## Problem 5.4.8 in Keener

The functional that we wish to minimize is $J[u]=\int_0^1 u''(x)^2dx$,
over all functions in $C^{(1)}[0,1]$ that have a piecewise continuous
second derivative and that have $u(x_j)$, where $ 0 = x_0 < x_1 <
\cdots < x_{n-1} < x_n=1 $, and $u'(0)$, $u'(1)$ given. Consider the
space $V$ of all admissible functions that vanish at the $x_j$'s and
have $0$ derivative at $x=0$ and $x=1$. The Frechet derivative of $J$
at $u$ is then
\[
\Delta_u J(z) =2\int_0^1 u''z''dx.
\]
If $u$ is an extremal for $J$, then $\int_0^1 u''z''dx =0$ for all
$z\in V$. Choose $z\in V$ such that $z$ is supported on the interval
$[x_j,x_{j+1}]$. Because $z\in V$, we have $z(x_j)=z(x_{j+1})=0$, In
addition, the continuity of $z'$ implies that
$z'(z_j)=z'(x_{j+1})=0$. We will now show that we can choose constants
$A,B,C,D$ so that the function $z$ that defined by
\[
z:= \left\{\begin{array}{cl} u - A - Bx - Cx^2 - Dx^3& x\in
[x_j,x_{j+1}] \\ 0& x \not\in
[x_j,x_{j+1}]\end{array}\right.
\]
is in $V$. We need to show that we can satisfy $z(x_j)=0$,
$z(x_{j+]})=0$, $z'(x_j)=0$, and $z'(x_{j+1})=0$. Applying these
conditions results in the system
\[
\begin{pmatrix}
1&0&0&0 \\
1&1&1&1 \\
0&1&0&0 \\
0&1&2&3
\end{pmatrix}
\begin{pmatrix}
A \\
B \\
C \\
D
\end{pmatrix}
=
\begin{pmatrix}
u(x_j)\\
u(x_{j+1})\\
u'(x_j)\\
u'(x_{j+1})
\end{pmatrix}.
\]
This system can be solved and so, with the constants from the
solution, we have that $z\in V$. Because $u$ is an extremal and $z$ is
supported on $[x_j,x_{j+1}]$, we see that $\int_{x_j}^{x_{j+1}}
u''z''dx =0$. Also, using integration by parts yields
\[ \int_{x_j}^{x_{j+1}} z''dx = 0 \quad \text{and}
\int_{x_j}^{x_{j+1}} x\,z''dx = 0, \]
and thus
\[
\int_{x_j}^{x_{j+1}} (\underbrace{u''- 2C -Dx}_{z''}) z''dx=
\int_{x_j}^{x_{j+1}}(z'')^2dx =0.
\]
It follows that $z''=0$ on $[x_j,x_{j+1}]$, and so $u''=2C+Dx$. From
this, we see that $u$ is a cubic on the interval $[x_j,x_{j+1}]$. The
rest of the problem is easily solved.
Updated 2/18/2014 (fjn)