User:Egm6321.f11.team2/HW3

= R3.1 - A More General N1-ODE That Can Be Rendered Exact by the IFM=

Given
where

Find
Show that $$-\frac{1}{N} (N_x - M_y)$$ is a function of x only, only if $$\displaystyle k_1 (y) = d_1$$ (a constant).

Show that $$\displaystyle [ a(x) y + k_2(x) ] + \bar{b} (x) y' = 0$$ is a special case of 3.1.1.

Solution
Using 3.1.2 and 3.1.3 to expand the expressions for N and M, then taking their derivatives yields

$$\displaystyle \begin{align} N (x, y) &= \bar{b} (x, y) c(y) \\&= c(y) \int^x b(s) ds + c(y) k_1(y) \\N_x &= c(y) b(x) \end{align} $$

$$ \begin{align} M (x, y) &= a(x) \bar{c} (x, y) \\&= a(x) \int^y c(s) ds + a(x) k_2(x) \\M_y &= a(x) c(y) \end{align} $$

Substituting these expressions into the 2nd Exactness Condition yields

$$ \begin{align} -\frac{1}{N} ( N_x - M_y ) &= - \frac{1}{ c(y) \int^x b(s) ds + c(y) \, k_1(y) } \cdot [ c(y) b(x) - a(x) c(y) ] \\ &= - \frac{1}{c(y) [ \int^x b(s) ds + k_1(y) ] } \cdot c(y) [ \left( b(x) - a(x) \right) ] \\ &= - \frac{1}{ \int^x b(s) ds + k_1(y) } \cdot [ b(x) - a(x) ] \end{align} $$

which is a function of both x and y unless k1 is a constant.

Consider the case $$\displaystyle c(y) = 1$$. Then

$$\bar{c} (x, y) = \int^y ds + k(x) = y + k(x)$$

so 3.1.1 reduces to

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

as required.

In fact, this is still the case if $$\displaystyle c(y)$$ is equal to any constant.

=R*3.2 - Integrating factor method =

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

Find
Show that the above equation is exact or can be made exact by IFM.

Find the integrating factor h.

Solution
In order to exact condition

First condition of exactness is that

$$ \displaystyle M(x,y) + N(x,y)y' = 0$$3.2.1

$$ \displaystyle (5x^3 +2)\left( \frac {1}{5}y^5 + \sin x + d_2\right) = M(x,y) $$

$$ \displaystyle \left( \frac {1}{3}x^3+d_1\right)(y^4)y' = N(x,y)y' $$

$$\underbrace{ \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)}_{\displaystyle\color{blue}{N(x,y)}}y' + \underbrace {(5x^3+2)\left( \frac{1}{5}y^5+ \sin x + d_2\right)}_{\displaystyle\color{blue}{M(x,y)}} = 0$$

Thus, this equation satisfies the first condition.

Second condition of exactness is that

$$\displaystyle M_y(x,y) = N_x(x,y)$$3.2.2

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

$$\displaystyle \frac {\partial M}{\partial y} = 5x^3y^4 + 2y^4 $$

$$\frac {\partial N}{\partial x} = x^2y^4 $$

$$ M_y \neq N_x $$

It shows that this equation does not satisfy the second condition.

Therefore, in order to make this equation to exact, integrating factor must be used.

$$ \displaystyle n(x) = - \frac {1}{N}(N_x-M_y) = - \frac {1}{\bar b (x) \cancel{c(y)}}\left[\cancel{c(y)} b(x)- a(x)\cancel{c(y)}\right] = -\frac {1}{\bar b (x)}[b(x)-a(x)]$$3.2.3

where

$$\displaystyle \bar b(x) = \frac {1}{3} x^3 + d_1$$

$$\displaystyle b(x) = x^2 $$

$$\displaystyle a(x) = 5x^3 + 2 $$

Let d1 = 0, $$\displaystyle n(x) = - \frac {1}{\frac {1}{3}x^3}[x^2 - 5x^3 - 2] $$

$$ \displaystyle h(x) = \exp [\int^x n(s)ds+k]$$3.2.4

$$h(x) = \exp [ \int^x - \frac {1}{\frac {1}{3}s^3}(s^2 - 5s^3 - 2)ds+k] $$

$$ h(x) = \exp [ \int^x \frac{3}{s^3}(5s^3 -s^2 + 2)ds + k ] = \exp [ \int^x 3(5 -\frac{1}{s} + \frac{2}{s^3})ds + k ] $$

$$ h(x) = \exp [ 15x - \frac{3}{x^2} - 3\ln (x) + k] = \frac {1}{x^3} \exp [15x - \frac{3}{x^2} + k] $$

=R*3.3 -First Integral =

Given
Consider a class of N1-ODEs of the form in (1)p.13-2:

$$\displaystyle \overline{b}(x,y)c(y)y'+a(x)\overline{c}(x,y)=0$$ (3.3.1)

and

$$\displaystyle a(x)=\sin x^{3}$$

$$\displaystyle b(x)=\cos x$$

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

---

Find
1. Find an N1-ODE of the form (1) that is either exact of can be made exact by IFM.

2. Find the first integral $$ \phi(x,y)=k$$

Solution
I wonder if (2)p.13-4 is right? if $$\displaystyle a(x)=\sin x^{3}$$ is right, I cannot integrate it. But Professor's latex code reads: a(x)= 5x^3+2.I hope Professor could find this problems. In this problems, I treat $$\displaystyle \sin x^{3}= (\sin x)^{3}$$ instead.

