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Starting with the assumption that the ground surface is smooth and there is no friction. Assuming that all parts are moving in the positive direction to the right. Taking a snap shot when s_{2}>s_{1} so that the spring k_{2} is in compression. Spring k_{1} is in compression by also assuming that s_{1}>u at this instance.
Any other assumptions will also lead to the same set of equations as long as they are used in consistent way when finding the forces in the springs.
Starting with drawing a free body diagram of each body showing all forces acting on them based on the above assumption, and then using F=ma to find the equation of motion of each body m_{1},m_{2}. The free body diagrams is shown below
Now F=ma is applied to each body to obtain the equation of motions. For mass m_{2} \begin{align*} m_{2}s_{2}^{\prime \prime } & =-k_{2}\left ( s_{2}-s_{1}\right ) \\ s_{2}^{\prime \prime } & =-\frac{k_{2}}{m_{2}}\left ( s_{2}-s_{1}\right ) \end{align*}
And for mass m_{1} \begin{align*} m_{1}s_{1}^{\prime \prime } & =k_{2}\left ( s_{2}-s_{1}\right ) -k_{1}\left ( s_{1}-u\right ) \\ s_{1}^{\prime \prime } & =\frac{k_{2}}{m_{1}}\left ( s_{2}-s_{1}\right ) -\frac{k_{1}}{m_{1}}\left ( s_{1}-u\right ) \end{align*}
Now the state space equations are found. \begin{align*} \begin{pmatrix} x_{1}=s_{1}\\ x_{2}=s_{2}\\ x_{3}=s_{1}^{\prime }\\ x_{4}=s_{2}^{\prime }\end{pmatrix} & \overset{\frac{d}{dt}}{\Longrightarrow }\begin{pmatrix} x_{1}^{\prime }=s_{1}^{\prime }=x_{3}\\ x_{2}^{\prime }=s_{2}^{\prime }=x_{4}\\ x_{3}^{\prime }=s_{1}^{\prime \prime }=\frac{k_{2}}{m_{1}}\left ( s_{2}-s_{1}\right ) -\frac{k_{1}}{m_{1}}\left ( s_{1}-u\right ) =\frac{k_{2}}{m_{1}}\left ( x_{2}-x_{1}\right ) -\frac{k_{1}}{m_{1}}\left ( x_{1}-u\right ) \\ x_{4}^{\prime }=s_{2}^{\prime \prime }=-\frac{k_{2}}{m_{2}}\left ( s_{2}-s_{1}\right ) =-\frac{k_{2}}{m_{2}}\left ( x_{2}-x_{1}\right ) \end{pmatrix} \\ & =\begin{pmatrix} x_{3}\\ x_{4}\\ x_{1}\left ( -\frac{k_{2}}{m_{1}}-\frac{k_{1}}{m_{1}}\right ) +\frac{k_{2}}{m_{1}}x_{2}+\frac{k_{1}}{m_{1}}u\\ \frac{k_{2}}{m_{2}}x_{1}-\frac{k_{2}}{m_{2}}x_{2}\end{pmatrix} \end{align*}
Hence\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\end{pmatrix} & =\overset{A\left ( n\times n\right ) }{\overbrace{\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -\left ( \frac{k_{2}}{m_{1}}+\frac{k_{1}}{m_{1}}\right ) & \frac{k_{2}}{m_{1}} & 0 & 0\\ \frac{k_{2}}{m_{2}} & -\frac{k_{2}}{m_{2}} & 0 & 0 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\overset{B\left ( n\times m\right ) }{\overbrace{\begin{pmatrix} 0\\ 0\\ \frac{k_{1}}{m_{1}}\\ 0 \end{pmatrix} }}u\left ( t\right ) \\\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} & =\overset{C\left ( r\times n\right ) }{\overbrace{\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\overset{D\left ( r\times m\right ) }{\overbrace{\begin{pmatrix} 0\\ 0 \end{pmatrix} }}u\left ( t\right ) \end{align*}
The above is in the form of x^{\prime }=Ax+Bu and y=Cx+Du where r=2 is number of outputs, m=1 is the number of input and n=4 is the number of states.
Using k_{1}=k_{2}=0.5,m_{1}=1,m_{2}=2 and x\left ( 0\right ) =0 now the unit step response for y_{1},y_{2} is found using Simulink. With the above values the system becomes\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\end{pmatrix} & =\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0.5 & 0 & 0\\ 0.25 & -0.25 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 0.5\\ 0 \end{pmatrix} u\left ( t\right ) \\\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} & =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\begin{pmatrix} 0\\ 0 \end{pmatrix} u\left ( t\right ) \end{align*}
Using simulink, state space block was used to implement the above. A step input source was used. Demux was used to send the y_{1} and y_{2} responses to two different time scopes. Simulation was set for 40 seconds to obtain long enough view of the response. The following figure shows the step response and the model used.
