User:Egm6321.f11.team4/HW4

= Problem R4.1: Direct Derivation of RTT = From [[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]

Given
A control volume, denoted by $$ \mathcal B_t $$. Its boundary is denoted by $$ \partial \mathcal B_t $$, as seen in figure below. This figure (shown below) was taken from Lecture Notes of Dr Vu-Quoc, for his course, EGM 6321 - Principles of Engineering Analysis - I. The outward normal to boundary is denoted by $$ \mathbf n $$



Find
Show the direct derivation of RTT, given by following equation,

$$\displaystyle \frac{D}{Dt} \int_{\mathcal B_t} \mathbf f (x,t) \, d \mathcal B_t = \int_{\mathcal B_t} \frac{\partial \mathbf f}{\partial t} \, d \mathcal B_t + \int_{\mathcal \partial \mathcal B_t} \mathbf n \cdot (\mathbf f \mathbf u) \, d (\partial \mathcal B_t) $$  (4.1)

Solution
- Solved on our own Consider the material property f(x,t) (given per unit volume), where x is the spatial variable. Principle of conservation of this material property governs that net rate of change of total amount of f(x,t) associated with material present within the control volume, $$ \mathcal B_t $$ (Term 1) is equal to instantaneous change of this property f(x,t) within the control volume, integrated over the complete domain (Term 2), and the net influx and outflux of the property f(x,t) happening at the surface (or boundary) of the control volume (Term 3). The above statement can be written mathematically as,

Term 1 = Term 2 + Term 3

These terms are given by,

Term 1: Net rate of change of total amount of f(x,t) associated with material currently present within control volume, (integrated over $$ \mathcal B_t $$)

$$\displaystyle \frac{D}{Dt} \int_{\mathcal B_t} \mathbf f (x,t) \, d \mathcal B_t $$

Term 2: Integration over the complete volume of the instantaneous change of the property f(x,t) associated with the material present instantaneously within the control volume,

$$\displaystyle \int_{\mathcal B_t} \frac{\partial \mathbf f}{\partial t} \, d \mathcal B_t $$

Term 3: Influx and outflux of the property f(x,t) happening at the surface (or boundary) of the control volume ($$ \partial \mathcal B_t $$), at a particular point (given by $$\displaystyle \mathbf n \cdot (\mathbf f \mathbf u) $$), and integrated over the complete boundary, $$ \partial \mathcal B_t $$,

$$\displaystyle \int_{\mathcal \partial \mathcal B_t} \mathbf n \cdot (\mathbf f \mathbf u) \, d (\partial \mathcal B_t) $$

Note in above term (Term 3), if $$\displaystyle (\mathbf f \mathbf u) $$ has same direction as n, then it is the outflux of the material, and it will add on the term 2 for conservation of total change of f(x,t) associated with material currently present in the control volume at given time.

However, when a fluid is coming into the control volume, it is not the material that we identified being associated with being present in the control volume at current time, and hence that has to be subtracted from term 2. This happens above since, in that case u has opposite direction to n, and hence we get negative sign with the dot product.

Add all the terms to get Eq. (4.1). Hence the above derivation is proven.

Author
Contributed by Ankush

= Problem R*4.2 Verifying Exactness of L2-ODE-VC = From [[media:Pea1.f11.mtg21.djvu|Mtg 21-1]]

Given
Given a L2-ODE-VC,

$$\displaystyle \sqrt{x} \,y'' + 2xy' + 3y = 0 $$  (2.1)

Find
Verify the exactness of Eq. (2.1)

Solution
- Solved on our own We can identify terms, f(x,y,p) and g(x,y,p) corresponding to the particular form for a L2-ODE to be exact (Eq. (1), p. 16-4 in the Lecture Notes).

$$\displaystyle \underbrace {\sqrt{x}}_{f(x,y,p)} \,y'' + \underbrace {2xy' + 3y}_{g(x,y,p)} = 0 $$  (2.2)

Therefore above equation satisfies first exactness condition. Now we will verify second exactness condition, which is Eqns. (1) - (2), p. 16-5 of the Lecture Notes, i.e.

$$ \displaystyle f_{xx} + 2pf_{xy} + p^2 f_{fyy} = g_{xp} + p g_{yp} - g_y $$                     (2.3)

$$ \displaystyle f_{xp} + pf_{yp} + 2 f_y = g_{pp} $$                     (2.4)

