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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?

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  • $\begingroup$ Something weird is going on for sure. When you say $k = y_{ss}$, do you mean that $k(s,t) = \frac{\partial^2 y(s,t)}{\partial s^2}$? In that case, I don't see how you arrive at the changed PDE you give, which is in fact exactly the same as the first one but with $k$ instead of $y$. $\endgroup$ Apr 3 '15 at 12:10
  • $\begingroup$ Ah no ignore me, I'm being silly. I see now the equation has to be the same but we can exploit the change of variables to better deal with the bc's. Now I understand what we're doing, I suspect what's happening is some kind of adjoint consideration - because the conditions at the clamped end apply to lower derivatives of $y$ than $k$ is we have to integrate $k$ up in some way to apply them. But integral boundary conditions are messy so if we take some kind of weighted average we can move the derivatives across (formally) by parts. Give me a while and I'll try and work out the details. $\endgroup$ Apr 3 '15 at 12:20
  • $\begingroup$ yes I meant that, yes the equations are equal but $y$ and $k$ have different physical meaning. I guessed that the second came from double derivation in respect of s of the first, but actually I'm not totally sure it doesn't come from a different formulation. $\endgroup$
    – tmms
    Apr 3 '15 at 12:24
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    $\begingroup$ I think if we multiply the equation by an arbitrary smooth function $\phi(s)$ that is zero at each boundary, integrate over the domain then do some integration by parts the correct boundary conditions should drop out. I'm trying to do it at the moment, but it'll take a while for me to think through properly. $\endgroup$ Apr 3 '15 at 12:27
  • $\begingroup$ Actually, we might need to put some non-trivial boundary conditions on $\phi$ to begin with - not sure yet. $\endgroup$ Apr 3 '15 at 12:28
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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.

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  • $\begingroup$ I thing I'm loosing something but if I put my boundary condition on $y$ in the last of your equation i get on $k $ these contions: $k_{ss}(L,t)=0,k_{sss}(L,t)=0,k(0,t)=0,k_{s}(0,t)=0$ that are kind of mirrored respect of my conditions on $k$ $k(L,t)=0\,k_{s}(L,t)=0,k_{ss}(0,t)=0,k_{sss}(0,t)=0$ $\endgroup$
    – tmms
    Apr 3 '15 at 15:05
  • $\begingroup$ Or it's that the last must be satisfy for every $k$ also at the boundary and so I should evince that my condition on $y$ are equivalent to my condition $k$, because using one or the other give me the same remaining terms? In this way I get it. Sorry too much steel and oil lately. $\endgroup$
    – tmms
    Apr 3 '15 at 15:22
  • $\begingroup$ Not sure exactly what you're trying to say but the 'arbitrary' choice of $k$ does indeed mean that each term in the sum of the last equation must be zero (in this case, as all the boundary conditions we're given fix that quantity to zero). For the ones on which the boundary conditions on $y$ or its derivatives are prescribed, this takes care of itself, but for the ones for which no aspect of $y$ is fixed, e.g. $y(L,t)$ we must therefore have that the term in $k$ is $0$, here $\frac{\partial^3 k(L,t)}{\partial s^3} = 0$. Is that clear? $\endgroup$ Apr 3 '15 at 16:53
  • $\begingroup$ yes that is clear, so to satisfy the last of your equations, $k$ must verify these condition: $k_{ss}(L,t)=0,k_{sss}(L,t)=0,k(0,t)=0,k_{s}(0,t)=0$, right? These are different from the conditions I expect when I'm solving for $k$ and not using it as test function. $\endgroup$
    – tmms
    Apr 3 '15 at 17:40
  • $\begingroup$ Ah, I see where the confusion is coming from. I was going off what you wrote after 'at the clamped end...' of your original question, where you say that the boundary conditions on $y$ at $s=0$ imply something about the boundary conditions on $k$ at $s=L$. Can you please check exactly what boundary conditions you were expecting $k$ to have? $\endgroup$ Apr 3 '15 at 17:48

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