MA 201: Partial Differential Equations Lecture - 5 · Charpit’s method It is a general method for...

Compatible Systems and Charpit’s MethodCharpit’s Method

Some Special Types of First-Order PDEs

MA 201: Partial Differential EquationsLecture - 5

IIT Guwahati MA201(2016):PDE



Definition (Compatible systems of first-order PDEs)A system of two first-order PDEs

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

andg(x , y , u, p, q) = 0 (2)

are said to be compatible if they have a common solution.

TheoremEquations (1) and (2) are compatible on a domain D if

(i) J = ∂(f ,g)∂(p,q) =

∣

∣

∣

∣

fp fqgp gq

∣

∣

∣

∣

6= 0 on D.

(ii) p and q can be explicitly solved from (1) and (2) as p = φ(x , y , u)and q = ψ(x , y , u). Further, the equation

du = φ(x , y , u)dx + ψ(x , y , u)dy

is integrable.




TheoremA necessary and sufficient condition for the integrability of the equationdu = φ(x , y , u)dx + ψ(x , y , u)dy is

[f , g ] ≡ ∂(f , g)

∂(x , p)+∂(f , g)

∂(y , q)+ p

∂(f , g)

∂(u, p)+ q

∂(f , g)

∂(u, q)= 0. (3)

In other words, equations (1) and (2) are compatible iff (3) holds.

ExampleShow that the equations

xp − yq = 0, xup + yuq = 2xy

are compatible and solve them.

Solution. Take f ≡ xp − yq = 0, g ≡ u(xp + yq)− 2xy = 0. Then

fx = p, fy = −q, fu = 0, fp = x , fq = −y ,

gx = up − 2y , gy = uq − 2x , gu = xp + yq, gp = ux , gq = uy .




Compute

J ≡ ∂(f , g)

∂(p, q)=

∣

∣

∣

∣

fp fqgp gq

∣

∣

∣

∣

=

∣

∣

∣

∣

x −yux uy

∣

∣

∣

∣

= uxy + uxy = 2uxy 6= 0

for x 6= 0, y 6= 0, u 6= 0. Further,

∂(f , g)

∂(x , p)=

∣

∣

∣

∣

fx fpgx gp

∣

∣

∣

∣

=

∣

∣

∣

∣

p xup − 2y ux

∣

∣

∣

∣

= uxp − x(up − 2y) = 2xy

∂(f , g)

∂(u, p)=

∣

∣

∣

∣

fu fpgu gp

∣

∣

∣

∣

=

∣

∣

∣

∣

0 xxp + yq ux

∣

∣

∣

∣

= 0− x(xp + yq) = −x2p − xyq

∂(f , g)

∂(y , q)=

∣

∣

∣

∣

fy fqgy gq

∣

∣

∣

∣

=

∣

∣

∣

∣

−q −yuq − 2x uy

∣

∣

∣

∣

= −quy + y(uq − 2x) = −2xy

∂(f , g)

∂(u, q)=

∣

∣

∣

∣

fu fqgu gq

∣

∣

∣

∣

=

∣

∣

∣

∣

0 −yxp + yq zy

∣

∣

∣

∣

= y(xp + yq) = y2q + xyp.




It is an easy exercise to verify that

[f , g ] ≡ ∂(f , g)

∂(x , p)+∂(f , g)

∂(y , q)+ p

∂(f , g)

∂(u, p)+ q

∂(f , g)

∂(u, q)

= 2xy − x2p2 − xypq − 2xy + y2q2 + xypq

= y2q2 − x2p2

= 0.

So the equations are compatible.• Next step is to determine p and q from the two equationsxp− yq = 0, u(xp+ yq) = 2xy . Using these two equations, we have

uxp + uyq − 2xy = 0 =⇒ xp + yq =2xy

u

=⇒ 2xp =2xy

u=⇒ p =

y

u= φ(x , y , u).

and

xp − yq = 0 =⇒ q =xp

y=

xy

yu

=⇒ q =x

u= ψ(x , y , u).




Substituting p and q in du = pdx + qdy , we get

udu = ydx + xdy = d(xy),

and hence integrating, we obtain

u2 = 2xy + k ,

where k is a constant.NOTE:

For the compatibility of f (x , y , u, p, q) = 0 and g(x , y , u, p, q) = 0, it isnot necessary that every solution of f (x , y , u, p, q) = 0 be a solution ofg(x , y , u, p, q) = 0 or vice-versa. For instance, the equations

f ≡ xp − yq − x = 0 (4)

g ≡ x2p + q − xu = 0 (5)

are compatible. They have common solutions u = x + c(1 + xy), where cis an arbitrary constant. Note that u = x(y + 1) is a solution of (4) butnot of (5).




