User:Egm6321.f11.team1/report3

= R3.1 Show N1-ODE satisfies certain condition=

Given
Show that the N1-ODE (1) p.13-2

$$ \underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0 $$

satisfies the condition (2) p,11-2 :

$$\frac{h_x}{h}=-\frac{1}{N}(N_x-M_y) \, =: \color{blue}{n(x)}$$

that an integrating factor h(x) can be found to render it exact, only if $$k_1(y)=d_1$$

Show that (1) p.13-2 includes (1) p.12-4

$$ M+Ny'=[a(x)\, y+k_2(x)] + \bar b(x) \,y'=0 $$

as a particular case.

From lecture notes [[media:pea1.f11.mtg13.djvu|Mtg 13]]

Solution
Part 1:

In this case,

$$N(x,y)= \bar b(x,y)c(y)$$,

$$M(x,y)= a(x) \bar c(x,y) $$

Thus,

$$\displaystyle N_x=c(y) b(x), $$

$$\displaystyle M_y= a(x)c(y) $$

Then (2) p.11-2 will be like:

$$\frac{h_x}{h}=-\frac{1}{N}(c(y) [b(x)-a(x)])$$


 * $$=\frac{a(x)-b(x)}{\bar b(x,y)}$$


 * $$=\frac{a(x)-b(x)}{\int^x b(s)ds+k_1(y)}$$

This term is only respect to x, and can be written as $$\displaystyle n(x)$$, only if $$\displaystyle k_1(y)=constant$$

Part 2:

(1)p.13-2:

$$ a(x) \bar c(x,y)+ \bar b(x,y)c(y) \,y'=0

$$

When functions b(x,y) and c(x,y) meet the following conditions,

b(x,y) is a function of x only:

$$b(x,y)=b(x)$$

$$c(x,y)=c$$ (c is a constant)

then(1)p.13-2 becomes like:

(2)p.12-4: $$[a(x)y+k_2(x)]+\bar b(x) \,y'=0

$$

which proves that (2) p.12-4 is a particular case of (1)p.13-2

=R*3.2 Show Exactness or determine IFM to make exact=

Given
1st order ODE:


 * $$ \left(\frac{1}{3}x^{3}+d_1\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}+sin(x)+d_2\right)=0$$ $$\displaystyle (1) p. 13-4

$$

From lecture notes [[media:pea1.f11.mtg13.djvu|Mtg 13]]

Find
Show that the above is exact or find the integrating factor $$ h $$ to make it exact

Solution
For the the 1st condition of exactness the given equation must be in the form of:


 * $$ N(x,y)y' + M(x,y) = 0 \!$$  $$\displaystyle (2) p. 7-6$$

[[media:pea1.f11.mtg7.djvu|Mtg 7 (b,x)]]

It has been shown in R3.1 that the integration functions must be constant in order to determine the integration factor. Thus the integration functions and constants will be set to zero for this problem, and $$\displaystyle (1) p. 13-4 $$, becomes:


 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

which is in the form of the first exactness condition $$\displaystyle (2) p. 7-6$$ so the 1st condition is satisfied.

For the 2nd exactness condition the partial derivatives must be equal per meeting lecture notes 9, [[media:pea1.f11.mtg9.djvu|Mtg 9]]


 * $$M_y = N_x\!$$  $$\displaystyle (2) p. 9-3$$.

It can be seen that:


 * $$M_y \neq N_x$$

An integrating factor will be found such that the given ODE can be integrated exactly.

The strategy is to get the following equation from lecture notes 13:


 * $$\underbrace{\bar b(x,y)c(y)}_{N(x,y)}\ y' + \underbrace{a(x) \bar c(x,y)}_{M(x,y)}=0        \!$$  $$\displaystyle (1) p. 13-2$$.

into the form of equtaion $$\displaystyle (2) p. 12-2$$


 * $$ y'+a_o(x)y=0 \!$$  $$\displaystyle (2) p. 12-2$$.

So first lets calculate the the inputs to $$\displaystyle (1) p. 13-2$$


 * $$\bar b(x)= \frac{x^3}{3}\!$$
 * $$\bar c(y)= \frac{y^5}{3}\!$$

We also know that


 * $$c(y)= y^4\!$$
 * $$a(x)= 5x^3+2\!$$

Plugging these values into $$\displaystyle (1) p. 13-2$$ yields:


 * $$ (\frac{x^3}{3} y^4) y'+ (5x^3+2) y^5=0\!$$

After some algebra to get in the form of $$\displaystyle (2) p. 12-2$$ the result is:


 * $$\underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y =0\!$$

From equation $$\displaystyle (3) p. 11-4$$ the integrating factor can be found as follows:


 * $$h = \exp\int^{x}a_0(s)ds \!$$

with the integration constant equal to zero

Substituting $$ a_0\!$$ yields


 * $$h = \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \!$$

Now the integral can be evaluated, which yields:


 * $$ h = \exp\left(3x - \frac{3}{5x^2}\right)\!$$

This is the integrating factor to make the ODE given exact.

=R*3.3 Find N1-ODE that is either exact or can be made exact=

Given

 * $$ a(x)=sinx^3 $$


 * $$ b(x)=cosx $$


 * $$ c(y)= \exp(2y) $$

1.Find an N1-ODE of the form (1) p.13-2


 * $$ \underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0 $$

$$\,(Eq.3.3.1)$$

that is either exact or can be made exact by IFM. (See R3.1)

2.Find the first integral


 * $$\phi(x,y)=k $$

From lecture notes [[media:pea1.f11.mtg13.djvu|Mtg 13]]

Solution
Part 1


 * $$\bar b(x)= \int\limits^x b(s)ds= \int\limits^x cos(s)ds= sin(x) $$


 * $$\bar c(y)= \int\limits^y c(s)ds= \int\limits^y e^{2s}ds= \frac{1}{2} e^{2y}$$

Now Eq 3.3.1 becomes:


 * $$sin(x)e^{2y}y'+\frac{1}{2}sin(x^3)e^{2y}=0$$

$$\,(Eq.3.3.2)$$

Eq 3.3.2 is the form of


 * $$M(x,y)+N(x,y)\,y'=0$$ and the first condition of Exactness is satisfied.

$$\,(Eq.3.3.3)$$

The condition below is not only necessary but also sufficient for Eq 3.2 to be an exact differencial equation.


 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$


 * $$\frac{\partial M}{\partial y}=\frac{\partial(\frac{1}{2}sin(x^3)\exp^2y)}{\partial y}=sin(x^3)\exp^2y$$

$$\,(Eq.3.3.4)$$


 * $$\frac{\partial N}{\partial x}=\frac{\partial(sin(x)\exp^2y)}{\partial x}=cos(x)\exp^2y$$

$$\,(Eq.3.3.5)$$

Eq 3.3.4 and Eq 3.3.5 reveals that


 * $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$

Hence the Eq.3.3.2 is not exact.

But it can be made exact by using Integrating factor method.Refer to Eq(2), page 11-2 of the lecture notes, and let's multyply Eq 3.3.2 by h(x) in order to have an exact differential equation.


 * $$h(x)=exp[\int^x(-\frac{1}{N}(N_x-M_y))dx]$$

We can rewrite Eq 3.3.2 by cancelling $$e^{2y}$$from both term as:


 * $$\underbrace{sin(x)}_{N(x,y)}{y'}+ \underbrace{\frac{1}{2}sin(x^3)}_{M(x,y)}=0 $$

Now:


 * $$ h(x)=exp[\int^x(-\frac{1}{sin(x)}(cos(x)-0))dx]=exp[\int^x(-\frac{cos(x)}{sin(x)})dx]=exp[-ln(sin(x))]=\frac{1}{sin(x)}$$

$$\,(Eq.3.3.6)$$

Now Exact differential equation is given by


 * $$h(x)[sin(x)e^{2y}y'+\frac{1}{2}sin(x^3)\frac{1}{sin(x)}=0]$$

That is


 * $$sin(x)\frac{1}{sin(x)}y'+\frac{1}{2}sin(x^3)\frac{1}{sin(x)}=0$$


 * $$y'+\frac{1}{2}{sin(x^3)}{sin(x)}=0$$

$$\,(Eq.3.3.7)$$

Part 2 We know that


 * $$M(x,y)=\phi_x(x,y)$$

$$\,(Eq.3.3.8)$$


 * $$N(x,y)=\phi_y(x,y)$$

$$\,(Eq.3.3.9)$$

Also,


 * $$\phi_x(x,y)=\frac{\partial\phi(x,y)}{\partial x}$$

$$\,(Eq.3.3.10)$$


 * $$\phi_y(x,y)=\frac{\partial\phi(x,y)}{\partial y}$$

$$\,(Eq.3.3.11)$$

By integrating Eq.3.3.10 and 3.3.11 with respect to x, y, respectively, we can have equations as given below.


