Symbolic Computation of Conservation Laws of Nonlinear PDEs in Multi-dimensions Willy Hereman

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Symbolic Computation of
Conservation Laws of Nonlinear
PDEs in Multi-dimensions
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, U.S.A.
[email protected]
http://www.mines.edu/fs home/whereman/
Centre de Recherches Mathématiques
Université de Montréal, Montréal, Canada
Tuesday, May 23, 2006, 3:30p.m.
Acknowledgements
Collaborators:
Mark Hickman (Christchurch, NZ), Bernard Deconinck
(Seattle), and Michael Colagrosso (CSM)
Ph.D. student Douglas Poole
Several undergraduate and Master students
Research supported in part by NSF
under Grant CCR-9901929
This presentation was made in TeXpower
Outline
•
Conservation Laws of PDEs in multi-dimensions
•
Example: Shallow water wave equations (Dellar)
•
Algorithmic Methods for conservation laws
•
Computer Demonstration
•
Tools:
• Euler operators (testing exactness)
• Calculus-based formulas for homotopy operator
? symbolic integration by parts
? inversion of the total divergence operator
•
Application to shallow water wave equations
•
Conclusions and Future Work
•
Software and Publications
Notations – Computations on the Jet Space
•
Independent variables x = (x, y, z)
•
Dependent variables u = (u(1) , u(2) , . . . , u(j) , . . . , u(N ) )
In examples: u = (u, v, θ, h, . . .)
∂k u
,
∂xk
∂ k+l u
,
∂xk y l
•
Partial derivatives ukx =
•
Differential functions
Example: f = uvvx + x2 u3x vx + ux vxx for u(x), v(x)
•
Total derivative (with respect to x)
ukx ly =
(1)
∂
Dx =
+
∂x
(1)
Mx
M
x
X
k=0
etc.
(2)
∂
u(k+1)x
+
∂ukx
M
x
X
k=0
∂
v(k+1)x
∂vkx
is the order of f in u (with respect to x), etc.
•
Example: f = uvvx + x2 u3x vx + ux vxx
(1)
(2)
Mx = 1 and Mx = 2
•
Total derivative with respect to x:
1
X
2
X
∂f
∂f
∂f
Dx f =
+
+
u(k+1)x
v(k+1)x
∂x k=0
∂ukx k=0
∂vkx
∂f
∂f
∂f
=
+ ux
+ u2x
∂x
∂u
∂ux
∂f
∂f
∂f
vx
+ v2x
+ v3x
∂v
∂vx
∂v2x
= 2xu3x vx + ux (vvx ) + uxx (3x2 u2x vx + vxx )
+vx (uvx ) + vxx (uv + x2 u3x ) + vxxx (ux )
Conservation Laws
•
Conservation law in (1 + 1) dimensions
Dt ρ + Dx J = 0 (on PDE)
conserved density ρ and flux J
•
Example: Korteweg-de Vries (KdV) equation
ut + uux + u3x = 0
•
Sample conservation law
Dt u3 − 3ux 2 +
3 4
Dx
u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x = 0
4
•
Key property: Dilation invariance
•
Example: KdV equation and its density-flux pairs
are invariant under the scaling symmetry
x t
(x, t, u) → ( , 3 , λ2 u),
λ λ
λ is arbitrary parameter.
•
Some density-flux pairs for the KdV equation:
2
u
ρ(1) = u
J (1) =
+ u2x
2
ρ(2) = u2
J (2)
2u3
=
+ 2uu2x − ux 2
3
ρ(3) = u3 − 3ux 2
J
(3)
3 4
= u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x
4
..
.
ρ(6) = u6 − 60 u3 ux 2 − 30 ux 4 + 108 u2 u2x 2
720
648
216
3
2
+
u2x −
uu3x +
u4x 2
7
7
7
..
.
•
Conservation law in (3 + 1) dimensions
Dt ρ + ∇ · J = Dt ρ + Dx J1 + Dy J2 + Dz J3 = 0 (on PDE)
conserved density ρ and flux J = (J1 , J2 , J3 )
•
Example: Shallow water wave (SWW) equations
[P. Dellar, Phys. Fluids 15 (2003) 292-297]
1
ut + (u·∇)u + 2 Ω × u + ∇(θh) − h∇θ = 0
2
θt + u·(∇θ) = 0
ht + ∇·(uh) = 0
where u(x, y, t), θ(x, y, t) and h(x, y, t)
•
In components:
1
ut + uux + vuy − 2 Ωv + hθx + θhx = 0
2
1
vt + uvx + vvy + 2 Ωu + hθy + θhy = 0
2
θt + uθx + vθy = 0
ht + hux + uhx + hvy + vhy = 0
•
SWW equations are invariant under
(x, y, t, u, v, h, θ, Ω) →
(λ−1 x, λ−1 y, λ−b t, λb−1 u, λb−1 v, λa h, λ2b−a−2 θ, λb Ω)
where W (h) = a and W (Ω) = b
(a, b ∈ Q)
•
First few densities-flux pairs of SWW system:


uh
(1)
(1)


ρ =h
J =
vh


uhθ
(2)
(2)

ρ = hθ
J =
vhθ


2
uhθ

ρ(3) = hθ2
J(3) = 
vhθ2


3 h + uv 2 h + 2uh2 θ
u

ρ(4) = (u2 + v 2 )h + h2 θ
J(4) = 
v 3 h + u2 vh + 2vh2 θ
ρ(5) = vx θ − uy θ + 2Ωθ


4Ωuθ − 2uuy θ + 2uvx θ − hθθy
1
(5)

J =2
4Ωvθ + 2vvx θ − 2vuy θ + hθθx
.
Computer Demonstration
Algorithmic Methods for Conservation Laws
•
Use Noether’s Theorem (Lagrangian formulation)
•
Direct methods (Anderson, Bluman, Anco, Wolf,
etc.) based on solving ODEs (or PDEs)
•
Strategy (linear algebra and variational calculus)
• Density is linear combination of scaling invariant
terms with undetermined coefficients
• Use variational derivative (Euler operator) to
compute the undetermined coefficients
• Use the homotopy operator to compute the flux
(invert Dx or Div)
• Work with linearly independent pieces in finite
dimensional spaces
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem (curl test): F = ∇ × G iff ∇ · F = 0
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem (curl test): F = ∇ × G iff ∇ · F = 0
The divergence annihilates curls!
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem (curl test): F = ∇ × G iff ∇ · F = 0
The divergence annihilates curls!
Question: How can one test that f = ∇ · F ?
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem (gradient test): F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem (curl test): F = ∇ × G iff ∇ · F = 0
The divergence annihilates curls!
Question: How can one test that f = ∇ · F ?
No theorem from vector calculus!
Tools from the Calculus of Variations
•
Definition: a differential function f is exact iff
f = Dx F
•
Theorem (exactness test): f = Dx F iff
(0)
Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
•
Definition: a differential function f is a divergence
if f = Div F
•
Theorem (divergence test): f = Div F iff
(0)
Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
Tools from the Calculus of Variations
•
Definition: a differential function f is exact iff
f = Dx F
•
Theorem (exactness test): f = Dx F iff
(0)
Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
•
Definition: a differential function f is a divergence
if f = Div F
•
Theorem (divergence test): f = Div F iff
(0)
Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
The Euler operator annihilates divergences!
Formulas for Euler operators (variational derivatives):
(j)
(0)
Lu(j) (x)
=
M
x
X
(−Dx )
k
∂
(j)
∂ukx
∂
k=0
∂
2 ∂
3 ∂
=
− Dx (j) + Dx (j) − Dx (j) + · · ·
(j)
∂u
∂ux
∂u
∂u
2x
(j)
Mx
(0,0)
Lu(j) (x,y)
=
3x
(j)
My
X X
kx
(−Dx ) (−Dy )
kx =0 ky =0
∂
ky
(j)
∂ukx x ky y
∂
∂
∂
= (j) − Dx (j) − Dy (j)
∂u
∂ux
∂uy
+ D2x
2 ∂
3 ∂
+Dx Dy (j) +Dy (j) −Dx (j)
(j)
∂u2x
∂uxy
∂u2y
∂u3x
∂
∂
···
(0,0,0)
Lu(j) (x,y,z)
(j)
=
(j)
(j)
My
M
M
x
z
X X X
kx
ky
(−Dx ) (−Dy ) (−Dz )
∂
kz
(j)
∂ukx x ky y kz z
kx =0 ky =0 kz =0
∂
∂
∂
∂
=
− Dx (j) − Dy (j) − Dz (j)
(j)
∂u
∂ux
∂uy
∂uz
+ D2x
∂
(j)
∂u2x
+ Dx Dy
+
∂
(j)
∂uxy
D2y
∂
(j)
∂u2y
+ Dx Dz
+
D2z
∂
(j)
∂uxz
∂
(j)
∂u2z
+ D y Dz
∂
(j)
∂uyz
−· · ·
Application: Testing Exactness
Consider, for example,
f = 3ux v 2 sin u − u3x sin u − 6vvx cos u + 2ux u2x cos u + 8vx v2x
for u(x) and v(x)
•
f is exact
•
After integration by parts (by hand):
Z
F = f dx = 4 vx2 + u2x cos u − 3 v 2 cos u
•
Exactness test with Euler operator:
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
∂f
∂f
2 ∂f
=
− Dx
+ Dx
∂u
∂ux
∂u2x
= 3ux v 2 cos u − u3x cos u + 6v vx sin u − 2ux u2x sin u
(0)
Lu(x) f
−Dx [3v 2 sin u − 3u2x sin u + 2u2x cos u] + D2x [2ux cos u]
= 3ux v 2 cos u − u3x cos u + 6v vx sin u − 2ux u2x sin u
−[3ux v 2 cos u + 6v vx sin u − 3u3x cos u − 6u u2x sin u
−2ux u2x sin u + 2u3x cos u]
+[−2u3x cos u − 6ux u2x sin u + 2u3x cos u] = 0
Similarly,
(0)
Lv(x) f
≡0
Inverting Dx and Div
Problem Statement
•
In 1D:
Example:
For u(x) and v(x)
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
Z
•
Find F =
•
Result (by hand):
f dx so, f = Dx F
F = 4 vx2 + u2x cos u − 3 v 2 cos u
Inverting Dx and Div
Problem Statement
•
In 1D:
Example:
For u(x) and v(x)
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
Z
•
Find F =
•
Result (by hand):
f dx so, f = Dx F
F = 4 vx2 + u2x cos u − 3 v 2 cos u
Mathematica cannot compute this integral!
•
In 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
•
In 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica cannot do this!
•
In 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica cannot do this!
Can this be done without integration by parts?
•
In 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica cannot do this!
Can this be done without integration by parts?
Can this be reduced to single integral in one variable?
•
In 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica cannot do this!
Can this be done without integration by parts?
Can this be reduced to single integral in one variable?
Yes! With the Homotopy operator
Tools from Differential Geometry
Higher Euler Operators
•
In 1D (with variable x):
(j)
(i)
Lu(j) (x)
=
M
x
X
k=i
k i
(−Dx )
k−i
∂
(j)
∂ukx
Examples for component u(x):
(1)
Lu(x)
(2)
Lu(x)
(3)
Lu(x)
∂
∂
2 ∂
3 ∂
=
− 2Dx
+ 3Dx
− 4Dx
+ ···
∂ux
∂u2x
∂u3x
∂u4x
∂
∂
2 ∂
3 ∂
=
− 3Dx
+ 6Dx
− 10Dx
+ ···
∂u2x
∂u3x
∂u4x
∂u5x
∂
∂
2 ∂
3 ∂
=
− 4Dx
+ 10Dx
− 20Dx
+ ···
∂u3x
∂u4x
∂u5x
∂u6x
•
In 2D (with variables x and y):
(j)
(j)
My
M
x
X
X
(ix ,iy )
Lu(j) (x,y)=
kx =ix ky =iy
k k ∂
x
y
kx −ix
ky −iy
(−Dx )
(−Dy )
(j)
ix iy
∂u
kx x k y y
Examples for component u(x, y):
∂
∂
∂
2 ∂
− 2Dx
− Dy
+ 3Dx
+ ···
∂ux
∂u2x
∂uxy
∂u3x
∂
∂
∂
(0,1)
2 ∂
Lu(x,y) =
− 2Dy
− Dx
+ 3Dy
+ ···
∂uy
∂u2y
∂uyx
∂u3y
∂
∂
∂
∂
(1,1)
2
Lu(x,y) =
− 2Dx
− 2Dy
+ 3Dx
+ ···
∂uxy
∂u2xy
∂ux2y
∂u3xy
∂
∂
∂
∂
(2,1)
2
Lu(x,y) =
− 3Dx
− 2Dy
+ 6Dx
+ ···
∂u2xy
∂u3xy
∂u2x2y
∂u4xy
(1,0)
Lu(x,y) =
•
In 3D (with variables x, y, and z):
(j)
(ix ,iy ,iz )
Lu(j) (x,y,z)
(j)
(j)
M
M
M
y
x
z
X
X X
=
kx =ix ky =iy kz =iz
(−Dx )
•
kx −ix
k k k x
y
z
ix iy iz
(−Dy )
ky −iy
(−Dz )
kz −iz
∂
(j)
∂ukx x ky y kz z
Theorem:
[Kruskal et al, J. Math. Phys. 11 (1970) 952-960]
(i)
f = Dxr F iff Lu(j) (x) f ≡ 0 for i = 0, 1, . . . , r−1,
j = 1, . . . , N
Integration by Parts with Homotopy Operator
•
Theorem (integration with homotopy operator):
• In 1D: If f is exact then
F = D−1
x f =
Z
f dx = Hu(x) f
• In 2D: If f is a divergence then
F = Div
−1
f =
(x)
(y)
(Hu(x,y) f, Hu(x,y) f )
• In 3D: If f is a divergence then
(x)
(y)
(z)
F = Div−1 f = (Hu(x,y,z) f, Hu(x,y,z) f, Hu(x,y,z) f )
Homotopy Operators
•
In 1D (with variable x):
Z
Hu(x) f =
N
1X
0
dλ
(Iu(j) f )[λu]
λ
j=1
with integrand
(j)
Iu(j) f =
MX
x −1
Dix
u
(j)
(i+1)
Lu(j) (x) f
i=0
N is the number of dependent variables and
(Iu(j) f )[λu] means that in Iu(j) f one replaces
u(x) → λu(x), ux (x) → λux (x), etc.
Example: u(x) = (u(1) (x), u(2) (x)) = (u(x), v(x)):
1
Z
Hu(x) f =
0
dλ
(Iu f + Iv f ) [λu]
λ
where
(1)
Iu f =
MX
x −1
Dix
Dix
(i+1)
u Lu(x) f
i=0
and
(2)
Iv f =
MX
x −1
i=0
(i+1)
v Lv(x) f
•
In 2D (with variables x and y):
(x)
Hu(x,y) f
Z
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
with
(j)
(x)
Iu(j) f
=
(j)
MX
y
x −1 M
X
ix =0 iy =0
1 + ix
1 + ix + iy
Analogous formulas for
!
ix iy
Dx Dy
(y)
Hu(x,y) f
and
(j) (1+ix ,iy )
u Lu(j) (x,y) f
(y)
Iu(j) f
•
In 3D (with variables x, y, and z):
(x)
Hu(x,y,z) f
Z
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
with
(x)
Iu(j) f =
(j)
(j)
(j)
MX
y Mz
x −1 M
X
X
ix =0 iy =0 iz =0
!
1 + ix
(j) (1+ix ,iy ,ix )
ix iy iz
Dx Dy Dz u Lu(j) (x,y,z) f
1 + ix + iy + iz
Analogous formulas for
(y)
Iu(j) f,
and
(z)
Iu(j) f
(y)
Hu(x,y,z) f,
(z)
Hu(x,y,z) f,
Simplified Formula for Homotopy Operator
Homotopy Operator in 1D (with variable x)
Z
Hu(x) f =
0
N
1X
dλ
(Iu(j) f )[λu]
λ
j=1
where
(j)
Iu(j) f
=
MX
x −1
(i+1)
Dix u(j) Lu(j) (x) f
i=0
(j)
=
MX
x −1
i=0
(j)
(j)
uix
M
x
X
k=i+1
(−Dx )
k−(i+1)
∂f
(j)
∂ukx
Application of Homotopy Operator in 1D
Example:
f = 3ux v 2 sin u − u3x sin u − 6vvx cos u + 2ux u2x cos u + 8vx v2x
•
Compute
Iu f
∂f
∂f
∂f
= u
+ ux
− uDx
∂ux
∂u2x
∂u2x
= 3uv 2 sin u − uu2x sin u + 2u2x cos u
•
Similarly,
Iv f
•
Finally,
∂f
∂f
∂f
= v
+ vx
− vDx
∂vx
∂v2x
∂v2x
= −6v 2 cos u + 8vx2
Z
F = Hu(x) f =
0
Z
=
0
1
1
dλ
(Iu f + Iv f ) [λu]
λ
3λ2 uv 2 sin(λu) − λ2 uu2x sin(λu) + 2λu2x cos(λu)
−6λv 2 cos(λu) + 8λvx2 dλ
= 4vx2 + u2x cos u − 3v 2 cos u
Why does this work?
Sketch of Derivation and Proof
(in 1D with variable x, and for one component u)
Definition: Degree operator M
M
X
∂f
∂f
∂f
∂f
∂f
Mf =
uix
= u +ux
+u2x
+· · ·+uM x
∂uix
∂u
∂ux
∂u2x
∂uM x
i=0
f is of order M in x
Example: f = up uqx ur3x (p, q, r non-negative integers)
3
X
∂f
g = Mf =
uix
= (p + q + r) up uqx ur3x
∂uix
i=0
Application of M computes the total degree
Theorem (Inverse operator)
M−1 g(u)
=
R1
0
g[λu] dλ
λ
Proof:
d
g[λu] =
dλ
M
X
∂g[λu] dλuix
i=0
∂λuix
dλ
M
X
1
∂g[λu]
1
=
uix
= Mg[λu]
λ i=0
∂uix
λ
Integrate both sides with respect to λ
Z 1
λ=1
d
g[λu] dλ = g[λu]λ=0 = g(u) − g(0)
0 dλ
Z 1
Z 1
dλ
dλ
=
Mg[λu]
=M
g[λu]
λ
λ
0
0
Assuming g(0) = 0,
−1
M
Z
g(u) =
0
1
dλ
g[λu]
λ
Example:
If g(u) = (p + q + r) up uqx ur3x , then
g[λu] = (p + q +
r)λp+q+r
q r
p
u ux u3x
Hence,
M−1 g =
Z
1
(p + q + r) λp+q+r−1 up uqx ur3x dλ
0
= up uqx ur3x
λ=1
p+q+r λ
= up uqx ur3x
λ=0
Theorem: If f is an exact differential function, then
Z
F = Dx−1 f = f dx = Hu(x) f
Proof: Multiply
(0)
Lu(x) f
M
X
∂f
=
(−Dx )
∂ukx
k=0
k
by u to restore the degree.
Split off u ∂f
. Integrate by parts.
∂u
∂f
Split off ux ∂u
. Repeat the process.
x
Lastly, split off uM x ∂u∂f .
Mx
Explicitly,
M
X
∂f
=u
(−Dx )
∂ukx
k=0


M
M
X
X
∂f
k−1 ∂f 
k−1 ∂f

= u
− Dx u
(−Dx )
+ ux
(−Dx )
∂u
∂ukx
∂ukx
k=1
k=1

M
X
∂f
∂f
∂f
k−1
= u
+ ux
− Dx u
(−Dx )
∂u
∂ux
∂ukx
k=1

M
M
X
X
k−2 ∂f 
k−2 ∂f
+ux
(−Dx )
+ u2x
(−Dx )
∂ukx
∂ukx
k=2
k=2
(0)
uLu(x) f
= ...
k
∂f
∂f
∂f
= u
+ ux
+ . . . + uM x
∂u
∂ux
∂uM x

M
M
X
X
k−1 ∂f
k−2 ∂f

−Dx u
(−Dx )
+ ux
(−Dx )
∂ukx
∂ukx
k=1
k=2

M
X
k−M ∂f 
+ . . . + u(M −1)x
(−Dx )
∂u
kx
k=M


M−1
M
M
X
X
X
∂f
k−(i+1) ∂f 

− Dx
uix
(−Dx )
=
uix
∂uix
∂ukx
i=0
i=0
k=i+1


M−1
M
X
X
∂f
k−(i+1)

= Mf − Dx 
uix
(−Dx )
∂ukx
i=0
k=i+1
= 0
So,

M−1
X
Mf = Dx 
i=0
uix
M
X
k=i+1

(−Dx )
k−(i+1)
∂f 
∂ukx
Apply M−1 and use M−1 Dx = Dx M−1 .


