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Vortex topology and global-Bernoulli counterexamples — formal worked fields

Example

For $κ = 1$ , $ρ_{0} = 2$ , $p_{0} = 0$ , and $r = (x^{2} + y^{2})^{1 / 2} > 0$ , the vortex is $u_{v} = (- y / r^{2}, x / r^{2}, 0)$ with pressure $p_{v} = - 1 / r^{2}$ . Its circle of radius $R = 2$ has circulation $2 π$ . Separately, for solid rotation take $Ω_{0} = 1$ , $ρ_{0} = 2$ , and $p_{0} = 0$ ; then $u_{s} = (- y, x, 0)$ and $p_{s} = r^{2}$ , and its Bernoulli quantity is $B = r^{2}$ . Thus circles of distinct radii have distinct Bernoulli values.

Example references

Vortex topology and global-Bernoulli counterexamples

Facts & Assumptions

Given: The vortex is defined only where $r > 0$ and the solid-rotation field is on all of $R^{3}$ . Both use the stated constant-density pressure multiplier.

[A1]

For the vortex use $κ = 1$ , $ρ_{0} = 2$ , $p_{0} = 0$ , $ν > 0$ . For solid rotation use $Ω_{0} = 1$ , $ρ_{0} = 2$ , $p_{0} = 0$ , $ν > 0$ .

[F1]

Vortex topology and global-Bernoulli counterexamples: the vortex has circulation $2 π κ$ and pressure $p_{0} - ρ_{0} κ^{2} / (2 r^{2})$ ; its nonzero circulation rules out a single-valued global potential. Solid rotation with pressure $p_{0} + ρ_{0} Ω_{0}^{2} r^{2} / 2$ is an exact steady field with $B = p_{0} / ρ_{0} + Ω_{0}^{2} r^{2}$ .

Verification

1.1givenA1F1algebra

The parameter substitutions in [F1] give $p_{v} = - 1 / r^{2}$ . Direct differentiation yields $\nabla p_{v} = 2 (x, y, 0) / r^{4}$ , so $ρ_{0}^{- 1} \nabla p_{v} = (x, y, 0) / r^{4}$ ; the vortex convection is $- (x, y, 0) / r^{4}$ and its Laplacian is zero. Along $γ_{2} (θ) = (2 \cos θ, 2 \sin θ, z_{0})$ , $u_{v} \cdot γ_{2}^{'} = 1$ , so $\oint_{γ_{2}} u_{v} \cdot d r = \int_{0}^{2 π} 1 d θ = 2 π \neq 0$ . A single-valued potential would have zero integral by the chain rule around the closed curve.

2.1givenA1F1algebra∎

For solid rotation the substituted pressure is $p_{s} = r^{2}$ and $∣ u_{s} ∣^{2} / 2 = r^{2} / 2$ , while $p_{s} / ρ_{0} = r^{2} / 2$ . Hence $B = r^{2}$ . Along a streamline $γ (s) = (x (s), y (s), z (s))$ of this selected field, $γ^{'} = (- y, x, 0)$ , so $\frac{d}{d s} (x^{2} + y^{2}) = 2 x (- y) + 2 y x = 0$ ; consequently $r$ and $B$ are constant along that curve, while different positive radii give different values.

Depends on

Vortex topology and global-Bernoulli counterexamples