Since

$$\displaystyle \overline{b}(x,y)=\int^{x}b(s)ds+k_{1}(y)$$

$$\displaystyle \overline{c}(x,y)=\int^{y}c(s)ds+k_{2}(x)$$

Substitute the given equations in the above two equations, we have

$$\displaystyle \overline{b}(x,y)=\int^{x}\cos sds+k_{1}(y)=\sin x+k_{1}(y)$$

$$\displaystyle \overline{c}(x,y)=\int^{y}\exp(2s)ds+k_{2}(x)=\frac{1}{2}\exp(2y)+k_{2}(x)$$ (3.3.2)

Let $$\displaystyle k_1=k_2=0$$ and substitute the above equations to (3.3.1)

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

hence

$$\displaystyle M(x,y)= \frac{1}{2}\exp(2y)\sin x^{3}$$ (3.3.3)

$$\displaystyle N(x,y)= \sin x\exp(2y)$$ (3.3.4)

It is easy to know that

$$\displaystyle M_y (x,y) \neq N_x (x,y)$$

Recall IFM

$$\displaystyle \frac{dh}{h}=\frac{M_{y}(x,y)-N_{x}(x,y)}{N(x,y)}=\frac{\exp(2y)(\sin x^{3}-\cos x)}{\sin x\exp(2y)}=\frac{\sin x^{3}-\cos x}{\sin x}$$ (3.3.5)

So, it can be made exact by IFM since the right hand side is a function of x only.

Since $$\displaystyle \exp(\int h(x)dx)=\exp(\frac{x}{2}-\frac{\sin2x}{4}-\ln(\sin x))$$

we have

$$\displaystyle \phi_{x}=M(x,y)\exp(\int h(x)dx)=\frac{1}{2}\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)\sin x$$

$$\displaystyle \phi_{y}=N(x,y)\exp(\int h(x)dx)=\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)$$

then

$$\displaystyle \phi=\frac{1}{2}\exp(\frac{x}{2}-\frac{\sin2x}{4}+2y)=k$$ (3.3.6)

=R*3.4- Construct a Class of N1-ODEs=

Given
A class of N1-ODES of the form is as follows:

Euler Integrating factor method (IFM) is written as

For (3.4.2) to be exact, 2nd exactness condition is applied to find h and (3.4.2) becomes

To solve (3.4.3), 2 cases is considered.

In the cases, the second case is defined as follows:

Case 2:suppose h(x,y), thus h is a function of y only, and then (3.4.3) becomes

Find
Construct a class of N1-ODEs,which is the counterpart of (3.4.1),and satisfies the condition (3.4.4) that an integrating factor h(y) can be found to render it exact.

Solution
This problem was solved without referring to past solutions.

Based on (3.4.4), consider:

Thus:

If  $$\displaystyle k_1(y)=d_1 $$ (constant), then (3.4.4) is satisfied as follows:

=R*3.5 Motion of a particle in the air. =

Given


We know that:

Find
Part 1 : Derive the [[media:pea1.f11.mtg14.djvu|equation of motion]]

$$\displaystyle m\frac{dv_x}{dt}=-kv^n\cos\alpha$$

$$\displaystyle m\frac{dv_y}{dt}=-kv^n\sin\alpha-mg$$

Part 2 : Particular case when $$\begin{align}k=0\end{align}$$, verify $$\begin{align}y(x)\end{align}$$ is a parabola

Part 3 : Consider $$k\ne0$$, $$\begin{align}v_{xo}=0\end{align}$$ for constant mass.

Part 3.1 : Consider $$k\ne0$$, $$\begin{align}v_{xo}=0\end{align}$$ for mass as function of time.

Solution
Part 1:

Newton's Second Law of Motion: Rate of change of momentum is directly proportion to force applied and this change takes place in the direction of force. Change in momentum in the X-direction: Using 3.5.4 we can write Putting the value of $$\displaystyle v $$ and $$\displaystyle \alpha $$ from Equation 3.5.1 and 3.5.2 into Equation 3.5.5, we get: Change in momentum in Y-direction: Using 3.5.7 we can write Putting the value of $$\displaystyle v $$ and $$\displaystyle \alpha $$ from Equation 3.5.1 and 3.5.2 into Equation 3.5.8, we get: Equation 3.5.6 and 3.5.9 forms a System of coupled N1-ODE.

Part 2:

When $$\displaystyle k=0 $$ Equation 3.5.6 reduces to: At $$\displaystyle t=0\,\,,v_x=v_{x0} $$

Therefore $$\displaystyle v_x=v_{x0} $$ Or Where x_0 is the initial X-coordinate of projectile.

When $$\displaystyle k=0 $$ Equation 3.5.9 reduces to: At $$\displaystyle t=0\,,v_y=v_{y0} \,; t=t\,,v_y=v_y $$

Integrating Equation 3.5.13 using the above boundary conditions we get: At $$\displaystyle t=0\,,y=y_0 \,; t=t\,,y=y $$

Integrating Equation 3.5.14 using the above boundary conditions we get: From Equation 3.5.12 Substituting the value of time from Equation 3.5.17 into equation 3.5.16 we get: Equation 3.5.18 is clearly an equation of parabola.

Part 3:

When $$\displaystyle k \ne 0 $$ and $$\displaystyle v_x=0 $$

The SC-N1-ODE reduces to: Equation 3.5.20 satisfies the first exactness condition but doesn't satisfy the second exactness condition. So we multiply it by the Integrating factor $$\displaystyle h(t,v_y) $$ to make it exact. It should satisfy the second exactness condition i.e Equation 3.5.21 becomes Mass is constant so $$\displaystyle\frac{\partial m}{\partial t}=0$$. Let's also consider that $$\displaystyle h_t(t,v_y)=0 $$. So equation 3.5.23 transforms to: Solving for Integrating factor from Equation 3.5.24 we get: Now Equation 3.5.21 becomes: Integrating both sides and solving for $$\displaystyle v_y $$ For n=1 Integrating both sides we get: At $$\displaystyle t=0, \, v_y=v_{y0} $$

Solving for $$\displaystyle v_y $$ using above condition we get: Solution for $$\displaystyle v_y $$ is possible when $$\displaystyle n=1 $$ but for higher value of $$\displaystyle n=2,3,4... $$ integration is not possible and solution cannot be found out.

Part 3.1

When mass is variable with respect to time:

Equation 3.5.23 is difficult to solve for Integrating Factor because it is a Partial Differential Equation containing terms $$\displaystyle m_t, h_v , h_t $$ Hence no definite solution is possible to find using Euler Integration factor Method.

=R*3.6 - Deriving the Equations of Motion for Coupled Pendulums=

Given


The equations of motion for the coupled pendulums shown above are

$$m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (1)

$$m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (2)

Find
1. Derive the equation of motion (1)-(2). 2. Write equations (1)-(2) in matrix form

$$\mathbf{\dot{x}}(t) = \mathbf{A}(t)\,\mathbf{x}(t) + \mathbf{B}(t) \,\mathbf{u}(t)$$ (3)

with

$$\mathbf{x} := \left \lfloor \theta_1 \ \dot \theta_1 \ \theta_2 \ \dot \theta_2 \right \rfloor ^T \in \mathbb R^{4 \times 1}$$

$$\mathbf{u} := \begin{Bmatrix} u_1l \\ u_2l \end{Bmatrix} \in \mathbb R^{2 \times 1}$$

Solution
This problem was solved without referring to past solutions.

Deriving the Equations of Motion
Using Newton's Second Law as it relates to rotation


 * $$\displaystyle \sum \tau := I \alpha$$ (4)

Where,


 * $$\displaystyle \tau $$ = torque about a singular principal axis.
 * $$\displaystyle I = r l^2$$ = Moment of Inertia.
 * $$\displaystyle \alpha = \ddot \theta$$ = angular acceleration

Using the definition of torque.


 * $$\displaystyle \tau := \mathbf r \ x \ \mathbf F = r F sin(\theta)$$(4)

Where r and F are the magnitudes of $$ \mathbf r$$ and $$ \mathbf F$$ respectively. We can use equation (4) to obtain the effect of each force, and then sum them up to equal the total torque.

Hooke's Law,


 * $$\displaystyle \mathbf F = -k x$$

where x is the displacement of the spring's end from its equilibrium position and k is the spring constant.


 * $$\displaystyle \tau_{s1} = -k a^2 (sin(\theta_1) - sin(\theta_2)) = -k a^2 (\theta_1 - \theta_2)$$

Since the spring force acting on the second pendulum is of equal magnitude and opposite direction of the spring force acting on the first pendulum the torque created by the spring for the second pendulum is,


 * $$\displaystyle \tau_{s2} = -k a^2 (\theta_2 - \theta_1)$$

The torque due to gravity,


 * $$\displaystyle \tau_{g1} = - m_1 g l sin(\theta_1) = - m_1 g l \theta_1$$
 * $$\displaystyle \tau_{g2} = - m_2 g l sin(\theta_2) = - m_2 g l \theta_2$$

The torque due to the applied forces,


 * $$\displaystyle \tau_{u1} = u_1 l cos(\theta_1) = u_1 l$$
 * $$\displaystyle \tau_{u2} = u_2 l cos(\theta_2) = u_2 l$$

Summing all the forces,


 * $$\displaystyle \sum \tau_1 = -k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l$$
 * $$\displaystyle \sum \tau_2 = -k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l$$

Using equation (1),


 * $$\displaystyle m_1 l^2 \ddot \theta_1 = -k a^2 (\theta_1 - \theta_2) - m_{1} g l \theta_1 + u_1 l $$ (5)
 * $$\displaystyle m_2 l^2 \ddot \theta_2 = -k a^2 (\theta_2 - \theta_1) - m_{2} g l \theta_2 + u_2 l $$ (6)

Which are equivalent to equations (1) and (2).

Write the Equations of Motion in Matrix Form
We need to find an equation that takes the form:

$$\begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} ... \ \ \ \\ \ ... \ \ \\ \ \ ... \ \\ \ \ \ ... \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} ... \ \\ \ \ \\ \ ... \\ \ \ \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7)

Reordering equations (5) and (6) we obtain,


 * $$\displaystyle \ddot \theta_1 = \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \theta_1 + \frac{k a^2}{m_1 l^2} \theta_2 + \frac{1}{m_1 l^2} u_1 l $$ (8)


 * $$\displaystyle \ddot \theta_2 = \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \theta_2 + \frac{k a^2}{m_2 l^2} \theta_1 + \frac{1}{m_2 l^2}u_2 l $$ (9)

Entering equations (8) and (9) into equation (7) we obtain:

$$\displaystyle \begin{Bmatrix} \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \end{Bmatrix} = \begin{Bmatrix} 0 \ \ 1 \ \ 0 \ \ 0 \\ \frac{-(k a^2 + m_{1} g l)}{m_1 l^2} \ 0 \ \frac{k a^2}{m_1 l^2} \ 0 \\ \ 0 \ \ 0 \ \ 0 \ \ 1 \\ \frac{k a^2}{m_2 l^2} \ 0 \ \frac{-(k a^2 + m_{2} g l)}{m_2 l^2} \ 0 \end{Bmatrix} \ \begin{Bmatrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \end{Bmatrix} + \begin{Bmatrix} 0 \ \ 0 \\ \frac{1}{m_1 l^2} \ 0 \\ 0 \ 0 \\ 0 \ \frac{1}{m_2 l^2} \end{Bmatrix} \ \begin{Bmatrix} u_1 l \\ u_2 l \end{Bmatrix}$$ (7)

= R*3.7 - Solving Linear Systems with the Integrating Factor Method =

Find
Show that the solution to 3.7.1 is

$$x(t) = \exp\{ a(t-t_0)\} \, x(t_0) + \int^t_{t_0} \exp\{a(t - \tau)\} \, b \, u(\tau) d\tau$$

and identify the integrating factor, the homogenous solution, and the particular solution.

