Afr. Mat. https://doi.org/10.1007/s13370-018-0542-z

Radiative heating of a glass plate: the semi-discrete problem (second revision) Luc Paquet1

Received: 24 July 2015 / Accepted: 13 January 2018 © African Mathematical Union and Springer-Verlag GmbH Deutschland, ein Teil von Springer Nature 2018

Abstract In a preceding paper (MSIA, 2012), we have studied the radiative heating of a glass plate. We have proved existence and uniqueness of the solution. Here, we want to study the semi-discrete problem and to prove a priori error estimates. Previously, in that purpose, we have to study the regularity of the solution to the exact problem and of its time derivative. A very important property, Proposition 13, is remarked concerning our elliptic projection, the milestone in deriving the a priori error estimates. A numerical test corroborating our theoretical a priori bounds is given. Keywords Planck function · Radiative heat flux in a glass plate · Nonlinear heat-conduction equation · Regularity of the exact solution and of its time derivative · Semi-discrete problem · Elliptic projection · A priori error estimates Mathematics Subject Classification 35K20 · 35K55 · 45K05 · 65M22 · 65M60 · 65M15

1 Statement of the problem and summary of our results The setting is the same as in our previous paper [11], but the problem we study here is the related semi-discrete problem, our purpose being to prove some a priori error estimate on the semi-discrete solution. Let us remind only the equations of the exact problem to which we have been led in [11] and some results that we will need in the proof of the a priori error estimates. We consider an infinite plane horizontal glass plate of thickness l, laid down on its lower face x g = 0, on a black sheet-metal maintained at absolute ambient temperature T = Ta . The x-axis is directed upward orthogonally to the glass plate so that the upper (resp. lower) face of the glass-plate has x = l (resp. x = 0) for equation. An infinite plane horizontal black sheet metal S, at absolute temperature u(t) at time t, placed above

B 1

Luc Paquet [email protected]; [email protected] LAMAV-EDP, EA 4015, Institut des Sciences et Techniques de Valenciennes (ISTV2), Univ. Valenciennes, EA 4015-LAMAV, FR CNRS 2956, F-59313 Valenciennes, France

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the glass plate, emits thermal radiation in every direction, whose spectral radiative intensity [10,13,17,18] at wavelength λ in the (dry) air gap between S and the glass plate is given by the famous Planck function: 2C1 B(T, λ) = with T = u(t), (1) C2 λ5 (e λT − 1) −2 m.◦ K [5, 0 where C1 = hc02 = 0.595531 10−16 W m2 /sr and C2 = hc k B = 1.438786 10 p. 98] (in [11], the temperature of the source of thermal radiation S at time t was denoted TS (t), but here we prefer to denote it u(t) like in [12]). After refraction at the interface {x = l} between air and glass, some part of the radiative energy emitted by the black source S, will be absorbed i.e. converted into heat in the glass producing in such a way an increase of the temperature T (x, t) in the glass plate. Assuming the dependent quantities, independent of the cartesian coordinates y and z, and denoting by t f the final time of the heating process, based on physical considerations, we have shown in [11], that the radiative heating of the glass plate, can be modelized by the following nonlinear initial boundary value problem in the unknown (absolute) temperature distribution in the glass plate T (x, t) along the thickness of the glass plate 0 ≤ x ≤ l for time 0 ≤ t ≤ t f : ⎧ 2 c p m g ∂∂tT (x, t) = kh ∂∂ xT2 (x, t) + ψ(T (x, t)) + h T,u (x, t), ⎪ ⎪ ⎪ ⎪ ⎪ 0 < x < l, 0 < t < t f , ⎪ ⎨ −kh ∂∂Tx (l, t) = h c (T (l, t) − Ta ) + ((T (l, t)) − (u(t))), (2) ⎪ ⎪ ⎪ 0 < t < t f , (B.C. at x = l), ⎪ ⎪ ⎪ ⎩ T (0, t) = Ta , 0 < t < t f , (B.C. at x = 0), T (x, 0) = T0 (x), ∀x ∈ [0, l], (I.C. at t = 0),

where c p , m g , kh are assumed to be positive constants named respectively heat capacity, mass density and thermal conductivity of the glass [17,18]. The quantities appearing in the righthand side of the above nonlinear parabolic integro-differential equation and in the boundary condition (B.C.) at x = l, are defined by the following functions: k=M 

ψ(T ) : = −

4πκk Bgk (T ), h T,u (x, t) :=

k=M 

k=1

2πκk h kT,u (x, t)

k=1

+∞ and (T ) := π λ B(T, λ)dλ,

(3)

λ0

where (the meaning of M and Bgk (T ) are explained below): l h kT,u (x, t) : =

G k (x, x  ) Bgk (T (x  , t)) d x  + Bgk (Ta ) E 2 (κk x)

0

+ Bgk (Ta ) 2 (κk (2l − x)) + Bgk (u(t))[E 2 (κk (l − x)) − 2 (κk (l − x))] l + κk



1 (κk (2l − x − x ))Bgk (T (x  , t)) d x  .

(4)

0

 Let us explain the meaning of these different quantities. In (4), ∀ x, x  ∈ [0, l]2 : y y



  1 1 G k (x, x  ) := κk E 1 (κk x − x  ), 1 (y) := e− μ ρg (μ) dμ , and 2 (y) := e− μ ρg (μ)dμ, 0

123

μ

0

Radiative heating of a glass plate: the semi-discrete…

∀y > 0. ρg (μ) (0 ≤ μ ≤ 1) denotes the reflectivity coefficient given by Fresnel’s relation [10, formula (2.96) p. 47]. E 1 (·) and E 2 (·) denote the exponential integral functions of order 1 and 2 respectively. In the definition of (T ) in (3), λ is a positive constant called the spectral hemispherical emittance [10, pp. 62–63]; like in [16, p. 70], we have supposed that the spectral hemispherical emittance is equal to the spectral hemispherical absorptance and is independent of the temperature, for wavelength λ of the electromagnetic wave spectrum belonging to [λ0 , +∞[ (λ0 ≈ 5 μm for glass, [16, p. 70]), wavelengths for which glass is opaque. On the other hand, the region of wavelengths in the electromagnetic spectrum where glass is semi-transparent has been decomposed into M disjoint bands [λk , λk+1 [ (k = 1, . . . , M) [3,8,15] and the constants κk appearing in the above formulas are the linear spectral absorpλ tion coefficients [7,10,13,17,18] in their respective band. B k (T ) := λkk+1 B(T, λ) dλ and Bgk (T ) := n 2g B k (T ) [11,15], where n g denotes the refractory index of the glass (n g ≈ 1.46). Let us also mention that ψ(T (x, t)) + h T,u (x, t) = − ∂q ∂ x (x, t) where q(x, t) denotes the total radiative heat flux at time t through the horizontal plane at level 0 < x < l [10, p. 276] in the positive x-direction. In the boundary condition at x = l (see (2)), h c > 0 denotes the convective heat transfer coefficient, the term h c (T (l, t) − Ta ) representing the conducto-convective flux density at the infinite surface {x = l} of the glass plate according to Newton’s law [17, p. 16], [18, p. 13–16]. We assume that the initial condition T0 (·) (see (2)) is a continuous positive function on the closed interval [0, l], as an absolute temperature is always positive in classical physics and T0 (·) is a datum. In order to obtain regularity on the solution T (., .) of the nonlinear initial boundary value problem (2), we will suppose moreover in the present paper, at least that the initial condition T0 (·) ∈ H 1 (]0, l[) and that the compatibility condition T0 (0) = Ta between the initial condition T0 and the inhomogeneous Dirichlet boundary condition T (0, .) = Ta (see (2)) on the lower face {x = 0} of the glass plate, is verified. We suppose given two different positive constants Tlow and Tup such that 0 < Tlow  T0 (x)  Tup , ∀x ∈ [0, l] (this implies that 0 < Tlow  Ta  Tup due to T0 (0) = Ta ) and that Tlow  u(t)  Tup , ∀t ∈ [0, t f ]. We have proved in [11] that for u ∈ Uad = {v ∈ H 1 (]0, t f [); Tlow ≤ v(t) ≤ Tup , ∀t ∈]0, t f [},

(5)

that the initial boundary value problem (2) possesses a unique bounded weak solution

∗ (6) Tu ∈ {T ∈ L 2 (0, t f ; H 1 (]0, l[)); T˙ ∈ L 2 (0, t f ; HL1 (]0, l[) )}

which is also continuous on Q := [0, l] × 0, t f , Q denoting the “cylinder” ]0, l[×]0, t f [. In (6), HL1 (]0, l[) denotes the closed subspace of H 1 (]0, l[) formed by those functions of H 1 (]0, l[) which vanish at the left extremity x = 0 of the interval ]0, l[. By a weak solution Tu of the initial boundary value problem (2), we mean that Tu belongs to 

∗  T ∈ L 2 (0, t f ; H 1 (]0, l[)); T˙ ∈ L 2 (0, t f ; HL1 (]0, l[) ) and satisfies ∀ t ∈]0, t f [: ⎧   l u ⎪ c p m g dT (·, t), ϕ 1 + kh 0 ∂∂Txu (x, t)ϕ  (x)d x ⎪ ⎪ dt 1 ∗ ⎪ HL (]0,l[) ,HL (]0,l[) ⎪ ⎪ ⎨ − l ψ(T (x, t))ϕ(x)d x −  l h (x, t)ϕ(x)d x 0

u

0 Tu ,u

− [(u(t)) − (Tu (l, t))] · ϕ(l) − [h c (Ta − Tu (l, t))] ϕ(l) = 0, ∀ϕ ∈ HL1 (]0, l[), ⎪ ⎪ ⎪  ⎪ t f [, ⎪ ⎪ Tu (0, t) = Ta , ∀ t ∈]0, ⎩ Tu (x, 0) = T0 (x), ∀ x ∈]0, l[. (7)

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We have written Tu to underline the dependence of the weak solution of (2) with respect to u. Using Stampacchia’s truncation method [1], we have also proved in [11], that the weak solution of the initial boundary value problem (2) Tu ≡ T (u) satisfies: Tlow  Tu (x, t)  Tup , ∀(x, t) ∈ Q.

(8)

Let us mention, that though an absolute temperature is always positive in classical physics, that for mathematical purposes only, we have extended the definition of the Planck’s function (1) to negative real numbers T by setting B(T, λ) = 0, if T ≤ 0; in that way for fixed λ > 0, the function T  → B(T, λ) is defined on the whole real line and is lipschitzian of Lipschitz constant C2Cλ14 (cf. Lemma 3.6 of [11]). 2 Now these things having been recalled, let us explain what are the results in the present paper. Firstly, we study the regularity of the exact solution Tu i.e. of the weak solution of our nonlinear initial boundary value problem (2). Under certain regularity hypotheses on the initial condition T0 and compatibility conditions, we prove that Tu ∈ H 2,1 (Q) (see 2,1 (Q) (see Corollary 8), regularity properties that we u Proposition 3), and that dT dt ∈ H will need for our a priori error estimates (68), (72) and (77) to be valid. We introduce the semi-discrete problem (47), its matrix differential form being given by (44) from which the existence and unicity of its solution Tu,h is nearly immediate (see Proposition 10). We introduce the elliptic projection T˜u,h (., t) of Tu (., t) by formula (49). If we would have a Dirichlet boundary condition at the point x = l, then the technique is well known [19, p. 7]. But in our case, we have the nonlinear Robin boundary condition (2)(ii) at the point x = l. The key property (see Proposition 13) of our elliptic projection T˜u,h (., t) of Tu (., t) (49) relative to our setting (2), which allows us to derive our a priori error estimates, is that T˜u,h (l, t) = Tu (l, t). After having written the error equation (48) and splitted the error e := Tu,h − Tu on Tu,h in e = θ + ρ where θ = Tu,h − T˜u,h and ρ = T˜u,h − Tu , we are on the road which error estimates in h 2 for    will lead us to the proofsof our a priori Tu,h (l, .) − Tu (l, .) 2 (Corollary 24), for Tu,h − Tu C([0,t ];L 2 (]0,l[)) (Theorem 25) L (]0,t f [) f    and in h for ∇x (Tu,h − Tu ) 2 2 2 (Proposition 27). A numerical test is given L (0,t f ;L (]0,l[) )

corroborating our theoretical bounds and we take also this opportunity to show the influence on the solution of our problem of the sum of terms ψ(T (x, t)) + h T,u (x, t), modelizing the participating medium character of the glass into the integro-differential equation (2)(i) .

2 Regularity results At the present time, we know only that Tu ∈ {T ∈ L 2 (0, t f ; H 1 (]0, l[)); T˙

∗ L 2 (0, t f ; HL1 (]0, l[) )}. But we need more on the exact solution Tu .

