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Task: Construct a vector field $v:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that all the circles that "touch" y-axis in the origin (i.e. derivative in $(0,0)$ is in form $(0,a)$ for some $a$) are field lines of this field, and this vector field has to be continuously differentiable.

My thoughts:

It could look like this (up to direction):

enter image description here

At first I consider the case $x>0$ (the right half of the plane). My idea was to parametrize each circle that "touches" y-axis as follows: $$\begin{align}x-r&=r\cdot \cos(t)\\y&=r\cdot \sin(t)\end{align}$$ and it implies

$$\begin{align}x&=r\cdot( \cos(t)+1)\\y&=r\cdot \sin(t)\end{align}$$

Also for each point $(x,y)$ in $\mathbb{R}^2$ ($x>0$) there is only one $r$ such that $(x-r)^2+y^2=r^2$:

$$\begin{align} (x-r)^2+y^2&=r^2\\ x^2-2xr+r^2+y^2&=r^2\\ x^2-2xr+y^2&=0 \end{align} $$

and thus $r=\frac{x^2+y^2}{2x}$.

If I want all those circles to be the field lines, I have to set the values of $v$ to be the derivatives of the curve: $$\frac{d}{dt}c(t)=\frac{d}{dt}(r\cdot(\cos(t)+1),r\cdot\sin(t))=(-r\sin(t),r\cos(t))$$

So the vector field $v$ could be $$v(x,y)=v(r\cdot(\cos(t)+1),r\cdot\sin(t))=(-r\sin(t),r\cos(t))=(-y,x-r)=(-y,x-\frac{x^2+y^2}{2x})=(-y,\frac{x^2-y^2}{2x})$$

This $v$ is actually the function plotted above. So the question is now whether this function is continuously differentiable or not. The critical point is when $x$ is close to $0$. If we take a look at partial derivative, then

$$\partial_xv=(0,\frac{x^2+y^2}{2x^2})$$

and it diverges when $x\rightarrow 0$, so it seems that this $v$ is not suitable, but I think the real $v$ should be somehow similar to the given $v$.

But these are only my considerations that for sure can be wrong.

I'd appreciate your help.

nakajuice
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    You can multiply your vector field by any scalar field and the field lines don't change. So, for example, you can take the vector field to be $f(x,y)v(x,y)$, where $f$ is a scalar field that goes to $0$ near the origin in such a way as to cancel out the singularity in $v$. (Looking at your definition of $v(x,y)$, the function $f(x,y)=x$ should work...) –  Jul 08 '13 at 22:43
  • This looks very much alike with a dipolar magnetic field, have you checked that? – Shuhao Cao Jul 08 '13 at 23:02

1 Answers1

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The circles centered at a point $(0,a)$ passing through the origin have equations $(x-a)^2+y^2=a^2.$ There is a standard way to "go backwards" from this and get a differential equation. One takes the implicit derivative, solves for $a$, and plugs the result back into the circle equation. This gives $$2xyy'+x^2-y^2=0.$$ Then the line field is obtained on solving for $y'$.

Note that the solved version $$y'=\frac{y^2-x^2}{2xy}$$ makes it clear the line field is singular only on the axes.

coffeemath
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