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Exact Controllability for the Equation

of the One Dimensional Plate

in Domains with Moving Boundary

Controlabilidad Exacta para la Ecuaci´

on de la Placa

Unidimensional en Dominios con Frontera M´

ovil

Cruz S. Q. de Caldas (

gmacruz@vm.uff.br

)

Juan Limaco (

juanbr@hotmail.com

)

Rioco K. Barreto (

rikaba@vm.uff.br

)

Departamento de Matem´atica Aplicada Universidade Federal Fluminense IMUFF

Rua M´ario Santos Braga s/no_,

CEP: 24020-140, Niter´oi, RJ, Brasil.

Pedro Gamboa (

pgamboa@dmm.im.ufrj.br

)

Instituto de Matem´atica Universidade Federal do Rio de Janeiro

Caixa Postal 68530, CEP 21945–970 Rio de Janeiro, RJ, Brasil.

Abstract

This work studies the problem of the exact controlability in the boundary of the equation utt+uxxxx = 0 in a domain with moving

boundary.

Key words and phrases: Exact control, moving boundary, HUM method.

Resumen

Este trabajo estudia el problema de la controlabilidad exacta en la frontera de la ecuaci´on utt+uxxxx = 0 en un dominio con frontera

m´ovil.

Palabras y frases clave: Control exacto, frontera m’ovil, m’etodo HUM.

1

Introduction

Letα: [0,+∞)→R_{andβ: [0,+∞)→}R_{be two functions of classC}3_.

Let us consider the noncylindrical domainQb, defined by:

b

Q=©(x, t)∈R2_{; α(t)< x < β(t), 0< t < T}ª_.

The lateral boundary ofQb is given by

[

0<t<T

[(α(t)×t)∪(β(t)×t)].

This work investigates the following problem of exact control: given T > T0,

for some fix T0>0 and inicial data

{uo_{, u}1_{} ∈L}2_{(α(0), β(0))×H}−2_{(α(0), β(0)),}

we find the controls g1∈L2(0, T), g2∈L2(0, T) , such that the solutionu

of the following system:

(I)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

utt+uxxxx= 0 in Qb

u(α(t), t) =u(β(t), t) = 0 in ]0, T[

ux(α(t), t) =g1(t), ux(β(t), t) =g2(t) in ]0, T[

u(x,0) =u0_{, u}

t(x,0) =u1 in ]α(0), β(0)[

verifies that u(x, T) = 0 ,ut(x, T) = 0 in ]α(T), β(T)[.

2

Notations and main results

Let us consider the real functions α(t) andβ(t) satisfying the following con-ditions:

(H1) α, β∈C3_{([0,+∞)R), withα}′_,α′′_,β′_,β′′

∈L1_(0,∞).

(H2) α(t) < β(t) for all t ≥ 0 and 0 < γ0 = inft≥0γ(t), where γ(t) =

β(t)−α(t).

Remark 2.1. _{It follows from (H1) and (H2) that α, α}′_{, β,andβ}′ _are

bounded. In fact,

||α(t)| − |α(0)|| ≤ |α(t)−α(0)|=

¯ ¯ ¯ ¯

Z t

0

α′_(s)ds ¯ ¯ ¯ ¯

≤

Z t

0 |

α′(s)|ds≤

Z ∞

0 |

α′(s)|ds.

Thus

|α(t)| ≤ |α(0)|+

Z ∞

0 |

α′_(s)

|ds.

hence,αis bounded. The proof is similar forα′_{,β, andβ}′_.

In what follows we use the notations:

γ1= sup

t≥0

γ(t), s0= sup

t≥0{|

α′(t)|,|β′(t)|},

l1=

Z ∞

0 |

γ′_(t)

|dt <∞, l2=

Z ∞

0 |

γ′′_(t)

|dt <∞,

l3=

Z ∞

0 |

α′_(t)

|dt <∞, l4=

Z ∞

0 |

α′′_(t)

|dt <∞.

