7 on partial differential equations

MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 16

Lecture 3 Quasilinear First-Order PDEs

A first order quasilinear PDE is of the form

a(x, y, z)∂z

∂x+ b(x, y, z)

∂z

∂y= c(x, y, z). (1)

Such equations occur in a variety of nonlinear wave propagation problems. Let us assume

that an integral surface z = z(x, y) of (1) can be found. Writing this integral surface in

implicit form as

F (x, y, z) = z(x, y)− z = 0.

Note that the gradient vector∇F = (zx, zy,−1) is normal to the integral surface F (x, y, z) =

0. The equation (1) may be written as

azx + bzy − c = (a, b, c) · (zx, zy,−1) = 0. (2)

This shows that the vector (a, b, c) and the gradient vector ∇F are orthogonal. In other

words, the vector (a, b, c) lies in the tangent plane of the integral surface z = z(x, y) at

each point in the (x, y, z)-space where ∇F = 0.

At each point (x, y, z), the vector (a, b, c) determines a direction in (x, y, z)-space is

called the characteristic direction. We can construct a family of curves that have the

characteristic direction at each point. If the parametric form of these curves is

x = x(t), y = y(t), and z = z(t), (3)

then we must have

dx

dt= a(x(t), y(t), z(t)),

dy

dt= b(x(t), y(t), z(t)),

dz

dt= c(x(t), y(t), z(t)), (4)

because (dx/dt, dy/dt, dz/dt) is the tangent vector along the curves. The solutions of (4)

are called the characteristic curves of the quasilinear equation (1).

We assume that a(x, y, z), b(x, y, z), and c(x, y, z) are sufficiently smooth and do not

all vanish at the same point. Then, the theory of ordinary differential equations ensures

that a unique characteristic curve passes through each point (x0, y0, z0). The IVP for

(1) requires that z(x, y) be specified on a given curve in (x, y)-space which determines a

curve C in (x, y, z)-space referred to as the initial curve. To solve this IVP, we pass a

characteristic curve through each point of the initial curve C. If these curves generate a

surface known as integral surface. This integral surface is the solution of the IVP.


REMARK 1. (i) The characteristic equations (4) for x and y are not, in general, uncoupled

from the equation for z and hence differ from those in the linear case (cf. Eq. (3) of Lecture

2).

(ii) The characteristics equations (4) can be expressed in the nonparametric form as

dx

a=dy

b=dz

c. (5)

Below, we shall describe a method for finding the general solution of (1). This method

is due to Lagrange hence it is usually referred to as the method of characteristics or the

method of Lagrange.

1 The method of characteristics

It is a method of solution of quasi-linear PDE which is stated in the following result.

THEOREM 2. The general solution of the quasi-linear PDE (1) is

F (u, v) = 0, (6)

where F is an arbitrary function and u(x, y, z) = c1 and v(x, y, z) = c2 form a solution of

the equationsdx

a=dy

b=dz

c. (7)

Proof. If u(x, y, z) = c1 and v(x, y, z) = c2 satisfy the equations (1) then the equations

uxdx+ uydy + uzdz = 0,

vxdx+ vydy + vzdz = 0

are compatible with (7). Thus, we must have

aux + buy + cuz = 0,

avx + bvy + cvz = 0.

Solving these equations for a, b and c, we obtain

a∂(u,v)∂(y,z)

=b

∂(u,v)∂(z,x)

=c

∂(u,v)∂(x,y)

. (8)

Differentiate F (u, v) = 0 with respect to x and y, respectively, to have

∂F

∂u

{∂u

∂x+∂u

∂z

∂z

∂x

}+∂F

∂v

{∂v

∂x+∂v

∂z

∂z

∂x

}= 0

∂F

∂u

{∂u

∂y+∂u

∂z

∂z

∂y

}+∂F

∂v

{∂v

∂y+∂v

∂z

∂z

∂y

}= 0.


Eliminating ∂F∂u and ∂F

∂v from these equations, we obtain

∂z

∂x

∂(u, v)

∂(y, z)+∂z

∂y

∂(u, v)

∂(z, x)=∂(u, v)

∂(x, y)(9)

In view of (8), the equation (9) yields

a∂z

∂x+ b

∂z

∂y= c.

Thus, we find that F (u, v) = 0 is a solution of the equation (1). This completes the

proof. �.

REMARK 3. • All integral surfaces of the equation (1) are generated by the integral

curves of the equations (4).

• All surfaces generated by integral curves of the equations (4) are integral surfaces of

the equation (1).

EXAMPLE 4. Find the general integral of xzx + yzy = z.

Solution. The associated system of equations are

dx

x=dy

y=dz

z.

From the first two relation we have

dx

x=dy

y=⇒ lnx = ln y + ln c1 =⇒

x

y= c1.

Similarly,

dz

z=dy

y=⇒ z

y= c2.

Take u1 =xy and u2 =

zy . The general integral is given by

F (x

y,z

y) = 0.

EXAMPLE 5. Find the general integral of the equation

z(x+ y)zx + z(x− y)zy = x2 + y2.

Solution. The characteristic equations are

dx

z(x+ y)=

dy

z(x− y)=

dz

x2 + y2.

Each of these ratio is equivalent to

ydx+ xdy − zdz

0=xdx− ydy − zdz

0.


Consequently, we have

d{xy − z2

2} = 0 and d{1

2(x2 − y2 − z2)} = 0.

Integrating we obtain two integrals

2xy − z2 = c1 and x2 − y2 − z2 = c2,

where c1 and c2 are arbitrary constants. Thus, the general solution is

F (2xy − z2, x2 − y2 − z2) = 0,

where F is an arbitrary function.

