Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(21): 2013
The Complete Solution of Non-Linear Second
Order Partial Differential Equations
Athera Nema Kthem
Kufa University College of Education .Department of Mathematics .
Abstract
The main aim in this paper is to find the complete solution of non-linear second
order partial
differential equations with constant coefficients which has the general form
f(Zxx ,Zxy ,Zyy ,Zx ,Zy ,Z) = 0 ,
but terms like ZxxZ ,ZyyZ ,ZxZxy ,Z2 ,…etc .We have found the substitution
,
transforms the above non-linear second order partial differential equation to the first
order ordinary
differential equation with two variables ,and we can solve by using methods for
finding the solution
of the first order differential equations.
‫الخالصة‬
‫الهدف الرئيسي لهذا البحث هو الحصول على الحل الكامل للمعادالت التفاضلية الجزئية غيرالخطية من الرتبة الثانية ذات‬
ZxxZ ,ZyyZ ,ZxZxy ,Z2
‫والتي حدودها تشبه‬
f(Zxx ,Zxy ,Zyy ,Zx ,Zy ,Z) = 0 , ‫المعامالت الثابتة والتي صيغتها العامة‬
‫ فوجدنا التعويض‬, ‫الخ‬...
‫يحول المعادلة اعاله الى معادلةة تفاضةلية اعتياديةة مةن الرتبةة االولةى بمتغيةرتن يوالتةي يم ةن ايجةاد حلهةا العةار باسةتخدار طةر حةل‬
. ‫معادالت الرتبة األولى‬
Introduction
The research Rehab [Rehab,2006] searched a new function u(x) such that the
assumption
,
gives the general solution of the second order ordinary differential equation ,which
has the form
y(x) + p(x)y +q(x)y = 0 ,
and the depends on the forms of p(x) and q(x) .
In our work we depend on the this idea ,so we search a new functions u(x) and v(y)
,such that the
assumption
,
gives the complete solution of non-linear second order partial differential equations
(PDEs) with
constant coefficients which have the general form
f(Zxx ,Zxy ,Zyy ,Zx ,Zy ,Z) = 0 ,
but terms like ZxxZ ,ZyyZ ,ZxZxy ,Z2 ,…etc .And we depend on the form of the terms
and the value
of coefficients on the above equations .
This assumption transforms the above non-linear second order partial differential
equations to the
first order ordinary differential equation with two variables x and y.
14
Definitions
1 – Bernoulli Equation [Earl and Phillip,1997]
The general form of this equation is written as
y + p(x)y = q(x) yn , n≠ 1
… (1)
where p(x) and q(x) are functions of x or constants .
For finding the solution of this equation see [Earl and Phillip,1997]
2 –Riccati Equation [Rehap, 2008]
The general form of the Riccati equation is written as
y = f(x) +g(x)y +h(x)y2
…(2)
where f(x) ,g(x) and h(x) are given functions of x or constants .
For finding the solution of this equation see [Rehap, 2008]
3 –Non-Linear Second Order Partial Differential Equations with
Constant Coefficients [Evans L .C .,1998]
The general form of non-linear partial differential equation with constant
coefficients is written
as
f(Zxx ,Zxy ,Zyy ,Zx ,Zy ,Z) = 0 ,
…(3)
which has at least one term contains Z or partial derivatives of higher than one
degree or
multiply together .
4 –Complete Solution [Polyanin,2002 ]
The solution of non-linear partial differential equation ,which contains only
arbitrary constants .
Description of The New Method
