Amelia Carolina Sparavigna

The Medieval Bastides, Their Urban Planning and Some Possible Astronomical Orientations

Keywords: Architecture, Urban planning, Archaeoastronomy, SunCalc.net Software, Photographer's Ephemeris Software, CalSky Software

Bastides are the fortified towns built in medieval Europe, in particular in France, during the thirteenth and fourteenth centuries. Many of these towns are displaying a regular planning based on a grid of parallel and perpendicular streets. Here we will show some examples of these layouts using the satellite images. The cases of the bastides of Sauveterre-De-Guyenne, Aigues-Mortes, Libourne and Pavie will be discussed in more detail, because they have possible astronomical orientations. Sauveterre-De-Guyenne has one of its axes oriented along the southern moonrise on a major lunar standstill. Probably this orientation was chosen, because the year of the foundation of the bastide (AD 1281), was a year of major lunar standstill. In the cases of Aigues-Mortes, Libourne and Pavie we can see alignments along the sunrise on solstices.

Several words exist that are coming from the Provencal verb "bastir", which is meaning to build, to construct. For what concerns the verbs, we have in Italian "imbastire", the same of the English "to baste", that is, to sew with long loose stitches in order to hold something in place temporarily. Words like "bastion" and "battlement" are coming from "bastir". Bastion derives from "bastillon", diminutive of Old French "bastille", which is a fortress, tower, or fortified building. [START_REF]Etymology on Line[END_REF]. Battlement is coming from Old French "bataillement", earlier "bastillement", a fortification, from the verb "bastillier", which means to fortify, to equip with battlements to have a "bastille" [START_REF]Etymology on Line[END_REF][START_REF] Vv | Wikipedia[END_REF]. Let us remember that a famous bastille existed. It was the Bastille, the Paris fortress used as prison, destroyed by revolutionaries on July 14, 1789. Another word is "bastide". Bastides are the fortified towns built in medieval France during the thirteenth and fourteenth centuries [START_REF] Vv | Wikipedia[END_REF]. Some of them were built under Raymond VII of Toulouse to replace villages destroyed during the Albigensian Crusade [START_REF] Vv | Wikipedia[END_REF]. It seems that almost 700 bastides were built between 1222 (Cordes-sur-Ciel, Tarn) and 1372 (La Bastide d'Anjou, Tarn) [START_REF] Randolph | The Bastides of southwest France[END_REF]. Most of these towns were founded in the southwest of France that, at the time, was a frontier region belonging partly to France and partly to the kings of England [START_REF] Vv | Medieval planned towns[END_REF]. As observed in [START_REF] Vv | Medieval planned towns[END_REF], the bastides were built to establish a more modern society in a part of Europe that was rather wild and inhospitable. These towns were used to collect the population scattered in villages into centers which were easy to control and defend. This new environment helped to develop the trade and other productive activities.

In [START_REF] Morris | History of Urban Form Before the Industrial Revolution[END_REF], the bastides are defined as the planned new towns (in France several places exist having the name of Villeneuve, that is of "new town"). However, they were not established in an arbitrary manner; most were set up on the sites of existing villages or at the intersections of routes. In this sense, the bastides were "new towns", due to the new layout they displayed. When compared to the medieval towns, they "were built in order to put some order into society, not to accommodate a rapidly growing population" [START_REF] Vv | Medieval planned towns[END_REF]. To remark the new order of the society, most bastides were laid out on a grid pattern, with a central square; the grid pattern may have been inspired by the model of the ancient Roman colonies. Actually, the Roman colonies were also "new towns" established by Rome in the conquered regions. Therefore, when allowed by the local environment, the bastides have a planning consisting of an almost regular grid of streets. The main roads in the grid are knowns as carreyras, or carriage ways, since they are wide enough for carts [START_REF] Vv | Medieval planned towns[END_REF]. These streets are crossing at a central square, generally surrounded by arcades [START_REF] Vv | Medieval planned towns[END_REF]. This square was used as the market place. Sometimes it possessed also a covered market hall [START_REF] Vv | Medieval planned towns[END_REF]. The bastides were built at a time of relative peace and prosperity. Before the start of the Hundred Years' War, the early bastides were not fortified [START_REF] Vv | Medieval planned towns[END_REF]; however once Anglo-French relations deteriorated, many bastides were fortified. One of the bastides, of which we can see well preserved walls is Aigues-Mortes, the medieval port of Crusaders on the edge of the Camargue. The bastides were discussed also by Francis Haverfield, British historian and archaeologist, in his book on the ancient town planning [START_REF] Haverfield | Ancient Town-Planning[END_REF]. This book is mainly devoted to the town planning of the Romans; the last chapter, entitled "The sequel", starts from a question. "What was the sequel to this long work of town-planning?" Haverfield pointed out two facts for substantiating his answer. "First, the Roman planning helped the towns of the Empire to take definite form, but when the Empire fell, it too met its end. Only here and there its vestiges lingered on in the streets of scattered cities like things of a former age. But, secondly, from this death it rose again, first in the thirteenth century, with ever-growing power to set the model for the city life of the modern world. ... Early in the thirteenth century men began to revive, with certain modifications, the rectangular planning which Rome had used. Perhaps copying Roman originals seen in northern Italy, Frederic Stupor Mundi now built on a chess-board pattern the Terra Nova which he founded in Sicily. Now, in 1231, Barcelonette was built with twenty square 'insulae' in south-eastern France. Now, too, the 'Bastides' and 'Villes Neuves' of southern France and towns like Aigues-Mortes (1240) were built on similar plans. Soon after, the chess-board pattern came to England and was used in Edwardian towns like Flint and Winchelsea ... It is unnecessary here to follow further the renaissance of town-planning. By intervals and revivals it continued to spread. In 1652 it reached Java, when the Dutch built Batavia. In 1682 it reached America, when Penn founded Philadelphia". In [START_REF] Haverfield | Ancient Town-Planning[END_REF], Haverfield illustrated the plan of a bastide by that of Sauveterre-De-Guyenne, near Bordeaux (AD 1281), as given in the Figure 1. Let us start from this bastide our discussion on the layout and related orientation of some of them. We will see that this bastide is very interesting, because in its orientation we find a link to the year of its foundation.

Figure 1:

The plan of the bastide of Sauveterre-De-Guyenne, near Bordeaux (AD 1281), illustrated by A.E. Brinckmann in [START_REF] Haverfield | Ancient Town-Planning[END_REF].

Sauveterre-De-Guyenne and the lunar standstill

This bastide was built in a village, the name of which was "Athala". The bastide was first named "Salva-Terra" (Salvation of the land), which later became Saubeterre and then Sauveterre [START_REF] Vv | Wikipedia[END_REF]. It was founded as an English Bastide in 1281 by King Edward I of England. As told in [START_REF] Vv | Wikipedia[END_REF], the king signed the "Charte des Coutumes de la Cité" in 1283, which protected the inhabitants and set the rules of life in the village community [START_REF] Vv | Wikipedia[END_REF]. Sauveterre-de-Guyenne maintains the typical Bastide 'grid' layout, with the entry points into the city guarded by four stone tower gates, and a central market square surrounded by stone houses having arcades on the ground floor [START_REF] Vv | Wikipedia[END_REF]. And in fact, if we use satellite maps, such as that provided by Wikipedia (Figure 2), we have the same layout depicted in [START_REF] Haverfield | Ancient Town-Planning[END_REF]. In the Figure 3, the northern gate is shown. As we have shown in several previous discussions [START_REF] Sparavigna | The Town Planning of Pompeii and Herculaneum Having Streets Aligned Along Sunrise on Summer Solstice[END_REF][START_REF] Sparavigna | Roman Towns Oriented to Sunrise and Sunset on Solstices[END_REF][START_REF] Sparavigna | Wien and the Winter Solstice[END_REF][START_REF] Sparavigna | I Castra Albana Orientati Verso Il Solstizio D' Estate (The Castra Albana Oriented to the Summer Solstice)[END_REF][START_REF] Sparavigna | The Walled Town of Alife and the Solstices[END_REF][START_REF] Sparavigna | Alignments of the Streets of Torino with Moonrise and Moonset Azimuths on Major and Minor Lunar Standstills[END_REF][START_REF] Sparavigna | Alcuni siti romani in Piemonte orientati coi lunistizi[END_REF][START_REF] Sparavigna | On the Orientation of the Roman Grumentum[END_REF][START_REF] Sparavigna | The Roman Saepinum Oriented to Moonrise on a Major Lunar Standstill[END_REF][START_REF] Sparavigna | Augusta Emerita and the Major Lunar Standstill of 24 BC[END_REF], the ancient towns could have been oriented according to the rising of the sun and of the moon. That is, we can find examples where the main street was aligned along the sunrise, in particular, on the summer or the winter solstice. However, also the northern or southern directions of the moonrise was relevant, in particular for the major lunar standstills. Let us note that, in this case, the moonrise directions cannot be confused with the sunrise directions [START_REF] Vv | Wikipedia[END_REF][START_REF]The moon has an apparent behavior, which is more complex than that of the sun. We have that the sunrise direction oscillates between the two solstitial positions during a year, whereas the moon does the same during a nodal period (about 27 days). Moreover[END_REF].

Since the planning of Sauveterre-De-Guyenne is not cardinally oriented, we can ask ourselves if other astronomical orientation, according to the sun or the moon, are present. In fact, if we apply the Photographer's Ephemeris software, we can see that one of the axes of the grid is along the southern moonrise on a major lunar standstill. In the Figure 4 we can see the result simulated for year 2025, a year of major lunar standstill. It is therefore possible that the bastide was founded with an astronomical orientation. To sustain this possibility, let us consider the example of Augusta Emerita, today Mérida, in Spain. This Roman town has the main street, the Decumanus, which is aligned along the northern moonrise on a lunar standstill. As we have discussed in [START_REF] Sparavigna | Augusta Emerita and the Major Lunar Standstill of 24 BC[END_REF], the year of the foundation of Augusta Emerita was 25 BC, close to a year of major lunar standstill (24 BC), as we can evidence using CalSky software. The moonrise directions on standstills for 25 BC and 24 BC are practically the same. Here, let us use CalSky for Sauveterre-De-Guyenne too. We obtain the following results for moonrise azimuths close to 133 degrees. [START_REF] Vv | Wikipedia[END_REF] These results indicate that the year AD 1281 was a year of major lunar standstill, and therefore, that an astronomical orientation of the town to remark the year of its foundation is possible too.

Aigues-Mortes and the sun

The name is attested in the Latinized form Aquae Mortuae in 1248. It is coming from the Occitan Aigas Mortas meaning "dead water", due to the presence of marshes and ponds present in the place [START_REF] Vv | Wikipedia[END_REF]. As told in [START_REF] Vv | Wikipedia[END_REF], the foundation of the city is said to have been by Gaius Marius, around 102 BC.

The region was under the rule of the monks of a local Benedictine abbey [START_REF] Vv | Wikipedia[END_REF]. In 1240, Louis IX, who wanted to have his own port for transporting his Crusaders, focused on the strategic position of the place. The king obtained the town and the surrounding lands from the monks of the abbey [START_REF] Vv | Wikipedia[END_REF].

Aigues-Mortes became the place from which the king departed for two Crusades: the Seventh Crusade in 1248 and the Eighth Crusade in 1270. The king died during this Crusade. In 1272, his son Philip III the Bold ordered to build the walls for protecting the town [START_REF] Vv | Wikipedia[END_REF], and we can see them in the Figure 5. It is interesting to note that this town has an orientation along the sunrise on the winter solstice. We can see it using the SunCalc.net software as shown in the Figure 6.

Libourne

This town was created as a bastide by Roger de Leybourne, English soldier and landowner, in 1270.

The original name of Leybourne changed to the present Libourne. Its original reason of existance was that of being a port to ship wine of the Dordogne valley to England [22]. We can see the bastide in the Figure 7. Also in the case of the bastide of Libourne, we can find an astronomical orientation. The main axis of the bastide is aligned along the sunrise on the summer solstice (Figure 8).

Bastide Saint Louis

This bastide is named after the king Louis IX, commonly known as Saint Louis. This bastide was built during his reign, with a regular grid about the central square, now the Place Carnot. As told in [23], «c'est en effet en 1247 que ce nouveau bourg est créé en rive gauche de l'Aude … Cette nouvelle ville dont le tracé se trouve plus près des rives d'Aude est à l'image des bastides du grand Sud Ouest». Today, the Bastide is delimited by the boulevards laid out in the XVIIIth and XIXth centuries on the site of the old moats [24]. We can see the bastide in the Figure 9. In this case, the town has an orientation according to the cardinal points. Let us consider in more detail the case of Pavie. Like Fleurance, also Pavie has the name of an Italian town. As told in [START_REF] Vv | Wikipedia[END_REF], la nouvelle bastide prend le nom officiel de la ville de Pavie en Italie. The town had maintained its original planning. Here we find a possible astronomical orientation along the sunrise on the summer solstice. We can evidence this orientation using SunCal.net software (see the Figure 14).

Villenueve-sur-Lot

Here an example of a Villenueve, that is of a "new town". It is Villenueve-sur-Lot. The area of this town was first populated by the Romans. In the 11th century, a Benedictine abbey was built there, with a small village of potters and farmers around the abbey [START_REF] Vv | Wikipedia[END_REF]. Between 1254 and 1263, Alphonse of Poitiers built a bastide. This is like those we have seen before, with a rectangular layout about a central square (Figure 15).

Conclusion

In this paper we have considered only a few of the existing bastides. In particular we have searched for possible astronomical orientations in the layout of those having a regular grid of parallel and perpendicular streets. We have found four cases of possible orientations along the sunrise on solstices and moonrise on a lunar standstill. Probably, the persons that planned them added these alignments in the layout of the bastides for symbolical reasons. Actually, the case of Sauveterre-De-Guyenne is very interesting and remarkable. The town has one of its axes oriented along the southern moonrise on a major lunar standstill, and the year of the foundation of the bastide, AD 1281, was a year of major lunar standstill. Therefore, it is possible that, in the case of Sauveterre-De-Guyenne, the alignment was decided as a symbolical reference to the year of the foundation.

Figure 2 :

 2 Figure 2: The bastide of Sauveterre-De-Guyenne in a Wikimapia image.

Figure 3 :

 3 Figure 3: One of the gates of Sauveterre-De-Guyenne in a Street View image of Google Earth.

Figure 4 :

 4 Figure 4: One of the axes of Sauveterre-De-Guyenne is oriented along the direction of the southern moonrise (pale blue line, azimuth 133 degrees from true north) on a major lunar standstill (Courtesy, Photographer's Ephemeris software). In the figure we can see also the directions of the moonset (dark blue line) and of sunrise and sunset (yellow and orange lines).

Figure 5 :

 5 Figure 5: The bastide of Aigues-Mortes in a Wikimapia image.

Figure 6 :

 6 Figure 6: The bastide and the sun. Thanks to the SunCalc.net software we can see the directions of sunrise (yellow) and sunset (red) on the satellite images. Here we see one of the streets aligned along the sunrise on the winter solstice. Also the southern part of the walls is oriented along this same direction.

Figure 7 :

 7 Figure 7: The bastide of Libourne in a Google Earth image.

Figure 8 :

 8 Figure 8: The bastide of Libourne. Using the SunCalc.net software we can see that the main axis of the bastide is aligned along the sunrise on the summer solstice.

Figure 9 :

 9 Figure 9: The bastide Saint Louis at Carcassonne in Google Earth.

Figure 10 :

 10 Figure 10: The bastide of Monpazier in Google Earth.

Figure 11 :Figure 12 :

 1112 Figure 11: The bastide of Mazères in Google Earth.

Figure 13 Figure 13 :

 1313 Figure13-a

Figure 14 :

 14 Figure 14: The bastide of Pavie. By using the SunCalc.net software we can see that the bastide is aligned along the sunrise on the summer solstice.

Figure 15 :

 15 Figure 15: The bastide of Villenueve-sur-Lot (Courtesy Google Earth).


Thi Minh

Nhat Vo
email: vtmnhat@gmail.com.

The local backward heat problem

Keywords: inverse problem, global backward, local backward, controllability, observation estimate, heat equation

In this paper, we study the local backward problem of a linear heat equation with time-dependent coefficients under the Dirichlet boundary condition. Precisely, we recover the initial data from the observation on a subdomain at some later time. Thanks to the " optimal filtering " method of Seidman, we can solve the global backward problem, which determines the solution at initial time from the known data on the whole domain. Then, by using a result of controllability at one point of time, we can connect local and global backward problem.

1 Introduction and main result

Our motivation

Inverse and ill-posed problems (see [I], [P], [K]) are the heart of scientific inquiry and technological development. They play a significant role in engineering applications, as well as several practical areas, such as image processing, mathematical finance, physics, etc. and, more recently, modeling in the life sciences. During the last ten years or so, there have been remarkable developments both in the mathematical theory and applications of inverse problems. Especially, in various industrial purposes, for example steel industries, glass and polymer forming and nuclear power station, the "backward heat problem", which recovers the temperature in the heating system from the observation at some later time keeps an important position. On the other hand, from the mathematical point of view, it is well-known to be an ill-posed problem in the sense of Hadamard (see [H]) due to the irreversibility of time. That is, there exists no solution from the given final data, and even if a solution exists, the small perturbations of the observation data may be dramatically scaled up in the solution. Hence, the interest of constructing some special regularization method is motivated. This topic has been studied extensively with lots of methods released such as: Tikhonov regularization [F], [M], [TS], [ZM], [MFH], Lavrentiev regularization [NT], [JSG], truncation method [NTT], [KT], [ZFM], filter method [S], [TKLT], [QW], the quasi-boundary value method [DB], [KT], [QTTT] and other methods [AE], [START_REF] Li | Solution of backward heat problem by Morozov discrepancy principle and conditional stability[END_REF], [START_REF] Liu | On stability and regularization for backward heat equation[END_REF], [HX], [TQKT], ... In [S], Seidman uses a so-called "optimal filtering" method in order to recover the solution at time t > 0 with an optimal result. The idea of improving his outcome to reconstruct the solution at time 0 is an interesting issue. Furthermore, the question that if we restrict our observation on a subregion inside the domain then how the local problem will be solved is also attractive.

Our problem

Let Ω be an open bounded domain in R n (n ≥ 1) with a boundary ∂Ω of class C 2 ; T be a fixed positive constant. Let p ∈ C 1 ([0; +∞)) such that 0 < p 1 ≤ p(t) ≤ p 2 , ∀t ∈ [0, +∞), where p 1 and p 2 are some positive real numbers. Let ω be a nonempty, open subdomain of Ω. We consider a linear heat equation with time-dependent coefficients, under the Dirichlet boundary condition with the state u ∈ C 1 ((0, T ); H 1 0 (Ω)):

∂ t up(t)∆u = 0 in Ω × (0, T ), u = 0 on ∂Ω × (0, T ).

(1.1)

Our target is constructing the initial solution u(•, 0) when the local measurement data of u(•, T ) on the subdomain ω is available. In practice, the data at time T is often measured by the physical instrument. Therefore, only a perturbed data f can be obtained. Let δ > 0 denote the noisy level with the following property

u(•, T ) -f L 2 (ω) ≤ δ. (1.2)
Moreover, in order to assure the convergence of the regularization approximation to the initial data u(•, 0), some a priori assumption on the exact solution is required u(•, 0) ∈ H 1 0 (Ω).

(1.3)

We will determine an approximate output g of the unknown exact solution u(•, 0) such that the error estimate e(δ) in u(•, 0)g L 2 (Ω) ≤ e(δ) (1.4) tends to 0 when δ tends to 0.

Relevant works

Now, we consider how our problem can be solved so far in the past. In fact, there are lots of papers on the global backward problem but the works on the local one are restricted. There has been a sizeable literature on the special case p ≡ 1 with various methods. From now on, we will denote by δ the noisy level.

1. In 1996 (see [S]), Seidman considers the heat equation which has the following form ∂ t u -∇a∇u + qu = 0 on Ω × (0, T ) with u = 0 on ∂Ω × (0, T ) (1.5)

where a and q belong to L ∞ (Ω). He succeeds in constructing the solution at a fixed time t ∈ (0, T ) from the observation f satisfying u(•, T )f L 2 (Ω) ≤ δ, under the assumption u(•, 0) ∈ L 2 (Ω). His strategy is using a "filter" with respect to the spectral decomposition of operator A : u → -∇a∇u + qu, which is defined as

F(t)e i = min 1, e -λ i (T -t) t T u(•, 0) L 2 (Ω) δ 1-t T e i (1.6)
where {λ i } i≥1 and {e i } i≥1 are respectively denoted by the eigenvalues and the corresponding eigenfunctions of the operator A. Then, he can get the optimal result, which is u(•, t)g t L 2 (Ω) ≤ δ t T u(•, 0)

1-t T L 2 (Ω) .
(1.7)

The regularization solution g t at time t is constructed as

g t := ∞ i=1 e λ i (T -t)
Ω f (x)e i (x)dx F(t)e i .

(1.8) 2. By generalizing the result of Seidman, in [TS], Tautenhahn and Schröter provide us a definition of the term "optimal method", in a sense the error of the estimate between the exact solution and the approximate one defined from the optimal method can not be greater than the best possible worst case error (see Definition 1.1, page 478). Their interest is finding the optimal results in different regularization methods for solving the backward heat equations. According to this sense, the result of Seidman is optimal.

3. In 2007, Trong et al. (see [TQKT]) improve the quasi-boundary value method to regularize the 1D backward heat equation. They succeed in recovering the initial data with the following error:

u(•, 0) -g L 2 (Ω) ≤ 4 √ 8C 4 √ T ln u(•, 0) L 2 (Ω) δ -1 4 (1.9)
where C is a positive constant depending on u(•, 0) H 1 0 (Ω) (see also [AP]).

The problem with case p ≡ constant is recently concerned, which can be mentioned in some following writings:

1. In 2013, Tuan et al. in [TQTT] consider the 1D backward heat equation with time-dependent coefficients. They use a so-called "modified method" to get the result below

u(•, t) -g t L 2 (Ω) ≤ 1 + u(•, 0) L 2 (Ω) δ u(•, 0) L 2 (Ω) p 2 1 t p 2 2 T
.

(1.10) 2. In 2014, Zhang, Fu and Ma (see [ZFM]) also study on the 1D backward heat equation with time-dependent coefficients, but use the truncation method. They can recover the solution at time t ∈ T (1 -p 1 p 2 ); T satisfying u(•, t)g t L 2 (Ω) ≤ u(•, 0)

1-t T L 2 (Ω) ((τ +1)δ) t T + u(•, 0) L 2 (Ω) τ -1 p 2 (T -t) p 1 T δ (p 2 -p 1 )T +p 2 t p 1 T
(1.11) for some constant τ > 1.

3. In 2016, Khanh and Tuan (in [KT]) solve an initial inverse problem for an inhomogeneous heat equation by using high frequency truncation method. Under the assumption that u(•, 0) ∈ H 1 0 (Ω), they can recover the initial data with the below error:

u(•, 0) -g L 2 (Ω) ≤ δ u(•, 0) L 2 (Ω) 1 2T ln u(•,0) L 2 (Ω) δ √ 2p 2 T + √ 2p 2 T u(•, 0) H 1 0 (Ω) ln u(•,0) L 2 (Ω) δ .
(1.12)

For the local inverse problem, we can pick up some of following works:

1. In 1995, Yamamoto in [Y] proposes a reconstruction formula for the spatial dependence of the source term in a wave equation ∂ tt u -∆u = f (x)σ(t), assuming σ(t) known, from local measurement using exact controllability.

2. In 2009, Li, Yamamoto and Zou in [LYZ] study the conditional stability of inverse problems. Here, the known data is observed in a subregion along a time period which may start at some point, possibly far away from the initial data.

3. In 2011, García and Takahashi (see [GT]) present some abstract results of a general connection between null-controllability and several inverse problems for a class of parabolic equations.

4. In 2013, García, Osses and Tapia in [GOT] succeed in determining the heat source from a single internal measurement of the solution, thanks to a family of null control functions.

1.4 Our method of solving the global backward problem (GBP) and the local backward problem (LBP)

Firstly, we deal with the global backward problem (GBP), which recovers the initial data from the observation on the whole domain at some later time τ > 0. Here, we assume that there exists solution of the (1.1) satisfying the a priori condition (1.3) and f be the known data on Ω at time τ such that u(•, τ ) -f L 2 (Ω) ≤ δ for some δ > 0. We will determine a function g which approximate the initial data. Our idea of constructing such function g is from the "optimal filtering method" of Seidman (see [S]): First, we define a continuous operator depending on a regularization parameter α

R α : L 2 (Ω) → L 2 (Ω) φ → ∞ i=1 min{e λ i τ 0 p(s)ds ; α} Ω φ(x)e i (x)dx e i ;
Then, the function R α f will be closed to the exact solution u(•, 0) in L 2 (Ω) where α is the minimizer of the problem min

α>0 u(•, 0) -R α f L 2 (Ω) .
Secondly, for the local backward problem (LBP), whose observation is measured on a subdomain, we need to use a tool of controllability to link with the (GBP). Precisely, we use the assertion about the existence of a sequence of control functions to get the information of solution on the whole domain Ω from the given data on the subdomain ω: For each i = 1, 2, ..., for any ε > 0, there exists h i ∈ L 2 (ω) such that the solution of

       ∂ t ϕ i -p(t)∆ϕ i = 0 in Ω × (0, 2T ) \ {T }, ϕ i = 0 on ∂Ω × (0, 2T ), ϕ i (•, 0) = e i in Ω, ϕ i (•, T ) = ϕ i (•, T -) + ½ ω h i in Ω (1.13) satisfies ϕ i (•, 2T ) L 2 (Ω) ≤ ε.
Here, ½ ω presents for the characteristic function on the region ω and ϕ i (T -) denotes the left limit of the function ϕ i at time T . Multiplying ∂ t ϕ ip(t)∆ϕ i = 0 by u(•, 2Tt) and using some computation technique, we can get the approximate solution f at time τ = 3T , which is

u(•, 3T ) -f L 2 (Ω) ≤ E(δ).
Here, f is computed by known data h i and f and E(δ) is a function of δ such that E(δ) → 0 when δ → 0. Lastly, applying the result of (GBP) with the information at 3T on the whole domain, the initial data of (1.1) is reconstructed.

Spectral theory

As a direct consequence of spectral theorem for compact, self-adjoint operators (see Theorem 9.16, page 225, [HN]), there exists a sequence of positive real eigenvalues of the operator -∆, which denoted by {λ i } i=1,2,... where

0 < λ 1 ≤ λ 2 ≤ λ 3 ≤ ...., λ i → ∞ as i → ∞.
(1.14)

Moreover, there exists an orthonormal basis {e i } i=1,2,... of L 2 (Ω), where e i ∈ H 1 0 (Ω) is an eigenfunction corresponding to λ i -∆e i = λ i e i in Ω, e i = 0 on ∂Ω.

(1.15)

When u 0 ∈ L 2 (Ω) and u 0 = ∞ i=1 a i e i with a i = Ω u 0 (x)e i (x)dx and ∞ i=1 |a i | 2 < ∞, then u(•, t) = ∞ i=1 a i e -λ i t 0 p(s)ds e i (1.16)
is the unique solution of

   ∂ t u -p(t)∆u = 0 in Ω × (0, T ), u = 0 on ∂Ω × (0, T ), u(•, 0) = u 0 in Ω.
(1.17)

Main result

Theorem 1.1. Let u be the solution of (1.1) with the a priori bound

(1.3). Let f ∈ L 2 (ω) and 0 < δ < u(•, 0) L 2 (Ω) satisfying u(•, T ) -f L 2 (ω) ≤ δ. (1.18)
There exists a function g ∈ L 2 (Ω) and a constant C = C(Ω, ω, p) > 0 such that the following estimate holds

u(•, 0) -g L 2 (Ω) ≤ Ce C T √ T u(•, 0) H 1 0 (Ω) ln u(•,0) L 2 (Ω) δ . (1.19) Remark 1.1. 1. When δ < De -D(T + 1 T ) u(•, 0) L 2 (Ω) for some positive constant D = D(Ω, ω, p), the approximate solution of the initial data satisfying (1.19) is constructed as below g := - ∞ i=1 min{e λ i 3T 0 p(s)ds , α}e -λ i 3T 2T p(s)ds ω h i (x)f (x)dx e i (1.20)
where {h i } i≥1 is a sequence of control functions (see Section 4) and α is the regularization parameter given by

α = A B -1 √ 3p 2 T u(•, 0) H 1 0 (Ω) K 1 e K 1 T u(•, 0) L 2 (Ω) 1-k 1 δ k 1 (1.21) with (i) A : [0; +∞) → √ e 2 ; +∞ x → e x 1 + 2x , (1.22) (ii) B : (0; +∞) → (0; +∞) x → √ xe x (1.23)
The existence of the function B -1 dues to the bijection property of the function B on (0, +∞),

(iii) K 1 = K 1 (Ω, ω, p) > 1 and k 1 = k 1 (Ω, ω, p) ∈ (0, 1)
. All these constants can be explicitly computed when Ω is convex or star-shaped with respect to some x 0 ∈ Ω.

The estimate (1.19) connects to the well-known following estimate

u(•, 0) L 2 (Ω) ≤ C 1 + T + 1 T u(•, 0) H 1 0 (Ω) ln u(•,0) L 2 (Ω) u(•,T ) L 2 (ω)
.

(1.24)

for some positive constant C = C(Ω, ω, p) (see Appendix for the proof ).

Outline

Section 2 will give us a result of the (GBP) (see Theorem 2.1), where the known data is observed on the whole domain. In section 3, we construct an observation estimate at one point of time for the parabolic equations with time-dependent coefficients (see Theorem 3.1 and Theorem 3.2). This is an important preliminary of the approximate controllability (see Theorem 4.1), which is studied in section 4. Lastly, combining the result of controllability and global backward, we get the proof of the Theorem 1.1, mentioned in section 5.

Global backward problem

First of all, we need to consider the special case, that is ω ≡ Ω. In [S], Seidman succeeds in recovering the solution at time t > 0 by an optimal filtering method, under the a priori condition u(•, 0) ∈ L 2 (Ω). Here we will use his method to recover the initial data at time 0 but with the stronger assumption u(•, 0) ∈ H 1 0 (Ω).

Theorem 2.1. Let u be the solution of (1.1) satisfying the a priori condition (1.3). Let f ∈ L 2 (Ω) and δ > 0 having the following property

u(•, T ) -f L 2 (Ω) ≤ δ. (2.1)
There exists a function g ∈ L 2 (Ω) such that for any ζ >

δ 2 2λ 1 p 2 T u(•,0) 2 L 2 (Ω)
, the following estimate holds

u(•, 0) -g L 2 (Ω) ≤ (1 + ζ)p 2 T u(•, 0) H 1 0 (Ω) ln √ 2ζλ 1 p 2 T u(•,0) L 2 (Ω) δ . (2.2) Remark 2.1. 1. When δ < u(•, 0) L 2 (Ω) e -λ 1 p 2 T
, the approximate solution of the initial data satisfying (2.2) is constructed as

g := ∞ i=1 min e λ i T 0 p(s)ds , ᾱ Ω f (x)e i (x)dxe i . (2.3)
Here, the regularization parameter ᾱ is given by

ᾱ := A B -1 √ p 2 T u(•, 0) H 1 0 (Ω) δ (2.4)
with A and B being respectively defined in (1.22) and (1.23).

2. When δ < u(•, 0) L 2 (Ω) , we can choose ζ = 1 2λ 1 p 2 T in order to get

u(•, 0) -g L 2 (Ω) ≤ p 2 T + 1 2λ 1 u(•, 0) H 1 0 (Ω) ln u(•,0) L 2 (Ω) δ .
(2.5)

This connects to the well-known following estimate

u(•, 0) L 2 (Ω) ≤ √ p 2 T u(•, 0) H 1 0 (Ω) ln u(•,0) L 2 (Ω) u(•,T ) L 2 (Ω)
.

(2.6)

Proof of Theorem 2.1

Proof. For the case δ ≥ u(•, 0) L 2 (Ω) e -λ 1 p 2 T , the estimate (2.2) holds with g = 0.

Indeed, combining with the fact that √ 2ζλ 1 p 2 T ≤ e ζλ 1 p 2 T ∀ζ > 0, we get

2ζλ 1 p 2 T u(•, 0) L 2 (Ω) δ ≤ e (1+ζ)λ 1 p 2 T . (2.7) It implies that 1 √ λ 1 ≤ (1 + ζ)p 2 T ln √ 2ζλ 1 p 2 T u(•,0) L 2 (Ω) δ . Hence u(•, 0) L 2 (Ω) ≤ u(•, 0) H 1 0 (Ω) √ λ 1 ≤ (1 + ζ)p 2 T u(•, 0) H 1 0 (Ω) ln √ 2λ 1 p 2 T u(•,0) L 2 (Ω) δ .
(2.8)

The main purpose concerns the case when

δ < u(•, 0) L 2 (Ω) e -λ 1 p 2 T .
(2.9)

In this case, we will determine the regularization solution at time 0 as follows: First of all, Step 1 will provide us the construction of a continuous operator R β depending on a parameter β, which will be chosen later. The regularization solution g is defined by applying this operator on the known-data f . Secondly, in Step 2, we compute the error between the exact solution and the approximate solution defined in Step 1. Lastly, by minimizing the error in Step 2 with respect to β, we can obtain the final result in Step 3.

Step 1: Construct the regularization solution. Let us define a continuous function R β depending on a positive parameter β, which will be chosen later:

R β : L 2 (Ω) → L 2 (Ω) f → ∞ i=1 min{e λ i T 0 p(s)ds ; β} Ω f (x)e i (x)dx e i
Put g := R β f . We will prove that such defined function g approximate the exact solution u(•, 0) with some suitable choice of β.

Step 2: Compute the error u(•, 0)g L 2 (Ω) . Put g T := R β u(•, T ), we will compute the error by using the following triangle inequality

u(•, 0) -g L 2 (Ω) ≤ u(•, 0) -g T L 2 (Ω) + g T -g L 2 (Ω) .
(2.10)

On one hand, we have

g -g T L 2 (Ω) = ∞ i=1 min{e λ i T 0 p(s)ds ; β} Ω f (x) -u(x, T ) dxe i L 2 (Ω) ≤ β f -u(•, T ) L 2 (Ω) ≤ βδ.
(2.11)

On the other hand

u(•, 0) -g T L 2 (Ω) = ∞ i=1 Ω u(x, 0)e i (x)dxe i - ∞ i=1 min{e λ i T 0 p(s)ds ; β}e -λ i T 0 p(s)ds Ω u(x, 0)e i (x)dxe i L 2 (Ω) = ∞ i=1 1 -min{e λ i T 0 p(s)ds ; β}e -λ i T 0 p(s)ds Ω u(x, 0)e i (x)dxe i L 2 (Ω) = e λ i T 0 p(s)ds >β 1 -βe -λ i T 0 p(s)ds Ω u(x, 0)e i (x)dxe i L 2 (Ω) = e λ i T 0 p(s)ds >β 1 -βe -λ i T 0 p(s)ds √ λ i λ i Ω u(x, 0)e i (x)dxe i L 2 (Ω) ≤ sup λ≥λ 1 1 -βe -λ T 0 p(s)ds √ λ u(•, 0) H 1 0 (Ω) ≤ sup λ≥λ 1 (1 -βe -λp 2 T ) √ λp 2 T p 2 T u(•, 0) H 1 0 (Ω) .
(2.12) Now, we solve the problem of finding sup

λ≥λ 1 (1-βe -λp 2 T ) √ λp 2 T . Define F : [λ 1 ; +∞) → (0; +∞) λ → (1 -βe -λp 2 T ) √ λp 2 T .
Obviously, F is differentiable and

F ′ (λ) = βp 2 T e -λp 2 T (1 + 2λp 2 T ) -p 2 T 2 √ λp 2 T λp 2 T . (2.13)
The equation Suppose the condition (2.15) is satisfied then there exists a unique λ ≥ λ 1 such that F ′ ( λ) = 0 and β = A( λp 2 T ). We can write λp 2 T = A -1 (β). On the other hand, the fact that F ′ (λ 1 ) > 0 leads us to the conclusion: the function F is strictly increasing on (λ 1 , λ) and strictly decreasing on ( λ, +∞). Consequently, F gets supremum at λ, i.e

F ′ (λ) = 0 is equivalent to β = e λp 2 T 1 + 2λp 2 T . ( 2
F ( λ) = sup λ≥λ 1 F (λ).
(2.16)

Step 3: Minimize the error with a suitable choice of β. Combining the two above steps, we get

u(•, 0) -g L 2 (Ω) ≤ βδ + (1 -βe -λp 2 T ) λp 2 T p 2 T u(•, 0) H 1 0 (Ω) = Θδe A -1 (β) + (1 -Θ) √ p 2 T u(•, 0) H 1 0 (Ω) A -1 (β) (2.17)
where Θ = βe -λp 2 T . Note that

ΘA + (1 -Θ)B ≥ min{A, B} ∀ A, B > 0, Θ ∈ (0, 1).
The equality occurs when and only when A = B. Hence, in order to minimize the right-hand side of (2.17), we will choose β such that

δe A -1 (β) = √ p 2 T u(•, 0) H 1 0 (Ω)
A -1 (β) .

(2.18)

The choice of

β = ᾱ := A B -1 √ p 2 T u(•,0) H 1 0 (Ω)
δ satisfies the condition (2.15) (dues to the assumption (2.9) on the smallness of δ) and the estimate (2.18). Therefore, we get the following estimate

u(•, 0) -g L 2 (Ω) ≤ √ p 2 T u(•, 0) H 1 0 (Ω) A -1 ( ᾱ) ≤ √ p 2 T u(•, 0) H 1 0 (Ω) B -1 √ p 2 T u(•,0) H 1 0 (Ω) δ . ( 2
√ p 2 T u(•, 0) H 1 0 (Ω) δ ≤ √ p 2 T u(•, 0) H 1 0 (Ω) u(•, 0) -g L 2 (Ω) e p 2 T u(•,0) 2 H 1 0 (Ω) u(•,0)-g 2 L 2 (Ω) .
(2.20)

Using the fact that

√ 2ζx ≤ e ζx 2 ∀ζ > 0 ∀x > 0, one obtains √ p 2 T u(•, 0) H 1 0 (Ω) δ ≤ 1 √ 2ζ e (1+ζ)p 2 T u(•,0) 2 H 1 0 (Ω) u(•,0)-g 2 L 2 (Ω)
.

(2.21)

It is equivalent to

(1 + ζ)p 2 T u(•, 0) 2 H 1 0 (Ω) u(•, 0) -g 2 L 2 (Ω) ≥ ln √ 2ζp 2 T u(•, 0) H 1 0 (Ω) δ . (2.22) With ζ > δ 2 2λ 1 p 2 T u(•,0) 2 L 2 (Ω)
, it deduces that

u(•, 0) -g L 2 (Ω) ≤ (1 + ζ)p 2 T u(•, 0) H 1 0 (Ω) ln √ 2ζλ 1 p 2 T u(•,0) L 2 (Ω) δ .
(2.23)

For the local case ω ⋐ Ω, it is required the existence of control functions on the subdomain at some point of time in order to link with a global result. This controllability problem has a sustainable connection with the observability one, which will be studied in the next Section.

Observability at one point of time

The issue on constructing an observation estimate is widely studied. It can be solved by global Carleman inequality, which is presented in [FI]; by using the estimate of Lebeau and Robbiano (see [LR]) or by transmutation (see [EZ]). Recently, Phung et al. provide a different method which is based on properties of the heat kernel with a parametric of order 0. In [START_REF] Phung | Quantitative unique continuation for the semilinear heat equation in a convex domain[END_REF] and [START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF], the authors work on a linear equation which has form

∂ t v -∆v + av + b∇v = 0 in Ω × (0, T ). Here, a ∈ L ∞ ((0, T ), L q (Ω)) with q ≥ 2 if n = 1 and q > n if n ≥ 2; b ∈ L ∞ (Ω × (0, T )) n
and Ω must be convex. Then, by using some geometrical techniques, Phung et. al improve their previous results by working on a general domain (i.e Ω is even convex or not). For the following form of linear equation

∂ t v -∆v + av = 0 in Ω × (0, T )
where a ∈ L ∞ (Ω × (0, T )), see [PWZ]. For the parabolic equations with space-time coefficients [BP]. Here, we also deal with the problem of determining an observation estimate in the general case of domain but for a linear heat equation with time-dependent coefficients

∂ t v -∇(A∇v) + av + b∇v = 0 in Ω × (0, T ) where a ∈ L ∞ (Ω × (0, T )), b ∈ L ∞ (Ω × (0, T )) n and A is a n × n symmetric positive- definite matrix with C 2 (Ω × [0, T ]) coefficients, see
∂ t v -p(t)∆v = 0 in Ω × (0, T )
where p ∈ C 1 (0, T ). In this section, we will study two results of observation estimates in two different geometrical cases: The general case (Theorem 3.1) and the special case (Theorem 3.2) when Ω is convex or star-shaped with respect to some x 0 such that B(x 0 , r)

:= {x; |x -x 0 | < r} ⊂ ω, 0 < r < R := max x∈Ω |x -x 0 |.
For the special case, we make a careful evaluation of the constants which can be explicitly computed. First of all, we state an observation result in general case of domain Ω.

Theorem 3.1. There exist constants K = K(Ω, ω, p) > 0 and µ = µ(Ω, ω, p) ∈ (0; 1) such that the solution of

   ∂ t v -p(t)∆v = 0 in Ω × (0, T ), v = 0 on ∂Ω × (0, T ), v(•, 0) ∈ L 2 (Ω), (3.1) satisfies v(•, T ) L 2 (Ω) ≤ Ke K T v(•, T ) µ L 2 (ω) v(•, 0) 1-µ L 2 (Ω) . (3.2)
Corollary 3.1. For any ε > 0, there exist positive constants c 1 and c 2 depending on Ω, ω and p such that the following estimate holds

v(•, T ) 2 L 2 (Ω) ≤ c 1 e c 1 T 1 ε c 2 v(•, T ) 2 L 2 (ω) + ε v(•, 0) 2 L 2 (Ω) . (3.3) Proof of Corollary 3.1 Proof. It implies from (3.2) in Theorem 3.1 that v(•, T ) 2 L 2 (Ω) ≤ K 2 e 2K T v(•, T ) 2µ L 2 (ω) v(•, 0) 2(1-µ) L 2 (Ω) .
Applying the Young's inequality ab

≤ a m m + b q q with a = K 1 µ e K µT v(•, T ) L 2 (ω) 1 ε 1-µ 2µ (1 -µ) 1-µ 2µ 2µ , b = ε 1 2 1 1 -µ 1 2 v(•, 0) L 2 (Ω) 2(1-µ) , m = 1 µ and q = 1 1 -µ , we get v(•, T ) 2 L 2 (Ω) ≤ µK 2 µ e 2K µT 1 ε 1-µ µ (1 -µ) 1-µ µ v(•, T ) 2 L 2 (ω) + ε v(•, 0) 2 L 2 (Ω) .
Therefore, we obtain the estimate (3.3) with

c 1 := max µK 2 µ (1 -µ) 1-µ µ , 2K µ and c 2 := (1 -µ) µ . (3.4)
Our next theorem will provide us an observation result in a special geometric case with specific constants.

(i) Ω is convex or star-shaped domain with respect to x 0 , (ii) R 2 < 2p 2 1 |p ′ |∞ if p ≡ constant where |p ′ | ∞ = sup t∈[0,T ] |p ′ (t)|; then the solution of (3.1) satisfies (3.2) with ω = {x; |x -x 0 | < r} where 0 < r < R, K = max 4 1+C 0 (1+S ℓ ) (1 + ℓ) n+2C 0 (1+S ℓ ) e 2C 1 (1+S ℓ ) e r 2 ℓ 4p 1 1 2(1+S ℓ ) , r 2 ℓ 4p 1 (1 + S ℓ ) and µ = 1 2(1 + S ℓ )
.

Here

C 0 := R 2 |p ′ | ∞ 2p 2 1 , C 1 := (2 + n) |p ′ | ∞ p 1 , ℓ :=        2 2+ξ R 2 e C 1 ξ ln 3 2 r 2 1 1-ξ -1 ∀ξ ∈ (0, 1) if C 0 = 0, 4R 2 e C 1 r 2 1-( 2 3 ) C 0 1 1-C 0 -1 if C 0 > 0 and S ℓ := e C 1    ln(1+ℓ) ln 3 2 if C 0 = 0, (1+ℓ) C 0 1-( 2 3 ) C 0 if C 0 > 0.
Remark 3.1. In the special case when p ≡ 1, the observation estimate (3.2) can be written as

v(•, T ) L 2 (Ω) ≤ 4(1 + ℓ) n e r 2 ℓ 4 ( 1+ 1 T ) v(•, T ) L 2 (ω) 1 2 1+ ln(1+ℓ) ln 3 2 v(•, 0) 1+2 ln(1+ℓ) ln 3 2 2 1+ ln(1+ℓ) ln 3 2 L 2 (Ω)
where ℓ :

= 2 2+ξ R 2 ξ ln 3 2 r 2 1 1-ξ -1 > 1 for any ξ ∈ (0, 1).
The interested readers can compare this result with Proposition 2.1 in [START_REF] Phung | Quantitative unique continuation for the semilinear heat equation in a convex domain[END_REF], Proposition 2.2 in [START_REF] Phung | An observability estimate for parabolic equations from a measurable set in time and its applications[END_REF] or Theorem 4.2 in [BP].

The main idea of the proof of both theorems is based on the logarithm convexity method (see [Ve]). In order to check a kind of logarithm convexity for a suitable functional, it requires that some boundary terms must be dropped or have a good sign. This is possible under the assumption (ii) in Theorem 3.2. But for the general case (Theorem 3.1), we need a local star-shaped assumption (to get a good sign of boundary terms) and a suitable cut-off function (to drop some boundary terms). Then, thanks to the covering argument and the propagation of smallness, we get the global desired result. First of all, we need some preliminary results in the first subsection. Then, the proof of Theorem 3.2 and Theorem 3.1 will be devoted in two next subsections, respectively.

Preliminary results

The strategy of the proof of Theorem 3.1 and Theorem 3.2 consists on choosing a suitable function whose logarithm can be a convex function and considering the differential inequalities associated to this function (see Lemma 3.1). Then by choosing a suitable weight function inspired by the heat kernel (see Corollary 3.2) and solving ODE inequalities (see Lemma 3.2), we obtain a Hölder type inequality (see Corollary 3.3). The localization process in the proof of general case makes appear the function F in Corollary 3.2, which will be treated due to the technical Lemma 3.3.

Lemma 3.1. Let ϑ be an open set in R n , x 0 ∈ ϑ, z ∈ H 1 (0, T ; H 1 0 (ϑ)) and φ ∈ C 2 (Ω × (0, T )). We define two functions from [0, T ] on (0, +∞) by y(t) := ϑ |z(x, t)| 2 e φ(x,t) dx, N(t) := p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx ϑ |z(x, t)| 2 e φ(x,t) dx .
With the notations Gφ := ∂ t φ + p(t)∆φ + p(t)|∇φ| 2 and w := ∂ t z -p(t)∆z, the following assertions hold for any times t > 0:

i/ y ′ (t) + 2N(t)y(t) = ϑ Gφ(x, t)|z(x, t)| 2 e φ(x,t) dx + 2 ϑ w(x, t)z(x, t)e φ(x,t) dx, ii/ N ′ (t) ≤ p ′ (t) p(t) N(t) + p(t) 2 y(t) ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x,t) dx + p(t) y(t) ϑ |∇z(x, t)| 2 Gφ(x, t)e φ(x,t) dx + 1 2y(t) ϑ |w(x, t)| 2 e φ(x,t) dx - 2p(t) 2 y(t) ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx - p(t) y(t) 2 ϑ Gφ(x, t)|z(x, t)| 2 e φ(x,t) dx ϑ |∇z(x, t)| 2 e φ(x,t) dx
where ν is the unit outward normal vector to ∂ϑ and ∇ 2 φ is the Hessian matrix of φ.

Proof of Lemma 3.1

Proof. First of all, we will prove the assertion i/.

We have ,t) dx.

y ′ (t) = 2 ϑ z(x, t)∂ t z(x, t)e φ(x,t) dx + ϑ |z(x, t)| 2 ∂ t φ(x, t)e φ(x
With w := ∂ t zp(t)∆z, one has ,t) dx.

y ′ (t) = 2 ϑ z(x, t)w(x, t)e φ(x,t) dx + 2p(t) ϑ z(x, t)∆z(x, t)e φ(x,t) dx + ϑ |z(x, t)| 2 ∂ t φ(x, t)e φ(x
(3.5)

Let us compute the second term of (3.5) by using integration by parts:

2p(t) ϑ z(x, t)∆z(x, t)e φ(x,t) dx = -2p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx -2p(t) ϑ z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx = -2p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx -p(t) ϑ ∇(|z(x, t)| 2 )∇φ(x, t)e φ(x,t) dx. (3.6)
We use the fact that 2z∇z = ∇(|z| 2 ) to get the second equality. Integrating by parts the second term in (3.6) gives

-p(t) ϑ ∇(|z(x, t)| 2 )∇φ(x, t)e φ(x,t) dx = p(t) ϑ |z(x, t)| 2 ∆φ(x, t)e φ(x,t) dx + p(t) ϑ |z(x, t)| 2 |∇φ(x, t)| 2 e φ(x,t) dx. (3.7)
Combining (3.5) and (3.7), we obtain: ,t) dx.

y ′ (t) = -2p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx + p(t) ϑ |z(x, t)| 2 ∆φ(x, t)e φ(x,t) dx +p(t) ϑ |z(x, t)| 2 |∇φ(x, t)| 2 e φ(x,t) dx + ϑ |z(x, t)| 2 ∂ t φ(x, t)e φ(x,t) dx +2 ϑ z(x, t)w(x, t)e φ(x
Thus, we can get the assertion i/. Now, we move to next step with the proof of assertion ii/.

Step 1:

Compute d dt p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx . d dt p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx = p ′ (t) ϑ |∇z(x, t)| 2 e φ(x,t) dx + 2p(t) ϑ ∇z(x, t)∂ t (∇z(x, t))e φ(x,t) dx +p(t) ϑ |∇z(x, t)| 2 ∂ t φ(x, t)e φ(x,t) dx = P 1 + P 2 + P 3 (3.8)
where P i (i = 1, 2, 3) is the i th term in the right-hand side of (3.8). For the second term P 2 , we use integration by parts, with the note that ∂ t z = 0 on ∂ϑ, to get: ,t) dx.

P 2 = 2p(t) ϑ ∇z(x, t)∇(∂ t z(x, t))e φ(x,t) dx = -2p(t) ϑ ∆z(x, t)∂ t z(x, t)e φ(x,t) dx -2p(t) ϑ ∇z(x, t)∂ t z(x, t)∇φ(x, t)e φ(x,t) dx = -2 ϑ |∂ t z(x, t)| 2 e φ(x,t) dx + 2 ϑ w(x, t)∂ t z(x, t)e φ(x,t) dx -2p(t) ϑ ∂ t z(x, t)∇z(x, t)∇φ(x, t)e φ(x
(3.9)

The last equality is implied from the fact: p(t)∆z = ∂ t zw. For the third term P 3 , since Gφ := ∂ t φ + p(t)∆φ + p(t)|∇φ| 2 , we get ,t) dx.

P 3 = p(t) ϑ |∇z(x, t)| 2 ∂ t φ(x, t)e φ(x,t) dx = p(t) ϑ |∇z(x, t)| 2 Gφ(x, t)e φ(x,t) dx -p(t) 2 ϑ |∇z(x, t)| 2 ∆φ(x, t)e φ(x,t) dx -p(t) 2 ϑ |∇z(x, t)| 2 |∇φ(x, t)| 2 e φ(x
(3.10)

Integrating by parts the second term in (3.10) gives ,t) dx.

-p(t) 2 ϑ |∇z(x, t)| 2 ∆φ(x, t)e φ(x,t) dx = p(t) 2 ϑ ∇(|∇z(x, t)| 2 )∇φ(x, t)e φ(x,t) dx + p(t) 2 ϑ |∇z(x, t)| 2 |∇φ(x, t)| 2 e φ(x,t) dx -p(t) 2 ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x
(3.11)

Now, we compute the first term in (3.11) by using standard summation notations ,t) dx.

p(t) 2 ϑ ∇(|∇z(x, t)| 2 )∇φ(x, t)e φ(x,t) dx = p(t) 2 ϑ ∂ i (|∂ j z(x, t)| 2 )∂ i φ(x, t)e φ(x,t) dx = 2p(t) 2 ϑ ∂ j z(x, t)∂ 2 ij z(x, t)∂ i φ(x, t)e φ(x,t) dx = -2p(t) 2 ϑ ∂ 2 jj z(x, t)∂ i z(x, t)∂ i φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∂ j z(x, t)∂ i z(x, t)∂ 2 ij φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∂ j z(x, t)∂ i z(x, t)∂ i φ(x, t)∂ j φ(x, t)e φ(x,t) dx +2p(t) 2 ∂ϑ ∂ j z(x, t)∂ i z(x, t)∂ i φ(x, t)ν j e φ(x
(3.12) Thus, we can write ,t) dx.

p(t) 2 ϑ ∇ |∇z(x, t)| 2 ∇φ(x, t)e φ(x,t) )dx = -2p(t) 2 ϑ ∆z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx -2p(t) 2 ϑ |∇z(x, t)∇φ(x, t)| 2 e φ(x,t) dx + 2p(t) 2 ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x
(3.13) Combining (3.10), (3.11) and (3.13), the third term P 3 in (3.8) can be computed as ,t) dx.

P 3 = p(t) ϑ |∇z(x, t)| 2 Gφ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∆z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx -2p(t) 2 ϑ |∇z(x, t)∇φ(x, t)| 2 e φ(x,t) +p(t) 2 ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x
(3.14) Thus, from above results (3.9) and (3.14), (3.8) can be written ,t) dx.

d dt p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx = p ′ (t) ϑ |∇z(x, t)| 2 e φ(x,t) dx -2 ϑ |∂ t z(x, t)| 2 e φ(x,t) dx + 2 ϑ w(x, t)∂ t z(x, t)e φ(x,t) dx -2p(t) ϑ ∂ t z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx + p(t) ϑ |∇z(x, t)| 2 Gφ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∆z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx -2p(t) 2 ϑ |∇z(x, t)∇φ(x, t)| 2 e φ(x,t) dx + p(t) 2 ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x,t) dx. (3.15) Since p(t)∆z = ∂ t z -w, one has -2p(t) 2 ϑ ∆z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx = -2p(t) ϑ ∂ t z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx + 2p(t) ϑ w(x, t)∇z(x, t)∇φ(x, t)e φ(x
Moreover, we also have ,t) dx.

-2 ϑ |∂ t z(x, t)| 2 e φ(x,t) dx + 2 ϑ w(x, t)∂ t z(x, t)e φ(x,t) dx -4p(t) ϑ ∂ t z(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx + 2p(t) ϑ w(x, t)∇z(x, t)∇φ(x, t)e φ(x,t) dx -2p(t) 2 ϑ |∇z(x, t)∇φ(x, t)| 2 e φ(x,t) dx = -2 ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) 2 e φ(x,t) dx + 1 2 ϑ |w(x, t)| 2 e φ(x
(3.16) Thus, (3.15) and (3.16) imply that ,t) dx.

d dt p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx = p ′ (t) ϑ |∇z(x, t)| 2 e φ(x,t) dx + p(t) ϑ |∇z(x, t)| 2 Gφe φ(x,t) dx -2p(t) 2 ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx + p(t) 2 ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x,t) dx -2 ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) 2 e φ(x,t) dx + 1 2 ϑ |w(x, t)| 2 e φ(x
(3.17)

Step 2: Compute y ′ (t)p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx. From the result i/, we have ,t) dx.

y ′ (t)p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx = -2 p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx 2 + 2p(t) ϑ z(x, t)w(x, t)e φ(x,t) dx ϑ |∇z(x, t)| 2 e φ(x,t) dx +p(t) ϑ Gφ(x, t)|z(x, t)| 2 e φ(x,t) dx ϑ |∇z(x, t)| 2 e φ(x,t) dx = 2A(-A + B) + p(t) ϑ Gφ(x, t)|z(x, t)| 2 e φ(x,t) dx ϑ |∇z(x, t)| 2 e φ(x
(3.18)

Here A := p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx and B := ϑ z(x, t)w(x, t)e φ(x,t) dx
Our target is making appear the term ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) -1 2 w(x, t). First of all, we compute A by integrating by parts (3.21)

A = p(t) ϑ |∇z(x, t)| 2 e φ(x,t) dx = -p(t) ϑ ∆z(x, t)z(x, t)e φ(x,t) dx -p(t) ϑ ∇z(x, t)z(x, t)∇φ(x, t)e φ(x,t) dx = ϑ w(x, t)z(x, t)e φ(x,t) dx - ϑ ∂ t z(x, t)z(x, t)e φ(x,t) dx -p(t) ϑ ∇z(x, t)z(x, t)∇φ(x, t)e φ(x,t) dx = - ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) z(x, t)e φ(x,t) dx + 1 2 ϑ w(x, t)z(x, t)e φ(x,t) dx. (3.19) Thus B -A = ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) z(x, t)e φ(x,t) dx + 1 2 ϑ w(x, t)z(x, t)e φ(
Step 3: Compute N ′ (t).

We have ,t) .

N ′ (t) = 1 y(t) 2 y(t) d dt p(t) ϑ |∇z(x, t)| 2 e φ(x,t) -y ′ (t)p(t) ϑ |∇z(x, t)| 2 e φ(x
The result (3.17 ,t) dx.

N ′ (t) = p ′ (t) p(t) N(t) + p(t) 2 y(t) ∂ϑ |∇z(x, t)| 2 ∂ ν φ(x, t)e φ(x,t) dx + p(t) y(t) ϑ |∇z(x, t)| 2 Gφe φ(x,t) dx - 2p(t) 2 y(t) ϑ ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx + 1 2y(t) ϑ |w(x, t)| 2 e φ(x,t) dx - 2 y(t) ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) 2 e φ(x,t) dx + 2 y(t) 2 ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) z(x, t)e φ(x,t) dx 2 - 1 2y(t) 2 ϑ w(x, t)z(x, t)e φ(x,t) dx 2 - p(t) y(t) 2 ϑ Gφ|z(x, t)| 2 e φ(x,t) dx ϑ |∇z(x, t)| 2 e φ(x
Thanks to Cauchy-Schwarz inequality:

ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) z(x, t)e φ(x,t) dx 2 ≤ ϑ ∂ t z(x, t) + p(t)∇z(x, t)∇φ(x, t) - 1 2 w(x, t) 2 e φ(x,t) dx ϑ |z(x, t)| 2 e φ(x,t) dx,
we receive the assertion ii/. Now, by choosing an explicit weight function e φ inspired from the heat kernel, we get the following result. Assume that ϑ be a convex domain or star-shaped with respect to x 0 . For any ρ > 0, with φ is chosen as below

φ(x, t) := -|x -x 0 | 2 4p(T )(T -t + ρ) - n 2 ln(T -t + ρ), (3.22)
we obtain two following estimates:

i/ |y ′ (t) + 2N(t)y(t)| ≤ C 0 T -t + ρ + C 1 y(t)+2 ϑ |w(x, t)z(x, t)|e φ(x,t) dx, (3.23) ii/ N ′ (t) ≤ 1 + C 0 T -t + ρ + C 1 N(t) + 1 2 ϑ |w(x, t)| 2 e φ(x,t) dx y(t) (3.24)
where

C 0 = |p ′ | ∞ R 2 2p 2 1 and C 1 = (2 + n) |p ′ | ∞ p 1 .
Proof of Corollary 3.2

Proof. Obviously, we can easily check the following properties of the function φ:

(1)

∂ t φ + p(T )∆φ + p(T )|∇φ| 2 = 0, (2) ∇φ = -(x-x 0 ) 2p(T )(T -t+ρ) , (3) ∆φ = -n 2p(T )(T -t+ρ) , ( 4
) ∇ 2 φ = -1 2p(T )(T -t+ρ)
I n where I n is the identity matrix of size n.

Remind that Gφ = ∂ t φ + p(t)∆φ + p(t)|∇φ| 2 . Thanks to properties (1), ( 2) and (3), we get

|Gφ| ≤ |p(t) -p(T )| ∆φ + |p(t) -p(T )| |∇φ| 2 ≤ n|p ′ | ∞ 2p(T ) + |p ′ | ∞ R 2 4p(T ) 2 1 T -t + ρ ≤ n|p ′ | ∞ 2p 1 + |p ′ | ∞ R 2 4p 2 1 1 T -t + ρ . (3.25)
Hence, from result i/ in Lemma 3.1, we get the assertion i/. Now, we turn to prove the assertion ii/. Thanks to the assumption that ϑ is star-shaped with respect to x 0 , one has

∂ ν φ = - (x -x 0 )ν 2p(T )(T -t + ρ) ≤ 0 ∀x ∈ ∂ϑ. (3.26)
Furthermore, property (4) implies ,t) dx.

Ω ∇z(x, t)∇ 2 φ(x, t)∇z(x, t)e φ(x,t) dx = -1 2p(T )(T -t + ρ) Ω |∇z(x, t)| 2 e φ(x
(3.27) Consequently, combining result ii/ in Lemma 3.1 with (3.25), (3.26) and (3.27), we get the assertion ii/. Now, the following lemma will solve the ODE inequalities getting from Corollary 3.2.

Lemma 3.2. Let ρ > 0, F ∈ C 0 ([0, T ]). Suppose two positive functions y, N ∈ C 1 ([0, T ]) satisfy the following conditions 1. |y ′ (t) + 2N(t)y(t)| ≤ C 0 T -t + ρ + C 1 + F (t) y(t), (3.28) 2. N ′ (t) ≤ 1 + C 0 T -t + ρ + C 1 N(t) + 1 2 F (t) (3.29)
where C 0 , C 1 > 0. Then for any 0 ≤ t 1 < t 2 < t 3 ≤ T , one has

(y(t 2 )) 1+M ≤ e G T -t 1 + ρ T -t 3 + ρ C 0 (1+M ) y(t 3 ) (y(t 1 )) M (3.30) with M = t 3 t 2 e C 1 s (T -s+h) 1+C 0 ds t 2 t 1 e C 1 s (T -s+h) 1+C 0 ds , G = (1 + M) (t 3 -t 1 ) t 3 t 1 F (s)ds + t 3 t 1 F (s)ds + (t 3 -t 1 )C 1 .
Proof of Lemma 3.2

Proof. From (3.29), we get:

N(t)(T -t + ρ) 1+C 0 e -C 1 t ′ ≤ 1 2 F (t)(T -t + ρ) 1+C 0 e -C 1 t . (3.31)
For t 1 < t < t 2 :

Integrating (3.31) over (t; t 2 ) gives

N(t) ≥ N(t 2 ) T -t 2 + ρ T -t + ρ 1+C 0 e -C 1 (t 2 -t) - 1 2 e C 1 t 1 T -t + ρ 1+C 0 t 2 t F (s)(T -s + ρ) 1+C 0 e -C 1 s ds. (3.32)
Using the fact that

(T -s + ρ) 1+C 0 e -C 1 s ≤ (T -t + ρ) 1+C 0 e -C 1 t ∀s ≥ t, one gets N(t) ≥ Q(t 2 ) e C 1 t (T -t + ρ) 1+C 0 - 1 2 t 2 t 1 F (s)ds (3.33)
where

Q(t 2 ) = e -C 1 t 2 (T -t 2 + ρ) 1+C 0 N(t 2 )
. From (3.28), we also have

y ′ (t) + 2N(t)y(t) ≤ C 0 T -t + ρ + C 1 + F (t) y(t).
(3.34)

Combining to (3.33), we obtain:

y ′ (t) + 2Q(t 2 ) e C 1 t (T -t + ρ) 1+C 0 - t 2 t 1 F (s)ds - C 0 T -t + ρ -C 1 -F (t) y(t) ≤ 0.
It is equivalent to

y(t)e 2Q(t 2 ) t 0 e C 1 s (T -s+ρ) 1+C 0 ds e -( t 2 t 1 F (s)ds+C 1) t (T -t + ρ) C 0 e -t 0 F (s)ds ′ ≤ 0.
(3.35)

Integrating (3.35) over (t 1 ; t 2 ), one has

y(t 1 ) ≥ y(t 2 )e 2Q(t 2 ) t 2 t 1 e C 1 s (T -s+ρ) 1+C 0 ds e -( t 2 t 1 F (s)ds+C 1) (t 2 -t 1 ) T -t 2 + ρ T -t 1 + ρ C 0 e -t 2 t 1 F (s)ds .
(3.36) For t 2 < t < t 3 : Integrating (3.31) over (t 2 ; t) gives

N(t) ≤ N(t 2 ) T -t 2 + ρ T -t + ρ 1+C 0 e -C 1 (t 2 -t) + 1 2 e C 1 t 1 T -t + ρ 1+C 0 t t 2 F (s)(T -s + ρ) 1+C 0 e -C 1 s ds. (3.37) Using the fact that (T -s + ρ) 1+C 0 e -C 1 s ≤ (T -t 2 + ρ) 1+C 0 e -C 1 t 2 ∀s ≥ t 2 , we obtain N(t) ≤ Q(t 2 ) 1 T -t + ρ 1+C 0 e C 1 t + 1 2 e C 1 (t-t 2 ) T -t 2 + ρ T -t + ρ 1+C 0 t 3 t 2 F (s)ds.(3.38)
From (3.28), we also have

y ′ (t) + 2N(t)y(t) ≥ - C 0 T -t + ρ + C 1 + F (t) y(t). (3.39)
It deduces from (3.38) and (3.39) that

y ′ (t) + 2Q(t 2 ) e C 1 t (T -t + ρ) 1+C 0 + e C 1 (t-t 2 ) T -t 2 + ρ T -t + ρ 1+C 0 t 3 t 2 F (s)ds ≥ - C 0 T -t + ρ + C 1 + F (t) y(t). (3.40)
It is equivalent to

y(t)e 2Q(t 2 ) t 0 e C 1 s (T -s+ρ) 1+C 0 ds+ t 3 t 2 F (s)ds t 0 e C 1 (s-t 2 ) ( T -t 2 +ρ T -s+ρ ) 1+C 0 ds+C 1 t+ t 0 F (s)ds 1 (T -t + ρ) C 0 ′ ≥ 0.
(3.41) Integrating (3.41) over (t 2 ; t 3 ) gives

y(t 2 ) ≤ y(t 3 )e 2Q(t 2 ) t 3 t 2 e C 1 s (T -s+ρ) 1+C 0 ds e t 3 t 2 F (s)ds t 3 t 2 e C 1 (s-t 2 ) ( T -t 2 +ρ T -s+ρ ) 1+C 0 ds × T -t 2 + ρ T -t 3 + ρ C 0 e C 1 (t 3 -t 2 ) e t 3 t 2 F (s)ds . (3.42)
Combining to (3.36), one gets

y(t 2 ) ≤ y(t 3 ) y(t 1 ) y(t 2 ) M T -t 1 + ρ T -t 2 + ρ M C 0 e ( t 2 t 1 F (s)ds+C 1) M (t 2 -t 1 ) e M t 2 t 1 F (s)ds ×e C 1 (t 3 -t 2 ) e t 3 t 2 F (s)ds t 3 t 2 e C 1 (s-t 2 ) ( T -t 2 +ρ T -s+ρ ) 1+C 0 ds e t 3 t 2 F (s)ds T -t 2 + ρ T -t 3 + ρ C 0 (3.43)
where

M = t 3 t 2 e C 1 s (T -s+ρ) 1+C 0 ds t 2 t 1 e C 1 s (T -s+ρ) 1+C 0 ds . (3.44)
We also have

t 3 t 2 e C 1 (s-t 2 ) T -t 2 + ρ T -s + ρ 1+C 0 ds = t 3 t 2 e C 1 s (T -s+ρ) 1+C 0 ds e C 1 t 2 (T -t 2 +ρ) 1+C 0 ≤ (t 2 -t 1 ) t 3 t 2 e C 1 s (T -s+ρ) 1+C 0 ds t 2 t 1 e C 1 s (T -s+ρ) 1+C 0 ds ≤ (t 2 -t 1 )M.
(3.45) Hence, it is deduced from (3.43) that

(y(t 2 )) 1+M ≤ y(t 3 ) (y(t 1 )) M e M (t 2 -t 1 ) t 2 t 1 F (s)ds e C 1 M (t 2 -t 1 ) e M t 2 t 1 F (s)ds ×e M (t 2 -t 1 ) t 3 t 2 F (s)ds e C 1 (t 3 -t 2 ) e t 3 t 2 F (s)ds × T -t 2 + ρ T -t 3 + ρ C 0 T -t 1 + ρ T -t 2 + ρ M C 0 ≤ y(t 3 ) (y(t 1 )) M e (1+M )[(t 3 -t 1 ) t 3 t 1 F (s)ds+ t 3 t 1 F (s)ds+(t 3 -t 1 )C 1] T -t 1 + ρ T -t 3 + ρ (1+M )C 0 . (3.46)
We move to an application of this Lemma with specific choice of time.

Corollary 3.3. Under the assumption of Lemma 3.2, for any ρ > 0 and ℓ > 1 such that ℓρ ≤ min{ 1 2 ; T 4 }, one has

(y(T -ℓρ)) 1+M ℓ ≤ e G ℓ (1 + 2ℓ) 2C 0 (1+M ℓ ) y(T ) (y(T -2ℓρ)) M ℓ (3.47)
where

M ℓ = T T -ℓρ e C 1 s (T -s+ρ) 1+C 0 ds T -ℓρ T -2ℓρ e C 1 s (T -s+ρ) 1+C 0 ds . (3.48)
and

G ℓ = (1 + M ℓ ) 2 T T -2ℓρ F (s)ds + C 1 . (3.49)
Moreover, the upper bound of M ℓ can be given as

M ℓ ≤ S ℓ := e C 1    ln(1+ℓ) ln 3 2 if C 0 = 0 (1+ℓ) C 0 1-( 2 3 ) C 0 if C 0 > 0 .
Proof of Corollary 3.3

Proof. Now, for ℓ > 1 and ℓρ < min{ 1 2 ; T 4 }, applying Lemma 3.2 for

t 1 = T -2ℓρ; t 2 = T -ℓρ; t 3 = T , we get (y(T -ℓρ)) 1+M ℓ ≤ y(T ) (y(T -2ℓρ)) M ℓ e (1+M ℓ )(2 T T -2ℓρ F (s)ds+C 1) (1 + 2ℓ) (1+M ℓ )C 0 (3.50) with M ℓ = T T -ℓρ e C 1 s (T -s+ρ) 1+C 0 ds T -ℓρ T -2ℓρ e C 1 s (T -s+ρ) 1+C 0 ds . If C 0 = 0 then M ℓ = T T -ℓρ e C 1 s (T -s+ρ) ds T -ℓρ T -2ℓρ e C 1 s (T -s+ρ) ds ≤ e C 1 ln(1 + ℓ) ln 1+2ℓ 1+ℓ ≤ e C 1 ln(1 + ℓ) ln 3 2 . If C 0 > 0 then M ℓ ≤ e 2ℓρC 1 (1 + ℓ) C 0 -1 (1+2ℓ) C 0 -(1+ℓ) C 0 (1+2ℓ) C 0 ≤ e C 1 (1 + ℓ) C 0 1 -2 3 C 0 . Lemma 3.3. Let x 0 ∈ Ω, ̺ > 0 and 0 < ǫ < ̺ 2 .
Let v be the solution of (3.1). Then there exist constants E 1 > 1, E 2 > 0 and E 3 > 0, which all depend on ̺ and ǫ, such that the following estimate holds

Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺) |v(x, t)| 2 dx ≤ E 1 e E 1 θ ∀ T 2 < T -E 2 θ ≤ t ≤ T .
Here

1 θ = ln E 3 e E 3 T Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx .
Proof of Lemma 3.3 In order to get a local estimate, we need to use a weight function e -|x-x 0 | 2 ( will be chosen later) and a cut-off function Ψ on B(x 0 , ̺). After finding an ODE inequality for

Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx (see
Step 1) and solving it (see Step 2), we can get the final result with a suitable choice of (see Step 3). Now, we start the proof with the definition of the following cut-off function.

Proof. Let Ψ ∈ C ∞ 0 (B(x 0 , ̺)) such that Ψ = 1 in B(x 0 , ̺ -ǫ), Ψ ∈ (0; 1) in B(x 0 , ̺).
Step 1: For < 1, find an ODE inequality for

Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx. We have d dt Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx = 2 Ω∩B(x 0 ,̺) |Ψ(x)| 2 v(x, t)∂ t v(x, t)e -|x-x 0 | 2 dx = 2p(t) Ω∩B(x 0 ,̺) |Ψ(x)| 2 v(x, t)∆v(x, t)e -|x-x 0 | 2 dx. (3.51)
By integrating by parts with the fact that Ψv = 0 on ∂(Ω ∩ B(x 0 , ̺)) one obtains

Ω∩B(x 0 ,̺) |Ψ(x)| 2 v(x, t)∆v(x, t)e -|x-x 0 | 2 dx = - Ω∩B(x 0 ,̺) ∇v(x, t)∇(|Ψ(x)| 2 v(x, t)e -|x-x 0 | 2 )dx = -2 Ω∩B(x 0 ,̺) Ψ(x)∇Ψ(x)∇v(x, t)v(x, t)e -|x-x 0 | 2 dx - Ω∩B(x 0 ,̺) |Ψ(x)| 2 |∇v(x, t)| 2 e -|x-x 0 | 2 dx +2 Ω∩B(x 0 ,̺) |Ψ(x)| 2 v(x, t)∇v(x, t)(x -x 0 ) 1 e -|x-x 0 | 2 dx := P 1 + P 2 + P 3 (3.52)
where P i (i = 1, 2, 3) is the i th term in the right-hand side of (3.52). Now, thanks to Cauchy-Schwarz inequality, we get

P 1 = -2 Ω∩B(x 0 ,̺) Ψ(x)∇Ψ(x)∇v(x, t)v(x, t)e -|x-x 0 | 2 dx ≤ 2 Ω∩B(x 0 ,̺) |Ψ(x)∇v(x, t)| 2 e -|x-x 0 | 2 dx 1 2 Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx 1 2
and

P 3 = 2 Ω∩B(x 0 ,̺) |Ψ(x)| 2 v(x, t)∇v(x, t)(x -x 0 ) 1 e -|x-x 0 | 2 dx ≤ 2 Ω∩B(x 0 ,̺) |Ψ(x)∇v(x, t)| 2 e -|x-x 0 | 2 dx 1 2 × Ω∩B(x 0 ,̺) Ψ(x)v(x, t) (x -x 0 ) 2 e -|x-x 0 | 2 dx 1 2
.

Thus

P 1 + P 3 ≤ 2 Ω∩B(x 0 ,̺) |Ψ(x)∇v(x, t)| 2 e -|x-x 0 | 2 dx 1 2 × Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx 1 2 + Ω∩B(x 0 ,̺) Ψ(x)v(x, t) (x -x 0 ) 2 e -|x-x 0 | 2 dx 1 2   . (3.53) It follows from 2ab ≤ a 2 + b 2 and (a + b) 2 ≤ 2a 2 + 2b 2 ∀a, b > 0 that P 1 + P 3 ≤ Ω∩B(x 0 ,̺) |Ψ(x)∇v(x, t)| 2 e -|x-x 0 | 2 dx + Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx 1 2 + Ω∩B(x 0 ,̺) Ψ(x)v(x, t) (x -x 0 ) 2 e -|x-x 0 | 2 dx 1 2   2 ≤ Ω∩B(x 0 ,̺) |Ψ(x)∇v(x, t)| 2 e -|x-x 0 | 2 dx + 2 Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx +2 Ω∩B(x 0 ,̺) Ψ(x)v(x, t) (x -x 0 ) 2 e -|x-x 0 | 2 dx (3.54)
Combining (3.51), (3.52) and (3.54), we can conclude

d dt Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx ≤ 4p(t) Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx +4p(t) Ω∩B(x 0 ,̺) Ψ(x)v(x, t) (x -x 0 ) 2 e -|x-x 0 | 2 dx ≤ 4p 2 Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx +4p 2 ̺ 2 2 Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx. (3.55) Moreover, due to ∇Ψ(•) = 0 in Ω ∩ B(x 0 , ̺ -ǫ), one has Ω∩B(x 0 ,̺) |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx = Ω∩{|x-x 0 |≥̺-ǫ} |v(x, t)∇Ψ(x)| 2 e -|x-x 0 | 2 dx ≤ |∇Ψ| ∞ e -(̺-ǫ) 2 Ω |v(x, t)| 2 dx ≤ |∇Ψ| ∞ e -(̺-ǫ) 2 Ω |v(x, 0)| 2 dx. (3.56) Thus d dt Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx ≤ 4p 2 ̺ 2 2 Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx + 4p 2 |∇Ψ| 2 ∞ e -(̺-ǫ) 2 Ω |v(x, 0)| 2 dx.
(3.57)

Step 2: Solve ODE inequality. It deduces from (3.57) that

d dt e -4p 2 ̺ 2 2 t Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx ≤ 4p 2 |∇Ψ| 2 ∞ e -4p 2 ̺ 2 2 t e -(̺-ǫ) 2 Ω |v(x, 0)| 2 dx. (3.58)
Integrating (3.58) over (t; T ) gives

Ω∩B(x 0 ,̺) |Ψ(x)v(x, T )| 2 e -|x-x 0 | 2 dx ≤ e 4p 2 ̺ 2 2 (T -t) Ω∩B(x 0 ,̺) |Ψ(x)v(x, t)| 2 e -|x-x 0 | 2 dx +4p 2 |∇Ψ| 2 ∞ e 4p 2 ̺ 2 2 T e -(̺-ǫ) 2 Ω |v(x, 0)| 2 dx T t e -4p 2 ̺ 2 2 s ds. (3.59) It implies from the fact Ψ(•) = 1 in B(x 0 , ̺ -ǫ) that Ω∩B(x 0 ,̺) |Ψ(x)v(x, T )| 2 e -|x-x 0 | 2 dx ≥ Ω∩B(x 0 ,̺-ǫ) |Ψ(x)v(x, T )| 2 e -|x-x 0 | 2 dx = Ω∩B(x 0 ,̺-ǫ) |v(x, T )| 2 e -|x-x 0 | 2 dx ≥ Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 e -|x-x 0 | 2 dx ≥ e -(̺-2ǫ) 2 Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx. (3.60)
Combining (3.59), (3.60) and the following estimate

T t e -4p 2 ̺ 2 2 s ds = 1 4p 2 ̺ 2 2 e -4p 2 ̺ 2 2 T -e -4p 2 ̺ 2 2 t ≤ 1 4p 2 ̺ 2 2 e -4p 2 ̺ 2 2 t ≤ 1 4p 2 ̺ 2 e -4p 2 ̺ 2 2 t (3.61) for < 1, we obtain Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx ≤ e 4p 2 ̺ 2 2 (T -t) e (̺-2ǫ) 2 Ω∩B(x 0 ,̺) |v(x, t)| 2 dx + |∇Ψ| 2 ∞ ̺ 2 e 4p 2 ̺ 2 2 (T -t) e (̺-2ǫ) 2 - (̺-ǫ) 2 Ω |v(x, 0)| 2 dx. (3.62) Let T 2 < T -η ≤ t ≤ T , it yields Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx ≤ e 4p 2 ̺ 2 η e (̺-2ǫ) 2 Ω∩B(x 0 ,̺) |v(x, t)| 2 dx + |∇Ψ| 2 ∞ ̺ 2 e 4p 2 ̺ 2 η e (̺-2ǫ) 2 -(̺-ǫ) 2 Ω |v(x, 0)| 2 dx. (3.63) We choose η = ǫ(2̺-3ǫ) 8p 2 ̺ 2 , that is 4p 2 ̺ 2 η = 1 2 ǫ(-3ǫ + 2̺) in order to get 4p 2 ̺ 2 η + (̺ -2ǫ) 2 -(̺ -ǫ) 2 < 0
Then, (3.63) becomes

Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx ≤ e (̺-ǫ) 2 +(̺-2ǫ) 2 2 Ω∩B(x 0 ,̺) |v(x, t)| 2 dx + |∇Ψ| 2 ∞ ̺ 2 e (̺-2ǫ) 2 -(̺-ǫ) 2 2 Ω |v(x, 0)| 2 dx.
(3.64)

Step 3: Choose for minimization problem. Now, for the purpose of minimizing the right-hand side of inequality (3.64), we choose such that

|∇Ψ| 2 ∞ ̺ 2 e (̺-2ǫ) 2 -(̺-ǫ) 2 2 Ω |v(x, 0)| 2 dx ≤ 1 2 Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx or e (̺-2ǫ) 2 2 ≤ e (̺-ǫ) 2 2 ̺ 2 2|∇Ψ| 2 ∞ Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx .
The choice of also satisfies the condition that < 1 and η < min{1, T 2 }. Such exists by choosing

= ǫ(2̺-3ǫ) 2 ln 2|∇Ψ| 2 ∞ ̺ 2 Ω |v(x,0)| 2 dx Ω∩B(x 0 ,̺-2ǫ) |v(x,T )| 2 dx + ǫ(2̺-3ǫ) 2 1 + η 1 + 2 T .
With this choice, it implies from (3.64) that

Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx ≤ ̺ 2 |∇Ψ| 2 ∞ e (̺-2ǫ) 2 Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺) |v(x, t)| 2 dx.

This is equivalent to

Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺) |v(x, t)| 2 dx ≤ ̺ 2 |∇Ψ| 2 ∞ e (̺-2ǫ) 2
. This completes the proof with

E 1 = max ̺ 2 |∇Ψ| 2 ∞ ; 2(̺-2ǫ) 2 ǫ(2̺-3ǫ) ; 1 > 1 and 1 θ = ln 2|∇Ψ| 2 ∞ ̺ 2 2 Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx + ǫ(2̺ -3ǫ) 2 1 + η 1 + 2 T = ln 2|∇Ψ| 2 ∞ ̺ 2 2 Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx e ǫ(2̺-3ǫ) 2 [1+η(1+ 2 T )]
Thus, there exists a constant E 3 > 0 such that

1 θ = ln E 3 e E 3 T Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,̺-2ǫ) |v(x, T )| 2 dx On the other hand, η = θ ǫ 2 (2̺-3ǫ) 2 16p 2 ̺ 2 . Hence, E 2 = ǫ 2 (2̺-3ǫ) 2 16p 2 ̺ 2 .
3.2 Proof of Theorem 3.2

Proof. Let x 0 ∈ Ω, ρ > 0 and ℓ > 1 such that ℓρ ≤ min{ 1 2 , T 4 }, Corollary 3.2 gives |y ′ (t) + 2N(t)y(t)| ≤ C 0 T -t + ρ + C 1 y(t) (3.65)
and

N ′ (t) ≤ 1 + C 0 T -t + ρ + C 1 N(t).
(3.66)

Here

y(t) = 1 (T -t + ρ) n 2 Ω |v(x, t)| 2 e -|x-x 0 | 2 4p(T )(T -t+ρ) dx, N(t) = p(t) (T -t + ρ) n 2 Ω |∇v(x, t)| 2 e -|x-x 0 | 2 4p(T )(T -t+ρ) dx y(t) , C 0 = |p ′ | ∞ R 2 2p 2 1 and C 1 = (2 + n) |p ′ | ∞ p 1 .
Now, thanks to Corollary 3.3, one has

y(T -ℓρ) 1+M ℓ ≤ e G ℓ (1 + 2ℓ) C 0 (1+M ℓ ) y(T )y(T -2ℓρ) M ℓ (3.67)
where

G ℓ = C 1 (1 + M ℓ )
and

M ℓ ≤ S ℓ = e C 1    ln(1+ℓ) ln 3 2 if C 0 = 0, (1+ℓ) C 0 1-( 2 3 ) C 0 if C 0 > 0.
Then (3.67) is equivalent to

Ω |v(x, T -ℓρ)| 2 dx 1+M ℓ ≤ ((1 + ℓ)ρ) (1+M ℓ ) n 2 ρ n 2 ((1 + 2ℓ)ρ) M ℓ n 2 e R 2 (1+M ℓ ) 4p(T )(1+ℓ)ρ e C 1 (1+M ℓ ) (1 + 2ℓ) C 0 (1+M ℓ ) × Ω |v(x, t)| 2 e -|x-x 0 | 2 4ρp(T ) dx Ω |v(x, T -2ℓρ)| 2 e -|x-x 0 | 2 4(1+ℓ)ρp(T ) dx M ℓ ≤ 2 C 0 (1+M ℓ ) (1 + ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )C 1 e R 2 (1+M ℓ ) 4p(T )(1+ℓ)ρ × Ω |v(x, t)| 2 e -|x-x 0 | 2 4ρp(T ) dx Ω |v(x, 0)| 2 dx M ℓ . (3.68)
On the other hand, for 0 < r < R such that B(x 0 , r) ⊂ ω, one has

Ω |v(x, T )| 2 e -|x-x 0 | 2 4ρp(T ) dx ≤ B(x 0 ,r) |v(x, T )| 2 dx + Ω∩{x;|x-x 0 |≥r} |v(x, T )| 2 e -|x-x 0 | 2 4ρp(T ) dx ≤ B(x 0 ,r) |v(x, T )| 2 dx + e -r 2 4ρp(T ) Ω |v(x, T )| 2 dx ≤ ω |v(x, T )| 2 dx + e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx.
(3.69)

Moreover, we also have

Ω |v(x, T )| 2 dx 1+M ℓ ≤ 2 C 0 (1+M ℓ ) (1 + ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )C 1 e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) × ω |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx M ℓ + e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx 1+M ℓ . (3.71)
With the notice that M ℓ ≤ S ℓ , we can write (3.71) as below, thanks to the energy estimate

Ω |v(x, 0)| 2 dx ≥ Ω |v(x, T )| 2 dx Ω |v(x, T )| 2 dx 1+S ℓ ≤ 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e R 2 (1+S ℓ ) 4(1+ℓ)ρp(T ) × ω |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx S ℓ + e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx 1+S ℓ . (3.72)
Now, in order to minimize the right-hand side of inequality (3.72), we will choose ℓ > 1 as

ℓ =        2 2+ξ R 2 e C 1 ξ ln 3 2 r 2 1 1-ξ -1 ∀ξ ∈ (0, 1) if C 0 = 0, 4R 2 e C 1 r 2 1-( 2 3 ) C 0 1 1-C 0 -1 if C 0 > 0.
The assumption (i) follows C 0 < 1. Hence, such choice of ℓ provides us

R 2 (1 + S ℓ ) 4(1 + ℓ)ρp(T ) ≤ r 2 8ρp(T )
.

Thus, (3.72) becomes:

Ω |v(x, T )| 2 dx 1+S ℓ ≤ 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e r 2 8p(T )h ω |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx S ℓ + 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e -r 2 8p(T )h Ω |v(x, 0)| 2 dx 1+S ℓ . (3.73)
This estimate is true for any ρ > 0 satisfying ρ < 1 ℓ min{ 1 2 ; T 4 }. For ρ ≥ 1 ℓ min{ 1 2 ; T 4 } which implies r 2 8ρp(T ) ≤ r 2 ℓ 4p(T ) 1 + 2 T , we can get the following estimate be true for any ρ > 0.

Ω |v(x, T )| 2 dx 1+S ℓ ≤ 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e r 2 8ρp(T ) ω |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx S ℓ + 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e r 2 ℓ 4p(T ) (1+ 2 T ) e -r 2 8ρp(T ) Ω |v(x, 0)| 2 dx 1+S ℓ . (3.74) Now, we choose ρ such that 2 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e r 2 ℓ 4p(T ) (1+ 2 T ) e -r 2 8ρp(T ) Ω |v(x, 0)| 2 dx 1+S ℓ = 1 2 Ω |v(x, T )| 2 dx 1+S ℓ that is e r 2 8ρp(T ) = 22 C 0 (1+S ℓ ) (1 + ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )C 1 e r 2 ℓ 4p(T ) ( 1+ 2 T ) Ω |v(x, 0)| 2 dx Ω |v(x, T )| 2 dx 1+S ℓ .
Therefore, we get

Ω |v(x, T )| 2 dx 2(1+S ℓ ) ≤ 4.4 C 0 (1+S ℓ ) (1 + ℓ) n+2C 0 (1+S ℓ ) e 2(1+S ℓ )C 1 e r 2 ℓ 4p(T ) (1+ 2 T ) × ω |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx 1+2S ℓ .
Thus, we can state

Ω |v(x, T )| 2 dx ≤ 4 1+C 0 (1+S ℓ ) (1 + ℓ) n+2C 0 (1+S ℓ ) e 2(1+S ℓ )C 1 e r 2 ℓ 4p(T ) ( 1+ 2 T ) ω |v(x, T )| 2 dx 1 2(1+S ℓ ) × Ω |v(x, 0)| 2 dx 1+2S ℓ 2(1+S ℓ )
.

(3.75)

This completes the proof.

Proof of Theorem 3.1

Let us move to the proof of Theorem 3.1 with the structure as: Thanks to Preliminary results in the previous subsection, we can get a Hölder type estimate in Step 1. Furthermore, thanks to the technical Lemma 3.3, we get an estimate for the term containing F (t), which is presented in Step 2. Step 3 will make appear a small ball, which is related to the presence of ω later, by using a splitting technique. Next, dealing with a minimization problem, Step 4 provides us a localized observation estimate. Due to propagation of smallness by constructing the sequence of balls, ω will appear in Step 5. Lastly, in Step 6, by using an adequate covering of Ω with a finite number of balls, we will get the desired result.

Proof.

Step 1: Get Hölder type inequality. Let x 0 ∈ Ω and R be small enough such that Ω ∩ B(x 0 ; R) is star-shaped with respect to x 0 . Such choice of x 0 and R will be mentioned in Step 6. Let

0 < ǫ < R 4 . Define ψ ∈ C 2 0 (B(x 0 ; R)) satisfying ψ = 1 in B(x 0 ; R -ǫ) and 0 < ψ(t) < 1 ∀t ∈ B(x 0 ; R). Then ψv ∈ H 1 ((0, T ); H 1 0 (Ω ∩ B(x 0 ; R)). For any ρ > 0 and ℓ > 1 such that ℓρ ≤ min{ 1 2 , T 4 }, Corollary 3.2 gives |y ′ (t) + 2N(t)y(t)| ≤ C 0 T -t + ρ + C 1 y(t) + 2 ϑ |w(x, t)z(x, t)|e φ(x,t) dx (3.76)
and

N ′ (t) ≤ 1 + C 0 T -t + ρ + C 1 N(t) + 1 2 ϑ |w(x, t)| 2 e φ(x,t) dx y(t) . (3.77) Here ϑ = Ω ∩ B(x 0 ; R), z = ψv, w := ∂ t z -∆z, y(t) = 1 (T -t + ρ) n 2 ϑ |z(x, t)| 2 e -|x-x 0 | 2 4p(T )(T -t+ρ) dx, N(t) = p(t) (T -t + ρ) n 2 ϑ |∇z(x, t)| 2 e -|x-x 0 | 2 4p(T )(T -t+ρ) dx y(t) , C 0 = |p ′ | ∞ R 2 2p 2 1 and C 1 = (2 + n) |p ′ | ∞ p 1 .
Thanks to the Cauchy-Schwarz inequality and the fact that 2ab ≤ a 2 + b 2 ∀a, b, one gets

2 ϑ |w(x, t)z(x, t)|e φ(x,t) dx ≤ 2 ϑ |w(x, t)| 2 e φ(x,t) dx 1 2 ϑ |z(x, t)| 2 e φ(x,t) dx 1 2 ≤ ϑ |w(x, t)| 2 e φ(x,t) dx + ϑ |z(x, t)| 2 e φ(x,t) dx, (3.78)
Thus, we can write (3.76) as below ,t) dx y(t) .

|y ′ (t) + 2N(t)y(t)| ≤ C 0 T -t + ρ + C 1 + 1 + F (t) y(t) (3.79) with F (t) := ϑ |w(x, t)| 2 e φ(x
Now, thanks to Corollary 3.3, one has

y(T -ℓρ) 1+M ℓ ≤ e G ℓ (1 + 2ℓ) C 0 (1+M ℓ ) y(T )y(T -2ℓρ) M ℓ (3.80)
where

G ℓ = (1 + M ℓ ) C 1 + 1 + 2 T T -2ℓρ F (s)ds and M ℓ ≤ S ℓ = e C 1 +1    ln(1+ℓ) ln 3 2 if C 0 = 0, (1+ℓ) C 0 1-( 2 3 ) C 0 if C 0 > 0.
Step 2:

Estimate T T -2ℓρ F (s)ds. Remind that F (s) = ϑ |w(x, s)| 2 e φ(x,s) dx ϑ |z(x, s)| 2 e φ(x,s) dx (3.81) with w(x, s) = ∂ t z(x, s) -p(s)∆z(x, s) = -p(s)v(x, s)∆ψ(x) -2p(s)∇v(x, s)∇ψ(x). (3.82) Note that ∇ψ = ∆ψ = 0 in Ω ∩ B(x 0 , R -ǫ), so ϑ |w(x, s)| 2 e φ(x,s) dx = p(s) 2 ϑ (v(x, s)∆ψ(x) + 2∇v(x, s))∇ψ(x)) 2 e φ(x,s) dx = p(s) 2 Ω∩{x;|x-x 0 |≥R-ǫ} (v(x, s)∆ψ(x) + 2∇v(x, s))∇ψ(x)) 2 e φ(x,s) dx. (3.83)
It implies from the fact that e φ(x,s) =

1 (T -s+ρ) n 2 e -|x-x 0 | 2 4p(T )(T -s+ρ) and (a+b) 2 ≤ 2(a 2 +b 2 ) ∀a, b that ϑ |w(x, s)| 2 e φ(x,s) dx ≤ 2p(s) 2 (T -s + ρ) n 2 e -(R-ǫ) 2 4p(T )(T -s+ρ) |∆ψ| 2 ∞ Ω |v(x, s)| 2 dx + 4|∇ψ| 2 ∞ Ω |∇v(x, s)| 2 dx .
(3.84)

Moreover, thanks to the following energy estimate

Ω |∇v(x, s)| 2 dx ≤ 1 2p 1 s Ω |v(x, 0)| 2 dx ∀s > 0,
we obtain:

ϑ |w(x, s)| 2 e φ(x,s) dx ≤ C 2 (T -s + ρ) n 2 e -(R-ǫ) 2 4p(T )(T -s+ρ) 1 + 1 2p 1 s Ω |v(x, 0)| 2 dx (3.85) where C 2 = 2p 2 2 max{|∆ψ| 2 ∞ , 4|∇ψ| 2 ∞ }. On the other hand, due to ψ = 1 on B(x 0 , R-ǫ), one gets ϑ |z(x, s)| 2 e φ(x,s) dx = Ω∩B(x 0 ,R-ǫ) |v(x, s)| 2 e φ(x,s) dx ≥ 1 (T -s + ρ) n 2 e -(R-2ǫ) 2 4p(T )(T -s+ρ) Ω∩B(x 0 ,R-2ǫ) |v(x, s)| 2 dx.(3.86)
Combining (3.81), (3.85) and (3.86) gives

F (s) ≤ C 2 e -ǫ(2R-3ǫ) 12ℓρp(T ) 1 + 1 2p 1 s Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,R-2ǫ) |v(x, s)| 2 dx (3.87) with s ∈ [T -2ℓρ, T ]. Now, apply Lemma 3.3 with ̺ = R-2ǫ, under condition 2ℓρ ≤ E 2 θ
for some E 2 > 0 depending on R and ǫ, there exists a constant

E 1 = E 1 (R, ǫ) > 1 such that Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,R-2ǫ) |v(x, s)| 2 dx ≤ E 1 e E 1 θ ∀T -ℓρ ≤ s ≤ T . (3.88)
Thus, from (3.87) and (3.88), one has

T T -2ℓρ F (s)ds ≤ E 1 e E 1 θ C 2 e -ǫ(2R-3ǫ) 12ℓρp 1 T T -2ℓρ 1 + 1 2p 1 s ds.
In order to get E 1 θ -ǫ(2R-3ǫ) 12ℓρp 1 < 0, we take 2ℓρ ≤ cE 2 θ with c = min ǫ(2R-3ǫ) 6p 1 E 1 E 2 ; 1 . On the other hand, due to

T T -2ℓρ 1 + 1 2p 1 s ds = 2ℓρ + 1 2p 1 ln T T -2ℓρ ≤ 1 + 1 2p 1 ln 2
, there exists a constant C 3 > 0 which does not depend on ℓ, ρ and T such that

T T -2ℓρ F (s)ds ≤ C 3 .
Step 3: Make appear a small ball. Remind that

y(t) = ϑ |z(x, t)| 2 1 (T -t + ρ) n 2 e -|x-x 0 | 2 4p(T )(T -t+ρ) .
From (3.80) in Step 1, we have

ϑ |z(x, T -ℓρ)| 2 dx 1+M ℓ ≤ (1 + 2ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )(C 1 +1+2C 3 ) e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) ϑ |z(x, t)| 2 e -|x-x 0 | 2 4ρp(T ) dx ϑ |z(x, T -2ℓρ)| 2 e -|x-x 0 | 2 4(1+2ℓ)ρp(T ) dx M ℓ . (3.89) From the fact that |z| ≤ |v| in Ω ∩ B(x 0 , R) and Ω |v(x, T -2ℓρ)| 2 dx ≤ Ω |v(x, 0)| 2 dx, we can write ϑ |z(x, T -ℓρ)| 2 dx 1+M ℓ ≤ (1 + 2ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )(C 1 +1+2C 3 ) e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) ϑ |v(x, t)| 2 e -|x-x 0 | 2 4ρp(T ) dx Ω |v(x, 0)| 2 dx M ℓ . (3.90)
On the other hand, thanks to ψ = 1 on B(x 0 , Rǫ), we obtain

ϑ |z(x, T -ℓρ)| 2 dx ≥ Ω∩B(x 0 ,R-2ǫ) |v(x, T -ℓρ)| 2 dx.
(3.91)

From inequality (3.88), one gets

Ω∩B(x 0 ,R-2ǫ) |v(x, T -ℓρ)| 2 dx ≥ 1 E 1 e E 1 θ Ω |v(x, 0)| 2 dx ≥ 1 E 1 e E 1 θ Ω |v(x, T )| 2 dx.
(3.92) Combining (3.90), (3.91) and (3.92), it yields

Ω |v(x, T )| 2 dx 1+M ℓ ≤ (E 1 e E 1 θ ) 1+M ℓ (1 + 2ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )(C 1 +1+2C 3 ) e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) ϑ |v(x, t)| 2 e -|x-x 0 | 2 4ρp(T ) dx Ω |v(x, 0)| 2 dx M ℓ . (3.93) Furthermore, for 0 < r ≤ R 2 such that B(x 0 , r) ⊂ Ω, one has ϑ |v(x, T )| 2 e -|x-x 0 | 2 4ρp(T ) dx ≤ B(x 0 ,r) |v(x, T )| 2 dx + Ω∩{x;|x-x 0 |≥r} |v(x, T )| 2 e -|x-x 0 | 2 4ρp(T ) dx ≤ B(x 0 ,r) |v(x, T )| 2 dx + e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx.
(3.94) Thus, we obtain that

Ω |v(x, T )| 2 dx 1+M ℓ ≤ (E 1 e E 1 θ ) 1+M ℓ (1 + 2ℓ) n 2 +C 0 (1+M ℓ ) e (1+M ℓ )(C 1 +1+2C 3 ) e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) ϑ |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx M ℓ + e R 2 (1+M ℓ ) 4(1+ℓ)ρp(T ) e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx 1+M ℓ . (3.95)
Notice that M ℓ ≤ S ℓ , therefore we can write (3.95) as below

Ω |v(x, T )| 2 dx 1+S ℓ ≤ (E 1 e E 1 θ ) 1+S ℓ (1 + 2ℓ) n 2 +C 0 (1+S ℓ ) e (1+S ℓ )(C 1 +1+2C 3 ) e R 2 (1+S ℓ ) 4(1+ℓ)ρp(T ) ϑ |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx S ℓ + e R 2 (1+S ℓ ) 4(1+ℓ)ρp(T ) e -r 2 4ρp(T ) Ω |v(x, 0)| 2 dx 1+S ℓ . (3.96)
Step 4: Choose suitable ℓ, ρ and solve a minimization problem. With R is small enough such that C 0 := R 2 |p ′ |∞ 2p 2 1 < 1, we will minimize the right-hand side of the estimate (3.96) by choosing ℓ as below

ℓ =        2 2+ξ R 2 e C 1 +1 ξ ln 3 2 r 2 1 1-ξ -1 ∀ξ ∈ (0, 1) if C 0 = 0; 4R 2 e C 1 +1
r 2 1-( 23 )

C 0 1 1-C 0 -1 if C 0 > 0 in order to get R 2 (1 + S ℓ ) 4(1 + ℓ)ρp(T ) ≤ r 2 8ρp(T )
.

With this choice of ℓ and the fact that E 1 > 1, we can conclude from (3.96) that: there exists a constant C 4 > 1 not depending on ρ such that

Ω |v(x, T )| 2 dx 1+S ℓ ≤ C 4 e C 4 θ e r 2 8p(T )ρ ϑ |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx S ℓ + C 4 e C 4 θ e -r 2 8p(T )ρ Ω |v(x, 0)| 2 dx 1+S ℓ .
This estimate is true for any ρ > 0 satisfying ρ ≤ 1 ℓ min{ 1 2 ; T 4 ; cE 2 θ}. For ρ > 1 ℓ min{ 1 2 ; T 4 ; cE 2 θ} which implies r 2 8ρp(T ) < r 2 ℓ 8p(T ) 2 + 4 T + 1 cE 2 θ , we can get the following estimate be true for any ρ > 0.

C 4 e C 4 θ e r 2 ℓ 8p(T ) 2+ 4 T + 1 cE 2 θ e -r 2 8ρp(T ) Ω |v(x, 0)| 2 dx 1+S ℓ = 1 2 Ω |v(x, T )| 2 dx 1+S ℓ ; that is e r 2 8ρp(T ) = 2C 4 e C 4 θ e r 2 ℓ 8p(T ) 2+ 4 T + 1 cE 2 θ Ω |v(x, 0)| 2 dx Ω |v(x, T )| 2 dx 1+S ℓ ; in order to get Ω |v(x, T )| 2 dx 2(1+S ℓ ) ≤ 4C 2 4 e 2C 4 θ e r 2 ℓ 8p(T ) 2+ 4 T + 1 cE 2 θ B(x 0 ,r) |v(x, T )| 2 dx Ω |v(x, 0)| 2 dx 1+2S ℓ .
Thus, there exists a constant C 5 not depending on T and θ such that

Ω |v(x, T )| 2 dx ≤ C 5 e C 5 θ e C 5 T B(x 0 ,r) |v(x, T )| 2 dx 1 2(1+S ℓ ) Ω |v(x, 0)| 2 dx 1+2S ℓ 2(1+S ℓ ) .
On the other hand, Lemma 3.3 says that there exists E 3 > 0 satisfying the following estimate

e 1 θ = E 3 e E 3 T Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx .
Hence, the following estimate holds

Ω |v(x, T )| 2 dx ≤ C 5 E C 5 3 e C 5 (E 3 +1) T Ω |v(x, 0)| 2 dx Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx C 5 × B(x 0 ,r) |v(x, T )| 2 dx 1 2(1+S ℓ ) Ω |v(x, 0)| 2 dx 1+2S ℓ 2(1+S ℓ ) . Using the fact that Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx ≤ Ω |v(x, T )| 2 dx, we obtain Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx 1+C 5 ≤ C 5 E C 5 3 e C 5 (E 3 +1) T B(x 0 ,r) |v(x, T )| 2 dx 1 2(1+S ℓ ) Ω |v(x, 0)| 2 dx C 5 + 1+2S ℓ 2(1+S ℓ ) .
Thus, there exist constants κ > 0 and σ ∈ (0, 1) satisfying the following estimate

Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx ≤ κe κ T B(x 0 ,r) |v(x, T )| 2 dx σ Ω |v(x, 0)| 2 dx 1-σ . (3.97)
Step 5 Make appear ω by propagation of smallness. Let r > 0 be small enough and x j ∈ Ω(j = 1, 2, ..., m)(m ∈ N), we can construct a sequence of balls {B(x j , r)} j∈1,m such that the following inclusions hold

1. B(x m , r) ∈ ω ; 2. B(x j , r) ⊂ B(x j+1 , 2r) ∀j = 1, 2, .., m -1; 3. B(x j , 2r) ⋐ Ω ∀j = 1, 2, ...m .
Then, thanks to (3.97), there exist σ 1 , κ 1 , σ m , κ m such that

Ω∩B(x 0 ,R-4ǫ) |v(x, T )| 2 dx ≤ κe κ T B(x 0 ,r) |v(x, T )| 2 dx σ Ω |v(x, 0)| 2 dx 1-σ ≤ κe κ T B(x 1 ,2r) |v(x, T )| 2 dx σ Ω |v(x, 0)| 2 dx 1-σ ≤ κe κ T κ 1 e κ 1 T B(x 1 ,r) |v(x, T )| 2 dx σ 1 Ω |v(x, 0)| 2 dx 1-σ 1 σ Ω |v(x, 0)| 2 dx 1-σ ≤ ... ≤ κ m e κm T B(xm,r) |v(x, T )| 2 dx σm Ω |v(x, 0)| 2 dx 1-σm ≤ κ m e κm T ω |v(x, T )| 2 dx σm Ω |v(x, 0)| 2 dx 1-σm . (3.98)
Thus, we already prove that if Ω ∩ B(x 0 , R) is star-shaped with respect to x 0 , then we obtain a local observation at one point of time, which has form (3.98). Now, in order to get the global result, we will cover Ω by a finite number of balls B(x 0 , R -4ǫ) satisfying assumption that Ω ∩ B(x 0 , R) is star-shaped with respect to x 0 .

Step 6 Cover Ω. We can see that Ω is covered by a finite number of balls B(x 0 , R -4ǫ) which have one of two following properties:

1. B(x 0 , R) ⊂ Ω; 2. B(x 0 , R) ∩ ∂Ω = ∅.
For the first case, obviously, the assumption that Ω ∩ B(x 0 , R) is star-shaped with respect to x 0 is satisfied because of the convexity of the ball B(x 0 , R). For the second one, we will use the result in [AEWZ] (see Theorem 8, page 2443), which says that Ω is locally star-shaped, i.e for each χ ∈ ∂Ω, there are x χ in Ω and R χ > 0 such that

χ ∈ B(x χ , R χ ) and Ω ∩ B(x χ , R χ ) is star-shaped with center x χ .
Thus, we can choose x 0 = x χ and R = R χ then Ω ∩ B(x 0 , R) is star-shaped with respect to x 0 .

Approximate controllability at one point of time

In [LR], Lebeau and Robbiano connect the controllability to an interpolation estimate for an elliptic system. Then, in [FI], Fursikov and Imanuvilov use a global Carleman inequality and a minimization technique to construct the control function. Recently, in [FCZ], Fernández-Cara and Zuazua establish a null controllability for semilinear heat equation. In [Vo], the author succeeds in computing a control function for the cubic semilinear heat equation in a constructive way. Those results are related to the controllability in L 2 (Ω × (0, T )). Here, we need to add a control at a fixed point of time, which is well studied in [PWX], but for case p ≡ 1. Now, our concern is approximate controllability at one point of time for linear heat equation with timedependent coefficients. This control will lead the given data at the initial time to an origin-center ball with a small radius at some later time. Denote ½ ω be the characteristic function on the region ω and ϕ(T -) be the left limit of the function ϕ at time T . Now, consider the following system.

       ∂ t ϕ -p(t)∆ϕ = 0 in Ω × (0, 2T ) \ {T }, ϕ = 0 on ∂Ω × (0, 2T ), ϕ(•, 0) = ϕ 0 in Ω, ϕ(•, T ) = ϕ(•, T -) + ½ ω h in Ω. (4.1)
The next theorem consists on the existence of the control function h at some fixed point of time T which leads the solution at final time 2T getting small. Theorem 4.1. Let ε > 0 and ϕ 0 ∈ L 2 (Ω), there exists a function h ∈ L 2 (ω) such that the solution of ( 4

.1) satisfies ϕ(•, 2T ) L 2 (Ω) ≤ ε ϕ 0 L 2 (Ω) . Moreover, there exist constants c 3 = c 3 (Ω, ω, p) > 0 and c 4 = c 4 (Ω, ω, p) > 0 such that h L 2 (ω) ≤ c 3 e c 3 T ε c 4 ϕ 0 L 2 (Ω) . (4.2)
Proof of Theorem 4.1

Proof.

Step 1: Define a functional which has a unique minimizer.

Let c 1 and c 2 be the constants from Corollary 3.1, put k := c 1 e c 1 T 1 ε 2c 2 . We consider the following functional

J(u 0 ) = k 2 2 u(•, T ) 2 L 2 (ω) + ε 2 2 u 0 2 L 2 (Ω) - Ω ϕ 0 (x)u(x, 2T )dx (4.3) where u(x, t) is the solution of    ∂ t u -p(t)∆u = 0 in Ω × (0, T ), u = 0 on ∂Ω × (0, T ), u(•, 0) = u 0 in Ω. (4.4)
Notice that J is a strictly convex, C 1 and coercive. Therefore, J has a unique minimizer

Φ 0 ∈ L 2 (Ω) such that J(Φ 0 ) = min u 0 ∈L 2 (Ω)
J(u 0 ). It implies that J ′ (Φ 0 )z 0 = 0 for any z 0 ∈ L 2 (Ω), i.e the following estimate holds for any z 0 :

k 2 Φ(•, T ) L 2 (ω) z(•, T ) L 2 (ω) + ε 2 Φ 0 L 2 (Ω) z 0 L 2 (Ω) - Ω ϕ 0 (x)z(x, 2T )dx = 0 (4.5)
where Φ(x, t) and z(x, t) are respectively the solution of (4.4) corresponding to Φ 0 := Φ(•, 0) and z 0 := z(•, 0).

Step 2: Construct a control function. On the other hand, multiplying ∂ t ϕp(t)∆ϕ = 0 by Φ(•, 2Tt) and integrating over Ω, we get

Ω ∂ t ϕ(x, t)Φ(x, 2T -t)dx -p(t) Ω ∆ϕ(x, t)Φ(x, 2T -t)dx = 0. (4.6)
Integrating by parts (4.6) two times with the fact that ϕ = Φ = 0 on ∂Ω, one has

Ω ∂ t ϕ(x, t)Φ(x, 2T -t)dx -p(t) Ω ϕ(x, t)∆Φ(x, 2T -t)dx = 0. (4.7) Since ∂ t Φ -p(t)∆Φ = 0, it yields Ω ∂ t ϕ(x, t)Φ(x, 2T -t)dx - Ω ϕ(x, t)∂ t Φ(x, 2T -t)dx = 0. (4.8) It implies that d dt Ω ϕ(x, t)Φ(x, 2T -t)dx = 0. (4.9)
Now, by integrating (4.9) over (0, T ), we obtain Integrating again (4.9) over (T, 2T ) forces

Ω ϕ(x, 2T )Φ(x, 0)dx = Ω ϕ(x, T )Φ(x, T )dx = Ω ϕ(x, T -)Φ(x, T )dx + Ω h(x)Φ(x, T )dx. (4.11)
Combining the above equality with (4.10), we conclude that

Ω ϕ(x, 2T )Φ 0 dx = Ω ϕ 0 (x)Φ(x, 2T )dx + Ω h(x)Φ(x, T )dx (4.12) that is - Ω h(x)Φ(x, T )dx + Ω ϕ(x, 2T )Φ 0 dx - Ω ϕ 0 (x)Φ(x, 2T )dx = 0. (4.13)
In addition, by choosing z 0 = Φ 0 in (4.5), it follows that

k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ 0 2 L 2 (Ω) - Ω ϕ 0 (x)Φ(x, 2T )dx = 0. (4.14)
Thus from (4.13) and (4.14), if we choose h(x) = -k 2 Φ(x, T ) then ϕ(x, 2T ) = ε 2 Φ 0 (x). Moreover, using the Cauchy-Schwarz inequality

Ω |v 0 (x)Φ(x, 2T )|dx ≤ Ω |ϕ 0 (x)| 2 dx 1 2 Ω |Φ(x, 2T )| 2 dx 1 2 , (4.15) we get k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ 0 2 L 2 (Ω) ≤ ϕ 0 L 2 (Ω) Φ(•, 2T ) L 2 (Ω) . (4.16)
On the other hand, Corollary 3.1 gives

Φ(•, T ) 2 L 2 (Ω) ≤ k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ(•, 0) 2 L 2 (Ω) . (4.17) Furthermore, the fact that Φ(•, 2T ) 2 L 2 (Ω) ≤ Φ(•, T ) 2 L 2 (Ω) provides us Φ(•, 2T ) 2 L 2 (Ω) ≤ k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ(•, 0) 2 L 2 (Ω) . (4.18)
Thus, combining (4.16) and (4.18), we conclude that

k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ 0 2 L 2 (Ω) ≤ ϕ 0 L 2 (Ω) k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ 0 2 L 2 (Ω) 1 2 . It implies that k 2 Φ(•, T ) 2 L 2 (ω) + ε 2 Φ 0 2 L 2 (Ω) ≤ ϕ 0 2 L 2 (Ω) . (4.19)
This is equivalent to

1 k 2 h 2 L 2 (ω) + 1 ε 2 ϕ(•, 2T ) 2 L 2 (Ω) ≤ ϕ 0 2 L 2 (Ω) . (4.20)
Thus, we get the desired estimate (4.2) with c 3 := max{ √ c 1 , c 1 2 } and c 4 := c 2 . This completes the proof.

5 The local backward -Proof of the Theorem 1.1

For the case ω ⋐ Ω, we need to use the controllability result at one point of time (Theorem 4.1) in order to get the information of the solution on the whole domain from the known data on the subdomain. In detail, the proof of Theorem 1.1 is structured as:

Step 1 will provide us the approximate data of u(•, 3T ) on whole domain dues to the controllability result; Then, by some computation technique, we make appear f in Step 2; Lastly, applying the global backward result in Theorem 2.1, Step 3 will complete the proof with the construction of the initial data.

Proof.

Step 1: Use controllability result to link the knowledge on the whole domain Ω and on the subdomain ω. Now, for each i = 1, 2, ..., Theorem 4.1 says that for any ε > 0, there exists h i ∈ L 2 (ω) such that the solution of        ∂ t ϕ ip(t)∆ϕ i = 0 in Ω × (0, 2T ) \ {T }, ϕ i = 0 on ∂Ω × (0, 2T ), ϕ i (•, 0) = e i in Ω, ϕ i (•, T ) = ϕ i (•, T -) + ½ ω h i in Ω (5.1) satisfies ϕ i (•, 2T ) L 2 (Ω) ≤ ε for any i ≥ 1. Recall that e i (i = 1, 2, ...) is the eigenfunction of Laplace operator. Moreover, there exist constants c 3 , c 4 > 0 such that the following estimate holds (5.7)

Step 2: Make appear f . Now, we will make appear f by using a triangle inequality u(•, 3T ) + (5.9)

The last inequality in (5.9) is followed from (5.2) and the assumption (2.1). Thus, from (5.7) and (5.9), we can conclude that u(•, 3T ) + (5.12)

Using the fact that e -x ≤ γ x γ ∀x > 0 ∀γ > 0 and λ i ≈ i

2 n
∀i ≥ 1 ( by the Weyl formula), we get:

∞ i=1 e -2λ i p 2 T ≤ γ 2p 2 T γ ∞ i=1 1 λ γ i ≤ γ 2p 2 T γ ∞ i=1 1 i 2 n γ .
When γ > n 2 , there exists a constant S > 0 such that for some positive constants K 1 = K 1 (Ω, ω, p) > 1 and k 1 = k 1 (Ω, ω, p) ∈ (0, 1).

Step 3: Apply the global backward result.

From the facts that h i ∈ L 2 (ω), z ∈ L 2 (ω) and ∞ i=1 e -λ i 3T 2T p(s)ds < ∞, one gets

- ∞ i=1
e -λ i 3T 2T p(s)ds ω h i (x)f (x)dxe i ∈ L 2 (Ω).

(5.14) Thus, Theorem 2.1 gives us the following estimate, where δ in (2.1) is replaced by

K 1 e K 1 T u(•, 0) 1-k 1 L 2 (Ω) δ k 1 u(•, 0) -g L 2 (Ω) ≤ (1 + ζ)p 2 T u(•, 0) H 1 0 (Ω) ln √ 2ζλ 1 p 2 T u(•,0) L 2 (Ω)
K 1 e (5.18)

This completes the proof. .

Appendix

(5.25)

  Theorem 3.2. Let x 0 ∈ Ω and R := max x∈Ω |x-x 0 |. Suppose all the following assumptions hold:

  ) in Step 1 and (3.21) in Step 2 provide us

  Corollary 3.2. Under the same assumption in Lemma 3.1, put R := max x∈ϑ |xx 0 |.

  choose ρ such as

  of the equation ∂ t ϕ i -p(t)∆ϕ i = 0 by u(•, 2T -t) and integrating over Ω, one getsd dt Ω ϕ i (x, t)u(x, 2Tt)dx = 0.(5.3) Integrating (5.3) over (0, T ) and (T, 2T ) respectively and using the fact thatϕ i (•, T ) = ϕ i (•, T -) + ½ ω h i , one has Ω ϕ i (x, 2T )u(x, 0)dx = Ω ϕ i (x, 0)u(x, 2T )dx + ω h i (x)u(x, T )dx. 0)e j (x)dxe jand ϕ i (•, 0) = e i in (5.4), we getΩ ϕ i (x, 2T )u(x, 0)dx = e -λ i 2T 0 p(s)ds Ω u(x, 0)e i (x)dx + ω h i (x)u(x, T )dx. (5.5)Now, multiplying both sides of (5.5) by e -λ i 3T 2T p(s)ds e i and take the sum from i = 1 to ∞,h i (x)u(x, T )dxe i L 2 (Ω) ≤ ε u(•, 0) L 2 (Ω) ∞ i=1 e -2λ i 3T 2T p(s)ds 1 2.

  first estimate in Step 1, let us estimate the second one.∞ i=1 e -λ i 3T 2T p(s)ds ω h i (x) (u(x, T )f ) dxe i L 2 (Ω) 2 (ω) u(•, T )f L 2 (ω)

  that the function x → ax+bx -s (a, b, s > 0) gets minimum at x 0 = sb a 1 s+1 . Hence, in order to minimize the right-hand side of (5.10) we choose ε such that

  0) 1-k 1 L 2 (Ω) δ k 1 (5.13)

.

  With the assumption that δ < u(•, 0) L 2 (Ω) that x ≤ e x ∀x > 0, there exists a constant C > 0 such that u(•, 0)g L 2 (Ω)

≥

  Proof of (1.24) in Remark 1.1 Let us remind the observation estimate (3.2) in Theorem 3.1 (see Section 3)u(•, T ) L 2 (Ω) ≤ K 2 e some K 2 = K 2 (Ω, ω, p) > 0 and k 2 = k 2 (Ω, ω, p) ∈ (0, 1). Combining to the known backward estimate below u(•, 0) L 2 (Ω) ≤ e p 2 T u(•,0) u(•, 0) L 2 (Ω) u(•, T ) L 2 (ω) λ 1 , there exists a constant C = max{ p 2 k 2 , K 2 k 2 λ 1 } > 0 such that u(•, 0) L 2 (Ω) u(•, T ) L 2 (ω)