Methods for solving boundary value problems in linear, second order, partial differential equations in two variables tend to be somewhat rigidly partitioned in some of the standard text-books. Problems for elliptic equations are sometimes solved by finding the fundamental solution which is defined as a solution with a given singularity at a certain point. Another approach is by way of Green's functions which are usually defined as solutions of the original homogeneous equations now made inhomogeneous by the introduction of adelta function on the right hand side. The Green's function coincides with the fundamental solution for elliptic equations but exhibits a totally different type of singularity for parabolic or hyperbolic equations. Boundary value problems for hyperbolic equations can often by solved by Riemann's method which depends on the existence of an auxiliary function called the Riemann or sometimes the Riemann-Green function. The main object of this paper is to show the close relationship between Riemann's method and the method of Green's functions. This not only serves to unify different methods of solution of boundary value problems but also provides an additional method of determining Riemann functions for given hyperbolic equations. Before establishing these relationships we shall survey the general approach to boundary value problems through the use of the Green's function.

1. The cardinal function is the interpolation function

which takes the values αr at the points α + rw. Its principal properties were discovered by Professor Whittaker, amongst others that

(A) When C (x) is analysed into periodic constituents by Fourier's integral-theorem, all constituents of period less than 2w are absent.

We identify the C*-envelopes of certain quotients of tensor algebras over C*-correspondences.

Let S be a subsemigroup of a semigroup Q. Then Q is a semigroup of left quotients of S if every element of Q can be written as a*b, where a lies in a group -class of Q and a* is the inverse of a in this group; in addition, we insist that every element of S satisfying a weak cancellation condition named square-cancellable lie in a subgroup of Q.

J. B. Fountain and M. Petrich gave an example of a semigroup having two non-isomorphic semigroups of left quotients. More positive results are available if we restrict the classes of semigroups from which the semigroups of left quotients may come. For example, a semigroup has at most one bisimple inverse ω-semigroup of left quotients. The crux of the matter is the restrictions to a semigroup S of Green's relations ℛ and ℒ in a semigroup of quotients of S. With this in mind we give necessary and sufficient conditions for two semigroups of left quotients of S to be isomorphic under an isomorphism fixing S pointwise.

The above result is then used to show that if R is a subring of rings Q1 and Q2 and the multiplicative subsemigroups of Q1 and Q2 are semigroups of left quotients of the multiplicative semigroup of R, then Ql and Q2 are isomorphic rings.

Let G be a torsion free group, F the free group generated by t. The equation r(t) = 1 is said to have a solution over G if there is a solution in some group that contains G. In this paper we generalize a result due to Klyachko who established the solution when the exponent sum of t is one.

The attempt to enumerate the possible distinct forms of knots of any order, though unsuccessful as yet, has led me to a number of curious results, some of which may perhaps be new. The general character of the methods employed will be obvious from an inspection of a few simple cases, and any one who has some practice in algebra may extend the results indefinitely.

§ 1. The roots of the cubic equation

are given by the formula

This solution is of little practical use when the roots of the cubic are all real and unequal, that is, when is negative (the-Irreducible Case of Cardan's Solution).

We prove two inequalities which relate the Lp modulus of continuity of n-th order, ωn(f,·)p, of an Hp function f with the p-th mean values of the n-th derivative f(n). Using these inequalities we extend classical results of Hardy and Littlewood [5], Gwiliam [4], Zygmund [13] and Taibleson [12] as well as a recent result of Oswald [6].

In this note we point out that certain algebraic-topological constructions are particular cases of one construction, namely double-negation sheafification. The principal cases we have in mind are concerned with booleanpowers, completions of boolean algebras, and maximal rings of quotients.We conjecture that several other constructions—particularly completion-type constructions—will turn out also to be examples of double-negation sheafification.

The modular group PSL(2, ℤ), which is isomorphic to a free product of a cyclicgroupof order 2 and a cyclic group of order 3, has many important homomorphic images. Inparticular, Macbeath [7] showed that PSL(2, q) is an image of the modular group if q ≠ 9. (Here, as usual, q is a prime power.) The extended modular group PGL(2, ℤ) contains PSL{2, ℤ) with index 2. It has a presentation

the subgroup PSL(2, ℤ) being generated by UV and VW.

The object of this note was to point out that in using the method of limits to find a geometrical maximum or minimum it is not correct to conduct all the reasoning at the final stage when the limit has been reached, and to call attention to the form of statement which lays stress on the fact that the reasoning should be based on the consideration of the quantities involved while they are yet finite. Examples were given from one or two well-known books for students where the fallacious method of proof is adopted.

One of the simplest three part boundary value problems is the electrostatic problem for the circular annulus and, at present, there seems to be no method available for obtaining the solution in a closed form. It has recently been shown by the author (1) and Cooke (2) that this problem can be reduced to the solution of a Fredholm integral equation of the second kind. The equation obtained in (1, 2) is fairly simple and is suitable for obtaining a numerical solution but, unfortunately, it cannot be solved iteratively to give a simple form of solution valid for small values of the ratio (inner radius/outer radius).