∈

Proposition 1 The right hand-side (x, t)  → ψ(Tu (x, t)) + h Tu ,u (x, t) of Eq. (2)(i) belongs to L 2 (Q). Proof As Tu ∈ L 2 (Q), ψ ◦ Tu ∈ L 2 (Q) by Corollary 3.3, p. 10 of [11]. By the same result h Tu ,u belongs also to L 2 (Q).   Proposition 2 The right-hand side t  −→ h c (Tu (l, t) − Ta ) + ((Tu (l, t)) − (u(t))) in the boundary condition at x = l of Eq. (2)(ii) belongs to L 2 (]0, t f [). Proof As Tu ∈ L 2 (0, t f ; H 1 (]0, l[)) and H 1 (]0, l[)) → C([0, l]), the mapping t  −→ Tu (l, t) belongs to L 2 (0, t f ). Using Corollary 3.4 p. 12 of [11], it follows that the function

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t  −→ (Tu (l, t)) belongs also to L 2 (0, t f ). By the same result  ◦ u ∈ L 2 (0, t f ). This proves the result.   Proposition 3 Tu ∈ H 2,1 (Q) := L 2 (]0, t f [; H 2 (]0, l[)) ∩ H 1 (]0, t f [; L 2 (]0, l[)) [9, p. 6]. Proof From the two previous propositions, it follows that Tu may be seen as the solution of: ⎧ ∂T ∂2T ⎪ ⎪ ∂t (x, t) = a ∂ x 2 (x, t) + f (x, t), ⎪ ⎪ ⎨ 0 < x < l, 0 < t < t f , ∂T (9) (l, t) = g(t), 0 < t < t f , (B.C. at x = l), ∂x ⎪ ⎪ ⎪ ⎪ ⎩ T (0, t) = Ta , 0 < t < t f , (B.C. at x = 0), T (x, 0) = T0 (x), ∀x ∈ [0, l], (I.C. at t = 0). where a ∈ R∗+ , f ∈ L 2 (Q), g ∈ L 2 (]0, t f [) and T0 ∈ H 1 (]0, l[) (see the Introduction). A fortiori f ∈ [H 1/2+2ε,1/4+ε (Q)] and g ∈ H −ε (]0, t f [), for ε ∈]0, 1/4[. Using the result at the bottom of page 78 of [9] with s = −1/4−ε, we obtain that Tu ∈ H 3/2−2ε,3/4−ε (Q) (multiplying Tu by a cutoff function depending only on the variable x, we reduce us respectively to the case of the Neumann boundary condition on the whole boundary {0, l} or respectively to the case of the Dirichlet boundary conditions on the whole boundary {0, l} and apply the cited result of [9] in these two cases). Using the “First Trace Theorem” p. 9 of [9], we deduce that Tu (l, .) ∈ H 1/2−ε (]0, t f [) and thus a fortiori to H 1/4 (]0, t f [). (.) being a Lipschitz continuous function [11, (3.13), p. 17], it follows that t f t f 0 0

|(Tu (l, t)) − (Tu (l, s))|2 ds ⊗ dt  |t − s|3/2

t f t f 0 0

|Tu (l, t) − Tu (l, s)|2 ds ⊗ dt < +∞, |t − s|3/2

H 1/4 (]0, t

which implies that (Tu (l, .)) ∈ f [). We have already seen in the proof of the previous proposition that  ◦ u ∈ L 2 (0, t f ). On the other hand u ∈ H 1 (]0, t f [), so that

∂B

 +∞ ∂ B d(◦u) 1   



dt (t) =  (u(t))u (t).  (T ) = π λ0 ελ ∂ T (T, λ)dλ and ∂ T (T, λ)  λ4 [11, p. 14]  1 imply that  (·) is a bounded function. Thus ◦u ∈ H (0, t f ) and a fortiori in H 1/4 (]0, t f [). In conclusion, we know now in (9) that f ∈ L 2 (Q) and that g ∈ H 1/4 (]0, t f [). Applying now Theorem 6.2, p. 37 of [9] with r = 0, we obtain that Tu ∈ H 2,1 (Q). What was to be proved.   Lemma 4 Let us consider the mapping f : Q → R : (x, t)  → ψ(Tu (x, t)) + h Tu ,u (x, t). Then

∂f ∂t

(10)

∈ L 2 (Q).

 k Proof Firstly ψ(Tu (x, t)) = − k=M k=1 4πκk Bg (Tu (x, t)). Thus

 ∂(ψ◦Tu ) (x, t) = − k=M k=1 ∂t d Bgk d Bgk ∂ Tu 4πκk dT (Tu (x, t)) ∂t (x, t). But, as we have proved in [11] page 14, dT (·) is bounded. u) By the previous proposition, we know that ∂∂tTu ∈ L 2 (Q). Thus ∂(ψ◦T ∈ L 2 (Q). Secondly, ∂t  ∂h Tu ,u k=M what can we say about ∂t . In view of formula h Tu ,u (x, t) := k=1 2πκk h kTu ,u (x, t) (see ∂h k (3)), it suffices to prove that ∂tTu ,u ∈ L 2 (Q). But ∂h kTu ,u ∂t

l (x, t) : = 0

G k (x, x  )

d Bgk dT

(Tu (x  , t))

∂ Tu  (x , t) d x  ∂t

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d Bgk

(u(t))u  (t)[E 2 (κk (l − x)) − 2 (κk (l − x))] dT l d Bgk ∂ Tu   + κk 1 (κk (2l − x − x )) (Tu (x  , t)) (x , t) d x  . dT ∂t

+

(11)

0

From that formula, page 23] and u  ∈ ∂h Tu ,u ∂t

d Bgk ∂ Tu 2 dT (·) bounded, ∂t ∈ L (Q), Corollary k ∂h L 2 (]0, t f [), follows that ∂tTu ,u ∈ L 2 (Q),

3.19, Corollary 3.20 of [11, ∀k = 1, . . . , M. Thus also

∈ L 2 (Q). The proof is complete.

 

Lemma 5 Let us suppose that T0 ∈ H 2 (]0, l[). Then, the following initial boundary value problem ⎧ ∂f ∂2v ⎪ c p m g ∂v ⎪ ∂t (x, t) = k h ∂ x 2 (x, t) + ∂t (x, t), ⎪ ⎪ ⎪ 0 < x < l, 0 < t < t f , ⎪ ⎪  ⎪ t ⎨ ∂v −kh ∂ x (l, t) = h c v(l, t) +  0 v(l, s)ds + T0 (l) v(l, t) −  (u(t))u  (t), ⎪ 0 < t < t f , (B.C. at x = l), ⎪ ⎪ ⎪ ⎪ v(0, t) = 0, 0 < t < t f , (B.C. at x = 0), ⎪ ⎪ ⎪ ⎩ v(x, 0) = 1 ( f (x, 0) + kh T  (x)), ∀ x ∈]0, l[, (I.C. at t = 0), 0 cpmg

(12)

possesses at least one weak solution. In (12), f denotes the function defined by formula (10). Proof Let us firstly precise what we mean by a weak solution of the initial boundary value problem (12). By a weak solution, we mean an element v ∈ {v ∈ L 2 (0, t f ; HL1 (]0, l[)); v˙ ∈

∗ L 2 (0, t f ; HL1 (]0, l[) )} such that ∀ t ∈]0, t f [, ∀ϕ ∈ HL1 (]0, l[) : ⎧ l d ⎪ v(·, t), ϕ H 1 (]0,l[)∗ ,H 1 (]0,l[) = −kh 0 ∂∂vx (x, t)ϕ  (x)d x c p m g dt ⎪ ⎪ L L   ⎪ t ⎨  l ∂f + 0 ∂t (x, t)ϕ(x)d x − h c v(l, t)ϕ(l) −  0 v(l, s)ds + T0 (l) v(l, t)ϕ(l) ⎪ + (u(t))u(t)ϕ(l), ˙ ⎪ ⎪ ⎪ ⎩ v(x, 0) = 1 ( f (x, 0) + k T  (x)), ∀ x ∈]0, l[. h 0 cpmg

(13)

The homogeneous Dirichlet boundary condition at x = 0, is expressed by the fact that v ∈ L 2 (0, t f ; HL1 (]0, l[)) and the initial condition at time t = 0 has sense due to the fact that

∗ {v ∈ L 2 (0, t f ; HL1 (]0, l[)); v˙ ∈ L 2 (0, t f ; HL1 (]0, l[) )} → C([0, t f ]; L 2 (]0, l[)). l We know also by the previous lemma that ∂∂tf ∈ L 2 (Q) so that 0 ∂∂tf (x, t)ϕ(x)d x exists ∀ t ∈ ]0, t f [. Also by Proposition 1 and Lemma 4, f and ddtf belong to the space L 2 (0, t f ; L 2 (]0, l[)) which implies that f (·, 0) ∈ L 2 (]0, l[). Thus the right-hand side in the initial condition of problem (13) belongs to L 2 (]0, l[). Let us apply Faedo–Galerkin’s method. Let us consider  w j j≥1 an orthonormal basis in L 2 (]0, l[) also orthogonal basis in HL1 (]0, l[). In particular, these basis functions are continuous functions on the closed interval [0, l] and thus bounded. Let mus consider the vectorial subspace generated by w1 , . . . , wm . Let us try to find vm (t) = j=1 (vm (t)| w j ) L 2 (]0,l[) w j solution of the following Cauchy problem for the nonlinear

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system of (integro-) differential equations ∀ j = 1, . . . , m:   ⎧ ∂f ⎪ c m ( v ˙ (t)|w ) + k (v (t)|w ) = (·, t)|w − hc 2 1 p g m j h m j j L (]0,l[) ⎪ HL (]0,l[) ∂t ⎨ L 2 (]0,l[)   t   ˙ j (l), ⎪ vm (t)(l)w j (l) −  0 vm (s)(l)ds + T0 (l) vm (t)(l)w j (l) +  (u(t))u(t)w ⎪ ⎩ 1  (vm (·, 0)|w j ) L 2 (]0,l[) = c p m g ( f (·, 0) + kh T0 (·)|w j ) L 2 (]0,l[) . (14) As is usual, to prove existence and uniqueness of a solution to a system of differential equations, we transform this system of differential equations in a system of nonlinear integral equations: t ⎧ c p m g (vm (t)|w j ) L 2 (]0,l[) + kh 0 (vm (s)|w j ) H 1 (]0,l[) ds = ( f (·, t)|w j ) L 2 (]0,l[) ⎪ ⎪   L  ⎪ t ⎪ t ⎨ −( f (·, 0)|w ) 2 v (s)(l)ds w (l) −  j L (]0,l[) − h c m j 0 0 vm (s)(l)ds  ⎪ ⎪ + T0 (l) w j (l) + (T0 (l))w j (l) + (u(t))w j (l) − (u(0))w j (l) ⎪ ⎪ ⎩ + c p m g (vm (0)|w j ) L 2 (]0,l[) , ∀ j = 1, . . . , m.

(15)

Let us note that c p m g (vm (0)|w j ) L 2 (]0,l[) = ( f (·, 0) + kh T0 (·)|w j ) L 2 (]0,l[) , ∀ j = 1, . . . , m. As vm (s) =

m

k=1 (vm (s)|

(16)

wk ) L 2 (]0,l[) wk ,

kh (vm (s)|w j ) H 1 (]0,l[) = kh L

m  (vm (s)|wk ) L 2 (]0,l[) (wk |w j ) H 1 (]0,l[) L

k=1

 2 = kh (vm (s)|w j ) L 2 (]0,l[) w j  H 1 (]0,l[) . L

 2 t In particular kh 0 (vm (s)|w j ) H 1 (]0,l[) ds = kh w j  H 1 (]0,l[) 0 (vm (s)|w j ) L 2 (]0,l[) ds. Let us t

( j)

L

L

set vm (t) := (vm (t)|w j ) L 2 (]0,l[) , ∀ j = 1, . . . , m. The above nonlinear integral equation may be rewritten: ⎛ ⎞  t (1) ⎞ ⎛ (1) w1 2H 1 (]0,l[) 0 vm (s)ds vm (t) L ⎜ ⎟ ⎟ ⎜. ⎟ .. ⎟ = −kh ⎜ . cpmg ⎜ ⎜ ⎟ ⎠ ⎝. ⎝. ⎠  t (m) (m) 2 wm  H 1 (]0,l[) 0 vm (s)ds vm (t) L ⎛ ⎞ ⎛ ⎞ ( f (t)|w1 ) L 2 (]0,l[) ( f (0)|w1 ) L 2 (]0,l[) ⎜ ⎟ ⎜ .. ⎟ + ⎝ ... ⎠ − ⎝. ⎠ ( f (t)|wm ) L 2 (]0,l[) ( f (0)|wm ) L 2 (]0,l[) ⎛ ⎞  w1 (l)  m   t ⎜ ⎟ (k) vm (s)ds · wk (l) ⎝ ... − hc ⎠ k=1 0 wm (l) ⎞ ⎛  w1 (l)  m   t ⎟ ⎜ (k) vm (s)ds · wk (l) + T0 (l) ⎝ ... − ⎠ + ( (T0 (l)) 0 k=1 wm (l)

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⎞ w1 (l) ⎟ ⎜ +  (u(t)) −  (u(0))) ⎝ ... ⎠ wm (l) ⎞ ⎛  ( f (·, 0) + kh T0 (·)|w1 ) L 2 (]0,l[) ⎟ ⎜ + ⎝ ... ⎠.  ( f (·, 0) + kh T0 (·)|wm ) L 2 (]0,l[) ⎛

(17)

Now, to prove that this system of m nonlinear scalar integral equations in the m unknowns ( j) vm (·), j = 1, . . . , m possesses one and only one solution, we proceed as in the proof of Cauchy’s theorem. Firstly, let us introduce the Banach space E = C([0, t f ]; Rm ) (1) (m) endowed with the sup norm. Let us define the mapping T : E → E : (vm , . . . , vm )  → (1) (m) (1) (m) T (vm , . . . , vm ), where T (vm , . . . , vm ) denotes the transposed of the right-hand side of Eq. (17). Now, we are going to prove that some power of the mapping T for the composition law is a contraction. Inequality (3.6) page 13 of [11] about Planck function, implies that  : R → R is a Lipschitz function. Thus there exists a positive constant C such that:





(1) (m) (1) (m) , . . . , vm ) (t) − T (v˜m , . . . , v˜m ) (t) m

T (vm ≤C

m t 

R





( j) ( j)

vm (s) − v˜m (s) ds.

