Initial value problems for diffusion equations with singular potential

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Initial value problems for diffusion equations with singular potential

Konstantinos Gkikas, Laurent Veron

To cite this version:

Konstantinos Gkikas, Laurent Veron. Initial value problems for diffusion equations with singular

potential. 2012. �hal-00736712v5�

potential

Konstantinos T. Gkikas and Laurent Veron

To Patrizia Pucci, with friendship and high esteem

Abstract. Let V be a nonnegative locally bounded function defined in Q

∞

:=

R

ⁿ

× (0, ∞ ). We study under what conditions on V and on a Radon measure µ in R

^d

does it exist a function which satisfies ∂

^t

u − ∆u + V u = 0 in Q

∞

and u(., 0) = µ. We prove the existence of a subcritical case in which any measure is admissible and a supercritical case where capacitary conditions are needed.

We obtain a general representation theorem of positive solutions when tV (x, t) is bounded and we prove the existence of an initial trace in the class of outer regular Borel measures.

Contents

1. Introduction 1

2. The subcritical case 6

3. The supercritical case 12

4. Initial trace 23

5. Appendix: the case of a bounded domain 28

References 30

1. Introduction

In this article we study the initial value problem for the heat equation (1.1) ∂ _t u − ∆u + V (x, t)u = 0 in Q _T := R ⁿ × (0, T )

u(., 0) = µ in R ⁿ ,

where V ∈ L

^∞

_loc (Q T ) is a nonnegative function and µ a Radon measure in R ⁿ . By a (weak) solution of (1.1) we mean a function u ∈ L ¹ _loc (Q _T ) such that V u ∈ L ¹ _loc (Q _T ), satisfying

(1.2) −

Z Z

Q

T

(∂ t φ + ∆φ) udxdt + Z Z

Q

T

V uφdxdt = Z

Ω

ζdµ

1991 Mathematics Subject Classification. 35K05, 35K67, 35K15, 35J70 .

Key words and phrases. Heat kernel Harnack inequality Radon measures initial trace representation.

1

for every function ζ ∈ C _c ^1,1;1 (Q _T ) which vanishes for t = T . Besides the singularity of the potential at t = 0, there are two main difficulties which appear for construct- ing weak solutions : the growth of the measure at infinity and the concentration of the measure near some points in R ⁿ . Diffusion equations with singular poten- tials depending only on x have been studied in connection with the stationnary equation (see e.g. [ 13 ]). The particular case of Hardy’s potentials v(x, t) = c | x |

⁻

² has been thorougly investigated since the early work of Baras and Goldstein [ 5 ], in connection with the problem of instantaneous blow-up. For time dependent sin- gular potentials most of the works are concentrated on the well posedness and the existence of a maximum principle; this is the case if V ∈ L

^∞

_t L x

ⁿ²

^,∞ , see e.g. [ 16 ]. In the case of time-singular potentials, a notion of non-autonomous Kato class have been introduced in [ 18 ] in order to prove that the evolution problem associated to the equation is well posed in L ¹ (R ⁿ ). This class is the extension to diffusion equations of the Kato’s class in Schr¨odinger operators. Other studies have been performed by probabilistic methods in order to analyze the L ^p − L ^q regularizing effect [12]. To our knowledge, no work dealing with the initial value problems with measure data for singular operators has already been published. We present here an extension to evolution equations of a series of questions raised and solved in the case of Schr¨odinger stationary equations in particular by [2], [3], [19], having in mind that one of the aim of this present work is to develop a framework adapted to the construction of the precise trace of solutions of semilinear heat equations. This aspect will appear in a forthcoming work [11]).

We denote by H (x, t) = _4πt ¹

ⁿ₂

e

^−|x|

2

4t

the Gaussian kernel in R ⁿ and by H[µ]

the corresponding heat potential of a measure µ ∈ M (R ⁿ ). Thus (1.3) H[µ](x, t) = _4πt ¹

ⁿ₂

Z

e

^−|x−y|

2 4t

dµ(y),

whenever this expression has a meaning: for example it is straightforward that if µ ∈ M (R ⁿ ) satisfies

(1.4) k µ k

^M_T

:=

Z

Rⁿ

e

^−|y|

2

4T

d | µ | (y) < ∞ ,

then (1.3) has a meaning as long as t < T , and let be M

_T

(R ⁿ ) the set of Radon measures in R ⁿ satisfying (1.4). If G ⊂ R ⁿ , let Q ^G _T be the cylinder G × (0, T ), B _R (x) the ball of center x and radius R and B _R = B _R (0). We prove

Theorem A Let the measure µ verifies (1.5)

Z Z

Q

^BR_T

H[ | µ | ](x, t)V (x, t)dxdt ≤ M R ∀ R > 0.

Then (1.1) admits a solution in Q T .

A measure which satisfies (1.5) is called an admissible measure and a measure for which there exists a solution to problem (1.1) is called a good measure. Notice that even when V = 0, uniqueness without any restriction on u is not true, however the next uniqueness result holds:

Theorem B Let u be a weak solution of (1.1) with µ = 0. If u satisfies (1.6)

Z Z

Q

T

(1 + V (x, t)) e

⁻

^λ

^|

^x

^|2

| u(x, t) | dxdt < ∞

for some λ > 0, then u = 0.

We denote by E ^ν (Q T ) the set of functions u ∈ L ¹ _loc (Q T ) for which (1.6) holds for some λ > 0. The general result we prove is the following.

Theorem C Let µ ∈ M (R ⁿ ) be an admissible measure satisfying (1.4). Then there exists a unique solution u µ ∈ E ^ν (Q T ) to problem (1.1). Furthermore

(1.7)

Z Z

Q

T

n 2T + V

| u | e

^{− |x|}

2

4(T−t)

dxdt ≤ Z

Rⁿ

e

^−|y|

2

4T

d | µ | (y).

We consider first the subcritical case, which means that any positive measure satisfying (1.4) is a good measure and we prove that such is the case if for any R > 0 there exist m _R > 0 such that

(1.8)

Z Z

Q

^BR_T

H (x − y, t)V (x, t)dxdt ≤ m R e

^−|y|

2 4T

.

Moreover we prove a stability result among the measures satisfying (1.4): if V verifies for all R > 0

(1.9) sup

y∈R

ⁿ

e

^|y|

2 4T

Z Z

E

H (x − y, t)V (x, t)dxdt → 0 when | E | → 0 , E Borel subset of Q ^B _T

^R

, then if { µ _k } is a sequence of Radon measures bounded in M

T

(R ⁿ ) which converges in the weak sense of measures to µ, then { (u µ

k

, V u _µ

_k

) } converges to (u µ , V u _µ ) in L ¹ _loc (Q _T ).

In the supercritical case, that is when not all measure in M

_T

(R ⁿ ) is a good measure, we develop a capacitary framework in order to characterize the good measures. We denote by M ^V (R ⁿ ) the set of Radon measures such that V H[µ] ∈ L ¹ (Q T ) and k µ k

M^V

:= k V H[µ] k L

¹

. If E ⊂ Q T is a Borel set, we set

(1.10) C V (E) = sup { µ(E) : µ ∈ M ^V

+ (R ⁿ ), µ(E ^c ) = 0, k µ k

M^V

≤ 1 } . This defines a capacity. If

(1.11) C _V

^∗

(E) = inf {k f k L

^∞

: ˇ H [f ](y) ≥ 1 ∀ y ∈ E } , where

(1.12) H ˇ [f ](y) =

Z Z

Q

T

H (x − y, t)V (x, t)f (x, t)dxdt = Z T

0

H[V f ](y, t)dt ∀ y ∈ R ⁿ , then C _V

^∗

(E) = C V (E) for any compact set. Denote by Z V the singular set of V , that is the largest set with zero C V capacity. Then

(1.13) Z _V = { x ∈ R ⁿ : Z Z

Q

T

H (x − y, t)V (y, t)dxdt = ∞} , and the following result characterizes the good measures.

Theorem D If µ is an admissible measure then µ(Z V ) = 0. If µ ∈ M

_T

(R ⁿ ) satisfies µ(Z V ) = 0, then it is a good measure. Furthermore µ is a positive good measure if and only if there exists an increasing sequence of positive admissible measures { µ k } which converges to µ in the weak ∗ topology.

Since many important applications deal with the nonlinear equation

(1.14) ∂ _t u − ∆u + | u | ^q

⁻

¹ u = 0 in Q

_∞

:= R ⁿ × (0, ∞ ),

where q > 1 and due to the fact that any solution defined in Q

_∞

satisfies (1.15) | u(x, t) | ^{q− 1} ≤ 1

t(q − 1) ∀ (x, t) ∈ Q

_∞

, we shall concentrate on potentials V which satisfy

(1.16) 0 ≤ V (x, t) ≤ C 1

t ∀ (x, t) ∈ Q T ,

for some C 1 > 0. For such potentials we prove the existence of a representation theorem for positive solutions of

(1.17) ∂ _t u − ∆u + V (x, t)u = 0 in Q _T .

If u is a positive solution of (1.1) in Q T with µ ∈ M ₊ (R ⁿ ), it is the increasing limit of the solutions u = u R of

(1.18)

∂ t u − ∆u + V (x, t)u = 0 in Q ^B _T

^R

u = 0 in ∂B R × (0, T ) u(., 0) = χ B

R

µ in B R ,

when R → ∞ , thus there exists a positive function H _V ∈ C(R ⁿ × R ⁿ × (0, T )) such that

(1.19) u(x, t) =

Z

R^N

H V (x, y, t)dµ(y).

Furthermore we show how to construct H V from V and we prove the following formula

(1.20) H V (x, y, t) = Z

R^N

e ^ψ(x,t) Γ(x, ξ, t)dµ y (ξ), where µ y is a Radon measure such that

(1.21) δ _y ≥ µ _y ,

(δ y is the Dirac measure concentrated at y), (1.22) ψ(x, t) =

Z T t

Z

Rⁿ

1

4π(s − t)

ⁿ₂

e

^−|x−y|

2

4(s−t)

V (y, s)dyds and Γ satisfies the following estimate

(1.23) c 1 t

⁻ⁿ²

e

⁻

^γ

^1|x−y|

2

t

≤ Γ(x, y, t) ≤ c 2 t

⁻ⁿ²

e

⁻

^γ

^2|x−y|

2 t

where A i , c i depends on T , d and V . Conversely, we first prove the following representation result

Theorem E Assume V satisfies (1.16). If u is a positive solution of (1.1) in Q _T , there exists a positive Radon measure µ in R ⁿ such that (1.19) holds.

If µ ∈ M

T

(R ⁿ ) is positive, we can define for any k > 0 the solution u k of (1.24) ∂ _t u − ∆u + V _k (x, t)u = 0 in Q _T

u(., 0) = µ in R ⁿ , where V k (x, t) = min { k, V (x, t) } , and

(1.25) u k (x, t) =

Z

R^N

H V

k

(x, y, t)dµ(y).

Moreover { H _V

_k

} and { v _k } decrease respectively to H _V and u

^∗

there holds

(1.26) u

^∗

(x, t) =

Z

R^N

H V (x, y, t)dµ(y).

However u

^∗

is not a solution of (1.1), but of a relaxed problem where µ is replaced by a smaller measure µ

^∗

called the reduced measure associated to µ. If we define the zero set of V by

(1.27) S ing V := { y ∈ R ^N : H V (x, y, t) = 0 } , we prove

Theorem F If

(1.28) lim sup

t→ 0

Z T t

Z

Rⁿ

1

4π(s − t)

ⁿ₂

e

^−|ξ−y|

2

4(s−t)

V (y, s)dyds = ∞ , then

ξ ∈ S ing V , i.e. H V (x, ξ, t) = 0, ∀ (x, t) ∈ R ⁿ × (0, ∞ ).

We note here that if V satisfies (1.28) then δ _ξ is not admissible measure and the reduced measure (δ ξ )

^∗

= µ _ξ associated to δ _ξ is zero.

Theorem G Assume V satisfies (1.15) and µ ∈ M

_T

(R ⁿ ). Then (i) supp(µ − µ

^∗

) ⊂ S ing V .

(ii) If µ( S ing V ) = 0, then µ

^∗

= 0.

(iii) S ing _V = Z _V .

The last section is devoted to the initial trace problem: to any positive solution u of (1.1) we can associate an open subset R (u) ⊂ R ⁿ which is the set of points y which possesses a neighborhood U such that

(1.29)

Z Z

Q

^U_T

V (x, t)u(x, t)dxdt < ∞ . There exists a positive Radon measure µ _u on R (u) such that (1.30) lim t

→

0

Z

Rⁿ

u(x, t)ζ(x)dx = Z

Rⁿ

ζdµ ∀ ζ ∈ C _c ( R (u)).

The set S (u) = R ⁿ \ R (u) is the set of points y such that for any open set U containing y, there holds

(1.31)

Z Z

Q

^U_T

V (x, t)u(x, t)dxdt = ∞ . If V satisfies (1.17), S (u) it has the property that

(1.32) lim sup _t

_→

₀

Z

U

u(x, t)dx = ∞ . Furthermore, if is satisfies (1.9), then S (u) = ∅ .

An alternative construction of the initial trace based on the sweeping method is also developed.

Precise definitions of the different notions used in the introduction will be given

in the next sections.

Aknowledgements This work has been prepared while the first author was visiting the Laboratory of Mathematics and Theoretical Physics, CNRS-UMR 7350, thanks to the support of a grant from R´egion Centre, in the framework of the program Cr´eation et propagations de singularit´es dans les ´equations non-lin´eaires.

2. The subcritical case

Let Q _T = R ⁿ × (0, T ]. In this section we consider the linear parabolic problem

(2.1) ∂ t u − ∆u + V u = 0 in Q T

u(., 0) = µ in R ⁿ × { 0 } , where V ∈ L ¹ _loc (Q T ) is nonnegative and µ is a Radon measure.

Definition 2.1. We say that µ ∈ M (R ⁿ ) is a good measure if problem (2.1) has a weak solution u i.e. there exists a function u ∈ L ¹ _loc (Q _T ), such that V u ∈ L ¹ _loc (Q _T ) which satisfies

(2.2) − Z Z

Q

T

u(∂ t φ + ∆φ)dxdt + Z Z

Q

T

V uφdxdt = Z

Rⁿ

φ(x, 0)dµ ∀ φ ∈ X (Q T ), where X (Q T ) is the space of test functions defined by

X (Q T ) = { φ ∈ C c (Q _T ), ∂ t φ + ∆φ ∈ L

^∞

_loc (Q

_∞

), φ(x, T ) = 0 }

Definition 2.2. Let H(x, t) be the heat kernel of heat equation in R ⁿ , we say that µ ∈ M (R ⁿ ) is an admissible measure if

(i)

|| V H[ | µ | ] || _L

¹

_(Q

^BR

T

) = Z Z

Q

^BR_T

Z

Rⁿ

H (x − y, t)d | µ(y) |

V (x, t)dxdt < M R,T

where M _R,T is a positive constant.

Definition 2.3. A function u(x, t) will be said to belong to the class E

V

(Q T ) if there exists λ > 0 such that

Z Z

Q

T

e

^−λ|x|2

| u(x, t) | (1 + V (x, t))dxdt < ∞ . A measure in R ⁿ belongs to the class M

_T

(R ⁿ ) if

k µ k

M_T

:=

Z

Rⁿ

e

^−|x|

2

4T

d | µ | < ∞ .

Lemma 2.4. There exists at most one weak solution of problem (2.1) in the class E

V

(Q T ).

Proof. Let u 1 and u 2 be two solutions in the class E

V

(Q T ) then w = u 1 − u 2 is a solution with initial data 0. Choose a standard mollifier ρ : B(0, 1) 7→ [0, 1] and define

w j (x, t) = j ⁿ Z

B

1 j

(x)

ρ(j(x − y))w(y, t)dy ≡ Z

B

1 j

(x)

ρ j (x − y)w(y, t)dy.

Then w j (., t) is C

^∞

and from the equation satisfied by w, it holds

∂ t w j − ∆w j + Z

B

1 j

(x)

V (y, t)ρ j (x − y)w(y)dy = 0,

where ∂ _t w _j is taken in the weak sense.

First we consider the case λ > 0 and t ≤ min { 16λ ¹ , T } = T

^′

. Set φ(x, t) = ξ(x, t)ζ(x), where ξ(x, t) = e

⁻

|x|2

4( 18λ−t)

and ζ ∈ C _c

^∞

(R ⁿ ). Given ε > 0 we define

g j = q w _j ² + ε.

Because ∂ t (g j φ) = √ ^∂ _w

^t

^w

2^j

j

+ε φ + g j ∂ t φ,, by a straightforward calculation we have Z

Rⁿ

h

g j φ(., s) i s=0 s=t dx =

Z Z

Q

t

w j

q w _j ² + ε

φ∆w j dxds

− Z Z

Q

t

w _j (x, s) q w _j ² (x, s) + ε

φ(x, s)



 Z

B

1 j

(x)

V (y, t)ρ j (x − y)w(y, s)dy



 dxds +

Z Z

Q

t

g _j φ _s dxds

= I 1 + I 2 + I 3 . By integration by parts, we obtain I 1 = −

Z Z

Q

t

|∇ w j | ² q

w ² _j (x, s) + ε

φdxds + Z Z

Q

t

|∇ w _j | ² w ² _j

(w _j ² (x, s) + ε)

³²

φdxds − Z Z

Q

t

w j

q

w ² _j + ε ∇ w j . ∇ φdxds

≤ − Z Z

Q

t

w j

q w ² _j + ε ∇ w j . ∇ φdxds

≤ − Z Z

Q

t

∇ g _j . ∇ φdxds

= −

Z Z

Q

t

ζ ∇ g _j . ∇ ξdxds − Z Z

Q

t

ξ ∇ g _j . ∇ ζdxds

= Z Z

Q

t

ζg _j ∆ξdxds + Z Z

Q

t

g _j ∇ ζ. ∇ ξdxds.

Since t ≤ T , there holds ξ |∇ g _j | ∈ L ¹ (Q T

^′

), ξg j ∈ L ¹ (Q T

^′

), | ∆ξ | g _j ∈ L ¹ (Q T

^′

),

∂ _s ξg _j ∈ L ¹ (Q T

^′

) and Z Z

Q

t

w _j (x, s) q w ² _j (x, s) + ε



 Z

B

1 j

(x)

V (y, t)ρ j (x − y)w(y, s)dy



 ξdxds < ∞ . The reason for which ξ |∇ g _j | ∈ L ¹ (Q T

^′

) follows from the next inequality

Z Z

Q

_T′

|∇ g _j | ξdxds = Z Z

Q

_T′

|∇ w _j | q ε + w _j ²

ξdxds

≤ Z Z

Q

_T′

e

⁻

|x|2 4( 18λ−t)



 Z

B

1 j

(x) |∇ ρ j (x − y) | w(y, s)dy



 dxds.

Since ∀ y ∈ B

¹

j

(x), we have | x | ² ≥ ( | y | ² − ¹ j ) ² = | y | ² + _j ¹

2

− 2

^|y|

_j ≥

^|y|

2

²

− (C − 1) _j ¹

2

,

for some positive constant C > 0 independent on j, y and x. Thus we have, using

the fact that e

⁻

^λ

^|

^y

^|2

w ∈ L ¹ (Q T ), Z Z

Q

_T′

|∇ g j | ξdxds ≤ C(j, λ) Z Z

Q

_T′

Z

B

1 j

(x)

e

⁻

|y|2 8( 18λ−t)

|∇ ρ j (x − y) | w(y, s)dydxds < ∞ . Also

Z Z

Q

t

w _j (x, s) q w ² _j (x, s) + ε

ξ



 Z

B

1 j

(x)

V (y, t)ρ j (x − y)w(y, s)dy



 dxds → j→∞

Z Z

Q

t

w ² (x, s)

p w ² (x, s) + ε ξV (y, t)dxds and Z

Rⁿ

q

w _j ² (x, s) + ε(ξ s + ∆ξ)dxds → ^j

→∞

Z

Rⁿ

p w ² (x, s) + ε(ξ s + ∆ξ)dxds.

We choose ζ _R = 1 in B _R , 0 ≤ ζ _R ≤ 1 in B R+1 \ B _R and 0 otherwise. Letting successively j → ∞ , R → ∞ and finally ε → 0, we derive

Z

Rⁿ

| w(x, t) | ξ(x, t)dx ≤ Z Z

Q

t

| w | (ξ s + ∆ξ)dxds − Z Z

Q

t

w(x, s)ξV (y, t)dxds.

Since

ξ _s + ∆ξ = − n 2( _8λ ¹ − s) ,

and V ≥ 0, we have w(x, t) = 0 ∀ (x, t) ∈ Q _T

^′

. If T

^′

= T this complete the proof for λ ≥ 0, otherwise the proof can be completed by a finite number of interations of the same argument on R ⁿ × (T

^′

, 2T

^′

), R ⁿ × (2T

^′

, 3T

^′

), etc. If λ = 0 we set ξ = 1

and the result follows by similar argument 2

Theorem 2.5. If µ ∈ M

T

(R ⁿ ) is an admissible measure, there exists a unique u = u _µ ∈ E

V

(Q T ) solution of (2.1). Furthermore the following estimate holds (2.3) n

2T Z Z

Q

T

| u | e

^{− |x|}

2

4(T−t)

dxds + Z Z

Q

T

| u | V e

^{− |x|}

2

4(T−t)

dxds ≤ Z

Rⁿ

e

^−|x|

2 4T

d | µ | . Proof. First we assume that µ ≥ 0. Let µ R = χ B

R

µ. It is well known that the heat kernel H ^B

^R

(x, y, t) in Ω = B R is increasing with respect to R and H ^B

^R

→ H, as R → ∞ in L ¹ (Q T ) for any T > 0. Thus µ R is an admissible measure in B R

and by Proposition 5.4, there exists a unique weak solution u R of problem 5.2 on Ω = B R . By (ii) of Proposition 5.5 we have

− Z Z

Q

T

| u R | (∂ t φ + ∆φ)dxdt + Z Z

Q

T

| u R | V φdxdt ≤ Z

B

R

φ(x, 0)d | µ R | . If we set φ _ε (x, t) = e

^{− |x|}

2

4(T+ε−t)

; ε > 0, then

∂ _t φ + ∆φ = − n

2(T + ε − t) e

^{− |x|}

2 4(T+ε−t)

, thus we have

Z Z

Q

T

| u R | n

2(T + ε − t) e

⁻

|x|2

4(T+ε−t)

dxdt+

Z Z

Q

T

| u R | V e

⁻

|x|2

4(T+ε−t)

dxdt ≤ Z

B

R

e

^−|x|

2 4T+4ε

dµ R ,

which implies n

2T + ε Z T

0

Z

B

R

| u _R | e

^{− |x|}

2

4(T+ε−t)

dxdt+

Z T 0

Z

B

R

| u _R | V e

^{− |x|}

2

4(T+ε−t)

dxdt ≤ Z

Rⁿ

e

^−|x|

2 4T+4ε

dµ _R .

Letting ε → 0, we derive n

2T Z Z

Q

T

| u R | e

⁻

|x|2

4(T−t)

dxdt+

Z Z

Q

T

| u R | V e

⁻

|x|2

4(T−t)

dxdt ≤ Z

Rⁿ

e

^−|x|

2 4T

dµ R ≤

Z

Rⁿ

e

^−|x|

2 4T

dµ.

Now by the maximum principle { u R } is increasing with respect to R and converges to some function u. By the above inequality u ∈ E

V

(Q T ) satisfies the estimate (2.5) and u is a weak solution of problem (2.1). By Lemma 2.4 it is unique. In the general case we write µ = µ ⁺ − µ

⁻

and the result follows by the above arguments and Lemma 2.4. In the sequel we shall denote by u _µ this unique solution. 2 Definition 2.6. A potential V is called subcritical in Q T if for any R > 0 there exists m _R > 0 such that

(2.4)

Z Z

Q

^BR_T

H (x − y, t)V (x, t)dxdt ≤ m R e

^−|y|

2

4T

∀ y ∈ R ⁿ . It is called strongly subcritical if moreover

(2.5) e

^|y|

2 4T

Z Z

E

H (x − y, t)V (x, t)dxdt → 0 when | E | → 0 , E Borel subset of Q ^B _T

^R

, uniformly with respect to y ∈ R ⁿ

Theorem 2.7. Assume V is subcritical. Then any measure in M _T (R ⁿ ) is admissible. Furthermore, if V is strongly subcritical and { µ k } is a sequence of measures uniformly bounded in M _T (R ^N ) which converges weakly to µ, then the corresponding solutions { u _µ

_k

} converge to u _µ in L ¹ _loc (Q _T ), and { V u _µ

_k

} converges to V u _µ in L ¹ _loc (Q _T ).

Proof. For the first statement we can assume µ ≥ 0 and there holds Z Z

Q

^BR_T

H (x − y, t)dµ(y)V (x, t)dxdt = Z

Rⁿ

Z Z

Q

^BR_T

H(t, x − y)V (x, t)dxdt

! dµ(y)

≤ m R

Z

Rⁿ

e

^−|y|

2 4T

dµ(y)

≤ m _R k µ k

^MT

.

Thus µ is admissible. For the second statement, we assume first that µ k ≥ 0. By lower semicontinuity µ ∈ M _T (R ^N ) and k V H[µ] k _L

¹

_(Q

^BR

T

) ≤ M R,T for any k. Since 0 ≤ u _µ

_k

≤ H[µ k ] and H[µ k ] → H[µ] in L ¹ _loc (Q _T ), the sequence { u _µ

_k

} is uniformly integrable and thus relatively compact in L ¹ _loc (Q _T ). Furthermore 0 ≤ V u _µ

_k

≤ V H[µ k ]. Let E ⊂ Q ^B _T

^R

be a Borel subset, then

Z Z

E

V H[µ k ]dxdt = Z

Rⁿ

Z Z

E

V H (x − y, t)dxdt

dµ k (y)

= Z

Rⁿ

e

^|y|

2 4T

Z Z

E

V (x)H (x − y, t)dxdt

e

^−|y|

2 4T

dµ k (y)

≤ ǫ( | E | ) k µ k k

M_T

,

where ǫ(r) → 0 as r → 0. Thus { (u µ

k

, V u _µ

_k

) } is locally compact in L ¹ _loc (Q _T ) and, using a diagonal sequence, there exist u ∈ L ¹ _loc (Q _T ) with V u ∈ L ¹ _loc (Q _T ) and a subsequence { k _j } such that { (u µ

_kj

, V u _µ

_kj

) } converges to (u µ , V u _µ ) a.e. and in L ¹ _loc (Q _T ). From the integral expression (2.2) satisfied by the u µ

k

, u is a weak solution of problem (2.1). Since the u µ

k

satisfy (2.3), the property holds for u, thereforeu = u µ is the unique solution of (2.1), which ends the proof. 2 .

As a variant of the above result which will be useful later on we have Proposition 2.8. Assume V satisfies

(2.6) e

^|y|

2 4T

Z Z

E

H(x − y, t)V (x, t + τ)dxdt → 0 when | E | → 0 , E Borel subset of Q ^B _T

^R

, uniformly with respect to y ∈ R ⁿ and τ ∈ [0, τ 0 ]. Let τ k > 0 with τ k → 0 and { µ k } be a sequence uniformly bounded in M _T (R ^N ) which converges weakly to µ. Then the solutions { u τ

k

,µ

k

} of

(2.7) ∂ t u − ∆u + V u = 0 on R ⁿ × (τ k , T ) u(., τ k ) = µ k on R ⁿ × { τ k }

(extended by 0 on (0, τ k )) converge to u _µ in L ¹ _loc (Q T ), and { V u _µ

_k

} converges to V u µ in L ¹ _loc (Q T ).

Condition (2.5) may be very difficult to verify and we give below a sufficient condition for it to hold.

Proposition 2.9. Assume V satisfies

(2.8) lim

λ

→

0 e

^|y|

2 4T

λ

⁻ⁿ

Z λ 0

Z

B

_λ2

(y)

V (x, t)dxdt = 0 uniformly with respect to y ∈ R ⁿ , then V is strongly subcritical.

Proof. Let E ⊂ Q ^B _T

^R

be a Borel set. For δ > 0, we define the weighted heat ball of amplitude δe

^−|y|

2 4T

by P _δ = P _δ (y, T ) =

(x, t) ∈ Q _T : H (x − y, t) ≥ δe

^−|y|

2 4T

.

By an straightforward computation, one sees that P δ (y, T ) ⊂ B

a

n

δ

⁻¹ⁿ

e

^|y|

2 4nT

(y) × [0, b n δ

⁻ⁿ²

e

^|y|

2

2nT

] := R δ (y, T ), for some a n , b n > 0. We write

Z Z

E

H (x − y, t)V (x, t)dxdt = Z Z

E

∩

P

δ

H (x − y, t)V (x, t)dxdt+

Z Z

E

∩

P

_δ^c

H (x − y, t)V (x, t)dxdt.

Then Z Z

E

∩

P

_δ^c

H(x − y, t)V (x, t)dxdt ≤ δe

^−|y|

2 4T

Z Z

E

V (x, t)dxdt,

and Z Z

E

∩

P

δ

H (x − y, t)V (x, t)dxdt ≤ Z δ

0

Z

{

(x,t)

∈

Q

^BR_T

:H(x

−

y,t)=τ e

^−|y|

2 4T }

V (x, t)dS τ (x, t)τ dτ

≤

"

τ Z τ

0

Z

{

(x,t)

∈Q^BRT

:H(x−y,t)=σe

^−|y|

2 4T }

V (x, t)dS σ (x, t)dσ

# τ=δ

τ=0

− Z δ

0

Z τ 0

Z

{

(x,t)

∈Q^BRT

:H(x−y,t)=σe

^−|y|

2 4T }

V (x, t)dS σ (x, t)dσdτ

≤ δ Z δ

0

Z

{

(x,t)

∈

Q

^BR_T

:H(x

−

y,t)=σe

^−|y|

2 4T }

V (x, t)dS σ (x, t)dσ.

The first integration by parts is justified since V ∈ L ¹ (Q ^B _T

^R

). Notice that δ

Z δ 0

Z

{

(x,t)

∈Q^BRT

:H(x−y,t)=σe

^−|y|

2 4T }

V (x, t)dS σ (x, t)dσ = δ Z Z

Q

^BR_T ∩Pδ

V (x, t)dxdt and

δ Z Z

Q

^BR_T ∩Pδ

V (x, t)dxdt ≤ δ Z Z

Q

^BR_T ∩Rδ

(y,T )

V (x, t)dxdt

≤ βr

⁻

ⁿ Z αr

0

Z

B

R∩

B

_(αr)2

(y)

V (x, t)dxdt,

for some α, β > 0 and if we have set r = δ

⁻¹ⁿ

. Notice also that B R ∩ B (αr)

²

(y) = ∅ if | y | ≥ R + (αr) ² , or, equivalently, if | y | ≥ R + α ² δ

⁻ⁿ²

.

(i) If | y | ≥ R + α ² , we fix δ such that 1 < δ, then e

^|y|

2 4T

Z Z

E

H (x − y, t)V (x, t)dxdt ≤ δ Z Z

E

V (x, t)dxdt, which can be made smaller than ǫ provided | E | is small enough.

(ii) If | y | < R + α ² , then e

^|y|

2 4T

Z Z

E

∩

P

_δ^c

H(x − y, t)V (x, t)dxdt ≤ e

^R2+α

4 2T

Z Z

E

∩

P

_δ^c

H(x − y, t)V (x, t)dxdt

≤ δe

^R2+α

4 2T

Z Z

E

V (x, t)dxdt.

Given ǫ > 0, we fix δ = r

⁻

ⁿ such that e

^R2+α

4 2T

Z Z

E∩P

δ

H(x − y, t)V (x, t)dxdt ≤ βe

^{R2 +α}

4 2T

r

⁻ⁿ

Z αr 0

Z

B

R∩B(αr)2

(y)

V (x, t)dxdt ≤ ǫ 2 , and then η > 0 such that | E | ≤ η implies

e

^|y|

2 4T

Z Z

E

∩

P

_δ^c

H (x − y, t)V (x, t)dxdt ≤ δe

^R2+α

4 2T

Z Z

E

V (x, t)dxdt ≤ ǫ 2 . Therefore

e

^|y|

2 4T

Z Z

E

H(x − y, t)V (x, t)dxdt ≤ ǫ,

which is (2.5). 2

Remark In Theorem 2.7 and Proposition 2.9, the assumption of uniformity with respect to y ∈ R ⁿ in (2.5), (2.6) and (2.8) can be replaced by uniformity with respect to y ∈ B _R

₀

if all the measures µ _k have their support in B _R

₀

. A extension of these assumptions, valid when the convergent measures µ _k have their support in a fixed compact set is to assume that V is locally strongly subcritical, which means that (2.5) holds uniformly with respect to y in a compact set. Similar extension holds for (2.8).

3. The supercritical case

3.1. Capacities. All the proofs in this subsection are similar to the ones of [19] and inspired by [9]; we omit them. We assume also that there exists a positive measure µ 0 such that H[µ 0 ]V ∈ L ¹ (Q T ).

Definition 3.1. If µ ∈ M ₊ (R ⁿ ) and f is a nonnegative measurable function defined in Ω such that

(t, x, y) 7→ H[µ](y, t)V (x, t)f (x, t) ∈ L ¹ (Q T × R ⁿ ; dxdt ⊗ dµ), we set

E (f, µ) = Z

Q

T

Z

Rⁿ

H (x − y, t)dµ(y)

V (x, t)f (x, t)dxdt.

If we put

H[f ˘ ](y) = Z

Q

T

H (x − y, t)V (x, t)f (x, t)dxdt,

then by Fubini’s Theorem, ˘ H[f ](y) < ∞ , µ − almost everywhere in R ⁿ and E (f, µ) =

Z

Rⁿ

Z

Q

T

H (x − y, t)V (x, t)f (x, t)dxdt

dµ(y).

Proposition 3.2. Let f be fixed. Then (a) y 7→ H[f ˘ ](y) is lower semicontinuous in R ⁿ .

(b) µ 7→ E (f, µ) is lower semicontinuous in M ₊ (R ⁿ ) in the weak* topology.

Definition 3.3. We denote by M ^V (R ⁿ ) the set of all measures µ on R ⁿ such that V H[ | µ | ] ∈ L ¹ (Q T ). If µ is such a measure, we set

|| µ ||

^MV

= Z

Q

T

Z

Rⁿ

H (x − y, t)d | µ | (y)

V (x, t)dxdt = || V H[ | µ | ] || ^L

¹

^(Q

T

) . If E ⊂ R ⁿ is a Borel set, we put

M ₊ (E) = { µ ∈ M ₊ (R ⁿ ) : µ(E ^c ) = 0 } and M ^V ₊ (E) = M ^V (R ⁿ ) ∩ M ₊ (E).

Definition 3.4. If E ⊂ R ⁿ is any borel subset we define the set function C V

by

C V (E) := sup { µ(E) : µ ∈ M ^V ₊ (E), || µ ||

^MV

≤ 1 } ; this is equivalent to,

C _V (E) := sup

µ(E)

|| µ ||

^MV

: µ ∈ M ^V

+ (E)

.

Proposition 3.5. The set function C _V satisfies C V (E) ≤ sup

y∈E

Z

Q

T

H (x − y, t)V (x, t)dxdt

−

1

∀ E ⊂ R ⁿ , E Borel.

Furthermore equality holds if E is compact. Finally,

C V (E 1 ∪ E 2 ) = sup { C V (E 1 ), C V (E 2 ) } ∀ E i ⊂ R ⁿ , E i Borel.

Definition 3.6. For any Borel E ⊂ R ⁿ , we set

C _V

^∗

(E) := inf {|| f || ^L

^∞

: ˘ H[f ](y) ≥ 1 ∀ y ∈ E } . Proposition 3.7. For any compact set E ⊂ R ⁿ ,

C _V

^∗

(E) = C _V (E).

3.2. The singular set of V . In this section we assume that V satisfies (1.16), although much weaker assumption could have been possible. We define the singular set of V , Z V by

(3.1) Z _V =

x ∈ R ⁿ : Z Z

Q

T

H (x − y, t)V (y, t)dydt = ∞

.

Since the function x 7→ f (x) = Z Z

Q

T

H (x − y, t)V (y, t)dydt is lower semicontinuous, it is a Borel function and Z V is a Borel set.

Lemma 3.8. If x ∈ Z V then for any r > 0, Z Z

Q

^Br_T ^(x)

H (x − y, t)V (y, t)dydt = ∞ .

Proof. We will prove it by contradiction, assuming that there exists r > 0, such

that Z Z

Q

^Br_T ^(x)

H(x − y, t)V (y, t)dy ≤ M.

Replacing H by its value, we derive Z Z

Q

T

H(x − y, t)V (y, t)dydt = Z Z

Q

^Br(x)_T

H(x − y, t)V (y, t)dydt + Z Z

Q

^Bc_T^r(x)

H (x − y, t)V (y, t)dydt

≤ M + C(n) Z T

0

t

⁻ⁿ⁺²²

e

^−r

2

4t

dt < ∞ .

Which is clearly a contradiction. 2

Lemma 3.9. If µ is an admissible positive measure then µ(Z V ) = 0.

Proof. Let K ⊂ Z _V be a compact set. In view of the above lemma there exists a R > 0 such that K ⊂ B _R and for each x ∈ K, we have

(3.2)

Z Z

Q

^B_T^2R

H(x − y, t)V (y)dy = ∞ and

(3.3)

Z Z

Q

^Bc_T^2R

H(x − y, t)V (y)dy < ∞ .

Now, µ _K = χ _K µ is an admissible measure and by Fubini theorem we have Z Z

Q

T

Z

Rⁿ

H (x − y, t)dµ K (y)

V (x, t)dxdt = Z

K

Z Z

Q

T

H (x − y, t)V (x, t)dxdtdµ(y)

= Z

K

Z Z

Q

^B_T^2R

H (x − y, t)V (x)dxdtdµ(y) +

Z

K

Z Z

Q

^Bc_T^2R

H (x − y, t)V (x)dxdtdµy.

By (3.3) the second integral above is finite and by (3.2) Z Z

Q

^B_T^2R

H(x − y, t)V (x)dxdt = ∞ ∀ y ∈ K.

It follows that µ(K) = 0. This implies µ(Z V ) = 0 by regularity. 2 Theorem 3.10. If µ ∈ M _T (R ⁿ ), µ ≥ 0 such that µ(Z V ) = 0, then µ is a good measure.

Proof. We set µ R = χ B

R

µ. By Proposition 5.8, since Z _V ^B

^R

⊂ Z V , µ R is a good measure in B _R with corresponding solution u ^R _µ . In view of Lemma 2.5, u ^R _µ satisfies

Z Z

Q

^BR_T

| u ^R _µ | n

4(T − t) e

^{− |x|}

2

4(T−t)

dxdt + Z Z

Q

^BR_T

| u ^R _µ | V e

^{− |x|}

2

4(T−t)

dxdt ≤ Z

B

R

e

^−|x|

2 4T

dµ.

Also { u ^R _µ } is an increasing function, thus converges to u µ . By the above estimate we have that u _µ belong to class E

V

(Q T ) and is a weak solution of (2.1). 2 Proposition 3.11. Let µ ∈ M ₊ (R ⁿ ). Then µ(Z V ) = 0 if and only if there exists an increasing sequence of positive admissible measures which converges to µ in the weak* topology.

Proof. The proof is similar as the one of [ 19 , Th 3.11] and we present it for the sake of completeness. First, we assume that µ(Z V ) = 0. Then we define the set

K _N =

x ∈ R ⁿ : Z

Q

T

H (x − y, t)V (y)dydt ≤ N

.

We note that Z _V ∩ K _N = ∅ . We set µ _n = χ _K

_N

µ then we have Z

Q

T

Z

Rⁿ

H (x − y, t)dµ n (y)

V (x, t)dxdt ≤ µ(K N ).

Thus µ _n is admissible, increasing with respect n. By the monotone theorem it follows that µ _n → χ _Z

_V^c

µ. Since µ(Z V ) = 0 the result follows in this direction.

For the other direction. Let { µ n } be an increasing sequence of positive admissible measure. Then by Lemma 3.9 we have that µ n (Z V ) = 0, ∀ n ≥ 1. Since µ n ≤ µ, there exist an increasing functions h n µ − integrable such that µ n = h n µ. Since

0 = µ n (Z V ) → µ(Z V ) the result follows. 2

3.3. Properties of positive solutions and representation formula. We first recall the construction of the kernel function for the operator w 7→ ∂ t w − ∆w + V w in Q T , always assuming that V satisfies (1.16). For δ > 0 and µ ∈ M _T , we denote by w _δ the solution of

(3.4) ∂ t w − ∆w + V δ w = 0, in Q T

w(., 0) = µ in R ⁿ .

where V _δ = V χ _Q

_δ,T

and Q _δ,T = (δ, T ) × R ⁿ . Then

(3.5) w _δ (x, t) =

Z

Rⁿ

H _V

_δ

(x, y, t)dµ(y)

Lemma 3.12. The mapping δ 7→ H V

δ

(x, y, t) is increasing and converges to H _V ∈ C(R ⁿ × R ⁿ × (0, T ])) when δ → 0. Furthermore there exists a function H _V ∈ C(R ⁿ × R ⁿ × (0, T ])) such that for any µ ∈ M _T (R ⁿ )

(3.6) lim

δ→ 0 w δ (x, t) = w(x, t) = Z

Rⁿ

H V (x, y, t)dµ(y).

Proof. Without loss of generality we can assume µ ≥ 0. By the maximum principle δ 7→ H _V

_δ

(x, y, t) is increasing and the result follows by the monotone

convergence theorem. 2

If R ⁿ is replaced by a smooth bounded domain Ω, we can consider the problem (3.7)

∂ _t w − ∆w + V _δ w = 0 in Q ^Ω _T

w = 0 in ∂ _l Q ^Ω _T := ∂Ω × (0, T ] w(., 0) = µ in Ω.

where V _δ

^′

= V χ _Q

^Ω

δ,T

and Q ^Ω _δ,T = (δ, T ) × Ω. Then

(3.8) w _δ (x, t) =

Z

Ω

H _V ^Ω

_δ

(x, y, t)dµ(y) The proof of the next result is straightforward.

Lemma 3.13. The mapping δ 7→ H _V ^Ω

_δ

(x, y, t) increases and converges to H _V ^Ω ∈ C(Ω × Ω × (0, T ])) when δ → 0. Furthermore There exists a fonction H _V ^Ω ∈ C(Ω × Ω × (0, T ])) such that for any µ ∈ M _b (Ω)

(3.9) lim

δ→ 0 w _δ (x, t) = w(x, t) = Z

Ω

H _V ^Ω (x, y, t)dµ(y).

Furthermore H _V ^Ω ≤ H _V ^Ω

^′

≤ H _V if Ω ⊂ Ω

^′

.

It is important to notice that the above results do not imply that w is a weak solution of problem (1.1). This question will be considered later on with the notion of reduced measure.

Lemma 3.14. Assume µ ∈ M ₊ (R ⁿ ) is a good measure and let u be a positive weak solution of problem (2.1). If Ω is a smooth bounded domain, then there exists a unique positive weak solution v of problem

(3.10)

∂ _t v − ∆v + V v = 0, in Q ^Ω _T , v = 0 on ∂ _l Q ^Ω _T v(., 0) = χ Ω µ in Ω.

Furthermore

(3.11) v(x, t) =

Z

Ω

H _V ^Ω (x, y, t)dµ(y).

Proof. Let { t _j }

^∞

j=1 be a sequence decreasing to 0, such that t _j < T, ∀ j ∈ N.

We consider the following problem (3.12)

∂ _t v − ∆v + V v = 0, in Ω × (t j , T ], v = 0 on ∂Ω × (t j , T ] v(., t j ) = u(., t j ) in Ω × { t _j } .

Since u, V u ∈ L ¹ (Q ^B _T

^R

) for any R > 0, t 7→ u(., t) is continuous with value in L ¹ _loc (R ⁿ ), therefore u(., t j ) ∈ L ¹ _loc (R ⁿ ) and there exists a unique solution v j to (3.12) (notice also that V ∈ L

^∞

(Q ^B _T

^R

)). By the maximum principle 0 ≤ v _j ≤ u and by standard parabolic estimates, we may assume that the sequence v _j converges locally uniformly in Ω × (0, T ] to a function v ≤ u. Also, if φ ∈ C ^1,1;1 (Q ^Ω _T ) vanishes on ∂ l Q ^Ω _T and satisfies φ(x, T ) = 0, we have

− Z T

t

j

Z

Ω

v j (∂ t φ+∆φ)dxdt+

Z T t

j

Z

Ω

V v j φdxdt+

Z

Ω

φ(x, T − t j )v j (x, T − t j )dx = Z

Ω

φ(x, 0)u(x, t j )dx, where in the above equality we have taken φ(., . − t _j ) as test function. Since φ(., T −

t _j ) → 0 uniformly and u(., t j ) → µ in the weak sense of measures, it follows by the dominated convergence theorem that

− Z Z

Q

^Ω_T

v(∂ t φ + ∆φ)dxdt + Z Z

Q

^Ω_T

V vφdxdt = Z

Ω

φ(y, 0)dµ(y),

thus v is a weak solution of problem (3.10). Uniqueness follows as in Lemma 2.4.

Finally, for δ > 0, we consider the solution w _δ of (3.7). Then it is expressed by (3.5). Furthermore

− Z Z

Q

^Ω_T

w δ (∂ t φ + ∆φ)dxdt + Z Z

Q

^Ω_T

V δ w δ φdxdt = Z

Ω

φ(x, 0)dµ(x),

The sequence w δ is decreasing, with limit w. Since w δ ≥ v, then w ≥ v. If we assume φ ≥ 0, it follows from dominated convergence and Fatou’s lemma that

− Z Z

Q

^Ω_T

w(∂ t φ + ∆φ)dxdt + Z Z

Q

^Ω_T

V wφdxdt ≤ Z

Ω

φ(x, 0)dµ(x),

Thus w is a subsolution for problem (3.10) for which we have comparison when

existence. Finally w = v and (3.11) holds. 2

Lemma 3.15. Assume µ ∈ M ₊ (R ⁿ ) is a good measure and let u be a positive weak solution of problem (2.1). Then for any (x, t) ∈ R ⁿ × (0, T ], we have

R lim

→∞

u R = u,

where { u _R } is the increasing sequence of the weak solutions of the problem (3.10) with Ω = B _R . Moreover, the convergence is uniform in any compact subset of R ⁿ × (0, T ] and we have the representation formula

u(x, t) = Z

Rⁿ

H V (x, y, t)dµ(y).

Proof. By the maximum principle, u _R ≤ u _R

^′

≤ u for any 0 < R ≤ R

^′

. Thus

u _R → w ≤ u. Also by standard parabolic estimates, this convergence is locally

uniformly. Now by dominated convergence theorem, it follows that w is a weak

solution of problem (2.1) with initial data µ. Now we set w e = u − w ≥ 0. Since w e satisfies in the weak sense

e

w _t − ∆ w e + V w e = 0 in Q _T w(x, t) e ≥ 0 in Q T

w(x, e 0) = 0 in R ⁿ , and V ≥ 0, it clearly satisfies

e

w t − ∆ w e ≤ 0 in Q T

w(x, t) e ≥ 0 in Q T

w(x, e 0) = 0 in R ⁿ ,

which implies w e = 0. By the previous lemme u R admits the representation u ^R (x, t) =

Z

B

R

H _V ^B

^R

(x, y, t)dµ(y).

Since { H _V ^B

^R

} is an increasing sequence and lim R

→∞

H _V ^B

^R

= H V , we have using again Fatou’s lemma as in the proof of Lemma 3.15

u(x, t) = lim

R→∞ u ^R (x, t) = lim

R→∞

Z

B

R

H _V ^B

^R

(x, y, t)dµ(y) = Z

Rⁿ

H V (x, y, t)dµ(y) 2 Lemma 3.16. Harnack inequality Let C 1 > 0 and V (x, t) be a potential satisfying (1.16) If u is a positive solution of (1.17), then the Harnack inequality is valid:

u(y, s) ≤ u(x, t) exp

C(n, C 1 )

| x − y | ² t − s + t

s + 1

, ∀ (y, s), (x, t) ∈ Q _T , s < t.

Proof. We extend V for t ≥ T by the value C 1 t

⁻

¹ . We consider the linear parabolic problem

(3.13) ∂ t u − ∆u + V u = 0, in R ⁿ × [1, ∞ ),

It is well known that, under the assumption (1.16), every positive solution u(x, t) of (3.13) satisfies the Harnack inequality

u(y, s) ≤ u(x, t) exp

C(n, C 1 )

| x − y | ² t − s + t

s + 1

, ∀ (x, t) ∈ R ⁿ × [1, ∞ ).

Set ˜ u(x, t) = u( _λ ^t

2

x

λ ). Then ˜ u satisfies u t − ∆u + 1

λ ² V ( t λ ²

x

λ )˜ u = 0, in R ⁿ × (0, ∞ ).

We note here that _λ ¹

2

V ( _λ ^t

2

x

λ ) ≤ C 1 , ∀ t ≥ λ ¹

²

, thus ˜ u satisfies the Harnack inequality e

u(y, s) ≤ u(x, t) exp e

C(n, C 1 )

| x − y | ² t − s + t

s + 1

, ∀ (x, t) ∈ R ⁿ × [ 1 λ ² , ∞ ).

By the last inequality and the definition of ˜ u we derive the desired result. 2 Next, we set

(3.14) S ing V (R ⁿ ) := { y ∈ R ⁿ : H V (x, y, t) = 0 }

If H _V (x, y, t) = 0 for some (x, t) ∈ Q _T , then H _V (x

^′

, y, t

^′

) = 0 for any (x

^′

, t

^′

) ∈ Q _T ,

t

^′

< t by Harnack inequality principle. We prove the Representation formula.

Theorem 3.17. Let u be a positive solution of (1.17). Then there exists a measure µ ∈ M ₊ (R ⁿ ) such that

u(x, t) = Z

Rⁿ

H V (x, y, t)dµ(y), and µ is concentrated on ( S ing V (R ⁿ )) ^c .

Proof. By Lemma 3.15 we have u(x, t) =

Z

Rⁿ

H V (x, y, t − s)u(y, s)dy for any s < t ≤ T.

We assume that s ≤ ^T 2 . By Harnack inequality on x 7→ H V (x, y, ^T ₂ ) Z

Rⁿ

H V (0, y, T

2 )u(y, s)dy ≤ c(n) Z

Rⁿ

H V (0, y, T − s, )u(y, s)dy = c(n)u(0, T ).

For any Borel set E, we define the measure ρ s by ρ _s (E) :=

Z

E

H _V (0, y, T

2 )u(y, s)dy ≤ Z

Rⁿ

H _V (0, y, T

2 )u(y, s)dy ≤ c(n)u(T, 0).

Thus there exists a decreasing sequence { s _j }

^∞

j=1 which converges to origin, such that the measure ρ s

j

converges in the weak* topology to a positive Radon measure ρ. Also we have the estimate ρ(R ⁿ ) ≤ C(n)u(0, T ). Now choose (x, t) ∈ Q T and j 0

large enough such that t > s j

0

. Let ε > 0, we set for any j ≥ j 0 , W j (y) = H _V (x, y, t − s _j )

H V (0, y, ^T ₂ ) + ε . For any R > 0 and | y | > R we have

W _j (y) ≤ 1

ε H _V (x, y, t − s _j ) ≤ 1

ε H (x − y, t − s _j ) < 1

ε C(x, R, t − s _j ), where lim R

→∞

C(x, R, t − s j ) = 0. We have also

Z

|

y

|≥

R

W j (y)dρ j ≤ 1

ε C(x, R, t − s j )c(n)u(T, 0).

For any | y | < R, we have by standard parabolic estimates that W j (y) → _H

_V

^H ₍

^VT

^(x,y,t)

2

,0,y)+ε

when j → ∞ , uniformly with respect to y. Thus by the above estimates it follows Z

Rⁿ

W _j (y)dρ j → Z

Rⁿ

H _V (x, y, t) H V (0, y, ^T ₂ ) + ε dρ.

For sufficiently large j we have Z

Rⁿ

H V (x, y, t − s j ) H V (0, y, ^T ₂ ) + ε dρ _s

_j

=

Z

Rⁿ

H V (x, y, t − s j ) H V (0, y, ^T ₂ ) + ε

H _V (0, y, T

2 ) + ε − ε

u(y, s j )dy

= u(x, t) − ε Z

Rⁿ

H V (x, y, t − s j , )

H _V (0, y, ^T ₂ ) + ε u(y, s j )dy.

Note that this is a consequence of the identity Z

Rⁿ

H V (x, y, t − s j )u(y, s j )dy = u(x, t).

Thus as before, we define d ρ e _j = H _V (x, y, t − s _j )u(y, s j )dy and thus there exists a subsequence, say { ρ e _j } converges in the weak* topology to a positive Radon measure e

ρ. Thus we have ε

Z

Rⁿ

H V (x, y, t − s j )

H _V (0, y, ^T ₂ ) + ε u(y, s j )dy = ε Z

Rⁿ

χ (

SingV

(

Rⁿ

))

^c

H V (x, y, t − s j )

H _V (0, y, ^T ₂ ) + ε u(s j , y)dy

→ ε Z

Rⁿ

χ (

SingV

(

Rⁿ

))

^c

1

H V ( ^T ₂ , 0, y) + ε d ρ. e Combining the above relations, we derive

(3.15) Z

Rⁿ

H _V (x, y, t)

H V (0, y, ^T ₂ ) + ε dρ = u(x, t) − ε Z

Rⁿ

χ ₍

_S

_ing

_V

₍

_Rⁿ

₎₎

^c

1

H V (0, y, ^T ₂ ) + ε d ρ. e Now, we have

ε lim

→

0 χ (

SingV

(

Rⁿ

))

^c

ε

H V ( ^T ₂ , 0, y) + ε = 0, and by Harnack inequality on the function x 7→ H V (x, y, t)

H V (x, y, t)

H _V (0, y, ^T ₂ ) + ε ≤ C(t, T ),

thus by dominated convergence theorem, we can let ε tend to 0 in (3.15) and obtain Z

Rⁿ

H V (x, y, t)

H V (0, y, ^T ₂ ) dρ = u(x, t).

And the result follows if we set

dµ = χ ₍

_S

_ing

_V

_(R

ⁿ

₎₎

^c

1

H V (0, y, ^T ₂ ) dρ.

2 In the next result we give a construction of H _V , with some estimates and a different proof of the existence of an initial measure for positive solutions of (1.16).

Theorem 3.18. Assume V satisfies (1.16) and u is a positive solution of (1.17) then there exists a positive Radon measure µ in R ⁿ such that

(3.16) u(x, t) =

Z

Rⁿ

ǫ ^ψ(x,t) Γ(x, y, t, 0)dµ(y) where

(3.17) ψ(x, t) =

Z T t

Z

Rⁿ

e

^−|x−y|

2 4(s−t)

4π(t − s) V (y, s)dyds and

(3.18) c 1

e

⁻

^γ

^1|x−y|

2 s−t

(t − s)

ⁿ²

≤ Γ(x, y, t, s) ≤ c 2

e

⁻

^γ

^2|x−y|

2 s−t

(t − s)

ⁿ²

for some positive constants c i and γ i , i = 1, 2.

Proof. Assuming that u is a positive solution of (1.17), we set u(x, t) = e ^ψ(x,t) v(x, t). Then

(3.19) ∂ _t v − ∆v − 2 ∇ ψ. ∇ v − |∇ ψ | ² v + (∂ t ψ − ∆ψ + V )v = 0.

We choose ψ as the solution of problem

(3.20) − ∂ t ψ − ∆ψ + V ψ = 0 in Q T

ψ(., T ) = 0 in R ⁿ .

Then ψ is expressed by (3.17). Furthermore, by standard computations, (3.21) (i) 0 ≤ ψ(x, t) ≤ c ln ^T _t

(ii) |∇ ψ(x, t) | ≤ c 1 (T ) + c 2 (T ) ln ^T _t The function v satisfies

(3.22) ∂ _t v − ∆v − 2 ∇ ψ. ∇ v − |∇ ψ | ² v = 0.

Then, by (3.21),

(3.23)

(i) 0 ≤

Z

Rⁿ

sup {| ψ(x, s) | ^q : x ∈ R ⁿ } ds ≤ M 1

(ii) 0 ≤

Z

Rⁿ

sup {|∇ ψ(x, s) | ^q : x ∈ R ⁿ } ds ≤ M 2

for any 1 ≤ q < ∞ for some M _i ∈ R + . This is the condition H in [4] with R 0 = ∞ and p = ∞ . Therefore there exists a kernel function Γ ∈ C(R ⁿ × R ⁿ × (0, T ) × (0, T )) which satisfies (3.18) and there exists also a positive Radon measure µ in R ⁿ such that

(3.24) v(x, t) =

Z

Rⁿ

Γ(x, y, t, 0)dµ(y).

Finally u verifies

(3.25) u(x, t) = e ^ψ(x,t) Z

Rⁿ

Γ(x, y, t, 0)dµ(y).

2 We recall that S ing V (R ⁿ ) := { y ∈ R ⁿ : H V (x, y, t) = 0 } .

Theorem 3.19. Let δ ξ be the Dirac measure concentrated at y and let V sat- isfies (1.16). Then

H V (x, ξ, t) = Z

Rⁿ

e ^ψ(x,t) Γ(x, y, t)dµ ξ (y), where µ _ξ is a positive Radon measure such that

δ _ξ ≥ µ _ξ ,

and ψ, Γ are the functions in (3.17) and (3.18) respectively.

Furthermore, if

lim sup

t

→

0 ψ(ξ, t) = lim sup

t

→

0

Z T t

Z

Rⁿ

1

4π(s − t)

ⁿ₂

e

^−|ξ−y|

2

4(s−t)

V (y, s)dyds = ∞ then

ξ ∈ S ing _V , i.e. H V (x, ξ, t) = 0, ∀ (x, t) ∈ R ⁿ × (0, ∞ ).

proof. First we note that H _V

_k

(x, ξ, t) is the solution of problem (1.24) with δ _ξ as initial data. Since H _V

_k

(x, ξ, t) ↓ H _V (x, ξ, t), we have by maximum principle , H(x, ξ, t) ≥ H _V (x, ξ, t). Now by Theorem 3.18, there exists a positive Radon measure µ _ξ in R ⁿ such that

(3.26) H V (x, , ξ, t) = Z

Rⁿ

ǫ ^ψ(x,t) Γ(x, y, t, 0)dµ ξ (y)

Let φ ∈ C 0 (R ⁿ ) then we have by the properties of Γ(x, ξ, t) (see [4]) and (3.26)

t lim

→

0

Z

Rⁿ

H V (x, ξ, t)φ(x)dx ≥ lim

t

→

0

Z

Rⁿ

Z

Rⁿ

Γ(x, y, t)φ(x)dxdµ ξ (y) = Z

Rⁿ

φ(y)µ ξ (y), That is

(3.27)

Z

Rⁿ

φ(x)dδ ξ (x) ≥ Z

Rⁿ

φ(x)dµ ξ (x) ⇒ δ _ξ ≥ µ _ξ , since φ is an abstract function in space C ₀ (R ⁿ ).

Also we have that there exist positive constants C ₁ , C ₂ such that (3.28) Γ(x, y, t) ≥ C 1 H(x, y, C 2 t).

Also we have

H(ξ, ξ, t) ≥ H V (ξ, ξ, t) = Z

Rⁿ

H V (ξ, y, t)dµ ξ (y) = Z

Rⁿ

e ^ψ(ξ,t) Γ(ξ, y, t)dµ ξ (y) (by (3.28)) ≥ C 1

Z

B(ξ,

√

C

2

t)

e ^ψ(t,ξ) H (ξ, y, C 2 t)dµ ξ (y) (By Harnack inequality) ≥ C(T, n, C 1 , C 2 )

Z

B(ξ,

√

C

2

t)

e ^ψ(ξ,t) H (ξ, ξ, C 2 t

2 )dµ ξ (y)

= C(T, n, C 1 , C 2 )e ^ψ(ξ,t) H (ξ, ξ, C 2 t

2 )µ ξ (B(ξ, p C 2 t)) Thus by the last inequality and the fact that

H (ξ, ξ, t)

H (ξ, ξ, ^C ₂

²

^t ) = C(C 2 , n) > 0, we have

C(T, n, C 1 , C 2 ) ≥ e ^ψ(t,ξ) µ ξ (B(ξ, p C 2 t)).

But limsup t

→

0 ψ(ξ, t) = ∞ which implies

t→ lim 0 µ _ξ (B(ξ, p

C 2 t) = µ _ξ ( { ξ } ) = 0.

Thus by (3.27) we have µ ξ ≡ 0, i.e. H V (x, ξ, t) = 0, ∀ (x, t) ∈ R ⁿ × (0, ∞ ). 2 3.4. Reduced measures. In this section we assume that V is nonnegative, but not necessarily satisfies (1.16), therefore we can construct H _V [µ] for µ ∈ M _T (R ⁿ ). Furthermore, if µ is nonnegative we can consider the solution u _k of the problem

(3.29) ∂ _t u − ∆u + V ^k u = 0, in Q _T u(., 0) = µ in R ⁿ , where V ^k = min { V, k } . Then there holds

u k (x, t) = Z

Rⁿ

H _V

^k

(t, x, y)dµ(y) = H _V

^k

[µ](x, t),