Non-linear equations are usually more difficult to solve than linear once .In this
search we
present a new method for solving their by search a new functions u(x) and v(y) such
that the
assumption
,
…(4)
represents the complete solution of equation (3) ,this assumption transforms to first
order ordinary
differential equation , by finding Zxx , Zyy , Zxy , Zx and Zy from equation (4) ,we get
,
and by substituting Zxx , Zyy , Zxy , Zx , Zy and Z into the equation (3) ,transforms to
the first order
ordinary differential equation with two functions u(x) and v(y) which has the general
form
f(u(x),v(y),u(x),v(y)) and we get
[f(u(x),v(y),u(x),v(y))]
=0
Since
≠0
15
Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(21): 2013
So
f(u(x),v(y),u(x),v(y)) = 0 ,
…(5)
the equation (5) is of the first order ordinary differential equation and contains two
dependent
functions u(x) and v(y) .
Note
In order to find the complete solution of non-linear second order partial differential
equation with
constant coefficients ,we will discuss special cases which depend on the form of terms
and the values
of coefficients .
The Complete
Differential
Equations
Solution
of
Non-Linear
Second
Order
Partial
To find the complete solution of the equation (3) we classify to many cases by
depending on the
form of terms and the values of Ai ,i= 1,…,n ,n N .
Case (1):- If the equation (3) has the form
A1(ZxxZ +Zx2) + A2(ZyyZ +Zy2) = 0 ,
we can be classified to many states by depending on the values of Ai , i=1,2
a – If A1 = 0,
so the PDE is given by
A2(ZyyZ +Zy2) = 0 ,
then the equation (5) becomes
A2(v(y) +2v2(y)) = 0 ,
and the complete solution of the above PDE is given by
Z(x,y) = B1 1(x) y-c1
,
where B1 and C1 are arbitrary constants and1(x) is an arbitrary function of x .
b -– If A2 = 0,
so the PDE is given by
A1(ZxxZ +Zx2) = 0 ,
then the equation (5) becomes
A1(u(x) +2u2(x)) = 0 ,
and the complete solution of the above PDE is given by
Z(x,y) = B2 2(y) x-c2
,
where B2 and C2 are arbitrary constants and 2(y) is an arbitrary function of y .
c –If A1 and A2 are not identically zero ,then the PDE is given by
A1(ZxxZ +Zx2) + A2(ZyyZ +Zy2) = 0 ,
then the equation (5) becomes
A1(u(x) +2u2(x)) + A2(v(y) +2v2(y)) = 0 ,
and the complete solution of the above PDE is given by
,
where di ,i= 1,…,4 and  are arbitrary constants .
Proof :a – since A2(v(y) +2v2(y)) = 0 , then
(v(y) +2v2(y)) = 0 ,
16
this equation is variable separable equation and the general solution is given by
,
where C1 is arbitrary constant .
Then
=
So the complete solution of the above PDE is given by
Z(x,y) = B1 1(x) y-c1
, B1= eg , 1(x) =
where B1 and C1 are arbitrary constants and 1(x) is an arbitrary function of x .
b – since A1(u(x) +2u2(x)) = 0 , then
(u(x) +2u2(x)) = 0 ,
then by the same method as in case (a) ,we get the complete solution to the above
PDE is given
by
Z(x,y) = B2 2(y) x-c2
, B2 = eg , 2(y) =
where B2 and C2 are arbitrary constants and 2(y) is an arbitrary function of y .
c – since A1(u(x) +2u2(x)) + A2(v(y) +2v2(y)) = 0 , then
A1(u(x) +2u2(x)) = - A2(v(y) -2v2(y)) = -2 ,
 u(x) +2u2(x) +
=0,
…(6)
v(y) +2v2(y) =0 ,
we get from equation (6)
…(7)
and


