Analytical Solutions to the Navier-Stokes Equations with Density-dependent Viscosity and with Pressure

Abstract

This article is the continued version of the analytical solutions for the pressureless Navier-Stokes equations with density-dependent viscosity in "M.W. Yuen, Analyitcal Solutions to the Navier-Stokes Equations, J. Math. Phys., 49 (2008) No. 11, 113102, 10pp". We are able to extend the similar solutions structure to the case with pressure under some restriction for $\gamma$ and $\theta$. Comment: 7 pages, Typos are corrected

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arXiv:0902.2740v3 [math-ph] 19 Feb 2009
Analytical Solutions to the Navier-Stokes
Equations with Density-dependent Viscosity and
with Pressure
Ling Hei Yeung∗
Department of Mathematics, The Hong Kong Baptist University,
Kowloon Tong, Hong Kong
Yuen Manwai†
Department of Applied Mathematics, The Hong Kong Polytechnic University,
Hung Hom, Kowloon, Hong Kong
Revised 20-Feb-2009
Abstract
This article is the continued version of the analytical solutions for the pressureless Navier-
Stokes equations with density-dependent viscosity [9].We are able to extend the similar
solutions structure to the case with pressure under some restriction to the constants γ and θ.
Key words: Navier-Stokes Equations, Analytical Solutions, Radial Symmetry, Density-
dependent Viscosity, With Pressure
∗E-mail address: lightisgood2005@yahoo.com.hk
†E-mail address: nevetsyuen@hotmail.com
1

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2
Yeung and Yuen
1 Introduction
The Navier-Stokes equations can be formulated in the following form:







ρt+∇ · (ρu) =0,
(ρu)t+∇ · (ρu ⊗ u) + δ∇P =vis(ρ,u).
(1)
As usual, ρ = ρ(x,t) and u(x,t) are the density, the velocity respectively. P = P(ρ) is the pressure.
We use a γ-law on the pressure, i.e.
P(ρ) = Kργ, (2)
with K > 0, which is a universal hypothesis. The constant γ = cP/cv≥ 1, where cpand cvare the
specific heats per unit mass under constant pressure and constant volume respectively, is the ratio
of the specific heats. γ is the adiabatic exponent in (2). In particular, the fluid is called isothermal
if γ = 1. It can be used for constructing models with non-degenerate isothermal fluid. δ can be
the constant 0 or 1. When δ = 0, we call the system is pressureless; when δ = 1, we call that it is
with pressure. And vis(ρ,u) is the viscosity function. When vis(ρ,u) = 0, the system (1) becomes
the Euler equations. For the detailed study of the Euler and Navier-Stokes equations, see [1] and
[3]. Here we consider the density-dependent viscosity function as follows:
vis(ρ,u).= ▽(µ(ρ) ▽ ·u).
where µ(ρ) is a density-dependent viscosity function, which is usually written as µ(ρ).= κρθwith
the constants κ, θ > 0. For the study of this kind of the above system, the readers may refer [6],
[8].
The Navier-Stokes equations with density-dependent viscosity in radial symmetry can be ex-
pressed by:







ρt+ uρr+ ρur+N−1
rρu = 0,
ρ(ut+ uur) + ∇Kργ= (κρθ)r
?N−1
ru + ur
?+ κρθ(urr+N−1
rur−N−1
r2 u),
(3)
Recently, Yuen’s results [9] showed that there exists a family of the analytical solutions for the
pressureless Navier-Stokes equations with density-dependent viscosity:

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Analytical Solutions to the Navier-Stokes Equations
3
for θ = 1:















ρ(t,r) =ey(r/a(t))
a(t)N , u(t,r) =
·a(t)
a(t)r,
··a(t) =λ
·a(t)
a(t)2,a(0) = a0> 0,
·a(0) = a1,
y(x) =
λ
2Nκx2+ α,
(4)
for θ ?= 1:























ρ(t,r) =







y(r/a(t))
a(t)N , for y(
r
a(t)) ≥ 0;
0, for y(
r
a(t)) < 0
, u(t,r) =
·a(t)
a(t)r,
··a(t) =
−λ·a(t)
a(t)Nθ−N+2, a(0) = a0> 0,
·a(0) = a1,
y(x) =
θ−1
?
1
2(θ − 1)−λ
Nκθx2+ αθ−1,
(5)
where α > 0.
In this article, we extend the results form the study of the analytical solutions in the Navier-
Stokes equations without pressure [9] to the case with pressure. The techniques of separation
method of self-similar solutions were also found to treat other similar systems in [2], [4], [5], [7],
[8] and [9].
Our main result is the following theorem:
Theorem 1 For the N-dimensional Navier-Stokes equations in radial symmetry (3), there exists
a family of solutions, those are:
for θ = γ = 1,







