AbstractIn this paper, after nearly 150 years of waiting, the Navier-Stokes equations in 3-D forincompressible fluid flow have been analytically solved. In fact, it is shown that these equationscan be solved in 4-dimensions or n-dimensions. The author has proposed and applied a new law, the law of definite ratio for fluid flow. This law states that in incompressible fluid flow, the otherterms of the fluid flow equation divide the gravity term in a definite ratio and each term utilizesgravity to function. The sum of the terms of the ratio is always unity. This law evolved from theauthor's earlier solutions of the Navier-Stokes equations. By applying the above law, the hithertounsolved magnetohydrodynamic equations were routinely solved. It is also shown that withoutgravity forces on earth, there will be no incompressible fluid flow on earth as is known (see p.23,p.13). The difficulty in solving the Navier-Stokes equations has been due to finding a logical way tosplit the equations. By using the most fundamental principle for dividing a quantity into parts, usingratios, all hidden flaws in splitting the equations have been eliminated. The resulting sub-equationswere readily integrable, and even, the nonlinear sub-equations were readily integrated. Thepreliminaries reveal how the ratio technique evolved as well as possible applications of the solutionmethod in mathematics, science, engineering, business, economics, finance, investment andpersonnel management decisions. The coverage is as follows. The x−direction Navier-Stokesequation will be linearized, solved, and the solution analyzed. The linearized equation represents,except for the numerical coefficient of the acceleration term, the linear part of the Navier-Stokesequation. This solution will be followed by the solution of the Euler equation of fluid flow. TheEuler equation represents the nonlinear part of the Navier-Stokes equation. The Euler equation wassolved in the author's previous paper. Following the Euler solution, the Navier-Stokes equation willbe solved, essentially by combining the solutions of the linearized equation and the Euler solution.For the Navier-Stokes equation, the linear part of the relation obtained from the integration of thelinear part of the equation satisfied the linear part of the equation; and the relation from theintegration of the non-linear part satisfied the non-linear part of the equation. For the linearizedequation, different terms of the equation were made subjects of the equation, and each such equationwas integrated by first splitting-up the equation, using ratio, into sub-equations. The integrationresults were combined. Four equations were integrated. The relations obtained using these terms assubjects of the equations were checked in the corresponding equations. Only the equation with thegravity term as subject of the equation satisfied its corresponding equation, and this only satisfactionled to the law of definite ratio for fluid flow, stated above. This equation which satisfied itscorresponding equation will be defined as the driver equation; and each of the other equations whichdid not satisfy its corresponding equation will be called a supporter equation. A supporter equationdoes not satisfy its corresponding equation completely but provides useful information about thedriver equation which is not apparent in the solution of the driver equation. The solutions revealedthe role of each term of the Navier-Stokes equations in fluid flow. Most importantly, the gravity termis the indispensable term in fluid flow, and it is involved in the parabolic as well as the forwardmotion of fluids. The pressure gradient term is also involved in the parabolic motion of fluids. Theviscosity terms are involved in parabolic, periodic and decreasingly exponential motion of fluids.As the viscosity increases, the periodicity increases. The variable acceleration term is also involvedin the periodic and decreasingly exponential motion of fluids. The convective acceleration term withx as the independent variable produces square root function behavior. The other convectiveacceleration terms produce fractional expressions containing square root functions.For a spin-off, the smooth solutions from above are specialized and extended to satisfy therequirements of the CMI Millennium Prize Problems, and thus prove the existence of smoothsolutions of the Navier-Stokes equations.
Solutions of the Navier-Stokes Equations
2
Options Option 1: Solutions of 3-D Linearized Navier-Stokes Equations 3 Option 2: Solutions of 4-D Linearized Navier-Stokes Equations 17 Option 3: Solutions of the Euler Equations 18 Option 4: Solutions of the Navier-Stokes Equations 20 Option 5: Solutions of 4-D Navier-Stokes Equations 22 Option 6: CMI Millennium Prize Problem Requirements 26 Option 7: Solutions of Magnetohydrodynamic Equations 28
The Navier-Stokes equations in three dimensions are three simultaneous equations in Cartesiancoordinates for the flow of incompressible fluids. The equations are presented below:
(N
µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂
µ∂∂
∂∂
∂∂
∂∂ ρ ρ
∂∂
( ) ( )
( ) (
)2
2
2
2
2
2
2
2
2
2
2
2
Vx
Vy
Vz
px
gVt
VVx
VVy
VVz
Vx
Vy
Vz
py
gVt
x x x xx
xy
xz
xx x
y y yy
y
+ + − + = + + +
+ + − + = + VVVx
VVy
VVz
Vx
Vy
Vz
pz
gVt
VVx
VVy
VVz
x z
z z z zx
zy
zz
z
yy
y y
z
y
z
∂∂
∂∂
∂∂
µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂
+ +
+ + − + = + + +
)
( ) ( )
)
)
(N
(N
2
2
2
2
2
2
Equation ( Nx ) will be the first equation to be solved; and based on its solution, one will be able towrite down the solutions for the other two equations, ( Ny), and ( Nz ).
Dimensional ConsistencyThe Navier-Stokes equations are dimensionally consistent as shown below:
µ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2
V
x
V
yVz
px
gVt
VVx
VVy
VVz
x x x xx
xx
xy
xz
x + + − + = + + +
Using MLT
M L T L T L T L T L T M L T L T L T L T( ) ( )− − − − − − − − − − − − − − − − − −+ + − + = + + +2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Using kg m s− −kg m s m s m s m s m s kg m s m s m s m s( (− − − − − − − − − − − − − − − − − −+ + − + = + + +2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
Solutions of the Navier-Stokes EquationsPreliminaries
3
Option 1Solution of the Linearized Navier-Stokes Equation
in the x-directionThe equation will be linearized by redefinition. The nine-term equation will be reduced to six terms.
Given: µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2Vx Vx Vx x Vx
xVx
yVx
zVx
x y zpx
gx tV
xV
yV
z+ + − + = + + + (A)
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ2
2
2
2
2
2Vx Vx Vx x Vx
xVx
yVx
zVx
x y zpx t
Vx
Vy
Vz
gx (B)
− + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx Vx Vx x Vx
x y zpx t
gx (C)
Plan: One will split-up equation (C) into five equations, solve them, and combine the solutions. Onsplitting-up the equations and proceeding to solve them, the non linear terms could be redefined andmade linear. This linearization is possible if the gravitational force term is the subject of theequation as in equation (B). After converting the non-linear terms to linear terms by redefinition,one will have only six terms as in equation (C). One will show logically how equation (C) wasobtained from equation (B), using a method which will be called the multiplier method.Three main steps are covered.In main Step 1, one shows how equation (C) was obtained from equation (B)In main Step 2, equation (C) will be split-up into five equations.In main Step 3, each equation will be solved.In main Step 4, the solutions from the five equations will be combined.In main Step 5, the combined relation will be checked in equation (C). for identity.
PreliminariesHere, one covers examples to illustrate the mathematical validity of how one splits-up equation (C).Let one think like a child - Albert Einstein. Actually, one can think like an eighth or a ninth grader. Suppose one performs the following operations:
Example 1: 10 20 25 55+ + = (1) 10 55 5510
55211= =× × (2)
20 55 552055
411= =× × (3)
25 55 552555
511= =× × (4)
Equations (2), (3), and (4) can be written as follows:
10 55= a (5) 20 55= b (6) 25 55= c (7) One will call a b, and c multipliers.
Above, a = 211, b = 4
11 , c = 511
Observe also that a b c+ + = 1( 2
114
115
111111 1+ + = = )
Solutions of the Navier-Stokes EquationsPreliminaries
4
Example 2: Addition of only two numbers 20 25 45+ = (8) 20 45 4520
4549= =× × (9)
25 45 452545
59= =× × (10)
Equations (9), and (10), can be written as follows: 20 45= a (11) 25 45= b (12) Rewrite (8) by transposition. If 20 45 25− = − Then 20 25= − d ( d is a multiplier) − = −45 25 f ( f is a multiplier)
Above, d = − = −2025
45 , f = −− =45
2595 ,
Observe also here that d f+ = 1 (− + = =45
95
55 1)
a b+ = 1 ( 49
59
99 1+ = = )
One can conclude that the sum of the multipliers is always 1.
More formally: Let A B C S+ + = , where A B C S, , and . are real numbers. (for the moment), and one excludes 0. Let a b c, , be respectively, multipliers of the sum S corresponding to A B C, , . Then A Sa= , B Sb= , C Sc= ; and a b c+ + = 1 To show that a b c+ + = 1, Sa Sb Sc S+ + = . S a b c S( )+ + = (factoring out the S) a b c+ + = 1. (Dividing both sides of the equation by S)
Solutions of the Navier-Stokes EquationsPreliminaries
5
Example 3: Solve the quadratic equation; 6 11 10 02x x+ − =Method 1 (a common and straightforward method)By factoring, 6 11 10 02x x+ − = ( )( )3 2 2 5 0x x− + = and solving, ( )3 2 0x − = or ( )2 5 0x + = x = 2
3 , x = − 52 . Solution set: { , }− 5
223
Method 2: One applies the discussion in Example 2 One will call this method the multiplier method.Step 1: From 6 11 10 02x x+ − = (1) 6 11 102x x+ = 6 102x a= ; (Here, a is a multiplier) 3 52x a= (2) 11 10x b= (Here, b is a multiplier) 11 10 1x a= −( ) ( )a b+ = 1 11 10 10x a= − x a= −10 10
Again, one obtains the same solution set { , }− 52
23 as by the factoring method.
About the multipliers
The values of the multipliers obtained were a = 154 3 3
4 or , b = − −2 114
34 or ; a b= =4
151115. .
It easy to understand, say, in 20 45 452045
49= =× × , that the multiplier 4
9 can be viewed as thefraction of the multiplicand, 45 .Later, one will learn that the multipliers are ratio terms as in Examples 5, 6 and 7, below.
Solutions of the Navier-Stokes EquationsPreliminaries
6
Example 4 Solve ax bx c2 0+ + = by completing the square and incorporating the multiplier method.
Step 1: From ax bx c2 0+ + = ax bx c2 + = − Let ax cd2 = − ; ( d is a multiplier) (1) Let bx cf= − ( f is a multiplier) (2) (and d f+ = 1)
ax bx cd cf2 + = − − (Adding equations (1) and (2)
x ba x c
a d ca f2 + = − −
x ba x b
aba
ca d f2
2 2
2 2+ + ( ) − ( ) = − +( )
(completing the square on the left-hand side))
x ba
ba
ca+( ) = ( ) −2 2
2 2 ( d f+ = 1) (3)
One's interest is in equations (1), (2) and (3).
Step 2 x ba
ba
ca+ = ± ( ) −2 2
2
x ba
ba
ca
x ba
ba
aca
b aca
x ba
b aca
x b b aca
+ = ± −
+ = ± −
= ± −
= − ± −
= − ± −
2 4
2 444
44
24
24
2
2
2
2
2 2
2
2
2
2
Example 5: A grandmother left $45,000 in her will to be divided between eight grandchildren, Betsy, Comfort, Elaine, Ingrid, Elizabeth, Maureen, Ramona, Marilyn, in
Check; Sum of shares = $45,000 Sum of the fractions = 1
Solutions of the Navier-Stokes EquationsPreliminaries
7
Example 6: Sir Isaac Newton left ρgx units in his will to be divided between −µ ∂∂2
2Vx
x , −µ ∂∂
2
2Vx
y,
−µ ∂∂2
2Vx
z,∂∂px
, ρ ∂∂Vxt
, ρ ∂∂Vxx
Vx , ρ ∂∂Vyy
Vx , ρ ∂∂Vzz
Vx in the ratio a b c d f h m n: : : : : : : .
where a b c d f h m n+ + + + + + + = 1. How much does each receive?
