# A direct approach to solving the hydrodynamic oseenlet

Aditya R. Pujari
International Centre for Theoretical Sciences - TIFR
Shivakote, Bengaluru (North) - 560 089, India.
<EMAIL_ADDRESS>
Alternate email<EMAIL_ADDRESS>
(November, 2021)

###### Abstract

We present a new direct approach to obtain the Green’s function of the Oseen
equations, also known as the oseenlet. This approach is different in that it
does not assume an initial form of the solution to begin with. The final
solution is expressed in a new, simplified form that resembles the Oseen-
Burger’s tensor of the stokeslet equations, from which the contributions due
to inertia are easily appreciable.

_Keywords_ Stokes flow $\cdot$ Oseen flow $\cdot$ Stokeslet $\cdot$ Oseenlet
$\cdot$ Point Force Singularities

##  1 Introduction

Green’s functions are useful mathematical constructs, especially in fluid
dynamics while describing low Reynolds number flows, such as Stokes flow and
Oseen flow. Stokes flow in particular has been extensively studied in
literature over the past several decades. Various methods have been devised to
obtain the Green’s function of Stokes flow—also known as Oseen-Burger’s
tensor, or the stokeslet [1]. Oseen flow, however, is a relatively new and
expanding subject, and literature on obtaining the Green’s function of Oseen
flow—such as that given by Pozrikidis [2]—is limited.

Here we present a new direct approach to obtain the Green’s function of the
Oseen equations, also popularly known as the oseenlet, using Fourier
transformations. This approach differs from that of Pozrikidis, in that it
does not assume a certain form of the solution to begin with. The motivation
to come up with a solution of the oseenlet equations using Fourier
transformations came from appreciating the elegance of the method of Zapryanov
and Tabakova to solve the stokeslet [3] (hereafter, referred to as the Z&T
method). Their method makes use of the Fourier transforms of the fundamental
solutions of known partial differential equations; as a result, it requires
little calculation and the solution is obtained within a page of simple
algebra. Intrigued by their method, a similar direct approach was devised to
solve the oseenlet.

##  2 The Oseen Equations and the Oseenlet

The Oseen equations are a linearized approximation of the Navier-Stokes
equations. We briefly derive them here for the steady case. Let $\mathbf{u}$
and $p$ denote the velocity and pressure of the fluid in the domain, which are
functions of the spatial variable $\mathbf{x}$. The Navier-Stokes equations
are given by,

$\rho\mathbf{u}(\mathbf{x})\cdot\mathbf{\nabla}\mathbf{u}(\mathbf{x})=-\mathbf{\nabla}p(\mathbf{x})+\mu\nabla^{2}\mathbf{u}(\mathbf{x})$
(1) $\mathbf{\nabla}\cdot\mathbf{u}(\mathbf{x})=0$ (2)

where $\rho$ is the density of the fluid and $\mu$ is the dynamic viscosity of
the fluid. Let us decompose $\mathbf{u}$ into a constant average flow
$\mathbf{U}$ and a perturbed flow $\mathbf{v}(\mathbf{x})$.

$\mathbf{u}=\mathbf{U}+\mathbf{v}(\mathbf{x})$ (3)

Rewriting the Navier-Stokes equations in terms of this decomposition we get,

$\rho(\mathbf{U}+\mathbf{v}(\mathbf{x}))\cdot\mathbf{\nabla}(\mathbf{U}+\mathbf{v}(\mathbf{x}))=-\mathbf{\nabla}p(\mathbf{x})+\mu\nabla^{2}(\mathbf{U}+\mathbf{v}(\mathbf{x}))$
(4) $\nabla\cdot(\mathbf{U}+\mathbf{v}(\mathbf{x}))=0.$ (5)

On expanding the terms, we see that the term
$\rho\mathbf{v}\cdot\nabla\mathbf{v}$ may be neglected as a higher order
$O(v^{2})$ term. This gives the Oseen approximation.

