= Solution
{c}
A $C^2$ piecewise cubic supported on $[0,ph]$ has the truncated-power representation with indices $0,\ldots,p$. Vanishing for $x\geq ph$ imposes
$$
\sum_{j=0}^pa_jj^m=0,\qquad m=0,1,2,3,
$$
after expanding its cubic exterior <polynomial>; powers of $h$ cancel. For $p\leq3$, the <Vandermonde matrix> columns are independent, forcing every coefficient to be zero. For $p=4$, coefficients proportional to $(1,-4,6,-4,1)$ satisfy these conditions because a fourth finite difference annihilates cubics.
Normalize to get the <Cardinal cubic B-spline>
$$
\boxed{N_0^3(x)=\frac1{6h^3}\sum_{j=0}^4(-1)^j\binom4j(x-jh)_+^3.}
$$
At its five support knots its values are $0,1/6,2/3,1/6,0$, so their sum is one. Thus \b[four intervals are necessary and sufficient].
For $h=1/2$, the five translates active in the interior of $[0,1]=[0,2h]$ are indexed $-3,-2,-1,0,1$. Write $s(x)=\sum_{j=-3}^1c_jN_j^3(x)$. Nodal <interpolation> gives
$$
\begin{aligned}
c_{-3}+4c_{-2}+c_{-1}&=6u(0),\\
c_{-2}+4c_{-1}+c_0&=6u(h),\\
c_{-1}+4c_0+c_1&=6u(2h).
\end{aligned}
$$
At the left endpoint the first <derivatives> of the three active <functions> are $-1/(2h),0,1/(2h)$, and their second <derivatives> are $1/h^2,-2/h^2,1/h^2$. The two remaining equations are
$$
\boxed{c_{-1}-c_{-3}=2h\,u'(0),\qquad
c_{-3}-2c_{-2}+c_{-1}=h^2u''(0).}
$$
Together these five equations form the requested system; insert $h=1/2$ if numerical coefficients are desired. The first, fourth and fifth determine the first three coefficients, and the other two then determine $c_0,c_1$, so the system is nonsingular.
The source only assumes $u\in C[0,1]$, which need not have either endpoint <derivative>. The Hermite conditions require those <derivatives> to exist, or to be supplied as separate <interpolation> data. The nodal conditions alone remain meaningful for every continuous $u$.
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