We list all required partial derivatives of f(x,y,p) and g(x,y,p) to verify Eq. (2.3) and (2.4) for checking 2nd exactness condition.

$$ \displaystyle g_y = 3, \, \, g_x = 2p, \, \, g_{xp} = 2, \, \, g_{yp} = 0, \, \, g_{p} = 2x, \, \, g_{pp} = 0, \, \, $$

$$ f_x = \frac{1}{2 \sqrt{x}}, \, \, f_{xx} = - \frac{1}{4 (x)^{\frac {3}{2}}}, \, \, f_{xy} = 0, \, \, f_{yy} = 0, \, \, f_{xp} = f_{yp} = f_y = 0 $$  (2.5)

Using above values we get for LHS of Eq. (2.3),

$$ \displaystyle f_{xx} + 2p \cancel{f_{xy}} + p^2 \cancel{f_{yy}} = - \frac{1}{4 (x)^{\frac {3}{2}}} $$                     LHS (2.3)

The RHS of Eq. (2.3) is given as,

$$ \displaystyle g_{xp} + p \cancel{g_{yp}} - g_y = 2 + 0 - 3 = -1 $$                     RHS (2.3)

Hence we clearly see that RHS and LHS of Eq. (2.3) do not match up thus violating the second exactness condition for L2-ODE.

For Eq. (2.4), LHS and RHS both come out to be zero, since all the terms are zero as can be seen from Eq. (2.5).

$$ \displaystyle f_{xp} + pf_{yp} + 2 f_y = 0 $$     LHS (2.4)

$$ \displaystyle g_{pp} = 0 $$     RHS (2.4)

Therefore, even though Eq. (2.4) is satisfied, but for we need both (2.3) and (2.4) to be satisfied for second exactness condition to be true for given L2-ODE. Hence we prove that Eq. (2.1) is not exact.

Author
Contributed by Ankush

= Problem R*4.3 Solve L2-ODE-VC via the Power Form = From [[media:Pea1.f11.mtg21.djvu|Mtg 21-3]]

Given
$$\displaystyle (x^my^n)[\sqrt{x}y''+2xy'+3y]=0$$

Find
$$\displaystyle \mathbf a)$$ Find $$\displaystyle m, n $$ to render the above equation, equation (1) p.21-3, exact. $$\displaystyle \mathbf b)$$ Show that the first integral is a L1-ODE-VC as shown below, equation (2) p.21-3. $$\displaystyle \phi(x,y,p)=xp+(2x^\frac{3}{2}-1)y=k $$ with $$\displaystyle p(x):=y'(x) $$. $$\displaystyle \mathbf c)$$ Solve equation (2) p.21-3 for $$\displaystyle y(x) $$.