Also ,we get from equation (7)


Then
=
=
=
So the complete solution of the original PDE is given by
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Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(21): 2013
,
where di ,i= 1,…,4 and  are arbitrary constants .
Case (2):- If the equation (3) has the form
A1ZxxZ + Zxy ( A2Zx + A3Zy ) + A4ZyyZx = 0 ,
we can be classified to many states by depending on the values of Ai , i=1,2
a – If A2 = A3 =A4 = 0 ,
so the PDE is given by
A1ZxxZ = 0 ,
then the equation (5) becomes
A1(u(x) +u2(x)) = 0 ,
and the complete solution of the above PDE is given by
Z(x,y) = B (y) (x - c) ,
where B and C are arbitrary constants and (y) is an arbitrary function of y .
b – If A3 =A4 = 0 ,
so the PDE is given by
A1ZxxZ + A2ZxyZx = 0 ,
then the equation (5) becomes
A1(u(x) +u2(x)) + A2u2(x) v(y) = 0 ,
and the complete solution of the above PDE is given by
Z(x,y) =
,
where D , K , C and  are arbitrary constants .
c – If A4 = 0 ,
so the PDE is given by
A1ZxxZ + Zxy (A2Zx + A3Zy) = 0 ,
then the equation (5) becomes
A1(u(x) +u2(x)) + A2u2(x) v(y) + A3v2(y)u(x) = 0 ,
so the complete solution of the above PDE is given by
Z(x,y) =
,
where Di , Bi , i = 1,2 and  are arbitrary constants .
d –If no any one of the coefficients identically zero then the PDE is given by
A1ZxxZ + Zxy (A2Zx + A3Zy) + A4ZyyZx = 0 ,
then the equation (5) becomes
A1(u(x) +u2(x)) + A2u2(x) v(y) + A3v2(y)u(x) +A4 (v(y) +v2(y))u(x) =
0,
so the complete solution of the above PDE is given by
Z(x,y) =
,
where Di , Bi , i = 1,2 and  are arbitrary constants .
Proof :a - since A1(u(x) +u2(x)) = 0 ,
then
(u(x) +u2(x)) = 0 ,
this equation is variable separable equation and the general solution is given by
,
18
where c is an arbitrary constant .
Then
=
= B (y) (x - c) ; B = eg , (y) =
where B and C are arbitrary constants and (y) is an arbitrary function of y .
b - since A1(u(x) + u2(x)) + A2 u2(x) v (y) = 0 , then
,
so
and

Where c is an arbitrary constant . Then
=
=
; D = eg
where D and  are arbitrary constants .
c –since
A1(u(x) +u2(x)) + A2u2(x) v(y) + A3v2(y)u(x) +A4 (v(y) +v2(y))u(x) =
0,
here we can't separate the variables in this equation ,so we suppose that v(y) = 
,where  is an
arbitrary constant . [Nada Z .A .,2006 ]
Then the last equation becomes
A1(u(x) +u2(x)) +A2u2(x) +A32u(x) = 0


;
B1 =
, B2 =
The last equation is similar to equation (1) ,then general solution is given by
Therefore
=
=
so the complete solution of the above PDE is given by
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Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(21): 2013
Z(x,y) =
,
where Di , Bi , i = 1,2 and  are arbitrary constants .
d – since A1(u(x) +u2(x)) + A2u2(x) v(y) + A3v2(y)u(x) +A4 (v(y) +v2(y))u(x) = 0
let v(y) =  ,where  is an arbitrary constant [Nada Z .A .,2006 ].Then the last
equation becomes
A1(u(x) +u2(x)) +A2u2(x) + (A32 + A42 ) u(x) = 0