Next, show that the solution to 3.7.2 is

$$x(t) = \left[\exp\int^t_{t_0} a(\tau) d\tau\right] x(t_0) + \int^t_{t_0}\left[\exp\int^t_\tau a(s) ds \right] b(\tau) u(\tau) d\tau$$

Solution
This problem was solved without referring to past solutions.

We first put 3.7.1 into the particular form, suppressing the functional dependence notation for now

$$0 = \underbrace{ a x + b u }_M + \underbrace{(- 1) }_N \dot x$$

In contrast to the normal notation for this course, in this problem t is the independent variable and x is the dependent variable. Thus, following the normal logic of multiplying the above equation through by h(t, x), requiring that $$\displaystyle (hM)_x = (hN)_t$$ and then assuming that h is a function of t only yields

Now,

$$ \begin{align} N_t &= 0 \\M_x &= a \end{align} $$

Plugging these values into 3.7.3 and then solving for h

$$ \begin{align} \frac{h_t}{h} &= -a \\ \int\frac{dh}{h} &= - \int^t_{t_0} a\, d\tau \\ \log h &= - a (t - t_0) \\ h &= \exp\{- a (t - t_0) \} \end{align} $$

This is the integrating factor.

We procede to solve for x(t) by multiplying 3.7.1 by the integrating factor and integrating

$$ \begin{align} \exp\{ -a (t-t_0) \} \, \dot{x} (t) &= \exp \{ -a (t-t_0) \} \, a\, x(t) + \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \exp\{ -a (t-t_0) \} \, \dot{x} (t) - a\, \exp \{ -a (t-t_0) \} \, x(t) &= \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \frac{d}{dt} \left( \exp\{ -a (t-t_0) \}\, x(t) \right) &= \exp \{ -a (t-t_0) \} \, b\, u(t) \\ \int^t_{t_0} d \left( \exp \{ -a (t-t_0) \}\, x(t) \right) &= \int^t_{t_0} \exp \{ -a (\tau-t_0) \}\, b \, u(\tau) \, d\tau \\ \exp \{ -a(t-t_0) \}\, x(t) - x(t_0) &= \int^t_{t_0} \exp \{ -a (\tau-t_0) \}\, b \, u(\tau) \, d\tau \\ x(t) &= \exp \{ a(t-t_0) \}\, x(t_0) + \int^t_{t_0} \exp \{ a(t-\tau) \} \, b \, u(\tau) \, d\tau \end{align} $$

The first term on the right-hand side is the homogenous solution (known in linear systems theory as the zero-input response) while the second term is the particular solution (know in linear systems theory as the zero-state response).

To solve 3.7.2, we simply note that, since we don't know the functional form of a(t), we cannot arrive at a closed-form representation of its integral. Thus, the integrating factor when a is a function of t is

$$h = \exp \{ - \int^t_{t_0} a(\tau) d\tau \}$$

The rest of the solution proceeds exactly the same as above, where we need to change the dummy integration variable in the particular solution since $$\tau$$ is being used in the outer integral, and of course we must add in the functional dependence of b on t.

=R*3.8 - Linear Time Variant System =

Given
$$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$3.8.1

Find
Generalize $$\displaystyle \mathbf {x}(t) = exp [{\mathbf {A}(t-t_0)}]\mathbf{x}(t_0) + \int^t_{t0} exp[{ \mathbf{A}(t-\tau)}]\mathbf{B}\mathbf{u}(\tau)d\tau $$ to the case of linear time varient system.

Verify that your expression is indded the solution for

$$\displaystyle \dot\mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Solution
$$\displaystyle \mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Let A(t)= A and B(t) = B Where $$\mathbf{x} $$ is a n x 1 vector, $$\mathbf{u}$$ is 'm x 1' vector, $$\mathbf{A}$$ is a 'n x n' matrix and $$\mathbf{B}$$ is a 'n x m' matrix.

The equation is following that

$$\displaystyle \dot x(t) = Ax(t) + Bu(t)$$

Rearrange the above equation. Thus,

$$\displaystyle \underbrace{1}_{N}\cdot \dot x(t) - \underbrace{Ax(t)}_{M} = Bt(u) $$

$$\displaystyle h(t) = e^{-\int \frac{1}{N}(N_t - M_x)dt}$$3.8.2

Where :

$$\displaystyle N =1, N_t = 0, M_x = - A$$ From the above equation, intergrating factor is that $$\displaystyle h(t) = e^{\int^t-Ads}=e^{-At}$$3.8.3

$$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = \mathbf{B}\mathbf{u}(t)e^{-At}$$

[LHS] : $$\displaystyle e^{-At}(\dot x(t)-Ax(t)) = [e^{-At}x(t)]'$$

Rearraing the eq

$$\displaystyle [e^{-At}x(t)]' = \mathbf{B}\mathbf{u}(t)e^{-At}$$3.8.4

Interval is $$\displaystyle t_0< \tau < t$$

The equation is following that

$$\displaystyle \int^t_{t_0} [e^{-A\tau}x(\tau)]'d\tau = \int^t_{t_0}Bu(\tau)e^{-A\tau}d\tau $$

$$\displaystyle e^{-At}x(t)-e^{-At_0}x(t_0) = \int^t_{t_0}e^{-A\tau}Bu(\tau)d\tau$$

Dividing by $$\displaystyle e^{-At}$$

And rearranging for x(t)

$$\begin{align} \displaystyle x(t)&=\frac {\int^t_{t_0}e^{-A\tau}\mathbf{B}\mathbf{u}(\tau)d\tau}{e^{-At}} + e^{A(t-t_0)}x(t_0) \\&={\int^t_{t_0}e^{A(t-\tau)}\mathbf{B}\mathbf{u}(\tau)d\tau} + e^{A(t-t_0)}x(t_0)\end{align} $$ SC-L1-ODE-VC can be based on the SC-L1-ODE-CC.

