MA 201: Partial Differential Equations

Lecture 4-6

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Method of Characteristicsfor

First Order Equations in two independent variables

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For a PDEF (x , y , z , p, q) = 0 (1)

the characteristic equations are

dx

dt= Fp (2)

dy

dt= Fq (3)

dz

dt= pFp + qFq (4)

dp

dt= −[Fx + pFz ] (5)

dq

dt= −[Fy + qFz ] (6)

A curve x = x(t), y = y(t), z = z(t) satisfying these equation is acharacteristic curve for (26).

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Example: Determine the characteristic curves of the equation

z = p2 − q2

and find the integral surface which passes through the parabola4z + x2 = 0, y = 0.

Solution: (a) Write the equation in the form F (x , y , z , p, q) = 0.

F (x , y , z , p, q) = z − p2 + q2 = 0. (7)

(b) Parametrize the given initial curve.

x0(s) = 2s, y0(s) = 0, z0(s) = −s2.

(c) Find initial values p0(s) and q0(s) from

F (x , y , z , p,q) = 0 anddz

ds= p

dx

ds+ q

dy

ds.

p0(s) = −s, q0(s) = ±√2s.

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(d) Write the characteristic equations.

Here F = z − p2 + q2. So, Fx = 0,Fy = 0,Fz = 1,Fp = −2p,Fq = 2q.Equations are

dx

dt= Fp = −2p (8)

dy

dt= Fq = 2q (9)

dz

dt= pFp + qFq = −2(p2 − q2) = −2z (10)

dp

dt= −[Fx + pFz ] = −p (11)

dq

dt= −[Fy + qFz ] = −q. (12)

(e) Solve the characteristic equations (ODE’s).

p = c1e−t , q = c2e

−t , x = 2c1e−t + c3, y = −2c2e

−t + c4, z = c5e−2t

(f) Find the constants satisfying the initial condition, i.e., at t = 0.

c1 = −s, c2 = ±√2s, c3 = 4s, c4 = ±2

√2s, c5 = −s2.

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(g) Write the characteristic curves passing through the initial curve.

x(s, t) = 2s(2− e−t) (13)

y(s, t) = ±[

2√2s(1− e−t)

]

(14)

z(s, t) = −s2e−2t (15)

For each value of s we get a characteristic curve.(h) Eliminate s and t to obtain the required surface.

4z + (x ±√2y)2 = 0.

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Compatible system of first-order equations

DefinitionIf every solution of the first-order pde

f (x , y , z , p, q) = 0 (16)

is also a solution of the pde

g(x , y , z , p, q) = 0 (17)

then the equations are said to be compatible.

TheoremA necessary condition that the two equations (16) and (17) arecompatible is

[f , g ] :=∂(f , g)

∂(x , p)+ p

∂(f , g)

∂(z , p)+∂(f , g)

∂(y , q)+ q

∂(f , g)

∂(z , q)= 0. (18)

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Proof. If

J ≡ ∂(f , g)

∂(p, q)6= 0, (19)

then one can solve (16) and (17) for p and q to obtain

p = φ(x , y , z), q = ψ(x , y , z). (20)

The equations (16) and (17) will be compatible only if the system (20) iscompletely integrable, i.e., the equation

φ dx + ψ dy − dz = 0 (21)

is integrable.

An equation of the formn

∑

i=1

Fidxi = 0 is called a Pfaffian

differential equation. A necessary condition that the Pfaffianequation is integrable is that ~X · curl ~X = 0, where~X = (F1,F2, . . . ,Fn).

Thus, (21) is integrable if φ(−ψz ) + ψ(φz )− (ψx − φy ) = 0, i.e.,

ψx + φψz = φy + ψφz . (22)

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Substituting from equations (20) into (16) and differentiating w.r.t. xand z , respectively, we get

fx + fpφx + fqψx = 0,

fz + fpφz + fqψz = 0,

from which it is deduced that

fx + φfz + fp(φx + φφz ) + fq(ψx + φψz ) = 0.

Similarly we can deduce from (17) that

gx + φgz + gp(φx + φφz ) + gq(ψx + φψz) = 0.

Solving these two equations, we find that

ψx + φψz =1

J

{

∂(f , g)

∂(x , p)+ φ

∂(f , g)

∂(z , p)

}

(23)

where J is as defined as in (19).

