Stabilization of a strong damped wave equation with acoustic boundary conditions. Nicolas Fourrier
U NIVERSITY
OF
V IRGINIA
[email protected]
September 11, 2012
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
1 / 42
Overview
1
Introduction
2
Theorems
3
Proof of the theorems
4
Numerical Calculations
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
2 / 42
Introduction
Overview
1
Introduction Presentation of the problem Mathematical model Preliminary definitions
2
Theorems
3
Proof of the theorems
4
Numerical Calculations
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
3 / 42
Introduction
Presentation of the problem
Physical Model
3-dimension bounded domain Ω. Boundary δΩ = Γ0 ∪ Γ1 Gas undergoes small irrotationnal perturbations away from Ω which are concentrated on a portion of the boundary Γ1 . Each point of the surface Γ1 acts like a spring in response to the excess pressure in the gas. There is no transverse tension between neighboring points of Γ1 , i.e., the springs are independent of each other. Dirichlet boundary conditions on Γ0 .
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
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Introduction
Presentation of the problem
Previous results
Wave equation Viscous-frictional dampers lead to analytic and exponentially stable semigroup in the presence of classical boundary conditions (Dirichlet, Robin) No longer true in the case of acoustic boundary conditions.
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Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Mathematical model
utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(1)
on Ω
cΩ is the coefficient associated with the viscous interior damper. kΓ is the boundary coefficient. α = 0 or 1, allows to add a viscous boundary damper. If α = 0, kΓ is only related to speed on the boundary. If α = 1, kΓ also determines the speed, but it also determines with cΩ the intensity of the damping.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
6 / 42
Introduction
Preliminary definitions
Definition of the operators Neumann Map: N
z = Ng ⇔
∆z = 0 δη z|Γ = g 1 z|Γ = 0 0
3
3 −ǫ
By elliptic theory, N ∈ L(L2 (Γ1 ), H 2 (Ω) ⊂ D(A 4
on Ω on Γ on Γ0
(2)
)) ∀ǫ > 0
Laplacian in the interior A : L2 (Ω) ⊃ D(A) → L2 (Ω) Let the operator A : L2 (Ω) ⊃ D(A) → L2 (Ω) be defined by: 2
A(u − Nδη u) = −∆u, D(A) = {z ∈ H (Ω), z|Γ = 0, δη z|Γ = 0} 0
1
(3)
Laplace-Beltrami on the boundary B : L2 (Γ1 ) ⊃ D(B) → L2 (Γ1 )
2
2
Bz = −∆Γ z, D(B) = {z ∈ H (Γ1 ), z|δΓ = 0} = H0 (Γ1 )
(4)
1
1 R with the weighted norm kzk 1 = kΓ kB 2 zk 2 . = kΓ Γ |∇Γ z|2 = kΓ kzk 1 L (Γ1 ) H (Γ1 ) 1 0
B2
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Introduction
Preliminary definitions
Energy space
Let (u, ut , u|Γ1 , u|Γ1 ) = (u1 , u2 , u3 , u4 ). The associated energy space is: 1
1
H = D(A 2 ) × L2 (Ω) × D(B 2 ) × L2 (Γ1 )
(5)
with associated norm: kukH = |∇u1 |2Ω + |u2 |2Ω + kΓ |∇Γ u3 |2Γ1 + |u4 |2Γ1 .