Solution −µ ∂∂2
2Vx
x's share of ρgx units = a gxρ units
−µ ∂∂2
2Vx
y's share of ρgx units = b gxρ units
−µ ∂∂2
2Vx
z's share of ρgx units= c gxρ units
∂∂px
's share of ρgx units = d gxρ units
ρ ∂∂Vxt
's share of ρgx units= f gxρ units
ρ ∂∂Vxx
Vx 's share of ρgx units = h gxρ units
ρ ∂∂Vyy
Vx 's share of ρgx units = m gxρ units
ρ ∂∂Vzz
Vx 's share of ρgx units= n gxρ units
Sum of shares = ρgx units Note: a b c d f h m n+ + + + + + + = 1
Example 7: The returns on investments A B C D, , , are in the ratio a b c d: : : . If the total return on these four investments is P dollars, what is the return on each of these investments? ( )a b c d+ + + = 1Solution Return on investment A aP= dollars Return on investment B bP= dollars Return on investment C cP= dollars Return on investment D dP= dollars
Check aP bP cP dP P+ + + = P a b c d P( )+ + + = a b c d+ + + = 1 (dividing both sides by P)The objective of presenting examples 1, 2, 3, 4, 5, 6, and 7 was to convince the reader that theprinciples to be used in splitting the Navier-Stokes equations are valid.In Examples 3 and 4, one could have used the quadratic formula directly to solve for x , withoutfinding a and b first. The objective was to convince the reader that the introduction of a and b didnot change the solution sets of the original equations.
For the rest of the coverage in this paper, a multiplier is the same as a ratio termThe multiplier method is the same as the ratio method.
Linearization of Non-Linear terms
8
Main Step 1Linearization of the Non-Linear Terms
Step 1: The main principle is to multiply the right side of the equation by the ratio terms This step is critical to the removal of the non-linearity of the equation. ρgx is to be divided by the terms on the left-hand--side of the equation in the ratio a b c d f h m n: : : : : : : ( a b c d f h m n+ + + + + + + = 1
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρ
2
2
2
2
2
2Vx Vx Vx x
x y zpx
Vxt
VxVxx
VyVxy
VzVxz
gx
nonlinear terms
all acceleration terms
6 744444 844444
1 24444444 34444444 (1)
Apply the principles involved in the ratio method covered in the preliminaries, to the nonlinear terms (the last three terms.)
Then ρ ∂∂ ρVVz
n gxzx = , where n is the ratio term corresponding to ρ ∂
∂VVzzx .
Vz
ngxzVx∂∂ = (2)
VdVdz ngxz
x = (One drops the partials symbol, since a single independent variable is involved)
dzdt
dVdz ngx V dz
dtx
z= = ( , by definition)
ddt ngxVx = (3)
Therefore, VVz
dVdt ngz
x xx
∂∂ = = (4)
Step 2: Similarly, Let ρ ∂∂ ρVVy
m gyx
x= ( m is the ratio term corresponding to ρ ∂∂VVyyx ) (5)
VdVdy mgy
xx= (One drops the partials symbol, since a single independent variable is involved)
dydt
dVdy mgx V
dydt
xy= = ( )
ddt mgxVx = (6)
Therefore, VdVdy
dVdt mgxy
x x= = (7)
Step 3: Let ρ ∂∂ ρVx
h gxVx
x= where h is the ratio term corresponding to ρ ∂∂Vxx
Vx .
VVx
hgxx
x∂∂ = (8)
VdVdx hgx
xx= (One drops the partials symbol, since a single independent variable is involved)
dxdt
dVdx hg V dx
dtx
x x= = ( )
ddt hgVx
x= (9) Therefore, VVx
dVdt hgx
x xx
∂∂ = = (10)
Linearization of Non-Linear terms
9
From equations (4), (7), (10), Vx
Vy
Vz
ddtx
Vxy
Vxz
Vx Vx∂∂
∂∂
∂∂= = = and
Vx
Vy
Vzx
Vxy
Vxz
Vx∂∂
∂∂
∂∂+ + = 3
ddtVx ( 11)
Thus, the ratio of the linear term ∂∂Vtx to the nonlinear sum V
xV
yV
zxVx
yVx
zVx∂
∂∂∂
∂∂+ + in
equation (1) is 1 to 3. Unquestionably, there is a ratio between the sum of the nonlinear
terms and the linear term ∂∂Vxt
. This ratio must be verified experimentally.
Note: One could have obtained equation (C) from equation (A) by redefining the nonlinear terms by carelessly disregarding the partial derivatives of the nonlinear terms in equation (1).However, the author did not do that, but logically, the terms became linearized. Note also that the above linearization is possible only if ρgx is the subject of the equation,and it will later be learned that a solution to the logically linearized Navier-Stokes equation isobtained only if ρgx is the subject of the equation.
Step 4: Substitute the right side of equation (11) for the nonlinear terms on the left- side of
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρ
2
2
2
2
2
2Vx Vx Vx x
x y zpx
Vxt
VxVxx
VyVxy
VzVxz
gx
nonlinear terms
all acceleration terms
6 744444 844444
1 24444444 34444444 (12)
Then one obtains − − − + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ
∂∂ ρ
∂∂ ρ
2
2
2
2
2
2 3Vx Vx Vx x
x y zpx
Vxt
Vxx
gx
all acceleration terms1 2444 3444
− − − + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ
2
2
2
2
2
2 4Vx Vx Vx x Vx
x y zpx t
gx (simplifying) (13)
Now, instead of solving equation (1), previous page, one will solve the following equation
− − − + + =KVx
KVy
KVz
px
Vt
gx x x xx
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4 (k = µ
ρ ) (14)
Main Step 2Step 5: In equation (14) divide gx by the terms on the left side in the ratio a b c d f: : : : .
− = − = − = = =KV
ag KV
bg KV
cgpx
dgVt
fgxx x
xy x
xz x x x
x∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4; ; ; ;
( a b c d f, , , , are the ratio terms and a b c d f+ + + + = 1).
As proportions: −
=−
=−
= = =K
Vx
ag
KV
yb
g KVz
cg
px
dg
Vt
fg
x x xx
x
x x x x
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
2
1 1 1
1
1
4
1; ; ; ;
One can view each of the ratio terms a b c d f, , , , as a fraction (a real number) of gx contributedby each expression on the left-hand side of equation (14) above
Solutions of the five sub-equations
10
Main Step 3Step 6: Solve the differential equations in Step 5.
Solutions of the five sub-equations
− =KV
agxx
∂∂2
2
kV
ag
V ak g
Vx
agk x C
Vag
k x C x C
xx
xx
x
x
∂∂
∂∂∂∂
2
22
2
12
1 2
1
2
= −
= −
= − +
= − + +
− =KVy
bgx∂∂2
2
KVy
bg
Vy
bk g
Vy
bgk y C
Vbg
k y C y C
x
x
x
x
∂∂
∂∂∂∂
2
2
2
2
3
22
3 42
= −
= −
= − +
= − + +
− =KVz
cgx∂∂2
2
KVz
cg
Vz
ck g
Vz
cgk z C
Vcgk z C z C
x
x
x
x
∂∂
∂∂∂∂
2
22
2
5
32
5 62
= −
= −
= − +
= − + +
1ρ∂∂px
dg=
1
7
ρ∂∂
∂∂ ρ
ρ
px
dg
px
d g
p d gx C
=
== +
4∂∂Vt
fgx =
∂∂Vt
fgx = 4
Vfg
tx4 4=
Main Step 4Step 7: One combines the above solutions
V V V V Vag
k x C x Cbg
k y C y Ccgk z C z C
fgt C
agk x C x
bgk y C y
cgk z C z
fgt C
agk x
bgk y
x x x x x= + + +
= − + + − + + − + + +
= − + − + − + +
= − − −
+
+
1 2 3 4
21 2
23 4
25 6 7
21
23
25 9
2 2
2 2 2 4
2 2 2 4
2 2cgcgk z C x C y C z
fgt C
agk x
bgk y
cgk z C x C y C z
fgt C
gk ax by cz C x C y C z
fgt C
Vg
ax by cz
x x
xx
2 4
2 2 2 4
2 4
2
21 3 5 9
2 2 21 3 5 9
2 2 21 3 5 9
2 2
+ + + +
= − − − + + + +
= − + + + + + +
= − + +
+
+
+( )
(ρµ
221 3 5 94)
( )
+ + + +
=
+C x C y C zfg
t C
P x d g x
x
xρ
V V V V V
V x y z tg
ax by cz C x C y C zfg
t C
P x d g x
x x x x x
xx x
x
= + + +
= − + + + + + +
=
+
1 2 3 4
2 2 21 3 5 92 4( , , , ) ( )
( )
ρµ
ρ
For V x tx ( , ), let y z= =0 0,
Then V x tg
ax C xfg
t Cxx x( , ) = − + + +ρµ2 4
21 9 P x d g xx( ) = ρ
V x V xg
ax C x Cx xx( , ) ( )0 2
0 210 9= = − + +ρ
µ
Checking in equation (C)
11
Main Step 5Checking in equation (C)
Step 8: Find the derivatives, using
Vg
ax by cz C x C y C zfg
t Cxx x= − + + + + + ++
ρµ2 4
2 2 21 3 5 9( )
P x d g xx( ) = ρ
∂∂
ρµ
Vx
gax Cx x= − +2 2 1( )
1. ∂∂
ρµ
2
2Vx
a gx x= −
4. ∂∂ ρpx
d g= ; 5. ∂∂Vt
fgx x= 4
∂∂
ρµ
Vy
gby Cx x= − +( ) 3
2. ∂∂
ρµ
2
2Vy
b gx x= −
∂∂
ρµ
Vz
gczx x= − ( )
3.∂∂
ρµ
2
2Vx
zc gx= − ;
Step 9: Substitute the derivatives from Step 8 in − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂∂ ρ( )
2
2
2
2
2
2 4Vx Vx Vx x Vx
x y zpx t
gx to check for identity (to determine if the relation obtained satisfies the original equation).
− + + + + =
− − − − + + =
+ + + + =
+
µ ∂∂
∂∂
∂∂
∂∂ ρ ∂∂ ρ
µ ρµ
ρµ
ρµ ρ ρ ρ
ρ ρ ρ ρ ρ ρ
( )
( )?
?
2
2
2
2
2
2 4
4 4
Vx
Vy
Vz
px
Vt
g
a g b g c gd g
fg g
a g b g c g d g fg g
ag
x x x xx
x x x
x x x x x x
x
x x x
bgbg cg dg fg g
g a b c d f g
g g a b c d f
g g
x x x x x
x x
x x
x x
(
Yes
+ + + =
+ + + + =
= + + + + =
=
?
?
?
?
