Optimal Strong Convergence of Finite Element Methods for One-Dimensional Stochastic Elliptic Equations with Fractional Noise

Abstract

We investigate the strong convergence order of piecewise linear finite element methods for a class of one-dimensional semilinear stochastic elliptic equations with additive fractional white noise. For the Hurst index $H\in (0,1)$, we approximate the fractional Brownian motion by two spectral expansions. We show that the resulting schemes are of order $H+1$ in the mean-square sense if the element size h is taken proportionally to the truncation parameters in the spectral approximations. Numerical results confirm our theoretical prediction.

Appendices

Verification of the Isometry of Wiener Integrals

In this section, we verify the itsometry (2.4). When $H>1/2$, we apply (2.3), stochastic Fubini’s theorem, and integration-by-parts twice to obtain that

$$\begin{aligned} \mathbb {E}[\left( \int _{\mathcal {D}} e_k \,dW_{H }(x)\right) ^2]= & {} \int _{\mathcal {D}}\int _{\mathcal {D}}\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) \mathbb {E}[W_{H}(x)W_{H}(y)]\,dx \,dy\\= & {} \frac{1}{2} \int _{\mathcal {D}}\int _{\mathcal {D}}\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) (\left| x\right| ^{2H} +\left| y\right| ^{2H} - \left| x-y\right| ^{2H} )\,dx \,dy\\= & {} -\frac{1}{2} \int _{\mathcal {D}}\int _{\mathcal {D}}\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) \left| x-y\right| ^{2H} \,dx \,dy\\= & {} - \frac{1}{2} \int _{\mathcal {D}}\int _{\mathcal {D}} e_k(x) e_k(y) \frac{d}{dx} \frac{d}{dy} \left| x-y\right| ^{2H} \,dx \,dy\\= & {} H(2H-1) \int _{\mathcal {D}}\int _{\mathcal {D}} {e_k(x) e_k(y) }{ \left| x-y\right| ^{2H-2}} \,dx \,dy. \end{aligned}$$

This is exactly the squared norm $\Vert e_k\Vert ^2_{|R_H|,1}$ in [24] (Equation 1.6.6., Page 19). When $H=1/2$, we may use the covariance function $\mathbb {E}[W_{H}(x)W_{H}(y)]= \min (x,y)$ and integration-by-parts to obtain the isometry.

For $0<H<1/2$, we can derive the isometry with more care. Applying (2.3), stochastic Fubini’s theorem, and integration-by-parts twice, we obtain that

$$\begin{aligned}&\mathbb {E}[\left( \int _{\mathcal {D}} e_k \,dW_{H }(x)\right) ^2]\\&\quad = -\frac{1}{2}\int _{\mathcal {D}}\int _{\mathcal {D}}\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) \left| x-y\right| ^{2H} \,dx \,dy\\&\quad = -\frac{1}{2}\int _0^1\int _0^y\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) (y-x)^{2H} \,dx \,dy\\&\qquad -\frac{1}{2}\int _0^1\int _y^1\frac{d}{dx} e_k(x)\frac{d}{dy} e_k(y) (x-y)^{2H} \,dx \,dy \\&\quad = -\frac{1}{2}\int _0^1\left( \int _0^y2H e_k(x)(y-x)^{2H-1}\,dx \right) \frac{d}{dy} e_k(y) \,dy\\&\qquad +\frac{1}{2}\int _0^1\left( \int _y^12H e_k(x)(x-y)^{2H-1}\,dx\right) \frac{d}{dy} e_k(y)\,dy\\&\quad =-H \int _0^1\left( \int _x^1 \frac{d}{dy} e_k(y)(y-x)^{2H-1}\,dy \right) e_k(x) \,dx\\&\qquad +H \int _0^1\left( \int _0^x \frac{d}{dy} e_k(y)(x-y)^{2H-1}\,dy\right) e_k(x)\,dx\\&\quad = H\Gamma (2H)\Big ((e_k,\,_xD_1^{2H}e_k)+(e_k,\,_0D_x^{2H}e_k)\Big )\\&\quad = H\Gamma (2H)\Big ((_0D_x^{H}e_k,\,_xD_1^{H}e_k)+(_xD_1^{H}e_k,\,_0D_x^{H}e_k)\Big )\\&\quad = \Gamma (2H+1)\cos (\pi H)\Vert _0D_x^{H}e_k\Vert _{L^2}^2. \end{aligned}$$

Here we have applied fractional integration by parts (see e.g., [37]) and used the formulation from [13] to obtain the last equality. Also, $\,_0D_x^{\alpha }u$ and $\,_xD_1^{\alpha }u$ denote the left- and right- Caputo fractional derivative with $0< \alpha <1$ respectively, defined by

$$\begin{aligned} (\,_0D_x^{\alpha }u)(x)=\frac{1}{\Gamma (\alpha )}\int _0^x\frac{u'(\tau )}{(x-\tau )^{1-\alpha }}\,d\tau ,(\,_xD_1^{\alpha }u)(x)=-\frac{1}{\Gamma (\alpha )}\int _x^1\frac{u'(\tau )}{(\tau -x)^{1-\alpha }}\,d\tau . \end{aligned}$$

Some Useful Statements

In order to estimate $\mathbb {E}[\left\| \zeta _2\right\| ^2]$ in Lemma 43, we need the following lemmas.

Lemma B1

([33]) For the Bessel function $J_{\nu }$, where $\nu >-1$, we have

$$\begin{aligned} J^2_{1+\nu }(z) + J^2_{\nu }(z)\approx & {} \frac{2}{\pi z} , \text{ for } \text{ large } \left| z\right| . \end{aligned}$$

(B.1)

$$\begin{aligned} 2\nu J_\nu (x)= & {} x J_{\nu +1}(x) +x J_{\nu -1}(x). \end{aligned}$$

(B.2)

Lemma B2

Let $H\ne \frac{1}{2}$ and $\alpha _n$ be the positive zeros of $J_{-H}$ and $\beta _n$ be the positive zeros of $J_{1-H}$.

There exists a positive constant C independent of n such that

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{4m^4} \bigg (\frac{1}{m\pi + \alpha _n} +\frac{1}{m\pi - \alpha _n}\bigg )^2\le & {} C \frac{1}{\alpha _n^4}, \end{aligned}$$

(B.3)

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{4m^4} \bigg (\frac{1}{m\pi + \beta _n} +\frac{1}{m\pi - \beta _n}\bigg )^2\le & {} C \frac{1}{\beta _n^4}. \end{aligned}$$

(B.4)

Proof

First, we observe that

$$\begin{aligned} \frac{\pi ^2}{4m^4} \bigg (\frac{1}{m\pi + \alpha _n} +\frac{1}{m\pi - \alpha _n}\bigg )^2= & {} \frac{ 1 }{m^2} \bigg (\frac{1}{m^2 - t^2} \bigg )^2 \\= & {} \frac{1}{t^2(m^2 - t^2)^2} - \frac{1}{t^4(m^2 - t^2)} + \frac{1}{t^4 m^2 }, \end{aligned}$$

where $t= \alpha _n/\pi$. According to (2.12), $\alpha _n,\beta _n\ne n\pi$ and thus t is not an integer when $H\ne \frac{1}{2}$.

By Formula 1.421.3 in [16] (Page 44), i.e.,

$\displaystyle \sum _{k=1}^\infty \frac{1}{y^2-k^2} =\frac{\pi }{2}\frac{\cot (\pi y)}{y} -\frac{1}{2y^2}$,

we have

$$\begin{aligned} -\sum _{m=1}^\infty \frac{1}{t^4(m^2 - t^2)}= & {} \frac{1}{t^4} \bigg ( \frac{\pi \cot (\alpha _n)}{2t} -\frac{1}{2t^2} \bigg ) \le \frac{\pi \cot (\alpha _n)}{2t^5}. \end{aligned}$$

(B.5)

By Formula 1.423 of [16] (Page 44), i.e., $\displaystyle \sum _{k=1}^\infty \frac{1}{(1-k^2l^2)^2}=\frac{\pi ^2}{4l^2} \csc ^2 (\frac{\pi }{l}) + \frac{\pi }{4l}\cot (\frac{\pi }{l}) -\frac{1}{2}$, we obtain that

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{t^2(m^2 - t^2)^2}= & {} \sum _{m=1}^\infty \frac{1}{t^6(m^2/t^2 - 1)^2} = \frac{1}{t^6} \bigg ( \frac{\pi ^2t^2}{4}\csc ^2(t\pi ) + \frac{t\pi }{4}\csc (t\pi )-\frac{1}{2} \bigg ). \qquad \nonumber \\ \end{aligned}$$

(B.6)

By the fact that $\sum _{m=1}^\infty \frac{1}{ m^2 } = \frac{\pi ^2}{6}$, (B.5) and (B.6), we then have

$$\begin{aligned}&\sum _{m=1}^\infty \frac{\pi ^2}{4m^4} \bigg (\frac{1}{m\pi + \alpha _n} +\frac{1}{m\pi - \alpha _n}\bigg )^2 \\&\quad = \sum _{m=1}^\infty \frac{1}{t^2(m^2 - t^2)^2} - \sum _{m=1}^\infty \frac{1}{t^4(m^2 - t^2)} + \sum _{m=1}^\infty \frac{1}{t^4 m^2 } \\&\quad \le \frac{1}{t^6} \bigg ( \frac{\pi ^2t^2}{4}\csc ^2(t\pi ) + \frac{t\pi }{4}\csc (t\pi )-\frac{1}{2} \bigg ) + \frac{\pi \cot (\alpha _n)}{2t^5} +\frac{1}{t^4} \frac{\pi ^2}{6} \\&\quad \le C\frac{1}{t^4} + C\frac{1}{t^5} \le C \frac{1}{\alpha _n^4}. \end{aligned}$$

Here we have applied the estimate of zeros of the Bessel function (2.12) that $\alpha _n$ is at the order of $n\pi$. We then have proved (B.3). Similarly, we have (B.4). $\square$