Solution Part A
Solved independently. $$\displaystyle \mathbf a)$$ In order to determine $$\displaystyle m $$ and $$\displaystyle n $$, we must analyze the two conditions of exactness for N2-ODEs. The first condition requires the N2-ODE to fit the following particular form of equation (2) p.16-4. $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ The second condition requires the N2-ODE to satisfy the following equations, (1)-(2) p.16-5. $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y $$ $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $$ First, we must determine the new $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ that include the Integrating Factor, $$\displaystyle h(x,y)=x^m y^n $$. Factor through by the integrating factor to yield the following: $$\displaystyle g(x,y,p)=2x^{m+1}y^np+3x^m y^{n+1} $$ $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ This satisfies the first exactness condition. Since $$\displaystyle f(x,y,p) $$ is not a function of $$\displaystyle p $$, the second equation of the second exactness condition significantly simplifies to $$\displaystyle \cancel {f_{xp}}+\cancel {pf_{yp}}+2f_y=\cancel {g_{pp}} $$ $$\displaystyle 2f_y=0 $$. Differentiate $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ with respect to $$\displaystyle y $$ to obtain $$\displaystyle f_y(x,y,p)=nx^{0.5+m}y^{n-1} $$. Plugging in $$\displaystyle f_y(x,y,p) $$ into $$\displaystyle 2f_y=0 $$ yields $$\displaystyle 2f_y=2nx^{0.5+m}y^{n-1}=0 $$. The only way to solve the problem is to set $$\displaystyle n=0 $$. Since we solved for $$\displaystyle n $$, we can simplify $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ to $$\displaystyle f(x,y,p)=x^{0.5+m} $$. $$\displaystyle g(x,y,p)=2x^{m+1} p+3x^m y $$. Now we turn our attention to the first equation of the second exactness condition after we simply. $$\displaystyle f_{xx}+\cancel {2pf_{xy}}+\cancel {p^2f_{yy}}=g_{xp}+\cancel {pg_{yp}}-g_y $$ Next, solve for the partial differentials of $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ with the results shown below. $$\displaystyle f_{xx}(x,y,p)=(m+0.5)(m-0.5)x^{m-0.5} $$. $$\displaystyle g_{xp}(x,y,p)=2(m+1)x^m $$. $$\displaystyle g_y(x,y,p)=3x^m $$. Plug these three equations back into the first equation of the second exactness condition to obtain $$\displaystyle (m+0.5)(m-0.5)x^{m-0.5}=2(m+1)x^m-3x^m $$. After a visual inspection, one can deduce the LHS to be zero due to no other like terms. $$\displaystyle 2(m+1)\cancel {x^m}-3 \cancel {x^m}=0 $$ The $$\displaystyle x^m $$ terms cancel out leaving a simple algegra problem. $$\displaystyle 2(m+1)-3=0 $$ So $$\displaystyle m=0.5 $$. The integrating factor for this L2-ODE-VC is $$\displaystyle h(x,y)=x^{0.5} $$.

Solution Part B
$$\displaystyle \mathbf b) $$ We can solve for the first integral using equation (3) p.16-5, $$\displaystyle \phi(x,y,p)=h(x,y)+\int f(x,y,p) dp $$. Since $$\displaystyle f(x,y,p)=x $$ we can simply equation (3) p.16-5 to $$\displaystyle \phi(x,y,p)=h(x,y)+xp $$. Solve this for the partials as shown below. $$\displaystyle \phi_x=h_x+p $$. $$\displaystyle \phi_y=h_y $$. Next, we use the first equation of the second exactness condition, $$\displaystyle g(x,y,p):=\phi_x+\phi_yp $$, to obtain the following: $$\displaystyle 3x^{0.5}y+2x^{1.5}p=h_x+p+h_yp $$. One can equate the following: $$\displaystyle h_x=3x^{0.5} $$. $$\displaystyle h_y+1=2x^{1.5} $$. Integrating these yields the following: $$\displaystyle h=2x^{1.5}y+k_1(y) $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2(x) $$. After differentiating the equation you obtain $$\displaystyle h_y=\cancel {2x^{1.5}}+k_1'(y)=\cancel {2x^{1.5}}-1 $$. $$\displaystyle h_x=\cancel {3x^{1.5}y}+k_x'(x)=\cancel {3x^{0.5}y} $$. Integrating yields the next result: $$\displaystyle h_1(y)=-y+k_1 $$. $$\displaystyle h_2(x)=k_2 $$. Plug these back into the integrating function equation to yield $$\displaystyle h=2x^{1.5}y-y+k_1 $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2 $$. These equation are the same. Now that we have solved for the integrating function equation, we can input this into the original equation for the final solution. $$\displaystyle \phi(x,y,p)=xp+(2x^{1.5}-1)y=k $$ with $$\displaystyle k=k_3-k_1 $$.

Solution Part C
$$\displaystyle \mathbf c) $$ Next we need to solve for $$\displaystyle y(x) $$ starting from the first integral equation above. The N1-ODE-VC can be rearranged to the following form: $$\displaystyle P(x)y'+Q(x)y=R(x) $$. $$\displaystyle xy'+(2x^{1.5}-1)y=k $$. Divide through by $$\displaystyle x $$ to obtain $$\displaystyle y'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)} $$. $$\displaystyle y'+(2x^{0.5}-\frac{1}{x})y=\frac{k}{x} $$. Using equation (3) p.11-4, we can solve for the integrating factor. Knowing $$\displaystyle a_0(x)=\frac{Q(x)}{P(x)} $$ and solving yields $$\displaystyle h(x)=\exp[\int^x a_0(s)ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x \frac{Q(s)}{P(s)} ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x (2s^{0.5}-\frac{1}{s})ds+k_1] $$ $$\displaystyle h(x)=\exp[\frac{4}{3}x^{1.5}-\ln{x}+k_1] $$ $$\displaystyle h(x)=\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1] $$. Next we use equation (1) p. 11-5 to solve for $$\displaystyle y(x) $$. $$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k_2\right] $$ $$\displaystyle b(x)=\frac{R(x)}{P(x)}=\frac{k}{x} $$ $$\displaystyle y(x)=\frac{1}{\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1]}*\left[\int^x (\frac{1}{s}\exp[{\frac{4}{3}s^{1.5}}+k_1])(\frac{k}{s})ds+k_2 \right]$$

