Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/59

Rh The existence of the minimum requires that while $$\delta \omega$$ is subject to the limitation that  and that the normal component of 8m at the bounding surface vanishes. To prove that the line-integral of $$u \omega$$ vanishes for any closed curve within the space, let us imagine the curve to be surrounded by an infinitely slender tube of normal section $$dz,$$ which may be either constant or variable. We may satisfy the equation $$\nabla. \delta \omega = 0$$ by making $$\delta \omega = 0$$ outside of the tube, and $$\delta \omega\,dz = \delta a \frac{d\rho}{ds}$$ within it, $$\delta a$$ denoting an arbitrary infinitesimal constant, $$\rho$$ the position-vector, and $$ds$$ an element of the length of the tube or closed curve. We have then whence We may express this result by saying that $$u \omega$$ is the derivative of a single-valued scalar function of position in space. (See No. 67.)

If for certain parts of the surface the normal component of $$\omega$$ is not given for each point, but only the surface-integral of $$\omega$$ for each such part, then the above reasoning will apply not only to closed curves, but also to curves commencing and ending in such a part of the surface. The primitive of $$u \omega$$ will then have a constant value in each such part. If the space extends to infinity and there is no special condition respecting the value of $$\omega$$ at infinite distances, the primitive of $$u \omega$$ will have a constant value at infinite distances within the space or within each separate continuous part of it.

If we except those cases in which the problem has no definite meaning because the data are such that the integral $$\int u \omega. \omega \, dv$$ must be infinite, it is evident that a minimum must always exist, and (on account of the quadratic form of the integral) that it is unique. That the conditions just found are sufficient to insure this minimum, is evident from the consideration that any allowable values of $$\delta \omega$$ may be made up of such values as we have supposed. Therefore, there will be one and only one vector function of position in space which satisfies these conditions together with those enumerated at the banning of this number.

b. In the second place, let the vector $$\omega$$ be subject to the conditions that $$\nabla \times \omega$$ is given throughout the space, and that the tangential component of $$\omega$$ is given at the bounding surface. The solution is that