- #1
John Creighto
- 495
- 2
I tried posting this to my blog but the preview function wasn't rendering the formula's correctly and once I posted it I couldn't edit it.
https://www.physicsforums.com/blog.php?b=1152
Therefor I'll post it here instead.
For simplicity let's consider a very simple ODE.
[tex]\dot{x_1}=a x_1^2[/tex]
We can approximate this first order system with a second order ODE as follows:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 \\
0 &
\frac{d f(x_1)}{d x_1}
\end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \end{array} \right]
[/tex]
Where
[tex]
x_2=\frac{dx_1}{dt}
[/tex]
Or in the simple case mentioned above we have:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 \\
0 & 2x_1(t_o) \end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \end{array} \right]
[/tex]
Using the matrix exponential the solution to the linear approximation of this stem as follows:
[tex]
\left[ \begin{array}{c}
x(t) \\
\dot{x}(t) \end{array} \right]
=
exp \left( \left[ \begin{array}{ccc}
0 & 1 \\
0 &
2x_1(t_o) \end{array} \right] (t-t_o) \right)
\left[ \begin{array}{c}
x(t_o)) \\
\dot{x}(t_o) \end{array} \right]
[/tex]
Where:
[tex]\dot{x}(t_o)=ax(t_o)^2[/tex]
Keep in mind the choice of using a second order approximation was somewhat arbitrary. We could of equally well, done a third order approximation as follows:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \\
\dot{x_3} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 2 a & 2 a x(t_o)
\end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \\
x_3 \end{array} \right]
[/tex]
Where
[tex]
x_2=\frac{dx_1}{dt}, x_3=\frac{d^2x_1}{dt^2}
[/tex]
and
[tex]\dot{x}(t_o)=ax(t_o)^2[/tex]
[tex]\ddot{x}(t_o)=2 a x(t_o) \dot{x}(t_o)=2a^2x(t_o)^3[/tex]Which can also be solved using the matrix exponential.
https://www.physicsforums.com/blog.php?b=1152
Therefor I'll post it here instead.
For simplicity let's consider a very simple ODE.
[tex]\dot{x_1}=a x_1^2[/tex]
We can approximate this first order system with a second order ODE as follows:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 \\
0 &
\frac{d f(x_1)}{d x_1}
\end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \end{array} \right]
[/tex]
Where
[tex]
x_2=\frac{dx_1}{dt}
[/tex]
Or in the simple case mentioned above we have:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 \\
0 & 2x_1(t_o) \end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \end{array} \right]
[/tex]
Using the matrix exponential the solution to the linear approximation of this stem as follows:
[tex]
\left[ \begin{array}{c}
x(t) \\
\dot{x}(t) \end{array} \right]
=
exp \left( \left[ \begin{array}{ccc}
0 & 1 \\
0 &
2x_1(t_o) \end{array} \right] (t-t_o) \right)
\left[ \begin{array}{c}
x(t_o)) \\
\dot{x}(t_o) \end{array} \right]
[/tex]
Where:
[tex]\dot{x}(t_o)=ax(t_o)^2[/tex]
Keep in mind the choice of using a second order approximation was somewhat arbitrary. We could of equally well, done a third order approximation as follows:
[tex]
\left[ \begin{array}{c}
\dot{x_1} \\
\dot{x_2} \\
\dot{x_3} \end{array} \right]
=
\left[ \begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 2 a & 2 a x(t_o)
\end{array} \right]
\left[ \begin{array}{c}
x_1 \\
x_2 \\
x_3 \end{array} \right]
[/tex]
Where
[tex]
x_2=\frac{dx_1}{dt}, x_3=\frac{d^2x_1}{dt^2}
[/tex]
and
[tex]\dot{x}(t_o)=ax(t_o)^2[/tex]
[tex]\ddot{x}(t_o)=2 a x(t_o) \dot{x}(t_o)=2a^2x(t_o)^3[/tex]Which can also be solved using the matrix exponential.
Last edited by a moderator: