tity. Now, it is easy to see that there are an infinite number of arcs which, multiplied respectively by their tangents, give the hl same definite product whence it follows that we can find for n or p an infinite number of different values. 322. If we denote by e,, €, €,, &c. the infinite number of arcs which satisfy the definite equation e tane = hl we can take k' for n any one of these arcs divided by l. The same would be the case with the quantity p; we must then take m2 = n2 + p2. If we gave to n and p other values, we could satisfy the differential equation, but not the condition relative to the surface. We can then find in this manner an infinite number of particular values of v, and as the sum of any collection of these values still satisfies the equation, we can form a more general value of v. Take successively for n and p all the possible values, namely, ,,, &c. Denoting by ɑ, ɑ,, ɑ,, &c., b1, b1, b, &c., constant coefficients, the value of v may be expressed by the following equation: v = (a ̧e−x √m2+ñ12 cos n ̧y +α ̧ ̄ ±√n ̧2+m2 cos n2y + &c.) b1 cos n ̧z +(a,ex √n2+",2 cos n ̧y+ae-√13 cos n ̧y + &c.) b ̧ cos në + (a ̧e ̄x √112+n ̧3 cos n ̧y +αe̟ ̄* ́ cos ny+&c.) b ̧ cos n ̧z + &c. 323. If we now suppose the distance x nothing, every point of the section A must preserve a constant temperature. It is therefore necessary that, on making x = 0, the value of v should be always the same, whatever value we may give to y or to z; provided these values are included between 0 and l. Now, on making x=0, we find v = (a ̧ cos n ̧y+a, cos n ̧y+a, cos n ̧y+&c.) × (b, cos n ̧z + b, cos ny+b ̧ cos ny+&c.). 1 Denoting by 1 the constant temperature of the end A, assume the two equations 1 = a, cos n ̧y + a, cos n ̧y + a, cos n ̧y + &c, 1 = b1 cos n1y + b, cos ny + b, cos ny + &c. 2 It is sufficient then to determine the coefficients a,, a,, a,, &c., whose number is infinite, so that the second member of the equation may be always equal to unity. This problem has already been solved in the case where the numbers n,, n2, n,, &c. form the series of odd numbers (Chap. III., Sec. II., Art. 177). Here n1, n, n ̧, &c. are incommensurable quantities given by an equation of infinitely high degree. 39 324. Writing down the equation 1 = a1cos n ̧y+a, cos n ̧y + α, cos n1y + &c., 1. multiply the two members of the equation by cos ny dy, and take the integral from y=0 to y=l. We thus determine the first coefficient ... The remaining coefficients may be determined in a similar manner. In general, if we multiply the two members of the equation by cos vy, and integrate it, we have corresponding to a single term of the second member, represented by a cos ny, the integral 1 1 a fcos ny cos by dy or a fcos (n − v) y dy + ', a focos (n + v) ydy, vy 2 h k Now, every value of n satisfies the equation n tan nl =; the same is the case with v, we have therefore or n tan vl = v tan vl; n sin nl cos il-v sin vl cos nl = 0. Thus the foregoing integral, which reduces to a n2 — ‚2 (n sin nl cos vl — v cos nl sin vl), is nothing, except only in the case where n = v. Taking then the we see that if we have n=v, it is equal to the quantity It follows from this that if in the equation 1 = a ̧ cos n1y + α, cos n ̧y + a ̧ cos ny + &c. we wish to determine the coefficient of a term of the second member denoted by a cos ny, we must multiply the two members by cos ny dy, and integrate from y=0 to y=l. We have the resulting equation cients a,, a, a,, &c. may be determined; the same is the case with b1, b, b,, &c., which are respectively the same as the former coefficients. 325. It is easy now to form the general value of v. 1st, it d'v d'v satisfies the equation + + =0; 2nd, it satisfies the two dx dy2+ dz2 dv dy conditions k + hv=0, and k dv + hv=0; 3rd, it gives a constant value to v when we make x = 0, whatever else the values of y and z may be, included between 0 and 7; hence it is the complete solution of the proposed problem. We have thus arrived at the equation 1 sin nl cos ny sin nl cos ny = + 4 ~ 2n ̧l + sin 2n ̧l ̄ ̄ 2n ̧l+ sin 2n ̧l or denoting by e,, €, e, &c. the arcs nl, nl, nl, &c. an equation which holds for all values of y included between O and l, and consequently for all those which are included between O and -7, when x = = 0. Substituting the known values of ɑ1, b1, α, b1, ɑ, b1, &c. in the general value of v, we have the following equation, which contains the solution of the proposed problem, + + &c. &c.) sin nl cos n sin nl cos ny + &c.) + &c...... (E). The quantities denoted by n,, ng, n,, &c. are infinite in number, and respectively equal to the quantities,,, &c.; the arcs, 1, € €, &c., are the roots of the definite equation 326. The solution expressed by the foregoing equation E is the only solution which belongs to the problem; it represents the d'v d'v d2v general integral of the equation + =0, in which the dx** dy3* dz2 arbitrary functions have been determined from the given conditions. It is easy to see that there can be no different solution. In fact, let us denote by (x, y, z) the value of v derived from the equation (E), it is evident that if we gave to the solid initial temperatures expressed by (x, y, z), no change could happen in the system of temperatures, provided that the section at the origin were retained at the constant temperature 1: for the equation d'v d'v d'v O being satisfied, the instantaneous variation of dx31 dy3 dz2 + + = the temperature is necessarily nothing. The same would not be the case, if after having given to each point within the solid whose co-ordinates are x, y, z the initial temperature (x, y, z), we gave to all points of the section at the origin the temperature 0. We see clearly, and without calculation, that in the latter case the state of the solid would change continually, and that the original heat which it contains would be dissipated little by little into the air, and into the cold mass which maintains the end at the temperature 0. This result depends on the form of the function (x, y, z), which becomes nothing when x has an infinite value as the problem supposes. A similar effect would exist if the initial temperatures instead of being + (x, y, z) were — ↓ (x, y, z) at all the internal points of the prism; provided the section at the origin be maintained always at the temperature 0. In each case, the initial temperatures would continually approach the constant temperature of the medium, which is 0; and the final temperatures would all be nul. 327. These preliminaries arranged, consider the movement of heat in two prisms exactly equal to that which was the subject of the problem. For the first solid suppose the initial temperatures to be + (x, y, z), and that the section at origin A is maintained at the fixed temperature 1. For the second solid suppose the initial temperatures to be (x, y, z), and that at the origin A all points of the section are maintained at the temperature 0. It is evident that in the first prism the system of temperatures cannot change, and that in the second this system varies continually up to that at which all the temperatures become nul. If now we make the two different states coincide in the same solid, the movement of heat is effected freely, as if each system alone existed. In the initial state formed of the two united systems, each point of the solid has zero temperature, except the points of the section A, in accordance with the hypothesis. Now the temperatures of the second system change more and more, and vanish entirely, whilst those of the first remain unchanged. Hence after an infinite time, the permanent system of temperatures becomes that represented by equation E, or v = (x, y, z). It must be remarked that this result depends on the condition relative to the initial state; it occurs whenever the initial heat |