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Vector Field

I already know that since that $0$ is a repulsor singular point over $x$-axis and $-x$ also works as a repulsor point in opposite direction. Can anyone help me out by finding the equation of this vector field given the phase portrait on picture? Thanks.

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  • $\begingroup$ There seem to be more information given on the paper. Could you provide also that information. It seems that the arrows are vertical at $y=-\mu x$, is that true? $\endgroup$ – mickep Feb 17 '15 at 19:31
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Let $A$ be the matrix of this vector field, so $$ \begin{bmatrix} \dot x \\ \dot y \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix} $$ There isn't enough on the photograph to tell so let me make some additional assumptions.

  1. $A \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix}1 \\ 0 \end{bmatrix}$. If that's not true, $A \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is at least a positive multiple of $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, but let's keep it simple.

  2. $A \begin{bmatrix} 1 \\ -\mu \end{bmatrix} = \begin{bmatrix}0 \\ -1 \end{bmatrix}$. Again, this is from the picture, modulo positive multiples.

If that's true then \begin{align*} A \begin{bmatrix} 0 \\ 1 \end{bmatrix} &= A \left(\begin{bmatrix} -1/\mu \\ 1 \end{bmatrix} + \begin{bmatrix} 1/\mu \\ 0 \end{bmatrix} \right)\\ &= \begin{bmatrix} 0 \\ 1/\mu \end{bmatrix} + \begin{bmatrix} 1/\mu \\ 0 \end{bmatrix} = \begin{bmatrix} 1/\mu \\ 1/\mu \end{bmatrix} \end{align*} So from this we know that $$ A = \begin{bmatrix} 1 & 1/\mu \\ 0 & 1/\mu \end{bmatrix} $$

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