TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS

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1 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS PROF. VLADIMIR VLADIMIROV 1. The Kinematics of Continuous Medium 1.1. Lagrangian and Eulerian Coordinates General Idea Fluid flow is represented mathematically by a continuous transformation of 3D Euclidean space into itself. Time, t is the parameter, describing the transformation x 1 x 1 A A τ τ x 2 x 2 x 3 x3 x 1, x 2, x 3 - the fixed rectangular coordinate system. x = (x 1, x 2, x 3 ) - The position of the particle A. Say A is a particle moving with fluid: at t = 0 it occupies the postion X = (X 1, X 2, X 3 ) at time t it has moved to the position x = (x 1, x 2, x 3 ) Then x is determined as a function of X and t: (1.1) x = ϕ( X, t) or x i = ϕ i (X i, t) 1

2 2 PROF. VLADIMIR VLADIMIROV If X is fixed while t varies then (1.1) gives the path of the particle A intially at X. We assume, that initially distinct points remain distinct throughout the entire motion; it means that the transformation (1.1) posesses the inverse (1.2) X = Φ( x, t) or Xi = Φ i (x i, t) It is also usually assumed that ϕ i and Φ i posses continuous derivatives up to the third order in all variables. The flow is completely determined by the transformation (1.1). Below we will consider the state of motion (say velocity u or density ρ) at given points x or X during the course of time: ρ = ρ( x, t), u( x, t) - Eulerian description ρ = ρ( X, t), u( X, t) - Lagrangian description x - spatial variables, i.e. Eulerian variables X - material variables, i.e. Lagrangian variables 1.2. Partial and Material Derivatives By means of (1.1), (1.2) any quantity F, which is a function of Eulerian variables ( x, t) is also a funtion of Lagrangian variables ( X, t) and conversely. To indicate a particular set of variables we write either Geometrically F = F( X, t) or F = F( x( X, t), t) F( X, t) is the value of F experienced at time t by the particle initially at X; F( x, t) is the value of F felt by the particle instantaneously at the position x. We shall use the symbols F F( x, t) df (1.3) t t F( X, t) t for the two possible time derivatives of F, obviously they are quite different quantites: df is called the material derivative of F; d df F t is the differentiation following the motion of the fluid; measures the rate of change of F following a particle; gives the rate of F apparent to a viewer stationed at the position x.

3 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS Velocity and Acceleration The velocity of a particle is given by the definition (1.4) u d x u i dx i ϕ( X, t). t As defined, u is a function of the material variables. However, one usually deals with the spatial form: u = u( x, t). In most problems it is sufficient to know u( x, t) rather than the actual motion (1.1). We have introduced the velocity field in terms of the motion (1.1). It is important to be able to proceed in the opposite direction; to detemine the motion (1.1) from u( x, t). This transistion is given by solving the system of ODE s with the intial condition x(0) = X. d x = u( x, t) The acceleration is the rate of change of velocity experienced by a moving particle: (1.5) a d u. So, the acceleration a( x, t) can be computed directly in terms of the velocity field u( x, t) or du i du i a = u t a i du i u i( X, t) t ( ) x( X, t), t = u i t + u i dx k x k u k + u. ( u) u t + ( u. ) u. Correspondently, the general formula for differentiation following the motion of the fluid is:

4 4 PROF. VLADIMIR VLADIMIROV (1.6) df F t df = F t + ( u. )F F = F( x, t) = F ( x( X, t), t ) = F X t + F x k ( X, t) F x k t t + u F k = F x k t + ( u. )F u k = Equation (1.3) may be interpreted as expressing, for an arbitrary quantity F = F( x, t), the time rate of cahnge of F apparent to a viewer situated on the moving particle instantaneously at the position x Jacobian and Euler s Formula The basic transformation between Eulerian x and Lagrangian X coordinates is given by (1.1) and (1.2). The Jacobian of the transformation is: J = (x ( ) 1, x 2, x 3 ) (X 1, X 2, X 3 ) = det xi X k From the assumption, that (1.1) possesses a differentiable inverse, (1.2), it follows that (1.7) 0 < J < (for invertibility J 0). Geometrically, J represents the dilation of an infintesimal volume as it follows the motion: (1.8) dx 1 dx 2 dx 3 = J dx 1 dx 2 dx 3 Euler s formula for J We shall use Euler s formula: (1.9) dj = J. u Proof. Let A ik be the cofactor of the element x i X k in the expression of the Jacobian determinant: The derivative of J is, x i X k A jk = J δ ij

5 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 5 dj = d ( ) xi A ik = u i A ik = u i X k X k x j x j X k A ik =J δ ij = u i x j δ ij J = u i x i J = J. u 1.5. The Transport Theorem If a fluid is assumed to be incompressible, that is to move without a change in volume (1.8), we get J 1 in (1.8) and (1.9) yields (1.10). u = 0 The Transport Theorem Let τ = τ(t) denote an arbitrary volume, which is moving with a fulid. Also let F( x, t) be a scalar or vector function of position. Consider the volume integral F dτ. τ(t) Its derivative is given by the important formula (1.11) ( ) d df F dτ = τ(t) τ(t) + F. u Proof. Consider the correspondence between the Eulerian coordinates and the Lagrangian coordinates (1.1), (1.2). In x 1, x 2, x 3 - a moving volume τ(t) In X 1, X 2, X 3 - a fixed volume τ 0 τ(t) τ 0 under transformations (1.1),(1.2) dτ

6 6 PROF. VLADIMIR VLADIMIROV Then, F dτ = F( X, t)j dτ 0 τ(t) τ 0 where dτ = J dτ 0 ; dτ = dx 1 dx 2 dx 3, dτ 0 = dx 1 dx 2 dx 3 (see (1.8)). Differentiation yields d F dτ = τ(t) = τ 0 τ(t) ( J F t + F J X t ( ) df + F. u dτ X ) dτ 0 where we have used (1.3) and (1.9). Another form of (1.11) can be obtained after transformation (1.12) which gives df + F(. u) = F t +.( uf) (1.12a) d F F dτ = τ(t) τ(t) t dτ + F( u. n) ds τ(t) n τ(t) τ(t) It should be emphasised that (1.11) and (1.12) express a kinematical theorem, independant of any meaning attached to F Equation of Continuity The density function ρ = ρ( x, t) > 0. The mass of the fluid occupying a region τ is: M = ρ dτ τ

7 Physics TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 7 We postulate the principle of conservation of mass. The mass of a fluid in material volume τ does not change as τ(t) moves with the fluid. (1.13) now from (1.13) and (1.11) it follows that τ(t) d ρ dτ = 0 τ(t) ( ) dρ + ρ(. u) and since τ(t) is an arbitrary volume, this implies: dτ = 0 dρ (1.14) + ρ(. u) = 0. This is the Eulerian form of the equation of continuity. We can express this in the alternative form (1.15) ρ t +.(ρ u) = 0. Now, multiplying (1.14) by J and using (1.9) we derive the Lagrangian form of the equation of continuity: d (1.16) (ρj ) = 0 ρj = ρ 0 where ρ = ρ 0 is the initial density distribution. For future use, let us also write that for an arbitrary function F( x, t) (1.16a) d ρf dτ = ρ df τ(t) τ(t) dτ To prove this equality one should use (1.9) and (1.14) The Rate of Strain Tensor (prove it!) Let us introduce a simple decomposition of the tensor u i x k, namely (1.17) where u i x k = e ik + Ω ik

8 8 PROF. VLADIMIR VLADIMIROV e ik = 1 2 ( ui + u ) k x k x i Ω ik = 1 2 ( uk u ) i x k x k The tensors e ik and Ω ik are respectively the symmetric and skew-symmetric part of u i x k. The deformation (rate of strain) tensor Let d x denote a material element of arc. It s rate of change during the fluid motion is given by the formula: or simply (1.18) From (1.18) we get d (dx i) = d ( ) xi dx k = u i dx k = u i dx j X k X k x j d (d x) = (d x. ) u. d (1.19) (dl2 ) = 2e ik dx i dx k where dl = d x. Thus the tensor e ik is a measure of the rate of change of the squared element of arc following a fluid motion. In a rigid motion dl const, whence a necessary condition that a motion be localy and instantaneously rigid is that e ik = 0 (see in homework). For this reason, e ik is called the deformation tensor (or the rate of strain tensor). The tensor e ik 1 3 e jjδ ik is also of interest, for its vanishing is the necessary and sufficient condition that the motion, locally and instantaneously preserves angles (prove it!). If e ik = 0 everwhere in the fluid, then the motion is rigid and (1.19a) u = 1 ω r + const. 2 where ω is the constant angular velocity of the motion. Equation (1.19a) can be derived 2 analytically as the integral of the system of first order partial differential equations e ik = Vorticity and the Potential of Velocity Let us consider the velocity field in the neighbourhood of a fixed point P. Denoting the evaluation of a quantity at the point P by a superscript, we have near P ( ) P u i = u P ui i + r k + O(r 2 ) r k

9 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 9 where r denotes the radius vector from P. Neglecting terms of order r 2 and using (1.17) we get (1.20) u i = u P i + e P ikr k + Ω P ikr k We must now interpret the various terms in this formula. The first term on the right represents a uniform translation of velocity u P. If we set e = e P ikr i r k then the second term can be written in the form (1.21) 1 2 e or 1 e. 2 x i This term represents a veloctiy field normal at each point to the quadric surface e = const which passes through that point. In this velocity field there are 3 mutually perpendicular directions which are suffering no instantaneous rotation (the axes of strian). The principal (or eigen-) values of e measure the rates of extension per unit length of fluid elements in these directions. r 2 n extension and rotation r 1 pure extension The final term in (1.20) may be written as (1.22) where ω = u is the vorticity vector. 1 2 ωp r

10 10 PROF. VLADIMIR VLADIMIROV The form (1.22) shows clearly that the final term in (1.20) represents a rigid motion of angular velocity 1 2 ωp. By considering the above results, the identity (1.17) can be fully interpreted. For an arbitrary motion, the velocity u in the neighbourhod of a fixed point P is given, up to terms of order r 2, by ( ) 1 (1.23) u = u P + 2 e ωp r where e = e ik r i r k is the rate of strain quadric and ω = u is the vorticity vector. Thus an arbitrary instantaneous state of continuous motion is at each translation, a dilation along three mutually perpendicular axes, and the rigid motion of these axes. The angular velocity of rotation is 1 2 ωp. This result amply esablished that ω represents the local and instantaneous rate of rotation of the fluid. If e ik = 0 at a point, it is apparent from (1.23) that the motion is locally and instantaneously a rotation, while if e ik = kδ ik the motion is a combination of pure expansion and rotation. On the other hand, if throughout a finite portion of fluid we have ω = 0, Ω ik = 0, then the relative motion of any element of that position consists of a pure deformation, and is called irrotational! In this case a is derivable from a potential ϕ( x, t): (1.24) u = ϕ (prove it!). Problem Prove, that the motions of irrotaional flows and potential flow coincide. Step 1: Let u = ϕ. Then ω = u = 0 u = ϕ u l 2 ϕ ω i = ε ikl = ε ikl 0 x k x k x l it is identically equal to zero, since it is the scalar product of a symmetric tensor 2 ϕ x k x l and antisymmetric tensor ε ikl. So u = ϕ ω = 0. Step 2: Let ω = u = 0. Then let x ϕ = u.d x x 0 the definition of the potential ϕ( x, t). Evidently ϕ = u. However, we have to prove that ϕ( x, t) is a single valued function. It means that the integral,

11 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 11 x x 0 u.d x does not depend on the path of integration. This statement is equivalent to the equality (1.24a) γ u.d x = 0 for any closed circuit γ. Using stokes theorem u.d x = ω. n ds γ S γ for any surface S γ based on γ. If ω = 0 then u.d x = 0 and ϕ( x, t) is a single γ valued function Streamlines, Vortex Lines and Trajectories A curve, everywhere tangent to a given continuous vector field is called a vector-line. In particular, the vecor-lines of the velocity field are called streamlines, and the vector-lines of the vorticity are called vortex-lines. Equations of Streamline (1.25) x = x(s); or d x ds = u( x(s), t ) ; t = const. dx u(x, y, z, t) = dy v(x, y, z, t) = dz w(x, y, z, t) ; t = const. Equation of a Vortex line (1.26) d x ds = ω( x(s), t ) ; t = const. We define a trajectory to be the curve traced out by a particle as time progresses.

12 12 PROF. VLADIMIR VLADIMIROV Equation of a Trajectory (or Pathline) (1.27) or d x = u( x(t), t ) dx u(x, y, z, t) = dy v(x, y, z, t) = dz w(x, y, z, t) = with suitable initial conditions. If u is independent of t, ( u t = 0) then the streamlines and trajectories coincide (prove it!). In this case the flow is called steady or stationary. 2. The Equations of Motion Our aim is to derive the equations, which describe the action of forces The Stress Principle (Cauchy, 1827) Upon any imagined closed surface S there exists a distribution of stress vector t whose resultant and moment (torque) are equivalent to those of the actual forces exerted by the material outside S upon that inside. t - surface force (stress); n - the outward unit normal to S. t t n t It is assumed that t = t( x, t, n), for all x S. Now we can set the fundamental principle of the dynamics of fluid motion - the principle of conservation of linear momentum:

13 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 13 The rate of change of linear momentum of a material volume τ(t) equals the resultant force on the volume (2.1) d ρ u dτ = ρf dτ + t ds τ(t) τ(t) τ(t) Here f( x, t) is the external force per unit mass. By means of our transport theorem, (1.11), and conservation of mass, (1.13), we can transform (2.1) to ρ d u τ(t) dτ = ρf dτ + t ds τ(t) τ(t) This equality is valid for any instant t and does not contain any differentiation of integrals over variable volume τ(t). Therefore here the integration over a moving volume can be replaced without loss of generality by integration over a fixed volume. (2.2) τ 0 ρ d u dτ = ρf dτ + t ds. τ 0 τ 0 A result of great importance follows from (2.2). Let l be the typical or characteristic size of τ 0 ; then l 3 will be the volume of τ 0. Dividing both sides of (2.2) by l 2, letting l 0 we obtain (2.3) lim l 2 t ds = 0 l 0 τ 0 that is, the stress forces in local equilibrium! ı x 3 j k Σ n x 2 Consider the tetrahedron with vertex at an arbitrary point x and with three of its faces parallel to the coordinate planes. The normals are: ı, j. k, n with n = (n 1, n 2, n 3 ) and with ı = (1, 0, 0), etc. The areas are n 1 Σ, n 2 Σ, n 3 Σ, Σ. Now let us consider (2.3). Since l 2 Σ we can write x 1 t ( n ) + n 1 t ( ı ) + n 2 t ( j ) + n 3 t ( k ) = 0 where t ( n ) t ( x, t, n ). Also by Newton s second law we have t ( ı ) = t ( ı ), t ( j ) = t ( j ), t ( k ) = t ( k ). Now we obtain, t ( n ) = n 1 t ( ı ) + n 2 t ( j ) + n 3 t ( k ) Therefore t may be expressed as a linear function of the components n i :

14 14 PROF. VLADIMIR VLADIMIROV (2.4) t i = T ik n k ; T ik = T ik ( x, t). The matrix (tensor) T ik is called the stress tensor. Using (2.4) we transform (2.2) to the form and since τ 0 is arbitrary τ 0 ρ du i dτ = τ 0 ( ρf i + T ) ik x k dτ (2.5) ρ du i = ρf i + T ik x k. This is the equation of motion discovered by Cachy (1828). As the result we have obtained four equations (2.5) and (1.14) (2.6) ρ du i = ρf i + T ik x k dρ + ρ u = 0 They relate to each other four quantities ρ, u and contain still undefined T ik. Tensor T ik (the stress tensor) can be expressed in terms of ρ, u by direct mechanical or thermodynamical assumptions. The various possibilities for this form the physical core of fluid mechanics Conservation of angular momentum Tensor T ik in (2.6) has nine independent components. However to guarantee the conservation of angular momentum, we have to accept that it is symmetric (2.7) T ik = T ki which is known as the Boltzmann postulate. Theorem 1 (Boltzmann postulate). For an arbitrary continuous medium satisfying (2.6), (2.7) we have (2.8) d ρ ( r u ) dτ = ρ ( r f ) dτ + r t ds τ τ τ where τ is an arbitrary material volume. Proof. Using (1.9), (1.3) we get

15 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 15 (2.9) d ρ ( r u ) dτ = τ τ = τ ( ρ r d u ) dτ ρ ( r f ) dτ + τ r t ds T 0 dτ where T 0i ǫ ikl T kl (prove it!). By virtue of (2.7) we have T 0 = 0 and (2.8) is proven. Conversely, if (2.8) holds for arbitrary volumes then T ik must be symmetric Perfect Fluid A perfect fluid is by definition a material for which (2.10) t = p n p is called the pressure: when p > 0 the vectors t acting on a closed surface tend to compress the fluid inside. Comparing t ( n ) = n 1 t ( ı ) +n 2 t ( j ) +n 3 t ( k ) and t ( n ) = p n we find p ( n ) = p ( ı ) = p ( j ) = p ( k ). That is, p is independent on n Also we can obtain p = p(x, t) (2.11) T ik = pδ ik The equations of motion (2.6) take Euler s (1755) form (2.12) ρ d u = ρ f p dp + ρ(. u) = 0 and usually are referred to as Euler s equations. This system contains four equations for five variables ρ, u, p, so it is not closed and it is not solvable. In order to make it closed, one should introduce one more equation, based on physical reasoning Barotropic Fluid A perfect fluid with ρ = ρ(p) is called the baratropic fluid. We will consider two basic cases i) Compressible Fluid (gas): with an invertible function ρ(p) p(ρ). Here the necessary conditions are ρ p 0, 0. From physical reasons one can see, that p ρ always c 2 p ρ > 0

16 16 PROF. VLADIMIR VLADIMIROV where c(ρ) is the Speed of Sound. (the increasing of pressure does correspond to the increasing of density). Equations (2.12) can be written as (2.13) u t + ( u. ) u = 1 p(ρ) = w = c2 ρ ρ ρ ρ t +.(ρ u) = 0 p c(ρ) ρ Here w = w(ρ) is a thermodynamic function, called the enthalpy. System (2.13) contains four equations for four unknown functions, so it is a closed system of equations. ii) Incompressible Fluid: It corresponds to the case ρ(p) = const. = ρ 0. Equations (2.12) can be written as (2.14) u t + ( u. ) u = 1 p + f ρ 0. u = 0 ρ = ρ 0 = const. This system of four equations also contains four unknown functions, so it is also closed Viscous Incompressible Fluid If the internal friction (viscosity) in a fluid is present, the expression for the stress tensor (2.11) should be generalized as (2.15) T ik = pδ ik + 2µe ik where e ik 1 2 ( ui + u ) k x k x i is the rate of deformation tensor, µ is a constant coefficient known as the coefficient of dynamic viscosity; the combination ν µ is called the coefficient of kinematic viscosity. ρ The form of viscous term (2.15) is also called the Newtonian viscosity. Substitution of (2.14) into (2.6) (in the case of incompressible fluid) gives the famous Navier-Stokes equations for a viscous incompressible fluid:

17 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 17 (2.16) u t + ( u. ) u = 1 p + ν u + f ρ 0. u = 0 2 x i x j This is also a closed system of equations which represent the most basic mathematical model in the modern fluid dynamic. Note. The main argument towards (2.14) is; the viscous part of stresses T ik should be proportional to the amplitude of deformations, which can be measured by e ik. In particular, the viscous stresses must be zero in the absence of deformations e ik 0 (which corresponds to the motion of fluid as a solid body, u = U + Ω r with constants U and Ω, see Homework) Boundary Conditions For the formulation of any particular problem of fluid motion the governing equations (2.6), (2.13), (2.14), or (2.16) must be complemented by the appropriate boundary conditions. There are many versions of such conditions, the particular one must be chosen on the basis of the physical and mathematical analysis of the problem. The non-leak boundary conditions If a fluid flow is bounded by solid walls, then the most evident condition requires that the fluid can not leak or penetrate into the boundary. For the stationary (fixed) boundary τ it means that (2.17) u n u. n = 0 on τ where n is the normal unit vector to τ. If the boundary is moving with the velocity U = U(t), then (2.17) should be modified as (2.18) ( u U ) n ( u U ). n = 0 or u. n = U n on τ In the remainder of this lecture we take U = 0; the conditions for U 0 can be restored in a straight forward way. It is proven that it all flow boundary τ is solid, then for the perfect fluid (2.17) gives us the full set of boundary data; any additional to (2.17) boundary condition will produce a mathematically controversial problem. So, for the perfect fluid the system of governing equations is usually written together with the boundary conditions as

18 18 PROF. VLADIMIR VLADIMIROV (2.19) d u = 1 ρ p(ρ) + f dp + ρ(. u) = 0 u. n = 0 on τ for the compressible baratropic fluid (2.13) and (2.20) for the inviscid incompressible fluid (2.14). The non-slip boundary condition d u = 1 p + f ρ 0. u = 0 ρ 0 = const. u. n = 0 on τ The non-leak boundary conditions (2.17), (2.18) leaves the tangential velocity component (2.21) u t = u n( n. u) on τ undetermined; u t can be found as a result of solving (2.19) or (2.20). In contrary, for a viscous fluid (2.16) the vector u t also must be zero. (2.22) u t = 0 on τ So, for viscous fluid both (2.17) and (2.18) must hold. Both these conditions can be unified into one (2.23) u = u n + u t = 0 on τ which is well known as the non-slip boundary condition. The physical reason for (2.22) is the friction on the molecular level (the attraction between the molecules of fluid and solid). Mathematically the doubled number of boundary conditions ((2.17) and (2.22) for viscous fluid in comparison with (2.17) only for an inviscid fluid) is required following to the doubling of spatial order of governing PDEs. Indeed, the Laplacian operator in the R.H.S. of (2.16) is of the second order, while without it (for ν = 0), the highest order of spatial derivative correspond to in the L.H.S. of (2.16). As the result, the governing equations for a viscous incompressible fluid are often written as the set of the Navier-Stokes equations along with the non-slip boundary conditions:

19 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 19 (2.24) d u = 1 p + ν u + f ρ 0. u = 0 ρ 0 = const. u = 0 on τ in τ The systems of governing equations (2.19), (2.20) and (2.24) represent the main result of this Lecture. In order to make the mathematical formulations complete, all these systems must be complemented by the initial data for the velocity field (2.25) u( x, 0) = u 0 ( x) in τ 2.7. Plane rotationally-symmetric and axisymmetric motions Example 2.1. A perfect incompressible fluid of constant density ρ = ρ 0 in the absence of external forcing is described by Euler s equations (2.14): (2.26) d u = 1 ρ 0 p. u = 0 In Cartesian coordinates (x, y, z) the components of velocity are u = (u, v, w). Write the component form of Euler s equation for: i) general three-dimensional (3D) motions; ii) plane (2D) motions. Solution: Taking the above cases separately we get i) In general 3D motions we describe u, v, w, p as (2.27) u = u(x, y, z, t) v = v(x, y, z, t) w = w(x, y, z, t) p = p(x, y, z, t) The component form of equations is (2.28) u t + uu k + vu y + wu z = 1 ρ 0 p x v t + uv x + vv y + wv z = 1 ρ 0 p y w t + uw x + vw y + ww z = 1 p z ρ 0 u x + v y + w z = 0

20 20 PROF. VLADIMIR VLADIMIROV ii) Plane (2D) motions: (2.29) u = u(x, y, t) v = v(x, y, t) w 0 p = p(x, y, t) (2.30) These equations reduce down to; u t + uu x + vu y = 1 p x ρ 0 v t + uv x + vv y = 1 ρ 0 p y u x + v y = 0 Example 2.2. A perfect incompressible fluid of a constant density ρ = ρ 0 in the absence of external forces is described by Euler s equations (2.15): (2.31) d u = 1 ρ 0 p. u = 0 In the cylindrical coordinates (r, ϕ, z) the components of velocity are u = (u, v, w). Write the component form of Euler s equation for i) general three-dimensional (3D) motions; ii) rotationally-symmetric motions; iii) axisymmetric motions. Solution: Again breaking the cases down separately we get. i) For general 3D motions (2.32) u = u(r, ϕ, z, t) v = v(r, ϕ, z, t) w = w(r, ϕ, z, t) p = p(r, ϕ, z, t) (2.33) In (2.31) operators d, p and. u must be expressed in the cylindrical coordinates. Euler s equations take the form, du {}}{ u t + ( u. )u v2 r = 1 ρ 0 p r dv {}}{ v t + ( u. )v + uv r = 1 rρ 0 p ϕ dw {}}{ w t + ( u. )w = 1 ρ 0 p z u r + u r + 1 r v ϕ + w z = 0

21 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 21 where = ( r, 1 r ϕ, ) z so, that u. = u r + v r ϕ + w z ii) Rotationaly-symmetric motions: Here all functions do not depend on ϕ (2.34) u = u(r, z, t) v = v(r, z, t) w = w(r, z, t) p = p(r, z, t) so in (2.33) we can put ϕ = 0 that leads to equations, (2.35) du {}}{ u t + uu r + wu z v2 r = 1 p r ρ 0 dv {}}{ v t + uv r + wv z + uv r = 0 dw {}}{ w t + uw r + ww z = 1 ρ 0 p z u r + u r + w z = 0 The second equation can be rewritten for γ = rv; (2.36) dγ = γ t + uγ r + wγ z = 0 That equation represents a special case of the conservation of the circulation of velocity, which we will consider later. iii) Axisymmetric motions: Here in addition to (2.34) also v 0, (2.37) v 0 u = u(r, z, t) w = w(r, z, t) p = p(r, z, t) so (2.35) reduces to

22 22 PROF. VLADIMIR VLADIMIROV (2.38) du {}}{ u t + uu r + wu z = 1 ρ 0 p r dw {}}{ w t + uw r + ww z = 1 ρ 0 p z u r + u r + w z = 0. The condition v 0 is natural from the view point of equation (2.36). Indeed if we take γ t=0 = 0, it follows that γ = 0 at any t. Example 2.3. Consider a plane steady rotationally-symmetric flow of an inviscid incompressible fluid of a constant density ρ 0 in a gap between two concentric cylinders of radii R 2 > R 1. Obtain; i) the general solution of this type: calculate its vorticity ω from Stokes s theorem and from the formula (2.38a) ω = 1 r (rv) r 1 r u ϕ, where u, v are r, ϕ-components of velocity. ii) the solution with constant vorticity ω = Ω 0 : does this solution contain any other arbitrary parameters? iii) the irrotational potential solution. Solution: y v r u ϕ τ x R 1 R 2

