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Convergence of finite element solutions of stochastic partial integro-differential equations driven by white noise

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Abstract

Numerical approximation of a stochastic partial integro-differential equation driven by a space-time white noise is studied by truncating a series representation of the noise, with finite element method for spatial discretization and convolution quadrature for time discretization. Sharp-order convergence of the numerical solutions is proved up to a logarithmic factor. Numerical examples are provided to support the theoretical analysis.

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Correspondence to Jilu Wang.

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The research of M. Gunzburger and J. Wang was supported in part by the USA National Science Foundation Grant DMS-1315259 and by the USA Air Force Office of Scientific Research Grant FA9550-15-1-0001. The research of B. Li was supported in part by the Hong Kong RGC Grant 15300817.

Appendices

Mild solution of (1.1)

In the case \(\alpha \in (1,2)\), the boundary condition \(\partial _t^{1-\alpha }\psi =0\) is equivalent to \(\psi =0\) on \(\partial \Omega \) (this can be checked by taking Laplace transform in time). Similarly, in the case \(\alpha \in (0,1]\), the boundary condition \(\partial _t^{1-\alpha }\psi =0\) is equivalent to \(\psi -\psi _0=0\) on \(\partial \Omega \times [0,\infty )\), where \(\psi _0=\psi (\cdot ,0)\) is the initial value in (1.1).

In the case \(\sigma =0\), the solution of the corresponding deterministic problem of (1.1) can be expressed by (via Laplace transform, cf. [23, (3.11) and line 4 of page 12] in the case \(\alpha \in (1,2)\))

$$\begin{aligned} \psi (\cdot ,t) = \left\{ \begin{array}{ll} \psi _0 + \int _0^t E(t-s) f(\cdot ,s)\mathrm{d}s &{}\quad \text {if}\,\,\,\alpha \in (0,1] , \\ E(t) \psi _0 + \int _0^t E(t-s) f(\cdot ,s)\mathrm{d}s &{}\quad \text {if}\,\,\,\alpha \in (1,2) , \end{array} \right. \end{aligned}$$
(A.1)

where the operator \(E(t):L^2(\mathcal {O})\rightarrow L^2(\mathcal {O})\) is given by

$$\begin{aligned} E(t) \phi :=\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{\alpha -1} (z^\alpha -\Delta )^{-1}\phi \, \mathrm{d}z \quad \forall \, \phi \in L^2(\mathcal {O}) \end{aligned}$$
(A.2)

with integration over a contour \(\varGamma _{\theta ,\kappa }\) on the complex plane.

Correspondingly, the mild solution of the stochastic problem (1.1) is defined as (cf. [18, Proposition 2.7] and [24])

$$\begin{aligned} \psi (\cdot ,t) = \left\{ \begin{array}{ll} \psi _0 + \int _0^t E(t-s) f(\cdot ,s)\mathrm{d}s +\sigma \int _0^t E(t-s)\mathrm{d}W(\cdot ,s) &{}\quad \text {if}\quad \alpha \in (0,1] , \\ E(t) \psi _0 + \int _0^t E(t-s) f(\cdot ,s)\mathrm{d}s +\sigma \int _0^t E(t-s)\mathrm{d}W(\cdot ,s) &{}\quad \text {if}\quad \alpha \in (1,2) . \end{array} \right. \end{aligned}$$
(A.3)

For any given initial data \(\psi _0\in L^2(\mathcal {O})\) and source \(f\in L^1(0,T;L^2(\mathcal {O}))\), the expression (A.3) defines a mild solution \(\psi \in C([0,T];L^2(\Omega ; L^2(\mathcal {O})))\). In the case \(\psi _0=f=0\) and \(\sigma \ne 0\), a simple proof of this result can be found in [12, Appendix]; in the case \(\sigma =0\) (\(\psi _0\) and f may not be zero), the result is a consequence of the boundedness of the operator \(E(t):L^2(\mathcal {O})\rightarrow L^2(\mathcal {O})\), i.e.,

$$\begin{aligned} \Vert E(t)v\Vert&\le C\int _{\varGamma _{\theta ,\kappa }} |e^{zt}| |z|^{\alpha -1} \Vert (z^\alpha -\Delta )^{-1}v\Vert |\mathrm{d}z| \nonumber \\&\le C\Vert v\Vert \int _\kappa ^{\infty } e^{-rt|\cos (\theta )|} r^{-1} \mathrm{d}r +C\Vert v\Vert \int _{-\theta }^\theta e^{\kappa t\cos (\varphi )} \mathrm{d}\varphi \nonumber \\&\le C\Vert v\Vert \qquad \forall \, v\in L^2(\mathcal {O}). \end{aligned}$$
(A.4)

Similarly, the discrete operator \(E^{(h)}(t):X_h\rightarrow X_h\) defined by

$$\begin{aligned} E^{(h)}(t) \phi :=\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{\alpha -1} (z^\alpha -\Delta _h )^{-1}\phi \, \mathrm{d}z \quad \forall \, \phi \in X_h , \end{aligned}$$
(A.5)

is also bounded on the finite element subspace \(X_h\), i.e.,

$$\begin{aligned} \Vert E^{(h)}(t)v\Vert \le C\Vert v\Vert \quad \forall \, v\in X_h , \end{aligned}$$
(A.6)

where the constant C is independent of the mesh size h.

Representation of the discrete solutions

For \(f=0\) we prove the following representation of the solutions of (4.2) and (2.23):

$$\begin{aligned} v^{(h)}(\cdot ,t_n)&= P_h\psi _0 + \frac{1}{2\pi \mathrm {i}}\int _{\varGamma _{\theta ,\kappa }} e^{zt_n}z^{-1} (z^\alpha - \Delta _h)^{-1} \Delta _hP_h\psi _0 \mathrm{d}z , \end{aligned}$$
(B.1)
$$\begin{aligned} v_n^{(h)}&=P_h\psi _0 + \frac{1}{2\pi i}\int _{\varGamma _{\theta ,\kappa }^{(\tau )}} e^{t_nz}e^{-z\tau } \delta (e^{-z\tau })^{-1} \big (\delta (e^{-z\tau })^{\alpha }-\Delta _h\big )^{-1} \Delta _h P_h\psi _0 \, \mathrm{d}z , \end{aligned}$$
(B.2)

which are used in (4.18) in estimating the error of temporal discretization.

In fact, (B.1) is a consequence of (A.3): replacing E(t) by \(E^{(h)}(t)\) and substituting \(\phi =P_h\psi _0\) yield

$$\begin{aligned} \begin{aligned} v^{(h)}(\cdot ,t_n)&=\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{\alpha -1} (z^\alpha -\Delta _h )^{-1}P_h\psi _0\, \mathrm{d}z \\&=\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{-1}(z^\alpha -\Delta _h+\Delta _h) (z^\alpha -\Delta _h )^{-1}P_h\psi _0\, \mathrm{d}z \\&=\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{-1} P_h\psi _0\, \mathrm{d}z + \frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{-1}(z^\alpha -\Delta _h )^{-1}\Delta _hP_h\psi _0\, \mathrm{d}z \\&=P_h\psi _0 + \frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{-1}(z^\alpha -\Delta _h )^{-1}\Delta _hP_h\psi _0\, \mathrm{d}z , \end{aligned} \end{aligned}$$

where we have used the identity \(\frac{1}{2\pi \mathrm{i}}\int _{\varGamma _{\theta ,\kappa }}e^{zt} z^{-1} \mathrm{d}z =1\) (i.e., the inverse Laplace transform of \(z^{-1}\) is 1).

It remains to prove (B.2). To this end, we rewrite (2.23) as

$$\begin{aligned} \bar{\partial }_\tau (v_n^{(h)}-P_h\psi _0) -\Delta _h \bar{\partial }_\tau ^{1-\alpha } (v_n^{(h)}-P_h\psi _0) = \Delta _h \bar{\partial }_\tau ^{1-\alpha } (P_h\psi _0)_n , \end{aligned}$$
(B.3)

where \(\bar{\partial }_\tau ^{1-\alpha } (P_h\psi _0)_n:=\frac{1}{\tau ^{1-\alpha }}\sum _{j=1}^n b_{n-j} P_h\psi _0\). Since we are only interested in the solutions \(v_n^{(h)}\), \(n=1,\dots ,N\), we define

$$\begin{aligned} \widetilde{v}_n^{(h)} =\left\{ \begin{array}{ll} v_n^{(h)} &{}\quad 1\le n\le N,\\ P_h\psi _0 &{}\quad n\ge N+1 , \end{array}\right. \end{aligned}$$

which satisfies the equation

$$\begin{aligned} \bar{\partial }_\tau (\widetilde{v}_n^{(h)}-P_h\psi _0) -\Delta _h \bar{\partial }_\tau ^{1-\alpha } (\widetilde{v}_n^{(h)}-P_h\psi _0) = \Delta _h \bar{\partial }_\tau ^{1-\alpha } (P_h\psi _0)_n + g_n , \end{aligned}$$
(B.4)

with \(g_n=0\) for \(1\le n\le N\). The right-hand side of (B.4) differs from (B.3) only for \(n\ge N+1\), that

$$\begin{aligned} \Vert g_n\Vert\le & {} \Vert \Delta _h \bar{\partial }_\tau ^{1-\alpha } (\widetilde{v}_n^{(h)}-P_h\psi _0)\Vert +\Vert \Delta _h \bar{\partial }_\tau ^{1-\alpha } (P_h\psi _0)_n \Vert \nonumber \\\le & {} \frac{1}{\tau ^{1-\alpha }}\sum _{j=1}^N |b_{n-j}|\Vert \widetilde{v}_j^{(h)}-P_h\psi _0\Vert +\frac{1}{\tau ^{1-\alpha }}\sum _{j=1}^n |b_{n-j}|\Vert P_h\psi _0\Vert \nonumber \\\le & {} C\tau ^{\alpha -1}\left( \sum _{j=1}^N |b_{n-j}|+\sum _{j=1}^n |b_{n-j}|\right) \nonumber \\\le & {} C\tau ^{\alpha -1}n^{\alpha -1}, \end{aligned}$$

as \(n\rightarrow \infty \). Thus \(\sum _{n=N+1}^\infty g_n\zeta ^n\) is an analytic function of \(\zeta \) for \(|\zeta |<1\).

By (2.8), summing up (B.4) times \(\zeta ^n\) for \(n=1,2,\dots \), yields

$$\begin{aligned} \bigg (\frac{1-\zeta }{\tau }-\bigg (\frac{1-\zeta }{\tau }\bigg )^{1-\alpha }\Delta _h \bigg ) \sum _{n=1}^\infty (\widetilde{v}_n^{(h)}-P_h\psi _0)\zeta ^n= \Delta _h \bigg (\frac{1-\zeta }{\tau }\bigg )^{1-\alpha } \frac{\zeta }{1-\zeta } P_h\psi _0 + \sum _{n=N+1}^\infty g_n\zeta ^n , \end{aligned}$$

which implies

$$\begin{aligned} \sum _{n=1}^\infty (\widetilde{v}_n^{(h)}-P_h\psi _0)\zeta ^n= & {} \bigg (\frac{1-\zeta }{\tau }\bigg )^{-1} \bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h \bigg ) ^{-1}\Delta _h P_h\psi _0 \frac{\zeta }{\tau } \\&+ \bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha -1}\bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h \bigg )^{-1}\sum _{n=N+1}^\infty g_n\zeta ^n . \end{aligned}$$

For \(\kappa >0\) and \(\varrho _\kappa =e^{-(\kappa +1)\tau } \in (0,1)\), the Cauchy integral formula implies that

$$\begin{aligned}&\widetilde{v}_n^{(h)}-P_h\psi _0 \nonumber \\&\quad = \frac{1}{2\pi i}\int _{|\zeta |=\varrho _\kappa } \zeta ^{-n-1}\sum _{n=1}^\infty (v_n^{(h)}-P_h\psi _0)\zeta ^n \mathrm{d}\zeta \nonumber \\&\quad = \frac{1}{2\pi i}\int _{|\zeta |=\varrho _\kappa } \zeta ^{-n} \bigg (\frac{1-\zeta }{\tau }\bigg )^{-1} \bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h\bigg )^{-1}\frac{1}{\tau }\Delta _h P_h\psi _0 \mathrm{d}\zeta \nonumber \\&\quad \quad + \frac{1}{2\pi i}\int _{|\zeta |=\varrho _\kappa } \bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha -1}\bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h \bigg )^{-1}\sum _{m=N+1}^\infty g_m\zeta ^{m-n-1} \mathrm{d}\zeta . \end{aligned}$$
(B.5)

For \(1\le n\le N\) the function \( \big (\frac{1-\zeta }{\tau }\big )^{\alpha -1}\big (\big (\frac{1-\zeta }{\tau }\big )^{\alpha }-\Delta _h \big )^{-1}\sum _{m=N+1}^\infty g_m\zeta ^{m-n-1} \) is analytic in \(|\zeta |<1\). Consequently, Cauchy’s integral theorem implies

$$\begin{aligned} \frac{1}{2\pi i}\int _{|\zeta |=\varrho _\kappa } \bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha -1}\bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h \bigg )^{-1}\sum _{m=N+1}^\infty g_m\zeta ^{m-n-1} \mathrm{d}\zeta =0 . \end{aligned}$$

Substituting this identity into (B.5) yields, for \(1\le n\le N\),

$$\begin{aligned}&\widetilde{v}_n^{(h)}-P_h\psi _0\nonumber \\&\quad =v_n^{(h)}-P_h\psi _0 \nonumber \\&\quad = \frac{1}{2\pi i}\int _{|\zeta |=\varrho _\kappa } \zeta ^{-n} \bigg (\frac{1-\zeta }{\tau }\bigg )^{-1} \bigg (\bigg (\frac{1-\zeta }{\tau }\bigg )^{\alpha }-\Delta _h\bigg )^{-1}\frac{1}{\tau }\Delta _h P_h\psi _0 \mathrm{d}\zeta \nonumber \\&\quad = \frac{1}{2\pi i}\int _{\varGamma ^\tau }e^{t_nz}e^{-z\tau } \bigg (\frac{1-e^{-\tau z}}{\tau }\bigg )^{-1} \bigg (\bigg (\frac{1-e^{-\tau z}}{\tau }\bigg )^{\alpha }-\Delta _h\bigg )^{-1} \Delta _h P_h\psi _0 \, \mathrm{d}z \nonumber \\&= \frac{1}{2\pi i}\int _{\varGamma ^\tau }e^{t_nz}e^{-z\tau } \delta (e^{-z\tau })^{-1} \big (\delta (e^{-z\tau })^{\alpha }-\Delta _h\big )^{-1} \Delta _h P_h\psi _0 \, \mathrm{d}z, \end{aligned}$$
(B.6)

where we have used the change of variable \(\zeta =e^{-z\tau }\), which converts the path of integration to the contour

$$\begin{aligned} \varGamma ^\tau =\left\{ z=\kappa +1+\mathrm {i} y: \, y\in {\mathbb R}\quad \text {and}\quad |y|\le {\pi }/{\tau } \right\} . \end{aligned}$$

The angle condition (3.4) and [3, Theorem 3.7.11] imply that the integrand on the right-hand side of (B.6) is analytic in the region

$$\begin{aligned} \Sigma _{\theta ,\kappa }^\tau = \Big \{z\in {\mathbb C} : |\mathrm{arg}(z)|\le \theta ,\quad |z|\ge \kappa ,\quad |\mathrm{Im}(z)|\le \frac{\pi }{\tau },\quad \mathrm{Re}(z)\le \kappa +1 \Big \} , \end{aligned}$$

enclosed by the four paths \(\varGamma ^\tau \), \(\varGamma _{\theta ,\kappa }^{(\tau )}\) and \({\mathbb R}\pm \mathrm {i}\pi /\tau \), where \(\varGamma _{\theta ,\kappa }^{(\tau )} \, =\left\{ z\in \varGamma _{\theta ,\kappa } : |\mathrm{Im}(z)|\right. \left. \le \frac{\pi }{\tau }\right\} \). Then Cauchy’s theorem allows us to deform the integration path from \(\varGamma ^\tau \) to \(\varGamma _{\theta ,\kappa }^{(\tau )}\) in the integral (B.6) (the integrals on \({\mathbb R}\pm \mathrm {i}\pi /\tau \) cancels each other). This yields the desired representation (B.2).

Some inequalities

In this appendix, we prove the following two inequalities:

$$\begin{aligned}&C_0^{\#}|z|\tau \le |1-e^{z\tau }|\le C_1^{\#}|z|\tau ,&\forall \, z\in \varGamma _{\theta ,\kappa }^{(\tau )}, \end{aligned}$$
(C.1)
$$\begin{aligned}&|1-e^{z\tau }|\le C|z|^{1/q}\tau ^{1/q},&\forall \, z\in \varGamma _{\theta ,\kappa },\quad 1\le q\le \infty , \end{aligned}$$
(C.2)

which have been used in (3.27), (4.26) and (4.32).

Proof of (C.1)

Note that

$$\begin{aligned} \varGamma _{\theta ,\kappa }^{(\tau )}&=\left\{ z\in \mathbb {C}: |z|=\kappa ,\, |\arg z|\le \theta \right\} \cup \left\{ z\in \mathbb {C}: z=\rho e^{\pm \mathrm{i}\theta }, \rho \ge \kappa , |\mathrm{Im}(z)|\le \frac{\pi }{\tau } \right\} \nonumber \\&=: \varGamma _{\theta ,\kappa }^{(\tau ),1}\cup \varGamma _{\theta ,\kappa }^{(\tau ),2}. \end{aligned}$$
(C.3)

For \(z\in \varGamma _{\theta ,\kappa }^{(\tau )}\) we have \(|z|\tau \le \pi /\sin (\theta )\). Since \(|z|\tau \) is bounded, the following Taylor expansion holds:

$$\begin{aligned}&1-e^{z\tau } =-z\tau +O(|z|^2\tau ^2) , \end{aligned}$$
(C.4)

which implies

$$\begin{aligned} |1-e^{z\tau }| \le C_1^{+}|z|\tau ,\quad \text {if}\,\,\,z\in \varGamma _{\theta ,\kappa }^{(\tau )}. \end{aligned}$$

This proves the right-half inequality of (C.1).

From (C.4) we also see that there exists a small constant \(\gamma \) such that

$$\begin{aligned} C_0^{+}|z|\tau \le |1-e^{z\tau }| ,\quad \text {if}\,\,\,z\in \varGamma _{\theta ,\kappa }^{(\tau )},\,\,\, |z|\tau <\gamma . \end{aligned}$$
(C.5)

If \(|z|\tau \ge \gamma \), then the following inequality holds for \(\theta \) satisfying the condition of Lemma 1:

$$\begin{aligned} \gamma \le |z|\tau \le \frac{\pi }{\sin (\theta )}\le \pi \sqrt{1+4/\pi ^2} \le \frac{3}{2}\pi . \end{aligned}$$

Since the function \(g(w):=|1-e^{w}|\) is not zero for \(\gamma \le |w| \le \frac{3}{2}\pi \), the function g(w) must have a positive minimum value \(\varpi \) for \(\gamma \le |w| \le \frac{3}{2}\pi \), i.e., \(g(w)\ge \varpi \). Consequently, we have

$$\begin{aligned} \begin{aligned}&\varpi \frac{\sin (\theta )}{\pi }|z|\tau \le \varpi \le |1-e^{z\tau }| ,&\text {if}\,\,\,z\in \varGamma _{\theta ,\kappa }^{(\tau )} ,\,\,\, |z|\tau \ge \gamma , \end{aligned} \end{aligned}$$
(C.6)

where we have used the inequality \(\frac{\sin (\theta )}{\pi }|z|\tau \le 1\) in the last inequality. Combining (C.5) and (C.6) yields (C.1).\(\square \)

Proof of (C.2)

If \(z\in \varGamma _{\theta ,\kappa }\) and \(|z|\tau \le \pi /\sin (\theta )\), then \(z\in \varGamma _{\theta ,\kappa }^{(\tau )}\). In this case, (C.1) implies

$$\begin{aligned}&|1-e^{z\tau }|\le C|z|\tau&\forall \, z\in \varGamma _{\theta ,\kappa } , \,\,\, |z|\tau \le \pi /\sin (\theta ),\\&|1-e^{z\tau }|\le C&\forall \, z\in \varGamma _{\theta ,\kappa } , \,\,\, |z|\tau \le \pi /\sin (\theta ) . \end{aligned}$$

The combination of the two inequalities above yields

$$\begin{aligned}&|1-e^{z\tau }|\le C|z|^{1/q}\tau ^{1/q}&\forall \, z\in \varGamma _{\theta ,\kappa } , \,\,\, |z|\tau \le \pi /\sin (\theta ). \end{aligned}$$
(C.7)

If \(z\in \varGamma _{\theta ,\kappa }\) and \(|z|\tau \ge \pi /\sin (\theta )\), then

$$\begin{aligned} |e^{z\tau }|=e^{-|z|\tau \cos (\theta )}\le e^{-\pi /\tan (\theta )}, \end{aligned}$$

which implies

$$\begin{aligned}&|1-e^{z\tau }|\le 1+e^{-\pi /\tan (\theta )} \le 2 \le 2 \bigg (\frac{\sin (\theta )}{\pi }\bigg )^{\frac{1}{q}} |z|^{1/q}\tau ^{1/q}&\forall \, z\in \varGamma _{\theta ,\kappa } , \quad |z|\tau \ge \pi /\sin (\theta ). \end{aligned}$$
(C.8)

Combining (C.7) and (C.8) yields (C.2).\(\square \)

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Gunzburger, M., Li, B. & Wang, J. Convergence of finite element solutions of stochastic partial integro-differential equations driven by white noise. Numer. Math. 141, 1043–1077 (2019). https://doi.org/10.1007/s00211-019-01028-8

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  • DOI: https://doi.org/10.1007/s00211-019-01028-8

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