User:Egm6321.f11.team4/HW3

=''' Problem R3.1 - Show Eq. (1) p.13-2 is exact only if $$k_1=d_1$$ (constant) and Eq. (1) p.12-4 is a particular equation '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-3]]

From [[media:Pea1.f11.mtg11.djvu|Mtg 11-2]]

From [[media:Pea1.f11.mtg12.djvu|Mtg 12-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$. Eq. (2) p.11-2 is $$\displaystyle n(x):=(-1/N)(N_x-M_y)$$. Eq. (1) p.12-4 is $$\displaystyle M+Ny'=[a(x)y+k_x(x)]+\bar b(x)y'=0$$.

Find
$$\displaystyle a)$$Show that the N1-ODE (1) p.13-2 satisfies the condition (2) p.11-2 that an integrating factor $$\displaystyle h(x)$$ can be found to render it exact, only if $$\displaystyle k_1(y)=d_1$$ (constant). $$\displaystyle b)$$Show that (1) p.13-2 includes (1) p.12-4 as a particular case.

Solution
Solved independently. $$\displaystyle a)$$Eq. (1) p. 13-2 can be written as $$\displaystyle N(x,y)y'+M(x,y)=0$$ which yields the following: $$\displaystyle N(x,y)=\bar b(x,y)c(y)$$ $$\displaystyle M(x,y)=a(x)\bar c(x,y)$$. These two equations yield the following when differentiated: $$\displaystyle N_x=\bar b(x,y)c_x(y)+\bar b_x(x,y)c(y)$$ $$\displaystyle M_y=a(x)\bar c_y(x,y)+a_y(x)\bar c(x,y)$$. Now we plug these equations into Eq. (2) p. 11-2 to obtain $$\displaystyle n(x)=\frac{-1}{\bar b(x,y)c(y)}*[\bar b_x(x,y)c(y)-a(x)\bar c_y(x,y)]$$ This simplifies to $$\displaystyle n(x)=\frac{a(x)\bar c_y(x,y)}{\bar b(x,y)c(y)}-\frac{\bar b_x(x,y)c(y)}{\bar b(x,y)c(y)}$$. Since $$\displaystyle \bar b_x(x,y)=b(x,y)$$ & $$\displaystyle \bar c_y(x,y)=c(x,y)$$ cancel out, the formula reduces to $$\displaystyle n(x)=\frac{a(x)}{\bar b(x,y)}-1$$. Rearranging the equation produces $$\displaystyle \bar b(x,y)=\frac{a(x)}{n(x)+1}$$. If you differentiate the above equation by $$\displaystyle y $$, you obtain $$\displaystyle \bar b_y(x,y)=0$$ because everything is a function of $$\displaystyle x $$ or a constant. This shows that $$\displaystyle k_1(y)=d_1$$ (constant) and not a function of $$\displaystyle y$$. $$\displaystyle b)$$In order to show Eq. (1) p. 12-4 is a particular case of Eq. (1) p. 13-2, we must make two assumptions. Those assumptions include $$\displaystyle \bar b(x,y)=\bar b(x)$$ and $$\displaystyle c(y)=C$$(constant). Working first with $$\displaystyle M(x,y) $$, we have $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ (left side is the general case and the right side is the particluar case). With $$\displaystyle c(y)=C $$, Eq. (1) p. 13-3 reduces to $$\displaystyle \bar c(x,y)=Cy+k_2(x)$$. Plug the above equation for $$\displaystyle \bar c(x,y) $$ into the $$\displaystyle a(x)\bar c(x,y)=a(x)y+k_2(x)$$ which yields $$\displaystyle a(x)[Cy+k_2(x)]=a(x)y+k_2(x)$$. The left side simplies becase $$\displaystyle C $$ is absorbed into $$\displaystyle a(x) $$ and $$\displaystyle a(x) $$ is absorbed into $$\displaystyle k_2(x) $$. Working last with $$\displaystyle N(x,y) $$, we have $$\displaystyle \bar b(x,y)c(y)=\bar b(x)$$ (left side is the general case and the right side is the particluar case). Since $$\displaystyle c(y)=C $$, it can be absorbed into $$\displaystyle \bar b(x,y)$$. Also, $$\displaystyle \bar b(x,y)=\bar b(x)$$ because it is not a function of $$\displaystyle y $$. This ultimately yields the particluar case, $$\displaystyle \bar b(x)$$.

Author
Contributed by Allen

=''' Problem R*3.2 - Show Eq. (1) p.13-4 is exact or can be made exact by the IFM. Find the integrating factor h '''=

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = 5x^3+2 $$ $$\displaystyle b(x) = x^2 $$ $$\displaystyle c(x) = y^4 $$ $$\displaystyle \bar b(x) = \frac{1}{3}x^3+d_1 $$ $$\displaystyle \bar c(x) = \frac{1}{5}y^5+\sin(x)+d_2 $$ So, after combining the above equations into Eq. (1) p.13-4 we obtain Eq. (1) p.13-4 which is $$\displaystyle (\frac{1}{3}x^3+d_1)y^4y'+(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)=0 $$.

Find
Show that Eq. (1) p. 13-4 either is exact or can be made exact by the IFM. Find the integrating factor $$\displaystyle h $$.

Solution
Solved independently. In order for Eq. (1) p. 13-4 to be exact, it must satisfy both conditions of exactness. The first exactness condition requires the formula to be in the following form: $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (1) p. 13-4 already satisfies the above requirement so it meets this condition. The second exactness condition requires the formula to satisfy the following condition: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)=(5x^3+2)*(\frac{1}{5}y^5+\sin x + d_2)$$ and $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, to obtain $$\displaystyle M_y(x,y)=5*(x^3+\frac{2}{5})y^4$$ and $$\displaystyle N_x(x,y)=x^2y^4$$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=(\frac{1}{3}x^3+d_1)y^4 $$, $$\displaystyle N_x(x,y)=x^2y^4 $$, and $$\displaystyle M_y(x,y)=(5x^3+2)y^4 $$ we have $$\displaystyle h(x)=\exp \int^x -[\frac{5s^3+s^2+2}{\frac{1}{3}s^3+d_1}]ds $$.

Author
Contributed by Allen

= Problem R*3.3 - Find an N1-ODE of the form (1) p. 13-2 and the First Integral, Phi =

From [[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]

Given
Eq. (1) p.13-2 is $$\displaystyle \bar b(x,y)c(y)y'+a(x)\bar c(x,y)=0$$ or $$\displaystyle N(x,y)y'+M(x,y)=0$$. Eq. (2) p.13-2 is $$\displaystyle \bar b(x,y) = \int^x b(s)ds+k_1(y) $$. Eq. (1) p.13-3 is $$\displaystyle \bar c(x,y)= \int^y c(s)ds+k_2(x) $$. $$\displaystyle a(x) = \sin(x^3) $$. $$\displaystyle b(x) = \cos(x) $$. $$\displaystyle c(x) = \exp(2y) $$.

Find
$$\displaystyle 1) $$ Find an N1-ODE of the form (1) p. 13-2 that is either exact or can be made exact by the IFM. $$\displaystyle 2) $$ Find the first integral $$ \phi (x,y)=k $$.

Solution
Solved independently. $$\displaystyle 1) $$ First solve for $$\displaystyle \bar b(x,y) $$ and $$\displaystyle \bar c(x,y) $$. $$\displaystyle \bar b(x,y) = \int^x \cos(s) ds $$ $$\displaystyle \bar b(x,y) = \sin(x) $$ $$\displaystyle \bar c(x,y) = \int^x \exp(2y) ds $$ $$\displaystyle \bar c(x,y) = \frac{1}{2}*\exp(2y) $$ Fill Eq. (1) p. 13-2 with the known values and you will obtain $$\displaystyle \sin(x)*\exp(2y)*y'+\frac{1}{2}*\sin(x^3)*\exp(2y)=0 $$ This equation simplifies to $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$. So, the above equation satisfies the first condition of exactness, $$\displaystyle N(x,y)y'+M(x,y)=0 $$. However, it fails to satisfy the second condition of exactness, as shown below. The second exactness condition requires the formula to satisfy the following form: $$\displaystyle M_y(x,y)=N_x(x,y)$$. Differentiate the following equations, $$\displaystyle M(x,y)= \frac{1}{2}*\sin(x^3)*\exp(2y) $$ and $$\displaystyle N(x,y)= \sin(x)*\exp(2y) $$, to obtain $$\displaystyle M_y(x,y)= \sin(x^3)*\exp(2y) $$ and $$\displaystyle N_x(x,y)= \cos(x)*\exp(2y) $$. Clearly, $$\displaystyle M_y(x,y)\ne N_x(x,y)$$. Since the equation is not exact, we must make it exact with the IFM. Using $$\displaystyle h(x)=\exp[\int^x n(s)ds+k] $$ and $$\displaystyle n(x)=\frac{-1}{N}(N_x-M_y) $$ with $$\displaystyle N(x,y)=\sin(x) $$, $$\displaystyle N_x(x,y)=\cos(x) $$, and $$\displaystyle M_y(x,y)=0 $$ we have $$\displaystyle h(x)=\exp[\int^x [\frac{-1}{\sin(s)}(\cos(s)-0)]ds] $$. This simplies to $$\displaystyle h(x)= \frac{1}{\sin(s)} $$. Multiply $$\displaystyle h(x)= \frac{1}{\sin(s)} $$ and $$\displaystyle \sin(x)*y'+\frac{1}{2}*\sin(x^3)=0 $$ which yields $$\displaystyle y'+\frac{1}{2}\frac{\sin(x^3)}{\sin(x)}=0 $$. $$\displaystyle 2) $$ Find $$ \phi (x,y)=k $$. $$ \phi_x(x,y)=M(x,y) $$ $$ \phi_y(x,y)=N(x,y) $$ Using the above equations, you can integrate with respect to $$\displaystyle x $$ or $$\displaystyle y $$ which yields $$\displaystyle \phi(x,y)=\frac{-1}{2x^2}\cos(x^3)+k_3 $$ $$\displaystyle \phi(x,y)=\sin(x)y+k_4 $$. Equate the equations above and solve for $$\displaystyle k=k_3+k_4 $$ which is $$\displaystyle \phi(x,y)=\sin(x)y+\frac{1}{2x^2}\cos(x^3)=k $$

Author
Contributed by Allen

= Problem 3.4: Construction of N1-ODEs = From [[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]

Given
Given a class of N1-ODEs of the form,

$$\displaystyle \underbrace{\bar b(x,y)c(y)}_{\color{blue}{N(x,y)}}\,y'+\underbrace{a(x)\bar c(x,y)}_{\color{blue}{M(x,y)}}=0$$ (4.1)

where, $$ \bar b(x,y) $$ and  $$ \bar c(x,y) $$ are given by,

$$ \bar b(x,y) := \int^x b(s)ds + k_1(y) $$ (4.2)

$$ \bar c(x,y) := \int^y c(s)ds + k_2(x) $$ (4.3)

Above, $$ \displaystyle a(x), b(x) $$ and $$ \displaystyle c(y) $$ are arbitrary functions.

In R3.1 (problems given above, on this page), it is shown that this N1-ODE (4.1) satisfies (2) p.11-2 (see lecture notes, XXXXX), i.e.

$$ \frac{1}{N}(N_x-M_y) \, =: \color{blue}{n(x)} $$ (4.4)

which means that the integrating factor, i.e. h, is a function of x alone.

Find
Find a class of N1-ODEs, which is counterpart of above equation, (4.1), that satisfies condition (1) p.11-3. This condition is given by,

$$ \frac{1}{M}(N_x-M_y) \, =: \color{blue}{m(y)} $$ (4.5)

This condition means that an integrating factor, h, which is a function of y alone, can be found for this new class of N1-ODEs that renders them exact.

