How to translate boundary conditions after rewrite pde applying derivatives? In this article (already mentioned in this question) the dynamics of a planar elastic beam with "cantilever constrains" (one clamped end and one free end) is modeled.
Using the Euler-Bernoulli Beam theory the PDE describing the dynamics of the beam should be:
$$ \gamma^2 \frac{\partial^4 y(s,t)}{\partial s^4} + \frac{\partial^2 y(s,t)}{\partial t^2}=0 $$
Where $y(s,t)$ in this formulation is the vertical displacement (expressed in a Cartesian reference) of the material point of the beam identified by the arch-lengh coordinate $s$ at time $t$.
The distributed load along the beam it's assumed to be zero.
The boundary conditions should be expressed as:


*

*At clamped end $s=0$:


$y(0,t)=0$
$\frac{\partial y(0,t)}{\partial s}=0 $


*

*At free end $s=L$:


$\frac{\partial^2 y(L,t)}{\partial s^2}=0 $ That correspond to zero bending moment on the last cross section.
$\frac{\partial^3 y(L,t)}{\partial s^3}=0 $ That correspond to zero shear forces on the last cross section.
In the mentioned article a curvature formulation is used that hold in the hypothesis of small $y_{s}$, in this case one can write $k=y_{ss}$, where $k(s,t)$ identify the curvature of the beam.
In this second formulation the previous PDE becomes:
$$ \gamma^2 \frac{\partial^4 k(s,t)}{\partial s^4} + \frac{\partial^2 k(s,t)}{\partial t^2}=0 $$
And its easier to treat in case of constant curvature initial conditions.
The equivalent boundary conditions used are:


*

*At free end $s=L$:


$\frac{\partial^2 y(L,t)}{\partial s^2}=0 \to k(L,t)0  $ 
$\frac{\partial^3 y(L,t)}{\partial s^3}=0 \to \frac{\partial k(L,t)}{\partial s}=0 $ 
And I'm fine with that.


*

*At clamped end $s=0$:


$y(0,t) \to  \frac{\partial^2 k(0,t)}{\partial^2 s}=0 $ (?)
$\frac{\partial y(0,t)}{\partial s}=0 \to  \frac{\partial^3 k(0,t)}{\partial^3 s}=0 $ (?)
Maybe I'm missing some trivial math step (I came from an engineering background) but these last transformation are not straight forward for me.
Why is it possible to use such conditions?
 A: Edit: this is answering the wrong question due to a miscommunication. 
Ok, I'm sure there are neater, more obvious ways of seeing why this has to be the case, but this works and the correct conditions do just drop out. It also doesn't use anywhere the form of $k$ in terms of $y$, which is kind of interesting and I haven't really thought through the ramifications of that. 
If you start by taking the weak formulation of the equation (that I alluded to in my comments - perhaps you haven't seen it before if you're an engineering student?), which amounts to taking the weighted average with a so-called test function over time and space. This has the form:
\begin{equation}
\int_{0}^{\infty} \int_0^L\gamma\frac{\partial^4 y}{\partial s^4}\phi(s,t) + \frac{\partial^2 y}{\partial t^2}\phi(s,t)\mathrm{d}s\mathrm{d}t = 0,
\end{equation}
and I've made no assumptions about $\phi$ whatsoever. In fact, given we have the strong form of the equation holds, it has to hold for any $\phi$ I could possible dream of. This being true, I can certainly pick that it has to hold for $\phi = k$. 
Now, the following relies heavily on $y$ and $k$ being smooth enough (no discontinuities in the first four derivatives) that the integration by parts formula holds - if the dependent variables can make sudden jumps, this is not necessarily the case. However,  we can appeal to physical intuition here to avoid having to do some fairly in depth analysis to show that this equation's solutions have this smoothness property - as long as the beam doesn't break, this should be fairly self-evident.
I'll break the integral up in to two parts, swap the order of integration on the second part, and integrate by parts until all the derivatives are on the $k$ term - what we end up with looks like
\begin{equation}
\int_0^\infty \gamma \left(\left[k\frac{\partial^3 y}{\partial s^3}\right]_0^L -\left[\frac{\partial k}{\partial s}\frac{\partial^2 y}{\partial s^2}\right]_0^L + \left[\frac{\partial^2 k}{\partial s^2}\frac{\partial y}{\partial s}\right]_0^L - \left[\frac{\partial^3 k}{\partial s^3}y\right]_0^L  + \int_0^L\frac{\partial^4 k}{\partial s^4}y\mathrm{d}s\right)\mathrm{d}t + \int_0^L\left(\left[k\frac{\partial y}{\partial t}\right]_0^\infty - \left[\frac{\partial k}{\partial t}y\right]_0^\infty + \int_0^\infty \frac{\partial^2 k}{\partial t^2}y\mathrm{d}t\right)\mathrm{d}s = 0.
\end{equation}
Now, the relevant parts are the first four terms of the first integrand, relating the boundary conditions of $y$ and $k$. The terms that remain under a double integral sum to zero as $k$ satisfies the same equation $y$ does (ah, using, albeit weakly, a relation between $k$ and $y$), so the boundary terms have to all cancel out for this new equation to hold.
Looking at the spatial boundary conditions, we get
\begin{equation}
\left. k(s,t)\frac{\partial^3 y(s,t)}{\partial s}\right|_{s=L} - \left. k(s,t)\frac{\partial^3 y(s,t)}{\partial s}\right|_{s=0} - \left.\frac{\partial k(s,t)}{\partial s}\frac{\partial^2 y(s,t)}{\partial s^2}\right|_{s=L} + \left.\frac{\partial k(s,t)}{\partial s}\frac{\partial^2 y(s,t)}{\partial s^2}\right|_{s=0} + \left.\frac{\partial^2 k(s,t)}{\partial s^2}\frac{\partial y(s,t)}{\partial s}\right|_{s=L} - \left.\frac{\partial^2 k(s,t)}{\partial s^2}\frac{\partial y(s,t)}{\partial s}\right|_{s=0} - \left.\frac{\partial^3 k(s,t)}{\partial s^3}y(s,t)\right|_{s=L} + \left.\frac{\partial^3 k(s,t)}{\partial s^3}y(s,t)\right|_{s=0} = 0.
\end{equation}
If we plug in the boundary conditions on $y$ that were given, the remaining terms summing to zero imposes exactly the boundary conditions on $k$ that you were after. Bit of a trek to get there, though, so there is perhaps a quicker way using $k = y_{ss}$ directly. 