ρ(t,r) =Ae
B(
r
a(t))2+C
a(t)N
, u(t,r) =
·a(t)
a(t)r,
¨ a(t) −2BK
a(t)+BNκ
·a(t)
a(t)2
= 0, a(0) = a0> 0,
·a(0) = a1,
(6)
where A ≥ 0, B and C are constants.
for θ = γ > 1,























ρ(t,r) =







y(
a(t)N, for y(
r
a(t))
r
a(t)) ≥ 0
0, for y(
r
a(t)) < 0
, u(t,r) =
·a(t)
a(t)r,
··
a(t)
a(t)N +
Kγ
a(t)θN+1−
Nκθ
a(t)θN+2 = 0, a(0) = a0> 0,
·a(t)
·a(0) = a1,
y(z) =
θ−1
?
1
2(θ − 1)z2+ αθ−1,
(7)
where a0, a1and α > 0 are constants;

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4
Yeung and Yuen
for
γ
2+1
2−
1
N= θ ≥ 1 −
1
N,























ρ(t,r) =







y(
a(t)N, for y(
r
a(t))
r
a(t)) ≥ 0
0, for y(
r
a(t)) < 0
, u(t,r) =
·a(t)
a(t)r
a(t) = σ(mt + n)s, 0 < s =
2
γN−N+2≤ 1
?
Kγ
sσγN+1y(z)γ−2−mNκθ
σθN+1y(z)θ−2?
˙ y(z) =(1−s)m2
σN−1 z, y(0) = α > 0
(8)
where m, n > 0, σ > 0 and α are constants.
2Separation Method of Self-Similar Solutions
Before we present the proof of Theorem 1, Lemmas 3 and 12 of [9] are quoted here.
Lemma 2 (Lemma 3 of [9]) For the equation of conservation of mass in radial symmetry:
ρt+ uρr+ ρur+N − 1
r
ρu = 0, (9)
there exist solutions,
ρ(t,r) =f(r/a(t))
a(t)N
, u(t,r) =
·a(t)
a(t)r, (10)
with the form f ≥ 0 ∈ C1and a(t) > 0 ∈ C1.
Lemma 3 (Lemma 12 of [9]) For the ordinary differential equation







·y(z)y(z)n− ξx = 0,
y(0) = α > 0,n ?= −1,
(11)
where ξ and n are constants,
we have the solution
y(z) =
n+1
?1
2(n + 1)ξz2+ αn+1, (12)
where the constant α > 0.
At this stage, we can show the proof of Theorem 1.
Proof. Our solutions (6), (7) and (8) fit the mass equation (3)1 by Lemma (2). Next, for the
equation (3)2, we plug our solutions to check that.

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Analytical Solutions to the Navier-Stokes Equations
5
For θ = γ = 1, we get
ρ(ut+ u · ur) + K [ρ]r− (κρ)r
?N − 1
r
a(t))2+C
r
u + ur
?
− κρr(urr+N − 1
r
ur−N − 1
r2
u) (13)
=AeB(
r
a(t))2+C
a(t)3
··a(t)
a(t)r + KAeB(
?
a(t)4
B
?−2r
·a(t)
a(t)2
a(t)
?
?
−ANκeB(
r
a(t))2+C
a(t)4
B
?−2r
a(t)
? ·a(t)
a(t)
(14)
=AeB(
r
a(t))2+Cr
a(t)4
¨ a(t) −2BK
a(t)
+BNκ
(15)
= 0, (16)
where the function a(t) is required to be
¨ a(t) −2BK
a(t)
+BNκ
·a(t)
a(t)2
= 0, a(0) = a0> 0,
·a(0) = a1, (17)
where a0and a1are constants.
For θ = γ > 1, we have:
ρ(ut+ u · ur) + K [ργ]r− (κρθ)r
?N − 1
r
a(t))
r
u + ur
?
− κρθ
r(urr+N − 1
r
ur−N − 1
r2
u) (18)
=
y(
r
a(t))
a(t)N
··a(t)
a(t)r +
Kθy(
r
a(t))θ−1˙ y(
a(t)γN+1
−
Nκθy(
r
a(t))θ−1˙ y(
a(t)
r
a(t))
θN+2
·a(t). (19)
By defining z := r/a(t), and requiring
y(z)z − y(z)θ−1˙ y(z) = 0,(20)
z − y(z)θ−2˙ y(z) = 0, (21)
(19) becomes
= y(z)z
?
··a(t)
a(t)N+
Kθ
a(t)θN+1−Nκθ
·a(t)
a(t)
θN+2
?
= 0, (22)
where the function a(t) is required to be
··a(t)
a(t)N+
Kγ
a(t)θN+1−Nκθ
·a(t)
θN+2= 0.
a(t)
(23)
Therefore, the equation (3)2is satisfied.
With n := θ − 2 and ξ := 1, in Lemma 3, (21) can be solved by
y(z) =
θ−1
?1
2(θ − 1)z2+ αθ−1, (24)