Charpit’s method

It is a general method for finding the general solution of a nonlinear PDEof first-order of the form

f (x , y , u, p, q) = 0. (6)

Basic Idea: To introduce another partial differential equation of the firstorder

g(x , y , u, p, q, a) = 0 (7)

which contains an arbitrary constant a and is such that(i) equations (6) and (7) can be solved for p and q to obtain

p = p(x , y , u, a), q = q(x , y , u, a).

(ii) the equation

du = p(x , y , u, a)dx + q(x , y , u, a)dy (8)

is integrable.




When such a function g is found, the solution

F (x , y , u, a, b) = 0

of (8) containing two arbitrary constants a and b will be the solution of(6).The compatibility of equations (6) and (7) yields

[f , g ] ≡ ∂(f , g)

∂(x , p)+∂(f , g)

∂(y , q)+ p

∂(f , g)

∂(u, p)+ q

∂(f , g)

∂(u, q)= 0.

Expanding it, we are led to the following linear PDE in g(x , y , u, p, q):

fp∂g

∂x+ fq

∂g

∂y+ (pfp + qfq)

∂g

∂u− (fx + pfu)

∂g

∂p− (fy + qfu)

∂g

∂q= 0. (9)




Now solve (9) to determine g by finding the integrals of the followingauxiliary equations:

dx

fp=

dy

fq=

du

pfp + qfq=

dp

−(fx + pfu)=

dq

−(fy + qfu)(10)

These equations are known as Charpit’s equations. Once an integralg(x , y , u, p, q, a) of this kind has been found, the problem reduces tosolving for p and q, and then integrating equation (8).Remarks.

• For finding integrals, all of Charpit’s equations (10) need not beused.

• p or q must occur in the solution obtained from (10).




ExampleFind a general solution of

p2x + q2y = u. (11)

Solution. To find a general solution, we proceed as follows:

• Step 1: (Computing fx , fy , fu, fp , fq).Set f ≡ p2x + q2y − u = 0. Then

fx = p2, fy = q2, fu = −1, fp = 2px , fq = 2qy ,

and hence,

pfp + qfq = 2p2x + 2q2y , −(fx + pfu) = −p2 + p,

−(fy + qfu) = −q2 + q.




• Step 2: (Writing Charpit’s equations and finding a solutiong(x , y , u, p, q, a)).The Charpit’s equations (or auxiliary) equations are:

dx

fp=

dy

fq=

du

pfp + qfq=

dp

−(fx + pfu)=

dq

−(fy + qfu)

=⇒ dx

2px=

dy

2qy=

du

2(p2x + q2y)=

dp

−p2 + p=

dq

−q2 + q

From which it follows that

p2dx + 2pxdp

2p3x + 2p2x − 2p3x=

q2dy + 2qydq

2q3y + 2q2y − 2q3y

=⇒ p2dx + 2pxdp

p2x=

q2dy + 2qydq

q2y

On integrating, we obtain

log(p2x) = log(q2y) + log a

=⇒ p2x = aq2y , (12)

where a is an arbitrary constant.




• Step 3: (Solving for p and q).Using (11) and (12), we find that

p2x + q2y = u, p2x = aq2y

=⇒ (aq2y) + q2y = u =⇒ q2y(1 + a) = u

=⇒ q2 =u

(1 + a)y=⇒ q =

[

u

(1 + a)y

]1/2

.

and

p2 = aq2y

x= a

u

(1 + a)y

y

x=

au

(1 + a)x

=⇒ p =

[

au

(1 + a)x

]1/2

.




• Step 4: (Writing du = p(x , y , u, a)dx + q(x , y , u, a)dy and findingits solution).Writing

du =

[

au

(1 + a)x

]1/2

dx +

[

u

(1 + a)y

]1/2

dy

=⇒(

1 + a

u

)1/2

du =( a

x

)1/2

dx +

(

1

y

)1/2

dy .

Integrate to have

[(1 + a)u]1/2

= (ax)1/2 + (y)1/2 + b

the general solution of equation (11).





• Equations involving only p and qIf the equation is of the form

f (p, q) = 0, (13)

then Charpit’s equations take the form

dx

fp=

dy

fq=

du

pfp + qfq=

dp

0=

dq

0

the last two are actually equivalent todp

dt= 0,

dq

dt= 0 and hence

an immediate solution is given by p = a, where a is an arbitraryconstant. Substituting p = a in (13), we obtain a relation

q = Q(a).

