\displaystyle 0 = 2(m+1)x^m-3x^m $$ $$\displaystyle 2(m+1)=3 $$ And therefore m=1/2 for 4.3.1 to be exact.

PART 2.

Now we know that $$\displaystyle \bar{g}(x,y,p)= 2x^{3/2}p+3x^{1/2}y$$ $$\displaystyle \bar{f}(x,y,p)=x$$ And $$\displaystyle \phi{x,y,p}=h(x,y)+\int f(x,y,p)dp +k_2= h(x,y) + xp +k_2$$ $$\displaystyle g(x,y,p)=\phi_x+\phi_yp$$ $$\displaystyle 2x^{3/2}p+3x^{1/2}y=h_x+p+h_yp$$ $$\displaystyle h=(2x^{3/2}-1)y$$ And therefore, $$\displaystyle \phi(x,y,p)=(2x^{3/2}-1)y + xp +k_2$$ $$\displaystyle \phi(x,y,p)=xp+(2x^{3/2}-1)y +k$$

PART 3.

The N1-ODE-VC:

$$\displaystyle \phi(x,y,p)=xp+(2x^{3/2}-1)y = k$$

so we can rewrite,

$$\displaystyle xy'+(2x^{3/2}-1)y=k $$

If we rearrange it, we get

$$\displaystyle y'+(2x^{1/2}-\frac{1}{x})y=\frac{k}{x} $$

$$\displaystyle a_1 y'+a_o y = b $$

We know that $$\displaystyle a_1(x)= 1 $$

From the equation (3) p.11-4, we can solve it by using integrating factor,

$$\displaystyle h(x)=\exp[\int^x a_0(s)ds+k_1] $$

We need to substitute $$\displaystyle a_0(x)=2x^{1/2}-\frac{1}{x} $$ into the h(x)equation

$$\displaystyle h(x)=\exp[\int^x (2s^{1/2}-\frac{1}{s})ds+k1] $$

$$\displaystyle h(x)=\exp[\frac{4}{3}x^{3/2}-\ln{x}+k1] $$

$$\displaystyle h(x)=\frac{1}{x}\exp[{\frac{4}{3}x^{3/2}}+k1] $$

From the equation (1) p. 11-5 to find out $$\displaystyle y(x) $$

$$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds + k2\right] $$

We know that $$\displaystyle b(x)=\frac{k}{x} $$

$$\displaystyle y(x)=\frac{1}{\frac{1}{x}\exp\left[{\frac{4}{3}x^{3/2}}+k1\right]}\left[\int^x \left(\frac{1}{s}\exp\left[{\frac{4}{3}s^{3/2}}+k1\right]\right)\left(\frac{k}{s}\right)ds+k2 \right]$$

Find
Show that 4.4.1 and 4.4.2 leads to 4.4.3.

Solution
We solved this problem on our own. By combining 4.4.1 and 4.4.2 we can see that, $$\displaystyle G=\phi_{x}+\phi_{y}y'+\phi_{p}y''=0 $$ This means $$\displaystyle\phi_p=P(x)$$and $$\displaystyle \phi= P(x)p+k_1 $$ Therefore, $$\displaystyle G=P'(x)p+\frac{\partial{k}}{\partial{dx}} + \frac{\partial{k}}{\partial{dy}}y' + P(x)y''=0 $$ And now group the terms by order, $$\displaystyle G=(\frac{\partial{k}}{\partial{dx}}) + (P'(x)+\frac{\partial{k}}{\partial{dy}})y' + P(x)y''=0 $$ By inspection, the terms are, $$\displaystyle R(x)y=\frac{\partial{k}}{\partial{dx}}$$ $$\displaystyle Q(x)=P'(x)+\frac{\partial{k}}{\partial{dy}}$$ And, $$\displaystyle \phi(x,y,p)=P(x)p+Q(x)y+k$$ $$\displaystyle \phi(x,y,p)=P(x)p+T(x)y+k$$

Find
1.Show equation 4.5.1 is exact

2.Find Φ

3.Slove for y(x)

Solution
We solved this problem on our own.

(1)Show equation 4.5.1 is exact:

So,the equation can be written as G=f(x,y,p){y}+g(x,y,p) It satifies 1st exactness condition. So,it satifies 2nd exactness condition. Thus,equation 4.5.1 is exact.

(2)Find Φ: Due to G(x,y,p,p') is exact We can suppose that: (3)Slove for y(x): We know how to slove all the L1-ODE:

Solution
We slove this according to the MTG22-p22-.

From the equation 4.6.1,we know:

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