=R*3.9 - System of Coupled L1-ODE-VC=

Given
The following equation (1),(2),(3)p.15-4

$$ \displaystyle \mathbf{\Phi}(t,t_{0})_{n\times n}=\exp(\mathbf{A}(t-t_{0}))$$ (3.9.1)

The fundamental or state transition matrix $$ \displaystyle \mathbf{\Phi}$$ is related to the integrating factor.

Properties of the state transition matrix $$ \displaystyle \mathbf{\Phi}$$

$$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n}$$ (3.9.2)

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$ <p style="text-align:center">(3.9.1)

Find
Verify that (3.9.1) satisfies (3.9.2)-(3.9.3)

Solution
$$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\frac{d}{dt}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n} $$

<p style="text-align:center">(3.9.4) According to the formula of exponentiation of a matrix

$$ \displaystyle \exp(\mathbf{A})=\sum_{k=0}^{\infty}\mathbf{A}^{k}$$

<p style="text-align:center">(3.9.5) and

$$ \displaystyle \exp(\mathbf{A}(t-t_{0}))=\sum_{k=0}^{\infty}\mathbf{A}^{k}(t-t_{0})^{k}$$ <p style="text-align:center">(3.9.6)

Rebember $$ \displaystyle (t-t_0)^0=1$$, and expand 3.9.6

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$

-- Remark Sophus Lie is the first one who found that the solutions of (3.9.2) form a sort of group, we then call it Lie Group. Lie group equips both geometric characteristic (differentiable manifold) and algebraic characteristic (group). =R3.10 Free Vibration of Coupled Pendulums =

Given
The matrix foam of linear time-variant systems is

The solution of generalized SC-L1-ODE-CC is

Coupled pendulums system is shown as follows:

Pendulums:

{| style="width:100%" border="0" $$\displaystyle a=0.3,l=1,k=0.2 $$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle m_{1}g=3,m_{2}g=6,g=10$$

No applied forces:

{| style="width:100%" border="0" $$\displaystyle u_{1}=u_{2}=0$$
 * style="width:95%" |
 * style="width:95%" |

Initial conditions:

{| style="width:100%" border="0" $$\displaystyle \theta_{1}(0)=0,\dot\theta_{1}(0)=-2 $$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle \theta_{2}(0)=0,\dot\theta_{2}(0)=1$$

Find
1. Use matlab's ode45 command to integrate the system, (3.10.3) and (3.10.4), in matrix form of (3.10.1) for $$\displaystyle t\in[0,7] $$

2. Use (3.10.2) to find the solution at the same time stations as in Q1.

3. Plot $$\displaystyle \theta_{1}(t) $$ and $$\displaystyle \theta_{2}(t) $$ from Q1 and Q2.

Solution
This problem was solved without referring to past solutions.

(3.10.3) and (3.10.4) are reordered as follows:

Finally we obtain the following matrix:

Mathlab codes and plot from Q1

Mathlab codes and plot from Q2

=R*3.11 Representing the solution of SC-L1-ODE-VC in the terms of State Transition Matrix using the properties of State Transition Matrix =

Find
Express Equation 3.11.1 in terms of [[media:pea1.f11.mtg16.djvu|State Transition Matrix]] $$\displaystyle \mathbf {X}(t)=\mathbf{\Phi}(t,t_0)\mathbf{X}(t_0)+\int^t_{t_0}\mathbf{\Phi}(t,\tau)\,\mathbf{B}(\tau)\,\mathbf{U}(\tau)\,d\tau $$ using properties of State Transition Matrix Equation 3.11.2 and 3.11.3.

Solution
Let,

Equation 3.11.1 becomes, Rearranging Equation 3.11.5 Clearly Equation 3.11.6 satisfies the first exactness condition but doesn't hold good for 2nd exactness condition so in order to make it exact we multiply it by an Integrating Factor $$\displaystyle h(t,{\Phi}) $$.

Equation 3.11.6 becomes Now applying the Second Exactness Condition to Equation 3.11.7. Since $$\displaystyle N(t,\phi)=1 $$.

$$\displaystyle \frac{\partial N}{\partial t}=0 $$ Let $$\displaystyle \frac{\partial h}{\partial \phi}=0 $$

Equation 3.11.9 becomes Since $$\displaystyle M(t,\phi)= -\mathbf {A}(t)\mathbf{\Phi}(t,t_0) $$

$$\displaystyle M_\phi= -\mathbf {A}(t) $$

Solving for integrating factor $$\displaystyle h(t) $$ from Equation 3.11.10.

$$\displaystyle \int \frac{dh}{h}= - \int \mathbf{A(t)}dt $$

Hence $$\displaystyle h(t)=\exp \{-\int \mathbf{A(t)}dt\}$$

Equation 3.11.7 transforms to : Clearly Equation 3.11.11 can be written as: Integrating Equation 3.11.12 within limits $$\displaystyle t_0-t $$ Solving 3.11.13 for $$\displaystyle \mathbf{\Phi}(t,{t_0}) $$ we get: Now from property 1 of State Transition Matrix i.e Equation 3.11.2, (Rearranging and integrating within the limits $$\displaystyle t_0-t $$) we get : Now using property 2 of State Transition Matrix $$\displaystyle\mathbf{\Phi}({t_0},{t_0})= \mathbf I $$ Now  $$\displaystyle \ln \,\mathbf I $$   is a zero matrix.