Solution
We note that in (4.1), $$ \displaystyle N_x = b(x)c(y) $$ and  $$ \displaystyle M_y = a(x)c(y) $$, so we get a common factor, $$  \displaystyle c(y) $$ in the numerator of (4.4) from both $$ \displaystyle N_x $$ and $$ \displaystyle M_y $$, which cancels out with c(y) from $$ \displaystyle N(x,y) $$ in the denominator. What remains after this canceling operation is a function of x alone if and only if $$ \displaystyle k_1(y) = d_1 $$ is a constant.

So, for construction of new class of N1-ODEs, we want $$ \displaystyle N_x $$ and  $$ \displaystyle M_y $$ to give a common factor that's a function of x alone, which will be cancelled by a factor in $$ \displaystyle M(x,y) $$, that will be a function of x alone. Therefore, let,

$$ M(x,y) := a(x) \bar c(x,y) $$ (4.6)

and,

$$ N(x,y) := \bar a(x,y) b(y) $$ (4.7)

Here,

$$ \bar a(x,y) := \int^x a(s)ds + k_1(y) $$ (4.8)

$$ \bar c(x,y) := \int^y c(s)ds + k_2(x) $$ (4.9)

and, now it can be verified that N1-ODE formed from $$ \displaystyle M(x,y) $$ and $$ \displaystyle N(x,y) $$ given by (4.6) and (4.7) respectively, with $$ \displaystyle \bar a(x,y) $$ and $$ \displaystyle \bar c(x,y)  $$ given by (4.8) and (4.9) respectively satisfies,

$$ \frac{h_y}{h} = \frac{1}{M}(N_x-M_y) \, = \frac{1}{\cancel{a(x)} \bar c(x,y)} (\cancel{a(x)}b(y) - \cancel{a(x)}c(y)) =: \color{blue}{m(y)} $$ (4.10)

only when $$ \displaystyle k_2(x) $$ in $$ \displaystyle \bar c(x,y) $$ is a constant.

Author
Contributed by Ankush

= Problem R*3.5 Derivation of Equations of Motion = From [[media:Pea1.f11.mtg14.djvu|Mtg 14-2]]

Given
A particle (e.g., a projectile) moving in the air, with initial velocities, $$ \displaystyle v_x(t=0) $$ and $$ \displaystyle v_y(t=0) $$ given by $$ \displaystyle v_{x0} $$ and $$ \displaystyle v_{y0} $$. The particle travels in air, with two forces acting on the particle, it's weight $$ \displaystyle mg $$, acting vertically downwards (in negative y direction), and $$ \displaystyle kv^n $$ the drag force provided by the fluid, i.e. air, which resists the motion of the particle. $$ \displaystyle m $$ and $$ \displaystyle g $$ are mass of the particle and gravitational acceleration, and $$ \displaystyle k $$ and $$ \displaystyle n $$ are constants.



Fig.5-1: A particle in motion in air, with weight and drag force provided by air to its motion. The initial velocities are also denoted as described above. Angle, $$ \displaystyle \alpha $$ is the angle that velocity of the particle, v makes with the x-axis at a particular time. This image was taken from

Find
1. Derive the equations of motion of the particle (shown above), i.e.

$$ m\frac{dv_x}{dt} = -kv^n\cos \alpha $$  (5.1)

$$ m\frac{dv_y}{dt} = -kv^n\sin \alpha - mg $$  (5.2)

$$ v^2 = (v_x)^2 + (v_y)^2 $$   (5.3)

$$ \tan \alpha\ = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{v_y}{v_x}  $$   (5.4)

2. For case k = 0, i.e. (no drag force), verify that y(x) is a parabola

3. For case k ≠ 0 and vx0 = 0 (i.e. particle is thrown vertically upwards, i.e. α = 90o), equation of motion (5.2) becomes,

$$ \displaystyle m \frac{dv_y}{dt} = - k(v_y)^n - mg $$  (5.5)

3.1 Find if Eq. (5.5) is exact? If not, find if it can be made exact using IFM [(Integration Factor Method)]?

Find $$ \displaystyle v_y(t) $$ and $$ \displaystyle y(t) $$ for constant $$ \displaystyle m $$.

3.2 For $$ \displaystyle m = m(t) $$, find $$ \displaystyle v_y(t) $$ and $$ \displaystyle y(t) $$. Mass as a function of time, t, is given in the following graph,



Fig. 5-2: Temporal variation of mass, m (in y-aixs) w.r.t. time, t (in x-axis). This graph is taken from

Solution
Solved independently.

Considering forces on the particle shown in Fig. 5-1, we can resolve the force, $$ \displaystyle kv^n $$ in horizontal and vertical directions (x and y respectively), therefore we will have,

$$ \displaystyle (kv^n)_x = -kv^n \cos \alpha $$  (5.6) $$ \displaystyle (kv^n)_y = -kv^n \sin \alpha $$ <p style="text-align:right"> (5.7)

Applying Newton's Second Law, which states that the net rate of change of momentum is governed by forces acting on a body, and resolving these forces along x and y directions, we get, (also assuming mass, m to be constant w.r.t. time or rate of change of mass w.r.t. time to be very low when this analysis is performed)

$$ m\frac{dv_x}{dt} = (kv^n)_x $$ <p style="text-align:right"> (5.8)

$$ m\frac{dv_y}{dt} = (kv^n)_y - mg $$ <p style="text-align:right"> (5.9)

Substituting (5.6) and (5.7) into (5.8) and (5.9) respectively, we will obtain the required equations of motion given by (5.1) and (5.2)

Also, since $$ \displaystyle v_x $$ and $$ \displaystyle  v_y $$ are components of velocity vector, v, (see Fig. 5-1) it is very easy to see that (5.3) is satisfied based on Pythagorus theorem. Also based on definition of tangent definition, and the triangle shown in Fig. 5-1, we can see that,

