Damped Wave Equation with Dynamic Boundary Conditions Approximate Solution Nicolas Fourrier
U NIVERSITY
OF
V IRGINIA
[email protected]
April 10, 2011
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P.D.E. Seminar
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Overview
1
Finite Element
2
Improvements of the approximation
3
Damped Wave Equation - 2D
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Finite Element
Overview
1
Finite Element Quick Recall Interpolation functions
2
Improvements of the approximation
3
Damped Wave Equation - 2D
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Finite Element
Quick Recall
Domain Definition - 2 Dimension Ω and Ωh Ω: convex domain with smooth boundary δΩ Th : partition of Ω into disjoint polygons. No vertex of any polygons lies on the interior of another polygon. Ωh : union of the polygons ⊂ Ω Size of the partition - h Let h denote the maximal length of the sides of the polygons of the partition Th . Thus h is a parameter which decreases as the partition is made finer. The elements of Th are essentially the same size and the angles each element is bounded below by a positive constant, indepently of h.
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P.D.E. Seminar
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Finite Element
Quick Recall
Finite Element Model
Finite dimensional subspace of H01 (Ω): Shr (Ω) Continuous functions on the closure Ω of Ω Piecewise polynomial of degree at most r − 1 in each polygon of Th Vanish outside Ωh
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P.D.E. Seminar
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Finite Element
Quick Recall
Finite Element Model
Strong form to Variational Form ∆u = ut on Ω
(1)
u = 0 on δΩ gives Z
Ω
Nicolas Fourrier (U.V.A.)
∇u∇vd Ω +
Z
Ω
ut vd Ω = 0 ∀v ∈ H01 (Ω)
P.D.E. Seminar
(2)
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Finite Element
Quick Recall
Finite Element Model On Shr (Ω) Let uje be the value of uhe at the j th node of the element e and Nje are the the interpolations function. u ≈ uh =
n X
uje Nje
(3)
j=1
Then the i th algebraic equation of the finite element model is obtained by substituting v = Nie n X
uje
j=1
Nicolas Fourrier (U.V.A.)
Z
Ωe
∇Nie ∇Nje d Ωe
+
P.D.E. Seminar
Z
Ωe
Nie Nje d Ωe
=0
April 10, 2011
(4)
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Finite Element
Interpolation functions
Construction of the interpolation functions 1-Dimensional Case One interpolation function per node per element. Two nodes per element. At the node, the corresponding shape function is 1, i.e., uhe = uje × 1 At the other node the shape function is 0.
N1e (x) = 1 − N2e (x) =
Nicolas Fourrier (U.V.A.)
x h
P.D.E. Seminar
x h
(5)
