國立高雄大學應用數學系
碩士論文
兩個索伯列夫不等式最佳常數逼近法的比較
The Comparison of Two Approcaches of Best Constant in
Sobolev Inequaility
研究生:吳家祥撰
指導教授:吳宗芳
兩個索伯列夫不等式最佳常數逼近法的比較
指導教授:吳宗芳教授 國立高雄大學應用數學系 學生:吳家祥 國立高雄大學應用數學系 摘要本論文文中,我們主要是研讀Talenti 所寫的 "Best Constant in Sobolev Inequality" athored 和 Agueh 所寫的 "A New ODE Approach To Sharp Sobolev Inequalities" 這兩篇 論文,並且研究這兩個工作的細節和比較這兩個關於索伯列夫不等式最佳常數逼近法的 差異。
在Talenti 的論文中,Talenti 的證明由兩個部分組成。第一步利用了球對稱重排去
簡化問題,找出問題所對應極致曲線及其對應的最佳函數,第二步Talenti 證明了在第
一部分給出的極致曲線的確給了我們最大值並找出索伯利夫不等式的最佳常數。 在Agueh 的論文中,他藉由研究 Talenti 和 另一個由 Cordero-Erausquin 提出的逼近
法之間的聯繫,找到了個變數變換幫助我們把從Talenti 獲得的極致曲線簡化成一個非
線性的常微分方程解出並由此獲得索伯列夫不等式最佳常數。
The Comparison of Two Approcaches of Best Constant in
Sobolev Inequaility
Advisor(s): Dr. Tsung-Fang Wu Department of Applied Mathematics
National University of Kaohsiung
Student: Chia-Hsiang Wu Department of Applied Mathematics
National University of Kaohsiung
ABSTRACT
In this thesis, we mainly study the paper with title "Best Constant in Sobolev Inequality" authored by Talenti and "A New ODE Approach To Sharp Sobolev Inequalities" authored by Agueh, elaborate the detail of the work and compare these two approaches of best constant in Sobolev Inequality.
In Talenti's approaches, Talenti proves the best constant for the Sobolev Inequality and the proof consists with twos step. In the first step, talents proofs are based on a spherically symmetric function which helped them reduce the problem, and find the extremal function and its solution. In the second step, Talenti shows that the extremal found in step 1 actually gives the maximum and finds the best constant in Sobolev Inequality.
In Agueh's approaches Agueh investigates the link between Talenti's approaches and another approach proposed by Cordero-Erausquin. We show that a strategic change of variable which helps to solve explicitly the nonlinear ordinary differential equation -- leading to the sharp constant and extremals in Sobolev Inequality obtained from the approach by Talenti.
Contents
1 Introduction 1 1.1 Background . . . 1 1.2 Sobolev Inequality . . . 1 1.3 Talenti’s Approach . . . 1 1.4 A ODE Approach . . . 3 2 Preknowledge 3 2.1 Lp space . . . . 3 2.2 Elementary Inequalities . . . 4 2.3 Weak Derivatives . . . 5 2.4 Wk;p Spaces . . . . 52.5 Lemmas and De…nitions . . . 5
2.6 Proof of Sobolev Inequaility . . . 6
3 Tanlti’s Approach[1976] 8 3.1 Spherically symmetric functions. . . 8
3.2 Extremal of J(u) . . . 9
3.3 Lagrange problem . . . 9
4 Agueh’s ODE Approach 13 4.1 Mass Transportation Approach . . . 13
4.2 An Optimization Problem . . . 14
1
Introduction
1.1
Background
Sobolev inequalities, also called Sobolev imbedding theorems, are very pop-ular among writers in partial di¤erential equations or in the calculus of vari-ations, and the best constant C and optimal functions in (1) are obtained independently with di¤erent approach by many authors. In this thesis, we will mainly study the approach for best constant C and optimal functions in (1) by Talenti and an ODE approaches proposed by Agueh.
1.2
Sobolev Inequality
Let u be a real valued fuction, de…ned on the whole m-dimensional euclidean space Rm;su¢ cenly smooth and decaying fast enough at in…nity. Moreover, let p be any number such that : 1 < p < m. Then:
kukq Ckrukp; (1) where q = mp=(m p) and C = p m1 1p p 1 m p 1 p1 m! (m2) 2 (np) (m + 1 np) !1 m ; (2) the equality holds in (1) if u has the form
u(x) = a + bjxjpp1 1 mp ; (3) where jxj = (x2 1+ ::: + x2m) 1
2 and a; b are postive constants.
1.3
Talenti’s Approach
Talenti’s Approach …rst concerned with the simplest Sobolev inequality kukq Ckrukp; (4) where C is the constant independent of u and we are interested in the smallest constant which is admissible in (4);
namely we will evaluate the expression C = sup kukq
krukp :
Following Talenti, the best constant C and extremals in (1) can be obtained from the solutions of the variational problem
sup ( I(u(x)) := kukq krukp : u2 C01(Rm) ) ; (5) or equivalently
supnkukq : u2 Cc1(Rm);krukp = 1o; (6) If u denotes the spherically symmetric rearrangement of u then by Cavalieri and Polya Szego Principle and let v depends on r = jxj the variational problem (5) reduce to the 1-dimensional variational problem
sup 8 < :J (u(x)) := R1 0 r m 1 jv(r)jqdr 1 q R1 0 rm 1jv(r)j p dr 1 p : v(r)2 K 9 = ;; (7) or equivalently sup Z 1 0 rm 1jv(r)jqdr : u 2 K; Z 1 0 rm 1jv(r)jpdr = C ; (8) where C > 0 is some constant, and K is the set of all nonnegative and nonincreasing functions, v : [0; 1) ! [0; 1); (v(r) = u(x); r = jxj) and deduce the best constant and extremals in the Sobolev inequality (1) just by changing conveniently r to jxj ;the maximizers of (7) or (8) are the solutions of the ODE
rm 1ju0(r)jp 1 0 Crm 1ju(r)jq 1 = 0; C > 0; (9) which verify certain limit conditions at 1 (as in u(r) 2 K). Note that (9) is nothing but the p-Laplacian (or Laplace’s if p = 2) equation,
div(jru(x)jp 2ru(x)) + Cu(x)q 1 = 0; (10) expressed in the radial form, u(r) = u(x); r = jxj.
Then Talenti proves that the solution of (9) is (3) and actually gives the maximum by using Hilbert invariant integral, and the Weierstrass E function.
1.4
A ODE Approach
Agueh shows that in fact, the mass transportation approach suggests a change of variable which helps to solve explicitly the nonlinear ordinary dif-ferential equation – leading to the sharp constant and extremals in (1) – obtained from the approach by Talenti without using the result by Bliss. Then, this leads to a new ODE approach to the sharp Sobolev inequalities
He observes that the Euler-Lagrange equation to (31) is the PDE (10)), whose radial form is the ODE (9) obtained in [1], where p represents here the Lagrange multiplier for the constraint kukq = 1, and from the result of mass transport theory [5] , we could say that solutions to (9) satisfy
rp0 p0 =
jxjp0
p0 = F
0(up0(x)) := H(u(x)) = H(u(r)); (11) for some constant . Therefore, (11) suggests the change of variable H(u(r)) =
rp0
p0 solve (9). Using this change of variable, we can rewrite (9) as the …rst
order nonlinear ODE in G0(t) = jH0(t)jp 2 H0(t) p 1 ; m = p q t qG00(t)+ p p 1(m 1)G 00(t) Z t 0 d G0( )+ (p 1)G 0(t)tq 1; 8t 2 (0; u(0)); (12) Setting G00(t)Rt 0 d
G0( ) = k (for some constant k), (12) reduces to the linear
…rst oder ODE in G0, m p p 1(m 1)k = Cp q t qG00(t) + (p 1)G0(t)tq 1;
whose solutions can be easily obtained in closed forms. Then from the so-lutions G0 to (12), we can derive a solution u(r) to (9) via the relations G0(t) = jH0(t)jp 2H0(t)
p 1 and (11).
Finally, we can compute the optimal function of extremals (3) and …nd the best constant (2) in (1)
The remainder of this paper is organized as follows. In Section 2 some preliminaries knowledge are presented. In Section 3 and Section 4 introduce two approaches respectively.
2
Preknowledge
2.1
L
pspace
De…nition 1 If E be a measurable set of Rm and p satis…es 0 < p < 1, then Lp(E) denotes the colletion of measurable f for which R
Ejfj p
that is
Lp(E) =ff : Z
E
jfjp < +1g; 0 < p < 1; and we shall write
kfkp;E = Z Ejfj p 1 p ; 0 < p <1; thus Lp(E) is the class of measurable f for which
kfkp;E is …nite
2.2
Elementary Inequalities
Theorem 2 (Young’s inequality)
Let (x) be continuous real-value and strictly increasing for x 0 and let (0) = 0: If x = (y) is the inverse of ; then for a; b > 0;
ab Z a 0 (x)dx + Z b 0 (y)dy:
Equality holds i¤ b = (a):
If (x) = x ; > 0; setting p = + 1 and p0 = 1 + 1 we obtain Young’s inequality becomes ab a p p + bp0 p0 , if a; b 0; 1 < p <1; and 1 p+ 1 p0 = 1 Theorem 3 (Hödler’s inequality)
If 1 p 1 and 1p + p10 = 1;then kfgk1 kfkpkgkq;that is Z E jfgj Z E jfjp 1 p Z E jgjp0 1 p0 ; (1 < p <1); Z Ejfgj (esssup E jfj) Z Ejgj : Theorem 4 (Minkowski’s inequality)
If 1 p 1;then kf + gkp kfkp+kgkp;that is Z E jf + gjp 1 p Z E jfjp 1 p + Z E jgjp 1 p ; (1 < p <1); esssup E jf + gj esssup E jfj + esssupE jgj :
2.3
Weak Derivatives
De…nition 5 Let F; f 2 L1loc( )(L1loc( ) :for each compact subset 0 it holds R 0ju(x)j dx < 18u 2 L1
loc( )) If for all fuctions g 2 C1
0 ( ) it holds that Z
F (x)g(x)dx = ( 1)j j Z
f (x)D g(x)dx:
Then F (x) is s called weak derivative of f (x) with respect to the multi-index :
2.4
W
k;pSpaces
De…nition 6 Let k 0 is a constant and 1 p 1, then the Sobolev space Wk;p is de…ned by
Wk;p =fu 2 Lp : D u2 Lp;8 with j j kg: This space is equipped with the norm
kukWk;p( ) =
X j j k
kD ukp:
2.5
Lemmas and De…nitions
De…nition 7 (Spherically symmetric rearrangement)
Let u be any smooth real valued function de…ned on the whole euclidean space Rm which decays fast enough at in…nity. Let u be the spherically symmetric rearrangement of u; that is u (x) = supft : (t)Cmjxj m g; where Cm = m 2= (1 + m
2) is the measure of the m-dimensional unit ball and
(t) = measfx 2 Rm : u(x) > tg:
Lemma 8 (Spherically Symmetric Rearrangement)Let u be any nonegative smooth real valued function de…ned on the whole euclidean space Rm which decays fast enough at in…nity. Let u be the spherically symmetric rearrange-ment of u: Then , for every exponent p 1; the following holds:
(i)R(u )pdx =R updx; (ii)R jru jpdx R jrujpdx;
De…nition 9 (Gâteaux derivative)
let f be a function on an open subset U of a Banach space X into the Banach space Y . We say f is Gâteaux di¤erentiable at x 2 U if there is bounded and linear operator T : X ! Y such that
Txv = lim t!0 f (x + tv) f (x) t = d dtf (x + tv)jt=0: Theorem 10 (Lebesgue’s Dominated Convergence Theorem)
Suppose fn : R ! [ 1; 1]are (Lebesgue) measurable functions such that the pointwise limit f (x) = limn!1fn exists. Assume there is an integrable g : R ! [0; 1] with jfn(x)j g for each x 2 R. Then f is integrable as is fn for each n, and
lim n!1 Z R fn = Z R lim n!1fn = Z R f:
Lemma 11 (Du Bois-Reymond lemma) If f 2 C0(a; b) and
Z b a
f (x) 0(x)dx = 0; for every (x)2 C01(a; b); then f c; c is a constant.
2.6
Proof of Sobolev Inequaility
Proof. We start from case p = 1 for u 2 C1
0(Rn) and …x index i 2 f1; 2; :::; t; :::; ng u(x) = Z xi 1 Diu(x1:::; t; :::; n)dt; ju(x)j = Z xi 1 Diu(x1:::; t; :::; n)dt Z xi 1 jDiuj dt Z 1 1jD iuj dt; so that ju(x)jnn1 ( Z 1 1jD iuj dt) n n 1 n Y i=1 ( Z 1 1jD iuj dt) 1 n 1: (13)
The inequality (13) now is tntetrated successuvely over each variable xi, i = 1; :::; n use generalized Hödler inequality then being applied after each integrantion Z 1 1 ::: Z 1 1ju(x)j n n+1dx 1:::dxn n Y i=1 Z 1 1 ::: Z 1 1jD iuj dx1:::dxn 1 n 1 ; then we have kukL n n 1 n Y i=1 Z jDiuj dx 1 n 1 n n X i=1 Z jDiuj dx 1 n n X i=1 Z jDuj dx = Z jDuj dx = krukL1:
For case 1 < p < 1; Let > 1 kjuj kL n n 1 Z jD juj j dx Z juj 1jDuj dx; by Hödler inequality kjuj kL n n 1 Z juj( 1)qdx 1 q Z jDujp 1 p ; (14) take q = p 1p ; = (n 1)pn p ; ( 1)q = np nn p , (14) becomes Z juj(nn 1)pp n n 1 dx n 1 n Z jujnpn pn p p 1 dx p 1 p Z jDujp 1 p ; Z jujnnppdx n p np Z jDujp 1 p ; kjujkn p np (n 1)p n p kDukp:
3
Tanlti’s Approach
In 1976 Talenti …rst concerned with the simplest Sobolev inequality
kukq Ckrukp; (15) where C is the constant independent of u and we are interested in the smallest constant which is admissible in (4);
namely we will evaluate the expression C = sup kukq
krukp
; (16)
where the sup is taken in the class of all (not identically zero) smooth func-tions u which decay rapidly at in…nity (e.g. compactly supported).