 * $$\int \phi_x(x,y)dx=\phi(x,y)=\int M(x,y)dx +k(y)$$

<p style="text-align:right">$$\,(Eq.3.3.12)$$


 * $$\int \phi_y(x,y)dy=\phi(x,y)=\int N(x,y)dy +k(x)$$

<p style="text-align:right">$$\,(Eq.3.3.13)$$

Now we can calculate $$k(y)$$ by differentiating Eq 3.3.12 with respect to y and equating to Eq 3.3.9=>


 * $$k'(y)=N(x,y)=1$$

Now integrate the above equation =>


 * $$k(y)=y+k1$$

Then$$\phi(x,y) $$ is obtained by substituting

$$k(y)$$ in Eq 3.3.12


 * $$\phi(x,y)=\int M(x,y)dx+y+k1$$

That is


 * $$\phi(x,y)=\frac{1}{2}\int \frac{sin(x^3)}{sinx}dx +y=k$$

= R3.4 Construct a class of N1-ODEs =

== Given == Consider a class of N1-ODEs of the form:


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

$$
 * $$\underbrace{\bar b(x,y)c(y)}_{\displaystyle{N(x,y)}}\,y'+\underbrace{a(x)\bar c(x,y)}_{\displaystyle{M(x,y)}}=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.4.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.4.1)
 * }
 * }

in which,


 * $$\bar b(x,y):=\int^x b(s)ds+k_1(y)$$


 * $$\bar c(x,y):=\int^y c(s)ds+k_2(x)$$

where $$a(x), b(x), c(y) \,$$ are arbitrary functions.

Equation 3.4.1satisfies the condition
 * $$\displaystyle \frac{h_x}{h}=-\frac{1}{N}\left(N_x-M_y\right)=:n(x)$$

that an integrating factor $$h(x)$$ can be found to render it exact, only if $$\displaystyle k_1(y)=constant$$.

== Find == Construct a class of N1-ODEs, which is the counterpart of Equation 3.x.x, and satisfies the condition
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle \frac{h_y}{h}=\frac{1}{M}\left(N_x-M_y\right)=:m(y)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.4.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.4.2)
 * }
 * }

that an integrating factor $$h(y)\,$$ can be found to render it exact.

Solution
The condition Equation (3.4.2) means that the intergrating factor only with respect y. According to this, suppose the N1-ODE we want has the form of:

with

$$\displaystyle \bar a(x,y)=\int^y a(s)ds+k_1(x)$$

$$\displaystyle \bar c(x,y)=\int^x c(s)ds+k_2(y)$$ The partial differential are:

$$\displaystyle M_y=a(y)\cdot c(x)$$

$$\displaystyle N_x=b(y)\cdot c(x)$$

$$\displaystyle \frac{h_y}{h}=\frac{1}{M}\left(N_x-M_y\right)=b(y)-a(a):=m(y)$$, which satisfies the condition Equation (3.4.2).

= R*3.5 Derive Equations for Projectile Motion =

Given
Schematic of motion of particle in air:


 * Egm6321.f11.team1.figure1.png

Particle velocity
 * $$v:=\parallel\boldsymbol v \parallel$$

Constants
 * $$k,n\in\mathbb R$$

Mass of particle = $$\displaystyle m$$

Acceleration of gravity = $$\displaystyle g$$

== Find ==

(1) Derive the equations of motion


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

$$
 * $$m\frac{dv_x}{dt}=-kv^n\cos\alpha$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.1)
 * }
 * }


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

$$
 * $$m\frac{dv_y}{dt}=-kv^n\sin\alpha-mg$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.2)
 * }
 * }


 * $$\displaystyle v^2=(v_x)^2+(v_y)^2$$


 * $$\tan\alpha=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{v_y}{v_x}$$

(2) For the particular case $$\displaystyle k=0$$: Verify that $$\displaystyle y(x)$$ is a parabola.

(3) Consider the case $$\displaystyle k \ne 0$$ and $$\displaystyle v_{x0}=0$$.


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

$$
 * $$m\frac{dv_y}{dt}=-k(v_y)^n-mg$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.3)
 * }
 * }


 * (3.1) Is Equation 3.5.3 either exact or can be made exact using Integrating Factor Method? Find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m$$ constant.


 * (3.2) Find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m=m(t)$$.

Part (1)
For a non-accelerating particle in motion, the sum of the forces is zero. Consider the forces in the x-direction:

$$\sum F_x=0=ma_x+kv^n\cos\alpha=m\frac{dv_x}{dt}+kv^n\cos\alpha$$

Now consider the forces in the y-direction.

$$\sum F_y=0=ma_y+kv^n\sin\alpha+mg=m\frac{dv_y}{dt}+kv^n\sin\alpha+mg$$

By given and definition of norm :

$$v=\parallel\boldsymbol v\parallel=\sqrt{(v_x)^2+(v_y)^2}$$

Thus it is clear that

Apply definition of slope to a differential portion.

$$\tan\alpha=\frac{\rm opposite}{\rm adjacent}=\frac{dy}{dx}$$

Now apply chain rule and definition of velocity.

$$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{v_y}{v_x}$$

Put it all together.

Part (2)
Consider Equation (3.5.1) and Equation (3.5.2) for the case of $$\displaystyle k=0:$$

$$m\frac{dv_x}{dt}=0$$

$$m\frac{dv_y}{dt}=0-mg$$

Which implies

$$\frac{dv_x}{dt}=0$$

$$\frac{dv_y}{dt}=-g$$

Integrate

$$\displaystyle v_x(t)=v_{x0}$$

$$\displaystyle v_y(t)=-gt+v_{y0}$$

Integrate again to calculate position.

$$\displaystyle x(t)=v_{x0}t+x_0$$

$$y(t)=-g\frac{t^2}{2}+v_{y0}t+y_0$$

Solve $$\displaystyle x(t)$$ for $$\displaystyle t$$ so that we can eliminate time dependency from $$\displaystyle y(t)$$.

$$t=\frac{x-x_0}{v_{x0}}$$

Entering this into $$\displaystyle y(t)$$ yields

This is the form of a paraboloid.

Section 3.1
To test Equation 3.5.3 for exactness, it must pass the two Exactness Conditions. Rearrange as such:

$$\left[k(v_y)^n+mg\right]+\left[m\right]\frac{dv_y}{dt}=0$$

If we define the following values:

$$\displaystyle M(v_y,t):=k(v_y)^n+mg$$

$$\displaystyle N(v_y,t):=m$$

Then we have

$$\displaystyle M(v_y,t)+N(v_y,t)v_y'=0$$

This satisfies the First Exactness Condition. The second is satisfied if

$$\frac{\partial M(v_y,t)}{\partial v_y}=\frac{\partial N(v_y,t)}{\partial t}$$

But (assuming mass $$\displaystyle m$$ is a function of time)

$$kn(v_y)^{n-1}\ne \frac{dm}{dt}$$

Thus the Second Exactness Condition is not satisfied. Let us try to transform the ODE with an integrating factor to make it exact. We need to find an $$\displaystyle h(v_y,t)$$ so that the following N1-ODE is exact:

$$h(v_y,t)\left[M(v_y,t)+N(v_y,t)v_y'\right]=0$$


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

$$
 * $$\left[h(v_y,t)k(v_y)^n+h(v_y,t)mg\right]+\left[h(v_y,t)m\right]\frac{dv_y}{dt}=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.4)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.5.4)
 * }
 * }

Attempting to find a $$\displaystyle h(v_y,t)$$ that makes Equation 3.5.4 exact is extremely difficult, as explained in R2.11. To put it simply, the equation for $$\displaystyle h(v_y,t)$$ is a non linear PDE, so it has varying coefficients. For this reason, I claim that there does not exist an Integrating Factor of this form. However, there are two cases in which it is possible to find an Integrating Factor. In both cases I will be pulling equations from the class lecture notes.