Then, integrating the expression

du = adx + Q(a)dy

we obtainu = ax + Q(a)y + b, (14)

where b is a constant. Thus, (14) is a general solution of (13).IIT Guwahati MA201(2016):PDE



Note: Instead of taking dpdt

= 0, we can take dqdt

= 0 ⇒ q = a. In someproblems, taking dq = 0 the amount of computation involved may bereduced considerably.

ExampleFind a general solution of the equation pq = 1.

Solution. If p = a then pq = 1 ⇒ q = 1a. In this case, Q(a) = 1/a.

From (14), we obtain a general solution as

u = ax +y

a+ b

=⇒ a2x + y − au = b,

where a and b are arbitrary constants.




• Equations not involving the independent variables

For equation of the type

f (u, p, q) = 0, (15)

Charpit’s equation becomes

dx

fp=

dy

fq=

du

pfp + qfq=

dp

−pfu=

dq

−qfu.

From the last two relations, we have

dp

−pfu=

dq

−qfu=⇒ dp

p=

dq

q

=⇒ p = aq, (16)

where a is an arbitrary constant. Solving (15) and (16) for p and q,we obtain

q = Q(a, u) =⇒ p = aQ(a, u).




Now

du = pdx + qdy

=⇒ du = aQ(a, u)dx + Q(a, u)dy

=⇒ du = Q(a, u) [adx + dy ] .

It gives general solution as∫

du

Q(a, u)= ax + y + b, (17)

where b is an arbitrary constant.

ExampleFind a general solution of the PDE p2u2 + q2 = 1.

Solution. Putting p = aq in the given PDE, we obtain

a2q2u2 + q2 = 1

=⇒ q2(1 + a2u2) = 1

=⇒ q = (1 + a2u2)−1/2.




Now,

p2 = (1− q2)/u2 =

(

1− 1

(1 + a2u2)

)(

1

u2

)

=⇒ p2 =a2

1 + a2u2

=⇒ p = a(1 + a2u2)−1/2.

Substituting p and q in du = pdx + qdy , we obtain

du = a(1 + a2u2)−1/2dx + (1 + a2u2)−1/2dy

=⇒ (1 + a2u2)1/2du = adx + dy

=⇒ 1

2a

{

au(1 + a2u2)1/2 − log[au + (1 + a2u2)1/2]}

= ax + y + b,

which is the general solution of the given PDE.




• Separable equations

A first-order PDE is separable if it can be written in the form

f (x , p) = g(y , q). (18)

For this type of equation, Charpit’s equations become

dx

fp=

dy

−gq=

du

pfp − qgq=

dp

−fx=

dq

gy.

From the last two relations, we obtain an ODE

dp

−fx=

dx

fp=⇒ dp

dx+

fxfp

= 0 (19)

which may be solved to yield p as a function of x and an arbitraryconstant a. Writing (19) in the form fpdp + fxdx = 0, we see thatits solution is f (x , p) = a. Similarly, we get g(y , q) = a. Determinep and q from the equation

f (x , p) = a, g(y , q) = a

and then use the relation du = pdx + qdy to determine a completeintegral.




ExampleFind a general solution of p2y(1 + x2) = qx2.

Solution. First we write the given PDE in the form

p2(1 + x2)

x2=

q

y(separable equation)

It follows that

p2(1 + x2)

x2= a2 =⇒ p =

ax√1 + x2

,

where a is an arbitrary constant. Similarly,

q

y= a2 =⇒ q = a2y .

Now, the relation du = pdx + qdy yields

du =ax√1 + x2

dx + a2ydy =⇒ u = a√

1 + x2 +a2y2

2+ b,

where a and b are arbitrary constants, a general solutionfor the givenPDE.




• Clairaut’s equation

A first-order PDE is said to be in Clairaut form if it can be written as

u = px + qy + f (p, q). (20)

Charpit’s equations take the form

dx

x + fp=

dy

y + fq=

du

px + qy + pfp + qfq=

dp

0=

dq

0.

Now, equivalently considering dpdt

= 0 =⇒ p = a, where a is anarbitrary constant.dqdt

= 0 =⇒ q = b, where b is an arbitrary constant.Substituting the values of p and q in (20), we obtain the requiredgeneral solution

u = ax + by + f (a, b).




ExampleFind a general solution of (p + q)(u − xp − yq) = 1.

Solution. The given PDE can be put in the form

u = xp + yq +1

p + q, (21)

which is of Clairaut’s type. Putting p = a and q = b in (21), a generalsolution is given by

u = ax + by +1

a+ b,

where a and b are arbitrary constants.


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