Notice that when (x, t) varies in Qb the point (y, t) with y = (x−α)/γ

varies in the cylinder Q=]0,1[×]0, T[. The application τ : Qb →Qgiven by

τ(x, t) = (y, t) is a diffeomorphism.

By the change of variables u(x, t) =υ(y, t) withy= (x−α)/γ we transform the operator

b

L u=utt+uxxxx inQb

into the operator

where

(1)

¯ ¯ ¯ ¯ ¯ ¯ ¯

a(t) = _γ41_(t); b(y, t) = (

α′

+γ′

y γ )

2_;

c(y, t) =−2(α′+_γγ′y); d(y, t) = −γα′′−yγγ′′+2_γ2γ′α′+2y(γ′)2.

So, the problem of the noncylindrical control (I) is transformed into the fol-lowing problem of cilyndrical control:

Given T > T0 ,T0>0 fix, and the initial dataυ0∈L2(0,1), υ1∈H−2(0,1),

find the controls f1 ∈L2(0, T), f2 ∈L2(0, T) , such that the solution v of

the system below

(II)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lυ= 0 in Q

υ(0, t) =υ(1, t) = 0 in ]0, T[

υy(0, t) =f1(t) ,υy(1, t) =f2(t) in ]0, T[

υ(y,0) =υ0_{(y) ,υ}

t(y,0) =υ1(y) in ]0,1[

verifiesυ(y, T) = 0, υt(y, T) = 0 in ]0,1[.

Our main results are:

Theorem 2.1_{Let us assume that the hypotheses (H1) and (H2) are satisfied.}

Then there exists T0>0,such that for T > T0 and the initial data{u0, u1} ∈

L2_{(α(0), β(0))×H}−2_{(α(0), β(0)), there exists a pair of controls {g}

1, g2} ∈

L2_{(0, T)×L}2_{(0, T)such that the ultra weak solution of (I) satisfies u(x, T) =}

ut(x, T) = 0for all α(T)< x < β(T).

Theorem 2.2_{Let us assume that the hypotheses (H1) and (H2) are satisfied.}

Then there existsT0>0,such that forT > T0 and the inicial data {υ0, υ1} ∈

L2_(0,1)×H−2_{(0,1). there exists a pair of controls{f}

1, f2} ∈[L2(0, T)]2such

that the ultra weak solution of (II) satisfies υ(y, T) = υt(y, T) = 0 for all

3

The problem of exact control in the cylinder

3.1

The Homogeneus Problem

Let us consider the operator R, given by:

Rw = wtt+a(t)wyyyy+b(y, t)wyy+c(y, t)wyt

+e(y, t)wy+g(y, t)w+h(y, t)wt,

where

a1≥a(t)≥a0>0 for allt≥0,

b, c, e, g, h∈W1,∞_{(0, T;L}∞_{(0,1)), c}

y ∈L∞(Q).

Our objetive in this section is to find a solution for the following problem:

(III)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Rw=G in Q

w(0, t) = 0, w(1, t) = 0 in ]0, T[

wy(0, t) =wy(1, t) = 0 in ]0, T[

w(0) =w0_{, w}

t(0) =w1 in ]0,1[.

Theorem 3.1_{(Strong solution) Givenw}0_∈H02_(0,1)∩H4_(0,1),w1_∈H02_(0,1),

G, Gt∈L1(0, T;L2(0,1)),then there exists a unique solutionwof (III) in the classw∈C([0, T];H2

0(0,1)∩H4(0,1))∩C1([0, T];H02(0,1)) verifyingRw=G

in L1_{(0, T;L}2_(0,1)).

Theorem 3.2_{(Weak solution) Given w}0 _{∈ H0}2_{(0,1), w}1 _{∈ L}2_{(0,1), G ∈}

L1_{(0, T;L}2_{(0,1)), then there exists a unique functionw:Q→R, solution of}

(III) in the classw∈C([0, T];H2

0(0,1))∩C1([0, T];L2(0,1))satisfying

(i) −(wt, zt)L2_(Q)+ (a(t)w_yy, z_yy)

L2(Q)+ (b wyy, z)L2 (Q)−(c wt, zy)L2 (Q)

+(e wy, z)_{L2 (Q)}+ (g wt, z)_{L2 (Q)}+ (h w, z)_{L2 (Q)} = (G, z)_{L2 (Q)},

for all z∈L2_{(0, T;H}2

0(0,1))with zt∈L2(Q), z(0) =z(T) = 0 and w(0) =

w0_{, w}

t(0) =w1.

(ii) Rw=Gin L1_{(0, T;H}−2_(0,1)).

(iii) E(t) =E(0) +1 2

Z t

0

a′_(s)

|wyy(s)|2ds−

Z t

0

−

It can be observed that the Theorems 3.1 and 3.2 are valid for the operators

L andL∗_{( adjoint ofL) given by}

Theorem 3.3_{Let bewa weak solution of the problem (III). Then there exists}

a positive constant c0,such that

where λ0 is Poincare’s constant. Then, from (H1)-(H2), we get

Next, we will obtain the Fundamental Identity which will allow us to have estimates forwyy(0, .) andwyy(1, .).

Lemma 3.4. _{If q∈C}3_{([0,1]) andw is the strong solution of problem (III),}

then wsatisfies the identity,

(6) 1

named Fundamental Identity.

Besides this, when q(y) =y−1

(7)

Proof. Multiplying (III)1byqwy and integrating inQ, we obtain :

(13)

Z T

0

Z 1

0

hqwwydydt=−1

2

Z T

0

Z 1

0

(hqy+hyq)|w|2dy dt.

From (8)-(13), we deduced (6).

As a consequence of the Lemma 3.4, we obtain the following result:

Theorem 3.5 _{If q ∈ C}3_{([0,1]) and {w}0_{, w}1_{, G} ∈ H0}2_(0,1)×L2_(0,1)×

L1_{(0, T;L}2_{(0,1)) then a weak solutionwof problem (III) verifies the}

inequal-ity

(14)

Z T

0

a(t)|wyy(1, t)|2dt+

Z T

0

a(t)|wyy(0, t)|2dt

≤C

(

E(0) + (

Z T

0 |

G(t)|dt)2

)

.

Proof. Consider F as being the vector space consisting of weak solutions of (III), with initial data{w0, w1} ∈H02(0,1)×L2(0,1)}andG∈L1(0, T;L2(0,1)).

We define the application

ξ:H2

0(0,1)×L2(0,1)×L1(0, T;L2(0,1))→F by ξ({w0, w1, G}) =w.

Observe thatξis bijective. We define also the aplication||.||ξ:F →R, given

by

kwkξ =|w0yy|+|w1|+|G|L1_(0,T;L2_(0,1)),that is a norm in F.

Observe also that (F,k.kξ) is a Banach space. Consider nowH as being the

vector space consisting of strong solutions of (III), with initial data

w0_∈H2

0(0,1)∩H4(0,1), w1∈H02(0,1), and G∈W1,1(0, T;L2(0,1)).

Note that H is dense in F. On the other side using Lemma 3.4, it results that,

|wyy(1, .)|L2_(0,T)≤c{|w_yy0 |+|w1|+|G|_L1_(0,T;L2_(0,1))}, for allw∈H

or

(15) |wyy(1, .)|L2_(0,T)≤ckwk_ξ, for allw∈H.

Analogously, we obtain

Motivated by (15) and (16), we define the continuous operators

φ0:H →L2(0, T), φ1:H →L2(0, T)

by

φ0(w) =wyy(0, .) and φ1(w) =wyy(1, .),

respectively. Since H is dense inF, there exist continuous operators

˜

φ0:F →L2(0, T), φ˜1:F →L2(0, T),

such that ˜φ0|H = φ0, ˜φ1|H = φ1. Therefore, if w ∈ F then there exists a

sequence (wm)∈H,such thatwm→winF. Then,

limφ0(wm) = lim ˜φ0(wm) = ˜φ0(w)

or

wmyy(0, .) = ˜φ0(wm)→φ˜0(w)inL2(0, T).