Therefore Equation 3.11.7 becomes: Similarly, since $$\displaystyle \tau $$ is a dummy integration variable: Now looking at the assumption we did in the beginning i.e Equation 3.11.4 And Using Equation 3.11.14, 3.11.18 , 3.11.19 , 3.11.20 and 3.11.21 we get:

=R*3.12 - Determining the SC-L1-ODE-CC Equation for the Roll Control of a Rocket=

Given


$$\delta$$ = aileron deflection (angle)

$$\phi$$ = roll angle

$$\omega$$ = roll angular velocity

$$Q$$ = aileron effectiveness proportional to inverse of rocket roll inertia

$$\tau$$ = roll time constant

$$\dot \phi = \omega$$ (1)

$$\dot \omega = -\frac{1}{\tau} \omega + \frac{Q}{\tau} \delta$$ (2)

$$\dot \delta = u$$ (3)

Find
Put equations (1)-(3) in the form of a SC-L1-ODE-CC (System of Coupled Linear First Order Differential Equations with constant coefficients) with form:

$$\mathbf{\dot{x}}(t) = \mathbf{A}(t)\,\mathbf{x}(t) + \mathbf{B}(t) \,\mathbf{u}(t)$$ (4)

with $$\mathbf{A}, \mathbf{B}$$ being constant matrices.

Solution
This problem was solved without referring to past solutions.

Equations (1)-(3) constitute a system of 3 coupled linear equations. Therefore

$$\mathbf{\dot x}:= \left \lfloor \ \dot \phi \ \dot \omega \ \dot \delta \ \right \rfloor ^T \in \mathbb R^{3 \times 1}$$

$$\mathbf{x}:= \left \lfloor \ \phi \ \omega \ \delta \ \right \rfloor ^T \in \mathbb R^{3 \times 1}$$

Since the aileron's angular velocity is the control,

$$\mathbf{u} := \dot \delta = u$$

Thus, placing equations (1)-(3) in the form of equation (4) yields

$$\begin{Bmatrix} \dot \phi \\ \dot \omega \\ \dot \delta \end{Bmatrix} = \begin{Bmatrix} 0 \ 1 \ 0 \\ 0 \ \frac{-1}{\tau} \ \frac{Q}{\tau} \\ 0 \ 0 \ 0\end{Bmatrix} \ \begin{Bmatrix} \phi \\ \omega \\ \delta\end{Bmatrix} + \begin{Bmatrix} 0 \\ 0 \\ 1\end{Bmatrix} \ u$$

= R*3.13 - Derivation of the 2nd Exactness Conditions for N2-ODEs =

Given
If an N2-ODE is exact, then it must be expressible in the form

$$ \begin{align} \frac{d\phi}{dx} &= \phi_x + \phi_y\, p + \phi_p\, p_x \\ &= g + f\, p_x \end{align} $$

where

$$ \begin{align} p &:= \frac{dy}{dx} \\ g &:= \phi_x + \phi_y\, p \\ f &:= \phi_p \end{align} $$

Furthermore, since we assume that $$\phi$$ is sufficiently smooth,

$$ \begin{align} \phi_{xy} &= \phi_{yx} \\ \phi_{yp} &= \phi_{py} \\ \phi_{xp} &= \phi_{px} \end{align} $$

Finally, we are given a particular equation,

Find
Derive the 2nd Exactness Conditions, given below

Also, show that 3.13.1 satisfies these conditions.

Solution
This problem was solved without referring to past solutions, though reference was made to the lecture notes from Meeting 17 for the second part.

3.13.3 is much easier to derive, so let's begin there

$$ \begin{align} g &= \phi_x + \phi_y\, p \\ g_p &= \phi_{xp} + \phi_{yp}\, p + \phi_y \\ &= f_x + f_y\, p + \phi_y \\ g_{pp} &= f_{xp} + p\, f_{yp} + f_y + \phi_{yp} \\ &= f_{xp} + p\, f_{yp} + 2 f_y \end{align} $$

In deriving 3.13.2, we will make much use of the continuity of the second derivatives of $$\phi$$. Much of this material is a restatement of that contained in the lecture notes from Meeting 17.

$$ \begin{align} \phi_{px} &= \phi_{xp} \\ f_x &= [ g - \phi_y\, p ]_p \\ f_x &= g_p - \phi_{yp}\, p - \phi_y \\ f_x &= g_p - f_y\, p - \phi_y \end{align} $$

where use was made of the definition of g in the second step to obtain an expression for $$\phi_x$$. We can rearrange this last expression to arrive at an expression for $$\phi_y$$

This expression will be useful later on.

$$ \begin{align} \phi_{py} &= \phi_{yp} \\ f_y &= [ \frac{1}{p} (g - \phi_x) ]_p \\ f_y &= - \frac{1}{p^2} (g - \phi_x) + \frac{1}{p} (g_p - \phi_{xp}) \end{align} $$

where again the definition of g was used in the second step, this time to obtain an expression for $$\phi_y$$. Solving this expression for $$\phi_x$$ yields

Finally, we make use of $$\phi_{xy} = \phi_{yx}$$. We first differentiate 3.13.4 with respect to x

$$ \begin{align} \phi_{y} &= g_p - f_y\, p - f_x \\ \phi_{yx} &= g_{px} - f_{yx}\, p - f_y\, p_x - f_{xx} \\ \phi_{yx} &= g_{xp} - f_{xy}\, p - f_y\, p_x - f_{xx} \end{align} $$

We now differentiate 3.13.5 with respect to y, noting that $$p_y = \partial^2 y\, / \partial x\, \partial y = \partial^2 y\, / \partial y\, \partial x = \partial\, 1 / \partial x = 0$$

$$ \begin{align} \phi_x = g - p (g_p - f_x) + p^2\,f_y \\ \phi_{xy} = g_y - p (g_{py} - f_{xy}) + p^2\, f_{xy} \\ \phi_{xy} = g_y - p (g_{yp} - f_{xy}) + p^2\, f_{xy} \end{align} $$

Equating these two expressions and collecting like terms yields

$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy} + f_y p_x = g_{xp} + p g_{yp} - g_y$$

Note that there is an extra term on the left hand side of this expression compared to the desired result.