23 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 23 i) A plane steady rotationally-symmetric solution in polar coordinates (r, ϕ) can be obtained from (2.34) after the elimination of the dependence on z and t; (2.39) u = u(r) v = v(r) w = 0 p = p(r) (2.40) Those functions must satisfy to the equations uu r v2 r = 1 p r ρ 0 uv r + uv r = 0 u r + u r = 0 and the non-leak boundary conditions in τ (2.41) u(r) = 0 at r = R 1 and r = R 2. The third equation in (2.40) is 1 r (ru) r = 0; the only solution to this is u(r) = Q with an arbitrary constant Q. Substitution to r the boundary condition (2.41) shows that the only possibility is Q = 0 and u 0. Taking it (u 0) into account, we see that the only equation which follows from (2.40) is (2.42) Therefore the required solution is v 2 r = 1 r V 2 (ξ) p r or p = ρ 0 ρ 0 ξ dξ (2.43) u 0 v = V (r) p = P(r) = ρ 0 r V 2 (ξ) ξ with an arbitrary function V (r). The vorticity of this flow is dξ (2.44) Ω(r) = 1 r (rv ) r = V r + V r. It can be obtained both from (2.38a) and from Stokes s theorem (see Homework). We can also introduce the circulation of velocity as: (2.45) Γ = Γ(r) = 2πrV (r)

24 24 PROF. VLADIMIR VLADIMIROV ii) Solution with constant vorticity Ω 0 according to (2.44) must satisfy the equation It follows that, 1 r (rv ) r = Ω 0 (rv ) r = Ω 0 r rv = 1 2 Ω 0r 2 + C (2.46) V = 1 2 Ω 0r + C r which contains two arbitrary constants Ω 0 and C. iii) The irrotational (potential) solution satisfies the equation 1 r (rv ) r = 0 and can be obtained from (2.46) after taking Ω 0 = 0 (2.47) V = Γ 2πr where we have taken C = Γ 2π (2.45). =const. This definition of Γ is in agreement with Example 2.4 (Lamb s form of Euler s equation). A perfect barotropic gas in the absence of external forcing is described by Euler s equations (2.13) (2.48) d u = w dp w = w(ρ) dw = ρ ρ t +.(ρ u) = 0 Show that the first equation in (2.48) can be written as (2.49) u t + ω u = ) (w + u2 2 where ω = u. Write the counterpart of equation (2.49) for the perfect incompressible fluid. Solution: The transformation is based on the use of the identity ( ) u 2 (2.50) ( u. ) u = ω u + 2 ω u. In order to prove it let us write ω u = u ω = u ( u). So, in components

25 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 25 ( ω u ) i = ǫ ikl ǫ lmn u k u n x m = (δ im δ kn δ in δ km )u k u n x m = u k u k x i + u k u i x k. or in vector form: ( ) u 2 ω u = + ( u. )u 2 and (2.50) is proven. Substitution of (2.50) into the first equation (2.48) gives the required form (2.49). For the perfect incompressible fluid the transformation (2.49) is the same, so the equation is ( p (2.51) u t + ω u = + u ) 2 ρ 0 2 Example 2.5 (Sound Waves. Linearization). Compressible fluid (gas) is described by (2.13): (2.52) u t + ( u. ) u = c2 ρ ρ ρ t +.(ρ u) = 0 p c = c(ρ) = ρ A particular solution of this system is the state of rest: u 0 ρ = ρ 0 = const. c 0 = c(ρ 0 ) Let us consider small deviations (perturbations) of the state of rest, which we denote with tildes, (2.53) u = 0 + ũ( x, t) ρ = ρ 0 + ρ 0 ( x, t) Substitution of (2.53) into (2.52) yields; ũ t + ũũ x = [c(ρ 0 + ρ)] 2 ρ 0 + ρ ρ t + [ (ρ 0 + ρ)ũ ] = 0 x ρ x

26 26 PROF. VLADIMIR VLADIMIROV Here for the sake of simplicity we consider one-dimensional problem (general 3D problem is given in the Homework). Now we use the lineraization procedure. It means that we keep only the linear (with respect to tilde variables) terms and neglect all their squares, cubes, etc. The result is: ũ t = c2 0 ρ x ρ 0 ρ t + ρ 0 ũ x = 0 now we can derive equations for ρ only or for ũ only. ρ tt c 2 0 ρ xx = 0 ũ tt c 2 0ũxx = 0 Both equations have the form (2.54) f tt c 2 0f xx = 0 c 2 0 = const. which is well known as the wave equations. It can be solved by introducing of new independent variables ξ x c 0 t and η x + c 0 t. In these variables (2.54) takes the form f ξη = 0 (prove it!). This has the general solution, f(ξ, η) = g(ξ) + h(η) with arbitrary funtions g(ξ) and h(η). In the original variables x, t this solution is f(x, t) = g(x c 0 t) + h(x + c 0 t) which represents the combination of the wave travelling to the right and the wave travelling to the left. The propagation of both waves takes place without any deformations of their profiles. The absence of the deformations is vitally important in our life: it allows us to hear the signal in the same form, as it has been generated! Example 2.6 (Archimede s Theorem). Now we consider an incompressible fluid, either viscous (2.16) or inviscid (2.14). For the state of rest ( u 0) both these systems of equations are reduced to (2.55) 0 = 1 ρ 0 p + g where g is the gravity field, representing in this case the force per unit mass f = g. This is the equations for hydrostatic equilibrium. In full, they can be written as p = ρ 0 g u = 0 In Cartesian coordinates (x, y, z), such that g = (0, 0, g) with g > 0 we get the system of Partial Differential Equations for pressure:

27 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 27 p x = 0 p y = 0 p z = ρg Solution of this system is (2.56) p = p(z) = ρ 0 ρgz where p 0 is a constant. The represents the distribution of pressure in hydrostatic equilibrium. Let us consider a solid volume V submerged into the fluid. x 3 z g n τ τ x 2 V x 1 y x The force acting on the solid is F = p u ds τ 0 (justify this!) The use of (2.21) and transformations yield: where we have; pu i ds = (p 0 ρgx 3 )n i ds = I 1 + I 2 τ τ I 1 = p 0 = 0 τ n i ds

