 Solution

In one dimension, $u\in W^{1,2}$ has an absolutely continuous bounded representative. For any smooth zero-endpoint test function $\eta$, the functions $u+t\eta$ are admissible and arbitrarily close to $u$ for small $t$. Local minimality forces the first variation to vanish. Differentiation under the integral is valid because $u,\eta$ are bounded and the derivative terms are integrable, giving the weak Euler-Lagrange equationIn distributional form this says $(2u^{\prime }{)}^{\prime }=4x{u}^3$. The fundamental theorem for weak derivatives, or the fundamental lemma applied after subtracting an antiderivative, gives a constant $C$ withThis is the integrated Euler-Lagrange equation. The boundary difference $u(1)-u(0)=2$ determinesThe weak identity extends to all zero-trace $W^{1,2}$ test functions by density.