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Swimming at low Reynolds number

Swimming at low Reynolds number

\[\newcommand{\p}{\partial} \newcommand{\eps}{\epsilon} \newcommand{\half}{\frac{1}{2}}\]

As we have argued, the fluid flow as \(Re \to 0\) is fixd completely by the boundary conditions at that time. The only way for an object to move forward at low Reynolds number through some kind of repetitive motion is if it executes a protocol which takes a loop in the space of possible shapes. So, a single paddle will not do; a flexible paddle might, or two independent hinges.

Oher option are available – fo example, a rotating corkscrew. We’ll look at a simpler one, which is a wave traveling along a flexible membrane.

Waving of a thin flexible sheet

We consider 2d flow again, with a membrane extended along teh \(x\) direction which is waving in the \(y\) direction. We ignore, for now, the finite size of his object. We wish to show tha this motion induces both oscillatory flow of the fluid and an additional steady flow along the \(x\) axis. We can then boost into te rest frame of the steady flow, so that the membrane itself is swimming in the \(-x\) direction.

We take the membrane shape to be:

\[ y = a \sin(kx - \omega t) \]

In particular we take the infintesimal pieces of the sheet to reside at fixed \(x\), and move up and down.

The exact boundary conditions will clearly be difficult to impose. We will take the limit tha the oscillations are small compared to the wavelength:

\[ \eps = ak \ll 1 \]

(actually the wavelength is \(\lambda = 2\pi/k\), but \(\epsilon\) is the parameter which will appear).

Once again, because this is 2d incompressible flow, we can describe the velociy in terms of a streamfunction:

\[\begin{split}& u_x = \p_y \psi\\ & u_y - - \p_x \psi\end{split}\]

Taking the curl of the slow/ceeping flow equation we find

\[ (\nabla^2)^2 \psi = 0 \]

Because the individual elements of the membrane are oscillating vertically, and the no-slip condition means that the fluid velocity at the membrane equals the membrane velocity

\[ {\vec v}_{membrane} = {\hat y} \frac{d}{dt} a \sin(k x - \omega t) = - a\omega \cos(kx - \omega t)\ , \]

we have

\[\begin{split}& v_x = \p_y \psi(x,y = a\sin(kx - \omega t)) = 0\\ &v_y = - \p_x \psi(x,y = a\sin(kx - \omega t)) = - a \omega \cos(kx - \omega t)\end{split}\]

These are obviously noninear boundary conditions. They are also time-dependent. However, \(x,t\) only appear explicitly in the equations in the form \(kx - \omega t\). If we solve the equations at \(t = 0\), we can find solutions at other times \(t\) by replacing \(x \to x - \omega t/k\). So we will do that.

At this point, we are going to nondimensionalize the problem. This will make the formulae simpler (fewer parameters floating around, and will isolate the dimensionless parameters which control the solution. We rescale

\[\begin{split}& k x = X\\ &k y = Y\\ &\frac{k}{\omega a} \psi = \Psi\end{split}\]

All of the capitalized quantities are now dimensionless. The form of the bulk equations for the steamfunction remain unchanged:

\[ \left(\p_X^2 + \p_Y^2\right)^2 \Psi = 0 \]

The boundary conditions become

\[\begin{split}& \p_Y \Psi(X,Y = \eps \sin X) = 0\\ & \p_X \Psi(X, Y = \eps \sin X) = \cos X\end{split}\]

We are going to try and solve these equations in a power series in \(\eps\). Looking at the structure of these equations, we see \(\Psi\) should start at \({\cal O}(1)\) in order that the second equation be solvable, so:

\[ \Psi = \Psi_0 + \eps \Psi_1 + \eps^2 \Psi_2 + \ldots \]

(unlike the book, I prefer to match the index of \(\Psi\) with the order of \(\eps\)). We can expand the boundary conditions in \(\eps\):

\[\begin{split}& \p_Y \Psi(X,0) + \eps \sin X \p_Y^2 \Psi(X,0) + \ldots = 0\\ & \p_X \Psi(X,0) + \eps \sin X \p_Y \p_X \Psi(X,0) + \ldots = \cos X\end{split}\]

Inserting the above expansion we have:

(61)\[\begin{split}& \p_Y \Psi_0(X,0) = 0\\ & \p_X \Psi_0(X,0) = \cos X\\ & \p_Y \Psi_1 + \sin X \p_Y^2 \Psi_0 = 0\\ & \p_X \Psi_1 + \sin X \p_Y \p_X \Psi_0 = 0\end{split}\]

At lowest order we are working on the UHP and the boundary conditions depend only on \(X\); thus we will try a separation-of-variables ansätz \(\Psi_0 = F(X) G(Y)\). The boundary conditions already guarantee that \(F(X) = \sin X\). The bulk equation for \(G\) is then

\[ \left(\p_Y^2 - 1\right)^2 G = 0 \]