M−1
M
X
X
−1
k−(i+1) ∂f 

f = Dx M
uix
(−Dx )
∂ukx
i=0
k=i+1
Apply Dx−1 and use the formula for M−1


Z 1 M−1
M
X
X
∂f
dλ
−1
k−(i+1)

 [λu]
F = Dx f =
uix
(−Dx )
∂ukx
λ
0
i=0
k=i+1
= Hu(x) f
using
Iu f
=
M−1
X
(i+1)
Dix u Lu(x) f
i=0
=
M−1
X
i=0
=
M−1
X
i=0


M X
k
i 
k−(i+1) ∂f 
Dx u
(−Dx )
∂ukx
i+1
k=i+1
uix
M
X
k=i+1
(−Dx )
k−(i+1)
∂f
∂ukx
The latter formula provides a fast algorithm to
compute the integrand Iu f
Simplified Formula for Homotopy Operator
Homotopy Operator in 2D (with variables x and y)
(x)
Hu(x,y) f
(y)
Hu(x,y) f
Z
=
N
1X
0 j=1
Z
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
(y)
(Iu(j) f )[λu]
dλ
λ
where
(j)
(j)
(x)
Iu(j) f
=
MX
y
x −1 M
X
(j)
ix +iy ix
uix x iy y
ix =0 iy =0
(−Dx )
M
x
X
(j)
My
X
kx =ix +1 ky =iy
kx −ix −1
−Dy
ky −iy
kx +ky −ix −iy −1
kx −ix −1
kx +ky kx
∂f
(j)
∂ukx x ky y
and
(j)
(y)
Iu(j) f
=
(j)
M
y −1
x
X MX
ix =0 iy =0
(−Dx )
(j)
M
x
X
(j)
My
X
ix +iy uix x iy y
iy
kx =ix ky =iy +1
kx −ix
−Dy
ky −iy −1
∂f
(j)
∂ukx x ky y
kx +ky −ix −iy −1
ky −iy −1
kx +ky ky
Simplified Formula for Homotopy Operator
Homotopy Operator in 3D (with variables x, y, and z)
(x)
Hu(x,y,z) f
(y)
Hu(x,y,z) f
(z)
Hu(x,y,z) f
Z
=
N
1X
0 j=1
Z
=
N
1X
0 j=1
Z
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
(y)
(Iu(j) f )[λu]
dλ
λ
(z)
(Iu(j) f )[λu]
dλ
λ
where
(x)
Iu(j) (x,y,z) f
(j)
(j)
=
(j)
MX
y Mz
x −1 M
X
X
ix +iy +iz iy +iz (j)
uix x iy y iz z
ix
iy
ix =0 iy =0 iz =0
(j)
(j)
Mx
My
(j)
Mz
X
X X
kx =ix +1 ky =iy kz =iz
(−Dx )
Integrands
kx −ix −1
kx +ky +kz −ix −iy −iz −1 ky +kz −iy −iz ky −iy
kx −ix −1
kx +ky +kz ky +kz kx
ky
−Dy
(y)
Iu(j) (x,y,z) f
ky −iy
and
(−Dz )
kz −iz
(z)
Iu(j) (x,y,z) f
∂f
(j)
∂ukx x ky y kz z
are similar
Application of Homotopy Operator in 2D
Example: f = ux vy − u2x vy − uy vx + uxy vx
Recall (by hand): F̃ = (uvy − ux vy , −uvx + ux vx )
•
Compute
Iu(x) f
∂f
∂f
∂f
= u
+ ux
− uDx
∂ux
∂u2x
∂u2x
∂f
1
∂f
1
+ uy
− uDy
2 ∂uxy
2
∂uxy
1
1
= uvy + uy vx − ux vy + uvxy
2
2
•
Similarly,
Iv(x) f
•
∂f
=v
= −uy v + uxy v
∂vx
Hence,
Z 1
dλ
(x)
(x)
(x)
F1 = Hu(x,y) f =
Iu f + Iv f [λu]
λ
0
Z 1 1
1
= λ uvy + uy vx − ux vy + uvxy − uy v + uxy v dλ
2
2
0
1
1
1
1
1
1
= uvy + uy vx − ux vy + uvxy − uy v + uxy v
2
4
2
4
2
2
•
Analogously,
Z 1
dλ
(y)
(y)
(y)
Iu f + Iv f [λu]
F2 = Hu(x,y) f =
λ
0
!
Z 1 1
1
=
λ −uvx − uv2x + ux vx + λ (ux v − u2x v) dλ
2
2
0
1
1
1
1
1
= − uvx − uv2x + ux vx + ux v − u2x v
2
4
4
2
2
•
So,

F=
1
uv
2 y
+ 14 uy vx − 12 ux vy + 14 uvxy − 12 uy v + 12 uxy v
− 12 uvx − 14 uv2x + 14 ux vx + 12 ux v − 12 u2x v


Let K = F̃−F then


1
1
1
1
1
1
uv
−
u
v
−
u
v
−
uv
+
u
v
−
uxy v
y
y x
x y
xy
y
2
4
2
4
2
2

K=
− 12 uvx + 14 uv2x + 34 ux vx − 12 ux v + 12 u2x v
then Div K = 0
•
Also, K = (Dy θ, −Dx θ) with θ = 12 uv − 14 uvx − 12 ux v
(curl in 2D)
Needed: Algorithm to remove curl terms!
Computation of Conservation Laws for SWW
Quick Recapitulation
•
Conservation law in (2 + 1) dimensions
Dt ρ + ∇ · J = Dt ρ + Dx J1 + Dy J2 = 0 (on PDE)
conserved density ρ and flux J = (J1 , J2 )
•
Example: Shallow water wave (SWW) equations
1
ut + uux + vuy − 2 Ωv + hθx + θhx = 0
2
1
vt + uvx + vvy + 2 Ωu + hθy + θhy = 0
2
θt + uθx + vθy = 0
ht + hux + uhx + hvy + vhy = 0
•
Typical density-flux pair:
ρ(5) = vx θ − uy θ + 2Ωθ


