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Degenerate parabolic stochastic partial differential equations

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M. Hofmanová / Stochastic Processes and their Applications 123 (2013) 4294–4336 4301<br />

Proof. In order to prove this claim, we denote by σ i the ith row of σ . Let us fix test functions<br />

ψ 1 ∈ C ∞ (T N ), ψ 2 ∈ Cc ∞(R) and define θ(ξ) = ξ<br />

−∞ ψ 2(ζ ) dζ . We denote by ⟨·, ·⟩ x the duality<br />

between the space of distributions over T N and C ∞ (T N ). It holds<br />

<br />

<br />

σ i v<br />

∇g, ψ 1 ψ 2 = − div(σ i ψ 1 ), ψ 2 (ξ) dξ = − div σ i <br />

ψ 1 , θ(v)<br />

x<br />

If the following was true<br />

= σ i ∇θ(v), ψ 1<br />

x .<br />

−∞<br />

σ i ∇θ(v) = θ ′ (v) σ i ∇v in D ′ (T N ), (12)<br />

we would obtain<br />

<br />

σ i ∇g, ψ 1 ψ 2<br />

<br />

=<br />

<br />

θ ′ (v) σ i ∇v, ψ 1<br />

x = σ i ∇v δ v=ξ , ψ 1 ψ 2<br />

<br />

and the proof would be complete.<br />

Hence it remains to verify (12). Towards this end, let us consider an approximation to the<br />

identity on T N , denoted by (ϱ τ ). To be more precise, let ˜ϱ ∈ Cc ∞(RN ) be a nonnegative<br />

symmetric function satisfying R N ˜ϱ = 1 and supp ˜ϱ ⊂ B(0, 1/2). This function can be easily<br />

extended to become Z N -periodic, and let this modification be denoted by ¯ϱ. Now it is correct to<br />

define ϱ = ¯ϱ ◦q −1 , where q denotes the quotient mapping q : R N → T N = R N /Z N , and finally<br />

<br />

ϱ τ (x) = 1<br />

τ N ϱ x<br />

.<br />

τ<br />

Since the identity (12) is fulfilled by any sufficiently regular v, let us consider v τ , the<br />

mollifications of v given by (ϱ τ ). We have<br />

σ i ∇θ(v τ ) −→ σ i ∇θ(v) in D ′ (T N ).<br />

In order to obtain convergence of the corresponding right hand sides, i.e.<br />

θ ′ (v τ ) σ i ∇v τ −→ θ ′ (v) σ i ∇v in D ′ (T N ),<br />

we employ similar arguments as in the commutation lemma of DiPerna and Lions (see<br />

[10, Lemma II.1]). Namely, since σ i (∇v) ∈ L 2 (T N ) it is approximated in L 2 (T N ) by its<br />

mollifications [σ i ∇v] τ . Consequently,<br />

θ ′ (v τ ) σ i ∇v τ −→ θ ′ (v) σ i ∇v in D ′ (T N ).<br />

Thus, it is enough to show that<br />

<br />

θ ′ (v τ ) σ i ∇v τ − σ i ∇v τ<br />

−→ 0 in D ′ (T N ). (13)<br />

It holds<br />

σ i (x)∇v τ (x) − σ i ∇v τ (x)<br />

<br />

<br />

= v(y) σ i <br />

(x)(∇ϱ τ )(x − y) dy + v(y)div y σ i (y)ϱ τ (x − y) dy<br />

T N T<br />

<br />

N<br />

= − v(y) σ i (y) − σ i (x) <br />

(∇ϱ τ )(x − y)dy + v(y)div σ i (y) ϱ τ (x − y)dy.<br />

T N T N<br />

x

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