1 Introduction

Main Problem. (Kazdan-Warner) Given (M, g) a compact Riemmanian manifold without boundaries, and a function eR 2 ^C•(M), is there any comformal metric eg=e^2ug, such that the scalar curvature with respect to g is exactly ee R?

In this note, we deal with the case of surfaces (n=dim M=2)

Observation. In the case dimM = 2, we have R = 2K and eR = 2 eK, where K is the sectional curvature (Gaussian curvature). By Gauss-Bonnet

theorem, _Z

MKdµ= Z

Me^2uKdµe =2pc(M), thus we have the necessary condition:

1. If c(M) < 0, then eK is negative at some point.

2. If c(M) = 0, then either eK⌘^{0 or e}K changes sign.

3. If c(M) > 0, then eK is positive at some point.

However these conditions are in general not sufficient.

1

Conformal changes of curvatures. Under local coordinate, we have eG^k_ij = ¹

2 eg^kl(^∂geil

∂x^j +^∂egjl

∂xⁱ

∂ge_ij

∂x^l)

=G^k_ij+¹

2(d_ik∂u

∂x^j +d_jk∂u

∂xⁱ g_ijg^kl ∂u

∂x^l), Re_ij = ^∂eG

tij

∂x^t

∂eG^t_it

∂x^j +eG^s_ijeG^t_st eG^s_iteG^t_sj

=R_ij (4^u)g_ij, (n=2) Re =g_e^ijRe_ij =e ^2u(R 24^u)

Thus Kazdan-Warner problem for surfaces is equivalent to solving the fol-lowing PDE on M.

4^{u K}+e^2uKe=0. (1)

2 Case I: c ( M ) < 0

We may try to study this case by sub- and sup- solution method.

Proposition 2.1. On a compact Riemannian manifold(M, g), consider the semilinear PDE4^u+ f(x, u) = 0, where f 2 ^C•(M⇥^R). If there exists f, y 2 ^C2(M), such that f^yand

4^f+ f(x, f) 0, 4^y+ f(x, y) ^0,

(We call f and y a sub-solution and a sup-solution for the PDE respec-tively.) Then there exists u 2 ^{C• s.t. f} ^yand (1) holds.

Proof. Since M is compact, we can choose A, c > 0 such that A < f  y<A and ct+ f(x, t)increasing in t2 [ ^{A, A}]. We rewrite (1) into

Lu =F(x, u),

where L is an elliptic operator defined by Lu , 4^u+cu, and F(x, u) , ct+ f(x, t). By maximum principle (looking at the minimum), we can see that L is a ”positive” operator, in the sense Lu 0 ) ^u 0, or equiva-lently, Lu₁ Lu2)^u1 u2.

On the other hand, we have the following result of Schauder estimate:

2

Proposition 2.2. As an operator L : C^2,a ! ^C0,a between Holder spaces (a2 (^{0, 1})), L has a compact inverse L ¹.

Consider two sequences of sub- and sup- solutions{^fk}^,{^yk}^{given by} f0 =f, f_k+1 =L ¹(F(x, f_k))

y₀ =y, y_k+1 =L ¹(F(x, y_k))

We inductively show that f < f1 < f2 < · · · < ^y2 < y1 < y. First, notice that Lf_k+1 = F(x, f_k) Lf_k, so f_k  ^fk+1 by positiveness of L.

Now if f_k ^yk, then

Lf_k+1 =F(x, f_k) ^F(x, y_k) = Ly_k+1, thus f_k+1^yk+1again by positiveness of L.

Thus we have pointwise convergence

{^fk} ! ^u,{^yk} ! ^{u, f} ^u^u ^y.

Since f_k, y_k, Lf_k, Ly_kare all bounded, using the following L^p-estimate, we can show that{^fk}^,{^yk}are bounded in the Sobolev space W^2,pfor p>n:

Proposition 2.3. For u 2 ^W2,p(M) with p > 1, we have the following inequality¹:

k^uk_W^2,p ^C(k^ukL^p +k4^ukL^p) for some constant C >0.

Hence by Sobolev embedding theorem², {^fk}^,{^yk} are also bounded in C^1,a with a = 1 ⁿ_p. As a result, {^fk}^,{^yk} converges in C^1,a to u, u respectively, and

Lu =F(x, u), Lu=F(x, u),

and then by elliptic regularity theorem, u, u are smooth solutions to (1).

1Theorem 9.13 of [2]

2Theorem 7.26 of [2]

3

Now we may apply the method to the problem (n=dim M=2) Corollary 2.4. In the case of c(M) < 0, if (1) has a sup-solution y in C², then it has a smooth solution.

Proof. We only have to show that there is also a sub-solution f such that fy. Choose an f 2 ^C•(M)such that

4^f =K K₀, where K0 = R

Kdµ/R

dµ is the mean value of K, which is negative by Gauss-Bonnet theorem. Note that R

(K K0)dµ = 0, so such f exists by Hodge decomposition. Now we consider f = f c for some constant c>0. We have

4^{f K}+ ˜Ke^2f = K0+ ˜Ke^{2 f 2c} >0

for c sufficient large, and we also take c large enough so that f c  ^y, then hence we find the f we want.

Theorem 2.5. In the case c(M) < 0, if eK  0, then (1) has a smooth solu-tion.

Proof. As previous, we only have to show that there is also a sup-solution yfor (1). Choose an f 2 ^C•(M)such that

4^f =K^e Ke0, where eK₀ = R _e

Kdµ/R

dµ < 0. Consider y = a f +b for some constant a, b>0. We have

4^{y K}+ ˜Ke^2y = (a eK0 K) + (e ^{2a f+2b} a)K^e <0 for a, b sufficient large, so we find the y we want.

For general eK, the problem remains unsolved.

4

3 Case II: c ( M ) = 0

For the case c(M) = 0, we have a complete criterion for eK that (1) has a smooth solution:

Theorem 3.1. For the case c(M) < 0, (1) has smooth solution if and only if one of the following condition holds:

(1) eK ⌘^{0, or}

(2) eK changes sign andR _e

Ke^{2 f}dµ<0,

where fR 2 ^C•(M) is a solution of4^f = K. (Note that such f exists since Kdµ=2pc(M) =0.)

Proof. Necessity. Suppose u is a solution of (1), then consider v = u f , then

4^v =4^{u K} = e^{2v+2 f}Ke (2) Times e^2von both side and integrate, we obtain

Z Kee ^{2 f}dµ=

Z e ^2v4^v=

Z 2e ^2v|r^v|^2dµ⁰

by Green’s identity. If the equality holds, then r^v ⌘ 0, so v must be a constant, which implies eK ⌘ 0 by (2). Thus if eK 6⌘ 0, then we must have R _e

Ke^{2 f}dµ<0, and also eK have to change sign by Gauss-Bonnet theorem.

Sufficiency. If eK ⌘ 0, then obviously we can take u = f , so we only have to deal with the second case. Consider the subset of W^2,p(M),

S , {^u 2^W2,p(M),^Z udµ=

Z Kee ^{2u+2 f}dµ=0}

It is nonempty since eK changes sign. We want to minimize inS the Dirich-let energy

J(u), ¹₂^Z |r^u|^2dµ

Suppose there is a minimizer u₀2 S, then by the theory of Lagrange mul-tiplier, u0is an emtreme of

Z 1

2|r^u|²+au+b eKe^{2u+2 f}. 5

Compute the variation in u gives

4^u0+a+2b eKe^{2u0+2 f} =0.

Integrating over M gives

aA= 2b^Z Kee ^{2u0+2 f}dµ=0,

where A is the area of M. Thus e ^2u04^u0+2b eKe^{2 f} =0. Again integrating over M gives

2b^Z Kee ^{2 f}dµ=

Z e ^2u04^u0dµ.

= 2^Z e ^2u0|r^u0|² ⁰ which implies b >0 by the condition.

Thus we have v₀ = u₀+¹

2log 2b is a weak solution of (2), i.e., u = u₀+ 1

2log 2b+ f is a weak solution of (1). By elliptic regularity, if we can show that e^u 2 ^Lp(M)for all p > 1, then u is a smooth solution. To do this we need the following Lemmas.

Lemma 3.2 (Trudinger). For any compact Riemannian surface(M, g), there exists b, C >0 such that any u 2 ^{W1,2satisfying}

Z udµ =0,^Z |r^u|^2dµ¹

hasR

e^bu2dµ<C.

Proof. We fix a partition of unity{(^Ui, f_i)}^k_i=1with each U_i diffeomorphic to a 2 dimensional Euclidean disc D, and let u_i =f_iu, thus u=Â_iu_i.

We first prove that on each U_i ⇠= D with standard Euclidean metric, there exists c0 >0 such that

k^uikp ^c0p_p

kr^uik2, 8^{p 2.}

6

For v2 ^C1c(D), we have the following identity, v(x) = ¹

2p Z

D4^v(y)·^log(|^{x y}|)^dy

= ¹ 2p

Z

Dr^v(y)· ^{x y}

|^{x y}|^2dy By Holder inequality(¹_p+¹_q =1),

|^v(x)|  _2p¹ Z

D

⇣|r^v(y)|²|^{x y}| ^q⌘

1p

|^{x y}| ^2q|r^v(y)|^{1 2pdy}

 _2p¹

✓_Z

D|r^v(y)|²|^{x y}| ^qdy

◆¹_p ✓_Z

D|^{x y}| ^qdy

◆¹₂ ✓_Z

D|r^v(y)|^2dy

◆¹₂ ¹_p

the middle term is controlled by Z

D|^{x y}| ^qdy  Z

2D|^y| ^qdy=2^{1 q}p(p+2). Taking pth power and integrate over x and we have

Z

D|^v(x)|^pdx^c1(p+2)^p2+1

✓_Z

D|r^v(y)|^2dy

◆^p₂

)

✓_Z

D|^v(x)|^pdx

◆¹_p

^c2p_p^✓Z

D|r^v(y)|^2dy

◆¹₂

since C¹_c is dense in W^1,2, we have on each(U_i, g), k^uikp ^c0p_p

kr^uik2, for some c₀depend on g.

Now we have

k^ukp

Â

^kuikpc0pp

Â

^kruik2

^c3p_p(kr^uk²+k^uk²)

by equivalence of sobolev norms. We further use the following estimates:

7

Proposition 3.3 (Poincare-Wirtinger). For any u 2 ^W1,2(M), there exists C >0 such that

k^{u u}0k2 ^Ckr^uk2, where u0=R

udµ/A is the average of u.

Since we have u₀=R

udµ =0, we obtain k^ukp ^c4p_p

kr^uk²

Now consider the Taylor expansion of e^bu2, with kr^uk ^{1, we have} Z e^bu2dµ=

Z

Â

k=1

1

k!(b|^u|²)^k+A



Â

k=1

1

k!(2kbc²₄)^k+A^C for b sufficiently small (Take p =2k.)

Lemma 3.4. There existsC, h >0 such that for any u2^W1,2, Z e^udµ^{C exp}(hkr^uk²2+ ¹

A

Z udµ)

Proof. We may assumer^u6⌘0, and let u₀=u _A¹ R

udµ. We write u0  ^{b✓ u0}

kr^uk2

◆2

+ ¹

4bkr^uk²2,

where b is the constant appeared in the last lemma. Note that u₀

kr^uk2 satis-fies the condition of the last lemma, thus taking exponent and integrating

over M gives _Z

e^u0dµ^{C exp}( ¹

4bkr^uk²2), and hence the original statement. (Take h = ¹

4b.)

Lemma 3.5. Foru 2^W1,2(M), we have e^u 2 ^{Lpfor all p} >1.

Proof. Replace u by pu in the above lemma.

8

We still need to show the existence of the minimizer of J. Suppose{^uk} is a minimizing sequence of J inS^{, i.e.,}

J(u_i) !^c0 = inf

u2SJ(u) Then by Poincare-Wirtinger inequality we have

k^uik2 ^Ckr^uik2 =2CJ(u_i)¹²

So {^ui} is bounded in W^1,2, and there is a subsequence {^ui} weakly con-verging to some u0 2 ^W1,2. We have J(u0)  ^c0by the weak semi-lower-continuity of J. On the other hand, we can show thatR _e

Ke^2udµ is a contin-uous functional of u in W^1,2-weak topology, and hence

Z Kee ^{2u0+2 f}dµ= lim

i!•

Z Kee ^{2ui+2 f}dµ=0

and alsoR

u0dµ = 0, hence u0 2 S ^{and J}(u0) c0 by the definition of c0. As a consequence, u₀ is a minimizer of J inS, and we complete the proof of the theorem.

Proof of the continuity ofR _e

Ke^2udµ. Suppose {^ui}weakly converges to u, then by Rellich–Kondrachov theorem, they are strongly converge to u in L^pfor p>1, thus

Z Ke(e^ui e^u)dµ= Z ₁

0

Z

(u_i u)Ke^{e u+t(ui u)}dµdt!⁰ as i !0, by Lemma 3.4. and Holder’s inequality.

9

4 Case III: c ( M ) > 0

Only partial results are known for c(M) > 0. In this case, M must be diffeomorphic to S² or RP². We first consider the case M = S² with the standard 2-sphere metric g (K ⌘^1).

Proposition 4.1. On(S², g), If f is a first eigenfunction of4^{, i.e.,} 4^f+2f=0,

then any solution u of (1) must satisfies Z

(r^Ke· r^f)e^2udµ =0

Note that f can be seen as a linear function on R³restricted to S². Proof. Multiply (1) byr^u· r^fand integrate over S², we have

Z

(r^u· r^f)4^udµ Z

(r^u· r^f)dµ+ Z

(r^u· r^f)Ke^{e 2u}dµ=0 we deal with these three terms respectively. Note that f_,ij = fg_ij, so we have

Z

(r^u· r^f)4^udµ= Z

r(r^u· r^f)· r^udµ

= ¹ 2

Z

r(|r^u|²)· r^fdµ+ Z

|r^u|^2fdµ

= ¹ 2

Z

|r^u|²(4^f+2f)dµ=0, Z

(r^u· r^f)dµ= Z

f4^udµ = Z

f(Ke^{e 2u} 1)dµ= Z

f eKe^2udµ, Z

(r^u· r^f)Ke^{e 2u}dµ= ¹ 2

Z

(r^e2u· r^f)Kdµ^e

= ¹ 2

Z

(r(^{Kee 2u})· r^f)dµ 1 2

Z

(r^Ke· r^f)e^2udµ

= ¹ 2

Z Kee ^2u4^{fdµ 1}₂ Z

(r^Ke· r^f)e^2udµ summing up the three terms then the proposition follows.

10

With this, consider eK = 1+efwith e > 0 sufficinet small so that eK > 0, then the above proposition says

e Z

|r^f|^2e2udµ =0,

which impliesr^f = 0, or f is a constant, contradiction. This shows that even eK>0 is not sufficient for (1) to have a solution.

Nevertheless, we have the following result:

Theorem 4.2. On(S², g), if eK(x) = K^e( x)for all x2 ^{S2, and e}K is positive somewhere, then there is a solution u 2 ^C•(S²) such that u(x) = u( x) for all x 2^S2.

Thus for standard RP², the condition that eK is positive somewhere is necessary and sufficient.

We need the fact that in this case (S² and symmetry) we can actually take the constant h in Lemma 3.4 to be a number close to 1/32p, that is, we can take b = 8p # with small # in Lemma 3.2. We first prove for b=4p #/2 if we drop the condition u( x) = u(x).

The first step is to symmetrize u into a radially symmetric function u^# using the following Polya-Szego inequality [4].

Lemma 4.3 (Weil-Polya-Szego). Let (M, g)be a Riemannian surface such that the sectional curvature of M is bounded from above by k, with B ⇢ M an open subset diffeomorphic to R² and with smooth boundary. For u2 ^W1,2(B)and u^#is a radially symmetric function on a geodesic ball B^# on S = k ^1/2S² (2-sphere of radius k ^1/2) monotone in latitude, such that

|^B#| = |^B|^and|(^u#) ¹((t, •))| = |^{u 1}(t, •)|^{for any t} 2 ^R, then we have u^#2 ^W1,2(B^#)and

Z

B^#|r^u#(x)|^2dµS  Z

B|r^u(x)|^2dµM. (and hence we can replace u with u^#if M=S²and k =1.)

Now we rewrite things into spherical coordinate. Let q, f be the longi-tude and altilongi-tude of the 2-sphere, then the metric can be written as

ds² =df²+cos q dq², 11

and the area element is

dµ=cos q dq df.

The conditions on u is then Z

|r^u|^2dµ= Z

(u²_q+cos ²q u²_f)cos q dq df

=2p^{Z p}

pu²_q cos q dq ^1, Z u dµ=2p^Z u cos q dq=0.

Parametrize q by t such that e ^t/2 =tan

✓q 2

p 4

◆

, and let

w(t) = (4p)^1/2u(q), r(t) = ¹

4sech²t 2 then the conditions become

Z

|r^u|^2dµ = Z _•

•w⁰(t)²dt ^1, Z u dµ =

Z _•

•w(t)r(t)dt =0, and we want a bound for

Z e^{(4p #/2)u2}dµ=4p^{Z •}

•e^{(1 #/8p))2w(t)2}r(t)dt.

Note that r(t)has the properties

r(t) < C0e ^|t|, ^{Z •}

•r(t)dt=1 for some constant C₀. By Cauchy’s inequality we have

(w(r) w(s))² =

✓_{Z r}

s w⁰(t)dt

◆2



✓_{Z r}

s w⁰(t)²dt

◆ ✓_{Z r}

s 1dt

◆

 |^{r s}|^. Write(1 #/8p)² =1 t, then

Z _•

•e^{(1 #/8p)2w(t)2}r(t)dt^{C0e(1 t)C1} Z _•

•e ^t|t|dt=2C0e^{(1 t)C1}t ¹ 12

if we have the following estimate

w(t)²<|^t| +^C1

for some constant C1. To show this inequality, we write r(s)(w(t) w(s))^r(s)|(^{t s})|^1/2 integrate over s and we have

w(t)  Z _•

•r(s)|(^{t s})|^{1/2 ds} (|^t| +^C1)^1/2 for some C₁ (R_•

•r(s)|(^{t s})|^1/2ds)² |^t|^{for all t.}

Remark. In fact the boundedness is also true for b = 4p, but that requires a longer argument controlling the case when

d,^{1 max}_r>s (w(r) w(s))²/|^{r s}| is small.

We return to the case b = 8p e with u( x) = u(x). In Lemma 4.3, we take B to be a hemisphere on S², and k=p

2, then the lemma implies Z

S|r^u#(x)|^2dµS  ¹₂ Z

S²|r^u(x)|^2dµ_S^2, where S = p¹

2S², and thus Z

S²|r^u#(p

2x)|^2dµ_S²  ¹₂ Z

S²|r^u(x)|^2dµ_S²

if we scale u^#to the standard 2-sphere. Now we can run the same estimate for the functionp

2u^#(p

2x), and we haveR

e^{(8p #)u2}dµ bounded.

Now consider the setS , {^u 2 ^W1,2(S²),R

udµ = 0,R _e

Ke^2udµ > 0}^. This is nonempty since eK is positive somewhere. We want to minimize

13

J(u) = ¹

2kr^uk²2 2p log^Z Kee ^2udµ Note that c0 ,^infu2S J(u) ^✓1

2

2p 8p #

◆

kr^uk²2+c1 • by Lemma 3.4 and h = ¹

32p 4#. A similar argument of case II shows that J has a minimizer u0 2 S. The theory of Lagrange multiplier and calculus of variation then gives

4^u0+R^{4p e}_e^Ke2u0

Ke^2u0dµ l =0

for some multiplier l. Integrate over S²gives 4p 4pl, hence l =1 and we have u =u0+¹₂log(_4p¹ R _e

Ke^2u0dµ)is a solution of (1).

在文檔中 1 Basic Morse Theory (頁 66-79)