Hence, u considered as a function of x, y, z, p, q satisfies the two partial differential equations (4), (5), both which are of the first order and second degree.

As u and venter symmetrically into the system (a), (b), &c., v will also satisfy two partial differential equations of the same form, viz. the equations

μ

Further, these two systems of equations constitute the complete system of equations resulting from the elimination of u from the five equations (a), (b), (c), &c.; for in their determination, no factor involving either the differential coefficients of u and v, or the quantities R, S, T, &c. has been rejected. directly or indirectly.

I am not aware that the above results of elimination have been noticed before.

3. PROP. II.

The system of partial differential equations

above obtained for the determination of u, viz.

admits of resolution into four systems, each consisting of two linear partial differential equations of the first order. Of these systems two only are relevant to the solution of the problem.

For, multiplying the second by an indeterminate quantity A, and adding the result to the first, we have

Now let us see if it is possible to determine λ so as to make the first member of the equation resolvable into linear factors. We cannot say à priori that such resolution is possible as we should be able to do if that member were homogeneous and of the second degree with respect to three instead of with respect to the four subject variables

dp dq

in the first member of (8) while those of (d)

dx

dy

appear, we are led to assume as the proposed equivalent of

that member an expression of the form

Multiplying the factors of this expression together and then equating the coefficients with those of the first member of (8) we have

values which will be found to satisfy (e) and (f) also, and which reduce (a) to the form

V2x2- SVX+RT- UV=0.

Supposing thus determined, the equation (8) becomes

T

{ B (du) + Vx (du) + v du} {a (de) + 1 (dy) + du) = 0.

R

}{

dp dx V

The result is a little simplified if we retain m in place of λ.

We thus find as the resolved form of the given equation

m being determined by the quadratic

m2- Sm+RT— UV = 0.

If m,, m, be the values of m thus found, we have

{R

{1 (du) + m, (du) + v du} {m, (d) + I (dy) + v dn) = 0,

dy

T

du

dul

dq

and these two equations are manifestly together equal to the system (7).

Now these equations can only be simultaneously satisfied by equating to 0, one factor in the first member of each; and the different combinations which are thus possible give rise to four binary systems of linear equations. Let us examine these systems separately.

If we simultaneously equate to 0 the two first factors of the left-hand members of the last two equations, we have the systems

a system which, when m, and m, are different, is reducible to

the system

It is clear that this cannot lead to a value of u satisfying the given differential equation (1),

because it takes no account of the forms of S, U, and T. Indeed if we actually eliminate

(derived from the assumed first integral u=ƒ(v) by making f(v) = c, and differentiating the result first with respect to x, then with respect to y), we find as the result

Again, if we equate to 0 the two last factors of the righthand members of (10), we have

which, if m, and m, are different, reduce to

0,

0,

And it is evident that neither are these equations consistent with the given equation (1), because they take no account of S, U, and R. The equation of the second degree to which they actually lead is

Vr + T (s2 — rt) = 0.

There remain then the two systems formed by combining the first factor of each one of the first members with the second factor of the other, viz.