Partial Differential Equations (MMB-452)

Partial Differential Equations(MMB-452)

UNIT 2

Dear Students !Two topics of this unit, namely: “Integral surfaces passing througha given curve” and “Compatible system of first order partial dif-ferential equations” have already been completed in each of the sections.Remaining portions of this unit are noted here in easy way. During thepreparation, if any student faces the difficulty to understand any topic, pleasecontact on my phone number (9358073197) for discussion.

Dr. Aftab AlamDeptt. of Mathematics

AMU

Cauchy Method of Characteristics for First or-

der PDEs

This method is used for finding the integral surface of first order partialdifferential equations

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

which passes through a given curve

x0 = x0(s), y0 = y0(s), z0 = z0(s). (2)

The curve (2) is called initial data curve for equation (1).

1

2

Integral Strip: A 5-tuple of real functions of the form(x(t), y(t), z(t), p(t), q(t)

), t ∈

I, (wherein I is an interval in real line) is called an integral strip of equation(1) if it satisfies the following ones:

(i)(x(t), y(t), z(t), p(t), q(t)

)= 0, ∀ t ∈ I

(ii) dzdt

= p(t)dxdt

+ q(t)dydt

.

The last condition is called strip condition.

Initial Strip: Along the curve (2), if we specify the functions p0(s) and q0(s)such that

(x0(s), y0(s), z0(s), p0(s), q0(s)

)becomes an integral strip then such

a strip is called an initial strip of Eq. (1). Notice that in this case, the initialstrip condition is defined as:

dz0ds

= p0(s)dx0ds

+ q0(s)dy0ds

.

Characteristic Equations: A system of five ordinary differential equations

dx

dt= Fp

dy

dt= Fq

dz

dt= pFp + qFq

dp

dt= −(Fx + pFz)

dq

dt= −(Fy + qFz) (3)

are called the characteristic equations of Eq. (1).

Fact: A general solution of characteristic equations (3) forms an integralstrip of Eq. (1).

Characteristic Curves: Let the general solution of system (3) of charac-teristic equations is

x = x(t), y = y(t), z = z(t), p = p(t), q = q(t) (4)

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Then first three functions

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

define a space curve in parameter ‘t’. This curve is called the characteristiccurve, or simply, the characteristics of Eq. (1).

Characteristic Strip: As the functions appearing in general solution (4) ofcharacteristic equations satisfy the Lipschitz condition, there exists a uniquesolution of system (3) for a given initial strip given by

x = x(t, s), y = y(t, s), z = z(t, s), p = p(t, s), q = q(t, s) (5)

such that at the initial value t = t0, it coincides with the initial strip. Such aunique solution containing two parametrs t and s is called the characteristicstrip of Eq. (1).Eliminating t and s from first three equations of (5), we obtain a relation be-tween x, y, z, which is the required integral suface of Eq. (1) passing throughthe curve (2).

Example 1. Find the characteristics of the equation pq = z and determinethe integral surface which passes through the straight line x = 1, z = y.

Solution. Given PDE is

F ≡ pq − z = 0. (6)

Parametric equation of initial data curve is

x0(s) = 1, y0(s) = s, z0(s) = s. (7)

Step I (Initial Strip): Let (x0, y0, z0, p0, q0) be the initial strips, then itsatisfies Eq. (6), i.e.,

p0q0 − z0 = 0 or p0q0 − s = 0

⇒ p0q0 = s. (8)

From Eq. (7), we get

dx0ds

= 0,dy0ds

= 1,dz0ds

= 1.

4

Hence, the initial strip condition is

dz0ds

= p0dx0ds

+ q0dy0ds

⇒ 1 = p00 + q01 or q0 = 1.

Putting this value of q0 in Eq. (8), we get

p0 = s.

Hence, the initial strips is

(x0, y0, z0, p0, q0) = (1, s, s, s, 1).

Step II (Characteristic Equations): From Eq. (6), we have

Fx = 0, Fy = 0, Fz = −1, Fp = q, Fq = p.