Different values of k_{1},k_{2} are used to see the effect on the responses. When the spring stiffness increased, the frequency of oscillation increased. For example, this is a simulation using k_{1}=0.5,k_{2}=10,m_{1}=1,m_{2}=2 and x\left ( 0\right ) =0. With the above values the system becomes
\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\end{pmatrix} & =\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -10.5 & 10 & 0 & 0\\ 5 & -5 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 0.5\\ 0 \end{pmatrix} u\left ( t\right ) \\\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} & =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\end{pmatrix} +\begin{pmatrix} 0\\ 0 \end{pmatrix} u\left ( t\right ) \end{align*}
The step response for y_{1},y_{2} is
Given\begin{align*} x^{\prime } & =Ax+Bu\\ y & =Cx+Du \end{align*}
Applying the Laplace transform to the above and using zero initial conditions gives the following result. In the following, X\left ( s\right ) is the Laplace transform of x\left ( t\right ) , Y\left ( s\right ) is the Laplace transform of y\left ( t\right ) , and U\left ( s\right ) is the Laplace transform of u\left ( t\right ) \begin{align*} sX\left ( s\right ) & =AX\left ( s\right ) +BU\left ( s\right ) \\ Y\left ( s\right ) & =CX\left ( s\right ) +DU\left ( s\right ) \end{align*}
The first equation above gives X\left ( s\right ) =\left ( sI-A\right ) ^{-1}BU\left ( s\right ) . Substituting this value of X\left ( s\right ) in the second equation above results in Y\left ( s\right ) =C\left ( sI-A\right ) ^{-1}BU\left ( s\right ) Where DU\left ( s\right ) was not used since D is zero matrix in this example. Therefore the system transfer function matrix is G\left ( s\right ) =\frac{Y\left ( s\right ) }{U\left ( s\right ) }=C\left ( sI-A\right ) ^{-1}B Using numerical values of A,B,C from part(c) gives G\left ( s\right ) =\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix} \left ( s\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} -\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0.5 & 0 & 0\\ 0.25 & -0.25 & 0 & 0 \end{pmatrix} \right ) ^{-1}\begin{pmatrix} 0\\ 0\\ 0.5\\ 0 \end{pmatrix} Hence G\left ( s\right ) is a 2\times 1 vector. The first entry is the transfer function between u and y_{1} and the second entry is the transfer function between u and y_{2}. The above is evaluated using Matlab syms as follows
From above, the transfer functions are \begin{align*} & \fbox{$H_1\left ( s\right ) =\frac{Y_1\left ( s\right ) }{U\left ( s\right ) }=\frac{4s^2+1}{8s^4+10s^2+1}$}\\ & \fbox{$H_2\left ( s\right ) =\frac{Y_2\left ( s\right ) }{U\left ( s\right ) }=\frac{1}{8s^4+10s^2+1}$} \end{align*}
Using H_{1}\left ( s\right ) above Y_{1}\left ( s\right ) \left ( 8s^{4}+10s^{2}+1\right ) =\left ( 4s^{2}+1\right ) U\left ( s\right ) And taking inverse Laplace transform gives (each s adds one derivative in time) \fbox{$8\frac{d^4y_1}{dt^4}+10\frac{d^2y_1}{dt^2}+y_1\left ( t\right ) =4\frac{d^2u}{dt^2}+u\left ( t\right ) $}
The kinetic energy is given by K=\frac{m}{2}\left ( \dot{r}^{2}+\left ( r\dot{\phi }\right ) ^{2}+\left ( r\dot{\theta }\cos \phi \right ) ^{2}\right ) And the potential energy is P=-\frac{km}{r} Where k is constant and m is mass of satellite. The Lagrangian is\begin{align*} L & =K-P\\ & =\frac{m}{2}\left ( \dot{r}^{2}+\left ( r\dot{\phi }\right ) ^{2}+\left ( r\dot{\theta }\cos \phi \right ) ^{2}\right ) +\frac{km}{r} \end{align*}
The equations of motions of the mass m~ in each degree of freedom r,\theta ,\phi are given by\begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r} & =u_{r}\left ( t\right ) \tag{1}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =u_{\theta }\left ( t\right ) \tag{2}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\phi }}-\frac{\partial L}{\partial \phi } & =u_{\phi }\left ( t\right ) \tag{3} \end{align}
Starting with (1) gives find \frac{\partial L}{\partial r}=mr\dot{\phi }^{2}+mr\left ( \dot{\theta }\cos \phi \right ) ^{2}-\frac{km}{r^{2}} and \frac{\partial L}{\partial \dot{r}}=m\dot{r}, hence (1) becomes m\ddot{r}-mr\left ( \dot{\phi }^{2}+\left ( \dot{\theta }\cos \phi \right ) ^{2}\right ) +\frac{km}{r^{2}}=u_{r}\left ( t\right )
Hence
