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Analytic Solution of Navier-Stokes Equation for Velocity
Vector Field v(g) and Scalar Pressure p(g) Functions of g
where, g=ax +by +cz +h(t).
Biruk Alemayehu Petros, birukoalex@gmail.com, Ethiopia/Bonga
May 5, 2023
Abstract
Analytic Solution of Navier-Stokes Equation for Velocity Vector Field v(g) and Scalar Pres-
sure p(g) Functions of gwhere, g=ax +by +cz +h(t) is solved from Navier Stokes equation.The
solution is found by assuming all components of velocity vector field and pressure are functions
of a linear function g.
1 Introduction
Definition. For a position vector r(in Cartesian coordinates), gradient operator ∇, Laplacian ∇2
, velocity vector field v, defined as
r=xi+yj+zk
∇=∂
∂x i+∂
∂y j+∂
∂z k.
∇2=∂2
∂x2+∂2
∂y2+∂2
∂z2
v=vxi+vyj+vzk
where x,y,z,vx,vy, and vzare the scalar components, the Navier Stokes equation is
ρ∂v
∂t + (v· ∇)v=−∇p+µ∇2v+f(1)
∇ · v= 0 (2)
where, density (ρ) describes the fluid mass per unit volume. For velocity vector field (v), ∂v
∂t is
its time derivative. Pressure (p) results from particles colliding with each other, and µrepresents
dynamic viscosity and fis the external force acting on the fluid.
1
2 Statement of the problem
Definition. For a constant a, b, c ∈R, let ∇g=∂g
∂x i+∂ g
∂y j+∂ g
∂z k=ai+bj+ckand ∇gis a unit
vector if a2+b2+c2= 1, i.e., a
√a2+b2+c2i+b
√a2+b2+c2j+c
√a2+b2+c2k
Assumptions. Assume density ρ= 1 and external applied force f= 0, and let the velocity
vector field vand scalar pressure pare functions of g(x, y, z, t), where gis a function of variables
x, y, z, and tfor a position vector rand function h(t) function of tonly as shown below
g(r, t) = r· ∇g+h(t)
Problem. Solution for Navier Stokes equation for velocity vector field vand scalar pressure p
under the given Assumptions.
3 Solve Navier Stokes equation for velocity vector field and
pressure under the given assuptions.
1. From Assumptions and equation(2)
∇ · v=d
dg (v· ∇g) (3)
Therefore, v· ∇ghas to be constant.
2. Simplify xcomponent of equation(1) using Assumptions
∂
∂t vx+ (v· ∇)vx=µ∇2vx−∂
∂x p(4)
∂
∂t g∂
∂g vx+ (v· ∇g)∂
∂g vx=µ∇2g∂
∂g vx+µ|
∇g|2∂2
∂g2vx−∂
∂g p∂
∂x g
∂
∂t g+ (v· ∇g)−
µ∇2g∂
∂g vx=µ
|∇g|2∂2
∂g2ux−∂
∂g p∂
∂x g
d
dt h(t)+(v· ∇g)∂
∂g vx=µ∂2
∂g2vx−∂
∂g p∂
∂x g
µ∂
∂g vx+kvx=ap +c0x(5)
where
k=−d
dt h(t)+(v· ∇g)
3. Solve equation(5)
vx(g) = e−k
µgc1x+a
µZek
µqp(q) + c0x
adq(6)
4. Simplify for y and z components of equation(1)
µ∂
∂g vy+kvy=ap +c0y(7)
µ∂
∂g vz+kvz=ap +c0z(8)
2
5. Solve equation(7) and (8) respectively
vy(g) = e−k
µgc1y+b
µZek
µqp(q) + c0y
bdq(9)
vz(g) = e−k
µgc1z+c
µZek
µqp(q) + c0z
cdq(10)
6. Multiplying awith equation(5), bwith equation(7) and cwith equation(8) and add
aµ∂
∂g vx+kvx=ap +c0x
bµ∂
∂g vy+kvy=bp +c0y