( )
( ) )1 1
Scrapwork∂∂
ρµ
∂∂
ρµ
∂∂
ρµ
2
2
2
2
2
2
Vx
Vx
Vx
xa g
yb g
zc g
x
x
x
= −
= −
= −
;
;
;
∂∂ ρpx
d g= ; ∂∂Vt
fgx x= 4
An identity is obtained and therefore, the solution of equation (C), p.96, is given by
V x y z tg
ax by cz C x C y C zfg
t C P x d g xx xx x( , , , ) ( ) ( )= − + + + + + + =+
ρµ ρ2 4
2 2 21 3 5 9;
Solutions Summary
12
Solution Summary for Vx , Vy and Vz
For Vx a b c d f+ + + + =1
µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2Vx Vx Vx x Vx
xVx
yVx
zVx
x y zpx
gx tV
xV
yV
z+ + − + = + + +
− − − + + =Kx
Ky
Kz
px t
gVx Vx Vx Vx
x∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4
V V V V Vag
k x C x Cbg
k y C y Ccgk z C z C
fgt C C
agk x C x
bgk y C y
cgk z C z
fgt C
agk x
bgk
x x x x x= + + +
= − + + − + + − + + + +
= − + − + − + +
= − −
+
+
1 2 3 4
21 2
23 4
25 6 7 8
21
23
25 9
2
2 2 2 4
2 2 2 4
2 2 yycgk z C x C y C z
fgt C
V x y z tg
ax by cz C x C y C zfg
t C
P x d gx
yx x
2 21 3 5 9
2 2 21 3 5 9
2 4
2 4
− + + + +
= − + + + + + +
=
+
+( , , , ) ( )
( )
ρµ
ρFor V x tx ( , ), let y z= =0 0,
Then V x tg
ax C xfg
t Cxx x( , ) ( )= − + + +ρµ2 4
21 9
For Vy h j m n q+ + + + =1
µ∂∂
∂∂
∂∂
∂∂ ρ ρ
∂∂
∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2
Vy Vy Vyy
Vyx
Vyy
Vyz
Vy
x y zpy
gt
Vx
Vy
Vz
+ + − + = + + +
− − − + + =KVx
KVy
KVz
py
Vt
gy y y yy
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4
Vhg
k x C xjg
k y C ymg
k z C zng
t
V x y z tg
hx jy mz C x C y C zqg
t C
P y n g y
y
y
y
y y y
y
y
y
= − + − + − + +
= − + + + + + + +
=
2 2 2 4
2 4
21
23
25
2 2 21 3 5( , , , ) ( )
( )
ρµ
ρ
For Vz r s u v w+ + + + =1
µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2Vz Vz Vz Vz
xVz
yVz
zVz
x y zpz
gt
Vx
Vy
Vzz+ + − + = + + +
− − − + + =k k kpz t
gVz
x
Vzy
Vzz
Vzz
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4
Vrg
k x C xsg
k y C yug
k z C zwg
t
V x y z tg
rx sy uz C x C y C zwg
t C
P z v g z
z
z
z
z z z z
z z
= − + − + − + +
= − + + + + + + +
=
2 2 2 4
2 4
21
23
25
2 2 21 3 5( , , , ) ( )
( )
ρµ
ρ
Discussion About Solutions
13
Discussion About SolutionsA solution to equation − + + + + =µ ∂
∂∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx Vx Vx Vx
xx y zpx t
g (C) is
V x y z tg
ax by cz C x C y C zfg
t C
P x d g x a b c d f
xx x
x
( , , , ) ( )
( ) (
= − + + + + + +
= =
+ρµ
ρ2 4
1
2 2 21 3 5 9
; + + + + ).
This relation gives an identity when checked in Equation (C) above.
One observes above that the most important insight of the above solution is the indispensability ofthe gravity term in incompressible fluid flow. Observe that if gravity, g , were zero, the first threeterms, the seventh term, and P x( ) would all be zero be. This result can be stated emphatically thatwithout gravity forces on earth, there will be no incompressible fluid flow on earth as is known.The above result will be the same when one covers the general case, Option 4.
The above parabolic solution is also encouraging. It reminds one of the parabolic curve obtainedwhen a stone is projected vertically upwards at an acute angle to the horizontal..The author also tried the following possible approaches: (D), (E) and (F), but none of the possiblesolutions completely satisfied the corresponding original equations (D), (E) or (F) .
µ ∂∂ µ ∂∂ µ ∂∂ ρ ρ ∂∂∂∂
2
2
2
2
2
2 4Vx Vx Vx
xVx
x y zg
tpx
+ + + − = ( D) (One uses the subject ∂∂px
K Vx
K Vy
K Vz
px
g Vt
x x x xx
4 4 41
4 42
2
2
2
2
2∂∂
∂∂
∂∂ ρ
∂∂
∂∂+ + − + = (E), (One uses the subject
∂∂Vtx
− − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
y z
gxt
px x
(F) (One uses subject ∂∂2
2Vx
x
Integration Results Summary
Case 1: − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx
Vy
Vz
px
Vt
gx x x xx (C)
V x y z t
gax by cz C x C y C z
fgt C
P x d g x a b c d f
x
x
x x( , , , ) ( )
( ) (
= − + + + + + +
= =
+ρµ
ρ2 4
1
2 2 23 5 91
; + + + + ) <----Solution
Case 2: µ ∂∂ µ ∂∂ µ ∂∂ ρ ρ ∂∂∂∂
2
2
2
2
2
2 4Vx Vx Vx
xVx
x y zg
tpx
+ + + − = ( D). (One uses the subject ∂∂px
V x y z t ax by cz C x C y C z
ft C
P x d g x
xx x
x
p( , , . ) ( )
( )
= + + + + + + − +
=
λµ λ λ
ρρ
2 41
2 2 251 3
Case 3: K Vx
K Vy
K Vz
px
g Vt
x x x xx
4 4 41
4 42
2
2
2
2
2∂∂
∂∂
∂∂ ρ
∂∂
∂∂+ + − + = (E). (One uses the subject
∂∂Vtx
V x y z t C x C x t C y C y t
C z C zt g
f t x C
P x x d g x
x x x y
z z
x
e e
e
yy
z
( , , , ) ( cos sin ) ( cos sin )
( cos sin )
( )
( ) ( )
( )
= + − + + −
+ + − + + +
= =
1 22
3 4
2
5 6
2
4 8
λ λ λ λ
λ λ λ
λ ρ
λ β λ ω
λ ε
Discussion About Solutions
14
Case 4: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
y z
gxt
px x
(F). (One uses the subject ∂∂2
2Vx
x
V x y z t A y B y Cex
Dex
E z F z He Lex g x
c Ax B A x B x t
x
aa
aa
bb
x bb x e
( , , , ) ( cos sin )
( cos sin(
( cos sin ) )
( ) )
( ) )
(
( /
= + + −
+ + +−
− + + + + −
+
λ λ
λ λ ρµ λ λ λ α
λ
λ λ
λ λ 22
2 1 1
222
2 3µρ
f x C x C
P x d g xx
+ +
=
)
( )
Note: Relations for equations with subjects gx and ∂∂px
are almost identical.
By comparing possible solutions for equations (C) and (D), λ ρx xg= − in relation for (D).
V x y z t ax by cz C x C y C zf
t C
P x d g x
xx x
x
p( , , , ) ( )
( )
= + + + + + + − +
=
λµ λ λ
ρρ
2 41
2 2 251 3
The comparative analysis of the possible solutions when checked in each corresponding equation ispresented in the table below.
EquationEquationSubject
Number of terms ofpossible solutions notsatisfying original equation
Case 1: − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx
Vy
Vz
px
Vt
gx x x xx gx
NoneCase 1 yields the solution
Case 2: µ ∂∂ µ ∂∂ µ ∂∂ ρ ρ ∂∂∂∂
2
2
2
2
2
2 4Vx Vx Vx
xVx
x y zg
tpx
+ + + − = ∂∂px One term
Case 3: K Vx
K Vy
K Vz
px
g Vt
x x x xx
4 4 41
4 42
2
2
2
2
2∂∂
∂∂
∂∂ ρ
∂∂
∂∂+ + − + =
∂∂Vtx At least 2 terms
Case 4: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
y z
gxt
px x
∂∂2
2Vx
x At least 2 terms
Outcome 1: With gx included and with gx as the subject of the equation.The solution is straightforward and the possible solution checks well in the original equation (C)Also, if gx or ρgx is not the subject of the equation, the linearization of the nonlinear terms couldnot be justified.
Discussion About Solutions
15
Outcome 2: With gx included but with ∂∂Vtx as the subject of the equation.
There are two problems when checking . 1. For ∂∂ ρ
∂∂
λρ
Vt
px
td
x = − → −14 4 ; 2. g V
tgt
fx
4 4= →∂∂
With d and f in the denominators, the multipliers sum a b c d f+ + + + = 1 is false.
Outcome 3 : With gx excluded, and ∂∂Vtx as the subject of the equation, there is one problem:
− = → −14 4ρ
∂∂
∂∂
λρ
px
Vt
td
x .With d in the denominator a b c d f+ + + + = 1 is false
Outcome 4 : With gx included, and ∂∂2
2Vx
x as the subject of the equation, there are at least, two
problems in the checking with the multipliers c f and in the denominators. Checking for a b c d f+ + + + = 1 is impossible.
Characteristic curves of the integration results
EquationsEquationSubject Curve characteristics
Case 1: − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx
Vy
Vz
px
Vt
gx x x xx gx
Parabolic and Inverted
Case 2: µ ∂∂ µ ∂∂ µ ∂∂ ρ ρ ∂∂∂∂
2
2
2
2
2
2 4Vx Vx Vx
xVx
x y zg
tpx
+ + + − = ∂∂px Parabolic
Case 3: K Vx
K Vy
K Vz
px
g Vt
x x x xx
4 4 41
4 42
2
2
2
2
2∂∂
∂∂
∂∂ ρ
∂∂
∂∂+ + − + =
∂∂Vtx Periodic and decreasingly
exponential
Case 4: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
y z
gxt
px x
∂∂2
2Vx
x Periodic, parabolic, andexponential
The following are possible interpretations of the roles of the terms based on the types of curvesproduced when using the terms as subjects of the equations.
1. gx and ∂∂px
are involved in the parabolic motion of fluids..
2. ∂∂Vtx and
∂∂2
2Vx
x are involved in the parabolic, periodic and decreasingly exponential motion.
3. gx is responsible for the forward motion.
Discussion About Solutions
16
Definitions and Classification of Equations
− − − + + =KVx
KVy
KVz
px
Vt
gx x x xx
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
2
2
2
2
2
21 4 (k = µ
ρ )
One may classify the equations involved in Option 1 according to the following:Driver Equation: A differential equation whose integral satisfies its corresponding equation.Supporter equation: A differential equation which contains the same terms as the driver equation but whose integral does not satisfy its corresponding equation but provides useful information about the driver equation.Note that the driver equation and a supporter equation differ only in the subject of the equation.
EquationEquationSubject Type of
equation
# of terms ofrelation notsatisfying original equation
Case 1: − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2 4Vx
Vy
Vz
px
Vt
gx x x xx gx
DriverEquation
None
Case 2: µ ∂∂ µ ∂∂ µ ∂∂ ρ ρ ∂∂∂∂
2
2
2
2
2
2 4Vx Vx Vx
xVx
x y zg
tpx
+ + + − = ∂∂px Supporter
equationOne term
Case 3: K Vx
K Vy
K Vz
px
g Vt
x x x xx
4 4 41
4 42
2
2
2
2
2∂∂
∂∂
∂∂ ρ
∂∂
∂∂+ + − + =
∂∂Vtx Supporter
equationAt least 2 terms
Case 4: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
y z
gxt
px x
∂∂2
2Vx
xSupporterequation At least 2 terms
Case 5: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1Vx Vx Vx Vx
x z
gxt
px y
∂∂
2
2Vx
y Supporterequation
At least 2 terms
Case 6: − − − + + =∂∂
∂∂
ρµ
ρµ∂∂ µ
∂∂
∂∂
2
2
2
2
2
24 1V
yVx
gx Vt
px
Vz
x x x x ∂∂2
2Vz
x Supporterequation
At least 2 terms
One can apply the above definitions in solving the magnetohydrodynamic equations.