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If we differentiate the given pair of equations w.r.t. y and z , respectively,we will obtain

φy + ψφz = − 1

J

{

∂(f , g)

∂(y , q)+ ψ

∂(f , g)

∂(z , q)

}

. (24)

so that, substituting from (23) and (24) into (22and replacing φ, ψ byp, q respectively, we see that the condition that the two

[f , g ] =∂(f , g)

∂(x , p)+ p

∂(f , g)

∂(z , p)+∂(f , g)

∂(y , q)+ q

∂(f , g)

∂(z , q)= 0. (25)

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Example: Show that the equations

xp = yq, z(xp + yq) = 2xy

are compatible and solve them.Solution: Here f = xp − yq, g = z(xp + yq) − 2xy , so that

∂(f , g)

∂(x , p)= 2xy ,

∂(f , g)

∂(z , p)= −x2p−xyq,

∂(f , g)

∂(y , q)= −2xy ,

∂(f , g)

∂(z , q)= xyp+y2q

which imply that [f , g ] = y2q2 − p2x2 = 0. Hence the given equationsare compatible.Solving the given equations for p and q we get p = y/z , q = x/z . Thus,dz = ∂z

∂x dx + ∂z∂y dy gives

z dz = y dx + x dy .

Hence the solutions arez2 = 2xy + c1.

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Charpit’s Method

For solvingf (x , y , z , p, q) = 0, (26)

findg(x , y , z , p, q, a) = 0 (27)

containing an arbitrary constant a such that

• The PDE (27) is compatible to (26).

• Equations (26) and (27) can be solved to give

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

• The equation

dz = p(x , y , z , a)dx + q(x , y , z , a)dy (28)

is integrable.

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Charpit’s Method

Such a g should satisfy

fp∂g

∂x+ fq

∂g

∂y+ (pfp + qfq)

∂g

∂z− (fx + pfz)

∂g

∂p− (fy + qfz )

∂g

∂q= 0. (29)

which is equivalent to

dx

fp=

dy

fq=

dz

pfp + qfq=

dp

−(fx + pfz)=

dq

−(fy + qfz). (30)

These equations are known as Charpit’s equations.Once an integral g(x , y , z , p, q, a) of this kind has been found, theproblem reduces to solving for p, q and then integrating the equation

dz = p(x , y , z , a)dx + q(x , y , z , a)dy (31)

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Example. Find a complete integral of the equation

p2x + q2y = z . (32)

Solution: The Charpit’s equations are:

dx

2px=

dy

2qy=

dz

2(p2x + q2y=

dp

p − p2=

dq

q − q2

from which it follows that

p2dx + 2pxdp

p2x=

q2dy + 2qydq

q2y

⇒ p2x = aq2y , a is a constant. (33)

Solving (32) and (33) for p and q, we get

p =

{

az

(1 + a)x)

}1/2

, q =

{

z

(1 + a)y)

}1/2

So, equation (28) in this case becomes

(

1 + a

z

)1/2

dz =( a

x

)1/2dx +

(

1

y

)1/2

dy

⇒ {(1 + a)z}1/2 = (ax)1/2 + y1/2 + b

which is the required complete integral of (32).

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Special types of first-order PDEs

We discuss now some special types of first order PDEs.

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Equations involving only p and q

Consider an equationf (p, q) = 0. (34)

Charpit’s equations reduce to

dx

fp=

dy

fq=

dz

pfp + qfq=

dp

0=

dq

0(35)

Then, dp = dq = 0. Suppose we choose p = a, a constant. Thecorresponding value of q is obtained from

f (a, q) = 0 (36)

so that q = Q(a), a constant. Thus, we get a solution (completeintegral) of the equation as

z = ax + Q(a)y + b (37)

where a, b are parameters.Note: at times assuming q = a and solving f (p, a) = 0 for p may reducethe work considerably.

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Example: Find the complete integral of pq = 1.

Solution: The equation involves only p and q, and so from Charpit’sequation dp = dq = 0. Putting p = a we get q = 1/a. The completeintegral is, therefore,

z = ax +y

a+ b

i.e.,a2x + y − az = c , where a, c are arbitrary constants.

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Equations not involving the independent variable

Let the pde be of the form

f (z , p, q) = 0. (38)

Then the Charpit’s equations take the form

dx

fp=

dy

fq=

dz

pfp + qfq=

dp

−pfz=

dq

−qfz. (39)

The last equation givesdp

p=

dq

q, which leads to the relation

p = aq (40)

where a is a constant. Solving (38) and (40), we obtain expressions forp, q from which a complete integral follows immediately.

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Example: Find a complete integral of p2z2 + q2 = 1.

Solution: The equation does not involve the independent variables.From Charpit’s equation, we have p = aq. Therefore, we have

q2(1 + a2z2) = 1

⇒ q = (1 + a2z2)−1/2, p = a(1 + a2z2)−1/2

Hencedz = pdx + qdy

⇒ (1 + a2z2)1/2dz = adx + dy .

which leads to the complete integral

az(1 + a2z2)1/2 − log[az + (1 + a2z2)1/2] = 2a(ax + y + b).