\fbox{$\ddot{r}=r\left ( \dot{\phi }^2+\left ( \dot{\theta }\cos \phi \right ) ^2\right ) -\frac{k}{r^2}+\frac{1}{m}u_r\left ( t\right ) $}
Similarly for (2), \frac{\partial L}{\partial \theta }=0 and \frac{\partial L}{\partial \dot{\theta }}=m\dot{\theta }r^{2}\cos ^{2}\phi , and \begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }} & =m\left ( \ddot{\theta }r^{2}\cos ^{2}\phi +2\dot{\theta }r\dot{r}\cos ^{2}\phi -2\dot{\theta }\dot{\phi }r^{2}\cos \phi \sin \phi \right ) \\ & =mr\cos \phi \left ( \ddot{\theta }r\cos \phi +2\dot{\theta }\dot{r}\cos \phi -2\dot{\theta }\dot{\phi }r\sin \phi \right ) \end{align*}
And (2) becomes\begin{align*} mr\cos \phi \left ( \ddot{\theta }r\cos \phi +2\dot{\theta }\dot{r}\cos \phi -2\dot{\theta }\dot{\phi }r\sin \phi \right ) & =u_{\theta }\left ( t\right ) \\ \ddot{\theta }r\cos \phi +2\dot{\theta }\left ( \dot{r}\cos \phi -\dot{\phi }r\sin \phi \right ) & =\frac{1}{mr\cos \phi }u_{\theta }\left ( t\right ) \\ \ddot{\theta }r\cos \phi & =-2\dot{\theta }\left ( \dot{r}\cos \phi -\dot{\phi }r\sin \phi \right ) +\frac{1}{mr\cos \phi }u_{\theta }\left ( t\right ) \end{align*}
Hence
\fbox{$\ddot{\theta }=-\frac{2\dot{\theta }\dot{r}}{r}+2\dot{\theta }\dot{\phi }\tan \phi +\frac{1}{mr^2\cos ^2\phi }u_\theta \left ( t\right ) $}
Similarly for (3), \frac{\partial L}{\partial \phi }=-mr^{2}\dot{\theta }^{2}\cos \phi \sin \phi and \frac{\partial L}{\partial \dot{\phi }}=mr^{2}\dot{\phi }, and \frac{d}{dt}\frac{\partial L}{\partial \dot{\phi }}=m\left ( 2r\dot{r}\dot{\phi }+r^{2}\ddot{\phi }\right ) hence (3) becomes\begin{align*} mr\left ( 2\dot{r}\dot{\phi }+r\left ( \ddot{\phi }+\dot{\theta }^{2}\cos \phi \sin \phi \right ) \right ) & =u_{\phi }\left ( t\right ) \\ 2\dot{r}\dot{\phi }+r\ddot{\phi }+r\dot{\theta }^{2}\cos \phi \sin \phi & =\frac{1}{mr}u_{\phi }\left ( t\right ) \end{align*}
Hence
\fbox{$\ddot{\phi }=-\frac{2\dot{r}\dot{\phi }}{r}-\dot{\theta }^2\cos \phi \sin \phi +\frac{1}{mr^2}u_\phi \left ( t\right ) $}
The state space becomes \begin{align*} \begin{pmatrix} x_{1}=r\\ x_{2}=\dot{r}\\ x_{3}=\theta \\ x_{4}=\dot{\theta }\\ x_{5}=\phi \\ x_{6}=\dot{\phi }\end{pmatrix} & \overset{\frac{d}{dt}}{\Longrightarrow }\begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\\ x_{5}^{\prime }\\ x_{6}^{\prime }\end{pmatrix} =\begin{pmatrix} \dot{r}\\ \ddot{r}\\ \dot{\theta }\\ \ddot{\theta }\\ \dot{\phi }\\ \ddot{\phi }\end{pmatrix} =\begin{pmatrix} \dot{r}\\ r\dot{\phi }^{2}+r\dot{\theta }^{2}\cos ^{2}\phi -\frac{k}{r^{2}}+\frac{1}{m}u_{r}\left ( t\right ) \\ \dot{\theta }\\ -\frac{2}{r}\dot{\theta }\dot{r}+2\dot{\theta }\dot{\phi }\tan \phi +\frac{1}{mr^{2}\cos ^{2}\phi }u_{\theta }\left ( t\right ) \\ \dot{\phi }\\ -\frac{2}{r}\dot{r}\dot{\phi }-\dot{\theta }^{2}\cos \phi \sin \phi +\frac{1}{mr^{2}}u_{\phi }\left ( t\right ) \end{pmatrix} \\ & =\begin{pmatrix} x_{2}\\ x_{1}\left ( x_{6}^{2}+\left ( x_{4}\cos x_{5}\right ) ^{2}\right ) -\frac{k}{x_{1}^{2}}+\frac{1}{m}u_{r}\left ( t\right ) \\ x_{4}\\ -\frac{2}{x_{1}}x_{4}x_{2}+2x_{4}x_{6}\tan x_{5}+\frac{1}{mx_{1}^{2}\cos ^{2}x_{5}}u_{\theta }\left ( t\right ) \\ x_{6}\\ -\frac{2}{x_{1}}x_{2}x_{6}-x_{4}^{2}\cos x_{5}\sin x_{5}+\frac{1}{mx_{1}^{2}}u_{\phi }\left ( t\right ) \end{pmatrix} \\ & =\begin{pmatrix} f_{1}\left ( x,u\right ) \\ f_{2}\left ( x,u\right ) \\ f_{3}\left ( x,u\right ) \\ f_{4}\left ( x,u\right ) \\ f_{5}\left ( x,u\right ) \\ f_{6}\left ( x,u\right ) \end{pmatrix} \end{align*}
The output equation is now found. y_{1}=r=x_{1},y_{2}=\theta =x_{3},y_{3}=\phi =x_{5}, hence\begin{pmatrix} y_{1}\\ y_{2}\\ y_{3}\end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\\ x_{6}\end{pmatrix}
Applying the values given to the state vector x results in x=\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5}\\ x_{6}\end{pmatrix} =\begin{pmatrix} r\\ \dot{r}\\ \theta \\ \dot{\theta }\\ \phi \\ \dot{\phi }\end{pmatrix} =\begin{pmatrix} r_{0}\\ 0\\ \omega t\\ \omega \\ 0\\ 0 \end{pmatrix} And \bar{u}=\begin{pmatrix} u_{r}\\ u_{\theta }\\ u_{\phi }\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} It is seen that the state vector x now has zero in all the components that can change the orbit from being in the equatorial orbit. Since the input u=0 then this state will not change. The satellite will remain in this orbit.
To obtain x^{\prime } at \left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) then f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) is evaluated using Taylor expansion and higher order terms are ignored. This results in the A,B matrices as follows f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) =\overset{0\text{ since equilibrium}}{\overbrace{f\left ( \bar{x},\bar{u}\right ) }}+\left . \frac{\partial f}{\partial x}\Delta x\right \vert _{x=\bar{x}}+\left . \frac{\partial f}{\partial u}\Delta u\right \vert _{u=\bar{u}}+H.O.T In matrix form, the above is f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) =\begin{pmatrix} \frac{\partial f_{1}}{\partial x_{1}}\Delta x_{1} & \frac{\partial f_{1}}{\partial x_{2}}\Delta x_{2} & \cdots & \cdots & \cdots & \frac{\partial f_{1}}{\partial x_{6}}\Delta x_{6}\\ \frac{\partial f_{2}}{\partial x_{1}}\Delta x_{1} & \frac{\partial f_{2}}{\partial x_{2}}\Delta x_{2} & \cdots & \cdots & \cdots & \frac{\partial f_{2}}{\partial x_{6}}\Delta x_{6}\\ \vdots & & & & & \vdots \\ \vdots & & & \ddots & & \vdots \\ \vdots & & & & & \vdots \\ \frac{\partial f_{6}}{\partial x_{1}}\Delta x_{1} & \frac{\partial f_{6}}{\partial x_{2}}\Delta x_{2} & \cdots & \cdots & \cdots & \frac{\partial f_{6}}{\partial x_{6}}\Delta x_{6}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }+\begin{pmatrix} \frac{\partial f_{1}}{\partial u_{1}}\Delta u_{1} & \frac{\partial f_{1}}{\partial u_{2}}\Delta u_{2} & \frac{\partial f_{1}}{\partial u_{3}}\Delta u_{3}\\ \frac{\partial f_{2}}{\partial u_{1}}\Delta u_{1} & \frac{\partial f_{2}}{\partial u_{2}}\Delta u_{2} & \frac{\partial f_{2}}{\partial u_{3}}\Delta u_{3}\\ \vdots & & \vdots \\ \vdots & \ddots & \vdots \\ \vdots & & \vdots \\ \frac{\partial f_{6}}{\partial u_{1}}\Delta u_{1} & \frac{\partial f_{6}}{\partial u_{2}}\Delta u_{2} & \frac{\partial f_{6}}{\partial u_{3}}\Delta u_{3}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) } Therefore f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) =\begin{pmatrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} & \cdots & \cdots & \cdots & \frac{\partial f_{1}}{\partial x_{6}}\\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{1}} & \cdots & \cdots & \cdots & \frac{\partial f_{2}}{\partial x_{6}}\\ \vdots & & & & & \vdots \\ \vdots & & & \ddots & & \vdots \\ \vdots & & & & & \vdots \\ \frac{\partial f_{6}}{\partial x_{1}} & \frac{\partial f_{6}}{\partial x_{2}} & \cdots & \cdots & \cdots & \frac{\partial f_{6}}{\partial x_{6}}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\begin{pmatrix} \Delta x_{1}\\ \Delta x_{2}\\ \Delta x_{3}\\ \Delta x_{4}\\ \Delta x_{5}\\ \Delta x_{6}\end{pmatrix} +\begin{pmatrix} \frac{\partial f_{1}}{\partial u_{1}} & \frac{\partial f_{1}}{\partial u_{2}} & \frac{\partial f_{1}}{\partial u_{3}}\\ \frac{\partial f_{2}}{\partial u_{1}} & \frac{\partial f_{2}}{\partial u_{2}} & \frac{\partial f_{2}}{\partial u_{3}}\\ \vdots & & \vdots \\ \vdots & \ddots & \vdots \\ \vdots & & \vdots \\ \frac{\partial f_{6}}{\partial u_{1}} & \frac{\partial f_{6}}{\partial u_{2}} & \frac{\partial f_{6}}{\partial u_{3}}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\begin{pmatrix} \Delta u_{1}\\ \Delta u_{2}\\ \Delta u_{3}\end{pmatrix} Each component in the above is now evaluated and A,B are evaluated at \bar{x}=\begin{pmatrix} r_{0} & 0 & \omega t & \omega & 0 & 0 \end{pmatrix} ^{T},\bar{u}=\begin{pmatrix} 0 & 0 & 0 \end{pmatrix} in order to obtain A,B. Since f_{1}\left ( x,u\right ) =x_{2}, then \frac{\partial f_{1}}{\partial x_{2}}=1 and all other values are zero. Since f_{2}\left ( x,u\right ) =x_{1}\left ( x_{6}^{2}+\left ( x_{4}\cos x_{5}\right ) ^{2}\right ) -\frac{k}{x_{1}^{2}}+\frac{1}{m}u_{r}\left ( t\right ) then \begin{align*} \frac{\partial f_{2}}{\partial x_{1}} & =\left ( x_{6}^{2}+\left ( x_{4}\cos x_{5}\right ) ^{2}\right ) +2\frac{k}{x_{1}^{3}}\\ \frac{\partial f_{2}}{\partial x_{2}} & =0\\ \frac{\partial f_{2}}{\partial x_{3}} & =0\\ \frac{\partial f_{2}}{\partial x_{4}} & =2x_{1}x_{4}\cos ^{2}x_{5}\\ \frac{\partial f_{2}}{\partial x_{5}} & =-2x_{1}x_{4}^{2}\cos x_{5}\sin x_{5}\\ \frac{\partial f_{2}}{\partial x_{6}} & =0 \end{align*}
Since f_{3}\left ( x,u\right ) =x_{4} then \frac{\partial f_{3}}{\partial x_{4}}=1 and all other components are zero. And since f_{4}\left ( x,u\right ) =-\frac{2}{x_{1}}x_{4}x_{2}+2x_{4}x_{6}\frac{\sin x_{5}}{\cos x_{5}}+\frac{1}{mx_{1}^{2}\cos ^{2}x_{5}}u_{\theta }\left ( t\right ) then\begin{align*} \frac{\partial f_{4}}{\partial x_{1}} & =0\\ \frac{\partial f_{4}}{\partial x_{2}} & =-\frac{2}{x_{1}}x_{4}\\ \frac{\partial f_{4}}{\partial x_{3}} & =0\\ \frac{\partial f_{4}}{\partial x_{4}} & =-\frac{2}{x_{1}}x_{2}+2x_{6}\frac{\sin x_{5}}{\cos x_{5}}\\ \frac{\partial f_{4}}{\partial x_{5}} & =2x_{4}x_{6}\sec ^{2}\left ( x_{5}\right ) +\frac{2u_{\theta }\sec ^{2}x_{5}\tan x_{5}}{mx_{1}^{2}}\\ \frac{\partial f_{2}}{\partial x_{6}} & =2x_{4}\tan x_{5} \end{align*}