The characteristic equations of Eq. (6) are

dx

dt= Fp = q

dy

dt= Fq = p

dz

dt= pFp + qFq = 2pq

dp

dt= −(Fx + pFz) = p

dq

dt= −(Fy + qFz) = q

or

dx = qdt (9)

dy = pdt (10)

dz = 2pqdt (11)

dp = pdt (12)

dq = qdt (13)

Step III (Integral Strips): Integrating (12) and (13), we get

p = c1et and q = c2e

t.

5

Hence, Eq. (9), (10) and (11), respectively, become

dx = c2etdt, dy = c1e

tdt, dz = 2c1c2e2tdt.

Integrating these equations, we get

x = c2et + c3, y = c1e

t + c4, z = c1c2e2t + c5.

Hence, the integral strip of Eq. (6) is(x(t), y(t), z(t), p(t), q(t)

), where

x(t) = c2et + c3, y(t) = c1e

t + c4, z(t) = c1c2e2t + c5, p(t) = c1e

t, q(t) = c2et.

(14)Step IV (Characteristic Strips): At the initial value t = 0,

x(0) = x0 = 1, y(0) = y0 = s, z(0) = z0 = s, p(0) = p0 = s, q(0) = q0 = 1

or

c2 + c3 = 1 (15)

c1 + c4 = s (16)

c1c2 + c5 = s (17)

c1 = s (18)

c2 = 1 (19)

From (18) and (19), we have c1 = s, c2 = 1.From (15), we get c3 = 1− c2 = 1− 1 = 0.From (16), we get c4 = s− c1 = s− s = 0.From (17), we get c4 = s− c1c2 = s− s = 0.Putting these values of constants in Eq. (14), we get the characteristic stripgiven by

x = et, y = set, z = se2t, p = set, q = et.

Step V (Final Solution): Eliminating t and s from first three equations,we get

z = se2t = z = et.set = xy

or z = xy

which is the required integral surface.

Example 2. Find the characteristics of the equation p2 + q2 = 2 and deter-mine the integral surface which passes through the straight line x = 0, z = y.

6

Solution. Given PDE is

F ≡ p2 + q2 − 2 = 0. (20)

Parametric equation of initial data curve is

x0(s) = 0, y0(s) = s, z0(s) = s. (21)

Step I (Initial Strip): Let (x0, y0, z0, p0, q0) be the initial strips, then itsatisfies Eq. (6), i.e.,

p20 + q20 − 2 = 0

⇒ p20 + q20 = 2. (22)

From Eq. (21), we get

dx0ds

= 0,dy0ds

= 1,dz0ds

= 1.

Hence, the initial strip condition is

dz0ds

= p0dx0ds

+ q0dy0ds

⇒ 1 = p00 + q01 or q0 = 1.

Putting this value of q0 in Eq. (22), we get

p20 + 1 = 2⇒ p0 = ±1.

Hence, the initial strips is

(x0, y0, z0, p0, q0) = (0, s, s,±1, 1).

Step II (Characteristic Equations): From Eq. (6), we have

Fx = 0, Fy = 0, Fz = 0, Fp = 2p, Fq = 2q.

The characteristic equations of Eq. (6) are

dx

dt= Fp = 2p

dy

dt= Fq = 2q

dz

dt= pFp + qFq = 2p2 + 2q2 = 2.2 = 4

(from Eq. (20)

)dp

dt= −(Fx + pFz) = 0

dq

dt= −(Fy + qFz) = 0

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or

dx = 2pdt (23)

dy = 2qdt (24)

dz = 2dt (25)

dp = 0 (26)

dq = 0 (27)

Step III (Integral Strips): Integrating (26) and (27), we get

p = c1 and q = c2.

Hence, Eq. (23), (24) and (25), respectively, become

dx = 2c1dt, dy = 2c2dt, dz = 2dt.


x = 2c1t+ c3, y = 2c2t+ c4, z = 2t+ c5.

Hence, the integral strip of Eq. (20) is(x(t), y(t), z(t), p(t), q(t)

), where

x = 2c1t+ c3, y = 2c2t+ c4, z = 2t+ c5, p = c1, q = c2. (28)

Step IV (Characteristic Strips): At the initial value t = 0,

x(0) = x0 = 0, y(0) = y0 = s, z(0) = z0 = s, p(0) = p0 = ±1, q(0) = q0 = 1

orc3 = 0, c4 = s, c5 = s, c1 = ±1, c2 = 1.

Putting these values of constants in Eq. (28), we get the characteristicstrip given by

x = ±2t, y = 2t+ s, z = 4t+ s, p = ±1, q = 1.

Step V (Final Solution): Eliminating t and s from first three equations,we get

z − y = (4t+ s)− (2t+ s) = 2t = ±xor z = y ± x

which is the required integral surface.

8

Semilinear Partial Differential Equations: Re-

duction to Canonical Forms

Usually, for the second order derivatives, we use the notations:

r =∂2z

∂x2, s =

∂2z

∂x∂y, t =

∂2z

∂y2.

A semilinear (or almost-linear) partial differential equation of second orderin two variables is of the form

R(x, y)r + S(x, y)s+ T (x, y)t = V (x, y, z, p, q). (29)

Here the coefficients R, S, T do not vanish simultaneously.

Definition 1. The function 4 defined by

4(x0, y0) = S2(x0, y0)− 4R(x0, y0)T (x0, y0)

is called the discriminant of Eq. (29) at a point (x0, y0).

We classify such equation according to the sign of the discriminant 4.

Definition 2. Eq. (29) is called• hyperbolic at (x0, y0) if 4(x0, y0) > 0,• parabolic at (x0, y0) if 4(x0, y0) = 0 and• elliptic at (x0, y0) if 4(x0, y0) < 0.Moreover, Eq. (29) is called hyperbolic (resp., parabolic or elliptic) in adomain D ⊆ R2 if it is hyperbolic (resp., parabolic or elliptic) at each (x, y) ∈D.

For illustration, consider the equation xr−t+q = sinx. The discriminantis 4 = S2−4RT = 4x and hence, the given equation is hyperbolic, parabolicand elliptic accordingly x <,=, > 0.

Definition 3. The two ordinary differential equations

dy

dx=S ±√4

2R

9

are called characteristic equations of Eq. (29).

The solutions of these characteristic equations are called the characteris-tic curves or the characteristic projections or simply, the characteristics ofEq. (29).

Clearly, a hyperbolic equation admits two distinct and real characteristiccurves while a parabolic equation has a single real characteristic curve. Incase of elliptic equation, there is no real characteristic curve, instead twocomplex conjugate characteristic curves are obtained.

Coordinate Transformation: We have to prove that with a making changeof independent variables, a semilinear partial differential equation reduced toa new semilinear equation. Consider the general transformation of indepen-dent variables given by

ξ = ξ(x, y) and η = η(x, y). (30)

Now, using the chain rule, we compute the partial derivatives of z regard-ing new variables ξ and η so that z = z(ξ, η).

p =∂z

∂x=∂z

∂ξ

∂ξ

∂x+∂z

∂η

∂η

∂x= ξxzξ + ηxzη (31)

q =∂z

∂y=∂z

∂ξ

∂ξ

∂y+∂z

∂η

∂η

∂y= ξyzξ + ηyzη (32)

From equations (31) and (32), we get respectively

∂

∂x= ξx

∂

∂ξ+ ηx

∂

∂η(33)

∂

∂y= ξy

∂

∂ξ+ ηy

∂

∂η(34)

Using (31) and (33), we get

r =∂p

∂x=

∂

∂x(ξxzξ + ηxzη)

= ξxxzξ + ξx∂

∂x(zξ) + ηxxzη + ηx

∂

∂x(zη)

= ξxxzξ + ξx

(ξx∂

∂ξ+ ηx

∂

∂η

)zξ + ηxxzη + ηx

(ξx∂

∂ξ+ ηx

∂

∂η

)zη

= ξ2xzξξ + 2ξxηxzξη + η2xzηη + ξxxzξ + ηxxzη.

