- Math 6750 and Potential Flows - Week 8

1 Vorticity and potential flow

The nonlinear term of the convective derivative can be expressed using a vector identity as 1 u · ∇u = ∇ |u|2 − u × (∇ × u). 2 We recall that the vorticity field is defined by ω = ∇ × u and that if the vorticity field is identically zero, then the fluid is irrotational. From Calculus, we know that a field is irrotational if and only if it is a potential flow, i.e.

ω = 0 ⇔ u = ∇Φ.

Definition 1 (Potential flow.). A potential flow is a flow that can be represented as u = ∇Φ, where Φ is a called the .

Given an irrotational vector field u defined on some fluid region D, we may define a potential φ: D −→ R as the line integral Z P φ(P ) = u · dx, O where O ∈ D is some arbitrary reference point. For simply-connected fluid region D, the potential φ is independent of the path between O and P . Consequently, φ is a well-defined function and we have u = ∇φ. If u is irrotational and incompressible, then the corresponding velocity potential φ must be harmonic since ∇ · u = ∇ · ∇φ = ∆φ = 0.

Definition 2 (.). The circulation of u on C (C simple, smooth, oriented, closed contour) is I ΓC = udγ. C Here the integral is understood as a line integral along C.

Remark 1. Stokes’ theorem applied to the velocity field yields Z ΓC = n · ωdS, S where S is the surface enclosed by C. The right-hand side is sometimes referred to as the flux of vorticity through S.

Example 1. If u is a potential flow, then ΓC = 0 on any well behaved closed curve. Remark 2. Since ∇ × ∇Φ = 0 for any field, the vorticity of a potential flow vanishes and irrotational flow are most always refer to as potential flow.

1 Example 2. Consider a rigid body rotation with angular velocity Ω. The flow field is given by u = (0, 0, Ω) × (x, y, z) = (−Ωy, Ωx, 0) and ω = ∇ × u = 2(0, 0, Ω) = 2Ω 6= 0. Example 3. Consider the two-dimensional irrotational with Ω = (0, 0, α/r2). Then  αy αx  u = − , , 0 r2 r2 and one can verify that ω = 0. Example 4. Consider the flow

u = (αx, −αy, 0).

One can verify that ω = 0 and there exists a velocity potential φ such that u = ∇φ, with α φ(x, y) = (x2 + y2), 2 up to some additive constant.

k Example 5. Consider the line vortex flow u = r eθ, with eθ the unit vector in polar coordinates defined by eθ = (− sin θ, cos θ, 0). Using the gradient operator in polar coordinates, the velocity potential φ, if it exists, satisfies 1 k ∂ φ = 0, ∂ φ = , ∂ φ = 0. r r θ r z Up to some additive constant, we find that φ = kθ. This doesn’t contradict the fact we 3 mentioned above since u is defined on R \{0} which is not simply-connected. For closed curves that don’t go around the origin, the line integral is zero. But for the unit circle centered at the origin, the line integral is I u · dx = 2πk. C 2 Inviscid vortical flow

A barotropic fluid is an example of a compressible fluid. Definition 3. A barotropic fluid is defined by specifying the as a given function of the density, i.e. p = p(ρ). Theorem 1 (Kelvin). Let C(t) be a simple closed material curve in an inviscid fluid with body force −ρ∇Φ (conservative). If either ρ =constant or the fluid is barotropic, then the circulation is invariant under the flow: D Γ = 0 (1) Dt C(t)

2 Proof. We parameterise C(t) as

C(t) = {x(s, t): 0 ≤ s ≤ 1}.

