"Standard methods of the calculus of variations" or "Do you read German?" I'm trying to understand an article of Reinsch (1967), on smoothing spline functions. The author uses some rules that unfortunately I couldn't find.
The minimized functional is:
$$
\int_{x_0}^{x_n} g''(x)^2 dx + p \left\lbrace \sum_{i=0}^n\left(\frac{g(x_i)-y_i}{\delta y_i}\right)^2 + z^2 -S \right\rbrace
$$
With this, the author deduces that:
$$
\forall i,\ f^{(3)}(x_i)_{-} - f^{(3)}(x_i)_{+} = 2p \frac{f(x_i)-y_i}{\delta y_i}
$$
where $f^{(3)}$ is undefined in each $x_i$, $f^{(3)}(x_i)_{-}$ is the inferior limit in $x_i$, and $f^{(3)}(x_i)_{+}$ the superior limit.
This is the point I don't get. I understand how to compute the derivative of a functional when it is formulated as an integral, but this one isn't. I guess there is some rules I missed.
The author cite a book (Variationsrechnung und ihre Anwendung in Physik und Technik, Funk, 1962), but unfortunately I can't read German, and I couldn't find any source to corroborate Reinsch in his reasoning.
What is the derivation rule I missed? Is there a source (in English or in French) to corroborate the author computation?
Thanks!
 A: Consider the functional
 $$
F(g)=\int_{x_{0}}^{x_{n}}(g^{\prime\prime}(x))^{2}dx+p\sum_{i=0}^{n}\left(
\frac{g(x_{i})-y_{i}}{\delta y_{i}}\right)  ^{2}
 $$
and let $f$ be a minimum over all function $g\in C^{2}([x_{0},x_{n}])$. Then
taking $f+th$, for $t\in\mathbb{R}$ and $h\in C^{2}([x_{0},x_{n}])$, you have
that
 $$
F(f+th)\geq F(f)
 $$
and so the one variable function $k(t)=F(f+th)$ has a minimum at $t=0$. Hence,
$k^{\prime}(0)=0$. So if we now differentiate under the integral sign, we get
\begin{align*}
k^{\prime}(t)  & =\frac{d}{dt}(F(f+ht))=\frac{d}{dt}\int_{x_{0}}^{x_{n}
}(f^{\prime\prime}(x)+th^{\prime\prime}(x))^{2}dx\\&\quad+p\frac{d}{dt}\sum_{i=0}
^{n}\left(  \frac{f(x_{i})+th(x_{i})-y_{i}}{\delta y_{i}}\right)  ^{2}\\
& =\int_{x_{0}}^{x_{n}}2(f^{\prime\prime}(x)+th^{\prime\prime}(x))h^{\prime
\prime}(x)\,dx\\&\quad+2p\sum_{i=0}^{n}\frac{(f(x_{i})+th(x_{i})-y_{i})h(x_{i}
)}{(\delta y_{i})^{2}}.
\end{align*}
Taking $t=0$ gives
 $$
0=k^{\prime}(0)=\int_{x_{0}}^{x_{n}}2f^{\prime\prime}(x)h^{\prime\prime
}(x)\,dx+2p\sum_{i=0}^{n}\frac{(f(x_{i})-y_{i})h(x_{i})}{(\delta y_{i})^{2}}.
 $$
This is true for all $h\in C^{2}([x_{0},x_{n}])$. We now play with $h$. Fix
$i$ and consider functions $h$ which are zero except on $(x_{i-1},x_{i})$.
Then
 $$
0=\int_{x_{i-1}}^{x_{i}}f^{\prime\prime}(x)h^{\prime\prime}(x)\,dx.
 $$
By Weyl's lemma, this implies that $f^{\prime\prime}$ has two derivatives in
$(x_{i-1},x_{i})$ and that $f^{\prime\prime\prime\prime}(x)=0$ in each
interval $(x_{i-1},x_{i})$. Thus, $f^{\prime\prime\prime}$ is constant in each
interval $(x_{i-1},x_{i})$ but can jump at each $x_{i}$. To find the
constants, fix $1<i<n$ and take $h\in C^{4}([x_{0},x_{n}])$ which is zero outside
of $(x_{i}-\delta,x_{i}+\delta)$, where $\delta<\min\{x_{i}-x_{i-1}%
,x_{i+1}-x_{i}\}$. Then
\begin{align*}
0  & =\int_{x_{i}-\delta}^{x_{i}+\delta}f^{\prime\prime}(x)h^{\prime\prime
}(x)\,dx+p\frac{(f(x_{i})-y_{i})h(x_{i})}{(\delta y_{i})^{2}}\\
& =\int_{x_{i}-\delta}^{x_{i}}f^{\prime\prime}(x)h^{\prime\prime}
(x)\,dx+\int_{x_{i}}^{x_{i}+\delta}f^{\prime\prime}(x)h^{\prime\prime
}(x)\,dx+p\frac{(f(x_{i})-y_{i})h(x_{i})}{(\delta y_{i})^{2}}.
\end{align*}
Integrating by parts twice both integrals and using the fact that
$f^{\prime\prime\prime}=0$ in each open interval and that $h$ and its
derivatives up to order 4 are zero at $x_{i}\pm\delta$ we get
\begin{align*}
\int_{x_{i}-\delta}^{x_{i}}&f^{\prime\prime}(x)h^{\prime\prime}(x)\,dx  
=-\int_{x_{i}-\delta}^{x_{i}}f^{\prime\prime\prime}(x)h^{\prime}
(x)\,dx+f^{\prime\prime}(x_{i})h^{\prime}(x_{i})-0\\
& =\int_{x_{i}-\delta}^{x_{i}}f^{\prime\prime\prime\prime}(x)h(x)\,dx+0-f_{-}
^{\prime\prime\prime}(x_{i})h(x_{i})+f^{\prime\prime}(x_{i})h^{\prime}
(x_{i})\\
& =0-f_{-}^{\prime\prime\prime}(x_{i})h(x_{i})+f^{\prime\prime}(x_{i}
)h^{\prime}(x_{i}).
\end{align*}
Similarly,
\begin{align*}
\int_{x_{i}}^{x_{i}+\delta}&f^{\prime\prime}(x)h^{\prime\prime}(x)\,dx  
=-\int_{x_{i}}^{x_{i}+\delta}f^{\prime\prime\prime}(x)h^{\prime}
(x)\,dx+0-f^{\prime\prime}(x_{i})h^{\prime}(x_{i})\\
& =\int_{x_{i}}^{x_{i}+\delta}f^{\prime\prime\prime\prime}(x)h(x)\,dx-0+f_{+}
^{\prime\prime\prime}(x_{i})h(x_{i})-f^{\prime\prime}(x_{i})h^{\prime}
(x_{i})\\
& =0+f_{+}^{\prime\prime\prime}(x_{i})h(x_{i})-f^{\prime\prime}(x_{i}
)h^{\prime}(x_{i}).
\end{align*}
Hence, if we combine the last three equations we get
\begin{align*}
0  & =-f_{-}^{\prime\prime\prime}(x_{i})h(x_{i})+f^{\prime\prime}
(x_{i})h^{\prime}(x_{i})+f_{+}^{\prime\prime\prime}(x_{i})h(x_{i}
)-f^{\prime\prime}(x_{i})h^{\prime}(x_{i})\\&\quad+2p\frac{(f(x_{i})-y_{i})h(x_{i}
)}{(\delta y_{i})^{2}}\\
& =-f_{-}^{\prime\prime\prime}(x_{i})h(x_{i})+f_{+}^{\prime\prime\prime}
(x_{i})h(x_{i})+2p\frac{(f(x_{i})-y_{i})h(x_{i})}{(\delta y_{i})^{2}}.
\end{align*}
Now take $h$ such that $h(x_{i})=1$ and you get
 $$
0=-f_{-}^{\prime\prime\prime}(x_{i})+f_{+}^{\prime\prime\prime}(x_{i}
)+2p\frac{f(x_{i})-y_{i}}{(\delta y_{i})^{2}}.
 $$
For $i=1$ and $i=n$ you do something similar.