28 28 PROF. VLADIMIR VLADIMIROV prove it! Also, So finally we have I 2 = ρg = ρg τ τ x 3 n i ds x 3 dτ x i = ρg 0 0 V 1 = ρg i V (2.57) F = ρ gv which gives us Archimedes s theorem: The buoyancy force is equal to the weight of the fluid in the submerged volume of the solid. Further exercises on this subject can be found in the homework! Examples of viscous flows will be given later. 3. Further elements of kinematics and dynamics 3.1. The momentum transfer equation The principle of conservation of linear momentum (2.1) may be transformed by (1.12a), the transport theorem, into the form (3.1) d ρu i dτ = ρf i dτ + (t i ρu i u k n k ) ds τ 0 τ 0 τ 0 expressed the rate of change of momentum of a fixed volume τ 0. (prove it!) Proof. We can prove this by using the transport theorem, (1.11). We transform equation (1.11) to give now taking F ρ u this gives (3.1a) d F dτ = F τ(t) τ(t) t dτ + F ( u. n ) ds τ(t) d ρ u ρ u dτ = τ(t) τ(t) t dτ + τ(t) ρ u ( u. n ) ds Now combining (3.1a) with the law for changing of momentum, (2.1), gives

29 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 29 (3.1b) τ(t) ρ u t dτ = τ(t) ρf dτ + τ(t) [ t ρ u ( u. n )] ds In this expression we moved the t-differentiation under the integral sign. It means that equality is valid at any instant t separately (t in τ(t) plays a part of a parameter). Therefore in (3.1b) the integration over a moving volume can be replaced (without the loss of generality) by integration over a fixed volume: (the fact that τ(t) is moving does not play any role in (3.1b)!) ρ u (3.1c) dτ = ρ τ 0 t [ f dτ + t ρ u ( u. n )] ds. τ 0 τ 0 The partial derivative on the left hand side can be pulled out of the integral sign. The result is the required expression (3.1) d ρ u dτ = τ 0 ρf + τ 0 τ 0 [ t ρ u ( u. n )] ds Because of the physical interpretation of the final term, (3.1) is known as the momentum transfer equation. The vector P i (3.2) P i t i ρu i u k n k = (T ik ρu i u k )n k is known as the flux of momentum through the fixed test surface of unit area and with the unit normal vector n. The first term in P represents the flux of momentum due to the surface force t; the second term ρu i u k n k is known as the convective flux of momentum (the minus sign appears since the positive influx of momentum into τ 0 takes place when u. n < 0). Example 3.1. Let us determine the force on an obstacle immersed in a steady flow. Suppose that the fluid occupies the entire exterior of some obstacle, and that the external force field is zero. Then if τ 0 denotes the surface of the obstacle and Σ denotes a control surface enclosing τ 0, we have the following formula for the force F acting on the obsticle: (3.3) F = t ds = τ 0 Σ [ t ρ u( u. n) ] ds If we can prescribe t, u and ρ at the remote surface Σ then we can calculate F. The values of functions at the remote surface or at infinity is often known. For the perfect fluid (2.10) we have (3.4) t i = pn i T ik = pδ ik P i = ( pδ ik ρu i u k )n k and (3.1)-(3.3) become slightly simpler.

30 30 PROF. VLADIMIR VLADIMIROV n τ 0 n f = 0 Σ 3.2. The energy transfer equation Let K denote the kinetic energy of a volume τ K 1 ρu i u k dτ. 2 τ Then for an arbitrary material volume τ(t) we have (3.5) dk = ρf i u i dτ + t. u ds T ik e ik dτ τ(t) τ(t) τ(t) (prove it!) where e ik is the rate-of-strain tensor (1.17). Proof. We show this by differentiating the kinetic energy,

31 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 31 where dk = d ρu i u i dτ τ(t) du i = ρu i dτ = = = = τ(t) τ(t) τ(t) τ(t) τ(t) ( u i ρf i + T ) ik dτ x k ( ) ui T ik u i ρu i f i dτ + T ik dτ τ(t) x k x k ρu i f i dτ + u i T ik n k ds 1 ( ui T ik + u ) k dτ τ(t) 2 τ(t) x k x i ρu i f i dτ + u i t i ds T ik e ik dτ τ(t) t i = T ik n k e ik = 1 2 τ(t) ( ui + u ) k x k x i It states that the rate of change of kinetic energy of a moving volume is equal to the rate at which the work is begin done on the volume by external forces, diminished by a dissipation term, involving the interaction of stress and deformation. For a perfect fluid (3.5) simplifies to (3.6) dk = ρ ( ) f. u dτ p ( u. n ) ds + p. u dτ τ(t) τ(t) τ(t) (prove it!) The last term is the rate at which work is done by the pressure in changing the volume of fluid elements. A simplification of (3.5) and (3.6) may be obtained if f = Φ, where Φ = Φ( x) is a time independent potential. In this case we can introduce the potential energy and (3.5) becomes (3.7) Π τ ρφ dτ d (K + Π) = t. u ds T ik e ik dτ τ τ It is interesting to consider separately the term (please prove it!)

32 32 PROF. VLADIMIR VLADIMIROV D T ik e ik dτ τ which represents the rate of energy dissipation in the volume τ. For a perfect fluid T ik = pδ ik and the integrand is T ik e ik = pe ii = p u i x i = p u. One can see, that it gives the last term in (3.6), which is equal to zero for an inviscid in-compresible fluid. For the viscous incompressible fluid (2.12) T ik = pδ ik + 2µe ik so (3.8) D = 2µ e ik e ik dτ. τ One can see that the viscous dissipation (3.8) is always negative, D < Convection of vorticity Consider perfect barotropic fluid, which can be either compressible or incompressible. The equations of motion, (2.12), are (3.9) where acceleration is ρ a = p Φ dρ + ρ. u = 0 a = d u = u t + ( u. ) u. There is a well known vector identity, ( ) 1 (3.10) ( u. ) u = ω u + 2 u2 so a = u t + ω u + ( 1 2 u2), ω u. Calculations of a yields a = ω + ( ω u) t (3.11) = d ω ( ω. ) u + ω(. u). From (3.11) and the equation for ρ, (3.9), we obtain

33 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 33 d Next, by virtue of a = (w + Φ) we obtain (3.12) ( ) ( ) ω ω = ρ ρ. u + 1 ρ a. d ( ) ( ) ω ω = ρ ρ. u This equation governs the convection of vorticity in a baratropic fluid and called Helmholtz s equation. It can be integrated explicitly using the Lagrangian coordinates. Let us introduce a vector field c( x, t) according to the definition (3.13) ω i ρc k x i X k this being possible since J 0. Substitution of (3.13) into (3.12) yields d c = 0 or c = c( X). Setting, in (3.13), t = 0 we get ω 0 = ρ 0 c ω 0 ω( x, 0) ρ 0 ρ( x, 0) that allow us to write the explicit expression for c which finally leads to c = ω 0( X) ρ 0 ( X) (3.14) ω i ρ = ω 0k ρ 0 x i X k. There are three important consequences from equation (3.14) 3.4. The Lagrange-Cauchy theorem on the persistence of irrotationality If a fluid particle or a portion of fluid (material volume) is initially in irrotational motion ( ω = 0), then it will retain this property throughout its entire history. In particular it means that any flow that appears from the state of rest or from a uniform motion is potential.