Then ! d d Z d Z 1 ∂x(s, t)  Γ(t) = u(x, t) dx = u(x(s, t), t) · ds dt dt C(t) dt 0 ∂s Z 1 Du ∂x(s, t) Z 1 ∂u(x(s, t), t) = · ds + u(x(s, t), t) · ds 0 Dt ∂x 0 ∂s Z 1 Du ∂x(s, t) Z 1 1 ∂ = · ds + {u(x(s, t), t) · u(x(s, t), t)} ds. 0 Dt ∂x 0 2 ∂s The second integral vanishes since C(t) is a closed curve. Rewriting the material derivative of u using Euler equation, we see that

d Z 1 p  ∂x(s, t) Γ(t) = −∇ + χ · ds dt 0 ρ ∂s Z p  = −∇ + χ · dx = 0, C(t) ρ since we are integrating a conservative vector field over a closed curve.  Remark 3. Kelvin’s theorem allows for a direct Lagrangian description of the dynamics of vorticity, even though vorticity is defined as an Eulerian variable. Remark 4. If a body is moving through an inviscid, incompressible fluid, or if a uniform flow of such a fluid passes around a body, then the vorticity far from the body will be zero. Then, from Kelvin’s theorem, the vorticity in the fluid will be zero everywhere. Remark 5. The theorem does not require the fluid region to be simply connected. The Euler equation enters the proof in evaluating the line integral around C, so the theorem continues to hold even if the viscous effect happens to be important elsewhere in the flow. Definition 4. A vortex line is a curve which is everywhere tangent to the vorticity ω(x, t). The vortex line which pass through some simple closed curve in space are said to form the boundary of a vortex tube. A consequence of the Kelvin’s theorem is the Helmholtz’s theorem. Theorem 2 (Helmholtz). 1. The strength of a vortex tube does not vary with time. A vortex tube is a collection of vortex line (a line whose tangent is parallel to the voracity) through a closed contour moving through the fluid.

2. Vortex lines move with the fluid. A vortex cannot end in the fluid. It must form a ring, end at a boundary or go to infinity.

3. Fluid elements initially free of vorticity remain fluid of vorticity.

3 Proof. Let us define a vortex surface as a surface such that ω is everywhere tangent to the surface. Suppose that at t = 0, the vortext line is the intersection of two vortex surfaces S1 and S2. On each of these surfaces,

ω(x, 0) · ni = 0 where ni is normal to Si, i=1,2.

We claim that both S1,S2 remain as vortex surfaces as time advances, i.e. as the points on Si move with the fluid, the surface they comprise at any time t is a vortex surface. Choose ∗ any arbitrary closed material curve C1(t) that encloses a surface S1 (t) ⊂ S1(t). It follows from Stokes theorem that Z Z u · dx = ω · n dS = 0. ∗ C1(0) S1 (0) ∗ This quantity remains zero by virtue of Kelvin’s circulation theorem. Since C1(t) and so S1 (t) was arbitrary, this shows that ω · n1 = 0 on S1(t) and an identical argument shows that ω · n2 = 0 on S2(t). We proceed to prove the second statement. Consider any two arbitrary cross sections A1,A2 of a vortex tube and consider the vortex tube bounded by these two cross sections. Since u is incompressible, ∇ · ω = ∇ · (∇ · u) = 0 and it follows from the theorem that Z Z Z 0 = ∇ · ω dx = ω · n dS = ω · n dS. V ∂V A1∪A2 Let C(t) be a closed material curve which lie on the wall of the vortex tube, enclosing a closed region A(t). Then Z Z ω(x, t) · n(x, t) dS = u(x, t) · dx. A(t) C(t) It follows from Kelvin’s circulation theorem that Γ remains constant as time proceeds.  Example 6 (Vorticity generation near a wall). Consider a one dimensional shear flow near a wall. The viscous shear stress is given by

τx = µ (∂xv − ∂yu) = −µωz.

Recall that for a Newtonian fluid and incompressible flow, the vorticity equation has the form (ν = µ/ρ) D ω = ω · ∇u + ν∇2ω. (2) Dt Remark 6. If the flow is steady, then in dimensionless form the vorticity equation is 1 u · ∇ω = ω · ∇u + ∇2ω. Re The left-hand side represents the advection of vorticity by u, and the second term on the right- hand side represents the transport of vorticity by diffusion. The first time on the right-hand side is the production term associated with the stretching of vortex lines.

4 We now state some other versions of the vorticity equation for inviscid flows. Theorem 3. For an inviscid fluid of constant density, the vorticity equation is D ω = ω · ∇u. (3) Dt For a barotropic inviscid fluid, the vorticity equation is D ω Dρ ω = ω · ∇u + . (4) Dt ρ Dt Proof. Assuming that the body force is a potential force, f = −∇Φ, the conservation of linear momentum equation can be written as Du 1 + ∇p + ∇Φ = 0. Dt ρ Taking the of the previous equation and using the fact that the curl of a gradient field is zero yield Du ∇ × = 0. Dt Expressing the nonlinear term with the vorticity and using the conservation of mass equation Dρ Dt + ρ∇ · u = 0, we find 1  0 = ∇ × ∂ u + ∇ × (u · ∇u) = ∂ ω + ∇ × ∇|u|2 − u × ω t t 2

= ∂tω − ω · ∇u + u · ∇ω − u∇ · ω + ω∇ · u Dω ω Dρ = − ω · ∇u − . Dt ρ Dt Here we used the vector identity ∇ × (a × b) = b · ∇a − a · ∇b + a∇ · b − b∇ · a and the facts that the divergence of vorticity is zero as well as the curl of a gradient field. This proves Eq. (4). For an incompressible fluid, the last term is zero and Eq. (3) follows.  Example 7. In two dimensions ω · ∇u = 0 and so the vorticity ω satisfies Dω = 0. Dt 3 Example of vortical flows

3.1 Rankine’s combined vortex This example models the bathtub vortex before the open funnel forms. It is a classical example of a flow with a free surface z = Z(r) in cylindrical coordinates. We assume that the density ρ is constant, that the pressure above the free surface is constant p0 and that the body force is gravitational. The vorticity is constant and the flow is a solid-body rotation in a vertical tube bounded by r = a, z < Z. The only nonzero velocity component is uθ. The detailed solution for uθ, p and Z can be found in An introduction to theoretical fluid mechanics by Stephen Childress (AMS Courant lecture notes).

5 3.2 Steady propagation of a vortex dipole We look at the two-dimensional flow of an inviscid fluid with constant density with no body force. Since u · ∇ω = 0, we introduce a ψ so that

2 2 2 (u, v) = (ψy, −ψx) ω = ∇ ψ ψy(∇ ψ)x − ψx(∇ ψ)y = 0.

Therefore, contours of constant ψ and of constant ω must agree and

∇2ψ = f(ψ).

Again, the detailed solution which involves Bessel function of order 1 can be found in An introduction to theoretical fluid mechanics by Stephen Childress (AMS Courant lecture notes).

3.3 Axisymmetric flow The simplest flows allowing vortex stretching is the axisymmetric Euler incompressible flows, i.e. the solution of the Euler’s equations in cylindrical coordinates under the assumption that all variables are independent of the polar angle θ. The incompressible Euler’s equations in cylindrical coordinates can be found at the end of any textbooks on fluid dynamics. Here, we assume that the flow is axisymmetric and further that the swirl velocity uθ vanishes. Then the equations simplify to 1 ∂ u + ∂ (ru ) = 0 (5) z z r r r 1 ∂ u + u ∂ u + u ∂ u + ∂ p = 0 (6) t z z z z r r z ρ z 1 ∂ u + u ∂ u + u ∂ u + ∂ p = 0 (7) t r z z r r r r ρ r

(ωz, ωr, ωθ) = (0, 0, ∂zur − ∂ruz) (8)

Since the only nonzero component of vorticity is ωθ, the vorticity equation becomes u ω ∂ ω + u ∂ ω + u ω − r θ = 0. (9) t θ z z z r θ r The last equation is equivalent to D ω D θ = 0 = ∂ + u ∂ + u ∂ . Dt r Dt t z z r r

In other words, ωθ/r is a material invariant of the flow (vortex rings move with the flow). Next, we introduce the stream function ψ (called the Stokes stream function) for the solenoidal velocity: 1 1 u = ∂ ψ u = − ∂ ψ. z r r r r z

Then, by Eq. (8), ωθ satisfies 1 ω = − L(ψ), θ r

6 1 where L is the differential operator L = ∂zz + ∂rr − r ∂r. In the steady cases, the vorticity equation has the form

1 1  1 ∂ ψ∂ = ∂ ψ∂ L(ψ) = 0. r r z r z r r2

Again, a family of solutions can be obtained by solving any equation of the form

L(ψ) = r2f(ψ), where f is an arbitrary function of ψ. The solution involves Bessel function of the first kind.

4 Inviscid and incompressible irrotational flow

Consider an incompressible and inviscid (no viscosity) flow satisfying

∇ · u = 0 (10) 1 ∂ u + u · ∇u = − ∇p + f (11) t ρ

Plugging the velocity potential into the continuity Eq. (10) shows that the velocity potential satisfies the Laplace equation ∇2Φ = 0. Therefore by solving the Laplace equation, the velocity field can be determined. Note that in doing so, the momentum equation Eq. (11) was not explicitly solved. This is possible, because the condition of irrationality has been used, which is justified by Kelvin’s theorem. Once the velocity is obtained from the Laplace equation with the adequate boundary conditions, the pressure can be obtained. This is usually done with the Bernoulli equation (the integrated version of the momentum equations).

Definition 5 (Stream function). A stream function is a function such that a two-dimensional flow u = uex + vey can be written as u = ∂yΨ, v = −∂xΨ. Remark 7. A flow field given by a stream function satisfies the continuity Eq. (10) automat- ically. A stream function can be defined for all two-dimensional flows both rotational and irrotational. To find the equation that a potential (irrotational) flow given by Ψ must satisfy, we use the vorticity equation. Because the flow is two-dimensional, the vorticity is purely in ez and ω = ∂xv − ∂yu. Therefore, the irrationality condition becomes

2 ω = ∂xv − ∂yu = ∇ Ψ = 0.

Thus, the stream function satisfies the Laplace equation as well.

Since an irrotational flow can be described both via Ψ and Φ, we have

u = ∂xΦ = ∂yΨ v = ∂yΦ = −∂xΨ.

7 That is Φ and Ψ satisfy the Cauchy-Riemann equations

∂xΦ = ∂yΨ ∂yΦ = −∂xΨ.

Recall that a complex function f(z) = f(x + iy) is analytic if the following limit exists

df f(z + ∆z) − f(z) = lim . dz ∆z→0 ∆z If f is analytic, then Re(f) and Im(f) satisfy the CR-equation. Conversely, if two functions g(x, y) and h(x, y) satisfy the CR-equation and their partial derivatives are continuous, then the function f = g + ih is analytic. In our case, if Φ, Ψ are C1 functions, therefore, it makes sense to define the complex potential F (x) as

F (z) = Φ(x, y) + iΨ(x, y) z = x + iy.

Choose a path along the real-axis. Then dF F (z + ∆z) − F (z) = lim dz ∆z→0 ∆z Φ(x + ∆x, y) + iΨ(x + ∆x, y) − Φ(x, y) − iΨ(x, y) = lim = ∂xΦ + i∂xΨ ∆x→0 ∆x = u − iv.

We recover the same relation if we choose a path along the imaginary-axis, in which ∆z = i∆y. A useful quantity is the magnitude of F 0(z) that is the flow speed:

dF 2 2 q = = |u + v | = |u|. dz

dF The complex velocity is given by W = dz . The disadvantage of the above complex approach is that it can’t be generalized to three-dimensional flow.

5 Two dimensional incompressible flows

The fundamental solution of the Laplace equation in two dimension is log(r), where r = px2 + y2. From , we also know that the real and imaginary parts of the analytical function F satisfy the Laplace equations. Next, we list special solutions of the potential two dimensional flow based on choices for F . We consider the natural choices F (z) = cz, c log(z), c/z.

5.1 Uniform flows The simplest choice for F (z) is F (z) = cz, where c is a (possibly complex) constant. If c is real, then u = c and v = 0. The flow is horizontal and uniform. Similarly, if c is purely imaginary u = 0 and v = −c the flow is vertical. If c = βe−iαz, then u = β cos α and v = β sin α. The flow is uniform at an angle α.

