Analytical Solution For Navier-Stokes Equations In Two Dimensions For Laminar Incompressible Flow

Abstract

The Navier-Stokes equations describing laminar flow of an incompressible fluid will be solved. Different group of general solutions for Navier stokes equations governing Laminar incompressible fluids will be derived.

Full text

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arXiv:physics/0609186v1 [physics.flu-dyn] 21 Sep 2006
Analytical Solution For Navier-Stokes Equations In Two
Dimensions For Laminar Incompressible Flow
Saeed Otarod∗1and Davar Otarod†
∗Department of Physics, Yasouj University, Yasouj, Iran
E-mail : sotarod@mail.yu.ac.ir, sotarod@yahoo.com
†Department of mechanical engineering, Yasouj University, Yasouj, Iran
Abstract
The Navier-Stokes equations describing laminar flow of an incom-
pressible fluid will be solved. Different group of general solutions for
Navier stokes equations governing Laminar incompressible fluids will
be derived.
Keywords: Applied mathematics , Fluid dynamics, Navier-Stokes equations
1 Introduction
Navier stokes equations for laminar flows, in different physical conditions,
have been solved in almost all fluid mechanics text books(1). In some phys-
ical conditions these equations have exact solutions, but in general these
equations are complicated nonlinear partial differential equations that can
not be solved easily. Therefore, the authors in most cases, in order to solve
these equations, have either used numerical methods, or they have appealed
to Physically acceptable simplifying considerations. In this way so many ef-
forts have been put on the recognition of physical parameters and terms that
could be neglected without affecting the system effectively.
Also, Navier stokes equations have a very wide application in stellar astro-
physics,[2],[3],[4].
1Corresponding Author
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Since 2000, The author has tried to somehow overcome the complexities gov-
erning the analytical solutions of nonlinear partial Differential equations and
in many respects these activities has been successful.[5],[6],[7],[8],[9],[10].
Since Navier stokes equations governing the incompressible fluids has a wide
application in technology and industry, in this article we try to find explicit
solutions of these equations for the laminar incompressible fluids in two di-
mensions. We have treated these set of equations in a general way without
restricting ourselves to any especial boundary value conditions, although from
the results found in sections 2 and 3, one can conclude that these results are
consistent with different, logical and acceptable, boundary conditions. How
to fit the results with a especial set of boundary conditions is beyond the
scope of this article. We will address this issue in our next efforts.
2Basic equations
Navier stokes equations in two dimensions in cartesian coordinates for lami-
nar incompressible fluids are;
u(x,y)∂u(x,y)
∂x
+v(x,y)∂u(x,y)
∂y
= −1
ρ
∂
∂xP(x,y)+ν(∂2u(x,y)
∂x2
+∂2u(x,y)
∂y2
)
(1)
u(x,y)∂v(x,y)
∂x
+v(x,y)∂v(x,y)
∂y
= −1
ρ
∂
∂yP(x,y)+ ν(∂2v(x,y)
∂x2
+∂2v(x,y)
∂y2
)
(2)
∂u(x,y)
∂x
+∂v(x,y)
∂y
= 0 (3)
Here, u(x,y) and v(x,y) are the x and y components of the velocity respec-
tively, and P(x,y) stands for the pressure. The two first equations are the
components of momentum equations, in the x and y direction. and obviously
the third equation is the continuity equation.
To solve the above set of equations, at first we suppose that, u and v, are
functions of a float function f(x,y), that is to say,
u = u(f(x,y)),v = v(f(x,y))(4)
However, this is a restricting assumption, it has no contradiction with logical
considerations based on the following discussions.
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1-What we do is similar to the application of separation method in solving
the problems. At the beginning, when we apply this method we do not
have any logical supportive argument behind that. We hope by writing the
function in the separated form, we will come to suitable results. Of course,
It is possible that it may not work and in fact in many cases it does not. If
it works we have succeeded, otherwise it would be put aside.