34 34 PROF. VLADIMIR VLADIMIROV 3.5. Vortex lines are material lines set of particles which composes a vortex line at one instant will continue to form a vortex line at later instants! The proof lies in the fact that a direction (material differential) d x once tangent to a vortex line is carried by the fluid so that it is always tangent to a vortex line. Indeed if at t = 0 then at any other instant d x = d X = ω 0 ds dx i = x i X k dx k = ds ω 0k =dx k x i X k = ρ 0 ρ ω ids and d x is tangent to a vortex-line. In other words, if d x is parrallel to ω at any instant, this property will be valid forever. Let dl be a material arc along a vortex line, then dl 0 = ω 0 ds and dl = ρ 0 ρ ωds where ω 0 = ω 0, ω = ω. Hence, we can write (3.15) ρ ω dl = ρ 0 ω 0 dl 0. This formula shows, that the stretching of vortex lines increases the vorticity (here is an open road to turbulence!). Plane flow In the plane flow ω is normal to the flow plane: (3.15a) ω x = ω y = 0 ω z = ω = v x u y = 0. hence ( ω. ) u = 0 and (3.12) gives (3.16) In other words ω ρ d ( ) ω ρ or ω ρ = f( X ). =const. following each particle. Similarly, in an axially-symmetrix flow ω r = ω z = 0 ω θ = ω = v z u r. Hence ω =const. following each particle (prove it!). rρ

35 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS Bernouliun theorems (1738) A first integral of the equations of motion (either for compressible fluid (3.10) or incompressible fluid (3.11)) is commonly called a Bernoulli equation. There are various forms which it can take, depending on the particular kinematical or dynamical assumptions made about the motion, though in all cases the basic expression (3.17) H 1 2 u2 + dp ρ + Φ is present. For incompressible fluid, (3.11), it takes the form (3.18) H 1 2 u2 + p ρ 0 + Φ. By means of (3.10) the first equation (either from (3.10) or (3.11)) takes the form u (3.18a) + ω u = H. t If we assume that the flow is steady (i.e. u = 0), then we arrive to the following important t result. Bernoulli theorem Consider the steady baratropic flow of a perfect fluid. Then if ω u 0 in the flow region we have (3.19) H const. On the other hand, if ω u 0 then there exists a set of surfaces (3.19a) H = const. each one covered by a network of vortex-lines and steamlines. In particular, H is constant on streamlines. The case (3.19) includes very important irrotational (or potential) flows, in which (3.20) ω 0 u = ϕ with a velocity potential ϕ = ϕ( x, t). In the case (3.20) one can derive a more general result. Equation (3.18a) takes the form ( ) ϕ t + H = 0 which can be integrated to obtain

36 36 PROF. VLADIMIR VLADIMIROV ϕ (3.21) t + H = f(t) where f(t) is an arbitrary function of time. Equation (3.21) is known as the Bernoulli theorem for irrotational flow; for steady flow it reduces to (3.19). Equations (3.19), (3.19a) and (3.20) constitute complete integrals of the equations of motion; they are of great importance in fluid dynamics The Stream function It is possible to introduce a stream function whenever the equation of continuity can be written as the sum of two derivatives. Plane flows of an incompressible fluid In this case we have u = u(x, y, t) v = v(x, y, t) w 0 ρ const. so the equation of continuity gives us ρ t +.(ρ u) = 0 (3.22). u = 0 or u x + v y = 0. It follows that the line integral u dy v dx from some fixed point (x 0, y 0 ) to the variable point (x, y) defines a (possibly mutiple-valued) function ψ = ψ(x, y, t) such that (3.23) u = ψ y = ψ y v = ψ x = ψ x. It is evident, that the curves ψ = const. are the streamlines of the field (prove it!); ψ is called the stream function. From the definition of vorticity ( (3.15a) and (3.23) ) we obtain the following important equation satisfied by ψ, namely (3.24) ψ = ψ xx + ψ yy = 2 ψ x ψ y 2 = ω. This equation serves to determine ψ when the vorticity magnitude is known. Now for the steady plane flow the equation of motion (3.18a) is equivalent to H = H(ψ) ω = H ψ (prove it!).

37 Thus any solution of the equation TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 37 (3.25) ψ = f(ψ) provides an example of steady two-dimensional flow; of course one must also take into account boundary conditions on ψ. In irrotational flow a velocity potential ϕ exists and (3.26) u = ϕ x = ψ y v = ϕ y = ψ x. The complex function (3.27) w = w(z, t) ϕ + iψ z x + iy is therefore analytic (prove it); this fact is very useful for solving particular problems. Axially symmetric flows In the cylindrical coordinates (z, r, θ) velocity components The equation of continuity takes the form: u = u(z, r, t) v = v(z, r, t) w 0. (3.28) r (rv) + z (ru) = 0 (prove it!) hence we can define a stream function ψ = ψ(r, z, t) such that (3.29) u = 1 r ψ r v = 1 r ψ x. The equation relating ψ to the vorticity is: 2 ψ (3.30) z + 2 ψ 2 r 1 ψ 2 r r = rω In steady flow H and ω are connected by (prove it!). (3.31) H = H(ψ) ω = rh (ψ) (prove it!). Thus any solution of the equation ψ zz + ψ rr 1 r ψ r = rf(ψ) provides an example of steady axially symmetric flow.

38 38 PROF. VLADIMIR VLADIMIROV Example 3.2 (Hill s spherical vortex). Show that, ψ = 1 2 Ar2 (a 2 z 2 r 2 ) ω = 5Ar H = 5Aψ + const. is an exact solution for an ideal incompressible steady axially symmetric flow. Draw the picture of streamlines (qualitatively) Lagrangian form of the equations of motion For the case of a perfect fluid (see section 2, equations (2.10)-(2.12)); it is simple to find equations satisfied by u, ρ and p as functions of the Lagrangian variables X α, t. Indeed, the first equation in (2.12) d u = f (p) can be transformed using d u = d2 x 2 and multiplying both sides by x i X α. The result is, (3.32) ( ) 2 x i t f xi 2 i = 1 X α ρ p X α where x i = x i ( X, t), p = p( X, t), f = f( X, t). This equation together with (1.16) (1.16) ρj = ρ 0 form the system of governing equations written in Lagrangian coordinates Kelvin s circulation theorem (1869) The circulation around any closed curve (circuit) in the fluid is defined by the integral dx γ