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

2. For particular case, k = 0, equations of motion simplify to,

$$ m\frac{dv_x}{dt} = 0 $$ <p style="text-align:right"> (5.10)

$$ \cancel m\frac{dv_y}{dt} = - \cancel mg $$ <p style="text-align:right"> (5.11)

Integrating above equations (5.10) and (5.11) we get,

$$ \displaystyle v_x = constant = k_2 $$

$$ \displaystyle v_y(t) = - gt + k_1 $$

Applying initial conditions (at t = 0) for velocities, $$ \displaystyle v_x $$ and $$ \displaystyle v_y $$, we get

$$ \displaystyle k_2 = v_{x0} $$

$$ \displaystyle k_1 = v_{y0} $$

Relating $$ \displaystyle v_x $$ and $$ \displaystyle v_y $$ to x and y respectively,

$$ \displaystyle v_x = \frac{dx}{dt} = v_{x0} $$

$$ \displaystyle v_y = \frac{dy}{dt} = -gt + v_{y0}$$

Integrating above two equations w.r.t. time, t, to obtain x and y, (Note initial location is origin, hence integration constants in both equations given below will be zero)

$$ \displaystyle x(t) = v_{x0} t $$ <p style="text-align:right"> (5.12)

$$ \displaystyle y(t) = -\frac{gt^2}{2} + v_{y0} t $$ <p style="text-align:right"> (5.13)

Substituting $$ \displaystyle t = \frac{x}{v_{x0}} $$ (from 5.12) into (5.13), we get the parabola form of y in terms of x,

$$ \displaystyle y(x) = -\frac{gx^2}{2(v_{x0})^2} + \frac{v_{y0}}{v_{x0}} x $$ <p style="text-align:right"> (5.14)

3. For k ≠ 0, and vx0 = 0, i.e. the particle is thrown vertically upwards, we have (following (5.9)),

$$ \displaystyle \underbrace {m}_{N(t,v_y)}\frac{dv_y}{dt} + \underbrace {k(v_y)^n + mg}_{M(t,v_y)} = 0 $$ <p style="text-align:right"> (5.15)

Above equation satisfies 1st exactness condition, as is seen above. Now we check if it satisfies 2nd exactness condition, to see if the equation is exact or not?

$$ \displaystyle \frac{\partial M}{\partial v_y} = nk (v_y)^{n-1} \ne \frac{\partial N}{\partial t} = 0 $$ <p style="text-align:right"> (5.16)

Therefore, we see from above that the 2nd exactness condition is not satisfied, therefore Eq. (5.15) is not in exact form.

To check if it can be put into exact form or not, we find that, it satisfies (1) p. 11-3 from the notes, i.e.

$$ \displaystyle \frac{1}{M}(N_t - M_{v_y}) = \frac{1}{(k{v_y}^n + mg)} (0 - nk{v_y}^{n-1}) = F(v_y) $$ <p style="text-align:right"> (5.17)

Therefore, equation (1), p.14-3 (in notes) or Eq. 5.15 can be made exact using IFM based on above equation (5.17), which states that the expression on LHS, is a function of $$ \displaystyle v_y $$ alone.

To find $$ \displaystyle v_y(t) $$ and y(t) for constant m case, we rearrange terms in (5.15) and integrate both sides,

$$ \displaystyle \int \frac{1}{(\frac{k}{m} {v_y}^n + g)} dv_y = - \int dt $$ <p style="text-align:right"> (5.18)

Above equation is solved using Wolfram Online Integrator, therefore, we have,

$$ \displaystyle v_y(t) = -\frac{gt}{_2F_1(1,\frac{1}{n}; 1 + \frac{1}{n}; - \frac{k}{mg} {v_y}^n)} + C_1 $$ <p style="text-align:right"> (5.19)

Note, $$ \displaystyle C_1 $$ above is integration constant, and is equal to vy0 due to the initial condition. Integration of above Eq. (5.19) w.r.t. t, can be performed to get y(t), but in its current form, with n, which can be any value, it is difficult to perform the integration manually. A numerical approach will suit for this purpose for some given n.

3.2 For m = m(t), linear function given in Fig. 5-2, we write the differential equation in following form,

$$ \frac{dv_y}{dt} = - \frac{k}{m} {v_y}^n - g $$ <p style="text-align:right"> (5.20)

Above equation is non homogenous ODE, so we can write the solution in terms of homogenous and particular solutions (and take n = 1), i.e. vy = (vy)H + (vy)P. For homogenous ODE, we solve

$$ \frac{dv_{yH}}{dt} = - \frac{k}{m} {v_{yH}}^n $$

Above equation can be solved for vyH for n =1 and we get

$$ v_{yH} = (t)^{-\frac{k t_1}{(m_1 - m_0)}} $$

For particular solution, we use,

$$ \frac{dv_{yP}}{dt} = - g $$

and we get

$$ v_{yP} = - g t $$

Now combining both homogenous and particular portion of vy, and applying initial condition at t = 0, to get the constant of integration to be vy0, we finally get vy to be (for n = 1)

$$ v_{y}(t) = (t)^{-\frac{k t_1}{(m_1 - m_0)}} - gt + v_{y0} $$

Integrating above equation w.r.t. t, we can get y(t), and applying initial condition that y(t = 0) = 0, we get

$$ y(t) = \frac{(t)^{-\frac{k t_1}{(m_1 - m_0)} + 1}}{-\frac{k t_1}{(m_1 - m_0)} + 1} - \frac{gt^2}{2} + v_{y0} t $$

Author
Contributed by Ankush

= Problem R*3.6 Coupled Pendulums = From [[media:Pea1.f11.mtg14.djvu|Mtg 14-6]]

Given
A system of Coupled pendulums shown below,



Fig. 6-1: 2 pendulums connected with a spring of spring constant, k, and point weights of both the pendulums being $$ \displaystyle m_1 $$ and $$ \displaystyle m_2 $$. The rods 1 and 2 (1 is the rod on left and 2 is the rod on right), make angles θ1 and θ2 with the vertical axis respectively. u1 and u2 are two control forces acting on the rods 1 and 2 respectively. Above figure was taken from.