CASE 1

Suppose $$\displaystyle h_t(v_y,t)=0$$, thus $$\displaystyle h$$ is a function of $$\displaystyle v_y$$ only. From the notes, we define

$$n(v_y):=-\frac{1}{N}(N_t-M_{v_y})=\frac{1}{m}\left[\frac{dm}{dt}-kn(v_y)^{n-1}\right]$$

Also from the same source,

CASE 2

Now suppose $$\displaystyle h_{v_y}(v_y,t)=0$$, thus $$\displaystyle h$$ is a function of $$\displaystyle t$$ only. From the notes cited above, we define

$$m(t):=-\frac{1}{M}(N_t-M_{v_y})=\frac{1}{k(v_y)^n-mg}\left[\frac{dm}{dt}-kn(v_y)^{n-1}\right]$$

From R*2.12 we know $$\displaystyle h(t)$$:

The next step of this problem is to find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m$$ constant. Solving the ODE directly is challenging due to the complexity $$\displaystyle (v_y)^n$$ poses. To counter this problem I will consider a few common cases.

$$\displaystyle n=0$$

This case represents vertical motion without air resistance. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-k-mg$$

$$v_y(t) = -\frac{k+mg}{m}t+v_{y0}$$

$$y(t)= -\frac{k+mg}{2m}t+^2v_{y0}t+y_0$$

$$\displaystyle n=1$$

This case represents vertical motion with a linear air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y-mg$$

This is a L1-ODE solvable via Integration Factor Method, using $$h=\int \exp\left(\frac{k}{m} \right )dt$$. Using this yields the solution

$$v_y(t)=C_1\exp^{-\frac{k}{m}t}-\frac{gm}{k}$$

$$y(t)=-C_1\frac{k}{m}\exp^{-\frac{k}{m}t}-\frac{gm}{k}t+C_2$$

$$\displaystyle n=2$$

This case represents vertical motion with a quadratic air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y^2-mg$$

This is a N1-ODE. I will choose to use Wolfram-Alpha to solve this equation.

$$v_y(t)=\frac{\sqrt g\sqrt m\tan(\frac{C_1\sqrt g\sqrt k m-\sqrt g\sqrt k t}{\sqrt m})}{\sqrt k}$$

$$y(t)=\frac{2m\log\left( \cos\left( \frac{\sqrt g\sqrt k(C_1m-t)}{\sqrt m}\right)\right)}{t-C_1m}+C_2$$

$$\displaystyle n=3$$

This case represents vertical motion with a cubic air resistance factor. Consider Equation (3.5.3).

$$m\frac{dv_y}{dt}=-kv_y^2-mg$$

This is another N1-ODE. I used Wolfram-Alpha to solve this equation, but it yielded an implicit equation for $$\displaystyle v_y(t)$$.



Note that an implicit form of $$\displaystyle v_y(t)$$ makes it impossible to solve for $$\displaystyle y(t)$$.

$$\displaystyle n\geq4$$

This case represents vertical motion with an resistance factor that is proportional to some power (greater than 3) of the velocity. Solving this via Wolfram-Alpha yields similarly complex implicit solutions as in the $$\displaystyle n=3$$ case. Thus any further investigation is non informative.

Section 3.2
I need to find $$\displaystyle v_y(t)$$ and $$\displaystyle y(t)$$ for $$\displaystyle m=m(t)$$, which is defined below.



$$\displaystyle n=0$$

This case represents vertical motion without air resistance. Consider Equation (3.5.3).

$$m(t)\frac{dv_y}{dt}=-k-m(t)g$$

$$\frac{dv_y}{dt}=-\frac{k}{m(t)}-g$$

$$dv_y=\left[-\frac{k}{m(t)}-g\right]$$

Use the equation $$\displaystyle m(t)=-\frac{m_0-m_1}{t_1}t-m_0$$ to define mass on the interval $$\displaystyle\left[0,t_1 \right ]$$. It is derived from simple slope-intercept equation. Elsewhere $$\displaystyle m(t)=m_1$$.

Note change to dummy integration variable $$\displaystyle s$$.

$$v_y(t)=\int_{0}^{t}\left[-\frac{k}{m(s)}-g \right ]ds$$

$$v_y(t)=\int_{0}^{t_1}\left[-\frac{k}{m(s)}\right]ds+\int_{t_1}^{t}\left[-\frac{k}{m_1}\right]ds+\int_{0}^{t}\left[-g\right ]ds$$

$$v_y(t)=\int_{0}^{t_1}\left[-\frac{k}{-\frac{m_0-m_1}{t_1}s-m_0}\right]ds-\frac{k}{m}(t-t_1)-gt+C_1$$

$$v_y(t)=\left[k\frac{t_1}{m_0-m_1}\ln\left(\frac{m_0-m_1}{t_1}s+m_0 \right )\right]_{0}^{t_1}-\frac{k}{m}(t-t_1)-gt+C_1$$

$$v_y(t)=k\frac{t_1}{m_0-m_1}\left[\ln(2m_0-m_1)-\ln(m_0)\right ]-\frac{k}{m_1}(t-t_1)-gt+C_1$$

$$v_y(t)=k\frac{t_1}{m_0-m_1}\ln\left(\frac{2m_0-m_1}{m_0}\right)-\left(\frac{k}{m_1}-g \right )t-\frac{kt_1}{m_1}+C_1$$

And thus it follows that

$$y(t)=k\frac{t_1}{m_0-m_1}\ln\left(\frac{2m_0-m_1}{m_0}\right)t-\left(\frac{k}{m_1}-g \right )\frac{t^2}{2}-\frac{kt_1}{m_1}t+C_1t+C_2$$

$$\displaystyle n\geq1$$

Further cases introduce two functions of $$\displaystyle t$$ (which are $$\displaystyle v_y(t)$$ and $$\displaystyle m(t)$$) that are incorporated into an ODE. These are feasible problems but beyond the scope of this report.

= R*3.6 Derive the equations of motion of the Double Pendulums =

Find
1.Derive the equations of motion
 * {| style="width:100%" border="0" align="left"

\displaystyle m_1l^2 \ddot{\theta_1} = -ka^2(\theta_1-\theta_2)-m_1gl\theta_1+ u_1l $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

\displaystyle m_2l^2 \ddot{\theta_2} = -ka^2(\theta_2-\theta_1)-m_2gl\theta_2+ u_2l $$ $$ 2.Write in matrix form
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Solution
For the left arm: 'M' is the torque relative to hinge point


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

\displaystyle \sum M=M_u+M_k+M_G $$ $$ 'I' is the moment of inertia '$$\alpha$$' is the angular acceleration
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.6.1 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \sum M=I\cdot\alpha $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle I_1=m_1l^2 $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle \alpha=\ddot{\theta_1} $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle M_u=u_1 \cdot l \cdot \cos \theta_1 \approx u_1 \cdot l $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle M_k=F_k \cdot a \cdot \cos \theta_1 =k(a \cdot \sin \theta_2 -a \cdot \sin \theta_1) \cdot a \cdot \cos \theta_1 \approx -ka^2(\theta_1 - \theta_2) $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle M_G=-m_1g \cdot l \cdot \sin \theta_1 \approx -m_1gl \theta_1 $$ Apply to equation 3.6.1:
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle m_1l^2 \ddot{\theta_1} = -ka^2(\theta_1-\theta_2)-m_1gl\theta_1+ u_1l $$ $$ Similarly, we can get the equation of motion for the right arm:
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.6.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle m_2l^2 \ddot{\theta_2} = -ka^2(\theta_2-\theta_1)-m_2gl\theta_2+ u_2l $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.6.3)
 * }
 * }