Since f_{5}\left ( x,u\right ) =x_{6} then \frac{\partial f_{5}}{\partial x_{6}}=1 and all other components are zero. Finally, since f_{6}\left ( x,u\right ) =-\frac{2}{x_{1}}x_{2}x_{6}-x_{4}^{2}\cos x_{5}\sin x_{5}+\frac{1}{mx_{1}^{2}}u_{\phi }\left ( t\right ) then\begin{align*} \frac{\partial f_{6}}{\partial x_{1}} & =-\frac{2}{mx_{1}^{3}}u_{\phi }\\ \frac{\partial f_{6}}{\partial x_{2}} & =-\frac{2}{x_{1}}x_{2}\\ \frac{\partial f_{6}}{\partial x_{3}} & =0\\ \frac{\partial f_{6}}{\partial x_{4}} & =-2x_{4}\cos x_{5}\sin x_{5}\\ \frac{\partial f_{6}}{\partial x_{5}} & =-x_{4}^{2}\cos ^{2}x_{5}+x_{4}^{2}\sin ^{2}x_{5}\\ \frac{\partial f_{6}}{\partial x_{6}} & =-\frac{2}{x_{1}}x_{2} \end{align*}
Therefore, the linearized A matrix is A=\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ \left ( x_{6}^{2}+\left ( x_{4}\cos x_{5}\right ) ^{2}\right ) +2\frac{k}{x_{1}^{3}} & 0 & 0 & 2x_{1}x_{4}\cos ^{2}x_{5} & -2x_{1}x_{4}^{2}\cos x_{5}\sin x_{5} & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & -\frac{2}{x_{1}}x_{4} & 0 & -\frac{2}{x_{1}}x_{2}+2x_{6}\frac{\sin x_{5}}{\cos x_{5}} & 2x_{4}x_{6}\sec ^{2}\left ( x_{5}\right ) +\frac{2u_{\theta }\sec ^{2}x_{5}\tan x_{5}}{mx_{1}^{2}} & 2x_{4}\tan x_{5}\\ 0 & 0 & 0 & 0 & 0 & 1\\ -\frac{2}{mx_{1}^{3}}u_{\phi } & -\frac{2}{x_{1}}x_{2} & 0 & -2x_{4}\cos x_{5}\sin x_{5} & -x_{4}^{2}\cos ^{2}x_{5}+x_{4}^{2}\sin ^{2}x_{5} & -\frac{2}{x_{1}}x_{2}\end{pmatrix} The above is evaluated at \bar{x}=\begin{pmatrix} r_{0} & 0 & \omega t & \omega & 0 & 0 \end{pmatrix} ^{T},\bar{u}=\begin{pmatrix} 0 & 0 & 0 \end{pmatrix} which results in\begin{align*} A & =\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ \left ( 0+\left ( \omega \cos 0\right ) ^{2}\right ) +2\frac{k}{r_{0}^{3}} & 0 & 0 & 2r_{0}\omega \cos ^{2}0 & -2r_{0}\omega ^{2}\cos 0\sin 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & -\frac{2}{r_{0}}\omega & 0 & -\frac{2}{r_{0}}\left ( 0\right ) +2\left ( 0\right ) \frac{\sin 0}{\cos 0} & 2\omega \left ( 0\right ) \sec ^{2}\left ( 0\right ) +\frac{2u_{\theta }\sec ^{2}\left ( 0\right ) \tan \left ( 0\right ) }{mr_{0}^{2}} & 2\omega \tan \left ( 0\right ) \\ 0 & 0 & 0 & 0 & 0 & 1\\ -\frac{2}{mr_{0}^{3}}\left ( 0\right ) & -\frac{2}{r_{0}}\left ( 0\right ) & 0 & -2\omega \cos 0\sin 0 & -\omega ^{2}\cos ^{2}0+\omega ^{2}\sin ^{2}0 & -\frac{2}{r_{0}}\left ( 0\right ) \end{pmatrix} \\ & =\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ \omega ^{2}+2\frac{k}{r_{0}^{3}} & 0 & 0 & 2r_{0}\omega & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & -\frac{2}{r_{0}}\omega & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & -\omega ^{2} & 0 \end{pmatrix} \end{align*}
Using r_{0}=1,m=1 and k=r_{0}^{3}\omega ^{2} then the above becomes \fbox{$A=\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0\\ 3\omega ^{2} & 0 & 0 & 2\omega & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & -2\omega & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & -\omega ^{2} & 0 \end{pmatrix} $} The B matrix is now found. Since f_{1}\left ( x,u\right ) =x_{2}, then \frac{\partial f_{1}}{\partial u_{i}}=0 for i=1\cdots 3. And since f_{2}\left ( x,u\right ) =x_{1}\left ( x_{6}^{2}+\left ( x_{4}\cos x_{5}\right ) ^{2}\right ) -\frac{k}{x_{1}^{2}}+\frac{1}{m}u_{r}\left ( t\right ) then \frac{\partial f_{2}}{\partial u_{r}}=\frac{1}{m} and the other two components are zero. Since f_{3}\left ( x,u\right ) =x_{4} then \frac{\partial f_{3}}{\partial u_{i}}=0 for i=1\cdots 3. And since f_{4}\left ( x,u\right ) =-\frac{2}{x_{1}}x_{4}x_{2}+2x_{4}x_{6}\frac{\sin x_{5}}{\cos x_{5}}+\frac{1}{mx_{1}^{2}\cos ^{2}x_{5}}u_{\theta }\left ( t\right ) then \frac{\partial f_{4}}{\partial u_{\theta }}=\frac{1}{mx_{1}^{2}\cos ^{2}x_{5}}and the other two components are zero. Since f_{5}\left ( x,u\right ) =x_{6} then \frac{\partial f_{5}}{\partial u_{i}}=0 for i=1\cdots 3. Finally, since f_{6}\left ( x,u\right ) =-\frac{2}{x_{1}}x_{2}x_{6}-x_{4}^{2}\cos x_{5}\sin x_{5}+\frac{1}{mx_{1}^{2}}u_{\phi }\left ( t\right ) then \frac{\partial f_{6}}{\partial u_{\phi }}=\frac{1}{mx_{1}^{2}} and the other two components are zero. Hence the B matrix becomes
B=\begin{pmatrix} 0 & 0 & 0\\ \frac{1}{m} & 0 & 0\\ 0 & 0 & 0\\ 0 & \frac{1}{mx_{1}^{2}\cos ^{2}x_{5}} & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1}{mx_{1}^{2}}\end{pmatrix} The above is evaluated at \bar{x}=\begin{pmatrix} r_{0} & 0 & \omega t & \omega & 0 & 0 \end{pmatrix} ^{T} which results in B=\begin{pmatrix} 0 & 0 & 0\\ \frac{1}{m} & 0 & 0\\ 0 & 0 & 0\\ 0 & \frac{1}{mr_{0}^{2}\cos ^{2}0} & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1}{mr_{0}^{2}}\end{pmatrix} =\begin{pmatrix} 0 & 0 & 0\\ \frac{1}{m} & 0 & 0\\ 0 & 0 & 0\\ 0 & \frac{1}{mr_{0}^{2}} & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1}{mr_{0}^{2}}\end{pmatrix} And using r_{0}=1,m=1 it reduces to \fbox{$B=\begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix} $}
The linearized A matrix found above is
This shows that due to zeros everywhere in the linkage between the states r,r^{\prime },\theta ,\theta ^{\prime } and the states \phi ,\phi ^{\prime }, these states are decoupled. What this means is that motion can be analyzed in the \phi ,\phi ^{\prime } states as its own system without having to carry along other terms from the other states. This simplifies both the analysis and design for these parts of the system since they are decoupled from each others. The above decoupling is also present in the B and C matrices.