Find
Derive the equations of motion for the two coupled pendulums,

$$ \displaystyle m_1 l^2 \ddot \theta_1 = -ka^2(\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l $$ <p style="text-align:right"> (6.1)

$$ \displaystyle m_2 l^2 \ddot \theta_2 = -ka^2(\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l $$ <p style="text-align:right"> (6.2)

Write the above equations in the matrix form (1) p. 14-4, i.e.

$$ \displaystyle \mathbf{\dot{x}}_{n \times 1}(t) = \mathbf{A}_{n \times n}(t) \, \mathbf{x}_{n \times 1}(t) + \mathbf{B}_{n \times m}(t) \, \mathbf{u}_{m \times 1}(t) $$ <p style="text-align:right"> (6.3)

where x and u are written as,

$$ \displaystyle \mathbf{x}:= \left \lfloor \theta_1 \ \dot \theta_1 \ \theta_2 \ \dot \theta_2 \right \rfloor^T \in \mathbb R^{4 \times 1} $$ <p style="text-align:right"> (6.4)

$$ \displaystyle \mathbf{u}:= \begin{Bmatrix} u_1| \\u_2| \end{Bmatrix} \in \mathbb R^{2 \times 1} $$ <p style="text-align:right"> (6.5)

Find $$ \displaystyle \mathbf{A} \in \mathbb R^{4 \times 4}, \ \mathbf{B} \in \mathbb R^{4 \times 2} $$

Solution
Solved independently.

Since we are considering linear system, we take θ1 and θ2 to be small angles. This assumption will be used in deriving the above equations of motion, (6.1) and (6.2)

Note that the term, $$ \displaystyle \ddot \theta $$ with subscripts 1 and 2 denotes angular acceleration of the rods, i.e.,

$$ \displaystyle \alpha_i = \ddot \theta_i $$ <p style="text-align:right"> (6.5)

Angular acceleration, αi, is related to the torque, Ti in following way,

$$ \displaystyle T_i = I_i \alpha_i = I_i \ddot \theta_i $$ <p style="text-align:right"> (6.6)

Considering net torque on the rod 1, by forces acting on it, we get three components of the Torque,

(a) $$ \displaystyle T_1a = - m_1 g l \sin \theta_1 = -m_1 g l \theta_1 $$ (Due to weight m1g)

(b) $$ \displaystyle T_1b =  k a (\theta_2 - \theta_1) a \cos \theta_1 = k a (\theta_2 - \theta_1) a $$  (Due to spring force, ka(θ2 - θ1))

(c) $$ \displaystyle T_1c = u_1 l \cos \theta_1 = u_1 l $$ (Due to control force u1)

Note, in above, due to θ1 and θ2 being small angles, we could assume sin(θ1) ~ θ1 and cos(θ1) ~ 1.

Combining all (a), (b) and (c) above, we will get,

$$ \displaystyle T_1 = I_1 \alpha_1 = m_1 l^2 \ddot \theta_1 = - k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l $$

Above equation is (6.1) which is derived using Newton's second Law of rotational motion.

Similarly we can also derive (6.2) by applying Newton's second law for rotational motion on the rod 2. Signs of -ka2(θ2 - θ1) is flipped since spring force will pull the rod 2 in -x-direction. Following the steps we will get,

$$ \displaystyle T_2 = I_2 \alpha_2 = m_2 l^2 \ddot \theta_2 = - k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l $$

For second part of the problem, we write $$ \displaystyle \dot {x} $$ as follows, (differentiating $$ \displaystyle x $$ in Eq. (6.4) w.r.t. t)

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

Equations (6.1) and (6.2) now denote rows 2nd and 4th of Eq. (6.3). Rows 1st and 3rd of Eq. (6.3) will have 1 in the 2nd and 4th columns respectively in matrix A. Other columns in these rows will be zero. For 2nd and 4th rows of matrix A, we denote them by unknown variables which we have to find in terms of known variables.

$$ \displaystyle \begin{bmatrix} \dot {\theta}_{1} \\ \ddot{ {\theta}_{1}} \\ \dot {\theta}_{2} \\ \ddot{{\theta}_{2}} \\ \end{bmatrix}

=

\underbrace{ \begin{bmatrix} 0 & 1 & 0 & 0 \\ {a}_{1} & {b}_{1} & {c}_{1} & {d}_{1} \\ 0 & 0 & 0 & 1 \\ {a}_{2} & {b}_{2} & {c}_{2} & {d}_{2} \\ \end{bmatrix} }_{\mathbf{A}(t) =4 \times 4}

\cdot

\begin{bmatrix} {\theta}_{1} \\ \dot{{\theta}_{1}} \\ {\theta}_{2} \\ \dot{{\theta}_{2}} \\ \end{bmatrix} +

\underbrace{ \begin{bmatrix} 0 & 0 \\ {e}_{1} & {f}_{1}  \\ 0 & 0 \\ {e}_{2} & {f}_{2} \\ \end{bmatrix} }_{\mathbf{B}(t) = 4 \times 2}

\cdot

\begin{bmatrix} {u}_{1}l \\ {u}_{2}l \\ \end{bmatrix}

$$<p style="text-align:right"> (6.7)

Above, terms ai - fi, i = 1, 2 are unknowns that can be found by comparing above equation to Eq. (6.1) and (6.2). E.g., a1 is the coefficient of θ1 in Eq. (6.1), which is $$ \displaystyle (-\frac{ka^2}{m_1 l^2} - \frac{g}{l}) $$. Similarly, b1 and d1 will be zero since there is no $$ \displaystyle \dot \theta_1 $$ and $$ \displaystyle \dot \theta_2 $$ in Eq. (6.1). c1 is the coefficient of θ2 in Eq. (6.2) and is thus equal to $$ \displaystyle \frac{ka^2}{m_1 l^2} $$. Note, in calculating all these coefficients above, we have considered dividing Eq. (6.1) and (6.2) by the coefficients of $$ \displaystyle \ddot \theta_1 $$ and $$ \displaystyle \ddot \theta_2 $$ respectively. Also e1 will be $$ \displaystyle \frac{1}{m_1 l^2} $$ and f1 should be zero.

Following the same analysis as highlighted in above paragraph, we can find variables a2 - f2 and thus final system in matrix form is written as,

$$ \displaystyle \begin{bmatrix} \dot {\theta}_{1} \\ \ddot{ {\theta}_{1}} \\ \dot {\theta}_{2} \\ \ddot{{\theta}_{2}} \\ \end{bmatrix}

=

\underbrace{ \begin{bmatrix} 0 & 1 & 0 & 0 \\ -\frac{k{a}^{2}-{m}_{1}lg}{{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_2gl}{{m}_{2}{l}^{2}} & 0 \\ \end{bmatrix} }_{\mathbf{A}(t) =4 \times 4}

\cdot

\begin{bmatrix} {\theta}_{1} \\ \dot{{\theta}_{1}} \\ {\theta}_{2} \\ \dot{{\theta}_{2}} \\ \end{bmatrix} +

\underbrace{ \begin{bmatrix} 0 & 0 \\ \frac{1}{{m}_{1}{l}^{2}} & 0  \\ 0 & 0 \\ 0 & \frac{1}{{m}_{2}{l}^{2}} \\ \end{bmatrix} }_{\mathbf{B}(t) = 4 \times 2}

\cdot

\begin{bmatrix} {u}_{1}l \\ {u}_{2}l \\ \end{bmatrix}

$$<p style="text-align:right"> (6.8)

Author
Contributed by Ankush

=''' Problem 3.7 - Use IFM to show that the solution of (2) is ... '''= From [[media:Pea1.f11.mtg15.djvu|Mtg 15-1]]

Find
Show that the solution of (7.1) is

Solution
Multiplying Integrating Factor h(t);

h is the function of t, so

Then, put (7.4) into (7.3)

We can rewrite <LHS> as

Then, integrate (7.5) with (7.6);

Author
Contributed by Chung

= Problem 3.8 - Generalize (2)p.15-2 to the case of linear time-variant system = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-3]]

Given
Linear Time Invariant Equation $$\displaystyle \dot x(t) = a(t)x(t) + b(t)u(t) $$

Solution
Integrating factor ;

Multiply integrating factor both side;

As a result,

Then,

Author
Contributed by Chung

= Problem 3.9 - Verify that (1) p.15-4 = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-4]]

Find
Verify 9.1 satisfies 9.2, 9.3

Author
Contributed by Chung

= Problem 3.10 - Free vibration of coupled pendulums = From [[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]

Given
The equations of motion of coupled pendulums: Parameters: $$\displaystyle \begin{align} \mbox{Pendulums: } &a=0.3, \mbox{ } l=1, \mbox{ } k = 0.2 \\ &m_1 g = 3, \mbox{ } m_2 g = 6 \end{align} $$ $$\displaystyle \mbox{No applied forces: } u_1=u_2=0 $$ $$\displaystyle \begin{align} \mbox{Ini}&\mbox{tial conditions: } \\ &\theta_1 (0)=0, \mbox{ } \dot{\theta_1}(0)=-2 \\ &\theta_2 (0)=0, \mbox{ } \dot{\theta_2}(0)=+1 \\ \end{align} $$

Find
1. Integrate the system in matrix form for $$\displaystyle t \in [0,7] $$ 2. Use (2) p.15-2 to find the solution 3. Plot the results

Integrate the system in matrix form
In order to represent the Eq. (10.1) in matrix form (1) From [[media:Pea1.f11.mtg14.djvu|p.14-4]], the following state variables are defined. Then, the Eq. (10.1) can be rewritten as following. Rearrange the terms in the Eq. (10.3), Finally, the follow matrix equation is obtained.

The following is a matlab code to integrate the system.
 * {| style="width:100%" border="0" align="left"

The following figure is the plot of the integration results.
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Use (2) p.15-2 to find the solution
Since no input is applied to the system, the second term in the Eq. (10.6) is zero. The matlab code shows the problem-solving using the Eq. (10.7).
 * {| style="width:100%" border="0" align="left"

The following figure is the plot of the results solved using the Eq. (10.7).
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * style="width:30%; padding:10px; border:1px solid #888888" align="center" |
 * }
 * }
 * }

Plot the results
Refer to Figure 10.1 and Figure 10.2 $$\displaystyle x_1 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_1(t) $$ $$\displaystyle x_3 $$ in Figure 10.1 and Figure 10.2 are $$\displaystyle \theta_2(t) $$

Author
Contributed by Shin

= Problem 3.11 SC-L1-ODE-CC with State Transition Matrix = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]

Solution
First, we rearrange the equation(11.3), we have Then, we substitute $$\displaystyle \mathbf{x}(t)$$ by $$\displaystyle \mathbf{\Phi}(t,t_0)$$, we can get Therefore, we can derive Euler integrating factor Thus we have So we get We integrate both sides of equation(11.9) in the interval $$\displaystyle t\geqslant{\tau}\geqslant t_{0}$$, then we have

Thus replace $$\mathbf{\Phi} (t,{{t}_{0}})$$ with $$\mathbf{x}(t)$$, and rearrange the equation, we can get

Author
Contributed by Kexin Ren

= Problem 3.12 Roll Control of Rocket = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]

Given
$$ \begin{align} &{\delta} = \text{Aileron angle/deflection}\\ &{\phi} = \text{Roll angle}\\ &{\omega} = \text{Roll angular velocity}\\ &Q = \text{Aileron effectiveness}\\ &{\tau} = \text{Roll time constant}\\ &u = \text{Command signal to Ailerons}\\ \end{align} $$

Find
Put equations(12.1~12.3) in the form of SC-L1-ODE-CC as in equation(12.4) with $$\displaystyle \mathbf{A, B}$$ being constant matrices.