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
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Introduction
Preliminary definitions
Operator A With the definitions of A, B and N, we set: 0 I 0 0 −A(I − Nδη ) −cΩ A(I − Nδη ) 0 0 A= 0 0 0 I −δη −cΩ δη −kΓ B −qB
1
1
1
(6)
1
D(A) ={[u1 , u2 , u3 , u4 ]T ∈ D(A 2 ) × D(A 2 ) × D(B 2 ) × D(B 2 ), such that (I − Nδη )u1 + cΩ (I + Nδη )u2 ∈ D(A), kΓ u3 + qu4 ∈ D(B), u1 |Γ1 = N ∗ Au1 = u3 } Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
9 / 42
Theorems
Overview
1
Introduction
2
Theorems Strong damping on the boundary No damping on the boundary Intermediate theorem
3
Proof of the theorems
4
Numerical Calculations
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
10 / 42
Theorems
Strong damping on the boundary
Having a strong damping on the boundary is equivalent to α = 1 in (1)
Theorem I With A defined in (6), generates an analytic C0 -semigroup {eAt }t≥0 on H. Also, {eAt }t≥0 is exponentially stable, i.e., ∃C, ω ≥ 0 such that keAt uk2H ≤ Ce−ωtkukH
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P.D.E. Seminar
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Theorems
No damping on the boundary
No damping on the boundary is equivalent to α = 0 in (1)
Theorem II Aα=0 given by A in (6), generates a C0 -semigroup of contractions {eAα=0 t }t≥0 on H. Also, {eAα=0 t }t≥0 is strongly stable, i.e., ∀u ∈ H lim , eAα=0 t u → 0 t→∞
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P.D.E. Seminar
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Theorems
Intermediate theorem
The following theorem will provide the necessary tool to obtain the stability results from Theorem I and Theorem II.
Theorem III With A defined in (6), σ(A) ∩ iR = ∅
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Proof of the theorems
Overview
1
Introduction
2
Theorems
3
Proof of the theorems Theorem III Theorem I Theorem II
4
Numerical Calculations
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem III
Statement
Theorem III With A defined in (6), σ(A) ∩ iR = ∅ Strategy This problem will give the stability results for both theorems. We will proceed in 3 steps: σc (A) ∩ iR = ∅ σp (A) ∩ iR = ∅ σr (A) ∩ iR = ∅
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Proof of the theorems
Theorem III
Adjoint of A
0 −I 0 0 A(I − Nδη ) −cΩ A(I − Nδη ) 0 0 A∗ = 0 0 0 −I δη −cΩ δη kΓ B −qB 1
1
1
(7)
1
D(A∗ ) ={[u1 , u2 , u3 , u4 ]T ∈ D(A 2 ) × D(A 2 ) × D(B 2 ) × D(B 2 ), such that (I − Nδη )u1 − cΩ (I + Nδη )u2 ∈ D(A), u3 − qu4 ∈ D(B), u1 |Γ1 = N ∗ Au1 = u3 and u2 |Γ1 = N ∗ Au2 = u4 }
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P.D.E. Seminar
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Proof of the theorems
Theorem III
Step 1: σc (A) ∩ iR = ∅ Theorem If λ ∈ σc (A), then A − λ does not have a closed range. Assumption With A as given in (6), assume that λ = ir ∈ σc (A) with r 6= 0. ∀f = [f1 , f2 , f3 , f4 ]T ∈ H, suppose that u = [u1, u2, u3, u4] ∈ D(A) such that: (ir − A)u = f
(8)
Recall that: ∗
u1 |Γ = N Au1 = u3 1 Then, this leads to a single equation: 2
2
∗
∗
∗
− r u1 + (1 + cΩ ir )Au1 − AN(r N Au1 − kΓ BN Au1 − qirBN Au1 ) = irf1 + cΩ Af1 + f2 + AN(f4 + irf3 + qBf3 )
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
(9)
17 / 42
Proof of the theorems
1
Theorem III
1
Let V = D(A 2 ) ∩ D(B 2 ) with associated norm: 1
1
2
2
kukV = |A 2 u|Ω + |B 2 u|Γ
(10)
1
Consider the left hand side of (9) and define the operators T , M, K : V → V ∗ , by T M K 1
1
1
=M+K = A + kΓ ANBN ∗ A + ir (cΩ A + qANBN ∗ A) = −r 2 (I + ANN ∗ A)
(11)
1
Define F : D(A 2 ) × D(A 2 ) × D(B 2 ) × D(B 2 ) → V ∗ by: F = (ir + cΩ A)
Nicolas Fourrier (U.V.A.)
I
AN(ir + qB)
P.D.E. Seminar
AN
(12)
September 11, 2012
18 / 42
Proof of the theorems
Theorem III
Compactness of K 1
By definition of A, N and N ∗ A, K is compact from D(A 2 ) into its dual.
M is boundedly invertible Given M, u ∈ V and v ∈ V ∗ , define its bilinear form M(., .), by:
M(u, v ) = (1 + cΩ ir )
1 1 A 2 u, A 2 v
Ω
+ (kΓ + qir )
1
1
B 2 u, B 2 v
Γ1
(13)
M(., .) is a coercive and bounded bilinear form. Therefore, by Lax-Milgram M is bounded invertible.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem III
T is invertible By the Fredholm’s alternative, since we can deduce the desired invertibility of T provided that T is injective. Suppose that Tu1 = 0, assume that r 6= 0 and take the duality pairing with respect to u1 ∈ V ∗ , then: 0 = ((1 + cΩ ir )Au1 , v )
2
1 1 [D(A 2 )]∗ ×D(A 2 )
∗
− r (I + ANN Au1 , v ∗
1 1 [D(A 2 )]∗ ×D(A 2 )
+ (kΓ + qir )ANBN Au1 , v
(14)
1 1 [D(A 2 )]∗ ×D(A 2 )
1 1 ∗ 1 1 ∗ 2 2 2 2 2 = (1 − r )|A 2 u1 |Ω + kΓ |B 2 N Au1 |Γ + ir cΩ |A 2 u1 |Ω + q|B 2 N Au1 |Γ 1 1 Taking the imaginary part in (14) implies u1 = 0. Thus T is injective.
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P.D.E. Seminar
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Proof of the theorems
Theorem III
End of the proof σp (A) ∩ iR = ∅ With A as given in (6), if for r ∈ R and r 6= 0, there exists u = [u1, u2, u3, u4]T ∈ D(A) such that: u1 u1 u 2 u 2 A u3 = ir u3 u4 u4
(15)
This turns to be equivalent to solve: Tu1 = 0. σr (A) ∩ iR = ∅ First, recall the theorem: If the eigenvalue λ ∈ C is in the residual spectrum of A, then λ ∈ σp (A∗ ) With A∗ as given in (7), if for r ∈ R and r 6= 0, there exists u = [u1, u2, u3, u4]T ∈ D(A∗ ) such that: u1 u1 u 2 u2 ∗ A = ir u 3 u3 u4 u4
(16)
This is also equivalent to solve: Tu1 = 0.
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P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem I
Statement
Theorem I With A defined in (6), generates an analytic C0 -semigroup {eAt }t ≥0 on H. Also, {eAt }t ≥0 is exponentially stable, i.e., ∃C, ω ≥ 0 such that ke
At
2
ukH ≤ Ce−ωtkukH
Strategy Introducing a new variable z = u + cΩ u t
(17)
we will to transform the system into the heat equation with General Wentzell Boundary Conditions (GWBC) with a perturbation.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem I
Heat Equation with GWBC Using z = u + cΩ ut , we obtain:
(
zt ∆z + δη z − kΓ ∆Γ z
= cΩ ∆z + cz − cu Ω Ω
(18)
=0
Theorem by Favini Consider the heat equation with GWBC: zt = ∆z in Ω
(19)
∆z + βδη z + γz − ζβ∆Γ = 0 on δΩ
(20)
Then the closure G2 of ∆ in X2 with domain 2
¯ z|δΩ satisfies (20)} D(G2 ) = {Z = (z|Ω , z|δΩ ) ∈ H (Ω), The operator G2 is self-adjoint and generates a cosine function and a quasicontraction (contraction, if γ ≥ 0) semigroup on X2 which is analytic in the right half plane.