Discussion About Solutions
17
Applications of the splitting technique in science, engineering, business fieldsThe approach used in solving the equations allows for how the terms interact with each other. The author has not seen this technique anywhere, but the results are revealing and promising.Fluid flow design considerations:
1. Maximize the role of gx forces, followed by; 2. ∂∂px
forces; then 3.∂∂Vtx
(Make gx happy by always providing a workable mgsinθ ) .For long distance flow design such as for water pipelines, water channels, oil pipelines. wheneverpossible, the design should facilitate and maximize the role of gravity forces, and if design is
impossible to facilitate the role of gravity forces, design for ∂∂px
to take over flow.
The performance of ∂∂2
2Vx
x should be studied further, since its role is the most complicated: periodic,
parabolic, and decreasingly exponential.Tornado Effect ReliefPerhaps, machines can be designed and built to chase and neutralize or minimize tornadoes duringtouch-downs. The energy in the tornado at touch-down can be harnessed for useful purposes.Business and economics applications.1. Figuratively, if gx is the president of a company, it will have good working relationships with allthe members of the board of directors, according to the solution of the Navier-Stokes equation. If gx
is present at a meeting gx must preside over the meeting for the best outcome.
2. If gx is absent from a meeting, let ∂∂px
preside over the meeting, and everything will workout well.
However, if gx is present, gx must preside over the meeting.
To apply the results of the solutions of the Navier-Stokes equations in other areas or fields, the
properties, characteristics and functions of gx , ∂∂px
, ∂∂vtx must be studied to determine analogous
terms in those areas of possible applications. Other areas of applications include investments choicedecisions, financial decisions, personnel management and family relationships.
Option 2Solutions of 4-D Navier-Stokes Equations (linearized)
One advantage of the pairing approach is that the above solution can easily be extended to anynumber of dimensions.
If one adds µ ∂∂2
2Vs
x and ρ ∂∂VVssx to the 3-D x−direction equation, one obtains the 4-D Navier--
Stokes equation − + + + + + + =µ ∂∂
∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ ∂∂ ρ( ) ( )
2
2
2
2
2
2
2
2 4Vx Vx Vx Vx Vx
sx
x y z spx t
VVs
gx
After linearization, − + + + + + =µ ∂∂
∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ( ) ( )2
2
2
2
2
2
2
2 5Vx Vx Vx Vx Vx
x y z spx t
gx and its solution is
. V x y z s t
gax by cz es C x C y C z C s
fgt C
P x d g x a b c d e f
x
x
x x( , , , , ) ( )
( )
= − + + + + + + + +
= + + + + + =
+ρµ
ρ2 5
1
2 2 2 21 3 5 7 9
( ) .
For n−dimensions one can repeat the above as many times as one wishes. Back to Options
Solutions of the Euler Equations
18
Option 3Solutions of the Euler Equations of Fluid flow
In the Navier-Stokes equation, if µ = 0, one obtains the Euler equation. From
µ ∂∂
∂∂
∂∂
∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂( ) ( )
2
2
2
2
2
2Vx Vx Vx Vx
xVx
yVx
zVx
x y zpx
gx tV
xV
yV
z+ + − + = + + + , one obtains
Euler equation : ( ) ( )µ ∂∂ ρ ρ ∂
∂∂∂
∂∂
∂∂= − + = + + +0 p
xg
tV
xV
yV
zx
Vxx
Vxy
Vxz
Vx or
ρ ∂∂
∂∂
∂∂
∂∂
∂∂ ρ( )
Vxx
xy
Vxz
Vx xxt
VVx
Vy
Vz
px
g+ + + + = <---driver equation.
Euler equation ( :µ ∂∂
∂∂
∂∂
∂∂ ρ
∂∂= + + + + =0 1)
Vxx
Vxy
Vxz
Vxt
Vx
Vy
Vz
px
gx <---driver equation
Split the equation using the ratio terms f h n q de e e e e, , , , , , and solve. ( )f h n q de e e e e+ + + + = 1
1. ∂∂Vt
f gxe x=
V f g tx e x4 =V fg tx x4 =
2. V Vx
h gx xx
e∂∂ =
VdVdx h g
V dV h g dxx x
x x x
xe
e
==
Vh g xx
xe
2
2 = or
V h g xx xe2 2=
V h g xx e x= ± 2
3. VVy
n gy xx
e∂∂ =
VdVdy n g
V dV n gV V n g V
Vn g
VV
V
y x
y x x
y x x y
xx
y
y
y
x
y
y
y
e
e
e
e
V
=
== +
= +
≠
dyy (
y
ψψ
)( )
6
0
4. V Vz
q gz xx
e∂∂ =
VdVdz q g
V dV q gV V q g V
Vq g
VV
VV
z x
z x x
z x x
xx
z
z
z
z
x
z z
z
e
e
e
e
=== +
= +
≠
dz;z (
z
ψψ
)( )
7
0
5. 1ρ∂∂px
d ge x=
1
7
ρ∂∂
∂∂ ρ
ρ
px
d g
px
d g
p d g x C
e x
e x
e x
=
== +
V x y z t f g t h g xn g
Vq g
VV
VV
V C
P x d g x f h n q d V
xx
y
x
z
y z
z
z
e ee e
e e e e e e
x xy
y
z
x yV
( , , , )( ) ( )
( ) ( ) ,
= ± + + + + +
= + + + + = ≠ ≠
2
1 0 0
y z
ψ ψ
ρ
Find the test derivatives to check in the original equation.
1. ∂∂Vt
f gxe x= 2. V h g xx xe
2 2= ; 2 2V h g xxx
xVx e
∂∂ = ;
∂∂Vx
hV
x egx
x
x
V= ≠, 0
3
0
.
∂∂Vy
n gV
V
x e x
y
y
=
≠4
0
. ∂∂Vz
q gV
V
x e x
z
z
=
≠
5. ∂∂ ρpx
d ge x=
∂∂
∂∂
∂∂
∂∂ ρ
∂∂ ψ ψ
ρ ρ
Vxx
Vxy
Vxz
Vxy z
x yx
zx
zx
tV
xV
yV
zpx
g V V
f g Vh g
Vn g
V Vq gV d g g
x y z
xx
x yxe
e e eeV
+ + + + =
+ + + + =
1
1
(Above, ( and ( are arbitrary functions)
) )
?
f g h g n g q g d g ge e e e ex x x x x x+ + + + = ?
g f h n q d gx xe e e e e( )?
+ + + + =
( g g f h n q dx x e e e e e( )?
)1 1= + + + + =
g gx x?= Yes
Solutions of the Euler Equations
19
The relation obtained satisfies the Euler equation. Therefore the solution to the Euler equation is
V x y z t fgt hg xngV
qgV
y VyVy
z VzVz
C
P x d g x V V
xx
y
x
z
x y z
x( , , , )( ) ( )
( ) ,
= ± + + + + +
= ≠ ≠
2
0 0
y z
arbitrary functions
ψ ψ
ρ
1 2444 3444
The above is the solution of the driver equation. There are 5 supporter equations which will not besolved here.Question: Has the Euler equation of fluid flow been solved for the first time?Note: So far as the solutions of the equations are concerned, one needs not have explicit expressions for Vx , Vy, and Vz .The impediment to solving the Euler equations has been due to how to obtain sub-equations fromthe six-term equation. The above solution was made possible after pairing the terms of the equationusing ratios (by way of multipliers). The author was encouraged by Lagrange's use of ratios andproportion in solving differential equations. One advantage of the pairing approach is that theabove solution can easily be extended to any number of dimensions.
Extra:Linearized Euler Equation: If one linearizes the Euler equation as was done in Option 1, oneobtains
4 1∂∂ ρ
∂∂
Vx xxt
px
g+ = ; whose solution is Vfg
t C P x d g xx xx= + =4 ; ( ) ρ . (see Option 1 results)
Results for the Euler equations are presented below: for Vx , Vy and Vz
For Vx : ∂∂ ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρp
xVt
VVx
VVy
VVz
gxx
xy
xz
xx+ + + + =)
V x y z t fg t hg xng yV
qg zV
y VyVy
z VzVz
P x d g x
V
x xz
y z
xx
y
xx
V
( , , , )( ) ( )
( )
,
= ± + + + + =
≠ ≠
2
0 0
ψ ψρ
arbitrary functions
;
1 2444 3444 x-direction
For Vy , ∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρ
∂∂ ρp
y tV
xV
yV
Vz
gVy
xVy
yVy
z yy+ + + + =
V x y z t g t gg x
Vg
VV
VV
V P y g y
V
yx z
x z
z
x z
y yy y x
x
zy
V
( , , , )( ) ( )
( )
,
= ± + + + + =
≠ ≠
λ λλ λ ψ ψ λ ρ5 7
6 842
0 0
yz
;
y-direction
For Vz : ∂∂ ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρp
z tV
xV
yV
zg
Vzx
Vzy
Vzz
Vzz+ + + + =
V x y z t g t g zg xV
g yV
VV
VV P z g z
V
zx
x y
zx
z
y
x
x
y y
yzz
z
V
( , , , )( ) ( )
( )
,
= ± + + + + =
≠ ≠
β ββ β
βψ ψ
ρ5 86 7
42
0 0
;
z-direction
Note:By comparison with Navier-Stokes equation and its relation, a relation to Euler equation can befound by deleting the Navier-Stokes relation resulting from the µ -terms. Back to Options
Solutions of the Navier-Stokes Equations (Original)
20
Option 4 Solutions of the Navier-Stokes Equations
(Original)As it was in Option 1 for solving these equations,the first step here is to split-up the equation intoeight sub-equations using the ratio method. One will solve only the driver equation, based on theexperience gained in solving the linearized equation. There are 8 supporter equations.
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ2
2
2
2
2
2Vx Vx Vx x
x y z xx y zpx
Vt
VVx
VVy
VVz
gx x x x
nonlinear terms6 744444 844444
(A)
− − − + + + + + = =Kx
Ky
KVz
px
Vt
VVx
VVy
VVz
g KVx Vx x
x y z xx x x x∂
∂∂∂
∂∂ ρ
∂∂
∂∂
∂∂
∂∂
∂∂
µρ
2
2
2
2
2
21 ( ) (B)
Step 1: Apply the ratio method to equation (B) to obtain the following equations:
1 1 52
2
2
2
2
2. . ; 2. ; 3. ; 4. ; − = − = − = = =KV
ag KV
bg KV
cgpx
dgVt
fgxx x
xy x
xz x x
xx
∂∂
∂∂
∂∂ ρ
∂∂
∂∂
6. ; V Vx
hgx xx∂
∂ = 7. ; VVy
qgy xx∂
∂ = 8. V Vz
ngz xx∂
∂ =
where a b c d f h n q, , , , , , , are the ratio terms and a b c d f h n q+ + + + + + + = 1
Step 2: Solve the differential equations in Step 1. Note that after splitting the equations, the equations can be solved using techniques of ordinary differential equations.One can view each of the ratio terms a b c d f h n q, , , , , , , as a fraction (a real number) of gx contributed by each expression on the left-hand side of equation (B) above.
