A Decision-Making Machine Learning Approach in Hermite Spectral Approximations of Partial Differential Equations

Abstract

The accuracy and effectiveness of Hermite spectral methods for the numerical discretization of partial differential equations on unbounded domains are strongly affected by the amplitude of the Gaussian weight function employed to describe the approximation space. This is particularly true if the problem is under-resolved, i.e., there are no enough degrees of freedom. The issue becomes even more crucial when the equation under study is time-dependent, forcing in this way the choice of Hermite functions where the corresponding weight depends on time. In order to adapt dynamically the approximation space, it is here proposed an automatic decision-making process that relies on machine learning techniques, such as deep neural networks and support vector machines. The algorithm is numerically tested with success on a simple 1D problem, but the main goal is its exportability in the context of more serious applications. As a matter of fact we also show at the end an application in the framework of plasma physics.

Appendix

We begin by collecting some basic relations concerning Hermite polynomials. We denote with a prime the derivative of a given Hermite polynomial with respect to the argument $\zeta$. Thus, the first and the second derivatives of the $\ell$-th Hermite polynomial, for $\ell \ge 2$, are given by:

$$\begin{aligned} H^{\prime }_{\ell }(\zeta )=2\ell H_{\ell -1}(\zeta ),\qquad \qquad H^{\prime \prime }_{\ell }(\zeta ) = 4\ell (\ell -1)H_{\ell -2}(\zeta ). \end{aligned}$$

(63)

In addition, we have: $H'_0(\zeta )=0, H'_1(\zeta )=2$, $H''_0(\zeta )=0, H''_1(\zeta )=0$, so that (63) formally holds for any $\ell \ge 0$. Useful recursive relations are:

$$\begin{aligned} 2\zeta H^{\prime }_{\ell }(\zeta )&= H^{\prime \prime }_{\ell }(\zeta ) + 2\ell H_{\ell }(\zeta ) = {\left\{ \begin{array}{ll} 2\ell H_{\ell }(\zeta ) , &{} \ell <2,\\ 4\ell (\ell -1)H_{\ell -2}(\zeta ) + 2\ell H_{\ell }(\zeta ) ,&{} \ell \ge 2. \end{array}\right. } \end{aligned}$$

(64)

$$\begin{aligned} \zeta H_{\ell }(\zeta )&= \frac{1}{4(\ell +1)}\,2\zeta H^{\prime }_{\ell +1}(\zeta ) = \frac{1}{4(\ell +1)}\Big [4(\ell +1)\ell H_{\ell -1}(\zeta ) + 2(\ell +1)H_{\ell +1}(\zeta )\Big ]\nonumber \\&= \ell H_{\ell -1}(\zeta ) + \frac{1}{2}H_{\ell +1}(\zeta ) , \qquad \ell \ge 1. \end{aligned}$$

(65)

For completeness, we note that for $\ell =0$ it holds $\zeta H_{0}(\zeta )=H_{1}(\zeta )/2$.

We now go through the computations relative to the three formulations (5), (15), (16). We start with the second one. We recall that $\alpha$ depends on t and we denote by $\alpha '$ its derivative. Concerning Hermite polynomials, the prime will continue to denote the derivative with respect to $\zeta$. We substitute the definitions (11),  (9) and we split the integral in three parts that will be computed separately:

$$\begin{aligned} \int _{\mathbbm {R}}\partial _tu_N\,\phi _{m}\,dx&= \int _{\mathbbm {R}} \frac{\partial _tw_{\alpha }(x,t)}{\sqrt{\pi }}\left[ \sum _{\ell =0}^{N} {{\widehat{u}}}_{\ell }(t)H_{\ell }\big (\alpha x\big )\right] \frac{\alpha }{2^m\,m!}H_{m}\big (\alpha x\big )\,dx\nonumber \\&\quad + \int _{\mathbbm {R}} \frac{w_{\alpha }(x,t)}{\sqrt{\pi }}\sum _{\ell =0}^{N}\bigg [ \Big (\partial _t{{\widehat{u}}}_{\ell }(t)\Big )H_{\ell }\big (\alpha x\big ) + {{\widehat{u}}}_{\ell }(t)\,\alpha 'xH^{\prime }_{\ell }\big (\alpha x\big ) \bigg ] \frac{\alpha }{2^m\,m!}H_{m}\big (\alpha x\big )\,dx\nonumber \\&= \frac{\alpha }{2^m\,m!\sqrt{\pi }}(-2\alpha \alpha ^{\prime }) \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\int _{\mathbbm {R}}x^2w_{\alpha }(x,t) H_{\ell }(\alpha x)H_{m}(\alpha x)\,dx\nonumber \\&\quad + \frac{1}{2^{m}\,m!\,\sqrt{\pi }} \sum _{\ell =0}^{N}\left[ \partial _t{{\widehat{u}}}_{\ell }(t)\int _{\mathbbm {R}}w_{\alpha }(x,t)H_{\ell }\big (\alpha x\big )H_{m}\big (\alpha x\big )\,\alpha \,dx\right. \nonumber \\&\quad \left. +{{\widehat{u}}}_{\ell }\alpha ^{\prime }\int _{\mathbbm {R}}w_{\alpha }(x,t)\,xH^{\prime }_{\ell } \big (\alpha x\big )\,H_{m}\big (\alpha x\big )\,\alpha \,dx\right] = \mathbf (I) + \mathbf (II) + \mathbf (III) . \end{aligned}$$

(66)

To compute (I), we substitute $\zeta =\alpha x$ and use the orthogonality properties of the Hermite polynomials to find that:

$$\begin{aligned} \mathbf (I)&= \frac{1}{2^m\,m!\sqrt{\pi }}\frac{-2\alpha ^{\prime }}{\alpha } \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\int _{\mathbbm {R}}(\alpha x)^2 \exp {\big (-(\alpha x)^2\big )}H_{\ell }(\alpha x)H_{m}(\alpha x)\alpha \,dx\\&= -\frac{\alpha ^{\prime }}{\alpha } \Big [(2m+1)\,{{\widehat{u}}}_{m}(t)+2(m+2)(m+1)\,{{\widehat{u}}}_{m+2}(t) +\frac{1}{2}{{\widehat{u}}}_{m-2}(t)\Big ]. \end{aligned}$$

To compute the term (II), we substitute $\zeta =\alpha x$ and use the orthogonality properties:

$$\begin{aligned} \mathbf (II) = \frac{1}{2^{m}\,m!\,\sqrt{\pi }}\sum _{\ell =0}^{N}\partial _t{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}\exp (-\zeta ^2)H_{\ell }(\zeta )H_{m}(\zeta )\,d\zeta = \partial _t{{\widehat{u}}}_{m}(t) . \end{aligned}$$

To compute term (III) we use (63) and the orthogonality of Hermite polynomials:

$$\begin{aligned} \mathbf (III)&= \frac{1}{2^{m}\,m!\,\sqrt{\pi }}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\frac{\alpha ^{\prime }}{\alpha } \int _{\mathbbm {R}}\exp (-\zeta ^2)\,\zeta H^{\prime }_{\ell }(\zeta )H_{m}(\zeta )\,d\zeta \\&= \frac{1}{2^{m}\,m!}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\frac{\alpha ^{\prime }}{\alpha }\Big [ 2\ell m\,2^{m-1}(m-1)!\,\delta _{\ell -1,m-1} + \ell \,2^{m+1}(m+1)!\,\delta _{\ell -1,m+1}\Big ]\\&= \frac{\alpha ^{\prime }}{\alpha }\Big [ m{{\widehat{u}}}_{m}(t) + 2(m+2)(m+1){{\widehat{u}}}_{m+2}(t) \Big ]. \end{aligned}$$

By collecting the results for $\mathbf (I) , \mathbf (II)$, and $\mathbf (III)$, we find out that:

$$\begin{aligned} \mathbf (I) + \mathbf (II) + \mathbf (III) = \partial _t{{\widehat{u}}}_{m}(t) + \frac{\alpha ^{\prime }}{\alpha }\Big [ -(m+1){{\widehat{u}}}_{m}(t) - \frac{1}{2}{{\widehat{u}}}_{m-2}(t) \Big ]. \end{aligned}$$

(67)

In a similar fashion, we split the integral of the second derivative of $u_N$ against the test function into three parts:

$$\begin{aligned} \int _{\mathbbm {R}}\partial _{xx}u_N\phi _{m}\,dx&= \int _{\mathbbm {R}}\partial _{xx}w_{\alpha }(x,t) \bigg [\frac{1}{\sqrt{\pi }}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)H_{\ell }(\alpha x)\bigg ]\, \bigg [\frac{\alpha }{2^m\,m!}H_{m}\big (\alpha x\big )\bigg ]\,dx\nonumber \\&\quad + \frac{1}{\sqrt{\pi }} \int _{\mathbbm {R}} \bigg [2\partial _xw_{\alpha }(x,t)\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\partial _xH_{\ell }(\alpha x)\bigg ] \bigg [\frac{\alpha }{2^m\,m!}H_{m}\big (\alpha x\big )\bigg ] \,dx\nonumber \\&\quad + \frac{1}{\sqrt{\pi }}\frac{\alpha }{2^m\,m!} \int _{\mathbbm {R}} \bigg [w_{\alpha }(x,t)\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\partial _{xx}H_{\ell }(\alpha x)\bigg ]\, \bigg [H_{m}\big (\alpha x\big )\bigg ]\,dx\nonumber \\&= \mathbf (A) + \mathbf (B) + \mathbf (C) . \end{aligned}$$

Using the properties of Hermite polynomials, we evaluate these terms as follows:

$$\begin{aligned} \mathbf (A)&= 2\alpha ^2\bigg [ 2m{{\widehat{u}}}_{m}(t) + 2(m+2)(m+1){{\widehat{u}}}_{m+2}(t) +\frac{1}{2}{{\widehat{u}}}_{m-2}(t) \bigg ],\\ \mathbf (B)&= -4\alpha ^2\Big [m{{\widehat{u}}}_{m}(t) + 2(m+2)(m+1){{\widehat{u}}}_{m+2}(t)\Big ],\\ \mathbf (C)&= 4(m+2)(m+1)\alpha ^2{{\widehat{u}}}_{m+2}(t). \end{aligned}$$

Putting all together we arrive at:

$$\begin{aligned} \mathbf (A) +\mathbf (B) +\mathbf (C)&= 2\alpha ^2\Big [ 2m{{\widehat{u}}}_{m}(t) + 2(m+2)(m+1){{\widehat{u}}}_{m+2}(t) +\frac{1}{2}{{\widehat{u}}}_{m-2}(t) \Big ] \nonumber \\&\quad -4\alpha ^2\Big [m{{\widehat{u}}}_{m}(t) + 2(m+2)(m+1){{\widehat{u}}}_{m+2}(t)\Big ] \nonumber \\&\quad +4(m+2)(m+1)\alpha ^2{{\widehat{u}}}_{m+2}(t) = \alpha ^2{{\widehat{u}}}_{m-2}(t). \end{aligned}$$

(68)

By equating (67) and (68) we finally obtain the scheme (14).

We then examine the scheme originating from (5). We must compute:

$$\begin{aligned} \int _{\mathbbm {R}}\frac{\partial u_N}{\partial x}\, \frac{\partial \phi _{m}}{\partial x}\,dx&= \frac{1}{\sqrt{\pi }}\,\frac{\alpha }{2^{m}\,m!} \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}\partial _x\Big (w_{\alpha }(x,t)H_{\ell }(\alpha x)\Big )\,\partial _xH_{m}(\alpha x)\,\,dx\nonumber \\&=\frac{1}{\sqrt{\pi }}\,\frac{\alpha }{2^{m}\,m!} \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\bigg [ \int _{\mathbbm {R}}\big (\partial _xw_{\alpha }(x,t)\big )H_{\ell } (\alpha x)\,\partial _xH_{m}(\alpha x)\,\,dx\nonumber \\&\quad +\int _{\mathbbm {R}}w_{\alpha }(x,t)\partial _xH_{\ell }\big (\alpha x)\,\partial _xH_{m} \big (\alpha x\big )\,\,dx\bigg ]\nonumber \\&=\frac{1}{\sqrt{\pi }}\,\frac{\alpha }{2^{m}\,m!} \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\bigg [ -2\alpha \int _{\mathbbm {R}}\big (\alpha x\big )w_{\alpha }(x,t) H_{\ell }(\alpha x)\,2mH_{m-1}(\alpha x)\,\alpha \,dx\nonumber \\&\quad +\alpha \int _{\mathbbm {R}}w_{\alpha }(x,t)H^{\prime }_{\ell }(\alpha x) \,H^{\prime }_{m}(\alpha x)\,\alpha \,dx\bigg ]\nonumber \\&=\frac{1}{\sqrt{\pi }}\,\frac{\alpha ^2}{2^{m}\,m!} \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\bigg [ -4m\int _{\mathbbm {R}}\zeta H_{\ell }(\zeta ) H_{m-1}(\zeta )e^{-\zeta ^2}\,d\zeta \nonumber \\&\quad +\int _{\mathbbm {R}}\exp {(-\zeta ^2)}\,H^{\prime }_{\ell }(\zeta )\,H^{\prime }_{m} (\zeta )e^{-\zeta ^2}\,d\zeta \bigg ]\nonumber \\&= \alpha ^2\Big ( -2m\,{{\widehat{u}}}_{m}(t)-\,{{\widehat{u}}}_{m-2}(t) \Big ) + \alpha ^2\Big ( 2m\,{{\widehat{u}}}_{m}(t) \Big ) = - \alpha ^2 {{\widehat{u}}}_{m-2}(t). \end{aligned}$$

(69)

By changing the sign of the last term in (69) we exactly get the same result in (68) which brings again to the scheme (14).

We finally examine the scheme originating from (16). The first integral is clearly equal to ${{\widehat{u}}}_{m}(t)$. Successively, we evaluate:

$$\begin{aligned} \int _{\mathbbm {R}}u_N\,\frac{\partial ^2\phi _{m}}{\partial x^2}\,\,dx&=\alpha ^2\int _{\mathbbm {R}}u_N\,\phi _{m-2}\,\,dx\\&= \alpha ^2\frac{1}{\sqrt{\pi }}\,\frac{\alpha }{2^{m-2}(m-2)!} \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}w_{\alpha }(x,t)H_{\ell }(\alpha x)\,H_{m-2}(\alpha x)\,\,dx\\&= \frac{\alpha ^2}{\sqrt{\pi }}\frac{1}{2^{m-2}(m-2)!}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}H_{\ell }(\zeta )H_{m-2}(\zeta )e^{-\zeta ^2}\,d\zeta \\&= \frac{\alpha ^2}{2^{m-2}(m-2)!\sqrt{\pi }}\sum _{\ell =0}^{N} {{\widehat{u}}}_{\ell }(t)\,2^{m-2}(m-2)!\sqrt{\pi }\delta _{\ell ,m-2} = \alpha ^2{{\widehat{u}}}_{m-2}(t). \end{aligned}$$

Regarding the last integral, we split it into two parts:

$$\begin{aligned} \int _{\mathbbm {R}}u_N\,\frac{\partial \phi _{m}}{\partial t}\,\,dx&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}w_{\alpha }(x,t)H_{\ell }(\alpha x)\,\partial _t\Big (\alpha H_{m}(\alpha x)\Big )\,dx\nonumber \\&=\frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!}\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \int _{\mathbbm {R}}w_{\alpha }(x,t)H_{\ell }(\alpha x)\, \Big (\alpha ^{\prime }H_{m}(\alpha x)+\alpha \alpha ^{\prime }xH^{\prime }_{m}(\alpha x)\Big )\,dx\nonumber \\&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!} \,\frac{\alpha ^{\prime }}{\alpha }\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \bigg [ \int _{\mathbbm {R}}H_{\ell }(\zeta )H_{m}(\zeta )e^{-\zeta ^2}\,d\zeta \nonumber \\&\quad + \int _{\mathbbm {R}}H_{\ell }(\zeta )\,\zeta H^{\prime }_{m}(\zeta )e^{-\zeta ^2}\,d\zeta \bigg ]\nonumber \\&= \mathbf (I) + \mathbf (II) . \end{aligned}$$

(70)

These are finally evaluated as follows:

$$\begin{aligned} \mathbf (I)&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!} \frac{\alpha ^{\prime }}{\alpha } \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t)\big (2^{m}\, m!\sqrt{\pi }\big )\delta _{\ell ,m} = \frac{\alpha ^{\prime }(t)}{\alpha }{{\widehat{u}}}_{m}.\nonumber \\ \mathbf (II)&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!} \,\frac{\alpha ^{\prime }}{\alpha }\sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \bigg [ \int _{\mathbbm {R}}H_{\ell }(\zeta )\,\zeta H^{\prime }_{m}(\zeta )e^{-\zeta ^2}\,d\zeta \bigg ]\nonumber \\&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!} \frac{\alpha ^{\prime }}{\alpha } \sum _{\ell =0}^{N}{{\widehat{u}}}_{\ell }(t) \big (2^{\ell }\,\ell !\sqrt{\pi }\big ) \bigg [ 2m(m-1)\delta _{m-2,\ell } + m\delta _{\ell ,m} \bigg ]\nonumber \\&= \frac{1}{\sqrt{\pi }}\,\frac{1}{2^{m}\,m!} \frac{\alpha ^{\prime }}{\alpha } \big (2^{m-2}\,(m-2)!\sqrt{\pi }\big ) \bigg [ 2m(m-1) {{\widehat{u}}}_{m-2}(t) + \big (2^{m}\,m!\sqrt{\pi }\big )\,m{{\widehat{u}}}_{m}(t) \bigg ]\nonumber \\&= \frac{\alpha ^{\prime }}{\alpha }\bigg [ \frac{1}{2}{{\widehat{u}}}_{m-2}(t) + m{{\widehat{u}}}_{m}(t) \bigg ]. \end{aligned}$$

(71)

The final result of all these computations is again the scheme (14).