$\rho\mathbf{U}\cdot\nabla\mathbf{U}+\rho\mathbf{U}\cdot\nabla\mathbf{v}(\mathbf{x})+\rho\mathbf{v}(\mathbf{x})\cdot\nabla\mathbf{U}+\rho\mathbf{v}(\mathbf{x})\cdot\nabla\mathbf{v}(\mathbf{x})=-\nabla
p+\mu\nabla^{2}\mathbf{U}+\mu\nabla^{2}\mathbf{v}+\mathbf{b}\delta(\mathbf{x})$
(6)

In the present case where we wish to describe the flow due to a point force
singularity, we obtain the equations of the oseenlet by adding the point force
term, given by,

$\rho\mathbf{U}\cdot\mathbf{\nabla}\mathbf{v}(\mathbf{x})=-\mathbf{\nabla}p(\mathbf{x})+\mu\nabla^{2}\mathbf{v}(\mathbf{x})+\mathbf{b}\delta(\mathbf{x})$
(7) $\mathbf{\nabla}\cdot\mathbf{v}(\mathbf{x})=0.$ (8)

Here $\mathbf{b}$ is the point force located at the origin and
$\delta(\mathbf{x})$ is the 3-dimensional delta function. The boundary
conditions are that $\mathbf{v}(\mathbf{x})\rightarrow 0$ and
$p(\mathbf{x})\rightarrow 0$ as $\mathbf{x}\rightarrow\infty$.

##  3 Solution

To solve the equations of the oseenlet, we begin, as in the Z&T method, by
taking the Fourier transform of eq. 7 and eq. 8. Refer to the appendix for the
definition of the Fourier transform.

$i\rho\mathbf{U}\cdot\mathbf{k}\,\mathbf{\hat{v}}=-i\mathbf{k}\hat{p}-\mu
k^{2}\mathbf{\hat{v}}+\mathbf{b}$ (9) $i\mathbf{k}\cdot\mathbf{\hat{v}}=0$
(10)

Taking the dot product of eq. 9 with $\mathbf{k}$, and using eq. 10 to
eliminate the two velocity terms, we rearrange to obtain the pressure solution
in k-space. This is equivalent to taking the divergence of eq. 7 and
eliminating the velocity terms due to the continuity equation (eq. 8). We are
able to do this trick owing to the linearity of the equation in $\mathbf{v}$;
had the inertial term not been linearized, it would not have been eliminated
on taking the divergence.

$\hat{p}=-i\frac{\mathbf{k}\cdot\mathbf{b}}{k^{2}}$ (11)

On inverting this, we obtain,

$p=-\frac{1}{4\pi}\nabla\left(\frac{1}{r}\right)\cdot\mathbf{b}=-\frac{1}{4\pi}\frac{\mathbf{x}}{r}\cdot\mathbf{b}.$
(12)

Here $r=|\mathbf{x}|$. The pressure solution is the same as that of the
stokeslet, as also pointed out by Pozrikidis. Re-substituting the pressure
solution into eq. 9 we obtain an expression for velocity in k-space.

$\displaystyle
i\rho\mathbf{U}\cdot\mathbf{k}\,\mathbf{\hat{v}}=-\frac{\mathbf{k}\mathbf{k}\cdot\mathbf{b}}{k^{2}}-\mu
k^{2}\mathbf{\hat{v}}+\mathbf{b}$ (13)
$\displaystyle\mathbf{\hat{v}}=\frac{1}{\frac{i\mathbf{U}\cdot\mathbf{k}}{\nu}+k^{2}}\left[\frac{\mathbf{b}}{\mu}-\frac{\mathbf{k}\mathbf{k}\cdot\mathbf{b}}{\mu
k^{2}}\right]$ (14)