Next, we shall discuss a method for solving a Cauchy problem for the first-order quasi-

linear PDE (1). The following theorem gives conditions under which a unique solution of

the initial value problem for (1) can be obtained.

THEOREM 6. Let a(x, y, z), b(x, y, z) and c(x, y, z) in (1) have continuous partial deriva-

tives with respect to x, y and z variables. Let the initial curve C be described parametrically

as

x = x(s), y = y(s), and z = z(x(s), y(s)).

The initial curve C has a continuous tangent vector and

J(s) =dy

dsa[x(s), y(s), z(s)]− dx

dsb(x(s), y(s), z(s)] = 0 (10)

on C. Then, there exists a unique solution z = z(x, y), defined in some neighborhood of

the initial curve C, satisfies (1) and the initial condition z(x(s), y(s)) = z(s).

Proof. The characteristic system (4) with initial conditions at t = 0 given as x =

x(s), y = y(s), and z = z(s) has a unique solution of the form

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

with continuous derivatives in s and t, and

x(s, 0) = x(s), y(s, 0) = y(s), z(s, 0) = z(s).

This follows from the existence and uniqueness theory for ODEs. The Jacobian of the

transformation x = x(s, t), y = y(s, t) at t = 0 is

J(s) = J(s, t)|t=0 =

∣∣∣∣∣ ∂x∂s

∂x∂t

∂y∂s

∂y∂t

∣∣∣∣∣t=0

=

[∂y

∂ta− ∂x

∂tb

]t=0

= 0. (11)


in view of (10). By the continuity assumption, the Jacobian J = 0 in a neighborhood

of the initial curve. Thus, by the implicit function theorem, we can solve for s and t as

functions of x and y near the initial curve. Then

z(s, t) = z(s(x, y), t(x, y)) = Z(x, y).

a solution of (1), which can be easily seen as

c =dz

dt=

∂z

∂x

dx

dt+∂z

∂y

dy

dt

= a∂z

∂x+ b

∂z

∂y,

where we have used (4). The uniqueness of the solution follows from the fact that any two

integral surfaces that contain the same initial curve must coincide along all the charac-

teristic curves passing through the initial curve. This is a consequence of the uniqueness

theorem for the IVP for (4). This completes our proof. �.

EXAMPLE 7. Consider the IVP:

∂z

∂y+ z

∂z

∂x= 0

z(x, 0) = f(x),

where f(x) is a given smooth function.

Solution. We solve this problem using the following steps.

Step 1. (Finding characteristic curves)

To solve the IVP, we parameterize the initial curve as

x = s, y = 0, z = f(s).

The characteristic equations are

dx

dt= z,

dy

dt= 1,

dz

dt= 0.

Let the solutions be denoted as x(s, t), t(s, t), and z(s, t). We immediately find that

x(s, t) = zt+ c1(s), y(s, t) = t+ c2(s), z(s, t) = c3(s),

where ci, i = 1, 2, 3 are constants to be determined using IC.

Step 2. (Applying IC) The initial conditions at s = 0 are given by

x(s, 0) = s, y(s, 0) = 0, z(s, 0) = f(s).


Using these condition, we obtain

x(s, t) = zt+ s, y(s, t) = t, z(s, t) = f(s).

Step 3. (Writing the parametric form of the solution)

The solutions are thus given by

x(s, t) = zt+ s = f(s)t+ s, y(s, t) = t, z(s, t) = f(s).

Step 4. (Expressing z(s, t) in terms of z(x, y)) Applying the condition (10), we find that

J(s) = −1 = 0, along the entire initial curve. We can immediately solve for s(x, y) and

t(x, y) to obtain

s(x, y) = x− tf(s), t(x, y) = y.

Since t = y and s = x− tf(s) = x− yz, the solution can also be given in implicit form as

z = f(x− yz).

EXAMPLE 8. Solve the following quasi-linear PDE:

zzx + yzy = x, (x, y) ∈ R2

subject to the initial condition

z(x, 1) = 2x, x ∈ R.

Solution. Here a(x, y, z) = z, b(x, y, z) = y, c(x, y, z) = x. The characteristics

equations are

dx

dt= z, x(s, 0) = s,

dy

dt= y, y(s, 0) = 1,

dz

dt= x, z(s, 0) = 2s.

On solving the above ODEs, we obtain

x(s, t) =s

2(3et − e−t), y(s, t) = et, z(s, t) =

s

2(3et + e−t).

Solving for (s, t) in terms of (x, y), we obtain

s(x, y) =2xy

3y2 − 1, t(x, y) = ln(y),

z(x, y) = z(s(x, y), t(x, y)) =(3y2 + 1)x

(3y2 − 1).


Note that the characteristics variables imply that y must be positive (y = et). In fact, the

solution z is valid only for 3y2 − 1 > 0, i.e., for y > 1√3> 0. Observe that the change of

variables is valid only where ∣∣∣∣∣ xs(s, t) xt(s, t)

ys(s, t) yt(s, t)

∣∣∣∣∣ = 0.

It is easy to verify that this condition leads to y = 1/√3.

Practice Problems

1. Find a solution of the PDE zx + zzy = 6x satisfying the condition z(0, y) = 3y.

2. Find the general integral of the PDE

(2xy − 1)zx + (z − 2x2)zy = 2(x− yz)

and also the particular integral which passes through the line x = 1, y = 0.

3. Solve zx + zzy = 2x, z(0, y) = f(y).

4. Find the solution of the equation zx + zzy = 1 with the data

x(s, 0) = 2s, y(s, 0) = s2, z(0, s2) = s.

5. Find the characteristics of the equation zxzy = z, and determine the integral surface

which passes through the parabola x = 0, y2 = z.

7 on partial differential equations

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