76.6. EXAMPLES 2581

Combining the two integrals over S yields∫S(bnt − (v,i (+)− v,i (−))ni)φ =

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

by assumption. Therefore, including f , we obtain∫Q−β (v)φ t + v,iφ ,i−

∫U

β (v(x,0))φ (x,0) =∫

Qf φ

which implies, using the initial condition∫U

β (v0)φ (x,0)+∫

Q

((β (v))′ φ + v,iφ ,i

)−∫

Uβ (v(x,0))φ (x,0) =

∫Q

f φ

Regard β as a maximal monotone graph and let α (t) ≡ β−1 (t) . Thus α is single valued.

It just has a horizontal place corresponding to the jump in β . Then let w = β (v) so thatv = α (w) .

α(w)

w

Then in terms of w, the above equals∫U

w0φ (x,0)+∫

Q

(w′φ +α (w),i φ ,i

)−∫

Uw(x,0)φ (x,0) =

∫Q

f φ

and so this simplifies tow′−∆(α (w)) = f , w(0) = w0

where α maps onto R and is monotone and satisfies

(α (r1)−α (r2))(r1− r2) ≥ 0, |α (r)| ≤ m |r| ,|α (r1)−α (r2)| ≤ m |r1− r2| ,α (r)r ≥ δ |r|2

for some δ ,m > 0. Then K−1 (α (w)) = u where u is the original dependent variable.Obviously, the original function k could have had more than one jump and you wouldhandle it the same way by defining β to be like K−1 except for having appropriate jumps atthe values of K (u) corresponding to the jumps in k. This explains the following example.

Example 76.6.2 It can be shown that the Stefan problem can be reduced to the considera-tion of an equation of the form

wt −∆(α (w)) = f , w(0) = w0

where α : L2(0,T,L2 (U)

)→ L2

(0,T,L2 (U)

)is monotone hemicontinuous and coercive,

α being a single valued function. Here f is the same which occurred in the original partialdifferential equation

ut −∑i

∂

∂xi

(k (u)

∂u∂xi

)= f