If the curl of some vector function = 0, Is it a must that this vector function is the gradient of some other scalar function?

Question

I know of course that If the curl of a vector function is equal to zero, then the vector function is the gradient of some other scalar function, but is this a must?

if so, please give mathematical proof.

Answer 1

It is rather sufficient to prove that the curl of a vector function $\mathbf{F}$ which is the gradient of a scalar-function $\phi$ is 0.

Let $\phi(x,y,z)$ be a scalar-function. Then its gradient will be $$\nabla\phi(x,y,z) = \frac{\partial \phi(x,y,z)}{\partial x}\hat{\mathbf{x}} + \frac{\partial \phi(x,y,z)}{\partial y}\hat{\mathbf{y}} + \frac{\partial \phi(x,y,z)}{\partial z}\hat{\mathbf{z}} \quad.$$ By assumption, $$\mathbf{F} = \nabla \phi$$. We have to show that $$\nabla \times \nabla \phi(x,y,z) = \mathbf{0}$$.

Since $\nabla \times \mathbf{h} = \text{vector}$, we can write the components of this vector by the rule of cross-product i.e. $$(\nabla \times \mathbf{h})_x = \nabla_y h_z - \nabla_z h_y$$. So, we can write $$\text{curl}\mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{ F_y}{\partial z}\right)\hat{\mathbf{x}} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{\mathbf{y}} +\left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}} $$. But we know $$F_x = \nabla\phi_x = \frac{\partial \phi(x,y,z)}{\partial x} \\\\\ F_y = \nabla\phi_y =\frac{\partial \phi(x,y,z)}{\partial y} \\\\\\\ F_z = \nabla \phi_z = \frac{\partial \phi(x,y,z)}{\partial z}$$. Substituting these in the definition of $\text{curl} \mathbf{F}$, we get $$\text{curl} \mathbf{F} = \nabla \times \nabla \phi = \left(\frac{\partial^2 \phi}{\partial y \partial z} - \frac{\partial^2 \phi}{\partial z \partial y}\right)\hat{\mathbf{x}} + \left(\frac{\partial^2 \phi}{\partial z \partial x} - \frac{\partial^2 \phi}{\partial x \partial z}\right)\hat{\mathbf{y}} + \left(\frac{\partial^2 \phi}{\partial x \partial y} - \frac{\partial^2 \phi}{\partial y \partial x}\right)\hat{\mathbf{z}} = \mathbf{0} \qquad .$$

Answer 2

If $X$ is a smooth vector field of the form $X=\nabla \phi$ then it follows immediately from the symmetry of partial derivatives that $\mathrm{curl}(X)=0$. The converse statement, however, that for any smooth vector field $X$ satisfying $\mathrm{curl}(X)=0$ there exists a $\phi$ such that $X=\nabla\phi$, which is called the Poincare Lemma, is trickier to establish, and depends in general on the topology of the underlying space.

Rather than get bogged down in a complete answer to this question, which would require advanced topology, I will give a proof of the Poincare Lemma for vector fields on $\mathbb{R}^3$, which extends easily to vector fields on $\mathbb{R}^n$ and to star-shaped subdomains of $\mathbb{R}^n$. I will then give a well-known example which shows that the Poincare Lemma does not hold on the domain $\mathbb{R}^2\setminus(0,0)$, to illustrate that the problem on general domains with "holes" is more complex.

Let X be a smooth vector field defined everywhere on $\mathbb{R}^3$. Note firstly that if there were to exist a $\phi$ such that $X=\nabla \phi$, then without loss of generality we could redefine $\phi$ by adding a constant so that $\phi(0,0,0)=0$, and we could then express $\phi$ in terms of $X$ as

\begin{align} \phi(x,y,z) &= \int_0^1 \frac{d}{ds} \phi(sx, sy, sz) ds \\ &= \int_0^1 \big( x \partial_1 \phi(sx, sy, sz) + y \partial_2 \phi(sx, sy, sz) + z \partial_3\phi(sx, sy, sz) \big) ds \\ &= \int_0^1 \big(x X_1(sx, sy, sz) + y X_2(sx, sy, sz) + zX_3(sx, sy, sz) \big) ds \end{align}

(think of this as evaluating the line integral $\int X \cdot dl$ along the ray from the origin to the point $(x,y.z)$). Motivated by this, we proceed by defining a function $\phi$ by $$ \phi(x,y,z) := \int_0^1 \big(x X_1(sx, sy, sz) + y X_2(sx, sy, sz) + zX_3(sx, sy, sz) \big) ds$$ and we claim that it satisfies $X=\nabla \phi$ as desired.

Indeed, we compute

\begin{align} \partial_1\phi (x,y,z) &= \int_0^1 \big(X_1(sx, sy, sz) + sx\partial_1 X_1(sx, sy, sz) + sy \partial_1X_2(sx, sy, sz) + sz \partial_1X_3(sx, sy, sz)\big) ds \\ &= \int_0^1 \frac{d}{ds} \big( s X_1(sx, sy, sz) \big) ds + \int_0^1 \Big( sy \big( \partial_1X_2(sx, sy, sz) - \partial_2X_1(sx, sy, sz)\big) \\ & \hskip150pt + sz \big( \partial_1 X_3(sx,sy,sz) - \partial_3X_1(sx, sy, sz) \big)\Big)ds \\ &= X_1(x,y,z) \end{align}

where the second integral on the second line vanishes because $\mathrm{curl}(X)=0$, and similarly one can compute $\partial_2 \phi=X_2$ and $\partial_3 \phi = X_3$, showing that $X=\nabla \phi$ as claimed.

Finally, to illustrate the importance of the underlying topology, consider the vector field $X(x,y) = \frac{1}{x^2 + y^2} (-y, x)$ on $\mathbb{R}^2\setminus(0,0)$. This satisfies $\mathrm{curl}(X)=0$ but there cannot exist any $\phi$ such that $X=\nabla\phi$. To see this, let $l(s) = (\cos s, \sin s)$ parameterise the unit circle and observe that $\oint X(l(s)) \cdot dl(s) = 2\pi$. But if we had $X=\nabla \phi$ for some $\phi$, then we would have $\oint X(l(s))\cdot dl(s) = \oint \nabla \phi(l(s)) \cdot dl(s) = \int_0^{2\pi} \frac{d}{ds} \big( \phi(l(s)) \big) ds = 0$, which is a contradiction.