No. 4 HU et al.: Constructions of Some Exact Vortex Superposition Solutions 259

Constructions of Some Exact Vortex Superposition Solutions 4

Xin HU and Yongnian HUANG (Department of Mechanics and Engineering Science, State Key Laboratory for Turbulence Research, Peking University, Beijing 100871, China)

Abstract: In theoretical ways to solve the N-S equation, we will confront many difficul- ties. For example, the complex mathematical calculation, the 3D velocity components and the nonlinear terms, etc. We find that from the basis of Stokes-Helmholtz decomposition and after some processing, we can get a method of constructing the vortex superposition solutions. We get a set of exact solutions in different coordinates and we hope these solutions will be helpful to a clearer understanding of the vortex in Fluid Mechanics. Keywords: exact solution, Stokes-Helmholtz decomposition, vorticity equation

Introduction

One of the most useful methods to solve the N-S equation is to decompose the velocity to two parts: ‘Eli = $) + uj2), where uil) = dp/&i and ~1~) = EijkZJQk/aZj. This is the famous Stokes-Helmholtz decomposition Ill. From this decomposition, we find that if we let the vector potential function \k have only one component, and let the vortex field w have only one nonzero component, we will simplify the vortex equation and get many different special solutions in different coordinate systems. For example, we consider three coordinate systems: rectangular, cylindrical and spherical coordinate system.

1 Rectangular Coordinate System (z~,Q,z~)

We only need to think about one situation as 9 = Q+!Z3 = @(xl, x2, t)e,, , where eZs is the unit vector of 53 axis. The vortex field has the form w =wgess, and ws = -A!@. We

From the vorticity equation group, we have

aw3 at + (21. v>

d’ll3 w3=w -+vAws, 3 ax3

au1 o 6212

zg=’ G= 0

So we get &3/6x3 = b(t), and after some direct derivations, we get

1 cp = -b(t)x; + k(t)xs + g(xl,x2),

2 Ag + b(t) = 0

And the velocity solution is:

ag a* ag aa! U’=&+,52, uz=a52-%, ‘113 = b(t)xs + k(t)

From the definition us = dxs/dt = b(t)53 + k(t), we can easily know this flow’s physical meaning: it is accelerated or decelerated according to the sign of b(t), while from any direction which is perpendicular to x3 axis, its movement is unrelated with the x3 coordinate.

4The paper was received on Oct.. 3, 1999

260 Communications in Nonlinear Science & Numerical Simulation Vol. 4, No. 4 (Dec. 1999)

Now, we show some special exact solutions. If w3 = wg(t), we will then get w3 = wO exp(J, b(t)dt). If we take

ijj=- 2(Aw; B) (Ax: + Bz;) + Qo

here, go and QO satisfies two-dimensional Laplace equation, we will get an exact solution as follows:

u1=-- a;$4t) B x’-A+B

--w,zaexp(~b(t)dt)+$+~

A u2=--- ~~/yW) x2-A+B --w,zlexp(lb(t)dt)+$-2 (1)

743 = b(t)23 + k(t)

If b(t) = X = const, A = B, a = B, k(t) = 0, w. = w*, and go, !Po = const, we will get the rotational-and-effusive flow (see [l], p. 115-116):

dzl 1 21l=-=--xx dt 2

1 - iw*x2 exp(M)

212 = $f = -:Xx2 - iw*xl exp(Xt) (2)

dZ3 = xx ‘1L3 = dt 3

The physical meaning of this kind of flow is that: If a vortex tube is stretched, its strength will increase; while if compressed, its strength will decrease.

If we take a different function of go, Qo, and we get the vorticity equation in another form, we will get a different solution from (1) and (2). For example, if let go = -b(t)xf/2, we will get

!!f? + (IL?) - b(t)xl)z + I$)$ = b(t)w3 + UAW, 1

Let

s(t) = ex:Ll b(t)dt), [I = s(t)xl, & = s(t)x2, T(t) = J

‘s(t)dt 0

then = vs(t)ACw**

Let w** =m(t)Q**. Because w** =-s(t;AcI**, we can conclude that dw**/L3T=vm(t)w**, thus,

a*** - = um(t)~** - s(t)A,Q** = m(t)***

aT So we can get the result of !l!**, then we can obtain the velocity solution.

2 Cylindrical Coordinate System (T, 0,~)

There are three different situations for discussion.

(1) * = Q(r, 0, %, where e, is the unit vector of z axis. We would have:

‘il =!f+lE iav a* ai T dr

u=---- r&J e T a0 ar’ u.z = 7&

No. 4 HU et al.: Constructions of Some Exact Vortex Superposition Solutions 261

The vortex field has the form w = w,e,, where w, = -A*. From the vorticity equation group, we have

aw, at+ur~+~$$=wZ~+“Aw 2’ w!3=o 2 az

aus= 1 wzaz

So we have du,/dz = 0, du,Jdz = 0, bu,/Bz = b(t), therefore

cp = ;b(t)z’ + k(t)z + g(r, 8, t), b(t) + Ag = 0

and $$+I+ + T$$ = b(t)w, + VAW,

We can solve g(r, 8, t) first, then we will obtain cp, by working out the vorticity equation, we will get \E, so ‘1~ can be obtained.

For example, the famous Lundgren Vortex121 can be got as follows. Firstly, we let k(t) = 0, g = -ar2/2, here, b(t) = 2a = const, then ~!l) = -ar, ue (‘) = 0, up) = 2az. Then we carry a transformation named Lundgren transformation, and can get its solution. In [3], Huang proceeded in a different way, and he got another kind of solution.