Using chain rule, and using \frac{\partial \zeta }{\partial x}=1 and \frac{\partial \zeta }{\partial t}=-v then1 \frac{\partial ^{2}\phi }{\partial x^{2}}=\frac{\partial }{\partial x}\left ( \frac{\partial \phi }{\partial x}\right ) =\frac{\partial }{\partial x}\left ( \frac{d\phi }{d\zeta }\frac{\partial \zeta }{\partial x}\right ) =\frac{\partial }{\partial x}\left ( \frac{d\phi }{d\zeta }\right ) =\frac{\partial }{\partial \zeta }\left ( \frac{d\phi }{d\zeta }\right ) \frac{\partial \zeta }{\partial x}=\frac{\partial }{\partial \zeta }\left ( \frac{d\phi }{d\zeta }\right ) =\frac{d^{2}\phi }{d\zeta ^{2}} And\begin{align*} \frac{\partial ^{2}\phi }{\partial t^{2}} & =\frac{\partial }{\partial t}\left ( \frac{\partial \phi }{\partial t}\right ) =\frac{\partial }{\partial t}\left ( \frac{d\phi }{d\zeta }\frac{\partial \zeta }{\partial t}\right ) =\frac{\partial }{\partial t}\left ( \frac{d\phi }{d\zeta }\left ( -v\right ) \right ) =\frac{\partial }{\partial \zeta }\left ( \frac{d\phi }{d\zeta }\left ( -v\right ) \right ) \frac{\partial \zeta }{\partial t}=\frac{\partial }{\partial \zeta }\left ( \frac{d\phi }{d\zeta }\left ( -v\right ) \right ) \left ( -v\right ) \\ & =v^{2}\frac{d^{2}\phi }{d\zeta ^{2}} \end{align*}
Hence the PDE becomes the ODE\begin{equation} \frac{d^{2}\phi }{d\zeta ^{2}}-v^{2}\frac{d^{2}\phi }{d\zeta ^{2}}=\sin \phi \left ( \zeta \right ) \nonumber \end{equation}
Hence the differential equation is
\begin{equation} \fbox{$\left ( 1-v^2\right ) \frac{d^2\phi }{d\zeta ^2}=\sin \phi \left ( \zeta \right ) $}\tag{1} \end{equation}
Let x_{1}=\phi ,x_{2}=\frac{d\phi }{d\zeta }, hence\begin{pmatrix} x_{1}=\phi \\ x_{2}=\frac{d\phi }{d\zeta }\end{pmatrix} \overset{\frac{d}{dt}}{\Longrightarrow }\begin{pmatrix} x_{1}^{\prime }=\frac{d\phi }{d\zeta }\\ x_{2}^{\prime }=\frac{d^{2}\phi }{d\zeta ^{2}}\end{pmatrix} =\begin{pmatrix} x_{2}\\ \frac{\sin x_{1}}{1-v^{2}}\end{pmatrix} Hence x^{\prime }=f\left ( x\right ) becomes\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\begin{pmatrix} x_{2}\\ \frac{\sin x_{1}}{1-v^{2}}\end{pmatrix} \\ & \equiv f\left ( x\right ) \end{align*}
Equation (1) is solved, and then the result is compared to the so called ”kink” solution provided in order to determine what the constant of integration are. The constant of integration will be x_{1}\left ( 0\right ) and x_{2}\left ( 0\right ) . Starting by multiplying both sides of (1) by \frac{d\phi }{d\zeta } results in \left ( 1-v^{2}\right ) \frac{d^{2}\phi }{d\zeta ^{2}}\frac{d\phi }{d\zeta }=\sin \phi \frac{d\phi }{d\zeta } Or \left ( 1-v^{2}\right ) \frac{d^{2}\phi }{d\zeta ^{2}}\frac{d\phi }{d\zeta }d\zeta =\sin \phi d\phi Now both sides are integrated. The RHS gives \int \sin \phi d\phi =-\cos \left ( \phi \right ) +c_{1} and the LHS gives \int \left ( 1-v^{2}\right ) \frac{d^{2}\phi }{d\zeta ^{2}}\frac{d\phi }{d\zeta }d\zeta =\left ( 1-v^{2}\right ) \frac{1}{2}\left ( \frac{d\phi }{d\zeta }\right ) ^{2} This is becuase if \frac{1}{2}\left ( \frac{d\phi }{d\zeta }\right ) ^{2} is differentiated, using chain rule, the result will be the integrand. Since differentiating \frac{1}{2}\left ( \frac{d\phi }{d\zeta }\right ) ^{2} w.r.t. \zeta gives \frac{d\phi }{d\zeta }\frac{d^{2}\phi }{d\zeta ^{2}} which is the integrand in the LHS. No need to introduced a new integration of constant again here as it can be absorbed with c_{1}. Therefore, the result after integration once is\begin{equation} \left ( 1-v^{2}\right ) \frac{1}{2}\left ( \frac{d\phi }{d\zeta }\right ) ^{2}=-\cos \left ( \phi \right ) +c_{1}\tag{2} \end{equation} To make some progress now, assuming the following initial conditions x_{1}\left ( 0\right ) =0=\phi and x_{2}\left ( 0\right ) =0=\frac{d\phi }{d\zeta }. Using these initial conditions results in \fbox{$c_1=1$} Equation (2) now becomes\begin{align} \frac{\left ( \frac{d\phi }{d\zeta }\right ) ^{2}}{1-\cos \left ( \phi \right ) } & =\frac{2}{\left ( 1-v^{2}\right ) }\nonumber \\ \frac{1}{\sqrt{1-\cos \left ( \phi \right ) }}\frac{d\phi }{d\zeta } & =\frac{\sqrt{2}}{\sqrt{1-v^{2}}}\tag{3} \end{align}
From trigonometric tables the relation \sin \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{2}} is used, therefore \sqrt{1-\cos \left ( \phi \right ) }=\pm \sqrt{2}\sin \frac{\phi }{2} and (3) becomes\begin{align*} \frac{1}{\pm \sqrt{2}\sin \frac{\phi }{2}}\frac{d\phi }{d\zeta } & =\frac{\sqrt{2}}{\sqrt{1-v^{2}}}\\ \pm \frac{d\phi }{\sin \frac{\phi }{2}} & =\frac{2}{\sqrt{1-v^{2}}}d\zeta \end{align*}
Doing integration again\begin{equation} \pm \int \frac{d\phi }{\sin \frac{\phi }{2}}=\frac{2}{\sqrt{1-v^{2}}}\int d\zeta \tag{4} \end{equation} From tables (or using substitutions) \int \frac{d\phi }{\sin \frac{\phi }{2}}=2\ln \left ( \tan \left ( \frac{\phi }{4}\right ) \right ) hence (4) becomes\begin{align*} \pm 2\ln \left ( \tan \left ( \frac{\phi }{4}\right ) \right ) & =\frac{2}{\sqrt{1-v^{2}}}\left ( \zeta -\zeta _{0}\right ) \\ \tan \left ( \frac{\phi }{4}\right ) & =\exp \left ( \pm \frac{\zeta -\zeta _{0}}{\sqrt{1-v^{2}}}\right ) \end{align*}
Therefore\begin{align*} \frac{\phi }{4} & =\arctan \left ( \pm \exp \left ( \frac{\zeta -\zeta _{0}}{\sqrt{1-v^{2}}}\right ) \right ) \\ \phi \left ( \zeta \right ) & =x_{1}=4\arctan \left ( \pm \exp \left ( \frac{\zeta -\zeta _{0}}{\sqrt{1-v^{2}}}\right ) \right ) \end{align*}
If \zeta _{0}=0 then \fbox{$x_1=4\arctan \left ( \pm \exp \left ( \frac{\zeta }{\sqrt{1-v^2}}\right ) \right ) $} This is the answer we are asked to show. Hence the initial conditions needed to obtain this answer are x_{1}\left ( 0\right ) =0 and x_{2}\left ( 0\right ) =0
The robotic arm coupled differential equations are\begin{align*} I\theta _{1}^{\prime \prime }+mgl\sin \theta _{1}+k\left ( \theta _{1}-\theta _{2}\right ) & =0\\ J\theta _{2}^{\prime \prime }-k\left ( \theta _{1}-\theta _{2}\right ) & =F\left ( t\right ) \end{align*}
Let\begin{align*} \begin{pmatrix} x_{1}=\theta _{1}\\ x_{2}=\theta _{2}\\ x_{3}=\theta _{1}^{\prime }\\ x_{4}=\theta _{2}^{\prime }\end{pmatrix} \overset{\frac{d}{dt}}{\Longrightarrow }\begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\\ x_{3}^{\prime }\\ x_{4}^{\prime }\end{pmatrix} & =\begin{pmatrix} \theta _{1}^{\prime }\\ \theta _{2}^{\prime }\\ \theta _{1}^{\prime \prime }\\ \theta _{2}^{\prime \prime }\end{pmatrix} =\begin{pmatrix} \theta _{1}^{\prime }\\ \theta _{2}^{\prime }\\ \frac{mgl}{I}\sin \theta _{1}+\frac{k}{I}\left ( \theta _{1}-\theta _{2}\right ) \\ \frac{k}{J}\left ( \theta _{1}-\theta _{2}\right ) +\frac{1}{J}F\left ( t\right ) \end{pmatrix} =\begin{pmatrix} x_{3}\\ x_{4}\\ -\frac{mgl}{I}\sin x_{1}-\frac{k}{I}\left ( x_{1}-x_{2}\right ) \\ \frac{k}{J}\left ( x_{1}-x_{2}\right ) +\frac{1}{J}F\left ( t\right ) \end{pmatrix} \\ & =\begin{pmatrix} f_{1}\left ( x,u\right ) \\ f_{2}\left ( x,u\right ) \\ f_{3}\left ( x,u\right ) \\ f_{4}\left ( x,u\right ) \end{pmatrix} \end{align*}
Where u, the input, is F\left ( t\right ) in this example, but the letter u is used since it is the common notation.
