Charpit’s method to find the complete integral∗

Attila Máté
Brooklyn College of the City University of New York
December 14, 2011

Contents
1 Description of the method 1

2 An example 3

3 Integrability 4

4 Integrability of Charpit’s equations 6

1 Description of the method

Consider a first order partial differential equation with two independent variables

(1) F (x, y, u, p, q) = 0,

where p = ∂u/∂x and q = ∂u/∂y, and we assume that Fp2 + Fq2 6= 0. The equation for the
characteristic strips for this equation are
dx dy dp dp du
= Fp , = Fq , = −Fx − pFu , = −Fy − qFu , = pFp + qFq
ds ds ds ds ds
(see [3, p. 2]). Eliminating the parameter s from these equations, one often writes them as

dx dy du dp dq
(2) = = = = .
Fp Fq pFp + qFq −Fx − pFu −Fy − qFu

These equations are called Lagrange–Charpit equations. In interpreting these equations, it is con-
venient to allow zero denominators. For example, if Fp = 0, these equations require that dx = 0;
that is, the denominator being zero just means that the numerator is also zero. Assume that from
equations (1) and (2) one can derive a new equation

(3) φ(x, y, u, p, q) = a,
∗ Written for the course Mathematics 4211 at Brooklyn College of CUNY.

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where a is an arbitrary constant. Assume that the Jacobian determinant of F and φ with respect
to p and q does not vanish, i.e., that
 
∂(F, φ) Fp Fq 
(4) J = det = = Fp φq − φp Fq 6= 0;
∂(p, q) φp φq 
this assumption implies that equations (1) and (3) can be solved locally for p and q. Let this solution
be
(5) p = p(x, y, u, a) and q = q(x, y, u, a).
To simplify the notation, we consider a fixed value of a, and we write p(x, y, u, a) = p(x, y, u) and
q(x, y, u, a) = q(x, y, u) for now. Then, to get u = u(x, y), we need to solve the system
(6) ux = p(x, y, u), uy = q(x, y, u)
of partial differential equations (for each value of the parameter a).
Several questions arise at this point: does this really simplify the problem, and in any case,
are these equations solvable. The second question arises because we must have uxy = uyx under
reasonable assumptions.1 Differentiating the first equation in (6) with respect to y and taking second
equation into account, we have
uxy = py + pu uy = py + pu q.
Similarly, differentiating the second equation with respect to x, we obtain
uyx = qx + qu ux = qx + qu p.
The equality of these mixed derivatives2 leads to the requirement
(7) py + pu q = qx + qu p.
Therefore, to make the above approach viable, we need to do two things: we need to show that
condition (7) is satisfied, and then to show that this condition is sufficient to solve the system of
equations given in (6). Solving the above system is described as integrating the system, and condition
(7) is called integrability condition, and a system satisfying this condition is an integrable system.
The system of equations given in (6) can also be written as
(8) du = p(x, y, u) dx + q(x, y, u) dy.
Here du denotes the total differential of u = u(x, y), and since du = ux dx + uy dy, the equality
meaning the equality of the coefficients in the differential expression, this equation is just another
way of writing the above system of equations.
Assuming that condition (7) is satisfied, and a function u = u(x, y) satisfying (6) exists (locally,
i.e., on an on open set containing the point (x0 , y0 )), given the value of u at a point (x0 , y0 ), such a u
can be determined as follows. Let x = x(t), y = y(t) be differentiable functions such that x(0) = x0
and y(0) = y0 , and write U (t) = U (x(t), y(t)). Then it follows from (6) (or, equivalently, from(8))
with the aid of the chain rule that
(9) U ′ (t) = p(x(t), y(t), U (t)) x′ (t) + q(x(t), y(t), U (t)) y ′ (t)
holds. Solving this differential equation for U with the initial condition U (0) = u(x0 , y0 ), we can
determine u(x(t), y(t)) = U (t) (locally, i.e., for small values of t, where equation (9) is solvable).
Charpit’s method is described in [2, §10-10, pp. 242–244] and in [1].
1 For example, this is the case if u has continuous second derivatives.
2 See [4] for a discussion of various conditions ensuring the equality of mixed partial derivatives.

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2 An example
Before discussing the theoretical issues raised at the end of the previous section, we will describe an
example of how the method can be used. Consider the equation

(10) p2 u + q 2 − 4 = 0.

This equation is given in [2, Problem 7, p. 244]; the solution is given in [2, p. 287], but it is not shown
how to arrive at this solution. The Lagrange–Charpit equations (see (2)) for the above equation can
be written as
dx dy du dp dq
= = 2 = = .
2pu 2q 2p u + 2q 2 −p3 −p2 q
The fourth equation here can be written as dp/p = dq/q, i.e., log |p| = log |q| + C, that is, q = ap
with a = ±e−C . Substituting this into equation (10), we obtain p2 u + a2 p2 − 4 = 0, i.e.
2
p = ±√ .
u + a2
For the sake of simplicity, we will consider only the + sign in the ± here. With this equation,
together with the equation q = ap obtained just before, equation (8) can be written as
2 2a
du = √ dx + √ dy.
u+a 2 u + a2
Taking x = t and y = λt with λ an arbitrary constant (cf. (9)), we obtain
2 2a
du = √ dt + √ λ dt,
u+a 2 u + a2
i.e.,
(u + a2 )1/2 du = 2(1 + aλ) dt.
Integrating both sides, we obtain
2
(u + a2 )3/2 = 2(1 + aλ)t + C ′ = 2x + 2ay + C ′ ,
3
where C ′ is an arbitrary constant; the second equation follows since we have x = t and y = λt.
Solving this for u, we obtain
u = (3x + 3ay + b)2/3 − a2 ,
where a and b are arbitrary constants (b = (3/2)C ′ ). This is a complete integral of (10).3
3 This deals with the case x 6= 0, since the equations x = t, y = λt does not allow the possibility of x = 0 and y 6= 0.

It would have been easy to deal with this case as well, by taking x = ξt and y = ηt with ξ and η being arbitrary
constants; the case x = 0 can also be dealt with by making x → 0 in the above solution, since the solution of the
given equation is continuous. This shows that the solution we obtained is also valid in case x = 0.
If we take the − sign in ± above instead of the + sign, we arrive at the same solution, since in one of the steps we
squared both sides of an equation.

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3 Integrability
Condition (7) ensures that equations (6) or, equivalently, (8) are solvable. This is expressed by the
following
Theorem 1. Let (x0 , y0 , u0 ) ∈ R3 be a point, and assume that the functions p and q have continuous
partial derivatives in a neighborhood of (x0 , y0 , u0 ). Assume, further, that

(11) py + pu q = q x + q u p

in this neighborhood. Then there is a unique function u in a neighborhood of (x0 , y0 ) in R2 such that
u(x0 , y0 ) = u0 and

(12) du = p(x, y, u) dx + q(x, y, u) dy

in a neighborhood of (x0 , y0 ).
This theorem has a far-reaching generalization for systems of first-order partial differential equa-
tions, called Frobenius’s theorem, usually formulated in terms of vector fields or differential 1-forms,
that plays an important role in differential geometry.
Proof. Consider the differential equations

(13) ux (x, y0 ) = p(x, y0 , u(x, y0 )), uy (x, y) = q(x, y, u(x, y)), uy (x0 , y0 ) = u0 .

It is clear that any function u satisfying equation (12) must also satisfy these equations. These
latter equations can be thought of ordinary differential equations: the first equation together with
the initial condition given by the third equation is to describe u(x, x0 ), and the second equation
for a fixed value of x given by the value of u(x, y0 ) determined by the first and third equations as
initial condition is to determine u(x, y) for arbitrary (x, y). It follows from the theory of ordinary
differential equations that these equations uniquely determine u(x, y) in a neighborhood of (x0 , y0 );
we need to show that the function so determined satisfies the requirements of the theorem.

