1916] HORIZONTAL LAYER OF FLUID 437 which with (5) and the initial and boundary conditions suffice for the solution of the problem. The boundary conditions are that w = 0, 0 = 0, when z = 0 or £ We now assume in the usual manner that the small quantities are proportional to so that (8), (5), (9) become nu =------, nv =--------, nw =------,—\-<y0, .........(ID p p p dz ' ilu + imv + dwfdz — 0, ........................(12) from which by elimination of u, v, is, we derive Tt fl^QD Iv Cv lAJ « /f >l \ TT-—--r^ = nw-y0................;........(14) I- + m2 dz" Having regard to the boundary conditions to be satisfied by w and 6, we now assume that these quantities are proportional to sinsz, where s = q^r/^ and q is an integer. Hence /3w + {n + K(l2 + m* + s2)}0=Q, ..................(15) n (I2 -I- m2 + s2) w - 7 (Z2 -I- m2) 6 = 0, ..................(16) and the equation determining n is the quadratic nz (I* + m? + s2) + UK (I2 + m2 + s2)2 + (By (I2 + m-) = 0.......(17) When K = 0, there is no conduction, so that each element of the fluid retains its temperature and density. If /3 be positive, the equilibrium is stable, and indicating vibrations about the condition of equilibrium. If, on the other hand, /3 be negative, say — /Q', „ _ ± V(ffy(ft + m2)} /1Q\ W-------TUT',-----T~i—5T~ .........................iiy' V{t2 + m? + s2} When n has the positive value, the corresponding disturbance increases exponentially with the time. For a given value of I2 + m2, the numerical values of n diminish without limit as s increases—that is, the more subdivisions there are along z. The greatest value corresponds with q = 1 or s = 7r/£. On the other hand, if s be given, j n \ increases from zero as I2 -f in2 increases from zero (great wavelengths along x and y} up to a finite limit when I2 + m2 is large (small wavelengths along x and y). This case of no conductivity falls within the scope dv 1 diff dw 1 d™ed force,