10


t =∂q

∂y=

∂

∂y(ξyzξ + ηyzη)

= ξyyzξ + ξy∂

∂y(zξ) + ηyyzη + ηy

∂

∂y(zη)

= ξyyzξ + ξy

(ξy∂

∂ξ+ ηy

∂

∂η

)zξ + ηyyzη + ηy

(ξy∂

∂ξ+ ηy

∂

∂η

)zη

= ξ2yzξξ + 2ξyηyzξη + η2yzηη + ξyyzξ + ηyyzη.


s =∂q

∂x=

∂

∂x(ξyzξ + ηyzη)

= ξxyzξ + ξy∂

∂x(zξ) + ηxyzη + ηy

∂

∂x(zη)

= ξxyzξ + ξy

(ξx∂

∂ξ+ ηx

∂

∂η

)zξ + ηxyzη + ηy

(ξx∂

∂ξ+ ηx

∂

∂η

)zη

= ξxξyzξξ + (ξxηy + ξyηx)zξη + ηxηyzηη + ξxyzξ + ηxyzη.

Collecting all these derivatives, we have

p = ξxzξ + ηxzη

q = ξyzξ + ηyzη

r = ξ2xzξξ + 2ξxηxzξη + η2xzηη + ξxxzξ + ηxxzη

t = ξ2yzξξ + 2ξyηyzξη + η2yzηη + ξyyzξ + ηyyzη

s = ξxξyzξξ + (ξxηy + ξyηx)zξη + ηxηyzηη + ξxyzξ + ηxyzη (35)

Under the transformation (30), Eq. (29) shall be reduced to a new semi-linear PDE, in which z is independent variable, while ξ and η are independentvariables.

Invertible Transformation: A coordinate transformation of independentvariables (x, y) to new variables (ξ, η) is called an invertible transforma-tion (or nonsingular transformation) if the Jacobian of the transformation

J :=∂(ξ, η)

∂(x, y)= ξxηy − ξyηx 6= 0.

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Fact: Any invertible transformation of independent variable does not modifythe types (hyperbolic/parabolic/elliptic) of equations.

Laplace Transformation and Canonical Forms: By a suitable choiceof variables (ξ, η), Eq. (29) can be transformed into a simpler form, whichis often called the canonical form or normal form of Eq. (29). Such newcoordinates (ξ, η), under which the original equation reduces to its canonicalform, are called the canonical coordinates, while such a transformation iscalled Laplace transformation. Notice that Laplace transformation must beconsidered as invertible. The choice of canonical coordinates depend uponweather the original partial differential equation (29) is hyperbolic, parabolicor elliptic.

Canonical Forms of Hyperbolic Equations: Let Eq. (29) be a hyper-bolic equation, whose characteristic curves are

u(x, y) = c1 and v(x, y) = c2.

Then the transformation

ξ = u(x, y) and η = v(x, y)

is invertible. Moreover, under this transformation Eq. (29) reduces to itscanonical form given by

zξη = Φ(ξ, η, z, zξ, zη).

Canonical Forms of Parabolic Equations: Let Eq. (29) be a parabolicequation, whose characteristic curve is

u(x, y) = c.

Assume that v(x, y) is arbitrarily chosen such that the Jacobian ∂(u,v)∂(x,y)

6= 0.Then under the invertible transformation


Eq. (29) reduces to its canonical form given by

zηη = Φ(ξ, η, z, zξ, zη).

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Remark 1. In practice, in parabolic case, we select the new variable v(x, y)in such a way that the functions u(x, y) and v(x, y) must be independent so

that ∂(u,v)∂(x,y)

6= 0.

Canonical Forms of Elliptic Equations: Let Eq. (29) be an ellipticequation, whose characteristic curves are

u(x, y) + iv(x, y) = c1 and u(x, y)− iv(x, y) = c2.

Then the transformation


is invertible. Moreover, under this transformation Eq. (29) reduces to itscanonical form given by

zξξ + zηη = Φ(ξ, η, z, zξ, zη).

Example 3. Reduce the equation

r = x2t

to a canonical form.