Here $\nu$ is the kinematic viscosity of the fluid. We see that in the k-space
velocity solution given by eq. 14, if $\mathbf{U}\rightarrow 0$, we obtain the
corresponding k-space velocity solution of the stokeslet equations
$\frac{\mathbf{b}}{\mu k^{2}}+\frac{\mathbf{k}\mathbf{k}\cdot\mathbf{b}}{\mu
k^{4}}$. Note that the stokeslet solution is symmetric in $\mathbf{k}$,
however, in the oseenlet the $\frac{i\mathbf{U}\cdot\mathbf{k}}{\nu}$ term
introduces an asymmetry in the velocity solution. We complete the square in
the denominator and see that,
$k^{2}+\frac{i\mathbf{U}\cdot\mathbf{k}}{\nu}=\left(\mathbf{k}+\frac{i\mathbf{U}}{2\nu}\right)^{2}+\left(\frac{|\mathbf{U}|}{2\nu}\right)^{2}$.
The completed square may be treated as a shift in k-space by the constant
$\frac{i\mathbf{U}}{2\nu}$. Thus, to invert the first term, we seek to find a
scalar function in this shifted k-space whose Fourier transform is the
fraction $\frac{1}{k^{2}+c^{2}}$, where $c$ is some constant. This leads us to
the Green’s function of the Helmholtz equation, given by:

$\nabla^{2}\Phi(\mathbf{x})-c^{2}\Phi(\mathbf{x})+\delta(\mathbf{x})=0.$ (15)

Thus we have,

$\hat{\Phi}(\mathbf{k})=\frac{1}{k^{2}+c^{2}}$ (16)

and,

$\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})=\frac{1}{k^{2}+\frac{i\mathbf{U}\cdot\mathbf{k}}{\nu}}.$
(17)

Here, $c$ is $\frac{|\mathbf{U}|}{2\nu}$. We know that the solution of the
Helmholtz equation is

$\mathcal{F}^{-1}\left\\{\hat{\Phi}(\mathbf{k})\right\\}=\frac{1}{4\pi
r}e^{-cr}$ (18)

from which, a shift in k-space corresponds to an exponential factor appearing
in x-space,

$\mathcal{F}^{-1}\left\\{\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})\right\\}=\frac{1}{4\pi
r}e^{-\frac{|\mathbf{U}|r}{2\nu}}e^{\frac{\mathbf{U}\cdot\mathbf{x}}{2\nu}}.$
(19)

Thus we have the inversion of the first term.

$\mathbf{v}_{1}=\frac{\mathbf{b}}{4\pi\mu
r}e^{\frac{\mathbf{U}\cdot\mathbf{x}}{2\nu}-\frac{|\mathbf{U}|r}{2\nu}}$ (20)

Before we move on to inverting the second term, it is notable that we have
naturally been graced by a dimensionless variable in the form of the exponent.
We choose to call this dimensionless variable as $\eta$ and expect to
encounter it again later while solving the second term. ($\eta$ is defined
with a negative sign to align with Pozrikidis’ notation.)

$\eta\,\,\dot{=}\,\,\frac{|\mathbf{U}|r}{2\nu}-\frac{\mathbf{U}\cdot\mathbf{x}}{2\nu}$
(21)

The second term in the velocity expression is not as straightforward as the
first. Let us first rewrite the second term in terms of $\hat{\Phi}$.

$\mathbf{v}_{2}=-\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})\frac{\mathbf{k}\mathbf{k}\cdot\mathbf{b}}{\mu
k^{2}}$ (22)

If, somehow, we are able to obtain the inversion of the function
$\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})/k^{2}$, then we would only
be left with calculating a dyadic differentiation after inversion. Let us call
this function $\hat{H}$. Thus,

$\displaystyle\hat{H}\,\,\dot{=}\,\,-\frac{\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})}{k^{2}}$
(23)
$\displaystyle\mathbf{\hat{v}}_{2}=\frac{\mathbf{k}\mathbf{k}\hat{H}\cdot\mathbf{b}}{\mu}$
(24)
$\displaystyle\mathbf{v}_{2}=-\frac{\mathbf{\nabla}\mathbf{\nabla}H\cdot\mathbf{b}}{\mu}.$
(25)

We simplify for $\hat{H}$,

$\displaystyle k^{2}\hat{H}=-\hat{\Phi}(\mathbf{k}+\tfrac{i\mathbf{U}}{2\nu})$
(27) $\displaystyle\nabla^{2}H=\frac{1}{4\pi r}e^{-\eta}.$ (28)

Since it is not possible to integrate $\nabla^{2}H$ directly, we work
backwards starting from $H$ instead. We anticipate that $H$ will be a function
of the dimensionless variable $\eta$ obtained above. With this assumption, we
proceed to differentiate $H$. From here, the solution follows in the footsteps
of Pozrikidis’ solution.

$H=H(\eta)$
$\mathbf{\nabla}H=\frac{|\mathbf{U}|}{2\nu}\left(\frac{\mathbf{x}}{r}-\mathbf{e}\right)\frac{dH}{d\eta}$
(29)

Here $\mathbf{e}$ is the unit vector in the direction of $\mathbf{U}$.
Calculating the Laplacian of $H$:

$\displaystyle\nabla^{2}H\equiv\mathbf{\nabla}\cdot\mathbf{\nabla}H=\frac{|\mathbf{U}|}{2\nu}\frac{d}{d\mathbf{x}}\left[\frac{dH}{d\eta}\left(\frac{\mathbf{x}}{r}-\mathbf{e}\right)\right]=\frac{|\mathbf{U}|}{\nu}\frac{1}{r}\left(\eta\frac{d^{2}H}{d\eta^{2}}+\frac{dH}{d\eta}\right).$
(30)

Equating the expression for $\nabla^{2}H$ in eq. 30 with that in eq. 28 and
integrating we get,

$\displaystyle\frac{1}{4\pi r}e^{-\eta}$
$\displaystyle=\frac{|U|}{\nu}\frac{1}{r}\frac{\partial}{\partial\eta}\left(\eta\frac{\partial
H}{\partial\eta}\right)$ (31)
$\displaystyle-\frac{1}{4\pi}\left.\frac{\nu}{|U|}e^{-\eta}\right|^{\eta}_{0}$
$\displaystyle=\eta\frac{\partial H}{\partial\eta}\;\;\;\;\;\;\;\text{(since
$\eta$ is always positive)}$ (32) $\displaystyle H$
$\displaystyle=\frac{1}{4\pi}\frac{\nu}{|U|}\int^{\eta}_{0}\frac{1-e^{-\eta^{\prime}}}{\eta^{\prime}}d\eta^{\prime}.$
(33)

Thus we have successfully obtained the function $H$. What remains is only the
dyadic differentiation of $H$ in order to calculate the second term.

$\displaystyle\mathbf{v}_{2}$
$\displaystyle=-\frac{\mathbf{b}\cdot\mathbf{\nabla}}{\mu}\mathbf{\nabla}H$
(34)
$\displaystyle=-\frac{\mathbf{b}\cdot\mathbf{\nabla}}{\mu}\left[\frac{\partial
H}{\partial\eta}\left(\frac{|\mathbf{U}|}{2\nu}\left(\frac{\mathbf{x}}{r}-\mathbf{e}\right)\right)\right]$
(35) $\displaystyle\mathbf{v}_{2}$
$\displaystyle=-\frac{\mathbf{b}\cdot\mathbf{\nabla}}{8\pi\mu}\left[\left(\frac{1-e^{-\eta}}{\eta}\right)\left(\frac{\mathbf{x}}{r}-\mathbf{e}\right)\right]$
(36)

Finally, we put together the two terms to obtain the oseenlet velocity
solution, which is in agreement with the solution obtained by Pozrikidis.

$\mathbf{v}(\mathbf{x})=\frac{1}{8\pi\mu}\left\\{\frac{2\mathbf{b}}{r}e^{-\eta}-\mathbf{b}\cdot\mathbf{\nabla}\left[\left(\frac{1-e^{-\eta}}{\eta}\right)\left(\frac{\mathbf{x}}{r}-\mathbf{e}\right)\right]\right\\}.$
(37)

##  4 Discussion

We continue to express the oseenlet solution in a more tractable format. On
carrying out the differentiation of the second term, we obtain:

$\mathbf{u}(\mathbf{x})=\frac{1}{8\pi\mu}\mathbb{O}\cdot\mathbf{b}$ (38)

where,

$\mathbb{O}\,\,\dot{=}\,\,\frac{2}{r}e^{-\eta}\mathbb{I}-\frac{U}{2\nu}\left(\frac{\eta
e^{-\eta}+e^{-\eta}-1}{\eta^{2}}\right)\left(\frac{\mathbf{x}}{r}-\frac{\mathbf{U}}{U}\right)\left(\frac{\mathbf{x}}{r}-\frac{\mathbf{U}}{U}\right)+\left(\frac{e^{-\eta}-1}{\eta}\right)\left(\frac{\mathbb{I}}{r}-\frac{\mathbf{x}\mathbf{x}}{r^{3}}\right).$
(39)

Here $\mathbb{I}$ is the identity tensor. The tensor may be rewritten in a
manner so as to resemble the Oseen-Burger’s tensor of the stokeslet:

$\mathbb{O}=\eta_{1}\frac{\mathbb{I}}{r}+\eta_{2}\frac{\mathbf{x}\mathbf{x}}{r^{3}}+\frac{U}{2\nu}\eta_{3}\mathbf{x}^{\dagger}\mathbf{x}^{\dagger}$
(40)

where the dimensionless weights $\eta_{1}$, $\eta_{2}$, $\eta_{3}$ and the
dimensionless unit vector $\mathbf{x}^{\dagger}$ are defined as,

$\displaystyle\eta_{1}\,\,\dot{=}\,\,\left(\frac{2\eta
e^{-\eta}+e^{-\eta}-1}{\eta}\right)$
$\displaystyle\eta_{2}\,\,\dot{=}\,\,\left(\frac{1-e^{-\eta}}{\eta}\right)$
$\displaystyle\eta_{3}\,\,\dot{=}\,\,\left(\frac{1-\eta
e^{-\eta}-e^{-\eta}}{\eta^{2}}\right)$ (41)
$\displaystyle\mathbf{x}^{\dagger}\,\,\dot{=}\,\,\left(\frac{\mathbf{x}}{r}-\frac{\mathbf{U}}{U}\right).$
(42)

$\eta$, acts as an _angular_ Reynolds number, i.e., the Reynolds number
weighted by a factor of $(1-\cos{\theta})/2$. This can be seen by manipulating
the definition of $\eta$, eq. 21:

$\eta\,\,\dot{=}\,\,\frac{|\mathbf{U}|r}{2\nu}-\frac{\mathbf{U}\cdot\mathbf{x}}{2\nu}=\frac{|\mathbf{U}|r}{2\nu}(1-\cos{\theta})=\dfrac{1}{2}Re_{r}(1-\cos{\theta}).$
(43)

Here $\theta$ is the angle made by the position vector with the mean flow
velocity direction $\mathbf{U}$ and the length scale for the calculation of
the Reynolds number $Re_{r}$ is taken to be magnitude of the position vector
$r$. $\eta_{i}$, $i=1,2,3$, is thus an _angular_ coefficient that weights each
of the terms. For a given Reynolds number $Re_{r}$, each $\eta_{i}$ is a
monotonically decreasing function of $\theta$. Similarly,
$\mathbf{x}^{\dagger}$ is the _skew_ vector describing the extent to which the
position vector is skewed from the mean flow velocity direction $\mathbf{U}$.