39 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 39 Γ γ u.d x. If now γ moves with the fluid, i.e. γ = γ(t), we may calculate the rate of change of the circulation around this moving circuit dγ = d u.d x. Kelvin s Theorem The baratropic flow of a perfect fluid under the action of a conservative extraneous force is circulation-preserving d (3.33) Γ = 0 where (3.33) holds for any material circuit γ(t). In order to prove (3.33) consider the transformation to Lagrangian coordinates (3.34) d u i dx i = d γ(t) = = γ 0 γ(t) u i ( X, t) x i dx k γ 0 X k ( ) x i u i u it + u i dx k X k X k du i dx i γ(t) γ(t) a i dx i where γ 0 = γ(0) and we also used ( ) u i u 2 u i dx k = d = 0. γ 0 X k γ 0 2 Now for barotropic flow we recall, from (2.13)-(2.14), that acceleration a is derivable from a potential, therefore the last integral in (3.34) vanishes and (3.33) is proven. One can see that for a potential flow we have the circulation to be Γ = γ u = ϕ u.d x = ( ϕ).d x = dϕ 0 γ γ

40 40 PROF. VLADIMIR VLADIMIROV if the potential function is single-valued. It is always true for a simply connected domain of flow, so the potential flows with nonzero circulation in this are forbidden. However it is not true for a multi-connected domain, where we can have a multi-valued potential ϕ( x, t) and, consequently, several different circulations around different components of boundary. Γ γ 1 γ 2 Γ 2 Γ 3 γ Γ 1 γ 3 4. Viscous Fluid 4.1. The Navier-Stokes equations The general equations for an incompressible continuous medium have been derived as (see (2.6)) (4.1) du i = f i + 1 T ik ρ 0 x k u i = 0 in τ. x i Here u = u( x, t) = (u, v, w) = (u 1, u 2, u 3 ) is the velocity field, x = (x, y, z) = (x 1, x 2, x 3 ) - Cartesian coordinates, ρ = ρ 0 = const. - the density of fluid, T ik ( x, t) - the stress tensor, f is the given external force (per unit mass). The stress tensor has been specified in eq2.12, as (4.2) T ik = pδ ik + 2µe ik e ik 1 2 ( ui + u ) k x k x i where p is pressure, µ is the coefficient of dynamic viscosity, e ik is the rate-of-strain (or rateof-deformation) tensor. The substitution of (4.2) into (4.1) produces the famous Navier- Stokes equations

41 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 41 (4.3) d u = f 1 ρ 0 p + ν u. u = 0 where ν µ ρ 0 is the coefficient of kinematic viscosity, τ is the region occupied by the fluid The Boundary Conditions For a solid boundary of flow we have introduced the non-slip boundary conditions (see (2.21) and (2.22)): (4.4) u = 0 on τ for a fixed boundary, and (4.4a) u = U on τ for a boundary moving with a velocity U. For the boundary condition (4.4) the stress vector t applied from the boundary τ to the fluid is generally non-zero (4.5) t i = T ik n k = pn i + 2µe ik n k 0 on τ where n is the external unit normal vector to τ. The total force acting on the fluid (from the wall) will appear as the surface integral F i = τ t i ds = τ T ik n k ds = pn i ds τ }{{} the force of pressure + 2µ e ik n k ds. τ }{{} viscous force Let us introduce also the free-surface (or the stress-free) boundary condition (4.6) t = 0 on τ or (4.6a) pn i + 2µe ik n k = 0 on τ which corresponds to the absence of surface forcing. In this situation u 0 on τ.

42 42 PROF. VLADIMIR VLADIMIROV 4.3. Unidirectional Flows u = (u, 0, 0) in Cartesian coordinates, (x, y, z). The full Navier-Stokes equations are: (4.7) ρ(u t + uu x + vu y + wu z ) = p x + µ(u xx + u yy + u zz ) ρ(v t + uv x + vv y + wv z ) = p y + µ(v xx + v yy + v zz ) ρ(w t + uw x + vw y + ww z ) = p z + µ(w xx + w yy + w zz ) u x + v y + w z = 0 for unidirectional flow they become ρ(u t + uu x ) = p x + ν(u xx + u yy + u zz ) 0 = p y u = u(y, z, t) ρu t = p x + µ(u yy + u zz ) 0 = p z u x = 0 p = p(x, t) since neither the first nor the last term depends on x we obtain p x = G(t) when function G > 0, the pressure gradient represents a uniform body force in the direction x > 0. Finally, the main equation for unidirectional flows is (4.8) ρu t = G(t) + µ(u yy + u zz ). In the case of steady flows u t = 0 and G = const, which implies (4.9) u yy + u zz = G µ. Equations (4.8), (4.9) are to be solved subject to the boundary conditions Tangential Stress on a Wall The Cauchy stress vector at the wall is t i = ( pδ ik + 2µe ik ) wall u k = pn i + 2µe ik n k. The first term in the right hand side is the normal force of pressure. To analyse the second term we introduce the local (at the wall) coordinates where the plane (x, z) is tangent to the wall, n is antiparallel to the y-axis n = (0, 1, 0). 2e ik = u i x k + u k x i = u x u y + v x u z + w x u y + v x v y v z + w y = u z + w x v z + w y w z 0 u y u z u y 0 0 u z 0 0

43 TOPICS IN MATHEMATICAL PHYSICS: FLUID DYNAMICS 43 y u(y, z) n x for the unidirectional flows, which implies z 2µe ik n k = µ 0 u y u z u y 0 0 u z = µ 0 u y 0 0 So, the tangential stress (applied to fluid) is directed against the flow (4.10) t 1 = µu y. If we wish to obtain the force acting on the wall we have to change sign: ( du (4.11) t 1 = µu y = µ dy) at the wall Formulae (4.10) and (4.11) were originally proposed by Newton, which implies the viscosity as it appears in the Navier-Stokes equations is called the Newtonian viscosity Poisenille Flow in a Tube Consider a flow in a long tube of circular section under the action of a difference between the pressures imposed at the two ends of the tube: Hagen (1839), Poisenille (1840). The flow is axisymmetric u = u(r), see (2.38), then equation (4.9) implies u = 1 r r (ru r) = G µ ( - in cylindrical coordinates, see Arfken) then the general solution is u(r) = G 4µ ( r2 + A log r + B).

44 44 PROF. VLADIMIR VLADIMIROV p 0 > p 1 r p 0 p 1 0 l x no singularity at r = 0 A = 0 u = 0 at r = a (non-slip), which implies (4.12) u = G 4µ (a2 r 2 ) the tangential stress at the wall (see (4.11) with y r): t 1 = µ du dr = 1 r=a 2 Ga So, the total frictional force on a length l of the tube is, ( ) 1 2πal 2 Ga = πa 2 (p 0 p 1 ). Such an expression was to be expected from the conservation of momentum (steady flow total force = 0). A quantity of practical importance is the flux of volume past any section of the tube: a Q = u2πr dr = πga4 0 8µ = πa4 (p 0 p 1 ). 8µl The law Qa 4 constant was established experimentally by Hagen and Poisneille Plane Poisenille Flow Consider the unidirectional flow in a gap between two parallel plates The flow is plane, which implies u = u(y). Now equation (4.9) implies the general solution µu yy = G