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I have the differential equation $$f'=c \times f$$ for $f: \mathbb{R} \to \mathbb{R}^3$ and constants $c \in \mathbb{R}^3$. How can I solve something like this?

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    $\begingroup$ Doesn't the cross product only exist when $n=3$ and $n=7$? $\endgroup$
    – mattos
    Feb 24 '18 at 13:51
  • $\begingroup$ Can you define what you mean by $\times : \mathbb{R}^n\times\mathbb{R}^n \rightarrow \mathbb{R}^n$? $\endgroup$
    – Alex Jones
    Feb 24 '18 at 13:59
  • $\begingroup$ Sorry I have mistakenly written $\mathbb{R}^n$ instead of $\mathbb{R}^3$. $\endgroup$
    – Dimtsol
    Feb 24 '18 at 14:06
  • $\begingroup$ Have you tried anything? Try letting $c=\begin{bmatrix}0\\0\\1\end{bmatrix}$ and expanding out the cross product. Perhaps you can solve the three equations. That would be a place to start at least. If you show some more work, I can post a complete answer. $\endgroup$
    – robjohn
    Feb 24 '18 at 18:33
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Note that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}|f(t)|^2 &=2f(t)\cdot\left(c\times f(t)\right)\\ &=0\tag1 \end{align} $$ so we have that $|f(t)|$ is constant. Thus, $f(t)$ lives on a sphere of radius $|f(0)|$.

Furthermore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}c\cdot f(t) &=c\cdot\left(c\times f(t)\right)\\ &=0\tag2 \end{align} $$ so we have that $c\cdot f(t)$ is constant. That is, $f(t)$ lives on the plane perpendicular to $c$ containing $f(0)$.

The speed is also constant: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}|f'(t)|^2 &=\frac{\mathrm{d}}{\mathrm{d}t}|c\times f(t)|^2\\ &=2\color{#C00}{(c\times f(t))}\cdot c\times\color{#C00}{(c\times f(t))}\\[4pt] &=0\tag3 \end{align} $$ Since the circumference of the circle is $\frac{2\pi}{|c|}|c\times f(0)|$ and the speed is $|c\times f(0)|$, the period is $\frac{2\pi}{|c|}$.

Therefore, the solution can be gotten geometrically since $f(t)$ circles the projection of $f(0)$ onto $c$ at a constant distance with a period $\frac{2\pi}{|c|}$.

$$f(t)=c\frac{c\cdot f(0)}{c\cdot c}+\left(f(0)-c\frac{c\cdot f(0)}{c\cdot c}\right)\cos\left(|c|\,t\right)+\frac{c}{|c|}\times f(0)\sin\left(|c|\,t\right)\tag4$$

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  • $\begingroup$ Since a complete answer is already given, I will uncomment my answer. $\endgroup$
    – robjohn
    Feb 25 '18 at 8:20
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Is it the typical cross product you are talking about? ($n=3$) If so, then compute $c \times f$ directly and try figuring a constant matrix $A$ for which your differential equation becomes $f' = Af (\equiv c\times f)$. Then you can solve this new equation by means of the eigendecomposition of $A$, if the latter has any. (Spoiler alert: $A$ is skew-symmetric and has zeroes over the main diagonal.)

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In this case it helps, a little to increase the order of the ODE, as $$ \ddot f=c\times \dot f=c\times(c\times f) = \langle c,f\rangle\, c -\|c\|^2f $$ which means that perpendicular to $c$ it is an oscillation equation, in the direction of $c$ it is constant. Taking the next derivative confirms that as $$ \dddot f+\|c\|^2\dot f = \langle c,c\times f\rangle\, c=0 $$ It has the general solution form $$ f(t)=A\cos(\|c\|\,t)+B\sin(\|c\|\,t)+C $$ where the constant vectors are bound by the original first order equation, $$ \begin{cases}A+C&=f(0),\\ \|c\|A&=-c×B,\\\|c\|B&=c×A,\\ c×C&=0\end{cases} \implies \begin{cases} B&=\frac1{\|c\|}c×f(0)\\ A&=\frac1{\|c\|^2}c×(c×f(0))=f(0)-\frac1{\|c\|^2}⟨c,f(0)⟩\,c\\ C&=\frac1{\|c\|^2}⟨c,f(0)⟩\,c \end{cases} $$ This all combines to give the Rodrigues formula $$ f(t)=\exp(t(c×))f(0) =f(0)\cos(\|c\|\,t)+\frac{c×f(0)}{\|c\|}\sin(\|c\|\,t)+\frac{\langle c,f(0)\rangle\,c}{\|c\|^2} (1-\cos(\|c\|\,t)). $$

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  • $\begingroup$ Up to the addition of a constant, this is correct. $\endgroup$
    – robjohn
    Feb 24 '18 at 23:50
  • $\begingroup$ @robjohn : After some consideration, no, it is not. The double cross product has a second term. $\endgroup$ Feb 25 '18 at 6:42
  • $\begingroup$ That is the constant. See my answer to see that $c\cdot f(t)$ is a constant. $\endgroup$
    – robjohn
    Feb 25 '18 at 8:17
  • $\begingroup$ Yes, that is the same idea, constant in the direction of $c$. I only leave my answer open for the link to the Rodriguez formula article. $\endgroup$ Feb 25 '18 at 8:20
  • $\begingroup$ I think that $\langle c,f(0)\rangle c(1-\cos(\|c\|\,t))$ needs to be divided by $\|c\|^2$ since $\|c\|\ne1$. $\endgroup$
    – robjohn
    Feb 25 '18 at 8:25

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