Lamb–Oseen vortex

In fluid dynamics, the Lamb–Oseen vortex models a line vortex that decays due to viscosity. This vortex is named after Horace Lamb and Carl Wilhelm Oseen.

Vector plot of the Lamb–Oseen vortex velocity field.

Evolution of a Lamb–Oseen vortex in air in real time. Free-floating test particles reveal the velocity and vorticity pattern. (scale: image is 20 cm wide)

Mathematical description

Oseen looked for a solution for the Navier–Stokes equations in cylindrical coordinates $(r,\theta ,z)$ with velocity components $(v_{r},v_{\theta },v_{z})$ of the form

$v_{r}=0,\quad v_{\theta }={\frac {\Gamma }{2\pi r}}g(r,t),\quad v_{z}=0.$

where $\Gamma$ is the circulation of the vortex core. Navier–Stokes equations lead to

${\frac {\partial g}{\partial t}}=\nu \left({\frac {\partial ^{2}g}{\partial r^{2}}}-{\frac {1}{r}}{\frac {\partial g}{\partial r}}\right)$

which, subject to the conditions that it is regular at $r=0$ and becomes unity as ⁠$r\rightarrow \infty$⁠, leads to

$g(r,t)=1-\mathrm {e} ^{-r^{2}/4\nu t},$

where $\nu$ is the kinematic viscosity of the fluid. At ⁠$t=0$⁠, we have a potential vortex (free or irrotational vortex) with concentrated vorticity at the $z$-axis; and this vorticity diffuses away as time passes.

The only non-zero vorticity component is in the $z$-direction, given by

$\omega _{z}(r,t)={\frac {\Gamma }{4\pi \nu t}}\mathrm {e} ^{-r^{2}/4\nu t}.$

The pressure field simply ensures the vortex rotates in the circumferential direction, providing the centripetal force

${\partial p \over \partial r}=\rho {v^{2} \over r},$

where $\rho$ is the constant density. Below are two methods by which Oseen's vortex can be derived.

Derivation of the Lamb-Oseen vortex in polar coordinates

We begin with the polar coordinate $(r,\theta )\in \mathbb {R} ^{2}$ vector field, ${\vec {v}}=\left(0,v_{\theta }\right)$ in which the azimuthal momentum equation reduces to the advectionless form, $${\frac {\partial v_{\theta }}{\partial t}}=\nu \left({\frac {\partial ^{2}v_{\theta }}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial v_{\theta }}{\partial r}}-{\frac {v_{\theta }}{r^{2}}}\right)$$ which is subject to the following boundary conditions: $${\begin{Bmatrix}v_{\theta }(0,t)=0,&v_{\theta }(r\to \infty ,t)=0\\v_{\theta }(r,0)={\frac {\Gamma }{2\pi r}},&v_{\theta }(r,t\to \infty )=0\end{Bmatrix}}$$ By the substitution, $v_{\theta }(r,t)={\frac {\Gamma }{2\pi r}}g(r,t)$, given some unknown $g:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$, we have,

$${\begin{aligned}{\frac {\partial v_{\theta }}{\partial t}}&=\nu \left({\frac {\partial ^{2}v_{\theta }}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial v_{\theta }}{\partial r}}-{\frac {v_{\theta }}{r^{2}}}\right)\\\iff \ {\frac {\partial }{\partial t}}\left({\frac {\Gamma }{2\pi r}}g(r,t)\right)&=\nu \left({\frac {\partial ^{2}}{\partial r^{2}}}\left({\frac {\Gamma }{2\pi r}}g(r,t)\right)+{\frac {1}{r}}{\frac {\partial }{\partial r}}\left({\frac {\Gamma }{2\pi r}}g(r,t)\right)-{\frac {1}{r^{2}}}{\frac {\Gamma }{2\pi r}}g(r,t)\right)\\\iff \ {\frac {\Gamma }{2\pi r}}{\frac {\partial g}{\partial t}}&=\nu \left(\left(2{\frac {\Gamma }{2\pi r^{3}}}g-{\frac {\Gamma }{2\pi r^{2}}}{\frac {\partial g}{\partial r}}-{\frac {\Gamma }{2\pi r^{2}}}{\frac {\partial g}{\partial r}}+{\frac {\Gamma }{2\pi r}}{\frac {\partial ^{2}g}{\partial r^{2}}}\right)+{\frac {1}{r}}\left(-{\frac {\Gamma }{2\pi r^{2}}}g+{\frac {\Gamma }{2\pi r}}{\frac {\partial g}{\partial r}}\right)-{\frac {1}{r^{2}}}{\frac {\Gamma }{2\pi r}}g\right)\\\iff \ {\frac {\partial g}{\partial t}}&=\nu \left(\left({\frac {2}{r^{2}}}g-{\frac {2}{r}}{\frac {\partial g}{\partial r}}+{\frac {\partial ^{2}g}{\partial r^{2}}}\right)+{\frac {1}{r}}\left(-{\frac {1}{r}}g+{\frac {\partial g}{\partial r}}\right)-{\frac {1}{r^{2}}}g\right)\\\therefore \ {\frac {\partial g}{\partial t}}&=\nu \left({\frac {\partial ^{2}g}{\partial r^{2}}}-{\frac {1}{r}}{\frac {\partial g}{\partial r}}\right)\end{aligned}}$$ We can invoke a self-similar solution, $g(r,t)=f(\eta )$ with $\eta =r/\varphi (t)$, and obtain the following derivatives using the chain rule: $${\begin{aligned}{\frac {\partial g}{\partial t}}&:={\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial t}}=-f'{\frac {r}{\varphi (t)^{2}}}\varphi '(t)\\{\frac {1}{r}}{\frac {\partial g}{\partial r}}&:={\frac {1}{r}}{\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial r}}={\frac {f'}{r\varphi (t)}}\\{\frac {\partial ^{2}g}{\partial r^{2}}}&:={\frac {\partial }{\partial r}}\left({\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial r}}\right)={\frac {\mathrm {d} f}{\mathrm {d} \eta }}\underbrace {{\frac {\partial }{\partial r}}\left({\frac {\partial \eta }{\partial r}}\right)} _{=0}+{\frac {\partial \eta }{\partial r}}{\frac {\partial }{\partial r}}\left({\frac {\mathrm {d} f}{\mathrm {d} \eta }}\right)={\frac {\partial \eta }{\partial r}}\cdot {\frac {\mathrm {d} ^{2}f}{\mathrm {d} \eta ^{2}}}{\frac {\partial \eta }{\partial r}}={\frac {f''}{\varphi (t)^{2}}}\end{aligned}}$$ By substitution, it follows that $${\begin{aligned}-f'{\frac {r}{\varphi ^{2}}}\varphi '=\nu \left({\frac {f''}{\varphi ^{2}}}-{\frac {f'}{r\varphi }}\right)\\\iff \ f''+\left(\eta {\frac {\varphi \varphi '}{\nu }}-{\frac {1}{\eta }}\right)f'=0\end{aligned}}$$ If ${\frac {\varphi \varphi '}{\nu }}=1$, then $\varphi (t)={\sqrt {2\nu t}}$, and hence, $${\begin{aligned}{\frac {\mathrm {d} ^{2}f}{\mathrm {d} \eta ^{2}}}&=-\left(\eta -{\frac {1}{\eta }}\right){\frac {\mathrm {d} f}{\mathrm {d} \eta }}\ \ {\text{, where,}}\ \ u(\eta )=f'(\eta )\\\iff \ {\frac {\mathrm {d} u}{\mathrm {d} \eta }}&=-\left(\eta -{\frac {1}{\eta }}\right)u\\\iff \ \int {\frac {\mathrm {d} u}{u}}&=-\int \left(\eta -{\frac {1}{\eta }}\right)\mathrm {d} \eta \\\iff \ \ln(u)&=-\left({\frac {\eta ^{2}}{2}}-\ln(\eta )\right)+C\\\iff \ u(\eta )&=Ae^{-{\frac {\eta ^{2}}{2}}+\ln(\eta )}\ \ \ {\text{, where,}}\ \ A=e^{C}\\\iff \ f'(\eta )&=A\eta \ e^{-{\frac {\eta ^{2}}{2}}}\\\iff \ f(\eta )&=A\int \eta \ e^{-{\frac {\eta ^{2}}{2}}}\mathrm {d} \eta \ \ {\text{, where,}}\ \ p=-{\frac {\eta ^{2}}{2}}\ \ ,\ \ \mathrm {d} p=-\eta \ \mathrm {d} \eta \\\iff \ f(\eta )&=A\int \eta \left(-{\frac {1}{\eta }}\right)e^{p}\mathrm {d} p\\\iff \ f(\eta )&=-Ae^{-{\frac {\eta ^{2}}{2}}}+B\\\therefore \ g(r,t)&=B-Ae^{-{\frac {r^{2}}{4\nu t}}}\end{aligned}}$$ The first boundary condition, $v_{\theta }(0,t)=0$, infers that $g(0,t)=0$, which holds if and only if $\displaystyle \lim _{r\to 0}{\frac {\Gamma }{2\pi r}}g(r,t)=0$. Verifying by L'Hôpital's rule, $${\begin{aligned}\displaystyle \lim _{r\to 0}{\frac {\Gamma }{2\pi r}}g(r,t)={\frac {\Gamma }{2\pi }}\cdot \displaystyle \lim _{r\to 0}{\frac {-Ae^{-{\frac {r^{2}}{4\nu t}}}+B}{r}}={\frac {\Gamma }{2\pi }}\cdot \displaystyle \lim _{r\to 0}{\frac {{\frac {\mathrm {d} }{\mathrm {d} r}}\left(-Ae^{-{\frac {r^{2}}{4\nu t}}}+B\right)}{{\frac {\mathrm {d} }{\mathrm {d} r}}(r)}}={\frac {\Gamma A}{2\pi }}\cdot \displaystyle \lim _{r\to 0}\left({\frac {r}{2\nu t}}e^{-{\frac {r^{2}}{4\nu t}}}\right)=0\end{aligned}}$$ The second boundary condition, $v_{\theta }(r\to \infty ,t)=0$, is satisfied by the irrotational vortex, $\displaystyle \lim _{r\to \infty }{\frac {\Gamma }{2\pi r}}=0$, for all $t$. Given the third boundary condition, $v_{\theta }(r,0)={\frac {\Gamma }{2\pi r}}$, it follows that $g(r,0)=B=1$. Likewise, $\forall r\in [0,\infty )$, the fourth boundary condition, $v_{\theta }(r,t\to \infty )=0$, implies that $\displaystyle \lim _{t\to \infty }{\frac {\Gamma }{2\pi r}}\left(B-Ae^{-{\frac {r^{2}}{4\nu t}}}\right)=0$, such that $B=A$. Therefore, the solution becomes $${\begin{aligned}v_{\theta }(r,t)={\frac {\Gamma }{2\pi r}}\left(1-e^{-{\frac {r^{2}}{4\nu t}}}\right).\end{aligned}}$$