cµ∂
∂g vz+kvz=cp +c0z
µ∇ · v+k(v· ∇g) = |∇g|2p+c0· ∇g(11)
where, c0=ic0x+jc0y+kc0z
7. Solve equation(11) for pressure pusing equation(2) and Assumptions
k(v· ∇g) = p+c0· ∇g
p
k=v· ∇g−c0· ∇g
k(12)
8. Collect equation(6),(9),(10) and substitute equation(12) to solve for velocity vector field
v(g) = e−k
µgc1+1
µZek
µq(p(q)∇g+c0)dq
v(g) = e−k
µgc1+p
k∇g+1
kc0
v(g) = e−k
µgc1+v· ∇g−c0· ∇g
k∇g+1
kc0(13)
where, c1=ic1x+jc1y+kc1z
4 Check validity of the solution
1. Validate equation(1)
∂v
∂t =−k
µe−k
µgd
dt h(t)c1
(v· ∇)v=−k
µe−k
µg(v· ∇g)c1
µ∇2v=kk
µe−k
µgc1
3
∇p= 0
∂v
∂t + (v· ∇)v=µ∇2v− ∇p
k=−d
dt h(t)+(v· ∇g)
therefore, equation(12) and (13) are valid solutions of equation(1) under the given Assump-
tions.
2. Validate equation(2)
v· ∇g=e−k
µgc1· ∇g+v· ∇g−c0· ∇g
k∇g· ∇g+1
kc0· ∇g
we conclude that c1· ∇gmust be zero for equation(2) to be valid
5 Sample solutions for different values of k
1. complex solution for k=√−1
v0(g) = e−√−1
µgc1+v0· ∇g−c0· ∇g
√−1∇g+1
√−1c0
p
√−1=v0· ∇g−c0· ∇g
√−1
2. complex solution for k=−√−1
v1(g) = e√−1
µgc1+v1· ∇g+c0· ∇g
√−1∇g−1
√−1c0
−p
√−1=v1· ∇g+c0· ∇g
√−1
3. real periodic solution for g=ax +by +cx
v2=e−t
µcos(g
µ)c2
p= 0
c2· ∇g= 0
4. real periodic solution g=ax +by +cx −(c0· ∇g)t
v3=e−t
µsin(g
µ)c3+(c0· ∇g)
2∇g+1
2c0
p=−c0· ∇g
c3· ∇g= 0
4
5. The sum of the solution is also solution g=ax +by +cx −(c0· ∇g)t
v=e−t
µsin(g
µ)c3+e−t
µcos(g
µ)c2+(c0· ∇g)
2∇g+1
2c0
p=−c0· ∇g
c2· ∇g= 0
c3· ∇g= 0
6. complete representation of periodic velocity vector field under a given Assumptions for g=
ax +by +cx −(c0· ∇g)t
v=(c0· ∇g)
2∇g+1
2c0+∞
X
n=0
ane−n2t
µcos(ng
µ)+∞
X
n=0
bne−n2t
µsin(ng
µ)
p=−c0· ∇g
an· ∇g= 0
bn· ∇g= 0
6 Conclusion
Based on the given assumption that all components of the velocity vector field and pressure are func-
tions of a linear function g(x, y, z, t), it can be concluded that solutions for both the velocity vector
field and pressure exist in three dimensions. This assumption simplifies the problem considerably,
making it easier to solve for these variables. It also provides a convenient and efficient way to analyze
fluid dynamics in 3Dsystems. Overall, this conclusion suggests that the proposed assumption has
important practical implications for researchers studying fluid mechanics and related fields.
References
[1] Charles, L. F. Existence And Smoothness Of The Navier–Stokes Equation, Clay mathematics
institute.
[2] Nakayama, Y. Introduction to Fluid Mechanics., Former Professor, Tokai University, Japan
[3] Peter, V.O’ N. Advanced Engineering Mathematic., University of Alabama Birmingham.
[4] Biruk ,A. P. Navier-Stokes Three Dimensional Equations Solutions Volume Three., Journal of
Mathematics Research.
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