We know that this method can also be applied to Burgers Vortex141. This time a/% = 0, Q = S(T, t), again let k(t) = 0, g = -ar2/2, then also use Lundgren transformation, see [l]. From Burgers Vortex, we can get a set of exact solutions, such as Sullivan Vortexl’], Long Vortex16T71, also see [l].

(2) \k = Q(T, z, t)e,, where eeis the unit vector of 0 axis. We would have:

ap a* 1 ap -- W= Y&-Y&l uE = T a01 Q =$+f;(TQ)

The vortex field has the form w = woes, where w0 = -A*. From the vorticity equation group, we have

au, au, Tg=o, w=o

(up + $$) + v(Awe - 3)

due 1 d’cp b(t) So we have de = ;w = 7, then

p = ib( + k(t)8 + s(T, z,t), b(t) T2+Ag=0

So we have (F + y)we + vAw,

We can solve g(r, z, t) first, then we will obtain ‘p, by working out the vorticity equation, we will get Q, so u can be obtained.

For example, if b(t) = 0, and w0 = CF, where c = const, then we will get the famous Hill Spherical Vortex (see [l]).

(3) * = ~(T,ht)% where e, is the unit vector of T axis. We would have:

262 Communications in Nonlinear Science & Numerical Simulation Vol. 4, No. 4 (Dec. 1999)

In order to let w = w,e,, we may let a2!P/a0 dz = 0, or \E = m(z, t)/r + m(8, t) + W(T, t). But from the vorticity equation, we will obtain m(z, t) = b(t)z/2 + k(t), and cp = -b(t)B/2 + h(z, t), therefore U, = 0, u0 = 0, and because At,o = 0, we have h(z,t) = p(t)z + q(t), so

anO% t) w? =dt) - F>

1a2n and w, = - - -, so vorticity equation turns into T a82

aw WT -$ = v(Aw, - -$

From those, we can solve !I! and Q, but the flow is now one-dimensional.

3 Spherical Coordinate System (R, 0, 4)

There are also three different situations for discussion.

(I) \k = tqR,e,t) e$, where e+ is the unit vector of 4 axis. We would have:

a(p+ 1 -J-(@sin@),

uR = aR Rsinede ug = ;g - $&(Rq), 1 ap --

“’ = Rsin8 ad

The vortex field w = w$e,+ From the vorticity equation, we can get the following equations:

We have au,/@ = b(t)/Rsine, therefore

So, from Acp = 0, we obtain

cp = ~WM2 + dR, 0, Q, b(t)

As + R2 sin2 e = 0

After solving the vorticity equation, we would solve u.

c2) * = we? 6 t)eRT where eR is the unit vector of R direction. We would have:

ap u =l!?E+ i a* i aik UR=dR’ 0

1 ap Rae R@T u47iZ&@--- R de

The vortex field w = WReR, then we can get the following equations:

8WR au, ~~+(U.V)WR=WR~R+V(AWR-~)

WR”O _ 8% 2~ awR v-w ---- R RaR ~2 de

Thus, we have ue/R = auo/aR, q/R = Bu+/aR. We will get wR = 0, this means that the flow is irrotational, we would get the value of Q, use the identity A(p = 0, we also can solve u.

No. 4 HU et al.: Constructions of Some Exact Vortex Superposition Solutions 263

(3) * = WC 0, $7 t) es, where es is the unit vector of 0 axis. We would have:

In order to let w = wOeg, we have to let

-&RI) = f(;;y), &(RI) = g(R, 4, t) sin 8

Then we have

SO

&l af aR=q + n($, t) sin 8 + ~(0, t)

wg = -$-&RQ) - R2s;n2e$ =-L(d2m + -.L$) Rsin9 dR2

From the vorticity equation, we can get

cp = ( sin2 f&(4, t) - g)$ + M(R, 4, t) + N(B, t)

while we have the vorticity equation

&e & +(u.V)wg=we ( 1 au, Eas+%-

u R2 sin2 8 > + UAW*

so we can solve wB first, then we will get Q and u.

4 Discussion

The method discussed above has not completely finished because we do not want to be halted by some difficult mathematical work. We just put forward a new idea, and some parts are simply touched, even not give complete result, but you can easily find some other solutions. And we will discuss more exact solutions to expand its examples in future.

Acknowledgment

This work is supported by the National Basic Research Projects “Nonlinear Science” and “The Several Key Problems of Fluid and Aerodynamics”.

References

[l] Wu Jiezhi, Ma Huiyang and Zhou Mingde, Introduction to Vorticity and Vortex Dynamics, 1993 (in Chinese)

[2] Lundgren, T. S., Strained spiral vortex model for turbulent fine structure, Phys. Fluids, 1982, 25: 2193-2203

[3] Huang Yongnian, Some kinds of typical exact solutions of 3D vortex, Mechanics and Engineering Applications, 1998, 7: 197-200 (in Chinese)

[4] Burgers, J., A mathematical model illustrating the theory of turbulence, Adv. Appl. Mech., 1948. 1: 197

[5] Sullivan, R. D., A two-cell solution of the Navier-Stokes equations, J. Aero. Space Sci., 1959, 26: 767

[6] Long, R., Vortex motion in a viscous fluid, J. of Meteorology, 1958, 15(l): 1088112 [7] Long, R., A vortex in an infinite viscous fluid, J. Fluid Mech., 1961, 11: 611