CHAPTER XV.

PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND ORDER.

1. THE general form of a partial differential equation of the second order is

F (x, y, z, p, q, r, s, t) = 0...............................

(1),

d'z t=

dx dy' dy3

It is only in particular cases that the equation admits of integration, and the most important is that in which the differential coefficients of the second order present themselves only in the first degree; the equation thus assuming the form

Rr + Ss + Tt= V .........

·(2), in which R, S, T and V are functions of x, y, z, Р and This equation we propose to consider. The most usual method of solution, due to Monge, consists in a certain procedure for discovering either one or two first integrals of the form

u and v being determinate functions of x, y, z, p, q, and ƒ an arbitrary functional symbol. From these first integrals, singly or in combination, the second integral involving two arbitrary functions is obtained by a subsequent integration.

An important remark must here be made. Monge's method involves the assumption that the equation (2) admits of a first integral of the form (3). Now this is not always the case. There exist primitive equations, involving two arbitrary functions, from which by proceeding to a second differentiation both functions may be eliminated and an equation of the form (2) obtained, but from which it is impossible to eliminate

one function only so as to lead to an intermediate equation of the form (3). Especially this happens if the primitive involve an arbitrary function and its derived function together. Thus the primitive

z = $(y + x) + ¥ (y − x) − x {p' (y + x) — ↓' (y− x)}...(4),

leads to the partial differential equation of the second order

but not through an intermediate equation of the form (3).

(5),

It is necessary therefore not only to explain Monge's method, but also to give some account of methods to be adopted when it fails.

2. It is not only not true that the equation (2) has necessarily a first integral of the form (3), but neither is the converse proposition true. We propose therefore, 1st, to inquire under what conditions an equation of the first order of the form (3) does lead to an equation of the second order of the form (2); 2ndly, to establish upon the results of this direct inquiry the inverse method of solution. And this procedure, though somewhat longer than that usually followed, is more simple, because exact and thorough.

PROP. 1. A partial differential equation of the first order of the form u= f (v)_can_only_lead to a partial differential equation of the second order of the form

when u and v are so related as to satisfy identically the condition

For, differentiating the equation u=f(v) with respect to x,

and observing that

dz dp
=r,
=P,
dx

dx

dq

=0, we have dx

In like manner differentiating u=f(v) with respect to y,

we have

(du

dx

+P dz

+r

dp

Eliminating f' (v) there results

du du du (dv dv dv dv

+ s

+9 + s +t
dz dp dq

dq

dy

On reduction it will be found that the only terms involving r, s, and t in a degree higher than the first will be those which contain rt and s2. The equation will in fact assume the form

Rr+Ss+ Tt+U (rt — s2) = V.......................... (9),

efficients it is unnecessary to examine.

Now this equation assumes the form (6) when the condition (7) is satisfied-and then only.

3. The proposition might also be proved in the following manner. Since u=f(v) we have du=f' (v) dv, an equation which, since f(v) is arbitrary, involves the two equations du= 0, dv=0. Hence

But dz = pdx+qdy, dp = rdx + sdy, dq=sdx+tdy. Whence on substitution

Whence eliminating dx and dy, we have the same result as before.

4. A consequence, which, though not affecting the present inquiry, is important, may here be noted. It is that it would be in vain to seek a first integral of the form u =ƒ (v) for any partial differential equation of the second order which is not of the form (9).

PROP. 2. To deduce when possible a first integral, of the form u =ƒ(v), for the partial differential equation (6).

By the last proposition u and v must satisfy the condition (7), which is expressible in the form,

Hence, if we represent each member of this equation by m,

we have

Substituting these values in (10), we have

and we are to remember that this system, being equivalent

to du = 0, dv=0 modified by the condition (7), can only have an integral system of the form,

a and b being arbitrary constants, and u and v connected by the condition (11).

Making dz = pdx + qdy in (13), we have

From these and from the equations

dp=rdx + sdy, dq=sdx+tdy

(16),

if we eliminate the differentials dx, dy, dp, dq, we shall necessarily obtain a result of the form (6). For in thus doing we only repeat the process of Art. 3, with the added condition (7).

To effect this elimination, we have from (16),

dp +mdq = (r + ms) dx + (s +mt) dy;

or, rdx+s (dy + mdx) + tmdy = dp + mdq ...... (17).

Now the system (15) enables us to determine the ratios of dy and dp + mdq to do, and these ratios substituted in (17), reduce it to the form (6).

But in order that it may be, not only of the form (6), but actually equivalent to (6), it is necessary and sufficient that we have

dx dy + mdx_mdy _ dp+mdq

......(18).

This system of relations among the differentials must thus include the equations (15). The same system (18), together with the equation dz = pdx+qdy, must therefore include the system (13). It must therefore in its final integral system include the equations u=a, v=b with their implied condition.