Linear time-variant systems
 * {| style="width:100%" border="0" align="left"

\displaystyle \dot{x}(t)=A(t)x(t)+B(t)u(t) $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle x:=\left \lfloor \theta_1 \dot \theta_1 \theta_2 \dot \theta_2 \right \rfloor ^T \in \mathbb {R} ^{4 \times 1} $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle u:= \begin{Bmatrix} u_1l\\u_2l \end{Bmatrix} \in \mathbb {R}^{2 \times 1} $$
 * <p style="text-align:right;">
 * }
 * }

to find A and B in the following format


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

\displaystyle \begin{bmatrix} \dot \theta_1\\ \ddot \theta_1\\ \dot \theta_2\\ \ddot \theta_2 \end{bmatrix} = \begin{bmatrix} & &  & \\ &  &  & \\ &  &  & \\ &  &  & \end{bmatrix} \begin{bmatrix} \theta_1\\ \dot \theta_1\\ \theta_2\\ \dot \theta_2 \end{bmatrix} + \begin{bmatrix} & \\ & \\ & \\ & \end{bmatrix} \begin{bmatrix} u_1l\\ u_2l \end{bmatrix} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.6.4)
 * }
 * }

finally, the matrix is:


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

\displaystyle \begin{bmatrix} \dot \theta_1\\ \ddot \theta_1\\ \dot \theta_2\\ \ddot \theta_2 \end{bmatrix} = \begin{bmatrix} 0&1 &0  &0 \\ \displaystyle -\frac{ka^2}{m_1l^2}-\frac{g}{l} &0 & \displaystyle \frac{ka^2}{m_1l^2}  &0 \\ 0 &0 &0  &1 \\ \displaystyle \frac{ka^2}{m_2l^2}&0  & \displaystyle -\frac{ka^2}{m_2l^2}-\frac{g}{l} &0 \end{bmatrix} \begin{bmatrix} \theta_1\\ \dot \theta_1\\ \theta_2\\ \dot \theta_2 \end{bmatrix} + \begin{bmatrix} 0&0 \\ \displaystyle \frac{1}{m_1l^2}&0 \\ 0&0\\ 0&\displaystyle \frac{1}{m_2l^2} \end{bmatrix} \begin{bmatrix} u_1l\\ u_2l \end{bmatrix} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.6.5)
 * }
 * }

= R*3.7 =

R*3.7 Question from meeting 15, page 2: Use IFM to show that the solution of (2) below is (3) underneath. $$\dot x(t)= ax(t) + bu(t)$$ $$x(t)=[\exp\{a(t-t_0)\}]x(t_0)+\int^t_{t_0}[\exp\{a(t-\tau)\}]\,b \,u(\tau)\, d \tau$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The solution requires an understanding of how the Euler Integrating Factor Method (IFM) is based on the 1st and 2nd exactness conditions. Please note that the equations below for the Euler IFM were derived in a previous homework assignment and noted again for clarity. The derivation builds to the solution. First Exactness Condition of N1-ODEs If an N1-ODE is exact then it has a particular form of M(x,y) + N(x,y)y' = 0 where $$ y' = \frac{dy}{dx}$$ <h3 style="color:red">The appended derivations to this point are from meeting 8 Second Exactness Condition of N1-ODEs Now, we must derive the 2nd exactness condition. A general ND-ODE is exact if for G the following relations exist: $$\frac{d\phi}{dx} = G$$ $$\phi(x,y) = k$$ $$\frac{d\phi(x,y(x))}{dx} = G(y',y,x) = 0$$ Let $$\phi(x,y)$$ be a smooth function, then for the 2nd exactness condition we have $$M_y(x,y) = \frac {\partial^2\phi(x,y)}{\partial x\partial y} = N_x(x,y) = \frac{\partial^2\phi(x,y)}{\partial y\partial x}$$ <h3 style="color:red">The appended derivations to this point are from meeting 9 If an N1-ODE satisfies the 1st exactness condition but not the 2nd, then the N1-ODE can be transformed by using the Euler Integrating Factor Method (IFM). Assume the conditions below. $$M(x,y) + N(x,y)y' = 0 \And M_y(x,y) \ne N_x(x,y)$$ The transformation of the given N1-ODE to a form that fulfills both exactness conditions begins by finding an integrating factor h(x,y). For condition 1, the substitution of h gives $$(hM) + (hN)y' = \bar{M} + \bar{N}y' = 0$$ The key is to derive a solution using $$\bar{M} = hM \And \bar{N} = hN$$ The derivations are as follows: $$\bar{M}_y = h_yM + hM_y = \bar{N}_x = h_xN + hN_x$$ So the Euler IFM gives the following solution using h(x,y) as the transformation. $$h_xN - h_yM + h(N_x - M_y) = 0 $$ <h3 style="color:red">The cumulative derivations to this point are from meetings 10-11 But in our case of this problem, the following equation is the direct basis for the derivation. The inhomogeneous form of this equation has a solution based on meeting 11. So the IFM takes a different derivation path. The updated solution process begins here for report 3. $$\underbrace{-a \,x}_{\displaystyle\color{blue}{M(t,x)}} + \underbrace{\color{blue}{1}}_{\displaystyle\color{blue}{N(t,x)}} \cdot dx/dt = bu(t) $$ The 1st order linear differential equation is a function of t with two variables x(t) and u(t). So let us rewrite the equation as $$-ax + x' = bu = M + Nx'$$ Note that M = -ax and N = 1, so that $$\frac{h_t}{h} = \frac{1}{\N}(N_t - M_x) = (0 - a) = n(s)$$ $$\int_{t_0}^t\frac{h_t}{h} = \int_{t_0}^t(-a)$$ The equation reduces to $$\ln h = -a(t-t_0)$$ Which can be exponentiated to give the integrating factor $$ h = \exp ({\ln h}) = \exp -a(t-t_0)$$ Once the integrating factor is found, the following identity is used to integrate for hx. $$\frac{h_t}{h} = -a$$ So rearrangement gives $$h_t = ha = \dot {h}$$ The substitution into the 1st exactness condition for the inhomogenous condition gives $$h\dot {x} + \dot {h} x = \dot{hx} = hbu$$ Integrate both sides and get $$\int_{t_0}^t{\dot {hx}} = \int_{t_0}^t{b*u}*dt$$ The use of the dummy variable tau and rearrangement with respect to the integration constant gives $$x(t)=[\exp\{a(t-t_0)\}]x(t_0)+\int^t_{t_0}[\exp\{a(t-\tau)\}]\,b \,u(\tau)\, d \tau$$ The first term on the left is the homogeneous solution, and the 2nd term on the right is the particular solution.

References for R3.7 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 9) [[media:pea1.f11.mtg9.djvu|Mtg 9]] 8 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 10) [[media:pea1.f11.mtg10.djvu|Mtg 10]] 8 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011

= R*3.8 =

R*3.8 Question from meeting 15, page 4: Generalize (2) p. 15-2 to the case of linear time-variant system (1) p. 14-4. See (1) p. 15-2. Verify that your expression is indeed the solution for (1) p.14-4. $$\mathbf{x}(t) = \left[\exp \{\mathbf{A}(t-t_0) \} \right] \, \mathbf{x}(t_0) + \int^t_{t_0} \left[\exp \{ \mathbf{A} (t-\tau) \} \right] \mathbf{B} \, \mathbf{u}(\tau) \, d\tau$$ $$\mathbf{\dot[x]}(t) = \mathbf{A}(t) \,\mathbf{x}(t) + \mathbf{B}(t)\, \mathbf{u}(t)$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The equation (2) from p. 15-2 was for the case of n = 1. But the solution can be generalized for all n by simply applying the same derivation to the equation in matrix form. $$\mathbf{\dot[x]}(t) = \mathbf{A}(t) \,\mathbf{x}(t) + \mathbf{B}(t)\, \mathbf{u}(t)$$

The derivation path is similar since the solution of a scalar linear 1st order differential equation is the same for a system of such equations.