1  4Ωuθ − 2uuy θ + 2uvx θ − hθθy 
(5)
J
=
2
4Ωvθ + 2vvx θ − 2vuy θ + hθθx
Algorithm
•
Step 1: Construct the form of the density
The SWW equations are invariant under the
scaling symmetries
(x, y, t, u, v, θ, h, Ω) → (λ−1 x, λ−1 y, λ−2 t, λu, λv, λθ, λh, λ2 Ω)
and
(x, y, t, u, v, θ, h, Ω) → (λ−1 x, λ−1 y, λ−2 t, λu, λv, λ2 θ, λ0 h, λ2 Ω)
Construct a candidate density, for example,
ρ = c1 Ωθ + c2 uy θ + c3 vy θ + c4 ux θ + c5 vx θ
which is scaling invariant under both symmetries
•
Step 2: Determine the constants ci
Compute E = −Dt ρ and remove time derivatives
∂ρ
∂ρ
∂ρ
∂ρ
∂ρ
E = −(
θt )
utx +
uty +
vtx +
vty +
∂ux
∂uy
∂vx
∂vy
∂θ
= c4 θ(uux + vuy − 2Ωv + 12 hθx + θhx )x
+ c2 θ(uux + vuy − 2Ωv + 12 hθx + θhx )y
+ c5 θ(uvx + vvy + 2Ωu + 12 hθy + θhy )x
+ c3 θ(uvx + vvy + 2Ωu + 12 hθy + θhy )y
+ (c1 Ω + c2 uy + c3 vy + c4 ux + c5 vx )(uθx + vθy )
Require that
(0,0)
(0,0)
(0,0)
(0,0)
Lu(x,y) E = Lv(x,y) E = Lθ(x,y) E = Lh(x,y) E ≡ 0
•
Solution: c1 = 2, c2 = −1, c3 = c4 = 0, c5 = 1 gives
ρ = 2Ωθ − uy θ + vx θ
•
Step 3: Compute the flux J
E = θ(ux vx + uv2x + vx vy + vvxy + 2Ωux
+ 12 θx hy − ux uy − uuxy − uy vy − u2y v
+2Ωvy − 12 θy hx )
+2Ωuθx + 2Ωvθy − uuy θx
−uy vθy + uvx θx + vvx θy
Apply the 2D homotopy operator:
(x)
(y)
J = (J1 , J2 ) = Div−1 E = (Hu(x,y) E, Hu(x,y) E)
Compute
Iu(x) E
∂E
∂E
∂E
1
∂E
1
∂E
=u
+ux
−uDx
+ uy
− uDy
∂ux
∂u2x
∂u2x 2 ∂uxy 2
∂uxy
1 2
= uvx θ + 2Ωuθ + u θy − uuy θ
2
Similarly, compute
1 2
= vvy θ + v θy + uvx θ
2
1 2
(x)
Iθ E =
θ hy + 2Ωuθ − uuy θ + uvx θ
2
1
(x)
Ih E = − θθy h
2
Iv(x) E
Next,
J1 =
=
(x)
Hu(x,y) E
Z 1
Iu(x) E
0
Z
=
0
1
dλ
(x)
(x)
(x)
+ Iv E + Iθ E + Ih E [λu]
λ
1 2
2
4λΩuθ + λ 3uvx θ + u θy − 2uuy θ + vvy θ
2
!
1 2
1 2
1
+ v θy + θ hy − θθy h
dλ
2
2
2
2
1
1 2
= 2Ωuθ− uuy θ+ uvx θ+ vvy θ+ u θy
3
3
6
1 2
1
1
+ v θy − hθθy + hy θ2
6
6
6
Analogously,
J2 =
(y)
Hu(x,y) E
2
1
1 2
1 2
= 2Ωvθ + vvx θ − vuy θ − uux θ − u θx − v θx
3
3
6
6
1
1
+ hθθx − hx θ2
6
6
Hence,


112Ωuθ−4uuy θ+6uvx θ+2vvy θ+u2 θy +v 2 θy −hθθy +hy θ2
J=
6 12Ωvθ+4vvx θ−6vuy θ−2uux θ−u2 θx −v 2 θx +hθθx −hx θ2
After removing the curl term (by hand)

J̃
(5)

1  4Ωuθ − 2uuy θ + 2uvx θ − hθθy 
=
2
4Ωvθ + 2vvx θ − 2vuy θ + hθθx
Needed: Algorithm to remove curl terms!
Conclusions and Future Work
•
Usefulness of the homotopy operator
•
−1
Integration by parts, D−1
,
and
Div
x
•
To do: Integration of non-exact differential
functions
Example: f = ux v + uvx + u2 u2x
R
R 2
f dx = uv + u u2x dx
•
To do: Integration of parametrized differential
functions
Example: f = aux v + buvx
R
f dx = uv if a = b
.
•
Algorithm to compute conservation laws in
multi-dimensions
•
To do: various PDEs (other than those of
evolution type)
•
To do: Full implementation in Mathematica
Implementation in Mathematica & Software
Codes are available via the Internet
URL: http://www.mines.edu/fs home/whereman/
and via anonymous FTP from mines.edu in directory
pub/papers/math cs dept/software/
Publications
1. W. Hereman, M. Colagrosso, R. Sayers, A.
Ringler, B. Deconinck, M. Nivala, and M. S.
Hickman, Continuous and Discrete Homotopy
Operators and the Computation of Conservation
Laws. In: Differential Equations with Symbolic
Computation, Eds.: D. Wang and Z. Zheng,
Birkhäuser Verlag, Basel (2005), Chapter 15, pp.
249-285.
2. W. Hereman, Symbolic computation of
conservation laws of nonlinear partial differential
equations in multi-dimensions, Int. J. Quan.
Chem. 106(1), 278-299 (2006).
3. W. Hereman, B. Deconinck, and L. D. Poole,
Continuous and discrete homotopy operators:
A theoretical approach made concrete, Math.
Comput. Simul. (2006) submitted.
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