Setting \(\Phi = (\p_Y^2 - 1) G\), we find

\[ \Phi = A' e^Y + B' e^{-Y} \]

is the most general solution. Now in solving

\[ \left(\p_Y^2 - 1\right)^2 G = \Phi = A' e^Y + B' e^{-Y} \]

a general solution takes the form

\[ G = C e^Y + D e^{-Y} + {\tilde G} \]

where \({\tilde \Psi}\) is a particular solution to:

\[ \left(\p_Y^2 - 1\right) {\tilde G} = A' e^Y + B' e^{-Y} \]

which can be:

\[ {\tilde G} = \half Y (A' e^Y + B' e^{-Y}) \]

Now for \(Y > 0\) we need \({\vec v} \to 0\) as \(Y \to \infty\) so we must have \(A' = C = 0\). Now we must impose the boundary conditions at \(Y =0\). Since \(F(X) = \sin x\), \(u_y = - \p_X \Psi_0 = \cos X\) is satisfied at \(y = 0\) if \(D = 1\). Finally \(\p_Y G = 0\), which means \(\half B' - 1 = 0\), and the solution for \(y > 0\) becomes

\[ G = (1 + Y) e^{-Y} \]

while for \(Y < 0\), we must have \(B' = D = 0\), and after the same arguments:

\[ G = (1 - Y) e^Y \]

Thus

\[\begin{split}\Psi & = (1 + Y) e^{-Y} \sin X\ \ {\rm for}\ Y > 0\\ & = (1 - Y) e^Y \sin X \ \ {\rm for}\ Y < 0\end{split}\]

Now we can sove for \(\Psi_1\) in (61). This still satisfies the equation \((\p_X^2 + \p_Y^2)^2 \Psi_1 = 0\). Le us start with \(Y \to 0^+\), for which:

\[\begin{split}& \p_Y \Psi_1 - \sin^2 X = 0\\ & \p_X \Psi_1 = 0\end{split}\]

We further write \(\sin^2 X = \half(1 - \cos 2 X)\). Now \(\Psi_1\) is the sum of two terms, one for which \(\p_Y \psi_{1a} = \half\) at \(Y = 0\), and one for which \(\p_Y \Psi_{1b} + \half \cos 2 X = 0\). The latter piece will again be solved as before with

\[ \Psi_{1b} = (E + F y) e^{-2Y} \cos 2X \]

To solve for \(\Psi_{1a}\) he boundary conditions suggest we consider constant \(x\) dependence in our separation-of-variables ans&umla;tz so that \(\p_Y^4 \Psi_{1a} = 0\). This is solved by the cubic

\[ \Psi_{1a} = H_0 + H_1 Y + H_2 Y^2 + H_3 Y^3 \]

Now we need to combine these. Using \(\p_X \Psi_1 = 0\) we find \(E = 0\). \(H_0\) has no physical meaning as only derivatives of \(\Psi\) are physical, and we can set it to zero. \(H_{2,3}\) will give components of he veocity scaling as
\(U_{X} = \p_Y \Psi = \sim 2 H_2 Y + 3 H_3 Y^2 \to \pm \infty\) as \(y \to \infty\) so we must set \(H_{2,3} = 0\). Finally, we want \(\p_Y \Psi_1 = \sin^2 X = \half(1 - \cos 2X)\) at \(Y = 0\), so we must have \(H_1 = \half\), \(F = - \half\), or

\[ \Psi_1 = \half \left(Y - Y e^{-2Y}\cos 2 X\right) \]

So that

\[ \Psi = (1 + Y) e^{-Y} \sin X + \eps \half \left(Y - Y e^{-2Y}\cos 2 X\right) \]

Redimensionalizing and shifing \(kx \to kx - \omega t\),

\[ \psi = \frac{\omega a}{k} \left[ \left(1 + ky\right) e^{-ky}\sin(kx - \omega t) + \half \eps ky \left(1 - e^{-2ky} \cos (2kx - 2\omega t)\right)\right] + \ldots \]

we can compute the velocity:

\[\begin{split}& u_x = \p_y \psi = - \eps \omega y e^{-k y} \sin(kx - \omega t) + \eps^2 c \left[\half + (ky - \half)e^{-2ky} \cos(2(kx - \omega t))\right] + \ldots\\ & u_y = - \p_x \psi = \eps c \left(1 + k y\right) e^{-ky} \cos(kx - \omega t) + \eps^2 c ky e^{-2k y} \sin(2kx - 2 \omega t) + \ldots\end{split}\]

Here \(c = \omega/k\); note further that the rescaling \(\Psi \to \psi\) means ta even though it appears that \(\Psi\) was nonvanishing as \(\eps \to 0\), \({\vec u}\) is still vanishing in this limit.

The upshot is that in addition to an oscillatory flow, we have a steady component

\[ {\vec u}_{steady} = \half \eps^2 c {\hat x} = \frac{k a^2 \omega}{2} {\hat x} \]

If we boost into the rest frame of the steady flow, each infinitesimal piece of the membrane will be moving as:

\[\begin{split}& x = x_0 - \frac{k a^2 \omega t}{2}\\ & y = a \sin\left[kx_0 - \left(\half k^2 a^2 + 1\right)\omega t\right]\end{split}\]

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