Lamb–Oseen vortex

Short description: Line 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

$v_r=0,v_{\theta }=\frac{\mathrm{\Gamma }}{2\pi r}g(r,t),v_z=0.$

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

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

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

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

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

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

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

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

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

Given the vorticity vector, $\mathit{\omega }=(0,0,{\omega }_z)$, the $z$-component of the linearized vorticity transport equation is $$\begin{array}{r}\frac{\partial {\omega }_z}{\partial t}=\nu (\frac{{\partial }^2{\omega }_z}{\partial r^2}+\frac{1}{r}\frac{\partial {\omega }_z}{\partial r}).\end{array}$$ 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 ${\omega }_z(r,t)=A\mathrm{Ei}(-\frac{r^2}{4\nu t})+B$ 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,

$$\begin{array}{rl}\frac{\partial {\omega }_z}{\partial t}& :=\frac{\partial }{\partial t}\left(\frac{f}{t}\right)=f\frac{\partial }{\partial t}\left(\frac{1}{t}\right)+\frac{\mathrm{d}f}{\mathrm{d}\eta }\frac{\partial \eta }{\partial t}=-\frac{1}{t^2}f-\frac{2\nu r}{t(4\nu t{)}^{3/2}}{f}^{\prime }\\ \frac{1}{r}\frac{\partial {\omega }_z}{\partial r}& :=\frac{1}{rt}\frac{\mathrm{d}f}{\mathrm{d}\eta }\frac{\partial \eta }{\partial r}=f^{\prime }\frac{1}{rt\sqrt{4\nu t}}\\ \frac{{\partial }^2{\omega }_z}{\partial r^2}& :=\frac{\partial }{\partial r}\left(\frac{1}{t}\frac{\mathrm{d}f}{\mathrm{d}\eta }\frac{\partial \eta }{\partial r}\right)=\frac{1}{t}\frac{\mathrm{d}f}{\mathrm{d}\eta }\underset{=0}{\underbrace{\frac{\partial }{\partial r}\left(\frac{\partial \eta }{\partial r}\right)}}+\frac{1}{t}\frac{\partial \eta }{\partial r}\frac{\partial }{\partial r}\left(\frac{\mathrm{d}f}{\mathrm{d}\eta }\right)=\frac{1}{t}\frac{\partial \eta }{\partial r}\cdot \frac{{\mathrm{d}}^{2}f}{\mathrm{d}{\eta }^2}\frac{\partial \eta }{\partial r}=f^{″}\frac{1}{4\nu t^2}\end{array}$$ yielding the Sturm–Liouville differential equation, $$\begin{array}{r}{f}^{″}+(\frac{1}{\eta }+2\eta )f^{\prime }+4f=0.\end{array}$$ Because this is a differential equation of the form, $$\begin{array}{rl}{y}^{″}+f(x)y^{\prime }+g(x)y& =0\\ \equiv x{y}^{″}+\frac{d}{dx}\left[(xf(x))y\right]& =0\text{, where,}g(x)=\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x}[xf(x)]\end{array}$$ for some function, $y(x)$, it admits the solution, $$\begin{array}{rl}y(x)& =C{e}^{-\int [f(x)-\frac{g(x)-f^{\prime }(x)}{f(x)}]\mathrm{d}x}\int e^{\int [f(x)-2\frac{g(x)-f^{\prime }(x)}{f(x)}]\mathrm{d}x}\mathrm{d}x\\ & =Cx{e}^{-\int f(x)\mathrm{d}x}\int \frac{1}{x^2}{e}^{\int f(x)\mathrm{d}x}\mathrm{d}x\end{array}$$ This can be checked by multiplying through by $\eta$ and using integration by parts as follows: $$\begin{array}{rl}0& =\eta (f^{″}+\frac{1}{\eta }{f}^{\prime }+2\eta f^{\prime }+4f)\\ ⟺C& =\int \eta