8 Example 8. Consider the uniform flow at an angle α extended from the x-axis, defined by u = U cos α and v = U sin α for some constant U > 0. Then dF = u − iv = U cos α − iU sin α = Ue−iα. dz Since F is independent of z, we may integrate once and find F = Ue−iαz = (U cos α − iU sin α)(x + iy) = Φ + iΨ which implies that Φ = U cos αx + U sin αy Ψ = −U sin αx + U cos αy. For α = 0, Φ = Ux and Ψ = Uy.

5.2 Sources, sinks and vortices We consider F (z) to be proportional to the principal branch of the complex logarithm, i.e. F (z) = c log(z). Writing z = reiθ and assuming that c is real, yield the velocity potential and stream function Φ = c log r Ψ = cθ. Here, the equipotential lines are circles and the streamlines are purely radial. Transforming into cylindrical coordinates, we find the velocity components c u = u = 0. r r θ

If c > 0, then ur points away from the origin. A velocity field as described above is called a source. If the streamlines point towards the origin it is called a sink. A source (or sink) is characterized by its strength given as Z 2π m = urrdθ = 2πc. 0 Expressing c in terms of the source/sink gives the general complex potential m F (z) = log(z − z ), 2π 0 where z0 is the location of the singularity. On the other hand, if c is purely imaginary, then the complex potential F (z) = ic log(z) yield the velocity potential and stream function Φ = cθ Ψ = −c log(r). The equipotential lines are now radial, while the streamlines are circles. The corresponding velocity field in cylindrical coordinates is c u = 0 u = . r θ r The resulting flow field is called a vortex and its strength is given by the circulation I Γ = u · dx = 2πc.

9 Example 9. Consider the line vortex Γ 1 u = e . 2π r θ Γ We have that Φ = θ. The stream function satisfies 2π 1 u = ∂θΨ and u = −∂ Ψ, r r θ r or Γ ∂ Ψ = 0 and − ∂ Ψ = . θ r 2πr Integrating once, we find that Γ Ψ = − ln r. 2π Consequently, Γ iΓ iΓ F = [θ − i ln r] = − [ln r + iθ] = − ln z 2π 2π 2π if we choose the principal branch of the complex logarithmic.

5.3 Doublet The final case is F (z) = c/z. The function 1/z has a singularity at the origin called a doublet. First, we show that the complex potential is the same one as the one associated by placing a source of strength m and a sink of source m on the real axis a distance 2ε apart. The complex velocity potential is then m m m 1 + ε/z  F (z) = log(z + ε) − log(z − ε) = log . 2π 2ε 2π 1 − ε/z Expanding around the small parameter ε/z, F (z) becomes

m  ε  ε2  F (z) = log 1 + 2 + O . 2π z z2 Further expanding the log, we find m  ε  ε2  F (z) = 2 + O . 2π z z2

Now, we let ε → 0 while simultaneously letting m → ∞ in such a way that limε→0(mε) = πµ. Then, the complex potential becomes F (z) = µ/z. One can then show that the associated stream function and velocity components are y µ µ Ψ = −µ u = − cos θ u = − sin θ. x2 + y2 R R2 θ R2 Streamlines are Ψ = constant, i.e. shifted circles along the y-axis. One can show that A A u = − cos θ, u = − sin θ. r r2 θ r2

10 5.4 General solutions via superposition Consider the problem of finding the flow generated by a vortex a distance from the origin on the x-axis in the half-space x ≥ 0. Consider the free space and another line vortex of strength −Γ at x = −d. By superposition,

iΓ iΓ iΓ z − d F (z) = F (z) + F (z) = − ln(z − d) + ln(z + d) = − ln + − 2π 2π 2π z + d    iΓ z − d z − d = − ln + iarg = Φ(r, θ) + iΨ(r, θ). 2π z + d z + d

Γ z − d z − d Consequently, the stream function is Ψ(r, θ) = − ln . The streamlines are = 2π z + d z + d c, i.e. these are circles. On the imaginary axis, observe that ψ = 0. One can check that the normal velocity component u = 0 on the boundary (the wall). For curved boundaries, we use the Mile-Thompson’s circle theorem to construct the complex potential.

