Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/52

36 If the region for which $$\nabla \times \omega = 0$$ is unlimited, these functions will be angle-valued. If the region is limited, but acyclic, the functions will still be single-valued and satisfy the condition $$\nabla u = \omega$$ within the same region. If the region is cyclic, we may determine functions satisfying the condition $$\nabla u = \omega$$ within the region, but they will not necessarily be single-valued. 68. If $$\omega$$ is any vector function of position in space, $$\nabla. \nabla \times \omega = 0.$$ This may be deduced directly from the definitions of No. 54.

The converse of this proposition will be proved hereafter.

69. If $$u$$ is any scalar function of position in space, we have by Nos. 52 and 54

70. Def.—If $$\omega$$ is any vector function of position in space, we may define $$\nabla. \nabla \omega$$ by the equation the expression $$\nabla. \nabla$$ being regarded, for the present at least, as a single operator when appKed to a vector. (It will be remembered that no meaning has been attributed to $$\nabla$$ before a vector.) It should be noticed that if that is, the operator $$\nabla. \nabla$$ applied to a vector affects separately its scalar components.

71. From the above definition with those of Nos, 52 and 54 we may easily obtain The effect of the operator $$\nabla. \nabla$$ is therefore independent of the directions of the axes used in its definition.

72. The expression $$-\tfrac{1}{6}a^2 \nabla. \nabla u$$, where $$a$$ is any infinitesimal scalar, evidently represents the excess of the value of the scalar function $$u$$