On a collocation point of view to reproducing kernel methods

Abstract

In this text, we discuss an interpolation (or a collocation) point of view to reproducing kernel methods to approximate solutions to some linear and non-linear functional equations. The proposed method allows us to avoid the usual process of orthogonalization, solving a system of algebraic equations, with a positive defined matrix in the linear case. This also helps us to understand some methods present in the literature on this subject. We include some examples to illustrate the proposed method and an appendix with algorithms implemented in the R language.

Appendix A: Complementary information to the numerical examples

We present the R language implementation of algorithm proposed in Theorem 2.5 adapted to Example 3.1.

We present the R language implementation of algorithm proposed in Theorem 2.5, jointly with Expression 2.5, to the nonlinear Riccati differential equation of Example 3.2. Graphs is produced as in the previous example.

We present the R language implementation of algorithm proposed in Theorem 2.5, jointly with Expression 2.5, to the nonlinear boundary value problem of Example 3.3.

We find the polynomial p(x) by using Lagrange’s interpolation formula.

1.1 A.1 On the reproducing kernel of some subspaces of $\mathcal {H}_K$

We include a lemma for the sake of completeness (please see Theorem 3.2 in Azarnavid and Parand (2018) to a particular version of this result).

Lemma A.1

Let $\mathcal {H}_K$ being a reproducing kernel Hilbert space, with reproducing kernel $K:X\times X\rightarrow \mathbb {R}$, and

$$\begin{aligned} l(u)= \sum _{i=0}^n\alpha _iu(x_i), \quad u\in \mathcal {H}_K, \end{aligned}$$

to some $x_0,x_1,\ldots ,x_n\in X$ and $\alpha _0,\alpha _1,\ldots ,\alpha _n\in \mathbb {R}$. If $l(l(K^x))\ne 0$, for all $x\in X$, then the subspace $W=\{u\in \mathcal {H}_K:\, l(u)=0\}$ is a reproducing kernel Hilbert space with kernel

$$\begin{aligned} K_1(x,y)=K(x,y)-\frac{l(K^x)l(K^y)}{l(l(K^x))},\qquad x,y\in X. \end{aligned}$$

Proof

Note that

$$\begin{aligned} l(K^x)=\sum _{i=0}^n\alpha _iK(x_i,x)=\sum _{i=0}^n\alpha _iK^{x_i}(x),\quad x\in X \end{aligned}$$

is in $\mathcal {H}_K$, and that

$$\begin{aligned} l(l(K^x))=\sum _{j=0}^n\alpha _j\sum _{i=0}^n\alpha _iK(x_i,x_j)=\sum _{i,j=0}^n\alpha _j\alpha _iK(x_i,x_j)\ge 0, \end{aligned}$$

does not depend of x.

Now let, $u\in W$, hence

$$\begin{aligned} \left\langle u,K_1^x\right\rangle _K= & {} \left\langle u,K^x-\frac{l(K^x)\left( \sum _{i=0}^n\alpha _iK^{x_i}\right) }{l(l(K^x))}\right\rangle _K \\= & {} u(x)-\left\langle u,\frac{l(K^x)\left( \sum _{i=0}^n\alpha _iK^{x_i}\right) }{l(l(K^x))}\right\rangle _K\\= & {} u(x)-\left\langle u,\sum _{i=0}^n\alpha _iK^{x_i}\right\rangle _K\frac{l(K^x)}{l(l(K^x))}\\= & {} u(x)-\left( \sum _{i=0}^n\alpha _iu(x_i)\right) \frac{l(K^x)}{l(l(K^x))}\\ {}= & {} u(x),\qquad x\in X. \end{aligned}$$

To finishes, note that

$$\begin{aligned} l(K_1^x)= & {} \sum _{j=0}^n\alpha _j\left( K(x,x_j)-\frac{l(K^x)\left( \sum _{i=0}^n\alpha _iK(x_i,x_j)\right) }{l(l(K^x))}\right) \\= & {} l(K^x)-\frac{l(K^x)\left( \sum _{j=0}^n\alpha _j\sum _{i=0}^n\alpha _iK(x_i,x_j)\right) }{l(l(K^x))}\\= & {} l(K^x)-l(K^x)\\ {}= & {} 0. \end{aligned}$$

This means that $K_1^x\in W$, for all $x\in X$. $\square$

Corollary A.2

Let $\mathcal {H}_K$ being a reproducing kernel Hilbert space, with reproducing kernel $K:X\times X\rightarrow \mathbb {R}$, and

$$\begin{aligned} l_j(u)= \sum _{i=0}^n\alpha _{ij}u(x_i), \quad u\in \mathcal {H}_K, \end{aligned}$$

to some $x_0,x_1,\ldots ,x_n\in X$ and $\alpha _{ij}\in \mathbb {R}$, to $1\le j\le m$. If $l_j(l_j(K_{j-1}^x))\ne 0$, then the subspace $W_j=\{u\in W_{j-1}:\, l_j(u)=0\}$ is a reproducing kernel Hilbert space with kernel $K_j$, where $W_0=\mathcal {H}_K$ and

$$\begin{aligned} K_1(x,y)= & {} K(x,y)-\frac{l_1(K^x)l_1(K^y)}{l_1(l_1(K^x))}\\ K_2(x,y)= & {} K_1(x,y)-\frac{l_2(K_1^x)l_2(K_1^y)}{l_2(l_2(K_1^x))}\\\vdots &\vdots&\vdots \\ K_m(x,y)= & {} K_{m-1}(x,y)-\frac{l_m(K_{m-1}^x)l_m(K_{m-1}^y)}{l_m(l_m(K_{m-1}^x))} \end{aligned}$$