Charpit for Nonlinear PDE

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Applied Mathematics and Computation 193 (2007) 446–454

www.elsevier.com/locate/amc

Charpit and Adomian for solving integral equations

Alice Gorguis

Mathematics Department, North Park University, 3225 W. Foster Avenue, Chicago, IL 60625, United States

Abstract

In this paper, we consider the nonlinear partial differential equations of first order involving two independent variables.Geometrically the complete solution exists in an infinite variety. Applying Charpit’s Method, we can get a two-parameterfamily of integral surfaces by passing integral surfaces through a suitably chosen two-parameter family of curves. Also,briefly, we will present an easier method of solution called Adomian decomposition method.� 2007 Elsevier Inc. All rights reserved.

Keywords: Nonlinear partial differential equations; Charpit’s Method; Adomian method; Complete integrals

1. Introduction

In this work, we will discus nonlinear partial differential equations of the first order, and various importanttypes of integrals possessed by them. We will consider the general method, due to Charpit, for solving non-linear equations involving two independent variables, and also will discuss the method of solving these equa-tions using the decomposition method, which can be used generally for all types of differential and integralequations. We will see how the decomposition method can be applied in a straightforward manner andhow it provides a rapidly convergent series.

2. Analysis of Charpit’s method

Consider a first order nonlinear, and homogeneous partial differential equation of the form,

0096-3

doi:10

E-m

F ðx; y; z; p; qÞ ¼ 0; ð1Þ
where, x, y are independent variables, and z is a dependent variable, p � oz
ox, q � ozoy.

To solve (1), we will determine a second equation of the same type,

Gðx; y; z; p; qÞ ¼ 0; ð2Þ
such that (1) and (2) can be solved for p and q, in terms of x, y, and z.
003/$ - see front matter � 2007 Elsevier Inc. All rights reserved.

.1016/j.amc.2007.03.078

ail address: [email protected]

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A. Gorguis / Applied Mathematics and Computation 193 (2007) 446–454 447

The resulting expression will make

pðx; y; zÞdxþ qðx; y; zÞdy � dz ¼ 0; ð3Þ
integrable. In order that (1) and (2) be solvable for p and q, these relations must be independent, and thereforetheir Jacobian
J ¼ oðF ;GÞoðp; qÞ ¼ F pGq � F qGp ð4Þ

cannot vanish identically. In what follows we will assume that J 5 0. The necessary and sufficient conditionthat the total differential equation

P ðx; y; zÞdxþ Qðx; y; zÞdy þ Rðx; y; zÞdz ¼ 0 ð5Þ
be integrable is that
PoQoz� oR

oy

� �þ Q

oRox� oP

oz

� �þ R

oPoy� oQ

ox

� �ð6Þ

vanish identically. Hence if (3) is to be integrable, where (P = p, Q = q, R = �1), we must have

poqoz� q

opoz� op

oyþ oq

ox¼ 0 ð7Þ

assuming that both p and q in (1) and (2) are functions of x, y, and z. Differentiating (1) partially with respectto x gives

oFoxþ oF

op� opoxþ oF

oq� oqox¼ 0: ð8Þ

Differentiating (2) partially with respect to x gives

oGoxþ oG

op� opoxþ oG

oq� oqox¼ 0: ð9Þ

Now, solving (8) and (9) for oqox, then substituting in Eq. (7) gives

oqox¼ F xGp � F pGx

J; J 6¼ 0: ð10Þ

In the same manner, we will differentiate (1) and (2) with respect to y and solve for

opoy¼ F qGy � F yGq

J; J 6¼ 0: ð11Þ

Also, differentiating (1) and (2) partially with respect to z gives

F z þ F popozþ F q

oqoz¼ 0;

Gz þ Gpopozþ Gq

oqoz¼ 0:

ð12Þ

Then solving (12) for opoz and oq

oz gives

opoz¼ F qGz � F zGq

J; J 6¼ 0;

oqoz¼ F zGp � F pGz

J; J 6¼ 0:

ð13Þ

Inserting the expressions for oqox,

opoy,

opoz and oq

oz in Eq. (7) and multiplying throughout by J we get

pðF zGp � F pGzÞ � qðF qGz � F zGqÞ � ðF qGy � F yGqÞ þ ðF xGp � F pGxÞ ¼ 0: ð14Þ

448 A. Gorguis / Applied Mathematics and Computation 193 (2007) 446–454

Rearranging (14) results in

ðF x þ pF zÞoGopþ ðF y þ qF zÞ

oGoq� ðpF p þ qF qÞ

oGoz� F p

oGox� F q

oGoy¼ 0: ð15Þ

Eq. (15) is a six-dimensional homogeneous linear partial differential equation for the determination of (G) as afunction of x, y, z, p, and q. The subsidiary equations for (15) are

dpF x þ pF z

¼ dqF y þ qF z

¼ dz�pF p � qF q

¼ dx�F p

¼ dy�F q

; dG ¼ 0: ð16Þ

Eq. (16) is called ‘‘Charpit subsidiary equation’’. Therefore, to solve Eq. (1) we have to form the subsidiaryequation (16) first, then try to find one integral of (16), the simpler the better, that contains p and q or both, say

uðx; y; z; p; qÞ ¼ a; ð17Þ
where a = arbitrary constant. This gives the desired relation (2), namely, G � u � a = 0. We then solve (1) and(17) simultaneously for p and q as a function of x, y, z, and a, then substitute in Eq. (3), and integrate to get
f ðx; y; z; a; bÞ ¼ 0; ð18Þ
where b is a second arbitrary constant. Eq. (18) with the two arbitrary constants will serve as a Complete Inte-gral for Eq. (1).
In what follows, we shall examine some equations of special form for which standard methods, obtainablefrom the general Charpit’s procedure, may be formulated.

2.1. Equations of the form F(p,q) = 0

Consider a nonlinear equation of the form

F ðp; qÞ ¼ 0; ð19Þ
containing p and q but none of the variables, x, y, and z.
Since Fx, Fy, and Fz = 0, the denominators of the first two ratios in the Charpits subsidiary equation (16)will vanish identically. Therefore, we have dp = 0 and dq = 0, we may thus use either value of p = a, togetherwith q obtained from F(a,q) = 0, or use q = a together with the value of p obtained from F(p,a) = 0, which-ever is more convenient.

2.2. Equations of the form F(z,p,q) = 0

Consider a nonlinear equation of the form

F ðz; p; qÞ ¼ 0; ð20Þ
in which the independent variables x and y do not appear.
Since Fx = 0 and Fy = 0, Charpits subsidiary equations (16) will reduce to

dppF z

¼ dqqF z

¼ dz�pF p � qF q

¼ dx�F p

¼ dy�F q

: ð21Þ

From the first two ratios, we get

q ¼ ap: ð22Þ
Combining (20) and (22), we can find p and q in terms of z,
p ¼ f ðzÞ; q ¼ ap ¼ af ðzÞ: ð23Þ
Therefore, dz = pdx + qdy yields
dzf ðzÞ ¼ dxþ ady; ð24ÞZ

dzf ðzÞ þ b ¼ xþ ay: ð25Þ


Thus it appears that, in a complete integral (25), z will be a function of the combination x + ay. We may usethis fact to formulate a method of solving an equation of type (20) by setting

z ¼ gðxþ ayÞ ¼ gðuÞ; ð26Þ
so that
p ¼ ogox¼ dg

du; q ¼ og

oy¼ a

dgdu: ð27Þ

Then the partial differential Eq. (20) is transformed into the following ordinary differential equation of firstorder:

F z;dzdu; a

dzdu

� �¼ 0: ð28Þ

The substitution u = x + ay has the Geometric interpretation of rotating the coordinate axes in the xy-plane through the angle arctana, and stretching the x and y-coordinates in the ratio

ffiffiffiffiffiffiffiffiffiffiffiffiffi1þ a2p

: 1. Since Eq.(25) represents a cylinder, the surfaces given by a complete integral of an equation of type (20) will thereforebe cylinders with elements at an angle arctana with the x-axes and parallel to the xy-plane.