The 1st order linear differential equation is a function of t with two matrices x(t) and u(t) and two varying coefficient matrices a(t) and b(t). So let us rewrite the equation as $$-\mathbf{A}\mathbf{X} + \mathbf{X}' = \mathbf{B}\mathbf{U} = \mathbf{M} + \mathbf{N}\mathbf{X}'$$ $$\frac{\mathbf{h_t}}{\mathbf{h}} = \frac{1}{\mathbf{N}}(\mathbf{N_t} - \mathbf{M_x}) = (0 - \mathbf{A}) = \mathbf{n(s)}$$ So rearrangement gives $$ x(t) = \left[\exp\int^t_(t_0)\mathbf{A}(\mathbf{\tau}d\tau\right]\mathbf{x}(t_0) + \int^t_{t_0}\left[exp \int^t_{\tau}\mathbf{A}(s)ds \right]\mathbf{b}(\mathbf{\tau})\mathbf{u(\tau)}\, d\tau $$

References for R3.8 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 14) [[media:pea1.f11.mtg14.djvu|Mtg 14]] 20 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011.

= R*3.9 = R*3.9 Question from meeting 15, page 5: Verify that (1) satisfies (2)-(3). The equations (1) to (3) are shown from top to bottom below. $$\mathbf{\Phi}(t,t_0) = \exp \{\mathbf{A}(t-t_0) \}$$ $$\displaystyle\frac{d}{dt} \mathbf{\Phi}(t,t_0) = \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$ $$\mathbf \Phi (t_0,t_0) = \mathbf I$$ From lecture notes [[media:pea1.f11.mtg15.djvu|Mtg 15]]

Solution
The derivation of a matrix is a direct derivation of a linear system. So it follows the general rules of derivation and matrix algebra. The derivative of an exponential function as a scalar is the same as that for a linear system in matrix algebraic notation. Therefore, the simple rules of derivation apply to prove that (1) satisfies (2) $$\frac {d}{dt} {\mathbf{\Phi}(t,t_0)} = \frac {d}{dt} \exp \{\mathbf{A}(t-t_0) \}$$ $$\frac {d}{dt} {\mathbf{\Phi}(t,t_0)} = \frac {d}{dt} [\mathbf{A}(t-t_0)] * \exp \{\mathbf{A} (t-t_0) \}$$ This simply reduces to show that (1) satisfies (2). $$\displaystyle\frac{d}{dt} \mathbf{\Phi}(t,t_0) = \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$ In addition, the identity matrix is derived from general matrix algebra. $${\mathbf{\Phi}(t_0,t_0)} = \exp \{\mathbf{A}(t_0-t_0) \} = \exp \{\mathbf{A}(0) \} $$ $${\mathbf{\Phi}(t_0,t_0)} = \exp {\begin{bmatrix} 1     & \cdots & 0      \\ \vdots & \ddots & \vdots \\ 0     & \cdots & 1 \end{bmatrix}} $$ Therefore, the substitutions give the identity matrix, which is a matrix with values of 1 along the diagonal and 0 elsewhere.

References for R3.9 King, A.C., J. Billingham, and S.R. Otto. Differential Equations: Linear, Nonlinear, Ordinary, Partial. New York: Cambridge University Press, 2003. Oulia, Masoud and David E. Stevens. Fundamentals of Engineering Exam, 2nd Edition. Boston: Barron's Educational Services Inc., 2008. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 15) [[media:pea1.f11.mtg15.djvu|Mtg 15]] 22 Sep 2011.

= R3.10 Free vibration of coupled pendulums =

== Given ==

The equations of motion of free vibration coupled pendulums are
 * {| style="width:100%" border="0" align="left"

$$
 * $$m_1l^2\ddot{\theta_1}=-ka^2(\theta_1-\theta_2)-m_1gl\theta_1+u_1l$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.10.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.10.1)
 * }
 * }
 * $$m_2l^2\ddot{\theta_2}=-ka^2(\theta_2-\theta_1)-m_2gl\theta_2+u_2l$$

A specific case of this coupled pendulums model:
 * Pendulums: $$a=0.3,\,l=1,\,k=0.2$$
 * $$m_1g=3,\,m_2g=6$$
 * No applied forces: $$\displaystyle u_1=u_2=0$$
 * Initial conditions:
 * $$\theta_1(0)=0,\dot\theta_1(0)=-2$$
 * $$\theta_2(0)=0,\dot\theta_2(0)=+1$$

1)
Use matlab's ode45 command to integrate the coupled pendulums system in matrix form for $$t\in [0,7]$$

2)
Find the solution of this coupled pendulums system by using
 * $$\mathbf{x}(t)=[exp{\mathbf{A}(t-t_0)}]\mathbf{x}(t_0)+\int_{t_0}^t[exp{\mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau$$

3)
Plot $$\displaystyle \theta_1(t)$$ and $$\displaystyle \theta_2(t)$$.

1)
Use matlab code ode45 to solve the coupled pendulums system, and plot the solution $$\displaystyle \theta_1$$ and $$\displaystyle \theta_2$$. Matlab code: hw3_problem10.m

=== 2) === The equations of motion Equation (3.10.1) can be written in matrix form
 * $$\mathbf{\dot{X}}(t)=\mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t)$$

The solution is:
 * $$\displaystyle \mathbf{x}(t)=[exp{\mathbf{A}(t-t_0)}]\mathbf{x}(t_0)+\int^t_{t_0}[exp{\mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u(\tau)}d\tau$$

In this case,
 * $$\mathbf{x}:=\left\lfloor\theta_1\,\dot\theta_1\,\theta_2\,\dot\theta_2\right\rfloor^T$$
 * $$\mathbf{u}:=\begin{Bmatrix}

u_1l\\ u_2l \end{Bmatrix}$$ The equation of motion is:
 * $$\begin{bmatrix}

\dot{\theta_1}\\ \ddot{\theta_1}\\ \dot{\theta_2}\\ \ddot{\theta_2} \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 & 0\\ -\frac{ka^2}{m_1l^2}-\frac{g}{l} & 0 & \frac{ka^2}{m_1l^2} & 0\\ 0 & 0 & 0 & 1\\ \frac{ka^2}{m_2l^2} & 0 & -\frac{ka^2}{m_2l^2}-\frac{g}{l} & 0 \end{bmatrix}\begin{bmatrix} \theta_1\\ \dot\theta_1\\ \theta_2\\ \dot\theta_2 \end{bmatrix}+\begin{bmatrix} 0 & 0\\ \frac{1}{m_1l^2} & 0\\ 0 & 0\\ 0 & \frac{1}{m_2l^2} \end{bmatrix}\begin{bmatrix} u_1l\\ u_2l \end{bmatrix}$$ Therefore, the solution is:
 * $$\displaystyle \mathbf{x}(t)=[exp{\mathbf{A}t}]\mathbf{x}(t_0)$$

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

\theta_1\\ \dot\theta_1\\ \theta_2\\ \dot\theta_2 \end{bmatrix}=exp{\begin{bmatrix} 0 & t & 0 & 0\\ -9.8588t & 0 & 0.0588t & 0\\ 0 & 0 & 0 & t\\ 0.0298t & 0 & -9.8298t & 0 \end{bmatrix}}\begin{bmatrix} 0\\ -2\\ 0\\ 1 \end{bmatrix} $$ $$ To calculate the matrix exponential, we use Matlab code expm. Matlab code: matrix_exp.m
 * $$\begin{bmatrix}
 * $$\begin{bmatrix}
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.10.2)
 * }
 * }

3)
To get the plot of Q1, we use the MATLAB code in 1). As shown, the red line is the solution of Q1 $$\displaystyle \theta_1$$ and the green line is $$\displaystyle \theta_2$$.