;
B1 =
, B2 =
Then by the same way as in the above case (b) ,we can prove the complete
solution of the above
PDE is given by
Z(x,y) =
,
where Di , Bi , i = 1,2 and  are arbitrary constants .
Case (3):- If the equation (3) has the form
Z(A1Zxx + A2Zyy) + Zx ( A3Zyy + A4Zy ) + Zy( A5Zxy + A6Zy ) = 0 ,
In this case we discuss only one when Ai ,i=1,…,6 are not identically zero because the
result if we can found ,they will be similar to the result in case (1) and case (2) .
So if Ai ,i=1,…,6 are not identically zero ,then the equation becomes
A1(u(x) +u2(x)) + A2(v(y) +v2(y))+A3 (v(y) +v2(y))u(x) + A4 u(x) v(y) +
A5v2(y)u(x) +A6v2(y) = 0
and the complete solution of the above PDE is given by
iIf B ≠
; A=
, B=
where di ,i=1,2 and  are arbitrary constants .
ii-
Z(x,y) =
If B =
; A=
, B=
where D and  are arbitrary constants .
Proof :- since
A1(u(x) +u2(x)) + A2(v(y) +v2(y))+A3 (v(y) +v2(y))u(x) + A4 u(x) v(y) + A5v2(y)u(x)
+A6v2(y) = 0
let v(y) =  ,where  is an arbitrary constant [Nada.,2006 ] .Then the last equation
becomes
A1(u(x) +u2(x)) + A22 + A32u(x) + A4 u(x) + A52u(x) +A62 = 0

Let A =
and B =
So the last equation becomes
This equation is like equation (2) and by the same way [Rehap, 2008]
20
i-If B ≠
,we get

 u(x) = b tan( f – bx ) –
, f = bc
then
so the complete solution of the above PDE is given by
where di ,i=1,2 and  are arbitrary constants .
ii-If B =
,we get


u(x) =
,
then
=
so the complete solution of the above PDE is given by
Z(x,y) =
If B =
; A=
, B=
g
where D = e , c and  are arbitrary constants .
Examples
Example (1) : To solve the PDE
(ZxxZ +Zx2) + (ZyyZ +Zy2) = 0 ,
since A1 = 1 and A2 =
then by using the formula
we get the complete solution of the above PDE which has the form
where di , i=1,…,4 and  are arbitrary constants .
Example (1) : To solve the PDE
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Journal of Babylon University/Pure and Applied Sciences/ No.(1)/ Vol.(21): 2013
A1ZxxZ + Zxy ( A2Zx + A3Zy ) + A4ZyyZx = 0 ,
since A1 = 1 , A2 = 1, A3 = -2 and A4 = 3 then by using the formula
Z(x,y) =
,
so the complete solution of the above PDE is given by
Z(x,y) =
,
where Di , i = 1,2 and  are arbitrary constants .
Example (3) : To solve the PDE
Z(Zxx +
since A1= 1 and A2=
the formula
Zyy) + Zx (2 Zyy +
, A3= 2 , A4=
Zy ) - Zy( 2Zxy + Zy ) = 0 ,
, A5= -2 , A6 =
and B ≠
then by using
we get the complete solution of the above PDE which has the form
and  are arbitrary constants .
Where C1 =
Example (4) : To solve the PDE
Z(Zxx +
since A1= 1 and A2=
the formula
Zyy) + Zx (2 Zyy + 3 Zy ) + Zy( -2Zxy + Zy ) = 0 ,
, A3= 2 , A4= 3 , A5= -2 , A6 = and B =
then by using
Z(x,y) =
so the complete solution of the above PDE is given by
Z(x,y) =
where D , c and  are arbitrary constants .
References
Earl D. R. and Phillip E. B. ,1997 ,Elementary Differential Equations ,New York .
Evans L. C. ,1998 ,Partial Differential Equations ,American Mathematical Society .
Nada Z. A. , 2006 , The Complete Solution of Linear Second Order Partial
Differential Equations , Msc , thesis , University of Kufa ,College of Education ,
Department of Mathematics .
Polyanin A. D. ,2002 ,Linear Partial Differential Equations for Engineers and
Scientists , Chapman .
Rehab A. K. ,2006 ,Solving Some Kinds of Linear Second Order Non-Homogeneous
Differential Equations with Variables Coefficients ,Msc , thesis , University of
Kufa , College of Education , Department of Mathematics .
Rehab A.K., 2008 ,The Complete Solution for a Special Types of Partial Differential
Equations ,Journal of Al – Qadisiyah , Vol.(33) , No.(3) .
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