Solutions of the eight sub-equations
12
2. − =kV
agxx
∂∂
kV
ag
V ak g
Vx
agk x C
Vag
k x C x C
xx
xx
x
x
∂∂
∂∂∂∂
2
22
2
12
1 2
1
2
= −
= −
= − +
= − + +
2. − =KVy
bgx∂∂2
2
KVy
bg
Vy
bk g
Vy
bgk y C
Vbg
k y C y C
x
x
x
x
∂∂
∂∂∂∂
2
2
2
2
3
22
3 42
= −
= −
= − +
= − + +
3. − =KVz
cgx∂∂2
2
KVz
cg
Vz
ck g
Vz
cgk z C
Vcgk z C z C
x
x
x
x
∂∂
∂∂∂∂
2
22
2
5
32
5 62
= −
= −
= − +
= − + +
4 1. ρ∂∂px
dg=
1
7
ρ∂∂
∂∂ ρ
ρ
px
dg
px
d g
p d gx C
=
== +
5. ∂∂Vt
fgx =
V fgtx4 =
6. V Vx
hgx xx∂
∂ =
V dVdx hg
V dV hg dxV
hg x
V hg x
x x
x x x
x
x
xx
x
==
=
= ±
2
225
7. VVy
ngy xx∂
∂ =
VdVdy ng
V dV ngV V ng V
VngV
VV
y x
y x x
y x x y
xx y y
x
y
y y
=
== +
= +
dyy (
y
ψψ
)( )
6
8. V Vz
qgz xx∂
∂ =
VdVdz qg
V dV qgV V qg V
VqgV
VV
z x
z x x
z x x z
xx
z
z z
z
x
z
=== +
= +
dz;z (
zψψ
)( )
7
Note:ψ y Vy( ), ψ z Vz( )are arbitraryfunctions,(integrationconstants)Vy ≠ 0
Vz ≠ 0
Solutions of the Navier-Stokes Equations (Original)
21
Step 3: One combines the above solutions
V x y z t V V V V V V Vag
k x C xbg
k y C ycg
k z C z fg t hg xng yV
qg zV
VV
VV
g
x x x x x x x x
xy z y
z z
z
x
x x xx
x x y y
( , , , )( ) ( )
= + + + + + +
= − + − + − + ± + + + +
−
+
1 2 3 4 5 6 7
21
23
252 2 2 2
ψ ψ
ρ22 22 2 2
1 3 5µψ ψ
( )( ) ( )
ax by cz C x C y C z fg t hg xng y qg z V V
x xx
y
x
z
z
zV V V Vy y
y
z+ + + + + + ± + + + +
relation for linear terms relation for non - linear terms
arbitrary functions
6 7444444444 8444444444
1 244 344
6 74444444 844444444
+
= + + + + + + + = ≠ ≠
C
P x d g x a b c d f h n qx V Vy z
9
1 0 0( ) ; ( ) ,ρ
Step 4: Find the test derivatives Test derivatives for the linear part Test derivatives for the non-linear part
∂∂ρµ
2
2Vxa g
x
x
=
−
∂∂ρµ
2
2Vyb g
x
x
=
−
∂∂ρµ
2
2Vx
zc gx
=
−
∂∂ρ
px
d gx
= ∂∂Vt
fg
x
x
=V hg xx x
2 2=
2 2V hg xxx
xVx
∂∂ =
∂∂Vx
hgx
x
xxV V= ≠, 0
∂∂Vy
ngV
x x
y= ∂
∂Vz
qgV
x x
z=
Step 5: Substitute the derivatives from Step 4 in equation (A) for the checking.
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ2
2
2
2
2
2Vx
Vy
Vz
px
Vt
VVx
VVy
VVz
gx x x xx y z x
x x x x (A)
− − − − + + + + + =
+ + + + + + + =
µ ρµ
ρµ
ρµ ρ ρ ρ ρ ρ ρ
ρ ρ ρ ρ ρ ρ ρ ρ ρ
( ) ( ) ( ) ( )?
?
a g b g c gd g f g V
hgV V
ngV V
qgV g
a g b g c g d g f g h g n g q g g
x x xx
x
x
x x x x
x x yx
yz
x
zx
x x x x xx
x x x x x
x x
x x
ag bg cg dg fg hg ng qg g
g a b c d f h n q g
g g a b c d f h n q
x x x x
Yes (
+ + + + + + +
+ + + + + + + =
= + + + + + + + =
=?
( )
( ) )
?
?1 1
Step 6: The linear part of the relation satisfies the linear part of the equation; and the non-linear part of the relation satisfies the non-linear part of the equation.(B) below is the solution.
Analogy for the Identity Checking Method: If one goes shopping with American dollars andJapanese yens (without any currency conversion) and after shopping, if one wants to check the costof the items purchased, one would check the cost of the items purchased with dollars against thereceipts for the dollars; and one would also check the cost of the items purchased with yens againstthe receipts for the yens purchase. However, if one converts one currency to the other, one wouldonly have to check the receipts for only a single currency, dollars or yens. This conversion case issimilar to the linearized equations, where there was no partitioning in identity checking.
Solutions of the Navier-Stokes Equations (Original)
22
Summary of solutions for Vy, Vz ( P y g yy( ) = λ ρ4 , P z g zz( ) = β ρ4 )
− + + + + + + ± + + + + +
= + + + + + + = ≠ ≠
ρµ
ψ ψ
ρ
gax by cz C x C y C z fg t hg x
ng yV
qg zV
VV
VV C
P x d g x a b c d h n q
xx x
x
y
x
z
y y
y
z z
z
y zx V V
2 2
1 0 0
2 2 21 3 5 9( )
( ) ( )
( ) ; ( ) ,
(B)
Vg
x y z C x C y C z g t gg x
Vg
VV
VV
VP y g y
yx
y
z
x
x
z z
z
y
yy y
y x
x zV V
= − + + + + + + ± + + + +
= ≠ ≠
ρµ λ λ λ λ λ
λ λ ψ ψ
λ ρ2 2
0 0
12
22
32
1 3 5 5 76 8
4
( )( ) ( )
( ) ,
yz
Vg
x y z C x C y C z g t g zg x g y
zx
x
x
y
x y
zz
z
y
x y
yz
zV V
VV
V
V
V V
= − + + + + + ± + + + +
≠ ≠
+ρµ β β β β β β β ψ ψ
2 2
0 0
1 2 3 5 86 72 2 2
1 3 5( )( ) ( )
,
Option 5Solutions of 4-D Navier-Stokes Equations
One advantage of the pairing approach is that the above solution can easily be extended to anynumber of dimensions.
If one adds µ ∂∂2
2Vs
x and ρ ∂∂VVssx to the 3-D x−direction equation, one obtains
the 4-D Navier-Stokes equation
− + + + + + + + + + =µ ∂∂
∂∂
∂∂
∂∂
∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ( )
2
2
2
2
2
2
2
2Vx
Vy
Vz
Vs
px
Vt
VVx
VVy
VVz
VVs
gx x x xx y z s
xx
x x x x
whose solution is given byV x y z s t
gax by cz es C x C y C z C s fg t hg x
ng yV
qg zV
rg sV
VV
VV
VV
x
xx x
x
y
x
z s
y y
y
z z
z
s s
s
x
( , , , , )
( )
( ) ( ) ( )
=
− + + + + + + + + ± + + + +
+ +
ρµ
ψ ψ ψ2 22 2 2 2
1 3 5 6
arbitrary functions1 244444 34444
+
= + + + + + + + + + = ≠ ≠ ≠
C
P x d g x a b c d e f h n q rx x y sV V V
9
1 0 0 0
( ) ( ) , , ,ρ
For n−dimensions one can repeat the above as many times as one wishes.
Extra: Two-term Linearization of the Navier-Stokes Equation (Equation contains one nonlinear term)
By linearization as in Option 1, if one replaces ρ ∂∂ ρ ∂
∂VVy
VVzy z
x x+ by 2ρ ∂∂Vt
x in
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ2
2
2
2
2
2Vx Vx Vx x
x y z xx y zpx
Vt
VVx
VVy
VVz
gx x x x one obtains
− + + + + + + =µ ∂∂
∂∂
∂∂
∂∂
∂∂ ρ ∂
∂ ρ ∂∂ ρ( ) ( )
2
2
2
2
2
2
2
2 3Vx Vx Vx Vx Vx
xx y z spx t
VVx
gxx , whose solution is
Vg
ax by cz C x C y C zfg t
hg x Cx x y z t xx
x( , , , ) ( )= − + + + + + + ± +ρµ2 3 22 2 2
1 3 5 6 Back to Options
Conclusion
23
ConclusionSince one began solving the Navier-Stokes equations by thinking like an eighth grader, and one wasable to find a ratio technique for splitting the equations and solving them, perhaps, it is appropriate,after a few months of aging, to think like a ninth grader in the conclusion. One will reverse thecoverage approach and begin from the general case and end with the special cases.
Solutions of the Navier--Stokes equations (general case)x−direction Navier-Stokes Equation (also driver equation)
− − − + + + + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ2
2
2
2
2
2Vx
Vy
Vz
px
Vt
VVx
VVy
VVz
gx x x xx y z x
x x x x x−direction
V x y z t
gax by cz C x C y C z fg t hg x
ng yV
qg zV
VV
VV
x
xx x
x
y
x
z
z
z
y y
y
z
( , , , )
( )( ) ( )
=
− + + + + + + ± + + + +ρµ
ψ ψ2 22 2 2
1 3 5
solution for linear terms
arbitrary functions
solution for non - linear terms6 7444444444 8444444444
1 244 344
6 7744444444 844444444
+
= + + + + + + = ≠ ≠
C
P x d g x a b c d h n q Vx Vy z
9
1 0 0( ) ; ( ) ,ρ
One observes above that the most important insight of the above solution is the indispensability ofthe gravity term in incompressible fluid flow. Observe that if gravity, g , were zero, the first threeterms, the 7th term, the 8th term, the 9th term, the 10th term and P x( ) would all be zero.This result can be stated emphatically that without gravity forces on earth, there will be noincompressible fluid flow on earth as is known. The above is a very important new insight, becausein posing problems on incompressible fluid flow, it is sometimes suggested that the gravity term iszero. Such a suggestion would guarantee a no solution to the problem, according to the abovesolution of the Navier-Stokes equation.The author proposed and applied a new law, the law of definite ratio for incompressible fluid flow.This law states that in incompressible fluid flow, the other terms of the fluid flow equation dividethe gravity term in a definite ratio and also each term utilizes gravity to function. This law wasapplied in splitting-up the Navier-Stokes equations. The resulting sub-equations were readilyintegrable, and even the nonlinear sub-equations were readily integrated.The x−direction Navier-Stokes equation was split-up into sub-equations using ratios. The sub-equations were solved and combined. The relation obtained from the integration of the linear part ofthe equation satisfied the linear part of the equation and the relation obtained from integrating thenonlinear part of the equation satisfied the nonlinear part of the equation. By solving algebraicallyand simultaneously for Vx , Vy and Vz , the ( )ng y Vx y and ( )qg z Vx z terms would be replaced byfractional terms containing square root functions. One may note that in checking the relationsobtained for integrating the equations for possible solutions, one needs not have explicitexpressions for Vx , Vy, and Vz , since these behave as constants in the checking process. The abovesolution is the solution to the driver equation. There are eight supporter equations (see below andsee also Option 1 solution, p110). Only the solution to the driver equation completely satisfies itscorresponding Navier-Stokes equation.. A supporter equation does not completely satisfy itscorresponding Navier-Stokes equation. The above x−direction solution is the solution everyone hasbeen waiting for, for nearly 150 years. It was obtained in two simple steps, namely, splitting theequation using ratios and integrating. The task for the future is to solve the equations for Vx , Vy
and Vz simultaneously. and algebraically. in order to replace two implicit terms of the solution.
Conclusion
24
Supporter Equations
1
2
2
2
2
2
2
2
2
2
2
2
2
2
.
.
− − − + + + + + =
− − − +
µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ρ ∂
∂
µ ∂∂ µ ∂∂ µ ∂∂∂∂
Vx
Vy
Vz
px
Vt
VVy
VVz
g VVx
Vx
Vy
Vz
px
x x x xy z x x
x x x x
x x x x
++ + + + =
− − − + + + + + =
−
ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ρ ∂∂
µ ∂∂ µ ∂∂ µ ∂∂ ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂
VVx
VVy
VVz
gVt
Vx
Vy
Vz
Vt
VVx
VVy
VVz
gpx
x y z x
x x xx y z x
x
x x x x
x x x x3
4
2
2
2
2
2
2.
