Finite Element Approximations of a Class of Nonlinear Stochastic Wave Equations with Multiplicative Noise

Abstract

Wave propagation problems have many applications in physics and engineering, and the stochastic effects are important in accurately modeling them due to the uncertainty of the media. This paper considers and analyzes a fully discrete finite element method for a class of nonlinear stochastic wave equations, where the diffusion term is globally Lipschitz continuous while the drift term is only assumed to satisfy weaker conditions as in Chow (Ann Appl Probab 12(1):361–381, 2002). The novelties of this paper are threefold. First, the error estimates cannot be directly obtained if the numerical scheme in primal form is used. An equivalent numerical scheme in mixed form is therefore utilized and several Hölder continuity results of the strong solution are proved, which are used to establish the error estimates in both $L^2$ norm and energy norms. Second, two types of discretization of the nonlinear term are proposed to establish the $L^2$ stability and energy stability results of the discrete solutions. These two types of discretization and proper test functions are designed to overcome the challenges arising from the stochastic scaling in time issues and the nonlinear interaction. These stability results play key roles in proving the probability of the set on which the error estimates hold approaches one. Third, higher moment stability results of the discrete solutions are proved based on an energy argument and the underlying energy decaying property of the method. Numerical experiments are also presented to show the stability results of the discrete solutions and the convergence rates in various norms.

A Proofs of Hölder Continuity

In this Appendix, we prove the Hölder continuity in time for the strong solution u in various norms in Sect. 2.

Proof of Lemma 3

The SPDE (1.1) leads to

$$\begin{aligned} u_t(t) - h_2=\int _0^t\varDelta ud\zeta +\int _0^tf(u)d\zeta +\int _0^tg(u)dW(\zeta ). \end{aligned}$$

(A.1)

Taking the square, the spatial integral, and the expectation on both sides of (A.1), and then using the triangle inequality, the Schwarz inequality, and Itô isometry, we obtain

$$\begin{aligned}&\mathbb {E}\big [ \Vert u_t(t)\Vert _{L^2}^2\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert h_2 \Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t \varDelta u(\zeta )d\zeta \bigr )^2dx\big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t f(u(\zeta ))d\zeta \bigr )^2dx\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t g(u(\zeta ))dW(\zeta )\bigr )^2dx\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert h_2 \Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _0^t\Vert f(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert g(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert h_2 \Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{2q}}^{2q}d\zeta \big ]+C, \end{aligned}$$

(A.2)

where (1.8) is used in the derivation of the last inequality. For any $s,t \in [0,T]$ with $s < t$, we have

$$\begin{aligned} \mathbb {E}\big [ \Vert u(t)-u(s) \Vert _{L^2}^2 \big ] =\mathbb {E}\big [ \Vert u_t(\xi )\Vert _{L^2}^2 \big ] (t-s)^2 \le C(t-s)^2, \end{aligned}$$

(A.3)

where $\xi \in (s,t)$ and

$$\begin{aligned} C&= C\mathbb {E}\big [ \Vert h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{2q}}^{2q}d\zeta \big ]+C. \end{aligned}$$

This finishes the proof of the lemma. $\square$

Proof of Lemma 4

Note that $v = u_t$. By (1.1), for any $s,t \in [0,T]$ with $s < t$, we have

$$\begin{aligned} v(t)-v(s)=\int _s^t\varDelta ud\zeta +\int _s^tf(u)d\zeta +\int _s^tg(u)dW(\zeta ). \end{aligned}$$

(A.4)

Taking the square, the spatial integral, and the expectation on both sides of (A.4), and then using the triangle inequality, the Schwarz inequality, and Itô isometry, we obtain

$$\begin{aligned}&\mathbb {E}\big [ \Vert v(t)-v(s) \Vert _{L^2}^2 \big ] \nonumber \\&\quad \le C\mathbb {E}\big [ \int _s^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ](t-s)\nonumber \\&\qquad +C\mathbb {E}\big [ \int _s^t\Vert f(u(\zeta ))\Vert _{L^2}^2d\zeta \big ](t-s)+C\mathbb {E}\big [\int _s^t\Vert g(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]\nonumber ,\\&\quad \le C\mathbb {E}\big [ \int _s^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ](t-s)\nonumber \\&\qquad +C\mathbb {E}\big [ \int _s^t\Vert u(\zeta )\Vert _{L^{2q}}^{2q}d\zeta \big ](t-s)+C\mathbb {E}\big [\int _s^t\Vert u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C(t-s)\nonumber ,\\&\quad \le C(t-s), \end{aligned}$$

(A.5)

where

$$\begin{aligned} C =C\mathbb {E}\big [ \int _s^t\Vert \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [ \int _s^t\Vert u(\zeta )\Vert _{L^{2q}}^{2q}d\zeta \big ]+C\sup _{s \le \zeta \le t}\mathbb {E}\big [\Vert u(\zeta )\Vert _{L^2}^2\big ]+C, \end{aligned}$$

and this finishes the proof of the lemma. $\square$

Proof of Lemma 5

From the Eq. (1.1), we get

$$\begin{aligned} u_t(t)-h_2=\int _0^t\varDelta ud\zeta +\int _0^tf(u)d\zeta +\int _0^tg(u)dW(\zeta ). \end{aligned}$$

(A.6)

Taking the gradient, the square, the spatial integral, and the expectation on both sides of (A.6), and then using the triangle inequality, the Schwarz inequality, and Itô isometry, we obtain

$$\begin{aligned}&\mathbb {E}\big [ \Vert \nabla u_t(t)\Vert _{L^2}^2\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla \varDelta u(\zeta )d\zeta \bigr )^2dx\big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla f(u(\zeta ))d\zeta \bigr )^2dx\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla g(u(\zeta ))dW(\zeta )\bigr )^2dx\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _0^t\Vert \nabla f(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert \nabla g(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert \nabla u(\zeta )\Vert _{L^4}^4d\zeta \big ]+C. \end{aligned}$$

(A.7)

Therefore, for any $s,t \in [0,T]$ with $s < t$, we have

$$\begin{aligned} \mathbb {E}\big [ \Vert \nabla (u(t)-u(s)) \Vert _{L^2}^2 \big ] = \mathbb {E}\big [ \Vert \nabla u_t(\xi )\Vert _{L^2}^2 \big ] (t-s)^2 \le C(t-s)^2, \end{aligned}$$

(A.8)

where $\xi \in (s,t)$ and

$$\begin{aligned} C&= C\mathbb {E}\big [ \Vert h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ] + C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ]\\&\quad +C\mathbb {E}\big [\int _0^t\Vert \nabla u(\zeta )\Vert _{L^4}^4d\zeta \big ]+C. \end{aligned}$$

This finishes the proof of the lemma. $\square$

Proof of Lemma 6

From the SPDE (1.1), for any $s,t \in [0,T]$ with $s < t$, we have

$$\begin{aligned} v(t)-v(s)=\int _s^t\varDelta ud\zeta +\int _s^tf(u)d\zeta +\int _s^tg(u)dW(\zeta ). \end{aligned}$$

(A.9)

Taking the gradient, the square, the spatial integral, and the expectation on both sides of (A.6), and then using the triangle inequality, the Schwarz inequality, and Itô isometry, we obtain

$$\begin{aligned}&\mathbb {E}\big [ \Vert \nabla (v(t)-v(s)) \Vert _{L^2}^2 \big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \int _s^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ](t-s)\nonumber \\&\qquad +C\mathbb {E}\big [ \int _s^t\Vert \nabla f(u(\zeta ))\Vert _{L^2}^2d\zeta \big ](t-s)+C\mathbb {E}\big [\int _s^t\Vert \nabla g(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]\nonumber ,\\&\quad \le C\mathbb {E}\big [ \int _s^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ](t-s)\nonumber \\&\qquad +C\mathbb {E}\big [ \int _s^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ](t-s)+C\mathbb {E}\big [\int _s^t\Vert \nabla u(\zeta )\Vert _{L^4}^4d\zeta \big ]+C(t-s)\nonumber ,\\&\quad \le C(t-s), \end{aligned}$$

(A.10)

where

$$\begin{aligned} C&=C\mathbb {E}\big [ \int _s^t\Vert \nabla \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [ \int _s^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ]\\&\quad +C\sup _{s \le \zeta \le t}\mathbb {E}\big [\Vert \nabla u(\zeta )\Vert _{L^4}^4\big ]+C. \end{aligned}$$

This finishes the proof of the lemma. $\square$

Proof of Lemma 7

Again, from the Eq. (1.1), we get

$$\begin{aligned} u_t(t) - h_2=\int _0^t\varDelta ud\zeta +\int _0^tf(u)d\zeta +\int _0^tg(u)dW(\zeta ). \end{aligned}$$

(A.11)

Taking the Hessian, the square, the spatial integral, and the expectation on both sides of (A.6), and then using the triangle inequality, the Schwarz inequality, and Itô isometry, we obtain

$$\begin{aligned}&\mathbb {E}\big [ \Vert \nabla ^2 u_t(t)\Vert _{L^2}^2\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla ^2 h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla ^2 \varDelta u(\zeta )d\zeta \bigr )^2dx\big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla ^2 f(u(\zeta ))d\zeta \bigr )^2dx\big ]+C\mathbb {E}\big [ \int _{\mathcal {D}}\bigl (\int _0^t\nabla ^2 g(u(\zeta ))dW(\zeta )\bigr )^2dx\big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla ^2 h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla ^2 \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _0^t\Vert \nabla ^2 f(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert \nabla ^2 g(u(\zeta ))\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\quad \le C\mathbb {E}\big [ \Vert \nabla ^2 h_2\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla ^2 \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert \nabla ^2 u(\zeta )\Vert _{L^4}^4d\zeta \big ]\nonumber \\&\qquad +C\mathbb {E}\big [\int _0^t\Vert \nabla u(\zeta )\Vert _{L^8}^8d\zeta \big ]+C. \end{aligned}$$

(A.12)

Therefore, for any $s,t \in [0,T]$ with $s < t$, we have

$$\begin{aligned} \mathbb {E}\big [ \Vert \nabla ^2 (u(t)-u(s)) \Vert _{L^2}^2 \big ]&= \mathbb {E}\big [ \Vert \nabla ^2 u_t(\xi )\Vert _{L^2}^2 \big ] (t-s)^2\nonumber \\&\le C(t-s)^2, \end{aligned}$$

(A.13)

where $\xi \in (s,t)$ and

$$\begin{aligned} C&= C\mathbb {E}\big [ \Vert \nabla ^2 u_t(0)\Vert _{L^2}^2\big ]+C\mathbb {E}\big [ \int _0^t\Vert \nabla ^2 \varDelta u(\zeta )\Vert _{L^2}^2d\zeta \big ]+C\mathbb {E}\big [ \int _0^t \Vert u(\zeta )\Vert _{L^{4(q-1)}}^{4(q-1)}d\zeta \big ]\\&\quad +C\mathbb {E}\big [\int _0^t\Vert \nabla ^2 u(\zeta )\Vert _{L^4}^4d\zeta \big ]+C\mathbb {E}\big [\int _0^t\Vert \nabla u(\zeta )\Vert _{L^8}^8d\zeta \big ]+C. \end{aligned}$$

This finishes the proof of the lemma. $\square$