# Can all conservative vector fields from $\mathbb{R}^2 \to \mathbb{R}^2$ be represented as complex functions?

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?

• 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. Dec 13 '13 at 23:51
• Right, of course. This answers my question nicely. Dec 14 '13 at 0:20
• 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, Aug 2 '15 at 3:05
• @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 $f\,d\overline{z}$. Aug 2 '15 at 10:49
• Awesome explanation, @DanielFischer - thanks so much and have a great night :-) Aug 4 '15 at 7:07

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 $\nabla \phi = (M,N)$.