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Consider a time-dependent family of curves given in polar coordinates, so described by $r(\theta,t)$, which is assumed to be a positive $2\pi$-periodic function of $\theta$ for each fixed $t$, with the property that the periodic extension is smooth. (In other words the induced function on $S^1$ is smooth.)

Suppose that the point on the curve at each $\theta$ moves with velocity vector $v(\theta,t) \nu(\theta,t)$. Here $v$ is a real-valued function (with the same hypotheses as $r$ except that it can have either sign), and $\nu(\theta,t)$ is the outward unit normal to the curve at $\theta$. (If it helps later, you may assume $v>0$.) In general the outward unit normal has both a radial and an angular component; the angular component vanishes if and only if $\partial_\theta r = 0$ at that point.

I am trying to come up with a PDE for the evolution of $r(\theta,t)$, of the form $\frac{\partial r}{\partial t}=\dots$.

It seems like there is a transport term in this PDE. To see that, consider that in a time interval of length $h$ the point $r \mathbf{e}_r(\theta)$ is approximately displaced to $(r+avh) \mathbf{e}_r(\theta) + bvh \mathbf{e}_\theta(\theta)$, where $\mathbf{e}_r$ and $\mathbf{e}_\theta$ are unit radial and angular vectors. Here $a$ and $b$ are the coefficients in the expansion of $\nu(\theta,t)$ into components. Explicitly, we can write

$$\nu(\theta,t)=\frac{1}{\sqrt{r^2+(\partial_\theta r)^2}} (r\cos(\theta)+\partial_\theta r \sin(\theta),r\sin(\theta) - \partial_\theta r \cos(\theta))^T$$

with $\mathbf{e}_r=(\cos(\theta),\sin(\theta))^T$ and $\mathbf{e}_\theta=(-\sin(\theta),\cos(\theta))^T$. Therefore $a=\frac{r}{\sqrt{r^2+(\partial_\theta r)^2}}$ and $b=-\frac{\partial_\theta r}{\sqrt{r^2+(\partial_\theta r)^2}}$.

If I try to push this a little further, it seems like the angle that is displaced to end up at $\theta$ is $\theta' \approx \theta - \frac{bvh}{r}$. The initial $r$ at $\theta'$ (before accounting for the motion in the radial direction) is approximately $r-\frac{bvh}{r} \partial_\theta r$. So it seems that we recover

$$\frac{\partial r}{\partial t} = v \cdot \left ( a - b \frac{\partial_\theta r}{r} \right )$$

which I guess can be simplified further by plugging in $a$ and $b$, to get

$$\frac{\partial r}{\partial t}=\frac{v}{\sqrt{r^2+(\partial_\theta r)^2}} \left ( r + \frac{(\partial_\theta r)^2}{r} \right ) \\ = \frac{v}{r\sqrt{r^2+(\partial_\theta r)^2}} \left ( r^2+(\partial_\theta r)^2 \right ) \\ = \frac{v \sqrt{r^2+(\partial_\theta r)^2}}{r}.$$

Is this correct? I can tell it is correct in the case of the circle but it seems simpler than I expected in the general situation.

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  • $\begingroup$ Since to first order every curve is a line, just check on a generic line $r=p\sec(\theta-\theta_0)$ with $v$ constant. That gives $r(\theta,t)=(p+tv)\sec(\theta-\theta_0)$ so you should end up with $\frac{\partial}{\partial t}r=v\sec(\theta-\theta_0)$, which it does. $\endgroup$ Oct 8 at 16:20
  • $\begingroup$ @user10354138 With $r(\theta,0)=\sec(\theta)$, say, you have $r^2+(\partial_\theta r)^2=\sec(\theta)^2 + \sec(\theta)^2 \tan(\theta)^2 = \sec(\theta)^2+\sec(\theta)^4-\sec(\theta)^2=\sec(\theta)^4$, then the square root gives $\sec(\theta)^2$, then the division by $r$ gives $\sec(\theta)$ back again. So the vertical line with constant velocity case is handled correctly also (a decent sign...) $\endgroup$
    – Ian
    Oct 8 at 16:24
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To do it more rigorously (read analysis-y), we have $S^1=\mathbb{R}/2\pi$, a smooth $$ (\rho,\psi)\colon\mathbb{R}\times S^1\to\mathbb{R}^+\times S^1 $$ such that:

  • $\psi(t,-)$ is a diffeomorphism $S^1\to S^1$ for all $t$, say orientation preserving, with inverse $\phi(t,-)$;
  • normal deformation flow (*) \begin{align*} (\partial_1\rho,\partial_1\psi) &=v(\psi)\frac{(\rho\partial_2\psi,-\frac1\rho\partial_2\rho)}{\sqrt{(\rho\partial_2\psi)^2+\rho^2(\frac1\rho\partial_2\rho)^2}}\\ &=\left( \frac{v(\psi)\rho\partial_2\psi}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}} , -\frac{v(\psi)\frac1\rho\partial_2\rho}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}\right) \end{align*}

Then $$ r(\theta,t):=\rho(t,\phi(t,\theta)) $$ so we calculate \begin{align*} \psi(t,\phi(t,\theta))=\theta &\Rightarrow \frac{\partial}{\partial t}\psi(t,\phi(t,\theta))=0,\quad \frac{\partial}{\partial \theta}\psi(t,\phi(t,\theta))=1 \\ &\Rightarrow \partial_1\psi(t,\phi(t,\theta))+\partial_2\psi(t,\phi(t,\theta))\partial_1\phi(t,\theta)=0,\quad \partial_2\psi(t,\phi(t,\theta))\partial_2\phi(t,\theta)=1\\ &\Rightarrow \partial_1\phi(t,\theta)=-\frac{\partial_1\psi(t,\phi(t,\theta))}{\partial_2\psi(t,\phi(t,\theta))},\quad \partial_2\phi(t,\theta)=\frac1{\partial_2\psi(t,\phi(t,\theta))} \end{align*} hence $$ \frac{\partial r}{\partial\theta}(\theta,t)=\frac{\partial}{\partial\theta}\rho(t,\phi(t,\theta))=\partial_2\rho(t,\phi(t,\theta))\cdot\partial_2\phi(t,\theta)=\left.\frac{\partial_2\rho}{\partial_2\psi}\right\rvert_{t,\phi(t,\theta)} $$ and \begin{align*} \frac{\partial}{\partial t}r(\theta,t) &=\partial_1\rho(t,\phi(t,\theta))+\partial_2\rho(t,\phi(t,\theta))\partial_1\phi(t,\theta)\\ &=\partial_1\rho(t,\phi(t,\theta))-\partial_2\rho(t,\phi(t,\theta))\frac{\partial_1\psi(t,\phi(t,\theta))}{\partial_2\psi(t,\phi(t,\theta))}\\ &=\left.\left(\partial_1\rho-\partial_1\psi\frac{\partial_2\rho}{\partial_2\psi}\right)\right\rvert_{(t,\phi(t,\theta))}\\ &=\left.\left(\frac{v(\psi)\rho\partial_2\psi}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}}+\frac{v(\psi)\frac1\rho\partial_2\rho}{\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}} \frac{\partial_2\rho}{\partial_2\psi}\right)\right\rvert_{(t,\phi(t,\theta))}\\ &=v(\theta)\left.\left( \frac{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}{\rho\partial_2\psi\sqrt{(\rho\partial_2\psi)^2+(\partial_2\rho)^2}} \right)\right\rvert_{(t,\phi(t,\theta))}\\ &=v(\theta)\left.\left( \frac1\rho\sqrt{\rho^2+\left(\frac{\partial_2\rho}{\partial_2\psi}\right)^2} \right)\right\rvert_{(t,\phi(t,\theta))}\\ &=\frac{v(\theta)}{r(\theta,t)}\,\sqrt{r(\theta,t)^2+\left(\frac{\partial r(\theta,t)}{\partial\theta}\right)^2}. \end{align*}

(*) The tangent (of the $S^1\to\mathbb{R}^+\times S^1$ at fixed $t$) is $(\partial_2\rho,\partial_2\psi)\in\mathbb{R}^2=T_\rho\mathbb{R}^+\times T_\psi S^1$, taking orthogonal with respect to metric $dr^2+r^2d\theta^2$ on $\mathbb{R}^+\times S^1$ at $(\rho,\psi)$ gives $\pm(\rho\partial_2\psi,-\frac1\rho\partial_2\rho)$. Since we assume $\psi(t,-)$ is orientation-preserving we choose the + sign to have the first component positive.

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