Analogously, we have

wmyy(1, .) = ˜φ1(wm)→φ˜1(w) in L2(0, T).

On the other hand, there exist w0

m ∈ H02(0,1)∩H4(0,1), w1m ∈ H02(0,1),

Gm∈W1,1(0, T;L2(0,1)), in the way that ξ({wm0, w1m, Gm}) =wm and

(17) w0

m→w0 in H02(0,1)

(18) w1

m→w1 in L2(0,1)

(19) Gm→G in L1(0, T;L2(0,1))

Besides this,

(20) wm→w in C([0, T];H02(0,1))

(21) wmt→wt in C([0, T];L2(0,1)) .

Then, wmyy(0, .)→wyy(0, .) in D′(0, T).

As L2_{(0, T)֒→D}′

(0, T) then φ˜0(w) =wyy(0, .). Therefore

(22) wmyy(0, .)→wyy(0, .) in L2(0,1).

Analogously, we have

(23) wmyy(1, .)→wyy(1, .) in L2(0,1).

3.3

Inverse Inequality

Theorem 3.6 _{There existsT0>0 such that forT > T0 the weak solution of}

problem (III) with G= 0verifies

(24)

2 in the identity (6) of the Lemma 3.4. we obtain

the following result

Then we have the following estimates:

Combining the above estimates, the proof of the Theorem 3.6 is completed

For the study of problem (II), we introduce the concept of ultra weak solution.

Definition 3.7 _{Let be v}0 _{∈ L}2_{(0,1) , v}1 _{∈ H}−2_{(0,1) , f = {f}

1, f2} ∈

[L2_(0,1)]2_{. Let us say thatv∈L}∞_{(0, T;L}2_{(0,1)) is an ultra weak solution of}

(II) if it satisfies

(25)

Then there exists a unique ultra weak solution of problem (II).

Proof. Define the applicationS:L1_{(0, T;L}2_(0,1))→Rby

From Theorem 3.5, it follows that

(26) |wt(0)|2+|wyy(0)|2≤C(

Z T

0 |

(27)

Z T

0 |

wyy(0, t)|2dt≤C(

Z T

0 |

G|dt)2,

(28)

Z T

0 |

wyy(1, t)|2dt≤C(

Z T

0 |

G|dt)2.

From (26)-(28), we obtain

(29) |hS, Gi| |G|L1_(0,T;L2_(0,1)) ≤

C{|v0|+|v1|H−2_(0,1)+|f₁|_L2_(0,T)+|f₂|_L2_(0,T)}

or

(30) |S|(L1_(0,T;L2_(0,1)))′ ≤C{|v0|+|v1|_H−2_(0,1)+|f₁|_L2_(0,T)+|f₂|_L2_(0,T)}.

From (30), we conclude thatS∈(L1_{(0, T;L}2_(0,1)))′_{. Therefore, by the Riez}

Theorem there exists a uniquev∈L∞_{(0, T;L}2_{(0,1)) such that}

hS, Gi=

Z T

0

(v, G)dt for allG∈L1_{(0, T;L}2_(0,1))

and also, |v|L∞_(0,T;L2_(0,1))=|S|_(L1_(0,T;L2_(0,1)))′_.

Then, vis the ultra weak solution of problem (II) and it satisfies :

(31) |v|L∞_(0,T;L2_(0,1)) ≤C{|v0|+|v1|_H−2_(0,1)+|f₁|_L2_(0,T)+|f₂|_L2_(0,T)}.

This completes the proof of the theorem.

Our objetive now is to know the regularity of the ultra weak solution of problem (II).

Lemma 3.9 _{Let v}0 _{∈ H0}2_{(0,1), v}1 _{∈ L}2_{(0,1), and {f1, f2} ∈ [H0}2_(0,1)]2_.

Then problem (II) admits a unique solution

v∈C([0, T];H2

0(0,1))∩C1([0, T];L2(0,1)).

Proof. Letv(y, t) =y2_(y−1)f

2(t) +y(y−1)2f1(t). Since f1, f2 ∈H02(0,1)

thenv∈H2

0(0, T;H01(0,1)∩H4(0,1)). Besides, we have

v(0, t) =v(1, t) = 0, vy(0, t) =f1, vy(1, t) =f2.

The solution of (II) will bev=bv+v where

b

is the solution of

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lv_b=−Lv in Q

b

v(0, t) =_bv(1, t) = 0 in ]0, T[

b

vy(0, t) =bvy(1, t) = 0 in ]0, T[

b

v(y,0) =v0_{, bv}

t(y,0) =v1, in ]0,1[.

This complete the proof of this lemma.

Theorem 3.10 _{If v}0 _{∈ L}2_{(0,1) , v}1 _{∈ H}−2_{(0,1) ,{f}

1, f2} ∈ [L2(0,1)]2.

Then the ultra weak solution v of (II) is such that v ∈C([0, T];L2_(0,1))∩

C1_{([0, T];H}−2_(0,1)).

Proof. Let be the sequences (v0

n) ∈ H02(0,1),(v1n) ∈ L2(0,1), {f1n, f2n} ∈

[H2

0(0,1)]2, such that

v0

n→v0in L2(0,1)

v1

n→v1in H−2(0,1)

(f1n, f2n)→(f1, f2) in [L2(0, T)]2.

Consider the problem

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lvn= 0 inQ

vn(0, t) =vn(1, t) = 0 in ]0, T[

vny(0, t) =f1n(t) ,vny(1, t) =f2n(t) in ]0, T[

vn(0) =v0n ,vnt(0) =vn1 in ]0,1[.

From Lemma 3.9, we deduce thatvn ∈C([0, T];H2(0,1))∩C1([0, T];L2(0,1)).

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lvn−Lvm= 0

(vn−vm)(0, t) = (vn−vm)(1, t) = 0

(vn−vm)y(0, t) =f1n(t)−f1m(t) , (vn−vm)y(1, t) =f2n(t)−f2m(t)

(vn−vm)(0) =vn0−vm0 , (vnt−vmt)(0) =v1n−vm1 .

From (31), it follows that

(32) |vn−vm|L∞_(0,T;L2_(0,1))≤C

³

|v0

n−v0m|+|v1n−vm1|H−2_(0,1)

+|f1n−f1m|L2_(0,T)+|f_2n−f_2m|_L2_(0,T)

´

.

Since (v0

n) , (vn1) , (f10n) , (f21n) are Cauchy sequences, it result from (32)

that (vn) is a Cauchy sequence in C([0, T];L2(0,1)). Then, there exists ev ∈

C([0, T];L2_{(0,1)), such that v}

n → vein C([0, T];L2(0,1)). It is proved that

e

v is the ultra weak solution of (II). By uniqueness, we conclude that ev = v

and so v ∈ C1([0, T];L2(0,1)). In analogous form it is verified that v ∈

C1_{([0, T];H}−2_(0,1)).

Now, the objetive is to prove the main result of exact control in the cylinder.

Proof of Theorem 2.2.

By the Theorems 3.2, 3.5 and 3.6, it is proved that for each pair {ϕ0_{, ϕ}1_{} ∈}

H2

0(0,1)×L2(0,1),there exists a unique solution ϕof the mixed problem

(IV)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

L∗_{ϕ= 0 in Q}

ϕ(0, t) =ϕ(1, t) = 0 ,ϕy(0, t) =ϕy(1, t) = 0 , in ]0, T[

ϕ(y,0) =ϕ0 _,ϕ

t(y,0) =ϕ1 in ]0,1[

such that ϕ ∈ C([0, T];H2

0(0,1))∩C1([0, T];L2(0,1)), ϕyy(0, .), ϕyy(1, .) ∈

L2_{(0, T).}

On the other hand we have

(33) (T−T0)C0E(0)≤

Z T

0

a(t)|ϕyy(1, t)|2dt+

Z T

0

a(t)|ϕyy(0, t)|2dt≤C1E(0).

