POISSON'S SOLUTION. 225 where ƒ and F are given functions. We therefore require the solution of (4) subject to (7). 217. The solution of (4), which was first obtained by Poisson, may be effected as follows. The symbolic solution is p=cosh (atV)x+sinh (atV) +, are functions of x, y, z; we therefore obtain where Χ and from (7) and We must now show how the operations denoted by the symbolic operators may be performed. With the point O as a centre, describe a sphere of radius r, and let a, ẞ, y be the coordinates of any point P on this sphere relatively to O; also let us temporarily denote d/dx, d/dy, d/dz by λ, μ, ν. where the integration extends over the surface of the sphere. If through the point a, B, y, a plane be drawn, whose direction cosines are proportional to λ, μ, v, and if p be the perpendicular from O on to this plane, it is known that λa + μß+vy = (x2 + μ2 + v2)3 p = Vp. Also if be the angle, which the radius drawn from 0 to the point P makes with p, then p=r cos 0, dS = 2πr2 sin ede; whence the integral But if d be the elementary solid angle subtended by dS at O, d=rd; whence putting rat, and restoring the values of λ, μ, v, we obtain Now the operation denoted by the exponential factor on the right-hand side of the last equation, can be performed by means of the symbolic form of Taylor's theorem; we thus obtain If l, m, n be the direction cosines of OP, we shall have a = lat, B = mat, y = nat; and therefore the portion of upon the initial velocities is which depends From the form of (8) it is at once seen, that the portion of depending on the initial displacements may be obtained by changing Finto f, and differentiating with respect to t; we thus obtain This equation determines the value of 4 at time t, at any point O of the medium whose coordinates are x, y, z, in terms of the initial values of 4 and 4. The portions of the displacements which depend upon the dilatation are obtained by differentiating (9) with respect to x, y, z. 218. If the initial disturbance is confined to a portion T of the medium, the double integrals in (9) will vanish, unless the sphere whose centre is 0 and whose radius is at cuts a portion of the space T. Hence if O be outside T, and if r1, r, be respectively the least and greatest values of the radius vector of any element of that space, there will be no dilatation at O until at=r1. The dilatation at O will then commence, and will last during an interval (r, r1)/a, and will then cease for ever. 3 219. If fi, f2, f, denote the initial values of u, v, w1, which are the portions of the displacements which depend upon the distortion, and if F1, F2, F, denote the initial values of u, v, wa, then since u2, v2, w2 each satisfy equations of the same form as (4) with b written for a, it follows that the values of these quantities PROPAGATION OF AN ARBITRARY DISTURBANCE. 227 at time t are determined by equations of the same form as (9). It must also be recollected that fi, fa, fs and also F1, F2, F, satisfy (3). If therefore we write for brevity F(at) for F(x + lat, y + mat, z +nat), the complete value of u will be where F1, F2, F, satisfy (3); and since our object is to find the values of u, v, w at any subsequent time in terms of the values of the initial displacements and velocities, we must proceed to eliminate the F's and f's from (10). It will however be sufficient to perform this operation for those parts of u, v, w which depend upon the initial velocities, for when this is done, the portions depending upon the initial displacements can be obtained by differentiating with respect to t and changing u, v, wo into 221. Let a, ẞ, y denote the coordinates of any point P relatively to 0; let OP=r, and let l, m, n be the direction cosines of OP; then at points on the surface of the sphere rat, we have a = lat, &c.; also if (11), .(12). = where Vd/da2 + d2/dẞ2 + d2/dy, dv is an element of the normal to the sphere r = at, and the upper or lower sign is to be taken, according as the volume integrals extend throughout the space external or internal to the sphere. = Putting = x, y, and applying the theorem to the space. outside the sphere r = at, we obtain 1 - [fx as = [] = d x as + = r Putting x = 1, and applying the theorem to the space inside the sphere r = at, we obtain Eliminating dx/dr between (13) and (14), we obtain t ..(14). 471aff V2xdadßdy (r<at) - 4+ [[]} ▼3xdadßdy (r> at) Απα Now the function x, and consequently the functions ử, v。, w。, when they occur in a triple integral, are functions of the position. of the point whose coordinates are x+a, y +ß, z+y; whence d/dx = d/da, and accordingly we may write d/da, &c. for d/dx, &c. Hence substituting the value of x from (16) in (15), integrating by parts, and observing that the two surface integrals which appear in the integration cancel one another, we obtain Integrating the right-hand side again by parts, it follows that PROPAGATION OF AN ARBITRARY DISTURBANCE. 229 if u be the portion of u which depends upon the initial velocity of dilatation, and let (%) at denote the value of q, at a distance at from O, then the surface integral also the triple integral can easily be shown to be equal to SSS (u。 - 3lq) r3 dadßdy; whence we finally obtain for the portion of u depending upon the initial rate of dilatation + 4 [[[(ù. – 31q.) r−3 dadßdy, (r> at).............(17). 4π 222. We must now find the portion of u due to the initial velocities of rotation. Applying Green's theorem to the space outside the sphere r=bt, by writing F, for x in (13), we obtain Since 1 - [] } dF, as - [[] r~1 ▼3 F1dadßdy, (r>bt) ... (18). == by (3), we have r dr dß |