Solution. HereR = 1, S = 0, T = −x2.

Step I (Classification): The discriminant of given equation is

4 = S2 − 4RT = 4x2 > 0

Therefore, the given equation is hyperbolic for x 6= 0.Notice that for x = 0, the given equation is parabolic and reduces to r = 0,which is already in canonical form.

Step II (Characteristics): The characteristic equations are

dy

dx=S ±√4

2R=±2x

2= ±x

which becomedy ± xdx = 0.

13

On integrating these equations, we get

y +1

2x2 = c1 and y − 1

2x2 = c2

which are the characteristic curves of given PDE.

Step III (Canonical Form): The canonical coordinates are

ξ = y +1

2x2 and η = y − 1

2x2. (36)

Thus, we have

ξx = x, ξy = 1, ξxx = 1, ξxy = 0, ξyy = 0

ηx = −x, ηy = 1, ηxx = −1, ηxy = 0, ηyy = 0.

Using above, Eq. (35) become

r = x2zξξ − 2x2zξη + x2zηη + zξ − zηt = zξξ + 2zξη + zηη.

Putting these values in given PDE, we get

x2zξξ − 2x2zξη + x2zηη + zξ − zη = x2(zξξ + 2zξη + zηη)

or 4x2zξη = zξ − zη.

Using (36), we have x2 = ξ − η and hence above equation reduces to

zξη =1

4(ξ − η)(zξ − zη)

which is the required canonical form of given equation.


r + x2t = 0


14

Solution. HereR = 1, S = 0, T = x2.


4 = S2 − 4RT = −4x2 < 0.

Therefore, the given equation is elliptic for x 6= 0.It may be noted that for x = 0, the given equation is parabolic and reducesto r = 0, which is already in canonical form.


dy

dx=S ±√4

2R=±2x

2= ±ix

which becomedy ± ixdx = 0.

On integrating these equations, we get

y + ix2

2= c1 and y − ix

2

2= c2



ξ = y and η =x2

2. (37)

Thus, we have

ξx = 0, ξy = 1, ξxx = ξxy = ξyy = 0

ηx = x, ηy = 0, ηxx = 1, ηxy = ηyy = 0.


r = x2zηη + zη

t = zξξ.


x2zηη + zη + x2zξξ = 0

15

or zξξ + zηη = − 1

x2zη.

Using (37), we have x2 = 2η and hence above equation reduces to

zξξ + zηη = − 1

2ηzη



x2r − 2xys+ y2t+ xp+ yq = 0


Solution. Here

R = x2, S = −2xy, T = y2.


4 = S2 − 4RT = 4x2y2 − 4x2y2 = 0

Therefore, the given equation is parabolic.

Step II (Characteristics): The characteristic equation is

dy

dx=

S

2R= −y

x

or xdy + ydx = 0.

Integrating above equation, we get

xy = c

which is the characteristic curve of given PDE.


ξ = xy and η = x. (38)

16

Notice that we have chosen η = x in such a way that ξ and η are inde-pendent. Now, we have

ξx = y, ξy = x, ξxx = 0, ξxy = 1, ξyy = 0

ηx = 1, ηy = 0, ηxx = ηxy = ηyy = 0.


p = yzξ + zη

q = xzξ

r = y2zξξ + 2yzξη + zηη

t = x2zξξ

s = xyzξξ + xzξη + zξ.


x2(y2zξξ+2yzξη+zηη)−2xy(xyzξξ+xzξη+zξ)+y2x2zξξ+x(yzξ+zη)+yxzξ = 0

or xzηη + zη = 0.

Using x = η (due to (38)), above equation reduces to

zηη = −1

ηzη


General Solution of Semilinear Equations

In general, there is not easy to obtain the general solution of a semilinearpartial differential equation. But some times the canonical form of an equa-tion is so simple such that the subsequent analysis of solving the equationis made easy. It turns out that the solution of a semilinear equation andthat of its canonical form share many exclusive qualitative properties. Inthis subsection, we illustrate several such examples.

