Problem setting:
Let $\Omega = [-1,1] \subset \mathbb{R}$ be an interval and consider the space of infinitely differentiable functions, that is $C^{\infty}$.
We successively define the sequence $f_k \in C^{\infty}, \ k \in \mathbb{N}$, by \begin{align} \|f_k''\|_{L_2(\Omega)} & = \inf_{g \ \in \ C^{\infty}}\|g''\|_{L_2(\Omega)} \\ & \mbox{subject to:} \ g \perp \ \mathrm{span}(f_1,\ldots,f_{k-1}), \ \|g\|_{L_2} = 1 \end{align} for the standard $L_2$ scalar product, such that \begin{align} \int_{\Omega} f_i \ f_j = \delta_{ij}. \end{align} Question: Is there a numerically feasible description of these functions, as well as their first and second derivatives? And what is the space spanned by these functions?

Comments on the problem: Quite obviously, $f_1$ and $f_2$ form the set of affine functions and I guess all further functions are unique.

By numerically feasible description I mean sorts of functions which are (numerically) evaluable with reasonable computational complexity at any one point (e.g. polynomials, trigonometrical functions, recursively defined functions, ...). Any further theoretical insight on these functions is helpful nonetheless.

If one replaces $\|\cdot''\|$ by $\|\cdot'\|$, then the $f_k$ are seemingly (ignoring their order and modulo scaling) given by the system \begin{align} 1, \quad\sin((2i + 1)\pi/2 \ x), \quad \cos((2i + 2)\pi/2 \ x), \quad i \in \mathbb{N}. \end{align} All these functions have derivative $0$ at the interval border.

For the original problem, by an SVD of a discretized version of the problem, one can get a notion of how these functions look like (the plots are pretty exact). Here are functions $f_3$ to $f_7$: f_3 to f_7 and $f_3$ to $f_{15}$: f_3 to f_15, from which one may observe several properties. For example, the second derivative at the border seems to be $0$.

Regarding the space $C^{\infty}$, I think it is unlikely, yet I am not sure, that the original phrasing of the problem is well-defined. In case of need, one may replace it by $C^2$ or even $H^2$ (the space of twice weakly differentiable functions). Note that the latter one is embedded into $C^1$.


2 Answers 2


The optimization problem for $f_3$ is $$ \min_g \int_{-1}^1 |g''(x)|^2dx \quad \text{s.t.}\quad \int_{-1}^1 g(x)dx = 0,\ \int_{-1}^1 x g(x)dx = 0,\ \int_{-1}^1 |g(x)|^2dx = 1. $$ Using Lagrange Multipliers to solve this optimization problem you form the Lagrangian $$ \mathcal{L}(g,\lambda_1,\lambda_2,\mu) = \int_{-1}^1 |g''(x)|^2dx + \lambda_1 \int_{-1}^1 g(x)dx + \lambda_2\int_{-1}^1 xg(x)dx + \mu\left(\int_{-1}^1|g(x)|^2dx-1\right). $$ Now derive $\mathcal{L}$ with respect to all variables and set them to zero to get the necessary (and, due to convexity of the problem, also sufficient) conditions for optimality.

The derivatives for $\lambda_1,\lambda_2$ and $\mu$ just give the constraints and for the derivative for the "function variable" just do a formal calculation to calculate the directional derivative $$ \lim_{t\to 0} \frac{\mathcal{L}(g+th,\dots) - \mathcal{L}(g,\dots)}{t}. $$ Using integration by parts this shows that $f_3$ has to solve the boundary value problem $$ g^{(4)}(x) = -\lambda_1 - \lambda_2 x - 2\mu g(x),\quad g^{(2)}(\pm 1) = 0,\quad g^{(3)}(\pm 1) = 0 $$ for some Lagrange multiplies $\lambda_{1/2},\mu$ and also fulfill the additional constraints. There is a closed form solution for this problem, but I did not push the calculation through…

Proceeding similarly gives $f_k$ as solution of $$ g^{(4)}(x) = -\lambda_1 - \lambda_2 x - \lambda_3 f_3(x) - \cdots - \lambda_{k-1}f_{k-1}(x) - 2\mu g(x),\quad g^{(2)}(\pm 1) = 0,\quad g^{(3)}(\pm 1) = 0 $$ for some Lagrange multiplies $\lambda_{l},\mu$, $l=1,\dots,k-1$. While it seem not too hard to write down $f_3$ explicitly, $f_4$ seems to be quite a mess.

  • $\begingroup$ Thank you, I assume $g^{(i)}$ is the $i$-th derivative of $g$. Unfortunately, I am not at all familiar with Lagrange Multipliers used on function spaces, so I will have to see if I can derive these conditions myself (in particular, how to turn it into a differential equation). Anyhow, I wonder about the following: The ODE is of degree four, hence gives $3$ parameters + $3$ Lagrange multipliers. Yet I only have $4$ boundary conditions. It seems like I am missing something here. $\endgroup$ Jan 4, 2018 at 16:18
  • $\begingroup$ Yes, this is the notation for the $i$-the derivative. I extended the answer with more details on Lagrange multipliers. Besides the boundary conditions there are also constraints to fulfill, so the problem is rather overdetermined, but I guess there should be a solution. $\endgroup$
    – Dirk
    Jan 5, 2018 at 8:35
  • 1
    $\begingroup$ I finally had time to write down the proof. Thank you again for your answer. $\endgroup$ Jan 19, 2018 at 13:33

Here is the proof$^\ast$ (and answer to my question) which I promised.

The functions $f_1$ and $f_2$ form the set of affine functions and we can define \begin{align} f_1(x) = \frac{1}{\sqrt{2}}, \quad f_2(x) = \sqrt{\frac{3}{2}} x. \end{align} All subsequent functions are given (in correct order and normalized) by \begin{align} \sqrt{1+\frac{\cos(\nu_{1+2i})^2}{\cosh(\nu_{1+2i})^2}} f_{1+2i} & = \cos(\nu_{1+2i}x)+\frac{\cos(\nu_{1+2i})}{\cosh(\nu_{1+2i})} \cosh(\nu_{1+2i}x) \\ \sqrt{1-\frac{\sin(\nu_{2+2i})^2}{\sinh(\nu_{2+2i})^2}} f_{2+2i} & = \sin(\nu_{2+2i}x)+\frac{\sin(\nu_{2+2i})}{\sinh(\nu_{2+2i})} \sinh(\nu_{2+2i}x), \quad i \in \mathbb{N} \end{align} where $\nu_{1+2i}$ is the $i$-th (positive) root of $\cos(\nu)\sinh(\nu)+\cosh(\nu)\sin(\nu)$ and $\nu_{2+2i}$ is the $i$-th (positive) root of $\cos(\nu)\sinh(\nu)-\cosh(\nu)\sin(\nu)$. Further more, we have \begin{align} \int_{-1}^1 f_i'' f_j'' = \delta_{ij} \nu_i^4. \end{align}

$^\ast$: At one point we assume $f_k$ to be sufficiently smooth. I guess this follows easily within a standard argument using calculus of variations, but every derivation I found so far, just makes the same assumption.


Using Lagrange Multipliers (this part is thanks to Dirk) in order to obtain $f_k$, $k > 3$, we form the Lagrangian \begin{align} \mathcal{L}^{(k)}(g,\lambda^{(k)},\mu^{(k)}) & = \int_{-1}^1 |g''|^2 + \sum_{i = 1}^{k-1} \lambda_i^{(k)} \int_{-1}^1 g f_i + \mu^{(k)} \left( \int_{-1}^1 |g|^2 - 1 \right). \end{align} The derivatives with respect to $\lambda_1,\ldots,\lambda_{k-1}$ give the orthogonality conditions, whereas $\mu$ gives the normalization. Variation of the function $g$ provides \begin{align} 2 \int_{-1}^1 g'' h'' + \sum_{i = 1}^{k-1} \lambda_i^{(k)} \int_{-1}^1 f_i h + 2 \mu^{(k)} \int_{-1}^1 g h = 0. \end{align} Using partial integration twice (at this points we actually assume $f$ to be sufficiently smooth) we get \begin{align} & 2 \int_{-1}^1 g^{(4)} h + 2 g^{(2)} h|_{-1}^1 - 2 g^{(3)} h|_{-1}^1 \\ & + \sum_{i = 1}^{k-1} \lambda_i^{(k)} \int_{-1}^1 f_i h + 2 \mu^{(k)} \int_{-1}^1 g h = 0. \end{align} Since $h \in C^\infty$ is arbitrary, we get the pointwise identity \begin{align} \mathcal{P}^{(k)}(g) := 2g^{(4)} + \sum_{i = 1}^{k-1} \lambda_i^{(k)}f_i + 2 \mu^{(k)} g = 0 \end{align} as well as the boundary conditions \begin{align} g^{(2)}(\pm 1) = 0, \quad g^{(3)}(\pm 1) = 0. \end{align} We inductively proof now that $\lambda^{(k)} \equiv 0$, $k \geq 3$. For $j < k$, by orthonormality conditions, we can simplify \begin{align} 0 = \int_{-1}^1 \mathcal{P}^{(k)}(g) f_j & = 2 \int_{-1}^1 g^{(4)} f_j + \sum_{i = 1}^{k-1} \lambda_i^{(k)} \int_{-1}^1 f_i f_j + 2 \int_{-1}^1 \mu^{(k)} g f_j \\ & = 2 \int_{-1}^1 g^{(4)} f_j + \lambda_j^{(k)} \end{align} Using the boundary conditions for both $f$ and $g$, this is equivalent to \begin{align} 0 = \int_{-1}^1 g f_j^{(4)} + \lambda_j^{(k)} \end{align} In the case of $k = 3$, both $f_1^{(4)} \equiv f_2^{(4)} \equiv 0$. Hence, $\lambda_j^{(3)} = 0$, $j = 1,2$. We remain with \begin{align} g^{(4)} + \mu^{(k)} g = 0 \end{align} In the $k$-th inductive step, we can hence assume that $f_j^{(4)} = -\mu^{(j)} f_j$. Again, by orthogonality, we conclude $\lambda_j^{(k)} = 0$ for (all) $k$. Thereby, $f_k$ is the one normalized solution of the simple differential equation \begin{align} f_k^{(4)} + \mu^{(k)} f_k = 0, \quad f_k^{(2)}(\pm 1) = 0, \ f_k^{(3)}(\pm 1) = 0. \end{align} Searching for its solutions, $f_k$ can be assumed to be either symmetric or skew-symmetric. If neither was the case, then due to the symmetry of the problem, $g(x) := f_k(-x)$ fulfills the same differential equation (for the same $\mu^{(k)}$). Hence, $g + f_k$ and $g - f_k$ are symmetric and skew-symmetric, respectively, fulfill the differential equation and span the same space as $g$ and $f_k$. For simplicity, we substitute $\nu = \sqrt[4]{-\mu^{(k)}}$. The unscaled symmetric and skew-symmetric solutions are given by \begin{align} \cos(\nu x) + a \cosh(\nu x), \quad \sin(\nu x) + b \sinh(\nu x), \end{align} respectively. In order to satisfy the boundary conditions, one calculates that \begin{align} \begin{pmatrix} -\cos(\nu) & \cosh(\nu) \\ \sin(\nu) & \sinh(\nu) \end{pmatrix} \begin{pmatrix} 1 \\ a \end{pmatrix} = 0, \quad \begin{pmatrix} -\sin(\nu) & \sinh(\nu) \\ -\cos(\nu) & \cosh(\nu) \end{pmatrix} \begin{pmatrix} 1 \\ b \end{pmatrix} = 0. \end{align} The points at which either matrix becomes singular, are the roots $r_c$ and $r_s$ of \begin{align} t_c := \cos(\nu)\sinh(\nu)+\cosh(\nu)\sin(\nu), \quad t_s := \cos(\nu)\sinh(\nu)-\cosh(\nu)\sin(\nu), \end{align} respectively. Both have infinitely many (single) roots which can be approximated by $-\pi/4+i\pi$ and $\pi/4+i\pi$, respectively, since for $k > 7$, these equal the true $i$-th roots $r_c^{(i)}$ and $r_s^{(i)}$ up to $20$ decimal places. Earlier values are easily found numerically. We therefor define $\nu_{1,2} = 0, \ \nu_{1+2i} = r^{(i)}_c$ and $\nu_{2+2i} = r^{(i)}_s$. After some calculus, we obtain \begin{align} \sqrt{1+\frac{\cos(\nu_{1+2i})^2}{\cosh(\nu_{1+2i})^2}} f_{1+2i} & = \cos(\nu_{1+2i}x)+\frac{\cos(\nu_{1+2i})}{\cosh(\nu_{1+2i})} \cosh(\nu_{1+2i}x) \\ \sqrt{1-\frac{\sin(\nu_{2+2i})^2}{\sinh(\nu_{2+2i})^2}} f_{2+2i} & = \sin(\nu_{2+2i}x)+\frac{\sin(\nu_{2+2i})}{\sinh(\nu_{2+2i})} \sinh(\nu_{2+2i}x), \quad i \in \mathbb{N} \end{align}

All these roots are distinct. The orthogonality conditions are easy to verify, considering \begin{align} \int_{-1}^1 f_i f_j = \frac{1}{{\nu_i}^4} \int_{-1}^1 f_i^{(4)} f_j = -\frac{1}{{\nu_i}^4} \int_{-1}^1 f_i f_j^{(4)} = \frac{\nu_j^4}{{\nu_i}^4} \int_{-1}^1 f_i f_j \end{align} since either $\nu_j \neq \nu_i$ or $\nu_j = 0$, $j = 1,2$, for $j < i$. One can further calculate \begin{align} \int_{-1}^1 f_i'' f_j'' = \delta_{ij} \nu_i^4. \end{align} The $f_k$ are hence in correct order and their second derivates are orthogonal as well.


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