The first two terms are similar to the two terms of the Oseen-Burger’s tensor
of the stokeslet solution, and differ only by the scalar weights. As
$U\rightarrow 0$, the scalar factors $\eta_{1},\eta_{2}\rightarrow 1$ and
$\eta_{3}\rightarrow 0$, appropriately recovering the stokeslet solution.

##  5 Conclusion

In this article, we discussed a new direct approach to obtain the Green’s
function of the oseen equations. Without beginning with an assumption of a
specific form of the solution, we made the use of Fourier transformation to
first obtain the solution in $k$-space. The denominator of the first term
naturally suggests using the known solution of the Helmholtz equation for
inversion. A non-dimensional parameter $\eta$, which behaves as an angular
Reynolds number, is naturally obtained in the exponent. To invert the second
term, we work backwards by performing an integration on the lines of
Pozrikidis’ solution.

The final solution is expressed in a new tractable format, similar to that of
the stokeslet. The presence of the $\eta_{i}$ factors is a direct result of
the inertia of the fluid. These $\eta_{i}$ factors along with the skew vector
$\mathbf{x}^{\dagger}$ are in turn responsible for the creation of fore-aft
asymmetry in the velocity field.

## Acknowledgements

The author acknowledges support of the Department of Atomic Energy, Government
of India, under project no. RTI4001. The author is grateful to Prof. Rama
Govindarajan for guidance and would also like to thank Anup Kumar and Divya
Jagannathan for fruitful discussions.

## Appendix A. Fourier Transforms

We define the Fourier transform as:

$\mathcal{F}\\{{f(\mathbf{x})}\\}\,\,\dot{=}\,\,\hat{f}(\mathbf{k})\,\,\dot{=}\,\,\int_{\mathbb{R}^{3}}d\mathbf{x}f(\mathbf{x})e^{-i\mathbf{k}\cdot\mathbf{x}}$
(A1)

and the inverse as

$\mathcal{F}^{-1}\\{{f(\mathbf{k})}\\}\,\,\dot{=}\,\,f(\mathbf{x})\,\,\dot{=}\,\,\frac{1}{(2\pi)^{3}}\int_{\mathbb{R}^{3}}d\mathbf{k}\hat{f}(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}.$
(A2)

We have made use of the Fourier transform of the Green’s function of the
Laplace equation while calculating the pressure solution.

$\nabla^{2}\phi(\mathbf{x})=-\delta(x)$ (A3)

where $\phi=\frac{1}{4\pi r}$. By integrating its solution, we can obtain the
corresponding Fourier space expression,

$\displaystyle\nabla^{2}\left(\frac{1}{4\pi r}\right)$
$\displaystyle=-\delta(\mathbf{x})=-\frac{1}{(2\pi^{3})}\int_{\mathbb{R}^{3}}d\mathbf{k}e^{i\mathbf{k}\cdot\mathbf{x}}$
(A4) $\displaystyle\nabla\left(\frac{1}{4\pi r}\right)$
$\displaystyle=-\frac{1}{(2\pi^{3})}\int_{\mathbb{R}^{3}}d\mathbf{k}\frac{i\mathbf{k}}{k^{2}}e^{i\mathbf{k}\cdot\mathbf{x}}$
(A5) $\displaystyle\left(\frac{1}{4\pi r}\right)$
$\displaystyle=\frac{1}{(2\pi^{3})}\int_{\mathbb{R}^{3}}d\mathbf{k}\frac{1}{k^{2}}e^{i\mathbf{k}\cdot\mathbf{x}}$
(A6)

## References

  * [1] M. Lisicki.  Four approaches to hydrodynamic green’s functions–the oseen tensors, 2013\.  Preprint at https://arxiv.org/abs/1312.6231.
  * [2] C. Pozrikidis.  Introduction to theoretical and computational fluid dynamics.  Oxford university press, 2011.
  * [3] Z. Zapryanov and S. Tabakova.  Dynamics of bubbles, drops and rigid particles, volume 50.  Springer Science & Business Media, 2013.