2.3. Equations of the form f(x, p) = g(y,q)

Consider a nonlinear partial differential equation of the form

f ðx; pÞ ¼ gðy; qÞ: ð29Þ
Letting F = f(x,p) � g(y,q), we will then have
F x ¼ fx; F y ¼ �gy ; F z ¼ 0; F y ¼ fp; F q ¼ �gq;

so that Charpit subsidiary equation (16) becomes

dpfx¼ dq�gy¼ dz�pfp þ qgq

¼ dx�fp¼ dy

gq: ð30Þ

From the first and fourth of these ratios, we have

fx dxþ fp dp ¼ 0; ð31Þ
and from the second and fifth
gy dy þ gq dq ¼ 0: ð32Þ

But the left sides of (31) and (32) are respectively, the total differentials of df and dq of the functions f(x,p)and g(y,q). Hence f(x,p) = constant and f(y,q) = constant, and by the original Eq. (29) these constants areequal. Therefore,

f ðx; pÞ ¼ a; gðy; qÞ ¼ a: ð33Þ

The procedure of solving an equation of type (29) is thus clear, we set each side of (29) equal to anarbitrary constant a, and solve the resulting pair of Eq. (33) for p in terms of x and a and for q in termsof y and a. We then insert these expressions for p and q in pdx + qdy � dz = 0, to get a total differentialequation

dz ¼ pðx; aÞdxþ qðy; aÞdy: ð34Þ
Then by integration we get
z ¼Z

pðx; aÞdxþZ

qðy; aÞdy þ b; ð35Þ

which is the complete integral of the problem (29).


2.4. Equations of the form z = xp + yq + f(p, q)

If we let

F � xp þ yqþ f ðp; qÞ � z ¼ 0; ð36Þ
then
F x ¼ p; F y ¼ q; F z ¼ �1; F p ¼ xþ fp; f q ¼ y þ fq; ð37Þ

so that

F x þ pF z ¼ 0 and F y þ qF z ¼ 0: ð38Þ

Consequently the Charpit subsidiary equation (16) yields dp = 0 and dq = 0, so that we may take eitherp = constant or q = constant in conjunction with the given equation.

3. Applications

In this section, we shall illustrate Charpit’s method through different examples.

Example 1. Find a complete integral of the nonlinear partial differential equation

q2 � 2qþ 3p ¼ 1: ð39Þ
Since the second denominator of the subsidiary equation (16) is Fy + qFz = 0, therefore we have dq = 0 andq = a. Then substituting q = a in Eq. (38), we get
a2 � 2aþ 3p ¼ 1: ð40Þ
Solving for p gives
p ¼ 1þ 2a� a2

3: ð41Þ

Substituting in

p dxþ qdy � dz ¼ 0; ð42Þ
then integrating and solving for z gives
3z ¼ ð1þ 2a� a2Þxþ 3ay þ b; ð43Þ
which is the required complete integral equation.
Example 2. Find a complete integral for the nonlinear partial differential equation

pqþ p þ q ¼ 0: ð44Þ
This equation is of the form F(p,q) = 0. From the first ratio of Charpit subsidiary equation, we get dp = 0 andp = a. Substitute this in (43) and solve for q to get
q ¼ �aaþ 1

: ð45Þ

Then substitute in

p dxþ qdy � dz ¼ 0; ð46Þ
which can be simply integrated and solved for z,
ðaþ 1Þz ¼ aðaþ 1Þx� ay þ b: ð47Þ

Eq. (47) is the complete integral equation.


