3 Resolution of Problem (1) in the Half Disc

Well-Posedness Results For A Third Boundary Value Problem For The Heat Equation In A Disc

Arezki Khelou…

Received 10 March 2014

Abstract

In this work we prove well-posedness results for the following one space linear second order parabolic equation@tu @²xu=f, set in a domain

= (t; x)2R²: r < t < r;'₁(t)< x < '₂(t)

ofR², where'_i(t) = ( 1)ⁱ r² t^{2 12}; i= 1;2and with lateral boundary con- ditions of Robin type. The right-hand sidef of the equation is taken in L²( ).

The method used is based on the approximation of the domain by a sequence of subdomains( n)_nwhich can be transformed into regular domains.

1 Introduction

Let =D(0; r)be the open disc centred at the origin of R² and with radius r >0, characterized by = (t; x)2R²: r < t < r;'₁(t)< x < '₂(t) ;where '₁ and'₂ are de…ned on[ r; r] by'_k(t) = ( 1)^k r² t²

1

2; k= 1;2. The lateral boundary of is de…ned by k = (t; '_k(t))2R²: r < t < r , k= 1,2. In , we consider the Robin type boundary value problem

@_tu @_x²u=f in ;

@_xu+ _kuj _k = 0, k= 1,2, (1)

where the coe¢ cients _k,k= 1,2are real numbers satisfying non-degeneracy assump- tions (to be made more precise later) and the right-hand side term f of the equation lies in L²( ), the space of square-integrable functions on with the measuredtdx.

The main di¢ culty related to this kind of problems is due to the fact that '₁ coincides with '₂ for t = r and for t = r, which prevents the domain to be transformed into a regular domain by means of a smooth transformation.

The case _k = 1, k = 1, 2, corresponding to Dirichlet boundary conditions is considered in [19]. We can …nd in [6] a study of the case _k = 0, k = 1, 2, corre- sponding to Neumann boundary conditions and in [23] an abstract study in the case

Mathematics Sub ject Classi…cations: 35K05, 35K20.

yLaboratoire des Mathématiques Appliquées, Faculté des Sciences Exactes, Université de Bejaia, 6000, Béjaia, Algeria. Lab. E.D.P.N.L., Ecole Normale Supérieure, 16050-Kouba, Algiers, Algeria

161

( ₁; ₂) = (1;0), corresponding to mixed (Dirichlet-Neumann) lateral boundary con- ditions. However, the boundary assumptions dealt with by the authors exclude our domain. Further references on the analysis of parabolic problems in non-cylindrical domains are: Labbas et al. [13, 14, 15], Khelou… et al. [8, 9, 10, 12], Degtyarev [5], Aref’ev and Bagirov [3, 4], Sadallah [20, 21, 22], Alkhutov [1, 2] and Paronetto [17].

In this work, we consider the case of Robin type boundary condition, namely, the case where _k 6= 0,k= 1,2, and we look for su¢ cient conditions (as weak as possible) on the lateral boundary of the domain and on the coe¢ cients _k, k = 1, 2 in order to obtain the maximal regularity of the solution in an anisotropic Hilbertian Sobolev space.

In previous works (see [7, 11]), we have studied the case where

= (t; x)2R²: 0< t < T; ₁(t)< x < ₂(t)

with the fundamental hypothesis ₁(0) = ₂(0)and we have proved that the solution uof Problem (1) is unique and has the optimal regularity, that is a solutionubelonging to the anisotropic Sobolev space

H^1;2( ) := u2H^1;2( ) : @xu+ _kuj k = 0,k= 1,2 with

H^1;2( ) = u2L²( ) :@tu; @xu; @_x²u2L²( ) ; under su¢ cient conditions on _k; k= 1;2, that are

0k(t) ( ₂(t) ₁(t)) ! 0 as t !0, k= 1;2.

Examples of functions satisfying this last condition are _k(t) = ( 1)^k r² t^{2 2+1} ; k= 1;2for all <0. However, the above condition is false in the case = 0corresponding to the class of domains considered in this article. So, the well-posedeness result which we will prove here can not be derived from [7] and [11]. In order to overcome this di¢ culty, we impose su¢ cient conditions on the lateral boundary of the domain and on the coe¢ cients _k,k= 1,2; that are,

1<0; ₂>0, (2)

( 1)^k _k t

2p

r² t² 0 a.e. t2] r; r[,k= 1;2; (3) and

1 (16 + 4 ²₁+ 4 ²₂)r+ (4j 1j+ 4j 2j)r²+ (8 + 4 ²₁+ 4 ²₂)r³ >0: (4) Then, our main result is following:

THEOREM 1. Under the hypothesis (2), (3) and (4), the heat operatorL=@_t @_x² is an isomorphism from H^1;2( ) intoL²( ).

It is not di¢ cult to prove the injectivity of the operatorL. Indeed, Ifuis a solution of Problem (1) with a null right-hand side, the calculations show that the inner product hLu; uiinL²( )gives

0 = X2 k=1

Z

k

( 1)^k _k t

2p

r² t² u²(t; '_k(t))dt+ Z

(@_xu)²dtdx.

The hypothesis (3) implies that@_xu= 0and consequently@_x²u= 0. Then, the equation of Problem (1) gives @_tu= 0. Thus, uis constant. The boundary conditions and the fact that _k 6= 0,k= 1,2imply thatu= 0in . So, in the sequel, we will be interested only by the question of the surjectivity of the operatorL.

The method used here is the domain decomposition method. More precisely, we divide into two parts

1=f(t; x)2 : r < t <0g and 2=f(t; x)2 : 0< t < rg.

So, we obtain two solutionsuk2H^1;2( k)in k,k= 1,2. Finally, we prove that the functionude…ned by

u:= u1 in 1; u2 in 2;

is the solution of problem (1) and has the optimal regularity, that is u 2 H^1;2( ).

The plan of this paper is as follows. In Section 2, we prove that Problem (1) admits a (unique) solution in the case of a "truncated" domain. Then, in Section 3, we approximate by a sequence ( _n) of such truncated domains and we establish an energy estimate which will allow us to pass to the limit and complete the proof of our main result.

2 Resolution of Problem (1) in a Truncated Disc

For each n2N , we de…ne

n:= (t; x)2R²: r < t < r 1

n;'₁(t)< x < '₂(t) :

THEOREM 2. Assume that _k and '_k, k = 1;2 verify assumptions (2) and (3) and letfn =fj n and

n;k= (t; '_k(t))2R²: r < t < r 1

n fork= 1; 2:

Then, for eachn2N , the problem

@tun @_x²un =fn2L²( n);

@xun+ _kunj n;k = 0,k= 1,2, (5) admits a (unique) solution un2H^1;2( n).

PROOF. We divide _n,n2N into two parts

=f(t; x)2 : r < t <0g and ⁺_n = (t; x)2 : 0< t < r 1 n . So, we have n= [ ⁺n [(f0g ]'₁(0); '₂(0)[).

LEMMA 1. Letf = fj and

k = (t; '_k(t))2R²: r < t <0 fork= 1; 2:

Then, the problem

@_tu @_x²u =f 2L²( );

@xu + _ku j _k = 0,k= 1,2, admits a (unique) solution u 2H^1;2( ).

PROOF. Since'₁is a decreasing function on] r;0[and'₂is an increasing function on] r;0[, then the result follows from [18].

Hereafter, we denote the trace u j_f0g ]'₁(0);'₂(0)[ by , which is in the Sobolev spaceH¹(f0g ]'₁(0); '₂(0)[)becauseu 2H^1;2( )(see [16]). Now, consider the following problem on ⁺_n,n2N

8>

<

>:

@tu⁺_n @_x²u⁺_n =f_n⁺2L²( ⁺_n);

u⁺_nj_f0g ]'₁(0);'₂(0)[= 2H¹(f0g ]'₁(0); '₂(0)[);

@xu⁺_n + _ku⁺_nj ⁺_n;k = 0,k= 1,2,

(6)

where ⁺_n;k= (t; '_k(t))2R²: 0< t < r _n¹ ,k= 1,2.

We use the following result, which is a consequence of Theorem 4.3 in [16] to solve Problem (6).

PROPOSITION 1. LetQbe the rectangle]0; T[ ]0;1[,f 2L²(Q)and 2H¹( ₀) with ₀=f0g ]0;1[. Then, the problem

8<

:

@tu @_x²u=f 2L²(Q); uj ₀= ;

@xu+ _kuj _k = 0,k= 1,2,

where ₁= ]0; T[ f0gand ₂= ]0; T[ f1gadmits a (unique) solutionu2H^1;2(Q).

REMARK 1. We have lies in H¹(f0g ]'₁(0); '₂(0)[), then @x is (only) in L²(f0g ]'₁(0); '₂(0)[) and its pointwise values should not make sense. So in the application of [[16] Theorem 4.3, Vol. 2], there are no compatibility conditions to satisfy.

Thanks to the transformation (t; x) 7! (t; y) = (t;('₂(t) '₁(t))x+'₁(t)); we deduce the following result:

PROPOSITION 2. For eachn2N , Problem (6) admits a unique solutionu⁺_n 2 H^1;2( ⁺_n).

So, the functionu_n 2H^1;2( _n),n2N de…ned by u_n := u in ;

u⁺_n in ⁺_n;

is the (unique) solution of Problem (5). This completes the proof of Theorem 2.

3 Resolution of Problem (1) in the Half Disc

In this section, we de…ne

+:= (t; x)2R²: 0< t < r;'₁(t)< x < '₂(t) and consider the following problem in ⁺

8<

:

@tu⁺ @²_xu⁺=f ⁺ 2L²( ⁺), u⁺j_f0g ]'₁(0);'₂(0)[= 0,

@xu⁺+ _ku⁺j ⁺_k = 0,k= 1,2,

(7)

where f⁺ =fj ⁺ and

+

k = (t; '_k(t))2R²: 0< t < r fork= 1; 2:

We assume that _k and'_k,k= 1;2verify assumptions (2), (3) and (4) and we denote f_n⁺ = f⁺j ⁺_n and u⁺_n 2H^1;2( ⁺_n) the solution of Problem (7) in ⁺_n. Such a solution exists by Proposition 2.

PROPOSITION 3. There exists a constantK >0independent ofnsuch that u⁺_{n H}1;2( ⁺n) K f_n⁺ _L2( ⁺n) K f⁺ _L2( ⁺),

where

u⁺_{n H}1;2( ⁺n)= r

u⁺n 2

L²( ⁺n)+ @tu⁺n 2

L²( ⁺n)+ @xu⁺n 2

L²( ⁺n)+ @_x²u⁺n 2 L²( ⁺n).

In order to prove Proposition 3, we need the following result LEMMA 2. We have the following estimations

(i) j'⁰_k(t)j('₂(t) '₁(t)) 2r fort2] r; r[andk= 1,2.

(ii) R'₂(t)

'₁(t)[@_x^ju⁺_n (s; x)]²ds ['₂(t) '₁(t)]²R'₂(t)

'₁(t)[@_x^j+1u⁺_n(s; x)]²dsforj= 0;1.

(iii) k@xu⁺_nk²L²( ⁺n) 4r² @_x²u⁺_n ²_L2( ⁺n).

PROOF OF PROPOSITION 3. We have

f_n^{+ 2}_L2( ⁺n) = h@tu⁺_n @_x²u⁺_n; @tu⁺_n @²_xu⁺_ni

= @tu⁺_n ²_L2( ⁺n)+ @²_xu⁺_n ²_L2( ⁺n) 2 Z

+n

@tu⁺_n:@_x²u⁺_ndtdx.

Let us consider the term 2R

+

n@_tu⁺_n:@²_xu⁺_ndtdx. We have

@_tu⁺_n:@_x²u⁺_n =@_x @_tu⁺_n:@_xu⁺_n 1

2@_t @_xu⁺_n ². Then

2 Z

+ n

@_tu⁺_n:@²_xu⁺_ndtdx = 2 Z

+ n

@_x @_tu⁺_n@_xu⁺_n dt dx+ Z

+ n

@_t @_xu⁺_n ²dt dx

= Z

@ ⁺n

h

@xu⁺_n ² t 2@tu⁺_n@xu⁺_n x

i d ,

with _t, _x are the components of the unit outward normal vector at@ ⁺_n. We shall rewrite the boundary integral making use of the boundary conditions. On the part of the boundary of ⁺_n where t = 0, we have u⁺_n = 0 and consequently @_xu⁺_n = 0.

The corresponding boundary integral vanishes. On the part of the boundary where t = r _n¹, we have x = 0 and t = 1. Accordingly the corresponding boundary integral R'₂(^r n¹)

'₁(^r n¹)(@xu⁺_n)² dx is nonnegative. On the parts of the boundary where x='_k(t),k= 1,2, we have

x= ( 1)^k q

1 + ('⁰_k)²(t)

, _t= ( 1)^k+1'⁰_k(t) q

1 + ('⁰_k)²(t)

and@_xu⁺_n(t; '_k(t)) + _ku⁺_n(t; '_k(t)) = 0:

Consequently, the corresponding boundary integrals In;k and Jn;k, k = 1;2 are the following:

In;k = ( 1)^k+1 Z r _n¹

0

'⁰_k(t) @xu⁺_n (t; '_k(t)) ²dt, k= 1,2,

J_n;k = ( 1)^k2 Z r _n¹

0 k@_tu⁺_n(t; '_k(t)):u⁺_n(t; '_k(t))dt,k= 1,2.

We have

2 Z

+ n

@tu⁺_n:@_x²u⁺_ndtdx jIn;1j jIn;2j jJn;1j jJn;2j. (8) It is the reason for which we look for an estimate of the type

jIn;1j+jIn;2j+jJn;1j+jJn;2j @_x²u⁺_n ²_L2( ⁺n);

where is a positive constant independent of n belonging to the interval ]0;1[. By introducing the function (t; x) = _'^{'2(t) x}

2(t) '₁(t) like in [18], we write forIn;1

jIn;1j =

Z r ¹_n 0

(Z '₂(t) '1(t)

'⁰₁(t)@x (t; x) @xu⁺_n(t; x) ² dx )

dt Z

+n

'⁰₁(t) ('₂(t) '₁(t)) @_x²u⁺_n ²dtdx+ 2 Z

+n

j'⁰₁j @xu⁺_n @²_xu⁺_n dtdx 2r @_x²u⁺_n ²+ @_x²u⁺_n ²+1Z

+ n

j'⁰₁j² @_xu⁺_n ²dtdx 2r+ +4r²

@_x²u⁺_n ²_L2( ⁺n) 7r @_x²u⁺_n ²_L2( ⁺n).

The last inequality is obtained by choosing =r. Similarly, we have jI_n;2j 7r @_x²u⁺_n ²_L2( ⁺n).

Let us now consider the terms Jn;k, k= 1, 2. By setting h(t) = (u⁺_n)²(t; '_k(t)), we obtain

J_n;k = ( 1)^k Z r _n¹

0 k:h

h⁰(t) '⁰_k(t)@_x u⁺_n ²(t; '_k(t))i dt

= ( 1)^k _k:h(t)^r

1 n

0 + ( 1)^k+1 Z r _n¹

0 k:'⁰_k(t)@_x u⁺_n ²(t; '_k(t))dt.

Condition (2) and the fact that (u⁺_n)²(0; '_k(0)) = 0 give ( 1)^k _k:h(t)^r

1 n

0 0: In

the sequel, we estimate the last boundary integral in the expression ofJn;k, namely Ln;k= ( 1)^k+1

Z r ¹_n

0 k:'⁰_k(t)@x u⁺_n ²(t; '_k(t))dt.

We have

@x u⁺_n ²(t; '₁(t))

= '₂(t) x

'₂(t) '₁(t)@_x u⁺_n ²(t; x)

x='₂(t)

x='₁(t)

=

Z '₂(t) '₁(t)

@_x '₂(t) x

'₂(t) '₁(t)@_x u⁺_n ²(t; x) dx

=

Z '₂(t) '₁(t)

1

'₂(t) '₁(t)@_x u⁺_n ²(t; x) '₂(t) x

'₂(t) '₁(t)@_x² u⁺_n ²(t; x) dx.

So, Ln;1=

Z

+n

1:'⁰₁(t)

'₂(t) '₁(t)@x u⁺_n ²(t; x) '₂(t) x

'₂(t) '₁(t) ¹:'⁰₁(t)@_x² u⁺_n ²(t; x) dtdx.

By using the equalities

@x u⁺_n ²(t; x) = 2@xu⁺_n(t; x)u⁺_n (t; x) and

@_x² u⁺_n ²(t; x) = 2@_x²u⁺_n (t; x)u⁺_n(t; x) + 2 @xu⁺_n(t; x) ²; we obtain

L_n;1 = Z

+ n

2 ₁:'⁰₁(t)

'₂(t) '₁(t)@_xu⁺_n(t; x)u⁺_n (t; x)dtdx Z

+n

'₂(t) x

'₂(t) '₁(t)2 ₁:'⁰₁(t)@_x²u⁺_n(t; x)u⁺_n(t; x)dtdx Z

+n

'₂(t) x

'₂(t) '₁(t)2 ₁:'⁰₁(t) @_xu⁺_n (t; x) ²dtdx

= An;1+Bn;1+Cn;1: Estimation of A_n;1,B_n;1 andC_n;1

a) We have

A_n;1= Z

+ n

2 ₁:'⁰₁(t)

'₂(t) '₁(t)@_xu⁺_n(t; x)u⁺_n (t; x)dtdx;

then jA_n;1j

Z

+ n

1[ ₁:'⁰₁(t)]² @_xu⁺_n (t; x) ²dtdx

+ Z

+n

1

['₂(t) '₁(t)]² u⁺_n (t; x) ²dtdx Z

+ n

1[ ₁:'⁰₁(t)]²['₂(t) '₁(t)]² @_x²u⁺_n(t; x) ²dtdx

+ Z

+n

['₂(t) '₁(t)]² @_x²u⁺_n(t; x) ²dtdx

2

14r²+ 4r² @_x²u⁺_n ²_L2( ⁺n) 4r³+ 4 ²₁r @_x²u⁺_n ²_L2( ⁺n). The last inequality is obtained by choosing =r.

b) We have B_n;1=

Z

+ n

'₂(t) x

'₂(t) '₁(t)2 ₁:'⁰₁(t)@²_xu⁺_n(t; x)u⁺_n (t; x)dtdx;

then

jBn;1j

2 1

Z

+n

j'⁰₁(t)j² u⁺_n(t; x) ²dtdx+ @_x²u⁺_n ²_L2( ⁺n)

2 1 sup

t2[0;r] j'⁰₁(t)j²['₂(t) '₁(t)]⁴ @²_xu⁺_n ²

L²( ⁺ⁿ) + @_x²u⁺_n ²_L2( ⁺n)

4 ²₁r⁴

+ @_x²u⁺_n ²_L2( ⁺n) 4 ²₁r³+r @_x²u⁺_n ²_L2( ⁺n): The last inequality is obtained by choosing =r.

c) We have C_n;1=

Z

+ n

'₂(t) x

'₂(t) '₁(t)2 ₁:'⁰₁(t) @_xu⁺_n(t; x) ²dtdx then

jC_n;1j 2j 1j Z

+ n

j'⁰₁(t)j j'₂(t) '₁(t)j² @_x²u⁺_n (t; x) ²dtdx 4j 1jr² @_x²u⁺_n ²_L2( ⁺n).

Consequently,

jLn;1j (4 + 4 ²₁)r³+ 4j 1jr²+ (1 + 4 ²₁)r @_x²u⁺_n ²_L2( ⁺n). Similarly, we can obtain

jLn;2j (4 + 4 ²₂)r³+ 4j 2jr²+ (1 + 4 ²₂)r @_x²u⁺_n ²_L2( ⁺n). Summing up the above estimates, we obtain

f_n^{+ 2}_L2( ⁺n) @_tu⁺_n ²_L2( ⁺n)+ @_x²u⁺_n ²_L2( ⁺n) jI_n;1j jI_n;2j jL_n;1j jL_n;2j

@_x²u⁺_n ²_L2( ⁺n) n

1 h

(16 + 4 ²₁+ 4 ²₂)r+ (4j 1j+ 4j 2j)r² +(8 + 4 ²₁+ 4 ²₂)r³io

+ @_tu⁺_n ²_L2( ⁺n):

Using the condition (4) and since kf_n⁺k²L²( ⁺ⁿ) kf⁺k²L²( ⁺), then Proposition 3 is proved.

THEOREM 3. Problem (7) admits a (unique) solutionu⁺2H^1;2( ⁺).

PROOF. The estimation of Proposition 3 shows that uf⁺n

L²( ⁺)+ @]tu⁺n

L²( ⁺)+ X2 i=1

@]_xⁱu⁺n

L²( ⁺) C f⁺ _L2( ⁺);

wheree:denotes the0 extension ofu⁺_n to ⁺. This means thatuf⁺n,@]tu⁺n;@]_xⁱu⁺n,i= 1;2 are bounded functions inL²( ⁺). The following compactness result is well known: A bounded sequence in a re‡exive Banach space (and in particular in a Hilbert space) is weakly convergent. So for a suitable increasing sequence of integers nk, k = 1;2; :::, there exists functionsu⁺; v⁺; v_i⁺; i= 1;2 inL²( ⁺)such that

ug⁺nk* u⁺;^@_tu⁺nk* v⁺;@^_xⁱu⁺nk* v⁺_i ; i= 1;2

weakly in L²( ⁺)as k! 1:Clearly, v⁺ =@_tu⁺; v_i⁺ =@_xⁱu⁺; i= 1;2 in the sense of distributions in ⁺ and so inL²( ⁺). Finally,u⁺ 2H^1;2( ⁺)and

@tu⁺ @_x²u⁺=f in ⁺:

On the other hand, the solution u⁺ satis…es the boundary conditions, since 8n2N ; u⁺ +

n =u⁺_n. REMARK 2. The functionu2H^1;2( ) de…ned by

u:= u in ;

u⁺ in ⁺, is the (unique) solution of Problem (1).

Acknowledgment. I want to thank the anonymous referee for a careful reading of the manuscript and for his/her helpful suggestions.

References

[1] Yu. A. Alkhutov, Lp-solubility of the Dirichlet problem for the heat equation in noncylindrical domains, (Russian) Mat. Sb., 193(2002), 3–40; translation in Sb.

Math., 193(2002), 1243–1279.

[2] Yu. A. Alkhutov, Lp-estimates of solutions of the Dirichlet problem for the heat equation in a ball, J. Math. Sci., 3(2007), 2021–2032.

[3] V. N. Aref’ev and L. A. Bagirov, Asymptotic behavior of solutions to the Dirich- let problem for parabolic equations in domains with singularities, Math. Notes, 59(1996), 10–17.

[4] V. N. Aref’ev and L. A. Bagirov, Solutions of the heat equation in domains with singularities, Math. Notes, 64(1998), 139–153.

[5] S. P. Degtyarev, The solvability of the …rst initial-boundary problem for parabolic and degenerate parabolic equations in domains with a conical point, Sb. Math., 201(2010), 999–1028.

[6] S. Hofmann and J. L. Lewis, TheL^pregularity problems for the heat equation in non-cylindrical domains, J. Funct. Anal., 220(2005), 1–54.

[7] A. Khelou… and B. K. Sadallah, Parabolic equations with Robin type boundary conditions in a non-rectangular domain, Electron. J. Di¤erential Equations 2010, No. 25, 14 pp..

[8] A. Khelou…, R. Labbas and B. K. Sadallah, On the resolution of a parabolic equation in a nonregular domain ofR³, Di¤er. Equ. Appl., 2(2010), 251–263.

[9] A. Khelou… and B. K. Sadallah, On the regularity of the heat equation solution in non-cylindrical domains: two approaches, Appl. Math. Comput., 218(2011), 1623–1633.

[10] A. Khelou…, Resolutions of parabolic equations in non-symmetric conical domains, Electron. J. Di¤erential Equations 2012, No. 116, 14 pp.

[11] A. Khelou…, Existence and uniqueness results for parabolic equations with Robin type boundary conditions in a non-regular domain of R³, Appl. Math. Comput., 220(2013), 756–769.

[12] A. Khelou… and B. K. Sadallah, Study of the heat equation in a symmetric conical type domain ofR^N+1, Math. Meth. Appl. Sci., (37)(2014), 1807–1818.

[13] R. Labbas, A. Medeghri and B. K. Sadallah, Sur une équation parabolique dans un domaine non cylindrique, C. R. Math. Acad. Sci. Paris 335(2002), 1017–1022.

[14] R. Labbas, A. Medeghri and B. K. Sadallah, On a parabolic equation in a trian- gular domain, Appl. Math. Comput., 130 (2002), 511–523.

[15] R. Labbas, A. Medeghri and B. K. Sadallah, An L^p approach for the study of degenerate parabolic equation, Electron. J. Di¤erential Equations 2005, No. 36, 20 pp.

[16] J. L. Lions and E. Magenes, Problèmes Aux Limites Non Homogènes Et Appli- cations, Vol. 2. (French) Travaux et Recherches Mathématiques, No. 18 Dunod, Paris 1968.

[17] F. Paronetto, An existence result for evolution equations in non-cylindrical do- mains, NoDEA Nonlinear Di¤erential Equations Appl., 20(2013), 1723–1740.

[18] B. K. Sadallah, Etude d’un problème 2m parabolique dans des domaines plan non rectangulaires, Boll. Un. Mat. Ital., 2(1983), 51–112.

[19] B. K. Sadallah, Existence de la solution de l’équation de la chaleur dans un disque, C. R. Acad. Sci. Paris Sér. I Math., 327(1998), 813–816.

[20] B. K. Sadallah, Regularity of a parabolic equation solution in a non-smooth and unbounded domain, J. Aust. Math. Soc., 84(2008), 265–276.

[21] B. K. Sadallah, A remark on a parabolic problem in a sectorial domain, Appl.

Math. E-Notes, 8(2008), 263–270.

[22] B. K. Sadallah, Study of a parabolic problem in a conical domain, Math. J.

Okayama Univ., 56(2014), 157–169.

[23] G. Savaré, Parabolic problems with mixed variable lateral conditions: an abstract approach, J. Math. Pures et Appl., 76(1997), 321–351.