Wave Equation with Dynamic B.C. Theorical and Numerical Approaches Nicolas Fourrier
U NIVERSITY
OF
V IRGINIA
[email protected]
October 4, 2011
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
1/1
Overview
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
2/1
Problem Definition
Overview
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
3/1
Problem Definition
Problem to solve
Problem to solve Wave Equation with dynamic boundary condition utt + cΩ ut = ∆u − kΩ u on Ω du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ
(1)
on Ω utt + cΩ ut = ∆u − kΩ u du − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 utt + cΓ ut + dη u=0 on Γ0
(2)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): c and k denote respectively dynamic and static coefficients Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary. Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
4/1
Problem Definition
Problem to solve
Problem to solve Wave Equation with dynamic boundary condition utt + cΩ ut = ∆u − kΩ u on Ω du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ
(1)
on Ω utt + cΩ ut = ∆u − kΩ u du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 u=0 on Γ0
(2)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): c and k denote respectively dynamic and static coefficients Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary. Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
4/1
Problem Definition
Problem to solve
Problem to solve Wave Equation with dynamic boundary condition utt + cΩ ut = ∆u − kΩ u on Ω du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ
(1)
on Ω utt + cΩ ut = ∆u − kΩ u du − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 utt + cΓ ut + dη u=0 on Γ0
(2)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): c and k denote respectively dynamic and static coefficients Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary. Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
4/1
Problem Definition
Problem to solve
Problem to solve Wave Equation with dynamic boundary condition utt + cΩ ut = ∆u − kΩ u on Ω du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ
(1)
on Ω utt + cΩ ut = ∆u − kΩ u du − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 utt + cΓ ut + dη u=0 on Γ0
(2)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): c and k denote respectively dynamic and static coefficients Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary. Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
4/1
Problem Definition
Importance of Dirichlet
Imaginary Axis
cΩ =2 cΓ =2 kΩ =1 kΓ =1
1.6i
Real Axis −1.1
½
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0
0
on Ω on Γ
Imaginary Axis
cΩ =2 cΓ =2 kΩ =1 kΓ =1
1.5i
Real Axis −0.6
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
Nicolas Fourrier (U.V.A.)
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
5/1
Problem Definition
Domain
No Dirichlet
Γ In this picture: 1
Ω: Interior
2
Γ: Boundary
3
Lx : Length
4
Ly : Width
Ω
Ly
Lx
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
6/1
Problem Definition
Domain
Dirichlet on Γ0
Γ In this picture: 1
Ω: Interior
2
Γ: Boundary
3
Lx : Length
4
Ly : Width
Ly
Ω
Γ
Γ
Γ0 Lx
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
7/1
Energy Identity
Overview
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
8/1
Energy Identity
Notation
Notation
hu, viΩ =
R
Ω uv
dΩ
|u|2Ω = hu, uiΩ Γ : Boundary on which dynamic BC are applied Γ0 : Dirichlet BC R hu, viΓ = Γ uv d Γ |u|2Γ = hu, uiΓ
The goal is to identify the energies and pull out an energy identity which will be used in the stabilization.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
9/1
Energy Identity
Introduction
Different steps
on Ω utt + cΩ ut = ∆u − kΩ u du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 u=0 on Γ0
(3)
Multiply by ut and integrate. ∂u ∂η
= kΓ ∆Γ u − utt − cΓ ut on Γ
On Γ0 , u = 0 Derive the boundary and interior energies. Obtain the energy identity.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
10 / 1
Energy Identity
Result
Result
Energies EΩ (t) = EΓ (t) =
1 2 1 2
2 + |∇u|2Ω + kΩ |u|2Ω |ut |Ω |ut |2Γ + kΓ |∇Γ u|2Γ
Energy Identity 1 d 2 dt
[EΩ (t) + EΓ (t)] = −cΩ |ut |2Ω − cΓ |ut |2Γ
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
11 / 1
Stabilization
Overview
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
12 / 1
Stabilization
Introduction
Procedure
on Ω utt + cΩ ut = ∆u − kΩ u du utt + cΓ ut + dη − kΓ ∆Γ u = 0 on Γ = Γ1,2,3 u=0 on Γ0
(4)
Multiply by u and integrate. ∂u ∂η
= kΓ ∆Γ u − utt − cΓ ut on Γ
u = 0 on Γ0 Set the left hand side to be the total energy. Set the right hand side to be the remaining terms. Estimate the right hand side.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
13 / 1
Stabilization
Integration
Integration by Parts Principal Equation Z
0
T
−|ut |2Ω
+
|∇u|2Ω
hu, ut iΩ |0T +
+
kΩ |u|2Ω dt
−
Z
T 0
∂u ,u ∂η
Γ
+
∂u , ut ∂η
dt =
Γ0
cΩ (|u(0)|2Ω − |u(T )|2Ω ) 2
Normal Term Z
T 0
∂u ,u ∂η
Z
T 0
dt =
Γ
−kΓ |∇Γ u|2Γ +|ut |2Γ dt −
cΓ (|u(T )|2Γ − |u(0)|2Γ ) − hu, ut iΓ |T0 2
It appears two unfriendly terms. Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
14 / 1
Stabilization
Use of the energy identity
Recall that the energy identity is: 1d [EΩ (t) + EΓ (t)] = −cΩ |ut |2Ω − cΓ |ut |2Γ 2 dt Using the energy identity allows to rewrite the two unfriendly terms: −
Z
T
0
|ut |2Ω +|ut |2Ω dt
Nicolas Fourrier (U.V.A.)
=
Z
T 0
cΩ cΓ 1 1 |ut |2Ω + |ut |2Γ dt+( + )(E (T )−E (0)) cΓ cΩ 2cΩ 2cΓ
P.D.E. Seminar
October 4, 2011
15 / 1
Stabilization
Z
0
T
Identifying the LHS and RHS
Z T cΩ cΓ 2 2 2 |ut |Ω + |∇u|Ω + kΩ |u|Ω dt + |ut |2Γ + kΓ |∇Γ u|2Γ dt cΓ c Ω 0 1 1 + )(E (0) − E (T )) = hu, ut iΩ |0T + hu, ut iΓ |0T + ( 2cΩ 2cΓ cΩ cΓ + (|u(0)|2Ω − |u(T )|2Ω ) + (|u(0)|2Γ − |u(T )|2Γ ) 2 2
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
16 / 1
Stabilization
Left Hand Side
Estimate the LHS
Left Hand Side Let c = max{ ccΩΓ , ccΩΓ } ≥ 1 T
cΩ |ut |2Ω + |∇u|2Ω + kΩ |u|2Ω dt + c c 0 cΓ Z T EΩ (t) + EΓ (t)dt ≥ Z
Z
T 0
cΓ |ut |2Γ + kΓ |∇Γ u|2Γ dt cΩ
0
=
Z
T
E (t)dt 0
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
17 / 1
Stabilization
Right Hand Side
Estimate the RHS I Recall 1 1 + )(E (0) − E (T )) + hu, ut iΩ |0T + hu, ut iΓ |0T 2cΩ 2cΓ cΓ cΩ + (|u(0)|2Ω − |u(T )|2Ω ) + (|u(0)|2Γ − |u(T )|2Γ ) 2 2
RHS = (
First estimate (
1 1 1 1 + )(E (0) − E (T )) ≤ ( + )E (0) 2cΩ 2cΓ cΩ cΓ
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
18 / 1
Stabilization
Right Hand Side
Estimate the RHS II
Second estimate Let k = max{ k1Ω , kDΓ }, and D be a Poincaré constant. hu, ut iΩ |0T + hu, ut iΓ |0T 1 D )EΩ (0) + (1 + )EΓ (0) ≤ 4 (1 + kΩ kΓ ≤ 4 [(1 + k)EΩ (0) + (1 + k)EΓ (0)] ≤ (4 + 4k)E (0)
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
19 / 1
Stabilization
Right Hand Side
Estimate the RHS III
Third estimate Γ Let ck = max{ ckΩΩ , Dc kΓ }
cΩ cΓ (|u(0)|2Ω − |u(T )|2Ω ) + (|u(0)|2Γ − |u(T )|2Γ ) 2 2 2cΩ 2DcΓ EΩ (0) + EΓ (0) ≤ kΩ kΓ ≤ 2ck E (0)
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
20 / 1
Stabilization
Energy Estimate
Recall Z
T
E (t)dt ≤ c × (LHS)
0
RHS ≤ E (0)(4 + 4k +
Final Estimate Z T
1 1 + + 2ck ) cΩ cΓ
E (t)dt ≤ E (0)c(4 + 4k +
0
c ck where k
1 1 + + 2ck ) cΩ cΓ
= max{ ccΩΓ , ccΩΓ } Γ = max{ kcΩΩ , Dc kΓ } 1 D = max{ kΩ , kΓ }
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
October 4, 2011
21 / 1
Numerical Results
0
Impact of kΩ
Maximum of eigenvalues real part in function of cΩ
−0.2
−0.4
−0.6
−0.8
−1
0
0.2
cΩ =2 cΓ =2 kΩ =0.4 kΓ =1
0.4
0.6
0.8
1
1.6
1.8
2
1.5i
Real Axis
Nicolas Fourrier (U.V.A.)
