Maximum of $y''$ for BVP with $y''''\leq0$. Consider the following boundary value problem for some $L>0$ and $w(x)\geq 0$:
$$\frac{d^4y}{dx^4}=-w(x)\,;\,\,\,y(0)=y(L)=0,\,y'(0)=y'(L)=0.$$
Here $w$ can be pathological: a step-function or an impulse function $w(x)=w_0\,\delta(x-a)$: anything that ensures that $y''$ is continuous (but not necessarily differentiable). From context (below) you can understand the kind of things that are allowed here.

Is it always the case that the maximum of $|y''|$ is at $x=0$ or $x=L$?

The fact that the second derivative of $y''$ is negative ensures that the maximum of $|y''|$ occurs at $x=0$, $x=L$, or where $\dfrac{d}{dx}(y''(x))=0$.
In context this is related to the maximum bending moment of fixed end beam. To that end I find with a point load at the midpoint, $w(x)=w_0\,\delta(x-L/2)$ we have:
$$|y''(0)|=|y''(L/2)|=|y''(L)|=\frac{w_0L}{8}.$$
But can we have $|y''(x_m)|>|y''(0)|,|y''(L)|$ for $0<x_m<L$.
I feel like I could ask this of an engineer (and have done so here), but would be interested in a more mathematical answer.
 A: Let $M:=y''$. Consider:
$$\int_0^LM(x)\,dx=\int_0^L\frac{d}{dx}(y'(x))\,dx=[y'(x)]_0^L=y'(L)-y'(0)=0-0=0.$$
Consider the graph of $M$ vs $x$ and take $x=0$, $x=L$ and $x=a$ such that $\max_{x\in[0,L]} M(x)$ occurs at $x=a$. Where $M_0:=M(0),\, M_a:=M(a),\,M_1:=M(L)$, assume that:
$$|M(a)|>|M_0|\text{ , and }|M(a)|>|M_1|.$$
It is the case that $M_0,M_1\leq 0$ so $M_a\geq 0$
Consider three points on the graph:
$$(0,M_0),\,(a,M_a),\,(L,M_1),$$
and draw a triangle with these vertices.
The sides are:
$$\ell_1(x)=\frac{M_a-M_0}{a}x+M_0\text{ for }0\leq x\leq a \text{, and}$$
$$\ell_2(x)=-\frac{M_a-M_1}{L-a}(x-L)+M_1\text{ for }a\leq x\leq L.$$
By concavity, on $[0,a]$, and decomposing $M=M_+-M_-$, we can see that (*this is a little iffy in terms of technicalities):
$$\int_0^aM(x)\,dx\geq \int_0^a\ell_1(x)\,dx\text{ and }\int_a^LM(x)\,dx\geq \int_a^L\ell_2(x)\,dx.$$
This implies that, because $\int_0^LM(x)\,dx=0$:
$$\begin{aligned}
\int_0^a\ell_1(x)\,dx+\int_a^L\ell_2(x)\,dx&\leq 0
\\\implies 2\left(\int_0^a\ell_1(x)\,dx+\int_a^L\ell_2(x)\,dx\right)&\leq 0,
\\ \implies M_0a+MaL+M_1(L-a)&\leq 0
\\\implies M_a&\leq \frac{a}{L}\cdot (-M_0)+\frac{L-a}{L}\cdot (-M_1),
\end{aligned}
$$
that is $M_a$ is less than a weighted average of $-M_0$ and $-M_1$, and therefore less than one (or both) of them.