(18)

j=1 0

Applying Cauchy–Schwarz inequality in Rm , we obtain:





(1) (m) (1) (m) , . . . , vm ) (t) − T (v˜m , . . . , v˜m ) (t)

T (vm √ ≤C m

t

Rm

√ |vm (s) − v˜m (s)|Rm ds ≤ C m t vm − v˜m  E .

(19)

0

This inequality implies that:   √  (1) (m) (1) (m)  , . . . , vm ) − T (v˜m , . . . , v˜m ) ≤ C m t f vm − v˜m  E . T (vm E

(20)

Iterating, we obtain ∀l ∈ N∗ that

l  √   C mt f  ◦l (1) (m) ◦l (1) (m)  vm − v˜m  E . T (vm , . . . , vm ) − T (v˜m , . . . , v˜m ) ≤ E l! √ +∞ (C √mt f )l Now eC mt f = l=1 and as the general term of a convergent series tends to zero, l! √ (C mt f )l tends to zero as l → +∞. Thus for l sufficiently large T ◦l : E → E is a contraction l!

(1)

(m)

and possesses thus one and only one fixed point (vm , . . . , vm ) ∈ E = C([0, t f ]; Rm ) which is also the unique fixed point of T : E → E. Thus, we know now that the Cauchy problem for the system of m nonlinear differential equations (14) possesses one and only one solution in the time interval [0, t f ]. Now, we have to pass to the limit as m → +∞. In that purpose, we are going to prove some energy estimate. As vm (t) is for a fixed t a linear combination of the functions w j , j = 1, . . . , m, it follows from (14) that:   ∂f c p m g (v˙m (t)|vm (t)) L 2 (]0,l[) + kh vm (t)2H 1 (]0,l[) = (·, t)|vm (t) ∂t L L 2 (]0,l[)

123

Radiative heating of a glass plate: the semi-discrete…

⎛ t 

⎝

−h c vm (t)(l) −  2

⎞ vm (s)(l)ds + T0 (l)⎠ vm (t)(l)2 +  (u(t))u(t)v ˙ m (t)(l).

0

As  (·) ≥ 0, it follows from the previous equality, the inequality: ∀ε > 0:     cpmg d l vm (t)2L 2 (]0,l[) + kh − l 1 + ε 2 vm (t)2H 1 (]0,l[) 2 dt 2 L  2



 ∂ f 1  1 2 ≤ 2 + 2 ( ◦ u) (t) . (·, t)  2 ε  ∂t ε

(21)

L (]0,l[)

Choosing in the previous inequality ε > 0 sufficiently small for l(1 + 2l )ε 2 to be smaller than k2h , we have in particular 2  

2 cpmg d ∂f 1  1

2  vm (t) L 2 (]0,l[) ≤ 2  (·, t) + 2 ( ◦ u) (t) .  2 dt ε ∂t ε L 2 (]0,l[) Integrating both sides from 0 to t, we obtain: ∀t ∈]0, t f [: cpmg 1 vm (t)2L 2 (]0,l[) ≤ 2 2 ε

2 t f  t f ∂f

2 

1  (·, t) dt + 2 ( ◦ u) (t) dt  ∂t  2 ε L (]0,l[) 0

0

 1   f (·, 0) + kh T  (·)2 2 + 0 L (]0,l[) 2c p m g  2 1 ∂f   ≤ 2 ε  ∂t  L 2 (Q)  1  2 1   f (·, 0) + kh T  (·)2 2 + 2  ∞ u2H 1 (]0,t [) + . 0 L (]0,l[) f ε 2c p m g This inequality shows that vm C([0,t f ];L 2 (]0,l[)) is bounded. Going back to the inequality (21) and integrating both sides from 0 to t f , we obtain: 

  t f  cpmg  l vm (t f )2 2 vm (t)2H 1 (]0,l[) dt + kh − l 1 + ε2 L (]0,l[) 2 2 L 

0

 1 ∂f ≤ 2 ε  ∂t

2    2 L

1 + 2 ε (Q)

t f



( ◦ u) (t) 2 dt + c p m g vm (0)2 2 . L (]0,l[) 2

0

As vm (0) L 2 (]0,l[) ≤

 1   f (·, 0) + kh T   2 0 L (]0,l[) cpmg

and thus bounded independently of m, it follows from the previous inequality that vm  L 2 (0,t f ;H 1 (]0,l[)) is also bounded independently of m. Now, we want to show that (v˙m )m≥1 L

is bounded in L 2 (0, t f ; HL1 (]0, l[)∗ ). Let w ∈ HL1 (]0, l[) such that w H 1 (]0,l[) ≤ 1. Let L us decompose w orthogonally into a part w˜ 1 ∈ span(w1 , . . . , wm ) and another part w˜ 2

123

L. Paquet

belonging to the orthogonal of the finite dimensional vector subspace span(w1 , . . . , wm ) in HL1 (]0, l[). This implies that w˜ 1  H 1 (]0,l[) ≤ 1. Now: L

v˙m (t), w H 1 (]0,l[)∗ ,H 1 (]0,l[) = (v˙m (t)|w) L 2 (]0,l[) L

L

= (v˙m (t)|w˜ 1 ) L 2 (]0,l[) + (v˙m (t)|w˜ 2 ) L 2 (]0,l[) .

+∞

wk ) H 1 (]0,l[) wk wk1 , k=m+1 (w| wk  1 L H (]0,l[) H (]0,l[)

But w˜ 2 =

L

this series being convergent in

L

HL1 (]0, l[) and thus a fortiori in L 2 (]0, l[) as HL1 (]0, l[) → L 2 (]0, l[). This implies that for j ∈ {1, . . . , m}:   +∞  wk (w j |w˜ 2 ) L 2 (]0,l[) = w| wk  H 1 (]0,l[) 1 k=m+1

L

HL (]0,l[)

(w j |wk ) L 2 (]0,l[) × = 0. wk  H 1 (]0,l[) L

As v˙m (t) is a linear combination of w1 , . . . , wm , it follows that (v˙m (t)|w˜ 2 ) L 2 (]0,l[) = 0. Thus v˙m (t), w H 1 (]0,l[)∗ ,H 1 (]0,l[) = (v˙m (t)|w˜ 1 ) L 2 (]0,l[) . By the first equation of (14), L

L

c p m g v˙m (t), w H 1 (]0,l[)∗ ,H 1 (]0,l[) = c p m g (v˙m (t)|w˜ 1 ) L 2 (]0,l[) L L   ∂f (·, t)|w˜ 1 = −kh (vm (t)|w˜ 1 ) H 1 (]0,l[) + − h c vm (t)(l)w˜ 1 (l) L ∂t L 2 (]0,l[) ⎛ t ⎞  ⎝ vm (s)(l)ds + T0 (l)⎠ vm (t)(l)w˜ 1 (l) +  (u(t))u(t) ˙ w˜ 1 (l) − 0

   ∂f   + vm (t) H 1 (]0,l[) + |u(t)| ˙ ≤C  ,  ∂t (·, t) 2 L L (]0,l[) for some constant C independent of m. Replacing w by −w, we obtain:   



 ∂f



 + vm (t) H 1 (]0,l[) + |u(t)| ˙ c p m g v˙m (t), w H 1 (]0,l[)∗ ,H 1 (]0,l[) ≤ C  ,  ∂t (·, t) 2 L L L L (]0,l[) for every w ∈ HL1 (]0, l[) such that w H 1 (]0,l[) ≤ 1. Thus L    ∂f   c p m g v˙m (t) H 1 (]0,l[)∗ ≤ C  + vm (t) H 1 (]0,l[) + |u(t)| ˙ .  ∂t (·, t) 2 L L L (]0,l[) Integrating both sides from 0 to t f , we obtain: ⎛ ⎞  2 t f t f t f ⎜ ∂ f  ⎟  v˙m (t)2H 1 (]0,l[)∗ dt ≤ C ⎝ + vm (t)2H 1 (]0,l[) dt + |u(t)| ˙ 2 dt ⎠ .  ∂t  2 L L L (Q)

0

0

0

Thus (v˙m )m≥1 is bounded in L 2 (0, t f ; HL1 (]0, l[)∗ ) independently of m as vm  L 2 (0,t f ;H 1 (]0,l[)) is bounded independently of m. Thus, there exists a subsequence L   vm l˜ of the sequence (vm )m≥1 which converges weakly to some element v in ˜ l≥1

L 2 (0, t f ; HL1 (]0, l[)), weakly star in L ∞ (0, t f ; L 2 (]0, l[)) and such that their derivatives

123

Radiative heating of a glass plate: the semi-discrete…

  v˙m l˜

˜ l≥1

converge weakly to v˙ in L 2 (0, t f ; HL1 (]0, l[)∗ ). Let ξ be an arbitrary function

belonging to the space L 2 (]0, t f [). ∀ j ∈ N∗ : t f

t f ˙ wj c p m g v(t),

c p m g (v˙m l˜ (t)|w j ) L 2 (]0,l[) ξ(t)dt → 0

! HL1 (]0,l[)∗ ,HL1 (]0,l[)

ξ(t)dt

0

as l˜ → +∞ as the mapping HL1 (]0, l[)∗ )

L (0, t f ; 2

t f c p m g χ(t), w j

→ R : χ →

! HL1 (]0,l[)∗ ,HL1 (]0,l[)

ξ(t)dt

0

is a continuous linear form on L 2 (0, t f ; HL1 (]0, l[)∗ ). t f

t f kh (vm l˜ (t)|w j ) H 1 (]0,l[) ξ(t)dt →

kh (v(t)|w j ) H 1 (]0,l[) ξ(t)dt

L

0

L

0

converges weakly to v in L 2 (0, t f ; HL1 (]0, l[)) and the mapping as the sequence (vm l˜ )l≥1 ˜ t f L (0, t f ; 2

HL1 (]0, l[))

→ R : χ →

kh (χ(t)|w j ) H 1 (]0,l[) ξ(t)dt L

0

converges is a continuous linear form on L 2 (0, t f ; HL1 (]0, l[)). As the sequence (vm l˜ )l≥1 ˜ 1 2 weakly to v in L (0, t f ; HL (]0, l[)), and the mapping L 2 (0, t f ; HL1 (]0, l[)) → L 2 (]0, t f [) : w  → w(., l) is linear and continuous, it follows that the sequence of functions vm l˜ (·)(l) converge weakly to v(·)(l) in L 2 (]0, t f [) as l˜ → +∞. This implies that t f t f − h c vm l˜ (t)(l)w j (l)ξ(t)dt → − h c v(t)(l)w j (l)ξ(t)dt 0

0

as l˜ → +∞. Let us show that: ⎞ ⎛ t t f  ⎝ −  vm l˜ (s)(l)ds + T0 (l)⎠ vm l˜ (t)(l)w j (l)ξ(t)dt → 0

0

⎛ t ⎞ t f  −  ⎝ v(s)(l)ds + T0 (l)⎠ v(t)(l)w j (l)ξ(t)dt 0

0

as l˜ → +∞. As for every t ∈]0, t f [: δl ⊗ χ]0,t[ ∈ HL1 (]0, l[)∗ ⊗ L 2 (]0, t f [)∗  L 2 (0, t f ; HL1 (]0, l[))∗

123

L. Paquet

  and as the sequence vm l˜ t 0

˜ l≥1

converges weakly to v in L 2 (0, t f ; HL1 (]0, l[)), it follows that

t vm l˜ (s)(l)ds → v(s)(l)ds as l˜ → +∞, ∀t ∈]0, t f [. Thus 0

⎛ t ⎛ t ⎞ ⎞    ⎝ vm l˜ (s)(l)ds + T0 (l)⎠ →  ⎝ v(s)(l)ds + T0 (l)⎠ 0

0

as l˜ → +∞, ∀t ∈]0, t f [. The sequence of functions vm l˜ (·)(l) converging weakly to v(·)(l) in L 2 (]0, t f [) as l˜ → +∞ is bounded in L 2 (]0, t f [). Thus:

⎞

t f ⎛ t



⎝ vm l˜ (s)(l)ds + T0 (l)⎠ vm l˜ (t)(l)w j (l)ξ(t)dt



0 0

⎞ ⎛ t

t f 



⎝ −  v(s)(l)ds + T0 (l)⎠ vm l˜ (t)(l)w j (l)ξ(t)dt



0 0    



  ≤ w j (l) sup vm l˜ (·)(l) 2 L (]0,t f [)

˜ l≥1

⎤1/2

⎛ t ⎞ ⎞ 2 ⎛ t

t f

 



⎥ ⎢ · ⎣

 ⎝ vm l˜ (s)(l)ds + T0 (l)⎠ −  ⎝ v(s)(l)ds + T0 (l)⎠

|ξ(t)|2 dt ⎦ ,



⎡

0

0

0

(22) → 0 as l˜ → +∞ by Lebesgue bounded convergence theorem. As  (·) is bounded, the function ⎛ t ⎞  ]0, t f [→ R : t  →  ⎝ v(s)(l)ds + T0 (l)⎠ 0

is bounded and thus the function

⎞ ⎛ t  ]0, t f [→ R : t  →  ⎝ v(s)(l)ds + T0 (l)⎠ ξ(t) 0

belongs to L 2 (]0, t f [). As the sequence of functions vm l˜ (·)(l) converges weakly to v(·)(l) in L 2 (]0, t f [) as l˜ → +∞, we have also that t f

⎛ ⎝

t

 0

v(s)(l)ds + T0 (l)⎠ ξ(t)vm l˜ (t)(l))dt

0

t f → 0

123

⎞

⎞ ⎛ t   ⎝ v(s)(l)ds + T0 (l)⎠ ξ(t)v(t)(l))dt 0

(23)

Radiative heating of a glass plate: the semi-discrete…

as l˜ → +∞. In conclusion, from (22) and (23) follows that: t f

⎞ ⎛ t   ⎝ vm ˜ (s)(l)ds + T0 (l)⎠ vm ˜ (t)(l)w j (l)ξ(t)dt

0

0

t f → 0

l

l

⎛

⎞

t

 ⎝ v(s)(l)ds + T0 (l)⎠ v(t)(l)w j (l)ξ(t)dt, 0

as l˜ → +∞. We are now in a position to pass to the limit as l˜ → +∞ in the first equation of (14) and we obtain: t f c p m g v(t), ˙ wj

t f

! HL1 (]0,l[)∗ ,HL1 (]0,l[)

ξ(t)dt +

0

kh (v(t)|w j ) H 1 (]0,l[) ξ(t)dt L

0

t f  = 0

∂f (·, t)|w j ∂t



t f L 2 (]0,l[)

ξ(t)dt −

h c v(t)(l)w j (l)ξ(t)dt 0

⎞ ⎛ t t f  t f ⎝ v(s)(l)ds + T0 (l)⎠ v(t)(l)w j (l)ξ(t)dt +  (u(t))u(t)w ˙ −  j (l)ξ(t)dt, 0

0

0

(24) converges weakly to v in ∀ j ∈ N∗ and ∀ξ ∈ L 2 (]0, t f [). As the sequence (vm l˜ )l≥1 ˜ L 2 (0, t f ; HL1 (]0, l[)), and their derivatives (v˙m l˜ )l≥1 weakly to v ˙ in ˜ 1 2 ∗ converges weakly to v(·, 0) in L 2 (]0, l[). L (0, t f ; HL (]0, l[) ), the sequence (vm l˜ (·, 0))l≥1 ˜ From, the second equation of (14) follows that v(., 0) = c p1m g ( f (·, 0) + kh T0 ). This fact and (24) prove the existence of at least one weak solution to the initial boundary value problem (12).   Lemma 6 The initial boundary value problem (12) possesses one and only one weak solution. Proof The existence of a solution has been proved in the preceding lemma. Thus it remains to prove uniqueness. Let us consider two weak solutions v and vˆ of the initial boundary value problem (12). Thus v and vˆ satisfies (13). In particular v(x, 0) = v(x, ˆ 0), ∀ x ∈]0, l[. We must prove that v = v. ˆ Considering v − v, ˆ we have: l  ! ∂ v − vˆ d v(·, t) − v(·, ˆ t), ϕ H 1 (]0,l[)∗ ,H 1 (]0,l[) = −kh cpmg (x, t)ϕ  (x)d x L L dt ∂x 0  ˆ t) ϕ(l) − h c v(l, t) − v(l, ⎡ ⎛ t ⎞ ⎞ ⎤ ⎛ t   − ⎣ ⎝ v(l, s)ds + T0 (l)⎠ v(l, t) −  ⎝ v(l, ˆ s)ds + T0 (l)⎠ v(l, ˆ t)⎦ ϕ(l). 0

0

123

L. Paquet

Taking ϕ = v(·, t) − v(·, ˆ t) for some fixed t ∈]0, t f [ in the previous equation, we obtain: 2 l    2 ∂ v − vˆ ˆ t) (v(x, t) − v(x, ˆ t)) d x + kh (x, t) d x + h c v(l, t) − v(l, ∂x 0 0 ⎛ t ⎞   2 = − ⎝ v(l, s)ds + T0 (l)⎠ v(l, t) − v(l, ˆ t)

cpmg d 2 dt

l

2

0

⎞ ⎛ t   − ⎝ v(l, s)ds + T0 (l)⎠ v(l, ˆ t) v(l, t) − v(l, ˆ t) 0

⎛ t ⎞   ˆ s)ds + T0 (l)⎠ v(l, ˆ t) v(l, t) − v(l, ˆ t) . + ⎝ v(l,

(25)

0

This equality can be rewritten: 2 l    2 ∂ v − vˆ ˆ t) (v(x, t) − v(x, ˆ t)) d x + kh (x, t) d x + h c v(l, t) − v(l, ∂x 0 0 ⎛ t ⎞   2 + ⎝ v(l, s)ds + T0 (l)⎠ v(l, t) − v(l, ˆ t)

cpmg d 2 dt

l

2

0

⎛ t ⎛ t ⎞ ⎞⎤    = ⎣ ⎝ v(l, ˆ s)ds + T0 (l)⎠ −  ⎝ v(l, s)ds + T0 (l)⎠⎦ v(l, ˆ t) v(l, t) − v(l, ˆ t) . ⎡

0

0

Now, it is easy to see that  is a Lipschitz function as  is bounded. Denoting by C the Lipschitz constant of  , we have: ⎡ ⎛ t ⎛ t ⎞ ⎞⎤    ⎣ ⎝ v(l, ˆ s)ds + T0 (l)⎠ −  ⎝ v(l, s)ds + T0 (l)⎠⎦ v(l, ˆ t) v(l, t) − v(l, ˆ t) 0

0

t











ˆ t) v(l, t) − v(l, ˆ t) . ≤ C (v(l, ˆ s) − v(l, s))ds

v(l,



0

Setting

C





= C v(l, ˆ t) , a = v(l, t) − v(l, ˆ t) , b =



t



ˆ s) − v(l, s))ds and applying

(v(l,

0

Young’s inequality: ab ≤ we have that:



t









ˆ t) v(l, t) − v(l, ˆ t) = C  ab ˆ s) − v(l, s))ds

v(l, C (v(l,



0

123

hc 2 C 2 a + b , 2C  2h c

Radiative heating of a glass plate: the semi-discrete…

⎞2 ⎛ t  2 (C  )2 hc  ⎝ (v(l, v(l, t) − v(l, ˆ t) + ≤ ˆ s) − v(l, s))ds ⎠ . 2 2h c 0

Thus: 2 l   2 ∂ v − vˆ hc  (v(x, t) − v(x, ˆ t)) d x + kh (x, t) d x + v(l, t) − v(l, ˆ t) ∂x 2 0 0 ⎞ ⎛ t   2 + ⎝ v(l, s)ds + T0 (l)⎠ v(l, t) − v(l, ˆ t)

cpmg d 2 dt

l

2

0

⎛ t ⎞2  t (C  )2 ⎝ (C  )2 ⎠ ≤ (v(l, ˆ s) − v(l, s))ds ≤ t (v(l, ˆ s) − v(l, s))2 ds. 2h c 2h c 0

0

Let us set ψ(t) =

t 0

cpmg d 2 dt

(v(l, ˆ s) − v(l, s))2 ds.

We have thus a fortiori the inequality:

l (v(x, t) − v(x, ˆ t))2 d x + 0

hc  C 2t ψ (t) ≤ v(l, ˆ t)2 ψ(t). 2 2h c

Let us integrate both sides of this inequality from 0 to t˜ for an arbitrary t˜ ∈]0, t f [. We obtain: cpmg 2

l

(v(x, t˜) − v(x, ˆ t˜))2 d x +

0

C2 hc ψ(t˜) ≤ 2 2h c

 t˜ t v(l, ˆ t)2 ψ(t)dt. 0

  t˜ ˆ t)2 ψ(t)dt, we obtain: Setting η t˜ = 0 t v(l, cpmg 2

l

(v(x, t˜) − v(x, ˆ t˜))2 d x +

0

C2  hc ψ(t˜) ≤ η t˜ . 2 2h c

A fortiori: hc C2  η t˜ , ∀t˜ ∈]0, t f [. ψ(t˜) ≤ 2 2h c Thus:  C2  ˆ t˜)2 η t˜ , ∀t˜ ∈]0, t f [. η t˜ ≤ 2 t˜v(l, hc   2 ˆ t˜)2 , we obtain η t˜ ≤ g(t˜)η t˜ , ∀t˜ ∈]0, t f [. We have also Setting g(t˜) := Ch 2 t˜v(l, c  η (0) = 0. Thus by Gronwall’s inequality [6, (ii), p. 624]: η t˜ = 0, ∀t˜ ∈]0, t f [ i.e.  t˜ t ˆ t)2 ψ(t)dt = 0. But ψ(t) = 0 (v(l, ˆ s) − v(l, s))2 ds. Thus: 0 t v(l, t v(l, ˆ t)

2

(v(l, ˆ s) − v(l, s))2 ds = 0, ∀ t ∈]0, t f [.

(26)

0

123

L. Paquet

By symmetry, we have also that: t v(l, t)

2

(v(l, ˆ s) − v(l, s))2 ds = 0, ∀ t ∈]0, t f [.

(27)

0

We want to prove that this implies v(l, ˆ s) = v(l, s), ∀ s ∈ [0, t f ].

(28)

Let us consider: t

t¯ := sup{t ∈ [0, t f ];

(v(l, ˆ s) − v(l, s))2 ds = 0}. 0

By continuity, we have of course that:  t¯ (v(l, ˆ s) − v(l, s))2 ds = 0. 0

∀ s

∈ [0, t¯], we have thus v(l, ˆ s) = v(l, s). If t¯ = t f , then (28) is proved. If t¯ < t f , t ˆ s) − v(l, s))2 ds > 0. By (26), (27): v(l, ˆ t) = 0 = v(l, t), ∀ t ∈]t¯, t f [. ∀t ∈]t¯, t f [: 0 (v(l, Thus (28) is also proved in that case. From (28) and (25) follows that ∀ t ∈]0, t f [: cpmg d 2 dt

l

2 l   ∂ v − vˆ (x, t) d x = 0. (v(x, t) − v(x, ˆ t)) d x + kh ∂x 2

0

0

The second term in the left-hand side of the previous equation is of course positive, so that:  d l ˆ t))2 d x ≤ 0, ∀ t ∈]0, t f [. Thus the absolutely continuous function: dt 0 (v(x, t) − v(x, l [0, t f ] → R+ : t  →

(v(x, t) − v(x, ˆ t))2 d x 0

is decreasing and null at t = 0. Being positive, this function is null ∀t ∈ [0, t f ] so that v(·, t) = v(·, ˆ t) in L 2 (]0, l[), ∀t ∈ [0, t f ]. Thus uniqueness is also proved.   Proposition 7 Under the regularity condition on the initial condition T0 that T0 ∈ H 2 (]0, l[), the compatibility condition T0 (0) = Ta at x = 0, and the compatibility condition between the initial condition and the boundary condition (29)(ii) at x = l: −kh T0 (l) = h c (T0 (l) − Ta ) + (T0 (l)) − (u(0)), the unique weak solution Tu of the initial boundary value problem (2) i.e. ⎧ 2 c p m g ∂∂tT (x, t) = kh ∂∂ xT2 (x, t) + ψ(T (x, t)) + h T,u (x, t), ⎪ ⎪ ⎪ ⎪ ⎪ 0 < x < l, 0 < t < t f , ⎪ ⎨ −kh ∂∂Tx (l, t) = h c (T (l, t) − Ta ) + ((T (l, t)) − (u(t))), ⎪ 0 < t < t f , (B.C. at x = l), ⎪ ⎪ ⎪ ⎪ T (0, t) = T ⎪ a , 0 < t < t f , (B.C. at x = 0), ⎩ T (x, 0) = T0 (x), ∀x ∈ [0, l], (I.C. at t = 0), satisfies the regularity property:

123

dTu dt

∈ L 2 (]0, t f [; H 1 (]0, l[)).

(29)

Radiative heating of a glass plate: the semi-discrete…

Proof Let us consider the weak solution v ∈ {w ∈ L 2 (0, t f ; HL1 (]0, l[)); w˙ ∈ L 2 (0, t f ; 1

∗ HL (]0, l[) )} of the initial boundary value problem (12); thus by (13) ∀ϕ ∈ HL1 (]0, l[) : ⎧ l ∂v l ⎪ ⎪ d  (x)d x + ∂ f (x, t)ϕ(x)d x ⎪ v(·, c t), ϕ m (x, t)ϕ 1 (]0,l[)∗ ,H 1 (]0,l[) = −k h p g ⎪ H dt ∂x ∂t ⎪ L L ⎪ 0 ⎨ 0  t  v(l, s)ds + T0 (l) v(l, t)ϕ(l) +  (u(t))u(t)ϕ(l) ˙ − h c v(l, t)ϕ(l) −  ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎩ and v(x, 0) = 1 ( f (x, 0) + kh T  (x)), ∀ x ∈]0, l[, 0

cpmg

(30) where f denotes the function f : Q → R : (x, t)  → ψ(Tu (x, t)) + h Tu ,u (x, t) ( 10). We know by the two preceding lemmas, that this initial boundary value problem possesses one and only one weak solution. Let us set: t T (x, t) :=

v(x, s)ds + T0 (x). 0

This implies that v(x, t) = ∂∂tT (x, t). Let us integrate both sides of (30) from 0 to t; we obtain: ⎧ l l l ⎪ c p m g 0 ∂∂tT (x, t)ϕ(x)d x − 0 f (x, 0)ϕ(x)d x + 0 kh T0 (x)ϕ  (x)d x − kh T0 (l)ϕ(l) ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ = −kh  l ∂ T (x, t)ϕ  (x)d x + kh  l T  (x)ϕ  (x)d x +  l f (x, t)ϕ(x)d x 0 ∂x

0 0

0

l ⎪ ⎪ − 0 f (x, 0)ϕ(x)d x − h c T (l, t)ϕ(l) + h c T0 (l)ϕ(l) ⎪ ⎪ ⎪ ⎪ ⎩ − (T (l, t))ϕ(l) + (T0 (l))ϕ(l) + (u(t))ϕ(l) − (u(0))ϕ(l).

Using the compatibility condition: −kh T0 (l) = h c (T0 (l) − Ta ) + (T0 (l)) − (u(0)), and f (x, t) = ψ(Tu (x, t)) + h Tu ,u (x, t), we obtain after simplifications ∀t ∈]0, t f [: ⎧ l l l c p m g 0 ∂∂tT (x, t)ϕ(x)d x + kh 0 ∂∂Tx (x, t)ϕ  (x)d x − 0 ψ(Tu (x, t))ϕ(x)d x ⎪ ⎪ ⎪ ⎪ l ⎪ ⎪ ⎪ ⎪ − 0 h Tu ,u (x, t)ϕ(x)d x − [(u(t)) − (T (l, t))] ϕ(l) ⎨ − [h c (Ta − T (l, t))] ϕ(l) = 0, ∀ϕ ∈ HL1 (]0, l[) ⎪ ⎪ ⎪ ⎪ ⎪ T (0, t) = Ta , 0 < t < t f , (B.C. at x = 0), ⎪ ⎪ ⎪ ⎩ T (x, 0) = T0 (x), ∀x ∈ [0, l], (I.C. at t = 0).

(31)

By (7) and (31), we have ∀t ∈]0, t f [: l ⎧ d Tu (·, t) − T (·, t), ϕ H 1 (]0,l[)∗ ,H 1 (]0,l[) + kh 0 ∂(T∂u x−T ) (x, t)ϕ  (x)d x c p m g dt ⎪ ⎪ L L ⎪ ⎪ ⎨ + [(Tu (l, t)) − (T (l, t))] ϕ(l) + [h c (Tu (l, t) − T (l, t))] ϕ(l) = 0, ∀ϕ ∈ HL1 (]0, l[) ⎪ ⎪ ⎪ ⎪ ⎩ (Tu − T ) (0, t) = 0, 0 < t < t f , (B.C. at x = 0), (Tu − T ) (x, 0) = 0, ∀ x ∈ [0, l], (I.C. at t = 0).

(32)

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L. Paquet

Considering ϕ = Tu (·, t) − T (·, t) ∈ HL1 (]0, l[) for some fixed t ∈]0, t f [ in (32), we obtain: d dt

l

(Tu (x, t) − T (x, t))2 d x ≤ 0, ∀ t ∈]0, t f [.

0

Thus, the absolutely continuous function l t →

(Tu (x, t) − T (x, t))2 d x 0

is a positive decreasing function with respect to the variable t. As this function takes the dT u value 0 for t = 0, this function is zero. Thus Tu = T which implies dT dt = dt = v ∈ 2 1   L (]0, t f [; H (]0, l[)). What was to be proved. Corollary 8 If we suppose moreover that T0 ∈ H 3 (]0, l[), then

dTu dt

∈ H 2,1 (Q).

Proof We keep the same notations as in the proof of the preceding proposition. By using formulas (10), (3) and (4) with t = 0, deriving with respect to x, using the fact that Bgk (T0 ) ∈ H 1 (]0, l[) and decomposing that function as the sum of an affine function defined on the interval ]0, l[ and a function in H˚ 1 (]0, l[), function that we extend by 0 as a function on the whole real line and using properties of the convolution operator, one can prove that d f (.,0) ∈ L 2 (]0, l[). Consequently c p1m g ( f (., 0) + kh T0 ) ∈ H 1 (]0, l[). As ∂∂tf ∈ L 2 (Q), by dx Lemma 4, ⎛ t ⎞  t  → h c v(l, t) +  ⎝ v(l, s)ds + T0 (l)⎠ v(l, t) −  (u(t))u  (t) 0

L 2 (]0, t

belongs to f [). Thus, the initial boundary value problem (12) is of the form (9) with u Ta replaced by 0 in the boundary condition at x = 0. Thus by Proposition 3, dT dt = v solution  of (30), i.e. weak solution of the initial boundary value problem (12), belongs to H 2,1 (Q). 

3 The semi-discrete problem In this section, we will suppose that the hypotheses of Corollary 8 are satisfied i.e. that T0 ∈ H 3 (]0, l[), that T0 (0) = Ta , and that −kh T0 (l) = h c (T0 (l) − Ta ) + (T0 (l)) − (u(0)). 2,1 (Q). In particular T and dTu belong to u This implies that Tu and dT u dt belong to H dt L 2 (]0, t f [; H 2 (]0, l[)) implying that Tu ∈ C([0, t f ]; H 2 (]0, l[)). Let us now define what is the semi-discrete problem corresponding to the exact problem (2). Given Nh ∈ N∗ , let us define the mesh x0 = 0 < x1 < x2 < · · · < x j < · · · < x Nh = l on the interval [0, l]. By ϕi (i = 0, . . . , Nh ), let us denote the continuous function on the interval [0, l], affine on each subinterval [x j−1 , x j ] ( j = 1, . . . , Nh ) which takes the value 1 at the node xi and 0 at all the other nodes x j , j  = i. Thus: ( x1 −x if x ∈ [x0 , x1 ], (33) ϕ0 (x) := x1 −x0 0 if x ∈]x1 , x Nh ],

123

Radiative heating of a glass plate: the semi-discrete…

ϕi (x) :=

for i = 1, . . . , Nh − 1, and

⎧ 0 ⎪ ⎪ ⎪ ⎨ x−xi−1

xi −xi−1 x −x

i+1 ⎪ ⎪ x −x ⎪ ⎩ 0i+1 i

)

ϕ Nh (x) :=

if x if x if x if x

∈ [0, xi−1 [, ∈ [xi−1 , xi [, ∈ [xi , xi+1 ], ∈]xi+1 , x Nh ],

(34)

if x ∈ [x0 , x Nh −1 ], if x ∈ [x Nh −1 , x Nh ].

0 x−x Nh −1 x Nh −x Nh −1

(35)

Every function ϕi (i = 0, . . . , Nh ) belongs to the space H 1 (]0, l[). Let us denote by X h the finite dimensional vector subspace of H 1 (]0, l[) generated by the functions ϕi (i = 0, . . . , Nh ). The semi-discrete problem is the following: find Tu,h (x j , ·) ∈ C 1 ([0, t f ]; R),

j = 1, . . . , Nh

satisfying the initial conditions Tu,h (x j , 0) = T0 (x j ), ∀ j = 1, . . . , Nh

(36)

such that the function Tu,h : [0, t f ] → X h : t  → Tu,h (·, t) = Ta ϕ0 +

Nh 

Tu,h (x j , t)ϕ j

(37)

j=1

satisfies on the time interval [0, t f ] the system of differential equations: cpmg

Nh  dTu,h j=1

Nh 

+

j=1

dt

l l (x j , t) ϕ j (x)ϕk (x)d x + Ta kh ϕ0 (x)ϕk (x)d x 0

l Tu,h (x j , t)kh

ϕ j (x)ϕk (x)d x −

0

Nh  + Tu,h (x j , t)ϕ j (x))ϕk (x)d x − j=1

0

l ψ(Ta ϕ0 (x) 0

l h 0

Ta ϕ0 +

Nh 

Tu,h (x j ,t)ϕ j ,u

(x, t)ϕk (x)d x

j=1

+ [(Tu,h (l, t)) − (u(t))]δk,Nh  + h c Tu,h (l, t) − Ta δk,Nh = 0, ∀k = 1, . . . , Nh .

(38)

A more “functional analytic” equivalent definition traced on the definition of what is a weak solution (7) of the exact problem (2) is given below by formula (47). Let us set α h (t) = (Tu,h (x1 , t), . . . , Tu,h (x Nh , t)).

(39)

Let us define the square matrix of order Nh , M h = (m hj,k )1≤ j,k≤Nh by m hj,k = l h h 0 ϕ j (x)ϕk (x)d x, ∀ j, k = 1, . . . , Nh and the related matrix A = c p m g M . We will h h need also in the following, the square matrix B = (b j,k )1≤ j,k≤Nh of order Nh by l bhj,k = kh 0 ϕ j (x)ϕk (x)d x, ∀ j, k = 1, . . . , Nh . These three square matrices of order Nh are tridiagonal, symmetric, positive definite and thus invertible. In the particular case of the

123

L. Paquet

equidistributed grid x j = j h, j = 0, 1, . . . , Nh on the interval [0, l] with h = computation shows that h = m i,i

l Nh ,

an easy

2h h h h h , m i,i+1 = m i+1,i = , ∀i = 1, . . . , Nh − 1, m hNh ,Nh = 3 6 3

and 2kh h −kh kh h , bi,i+1 = bi+1,i , ∀i = 1, . . . , Nh − 1, bhNh ,Nh = . = h h h Now, let us introduce the nonlinear function h = bi,i

* f h (α h ) = ( * f 1h (α h ), . . . , * f Nhh (α h )) f h : R Nh → R Nh : α h = (α1 , . . . , α Nh )  → * defined as follows: * f kh (α h ) =

l ψ(Ta ϕ0 (x) + l

M  k ∗ =1

+

α j ϕ j (x))ϕk (x)d x

j=1

0

+

Nh 

2π κk ∗ 0

l

M  k ∗ =1

2π κk2∗

⎢ ⎣ ⎡

l

∗

G k ∗ (x, x  )Bgk (Ta ϕ0 (x  ) +

Nh 

⎤ ⎥ α j ϕ j (x  ))d x  ⎦ ϕk (x)d x

j=1

0

l

Nh

0

j=1

⎤

 ⎢ ⎥  k∗  α j ϕ j (x  ))d x  ⎦ ϕk (x)d x ⎣ 1 (κk ∗ (2l − x − x ))Bg (Ta ϕ0 (x ) +

0





⎡

− α Nh δk,Nh − h c α Nh δk,Nh , ∀k = 1, . . . , Nh .

(40)

Lemma 9 The nonlinear mapping * f h (α h ) f h : R Nh → R Nh : α h = (α1 , . . . , α Nh )  → * is lipschitzian. Proof By [11, p. 14], the first order derivative of the Planck function with respect to the absolute temperature T satisfies the bounds: 0 ≤ ∂∂ TB (T, λ) ≤ C2Cλ14 . This inequality implies 2  ∗  k  and  are bounded which in turn implies that the first that the first order derivatives Bg order partial derivatives that * f h is lipschitzian.

∂* fh ∂α j

( j = 1, . . . , Nh ) are also bounded. From this last fact follows  

We also define the nonlinear function * g h : R → R Nh (* g1 (u), . . . , * g Nh (u)) as follows: * gk (u) =

l

M 

∗

[E 2 (κk ∗ (l − x)) − 2 (κk ∗ (l − x))]ϕk (x)d x Bgk (u)

2πκk ∗

k ∗ =1

: u → * g h (u) =

0

+ (u) δk,Nh , ∀k = 1, . . . , Nh , and the vector

Ch

=

(ckh )1≤k≤Nh l

ckh

= −kh Ta 0

123

of

R Nh

ϕ0 (x))ϕk (x)d x

(41)

by +

M  k ∗ =1

l

∗ 2πκk ∗ Bgk (Ta )

E 2 (κk ∗ x)ϕk (x)d x 0

Radiative heating of a glass plate: the semi-discrete…

+

M 

k ∗ =1

l

2πκ

k∗

∗ Bgk (Ta )

2 (κk ∗ (2l − x))ϕk (x)d x + 0

+h c Ta δk,Nh , ∀k = 1, . . . , Nh .

(42)

It is easy to see that the nonlinear function * g h : R → R Nh is Lipschitz-continuous. Let us still introduce the initial condition, the vector T0h := (T0 (x j ))1≤ j≤Nh

(43)

R Nh .

Having introduced these notations, in particular (39) and (43), we may rewrite the of semi-discrete problem (38) in the following matrix differential form: ( h h A α˙ (t) + B h α h (t) = f˜h (α h (t)) + g˜ h (u(t)) + C h , ∀t ∈ [0, t f ], (44) α h (0) = T0h . Proposition 10 The nonlinear Cauchy problem (44) possesses one and only one solution. Proof Let us set F : [0, t f ] × R Nh → R Nh : (t, y)  → (Ah )−1 [−B h y + f˜h (y) + g˜ h (u(t)) + C h ]. F is a continuous function, which by Lemma 9 is moreover lipschitzian with respect to the y variable uniformly in t. Applying the Cauchy–Lipschitz Theorem [4, p. 65], the result follows.   As an immediate corollary, we have: Corollary 11 The semi-discrete problem defined by the conditions (36), (37) and (38) possesses one and only one solution. Let us set: Sh = span < ϕ1 , . . . , ϕ Nh > . It is a vectorial subspace of dimension Nh of Tu,h (t) = Ta +

HL1 (]0, l[).

(45)

Formula (37) may be rewritten:

Nh  (Tu,h (x j , t) − Ta )ϕ j ,

(46)

j=1

which shows us that ∀t ∈ [0, t f ], Tu,h (t) ∈ Ta + Sh . The semi-discrete problem defined by the conditions (36), (37) and (38) may be rewritten in the following form: find Tu,h ∈ C 1 ([0, t f ]; Ta + Sh ) such that ∀t ∈ [0, t f ]: ⎧  l ∂T  l ∂T  c p m g 0 ∂tu,h (x, t)χ(x)d x + kh 0 ∂u,h ⎪ x (x, t)χ (x)d x ⎪   ⎪ l l ⎪ ⎨ − 0 ψ(Tu,h (x, t))χ(x)d x − 0 h Tu,h (.,t),u (x, t)χ(x)d x

 + (Tu,h (l, t)) − (u(t)) χ(l) + h c Tu,h (l, t) − Ta χ(l) = 0, ∀χ ∈ Sh , (47) ⎪ ⎪ ⎪ ⎪ ⎩ Tu,h (0, t) = Ta , ∀t ∈ [0, t f ], Tu,h (., 0) = Ih (T0 ). In formula (47), Ih denotes the Lagrange interpolation operator at the Nh + 1 points x0 = 0, x1 , x2 , . . . , x Nh = l by continuous functions, which are polynomials of degree at most 1 in each interval ]x j , x j+1 [, ∀ j = 0, . . . , Nh − 1. Our purpose is now to establish, an a priori error estimate. In that purpose, let us firstly write the error equation that is the equation for the error e := Tu,h − Tu :

123

L. Paquet

Proposition 12 The error e := Tu,h − Tu on Tu,h satisfies for ∀ t ∈]0, t f [: ⎧ l d ⎪ e(·, t), χ H 1 (]0,l[)∗ ,H 1 (]0,l[) + kh 0 ∂∂ex (x, t)χ  (x)d x c p m g dt ⎪ L L ⎪ l ⎪ ⎪ ⎪ u (x, t) + e(x, t)) − ψ(Tu (x, t))] χ(x)d x ⎪ ⎨ −0l [ψ(T

− 0 h Tu (.,t)+e(.,t),u (x, t) − h Tu (.,t),u (x, t) χ(x)d x ⎪ ⎪ + [(Tu (l, t) + e(l, t)) − (Tu (l, t))] χ(l) + h c e(l, t)χ(l) = 0, ∀χ ∈ Sh , ⎪ ⎪ ⎪ ⎪ e(0, t) = 0, ∀t ∈ [0, t f ], ⎪ ⎩ e(., 0) = Ih (T0 ) − T0 .