3.1
Spherically symmetric functions.
If u depends on r = jxj only, the ratio ((16) becomes 2 m1 1 2 h (m 2) i1 m J (u); where J (u) = R1 0 r m 1 jujqdr 1 q R1 0 r m 1ju0jpdr p1 ;
Lemma 12 Let m; p; q be real numbers such that 1 < p < m; q = mp
m p:
Let u be any real valued function of a real variable r, which is su¢ eiently smooth on the hall-line ]0; +1[ and which is such that
Z 1 0
rm 1ju0jpdr < +1; u(r)! 0 if r ! 1: Then
J (u) J ( ); where is any function of J the form
3.2
Extremal of J(u)
In this part we will show that the extremal of J(u) is equal to (9). Proof. From the de…nition of Gâteaux derivative we have
d dtJ (u + tv) = J (u) 1 q R1 0 r m 1q ju + tvjq 1sgn(u + tv)vdr R1 0 r m 1ju + tvjqdr J (u) R1 0 r m 1 ju0+ tv0jp 1 sgn(u0+ tv0)v0dr R1 0 r m 1ju0jpdr ; d dtJ (u + tv)jt=0 = J (u) R1 0 r m 1 jujq 1sgn(u)vdr R1 0 rm 1juj q dr R1 0 r m 1 ju0jp 1sgn(u0)v0dr R1 0 rm 1ju0j p dr :
Let J0(u)(v) = 0 we have Z 1 0 rm 1jujq 1sgn(u)vdr Z 1 0 rm 1ju0jp 1sgn(u0)v0dr = 0; = R1 0 r m 1 ju0jpdr R1 0 rm 1juj q dr; by integrations by parts we have
Z 1 0 rm 1jujq 1sgn(u)vdr + Z 1 0 rm 1ju0jp 1sgn(u0) 0vdr = 0 and by Du Bois-Reymond lemma we have the extremal of di¤eretail equations of the form
rm 1ju0jp 1sgn(u0) 0 + rm 1jujq 1sgn(u) = 0; is a postive constant. (17)
In this approach Talenti does not calculate the general case of exactly, Talnti only derive the general case from the case p = 2 of (17),which is a particularly simple case of Emden-Fowler equations and all its solutions can be obtained with quadratures.
3.3
Lagrange problem
In this part, we will show that the extremals we have found in the previous actually give the maximum. For convenience, we purpose to put our problem J (u) = maximum, (18)
in the form of a lagrange problem, namely: 8 > > > > < > > > > : R1 0 r m 1 ju1j q dr =maximum where u0 2(r) = rm 1ju01(r)j p and u2(0) = 0; u1(1) = 0; u2(1) = 0 (19)
Clearly the problem (18) is equivalent to the Lagrange problem (19). In fact,if u is a solution of (18), then the pair
u1 = u (r) R1 0 t m 1 ju0(t)jp dt 1=p u2(r) = Rr 0 t m 1 ju0 1(t)j p dt
is a solution of (19). Conversely, if a pair(u1; u2) is a solution of (19), then u = u1is a solution of (18)
A set of extremals of the Lagrange problem is easily obtained from the results of the previous subsection; namely:
( '1(r) = a 1 + brp0 1 m=p '2(r) =R0rtm 1 j'0 1(t)j p dt (20) More explicitly '2(r) = rm p'1(r)pf br p0 1 + brp0 ; (21)
where f ( ) is tbe function f ( ) = 1 p0 m p p 1 p Z 1 0 (1 t)m=p0(1 t) mdt: (22) We claim the two-parameter family of extremals in the …rst octant of R3. In other words, the paths r ! (r; '1; '2) are the trajectories of a smooth vector …eld X, de…ned in the …rst octant of R: Thus exactly one such path passes through any point of the …rst octant of and X (r; u1; u2)is the slope at the point(r; u1; u2) of the path passing through this point. The components of the vector …eld X are given by
8 < : X0(r; u1; u2) = 1 X1(r; u1; u2) = m pp 1 (u1=r) X2(r; u1; u2) = rm 1jX1(r; '1; '2)j p (23)
where is the root of the equation
We prove the following: there exists an exact di¤erential dW such that the integral R dW, along any path r ! (r; u1; u2) which satis…es the constraint u0 2(r) = rm 1ju01(r)j p is R01rm 1 ju1j q
drand equality holds when the path is an extremal. To see this, we consider a C2 real valued function W de…ned in the …rst octant of R3 and enjoying the following property: For every point (r; u1; u2) of the …rst octant of R3, the function
( 0; 1; 2)! rm 1uq1 0 @W (r; u1; u2) @r 0 @W (r; u1; u2) @u1 1 @W (r; u1; u2) @u2 2 restricted to the cone of all directions issuing from the point (r; u1; u2) such that 0 > 0 and p 1 0 2 = r m 1 j 1j p has a critical point at X (r; u1; u2) :
From the Lagrange multipliers rule and the de…nition (23) we obtain the relations 8 < : @W @r = r m 1uq 1+ (p 1) rm 1jX1jp @W @u1 = pr m 1 jX1jp 1 @W @u2 = (25) where ~is some continuously di¤erentiable function to be determined.
To …nd , we write down the compatibility conditions for the system (25). 0 @ 1 0 (p 1) (rm 1 jX1j p ) 0 1 p rm 1 jX1j p 1 p rm 1jX1j p 1 (p 1) (rm 1jX1j p ) 0 1 A 0 @ @ @r @ @u1 @ @u2 1 A = qrm 1uq 11 0 @ 00 1 1 A p 0 @ X1@u@ 2 @ @u2 @ @r + X1 @ @u1 1 A rm 1 jX1jp 1 : (26) since the matrix on the left-hand side of (26) has rank 2, we must impose orthogonality between the right-hand side of (26) and the eigenvectors of the transposed matrix; i.e.
p @ @X r m 1 jX1jp 1 = qrm 1uq 11 (27) where @ @X = @ @r + X1 @ @u1 + rm 1jX1jp @ @u2 is the derivative in the direction of the vector …eld X.
This gives the only possible value of X. A more explicit formula is = (p 1) p 1 (m p)p rpuq p 1 p 1 (1 ); (28)
where is the root of the equation (24). The formula (28) follows from (27) because @ @X r m 1 jX1jp 1 = m m p p 1 p 1 rm p 1up 11 p 1(1 ) ; (29)
where is as before and we can also cheek equation (29) by a direct compu-tation.
We can prove that the function , de…ned by (28), is actually a solution to the system (26). As the matrix at the left-hand side of (26) has rank 2, it is enough to verify two equations only. Disregarding (26.3) and combining (26.1) and (26.2) we have to check the following pair
@ @X = 0 @ @u1 = p @ @u2 r m 1 jX1j p 1 (30)
Equation (30.1) is trivial since its characteristic lines are precisely the mem-bers of X and the function as de…ned by (28) is constant. Indeed, we can write (28) as =hu1(1 ) 1 m=piq p 1 r p0 1 p ;
so the pair of equations u1(1 ) 1 m=p
and 1 r p0 is constant.
The procedure just described gives us a smooth solution W to the system (25) when we use as de…ned by ((28). Let us show that the di¤erential dW has the property stated at the beginning; this will follow from the fact that
is positive. The di¤erence E (r; u1; u2; 0; 1; 2) = rm 1uq1 0 @W (r; u1; u2) @r 0 @W (r; u1; u2) @u1 1 @W (r; u1; u2) @u2 2 ;
which we have already considered, is essentially the Weierstruss excess func-tion.
If the direction ( 0; 1; 2)is so restricted that
0 > 0 and p 1 0 2 = r m 1 j 1j p ; then we have from (25) the equation
E (r; u1; u2; 0; 1; 2) = 0r m 1 j 1= 0j p + pjX1j p 1 ( 1= 0) + (p 1)jX1j p : Clearly the expression in brackets is always 0and vanishes if (and only if) the direction ( 0; 1; 2) is parallel to X (r; u1; u2). As the factor in front is 0, we obtain that E 0the pair (u1; u2)give the maximum is our desired result.
4
Agueh’s ODE Approach
Agueh proposed a new variational approach to …nd the best constants and op-timal functions of the Sobolev inequalities. Agueh shows that the extremals for these inequalities are obtained via solutions of a nonlinear ordinary dif-ferential equation that we solve explicitly using a strategic change of variable In this approach, Agueh investigate the link between Tanlti’s approach and a mass transportation approach presented by Cordero-Erausquin et al. Agueh show that in fact, the mass transportation approach suggests a change of variable which helps to solve explicitly the nonlinear ordinary di¤erential equation – leading to the sharp constant and extremals obtained from the approach by Talenti.
4.1
Mass Transportation Approach
Cordero-Erausquin et al [5] presented another proof of (1) using a mass trans-portation approach. In [[5]], the authors established a duality between the functional E(u) = RRmjru(x)j
p
and a certain energy functional on prob-ability densities on Rm, from which they derived the sharp constant and extremals in (1).
We observe that the Euler-Lagrange equation to the variational problem inf E(u(x)) := Z Rmjruj p dx : u2 W1;p(Rm);kukq = 1 ; (31) where W1;p(Rm) :=fu 2 Lp(Rm) :ru 2 Lp(Rm)g : From mass transport theory give us the result
supn HF jxjp0=p0( 1) : 1 2 P (R m)o = inf HF +nPF ( 0) + 1 p Z Rm 0jrF 0( 0)j p dx : 0 22 P (Rm) ; and an optimal function 0 = 1 = 1 in both problems is a solution to the equation r F0( 1) + jxj p0 p0 ! = 0;
where 2 P (Rm) is in a space of probability densities on Rm; F (x) : [0;1) ! R such that F (0) = 0; x 7! xnF (x n) ;and HF ( ) =R F ( ) dx
By choosing F (x) = nx1 1=n; 1= uq1(x) we have inf E(u(x)) := Z Rmjruj p dx : u2 W1;p(Rm);kukq = 1 = p n p (n 1) p HF jxjp0=p0( 1) ; and u1 de…ned by r F0(uq1(x)) + jxj p0 p0 ! = 0; 1:= uq1(x)2 P (Rm); (32) is a minimizer of (31)
4.2
An Optimization Problem
Theorem 13 Let 1 < p < n and set q = (n p)np . Then the variational problem
inf E(u(x)) := Z
Rmjruj p
dx : u2 W1;p(Rm);kukq = 1 ; (33) has a solution u1 which is nonnegative, spherically symmetric and nonin-creasing. Moreover, u1 is a solution of the PDE
div(jru(x)jp 2ru(x)) + Cu(x)q 1 = 0; (34) for a well-chosen C > 0 (such that kukq = 1). Conversely, any nonnega-tive spherically symmetric and nonincreasing solution u 2 W1;p
(Rm) of (34) which satis…es the condition kukq = 1 solves (33).
Furthermore, the Sobolev inequality
kukq Ckrukp; u2 W 1;p
(Rm) holds, with the best constant
C = 1 (pE(u1))1=p: Proof. Let u 2 W1;p
(RN),
kukq = 1 , and u is nonnegative, radially sym-metric and nonincreasing solution of juj,such that the level sets of u has the same measure as the level sets of juj and they are balls centered at the origin. by lemma 8 we have ku kq =kukq = 1; ku kp =kukp; Z jru jpdx Z jrujpdx;
so that u is also admissible in (33) and E(u ) E(u). Therefore we can assume without loss of generality that the test functions in (33) are nonnega-tive, radially symmetric and nonincreasing Let E1 denote the value of (33). Clearly E1 is …nite because E(u) is nonnegative. Now, let (uk)k be a mini-mizing sequence in (33). We have that uk(x)=vk(r); r = jxj and kukkq = 1 for some nonincreasing function vk : [0;1) !.[0; 1):Then vk(r) ! 0 as r ! 1. And since vk is nonincreasing, we can assume without loss of gen-erality that
vk(r) 1;8r 0; (35) otherwise we replace vk(r) by ~vk(r) = vk(r + r0) where r0 is such that vk(r) 1;8r r0.By de…nition, (uk)k is bounded in W1;p(RN), and then up to a subsequence, it converges weakly to a function u1 2 W1;p(RN) which obviously is nonnegative, radially symmetric and nonincreasing (because the uk are). Then we can write that u1(x) = v1(r); r = jxj,for some nonin-creasing function v1 : [0;1) !.[0; 1). By weak lower-semicontinuity,we have that
E(u1) lim inf
k!1 E(uk) = E1;
It remains to show that ku1kq = 1. For that, we will prove that (uk)k con-verges strongly to u1 in Lq
(Rn). Indeed, since the u
k are radial and bounded in Lp
(Rn), there exists a constant C > 0 such that C kukk p p = n!n Z 1 0 rn 1vk(r)pdr;8k;
where !n denotes the Lebesgue measure of the unit ball in RN. Hence for all R 1, we have C n!n Z R 0 rn 1vk(r)pdr n!n Z R 0 rn 1vk(R)pdr = !nRnvk(R)p; i.e. vk(r) (C=!n) 1 pr n p;8r 1: (36) Set h(r) := min(1; r np); := max(1; (C=! n) 1 p); (37a) we have that Z 1 0 rn 1h(r)qdr = Z 1 0 rn 1h(r)qdr + Z 1 1 rn 1h(r)qdr p Z 1 0 rn 1dr + p Z 1 1 rn(1pq) 1dr < 1;
because n(1 qp ) 1 = n pnp < 0. Then rn 1h(r)q
2 L1((0;
1)) and from (35)- (37a), rn 1vk(r)q rn 1h(r)q;8k. Hence, the dominated convergence theorem gives that
1 = lim k!1kukk q Lq(Rn)= n!n lim k!1 Z 1 0 rn 1vk(r)qdr = n!n Z 1 0 lim k!1r n 1v k(r)qdr = n!n Z 1 0 rn 1v1(r)qdr = Z Rn u1(x)qdx =ku1kqLq(Rn):
This shows that u1 is a minimizer of (33)
Clearly u1 is a solution of the Euler-Lagrange equation (34) to the variational problem (33), where is the Lagrange multiplier for the constraint jjujjq = 1.
4.3
Sharp Sobolev Inequalities
Here we will compute the extremal fuction of Sobolev inequalities. Proof. Let u 2 W1;p
(RN) be a nonnegative, radially symmetric and non-increasing solution such that kukq = 1:Hence, there is a nonincreasing v : [0;1) ! [0; 1) such that limr!1v(r) = 0; limr!1v0(r) = 0 and
v(r) = u(x); r =jxj ; (38) and the extremal fuction
rm 1jv0(r)jp 1 0 rm 1jv(r)jq 1 = 0; read as (m 1) r r m 2 jv; (r)jp 1+ (p 1)jv0(r)jp 2v00(r) + v(r)q 1 = 0: (39) To solve 39 we consider the change of variable
H(v(r)) = r p0 p0 ; 1 p + 1 p0 = 1; (40) where H is C2 f unction on (0; v(0)), we have H0(t) 0,for all t 2 (0; v(0)); limt!1H(t) =1; limt!1H0(t) = 1 Di¤erentiating (40) H(v(r)) = r p0 p0 ; H0(v(r))v0(r) = rp0 1;
r = [H0(v(r))v0(r)]p011 = [H0(v(r))v0(r)]p 1: (41)
Di¤erentiating again (41) with respect to r, we have
1 = (p 1)[H0(v(r))v0(r)]p 2[H00(v(r))v0(r)v0(r) + H0(v(r))v00(r)]; 1 p 1 = H 00(v(r))H0(v(r))p 2 v0(r)p+ H0(v(r))p 1v0(r)p 2v00(r); 1 p 1 = H 00(v(r))jH0(v(r))jp 2 jv0(r)jp +jH0(v(r))jp 2H0(v(r))jv0(r)jp 2v00(r); (42) Now we express jv0(r)jp 2 v00(r) and jv0(r)jp
in terms of v and H(v): From (39) jv0(r)jp 2v00(r) = 1 p 1 v(r) q 1 m 1 r jv 0(r)jp 2 v0(r) = 1 p 1 m 1 r jv 0(r)jp 2 v0(r) + v(r)q 1 ; (43) and from (41) r = [H0(v(r))v0(r)]p 1= H00(v(r))jH0(v(r))jp 2jv0(r)jp 2v0(r) =) jv 0(r)jp 2 v0(r) r = 1 jH0(v(r))jp 2H0(v(r)): (44) Combing (43) and (44) we have
jv0jp 2v00 = 1
p 1[(m 1)
1
jH0(v)jp 2H0(v) + v
q 1] (45)
we multipy (45) by v0 and intrgate over (r; 1) we have Z 1 r jv0jp 2v00v0dr = 1 p 1 Z 1 r [(m 1) 1 jH0(v)jp 2H0(v)+ v q 1]v0dr: Noticing that d dr( jv0jp p ) =jv0j p 1 v00sgn(v0) =jv0jp 2 v00v0 Z 1 r d dr( jv0jp p )dr = jv0jp p = 1 p 1 Z 1 r (m 1) v 0 jH0(v)jp 2H0(v)+ Z 1 r vq 1v0dr =) jv0jp = ( p p 1)(qv q Z v(r) 0 (m 1) d jH0( )jp 2H0( )); (46)
where we substution = v(r) in the intrgral term.We insert (45) and (46) into (42) to obtain the ODE in H;
1 p 1 = jH 0(v(r))jp 2 H00(v(r))jv0(r)jp+jH0(v(r))jp 2H0(v(r))jv0(r)jp 2v00(r) = jH0(v(r))jp 2H00(v(r))( p p 1)(qv q Z v(r) 0 (m 1) d jH0( )jp 2H0( )) +jH0(v(r))jp 2H0(v(r)) 1 p 1 m 1 jH0(v(r))jp 2H0(v(r))+ v(r) q 1 ; 1 p 1 = jH 0(v(r)) jp 2H00(v(r))( p p 1)qv q jH0(v(r))jp 2H00(v(r))( p p 1) Z v(r) 0 (m 1) d jH0( )jp 2H0( )) +jH0(v(r))jp 2H0(v(r))( 1 p 1) m 1 jH0(v(r))jp 2H0(v(r)) +jH0(v(r))jp 2H0(v(r))( 1 p 1)Cv(r) q 1; 1 = jH0(v(r))jp 2H00(v(r))p q v q+ m 1 + jH0(v(r))jp 2H0(v(r))Cv(r)q 1 jH0(v(r))jp 2H00(v(r))p Z v(r) 0 (m 1) d jH0( )jp 2H0( )); m = jH0(v(r))jp 2H00(v(r))p q v q+ jH0(v(r))jp 2H0(v(r)) v(r)q 1 jH0(v(r))jp 2H00(v(r))p Z v(r) 0 (m 1) d jH0( )jp 2H0( )); (47) For convenience,we set
G0(t) = jH 0(t)jp 2 H0(t) p 1 ; G 00(t) =jH0(t)jp 2 H00(t); (48)
and rewrite (47) as the nonlinear ODE in G0
m = p q t qG00(t)+ p p 1(m 1)G 00(t) Z t 0 d G0( )+ (p 1)G 0(t)tq 1; 8t 2 (0; v(0)); (49)
To slove (49),we write as the linear ODE m p p 1(m 1)k = p q t qG00(t) + (p 1)G0(t)tq 1; m +p 1p (m 1)k = p qt q G00(t) + (p 1)G0(t)tq 1; (50) where G00(t) Z t 0 d G0( ) = k; (51) By slove 51 we have G0(t) = Btk+1k ; (52)
for some arbitrary constant B > 0;then 50 become m + p0(m 1)k = p qt q( B) k k + 1t 1 k+1 + (p 1)( B)t k k+1tq 1; m + p0(m 1)k B = p q k k + 1t q k+11 + (p 1)tq k+11 ; m + p0(m 1)k B = ( pk q(k + 1) + (p 1))t q k+11 ; (53) if 53 holds i¤ m + p0(m 1)k B = 0 and pk q(k + 1) + (p 1) = 0; (54) or q 1 k + 1 = 0 and m + p0(m 1)k B = pk q(k + 1) + (p 1); (55) if 54 holds m + p0(m 1)k = 0 =) k = m m 1p 0; pk q(k + 1) + (p 1) = 0 =) k k + 1 = (1 p) q p = q p0 ; and 52 become G0(t) = Bt p0q; (56)
for some arbitrary constant B > 0: Combing (56) and (48) we have H0(t) = ((p 1)B)p11t
n m p;
is a solution of (47) such that H(t) = n p n ((p 1)B) 1 p 1t p m p D;
for some constant D ,Then we solve v in (40)
v(r) = a (b + rp0)(m p)=p; a := " p0(m p)((p 1)B)p11 n #m p p ; b := Dp: (57) We substitute (57) to (39) to obtain (57) solves (39), where b is given by
b : a
m(p 1)+p m p
m(a(m p)p 1 )p 1;
and a > 0 is arbitrary.
and the other case if 55 holds, and 52 become G0(t) = m p
Cp(p 1)t
1 q; (58)
for B = Cp(p 1)n p : and k = 1q 1 Combing (58) and (48) we have
H0(t) = (m p Cp ) 1 p 1t 1 q p 1;
is a solution of (47) such that H(t) = (m p)(p 1) p2 ( m p Cp ) 1 p 1t p2 (m p)(p 1) D:
for some constant D ,Then we solve v in (40) v(r) = C p m p2 (p m) p m p (b + rp0 ) (m p)(p 1) p2 ; ; b := Dp0: (59)
We substitute (59) to (39) we can see that (59) does not solve (39). Therefore (57) is the only solution Best Constant.
References
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