Symbolic Computation of
Conservation Laws of Nonlinear
PDEs in N+1 Dimensions
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, U.S.A.
whereman@mines.edu
http://www.mines.edu/fs home/whereman/
Department of Mathematics and Statistics
University of Canterbury
Christchurch, New Zealand
Thursday, October 25, 2007, 3:10p.m.
Acknowledgements – Collaborators
Mark Hickman (University of Canterbury, NZ)
Bernard Deconinck (University of Washington)
Michael Colagrosso (CSM)
Ph.D. students: Douglas Poole (CSM) and
Michael Nivala (University of Washington)
Several undergraduate and graduate students
Research supported in part by NSF
under Grant CCR-9901929
This presentation was made in TeXpower
Outline
•
Conservation laws of PDEs in n+1 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
Notation – 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)
Here, Mx = 1 and Mx = 2
•
Total derivative with respect to x:
1
X
2
X
∂f
∂f
∂f
Dx f =
+
v(k+1)x
+
u(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
•
Example of a 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
Additional Examples
•
Example: Kadomtsev-Petviashvili (KP) equation
(ut + αuux + u3x )x + σ 2 u2y = 0
parameter α ∈ IR and σ 2 = ±1.
The equation be written as a conservation law
Dt (ux ) + Dx (αuux + u3x ) + Dy (σ 2 uy ) = 0.
Exchange y and t and set ut = v.
ut = v
vt = −σ 2 (uxy + αu2x + αuu2x + u4x )
•
Examples of conservation laws explicitly
dependent on t, x, y.
2
Dt (xux ) +Dx 3u − u2x − 6xuux + xu3x +Dy αxuy = 0
Dt (yux ) +Dx (αyuux + yu3x ) +Dy −σ 2 u + σ 2 yuy = 0
1 √ 2 √
σ2y2
σ2y2
α tu + tu2x + √ ut + √ u3x
2
4 t
4 t
!
√
√
√
ασ 2 y 2
+ √ uux − x tut − αx tuux − x tu3x
4 t
!
√
yu
y 2 uy
+Dy − √ + √ + x tuy = 0
2 t
4 t
√ tu + Dx
Dt
•
Example: Potential KP equation
Replace u by ux and integrate with respect to x.
uxt + αux u2x + u4x + σ 2 u2y = 0
•
Examples of conservation laws
(not explicitly dependent on x, y, t).
1 2
Dt (ux ) + Dx
αux + u3x + Dy (σ 2 uy ) = 0
2
2 3
2
αux − u22x + 2ux u3x − σ 2 u2y
Dt (ux ) + Dx
3
+Dy 2σ 2 ux uy = 0
Dt ux uy + Dx αu2x uy + ut uy + 2u3x uy − 2u2x uxy
1 3
2 2
+Dy σ uy − ux − ut ux + u22x = 0
3
Dt 2αuux u2x + 3uu4x − 3σ 2 u2y + Dx 2αut u2x + 3u2t
−2αuux utx − 3utx u2x + 3ut u3x + 3ux ut 2x − 3uut 3x )
+Dy 6σ 2 ut uy = 0
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.
.
Computer Demonstration
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!
Formula for Euler operator (variational derivative)
in 1D:
(j)
(0)
Lu(j) (x)
=
M
x
X
(−Dx )
k=0
k
∂
(j)
∂ukx
∂
∂
2 ∂
3 ∂
=
− Dx (j) + Dx (j) − Dx (j) + · · ·
(j)
∂u
∂ux
∂u
∂u
2x
where j = 1, 2, . . . , N.
3x
Formula for Euler operator in 2D:
(j)
Mx
(0,0)
Lu(j) (x,y)
=
(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
where j = 1, 2, . . . , N.
∂
···
Formula for Euler operator in 3D:
(0,0,0)
Lu(j) (x,y,z)
(j)
=
(j)
(j)
M
M
M
y
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
+ D x Dy
+
∂
(j)
∂uxy
D2y
∂
(j)
∂u2y
+ Dx Dz
where j = 1, 2, . . . , N.
+
D2z
∂
(j)
∂uxz
∂
(j)
∂u2z
+ Dy Dz
∂
(j)
∂uyz
−· · ·
Application: Testing Exactness
Example:
f = 3ux v 2 sin u − u3x sin u − 6vvx cos u + 2ux u2x cos u + 8vx v2x
where 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:
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
•
where u(x) and v(x).
Z
Find F = f dx. So, f = Dx F
Inverting Dx and Div
Problem Statement in 1D
•
Example:
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
•
where u(x) and v(x).
Z
Find F = f dx. So, f = Dx F
•
Result (by hand):
F = 4 vx2 + u2x cos u − 3 v 2 cos u
Inverting Dx and Div
Problem Statement in 1D
•
Example:
f = 3ux v 2 sin u−u3x sin u−6vvx cos u+2ux u2x cos u+8vx v2x
•
where u(x) and v(x).
Z
Find F = f dx. So, f = Dx F
•
Result (by hand):
F = 4 vx2 + u2x cos u − 3 v 2 cos u
Mathematica cannot compute this integral!
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
Find F = Div−1 f so, f = Div F
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
Find F = Div−1 f so, f = Div F
•
Result (by hand):
F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica cannot do this!
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
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?
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
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?
Problem Statement in 2D
•
Example:
f = ux vy − u2x vy − uy vx + uxy vx
where u(x, y) and v(x, y).
•
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 (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 (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)
i x 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 (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
kx −ix
(−Dx )
•
k k k x
y
z
i x i y 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
Using the 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
F = Div
−1
f =
(x)
(y)
(z)
(Hu(x,y,z) f, Hu(x,y,z) f, Hu(x,y,z) f )
•
Homotopy Operator in 1D (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
(j)
u
(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 → λu, ux → λux , etc.
More general: u → λ(u − u0 ) + u0 ,
ux → λ(ux − ux 0 ) + ux 0 , 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
Simplified Formula in 1D (variable x)
Z
Hu(x) f =
N
1X
0
dλ
(Iu(j) f )[λu]
λ
j=1
where
(j)
Iu(j) f
=
MX
x −1
Dix
(j)
u
(i+1)
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
+· · ·+uM x
= u +ux
+u2x
Mf =
uix
∂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
M
X
(0)
k ∂f
Lu(x) f =
(−Dx )
∂ukx
k=0
by u to restore the degree.
Split off u ∂f
.
∂u
∂f
Split off ux ∂u
.
x
Integrate by parts.
Repeat the process.
Lastly, split off uM x ∂u∂f .
Mx
Explicitly,
M
X
(0)
k ∂f
uLu(x) f = u
(−Dx )
∂ukx
k=0


M
M
X
X
∂f
k−1 ∂f
k−1 ∂f 

+ ux
(−Dx )
= u
− Dx u
(−Dx )
∂u
∂ukx
∂ukx
k=1
k=1

M
X
∂f
∂f
∂f
k−1
= u
(−Dx )
+ ux
− Dx u
∂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
= ...
∂f
∂f
∂f
= u
+ ux
+ . . . + uM x
∂u
∂ux
∂uM x

M
M
X
X
k−1 ∂f
k−2 ∂f

(−Dx )
−Dx u
+ ux
(−Dx )
∂ukx
∂ukx
k=1
k=2

M
X
k−M ∂f 
+ . . . + u(M −1)x
(−Dx )
∂u
kx
k=M


M
M−1
M
X
X
X
∂f
k−(i+1) ∂f 

=
uix
− Dx
uix
(−Dx )
∂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
dλ
∂f
−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
Dix
(i+1)
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 )
i
+
1
∂u
kx
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.
•
Homotopy Operator in 2D (variables x and y):
(x)
Hu(x,y) f
Z
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
with
(j)
Mx −1
(x)
Iu(j) f =
(j)
My
X X
ix =0 iy =0
1 + ix
1 + ix + iy
!
(y)
i
Dixx Dyy
(1+ix ,iy )
u(j) Lu(j) (x,y)
f
(y)
Analogous formulas for Hu(x,y) f and Iu(j) f.
Simplified Formula in 2D (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)
(x)
Iu(j) f
=
(j)
(j)
MX
y
x −1 M
X
M
x
X
(j)
My
X
ix +iy uix x iy y
ix
ix =0 iy =0
kx =ix +1 ky =iy
(−Dx )
kx −ix −1
−Dy
ky −iy
∂f
(j)
∂ukx x ky y
kx +ky −ix −iy −1
kx −ix −1
kx +ky kx
and
(j)
(y)
Iu(j) f
=
(j)
M
y −1
x
X MX
(j)
ix +iy iy
uix x iy y
ix =0 iy =0
(−Dx )
kx −ix
M
x
X
(j)
My
X
kx =ix ky =iy +1
−Dy
ky −iy −1
∂f
(j)
∂ukx x ky y
kx +ky −ix −iy −1
ky −iy −1
kx +ky ky
Application of Homotopy Operator in 2D
Example: f = ux vy − u2x vy − uy vx + uxy vx
By hand: F̃ = (uvy − ux vy , −uvx + ux vx )
•
Compute
Iu(x) f
∂f
∂f
∂f
= u
+ ux
− uDx
∂ux
∂u2x
∂u2x
1
∂f
1
∂f
+ 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)
Iu f + Iv f [λu]
F1 = Hu(x,y) f =
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)
F2 = Hu(x,y) f =
Iu f + Iv f [λu]
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
uvy
2

+ 41 uy vx − 12 ux vy + 14 uvxy − 12 uy v + 21 uxy v
− 12 uvx − 14 uv2x + 41 ux vx + 21 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 + 43 ux vx − 21 ux v + 21 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: Fast algorithm to remove curl terms
(and strategy to avoid curl terms)
•
Homotopy Operator in 3D (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
ix iy iz
(j) (1+ix ,iy ,ix )
Dx Dy Dz u Lu(j) (x,y,z) f
1 + i x + 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 in 3D (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)
Mz
(j)
Mx
My
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 kx −ix −1
ky −iy
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.
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 = −(
utx +
uty +
vtx +
vty +
θt )
∂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 + 21 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
+ 21 θx hy − ux uy − uuxy − uy vy − u2y v
+2Ωvy − 21 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
dλ
+ v θy + θ hy − θθy h
2
2
2
2
1
1 2
= 2Ωuθ− uuy θ+ uvx θ+ vvy θ+ u θy
3
3
6
1
1
1 2
+ 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


1  4Ωuθ − 2uuy θ + 2uvx θ − hθθy 
(5)
J̃ =
2
4Ωvθ + 2vvx θ − 2vuy θ + hθθx
Needed: Fast algorithm to remove curl terms
(and strategy to avoid curl terms)
Conclusions and Future Work
•
Scope and limitations of the homotopy operator.
−1
Integration by parts, D−1
,
and
Div
.
x
•
Integration of non-exact expressions.
Example: f = ux v + uvx + u2 u2x
R
R 2
f dx = uv + u u2x dx.
•
Integration of parametrized differential functions.
Example: f = aux v + buvx
R
f dx = uv if a = b.
•
Broader class of PDEs (other than those of
evolution type).
•
Full implementation in Mathematica.
Software packages in Mathematica
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. 74(4-5), 352-360 (2007).