User:Egm6321.f10.team5.abc/숙제3

Problem 1

 * 1) Derive the equation of motion.
 * 2) Verify $$y(x)$$ is parabola for the particular case of $$k = 0$$.
 * 3) First, consider $$k \ne 0$$, $${v_x}_0 = 0$$, then $$m \frac {d v_y} {d t} = - k (v_y)^n - mg$$. Second, find $$v_y(t)$$ and $$y(t)$$ for $$m = \mbox{constant}.$$

Solution 1

 * 1) Using the Newton's second law, $$\Sigma \vec F = m \vec a$$, we derive the equations of motion. Note that the only forces acting on the particles are gravity, $$m g$$, and air resistance, $$k v^n$$. First, we apply the Newton's second law in x-direction. $$\Sigma F_x = m a_x$$ $$- k v^n \cos \alpha = m \frac {d v_x} {d t}$$ Now, we apply the Newton's second law in y-direction. $$\Sigma F_y = m a_y$$ $$- k v^n \sin \alpha - m g = m \frac {d v_y} {d t}$$
 * 2) If we let $$k = 0$$, the equations of motion become $$m \frac {d v_x} {d t} = 0 \quad \mbox{or} \quad \frac {d v_x} {d t} = 0$$ $$m \frac {d v_y} {d t} = - mg \quad \mbox{or} \quad \frac {d v_y} {d t} = - g$$ We can express the above equations like following $$\frac {d^2 x} {d t^2} = 0$$ $$\frac {d^2 y} {d t^2} = - g$$ Take indefinite integral twice for each equation $$x = at + b$$ $$y = - g \frac {t^2} 2 + c t + d$$ Write $$t$$ in terms of $$x$$ $$t = \frac {x-b} a$$ Substitute this expression into the equation of $$y$$. $$y = - \frac g 2 (\frac {x-b} a)^2 + c (\frac {x-b} a) +d$$ Since the last equation is in the form of $$y = a x^2 + b x + c$$, $$y(x)$$ which is a polynomial of degree $$2$$ in $$x$$ is a parabola with a vertical axis.
 * 3) First, if $$v_{x0} = 0$$, $$v$$ moves in y-direction only meaning that $$v_x = 0$$. Then $$v^2 = (v_x)^2 + (v_y)^2$$ becomes $$v^2 = (v_y)^2$$. We have $$v = v_y$$ since $$v \geq 0$$ and $$v_y \geq 0$$. The angle becomes $$\alpha = 90^\circ$$. As a result, the equation   $$m \frac {d v_y} {d t} = - k v^n \sin \alpha - m g$$  transforms into $$m \frac {d v_y} {d t} = - k (v_y)^n - mg$$ Second, we arrange the equation to be expressed as   $$(k (v_y)^n + mg) + m \frac {d v_y} {d t} = 0$$  Arranging the last equation and dividing each term by $$m$$ to be  $$\frac {d v_y} {d t} = - \frac k m (v_y)^n - g$$  $$\frac {d v_y} {- \frac k m (v_y)^n - g} = d t$$  Taking the integral on each side gives $$- \frac {v_y} g {~}_2 F_1(1,\frac 1 n;1+\frac 1 n;-\frac k {g m} v_y^n) = t$$ Third, we can derive that  $$m = \frac {m_1 - m_0} {t_1} t + m_0 \quad \mbox{for} \quad t_0 \leq t \leq t_1$$  $$m = m_1 \quad \mbox{for} \quad t \geq t_1$$  Now express $$m \frac {d v_y} {d t} = - k (v_y)^n - mg$$ with changing mass.  $$(\frac {m_1 - m_0} {t_1} t + m_0) \frac {d v_y} {d t} = - k (v_y)^n - (\frac {m_1 - m_0} {t_1} t + m_0) g \quad \mbox{for} \quad t_0 \leq t \leq t_1$$  $$m_1 \frac {d v_y} {d t} = - k (v_y)^n - m_1 g \quad \mbox{for} \quad t_0 \leq t \leq t_1$$  Arrange the both equations to be in the form of $$M(v_y,t) + N(v_y,t) \frac {d v_y} {d t} = k$$  $$k (v_y)^n + g (\frac {m_1 - m_0} {t_1}) t + (\frac {m_1 - m_0} {t_1} t + m_0) \frac {d v_y} {d t} = - m_0 g \quad \mbox{for} \quad t_0 \leq t \leq t_1$$  $$k (v_y)^n + m_1 \frac {d v_y} {d t} = - m_1 g \quad \mbox{for} \quad t_0 \leq t \leq t_1$$  Because the both equations satisfy the first condition of exactness, $$m \frac {d v_y} {d t} = - k (v_y)^n - mg$$ can be made exact by integral factor method.

