/UQNO

Error estimate of functional type for neural operators

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We consider elliptic equation in the domain $\Gamma \subseteq [0, 1]^{D}$

$$\begin{equation} \begin{split} -\sum_{ij=1}^{D}\frac{\partial}{\partial x_i}\left(a_{ij}(x) \frac{\partial u}{\partial x_j} \right) + b^{2}(x) u(x) = f(x),\,\left.u\right|_{\partial\Gamma} = 0,\,a_{ij}(x) \geq c > 0. \end{split} \end{equation}$$

First, we define random trigonometric polynomials

$$\mathcal{P}(N_1, N_2, \alpha) = \left\{f(x) = \mathcal{R}\left(\sum_{m=0}^{N_1}\sum_{n=0}^{N_2}\frac{c_{mn}\exp\left(2\pi i(mx_1 + nx_2)\right)}{(1+m+n)^\alpha}\right):\mathcal{R}(c), \mathcal{I}(c)\simeq \mathcal{N}(0, I)\right\}.$$

For the first equation, we use Cholesky factorization to define matrix $a$ and random trigonometric polynomials for $b$ and $f$:

$$\begin{equation} \begin{split} &a(x) = \begin{pmatrix} \alpha(x) & 0 \\\ \gamma(x) & \beta(x) \end{pmatrix} \begin{pmatrix} \alpha(x) & \gamma(x) \\\ 0 & \beta(x) \end{pmatrix},\\\ &\,\alpha(x),\beta(x) \simeq 0.1\mathcal{P}(5, 5, 2) + 1;\,\gamma(x),\,b(x),\,f(x) \simeq \mathcal{P}(5, 5, 2). \end{split} \end{equation}$$

The second equation has a discontinuous scalar diffusion coefficient:

$$\begin{equation} \begin{split} a(x) = \alpha(x)I,\,\alpha(x) = \begin{cases} 10,\,p_1(x) \geq 0;\\\ 1,\, ~~p_1(x) < 0, \end{cases}\,b(x) = 0,\,f(x) = 1,\,p_1(x) \simeq \mathcal{P}(5, 5, 2). \end{split} \end{equation}$$

Third equation is similar to the second one but with more diverse $b$ and $f$:

$$\begin{equation} \begin{split} a(x) = \alpha(x)I,\,\alpha(x) = \begin{cases} 10,\,p_1(x) \geq 0;\\\ 1,\, ~~p_1(x) < 0, \end{cases}\,b(x),\,f(x),\,p_1(x) \simeq \mathcal{P}(5, 5, 2). \end{split} \end{equation}$$

Last equation is similar to the first one but with $b=0$:

$$\begin{equation} \begin{split} &a(x) = \begin{pmatrix} \alpha(x) & 0 \\\ \gamma(x) & \beta(x) \end{pmatrix} \begin{pmatrix} \alpha(x) & \gamma(x) \\\ 0 & \beta(x) \end{pmatrix},\\\ &\,\alpha(x),\beta(x) \simeq 0.1\mathcal{P}(5, 5, 2) + 1;\,\gamma(x), f(x) \simeq \mathcal{P}(5, 5, 2);\,b(x) = 0. \end{split} \end{equation}$$

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