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In the context of a school project, I'm trying to redo the derivation of the Korteweg-de Vries equation. At the point where the normal free surface condition is combined with the effect of surface tension, I'm lost.

(Notation: y = function describing the heigth of the water wave at a point x and a time t, p = atmospheric pressure, T= surface tension, p' = pressure just below water surface)

Korteweg and de Vries state in their paper that $p' = p - T\frac{\partial^2 y}{\partial x^2}$. I was surprised by this statement, as I recall from some class that the Young-Laplace equation states that $\Delta p = p'-p = T(\frac{1}{R_x})$, $R_x$ being the radius of curvature in the $x$-direction (I'm doing 1D so there is no other direction). I thought that $1/R_x$ was then given by $\frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}$, but I don't see why that would be equal to $\frac{\partial^2 y}{\partial x^2}$ in this case.

I'd really appreciate it if someone could shine a light on this matter.

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The exact formula for the radius of curvature is $$ \frac{1}{R} = \frac{y''}{(1+y'^2)^{3/2}}. \tag{1}$$

Exactly at the minimum or maximum you have $y'=0$. Near the minimum/maximum you still have a very small $y'$, i.e. $|y'| \ll 1$. Therefore you can use the approximation $1+y'^2 \approx 1$. Then equation (1) from above can be simplified to $$ \frac{1}{R} \approx y'' \tag{2}$$

Note that equation (2) is not an exact, but an approximate equation, only valid where $|y'| \ll 1$.

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  • $\begingroup$ I'm confused as to why we're looking at or in the neighbourhood of the minimum, and not just everywhere on the surface. $\endgroup$
    – Linde
    Oct 13, 2019 at 19:38
  • $\begingroup$ @Linde It is just because it simplifies the math very much, so that it is easier to draw more conclusions from this result. But of course you need be to be aware that any further conclusions are true only near the minimum of the surface. $\endgroup$
    – Thomas Fritsch
    Oct 13, 2019 at 19:45
  • $\begingroup$ But, the derivation of de KdV-equation, which describes the whole surface, seems to be partly built on this approximation? Without it, the free surface condition becomes much more complicated, which leads to a much more complicated equation than de Korteweg-de Vries one $\endgroup$
    – Linde
    Oct 13, 2019 at 19:50
  • $\begingroup$ @Linde I have rewritten my answer to be more exactly. For waves with small enough amplitudes and/or long enough wavelengths the condition $|y'| \ll 1$ may still be valid for the whole surface. $\endgroup$
    – Thomas Fritsch
    Oct 13, 2019 at 20:10
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The formula for the radius of curvature is $$ \kappa = \frac{y''}{(1+y'^2)^{3/2}} = \frac 1 R, $$

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  • $\begingroup$ I know, I accidentally wrote R instead of 1/R. $\endgroup$
    – Linde
    Oct 13, 2019 at 19:18

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