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Separable equations

We say that a first-order pde is separable, if it can be written in the form

f (x , p) = g(y , q) (41)

For such equations, Charpit’s equations become

dx

fp=

dy

−gq=

dz

pfp − qgq=

dp

−fx=

dq

−gy. (42)

We havedx

fp=

dp

−fx, i.e., df = fxdx + fpdp = 0, i.e., f (x , p) = a, a

constant. Similarly, g(y , q) = a. Now, solve them for p and q to getp = r(x , a), q = s(y , a). Solve

dz = p dx + q dy , i.e., dz = r(x , a)dx + s(y , a)dy

to get a complete integral.

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Example. Determine the complete integral of

p2y(1 + x2) = qx2.

Solution: Given thatp2y(1 + x2) = qx2.

⇒ p2(1 + x2)

x2=

q

y,

so that p =ax√1 + x2

, q = a2y .

Hence a complete integral is

z = a√

1 + x2 +1

2a2y2 + b,

where a and b are constants.

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Clairaut’s Equation

A first-order PDE is said to be of Clairaut type if it can be written in theform

z = px + qy + g(p, q) (43)

The corresponding Charpit’s equations are

dx

x + gp=

dy

y + gq=

dz

px + qy + pgp + qgq=

dp

0=

dq

0

⇒ p = a, q = b.

Using these values in (43), we get the complete integral

z = ax + by + g(a, b) (44)

as is readily verified by direct differentiation.

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Example Find a complete integral of

(p + q)(z − xp − yq) = 1.

Solution: The given equation can be written in the form

z = xp + yq +1

p + q,

i.e., the equation is a Clairaut’s equation. The complete integral is

z = ax + by +1

a+ b.

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Solution/Integral Surface of First Order PDE

Passing through a Given Curve

Refer: Sneddon, Chapter 2, Section 12

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Suppose a complete integral (a two parameter family of surfaces)

f (x , y , z , a, b) = 0 (45)

is known for the PDEF (x , y , z , p, q) = 0. (46)

Can we use the complete integral for finding an integral surface passingthrough a given curve C given by

x = x(t), y = y(t), z = z(t) (47)

We expect such a surface can be

1 A particular case of the complete integral f (x , y , z , a, b) = 0obtained by giving a or b particular values,

2 A particular case of the general integral which is the envelope of aone parameter subsystem of f (x , y , z , a, b) = 0,

3 The envelope of the two parameter system f (x , y , z , a, b) = 0.

Most frequently one finds a surface of type (2).

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Suppose E is an integral surface of type (2) containing C . Then E is theenvelope of a one parameter subfamily each of which touches C at apoint. A one-parameter subfamily of (45) is obtained by puttingb = φ(a).For each t, the corresponding point on C is on a member of (45), and so

f (x(t), y(t), z(t), a, b) = 0. (48)

Because the point touches the surface, it implies that

∂

∂t

(

f (x(t), y(t), z(t), a, b))

= 0. (49)

This implies, (48) has two equal roots for t. Eliminating t from (48) and(49) we get relation

ψ(a, b). (50)

Each solution b = φ(a) of (50) gives a one parameter subfamily of (45)whose envelope gives a required integral surface.

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Find a complete integral of the PDE (p2 + q2)x = pz and deduce thatthe solution which passes through the curve x = 0, z2 = 4y .

Step 1: Find solution of (p2 + q2)x = pz using Charpit’s Method.Using Charpit’s method, we find the complete integral of the given PDEas

z2 = a2x2 + (ay + b)2 (51)

where a and b are real parameters.

Step 2: Find one Parameter Subsystem of (51) passing through givencurve,

Step 3: Find Envelope of One Parameter Subsystem of (51)

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Jacobi’s Method

Refer: Sneddon, Chapter 2, Section 13

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Let u(x , y , z) = 0 be a solution of F (x , y , z , p, q) = 0.Then ux + puz = 0 gives p = −ux/uz . Similarly, q = −uy/uz . Puttingthese in the equation, we get a new equation

f (x , y , z , ux , uy , uz) = 0.

An auxiliary equation is obtained as follows:

dx

fux=

dy

fuy=

dz

fuz=

du1

−fx=

duy

−fy=

duz

−fz

Solve these equations to obtain ux , uy , uz and get a complete solutionsfrom

du = uxdx + uydy + uzdz

involving three parameters a, b, c . For each value of one of theparameters, we get a complete solution.

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Example - Jacobi’s Method

Using Jacobi’s method, find a complete integral of the equation

p2x + q2y = z . (52)

Step 1: Convert (52) into the form f (x , y , z , ux , uy , uz) = 0

Step 2: Solving the PDE f (x , y , z , ux , uy , uz) = 0 by Jacobi’s method.

Step 3: Getting the Solution of (52) from the solution of f = 0.

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