Applying the definition of g and f to 3.13.1 yields

$$ \begin{align} g &= x (y\,')^2 + y y\,' \\ &= x p^2 + y p \\ f &= xy \end{align} $$

Performing the necessary derivatives is simple

$$ \begin{align} g_p &= 2xp + y \\ g_{pp} &= 2x \\ g_y &= p \\ g_{yp} &= 1 \\ g_{xp} &= 2p \\ \\ f_{xx} &= f_{yy} = f_{xp} = f_{yp} = 0 \\ f_{xy} &= 1 \\ f_y &= x \end{align} $$

The exactness conditions are

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

=R*3.14 - Find solution for h(x,y) =

Given
$$\displaystyle g=\cancel{xp^2}+\cancel{yp} = \phi_x + \phi_y p = \underbrace{(h_x + \cancel{yp})}_{\phi_x}+\underbrace{(h_y+\cancel{xp})}_{\phi_y}p $$

$$\displaystyle p(x)= y'(x) $$

Find
$$\displaystyle h_x + h_y p = 0$$<p style="text-align: center;">3.14.1

Find h(x,y).

Solution
$$\displaystyle h_x + h_y p = 0 $$

where

$$\displaystyle h_x = \frac {\partial h}{\partial x}$$

$$\displaystyle h_y = \frac {\partial h}{\partial y}$$

$$\displaystyle p = \frac {dy}{dx} $$

The eq 3.14.1 is following that

$$ \displaystyle \frac {\partial h(x,y)}{\partial x}+\frac {\partial h(x,y)}{\partial y} \frac {dy}{dx} = \frac {d}{dx} h(x,y) = 0$$

Since $$\displaystyle \frac {d}{dx} h(x,y)$$ is 0, h(x,y) is only the function of y.

$$\displaystyle h(x,y) = f(y) $$ It shows that

$$\displaystyle h_x = 0 $$ Thus, the eq 3.14.1 is $$\displaystyle h_y \cdot p = 0 $$<p style="text-align: center;">3.14.2

In order to satisfy the eq 3.14.2,

1.$$\displaystyle h_y = 0 $$ or

2.$$\displaystyle p = \frac {dy}{dx} = 0 $$

If the second one is possible, the entire equation will be zero.It is trivial.

Thus, the possible solution is the first one.

$$\displaystyle h_x = 0, h_y = 0$$

Therefore, we can conclude that

h = constant = k

=R*3.15 - Second Exact condition=

Given
Given the following equations (1)p.18-3

$$ \displaystyle (15p^{4}\cos x^{2})y''+(6xy^{2})y'+[-6xy^{5}\sin x^{2}+2y^{3}]=0$$ <p style="text-align:center">(3.15.1)

$$ \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

<p style="text-align:center">(3.15.2) $$ \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

<p style="text-align:center">(3.15.3)

Find
Verify whether (3.15.1) satisfies the 2nd exactness condition (3.15.2) and (3.15.3).

Solution
According to Prof. Loc Vu-Quoc's lecture note, we know that

$$ \displaystyle \phi_{x}=3p^{5}(-\sin x^{2})(2x)+2y^{3}$$

<p style="text-align:center">(3.15.4)

$$ \displaystyle \phi_{y}=6xy^{2}$$

<p style="text-align:center">(3.15.5)

$$ \displaystyle f:=\phi_{p}=15p^{4}\cos x^{2}$$ <p style="text-align:center">(3.15.6)

$$ \displaystyle g:=\phi_{y}y'+\phi_{x}=(6xy^{2})y'+[-6xp^{5}\sin x^{2}+2y^{3}]$$ <p style="text-align:center">(3.15.7)

Hence

$$ \displaystyle f_{xx}=(\phi_{p})_{xx}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$ <p style="text-align:center">(3.15.8)

$$ \displaystyle f_{xp}=-120xp^{3}\sin x^{2},\, f_{xy}=0,\, f_{yy}=0,\, f_{y}=0,\, f_{yp}=0$$ <p style="text-align:center">(3.15.9) $$ \displaystyle g_{y}=(12xy)y'+6y^{2},\, g_{yp}=0$$ <p style="text-align:center">(3.15.10) $$ \displaystyle g_{xp}=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2},\,g_{pp}=-120xp^{3}\sin x^{2}$$ <p style="text-align:center">(3.15.11)

Substitute (3.15.8)-(3.15.11) to (3.15.2)

$$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+2p\times0+p^{2}\times0$$

$$ \displaystyle LHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}$$ <p style="text-align:center">(3.15.12) $$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}+p\times0-((12xy)y'+6y^{2})$$

$$ \displaystyle RHS=-30p^{4}\sin x^{2}-60x^{2}p^{4}\cos x^{2}-((12xy)y'+6y^{2})$$ <p style="text-align:center">(3.15.13)

Thus $$ \displaystyle LHS\neq RHS$$, then (3.15.11) does not satisfy (3.15.2), but we can easily find that (3.15.1) does satisfy (3.15.2)

=R*3.16- Completion of Second Choice=

Given
From (3.16.1) and (3.16.2), we get

For the first choice of the solution, assume that

The process to find $$\displaystyle \phi(x,y,p)$$ is as follows:

Find
Finish the second choice assuming that

Solution
This problem was solved without referring to past solutions.

(3.16.10) can be rewritten as follows:

And then the both sides of (3.16.11) are integrated as follows:

The above integration results in

The result is exactly same with (3.16.5).

