On a doubly nonlinear PDE with stochastic perturbation

Abstract

We consider a doubly nonlinear evolution equation with multiplicative noise and show existence and pathwise uniqueness of a strong solution. Using a semi-implicit time discretization we get approximate solutions with monotonicity arguments. We establish a-priori estimates for the approximate solutions and show tightness of the sequence of image measures induced by the sequence of approximate solutions. As a consequence of the theorems of Prokhorov and Skorokhod we get a.s. convergence of a subsequence on a new probability space which allows to show the existence of martingale solutions. Pathwise uniqueness is obtained by an $L^1$-method. Using this result, we are able to show existence and uniqueness of strong solutions.

Appendix

Theorem 7.1

Let $a: D \times \mathbb {R}^d \rightarrow \mathbb {R}^d$ and $b: \mathbb {R} \rightarrow \mathbb {R}$ as in Sect. 2. Then the operator $A: W_0^{1,p}(D) \rightarrow W^{-1,p'}(D)$, $Au = -div ~ a( \cdot , \nabla (b^{-1}(u)))$ is pseudomonotone.

Proof

Let $u_n \rightharpoonup u$ in $W_0^{1,p}(D)$ and $\limsup \limits _{n \rightarrow \infty } \langle Au_n,u_n - u\rangle \le 0$.

Then we have

$$\begin{aligned}&\langle Au_n,u_n - u\rangle = \int _D a(x, \nabla (b^{-1}(u_n))) \cdot \nabla (u_n - u)~dx \\&\quad =\int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla u_n) - a(x, (b^{-1})'(u_n)\nabla u) \bigg ) \cdot \nabla (u_n - u)~dx \\&\qquad + \int _D a(x, (b^{-1})'(u_n) \nabla u) \cdot \nabla (u_n - u)~dx \\&\quad =\int _D \frac{1}{(b^{-1})'(u_n)}\bigg ( a(x, (b^{-1})'(u_n) \nabla u_n) - a(x, (b^{-1})'(u_n)\nabla u) \bigg ) \cdot \\&\qquad \cdot ((b^{-1})'(u_n)\nabla u_n - (b^{-1})'(u_n) \nabla u) + \int _D a(x, (b^{-1})'(u_n) \nabla u) \cdot \nabla (u_n - u)~dx \\&\quad \ge \int _D a(x, (b^{-1})'(u_n) \nabla u) \cdot \nabla (u_n - u)~dx \rightarrow 0. \end{aligned}$$

Here we used that $(b^{-1})'>0$, a is monotone and $(b^{-1})'(u_n) \rightarrow (b^{-1})'(u)$ in $L^p(D)$.

Therefore we obtain $\liminf \limits _{n \rightarrow \infty } \langle Au_n, u_n - u \rangle \ge 0$.

By using the assumption we may conclude $\lim \limits _{n \rightarrow \infty } \langle Au_n, u_n - u \rangle =0$.

Now let $w \in W_0^{1,p}(D)$ and set $z=u + t(w-u)$, $t>0$. It follows that $z \rightarrow u$ in $W_0^{1,p}(D)$ for $t \rightarrow 0^+$. We obtain:

$$\begin{aligned}&\langle Au_n - Az, u_n-z \rangle \\&\quad = \int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla u_n) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u_n - z) ~dx\\&\quad = \int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla u_n) - a(x, (b^{-1})'(u_n)\nabla z) \bigg ) \cdot \nabla (u_n - z) ~dx\\&\qquad + \int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u_n - z)~dx\\&\quad \ge \int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u_n - z)~dx. \end{aligned}$$

Therefore it follows

$$\begin{aligned} t \langle Au_n, u-w \rangle&\ge - \langle Au_n, u_n - u\rangle + t \langle Az, u-w \rangle + \langle Az, u_n - u \rangle \\&\quad + \int _D \bigg ( a(x, (b^{-1})'(u_n) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u_n - z)~dx. \end{aligned}$$

Now we take the limit inferior on both sides of the inequality. Since a is monotone we may conclude

$$\begin{aligned}&t \liminf \limits _{n \rightarrow \infty } \langle Au_n, u_n - w \rangle \\&\quad \ge t \langle Az, u-w \rangle + \int _D \bigg ( a(x, (b^{-1})'(u) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u - z)~dx \\&\quad = t \langle Az, u-w \rangle + t \int _D \bigg ( a(x, (b^{-1})'(u) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u - w) ~dx \end{aligned}$$

We divide by t and get

$$\begin{aligned}&\liminf \limits _{n \rightarrow \infty } \langle Au_n, u_n - w \rangle \\&\quad \ge \langle Az, u-w \rangle + \int _D \bigg ( a(x, (b^{-1})'(u) \nabla z) - a(x, (b^{-1})'(z)\nabla z) \bigg ) \cdot \nabla (u - w)~dx \end{aligned}$$

By passing to the limit $t \rightarrow 0^+$ we get:

$$\begin{aligned}&\liminf \limits _{n \rightarrow \infty } \langle Au_n, u_n - w \rangle \\&\quad \ge \langle Au, u-w \rangle + \int _D \bigg ( a(x, (b^{-1})'(u) \nabla u) - a(x, (b^{-1})'(u)\nabla u) \bigg ) \cdot \nabla (u - w) ~dx\\&\quad = \langle Au, u-w\rangle . \end{aligned}$$

Hence, A is pseudomonotone.