145
~ r h=-jGh
dt 1n
"I
(1.3.1a)
-~ (meas(D) + IIGllu(E+»):::; IlvI1200( . 2(ro)2) +2sup r hlogh
(1.3.1b)
e
L
O,T,L"
t 1n
+ Ilvll~2(O,T;I') (B - 2Ci(llwll~2 (o,T;L'(n)2) + 2Dll w ilLoo (o,T;U(n)2))
- Cllvll 00 ( . 2(ro)2)) + 2 r Ghlogh:::; Cst
L
O,T,L"
1E+
K - CIIvIIL2(n)2 > 0
(1.3.1c)
h?,O
(I.3.1d)
Proof.-.
The result (I.3.1a) comes directly from the integration of the mass equation.
The result (I.3.1d) is a classical result: If we consider the path (7) defined by the
equation : = U where the data is x(t = T) = z if T =1= 0 or x(t = 0) = Xo if T = O. The
solution of this problem is noted x( t, T, z) and it verifies:
x( T, T, z) = z if T =1= 0
x(O, O,z) = Xo itT = 0
We solve the mass equation along the curves x(t, T, z) with the condition h(xo) fixed
or h(z) fixed. These data are given by the initial conditions ho(x) or the boundary condition
h=J1, on 2::-.
lU
b'
dh
hd'
.
"eo tam: dt =- IVU,l.e.
dh(t,x(t))
dt
= -h(t,x(t)) (divu) (t,x(t))
Then the solution of this problem (H) is
h(t,X(t,T,Z)) = C.e- !:tdivu)(€,x(€,r,z»)d€
where C is a constant obtained by :
h(T,X(T,T,Z))=J1,?'O itT=I=OandZE"Yh(O,x(O,O,z)) = ho ?' 0 itT = 0 and z En
That proves that C is a positive constant and then h is positive.
To get (I.3.1b) the difficulty comes from the mass equation; we have not div u = 0
as in the Stokes problem, and the only estimate onto his h bounded into L'(D).
When we change the test functions in the weak problem by v, we have two terms that
are not necessarily bounded: