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.
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 $[0,1]$,