5
$\begingroup$

Considering that such a vector field $(M,N)$ is conservative iff for $M,N$ differentiable, $\frac{\partial{N}}{\partial{x}} = \frac{\partial{M}}{\partial{y}}$, we only have one of the two Cauchy-Riemann equations satisfied. Are there then examples of conservative vector fields $\mathbb{R}^2 \to \mathbb{R}^2$ that are not complex-differentiable?

$\endgroup$
5
  • 8
    $\begingroup$ Whenever $M$ depends only on $x$ and $N$ only on $y$, you have $\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}$. That is rarely holomorphic. $\endgroup$ Commented Dec 13, 2013 at 23:51
  • $\begingroup$ Right, of course. This answers my question nicely. $\endgroup$ Commented Dec 14, 2013 at 0:20
  • $\begingroup$ Hi @DanielFischer, I'm trying to show that this is true in the other direction: that all holomorphic functions on a simply connected domain are curl-free and conservative, and integrate to zero on any closed curve that is contained in the domain -- which agrees with the Cauchy-Goursat theorem. However, I can only verify that the conjugate of f is conservative, using the C-R equations. What can we say about f itself? Thanks, $\endgroup$
    – User001
    Commented Aug 2, 2015 at 3:05
  • 3
    $\begingroup$ @LebronJames Generally, if you associate the field $(u,v)$ to the function $f = u+iv$, holomorphic functions don't give conservative fields. The line integrals you consider for the Cauchy theorem are different from the line integrals for vector fields on $\mathbb{R}^2$. In the Cauchy theorem, you consider $f\,dz$, which becomes $(u+iv)(dx+idy) = (u\,dx - v\,dy) + i(v\,dx + u\,dy)$. For vector fields $(a,b)$ on $\mathbb{R}^2$, you consider $a\,dx + b\,dy$ or $a\,dy - b\,dx$, so you have a sign-flip in the second component. Roughly, for vector fields you look at $f\,d\overline{z}$. $\endgroup$ Commented Aug 2, 2015 at 10:49
  • $\begingroup$ Awesome explanation, @DanielFischer - thanks so much and have a great night :-) $\endgroup$
    – User001
    Commented Aug 4, 2015 at 7:07

1 Answer 1

0
$\begingroup$

If a scalar function $\phi(x,y)$ on $\mathbb{R}^{2}$ is twice continuously differentiable, then $\nabla \phi = (\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y})$ is a conservative vector field because $\frac{\partial^{2}\phi}{\partial x\partial y} =\frac{\partial^{2}\phi}{\partial y \partial x}$. (That's not much of a restriction on $\phi$, and certainly not enough to guarantee that $\phi$ is the real part of a holomorphic function, for example.)

Conversely, given $(M,N)$ as you describe and continuously differentiable, then you can integrate $$ \phi(x,y) = \int_{(0,0)}^{(x,y)}M(x',y')dx' + N(x',y')dy'. $$ The above is well-defined because the smooth path you choose from $(0,0)$ to $(x,y)$ doesn't matter, which follows from Green's Theorem: $$ \oint_{C}Mdx+Ndy = \int\int_{R} \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\,dx dy = 0, $$ where $C$ is a simple closed positively oriented curve enclosing a region $R$. Because you can choose any smooth path from $(0,0)$ to $(x,y)$, it is not hard to show that $\nabla \phi = (M,N)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .