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") and require the test function if to be nonnegative. To define the generalized super-solutions(subsolutions) of the Cauchy problem (1.14), (1.19), it suffices to replace " = " in (1.23) by " < " ( " > " ) and require f to be nonnegative, or replace " = " in (1.20) and (1.22) by " < " ( " > " ) and require f>0,h> 0. Similarly, to define the generalized super-solutions(sub-solutions) of the boundary value problem (1.14), (1.27), (1.28), it suffices to replace " = " in (1.29) by " < " ( " > " ) and require tp > 0 , or replace " = " in (1.30) and (1.31) by " < " ( " > " ) and " > " ( " < " ) respectively and require f > 0. Remark 1.1.2
If u is a generalized solution of the Cauchy problem
11
Introduction
(1.14), (1.19) in Definition 1.1.3, then for any r <E (0,T) there holds
//
[u—+
JJQT
\
A(u)A wherefc= I m — 1 + — I
0 is an arbitrary constant. Its initial value is !
, l/(m-l)
x m — -c(±ca;) 1 +
In fact, the function (1.38) is a classical solution of (1.17) on QT0\ this can be checked immediately.
Existence
and Uniqueness of Solutions:
One Dimensional
Case
13
For any fixed 0 < t < T, the function (1.37) has a compact support \x\2
0} and both Bm(x,t)
and ——B™(x,t) ux%
(i = 1, 2, • • • , N) vanish at the lateral boundary of the domain: 2 ]X]
_ 2mNt2k/N ~ jfe(m-l) '
Similarly, we can check that the function (1.39) is a generalized solution of the equation (1.17) in one dimension. 1.2
Existence and Uniqueness of Solutions: One Dimensional Case
In this section, we study the Cauchy problem for the filtration equation (1.14) in one dimension: du _ dt ~
d2A(u) dx* '
[Z l)
-
u(x,0) = u0(x),
(2.2)
where uQ(x) > 0 is a locally integrable function on R and A(s) £ C 1 [0, oo) satisfies the conditions A(s) > 0, A'(s) > 0, for s > 0, A(0) = A'(Q) = 0.
(2.3)
Denote Q T = R x ( 0 , T ) . 1.2.1
Uniqueness
of
solutions
We first discuss the uniqueness of generalized solutions of the Cauchy problem. Theorem 1.2.1
The Cauchy problem (2.1), (2.2) has at most one gendA(u) eralized solution u bounded together with the weak derivative —-^—^. ox
14
Newtonian
Filtration
Equations
dA(u) First notice that if both u and — - — are bounded, then by apax proximation, the integral identity in the definition of generalized solutions (Definition 1.1.3) holds for any ip G W1'°°(QT) vanishing when |a;| is large enough and t = T. Such functions will be also called test functions. Now let ui, u2 be generalized solutions of (2.1), (2.2). Then ui and u2 satisfy (1.26). Hence we have Proof.
//S>-»WU(^-^)- ™ for any test function ip. If oo
lim Iln = / / "^•°°
{A(ui) - A(u2))(ui
-
u2)dxdt.
JJQT
Furthermore we can prove lim I2n = 0.
(2.7)
16
Newtonian
Filtration
Equations
dAtuA In fact, from the boundedness of ui and — and the uniform bounded edness of ct'n(x), we have \L2n|
< C (J
\A{Ul) -
A{u2)\dxdt 1/2
-ooin(2.6) yields //
{A(ui) - A(u2))(ui
- u2)dxdt = 0,
JJQT
and hence u\ = u2 a.e. on QT- This completes the proof of our theorem. 1.2.2
Existence
of
solutions
Next we discuss the existence of generalized solutions of the Cauchy problem (2.1), (2.2). • Theorem 1.2.2 Assume thatuo is a nonnegative, continuous and bounded function on K, A(uo) satisfies the Lipschitz condition, A(s) is appropriately smooth and A(s) —> +oo as s —>• +oo. Then for any T > 0 the Cauchy problem (2.1), (2.2) admits a continuous and bounded generalized solution dA(u) u on QT such that — - — is bounded. Moreover, the solution u is classical ox in the domain {(x,t) G Qr,u(x,t) > 0}. Proof. {VQ(X)}
Denote Vo = A(uo) and choose a sequence of smooth functions such that {VQ (X)} uniformly converges to VQ(X) as n —> oo and
!*
^ -» from which it follows by letting n -> oo and noticing that for large n, wn(x) = VQ(X) and wn(x) converges uniformly to VQ = A(UQ) on any finite interval, that u satisfies (1-24), i.e u is a generalized solution of (2.1), (2.2). Finally, we prove that the solution u is classical in {(x, t) € QT, U(X, t) > 0}. Let (x0, t0) £ QT, U(XQ, to) > 0. Then there is a neighborhood U C QT of (xo,io) and constant ao > 0 such that u(x,t) > ao > 0,
for (x,t) G U.
Hence un(x, t) > — > 0
for (x, t) E.U and large n.
This means that for large n the solution un satisfies dun
d (
dun
with a(x,t) = A'(un) being uniformly parabolic on U. From the standard theory for parabolic equations, it follows that for large n, un is uniformly bounded and equi-continuous in C2(U). Thus u eC2(U) and satisfies (2.1) in the classical sense. The proof of Theorem 1.2.2 is complete. •
Existence and Uniqueness of Solutions:
1.2.3
Comparison
One Dimensional
Case
21
theorems
First we have Theorem 1.2.3 Assume the conditions in Theorem 1.2.1 and Theorem 1.2.2. Let Ui and u2 be bounded generalized solutions of the equation (2.1) dA(ui) with bounded weak derivatives — and initial data uoi(x) (i = 1,2). / / ox u oi( ; c ) < UQ2(X) a.e. on K, then u\ < u2 a.e. on QTProof. As in the proof Theorem 1.2.2, we construct a sequence of approximate smooth functions {u^t{x)} (i = 1,2) of {uot(x)} (i = 1,2). However, an additional condition UQ^X) < UQ2(X) is required. Then, by the comparison theorem for classical solutions, we get u™(x,t) < u2(x,t) on Gn. Letting n —> oo and using the uniqueness of generalized solutions yield iii < u2 a.e on QT and complete the proof of our theorem. • Proving comparison theorem in this way seems to be too roundabout. Many approaches applied to prove uniqueness theorem are also adapted in establishing comparison theorem. The proof of Theorem 1.3.1 is an example in this respect. As we will point out in Remark 1.3.1, we can prove Theorem 1.2.4 Let ut G L1(QT) D L°°(Q r ) (i = 1,2) be generalized solutions of (2.1) with initial data Uoi, (i = 1, 2). Ifuoi(x) < uo2{x) a.e on R, then u\ 0)
(2.19)
and other appropriate conditions, where m, M, p are positive constants. It is indicated in [KA3] that if the condition (2.19) is replaced by dA{x,u0{x)) du
^ A/f /n t 2J < M i ( l + z ),
(2.20)
then one can not obtain the global solution in general. The pressure solution of degree two given in §1.1.1 also shows this matter. In fact, in case A(x,t,u) = um (m > 1), the condition (2.20) becomes / A/f \
1
/(m_1)
(l + z 2 ) 1 / ( m - 1 } .
«o(s)0)
and proved the existence and uniqueness of the so-called source-type solutions under appropriate conditions on A(u). 2.Extension to equations of other form. Kalashnikov [KA2] studied equations with absorption term du
d2A(u)
,, ,
where A(u) and ip(u) satisfy A'(u) > 0, V>'(«) > 0, A'(0)>0,ij(u)>0,
for u > 0, toiu>0.
He proved the existence and uniqueness of generalized solutions of the Cauchy problem under some additional conditions on A(u) and I/J(U), the solution obtained is Holder continuous.
Existence and Uniqueness of Solutions:
One Dimensional
Case
25
Kershner [KE1] has studied this kind of equations too. His method can be applied to more general equations d2A(u) dx2
du dt~
Wju) dx
+
+1{U)
-
In another paper, published a little early [KE2], he studied the first boundary value problem for equations of the special form du _ d2um dt dx2
dun _ dx
j
with some positive constants m, n, I. Gilding and Peletier [GP2] considered the Cauchy problem for the equation d2um
8u
m=-dx^
+
dun
.
.,
n
^x- (™>L»>o)
and proved that it admits at most one generalized solution whenever
n > i ( m + l) and it admits a generalized solution if UQ > 0 is bounded and continuous with uam Lipschitz continuous. Soon afterwards these results were extended by Gilding [GI] to more general equations du
m
=
d ( . ,du\ aiu)
dx-{
,.
.du
+b{u)
dx-)
dx-'
where a(u), b(u) are continuous and a(u) > 0 for u > 0, a(0) = 0. He proved the uniqueness for the Cauchy problem under the condition b2(u) = 0(a(u))
(u -»• 0+)
(2.21)
and proved the existence under the conditions that a'(u) and b'(u) are locally integrable , ua'{u),ub'{u) € Lx(0,1) and (x) ruu0ayx)
A(u0(x)) = / Jo
/
\
1
a{r)d
is Lipschitz continuous. It should be pointed out that as a condition to ensure the uniqueness of generalized solutions, (2.21) is unnatural. Wu Dequan [WD] and Dong Guangchang, Ye Qixiao [DY] have ever made efforts to improve this condition. Chen Yazhe [CHI] removed any kind of conditions in which b(u)
26
Newtonian
Filtration
Equations
is controlled by a(u) and established the uniqueness by assuming a(u) to satisfy the following condition: there exist constants 6, m > 1 such that 1 fm\
a(ui) < —,—r < m a.[u2)
m \u2J
for any 0 < u\ < u2 < 6. A substantial progress in the study of uniqueness was made by Zhao Junning [ZHl] who did not require a{u) and b{u) to have any relation and only assumed a(u) > 0 to satisfy the condition that the set E = {s,a{s) = 0} does not include any interior point; uniqueness was proved in L°°(QT)The uniqueness of solutions for equations without any additional assumption except a(u) > 0 is much more difficult to prove. Vol'pert and Hudjaev [VH1] tried to do that in the class of BV functions; however, as pointed out in [WZQl], their proof is incorrect. Based on a deep investigation of functions in BV and in a more general class BVX, Wu Zhuoqun and Yin Jingxue [WYl] completed the proof of uniqueness of solutions in BV. (For details see Chapter 3).
1.3
Existence and Uniqueness of Solutions: Higher Dimensional Case
We are ready to turn to the filtration equation in higher dimension. We will concentrate our attention on its typical case, i.e, equation of the form du
Aum
dt N
with constant m > 1, where A = ^
(3.1)
d2
- ^ , and only discuss the Cauchy
i=i
problem, whose initial value condition is u(a;)0) = uo(a:). 1.3.1
Comparison
theorem
and uniqueness
(3-2) of
solutions
Denote QT = RN x (0,T). Theorem 1.3.1 Let u{ G L1(QT) n L°°{QT) (i = 1,2) be generalized solutions of the Cauchy problem for (3.1) with initial data uoi(x), (i = 1,2). 7/0 / /,
w\Vu™\2dxdT
w(x)\Vu™(x,t)\2dx
+i [
with small e > 0. Hence (m+l)/2>
Jt JWL" \
-VQ^W}
and use (,(x,t) to cut off the function p2, namely to consider the function
42
Newtonian
Filtration
Equations
z = £ 2 p 2 instead of p2. Multiplying (4.5) by £ 2 , we can check that z satisfies d2z\
1 fdz 2\dt-m^dx2Jm.io
—
2 \
0,a > 1) we further obtain
K( a ; ,i)- u „( 2 /, S )r- i k},Bk,P
= {x € Bp;w(x,t)
< k}.
Multiplying (5.3) by (2(w — k)+ and integrating with respect to x over Bp yield (2(w-k)+?^-dx
/
vt
JB„
C2(w-k)+Awdx.
= f JBP
It is clear that
°t
JB„
ft
[ C2!" / JBP a
Aw-k)+
&{k + T)TdTdx
ut Jo p{w-k)+
/•
— / C2 / "t JB0 JO =
(2Xk{w-k)dx,
-£. / 0 1
&{k + T)Tdrdx
JAklP(t)
where Xk(s)=
[' &{k+T)TdT. Jo
Integrating by parts gives (2(w -
/
k)+Awdx
JBP
=
C2V(w - k)+Wwdx - 2 J C(w - fc)+VC • Vwdx
-f JBP
=
- [ JAkJt)
JB0 2
2
< |VH -2/" JAkJt)
C(w - fc)VC • Vwdx.
(5.4)
Regularity of Solutions:
Higher Dimensional
51
Case
Therefore we have d
(*)
0} and C is a constant depending only on N. Proof.
By Holder's inequality we have / \w\2dx < (mesS)1-^-2^
\w\2NKN-Vdx\
( /
Using the embedding theorem yields /
f
(N-2)/2N
\Vw\2NKN-Qdx\ ,(N~2)/2N
f
/
k} and P is a constant depending
Consider the following nonnegative function w(x) = 0
for x G
Bp\AktP,
w(x) = w — k
for x G
Ak!P\A\tP,
w(x) = X — k
for x G A\tP.
Regularity
of Solutions:
Higher Dimensional
53
Case
For almost all s, y £ Bp, we have _ _ w(y)—w(x)=
r\v-*\ QW / — (x + ru)dr, Jo Of where (r,u>) is the spherical coordinates centered at x. Integrating this formula with respect toj/G Bp\AktP and noting that w(y) = 0, we obtain mes
(Bp\Ak,p)w(x
-_f
f
JBP\AK,P
\v-*\ gw
^L (x +
JO
roj)drdy.
dr
Now we estimate the integral on the right hand side:
j
JBp\AkyP
<
~>0,0-wr-
The proof is easy. We leave it to the readers. We need also the following iteration lemma which can be proved by induction. Lemma 1.5.4 bers satisfying
Let {yn} (n = 0,1,2, • • •) be a sequence of positive num-
yn+1
n
,,i+ < cb,ny,
a
where c> 0, 6 > 1, a > 0 are constants. If y0 then lim yn = 0. n—>oo
mes(B / , 1 _ 0 . l P l \yl f c + / 3 h i / , 1 _ o . l P l (t))
>
(l-a1)wKjvpf-(l-61)(l-f71)^wpf
=
h(l-CT^KNP?,
and (i) is proved by choosing b = &i(l — a\)N. (ii) can be proved similarly.
•
Lemma 1.5.6 For any 6\ > 0, there exists s(#i) > 0 such that (i) if fi > k > /J./2, h = fi — k and 1 - aAp2) < -Kjvpf,
mesAk,Pl(to then (° Jtn—aAp2
mesA^h/2,+liPl(t)dtBP+"/2,pAt)^oKlfpir>
t € [ t o — a ^ P >'o—"^4p ]
^5'14)
-s
then s+2 _£
w < 2' + V or /•to
mesBka+uPl
< 0i£ f c 3 + 2 pf + 2 ,
(5.15)
7t0-aBfca+2
w/iere A;s = p, + u>/2s, Bka = r0,
which is possible from (5.19), where k) = fl + —r. Then similar to the 2' derivation from (5.16) to (5.18), we can verify that for s > I > ro,
221+2
0, there exists 6\ = 61(02) > 0 such that (i) if pb> k > p,/2, h = p, — k > 0 and
J 2
mesAktPl{t)dt T > t0 - ap2Bk_h/2p2 and then integrate (5.9) over [r, t]. We easily see that Xk(-^)raesBk_h/2iP2(t)
( # 7 - {Tp^^
-
w1
+
xk(h)mesBk,Pl(T)
(5 25)
-
l mesBk pAt)dL
)l
'
From (5.23), we have, in particular, ft0—ap
Bk_h/i
mes Bk,Pl(t)dt
2
Jtrto-ap
0iBk_h/2pN+2 l
we can conclude the existence of 9\ satisfying (5.24). If mesB k t P l (to — ap2Bk_h/2) = 0, then we take T — to — ap2Bk^h/2 (5.25) and assert that (5.24) holds for all i e [to - ap2Bk_h/2,to\-
in •
Lemma 1.5.8 There exists 62 > 0 such that (i) if p> k > p/2, h = p — k > 0 and max te[t0-aAp2,t0]
mesAklP2(t)
pe, max
mesBkiP2(t)
te[to-aBkp2,
then for any t £ [to —
< 62P2 ,
(5.28)
t0]
aBkp2/4,to],
mesBk„h/2tP2(t)=0; (5.29) 2 in addition, if mesBktP2{to — aBkp ) = 0, then for any t £ [to — aBkp2, to], (5.29) holds. Proof. We only prove (ii); the proof of (i) is similar. Let h h j = k ~ 2 + 2I+T'
k
tj =t
°~
Pi = Ps + —TTi—.
N =
Q(x) = C(x; Pj,Pj+i),
1 4aBkP
m a x
2
3 ,_, o ~ ^p^aBkP >
, m e s Bkj ,Pj (t),
Ij{t) = /
w)(2dx.
Xkj(kj -
J
Bkj,Pj(t)
Since kj > fc/2, by Lemma 1.5.3, we have Xkj(kj -w) < m$'(kj)(kj
- w)2 < 2mBk(kj
- w)2.
Hence Ij(t)Bkj,Pj(t)
Using Lemma 1.5.1, we further derive /,-(*) < CBkp)/N
f
\Vw[2(2dx + CBkp)/N
•
JBkj,Pj(t)
^H.
(5.31)
(Pj-Pj+1)
On the other hand, (5.4)implies
m +U 2
JBkj,Pj(t)
Given t G [tj+i,t0].
Ww\2C]dx < 7 (
h
\{Pj-Pj+l)
]
+ l) ^. J
(5-32)
63
Regularity of Solutions: Higher Dimensional Case (a) If I'jit) > 0, then from (5.32), we have / \Vw\2C]dx < 2 7 ( 'Bki,Pj(t)
k2
+1)
H.
Substituting it into (5.31) yields /,(«) < CBtf™
((27 + »
i
^
~ + 27) •
(5-33)
(b) If IAt) < 0 and for some r e [tj,t], I'J(T) — 0, then we can take r such that Jj(s) < 0 for s € (r,t] and J,-(f) < ^-(r). Since (5.31), (5.32) imply (5.33) for t = r, (5.33) holds for £ too. (c) If for any r G [£j, i], 7j(r) < 0, then from (5.32) we have \ 1
f f \Vw\\]dxdt hi JBkj,Pj(t)
< Ij{tj) + 7 (* - *,-) ( 7 ^—T2 + l ) H\{Pj ~ Pj+i) J
Integrating (5.31) over [tj,t] and using the above inequality yield / ' 7,(r)dr < CBktfN 7tj
hl^)
L
+ ( 2 7 + l ) ^ ( * " * j ) ^ + 2 7 (* - i * ) ^ ' \Pj - Pj+i)
Substituting (5.30) into this inequality and noting that IJ(T) is increasing on [tj,t], we derive that for t € [t J+ i,io], « « ) < C B ^ » « ( ± r ^ v*j+i-*i
+ 7 ei±lSl + 2
(Pj-Pj+ir
7). )
Thus in all cases (5.34) holds for t G [t,+i,io]On the other hand, using (5.3) we have Ii(t) > Xkj(kj - kj+1 )mesB k . + 1 iPj+1 (t) >
^Bkikj - kj+1)2mesBkj+uPj+1
(t),
which combining with (5.34) yields ^ ori 2/JV+l920+l) ( ±™Bk 27 + l 27\ /ij+i < 2G/i/ 2w^' I —+ —+ — . 3 ytj+i-tj (Pj-Pj+1)2 h2j
(5.34,
64
Newtonian
Filtration
Equations
Hence from the definition of tj and pj, we see that the above inequality can be simplified as C12^p2/N+1 M+ ni 1
?
where C\ is a constant depending only on N, m, 7, T. If we set yj = p,jj pN, then the above inequality can be written as
yj+1oo
62 is chosen small enough and complete the proof of (5.29). If mesBk tP2 (to — aBkp2) = 0, then we take tj = to — aB^p2 instead 1 3 of tj = to — -aBkp2 — . „abkp2. The desired conclusion can be easily obtained. • Lemma 1.5.9 There exists s > 0 such that (i) if fi> k> n/2, h = p, — k > 0 and mesAk,Pl(to
- aAp2)
1 - 2"1-(S+3)AS
Nc = 4m3(71-1/"\
and let P°
Pi = —r, if
Qi = QPl = {(a;,i);|a;-a;o| < pi,t0 -a&(pf)pf fii = sup w(x,t),
Jli = inf w(x,t)
Qi
Ljt= in-fit,
^ ( } y-i/m> 16m
3
^1™
\2CT)*J
~
>x
32m3C1-1/m
Thus Qi+i C Qp,/4 and hence u>i+i < Cpf+1. If (5.43) does not hold, then there exists r € [to—aBpf,to — aApf], such 1 that mesB^ i+Wj / 2 ,p li ('r) > -KNp^ and hence 2'
m e s ^ + W j / 2 , P l i < 2 K ivPu or mesA w _ W i / 2 i P l l (r) < -KNp%.
(5.44)
Now we divide the interval [r, to] into K equal parts, such that ^aApf 0 are nonnegative, bounded and Holder continuous, then by a similar argument we can prove that the corresponding generalized solution is Holder continuous down to t = 0.
1.6
1.6.1
Properties of the Free Boundary: Case Finite propagation
of
One Dimensional
disturbances
Consider the equation 8u dt ~
d2A(u) dx2
(6.1)
with initial data u(x,0) = uo(x)
(6.2)
i £ l
We always assume that A(u) e C^O, oo) n C 2 (0, oo), A(u) > 0, A'(u) > 0, A"(u) > 0
for u > 0,
,4(0) = A'(0) = 0, and UQ is nonnegative, bounded and continuous on R with A(uo) satisfying the Lipschitz condition. By virtue of Theorem 1.2.1 and Theorem 1.2.2, the Cauchy problem (6.1), (6.2) admits exactly one nonnegative, bounded and continuous generalized solution u on Q = R x (0, oo) with bounded weak derivative ^—-. ox In §1.4, we have discussed the regularity of generalized solutions for the equation du _ ~dt~
d2um dx2
(6.3)
with m > 1 which corresponds to the slow diffusion. In physics, slow diffusion should imply that the speed of propagation of disturbances is finite. The mathematical description of this fact is that if suppwo is bounded, then for any t > 0, swpp u(x,t) is also bounded. We have the following general result.
72
Newtonian
Theorem 1.6.1
Filtration
Equations
Assume that for any u > 0, tf («) = / -±lds s Jo
< +oo.
(6.4)
Let u be a generalized solution of the Cauchy problem (6.1), (6.2) on Q. If suppuo(x) is bounded, then for any t > 0, supp u(-,t) is also bounded. Proof.
Consider the function of the form u(x,t) = * _ 1 ( c ( c i + ii) - x ) + ) .
It is easy to verify that for any £i > 0, c > 0, u(x, t) is a generalized solution of (6.1) with initial data y~1(c(ct1-x)+).
u(x,0) = Let XQ = sup{suppuo(:r)}. Then
u(x, 0) = uo(x) = 0 < u(x, 0)
for x >
XQ.
Besides, since ty and ^ _ 1 are increasing, we have u{x,t)
=y-1(c{c{t
+
t1)-x0)+)
> ^ _ 1 ( c ( c t i - x0)+) >M = supw, 0 provided ct\ — x0 > 0 and c(cti — xo) > * ( M ) . Hence u(x0, t) <M < u(x0,t)
for t > 0.
Now we apply the comparison theorem on GT = (^0;°°) Theorem 1.2.4 and Remark 1.2.2) and then obtain u(x,t) < u(x,t),
x
(0, T) (see
on GT,
from which it follows that u{x,t) = 0 when x > Xt — c(t + ti), since u(x, t) = 0 for x > Xt = c(t +1\). Notice that in applying the comparison theorem on GT, we need to check u e L1(GT) (see Remark 1.2.2). From Definition 1.1.3 and Remark 1.1.2, for any T G (0,T) and
£)**• In particular, we take
oo
Theorem 1.6.2 means that in case of slow diffusion, disturbances will be propagated to infinite scope, although the speed of propagation is finite. Before proving Theorem 1.6.2, we first prove the following proposition which is also very useful in the sequel. Proposition 1.6.1 Assume that m > 1 and u is the generalized solution of the Cauchy problem (6.3), (6.2)on Q = R x (0, oo). Then ^ > - - , dt ~ t ' dv ^ (m - l)kv dt~ t
(6.6) ' . . [ ' y
in the sense of distributions. Remark 1.6.2 As will be seen from the proof stated below, this proposition is valid for any initial data UQ G L°°(M.N). Proof of Proposition 1.6.1. From the proof of Theorem 1.3.5 and point 4 in §1.3.3, the given generalized solution u can be obtained as the limit of a sequence of classical solutions which are positive, and uniformly bounded on Q and whose derivatives up to second order are bounded. To prove the proposition, we may simply suppose that u is just its approximate smooth solution. We first verify (6.8). Notice that v satisfies dv
,
. d2v
dt-^-1)Vo^
+
(dv^
[9-X
(6 9)
-
Properties of the Free Boundary:
Lw = -
- (m - l)v^
- 2m-
One Dimensional
75
Case
• - - (m + l)W2 = 0.
(6.10)
~ k Clearly the function w = — also satisfies the equation (6.10) on Q. Since w = -r—^ is bounded, we have ox* k wix.e) > — £
provided e > 0 is small enough. Thus we may use the comparison theorem on Rx (e, oo) to assert ~ k w(x,t) > w(x,t) = —
for (x,t) e R x (e, oo),
from which (6.8) follows by virtue of the arbitrariness of e. (6.7) follows from (6.8) and (6.9). To verify (6.6), it suffices to note that du
/m-l\
1 / ( m - 1 )
(82V
1
at
\
J
\ dx2
m — 1
m d2v
v(2-m)/(m-l)
\dxj
ku
^Udx~2^---
Proof of Theorem
( OV \ ^
• 1.6.2.
From (6.6) we have
!(**«>> o. From this and the continuity of u, it is easy to see that for any x &R, tku is increasing in t. Therefore fi(ii) C ft(£2), f o r £ i < £ 2 , in other words, (—l)Xi(t) {i = 1,2) is increasing. Next we prove (6.5). For simplicity we suppose that x\ < 0 < Xi, uo(0) > 0. Then there exist 6 > 0, £o > 0 such that UQ(X)
>
£Q
for |a;| < 6.
76
Newtonian
Filtration
Equations
Consider the function BTtL(x, t) = l}lm-xBm{x,
L{t + T))
{L,T > 0)
where Bm(x,t) is the Barenblatt solution of (6.3) (see (1.37)). It is easy to check by direct calculation that for any L, r > 0, BTIL{X, t) is a generalized solution of (6.3) with initial data Ll/(m-^B(x,LT)
=
BT,L(X,0)
m - 1 x> * X l / ( m _ 1 ) 2 2m(m+l)(Lr) /(m+i)y+y
LV(«-D / / (Lr)V("»+i) ^ Obviously
L 2/(m
0 0 such that L 2/(m
2
-l)
=
l/(m+l)
?m(m+_l)
2 / ( m + 1 )
=
^
m—1 It is easy to see that this is possible. For such L, r > 0, we have BTiL(x,0)L(x,t)
< u(x,t).
In particular
Q(t)
D{X:\X\>
0 for u > 0 and c(0) = 0. Similar to the argument stated in §1.1.2, we can prove the existence and uniqueness of generalized solutions (which can be defined in an obvious way) of the Cauchy problem for (6.11). Also the comparison theorem is valid (cf. [KA2]). In the sequel, we will use these results without proof. Theorem 1.6.3 Let u be a generalized solution of the Cauchy problem (6.11), (6.2) on Q. Ifu0{x) = 0 for \x\ > X > 0 and f ( f
*(£Kj
where &(v) = c($(v)), $(v) = A~1(v), u(x,t) = 0 for \x\ >X,t>0.
dv X such that
Proof. To prove that for some X\ > X, u(x,t) = 0 for x > X, t > 0, it suffices to construct a generalized solution w(x,t) on the domain Gx = {(x, t); x > X, t > 0} such that u(x, t) < w(x, t)
for (x, t) e Gx
(6.13)
and w(x, t) = 0 for x > Xi, t > 0. We try to seek such generalized solution w(x,t) among functions which depend only on x. First we require w(x) to satisfy (6.11) on G x \ G x 1 5 namely, for X < x < X\, d2 •^A(w(x))
fM
= cmj(x)))
= c(w(x))
= ^(j(x)),
(6.14)
78
Newtonian
Filtration
Equations
where j(x) = A(w(x)) and Xi > X is a constant to be determined later. Denote the inverse function of j(s) by J(v). Then dj(x) dx 2 d j(x) dx2
1 J'(v)' _ J"(v) (J'(v))2
dj(x) dx
J"{v) {J'(v))3
( 1 \2(J'{v)y
and (6.14) turns out to be
1 2{J'{v))2
V = *(v).
Integrating this equality yields J(u) =
i, W o *(0 7
d?7
'
(6 15)
'
Prom the above analysis, it is natural to define J(v) by (6.15) and then to consider its inverse function j(x). The condition (6.12) ensures the definition of J(v) for all v > 0. Since J(v) is increasing and J(+oo) = +00, j(x) is well-defined for all x > 0. j ( s ) is a solution of (6.14) for x > 0; so is j(X\ — x) for any X\ and a; < Xi. Hence for any Xi > X, <J>(j(Xi — x)) is a (classical) solution for x < X\. We choose X\ such that u{x,t)<w{X).
(6.16)
This is possible; for example, we may take X\ = X + J(A(M)), M = sup u. Now we define ( $(j(Xi - x)) w(x,t) — w(x) = < [ 0
where
for X < x < Xlt fora;>Xi.
n, dA(w) , ,r . . , , , bince w and — equal zero at x = X1, it is easy to check that a n s a generalized solution of (6.11) on Gx- Using the comparison theorem for u and w on Gx and noticing (6.16) and that uo(x) = 0 for x > X, we arrive at (6.15) and that u(x,t) = 0 for x > X, t > 0. Similarly we can prove that u(x, t) = 0 for x < —X, t > 0. •
Properties of the Free Boundary:
Remark 1.6.3
One Dimensional
79
Case
For equations of the form
d2um
du
the condition (6.12) becomes 1/2
(m + n\ \
mc J
-i
f1
vHm+n)/{2m)dv
J0
< +oo
which is equivalent to n < m. Theorem 1.6.4 Letu(x,t) be a generalized solution of the Cauchy problem (6.11), (6.2) on Q = R x (0, oo). If
t^ T. In this case we will say that extinction occurs for the solution u(x, t) at the time t = T. Proof.
To prove our theorem, we first choose a function w(t) such that ^ p
= -c(tu(t))
forO 0 to be determined later. Integrating (6.18) and using (6.19) we obtain
I
w(t)
dv -rr=T-t,
for 0 < t < T.
(6.20)
Here we notice that the condition (6.17) ensures that the integral / -u Jo
c(y)
is defined for any v > 0 and is increasing in v. Extend the function w(t) defined by (6.20) to (0, oo) with w(t) = 0 for t € [T, oo). Then it is easy to check that w{t) is a generalized solution of (6.11) on (0,oo). If we have u{x,0) =u0{x)
<w(0),
(6.21)
Newtonian
80
Filtration
Equations
then the comparison theorem gives u(x,t) < w(t)
for (x,t) G Q
and hence u(x, t) = 0 for x G R, t > T. Since
Jo
c(y) /•M
for (6.21) to be held, it suffices to take T = The proof is thus completed.
Jo
j
—r^where M > c
iy)
svpuoix). •
Remark 1.6.4 Kalashnikov proved in [KAl] that for the generalized solution of (6.11)to have the property of extinction, the condition (6.17) is also necessary. 1.6.3
Differential
equation
on the free
boundary
In what follows we will further investigate the properties of free boundaries x = d(t) (i = 1,2) of the generalized solution u of the equation (6.3). Naturally we expect the free boundaries to move with the local velocity. Set m
,m— 1
m —1 Then, in view of the equation of state, which we have used to derive the equation (6.3), v is essentially the pressure, and, by Darcy's law, we expect
(x,t)gn x-Ki(t)
where Q = {(x,t) almost true.
G Q;u(x,t)
Theorem 1.6.5
The limits
vx((i(t),t)
=
lim
OX
> 0}. The next result shows that this is
(*,t)en
^
^
OX
(t = l,2),
Properties of the Free Boundary:
exist for allt>0
One Dimensional
Case
81
and
C(t + 0) = -vx(d(t),t)
(i = l,2).
Proof. FVom Theorem 1.2.2, u e C°°(0). By Proposition 1.6.1, for any T > 0, there exists a constant /? depending only on r such that 0^2 >~P
for {x,t) e£l,
t>r
in the sense of distributions. Hence the function f(x,t) rj2 £
= v(x,t) + f3x2
t) £
satisfies 7-—^ > 0. This means that —— is an increasing function of x for oxz ox each fixed t > r. Since for any t > r, u has compact support, it follows from Lemma 1.4.1 that for any finite T > r , — and —— are bounded ox ox df(x t); on n n {{x,t);t e [r,T]}. Thus the limits lim ^ ' (i = 1,2) and (x,t)en dv(x i) lim ——-— (i — 1,2) exist for any t G [r,T\. Since T, r are arbitrary, (a;,t)6fi
these limits exist for any t > 0. We are ready to prove the rest part of the theorem for any t. For simplicity we take t = 0 and treat £ 2 (i) only. Set a = £2(0), vx(a, 0) = a. Then either a = 0 or a < 0. Case 1. a < 0. We will show that for any sufficiently small e > 0, there exists a 6 > 0 depending only on e such that
C2(Ai) -a At
+a < s
(6.22)
dv(x 0) whenever 0 < At < S. Since lim — ' = a, it follows that for any 0 < e < —a, there exists a do > 0 such that a-
dv(x,0) e < — ^ — - < a + e,
. . (a-60 < x < a).
Using the mean value theorem we get (a + e)(x - a) < v(x, 0) < (a - e)(x - a)
82
Newtonian
Filtration
Equations
and hence u(x,0) > u(x,0)
a — So,0 < t < T
and from this it is easy to see that a - (a + e)t < (2(t) -YPC'{t
(6.34)
+ 0) + O(h)
at any point t = to > 0, where
Mt) = at Now, in any interval
+ h)
-^-hCit h?/2
* ;
+ 0)
(*>o).
and for any h £ ( 0, - ), $h(t) G £ c
since by Proposition 1.6.2, ((t) is Lipschitz continuous on S, - . Also, for any 0 < s < T < 00, rT
2
/ rT+h
j $h(t)dt= -^u n
rs+h
c(t)dt-j
/»T+/i
= V]T
\
at)dt-h(c(T)-as))\ o
ps-\-h
(C(*)-C(T))dt-^jf
(C(t)-CW)dt 0 is a universal constant independent of h. Since, by (6.34), £(£) is Lipschitz continuous, $h(t)>-C,
for
te(s,T),
Properties of the Free Boundary:
One Dimensional
Case
87
we conclude that
I
\$h(t)\dt• 0 such that $hn weakly converges to a measure /L*O- Using (6.34) we then have $/in ( 0, £»'(* ± 0) eiriste and (-l)*Ci(* - 0) < (-1)*C*(* + 0).
(6.36)
Furthermore, for any 5 > 0, there exists P > 0 suc/i f/iat /or i 2 > *i > (-l)*C,'(*i + 0)eyPtl,
(6-37)
which implies that there exists a constant t* > 0 suc/i that £'(£) is strictly increasing for t > t* and Q(t) = x, for 0 0, which implies (6.37). If £(£) =const. on the interval 0 < si < t < s2, then ('(t) = 0 on this interval. Using(6.37) we deduce ('(£) = 0 on 0 < t < si. This completes the proof. Now we are ready to discuss the continuous differentiability of Q(t) for t > t*. We will treat C,{t) = (2(t) only. Let (xo,to) = (C(^o)i*o) with io > *2 a n d Ng be the intersection of a ^-neighborhood of (x0,t0) and Q, = {{x,t) 6 Q;u(x,t) > 0}. Take 6 > 0 so small that N25 stays away from t = t\ and from the free boundary x = Ci(i). Denote by d(x,t) the distance from (x,t) to the free boundary. • Lemma 1.6.1 m, 5 such that
There exist positive constants C\, C2 depending only on
C : < ^ < C d(x,t)
2
inN6.
Proof. By Corollary 1.6.1 and our assumption on Ng, there exist constants c > 0,C > 0, depending only on 6, such that for all t with (C(£), t) € dN5, c < C(t ± 0) < C.
(6.38)
Hence there exist constants C\ > 0, C2 > 0 depending only on S , such that Ci < , ^ l * * , < C 2 ,
for^eiV*.
Since t>(a:, t) is Lipschitz continuous, v(a;,t) = v(x,i) - v(£(t),t)
< C'\x - C(t)|
—2P (see (6.29)), using the Taylor formula, Theorem ox*
Properties of the Free Boundary:
One Dimensional
Case
89
1.6.5 and (6.38), we see that for 6 > 0 small enough, > \vx(C{t),t)\\x - C(*)l - P\x - C(*)|2
v(x,t)
= |C'(t + 0)||x - C(*)| - P\x - CWI2 > c'\x - C(t)|
>
dd(x,t)
with a constant Ci > 0 depending only on S . The rest part of the lemma can be proved similarly. • L e m m a 1.6.2
There exists a constant C depending only on 5, such that
dv 0, define vs(x, t) = ri^v^x
-xo,r]t-
t0).
Oi
Fix a £ (0, to)- Then v is defined in the —neighborhood of the origin and V v is also a generalized solution of the equation (6.39). (This means that \ V("»-l) m _ l vo J is a generalized solution of (6.3)). By Proposition 1.6.1, for t>
--, V d2v„ d2Vr,, l ^ =^ ^
+
xo,Vt
+
. r]k rik to)>-^>-^-v.
, . (6.40)
From the proof of Theorem 1.4.1, we see that vn G C 1 ' 1 / 2 with the Holder coefficient independent of 77. Therefore there exists a subsequence {vnn } with T]n —*• 0 such that vVn —> v in M2 uniformly on compact sets and v is a generalized solution of (6.39). We may simply suppose that vv —> v in R2 as 77 —>• 0 uniformly on compact sets. It is easily seen that all of the derivatives of v„ converge to the corresponding derivatives of v.
Properties of the Free Boundary:
One Dimensional
Case
91
In view of our hypothesis, for each fixed x < 0, — (z,0) = lim ——{x,0) = -z-{x0,t0) dx 17-+0 ox ox
= -b;
and for each fixed x > 0, v(x,Q) = lim Vr,(xo,0) = lim -v(rjx + x0,t0) n—>0
= 0.
T;-+0 T]
Thus —bx
for a; < 0
0
for x > 0.
v(x,0)
Moreover, (j
v„{x,t)
< — \r]x + xo -C(vt + to)\ < -{r]\x\ + \x0 - Civt + to)|) < C(\x\ + 2bt) V
provided i £ l and 0 < t < e with e > 0 small enough. Therefore we can use a uniqueness theorem in [KA3] mentioned in §1.2.4 to conclude that for ieR,0 0. Thus for x < bt, t < 0, dv(x, t) dv(bt, t) dx ~ dx v{x,t)
=v(bt,t)+
—{y,t)dy
bt dx Jbt
> v(x, t) - b(x - bt) = v{bt, t) + Lb(x, t) from which (6.44) follows. Now we prove that (6.41) holds for t < 0. Consider a fixed point (x,t) with x < bt. Let S be a rectangle with center (x, t) and boundaries parallel to the coordinate axis, such that (x,t) G S implies that x < bt and S contains points with t > 0. In S, v — Lb > 0 and achieves its minimum value, 0, on S D {t > 0}. By the strong minimum principle, we must have v = Lb in S. Therefore (6.41) holds for x < bt, t < 0. Suppose that v(x, t) > 0 for some (x, t) with x > bt, t < 0. Then from Theorem 1.6.2, we have v(x, t) > 0 for all t > t. However this contradicts the fact that the line x = x must intersect the line x = bt for some t > t and v = Lb = 0 at that intersection. Therefore (6.41) also holds for x > bt, t 0. Let rj = t - to ^ 0. Since v\v\ -» v = Lb as |7y| ->• 0, given e > 0 there exists a % = %(£) > 0 s u c h that —-v(?7a + a:o,7? + to)-i6(asgnr?,sgnr/)
<e
H X — Xc\
provided \rj\ = \t — t0\ G (0,770), - — — = a. Namely t — to \v(x,t) - Lb(x -x0,t-
t0)\ < e|t — *o|-
The proof of Lemma 1.6.3 is complete. • T h e o r e m 1.6.6 £*(£) (i = 1,2) is continuously differentiable at any t ^ t* (i — 1,2), where t* (i = 1,2) is defined in Corollary 1.6.1. Proof. We will prove the assertion only for £(i) = (^2(*)• Since a convex function which is differentiable everywhere on an interval is necessarily continuously differentiable there, by Proposition 1.6.3, it suffices to prove that £(£) is differentiable everywhere at any t = to > t^, i.e C(to - 0) = C'(*o + 0) = b.
(6.45)
If b = 0, then by Corollary 1.6.1, (6.45) is trivial. We suppose b > 0. Denote x0 = C(£o)- From the definition of t\, there must be a r e (*2,*o) such that C'(T + 0) > 0. First we prove that for any a > b, there exists e > 0 such that C(t)> x0+a(t-t0),
t0-e b > 0. Therefore (6.46) holds and we conclude that hm — < a. t->t0-o to — t Since a > 6 is arbitrary, it follows that hm ^ ^ < 6 . t-tto-0 to — t Next we prove that for any c < b, there exists T] > 0 such that C(i) < z 0 + c ( t - i o ) ,
for t0-r] X0 - fofoBy the mean value theorem there exists x G (XQ — br\ni x§ — cr)n), such that