And, this is the MATLAB code for plot the solution of Q2. Matlab code: plot2.m

As shown, the red is $$\displaystyle \theta_1$$, and the blue line is $$\displaystyle \theta_2$$.

=*R3.11 SC_L1_ODE_CC=

Given
Use (2)-(3) p.15-4 together with (1) p.14-4 to show (3) p.15-3. (2) p.15-4


 * $$\displaystyle \frac{d}{dt} \mathbf{\Phi}(t,t_0)= \mathbf{A} \, \mathbf{\Phi}(t,t_0)$$

<p style="text-align:right">$$\,(Eq.11.1)$$

(3) p.15-4


 * $$\mathbf \Phi (t_0,t_0)= \mathbf I$$

<p style="text-align:right">$$\,(Eq.11.2)$$

(1) p.14-4


 * $$\mathbf{\dot{x}}(t)= \mathbf{A}(t) \, \mathbf{x}(t) + \mathbf{B}(t) \, \mathbf{u}(t)$$

<p style="text-align:right">$$\,(Eq.11.3)$$

(3)p.15-3


 * $$ \mathbf{x}(t)=\mathbf{\Phi}(t,t_0) \mathbf{x}(t_0) + \int^t_{t_0} \mathbf{\Phi}(t, \tau) \, \mathbf{B}(\tau) \, \mathbf{u}{\tau} \, d\tau $$

<p style="text-align:right">$$\,(Eq.11.4)$$

Solution
Substituting$$\mathbf{x}(t)$$ in Eq11.1 by $$\mathbf{\phi}(t,t_0)$$ <p style="text-align:right">$$\,(Eq.11.5)$$

to obtain the differential equation of the state transition matrix, we get:


 * $$ \dot\phi (t,t_0)=A \mathbf{\phi}(t,t_0) + B u(t)$$

Re-arranging equation 11.5 in the following form:


 * $$\underbrace{1}_{N}\dot\phi - \underbrace{A \phi(t,t_0)}_{M}=Bu(t)$$

<p style="text-align:right">$$\,(Eq.11.6)$$

It is clear that this equation complies with the first condition of exactness.

So to make this equation exact we will apply the Euler Integrating Factor Method.

It has been previously shown that in order to meet the second condition of exactness we can multiply equation 11.6 by a function h to obtain the following:


 * $$\underbrace{h1}_{N}\dot\phi - \underbrace{hA \phi}_{M}=hBu$$

Now the second condition of exactness may be applied, where x is the dependent variable and t is the independent variable.


 * $$\underbrace{(h \frac{\partial M}{\partial{\phi}}+M \frac{\partial h}{\partial{\phi}})}_{\frac{\partial M}{\partial{\phi}}}=\underbrace{(h \frac{\partial N}{\partial{t}}+N \frac{\partial h}{\partial{t}})}_{\frac{\partial N}{\partial{t}}}$$

Now for the special case where $$h_x$$ equals zero we obtain the following:


 * $$hM_\phi=hN_t+Nh_t$$

Which may be rearranged to obtain:


 * $$hN_t+Nh_t-hM_\phi=0$$

We can now solve for h as a function of t.


 * $$h(t)=e^{-\int^t_{t_0}\frac{1}{N}(N_t-M_\phi)dt}$$

Where:
 * $$N=1$$


 * $$N_t=\frac{\partial 1}{\partial t}=0$$


 * $$M=-A \phi$$


 * $$M_\phi=-A$$


 * $$h(t)=e^{\int^t-Ads}=e^{-At}$$

<p style="text-align:right">$$\,(Eq.11.7)$$

Substituting 11.7 in the equation 11.6 we get


 * $$\underbrace{e^{-At} \dot\phi(t,t_0)}_{f'(x)\cdot g(x)}-\underbrace{e^{-At} \dot\phi(t,t_0)}_{f(x)\cdot g'(x)}=e^{-At} Bu(t)

$$ <p style="text-align:right">$$\,(Eq.11.8)$$

We should recognize equation 11.8 as the derivative of $$ (h\phi)$$ , by using the relation given in equation 11.2:


 * $$\frac{d(h\phi)}{dt}=hBu$$

Therefore, to find the solution we integrate both sides in the interval
 * $$t_0 \le \tau \le t$$


 * $$\int^t_{t_0} \frac{d(e^{-A\tau}\phi)}{d\tau}d\tau=\int^t_{t_0} e^{-A\tau}Bu(\tau)d\tau

$$ <p style="text-align:right">$$\,(Eq.11.9)$$

Equation 11.9 yields:


 * $$e^{-A(t-t_0)}\phi(t,t_0)-e^{-A(t_0-t_0)}\phi(t_0,t_0)+\int^t_{t_0}e^{-A\tau}Bu(\tau)d\tau$$

<p style="text-align:right">$$\,(Eq.11.10)$$

Re-arranging terms in equation 11.10 we obtain:


 * $$\phi(t,t_0)=e^{A(t-t_0)} \phi(t_0,t_0)+\int^t_{t_0} e^{A(t-\tau)} Bu(\tau)d\tau$$

<p style="text-align:right">$$\,(Eq.11.11)$$

Substituting $$\phi(t,t_0)$$ in the equation 11.11 by $$x(t)$$ we obtain the differential equation in the form of equation 11.4:


 * $$x(t)=e^{A(t-t_0)}x(t_0)+\int^t_{t_0}e^{A(t-\tau)}bu(\tau)d\tau$$

=R*3.12 Linear time-variant system, Rocket Roll Control in matrix notation=

Given

 * $$ \displaystyle \dot{\phi}=\omega $$ <p style="text-align:right;">$$\displaystyle (1) p. 16-2

$$


 * $$ \displaystyle \dot{\omega}=-\frac{1}{\tau }\omega+\frac{Q}{\tau }\delta $$<p style="text-align:right;">$$\displaystyle (2) p. 16-2

$$


 * $$ \displaystyle \dot{\delta}=u $$ <p style="text-align:right;">$$\displaystyle (3) p. 16-2

$$

See lecture notes [[media:pea1.f11.mtg16.djvu|Mtg 16 (b)]] for problem description.


 * $$ \displaystyle \mathbf{\dot{x}}(t)=\mathbf{A}(t)\, \mathbf{x}(t)+\mathbf{B}(t)\, \mathbf{u}(t) $$<p style="text-align:right;">$$\displaystyle (1) p. 14-4

$$

Reference lecture notes[[media:pea1.f11.mtg14.djvu|Mtg 14]] for form of equation.

Find
Put (1-3)p.16-2 in the form of (1)p.14-4

Solve

 * $$ \displaystyle \begin{bmatrix}

\dot{\phi }\\ \dot{\omega }\\ \dot{\delta } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi \\ \omega \\ \delta \end{bmatrix}

+ \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \mathbf {u}(t) $$

= R*3.13 Second exactness condition =

Given

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

\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

\displaystyle g(x,y,p):=\phi_x (x,y,p)+\phi_y (x,y,p)p $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


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

\displaystyle f(x,y,p):=\phi_p (x,y,p) $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Find
1.Derive the 2nd relation in the 2nd exactness condition 2.Derive the 1st relation in the 2nd exactness condition 3.Verify that $$x(y')^2+yy'+(xy)y''=0$$satisfies the 2nd exactness condition

1.Derive the 2nd relation in the 2nd exactness condition

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

\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.1 )
 * }
 * }


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

\displaystyle g(x,y,p):=\phi_x (x,y,p)+\phi_y (x,y,p)p $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.13.2)
 * }
 * }


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

\displaystyle f(x,y,p):=\phi_p (x,y,p) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.3 )
 * }
 * }


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

\displaystyle p:=y' $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle g_p = \phi_{xp} + \phi_{yp} \cdot p +\phi y $$ for
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle f=\phi_p $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle g_p = f_x+f_y \cdot p + \phi y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.4 )
 * }
 * }

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

\displaystyle g_{pp}=f_{xp} + pf_{yp}+f_y+\phi_{yp} $$
 * <p style="text-align:right;">
 * }
 * }