.
µµ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂ ρ µ ∂∂2
2
2
2
2
2Vy
Vz
px
Vt
VVx
VVy
VVz
gVx
x x xx y z x
xx x x x− + + + + + + = −
Explicit Functions for Vx , Vy, and Vz ,For explicit functions for Vx , Vy, and Vz , one has to solve (algebraically) the simultaneous system
of solutions for Vx , Vy, and Vz .
System of Navier Stokes relations to solve for simultaneously −=
− + + + + + + ± + + + +
=
− +
V V VV
gax by cz C x C y C z fg t hg x V V qg V V ng V V
V VV
gx
x y z
x
z x x z
y z
y
xx x y z z y y y
y
, ,
( ( ) [ ( ( )
( (
) z )] [ y ]
(algebraically).
ρµ ψ ψ
ρµ λ λ
2 2
2
2 2 21 3 5
12
222
32
1 3 5 5 7 8 6
12
22
32
1 3 5
2
2
y z C x C y C z g t g g g x
V VV
gx y z C x C y C z
y y x z z x xV V V V V Vz y x y z
x z
z
z
+ + + + + ± + + + +
=
− + + + + +
λ λ λ λ ψ λ ψ
ρµ β β β
) [ ( ) [ ( )]
( ( )
y ) z ]
++ ± + + + +β β β ψ β ψ5 826 7
g t g z g x g y
V V
z z x y
x y
V V V V V Vz x x y z y y x) [ ( )] [ ( )]
Special Cases of the Navier-Stokes Equations1. Linearized Navier--Stokes equationsOne may note that there are six linear terms and three nonlinear terms in the Navier-Stokesequation. The linearized case was covered before the general case, and the experience gained in thelinearized case guided one to solve the general case efficiently. In particular, the gravity term mustbe the subject of the equation for a solution. When the gravity term was the subject of the equation,the equation was called the driver equation. A splitting technique was applied to the linearizedNavier-Stokes equations (Option 1). Twenty sub-equations were solved. (Four sets of equationswith different equation subjects). The integration relations of one of the sets satisfied the linearizedNavier-Stokes equation; and this set was from the equation with gx as the subject of the equation.In addition to finding a solution, the results of the integration revealed the roles of the terms of theNavier-Stokes equations in fluid flow. In particular, the gravity forces and ∂ ∂p x are involved
mainly in the parabolic as well as the forward motion of fluids; ∂ ∂V tx and ∂ ∂2 2Vx x are involvedin the periodic motion of fluids, and one may infer that as µ increases, the periodicity increases.One should determine experimentally, if the ratio of the linear term ∂ ∂V tx to the nonlinear sumV V x V V y V V zx x y x z x( ) ( ) ( )∂ ∂ ∂ ∂ ∂ ∂+ + is 1 to 3.
Conclusion
25
V x y z t
gax by cz C x C y C z
fgt C P x d g xx
x xx( , , , ) ( ) ( )= − + + + + + + +
−
=ρµ ρ2 4
2 2 21 3 5 9 ;
Solution to linearized Navier Stokes equation6 74444444444 84444444444
− − − + + =µ ∂∂ µ ∂∂ µ ∂∂∂∂ ρ ∂∂ ρ
2
2
2
2
2
2 4Vx Vx Vx x Vx
x y zpx t
gx
Linearized Equation6 7444444444 8444444444
2. Solutions of the Euler equationSince one has solved the Navier-Stokes equation, one has also solved the Euler equation.
Euler equation ( = 0) : µ ∂∂
∂∂
∂∂
∂∂ ρ
∂∂
Vxx
Vxy
Vxz
Vxt
Vx
Vy
Vz
px
gx+ + + + =1
V x y z t f g t h g xn g
Vq g
VV
VV
V C
P x d g x f h n q d V
xx x
z
y y
y
z z
z
y z
e ee e
e e e e e e
x xy
x V
( , , , )( ) ( )
( ) ( ) ,
= ± + + + + +
= + + + + = ≠ ≠
2
1 0 0
y z
arbitrary functions
ψ ψ
ρ
1 244 344 x-direction
A Euler solution system to solve for
) z )] [ y ]
y ) z ]
V V V
Vf g t h g x V V q g V V n g V V
V V
Vg t g V V g V V
x z
xz z
y z
yy y z y x
y
e x e x y e x z z y e x y y
x z z
, ,
( [ ( ( )
( [ ( ) [
=± + + + +
=± + + +
2
25 7 8
ψ ψ
λ λ λ ψ λ66
5 8 6 72
g x V V
V V
Vg t g z V V g x V V g y V V
V V
y z
x z
zx y
x y
x x
z z z x x y z y y x
+
=± + + + +
ψ
β β ψ β ψβ
( )]
( [ ( )] [ ( )]
)
Overall Conclusion The author was encouraged by Lagrange's use of ratios and proportions in solving differentialequations. However the use of ratios in this paper is much more direct. One very interesting fact isthat after using ratios to split the equation with the gravity term as the subject of the equation, theintegration was straightforward. The author believes that if the ratio or proportion method ofsplitting the equations could not yield the solution, no other method can even come close, since useof ratios is the most fundamental principle in the division of any quantity into parts.Finally, in fluid flow, the indispensable term or factor is gravity, according to the above solutions.For any fluid flow design, one should always maximize the role of gravity for cost-effectiveness,durability, and dependability. Perhaps, Newton's law for fluid flow should read "Sum of everythingelse equals ρg" ; and this would imply the proposed new law that the other terms divide the gravityterm in a definite ratio, and also that each term utilizes gravity force to function in fluid flow.
Determining the ratio termsIn applications, the ratio terms a b c d f h n q, , , , , , , and others may perhaps be determined usinginformation such as initial and boundary conditions or may have to be determined experimentally.The author came to the experimental determination conclusion after referring to Example 5, page 6..The question is how did the grandmother determine the terms of the ratio for her grandchildren?Note that so far as the general solutions of the N-S equations are concerned one needs not find thespecific values of the ratio terms. Back to Options
CMI Millennium Prize Problem Requirements
26
Option 6Spin-off: CMI Millennium Prize Problem Requirements
Proof 1For the linearized Navier-Stokes equations
Proof of the existence of solutions of the Navier-Stokes equations
Since from page 13, it has been shown that the smooth equations given by
V x y z tg
ax by cz C x C y C zfg
t C P x d g xxx x
x( , , , ) ( ) ( )= − + + + + + + + =ρµ ρ2 4
2 2 21 3 5 9 ; are solutions
of the linearized equation, − + + + + =µ ∂∂
∂∂
∂∂
∂∂ ρ ∂∂ ρ( )
2
2
2
2
2
2 4Vx Vx Vx x Vx
x y zpx t
gx , it has been shown that
smooth solutions to the above differential equation exist. and the proof is complete.
From, above, if y z= =0 0, , V x tg
ax C xfg
t Cxx x( , ) = − + + +ρµ2 4
21 9 ; P x d gx C( ) = +ρ 10
Therefore, V x V xg
ax C x Cx xx( , ) ( )0 2
0 210 9= = − + +ρ
µFinding P x t( , )
1. V x tg
ax C xfg
t Cxx x( , ) ( )= − + + +ρµ2 4
21 9 ; P x d g xx( ) = ρ 2.
∂∂ ρpx
d g= ;
Required: To find P x t( , ) (that is, find a formula for P in terms of x and t)
dpdt
dpdx
dxdt=
dpdt
dpdx Vx= ( dx
dt Vx= )
dpdt d g
gax C x
fgt C
dpdx d g
dpdt
ad gx C d g x
d fgt C d g
P x tad g
x C d g xd fg
t
xx x
x
x x
x
x x
x x
= − + + +
=
= − + + +
= − + + +∫
ρ ρµ ρ
ρµ ρ ρ ρ
ρµ ρ ρ
2 4
2 4
2 4
21
21 9
21
9
2 2
2 2
2
2
( ) )
( , )
(
CC d g dt
P x tad g
x t C d g xtd fg
t C d g t C
d ga g
x t C xtfg
t C t C
x
x x
xx x
x x
9
2 29 10
21
210
2 2
1
2
9
2 8
2 8
ρ
ρµ ρ ρ ρ
ρ ρµ
= − + + + +
= − + + +
+
( , )
For the corresponding coverage for the original Navier-Stokes equation, see the next page
CMI Millennium Prize Problem Requirements
27
Proof 2For the Non-linearized Navier-Stokes equations (Original Equations)
Proof of the existence of solutions of the Navier-Stokes equationsFrom page 23, if y z= =0 0, in
Vg
ax by cz C x C y C z fg t hg xng yV
qg zV
VV
VVx x y z t
y z
xx x
x x y y
y
z z
z( , , , ) ( )
( ) ( )= − + + + + + + ± + + + +ρµ
ψ ψ2 22 2 2
1 3 5
continued
Solution to Linear part
solution of Euler equationl
{
6 7444444444 8444444444
1 24444444 334444444
P x d g xx( ) = ρone obtains
V x tg
ax C x fg t hg x C P x d g xxx
x x x( , ) ( ) ;= − + + ± + =ρµ ρ2 22
1 9 ;
V x V xg
ax C x hg x C P x d g xx xx
x x( , ) ( ) ( ) ;0 2 20 21 9= = − + ± + =ρ
µ ρ ;
Since previously, from p.113, it has been shown that the smooth equations given by
V x tg
ax C x fg t hg x C P x d g xxx
x x x( , ) ( ) ;= − + + ± + =ρµ ρ2 22
1 9 ; are solutions of
− + + + =µ ∂∂∂∂ ρ ∂∂ ρ ∂
∂ ρ2
2Vx
px
Vt
VVx
gx xx x
x x (deleting the x y− −and terms of (A)), p.112, one has
shown that smooth solutions to the above differential equation exist, and the proof is complete.Finding P x t( , ):
1. V x tg
ax C x fg t hg x C P x d g xxx
x x x( , ) ( ) ;= − + + ± + =ρµ ρ2 22
1 9; 2. ∂∂ ρpx
d g= ;
dpdt
dpdx
dxdt=
dpdt
dpdx Vx= ( dx
dt Vx= )
dpdt d g
gax C x hg x fg t C
dpdx d gx
xx x x= +
=− ± + +ρρµ ρ2 22
1 9( ) ) (
P x t d gg
ax C x hg x fg t C dtxx
x x( , ) ( )= +
∫ − ± + +ρρµ2 22
1 9
P x t d ga g
x t C xt hg x tfg
t C t Cxx
xx( , ) = − + ± ( ) + +
+ρ ρ
µ2 2 22
12
9 10
Back to Options
Magnetohydrodynamic Equations
28
Option 7Solutions of the Magnetohydrodynamic Equations
This system consists of four equations and one is to solve for V V V B B Bx y z x y z, , , , , ,
Magnetohydrodynamic Equations
1. < - - continuity equation
2.
Navier Stokes
Lorentz force
∂∂
∂∂
∂∂
ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂
∂∂ µ ρ
Vx
Vy
Vz
Vt
VVx
VVy
VVz
px
B B g
x y z
xx
xy
xz
xx
+ + =
+ + +
−
= − + ∇ × × +
0
16 74444444 84444444 6 744444 8444
( )
444
3
4 0
0
2
22
22
22
. ( )
( ) ( )
(
.
magnetic diffusivity)
ρ ∂∂ η
ρ ∂∂ η ∂∂
∂∂
∂∂
η
∂∂
∂∂
∂∂
Bt
V B B
Bt
V B Bx
By
Bz
BBx
By
Bz
x y z
= ∇ × × + ∇
= ∇ × × + + +
=
∇ • =
+ + =
Step 1:
1. If is constant : (for incompressible fluid)
< - - continuity equation
2.