(V)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lψ= 0 in Q

ψ(0, t) =ψ(1, t) = 0, ψy(0, t) =−ϕyy(0, t),

ψy(1, t) =ϕyy(1, t) in ]0, T[

ψ(y, T) = 0, ψt(y, T) = 0 in ]0,1[.

Considering ξ = T −t e ψ(y, t) = ψb(y, ξ) and using the Theorem 3.10, it is shown that there exists an ultra weak solution ψ of (V), such that ψ ∈

C([0, T];L2_(0,1)∩C1_{([0, T];H}−2_(0,1)).

From the definition of the ultra weak solution we have the following identity :

(34)

Z T

0

a(t)|ϕyy(0, t)|2dt+

Z T

0

a(t)|ϕyy(1, t)|2dt=−hψ(0), ϕ1i

−hψt(0), ϕ0i − 2h

α′_{(0) +γ}′_(0)y

γ(0) ψy(0), ϕ

0

i.

The identity (34) suggests the following operator :

Λ :H02(0,1)×L2(0,1)→H−2(0,1)×L2(0,1),such that

Λ{ϕ0, ϕ1}={ψt(0)−2

α′_{(0) +γ}′_(0)y

γ(0) ψy(0), ψ(0)}.

Then

(35) hΛ{ϕ0, ϕ1},{ϕ0, ϕ1}i=

Z T

0

a(t)|ϕyy(1, t)|2dt+

Z T

0

a(t)|ϕyy(0, t)|2dt.

From (H2), Remark 2.1 and remembering thata(t) = _γ41_(t),we have (γ1)−4≤

a(t)≤(γ0)−4. From (31), we obtain

(36) (γ0)−4{

Z T

0 |

ϕyy(1, t)|2dt+

Z T

0 |

ϕyy(0, t)|2dt} ≤ hΛ{ϕ0, ϕ1},{ϕ0, ϕ1}i

≤(γ1)−4{

Z T

0 |

ϕyy(1, t)|2dt+

Z T

0 |

From (35) and (36) we conclude that Λ is an injective selfadjoint operator . Therefore, Λ is an isomorphism ofH2

0(0,1)×L2(0,1) overH−2(0,1)×L2(0,1).

Finally, considering f1(t) =−ϕyy(0, t) andf2(t) =ϕyy(1, t), 0< t < T, and

initial data {v0_{, v}1_{} ∈ L}2_(0,1)×H−2_{(0,1), problem (II) has an ultra weak}

solutionv.

On the other side, considering problem (II) withv0_{=ψ(0) ,v}1_=ψ

t(0) where

ψis ultra weak solution of (V), it results thatψis also ultra weak solution of (II). By uniqueness, it can be concluded thatv=ψ.So from the system (V), we obtainv(y, T) = 0 ,vt(y, T) = 0,which proves the Theorem 2.2.

4

The problem of exact controllability in

noncylindrical domains

4.1

Weak solutions.

Let us consider the following problem :

(VI)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

θtt+θxxxx=bh in Qb

θ(α(t), t) =θ(β(t), t) = 0 in ]0, T[

θx(α(t), t) =θx(β(t), t) = 0 in ]0, T[

θ(x,0) =θ0_{, θ}

t(x,0) =θ1 in ]α(0), β(0)[.

Definition 4.2 _{Let θ}0 _{∈ H0}2_{(α(0), β(0)), θ}1 _{∈ L}2_{(α(0), β(0)), and} b_{h ∈}

L1_{(0, T;L}2_{(α(t), β(t))). We say that θ is a weak solution of problem (VI)}

ifθ∈C([0, T];H2

0(α(t), β(t))),θt∈C([0, T];L2(α(t), β(t))) and satisfies

−

Z T

0

(θt, ψt)L2_{(α(t),β(t))}dt+

Z T

0

(θxx, ψxx)L2_{(α(t),β(t))}dt

=

Z T

0

(bh, ψ)L2_{(α(t),β(t))}dt.

for all ψ ∈ L2_{(0, T;H}2

0(α(t), β(t))) and ψt ∈ L2(Qb) with ψ(0) = ψ(T) = 0

and initial data θ(0) =θ0_{, θ}

t(0) =θ1.

(37) u(x, t) =v(x−α γ , t)

(38) θ(x, t) =_γ1w(x−α γ , t)

(39) g1(t) =_γ1f1(t)

(40) g2(t) =_γ1f2(t).

From (38) we conclude that problem (VI) is equivalent to the problem (III). Therefore by Theorem 3.2 we guarantee the existence of weak solutions of the problem (VI).

It can be noted that the previus results are also valid when we replace the initial conditions of problem (VI) byθ(x, T) =θ0_,θ

t(x, T) =θ1.

4.2

Ultra weak solutions

Definition 4.3_{Given (u}0_{, u}1_)∈L2_{(α(0), β(0))×H}−2_{(α(0), β(0)), (g} 1, g2)∈

[L2_{(0, T)]}2_{,we say thatu∈L}∞_{(0, T;L}2_{(α(t), β(t))) is an ultra weak solution}

of problem (I) if

Z T

0

(u,bh)L2_{(α(t),β(t))}dt=hu1, θ(0)i_H−2_×H2 0−(u

0_{, θ}

t(0))L2_{(α(0),β(0))}

+

Z T

0

g1(t)θxx(α(t), t)dt−

Z T

0

g2(t)θxx(β(t), t)dt.

for allbh∈L1_{(0, T;L}2_{(α(t), β(t)) whereθis a weak solution of problem (IV) with}

θ(x, T) = 0, θt(x, T) = 0 in the place of (IV)4.

Using (37) we deduce that the ultra weak solution of problem (I) is equivalent to that of problem (II).

4.3

Proof of Theorem 2.1

As problems (I) e (II) are equivalent, like (III) and (IV), we can consider the following isomorphisms:

G1:L2(0,1)×H−2(0,1)→L2(α(0), β(0))×H−2(α(0), β(0)), such that

G1{v0, v1}={u0, u1}, and

G2:H02(0,1)×L2(0,1) →H02(α(0), β(0))×L2(α(0), β(0)), such that

G2{w0, w1}={θ0, θ1}.

Let us consider also the following isomorphisms :

σ: L2_(0,1)×H−2_(0,1)→H−2_(0,1))×L2_{(0,1), such that}

σ{v0_{, v}1_}={v1₋₂α′

(0)+γ′

(0)y

Λ : H2

0(0,1)×L2(0,1) →H−2(0,1)×L2(0,1), such that

Λ{w0_{, w}1_}={v1₋2α′

(0)+γ′

(0)y

γ(0) vy0,−v0}.

Therefore, the composed function Λ1=G2◦Λ−1◦σ◦G−11is an isomorphism

ofL2_{(α(0), β(0))×H}−2_{(α(0), β(0)) inH}2

0(α(0), β(0))×L2(α(0), β(0)).

By that, for the initial data {u0_{, u}1_{} ∈L}2_{(α(0), β(0))×H}−2_{(α(0), β(0)), we}

determine{θ0_{, θ}1_{}. Then with these data we find a weak solution to problem}

(IV). In a similar way with {w0_{, w}1_{} = G}−1

2 {θ0, θ1} , we find the weak

solutionwto problem (III).

Therefore we determine the ultra weak solution of the problem below:

(VII)

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Lv= 0 in Q

v(0, t) =v(1, t) = 0 in ]0, T[

vy(0, t) =−wyy(0, t) ,vy(1, t) =wyy(1, t) in ]0, T[

v(0) =v0 _,v

t(0) =v1 in ]0,1[

where{v0_{, v}1_}and{w0_{, w}1_{}are related by the aplication Λ. Then by Theorem}

2.2, v(T) =vt(T) = 0.

Finally, from (37) we obtain thatusatisfies the condition u(T) =ut(T) = 0.

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