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P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem I
Perturbation Theory
Let U = (z, u|Ω , u|Γ )T and set β = cΩ , ζ = kΓ , then one can rewrite (17) and (18) as follows: 1 u| z|Ω − 2Ω G 2 cΩ z c2 c zt u|Ω Ω Ω − c z|Ω U t = u t |Ω = Ω + c Ω u|Γ u t |Γ z|Γ 1 − c 1 1
Ω
(21)
cΩ
= (B + P)U
Note that the second equation is simply an ordinary differential equation. We can let u to be defined on H 1 (Ω). ¯ × H 1 (Ω)} is analytic on X2 × H 1 (Ω). Then A with domain D(A) = {H 2 (Ω) It is enough to show that B is relatively A-bounded. z|
u|
But | 2Ω − 2Ω |B ≤ C(kzkH 1 (Ω) + kukH 1 (Ω) ) ≤ C(kzkX + kukH 1 (Ω) ) = CkU kA 2 c c Ω Ω Therefore B is relatively A-bounded and thus analytic.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem I
Perturbation Theorem Let the operator (A, D(A)) generate an analytic semigroup T (z) on a Banach space X. Then there exists a constant α such that (A + B, D(A + B)) generates an analytic semigroup for every A-bounded operator B having A-bound α0 < α. Conclusion Then A generates an analytic semigroup {eAt }t ≥0 on H. Note that the proof also holds for kΓ = 0. Since σ(A) ∩ iR = ∅ by Theorem III, then {eAt }t ≥0 is exponentially stable by the stability theorem which achieves the proof theorem Theorem I.
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P.D.E. Seminar
September 11, 2012
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Proof of the theorems
Theorem II
Statement
Theorem II Aα=0 given by A in (6), generates a C0 -semigroup of contractions {eAα=0 t }t ≥0 on H. Also, {eAα=0 t }t ≥0 is strongly stable, i.e., A t ∀u ∈ H lim , e α=0 u → 0 t →∞
Strategy Using Lumer-Phillips theorem, we need to show that Aα=0 is maximal dissipative.
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Proof of the theorems
Theorem II
Dissipativity
∀u ∈ D(Aα=0 ), take the inner product:
< Aα=0 U , V > = (u2 , u1 )H 1 (Ω) + (∆u1 + cΩ ∆u2 , u2 )Ω + (u4 , u3 )H 1 (Γ ,k dS) + −δη (u1 + cΩ u2 ) + kΓ ∆Γ u3 , u4 Γ 1 1 Ω
= (∇u2 , ∇u1 )Ω − (∇u1 + cΩ ∇u2 , ∇u2 )Ω + δη (u1 + cΩ u2 ), u2 |δΩ Γ 1
+ kΓ h∇Γ u4 , ∇Γ u3 iΓ − δη (u1 + cΩ u2 ), u4 Γ − hkΓ ∇Γ u3 , ∇Γ u4 iΓ 1
1
1
2
= −cΩ |∇u2 |Ω < 0
(22)
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P.D.E. Seminar
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Proof of the theorems
Theorem II
Range Condition I
if λ ∈ C, Reλ > 0 and F = f1
f2
f3
f4 T ∈ H is given, then the elliptic equation: f1 u1 u2 = f2 (λ − Aα=0 ) f3 u3 f4 u4
is satisfied for some u1
u2
u3
(23)
u4 T ∈ D(Aα=0 )
Then the strategy is to create a bilinear form in order to eventually use Lax-Milgram. After multiplying by V = (v1 , v2 , v3 , v4 ) ∈ H, integrating over Ω and simplifying, we get: D E 2 2 + (∇u1 , ∇v1 )Ω + (cΩ λ∇u1 , ∇v1 )Ω + λ u3 , v3 λ u 1 , v1 Ω
Γ1
+ hkΓ ∇Γ u3 , ∇Γ v3 iΓ
1
(24)
= (f2 + λf1 , v1 )Ω + (cΩ λ∇f1 , ∇v1 )Ω + hf4 + λf3 , v3 iΓ 1
Let L(U , V ) be the left hand side and F (V ) be the right hand side.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
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Proof of the theorems
Theorem II
Range Condition II
Lax-Milgram conditions
2
|L(U , V )| ≤ max {λ + cΩ λ + kΓ , 1}kU kH kV kH |F (V )| ≤ max {λ, 1}kF kH kV kH ReL(U , U ) =
D E 2 2 + (∇u1 , ∇u1 )Ω + (cΩ λ∇u1 , ∇u1 )Ω + λ u3 , u3 λ u1 , u1 Ω
(25)
Γ1
+ hkΓ ∇Γ u3 , ∇Γ u3 iΓ
1
(26)
2
2 2 2 2 2 ≥ Reλ |u1 |Ω + |∇u1 |Ω + Reλ |u3 |Γ + |kΓ ∇Γ u3 |Γ 1 1
Conclusion L(U , V ) defined a bounded (25), coercive (26) bilinear form. So Aα=0 generates a semigroup of contraction {eAt }t ≥0 . By the stability theorem from Arendt-Batty, Theorem III shows the strong stability for {eAt }t ≥0 .