Example 3. Solve the nonlinear equation for a complete integral

p2 þ q2 ¼ z4: ð48Þ
To solve this example which is of the type F(z,p,q), we chose z = g(x + ay) = g(u), where u = x + ay. Thisgives
p ¼ ogox¼ dz

du; ð49Þ

q ¼ ogoy¼ a

dzdu: ð50Þ

By substitution of (52) and (53) into the original problem(51), we get

dzdu

� �2

þ a2 dzdu

� �2

¼ z4; ð51Þ

dzdu

� �2

ð1þ a2Þ ¼ z4; ð52Þ

dzdu

� �2

¼ z4

1þ a; ð53Þ

dzdu¼ z2ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ a2p ; ð54Þ

dzz2¼ z2ffiffiffiffiffiffiffiffiffiffiffiffiffi

1þ a2p : ð55Þ

Integrating both sides of Eq. (58) gives

z2ðxþ ay þ bffiffiffiffiffiffiffiffiffiffiffi1þ ap

Þ2 ¼ 1þ a2: ð56Þ
Replacing u with x + ay in Eq. (59) gives the desired complete integral
z2ðxþ ay þ bÞ2 ¼ ð1þ a2Þ: ð57Þ

Example 4. Find the complete integral of the given nonlinear partial differential equation

yp � x2q2 ¼ x2y: ð58Þ
Rewriting the equation into the form f(x,p) = g(y,q), we get
px2¼ q2 þ y

y¼ a: ð59Þ

Solving for p and q separately to get

p ¼ ax2 ð60Þ
and ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip
q ¼ � yða� 1Þ: ð61Þ
Substituting both Eqs. (53) and (54) into
dz ¼ p dxþ p dy; ð62Þ
and integrating gives the complete integral equation
ð3z� ax3 � 3bÞ2 ¼ 4ða� 1Þy3: ð63Þ

Example 5. Solve for the integral equation

z ¼ xp þ yq� 2p2 � 3q2: ð64Þ


This equation is of the form z = xp + yq + f(p,q).To solve this type of equations, let F � xp + yq � 2p2 � 3q2 � z = 0.We have Fx = p, Fy = , Fz = �1, Fp = x � 4p , Fq = y � 6q.Whence Fx + pFz = 0 and Fy + qFz = 0. Consequently Charpit’s subsidiary equation (16) yields dp = 0,

p = a, and dq = 0, q = b.Substituting in (57) gives the complete integral

z ¼ axþ by � 2a2 � 3b2: ð65Þ
Differentiating with respect to a and b gives
a ¼ x4

and b ¼ y6: ð66Þ

Substitute in Eq. (58) and simplify to get the integral equation

24z ¼ 3x2 þ 2y2: ð67Þ
Notice that one can easily verify that Eq. (67) satisfies Eq. (65), and consequently Eq. (67) is a singular inte-
gral equation.

Even though Charpit’s method seems to be simple, but that is after the subsidiary equation is formed, whichrequires a lengthy step. Next, we will see how Adomian method can prove that the problems of nonlinear dif-ferential equations can be solved without using any helping substituting equation.

3.1. Adomian’s method and its application

In this section we will give a general description to the Adomian decomposition method applied to a firstorder nonlinear partial differential equation, then it will be followed by an example to illustrate the method.

Consider the nonlinear partial differential equation of the form

uxðx; yÞ þ uyðx; yÞ þ Ruðx; yÞ þ F ðuðx; yÞÞ ¼ 0: ð68Þ
In operator form, Eq. (68) can be written as
Lxuðx; yÞ þ Lyuðx; yÞ þ Ruðx; yÞ þ F ðuðx; yÞÞ ¼ 0; ð69Þ
where Lx is the highest order differential in x, Ly is the highest order in y, R contains the remaining terms oflower derivatives and F(u(x,y)) is an analytic nonlinear term.
The solutions for u(x,y) obtained from the operator equations Lx and Ly are called partial solutions, thesepartial solutions are equivalent and each converges to the exact solution [2]. However, to solve the problem, adecision on selecting one of the two operators should be made, that is the one which minimizes the size ofcomputation, and has the best known condition, to accelerate the evaluation of the components of thesolution.

