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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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