we can regard f(x) as a function of the two variables x and y. The function f(a) will then be a function of a and y. We shall now regard this function f(x, y) as a function of the variable y, and we then conclude from the same theorem (B), Article 404, that 1 ƒ (a, y) = 2 [ƒ (z, B) [dq cos (gy – 9P). — 28). We have therefore, for the purpose of expressing any function whatever of the two variables x and y, the following equation We form in the same manner the equation which belongs to functions of three variables, namely, fdp cos (px − p) fdq cos (qy — 98) [dr cos (rz — ry)....(BBB), 81 each of the integrals being taken between the limits -∞ and +∞. It is evident that the same proposition extends to functions which include any number whatever of variables. It remains to show how this proportion is applicable to the discovery of the integrals of equations which contain more than two variables. 409. For example, the differential equation being d'v = d'v d'v ..(c), we wish to ascertain the value of v as a function of (x, y, t), such that; 1st, on supposing t = 0, v or f(x, y, t) becomes an arbitrary function (x, y) of x and y; 2nd, on making t=0 in the value dv of or f'(x, y, t), we find a second entirely arbitrary function. dt' ¥ (x, y). From the form of the differential equation (c) we can infer that the value of which satisfies this equation and the two preceding conditions is necessarily the general integral. To discover this integral, we first give to v the particular value or v = cos mt cos px cos qy. The substitution of v gives the condition m = √p2+q. It is no less evident that we may write =fda faß F (1, B) fdp cos (px − p) fdq cos (qy − q) cos t√p*+q*, whatever be the quantities p, q, a, B and F (x, B), which contain neither x, y, nor t. In fact this value of t is merely the sum of particular values. If we suppose t=0, v necessarily becomes (x, y). We have therefore Þ (x, y) = fdz fdß F (2, ß) fdp cos (px − px) fdq cos (qy —98). Thus the problem is reduced to determining F (a, ẞ), so that the result of the indicated integrations may be (x, y). Now, on comparing the last equation with equation (BB), we find 2 +00 dx + (x, y) = ( 21 ) * [ ̃^^dz [**d,3 4 (1, 6) [_dp cos (px − pz) -∞ 81 Hence the integral may be expressed thus: v=(-1)* fd. fd8$ (1,8) [dpcos (pr-p1) [dy cos (qy-98) cost.√ p2+qʻ. fdq We thus obtain a first part u of the integral; and, denoting by W the second part, which ought to contain the other arbitrary function (x, y), we have F. H. v = u + W, 27 and we must take W to be the integral fudt, changing only & into. In fact, u becomes equal to p(x, y), when t is made = 0; and at the same time W becomes nothing, since the integration, with respect to t, changes the cosine into a sine. dv Further, if we take the value of and make t = 0, the first dt Thus the equation part, which then contains a sine, becomes nothing, and the second part becomes equal to (x, y). v=u+W is the complete integral of the proposed equation. We could form in the same manner the integral of the equation 410. Let the proposed equation be + + =0; it is d'v d'v I'v required to express v as a function f(x, y, z), such that, 1st, f(x, y, 0) may be an arbitrary function (x, y); 2nd, that on d making z = 0 in the function f(x, y, z) we may find a second dz arbitrary function (x, y). It evidently follows, from the form of the differential equation, that the function thus determined will be the complete integral of the proposed equation. To discover this equation we may remark first that the equation is satisfied by writing v = cos px cos qy emz, the exponents and q being any numbers whatever, and the value of m being ± √ p2 + y2. We might then also write v = cos (px − p1) cos (qy − qB) (e2√μ2+y2 + e−z√p2+q*), or dq = faz faß F (1, B) fdp faq cos ( px − px) cos (qy − q8) (e2Vp2+q2 + e−z√p*+q*). If z be made equal to 0, we have, to determine F(x, B), the following condition dx Þ (x, y) = faz fdß F (1, ß) fdp fdq cos (px — pa) cos (qy — qB); and, on comparing with the equation (BB), we see that we have then, as the expression of the first part of the integral, u = 2 = ('-')" [dz [d8 4 (z, 8) [dp cos (pz-pa) fdq cos (qy-28). (e2√p2+q2 + e-z√p2+q2). The value of u reduces to p(x, y) when z=0, and the same du substitution makes the value of dx nothing. We might also integrate the value of u with respect to z, and give to the integral the following form in which is a new arbitrary function: W= ́= (21)* faz fdß 4 (x, 8) fdp da d3 (a, B) [dp cos (px — px) fdq cos (qy — qB) The value of W becomes nothing when z = 0, and the same. substitution makes the function d W equal to y(x, y). Hence the general integral of the proposed equation is v=u+ W. it is required to determine v as a function f(x, y, t), which satisfies the proposed equation (e) and the two following conditions: namely, 1st, the substitution t = 0 in f(x, y, t) must give an arbitrary function (x, y); 2nd, the same substitution in dtf(x, y, t) must give a second arbitrary function (x, y). d It evidently follows from the form of equation (e), and from the principles which we have explained above, that the function v, when determined so as to satisfy the preceding conditions, will be the complete integral of the proposed equation. To discover this function we write first, whence we derive v = cos px cos qy cos mt, or or v = cos (px − p2) cos (qy — qß) cos (p3t + q3t), v=fd1fdß F(a, B) fdp fdq cos (px — pa) cos (qy — q8) cos (p3t + q't). When we make t=0, we must have v = (x, y); which serves to determine the function F (a, B). If we compare this with the general equation (BB), we find that, when the integrals are taken 2 between infinite limits, the value of F (a, ß) is (-1)* (2, 3). We have therefore, as the expression of the first part u of the integral, น = (21)* fda fåß 4 (1, 3) fdp fdq cos (px — p2) cos (qy − 98) cos (p't + q't). Integrating the value of u with respect to t, the second arbitrary function being denoted by y, we shall find the other part W of the integral to be expressed thus: |