and substituting for ƒ (2) from the given primitive, we have and replacing 2 xy F′(xy) — F (xy) by 4(xy), we have 104.] Thus far the dependent variable has been explicitly involved if however all the variables are implicitly involved, a process similar to the preceding may be employed, if we consider one of the variables to be a function of, and thus dependent on the other two, and on this supposition calculate its partial variations due to the variations of the others. Some cases of this kind have already occurred in Art. 52 and 53, and some functions, the subjects of which are in particular combinations, have in them been eliminated. Ex. 1. Eliminate the function from x2 + y2+z2 = f(lx+my+nz). Let z be a variable dependent on x and y, which are two independent variables; .. 2x+2z (dz) = {1 And if the primitive is in the form F(x, y, z) = 0, the resulting partial differential equation is F that they are the functional factors in the derived-functions of F, according as the first or second subject varies. du dx du dy Ex. 3. u = f(ax2+by3 + cz1)+(cos lx + cos my + cos nz). (du) =2ax f'(ax2 + by3 + cz1) —l sin la d'(cos lx + cos my + cos nz), = 3 by3 f'(ax2+by3+cz1)—m sin my p'(cos lx + cos my + cos nz), = 4 cz3 f' (ax2+by3 + cz1) — n sin nz p′(cos lx + cos my + cos nz) ; therefore by Lagrange's method of cross-multiplication, see Preliminary Proposition II, we have du dx {4cmz3 sin my-3b ny2 sin nz}+ (dy) { 2 anx sin nz — 4c 123 sin læ} + (du) {3b ly3 sin lx — 2 amx sin my} = 0, an equation independent of the arbitrary functions, and there PRICE, VOL. I. Аа fore expressive of the properties of such functions, whatever is their specific form. Ex. 4. Let f' and : then z = f(x+ay) +$(x−ay). ', f" and 4" stand in their usual relation to ƒ and (dz) = ƒ'(x + ay) + 4′(x−ay), dx dz dy = a f'(x + ay) — a p′(x — ay), (d22) = ƒ'′′ (x+ay) +$′′ (x— ay), 105.] In general, for determining to what order of differentiation we must proceed to eliminate any number of arbitrary functions from an expression containing two variables in given combinations, for we will not enter upon the more general case, let the following considerations suffice. Suppose u = 0 to comprise m arbitrary functions of x and y, then it is plain that each successive differentiation introduces m other arbitrary functions, which are the derived of the given functions; so that by proceeding to the nth order of differentiation, we have (n+1)m different arbitrary functions: but as the original equation u = 0 gives one relation amongst these functions, so do give us other relations; and therefore by means of n differentiations we have the number of relations equal to In order that we may be able to eliminate all these, we must evidently have the number of relations greater than the number of unknown quantities, that is, that is, (n + 1)(n+2) > (n + 1) m; n+2 > 2m, n>2m-2; that is, n, which expresses the order of differentiation, must = 2m-1 at least; and we shall then have a sufficient number of equations to eliminate the arbitrary functions from. Thus, if the original equation involve but one arbitrary function, m = 1, and we need differentiate but once; if it involves two arbitrary functions, we must in the general case differentiate thrice, and so on. An example is subjoined in which three differentiations are required: u = f(x + y) + xy (xy). (du) = f'(x + y) + y + (x − y) + xy p′(x − y), dx (du) dy = f'(x+y)+x+(x − y) — xy p′(x-y); (du) - - (dy) = (y — x) p (x − y) + 2 ay 4′(x − y). dx - (dx dy) = (y—x) '(x—y) — 4 (x − y) d2u dx dy dy2 + 2y p′(x−y) + 2 xy p′′ (x − y), =(y-x) (x−y)+4(x-y) (du) – (džu) = 2 (x + y) 4' (x —y). dy2 And differentiating again, ;) = 2.4′(x − y) + 2(x + y) p′′ (x − y), d3u a) — (dy y ) = 2 p′(x − y) − 2 (x + y) $′′ (x − y), (dx2 dy) – (dx dy3) – (day) = 44′(x−y); 1 + d2u d2u $'(x − y) = d3u +( dys) = 2(x+y) dx2 dy2 d3u d3u SECTION 8.-Transformation of partial differential expressions into their equivalents in terms of new variables. 106.] In a previous section of the present chapter we have investigated the process by which an equation involving total differential expressions may be changed into its equivalent in terms of new variables, and we explained the modifications which such expressions underwent according as one or another of the variables is equicrescent. In the last section the elimination of arbitrary functions and other processes have given rise to equations containing partial differential quantities; and for purposes of simplification or for other reasons it is often necessary to transform them into their equivalents in terms of new variables. The process of this transformation I proceed to explain; but I shall take only simple cases, as they will be sufficient to unfold the principles; and the more complicated will be discussed where they arise in the second volume of our work. Let us first take the most simple case: that, viz., of a differ ential expression involving (du), (du), dx, dy ; dx, dy; in which accordingly the first partial derived-functions and differentials are involved, and wherein are none of a higher order. Now let it be observed that as a relation between (du), (du), the vari ables x and y, and certain constants is given by the given equation, so is there of necessity a relation of the form r(u, x, y) = 0, whether such a relation can be found or not. Let us suppose that the variables x and y are connected with two new variables r and 0 by means of equations of the form x = 1 (r, 0) y = 42 (r, 0) } ; (178) and that it is required to transform the given differential equation into its equivalent in terms of the new variables. If the primitive function is of the form u = f(x, y), then after the substitutions of the preceding values for x and y, we shall have |