Solution
Integrate both sides of equation(12.3), we have

So we can find equation(12.2) become

Then we integrate both sides of equation(12.6)

Similar to equation(12.1), we get

So equation(12.7) becomes

Therefore, we can get

where,

Author
Contributed by Kexin Ren

= Problem 3.13 - The 2nd exactness condition = From [[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]

Given
2nd exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg16.djvu|16-5]] And, the N2-ODE equation

Find
1. Derive the Eq. (13.2), i.e., the 2nd relation in the 2nd exactness condition. 2. Derive the Eq. (13.1), i.e., the 1st relation in the 2nd exactness condition. 3. Verify that the Eq. (13.3) satisfies the 2nd exactness condition.

Derive the 2nd relation in the 2nd exactness condition.
From the equation (3), (4) in the class note [[media:Pea1.f11.mtg16.djvu|16-4]], g(x,y,p) and f(x,y,p) are as below. Differentiating the Eq. (13.4) with respect to p, Differentiating the Eq. (13.6) with respect to p, Rewriting the Eq. (13.7) using the symmetric property of mixed 2nd partial derivatives of the function $$\displaystyle \phi(x,y,p) $$ as below, Substituting the Eq. (13.5) into the Eq. (13.8),

- The Eq. (13.9) shows the derivation of the 2nd relation in the 2nd exactness condition.

Derive the 1st relation in the 2nd exactness condition.
To solve this problem, 3 relations (Eq. 13.10) for the symmetry of mixed 2nd partial derivatives are exploited. From the first relation in the Eq. (13.10), Express $$\displaystyle \phi_y $$ the Eq. (13.12) in terms of $$\displaystyle f, g, p $$, From the second relation in the Eq. (13.10), Rewrite the Eq. (13.14) using the Eq. (13.4), (13.5), Express $$\displaystyle \phi_x $$ the Eq. (13.15) in terms of $$\displaystyle f, g, p $$, Take the partial derivative of the Eq. (13.13) with respect to x, Take the partial derivative of the Eq. (13.16) with respect to y, Use the third relation in the Eq. (13.10), Move the terms with f to the LHS and move the terms with g to the RHS, - The Eq. (13.20) shows the derivation of the 1st relation in the 2nd exactness condition.

Verify that the Eq. (13.3) satisfies the 2nd exactness condition.
From the 1st exactness condition for N2-ODEs in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], The 1st relation in the 2nd exactness condition is satisfied as below. Also, the 2nd relation in the 2nd exactness condition is satisfied as below.

Author
Contributed by Shin

= Problem 3.14 - Find h(x,y) = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|17-1]]. where $$\displaystyle p(x) := y^{\prime}(x) $$. The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-1]]

Find
Find $$\displaystyle h(x,y) $$ without assuming a priority that h=constant.

Solution
The Eq. (14.2) can be rewritten in the following form. Since y is the function of x, the Eq (14.3) is the partial derivative of the function h(x,y) with respect to x. So, the LHS term in the Eq. (14.3) is rewritten as follows.

Then, the following is true from the Eq. (14.3), (14.4).

The Eq. (14.5) implies that h(x,y) is the function of only y.

Substituting the Eq. (14.5) into the Eq. (14.2) would result the following. It is concluded that $$\displaystyle h_y = 0 $$ because if $$\displaystyle y^{\prime} = 0 $$ the solution would be trivial.

Author
Contributed by Shin

= Problem 3.15 - Verification of the 2nd exactness condition = From [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]

Given
The equation (1) in the class note [[media:Pea1.f11.mtg18.djvu|18-3]].

2nd exactness condition for N2-ODEs

Find
Verify whether the Eq. (15.1) satisfies the 2nd exactness condition (Eq. 15.2).

Solution
As proven in the class note [[media:Pea1.f11.mtg18.djvu|Mtg 18-3]], the Eq. (15.1) satisfies the 1st exactness condition for N2-ODEs.

The first condition (Eq. 15.2)
Computing each term in the Eq. (15.2) $$\displaystyle f_{xx} :$$ $$\displaystyle 2pf_{xy} :$$ $$\displaystyle f_{xy} = 0$$, since $$\displaystyle f_x$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle p^2 f_{yy} :$$ $$\displaystyle f_{y} = 0, f_{yy} = 0 $$, since $$\displaystyle f$$ is a function of $$\displaystyle x$$ and $$\displaystyle y^{\prime}$$. $$\displaystyle g_{xp} :$$ $$\displaystyle g_{yp} :$$ $$\displaystyle g_{y} :$$ Substituting the Eq. (15.5)-(15.8) into the LHS of the Eq. (15.2), Substituting the Eq. (15.9)-(15.12) into the RHS of the Eq. (15.2), - Since the Eq. (15.13) and the Eq. (15.14) are the same equations, the first condition is satisfied.

The second condition (Eq. 15.3)
Computing each term in the Eq. (15.3) $$\displaystyle f_{xp} :$$ $$\displaystyle f_{yp} :$$ $$\displaystyle f_{y} :$$ $$\displaystyle g_{pp} :$$ Substituting the Eq. (15.13)-(15.15) into the LHS of the Eq. (15.3), Substituting the Eq. (15.16) into the RHS of the Eq. (15.3), - Since the Eq. (15.17) and the Eq. (15.18) are the same equations, the second condition is satisfied.