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

\displaystyle f=\phi_p $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle g_{pp}=f_{xp}+pf_{yp}+2f_y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.5 )
 * }
 * }

2.Derive the 1st relation in the 2nd exactness condition
For we know that:
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi_{xy}= \phi_{yx} $$
 * <p style="text-align:right;">
 * }
 * }

From equation 3.13.4 we can get:
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi_y = g_p-f_x-pf_y $$
 * <p style="text-align:right;">
 * }
 * }

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

\displaystyle \phi_x = g-\phi_y \cdot p =g- pg_p + pf_x + p^2f_y $$ then
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi_{xy}=g_y-p_yg_p+p_yf_x-pg_{py}+pf_{xy}+2pp_yf_y+p^2f_{yy} $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle \phi_{yx}=gp_x - pf_{xy}-f_yp_x-f_{xx} $$ apply $$\phi_{xy}=\phi_{yx}$$
 * <p style="text-align:right;">
 * }
 * }

and $$p_x=p_y=0$$ We get:
 * {| style="width:100%" border="0" align="left"

\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.6 )
 * }
 * }

3.Verify that $$x(y')^2+yy'+(xy)y''=0$$satisfies the 2nd exactness condition

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

\displaystyle x(y')^2+yy'+(xy)y''=0 $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle g(x,y,y')=x(y')^2+yy' $$
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle f(x,y,y')=xy $$
 * <p style="text-align:right;">
 * }
 * }

1.Verify the first relation
 * {| style="width:100%" border="0" align="left"

\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=0+2p+0=2p $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.7 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle g_{xp}+pg_{yp}-g_y=2p+p-p=2p $$ $$ so Eq7=Eq8, satisfied 2.Verify the second relation
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.13.8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle f_{xp}+pf_{yp}+2f_y=0+0+2x=2x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.13.9 )
 * }
 * }


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

\displaystyle g_{pp}=2x $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.13.10)
 * }
 * }

so Eq9=Eq10, satisfied

=R*3.14 Discuss the search for the solution with out making h constant=

Given

 * $$ {h}_{x}+{h}_{y}\cdot P =0$$ <p style="text-align:right;">$$\displaystyle (1) p. 17-1

$$

Find
$$ h(x,y) $$ with out assuming that h=constant, and discuss the search for the solution

Solution
This problem is assigned in [[media:pea1.f11.mtg18.djvu|Mtg 18 (b)]]

As a starting point I looked at this:

This solution is based on http://en.wikiversity.org/w/index.php?title=User:Egm6321.f10.team03/Hwk3&oldid=621841#Problem_8-_Discuss_the_search_for_h

To find $$ h(x,y)$$ solve equation $$ (1) p. 17-1 $$ for $$\displaystyle {h}_{x}$$ as shown below


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


 * style="width:95%" |
 * $$\displaystyle
 * $$\displaystyle

{h}_{x} = - {h}_{y}P $$ $$
 * <p style="text-align:right;">$$\displaystyle (3.14.1)
 * }

Use the 2nd exactness condition, which states


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


 * style="width:95%" |
 * $$\displaystyle
 * $$\displaystyle

{h}_{xy} = {h}_{yx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (3.14.2)
 * }

to determine $$ h(x,y)$$. Differentiate both sides of $$\displaystyle (3.14.1)$$ to determine $$ {h}_{xy}$$ using the chain rule (http://en.wikipedia.org/wiki/Chain_rule)


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


 * style="width:95%" |
 * style="width:95%" |

{h}_{x}= - {h}_{y} \underbrace_{:= {y}^{'}}

\Rightarrow

{h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} $$ $$ Substituting $$ (3.14.3) $$ into $$ (3.14.2)$$ yields:
 * <p style="text-align:right;">$$\displaystyle (3.14.3)
 * }


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


 * style="width:95%" |
 * style="width:95%" |

{h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} \neq {h}_{yx} $$ $$
 * <p style="text-align:right;">$$\displaystyle (3.14.4)
 * }

By comparimg $$\displaystyle(3.14.1)$$ to $$\displaystyle(3.14.4)$$, and realizing $$\displaystyle y' = P $$ ,it can be see that the last term must be zero to satify the given equation.

Since $$\displaystyle(3.14.1)$$ is just $$ \displaystyle (1) p. 17-1 $$ rearranged, in order for $$ (1) p. 17-1 $$ to be true, $$\displaystyle {h}_{x} = {h}_{y} = 0$$.

The next step is to integrate both sides to get $$\displaystyle h(x,y) = \rm constant$$. When integrating $$\displaystyle {h}_{x} $$ and $$\displaystyle {h}_{y} $$ it must be considered that in general a function of of integration is also possible.

For reference a similar answer was obtained by Team 6 Fall 2010, for convienience that solution is shown in its entirety with a slight adaptation of equation numbering only for completeness from http://en.wikiversity.org/w/index.php?title=User:EGM6321.f10.team6.cook/hw3&oldid=621925#solution_of_h.28x.2Cy.29

we can get eqn(3.14.6) using definition of $$\begin{align} g(x,y,y') \end{align}$$

where $$\begin{align} P:=y' \end{align}$$


 * without assuming h=const. find the solution of eqn(3.14.6)

We can rewrite the eqn(3.14.5) as (3.14.6)

We are familiar with this equation, as we learned already. Total derivative - Egm6321.f10_HW1_prob#1_team6

As $$\begin{align} \frac{d}{dx}h(x,y)=0 \end{align}$$, we know that $$\begin{align} h(x,y)=f(y) \end{align}$$ only.

It means $$\begin{align} {h}_{x}=0 \end{align}$$

Hence, eqn(3.14.6) becomes,

There are two possible solutions.

1) $$\begin{align} P=y'=0 \end{align}$$

2) $$\begin{align} {h}_{y}=0 \end{align}$$

If 1) were satisfied, whole problems became zero, which is trivial. We can conclude that 2) is the solution.

As $$\begin{align} {h}_{x}=0 \end{align}$$ and $$\begin{align} {h}_{y}=0 \end{align}$$,

Based on the first clarified solution which has been improved for clarity and correctness and the 2nd solution which was copied almost directly, it appears that the search for $$\displaystyle h $$ with out assuming $$\displaystyle h $$ is constant yields the result that $$\displaystyle h $$ is indeed constant to satisfy the exactness conditions.

= R*3.15 Verify whether the equation satifies the 2nd exactness condition =

== Given == Consider the particular form of exact N2-ODE:
 * $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0$$

with
 * $$g(x,y,p)=:\phi_x+\phi_yy'$$


 * $$=\phi_x(x,y,p)+\phi_y(x,y,p)p$$


 * $$f(x,y,p):=\phi_p(x,y,p)$$

The 2nd exactness condition for N2-ODEs are
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp}$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.2)
 * }
 * }

== Find == There is an equation that
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle (15p^4cosx^2)y''+(6xy^2)y'+[-6xp^5sinx^2+2y^3]=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.15.3)
 * }
 * }

Verify whether it satisfies the 2nd exactness condition.

== Solution ==

As for the Equation (3.15.3)

$$\displaystyle f(x,y,p)=15p^4cosx^2=\phi_p(x,y,p)$$

$$\displaystyle g(x,y,p)=\underbrace{(6xy^2)}_{\phi_y(x,y,p)}y'+\underbrace{[-6xp^5sinx^2+2y^3]}_{\phi_x(x,y,p)} $$

And then we got the partial derivative terms:


 * $$\displaystyle f_x=15p^4(-sinx^2)\cdot 2x=-30xp^4sinx^2$$


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


 * $$\displaystyle f_{xx}=-30p^4sinx^2-60x^2p^4cosx^2$$


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


 * $$\displaystyle f_{xp}=-120xp^3sinx^2$$


 * $$\displaystyle g_x=6y^2p-6p^5sinx^2-12x^2p^5cosx^2$$


 * $$\displaystyle g_{xp}=-30p^4sinx^2-60x^2p^4cosx^2+6y^2$$


 * $$\displaystyle g_y=12xyp+6y^2$$


 * $$\displaystyle g_{yp}=12xy$$


 * $$\displaystyle g_p=-30xp^4sinx^2+6xy^2$$


 * $$\displaystyle g_{pp}=-120xp^3sinx^2$$

To verity whether it satisfies the 2nd exactness condition, Substituting them into Equation (3.15.1):

Therefore,

which means it satisfies the condition Equation (3.15.1). Substituting those values into Equation (3.15.2):

Therefore,

which means it satisfies the condition Equation (3.15.2).