Lorentz forceNavier - Stokes
ρ∂∂
∂∂
∂∂
ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂
∂∂ µ ρ
Vx
Vy
Vz
Vt
VVx
VVy
VVz
px
B B g
x y z
xx
xy
xz
xx
+ + =
+ + + = − + ∇ × × +
0
16 74444444 84444444 6 74444
( )
44 844444
ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂
∂∂ µ
∂∂
∂∂
∂∂
∂∂ ρV
tV
Vx
VVy
VVz
px
BBz
Bx
BBx
By
gxx
xy
xz
xz
x zy
y xx+ + + = − + − − − +1 ( ( ) ( )
ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂
∂∂ µ
∂∂
∂∂
∂∂
∂∂ ρV
tV
Vx
VVy
VVz
px
BBz
BBx
BBx
BBy
gxx
xy
xz
xz
xz
zy
yy
xx+ + + = − + − − + + 1 ( )
3 2
22
22
22
. ( )
( ) ( ) ( )
ρ ∂∂ η
ρ ∂∂∂∂
∂∂ η ∂
∂∂∂
∂∂
Bt
V B B
Bt y
V B V Bz
V B V B Bx
By
Bzx y y x z x x z
= ∇ × × + ∇
= − − − + + +
ρ ∂∂∂∂
∂∂
∂∂
∂∂ η ∂∂ η ∂∂ η ∂∂
Bt y
V By
V Bz
V Bz
V BB
xB
yBzx y y x z x x z
x x x= − − + + + +2
2
2
2
2
2
4 0
0
. ∇• =
+ + =
BBx
By
Bz
x y z∂∂
∂∂
∂∂
Magnetohydrodynamic Equations
29
Step 2:After the "vector juggling" one obtains the following system of equations which one will solve.
1 0
2 1 1 1 1
3
.
.
.
∂∂
∂∂
∂∂
ρ ∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂
∂∂ µ
∂∂ µ
∂∂ µ
∂∂ µ
∂∂ ρ
ρ∂∂
Vx
Vy
Vz
Vt
VVx
VVy
VVz
px
BBz
BBx
BBx
BBy
g
Bt
V
x y z
xx
xy
xz
xz
xz
zy
yy
xx
x
+ + =
+ + + + − + + − =
− xx yx
yx
x zx
xz
xz
zx x x x
x y z
By
BVy
VBy
BVy
VBz
BVz
VBz
BVz
Bx
By
Bz
Bx
By
Bz
y y∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
η∂∂
η∂η∂
η∂η∂
∂∂
∂∂
∂∂
− + + + + − − − − − =
+ + =
2
2
2
2
2
2 0
4 0.
At a glance, and from the experience gained in solving the Navier-Stokes equations, one can identifyequation (2) as the driver equation, since it contains the gravity term, and the gravity term is thesubject of the equation. However, since the system of equations is to be solved simultaneously andthere is only a single "driver", the gravity term, all the terms in the system of equations will beplaced in the driver equation, Equation 2. As suggested by Albert Einstein, Friedrich Nietzsche, andPablo Picasso, one will think like a child at the next step.Step 3: Thinking like a ninth grader, one will apply the following axiom: If and a b= c d a c b d= + = +, then ; and therefore, add the left sides and add the right sides of the above equations . That is, ( ) ( ) ( ) ( )1 2 3 4+ + + = ρgx
∂∂
∂∂
∂∂ ρ ∂∂ ρ ∂
∂ ρ ∂∂ ρ ∂
∂∂∂ µ
∂∂ µ
∂∂ µ
∂∂
µ∂∂
ρ∂∂
∂∂
∂
Vx
Vy
Vz
Vt
VVx
VVy
VVz
px
BBz
BBx
BBx
BBy
Bt
VBy
BV
x y z xx
xy
xz
xz
xz
zy
y
yx x
x yy
+ + + + + + + − + +
+ − −
−1 1 1
1 xxy
xx z
xx
zx
zz
x
x x x x y zx
yV
By
BVy
VBz
BVz
VBz
BVz
Bx
By
Bz
Bx
By
Bz
g
y
∂∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
η∂∂
η∂η∂
η∂η∂
∂∂
∂∂
∂∂ ρ
+ + + + − −
− − + + + =
−2
2
2
2
2
2 (Three lines per equation)
Step 4: Writing all the linear terms first
∂∂
∂∂
∂∂ ρ ∂∂
∂∂
ρ∂∂
η∂∂
η∂η∂
η∂η∂
∂∂
∂∂
∂∂
ρ ∂∂ ρ ∂
∂ ρ ∂∂ µ
∂
Vx
Vy
Vz
Vt
px
Bt
Bx
By
Bz
Bx
By
Bz
VVx
VVy
VVz
BB
x y z x x x x x x y z
xx
yx
zx
z
+ + + + + − − − + + +
+ + + −
2
2
2
2
2
2
1 xxz
zy
yy
xx y
x
yx
x zx
xz
xz
zx
x
zB
Bx
BBx
BBy
VBy
BVy
VBy
BVy
VBz
BVz
VBz
BVz
g
y
y
∂ µ∂∂ µ
∂∂ µ
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂ ρ
+ + − −
+ + + + − − =
−1 1 1
(Three lines per equation)
(Since all the terms are now in the same driver equation, let ρgx "drive them" simultaneously.)Step 5: Solve the above 28-term equation using the ratio method. (27 ratio terms) The ratio terms to be used are respectively the following: (Sum of the ratio terms = 1)β β β ω ω ω ω ω ω λ λ λ λ λ λ λ λ λ1 2 3 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9, , , , , , , , , , , , , , , , ,, , , , , , , , , a b c d f m q r s
1 1
1
1 16
.
∂∂ β ρ
β ρβ ρ
Vx
g
dVdx g
V g x C
x
x
x
x
x x
=
== +
2 2
2
2 17
.
∂∂ β ρ
β ρ
β ρ
Vy
g
dVdy g
V g y C
y
y
x
x
y x
=
=
= +
3 3
3
3 18
.
∂∂ β ρ
β ρβ ρ
Vz
g
dVdz g
V g z C
z
z
x
x
z x
=
== +
4
1
. ρ ∂∂ ρ∂∂
Vt
a g
Vt
ag
V ag t C
xx
xx
x x
=
== +
Magnetohydrodynamic Equations
30
5.
( )
+ C
∂∂ ρ
ρρ
px
b g
dpdx b g
P x b g x
x
x
x
=
==
6
1
.
ρ ∂∂ ρ∂∂
Bt
c g
Bt
cg
dBdt cg
B cg t C
xx
xx
xx
x x b
=
=
== +
7
2
2
22
2
2
2
2 3
.
− =
= −
= − +
= − + +
η ∂∂ ρρηρηρη
Bx
d g
d Bdx
d g
dBdx
d g xC
Bd g x
C x C
xx
x x
x x
xx
8
2
2
2
2
2
4
2
4 5
.
− =
= −
= − +
= − + +
η ∂∂ ρ
ρηρηρη
By
f g
d Bdy
f g
dBdy
f g yC
Bf g y
C y C
xx
x x
x x
xx
9
2
2
22
2
2
6
6 7
.
− =
= −
= − +
= − + +
η ∂∂ ρρηρηρη
Bz
m g
d Bdz
m g
dBdz
m g zC
Bm g z
C x C
xx
x x
x x
xx
10
19
.∂∂ ρ
ρρ
Bx
q g
dBdx q g
B q g x C
xx
x
x x
x
=
== +
11
20
.∂∂ ρ
ρρ
By
r g
dBdy r g
B r g y C
yx
x
y x
y
=
=
= +
12
21
.∂∂ ρ
ρρ
Bz
s g
dBdz s g
B s g z C
zx
x
z x
z
=
== +
13
22
2
1
1
12
12
1
1 2
.
ρ ∂∂ ω ρ
ωω
ω
ωω
VVx
g
VdVdx g
V dV g dx
Vg x
V g xV g x C
xx
x
xx
x
x x x
xx
x x
x x
=
==
=
== ± +
14.
((
ρ ∂∂ ω ρ
ωω ψ
ω ψ
VVy
g
V dV g dyV V g y V
Vg y
V
yx
x
y x x
y x x y
xx
y
y
y
y
y
yV
V
V
=
== +
= +
≠
2
2
2
2
0
))
15.
((
ρ ∂∂ ω ρ
ωωω ψ
ω ψ
VVz
g
VdVdz g
V dV g dzV V g z V
Vg z
VV
V
zx
x
zx
x
z x x
z x x z
xx
z
z
z
z
z
z
V
=
=== +
= +
≠
3
3
3
3
3
0
))
((
16
0
4
4
4
4
BBz
g
B dB g dzB B g z B
Bg z
BB
B
zx
x
z x x
z x x z
xx
z
z
z
z
z
z
B
∂∂ ω µρ
ω µρω µρ ψ
ω µρ ψ
= −= −= − +
= − +
≠
))
17
22
2
5
5
5
5
5
5
2
2
.
BBx
g
BdBdx g
B dB g dx
Bg x
B g x
B g x C
zz
x
zz
x
z z x
zx
z x
z x
∂∂ ω µρ
ω µρω µρ
ω µρ
ω µρω µρ
=
==
=
== ± +
Magnetohydrodynamic Equations
31
18
22
2
6
6
6
6
6
6
2
2
.
BBx
g
BdBdx g
B dB g dx
Bg x
B g x
B g x C
yy
x
yy
x
y y x
yx
y x
y x
∂∂ ω µρ
ω µρω µρ
ω µρ
ω µρω µρ
=
==
=
== ± +
((
19
1
0
1
1
1
1
1
.
))
− =
= −
= −= − +
= − +
≠
µ∂∂ λ ρ
λ µρ
λ µρλ µρ ψλ µρ ψ
BBy
g
BdBdy g
B dB g dyB B g y B
Bg y
BB
BB
yx
x
yx
x
y x x
y x x y y
xx
y
y y
y
y
20
0
2
2
2
2
2
− =
= −
= −= − +
= − +
≠
VBy
g
VdBdy g
V dB g dyV B g y V
Bg y
VV
VV
xy
x
xy
x
x y x
x y x x x
yx
x
x x
x
x
∂∂ λ ρ
λ ρ
λ ρλ ρ ψλ ρ ψ
((
))
21
0
3
3
3
3
3
.
))
− =
= −
= −= − +
= − +
≠
BVy
g
BdVdy g
B dV g dyB V g y B
Vg y
BB
BB
yx
x
yx
x
y x x
y x x y y
xx
y
y y
y
y
∂∂ λ ρ
λ ρ
λ ρλ ρ ψλ ρ ψ
((
22
0
4
4
4
4
4
.
))
VBy
g
VdBdy g
V dB g dyV B g y V
Bg y
VV
VV
yx
x
yx
x
y x x
y x x y y
xx
y
y y
y
y
∂∂ λ ρ
λ ρ
λ ρλ ρ ψλ ρ ψ
=
=
== +
= +
≠
((
23
0
5
5
5
5
5
.
))
BVy
g
BdVdy g
B dV g dyB V g y B
Vg y
BB
BB
xy
x
xy
x
x y x
x y x x x
yx
x
x x
x
x
∂∂ λ ρ
λ ρ
λ ρλ ρ ψλ ρ ψ
=
=
== +
= +
≠
((
24
6
0
6
6
6
6
.