Author
Egm6341.f11.team4.allen 04:50, 19 October 2011 (UTC)

= Problem R*4.4 - Derivation of a form of exact L2-ODE-VC = From [[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]

Given
The equation (2),(3) in the class note [[media:Pea1.f11.mtg13.djvu|21-4]]. The equation (1) in the class note [[media:Pea1.f11.mtg13.djvu|21-5]].

Find
Show that (2),(3) in the class note [[media:Pea1.f11.mtg13.djvu|21-4]] leads to the equation (1) in the class note [[media:Pea1.f11.mtg13.djvu|21-5]].

Solution
- Solved on our own The first exactness condition for L2 ODE VC: where $$\displaystyle p = y^{\prime} $$. By comparing the Eq. (4.4) and the Eq. (4.1), the following is obtained. Integrate the Eq. (4.5), By differentiating the Eq. (4.6) with respect to x and y, the following is obtained. Substituting the Eq. (4.7) into the Eq. (4.4) yields the following. From the last term of the Eq. (4.8), the following is true. Integrate the Eq. (4.9) with respect to $$\displaystyle x $$. Substitute the Eq. (4.10) into the Eq. (4.6). Take the partial derivative of the Eq. (4.11) with respect to $$\displaystyle y $$. Also, the following can be proven to be true using the problem statement. Hence, by comparing the Eq. (4.12) and the Eq. (4.13), the following can be concluded. Therefore, the following is justified.

Author
Contributed by Shin

=Problem 4.5 - Solve a L2-ODE-VC=

Given
A L2-ODE-VC:

Find
1. Show equation (5.1) is exact. 2. Find $$\displaystyle \phi $$ for (5.1). 3. Solve for $$\displaystyle y(x) $$.

Solution
- Solved on our own

1. Show Exactness
First exactness condition requires the equation to be in the form of: $$\displaystyle g(x,y,p)+f(x,y,p)y''=0 $$, where $$\displaystyle p=y' $$ , $$\displaystyle g(x,y,p)= \phi_x+ \phi_y p $$ and $$\displaystyle f(x,y,p) = \phi_p $$. In equation (5.1),

So (5.1) satisfies the first exactness condition. The second exactness condition requires the following two equations to be satisfied:

In (5.1),

(5.4) is satisfied.

(5.5) is satisfied.

So (5.1) satisfies the second exactness condition. Then equation is exact.

2. Find $$\displaystyle \phi $$
Since (5.1) is exact, then:

So:

Compare (5.11) and (5.2), assume $$\displaystyle h_x = 2xy $$.Then:

So $$\displaystyle h_y = x^2 +k_{1}^{'} (y) $$. Then (5.11) becomes:

We get:

So

3. Solve for $$\displaystyle y(x) $$
Equation (5.16) is a L1-ODE-VC. It can be transformed into:

In order to solve (5.17), we need to find a $$\displaystyle \bar{h}(x,y) $$ to make (5.17) exact. Assume $$\displaystyle \bar{h}_y=0 $$.So:

Solve for $$\displaystyle \bar{h} $$:

And:

where $$\displaystyle k_2, k_3 $$ are constants.

Author
Contributed by YuChen

=Problem 4.6 - Show the Equivalence=

Given
When $$\displaystyle n=2$$, we have:

Find
Use (6.1) to show that:

Solution
- Solved on our own Since:

So (6.1) becomes:

We get:

The LHS of (6.9) is:

And the RHS of (6.9) is:

From equations (6.10) (6.11) (6.12) (6.13) we can tell that:

Author
Contributed by YuChen

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