Problem 2

 * 1) Derive the following equations. $$m_1 l^2 \ddot \theta_1 = - k a^2 (\theta_1 - \theta_2) - m_1 g l \theta_1 + u_1 l$$ $$m_2 l^2 \ddot \theta_2 = - k a^2 (\theta_2 - \theta_1) - m_2 g l \theta_2 + u_2 l$$
 * 2) Write the above equations in the form of $$\dot x(t) = A(t) x(t) + B(t) u(t)$$. $$ x := \left[ {\begin{matrix}  \theta_1 & {\dot \theta_1} & \theta_2 & {\dot \theta_2} \\ \end{matrix} } \right]^T $$ Find A, B, and u.

Solution 2

 * 1) We use Newtons's second law for Rotation which is $$\Sigma \tau = I \alpha$$ for the derivation.  $$\Sigma \tau_1 = I_1 \alpha_1$$ We note that there are three torques applied on the particle 1; gravitational torque, controlling torque, and spring torque. Gravitational torque is $$\tau_g = \vec l \times \vec {F_g} = - l \sin \theta_1 m_1 g$$ Controlling torque is $$\tau_c = \vec l \times \vec {F_c} = l \cos \theta_1 u_1$$ Spring torque is $$\tau_s = \vec a \times \vec {F_s} = - a \cos \theta_1 k a (\sin \theta_1 - \sin \theta_2)$$ Moment of inertia for the point mass is $$I_1 = m_1 l^2$$ Angular acceleration for the point mass is $$\alpha_1 = \ddot \theta_1$$ We set up the equation of motion for the first particle. $$m_1 l^2 \ddot \theta_1\ = \tau_a + \tau_g + \tau_c = - a \cos \theta_1 k a (\sin \theta_1 - \sin \theta_2) - l \sin \theta_1 m_1 g + l \cos \theta_1 u_1$$ Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as $$m_1 l^2 \ddot \theta_1\ = - k a^2 (\theta_1 - \theta_2) - l \theta_1 m_1 g + l u_1$$ We use the same techniques to derive the equation of motion for second particle. Gravitational torque is $$\tau_g = \vec l \times \vec {F_g} = - l \sin \theta_2 m_2 g$$ Controlling torque is $$\tau_c = \vec l \times \vec {F_c} = l \cos \theta_2 u_2$$ Spring torque is $$\tau_s = \vec a \times \vec {F_s} = - a \cos \theta_2 k a (\sin \theta_2 - \sin \theta_1)$$ Moment of inertia for the point mass is $$I_2 = m_2 l^2$$ Angular acceleration for the point mass is $$\alpha_2 = \ddot \theta_2$$ We set up the equation of motion for the second particle. $$m_2 l^2 \ddot \theta_2\ = \tau_a + \tau_g + \tau_c = - a \cos \theta_2 k a (\sin \theta_2 - \sin \theta_1) - l \sin \theta_2 m_2 g + l \cos \theta_2 u_2$$ Apply the small-angle approximation which states that $$\sin x \approx x$$ and $$\cos x \approx 1$$ to simplify equation as $$m_2 l^2 \ddot \theta_2\ = - k a^2 (\theta_2 - \theta_1) - l \theta_2 m_2 g + l u_2$$
 * 2) We write $$\dot x(t) = A(t) x(t) + B(t) u(t)$$ in matrix form.  $$ \left[ {\begin{matrix}  \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \\ \end{matrix} } \right] = \left[ {\begin{matrix} A_{11} & A_{12} & A_{13} & A_{14} \\ A_{21} & A_{22} & A_{23} & A_{24} \\ A_{31} & A_{32} & A_{33} & A_{34} \\ A_{41} & A_{42} & A_{43} & A_{44} \\ \end{matrix} } \right] \left[ {\begin{matrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \\ \end{matrix} } \right] + \left[ {\begin{matrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ B_{31} & B_{32} \\ B_{41} & B_{42} \\ \end{matrix} } \right] \left[ {\begin{matrix} u_1 l \\ u_2 l \\ \end{matrix} } \right] $$ We divide the both side of $$m_1 