Alternative derivation of the Lamb-Oseen vortex by the vorticity equation

Given the vorticity vector, ${\boldsymbol {\omega }}=\left(0,0,\omega _{z}\right)$, the $z$-component of the linearized vorticity transport equation is $${\begin{aligned}{\frac {\partial \omega _{z}}{\partial t}}=\nu \left({\frac {\partial ^{2}\omega _{z}}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial \omega _{z}}{\partial r}}\right).\end{aligned}}$$ If one insists on invoking the similarity solution, $\omega _{z}(r,t)=f(\eta )$ with $\eta =r/{\sqrt {2\nu t}}$ as done previously, one would exclusively obtain $\omega _{z}(r,t)=A\operatorname {Ei} \left(-{\frac {r^{2}}{4\nu t}}\right)+B$ for arbitrary constants, $A$ and $B$. To obtain the complete solution to $\omega _{z}$, one must introduce the similarity solution, $\omega _{z}(r,t)=f(\eta )/t$ with $\eta =r/{\sqrt {4\nu t}}$. By the product rule and the chain rule, each derivative becomes,

$${\begin{aligned}{\frac {\partial \omega _{z}}{\partial t}}&:={\frac {\partial }{\partial t}}\left({\frac {f}{t}}\right)=f{\frac {\partial }{\partial t}}\left({\frac {1}{t}}\right)+{\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial t}}=-{\frac {1}{t^{2}}}f-{\frac {2\nu r}{t(4\nu t)^{3/2}}}f'\\{\frac {1}{r}}{\frac {\partial \omega _{z}}{\partial r}}&:={\frac {1}{rt}}{\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial r}}=f'{\frac {1}{rt{\sqrt {4\nu t}}}}\\{\frac {\partial ^{2}\omega _{z}}{\partial r^{2}}}&:={\frac {\partial }{\partial r}}\left({\frac {1}{t}}{\frac {\mathrm {d} f}{\mathrm {d} \eta }}{\frac {\partial \eta }{\partial r}}\right)={\frac {1}{t}}{\frac {\mathrm {d} f}{\mathrm {d} \eta }}\underbrace {{\frac {\partial }{\partial r}}\left({\frac {\partial \eta }{\partial r}}\right)} _{=0}+{\frac {1}{t}}{\frac {\partial \eta }{\partial r}}{\frac {\partial }{\partial r}}\left({\frac {\mathrm {d} f}{\mathrm {d} \eta }}\right)={\frac {1}{t}}{\frac {\partial \eta }{\partial r}}\cdot {\frac {\mathrm {d} ^{2}f}{\mathrm {d} \eta ^{2}}}{\frac {\partial \eta }{\partial r}}=f''{\frac {1}{4\nu t^{2}}}\end{aligned}}$$ yielding the Sturm–Liouville differential equation, $${\begin{aligned}f''+\left({\frac {1}{\eta }}+2\eta \right)f'+4f=0.\end{aligned}}$$ Because this is a differential equation of the form, $${\begin{aligned}y''+f(x)y'+g(x)y&=0\\\equiv \ xy''+{\frac {d}{dx}}\left[\left(xf(x)\right)y\right]&=0\ \ \ {\text{, where,}}\ \ g(x)={\frac {1}{x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left[xf(x)\right]\end{aligned}}$$ for some function, $y(x)$, it admits the solution, $${\begin{aligned}y(x)&=Ce^{-\int \left[f(x)-{\frac {g(x)-f'(x)}{f(x)}}\right]\mathrm {d} x}\int e^{\int \left[f(x)-2{\frac {g(x)-f'(x)}{f(x)}}\right]\mathrm {d} x}\mathrm {d} x\\&=Cxe^{-\int f(x)\mathrm {d} x}\int {\frac {1}{x^{2}}}e^{\int f(x)\mathrm {d} x}\mathrm {d} x\end{aligned}}$$ This can be checked by multiplying through by $\eta$ and using integration by parts as follows: $${\begin{aligned}0&=\eta \left(f''+{\frac {1}{\eta }}f'+2\eta f'+4f\right)\\\iff \ C&=\int \eta f''d\eta +\int f'd\eta +\int 2\eta ^{2}f'd\eta +\int 4\eta fd\eta \\\iff \ C&=\left(\eta f'-\int f'd\eta \right)+\int f'd\eta +\left(2\eta ^{2}f-\int 4\eta fd\eta \right)+\int 4\eta fd\eta \\\iff \ C&=\eta f'+2\eta ^{2}f\\\iff \ f'+2\eta f&={\frac {C}{\eta }}\ \ \ {\text{, with an integrating factor,}}\ \ p(\eta )=e^{\int 2\eta d\eta }=e^{\eta ^{2}}\\\iff \ e^{\eta ^{2}}f'+2\eta e^{\eta ^{2}}f&=C{\frac {e^{\eta ^{2}}}{\eta }}\\\iff \ {\frac {d}{d\eta }}\left(e^{\eta ^{2}}f\right)&=C{\frac {e^{\eta ^{2}}}{\eta }}\\\iff \ e^{\eta ^{2}}f&=C\int {\frac {e^{\eta ^{2}}}{\eta }}d\eta \\\therefore \ f(\eta )&=C\operatorname {Ei} (\eta ^{2})e^{-\eta ^{2}}+C_{1}e^{-\eta ^{2}}\end{aligned}}$$ Because $\displaystyle \lim _{\eta \to 0}\operatorname {Ei} (\eta ^{2})$ is not well defined, $C=0$. Substituting $\eta$ and $f(\eta )$ into $\omega _{z}(r,t)=f(\eta )/t$ yields, $${\begin{aligned}\omega _{z}(r,t)={\frac {C_{1}}{t}}e^{-{\frac {r^{2}}{4\nu t}}}\end{aligned}}$$ The circulation flux integral over a surface, $S=\left\{(r,\theta )\in \mathbb {R} ^{2}\ |\ 0\leq r\leq \infty ,\ 0\leq \theta \leq 2\pi \right\}$, may be solved for constant circulation, $\Gamma$. $${\displaystyle {\begin{aligned}\Gamma$$ :=\iint _{S}{\boldsymbol {\omega }}\cdot \mathrm {d} {\boldsymbol {S}}&=\int _{0}^{2\pi }\int _{0}^{\infty }\omega _{z}\cdot r\mathrm {d} r\mathrm {d} \theta \\&=2\pi \int _{0}^{\infty }{\frac {C_{1}}{t}}e^{-{\frac {r^{2}}{4\nu t}}}\cdot r\mathrm {d} r\\&=4\pi \nu C_{1}\int _{0}^{\infty }e^{-u}\mathrm {d} u\ \ \ {\text{where,}}\ \ \ u={\frac {r^{2}}{4\nu t}}\\\therefore C_{1}&={\frac {\Gamma }{4\pi \nu }}.\end{aligned}}} Since, $$\omega _{z}={\frac {\partial v_{\theta }}{\partial r}}+{\frac {v_{\theta }}{r}}-{\frac {1}{r}}{\frac {\partial v_{r}}{\partial \theta }}$$ bearing in mind that $v_{r}=0$, we can use the integrating factor, $e^{\int {\frac {1}{r}}dr}=r$, to isolate $v_{\theta }$. $${\begin{aligned}v_{\theta }(r,t)&={\frac {1}{r}}\int r\omega _{z}\mathrm {d} r\\&={\frac {\Gamma }{4\pi \nu t}}{\frac {1}{r}}\int re^{-{\frac {r^{2}}{4\nu t}}}\mathrm {d} r\\&={\frac {\Gamma }{4\pi \nu t}}{\frac {1}{r}}\int re^{u}\left(-{\frac {2\nu t}{r}}\right)\mathrm {d} u\ \ \ {\text{, where,}}\ \ u=-{\frac {r^{2}}{4\nu t}}\\&={\frac {\Gamma }{2\pi r}}\left(C_{2}-e^{-{\frac {r^{2}}{4\nu t}}}\right)\\\end{aligned}}$$ The aforementioned boundary conditions on $v_{\theta }$ yields the Lamb-Oseen vortex.

Generalized Oseen vortex

The generalized Oseen vortex may be obtained by looking for solutions of the form

$v_{r}=-\gamma (t)r,\quad v_{\theta }={\frac {\Gamma }{2\pi r}}g(r,t),\quad v_{z}=2\gamma (t)z$

that leads to the equation

${\frac {\partial g}{\partial t}}-\gamma r{\frac {\partial g}{\partial r}}=\nu \left({\frac {\partial ^{2}g}{\partial r^{2}}}-{\frac {1}{r}}{\frac {\partial g}{\partial r}}\right).$

Self-similar solution exists for the coordinate ⁠$\eta =r/\varphi (t)$⁠, provided ⁠$\varphi \varphi '+\gamma \varphi ^{2}=a$⁠, where $a$ is a constant, in which case ⁠$g=1-\mathrm {e} ^{-a\eta ^{2}/2\nu }$⁠. The solution for $\varphi (t)$ may be written according to Rott (1958) as

$\varphi ^{2}=2a\exp \left(-2\int _{0}^{t}\gamma (s)\,\mathrm {d} s\right)\int _{c}^{t}\exp \left(2\int _{0}^{u}\gamma (s)\,\mathrm {d} s\right)\,\mathrm {d} u,$

where $c$ is an arbitrary constant. For ⁠$\gamma =0$⁠, the classical Lamb–Oseen vortex is recovered. The case $\gamma =k$ corresponds to the axisymmetric stagnation point flow, where $k$ is a constant. When ⁠$c=-\infty$⁠, ⁠$\varphi ^{2}=a/k$⁠, a Burgers vortex is a obtained. For arbitrary ⁠$c$⁠, the solution becomes ⁠$\varphi ^{2}=a(1+\beta \mathrm {e} ^{-2kt})/k$⁠, where $\beta$ is an arbitrary constant. As ⁠$t\rightarrow \infty$⁠, Burgers vortex is recovered.

See also

• Rankine vortex and Kaufmann (Scully) vortex – Common simplified approximations for a viscous vortex.