# Function on $\mathbb{R}^{2}-\{0\}$.

Does there exist any compactly supported function $f= (f_1,f_2): \mathbb R^2-\{0\}\to \mathbb R^2$ such that $$\frac{\partial}{\partial x_2}f_1=\frac{\partial}{\partial x_1}f_2.$$

Also there does not exists any function $F: U(\subset \mathbb R^2-\{0\})\to \mathbb R$ such that $$\frac{\partial}{\partial x_1}F= f_1\text{ and } \frac{\partial}{\partial x_2}F= f_2.$$

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Pick your favorite $C^2$ compactly-supported function $F$. Then equality of mixed partials implies... –  user7530 Jan 29 '13 at 6:04

The answer is false, every compact supported $f$ that satisfies $\nabla \wedge f = 0$ is a gradient.

Identify $\mathbb{R}^2$ as $S^2−\{p\}$, the sphere with a point $p$ removed. Let $q$ be the point on $S^2$ that corresponds to $0$ in $\mathbb{R}^2$. Any compact supported vector field $f$ on $\mathbb{R}^2 - \{0\} \approx S^2−\{p,q\}$ vanishes identically on some neighborhood of $p$. As a result, one can extend its definition over the whole of $S^2−\{q\}$.

If $f$ is irrotational, i.e., $\nabla \wedge f = 0$, so is this extension to $S^2−\{q\}$. Since $S^2−\{q\}$ is simply connected, we can find an $F$ on $S^2−\{q\}$ such that $f = \nabla F$. Pull this back to $\mathbb{R}^2 - \{0\}$, and we see any compact supported irrotational vector field on the punctured plane is also a gradient.

Let's look at the problem from another angle without leaving $\mathbb{R}^2$.

For any path or closed loop $\gamma$ in $\mathbb{R}^2 - \{0\}$, let $I_{\gamma}(f)$ be the line integral: $$I_{\gamma}(f) = \int_{\gamma} ( f_1 dx_1 + f_2 dx_2 )$$

When $\nabla \wedge f = 0$ and $\gamma$ is a closed loop, Stokes' theorem tells us $I_{\gamma}(f)$ is invariant under continuous deformation of $\gamma$. For $f$ with compact support, we can evaluate $I_{\gamma}(f)$ by deforming $\gamma$ to another loop $\gamma_{\infty}$ outside of $f$'s support. We get: $$I_{\gamma}(f) = I_{\gamma_{\infty}}(f) = I_{\gamma_{\infty}}(0) = 0$$

Pick an arbitrary point $c \in \mathbb{R}^2-\{0\}$. For any $x \in \mathbb{R}^2-\{0\}$, let $\gamma_i : [0,1] \to \mathbb{R}^2-\{0\}$, $i = 1, 2$ be any two paths that satisfy: $\gamma_i(0) = c, \gamma_i(1) = x$. Join $\gamma_1$ and $\gamma_2$ to form the closed loop $\gamma_{12} : [0,1] \to \mathbb{R}^2-\{0\}$: \begin{align} \gamma_{12}(t) &= \gamma_1(2t) & \text{ for } & t \in [0,1/2]\\ &= \gamma_2(1-2t) & & t \in [1/2,1] \end{align} We have: $$I_{\gamma_1}(f) - I_{\gamma_2}(f) = I_{\gamma_{12}}(f) = 0$$ This means that $I_{\gamma_i}(f)$ is independent of the choice of $\gamma_i$ and can be viewed as a function $F(c,x)$ of $c$ and $x$ only. Finally, $F$ satisfies: $\nabla_{x} F(c,x) = f(x)$ and $\nabla_{c} F(c,x) = -f(c)$.

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Thanks for the answer.. :) –  Junu Jan 29 '13 at 19:04

Your first equation reduces to $\nabla \wedge f = 0$. There are plenty of curlless vector fields, so there can exist such a vector field. This condition says that $f$ is closed.

Your second equations reduce to

$$e_1 \cdot \nabla F = f_1, \quad e_2 \cdot \nabla F = f_2$$

$F$ must be a scalar field for this to be the case, so what you're really saying is that you want,

$$\nabla F = f$$

This may be the case, and by construction it would give $\nabla \wedge f = 0$. This condition says that $f$ is exact.

Without more information (in particular, the divergence), you can't reconstruct the original vector field $f$. Since its domain is the punctured plane, it's possible $f$ is closed but not exact--an easy example is $f = e_\theta$. Of course, the reason this is so is because if you could take the curl at the origin, you would get a nonzero result, even though $f$ would be well-defined (the zero vector) at that point.

Of interest may be Poincare's lemma, which says that closed forms are also exact if, for any two points $p, q$ in the domain, the line segment connecting them is also entirely in the domain (this is not true for the punctured plane).

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thanks for the answer.. but condition on $f$ is, it is compactly supported... I can take $f(x_1,x_2)= (\frac{-x_2}{x_1^2+x_2^2}, \frac{x_1}{x_1^2+x_2^2})$ for this No $F$ exits.. But $f$ is not compactly supported.. So i want compactly supported $f$. –  Junu Jan 29 '13 at 6:34