l^2 \ddot \theta_1\ = - k a^2 (\theta_1 - \theta_2) - l \theta_1 m_1 g + l u_1$$ by $$m_1 l^2$$ to get $$\ddot \theta_1\ = - \frac {k a^2} {m_1 l^2} (\theta_1 - \theta_2) - \frac 1 l \theta_1 g + l u_1 = - \theta_1 (\frac {k a^2} {m_1 l^2} + \frac g l) + \theta_2 \frac {k a^2} {m_1 l^2} + l u_1$$ Similarly, we divide the both side of $$m_2 l^2 \ddot \theta_2\ = - k a^2 (\theta_2 - \theta_1) - l \theta_2 m_2 g + l u_2$$ by $$m_2 l^2$$ to get $$\ddot \theta_2\ = - \frac {k a^2} {m_2 l^2} (\theta_2 - \theta_1) - \frac 1 l \theta_2 g + l u_2 = \theta_1 \frac {k a^2} {m_2 l^2} - \theta_2 (\frac {k a^2} {m_2 l^2} + \frac g l) + l u_2$$ We can easily determine that the all the components on row 1 and row 3 of matrix A and matrix B are null, so we write that $$ \left[ {\begin{matrix}  \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 0 & 0 & 0 & 0 \\ A_{21} & A_{22} & A_{23} & A_{24} \\ 0 & 0 & 0 & 0 \\ A_{41} & A_{42} & A_{43} & A_{44} \\ \end{matrix} } \right] \left[ {\begin{matrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \\ \end{matrix} } \right] + \left[ {\begin{matrix} 0 & 0 \\ B_{21} & B_{22} \\ 0 & 0 \\ B_{41} & B_{42} \\ \end{matrix} } \right] \left[ {\begin{matrix} u_1 l \\ u_2 l \\ \end{matrix} } \right] $$ Now we plug in the expressions of $$\ddot \theta_1$$ and $$\ddot \theta_2$$ into the matrix. $$ \left[ {\begin{matrix}  \dot \theta_1 \\ \ddot \theta_1 \\ \dot \theta_2 \\ \ddot \theta_2 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 0 & 0 & 0 & 0 \\ - \frac {k a^2} {m_1 l^2} - \frac g l & 0 & \frac {k a^2} {m_1 l^2} & 0 \\ 0 & 0 & 0 & 0 \\ \frac {k a^2} {m_2 l^2} & 0 & - \frac {k a^2} {m_2 l^2} - \frac g l & 0 \\ \end{matrix} } \right] \left[ {\begin{matrix} \theta_1 \\ \dot \theta_1 \\ \theta_2 \\ \dot \theta_2 \\ \end{matrix} } \right] + \left[ {\begin{matrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{matrix} } \right] \left[ {\begin{matrix} u_1 l \\ u_2 l \\ \end{matrix} } \right] $$

Problem 3
Solve $$\dot x(t) = a(t) x(t) + b(t) u(t)$$ for the case where $$a(t)$$ and $$b(t)$$ are constant coefficients while u(t) is prescribed.

Solution 3
We state that $$a(t) = a$$ and $$b(t) = b$$ for this case. $$\dot x(t) = a x(t) + b u(t)$$ $$-a x(t) + \dot x(t) = b u(t)$$ $$h(t) = \exp [- \int^t \frac 1 1 (0 + a) ds] = \exp (-at) $$ So we have the equation in the form of $$hx' + h'x = hk$$. $$-\exp (-at) a x(t) + \exp (-at) \dot x(t) = \exp (-at) b u(t)$$ Note that we cannot use the equation of $$y(x) = \frac 1 {h(x)} \int^x h(s) d(s) ds$$ since the lower limit is bounded as $$t_0$$. Deriving an equation for the lower bound $$y(x) h(x) - y(x_0) h(x_0) = \int^x_{} h(s) d(s) ds$$ $$y(x) = \frac 1 {h(x)} \left[y(x_0) h(x_0) + \int^x_{} h(s) d(s) ds\right]$$ Apply the modified equation given above to our problem. $$x(t) = \frac 1 {\exp (-at)} \left[ x(t_0) \exp(-at_0) + \int^t_{t_0} \exp(-a\tau) b u(\tau) d\tau\right]$$ It is possible to simplify the equation further as $$x(t) = x(t_0) \exp[a(t-t_0)] + \int^t_{t_0} \exp[a(t-\tau)] b u(\tau) d\tau$$

Problem 4
Expand the Taylor series of $$\sum_{k=0}^\infty \frac {x^k} {k !}$$.

Solution 4
We derive the Taylor series for the exponential function $$f(x) = \exp x$$ at $$a=0$$. $$\sum_{k=0}^\infty \frac {f^{(n)} (a)} {n !} (x - a)^n = \exp (0) + \frac {\exp (0)} {1 !} (x - 0) + \frac {\exp (0)} {2 !} (x - 0)^2 + \frac {\exp (0)} {3 !} (x - 0)^3 + \cdot \cdot \cdot$$ $$= 1 + \frac x 1 + \frac {x^2} {2 !} + \frac {x^3} {3 !} + \cdot \cdot \cdot = \sum_{k=0}^\infty \frac {x^k} {k !} = \exp x$$

Problem 5
Transform generalized $$x(t) = [\exp \underline A (t - t_0)] x(t_0) + \int^t_{t_0} [\exp \underline A (t - \tau)] \underline B \underline u(\tau) d\tau$$ to SC-L1-ODE-VC.

Solution 5
If we transform the equation from SC-L1-ODE-CC to SC-L1-ODE-VC, the following two operations have to be applied. $$\underline A (t - t_0) \rightarrow \int_{t_0}^t \underline A (\tau) d\tau$$ $$\underline A (t - \tau) \rightarrow \int_\tau^t \underline A (s) ds$$ Now we express the transformed equation. $$\underline x(t) = [\exp \int_{t_0}^t \underline A (\tau) d\tau] \underline x(t_0) + \int_{t_0}^t [\exp \int_\tau^t \underline A (s) ds] \underline B \underline u(\tau) d\tau$$

Problem 6
Use $$\frac d {dt} \underline \phi (t,t_0) = \underline A \underline \phi (t,t_0)$$ and $$\underline \phi (t_0,t_0) = \underline 1$$ along with $$\dot {\underline x}(t) = \underline A \underline x(t) + \underline B \underline u(t)$$ to obtain $$\underline x(t) = [\exp \underline A (t - t_0)] \underline x (t_0) + \int_{t_0}^t [\exp \underline A (t - \tau)] \underline B \underline u (\tau) d\tau$$.

Solution 6
We can derive from $$\frac d {dt} \underline \phi (t,t_0) = \underline A \underline \phi (t,t_0)$$ that $$\underline A = \frac {\frac d {dt} \underline \phi (t,t_0)} {\underline \phi (t,t_0)}$$ $$\dot {\underline x}(t) = \frac {\frac d {dt} \underline \phi (t,t_0)} {\underline \phi (t,t_0)} \underline x(t) + \underline B \underline u(t)$$ $$\dot {\underline x}(t) {\underline \phi (t,t_0)} = {\frac d {dt} \underline \phi (t,t_0)} \underline x(t) + \underline B \underline u(t) {\underline \phi (t,t_0)}$$ $$\dot {\underline x}(t) {\underline \phi (t,t_0)} - {\frac d {dt} \underline \phi (t,t_0)} \underline x(t) = \underline B \underline u(t) {\underline \phi (t,t_0)}$$ $$[\underline \phi (t,t_0)]^2 [\frac {\underline x(t)} {\underline \phi (t,t_0)}]' = \underline B \underline u(t) {\underline \phi (t,t_0)}$$ $$[\underline \phi (t,t_0)]^2 \int_{t_0}^t [\frac {\underline x(t)} {\underline \phi (t,t_0)}]' dt = \int_{t_0}^t \underline B \underline u(\tau) {\underline \phi (\tau,t_0)} d\tau$$ $$\underline x(t) \underline \phi (t,t_0) - \underline x(t_0) [\underline \phi (t,t_0)]^2 = \int_{t_0}^t \underline B \underline u(\tau) {\underline \phi (\tau,t_0)} d\tau$$ $$\underline x(t) = \underline x(t_0) \underline \phi (t,t_0) + \frac 1 {\underline \phi (t,t_0)} \int_{t_0}^t \underline B \underline u(\tau) {\underline \phi (\tau,t_0)} d\tau$$ $$\underline x(t) = \underline x(t_0) \underline \phi (t,t_0) + \int_{t_0}^t \underline B \underline u(\tau) {\underline \phi (t,\tau)} d\tau$$

