Societies and Academies

Abstract

LONDON. Royal Society. January 13.—Sir Archibald Geikie, K.C.B., president, in the chair.—Sir Edward Thorpe and A. G. Francis: The atomic weight of strontium.—L. F. Richardson: The approximate arithmetical solution by finite differences of physical problems involving differential equations, with an application to the stresses in a masonry dam. In order to deal with irregular boundaries, analysis is replaced by arithmetic, continuous functions are represented by tables of numbers, differentials by central differences. Then problems fall into two classes. (A) The relation between the equation obtaining throughout the body and the boundary condition is such that the integral can be stepped out from a boundary. This class includes equations of all orders and degrees. It has been treated by arithmetical differences by Runge, W. F. Sheppard, Karl Heun, W. Kutta, and Richard Ganz. Examples of a specially simple method are given. (B) The integral must be determined with reference to the boundary as a whole, as in Dirichlet's problem. The method given has only been worked out for a limited group of linear equations, namely, for those in connection with which a function analogous to potential energy exists, which is a complete minimum when and only when the difference equations are satisfied. Under this condition the difference between the integral φu and a function φ1 of Ihe independents, having the correct boundary conditions hut otherwise arbitrary, can be expanded in the form φ1 - φu = σAkPk where the A1s … An are constants and P1 … Pn are “principal modes of oscillation” defined by D′Pk = λk2Pk where D′φu,=0 is the difference equation to he integrated and λ2 is a constant. Now we start with the (able of numbers φ1 and calculate D′φ1. Then as D′φu =o we have D′φ1,=D′(φ1 - φ1)=σAk.λk2Pk. Multiplying both sides by some number α-1 and subtracting from φ1 and altering the boundary numbers so that the boundary condition is still satisfied, we have a new table which may be called φ2; and φ2-φu=σAk(1-α1-1λk2. Repeating the process with α2 … αm we get: φm+1-φu=σAk(1-α2-1λk2 … (I-αm-1λk2)Pk. Now a function I exists such that SIPk2=I, SIPlPk=o where S denotes a summation throughout the region. Therefore: