5 Harnack inequality for the heat equation

4P_w; 2d₀ 1

2

; 2 (1 ) d₀ 1

2)

= max (

4P_w; 2d₀ 1

2) :

Note that Pw is a constant which depends on d0 and P2, we conclude that C is a constant which depends on , d0 and P2. This …nishes the proof.

5 Harnack inequality for the heat equation

Lemma 5.1. Let fR g _2(0;1] be collection of measurable subsets of M R such that R ⁰ R if ⁰ . Fix m > 0, K > 0, 0 2 ¹₂; 1 and 0 < p₁ < p₀ 1. Let u be positive measurable on R1. If

Z

R₀

u^p0

!_p0¹

K 1

( ⁰)^mjR¹j

1 p

1

p0 Z

R

u^p

1 p

;

holds for all , ⁰ and p such that ¹₂ ⁰ < 1 and 0 < p p₁ < p₀, and if, moreover, for any 2 (0; 1),

jf(x; t) 2 R : ln u > gj KjR¹j ¹

holds for all > 0. Then there exists a constant C0 = C0(m; K; 0; p0; p1) > 0 such that Z

R₀

u^p0

!_p0¹

C₀jR¹j^p01 :

Proof. De…ne : [ 0; 1]! R by

( ) = ln 1 jR¹j

Z

R

u^p0

1 p0 ;

for any 2 [ ⁰; 1]. Now, for a given 2 [ ⁰; 1], we let 8<

:

R ;1=n

(x; t)2 R¹ : ln u > ^{( )}₂ o R _;2=n

(x; t)2 R¹ : ln u ^{( )}₂ o ; then R = R ;1[ R ^;2 and, for any (x; t) 2 R ^;2,

u (x; t) e ^{( )2} : (1)

So, in this case, for any 0 < p < p0, applying Hölder’s inequality and by the de…nition of ,

Note that, by second assumption, we have

jR ^;1j = (x; t)2 R¹ : ln u > ( ) which is equivalent to

1

Solving

e ^{( )} 2K ( )

1 p

1 p0

= e ^{( )2} ; (4)

we get

2K ( )

1 p

1

p0 = e ^{( )2} and thus

1 p

1

p₀ = ( ) 2 ln ^{( )}_2K

: (5)

Note that, if ^{( )}_2K 1, then 2K ( ) = ln _jR¹

1j

R

R u^p0

1

p0 which implies that R

R u^p0

1 p0

jR¹j^p01 e^2K. This satis…es the conclusion of this theorem. Thus, it is enough to analyze the case of ^{( )}_2K > 1. In this case, we have ln ^{( )}_2K > 0. Since ( ) is always positive, the solution p to (5) can be choosen so that p < p0. Also, since ln ^{( )}_2K tends to 0⁺as K tends to ^{( )}₂ , the solution p to (5) can be choosen so that p p₁ < p₀ by enlarging K to be large enough. Therefore, (3) also holds for such p. Pluging (4) into (3), we obtain that

1 jR¹j

Z

R

u^p e ^{( )2 p}+ e ^{( )2 p}

= 2 e ^{( )2}

p

which is equivalent to Z

R

u^p 2jR¹j e ^{( )2 p}: (6)

Now, for any 0 0 < , if ( ⁰) > 2K ( ⁰) ^2m, then, since

( ⁰) = ln 1 jR¹j

Z

R₀

u^p0

!_p0¹

ln 1 jR¹j

Z

R

u^p0

1 p0

= ( ) ; we have

( ) > 2K ( ⁰) ^2m (7)

and thus, by the …rst assumption,

( ⁰) = ln 1 jR¹j

Z

R₀

u^p0

!_p0¹

= lnjR¹j ^p01 Z

R₀

u^p0

!_p0¹

lnjR¹j ^p01 K 1 ( ⁰)^mjR¹j

1 p

1

p0 Z

R₀

u^p0

!¹_p

lnjR¹j ^p01 K 1 ( ⁰)^mjR¹j

1 p

1 p0

2jR¹j e ^{( )2}

p ¹_p

(by (6))

= 1

p₀ lnjR¹j + ln K + 1 p

1

p₀ ln ( ⁰) ^mjR¹j ¹+1

pln 2jR¹j e ^{( )2}

p

= ln K + 1 p

1

p₀ ln ( ⁰) ^m+ 1

pln 2 + ( ) 2

= ln K + ( ) 2 ln ^{( )}_2K

ln ( ⁰) ^m+1

pln 2 + ( )

2 (by (5))

ln K + ( ) 2 ln ^{( )}_2K

ln ( ) 2K

1 2

+ 1

pln 2 + ( )

2 (by (7))

= ln K + ( ) 4 +1

pln 2 + ( ) 2

= 3

4 ( ) + ln K + 1 pln 2:

That is,

( ⁰) 3

4 ( ) + ln 2^1pK:

Note that the last inequality holds when ( ⁰) > 2K ( ⁰) ^2m. Since ( ⁰) may by less or equal to 2K ( ⁰) ^2m, together with the last inequality, we arrive at

( ⁰) 3

4 ( ) + ln K + 1

pln 2 + 2K ( ⁰) ^2m: (8)

Recall that ¹_{p p}¹

0 = ^{( )}

2 ln( ^{( )}_2K) and 0 < 1, we have 1

p = ( )

2 ln ^{( )}_2K + 1

p₀

< (1) 2 ln ⁽_2K⁰⁾

+ 1 p₀: So, setting

L = (1) 2 ln ⁽_2K⁰⁾

+ 1 p₀;

(8) becomes

( ⁰) 3

4 ( ) + ln K + L ln 2 + 2K ( ⁰) ^2m

= 3

4 ( ) + ln 2^LK + 2K ( ⁰) ^2m: (9) Here we should note that L is independent with and ⁰.

Finally, iterating on (9) by choosing some > 1 so that ³₄ ^2m < 1 and letting i+1 =

i+ ^{i 1}(1 ₀) for all integer i 0, we get ( ₀) 3

4 ( ₁) + ln 2^LK + 2K ¹(1 ₀) ^2m

= 3

4 ( ₁) + ln 2^LK + 2K ^2m(1 ₀) ^2m 3

4 3

4 ( ₂) + ln 2^LK + 2K ²(1 ₀) ^2m + ln 2^LK + 2K ^2m(1 ₀) ^2m

= 3 4

3

4 ( ₂) + ln 2^LK + 2K ^4m(1 ₀) ^2m + ln 2^LK + 2K ^2m(1 ₀) ^2m

= 3

4

2

( ₂) + 1 + 3

4 ln 2^LK + 2K (1 ₀) ^{2m 2m}+3 4

4m

...

= 3

4

k

( _k) + ln 2^LK Xk 1

i=0

3 4

i

+ 2K (1 ₀) ^2m Xk 1

i=0

2m 3 4

2m i

:

Letting k ! 1, since ³4

2m< 1, we obtain that

( ₀) 4 ln 2^LK + 8K (1 ₀) ^2m: Recall that

( ₀) = ln 1 jR¹j

Z

R₀

u^p0

!_p0¹

; the last inequality becomes

ln 1 jR¹j

Z

R₀

u^p0

!¹

p0

4 ln 2^LK + 8K (1 ₀) ^2m

= ln 2^4LK⁴e^{8K(1 0) 2m} which is equivalent to

Z

R₀

u^p0

!_p0¹

C₀jR¹j^p01 ;

where C0 = 2^4LK⁴e^{8K(1 0) 2m}. Since L depends only on 0, p0 and K, we conclude that C0

is a constant depends nothing but m, K, 0, p0 and p1. Therefore we get the proof.

Theorem 5.1. Let E be a ball of center x0 2 M and raius r > 0 in M. Let t⁰ > 2r² and u be a positive solution to the Dirichlet heat equation on E [t₀ 2r²; t₀]. If M has both VDC and WPI, then there exists a constant CH = C_H (d₀; P₂) > 0 such that

sup

Q

u C_Hinf

Q+

u;

where

Q₊= ¹₂E t₀ ¹₂r²; t₀

Q = ¹₂E t₀ 2r²; t₀ ³₂r² ;

d₀ and P2 are controlling constants in VDC and WPI, respectively. The inequality in this theorem is called the Harnack inequality.

Proof. Let a be the constant in Theorem 4.3 (see page 74). Since u is a positive solution (to the Dirichlet heat equation on E [t₀ 2r²; t₀]), both e ^au and e ^au ¹ are positive solution. Let t1 = t₀ r², then Q can be rewritten as Q = ¹₂E t₁ r²; t₁ ¹₂r² . Also, for any 0 < " < 1, we set

Q_+;" = ¹₂ + " E t₀ ¹₂ + " r²; t₀

Q _;" = ¹₂ + " E t₁ r²; t₀ ¹₂ + " r² :

Now, since e ^au is a positive solution on E [t₀ 2r²; t₀], e ^au is a supersolution on E [t₁ r²; t₁]. By Theorem 4.2 and Theorem 4.3, we know that e ^au saties…es the conditions in Lemma 5.1 and thus

"Z

Q ;"

e ^au ^p0

#_p0¹

C0; jEj r²

1 p0

or, equivalently,

Z

Q ;"

u^p0

!_p0¹

e^aC_0; jEj r²

1

p0 : (1)

Thus

sup

Q

u CRM

1 jEj r²

1

p0 Z

Q ;"

u^p0

!_p0¹

(by (4.2))

C_RM 1

jEj r²

1

p0 e^aC_0; jEj r²

1

p0 (by (1))

= C_RMC_0; e^a: (2)

On the other hand, since e ^au ¹ is a positive solution on E [t₀ 2r²; t₀], e ^au ¹ is a subsolution on E [t₀ r²; t₀]. By Theorem 4.1 and Theorem 4.3, we know that e ^au ¹ saties…es the conditions in Lemma 5.1 and thus

"Z

Q+;"

e ^au ^{1 p0}

#¹

p0

C_0;+ jEj r²

1 p0

or, equivalently,

"Z

Q+;"

u ^{1 p0}

#_p0¹

e^aC_0;+ jEj r²

1

p0 : (3)

Note that u ¹ is a positive solution since u is a positive solution, so u ¹ is a subsolution and thus, by Theorem 4.1, we have

sup

Q+

u ¹ C_M 1

jEj r²

1 p0 "Z

Q+;"

u ^{1 p0}

#_p0¹

which is equivalent to

"Z

Q+;"

u ^{1 p0}

#_p0¹

C_M¹ jEj r²

1 p0 sup

Q+

u ¹: (4)

So, by (3) and (4), we get

C_M¹ jEj r²

1 p0 sup

Q+

u ¹ e^aC_0;+ jEj r²

1 p0

or, equivalently,

sup

Q+

u ¹ C_MC_0;+e^a:

Note that sup_Q+u ¹ = inf_Q+u ¹, thus the last inequality becomes inf

Q+

u C_M¹C_0;+¹e ^a which is equivalent to

e^a C_MC_0;+e^2ainf

Q+

u: (5)

Finally, plugging (5) into (2), we obtain that sup

Q

u C_RMC_0; C_MC_0;+e^2ainf

Q+

u

= C_Hinf

Q+

u;

where CH = C_RMC_MC_0; C_0;+e^2a. Since CH is contant depends only on d0 and P2, we get the proof.

References

[1] A. Grigor’yan, The heat equation on non-compact Riemannian manifolds, (in Russian) Matem. Sbornik, 182 (1991) no.1, 55-87. Engl. transl.: Math. USSR Sb., 72 (1992) no.1, 47-77.

[2] A. Grigor’yan, Heat Kernel and Analysis on Manifolds, Providence, R.I.: American Mathematical Society, 2009.

[3] A. Grigor’yan, Heat Kernels on Weighted Manifolds and Applications, Contemporary Mathematics, 398 (2003) 143-172.

[4] A. Young, Eigenvalues and the Heat Kernel, unpublished, 2003.

[5] C. J. Sung, Lecture Notes on Heat Kernels, unpublished, 1998.

[6] D. Jerison, The Poincaré inequality for vector …elds satisfying Hörmander’s condition, Duke Math. J. vol. 53, 1986, 503-523.

[7] E. B. Davies, Heat kernels and spectral theory, Cambridge University Press, 1989.

[8] E. M. Stein, Singular integrals and di¤erentiability properties of functions, Princeton, N.J.: Princeton University Press, 1970.

[9] I. Chavel, Eigenvalues in Riemannian Geometry, Pure and Applied Mathematics, vol.

115, Academic Press Inc., Orlando, FL, 1984.

[10] I. Chavel, Riemannian Geometry - A Modern Introduction, Cambridge Tracts in Math-ematics, vol. 108, Cambridge University Press, Cambridge, 1993.

[11] J. Moser, A Harnack inequality for parabolic di¤erential equations, Comm. Pure Appl.

Math., 17 (1964) 101-134. Correction: Comm. Pure Appl. Math., 20 (1967) 231-236.

[12] Kensuke Onishi, Jin-ichi Itoh, Estimation of the necessary number of points in Rie-mannian Voronoi diagram, 15th Canadian Conference on Computational Geometry, 2003.

[13] L. Salo¤-Coste, Aspects of Sobolev inequalities, LMS Lecture Notes Series 289, Cam-bridge Univ. Press, 2002.

[14] L. Salo¤ -Coste, A note on Poincaré, Sobolev, and Harnack inequalities. Internat. Math.

Res. Notices 1992, no. 2, 27-38.

[15] N. Th. Varopoulos, Hardy-Littlewood theory for semigroups, J. Funct. Anal., 63 (1985) no.2, 240-260.

[16] P. Li, R. Schoen, Lp and mean value properties of subharmonic functions on Rie-mannian manifolds, Acta Math., 153 (1984) 279-301.

[17] P. Li, L. Karp, The heat equation on complete Riemannian manifolds, unpublished, 1983.

[18] P. Li, Harmonic functions and applications to complete manifolds, unpublished, 2004.

[19] P. Li, Harmonic functions on complete Riemannian Manifolds. “Handbook of Geomet-ric Analysis, No. 1” Advanced Lectures in Mathematics, Vol. 7 (2008), International Press.

[20] P. Li, Lecture Notes on Geometric Analysis, Lecture Notes Series No. 6 - Research Insti-tute of Mathematics and Global Analysis Research Center, Seoul National University, Seoul, 1993.

[21] P. Li, Geometric Analysis, Cambridge Studies in Advanced Mathematics, vol. 134, 2012.

[22] Pierre H. Bérard, Spectral Geometry: Direct and Inverse Problems, Springer, 1986.

[23] Qi S. Zhang, Sobolev Inequalities, Heat Kernels under Ricci Flow, and the Poincaré Conjecture. CRC Press, 2010.

[24] Richard Schoen, Shing-Tung Yau, Lectures on Di¤erential Geometry, International Press, 1994.

在文檔中 在完備黎曼流形上針對熱方程的哈納克不等式 (頁 85-93)