Problem 7
Put $$\dot \phi = \omega$$, $$\dot \omega = - \frac 1 \tau \omega + \frac Q \tau \delta$$, and $$\dot \delta = u$$ in the form of $$\dot {\underline x}(t) = \underline A \underline x(t) + \underline B \underline u(t)$$.

Solution 7
$$\dot {\underline \phi}(t) = \underline \omega(t)$$ $$\dot {\underline \omega}(t) = - \frac 1 \tau \underline \omega(t) + \frac Q \tau \underline \delta(t)$$ $$\dot {\underline \delta}(t) = \underline u(t)$$

Problem 8
Without assuming prior that $$h$$ is a constant, discuss the search for the solution of $$h_x + h_y P = 0$$.

Solution 8
not available yet

Problem 9
Show that $$15 p^4 \cos x^2 y'' + 6 x y^2 y' - 6 x p^5 \sin x^2 + 2 y^3$$ satisfies the second condition of exactness for N2-ODE.

Solution 9
We are already given that $$f = 15 p^4 \cos x^2$$ $$g = (6 x y^2) p - 6 x p^5 \sin x^2 + 2 y^3$$ We derive the follwing terms $$f_x = - 30 p^3 x \sin x^2$$ $$f_y = 0$$ $$f_{xx} = - 30 p^3 \sin x^2 - 60 x^2 p^4 \cos x^2$$ $$f_{xy} = 0$$ $$f_{yy} = 0$$ $$g_x = 6 y^2 p - 6 p^5 \sin x^2\ - 12 x^2 p^5 \cos x^2$$ $$g_{xp} = 6 y^2 - 30 p^4 \sin x^2 - 60 x^2 p^4 \cos x^2$$ $$g_y = 12 x y p + 6 y^2$$ $$g_{yp} = 12 x y$$ $$f_{xp} = - 120 p^3 x \sin x^2$$ $$f_{yp} = 0$$ $$g_p = -30 x p^4 \sin x^2$$ $$g_{pp} = - 120 x p^3 \sin x^2$$ Now we test the second condition of exactness. $$f_{xx} + 2 p f_{xy} + p^2 f_{yy} = - 30 p^3 \sin x^2 - 60 x^2 p^4 \cos x^2 + 0 + 0 = - 30 p^3 \sin x^2 - 60 x^2 p^4 \cos x^2$$ $$g_{xp} + p g_{yp} - g_y = 6 y^2 - 30 p^4 \sin x^2 - 60 x^2 p^4 \cos x^2 + 12 x y p - 12 x y p - 6 y^2 = - 30 p^4 \sin x^2 - 60 x^2 p^4 \cos x^2$$ $$f_{xp} + p f_{yp} + 2 f_y = - 120 p^3 x \sin x^2 + 0 + 0 = - 120 p^3 x \sin x^2$$ It is verified that $$f_{xp} + p f_{yp} + 2 f_y = g_{pp}$$ and $$f_{xx} + 2 p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y$$. Thus, this differential equation is exact.

Problem 10
Assume $$h_y y' = 2 y^3$$ and solve $$(6 x y^2) y' + 2 y^3 = h_y y' + h_x$$.

Solution 10
Substitute in the expression of $$h_y y'$$ into the equation. $$(6 x y^2) y' + 2 y^3 = 2 y^3 + h_x$$ So we have $$h_x = 6 x y^2 y'$$. $$h(x,y) = 3 x^2 y^2 y' + k_1$$ $$\phi(x,y,p) = (3 x^2 y^2 y' + k_1) + 3 p^5 \cos x^2 = k_2$$ Arrange the equation and obtain that $$\phi(x,y,p) = 3 x^2 y^2 y' + 3 p^5 \cos x^2 = k_2 - k_1$$