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!
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.