2-Here,What we do is to collect all the variables into a single float variable
f. In fact, it is Some sort of variable change or transformation.
3- Any way u and v, some how are related to each other, in advance we do not
know this relationship, one possibility is that, they both are the function of
the same function. this is the case, in many physical situations, for example
in electromagnetic radiation all components of electrical and magnetic fields
all are functions of K.r + ωt and in harmonic oscillation the oscillations in
all directions are all functions of ωt .Therefore it is reasonable to think that
the two components of the velocity are the functions of a common function
like f(x,t).
Now according to the above considerations equation(3) will be written as;
du
df
∂f
∂x+dv
df
∂f
∂y= 0(5)
Since f is a float and arbitrary function, so far as it fulfills the conditions
expressed by the equations 1- 4,any properties can be attributed to that.
Obviously at first we prefer to chose f in such a way that the equations
could be solved as easily as possible. We do have so Many options and as
the first choice we assume
β∂f
∂x= α∂f
∂y
(6)
with this assumption equation(5) will be simplified to
αdu
df+ βdv
df= 0 (7)
which immediately results in
αu(f) + βv(f) = c(8)
As we mentioned this is not the only possible choice and for example we may
choose f to be
∂f
∂x= f∂f
∂y
(9)
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or
∂f
∂x= ef∂f
∂y
(10)
What choice is the most suitable one, it depends on the boundary conditions
that are to be satisfied, at this point let us concentrate on equations (6) and
(7) and look for the consequences. From elementary algebra The solution to
equation(6) is ,
f = f(αx + βy) (11)
That is to say f can be any arbitrary function of (αx+βy). From equations(7)
and (8) we have
∂v
∂x= −α
β
∂u
∂x,
∂2v
∂x2= −α
β
∂2u
∂x2
(12)
Substituting results of (11) into equation(2), we will come to
u(x,y)∂u(x,y)
∂x
+v(x,y)∂u(x,y)
∂y
=β
α
1
ρ
∂
∂yP(x,y)+ν(∂2u(x,y)
∂x2
+∂2u(x,y)
∂y2
)
(13)
subtracting equation(1) from equation(12) will result in
1
ρ(α∂P
∂x+ β∂P
∂y) = 0(14)
The general solution to this equation is P = P(βx − αy), that is, P can be
any function of (βx − αy).
Now if we rewrite equation(1) in this way;
u(x,y)∂u(x,y)
∂x
+v(x,y)∂u(x,y)
∂y
−ν(∂2u(x,y)
∂x2
+∂2u(x,y)
∂y2
) = −1
ρ
∂
∂xP(x,y)
(15)
it will be seen that, while according to our first assumption the left hand side
of the above equation is a function of f(αx + βy) the right hand side has
to be a function of (βx − αy), but these functions are linearly independent.
This can not happen unless
is true for y dependence of P; that is,∂P
we find that our original assumptionu = u(f) and v = v(f) is only consistent
with those physical situations, at which
∂P
∂x= 0 or
∂p
∂x= constant. The same argument
∂y= 0 or∂P
∂y= constant. In this way
p = constant(16)
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or
p = βx − αy (17)
So far we have derived two answer for P, and each will lead to a different
result. If we follow our analysis by taking into account the result of equation
(15) and substituting that into the equations (1) and (2), we will see that
both of these equations are identical. Therefore we need only one of them to
be solved. As a result equation(1) will be written as
u(x,y)∂u(x,y)
∂x
+ v(x,y)∂u(x,y)
∂y
= ν(∂2u(x,y)
∂x2
+∂2u(x,y)
∂y2
) (18)
since u = u(f),the following results are valid
∂u
∂x=∂f
∂x
du
df
∂u
∂y=∂f
∂y
du
df
(19)
∂2u
∂x2=d2u
df2(∂f
∂x)2+∂2f
∂x2
du
df
∂2u
∂y2=d2u
df2(∂f
∂y)2+∂2f
∂y2
du
df
(20)
From equations(6), (8), (19) and (20) will will have
c
β
du
df
∂f
∂y= ν(d2u
df2[(∂f
∂x)2+ (∂f
∂y)2] +du
df[∂2f
∂x2+∂2f
∂y2])(21)
Up to this point u has been separated from other parameters. To be able to
solve for u we have to decide about f. In the next two subsections we will
try two different cases.
2.1
f = αx + βy
As we said f can be any function of αx+βy and each f will give results that
are consistent with an especial boundary conditions. The simplest choice will
be
f = αx + βy (22)
For this choice, equation(20) will be reduced to
c
ν(α2+ β2)
d2u
df2=d2u
df2
(23)
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