Example 6. Find the general solution of the equation

y2r − 2xys+ x2t =y2

xp+

x2

yq.

17

Solution. Here

R = y2, S = −2xy, T = x2.


4 = S2 − 4RT = 4x2y2 − 4x2y2 = 0.



dy

dx=

S

2R= −x

y

or xdx+ ydy = 0.

On integrating, we get

x2 + y2 = c



ξ = x2 + y2 and η = y2.

Using these coordinates, the given equation can be reduced to its canonicalform given by

zηη = 0.

Step IV (General Solution): Integrating twice partially w.r.t. η, weobtain

z = ηφ(ξ) + ψ(ξ).

Putting the values of ξ and η in above equation, we get

z = y2φ(x2 + y2) + ψ(x2 + y2)

which is the required general solution.


xyr − (x2 − y2)s− xyt+ py − qx = 2(x2 − y2), x 6= 0.

18

Solution. Here

R = xy, S = −(x2 − y2), T = −xy.


4 = S2 − 4RT = (x2 − y2)2 + 4x2y2 = (x2 + y2)2 > 0 as x 6= 0

Therefore, the given equation is hyperbolic.


dy

dx=S ±√4

2R=−(x2 − y2)± (x2 + y2)

2xy

which becomedy

dx=y

xand

dy

dx= −x

y

or xdy − ydx = 0 and xdx+ ydy = 0.


y

x= c1 and x2 + y2 = c2



ξ =y

xand η = x2 + y2.


zξη =ξ2 − 1

(ξ2 + 1)2.

Step IV (General Solution): Integrating above equation partially w.r.t.

19

ξ, we obtain

zη =

∫ξ2 − 1

(ξ2 + 1)2dξ + φ(η)

=

∫ 1− 1ξ2(

ξ + 1ξ

)2dξ + φ(η)

=

∫dτ

τ 2+ φ(η), (Putting ξ +

1

ξ= τ)

= −1

τ+ φ(η) = − ξ

ξ2 + 1+ φ(η).

Integrating now partially w.r.t. η, we get

z = − ξη

ξ2 + 1+

∫φ(η)dη + ψ(ξ) = − ξη

ξ2 + 1+ θ(η) + ψ(ξ)


z = θ(x2 + y2) + ψ(yx

)− xy



yr + 3ys+ 3p = 0, y 6= 0

Solution. HereR = y, S = 3y, T = 0.


4 = S2 − 4RT = 9y2 > 0 as y 6= 0.

Therefore, the given equation is hyperbolic.


dy

dx=S ±√4

2R=

3y ± 3y

2y= 3, 0

20

which becomedy = 0 and dy − 3dx = 0.


y = c1 and y − 3x = c2



ξ = y and η = y − 3x.


zξη +1

ξzη = 0.

Step IV (General Solution): Setting zη = w, above equation reduces to

∂w

∂ξ+w

ξ= 0

or∂w

w+∂ξ

ξ= 0.

Integrating partially, we obtain

wξ = φ(η)

or zη =1

ξφ(η)

which on integrating w.r.t. η gives rise

z =1

ξ

∫φ(η)dη + ψ(ξ)

or z =1

ξθ(η) + ψ(ξ).


z =1

yθ(y − 3x) + ψ(y)


21


x2r − 2xys+ y2t+ xp+ yq = 0.

Solution. HereR = x2, S = −2xy, T = y2.


4 = S2 − 4RT = 4x2y2 − 4x2y2 = 0



dy

dx=

S

2R= −y

x

or xdy + ydx = 0

which on integrating, gives rise

xy = c



ξ = xy and η = x.


zηη +1

ηzη = 0.

Step IV (General Solution): Setting zη = w, above equation reduces to

∂w

∂η+w

η= 0

or∂w

w+∂η

η= 0.

22

Integrating partially, we obtain

wη = φ(ξ)

or zη =1

ηφ(ξ)

which on integrating w.r.t. η gives rise

z = φ(ξ) log η + ψ(ξ).


z = φ(xy) log x+ ψ(xy)


Partial Differential Equations (MMB-452)

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