Author
Contributed by Shin

=Problem 3.16 - Solve the Second Choice of Solution=

Given
Remove the cancelled terms:

The first choice for solution assumes:

thus:

so:

So we get:

Then:

and:

Find
Finish the other solution assuming:

Solution
Equation (16.9) is actually:

Integrating both sides yields:

Take derivative of equation (16.12) with respect to $$\,x$$ gives:

It can be seen from (16.2) and (16.9) that $$\,{{h}_{x}}$$ is also equal to:

So we get:

Substitute $$\displaystyle y' $$ for $$\displaystyle \frac{y}{3x}$$ in equation (16.9) yields:

That means:

So we have:

and:

which is the same as the first choice of solution.

Author
Contributed by YuChen

=Problem 3.17 - Prove the Simple Form of Reynolds Transport Theorem=

Given
The Reynolds transport theorem:

Let:

And the conservation of mass equation:

Find
Prove the simple form of the Reynolds transport theorem:

Solution
Replace $$\displaystyle \mathbf f(x,t) $$ with $$\displaystyle \rho (x,t) \mathbf u (x,t) $$ in equation (17.1) yields:

As equation (17.3) shows $$\displaystyle \frac{\partial \rho}{\partial t}+{\rm div}(\rho \mathbf u) = 0$$:

Author
Contributed by YuChen

=Problem 3.18 - Obtain 1-dimensional Reynolds Transport Theorem from n-dimensional Reynolds Transport Theorem=

Given
N-dimensional RTT:

Find
Obtain 1-dimensional RTT:

Solution
In 1-D case, $$\displaystyle \mathcal B_t $$ is actually $$\displaystyle s $$. So:

and

From equations (17.1) and (18.1) we know that $$\displaystyle \int_{\partial \mathcal B_t} \mathbf n (\mathbf f \mathbf u) \, d \partial \mathcal B_t = \int_{\mathcal B_t} {\rm div}(\mathbf f \mathbf u) \, d \mathcal B_t$$, so:

Since in 1-D case $$\displaystyle \mathbf u=u= \frac{ds}{dt}$$, then:

Add equations (18.3), (18.4) and (18.6) will get:

Author
Contributed by YuChen

=Contributing Members=

{| cellspacing=0 align=center cellpadding=5px width=60% style="background: white; border: 2px solid black;"
 * colspan="7" style="background: yellow;border-bottom:2px solid gray" | Team Contribution Table
 * Problem Number||Lecture(Mtg)||Assigned To||Solved By||Typed By||Signature(Author)
 * R3.1||[[media:Pea1.f11.mtg13.djvu|Mtg 13-3]]|| Allen || Allen || Allen || Allen 03:51, 4 October 2011 (UTC)]]
 * R*3.2||[[media:Pea1.f11.mtg13.djvu|Mtg 13-4]]|| Allen || Allen || Allen || Allen 04:50, 5 October 2011 (UTC)
 * R*3.3||[[media:Pea1.f11.mtg13.djvu|Mtg 13-4,5]]|| Allen || Allen || Allen || Allen 04:00, 5 October 2011 (UTC)]]
 * R3.4||[[media:Pea1.f11.mtg13.djvu|Mtg 13-5]]|| Ankush || Ankush || Ankush || Ankush 10:51, 4 October 2011 (UTC)
 * R*3.5||[[media:Pea1.f11.mtg14.djvu|Mtg 14-2]]|| Ankush || Ankush || Ankush || Ankush 20:51, 4 October 2011 (UTC)
 * R*3.6||[[media:Pea1.f11.mtg14.djvu|Mtg 14-6]]|| Ankush || Ankush || Ankush || Ankush 10:51, 5 October 2011 (UTC)
 * R*3.7||[[media:Pea1.f11.mtg15.djvu|Mtg 15-1]]|| Chung || Chung .|| Chung || Chung  3 Oct. 2011 at 23:36(UTC)
 * R*3.8||[[media:Pea1.f11.mtg15.djvu|Mtg 15-3]]|| Chung || Chung || Chung || Chung  2 Oct. 2011 at 21:55(UTC)
 * R*3.9||[[media:Pea1.f11.mtg15.djvu|Mtg 15-4]]|| Chung || Chung || Chung || Chung  2 Oct. 2011 at 21:26(UTC)
 * R3.10||[[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]|| Shin || Shin || Shin || Shin 5 Oct. 2011 at 17:32(UTC)
 * R*3.11||[[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 01:52(UTC)
 * R*3.12||[[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 02:45(UTC)
 * R*3.13||[[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]|| Shin || Shin || Shin || Shin 30 Sep. 2011 at 03:54(UTC)
 * R*3.14||[[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 18:00(UTC)
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R3.10||[[media:Pea1.f11.mtg15.djvu|Mtg 15-5]]|| Shin || Shin || Shin || Shin 5 Oct. 2011 at 17:32(UTC)
 * R*3.11||[[media:Pea1.f11.mtg16.djvu|Mtg 16-1]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 01:52(UTC)
 * R*3.12||[[media:Pea1.f11.mtg16.djvu|Mtg 16-3]]|| Ren || Ren || Ren || Ren 4 Oct. 2011 at 02:45(UTC)
 * R*3.13||[[media:Pea1.f11.mtg16.djvu|Mtg 16-6]]|| Shin || Shin || Shin || Shin 30 Sep. 2011 at 03:54(UTC)
 * R*3.14||[[media:Pea1.f11.mtg18.djvu|Mtg 18-1]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 18:00(UTC)
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R*3.15||[[media:Pea1.f11.mtg18.djvu|Mtg 18-3]]|| Shin || Shin || Shin || Shin 29 Sep. 2011 at 20:18(UTC)
 * R*3.16||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-2]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 20:00(UTC)
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * R*3.17||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-10]]|| YuChen || YuChen || YuChen || YuChen 04 Oct. 2011 at 22:15(UTC)
 * R*3.18||[[media:Pea1.f11.mtg19-20.djvu|Mtg 19-12]]|| YuChen || YuChen || YuChen || YuChen 05 Oct. 2011 at 01:25(UTC)
 * -|}
 * -|}