How Much Can One Learn a Partial Differential Equation from Its Solution?

Abstract

In this work, we study the problem of learning a partial differential equation (PDE) from its solution data. PDEs of various types are used to illustrate how much the solution data can reveal the PDE operator depending on the underlying operator and initial data. A data-driven and data-adaptive approach based on local regression and global consistency is proposed for stable PDE identification. Numerical experiments are provided to verify our analysis and demonstrate the performance of the proposed algorithms.

Appendix A. Proof of (4.20) and (4.21)

In the following, we assume $N > 1$. Recalling that the variance of a random variable X can be expressed as ${\mathbb {E}}[X^2]-({\mathbb {E}}[X])^2$, we get

$$\begin{aligned} \sum _{m=1}^N{\mathbb {E}}\left[ (\zeta _m-\frac{1}{N}\sum _{n=1}^N\zeta _n)^2\right]&=\sum _{m=1}^N{\mathbb {E}}\left[ \zeta _m^2\right] -\frac{1}{N}{\mathbb {E}}\left[ (\sum _{n=1}^N\zeta _n)^2\right] \\&=\frac{N(B-1)}{B}\sigma ^2+\sum _{m=1}^N\mu _m^2-\frac{B-1}{B}\sigma ^2-\frac{1}{N}\left( \sum _{n=1}^N\mu _n\right) ^2\\&=\frac{(N-1)(B-1)}{B}\sigma ^2+\sum _{m=1}^N\mu _m^2-\frac{1}{N}\left( \sum _{n=1}^N\mu _n\right) ^2. \end{aligned}$$

Hence by defining the estimator:

$$\begin{aligned} {\widehat{\sigma }}^2 = \frac{B\sum _{m=1}^N\left( \zeta _m-\frac{1}{N}\sum _{n=1}^N\zeta _n\right) ^2}{(N-1)(B-1)} \end{aligned}$$

and noting the Lipschitz assumption, we have

$$\begin{aligned} {\mathbb {E}}[{\widehat{\sigma }}^2-\sigma ^2]&=\frac{B\left( \sum _{m=1}^N\mu _m^2-\frac{1}{N}(\sum _{n=1}^N\mu _n)^2\right) }{(N-1)(B-1)}\\&\le \frac{B\sum _{m=1}^N\mu _m^2}{(N-1)(B-1)}\le \frac{DNBL^2R^2}{(N-1)(B-1)}. \end{aligned}$$

As for the variance of the estimator, we notice that since $\zeta _n$, $n=1,\dots , N$ are independent Gaussian random variables, if we denote $S:=\sqrt{\frac{B-1}{B}}\sigma$, $\sum _{n=1}^N\zeta _n^2/S^2$ has a non-central Chi-squared distribution whose mean is $N+\sum _{n=1}^N\mu _n^2/S^2$, and variance is $2(N+2\sum _{n=1}^N\mu _n^2/S^2)$; and $(\sum _{n=1}^N\zeta _n)^2/(NS^2)$ also has a non-central Chi-squared distribution whose mean is $1+(\sum _{n=1}^N\mu _n)^2/(NS^2)$, and variance is $2(1+2(\sum _{n=1}^N\mu _n)^2/(NS^2))$. First, we compute the covariance

$$\begin{aligned}&\text {Cov}\left( \sum _{n=1}^N\zeta _n^2,\left( \sum _{n=1}^N\zeta _n\right) ^2\right) ={\mathbb {E}}\left[ \sum _{n=1}^N\zeta _n^2(\sum _{m=1}^N\zeta _m)^2\right] -{\mathbb {E}}\left[ \sum _{n=1}^N\zeta _n^2\right] {\mathbb {E}} \left[ \left( \sum _{n=1}^N\zeta _n\right) ^2\right] \\&\quad ={\mathbb {E}}\left[ \sum _{n=1}^N\zeta _n^2\left( \sum _{m=1}^N\zeta _m\right) ^2\right] -\left( NS^2+\sum _{n=1}^N \mu _n^2\right) \left( NS^2 +\left( \sum _{n=1}^N\mu _n\right) ^2 \right) . \end{aligned}$$

Focusing on the first term, we have

$$\begin{aligned}&{\mathbb {E}}\left[ \sum _{n=1}^N\zeta _n^2\left( \sum _{m=1}^N\zeta _m\right) ^2\right] = \sum _{n=1}^N{\mathbb {E}}\left[ \zeta _n^2\sum _{m=1}^N\zeta _m^2\right] +\sum _{n=1}^N{\mathbb {E}}\left[ \zeta _n^2\sum _{i\ne j}\zeta _i\zeta _j\right] \\&\quad =\sum _{n=1}^N{\mathbb {E}}[\zeta _n^4]+\sum _{n=1}^N\left( {\mathbb {E}}[\zeta _n^2]\sum _{m=1,m\ne n}^N{\mathbb {E}}[\zeta _m^2]\right) +2\sum _{n=1}^N\left( {\mathbb {E}}[\zeta _n^3]\sum _{m=1,m\ne n}^N{\mathbb {E}}[\zeta _m]\right) \\&\qquad + \sum _{n=1}^N\left( {\mathbb {E}}[\zeta _n^2]\sum _{i\ne n,j\ne n,i\ne j}{\mathbb {E}}[\zeta _i]{\mathbb {E}}[\zeta _j]\right) \\&\quad =\sum _{n=1}^N(\mu _n^4+6\mu _n^2S^2+3S^4) +\sum _{n=1}^N\left( (\mu _n^2+S^2)\sum _{m=1,m\ne n}^N(\mu _m^2+S^2)\right) \\&\qquad + 2\sum _{n=1}^N\left( (\mu _n^3+3\mu _nS^2)\sum _{m=1,m\ne n}^N\mu _m\right) +\sum _{n=1}^N\left( (\mu _n^2+S^2)\sum _{i\ne n,j\ne n,i\ne j}\mu _i\mu _j\right) \\&\quad =\sum _{n=1}^N\mu _n^2\left( \sum _{n=1}^N\mu _n\right) ^2+\left( 2(N-1)\sum _{n=1}^N\mu _n^2\right. \\&\left. \qquad +6\left( \sum _{n=1}^N\mu _n\right) ^2 +(N-2)\sum _{n\ne m}\mu _n\mu _m\right) S^2 + N(N+2)S^4. \end{aligned}$$

Hence, we have

$$\begin{aligned}&\text {Cov}\left( \sum _{n=1}^N\zeta _n^2,\left( \sum _{n=1}^N\zeta _n\right) ^2\right) \\&\quad =\left( (N-2)\sum _{n=1}^N\mu _n^2+(6-N)\left( \sum _{n=1}^N\mu _n\right) ^2+(N-2)\sum _{n\ne m}\mu _n\mu _m\right) S^2+2NS^4\\&\quad =4\left( \sum _{n=1}^N\mu _n\right) ^2S^2+2NS^4 . \end{aligned}$$

Now we note that

$$\begin{aligned}&\text {Var}\left[ \sum _{m=1}^N\left( \zeta _m-\frac{1}{N}\sum _{n=1}^N\zeta _n\right) ^2\right] \\&\quad =\text {Var}\left[ \sum _{m=1}^N\zeta ^2_m\right] +\frac{1}{N^2}\text {Var}\left[ (\sum _{n=1}^N\zeta _n)^2\right] \\&\qquad -\frac{2}{N}\text {Cov}\left( \sum _{m=1}^N\zeta _m^2,\left( \sum _{m=1}^N\zeta _m\right) ^2\right) \\&\quad =2\left( NS^4+2\sum _{n=1}^N\mu _n^2S^2\right) +2S^4\left( 1+2\left( \sum _{n=1}^N\mu _n\right) ^2/(NS^2)\right) \\&\qquad -\frac{8}{N}\left( \sum _{n=1}^N\mu _n\right) ^2S^2-4S^4. \end{aligned}$$

After simplification, we get

$$\begin{aligned}&\text {Var}\left[ \sum _{m=1}^N\left( \zeta _m-\frac{1}{N}\sum _{n=1}^N\zeta _n\right) ^2\right] \\&\quad =2(N-1)S^4+4\left( \sum _{n=1}^N\mu _n^2-\frac{\left( \sum _{n=1}^N\mu _n\right) ^2}{N}\right) S^2. \end{aligned}$$

Considering the Lipschitz assumption, we obtain

$$\begin{aligned}&\text {Var}\left[ \sum _{m=1}^N\left( \zeta _m-\frac{1}{N}\sum _{n=1}^N\zeta _n\right) ^2\right] \le 2(N-1)S^4+4NDL^2R^2S^2. \end{aligned}$$

Therefore, denoting $\gamma =4DL^2R^2$, then we get

$$\begin{aligned} \text {Var}[{\widehat{\sigma }}^2]&\le \frac{2(N-1)(B-1)^2\sigma ^4+\gamma N B(B-1)\sigma ^2}{(N-1)^2(B-1)^2}\\&=\frac{2\sigma ^4}{N-1}+\frac{NB\sigma ^2\gamma }{(N-1)^2(B-1)}. \end{aligned}$$