76.6. EXAMPLES 2579

Also note that from the integral equation,

(−∆)−1(

u(t)−u0−∫ t

0ΦdW

)+∫ t

0u |u|p−2 ds = (−∆)−1

∫ t

0f ds

and so, since (−∆) is the Riesz map on H10 (U) , the integral equation above shows that off

a set of measure zero,∫ t

0u |u|p−2 ds = (−∆)−1

(∫ t

0f ds−

(u(t)−u0−

∫ t

0ΦdW

))∈ L2 (0,T,H1

0 (U)∩H2 (U))

by elliptic regularity results. Without that stochastic integral, one could assert that |u|p−2

2 u∈L2(0,T,H1

0 (U)). This is shown in [23]. However, it appears that no such condition can

be obtained here because of the nowhere differentiability of the stochastic integral, even ifmore is assumed on u0 and Φ.

Also in this reference is a treatment of the Stefan problem. The Stefan problem involvesa partial differential equation

ut −∑i

∂

∂xi

(k (u)

∂u∂xi

)= f , on U× [0,T ]≡ Q

for (x, t) /∈ S where u is the temperature and k (u) has a jump at σ and S is given by u(x, t) =σ . It is assumed that 0 < k1 ≤ k (r)≤ k2 < ∞ for all r ∈R. For example, its graph could beof the form

σ u

k(u)

On S there is a jump condition which is assumed to hold. Namely

bnt − (k (u+)u,i (+)− k (u−)u,i (−))ni = 0

where the sum is taken over repeated indices and b > 0. u(+) is the “limit” as (x′, t ′)→(x, t) ∈ S where (x′, t ′) ∈ S+,u(−) defined similarly. Also n will denote the unit normalwhich goes from S+ ≡ {(x, t) : u(x, t)> σ} toward S− ≡ {(x, t) : u(x, t)< σ}.

n =(nt ,nx1 , · · · ,nxn)

In addition, there is an initial condition and boundary conditions

u(x,0) = u0 (x) /∈ S, u(x, t) = 0 on ∂U.

The idea is to obtain a variational formulation of this thing. To do this, let K (r)≡∫ r

0 k (s)ds.Thus in the case of the above picture, the graph of K (r) would look like

σ

K(u)

u