 Solution

The trial space must be a subspace of $H_0^1(0,1)$, not merely of $L^2(0,1)$. The printed assumptions on the basis do not ensure square-integrable weak derivatives, and boundary traces are not defined for arbitrary L2 space functions. For example, $\phi (x)=x^{1/4}(1-x)$ is continuous, vanishes at both endpoints and belongs to $L^2$, but $|{\phi }^{\prime }(x){|}^2\sim x^{-3/2}/16$ is not integrable at zero. Assume the basis belongs to the zero-boundary Sobolev space, as it does for the hat functions in part (c).

Write $u_H={\sum }_{j=1}^M{c}_j{\phi }_j$. Differentiating the restricted energy with respect to $c_i$, or imposing the Galerkin method with test function ${\phi }_i$, givesThe derivative term is the usual diffusion stiffness matrix and the second is the mass matrix. The full matrix is real symmetric, and for every nonzero coefficient vector $c$,because a basis is linearly independent. Thus it is a positive-definite matrix, so the restricted Ritz method has a unique solution and minimum.