The Picard–Lindelöf theorem states the following: Let (x0 , y0 ) be a point and let f be a continuous
function on the set
S = {(x, y) ∈ R2 : |x − x0 | ≤ a, |y − y0 | ≤ b},
where a and b are positive real numbers. Assume that there is a real number L such that

|f (x, y1 ) − f (x, y2 )| ≤ L|y1 − y2 |

(such a condition is called a Lipschitz condition) for all points (x, y1 ) and (x, y2 ) in S. Let

M = max{|f (x, y)| : (x, y) ∈ S}.

Then the differential equation

dg(x)
= f (x, g(x))
dx
with initial condition g(x0 ) = y0 has a unique solution for g in the interval (x0 − h, x0 + h), where h =
min(a, b/M ).
This theorem ensures the uniqueness and the existence of a function u(x, y) satisfying equations (13)
in a neighborhood of the point (x0 , y0 ). The theorem is discussed in many standard books on ordinary
differential equations.

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 To complete the proof of the theorem, we need to show that the function u(x, y) so defined satisfies
the first equation in (6); the second equation there is satisfied, since that equation is required in
(13) used to define u(x, y). Integrating the equations in (13), we obtain
Z x Z y
u(x, y) = u(x0 , y0 ) + p(ξ, y0 , u(ξ, x0 )) dξ + q(x, η, u(x, η)) dη.
x0 y0

Differentiating this with respect to x, we obtain, by the Fundamental Theorem of Calculus and by
the interchangeability of differentiation and integration, that
Z y
ux (x, y) = p(x, y0 , u(x, y0 )) + (qx (x, η, u(x, η)) + ux (x, η)qu (x, η, u(x, η)) dη
y0
Z y
= p(x, y0 , u(x, y0 )) + (qx (x, η, u(x, η)) + p(x, η, u(x, η)qu (x, η, u(x, η)) dη
y0
Z y
+ (ux (x, η) − p(x, η, u(x, η))qu (x, η, u(x, η)) dη
y0
Z y
= p(x, y0 , u(x, y0 )) + (py (x, η, u(x, η)) + q(x, η, u(x, η))pu (x, η, u(x, η)) dη
y0
Z y
+ (ux (x, η) − p(x, η, u(x, η))qu (x, η, u(x, η)) dη,
y0

where the last equation was obtained by (11). Using the second equation in (13), we can further
write
Z y
ux (x, y) = p(x, y0 , u(x, y0 )) + (py (x, η, u(x, η)) + uy (x, η)pu (x, η, u(x, η)) dη
y0
Z y
+ (ux (x, η) − p(x, η, u(x, η))qu (x, η, u(x, η)) dη
y0
= p(x0 , y0 , u(x, y0 )) + p(x, y, u(x, y)) − p(x, y0 , u(x, y0 ))
Z y
+ (ux (x, η) − p(x, η, u(x, η))qu (x, η, u(x, η)) dη
y0
Z y
= p(x, y, u(x, y)) + (ux (x, η) − p(x, η, u(x, η))qu (x, η, u(x, η)) dη,
y0

where the second equation used the Newton-Leibniz formula to evaluate a definite integral. There-
fore, we have
Z y
(14) |ux (x, y) − p(x, y, u(x, y))| ≤ |ux (x, η) − p(x, η, u(x, η)||qu (x, η, u(x, η)| dη.
y0

Let U be a neighborhood of (x0 , y0 ) in which the function u(x, y) described in (13) is defined
and in which the functions ux (x, y), p(x, y, u(x, y)), and pu (x, y, u(x, y)) are bounded. Let M > 0
be such that
M ≥ |pu (x, y, u(x, y)| for (x, y) ∈ U.
Let δ with 0 < δ ≤ 1/(2M ) be such that

{(x, y)| : |x − x0 | < δ, |y − y0 | < δ} ⊂ U.

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Let

(15) A = sup{|ux (x, y) − p(x, y, u(x, y))| : |x − x0 | < δ, |y − y0 | < δ}.

Using (14), for (x, y) with |x − x0 | < δ and |y − y0 | < δ we obtain

|ux (x, y) − p(x, y, u(x, y))| ≤ AM |y − y0 | ≤ AM δ ≤ A/2.

This implies that

sup{|ux (x, y) − p(x, y, u(x, y))| : |x − x0 | < δ, |y − y0 | < δ} ≤ A/2,

contradicting (15) unless A = 0. This shows the first equation in (6) is satisfied in a neighborhood
of (x0 , y0 ), completing the proof.

4 Integrability of Charpit’s equations

In order to show that the integrability condition (7) to solve equations (6) (or, equivalently, (8)) is
satisfied, we need the following
Lemma 1. The function φ given in (3) satisfies the equation

(16) Fp φx + Fq φy + (pFp + qFq )φu − (Fx + pFu )φp − (Fy + qFu )φq = 0.

The above equation can also be written as

(17) (Fp φx − φp Fx ) + (Fq φy − φq Fy ) + p(Fp φu − φp Fu ) + q(Fq φu − φq Fu ) = 0,

and so it is clear that this equation is satisfied if the F and φ are the same functions. On the other
hand, equation (4) ensures that F and φ are not the same.
Proof. This equation means that the gradient vector (φx , φy , φu , φp , φp , φq ) is orthogonal to the
tangent vector (Fp , Fq , pFp + qFq , −Fx − pFu , −F y − qFu ) of the characteristic curves described by
equations (2).4 As equation (3) is a consequence of equations (2) and (1), it must be compatible
with characteristic curves.5 The gradient of the surface φ(x, y, u, p, q) = a must be orthogonal to
any curve contained in this surface; hence, a fortiori ,6 it must be orthogonal to the characteristic
curves contained in it.
Now, consider equations (1) and (3) with p and q as in (5). That is, suppressing the parameter
a, we have

F (x, y, u, p(x, y, u), q(x, y, u)) = 0 and φ(x, y, u, p(x, y, u), q(x, y, u)) = 0.

Differentiating these with respect to x we obtain

Fx + Fp px + Fq qx = 0 and φx + φp px + φq qx = 0.
4 The characteristic curves are in R5 . If one talks about these “curves” as objects in = R3 , it is better to call them

characteristic strips.
5 That is, if the surface φ(x, y, u, p, q) = a contains a point of a characteristic curve, it must contain the whole

characteristic curve (or, at least, a local piece of it, since these arguments only work locally). This is because
equations (2) describe the characteristic curves, and equation (1) must also be compatible with these characteristic
curves.
6 Even more so. The phrase “a fortiori” is often used in mathematical writing.

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Using the notation J = Fp φq − Fq φp (cf. (4)), multiplying the first equation by φp , the second one
by Fp , and subtracting the equations, we obtain

(18) Fx φp − φx Fp − Jqx = 0.

Similarly, differentiating the above equations with respect to y, we obtain

Fy + Fp py + Fq qy = 0 and φy + φp py + φq qy = 0.

Multiplying the first equation by φq and the second one by Fq , we obtain

(19) Fy φq − φy Fq + Jpy = 0.

Next, differentiating the above equations with respect to u, we obtain

Fu + Fp pu + Fq qu = 0 and φu + φp pu + φq qu = 0.

Multiplying the first equation by φp and the second one by Fp , we obtain

(20) Fu φp − φu Fp − Jqu = 0.

Similarly, multiplying the first equation by φq and the second one by Fq , we obtain

(21) Fu φq − φu Fq + Jpu = 0.

Adding equations (18), (19), p times (20), and q times (21) to the equation (17), we obtain

(−qx + py − pqu + qpu )J = 0.

Noting that J 6= 0 according to (4), equation (7) follows.

References
[1] M. Delgado. The Lagrange–Charpit method. SIAM Review, 39:298–304, 1997. Stable URL:
http://www.jstor.org/stable/2133111.

[2] L. E. Ford. Differential Equations, 2nd ed. McGraw Hill, New York, Toronto, London, 1962.

[3] Attila Máté. First order partial differential equations.

http://www.sci.brooklyn.cuny.edu/~mate/misc/partial_diffeqs/first_ord_partdiff_eq.pdf,
September 2011.

[4] Attila Máté. On the equality of mixed partial derivatives.

http://www.sci.brooklyn.cuny.edu/~mate/misc/mixedpartial.pdf, September 2011.