Converting scalar ODE to coupled system I'm currently battling the following problem:
\begin{align}
u^{(iv)} (x) &= f(x)\quad\text{on }(0,1)\\
u(0) = u'(0) &= 0\\
u''(1) = u'''(1) &= 0
\end{align}
which is, as I've understood, a beam equation for a cantilevered beam.
I am given the hint to define a new variable $p:=u''$ and to reformulate
the first equation  as a coupled system of second order ODEs.
Now the first idea was to just use the substitution, thus arriving at:
\begin{align*}
p(x) &= u''(x) \\
p'' (x) &= f(x)\\
u(0) = u'(0) &= 0\\
p(1) = p'(1) &= 0
\end{align*}
where in the first two equations $x\in(0,1)$ is understood. This however, is not what we're looking for.
Furthermore using the fundamental theorem of calculus I can gain
\begin{align*}
u'(x) &= u'(0) + \int_0^x u''(s)ds = \int_0^x p(s)ds\\
u(x) &= u(0) + \int_0^x u'(s)ds = \int_0^x u'(s)ds\\
&= \int_0^x \int_0^s p(y)dy ds\quad .
\end{align*}
I would like to ask for a hint (and hopefully not one applied with a sledgehammer)
where I'm going in the wrong direction.
N.B.
The new variable looks awfully like a Legendre transform, but there I again don't see where I
should go.
 A: Since $p(1)=p'(1)=0$, one can use the trick you applied to $u$ on $(0,x)$, but on $(y,1)$ for $p$. This yields $$p(y)=\int_y^1\int_z^1f(t)\mathrm dt\mathrm dz,$$ hence $$u(x)=\int_0^x\mathrm ds\int_0^s\mathrm dy\int_y^1\mathrm dz\int_z^1f(t)\mathrm dt,$$ and the task is to transform this into a single integral. An approach which is less error-prone than some others, is to use indicator functions, that is, to rewrite $u(x)$ as $$u(x)=\int_{[0,1]^4}f(t)\mathbf 1_{y\lt s\lt x}\mathbf 1_{y\lt z\lt t}\mathrm ds\mathrm dy\mathrm dz\mathrm dt=\int_0^1f(t)g_x(t)\mathrm dt,$$ where, for every $t$ and $x$ in $[0,1]$, $$g_x(t)=\int_{[0,1]^3}\mathbf 1_{y\lt s\lt x}\mathbf 1_{y\lt z\lt t}\mathrm ds\mathrm dy\mathrm dz=\int_0^x\int_0^t\min(s,z)\mathrm dz\mathrm ds.$$ 
For every $(t,s)$ such that $u=\min(s,t)$, $$\int_0^t\min(s,z)\mathrm dz=\int_0^uz\mathrm dz+\int_u^ts\mathrm dz=\tfrac12u^2+s(t-u),$$ hence $$g_x(t)=\int_0^x\left(\tfrac12\min(s,t)^2+s(t-\min(s,t))\right)\mathrm ds,$$ that is, considering $v=\min(x,t)$, $$g_x(t)=\int_0^v\left(\tfrac12s^2+s(t-s)\right)\mathrm ds+\int_v^x\left(\tfrac12t^2+s(t-t)\right)\mathrm ds=\int_0^v\left(st-\tfrac12s^2\right)\mathrm ds+\tfrac12t^2\int_v^x\mathrm ds,$$
or, equivalently, $$g_x(t)=\tfrac12tv^2-\tfrac16v^3+\tfrac12t^2(x-v)=\tfrac12txv-\tfrac16v^3.$$ Finally, for every $x$ in $[0,1]$,

$$u(x)=\tfrac12x\int_0^xt^2f(t)\mathrm dt-\tfrac16\int_0^xt^3f(t)\mathrm dt+\tfrac12x^2\int_x^1tf(t)\mathrm dt-\tfrac16x^3\int_x^1f(t)\mathrm dt.$$