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Numerical Calculations
Overview
1
Introduction
2
Theorems
3
Proof of the theorems
4
Numerical Calculations Spectrum of a classical strong damped wave equation Spectrums of the strong damped equation with Acoustic Boundary Conditi Solution’s behavior
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
30 / 42
Numerical Calculations
Spectrum of a classical strong damped wave equation
Spectral result In the paper, Proof of extensions of two conjecteures o structural damping for elastic systems, Pacific Journal of Mathematics - S. Chen and R. Triggiani, the eigenvalues for a strong damped wave equation are given: q +,− −1 α 2 λn = µn −cΩ ± cΩ − µn (27) ∞ where {µ}n=1 are the eigenvalues of ∆in 2D. The following picture shows the first eigenvalues λ+,− n .
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Numerical Calculations
Spectrum of a classical strong damped wave equation
Imaginary Axis
3i
Real Axis −24
Nicolas Fourrier (U.V.A.)
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0
September 11, 2012
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Numerical Calculations
21
22
11 16
23
12 17
7 11
13
8
14
9
1
10 15
6
8
2 2
20
14
Γ0 7
25
19
13
5
1
24
18
12
6
Spectrums of the strong damped equation with Acoustic Boundary
9
3 3
10
4 4
5
Γ
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Numerical Calculations
Spectrums of the strong damped equation with Acoustic Boundary
In the next two pictures, we will observe that the equation (1) with α = 1, i.e., with a damping on the boundary presents a very similar spectrum to the previous case. However, when there is no more damping on the boundary (α = 0), this leads to a new configuration.
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Numerical Calculations
Spectrums of the strong damped equation with Acoustic Boundary
Imaginary Axis
cΩ =0.15 cΓ =1 kΩ =1 kΓ =1
4i
α =1
Real Axis −31
utt − cΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (cΩ αut + u) = 0 u + δ(cΩδη tt u=0
Nicolas Fourrier (U.V.A.)
0
on Ω on Γ1 on Γ0
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Numerical Calculations
Spectrums of the strong damped equation with Acoustic Boundary
Imaginary Axis
cΩ =0.15 cΓ =1 kΩ =1 kΓ =1
11i
α =0
Real Axis −30
utt − cΩ ∆ut − ∆u = 0 ut +u) − kΓ ∆Γ (cΩ αut + u) = 0 u + δ(cΩδη tt u=0
Nicolas Fourrier (U.V.A.)
0
on Ω on Γ1 on Γ0
P.D.E. Seminar
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Numerical Calculations
Solution’s behavior
Model utt − ∆u − cΩ ∆ut = 0 utt − δη (u + cΩ ut ) − kΓ ∆Γ (cΩ αut + u) = 0 u=0 Initial conditions u(0, x) = u0 , ut (0, x) = u1
in Ω in Γ1 in Γ0
(28)
on Ω
Solution with α = 1 Solution with α = 0 Comparaison between these two cases. Comparaison of the L2 and H 1 norm.
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Numerical Calculations
Nicolas Fourrier (U.V.A.)
Solution’s behavior
P.D.E. Seminar
September 11, 2012
38 / 42
Numerical Calculations
Nicolas Fourrier (U.V.A.)
Solution’s behavior
P.D.E. Seminar
September 11, 2012
39 / 42
Numerical Calculations
Nicolas Fourrier (U.V.A.)
Solution’s behavior
P.D.E. Seminar
September 11, 2012
40 / 42
Numerical Calculations
Nicolas Fourrier (U.V.A.)
Solution’s behavior
P.D.E. Seminar
−−−− |uα = 1|L2(Ω)
−−−− |uα = 0|L2(Ω)
..... |uα = 1|H1(Ω)
..... |uα = 0|H1(Ω)
September 11, 2012
41 / 42
Numerical Calculations
Solution’s behavior
Lili Rose
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
September 11, 2012
42 / 42