Theorem 4 (Milne-Thompson’s Circle Theorem). Let f(z) be a complex function. Suppose that all singularities of f(z) lie in |z| > a. Then the function

a2  F (z) = f(z) + f z¯ is the complex potential of a flow with

1. the same singularities as f in |z| > a

2. |z| = a is a streamline of w(z).

 2  2 a a Proof. The singularities of f are in > a, i.e. in |z| < a. On the circle |z| = a, z¯ z¯ observe that a2 a2z a2z = = = z. z¯ |z|2 a2 This means that the complex function w(z) restricted to the circle |z| = a is

F = f(z) + f(z) = 2Re (f(z)) . |z|=a

In particular, the imaginary part which is the streamline is zero at the circle |z| = a.  Since the governing equations are linear, more general solutions can be obtained through the principle of superposition.

11 5.5 Circular cylinder We consider the flow around a circular cylinder that is transverse to the flow. In general, we don’t assume that no slip is satisfied on the body (at first).

Consider the uniform flow past a cylinder with radius a > 0. In the absence of the cylinder, the flow is uniform and the complex potential is f(z) = Uz and the only singularity is at z = ∞? By the circle theorem,

a2  Ua2 F (z) = f(z) + f = Uz + z¯ z has the same singularities outside the circle and |z| = a is a streamline. We claim that the circulation of the corresponding flow around the cylinder is 0. Use polar coordinates.

Ua2  a2 a2  F (r, θ) = Ureiθ + e−iθ = U r cos θ + ri sin θ + cos θ − i sin θ r r r

 a2   a2  which gives Φ(r, θ) = U cos θ r + and Ψ(r, θ) = U sin θ r − . The velocity compo- r r nents are 1  a2  1  a2  u = ∂ ψ = ∂ φ = U cos θ 1 − u = ∂ ψ = ∂ φ = −U sin θ 1 + r r θ r r2 θ r r θ r2

The circulation around a circle γ = Rer is I Z 2π Z 2π

Γ = u · γ = u(γ) · γ˙ dθ = (urer + uθeθ) · Reθ dθ C 0 0 r=R Z 2π  2  Z 2π a = uθ R dθ = −U 1 + 2 R sin θ dθ = 0 0 r=R r 0 Remark 8. There is no hydrodynamic force on the cylinder and therefore there can’t be any on it. This is a consequence of the absence of viscosity.

Definition 6 (Stagnation point). A stagnation point is a point where all the components of velocity vanish.

Remark 9. Check the boundary conditions on the circle |z| = a. We know by construction that Ψ = 0 on the circle. Moreover

Φ = 2U cos θ, ur = 0, uθ = −2U sin θ 6= 0. r=a r=a r=a The second one is no penetration, and the last one is the slip condition. There are two stagnation points at θ = 0, π in which uθ = 0 on the circle. Otherwise, there is slip on the body.

12 Now, we impose the additional constraint of nonzero circulation around the cylinder. By , we add a vortex of strength Γ at the origin. The corresponding complex potential is  a2  iΓ F (z) = U z + − ln z. z 2π Computing F 0(z) gives  a2  iΓ 1 W = F 0(z) = U 1 − − z2 2π z and as |z| −→ ∞, w0(z) −→ U. The velocity potential and stream function are

 a2 a2  iΓ iΓ F (r, θ) = U r cos θ + ir sin θ + cos θ − i sin θ − ln r − iθ r r 2π 2π which implies

 a2  Γ  a2  Γ Φ(r, θ) = U cos θ r + + θ Ψ(r, θ) = −U sin θ r − − ln r. r2 2π r 2π

The cylinder remains a streamline since

Γ Ψ = − ln a. r=a 2π To obtain Ψ = 0 on the cylinder, we substract the corresponding constant from the contribution of stream function in F (z)

 a2  iΓ iΓ  a2  iΓ z  F (z) = U z + − ln z + ln a = U z + − ln . z 2π 2π z 2π a

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