(48)

Proof Equation (48)(i) follows by taking the difference between Eqs. (47)(i) and (7)(i) which ¯ which implies are both valid for every χ ∈ Sh . We have proved in [11], that Tu ∈ C( Q) that Tu (0, .) ∈ C([0, t f ]). Thus Tu (0, t) = Ta for all t ∈ [0, t f ]. Taking the difference between Eqs. (47)(ii) and (7)(ii) , we obtain e(0, t) = 0, ∀t ∈ [0, t f ] (not only for almost every t ∈ [0, t f ]). Finally, taking the difference between Eqs. (47)(iii) and (7)(iii) , we obtain e(., 0) = Ih (T0 ) − T0 (this equality being true at every point x ∈ [0, l], not only almost everywhere, as both sides are continuous functions).   For an arbitrary t fixed in the interval [0, t f ], let us introduce the elliptic projection T˜u,h (., t) of Tu (., t) defined by: T˜u,h (., t) ∈ Ta + Sh , and satisfies: l 0

 ∂ T˜u,h ∂ Tu (x, t)χ  (x)d x = (x, t)χ  (x)d x, ∀χ ∈ Sh . ∂x ∂x l

(49)

0

In particular T˜u,h (0, t) = Ta . Also Tu (0, t) = Ta . Thus Tu (., t) − Ta and T˜u,h (., t) − Ta both belong to HL1 (]0, l[). Equation (49) may be rewritten in the equivalent form:    l ∂ T˜u,h − Ta l ∂ (Tu − Ta )  (x, t)χ (x)d x = (x, t)χ  (x)d x, ∀χ ∈ Sh . (50) ∂x ∂x 0

0

Equation (50), shows us that T˜u,h (., t) − Ta is the orthogonal projection of Tu (., t) − Ta onto Sh ⊂ HL1 (]0, l[) for the scalar product of HL1 (]0, l[). As already said in the introduction, if we would have a Dirichlet boundary condition at the point x = l, then the technique of elliptic projection is well known [19, p. 7]. But in our case, we have the nonlinear Robin boundary condition (2)(ii) at the point x = l. The key property of our elliptic projection (49), which will allow us to prove optimal a priori error estimates is that T˜u,h (l, t) = Tu (l, t), a property that we now establish: Proposition 13 Equation (49) is equivalent to T˜u,h (l, t) = Tu (l, t) and l 0

 ∂ T˜u,h ∂ Tu (x, t)χ  (x)d x = (x, t)χ  (x)d x, ∀χ ∈ S˚h , ∂x ∂x l

0

where S˚h := {χ ∈ Sh ; χ(l) = 0}. Proof Let us consider the function χ :]0, l[→ R : x  →

123

x . l

Radiative heating of a glass plate: the semi-discrete…

This function χ ∈ Sh and χ(l) = 1. Also χ  (x) = function χ gives us:

1 l,

∀x ∈]0, l[. Equation (49) with that

1 ˜ 1 (Tu,h (l, t) − Ta ) = (Tu (l, t) − Ta ) l l and thus T˜u,h (l, t) = Tu (l, t). The equivalence with Eq. (49) follows from the fact that the finite dimensional !vectorial subspace Sh defined by ! formula (45) is also equal to span ϕ1 , . . . , ϕ Nh −1 , χ and also that span ϕ1 , . . . , ϕ Nh −1 = S˚h .   Remark 14 The above definition and properties of the elliptic projection of Tu (., t) extends to an arbitrary function T ∈ H 1 (]0, l[) by replacing Ta by T (0). In particular, we will speak u in the next corollary of the elliptic projection of dT dt (., t) for almost every t ∈ [0, t f ]; in that dTu case dt (., t)(0) = 0 and Ta must be replaced by 0. Now, we split the error e := Tu,h − Tu on Tu,h , into two terms: e =θ +ρ where:

(

(51)

θ = Tu,h − T˜u,h , ρ = T˜u,h − Tu .

(52)

θ verifies the following equations: Proposition 15 For almost every t ∈ [0, t f ], we have ∀χ ∈ Sh :  l ∂θ l ⎧  ⎪ ⎪ c p m g θt (·, t), χ HL1 (]0,l[)∗ ,HL1 (]0,l[) + kh 0 ∂ x (x, t)χ (x)d x − 0 [ψ(Tu,h (x, t)) ⎪ ⎪ l ⎪ ⎪ −ψ(T˜u,h (x, t))]χ(x)d x − + 0 [h Tu,h (.,t),u (x, t)− ⎪ , ⎪ ⎪ ⎪ ˜u,h (l, t)) χ(l) + h c θ (l, t)χ(l) h (x, t)]χ(x)d x + (T (l, t)) − ( T ⎪ u,h ˜ ⎪ T (.,t),u ⎪ l ⎨ u,h = −c p m g ρt (·, t), χ H 1 (]0,l[)∗ ,H 1 (]0,l[) + 0 [ψ(T˜u,h (x, t)) L L + , l ⎪ ⎪ ⎪ h (x, t))]χ(x)d x + (x, t) − h (x, t) χ(x)d x, −ψ(T u T (.,t),u ⎪ u T˜u,h (.,t),u 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ θ (0, t) = 0, ∀t ∈ [0, t f ], ⎪ ⎩ θ (., 0) = Tu,h (·, 0) − T˜u,h (·, 0).

(53)

Proof The first equation follows by splitting the terms in the error equation (48) using the l decomposition e = θ +ρ of the error e, from Eq. (49) which implies that 0 ∂∂ρx (x, t)χ  (x)d x = 0, ∀χ ∈ Sh and T˜u,h (l, t) = Tu (l, t). The second one follows from the fact that Tu,h − Ta ∈ C([0, t f ]; Sh ) and also T˜u,h − Ta due to the fact that Tu ∈ C([0, t f ]; H 1 (]0, l[)) and Eq. (50). Thus θ (0, t) = Tu,h (0, t) − T˜u,h (0, t) is null for every t ∈ [0, t f ]. The third equation has   sense due to the continuity of Tu,h and T˜u,h and is trivial. Corollary 16 For almost every t ∈]0, t f [, θ (., t) verifies the following inequality: cpmg d θ (·, t)2L 2 (]0,l[) + kh ∇x θ (·, t)2L 2 (]0,l[)2 + h c θ (l, t)2 ≤ c p m g 2 dt · ρt (·, t) L 2 (]0,l[) θ (·, t) L 2 (]0,l[) + (L h + L ψ ) ρ(·, t) L 2 (]0,l[) θ (·, t) L 2 (]0,l[) +L h θ (·, t)2L 2 (]0,l[) ,

(54)

where L ψ and L h denote the Lipschitz constants of ψ and of the mapping L 2 (]0, l[) → L 2 (]0, l[) : T  → h T,u .

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L. Paquet

   dTu dt (., t) and that being dTu d ˜ Tu,h (., t) belongs to the orthogonal projection of dt (., t) onto Sh , the mapping t  → dt d L 2 (]0, t f [; Sh ). Also, t  → dt Tu,h (., t) belongs to L 2 (]0, t f [; Sh ). Thus t  → dθ dt (., t) belongs dρ 2 2 1 to L (]0, t f [; Sh ). Also t  → dt (., t) belongs to L (]0, t f [; H (]0, l[)). Putting χ = θ (·, t) in equality (53), we obtain: l cpmg d ∂θ θ (·, t)2L 2 (]0,l[) + kh (x, t)2 d x + h c θ (l, t)2 2 dt ∂x Proof Firstly, let us observe that by Eq. (50),

d ˜ dt Tu,h (., t)

=

0

l + , + , ψ(Tu,h (x, t)) − ψ(T˜u,h (x, t)) θ (x, t)d x + (Tu,h (l, t)) − (T˜u,h (l, t)) − 0



l

 + ,



h Tu,h (.,t),u (x, t) − h T˜u,h (.,t),u (x, t) θ (x, t)d x

·θ (l, t) ≤



0

l

 +

,



+

h T˜u,h (.,t),u (x, t) − h Tu (.,t),u (x, t) θ (x, t)d x



0

+ c p m g ρt (·, t) L 2 (]0,l[) θ (·, t) L 2 (]0,l[)

l

 +

,



ψ(T˜u,h (x, t)) − ψ(Tu (x, t)) θ (x, t)d x

. +





(55)

0

 k Now, ψ(T ) := − k=M k=1 4πκk Bg (T ) and thus the mapping T  −→ −ψ(T ) is an increasing function , + , + of T . Thus − ψ(Tu,h (x, t)) − ψ(T˜u,h (x, t)) θ (x, t) = (−ψ)(Tu,h (x, t)) − (−ψ)(T˜u,h (x, t)) · (Tu,h (x, t) − T˜u,h (x, t))  +∞ is always positive. In the same way, as (T ) := π λ0 λ B(T, λ)dλ, it is also an increasing function of T , which implies the positivity of the term + , + , (Tu,h (l, t)) − (T˜u,h (l, t)) θ (l, t) = (Tu,h (l, t)) − (T˜u,h (l, t)) · (Tu,h (l, t) − T˜u,h (l, t)). We are thus allowed to drop the last two terms in the left hand side of inequality (55). Now by Corollary 3.7 page 14 and Lemma 3.18 page 22 of [11] and formula (4), it follows that the mapping L 2 (]0, l[) → L 2 (]0, l[) : T  → h T,u is lipschitzian. Calling its Lipschitz constant L h , it follows that:

l

 +

,





h (x, t) − h (x, t) θ (x, t)d x Tu,h (.,t),u T˜u,h (.,t),u





0     θ (·, t) L 2 (]0,l[) ≤ h Tu,h (.,t),u (·, t) − h T˜u,h (.,t),u (·, t) 2 L (]0,l[)     θ (·, t) L 2 (]0,l[) = L h θ (·, t)2L 2 (]0,l[) . (56) ≤ L h Tu,h (·, t) − T˜u,h (·, t) 2 L (]0,l[)

123

Radiative heating of a glass plate: the semi-discrete…

Proceeding analogously, we also have the bound:

l

 +

,



h T˜u,h (.,t),u (x, t) − h Tu (.,t),u (x, t) θ (x, t)d x





0

≤ L h ρ(·, t) L 2 (]0,l[) θ (·, t) L 2 (]0,l[) .

(57)

By Lemma 3.6 page 13 of [11] and formula (3), it follows that the mapping R → R : T  −→ ψ(T ) is Lipschitzian. Calling its Lipschitz constant L ψ , it follows that:

l

 +

,



˜ ψ(Tu,h (x, t)) − ψ(Tu (x, t)) θ (x, t)d x

≤ L ψ ρ(·, t) L 2 (]0,l[) θ (·, t) L 2 (]0,l[) .



0 (58) From (55) and inequalities (56) to (58), follows inequality (54). What was to be proved.   We can not a priori absorb the term L h θ (·, t)2L 2 (]0,l[) in the right-hand side of inequality (54) of our previous corollary, by a fraction of a term in the left-hand side. To remedy to that, we multiply both sides of inequality (54), by exp(2λt) for λ = − c pLmh g . Let us set: ˜ t) := exp(λt)θ (·, t) and ρ(·, θ(·, ˜ t) := exp(λt)ρ(·, t).

(59)

Proposition 17 The following bound holds: 2   ˜ t) 2 c p m g θ(·,

L (]0,l[)

t  2   + kh ∇x θ˜ (·, s) 2

L (]0,l[)2

0

2    ≤ c p m g θ˜ (·, 0)

L 2 (]0,l[)

+2

2

(L b C P O ) kh

t



+2

cpmgCP O kh

ds + h c

θ˜ (l, s)2 ds

0

2  t

ρ˜t (·, s)2L 2 (]0,l[) ds 0

t

ρ(·, ˜ s)2L 2 (]0,l[) ds,

(60)

0

where L b := 2L h + L ψ . Proof Multiplying both sides of inequality (54) by exp(2λt), and using exp(λt)θt (·, t) = ∂t (exp(λt)θ (·, t)) − λ exp(λt)θ (·, t), as well as exp(λt)ρt (·, t) = ∂t (exp(λt)ρ(·, t)) − λ exp(λt)ρ(·, t), we obtain:

2 2  cpmg d     ˜ + kh ∇x θ˜ (·, t) 2 + h c θ˜ (l, t)2 θ (·, t) 2 L (]0,l[) L (]0,l[)2 2 dt    ˜ t) ≤ c p m g ρ˜t (·, t) L 2 (]0,l[) θ(·, + (L h + L ψ + |λ| c p m g ) ρ(·, ˜ t) L 2 (]0,l[)     · θ˜ (·, t)

L 2 (]0,l[)

L 2 (]0,l[)

2   ˜ t) 2 + (L h + λc p m g ) θ(·,

L (]0,l[)

.

123

L. Paquet

Taking λ = − c pLmh g , we have thus L h + λc p m g = 0. The previous inequality becomes: 2 2  cpmg d  ˜    ˜ t)2 + kh ∇x θ˜ (·, t) 2 + h c θ(l, θ (·, t) 2 L (]0,l[) L (]0,l[)2 2 dt         ˜ t) L 2 (]0,l[) θ˜ (·, t) ≤ c p m g ρ˜t (·,t) L 2 (]0,l[) θ˜ (·, t) + L b ρ(·, L 2 (]0,l[)

L 2 (]0,l[)

. (61)

Integrating both sides of the previous inequality from 0 to t, we obtain: t  t 2 2 cpmg     ˜ 2 + kh ∇x θ˜ (·, s) 2 ds + h θ (·, t) 2 c θ˜ (l, s) ds L (]0,l[) L (]0,l[)2 2 0

0

t  2  cpmg     ˜ + c p m g ρ˜t (·, s) L 2 (]0,l[) θ˜ (·, s) 2 ds ≤ θ (·, 0) 2 L (]0,l[) L (]0,l[) 2 0

t +L b

   ˜ ρ(·, s) ˜ s) L 2 (]0,l[) θ(·,

L 2 (]0,l[)

ds.

(62)

0 2

Now, let us apply “Cauchy’s inequality with δ”: ∀a ∈ R+ , ∀b ∈ R+ , ∀δ ∈ R∗+ : ab ≤ δa 2 + b4δ to bound the products in the right-hand side of the previous inequality. Firstly, let us denote by C P O the Poincaré constant i.e. ∀ϕ ∈ HL1 (]0, l[) : ϕ L 2 (]0,l[) ≤ C P O ∇x ϕ L 2 (]0,l[)2 (it is easy to see that C P O = π2 l). (1◦ ) ∀ε > 0 : t

    ρ˜t (·, s) L 2 (]0,l[) θ˜ (·, s)

cpmg 0

⎛

≤ cpmg ⎝

t

L 2 (]0,l[)

ds

⎞1/2 ⎛ t   2  ˜ 2 ⎠ ⎝ ρ˜t (·, s) L 2 (]0,l[) ds θ (·, s) 2

L (]0,l[)

0

L

2  t  cpmg ρ˜t (·, s)2L 2 (]0,l[) ds ds + (]0,l[) 4ε

0

0

t  2   ≤ εC 2P O ∇x θ˜ (·, s) 2 L

 2  t cpmg ρ˜t (·, s)2L 2 (]0,l[) ds. ds + (]0,l[)2 4ε

0 kh . 4C 2P O

t cpmg

0

Thus:

    ρ˜t (·, s) L 2 (]0,l[) θ˜ (·, s)

0

kh ≤ 4

t  2   ˜ ∇x θ(·, s) 2 L

0

123

ds ⎠

0

t  2   ≤ ε θ˜ (·, s) 2

Let us take ε =

⎞1/2

L 2 (]0,l[)



ds

cpmgCP O ds + (]0,l[)2 kh

2  t ρ˜t (·, s)2L 2 (]0,l[) ds. 0

(63)

Radiative heating of a glass plate: the semi-discrete…

(2◦ ) ∀˜ε > 0 : t Lb

    ρ(·, ˜ s) L 2 (]0,l[) θ˜ (·, s)

0

⎛

≤ Lb ⎝

t

L 2 (]0,l[)

ds

⎞1/2 ⎛ t   2  ⎝  θ˜ (·, s) 2

ρ(·, ˜ s)2L 2 (]0,l[) ds ⎠

L (]0,l[)

0 t 2

L 2 (]0,l[)

ds +

0

t  2   2 ≤ ε˜ C P O ∇x θ˜ (·, s) 2 L

0

Let us take ε˜ = t Lb

ds ⎠

0

t  2   ≤ ε˜ θ˜ (·, s)

kh . 4C 2P O

⎞1/2

Lb 4˜ε

ρ(·, ˜ s)2L 2 (]0,l[) ds

0

L 2b ds + (]0,l[)2 4˜ε

t ρ(·, ˜ s)2L 2 (]0,l[) ds. 0

Thus:

    ρ(·, ˜ s) L 2 (]0,l[) θ˜ (·, s)

L 2 (]0,l[)

ds

0

kh ≤ 4

t  2   ˜ ∇x θ (·, s) 2 L

0

(L b C P O )2 ds + 2 (]0,l[) kh

t ρ(·, ˜ s)2L 2 (]0,l[) ds.

(64)

0

Multiplying both sides of inequality (62) by 2 and using inequalities (63) to (64), we obtain: 2   ˜ t) 2 c p m g θ(·,

L (]0,l[)

+ kh

t  2   ˜ ∇x θ (·, s) 2

L (]0,l[)

0

2    ≤ c p m g θ˜ (·, 0) 2

L (]0,l[)

+2

(L b C P O ) kh

t 2

t ds + h c 2



+2

cpmgCP O kh

2  t

θ˜ (l, s)2 ds

0

ρ˜t (·, s)2L 2 (]0,l[) ds 0

ρ(·, ˜ s)2L 2 (]0,l[) ds.

0

  2  t  ˜ In view of the previous proposition, we must bound θ(·,0) , 0 ρ˜t (·, s)2L 2 (]0,l[) ds  2 L (]0,l[) t ˜ s)2L 2 (]0,l[) ds. Firstly: and 0 ρ(·,     θ (·, 0) L 2 (]0,l[) = Tu,h (·, 0) − T˜u,h (·, 0) 2 L (]0,l[)         ≤ Tu,h (·, 0)− Tu (·, 0) L 2 (]0,l[) + Tu (·, 0)− T˜u,h (·, 0) 2 . (65) What was to be proved.

L (]0,l[)

Lemma 18 ∀t ∈ [0, t f ]:    ˜ Tu,h (·, t) − Tu (·, t)

HL1 (]0,l[)

 h Tu (·, t) H 2 (]0,l[) .

123

L. Paquet

Proof Equation (50), shows us that T˜u,h (., t)−Ta is the orthogonal projection of Tu (., t)−Ta onto Sh ⊂ HL1 (]0, l[) for the scalar product of HL1 (]0, l[). Thus:        ˜  = (Tu (·, t) − Ta ) − T˜u,h (·, t) − Ta  1 Tu,h (·, t) − Tu (·, t) 1 HL (]0,l[)

HL (]0,l[)

= inf (Tu (·, t) − Ta ) − χ H 1 (]0,l[) = χ ∈Sh

L

inf

χ ∈Ta +Sh

Tu (·, t) − χ H 1 (]0,l[) L

≤ Tu (·, t) − Ih Tu (·, t) H 1 (]0,l[)  h Tu (·, t) H 2 (]0,l[) . L

  But, in view of (65), what we need (at least for t = 0) is a bound in the L 2 (]0, l[)-norm of T˜u,h (·, 0) − Tu (·, 0). We will thus apply a duality argument (like in [19, p. 5]): Proposition 19 The following bound holds ∀t ∈ [0, t f ]:     ρ(·, t) L 2 (]0,l[) = T˜u,h (·, t) − Tu (·, t) 2  h 2 Tu (·, t) H 2 (]0,l[) . L (]0,l[)

In the right hand side of the above inequality, h means max j=1,...,Nh (x j − x j−1 ). Proof Let us consider the two-points boundary value problem in the interval ]0, l[: given ϕ ∈ L 2 (]0, l[), find ψ ∈ H 2 (]0, l[) such that ⎧ ⎨ −ψ  (x) = ϕ(x), ∀x ∈]0, l[, ψ(0) = 0, ⎩ ψ(l) = 0. By the closed graph theorem, there exists a constant C > 0, such that ψ H 2 (]0,l[) ≤ C ϕ L 2 (]0,l[) . Using T˜u,h (0, t) = Tu (0, t) = Ta and Proposition 13 which tells us that T˜u,h (l, t) = Tu (l, t), we have: (T˜u,h (·, t) − Tu (·, t)|ϕ) L 2 (]0,l[) = −(T˜u,h (·, t) − Tu (·, t)|ψ  ) L 2 (]0,l[)

l  ˜ ∂ Tu,h = −(T˜u,h (l, t) − Tu (l, t))ψ (l) + (T˜u,h (0, t) − Tu (0, t))ψ (0) + (x, t) ∂x 



0

  l  ˜ ∂ Tu ∂ Tu,h ∂ Tu  (x, t) ψ (x)d x = (x, t) − (x, t) ψ  (x)d x − ∂x ∂x ∂x 0

 l  ˜ ∂ Tu,h ∂ Tu = (x, t) − (x, t) (ψ  (x) − χ  (x))d x, ∀χ ∈ Sh , ∂x ∂x 0

by Eq. (49). Thus using the preceding lemma: 







˜

(Tu,h (·, t) − Tu (·, t)|ϕ) L 2 (]0,l[) ≤ T˜u,h (·, t) − Tu (·, t)

H 1 (]0,l[)

· inf ψ − χ H 1 (]0,l[) ≤ h Tu (·, t) H 2 (]0,l[) inf ψ − χ H 1 (]0,l[) χ ∈Sh

≤ h Tu (·, t) H 2 (]0,l[) ψ − Ih ψ H 1 (]0,l[)  h 2 Tu (·, t) H 2 (]0,l[) ψ H 2 (]0,l[)

123

χ ∈Sh

Radiative heating of a glass plate: the semi-discrete…

 h 2 Tu (·, t) H 2 (]0,l[) ϕ L 2 (]0,l[) . Taking the supremum over all ϕ ∈ L 2 (]0, l[) such that ϕ L 2 (]0,l[) ≤ 1, we obtain:   ˜   h 2 Tu (·, t) H 2 (]0,l[) . Tu,h (·, t) − Tu (·, t) 2 L (]0,l[)

 

What was to be proved.     Corollary 20 Assuming that Tu (·, 0) ∈ H 2 (]0, l[): θ˜ (·, 0)

L 2 (]0,l[)

 h 2 Tu (·, 0) H 2 (]0,l[) .

Proof By inequality (65) and the previous proposition:        ˜   ≤ Tu,h (·, 0) − Tu (·, 0) L 2 (]0,l[) + Tu (·, 0) − T˜u,h (·, 0) 2 θ (·, 0) 2 L (]0,l[) L (]0,l[)   2    Tu,h (·, 0) − Tu (·, 0) L 2 (]0,l[) + h Tu (·, 0) H 2 (]0,l[) = Ih (Tu (·, 0)) − Tu (·, 0) L 2 (]0,l[) + h 2 Tu (·, 0) H 2 (]0,l[)  h 2 Tu (·, 0) H 2 (]0,l[) .   Proposition 21

t 0

ρ(·, ˜ s)2L 2 (]0,l[) ds  h 4

t 0

Tu (·, s)2H 2 (]0,l[) ds.

t Proof ρ(·, ˜ s) := exp(λs)ρ(·, s) and λ is negative. Thus: 0 ρ(·, ˜ s)2L 2 (]0,l[) ds ≤ t  t 2 2 ˜ 0 ρ(·, s) L 2 (]0,l[) ds. Thus it suffices to bound 0 ρ(·, s) L 2 (]0,l[) ds. ρ(·, s) := Tu,h (·, s)− Tu (·, s). By the preceding proposition ρ(·, s) L 2 (]0,l[)  h 2 Tu (·, s) H 2 (]0,l[) . Thus: t

t ρ(·, s)2L 2 (]0,l[) ds

h

0

Tu (·, s)2H 2 (]0,l[) ds.

4 0

 

What was to be proved. It remains thus to bound Proposition 22

t 0

t 0

ρ˜t (·, s)2L 2 (]0,l[) ds.

ρ˜t (·, s)2L 2 (]0,l[) ds  h 4

2 t   dTu  0  dt (·, s)

H 2 (]0,l[)

ds + h 4

t 0

Tu (·, s)2H 2 (]0,l[) ds.

Proof ∂s (exp(λs)ρ)(., s) = exp(λs) (∂s ρ) (., s) + λ exp(λs)ρ(., s). Thus: ρ˜t (·, s) = exp(λs)ρt (., s) + λ exp(λs)ρ(., s), ˜ s), = exp(λs)ρt (., s) + λρ(., which implies: ρ˜t (·, s) L 2 (]0,l[) ≤ exp(λs) ρt (·, s) L 2 (]0,l[) + |λ| ρ(·, ˜ s) L 2 (]0,l[) ≤ ρt (·, s) L 2 (]0,l[) + |λ| ρ(·, ˜ s) L 2 (]0,l[) ,

123

L. Paquet

as λ is negative (see (59). By Proposition 21: ∀s ∈ [0, t f ]: t

t ρ(·, ˜ s)2L 2 (]0,l[) ds

h

0

Tu (·, s)2H 2 (]0,l[) ds.

4 0

t

d T˜

u ρt (·, s)2L 2 (]0,l[) ds. ρt (·, s) = dtu,h (·, s)− dT dt (·, s). As, we have    d ˜ d ˜ u already observed, by Eq. (50): dt Tu,h (., t) = dT dt (., t) . In particular, dt Tu,h (., t) is the

It suffices thus to bound

orthogonal projection of

0

dTu 1 1 dt (., t) onto Sh ⊂ HL (]0, l[) for the scalar product of HL (]0, l[). d T˜u,h dTu dTu ˜ dt (l, t) = dt (l, t). Replacing Tu (·, t) by dt (., t) and Tu,h (·, t)

Also by Proposition 13,    d ˜  u Tu,h (., t) = dT by dt dt (., t) , in the estimate stated in Proposition 19, we obtain ∀ t ∈ [0, t f ]        d 2  dTu  T˜u,h (·, t) − dTu (., t) h  . (., t)  2  2  dt dt dt L (]0,l[) H (]0,l[) This estimate implies that: t ρt (·, s)2L 2 (]0,l[) ds 0

2 t   d ˜u,h (·, s) − dTu (., s) =  ds T  2  dt dt L (]0,l[) 0

2 t    dTu   (·, s) ds, ≤h  2  dt H (]0,l[) 4

(66)

0

 

from which the result now follows. Collecting the above results, we obtain the following estimate on θ = Tu,h − T˜u,h : Proposition 23 The following estimate holds for θ = Tu,h − T˜u,h : t c p m g θ (·, t)2L 2 (]0,l[)

t ∇x θ (·, s)2L 2 (]0,l[)2

+ kh 0

ds + h c

θ (l, s)2 ds 0

2 t  t  dTu  4 2 4 4   Tu (·, s)2H 2 (]0,l[) ds.  h Tu (·, 0) H 2 (]0,l[) + h ds + h  dt (·, s) 2 H (]0,l[) 0

0

(67) Proof This result follows from Proposition 17, Corollary 20, Proposition 22 and Proposition 21. Corollary 24 For the value of the solution Tu,h of the semi-discrete problem at the point x = l, we have the following a priori error estimate:      dTu    2 Tu,h (l, .) − Tu (l, .) 2   h Tu (·, 0) H 2 (]0,l[) +  . (68)  dt  2 L (]0,t f [) L (0,t f ;H 2 (]0,l[)

123

Radiative heating of a glass plate: the semi-discrete…

Proof As T˜u,h (l, .) = Tu (l, .) by Proposition 13, it follows that: Tu,h (l, .) − Tu (l, .) = Tu,h (l, .) − T˜u,h (l, .) = θ (l, .). Thus by (67), it follows   Tu,h (l, .) − Tu (l, .) 2 L (]0,t

f [)

 h 2 (Tu (·, 0) H 2 (]0,l[) + Tu  L 2 (0,t f ;H 2 (]0,l[)     dTu    + . dt  L 2 (0,t f ;H 2 (]0,l[)

(69)

But for every t ∈ [0, t f ]:

   dTu   Tu (·, t) H 2 (]0,l[)  Tu (·, 0) H 2 (]0,l[) +   dt  2 L (0,t f ;H 2 (]0,l[))

(70)

which implies

   dTu   Tu  L 2 (0,t f ;H 2 (]0,l[))  Tu (·, 0) H 2 (]0,l[) +  .  dt  2 L (0,t f ;H 2 (]0,l[))

(71)  

By inequalities (69) and (71), the result follows.

We are now ready to state our a priori error estimate for the solution Tu,h of the semidiscrete problem: Theorem 25 For the solution Tu,h of the semi-discrete problem, we have the following a priori error estimate:      dTu    2  Tu,h − Tu  . (72)  h Tu (·, 0) H 2 (]0,l[) +   dt  2 C([0,t f ];L 2 (]0,l[)) 2 L (0,t f ;H (]0,l[))

Proof As we have explained at the beginning of this section, Tu ∈ C([0, t f ]; H 2 (]0, l[)). As T˜u,h (·, t) − Ta is the orthogonal projection of Tu (·, t) − Ta onto Sh in HL1 (]0, l[), T˜u,h ∈ C([0, t f ]; H 1 (]0, l[)). A fortiori T˜u,h ∈ C([0, t f ]; L 2 (]0, l[)). Using the previous proposition, we obtain:      2 Tu (·, 0) H 2 (]0,l[) + Tu  L 2 (0,t f ;H 2 (]0,l[))  h Tu,h − T˜u,h  C([0,t f ];L 2 (]0,l[))     dTu    + . (73) dt  L 2 (0,t f ;H 2 (]0,l[)) By Proposition 19:   ˜  Tu,h (·, t) − Tu (·, t)

L 2 (]0,l[)

By inequality (74) and (70) follows:   ˜  Tu,h − Tu 

C([0,t f ];L 2 (]0,l[))

h

2

 h 2 Tu (·, t) H 2 (]0,l[) , ∀t ∈ [0, t f ].

(74)



    dTu   Tu (·, 0) H 2 (]0,l[) +  . (75)  dt  2 L (0,t f ;H 2 (]0,l[))

By inequalities (75), (73), and the triangular inequality follows:    2 Tu − Tu,h  Tu (·, 0) H 2 (]0,l[) + Tu  L 2 (0,t f ;H 2 (]0,l[))  h C([0,t ];L 2 (]0,l[)) f

123

L. Paquet

    dTu   + .  dt  2 L (0,t f ;H 2 (]0,l[))

(76)  

Using inequality (71), the result follows.

Now, we want also to give some a priori error estimate for the gradient ∇x Tu,h . Firstly a lemma: t t Lemma 26 0 ρ(·, ˜ s)2H 1 (]0,l[) ds  h 2 0 Tu (·, s)2H 2 (]0,l[) ds. Proof Equation (50), shows us that T˜u,h (., t)−Ta is the orthogonal projection of Tu (., t)−Ta onto Sh ⊂ HL1 (]0, l[) for the scalar product of HL1 (]0, l[). Thus:     ρ(·, s) H 1 (]0,l[)  Tu (·, s) − T˜u,h (·, s)

HL1 (]0,l[)

=

inf

χ ∈Ta +Sh

Tu (·, s) − χ  H 1 (]0,l[) L

  ≤ Tu (·, s) − Ih Tu (·, s) H 1 (]0,l[)  h Tu (·, s)  L 2 (]0,l[)  h Tu (·, s) H 2 (]0,l[) . L

As λ is negative, it follows from the previous inequality: t

t ρ(·, ˜ s)2H 1 (]0,l[) ds

=

0

t exp(λs)ρ(·, s)2H 1 (]0,l[) ds

0

t ρ(·, s)2H 1 (]0,l[) ds

≤

exp(2λs) 0

t · ρ(·, s)2H 1 (]0,l[) ds

=

0

h

Tu (·, s)2H 2 (]0,l[) ds.

2 0

  Proposition 27 For the gradient ∇x Tu,h , we have the following a priori error estimate:      dTu    ∇x (Tu,h − Tu ) 2   h Tu (·, 0) H 2 (]0,l[) +  .  dt  2 L (0,t f ;L 2 (]0,l[)2 ) L (0,t f ;H 2 (]0,l[)) (77) Proof From inequality (67) follows that:        h 2 Tu (·, 0) H 2 (]0,l[) + Tu  L 2 (0,t f ;H 2 (]0,l[)) ∇x Tu,h − T˜u,h  2 2 2 L (0,t f ;L (]0,l[) )     dTu    + . (78) dt  L 2 (0,t f ;H 2 (]0,l[)) On the other hand, from Lemma 26, it follows that:       h Tu  L 2 (0,t f ;H 2 (]0,l[)) . ∇x T˜u,h − Tu  2 2 2 L (0,t f ;L (]0,l[) )

(79)

By inequalities (78), (79) and the triangle inequality follows that:    ∇x (Tu,h − Tu ) 2 Tu (·, 0) H 2 (]0,l[) + Tu  L 2 (0,t f ;H 2 (]0,l[))  h 2 2 L (0,t f ;L (]0,l[) )     dTu   + .  dt  2 L (0,t f ;H 2 (]0,l[)) Using inequality (71) the result follows.

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Radiative heating of a glass plate: the semi-discrete… Temperature along the thickness of the Glass Plate, Ts=1500°C, Ta=T0=25°C, hc= 5.0, kh= 0.8 400 temperature temperature temperature temperature temperature

Temperature in degree Celsius

350

300

on the lower face at the quarter of the thickness in the middle plane at the three−quarter of the thickness on the upper face

250

200

150

100

50

0

0

10

20

30

40

50

60

70

80

90

100

time in seconds

Fig. 1 Temperatures in function of time

4 Numerical Tests Using the P1 −finite element in space with a regular grid on the interval [0, l] and the implicit Euler scheme in time, with prediction-correction on the nonlinear terms in (40), we have computed the temperature along the thickness in the glass plate. Moreover, to avoid the computation of double integrals with logarithmic singularities at each time iteration step in (40) very costly in computational time, we have turned back to the origin of these terms ∗ coming from the integration of the radiative intensities I k (x, t, μ) (k ∗ = 1, . . . , M) with respect to the cosine direction μ on the interval [−1, 1] (see formula (2.18) p. 9 of [11]), by performing rather a composite midpoint numerical integration in μ like in Siedow’s program based on their publication [15]. M has been chosen to be 30 like in that program also. The numerical results shown in Fig. 1 below, have been obtained by decomposing the interval [0, l] (the thickness l of the plate is taken equal to 6.1 millimeters like in [15]) into nss = 64 subintervals of equal length and by taking as time step τ = 0.1 second. The temperature u (denoted TS in [11] but u in [12]) of the infinite black radiative source placed above the glass plate has been chosen equal to 1500 ◦ C. The ambient air temperature has been taken equal to 25◦ C and also the initial temperature T0 of the glass plate.1 The heat transfer coefficient h c at the upper face of the glass plate has been chosen equal to 5.0 W/(m2 × ◦ K) (in [15], it is 280.5 W/(m2 × ◦ K) but it concerns tempering.2 ) The heat conductivity of the glass plate kh has been chosen equal to 0.8 W/(m × ◦ K) in conformity with [18, p. 449]. The graph of the temperature for x = 0, 4l , 2l , 3l4 , l with respect to time is shown in Fig. 1. If we do not take into account the terms ψ(T (x, t))+ h T,u (x, t) modelizing the participating medium character of the glass3 into Eq. (2)(i) , instead of T (l, 100) = 386.29 ◦ C we obtain T (l, 100) = 393.58 ◦ C, instead of T ( 3l4 , 100) = 295.71 ◦ C we obtain T ( 3l4 , 100) = 300.50 ◦ C, instead of T ( l , 100) = 206.38 ◦ C we obtain T ( l , 100) = 208.17 ◦ C, and instead of 2 2 T ( 4l , 100) = 116.12 ◦ C we obtain T ( 4l , 100) = 116.42 ◦ C. 1 For convenience the temperatures are cited in ◦ C; the corresponding absolute temperatures in ◦ K are obtained by adding 273.15 to the temperatures cited in ◦ C. 2 I take here the opportunity to thank my colleague F. Béchet from the Tempo laboratory in our university,

who has made me that remark. 3 Semi-transparent medium

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L. Paquet 400 Temperature along the thickness at t=5 s Temperature along the thickness at t=10 s Temperature along the thickness at t=100 s

Temperature in Celsius degrees

350 300 250 200 150 100 50 0

0

1

2

3

4

5

x−axis

6

7 −3

x 10

Fig. 2 Temperatures in ◦ C at 5, 10, and 100 s with respect to x

In Fig. 2, we have plotted the distribution of the temperature at the different times: 5, 10, 100 seconds along the thickness x (0 ≤ x ≤ l) of the glass plate, having chosen l l to make the computation h = nss = 1024 and as time step τ = 0.01 second. To confront our error estimate (72) with a numerical experiment, we have computed the temperature along the thickness of the glass plate at the fixed time t = 10 seconds choosing a small fixed time step τ = 0.01 second. The numerical results have been compared after one thousand of time steps for more and more refined meshes on the space interval [0, l]: l l l l l l h = 8l , 16 , 32 , 64 , 128 , 256 , taking as reference the solution computed with h = 1024 whose graph has been plotted previously in green in Fig. 2. We observe quadratic behavior of the relative error measured in the L 2 (]0, l[) norm with l respect to h = nss in accordance with our a priori error estimate (72) in Theorem 25 (nss denotes the number of subintervals [(i − 1)h, i h], i = 1, . . . , nss, into which the interval [0, l] has been decomposed): In blue in Fig. 3, we have plotted the six points, results of our numerical computations:  nss

Log10

sol−sol_r e f 2 sol_r e f 2



−3.0402 −3.6421 −4.2443 −4.8470 −5.4515 −6.0637

8 = 23 16 = 24 32 = 25 64 = 26 128 = 27 256 = 28

In red, on the same figure, we have plotted the regression line fitting at best these six points in the sense of mean-squares. The equation of that straight line is the following:  Log10

123

sol − sol_r e f 2 sol_r e f 2

 = −1.2248 − 0.60424 × Log2 (nss)

Radiative heating of a glass plate: the semi-discrete… Log10(relative error) / Log2(number of subintervals) 1 points linear regression

Log

10

(||sol−sol ref||2 / ||sol ref||2)

0 −1 −2 −3 −4 −5 −6 −7 −8

2

3

4

5

6 Log (nss)

7

8

9

10

2

Fig. 3 Relative error at time t = 10 s in function of nss Relative error on the top face / log2(number of subintervals) 1 points linear regression

0 (estimated relative error)

−1 −2 −3

Log

10

−4 −5 −6 −7 −8

2

3

4

5

6 Log2 (nss)

7

8

9

10

Fig. 4 Relative error on the temperature of the upper face

which implies that: sol − sol_r e f 2 = 0.059592 × nss −2.00772 . sol_r e f 2 Remark 28 For the temperature on the upper face of the glass plate at the fixed time t = 10 seconds, we have also observed quadratic convergence with respect to h in accordance with our a priori error estimate in Corollary 24: The equation of the regression line (see Fig. 4) for the relative error is: Log10 (relative_error) = −1.34 − 0.609 × Log2 (nss)

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L. Paquet

which implies that: relative_error = 0.0457 × nss −2.02 .

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