)
)
VBz
g
VdBdz g
V dB g dzV B g z V
Bg z
VV
VV
zx
x
zx
x
z x x
z x x z z
xx
z
z z
z
z
∂∂ λ ρ
λ ρλ ρλ ρ ψλ ρ ψ
=
=== +
= +
≠
(
(
25
0
7
7
7
7
7
.
))
BVz
g
BdVdz g
B dV g dzB V g z B
Vg z
BB
BB
xz
x
xz
x
x z x
x z x x x
zx
x
x x
x
x
∂∂ λ ρ
λ ρλ ρλ ρ ψλ ρ ψ
=
=== +
= +
≠
(
(
26
0
8
8
8
8
8
− =
= −= −= − +
= − +
≠
VBz
g
VdBdz g
V dB g dzV B g z V
Bg z
VV
VV
xz
x
xz
x
x z x
x z x x x
zx
x
x x
x
x
∂∂ λ ρ
λ ρλ ρλ ρ ψλ ρ ψ
((
))
27
0
9
9
9
9
9
.
))
− =
= −= −= − +
= − +
≠
BVz
g
BdVdz g
B dV g dzB V g z B
Vg z
BB
BB
zx
x
zx
x
z x x
z x x z z
xx
z
z z
z
z
∂∂ λ ρ
λ ρλ ρλ ρ ψλ ρ ψ
((
Magnetohydrodynamic Equations
32
Step 6: One collects the integrals of the sub-equations, above, for V V V B B Bx y z x y z, , , , , ,
V x y z t
g x ag t g xg y
Vg y
Bg z
Vg z
BV
VB
BV
VB
x
x x xx
y
x
y
x
z
x
z
z z
z
y y
y
y y
y
z z
( , , , )) ) ) )
=
+ ± + − + − + + + +
(sum of integrals from sub - equations #1, # 4,#13,#14,#15,# 21,# 27)( ( ( ( β ρ ω ω λ ρ ω λ ρ ψ ψ ψ ψ
1 12 3 3 92 BB C
z
arbitrary functions1 2444444 3444444
+ 1 ;
(integral from sub equation #5)+
−=P x b g x Cx( ) ρ 2
(sum of integrals from sub equations # 2,# 23)(
arbitrary function
−
= + + +V y g yg y
BB
B Cy xx
x
x x
x( )
)β ρ λ ρ ψ2
53
124 34
(sum of integrals from sub equations #3, # 25)(
arbitrary function
−
= + + +V z g zg z
BB
B Cz xx
x
x x
x( )
)β ρ λ ρ ψ3
74
124 34
(sum of integrals from sub - equations #6, # 7, #8, # 9, #10, #16,#19, # 22, # 24)
B x y z t
Bg
dx fy mz q g x C x C y C z cg tg y
Bg y
Vg z
B
g zV
x
xx
x xx
y
x
y
x
z
x
( , , , )
( )
=
= − + + + + + + + + − + − +ρη ρ λ µρ λ ρ ω µρ
λ ρ2
2 2 2 1 4 4
6
2 4 6
zz
z
z
y y
y
y y
y
z z
z
BB
BB
VV
VV Cz
s
+ + + + +ψ ψ ψ ψ( ( ( (
arbitrary function
) ) ) )
1 2444444 34444447
(sum of integrals from sub equations #11,#18,# 20)(
arbitrary function
−
= ± − + +B r g y g xg y
VV
V Cy x xx
x
x x
xρ ω µρ λ ρ ψ
2 62
8)
124 34
(sum of integrals from sub equations #12,#17,# 26)(
arbitrary function
−
= ± − + +B s g z g xg z
VV
V Cz x xx
x
x x
xρ ω µρ λ ρ ψ
2 58
21)
124 34
Magnetohydrodynamic Equations
33
Step 7: Find the test derivatives for the linear part
1
1
.
( )∂∂ β ρVx
gxx=
2
2
.
( )∂∂ β ρVy
gyx=
3. ∂∂ β ρVz
gzx= ( )3
4. ∂∂Vt
agxx= ( )
5.
( )
∂∂ ρpx
b gx=
6.
( )dBdt cgx
x=
72
2
.∂∂
ρη
Bx
d gx x= −
82
2
.∂∂
ρη
By
f gx x= −
92
2
.∂∂
ρη
Bz
m gx x= −
10.∂∂ ρBx
q gxx=
11.∂∂ ρBy
r gyx=
12.∂∂ ρBz
s gzx=
Test derivatives for the nonlinear part
13
1
.∂∂
ωVx
gV
x x
x=
14. ∂∂
ωVy
gV
x x
y= 2
15. ∂∂
ωVz
gV
x x
z= 3
16
4∂∂
ω µρBz
gB
x x
z
= −
17
5
.∂∂
ω µρBx
gB
z x
z=
18
6
.∂∂
ω µρBx
gB
y x
y=
19
1
.∂∂
λ µρBy
gB
x x
y= −
20
2
.∂∂
λ ρBy
gV
y x
x= −
21
3
.∂∂
λ ρVy
gB
x x
y= −
22
4
.∂∂
λ ρBy
gV
x x
y=
23
5
.∂∂
λ ρVy
gB
y x
x=
24
6
.∂∂
λ ρBz
gV
x x
z=
25
7
.∂∂
λ ρVz
gB
z x
x=
26
8
.∂∂
λ ρBz
gV
z x
x= −
27
9
.∂∂
λ ρVz
gB
x x
z= −
Step 8: Substitute the above test derivatives respectively in the following 28-term equation
g g g a g b g c g d g f g m g q g r g s g gg g g g g g g g
x x x x x x x x x x x x x
x x x x x x x x
+ + + + + + + + + + ++ + + + + + + + +ωω ρ ω ρ
λ ρ λ ρ λ ρ λ ρ ρ2 3
6 7 8 9
g g
g g g g g
x x
x x x x x
+
+ + + + =
?
(Three lines per equation)
Magnetohydrodynamic Equations
34
β β β ω ω ω
ω λ λ λ λ λ ω ω λ λ λ λ1 2 3 1 3 5
6 1 2 3 4 5 2 3 6 7 8 9
g g g ag bg cg dg fg mg qg rg sg g g g
g g g g g g g g g g g
x x x x x x x x x x x x x x x
x x x x x x x x x x x
+ + + + + + + + + + + + +
+ + + + + + + + + + + + gg gx x=
? (2 lines)
g a b c d f m q r s
g
x
x
(?
)
β β β ω ω ω λ λ λ ω ω λ λ
ω λ λ λ λ1 2 3 1 3 5 3 4 5 2 3 6 7
6 1 2 8 9
+ + + + + + + + + + + + + + + + + + + + +
+ + + + + =
(Two lines per equation)
g gx x( )?
1 = (Sum of the ratio terms = 1)
g gx x=?
Yes
Since an identity is obtained, the solutions to the 28-term equation are as follows
V x y z t
g x ag t g xg y
Vg y
Bg z
Vg z
BV
VB
BV
VB
x
x x xx
y
x
y
x
z
x
z
z z
z
y y
y
y y
y
z z
( , , , )) ) ) )
=
+ ± + − + − + + + +
(sum of integrals from sub - equations #1, # 4,#13,#14,#15,# 21,# 27) ( ( ( ( β ρ ω ω λ ρ ω λ ρ ψ ψ ψ ψ
1 12 3 3 92 BB C
z
arbitrary functions1 2444444 3444444
+ 1 ;
(integral from sub equation #5)+
−=P x b g x Cx( ) ρ 2
(sum of integrals from sub equations # 2,# 23)(
arbitrary function
−
= + + +V g yg y
BB
B Cy xx
x
x x
xβ ρ λ ρ ψ
25
3)
124 34
(sum of integrals from sub equations #3, # 25)(
arbitrary function
−
= + + +V g zg z
BB
B Cz xx
x
x x
xβ ρ λ ρ ψ
37
4)
124 34
(sum of integrals from sub - equations #6, # 7, #8, # 9, #10, #16,#19, # 22, # 24)
B x y z t
Bg
dx fy mz q g x C x C y C z cg tg y
Bg y
Vg z
B
g zV
x
xx
x xx
y
x
y
x
z
x
( , , , )
( )
=
= − + + + + + + + + − + − +ρη ρ λ µρ λ ρ ω µρ
λ ρ2
2 2 2 1 4 4
6
2 4 6
zz
z
z
y y
y
y y
y
z z
z
BB
BB
VV
VV Cz
s
+ + + + +ψ ψ ψ ψ( ( ( (
arbitrary function
) ) ) )
1 2444444 34444447
(sum of integrals from sub equations #11,#18,# 20)(
arbitrary function
−
= ± − + +B r g y g xg y
VV
V Cy x xx
x
x x
xρ ω µρ λ ρ ψ
2 62
8)
124 34
(sum of integrals from sub equations #12,#17,# 26)(
arbitrary function
−
= ± − + +B s g z g xg z
VV
V Cz x xx
x
x x
xρ ω µρ λ ρ ψ
2 58
21)
124 34
Magnetohydrodynamic Equations
35
Supporter Equation ContributionsNote above that there are 28 terms in the driver equation, and 27 supporter equations, Each supporterequation provides useful information about the driver equation. The more of these supporterequations that are integrated, the more the information one obtains about the driver equation.However, without solving a supporter equation, one can sometimes write down some characteristicsof the integral of the supporter equation by referring to the subjects of the supporter equations of theNavier-Stokes equations. For example, if one uses ( )η∂ ∂2 2B xx as the subject of a supporterequation here, the curve for the integral obtained would be parabolic, periodic, and decreasinglyexponential.
Determining the ratio termsIn applications, the ratio terms β β β ω ω1 2 3 1 2, , , , ,, , , , , , , , , a b c d f m q r s and others may perhapsbe determined using initial and boundary conditions, or may have to be determined experimentally.Note that so far as the general solutions of the equations are concerned, one needs not find thespecific values of the ratio terms.
Comparison of Solutions of Navier-Stokes Equationsand
Solutions of Magnetohydrodynamic Equations Navier-Stokes x−direction solution
Vg
ax by cz C x C y C z fg hgxngyV
qgzV V V
P x d g x
x x y z t x
y z
y y
y
z z
z
V V
x
( , , , ) ( )( ) ( )
( )
= − + + + + + + ± + + + +
=
ρµ
ψ ψ
ρ
2 22 2 21 3 5
arbitrary functions
1 244 344
For magnetohydrodynamic solutions, see previous page1. Vx for MHD system looks like the Vx for the Euler solution.2. P x( )) for N-S and MHD equations are the same.3. Vy and Vz for MHD are different from those of N-S equations.
4. Bx is parabolic and resembles Vx for N-S, except for the absence of the square root function.5. By and Bz resemble the Euler solution.
ConclusionThe author proposes a law of definite ratio. This law states that in magnetohydrodynamics, all theother terms in the system of equations divide the gravity term in a definite ratio and each termutilizes gravity to function. As in the case of incompressible fluid flow, one can add that withoutgravity forces, there would be no magnetohydrodynamics on earth as is known, according to thesolutions of the magnetohydrodynamic equations. Back to Options
References:For paper edition of the above paper, see Chapter 11 of the book entitled "Power of Ratios"by A. A. Frempong, published by Yellowtextbooks.com.Without using ratios or proportion, the author would never be able to split-up the Navier-Stokesequations into sub-equations which were readily integrable. The impediment to solving the Navier-Stokes equations for over 150 years (whether linearized or non-linearized ) has been due to findinga way to split-up the equations. Since ratios were the key to splitting the Navier-Stokes equations,and also splitting the 28-term system of magnetohydrodynamic equations, and solving them, thesolutions have also been published in the " Power of Ratios" book which covers definition of ratioand applications of ratio in mathematics, science, engineering, economics and business fields.