f^{″}d\eta +\int f^{\prime }d\eta +\int 2{\eta }^2{f}^{\prime }d\eta +\int 4\eta fd\eta \\ ⟺C& =(\eta f^{\prime }-\int f^{\prime }d\eta )+\int f^{\prime }d\eta +(2{\eta }^{2}f-\int 4\eta fd\eta )+\int 4\eta fd\eta \\ ⟺C& =\eta f^{\prime }+2{\eta }^{2}f\\ ⟺f^{\prime }+2\eta f& =\frac{C}{\eta }\text{, with an integrating factor,}p(\eta )=e^{\int 2\eta d\eta }=e^{{\eta }^2}\\ ⟺e^{{\eta }^2}{f}^{\prime }+2\eta e^{{\eta }^2}f& =C\frac{e^{{\eta }^2}}{\eta }\\ ⟺\frac{d}{d\eta }\left(e^{{\eta }^2}f\right)& =C\frac{e^{{\eta }^2}}{\eta }\\ ⟺e^{{\eta }^2}f& =C\int \frac{e^{{\eta }^2}}{\eta }d\eta \\ \therefore f(\eta )& =C\mathrm{Ei}({\eta }^2)e^{-{\eta }^2}+C_1{e}^{-{\eta }^2}\end{array}$$ Because $\underset{\eta \to 0}{\lim }\mathrm{Ei}({\eta }^2)$ is not well defined, $C=0$. Substituting $\eta$ and $f(\eta )$ into ${\omega }_z(r,t)=f(\eta )/t$ yields, $$\begin{array}{r}{\omega }_z(r,t)=\frac{C_1}{t}{e}^{-\frac{r^2}{4\nu t}}\end{array}$$ The circulation flux integral over a surface, $S=\{(r,\theta )\in {\mathbb{ℝ}}^2|0\le r\le \mathrm{\infty },0\le \theta \le 2\pi \}$, may be solved for constant circulation, $\mathrm{\Gamma }$. $$\begin{array}{rl}\mathrm{\Gamma }:=\underset{S}{\iint }\mathit{\omega }\cdot \mathrm{d}\mathit{S}& =\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\omega }_z\cdot r\mathrm{d}r\mathrm{d}\theta \\ & =2\pi \underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{C_1}{t}{e}^{-\frac{r^2}{4\nu t}}\cdot r\mathrm{d}r\\ & =4\pi \nu C_1\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-u}\mathrm{d}u\text{where,}u=\frac{r^2}{4\nu t}\\ \therefore C_1& =\frac{\mathrm{\Gamma }}{4\pi \nu }.\end{array}$$ 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 $v_{\theta }$. $$\begin{array}{rl}{v}_{\theta }(r,t)& =\frac{1}{r}\int r{\omega }_z\mathrm{d}r\\ & =\frac{\mathrm{\Gamma }}{4\pi \nu t}\frac{1}{r}\int r{e}^{-\frac{r^2}{4\nu t}}\mathrm{d}r\\ & =\frac{\mathrm{\Gamma }}{4\pi \nu t}\frac{1}{r}\int r{e}^u(-\frac{2\nu t}{r})\mathrm{d}u\text{, where,}u=-\frac{r^2}{4\nu t}\\ & =\frac{\mathrm{\Gamma }}{2\pi r}(C_2-e^{-\frac{r^2}{4\nu t}})\\ \end{array}$$ 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

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

that leads to the equation

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

Self-similar solution exists for the coordinate $\eta =r/\phi (t)$, provided $\phi {\phi }^{\prime }+\gamma {\phi }^2=a$, where $a$ is a constant, in which case $g=1-{\mathrm{e}}^{-a{\eta }^2/2\nu }$. The solution for $\phi (t)$ may be written according to Rott (1958) as

${\phi }^2=2a\exp (-2\underset{0}{\overset{t}{\int }}\gamma (s)\mathrm{d}s)\underset{c}{\overset{t}{\int }}\exp (2\underset{0}{\overset{u}{\int }}\gamma (s)\mathrm{d}s)\mathrm{d}u,$

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 $c=-\mathrm{\infty }$, ${\phi }^2=a/k$, a Burgers vortex is a obtained. For arbitrary $c$, the solution becomes ${\phi }^2=a(1+\beta {\mathrm{e}}^{-2kt})/k$, where $\beta$ is an arbitrary constant. As $t\to \mathrm{\infty }$, Burgers vortex is recovered.

See also

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