= R*3.16 Finish the story =

Given

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

\displaystyle \phi (x,y,p)=3p^5 \cos x^2 +2xy^3 =k $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _x = 3p^5(-\sin x^2)(2x)+2y^3 $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _y = 6xy^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _p = 15p^4 \cos x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle h_yy'+h_x=(6xy^2)y' + 2y^3 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.0 )
 * }
 * }

Find
$$ \displaystyle \phi (x,y,p) $$ using reverse engineering

Solution

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

\displaystyle \phi (x,y,p)=3p^5 \cos x^2 +2xy^3 =k $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.16.1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _x = 3p^5(-\sin x^2)(2x)+2y^3 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _y = 6xy^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.16.3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi _p = 15p^4 \cos x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.4 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle f= \phi _p =15p^4 \cos x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.5 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle g= \phi_yp+\phi _x = (6xy^2)y'+[-6xp^5 \sin x^2 +2y^3] $$ $$ From equation 6, we get:
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.6 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi = h(x,y)+3p^5\cos x^2 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.7 )
 * }
 * }

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

\displaystyle g=(h_y+0)y' + [h_x + 3p^5(-sinx^2)(2x)]=(6xy^2)y' - 6xp^5 \sin x^2 + 2y^3 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\; 3.16.8)
 * }
 * }

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

\displaystyle h_yy'+h_x=(6xy^2)y' + 2y^3 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.16.9 )
 * }
 * }

We have solved the problem assumed that $$\displaystyle h_x=2y^3$$

Now we assume that $$\displaystyle h_yy'=2y^3$$ so:


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

\displaystyle h_x=(6xy^2)y'=6y^2y' \cdot x $$
 * <p style="text-align:right;">
 * }
 * }

integrating h:


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

\displaystyle h(x,y)=3x^2 y^2 y' + k_2(y) $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle h_y=6x^2 y y'+{k_2}'(y) $$
 * <p style="text-align:right;">
 * }
 * }

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

\displaystyle h_y=\frac{2y^3}{y'} $$
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle {k_2}'(y)=\frac{2y^3}{y'}-6x^2y \cdot y' $$
 * <p style="text-align:right;">
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle k_2(y)=\frac{y^4}{2y'}-3x^2y'y^2+k_3 $$ so:
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle h(x,y)=\frac{y^4}{2y'}+k_3 $$
 * <p style="text-align:right;">
 * }
 * }


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

\displaystyle \phi (x,y,p)=3p^5\cos x^2 + \frac{y^4}{2y'}=k $$
 * <p style="text-align:right;">
 * }
 * }

= R*3.17 Derive Equation via RTT =

Given
Component form of gradient/divergence property:


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

$$
 * $$\frac{\partial\rho}{\partial x_i}u_i+\rho\,\frac{\partial u_i}{\partial x_i}=\frac{\partial}{\partial x_i}(\rho u_i)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.1)
 * }
 * }

Conservation of mass:


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

$$
 * $$\frac{\partial\rho}{\partial t}+{\rm div}(\rho\mathbf u)=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.2)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.2)
 * }
 * }

Reynolds Transport Theorem:


 * $$\frac{D}{Dt}\int_{\beta_t}f(x,t)\,d\beta_t=\int_{\beta_t}\left[\frac{\partial f}{\partial t}+{\rm div}(f\mathbf u)\right]\,d\beta_t$$

== Find ==

Show:


 * $$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\rho\frac{D\mathbf u}{Dt}\,d\beta_t$$

Solution
Apply $$\displaystyle f(x,t)=\rho(x,t)\,u_i(x,t)$$ to Reynolds Transport Theorem.

$$\frac{D}{Dt}\int_{\beta_t}\rho(x,t)\,u_i(x,t)\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\left(\rho(x,t)\,u_i(x,t)\right)}{\partial t}+{\rm div}(\rho(x,t)\,u_i(x,t)\mathbf u)\right]\,d\beta_t$$

$$\frac{D}{Dt}\int_{\beta_t}\rho\,u_i\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\left(\rho u_i\right)}{\partial t}+{\rm div}(\rho\mathbf u\mathbf u)\right]\,d\beta_t$$

Now apply Equation 3.17.1 to the first term under the integral on the right hand side:

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\left(\frac{\partial\rho}{\partial t}u_i+\rho\frac{\partial u_i}{\partial t}\right)+{\rm div}(\rho\mathbf u\mathbf u)\right]\,d\beta_t$$

Group terms:

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\rho}{\partial t}u_i+\left(\rho\frac{\partial \mathbf u}{\partial t}+{\rm div}(\rho\mathbf u\mathbf u)\right)\right]\,d\beta_t$$

Via application of Equation 3.17.2 we can eliminate the terms under the parentheses.

$$\frac{D}{Dt}\int_{\beta_t}\rho\mathbf u\,d\beta_t=\int_{\beta_t}\left[\frac{\partial\rho}{\partial t}u_i\right]\,d\beta_t$$

To complete the problem, we can use the definition of material time derivative:

$$\frac {D u_i(x,t)}{Dt}=\left.\frac{d u_i(x,t)}{dt}\right|_{X \rm fixed}=\frac{\partial u_i}{\partial t}$$

= R*3.18 Relate RTT to the 1D Case =

Given
Adjusted Reynold's Transport Theorem:


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

$$
 * $$\frac{D}{Dt}\int_{\beta_t}f(x,t)\,d\beta_t=\int_{\beta_t}\frac{\partial f}{\partial t}\,d\beta_t+\int_{\partial\beta_t}\mathbf n\cdot(f\mathbf u)\,d(\partial\beta_t)$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.3)
 * <p style="text-align:right;">$$\displaystyle (Equation\;3.17.3)
 * }
 * }

== Find ==

Obtain the following by means of Equation 3.17.3.


 * $$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=f(t,b(t))\,\frac{db(t)}{dt}-f(t,a(t))\frac{da(t)}{dt}+\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}ds$$

Solution
Let us choose our function $$\displaystyle f$$ to be a function of $$\displaystyle s$$ and $$\displaystyle t$$, so that $$\displaystyle f=f(t,s)$$. Let $$\displaystyle\mathcal B_t$$ be denoted by the interval between $$\displaystyle s=a(t)$$ and $$\displaystyle s=b(t)$$ so that $$\displaystyle \mathcal B_t \mapsto s$$. This has the following effect on Equation 3.17.3:

$$\frac{D}{Dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\int_{\partial s}f(s,t)\mathbf n\cdot\mathbf u\,d(\partial s)$$

Since the left hand side does not depend on $$\displaystyle x$$, the definition of material time derivative simplifies to $$\frac{D}{Dt}=\frac{d}{dt}.$$ Also the velocity of the material point, $$\displaystyle\mathbf u$$, is in the same direction of $$\displaystyle\mathbf n$$, and will be a derivative of $$\displaystyle s$$ rather than $$\displaystyle x.$$

$$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\int_{\partial s}f(s,t)\frac{\partial s}{\partial t}\,d(\partial s)$$

$$\frac{d}{dt}\int^{s=b(t)}_{s=a(t)}f(t,s)\,ds=\int^{s=b(t)}_{s=a(t)}\frac{\partial f(t,s)}{\partial t}\,ds+\left.\left[f(t,s)\frac{\partial s}{\partial t} \right ]\right|^{s=b(t)}_{s=a(t)}$$

Note we can transform the partial derivatives to regular derivatives since $$\displaystyle s$$ only depends on $$\displaystyle t$$. The next step of evaluating the integration at the bounds completes the proof.

= References =

= Team Work Distribution =

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