Let equilibrium point be \left ( \bar{x},\bar{u}\right ) =\left ( 0,0\right ) . The system is now linearized around this point.\begin{align*} f\left ( \bar{x}+\Delta x,\bar{u}+\Delta u\right ) & =\begin{pmatrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} & \frac{\partial f_{1}}{\partial x_{3}} & \frac{\partial f_{1}}{\partial x_{4}}\\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}} & \frac{\partial f_{2}}{\partial x_{3}} & \frac{\partial f_{2}}{\partial x_{4}}\\ \frac{\partial f_{3}}{\partial x_{1}} & \frac{\partial f_{3}}{\partial x_{2}} & \frac{\partial f_{3}}{\partial x_{3}} & \frac{\partial f_{3}}{\partial x_{4}}\\ \frac{\partial f_{4}}{\partial x_{1}} & \frac{\partial f_{4}}{\partial x_{2}} & \frac{\partial f_{4}}{\partial x_{3}} & \frac{\partial f_{4}}{\partial x_{4}}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\begin{pmatrix} \Delta x_{1}\\ \Delta x_{2}\\ \Delta x_{3}\\ \Delta x_{4}\end{pmatrix} +\begin{pmatrix} \frac{\partial f_{1}}{\partial u}\\ \frac{\partial f_{2}}{\partial u}\\ \frac{\partial f_{2}}{\partial u}\\ \frac{\partial f_{2}}{\partial u}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\Delta u\\ & =\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -\frac{mgl}{I}\cos x_{1}-\frac{k}{I} & \frac{k}{I} & 0 & 0\\ \frac{k}{J} & -\frac{k}{J} & 0 & 0 \end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\begin{pmatrix} \Delta x_{1}\\ \Delta x_{2}\\ \Delta x_{3}\\ \Delta x_{4}\end{pmatrix} +\begin{pmatrix} 0\\ 0\\ 0\\ \frac{1}{J}\end{pmatrix} _{\left ( \bar{x},\bar{u}\right ) }\Delta u \end{align*}
The above is evaluated at \left ( \bar{x},\bar{u}\right ) =\left ( 0,0\right ) giving\begin{align*} A & =\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -\frac{mgl}{I}-\frac{k}{I} & \frac{k}{I} & 0 & 0\\ \frac{k}{J} & -\frac{k}{J} & 0 & 0 \end{pmatrix} \\ B & =\begin{pmatrix} 0\\ 0\\ 0\\ \frac{1}{J}\end{pmatrix} \end{align*}
Using values I=J=mgl=k=1, the A matrix becomes \fbox{$A=\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -2 & 1 & 0 & 0\\ 1 & -1 & 0 & 0 \end{pmatrix} $} The real part of the eigenvalues of this matrix are
There is no positive real part. The real part of the eigenvalues are effectively zero. They are pure complex conjugate values. Hence the system is stable (sometimes also called marginally stable in this case).
\begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} =\begin{pmatrix} -\frac{t}{1+t^{2}} & 1\\ 0 & \frac{-4t}{1+t^{2}}\end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} With initial conditions \begin{pmatrix} x_{1}\left ( 0\right ) \\ x_{2}\left ( 0\right ) \end{pmatrix} =\begin{pmatrix} x_{10}\\ x_{20}\end{pmatrix}
x_{2}\left ( t\right ) is first solved since it does not depend on x_{1}\left ( t\right ) and then the solution is used to solve for x_{1}\left ( t\right ) . The differential equation for x_{2}\left ( t\right ) is\begin{align*} \frac{dx_{2}}{dt} & =\frac{-4t}{1+t^{2}}x_{2}\\ \frac{dx_{2}}{x_{2}} & =\frac{-4t}{1+t^{2}}dt \end{align*}
Integrating\begin{align*} \ln x_{2} & =\ln \left ( \frac{1}{\left ( 1+t^{2}\right ) ^{2}}\right ) +c\\ x_{2} & =c\frac{1}{\left ( 1+t^{2}\right ) ^{2}} \end{align*}
When t=0, x_{2}\left ( 0\right ) =x_{20}, hence c=x_{20} and \fbox{$x_2\left ( t\right ) =\frac{x_{20}}{\left ( 1+t^2\right ) ^2}$} Now since x_{2}\left ( t\right ) was found, it is used to obtain x_{1}\left ( t\right ) . The differential equation for x_{1}\left ( t\right ) is\begin{align*} \frac{dx_{1}}{dt} & =-\frac{t}{1+t^{2}}x_{1}+x_{2}\\ \frac{dx_{1}}{dt}+\frac{t}{1+t^{2}}x_{1} & =x_{20}\frac{1}{\left ( 1+t^{2}\right ) ^{2}} \end{align*}
The integrating factor is I=e^{\int \frac{t}{1+t^{2}}dt}=\sqrt{\left ( 1+t^{2}\right ) }, hence the solution to the above is d\left ( Ix_{1}\right ) =x_{20}\frac{I}{\left ( 1+t^{2}\right ) ^{2}} Integrating\begin{align*} Ix_{1} & =\int x_{20}\frac{I}{\left ( 1+t^{2}\right ) ^{2}}dt\\ & =x_{20}\int \frac{\sqrt{\left ( 1+t^{2}\right ) }}{\left ( 1+t^{2}\right ) ^{2}}dt\\ & =x_{20}\int \frac{1}{\left ( 1+t^{2}\right ) ^{3/2}}dt\\ & =x_{20}\frac{t}{\sqrt{1+t^{2}}}+c_{2} \end{align*}
Hence, dividing by I gives the final solution x_{1}\left ( t\right ) =x_{20}\frac{t}{1+t^{2}}+\frac{c_{2}}{\sqrt{\left ( 1+t^{2}\right ) }} When t=0,x_{1}\left ( 0\right ) =x_{10}, hence c_{2}=x_{10} and the solution becomes x_{1}\left ( t\right ) =x_{20}\frac{t}{1+t^{2}}+\frac{x_{10}}{\sqrt{\left ( 1+t^{2}\right ) }} Now we are asked to let x_{10}=1, hence x_{1}\left ( t\right ) becomes \fbox{$x_1\left ( t\right ) =x_{20}\frac{t}{1+t^2}+\frac{1}{\sqrt{\left ( 1+t^2\right ) }}$} Now taking the limit of x_{1}\left ( t\right ) _{t\rightarrow \infty } gives x_{1}\left ( t\right ) \rightarrow 0
Using Matlab syms, these can be solved as follows
