Given a 3D vector field $\vec{F}=(P,Q,R)$, I know that if $\boldsymbol\nabla \times \vec{F}=\mathbf{0}$, then it is a conservative field.
Moreover, I made an observation that if a vector field is conservative, then $\frac{\partial^2{P}}{\partial{y}\partial{z}} = \frac{\partial^2{Q}}{\partial{x}\partial{z}} = \frac{\partial^2{R}}{\partial{x}\partial{y}}$
Therefore, I am wondering is it true the other way around? i.e. $\frac{\partial^2{P}}{\partial{y}\partial{z}} = \frac{\partial^2{Q}}{\partial{x}\partial{z}} = \frac{\partial^2{R}}{\partial{x}\partial{y}} \implies$ vector field being conservative.
Thank you!
Unfortunately, this is not true in general. Consider the vortex vector field $$\mathbf V(x,y,z) = \bigg \langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, 0 \bigg \rangle.$$ We have that $\nabla \times \mathbf V = 0 \mathbf i + 0 \mathbf j + \big(\frac{-x^2 + y^2}{(x^2 + y^2)^2} - \frac{-x^2 + y^2}{(x^2 + y^2)^2} \big) \mathbf k = \mathbf 0;$ however, it is also true that $$\int_{\mathcal C} \mathbf V \cdot d \mathbf r = \int_0^{2 \pi} \bigg \langle \frac{-r \sin t}{r^2}, \frac{r \cos t}{r^2}, 0 \bigg \rangle \cdot \langle -r \sin t, r \cos t, 0 \rangle \, dt = 2 \pi,$$ where $\mathcal C$ is parametrized by $\mathbf r(t) = \langle r \cos t, r \sin t, 0 \rangle,$ i.e., $\mathcal C$ is a circle of radius $r > 0$ centered at the origin in the $xy$-plane. One can easily verify that your condition on the partials holds.