Suppose the best operator that meets the conditions is Lx, then Eq. (69) can be written as

Lxuðx; yÞ ¼ �Lyuðx; yÞ � Ruðx; yÞ � F ðuðx; yÞÞ: ð70Þ
Assuming that the operator is invertible, the inverse operator can be applied,
L�1x Lxuðx; yÞ ¼ �L�1

x Lyuðx; yÞ � RL�1x uðx; yÞ � L�1

x F ðuðx; yÞÞ; ð71Þ
where Lx ¼ o
ox and L�1x ¼

R x0ð�Þdx, then Eq. (71) can be written as

uðx; yÞ ¼ uð0; yÞ � L�1x Lyuðx; yÞ � L�1

x fRuðx; yÞ � F ðuðx; yÞÞg: ð72Þ

In the series form, the solution u(x,y) is

uðx; yÞ ¼X1n¼0

unðx; yÞ: ð73Þ


Applying (73) to Eq. (71) gives

X1n¼0

unðx; yÞ ¼ uð0; yÞ � L�1x Ly

X1n¼0

unðx; yÞ !

� L�1x R

X1n¼0

unx; y

!� L�1

x

X1n¼0

An

!; ð74Þ

where

F ðunðx; yÞÞ ¼X1n¼0

An: ð75Þ

The components un(x,y), n P 0 of the solution u(x,y) can be recursively determined,

u0ðx; yÞ ¼ uð0; yÞ;ukþ1ðx; yÞ ¼ �L�1

x Lyuk � L�1x Ruk � L�1

x ðAkÞ; k P 0:ð76Þ

Then, using the Algorithm for the Adomian polynomials An [1] for the nonlinear term F(u(x,y)), the com-ponents can be identified by

u0ðx; yÞ ¼ uð0; yÞ;u1ðx; yÞ ¼ �L�1

x Lyu0ðx; yÞ � L�1x Ru0ðx; yÞ � L�1

x A0;

u2ðx; yÞ ¼ �L�1x Lyu1ðx; yÞ � L�1

x Ru1ðx; yÞ � L�1x A1;

::::

ð77Þ

where each component can be determined by using the preceding component. Then the solution in series formwill be obtained as

X1n¼0

unðx; yÞ ¼ u0 þ u1 þ u2 þ � � � ð78Þ

Example. Solve the nonlinear, non-homogeneous first order partial differential equation given as

q2 � 2qþ 3p ¼ 0; ð79Þ
where q ¼ o
oy, and p ¼ oox.

Then Eq. (79) can be written as

u2y � 2uy þ ux ¼ 0: ð80Þ

Then Eq. (80) in operator form will be

3Lxu ¼ 2Lyu� Lyu; ð81Þ
where Lx ¼ o
ox ; Ly ¼ ooy. Then applying the inverse operator L�1

x to both sides of Eq. (81), we get

uðx; yÞ ¼ uð0; yÞ þ 2=3L�1x ðLyuÞ � 1=3L�1

x ðLyðu2ÞÞ: ð82Þ
where L�1
x ¼R x

0ð�Þdx.

The decomposition method identifies the unknown function u(x,y) as an infinite number of components,un(x,y), n P 0 given by

uðx; yÞ ¼X1n¼0

unðx; yÞ: ð83Þ

In terms of (83), Eq. (82) will be written as

u0 þ u1 þ u2 þ � � � ¼ uð0; yÞ þ 2=3L�1x fLyðu0 þ u1 þ u2 þ � � �Þg � 1=3L�1

x fLyðu0 þ u1 þ u2 þ � � �Þg: ð84Þ


4. Discussion

The main goal of this work was to propose Charpit’s method for solving the integral equations in a verysimple and nice way. Charpit’s method has worked effectively to handle the presented standard models. Butusing Adomian Decomposition Method to solve theses problems is even easier.

References

[1] D. Zwillinger, Hand Book of Differential Equations, Academic Press, Inc., New York, 1989.[2] Wazwaz, Partial Differential Equations, Methods and Applications, Balkema Publishers, 2002.

Further reading

[1] L.R. Ford, Differential Equations, McGraw and Hill, 1955.[2] F.H. Miller, Partial Differential Equations, John Willy and Sons, Inc., 1956.[3] C.R. Chester, Techniques in Partial Differential Equations, McGraw-Hill, New York, 1970, p. 212 and Chapter 15 (pp. 315–337).

Charpit for Nonlinear PDE

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