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Finite Element
Interpolation functions
1 0.8 0.6 0.4 0.2 0 −1
0
1 Nicolas Fourrier (U.V.A.)
−1
−0.5
P.D.E. Seminar
0
0.5 April 10, 2011
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Finite Element
Interpolation functions
Finite Element Model
2-Dimensional Case Similarly, we can build the shape functions for square elements using the above: x y )(1 − ) h h x y N2e (x) = (1 − ) h h x y e N3 (x) = (1 − ) h h xy e N4 (x) = hh N1e (x) = (1 −
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
(6)
April 10, 2011
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Finite Element
Interpolation functions
1
0.8
0.6
0.4
0.2
0 2 1.5 1 0.5 0 −0.5
2 1.5 1
−1 0.5 0 −1.5
−0.5 −1 −1.5
−2 −2
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P.D.E. Seminar
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Finite Element
Interpolation functions
Mass and stiffness matrix Now we are able to compute the integrals a : K →
Z
Ωe
M→
∇Nie ∇Nje d Ωe =
Z
h
Z
Z
Nie Nje d Ωe =
Z
h
Z
=
Z
hZ
Ωe
0
0
0
a
h 0 h 0 h 0
∇Nie (x, y)∇Nje (x, y)dx dy Nie (x, y)Nje (x, y)dx dy
(7)
x y xy (1 − ) dx dy hh h h for the entry (3,4) of M
We assume that each side of the element is the same
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P.D.E. Seminar
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Finite Element
Interpolation functions
Mass and stiffness matrix II
This provides the element mass (M) and stiffness (K) matrices for each element. Then, incorporating these element matrices into the global stiffness and mass matrices allows to solve the problem. 4 2 1 2 4 −1 −2 −1 1 −1 4 −1 −2 h2 2 4 3 1 (8) K = M= 36 1 2 4 2 6 −2 −1 4 −1 2 1 2 4 −1 −2 −1 4
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P.D.E. Seminar
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Improvements of the approximation
Overview
1
Finite Element
2
Improvements of the approximation Mesh refinement vs. Higher order shape function Example Lobatto Polynomials Chebyshev Points 1-D shape functions using the Chebychev points 2-D shape function
3
Damped Wave Equation - 2D
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Improvements of the approximation
Mesh refinement vs. Higher order shape function
Estimate Error estimate for the finite methods are of the form: |u − uh |L2 (Ω) ≤ Chr Mesh refinement More natural.
Higher order
No additional work is necessary.
Lot of work necessary.
Computation time increase dramatically.
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(9)
P.D.E. Seminar
Much better solution for smooth solution.
April 10, 2011
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Improvements of the approximation
Example
1.2
1
½
−∆u(x) = u=0
π2 πx 4 cos( 2 )
in Ω = [−1; 1] on δΩ
p−refinement
0.8
0.6
exact solution 0.4
h−refinement 0.2
0
−0.2 −1.5
Nicolas Fourrier (U.V.A.)
−1
−0.5
0
P.D.E. Seminar
0.5
1
April 10, 2011
1.5
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Improvements of the approximation
Lobatto Polynomials
Non Uniform distributed nodes
Recall that shape functions are polynomials of degree at most r − 1 in space and distribute the solution uh computed at its corresponding node through the whole element. It is common to use uniformly distributed nodes, but it has been shown that the non-uniform distribution (with high density toward the endpoints) gives the least error in the L2 -norm. The Gauss-Lobatto points (and Chebyshev points) are a way to define this non-uniform distribution. The drawback is that there is no explicit formula to determine them. The following picture shows the GaussLobatto points on a 1-dimension element [−1, 1].
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Improvements of the approximation
16
Chebyshev Points
Chebyshev points for p = 1, 2, ...15
14
12
10
8
6
4
2
0
−1
−0.8
Nicolas Fourrier (U.V.A.)
−0.6
−0.4
−0.2
0
P.D.E. Seminar
0.2
0.4
0.6
0.8
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1
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Improvements of the approximation
1-D shape functions using the Chebychev points
1
(7)
(7)
N1 (x)
N5 (x)
0
−1 (7) (7)
l0 l1
Nicolas Fourrier (U.V.A.)
(7)
l2
(7)
l3
P.D.E. Seminar
1 (7)
l4
(7)
l5
(7) (7)
l6 l7
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Improvements of the approximation
2-D shape function
1.2
1
0.8
0.6
0.4
0.2
−1
0
−0.8 −0.6 −0.4
−0.2
−0.2 0 −0.4 −1
0.2 −0.8
0.4 −0.6
Nicolas Fourrier (U.V.A.)
−0.4
−0.2
0.6 0
0.2
0.4
P.D.E. Seminar
0.8 0.6
0.8
1
1
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Damped Wave Equation - 2D
Overview
1
Finite Element
2
Improvements of the approximation
3
Damped Wave Equation - 2D Domain Problem definition Pictures & Movies
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
21
Domain
22
11 16
23
12 17
7 11
8
14
9
1
10
9
3 3
15
6
8
2 2
20
14
Γ0 7
25
19
13
5
1
13 18
12
6
24
10
4 4
5
Γ
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
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Damped Wave Equation - 2D
Problem definition
Problem definition
on Ω utt + cΩ ut = ∆u − kΩ u du − kΓ ∆Γ u = 0 on Γ = Exterior Boundary utt + cΓ ut + dη u=0 on Γ0 =Interior Boundary
(10)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): Today, kΩ = 0 and kΓ = 1, so that we can focus on the dynamic coefficients cΩ and cΓ . Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
Problem definition
Problem definition
on Ω utt + cΩ ut = ∆u − kΩ u du − kΓ ∆Γ u = 0 on Γ = Exterior Boundary utt + cΓ ut + dη u=0 on Γ0 =Interior Boundary
(10)
There are four damping coefficients (cΩ , kΩ , cΓ , kΓ ): Today, kΩ = 0 and kΓ = 1, so that we can focus on the dynamic coefficients cΩ and cΓ . Subcript Ω and Γ denote coefficients acting respectively on the interior and on the boundary.
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
Pictures & Movies
Next picture
General behavior of the solution Main result, boundary & interior damping Impact of the boundary damping on the solution Impact of the interior damping on the solution
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
24 / 32
Damped Wave Equation - 2D
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
Pictures & Movies
on Ω on Γ on Γ0
1
cΩ =0 cΓ =1 kΩ =0
0.5 0 −0.5
kΓ =1 1 1 0.8
0.5
0.6 0.4 0
Nicolas Fourrier (U.V.A.)
0.2 0
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
Pictures & Movies
Next picture
General behavior of the solution Main result, boundary & interior damping First, let’s observe the spectrum. Impact of the boundary damping on the solution Impact of the interior damping on the solution
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
Pictures & Movies
Imaginary Axis
cΩ =2 cΓ =2 kΩ =0 kΓ =1
8.5i
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 Γ on Γ0
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
Pictures & Movies
on Ω on Γ on Γ0
0.4
cΩ =2 0.2 0
cΓ =2 kΩ =0
−0.2 −0.4 −0.6
kΓ =1 −0.8 1 1 0.8
0.5
0.6 0.4 0
Nicolas Fourrier (U.V.A.)
0.2 0
P.D.E. Seminar
April 10, 2011
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Damped Wave Equation - 2D
Pictures & Movies
Next picture
General behavior of the solution Main result, boundary & interior damping Impact of the boundary damping on the solution Impact of the interior damping on the solution
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
29 / 32
Damped Wave Equation - 2D
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
0.4
cΩ =2
0.4
0.2
cΩ =2
0 −0.2
cΓ =2
Pictures & Movies
cΓ =0
−0.4
kΩ =0 kΓ =1
on Ω on Γ on Γ0
0.2 0 −0.2 −0.4
kΩ =0
−0.6 −0.8
kΓ =1
1
−0.6 −0.8 1
1 0.5 0
Nicolas Fourrier (U.V.A.)
1 0.5
0.5
0.5 0
0
P.D.E. Seminar
0
April 10, 2011
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Damped Wave Equation - 2D
Pictures & Movies
Next picture
General behavior of the solution Main result, boundary & interior damping Impact of the boundary damping on the solution Impact of the interior damping on the solution
Nicolas Fourrier (U.V.A.)
P.D.E. Seminar
April 10, 2011
31 / 32
Damped Wave Equation - 2D
utt + cΩ ut = ∆u − kΩ u utt + cΓ ut + du dη − kΓ ∆Γ u = 0 u=0
0.4
cΩ =2
0.4
0.2
cΩ =0
0 −0.2
cΓ =2
Pictures & Movies
cΓ =2
−0.4
kΩ =0 kΓ =1
on Ω on Γ on Γ0
0.2 0 −0.2 −0.4
kΩ =0
−0.6 −0.8
kΓ =1
1
−0.6 −0.8 1
1 0.5 0
Nicolas Fourrier (U.V.A.)
1 0.5
0.5
0.5 0
0
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0
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