Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 3 / 11C (Vector Calculus)

Use the Einstein summation convention and the contraction of two Levi-Civita symbols. For the first cross product term,The last term cancels $(F\cdot \nabla )G_i$. Interchanging $F,G$ gives the analogous cancellation with $(G\cdot \nabla )F_i$. The surviving terms are $F_j{\partial }_i{G}_j+G_j{\partial }_i{F}_j={\partial }_i(F_j{G}_j)$ by the product rule. This proves the dot-product gradient identityThe proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.

Assume the continuously differentiable irrotational vector field is defined on all of ${\mathbb{R}}^3$, as in this vector calculus setting. A constructive vector-field potential with the requested negative-gradient convention isSince zero curl means ${\partial }_i{E}_j={\partial }_j{E}_i$, differentiation under the integral givesThus $E=-\nabla \varphi$. This is the Poincare lemma in its straight-line form, and works on any star-shaped set about the integration base point. Zero curl on a domain with holes would not by itself guarantee a global vector-field potential; the full-space hypothesis matters.

For the given field, the second component requires ${\varphi }_y=y{e}^{-x^{2}z}$. Integration in $y$ gives $\varphi =\frac{1}{2}{y}^2{e}^{-x^{2}z}+C(x,z)$. Its $x$ derivative is $-x{y}^{2}z{e}^{-x^{2}z}+C_x$, and its $z$ derivative is $-\frac{1}{2}{x}^2{y}^2{e}^{-x^{2}z}+C_z$. Comparison with the other two components forces $C_x=C_z=0$. HenceIts negative gradient is exactly the given field. Because this vector-field potential is smooth everywhere, commutation of mixed partial derivatives gives $\nabla \times E=-\nabla \times \nabla \varphi =0$, verifying irrotationality. All such vector-field potentials differ by a constant on the connected domain.