a non-stiff boundary integral method for internal waves
TRANSCRIPT
A non-stiff boundary integral method for internal waves
NJIT, 4 June 2013
Oleksiy Varfolomiyevadvisor Michael Siegel
Wednesday, June 19, 13
Motivation
Develop a model and a numerical method that can be efficiently applied to study the motion of internal waves for doubly periodic interfacial flows with surface tension.
Wednesday, June 19, 13
Outline
•Model Description
•Linear Stability Analysis
•Discretization
•Numerical Experiment
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model Description
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model DescriptionEvolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in 3D.
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model Description
Motion of the fluids is driven by
Evolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in 3D.
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model Description
➡ Gravity
Motion of the fluids is driven by
Evolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in 3D.
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model Description
➡ Gravity
➡ Surface Tension
Motion of the fluids is driven by
Evolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in 3D.
Wednesday, June 19, 13
⇤! p(�) : p(�) = minq⇥Pk
max�⇥[�min,�max]
����1⌅�� q(�)
���� (41)
We used Wolfram Mathematica intrinsic function MiniMaxApproximation toobtain p(�). The next figure shows the approximation error
10 100 1000 104
10!15
10!14
10!13
10!12
10!11
To obtain grid steps we rewrite obtained approximation of the impedancefunction in the form of continued fraction (12). We proceed with Euclidean typealgorithm with 2k polynomial divisions, i.e.
p(�) =ck�1�k�1 + ck�2�k�2 + · · ·+ c0dk�k + dk�1�k�1 + · · ·+ d0
(42)
=1
dk�k+dk�1�k�1+···+d0ck�1�k�1+ck�2�k�2+···+c0
=1
dkck�1
�+
✓dk�1�
dkck�2ck�1
◆�k�1+···+
✓d1�
dkc0ck�1
◆�+d0
ck�1�k�1+ck�2�k�2+···+c0
,
ˆ d
Model Description
➡ Gravity
➡ Surface Tension
➡ Prescribed far-field pressure gradient
Motion of the fluids is driven by
Evolution of the interface between two immiscible, inviscid, incompressible, irrotational fluids of different density in 3D.
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
The interface S is parametrized by
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
The interface S is parametrized by
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
The interface S is parametrized by
The fluid velocities are governed by the Bernoulli’s law
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
@�i
@t�r�i ·Xt +
1
2|r�i|2 +
pi⇢i
+ gz = 0 in Di
The interface S is parametrized by
The fluid velocities are governed by the Bernoulli’s law
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
@�i
@t�r�i ·Xt +
1
2|r�i|2 +
pi⇢i
+ gz = 0 in Di
The interface S is parametrized by
The evolution equation for the free surface S
The fluid velocities are governed by the Bernoulli’s law
Wednesday, June 19, 13
Governing Equations
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
@�i
@t�r�i ·Xt +
1
2|r�i|2 +
pi⇢i
+ gz = 0 in Di
The interface S is parametrized by
The evolution equation for the free surface S
The fluid velocities are governed by the Bernoulli’s law
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Laplace-Young boundary condition
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Laplace-Young boundary condition
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Far-field boundary conditions
Laplace-Young boundary condition
Wednesday, June 19, 13
Boundary Conditions
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Kinematic boundary condition
Far-field boundary conditions
Laplace-Young boundary condition
Wednesday, June 19, 13
1 Model problem
Consider the following boundary value problem: Laplace equation on a semi-infinite strip with the given Neumann data on the left boundary and Dirichletdata elsewhere
�⌅2w(x, y)
⌅y2� ⌅2w(x, y)
⌅x2= 0, (x, y) ⌅ [0,⇤)⇥ [0, 1], (1)
⌅w
⌅x(0, y) = �⇤(y), y ⌅ [0, 1], (2)
w|x=⇥ = 0, (3)
w(x, 0) = 0, w(x, 1) = 0, x ⌅ [0,⇤). (4)
Assume
⇤(y) =m�
i=1
ai sin(i⇥y) (5)
The above problem can be studied as a problem
Aw(x)� d2w(x)
dx2= 0, x ⌅ [0,⇤) (6)
dw
dx(x = 0) = �⇤ (7)
w|x=⇥ = 0, (8)
where A = � �2
�y2 is defined on span{sin (⇥y), . . . , sin (m⇥y)}, spA =
{⇥2, . . . , (m⇥)2}. We can obtain Dirichlet data on the left boundary using
Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = d2w(x)dx2 , therefore
�dw(x)dx = A
12w(x) and now we can use given in (7) Neumann data to get at
x = 0w(0) = f(A)⇤,
here f(�) = �� 12 is impedance function.
2
Wednesday, June 19, 13
Gravity driven flow (Rayleigh-Taylor Instability)
Surface tension interface relaxation
−20
24
68
−20
24
68
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
Numerical Solution
Solution z at T=1 (Implicit method, A=1, dt = 0.1, N=32)
−20
24
68
−20
24
68
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
Numerical Solution
Solution z at T=5 (Implicit method, A=0.5, dt = 0.1, N=32)
Wednesday, June 19, 13
Numerical Results
Max interface height for lin & num soln. Explicit method, N=32, A=0.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
t
max of the lin and num solution zLin and z
lin solnnum soln
Max interface height for lin & num soln. Explicit method, N=32, A=1.
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
t
max of the lin and num solution zLin and z
lin solnnum soln
Wednesday, June 19, 13
Stability ChartLargest stable time step
for the explicit and implicit methods.
4t ⇠ (4x)32
Wednesday, June 19, 13
Conclusions✦ We have developed a non-stiff boundary integral method for
3D internal waves
✦ The algorithm is effective at eliminating the third order t-step constraint that plagues explicit methods
✦ Efficient algorithm for calculating the Birkhoff-Rott integral for a doubly-periodic surface. This algorithm is based on Ewald summation, computes the integral in O(N log N ) operations per time step
✦ Presented method is useful for computing the motion of doubly-periodic fluid interfaces with surface tension in 3D flow
Wednesday, June 19, 13