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)