1.4
Imaginary Axis
cΩ =2 cΓ =2 kΩ =1.4 kΓ =1
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
1.2
−1.1
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
22 / 1
Numerical Results
Impact of kΓ
−0.2 −0.3
Maximum of eigenvalues real part in function of kΓ
−0.4 −0.5 −0.6 −0.7 −0.8 −0.9 −1
0
0.5
cΩ =2 cΓ =2 kΩ =1 kΓ =0.4
1
1.5
2
3.5
4
2.5i
Real Axis
Nicolas Fourrier (U.V.A.)
3
Imaginary Axis
cΩ =2 cΓ =2 kΩ =1 kΓ =2.8
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
2.5
−0.88
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
23 / 1
Numerical Results
Impact of cΩ
0
Maximum of eigenvalues real part in function of cΩ −0.2
−0.4
−0.6
−0.8
−1
0
0.5
cΩ =0.9 cΓ =2 kΩ =1 kΓ =1
cΩ =1.9 cΓ =2 kΩ =1 kΓ =1
1
2
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
2.5
3
3.5
4
Imaginary Axis
cΩ =3.5 cΓ =2 kΩ =1 kΓ =1
1.5i
Real Axis
Nicolas Fourrier (U.V.A.)
1.5
−1.9
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
24 / 1
Numerical Results
Impact of cΓ
−0.4
Maximum of eigenvalues real part in function of cΓ
−0.5 −0.6 −0.7 −0.8 −0.9 −1 −1.1
0
0.5
cΩ =2 cΓ =0.9 kΩ =1 kΓ =1
cΩ =2 cΓ =2.1 kΩ =1 kΓ =1
1
2
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
2.5
3
3.5
4
Imaginary Axis
cΩ =2 cΓ =2.9 kΩ =1 kΓ =1
Real Axis
Nicolas Fourrier (U.V.A.)
1.5
1.5i
−1.2
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
25 / 1
Numerical Results
Impact of kΓ on cΓ
−0.4 −0.5 −0.6
Maximum of eigenvalues real part in function of cΓ
−0.7 −0.8 −0.9 −1 −1.1 0.5
cΩ =2 cΓ =1 kΩ =1 kΓ =2.5
1
cΩ =2 cΓ =2 kΩ =1 kΓ =2.5
1.5
2.5
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
3
3.5
4
4.5
Imaginary Axis
cΩ =2 cΓ =4.2 kΩ =1 kΓ =2.5
Real Axis
Nicolas Fourrier (U.V.A.)
2
2.3i
−1.5
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
26 / 1
c Impact of c Γ
Numerical Results
Ω
0
Maximum of eigenvalues real part in function of cΩ
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
cΩ =0.6 cΓ =0.6 kΩ =1 kΓ =1
cΩ =2 cΓ =2 kΩ =1 kΓ =1
1
1.5
2.5
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
3
3.5
4
4.5
Imaginary Axis
cΩ =3 cΓ =3 kΩ =1 kΓ =1
1.6i
Real Axis
Nicolas Fourrier (U.V.A.)
2
−1.6
0
on Ω on Γ2,3,4 on Γ1
P.D.E. Seminar
October 4, 2011
27 / 1
