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A one-dimensional Cubic Hermite finite element represents a cubic polynomial on $[a,b]$ by its value and first derivative at each endpoint. Put $h=b-a>0$ and $\xi =(x-a)/h$. Its four reference shape functions areThe interpolating polynomial isThus $p(a)=y_a$, $p^{\prime }(a)=s_a$, $p(b)=y_b$ and $p^{\prime }(b)=s_b$. The factors $h$ convert reference-coordinate derivatives to physical derivatives. These conditions determine the cubic uniquely: a difference with zero endpoint values and slopes would have two double roots, impossible for a nonzero cubic.

Sharing both value and slope between adjacent intervals produces globally $C^1$ piecewise cubics in Sobolev space $H^2$. With the required endpoint constraints, they form a conforming finite element space for fourth-order variational problems, such as a bending problem whose energy contains $\int |u^{\prime \prime }{|}^2$. Ordinary continuous piecewise-linear elements have slope jumps and do not have this $H^2$ conformity.

Choose a mesh of $[-1,1]$ and the conforming space of Cubic Hermite finite elements: piecewise cubic functions that are globally $C^1$ and satisfy $v(\pm 1)=v^{\prime }(\pm 1)=0$. Let ${\varphi }_1,\dots ,{\varphi }_n$ be its nodal value-and-slope basis. For $u_n={\sum }_k{a}_k{\varphi }_k$, the Ritz method imposesHence the coefficient vector solvesThe stiffness matrix is symmetric positive definite by part a.