Thus, the next process and resulted $$\displaystyle \phi(x,y,p)$$ would be same with the above given functions from (3.16.6) to (3.16.9).

=R*3.17- Application of Reynold's Transport Theorem =

Given
Considering: in [[media:pea1.f11.mtg19-20.djvu|Reynold's Transport Theorem]] Where:

$$\displaystyle \mathbf u = \mathbf u (x_i,t):= u_i(x_i,t) $$

Find
Show: Using Equations 3.17.4 and 3.17.5: Conservation of Mass:

Solution
Putting the value of $$\displaystyle f(x,t)$$ i.e Equation 3.17.1 in Reynold's Transport Theorem, Equation 3.17.2 we get: Partially differentiating within the integral on the right hand side we get: Rearranging terms: Since: Taking the Material Time Derivative of Equation 3.17.9 : Term $$\displaystyle \frac{\partial x_i}{\partial t}= u_i $$ in Equation 3.17.10.

We see that the first bracket of of right hand side in Equation 3.17.8 is nothing but $$\displaystyle \frac {D\mathbf u}{Dt} $$ of Equation 3.17.10 and second bracket is zero by Conservation of Mass (Equation 3.17.5).

Hence Equation 3.17.8 reduces to:

=R*3.18 - Deriving the 1-Dimensional Case from Reynolds Transport Theorem=

Given
The alternate version of Reynolds Transport Theorem is:

$$\frac{D}{Dt} \int_{\mathcal B_t}f(x,t) \ d \mathcal B_t = \int_{\mathcal B_t} \frac{\partial f(x,t)}{\partial t} \ d \mathcal B_T + \int_{\partial \mathcal B_t} \mathbf n \cdot (f(x,t) \mathbf u(x,t)) \ d (\partial \mathcal B_t)$$ (1)

$$\mathcal B_t$$ = region of integration $$\displaystyle x$$ = a point in the region $$\mathbf u(x,t)$$ = the velocity of x at time t $$\mathbf n$$ = outward unit normal to the boudary

A particular case of equation (1) for 1-Dimension is:

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)} f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + f(b(t),t) \ \frac{db(t)}{dt} - f(a(t),t) \ \frac{da(t)}{dt}$$ (2)

Find
Obtain particular case of equation (2) from the general case of equation (1)

Solution
This problem was solved without referring to past solutions.

Various terms simplify in the 1-dimensional case, since all measurements are now 1-dimensional:

$$\mathcal B_t = s$$ $$\displaystyle f(x,t) = f(s,t)$$ $$\mathbf u(x,t) = u(s,t)$$ $$\mathbf n = 1$$

$$\displaystyle \mathbf n$$ becomes 1 because it is a unit vector. Since a unit vector must have a magnitude of 1, and it only has 1 dimension, $$\displaystyle \mathbf n$$ must equal 1. The dot product also becomes a multiplication in the 1-dimensional case. Substituting these new values into equation (1) we obtain,

$$\frac{d}{dt} \int_{s}f(s,t) \ ds = \int_{s} \frac{\partial f(s,t)}{\partial t} \ ds + \int_{\partial s} 1 \times (f(s,t) u(s,t)) \ d (\partial s)$$ (3)

Where s is defined over the interval $$ \left [ a(t),b(t) \right ]$$ making equation (3):

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)}f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + \int_{\partial s} f(s,t) u(s,t) \ d (\partial s)$$ (4)

$$\displaystyle \partial s$$ is the boundary of s. The boundary of s is only defined at two points, a(t) and b(t). Because of this the last term in equation (4) becomes:

$$\int_{\partial s} f(s,t) u(s,t) \ d (\partial s) = f(b(t),t)u(b(t),t) - f(a(t),t)u(a(t),t)$$ (5)

Since u(b(t),t) is the velocity of b(t) at time t

$$u(b(t),t) = \frac{db(t)}{dt}$$

Substituting for the last term we obtain equation (2).

$$\frac{d}{dt} \int^{s=b(t)}_{s=a(t)}f(s,t) \ ds = \int^{s=b(t)}_{s=a(t)} \frac{\partial f(s,t)}{\partial t} \ ds + f(b(t),t) \frac{db(t)}{dt} - f(a(t),t) \frac{da(t)}{dt}$$ (6)

=References=

1. Malvern 1969, Introduction to the mechanics of a continuous medium, p.211

=Contributing Members(need updating)=

Egm6321.f11.team2.rho I solved R*3.4, R*3.10, and R*3.16 and read the proof of R*3.2, and  R*3.3. 20:12, 5 October 2011 (UTC) Egm6321.f11.team2.Xia I solved R*3.3, R*3.9, and R*3.15 and read the proof of R*3.1, R*3.2. and R*3.5   14:21, 3 October 2011 (UTC) Egm6321.f11.team2.kim I solved R*3.2, R*3.8, and R*3.14 and read the proof of R*3.1 and R*3.18. 02:12, 4 October 2011 (UTC) Egm6321.f11.team2.epps I solved R3.1, R3.7, and R3.13, reviewed R3.18 and R3.17, and made a small correction to the A matrix in R3.6.) Egm6321.f11.team2.epps 01:53, 5 October 2011 (UTC) Egm6321.f11.team2.johri.y 04:53, 5 October 2011 (UTC) I solved R*3.5, R*3.11 and R*3.17 and read the proof of R 3.1, R*3.4, R*3.6, R*3.7, R*3.12, R*3.13, R*3.14, R*3.16 and R*3.18 20:29, 5 October 2011 (UTC) Egm6321.f11.team2.wise 22:00, 5 October 2011 (UTC) I solved R*3.6, R*3.12,R*3.18; and read the proof of R*3.16 and R*3.17