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/
Joint Meeting of the AMS and NZMS
Victoria University of Wellington
Wellington, New Zealand
Thursday, December 13, 2007, 10:15a.m.
Acknowledgements – Collaborators
Mark Hickman (University of Canterbury, NZ)
Ph.D. student: Douglas Poole
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)
•
Computer demonstration
•
Tools:
• The Euler operator (testing exactness)
• The homotopy operator (invert Dx and Div)
•
Application to shallow water wave equations
•
Additional examples
•
Conclusions and future work
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.
Conservation Laws
•
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
Strategy to Compute Conservation Laws
• Density is linear combination of scaling invariant
terms with undetermined coefficients.
• Compute Dt ρ with total derivative operator.
• Use variational derivative (Euler operator) to
compute the undetermined coefficients.
• Use homotopy operator to compute flux J
(invert Dx or Div).
• Use linear algebra and variational calculus
(algorithmic).
• Work with linearly independent pieces in finite
dimensional spaces.
.
Computer Demonstration
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 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.
∂
···
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
(0)
Lu(x) f
∂f
∂f
2 ∂f
=
− Dx
+ Dx
≡0
∂u
∂ux
∂u2x
(0)
Lv(x) f
∂f
∂f
2 ∂f
=
− Dx
+ Dx
≡ 0.
∂v
∂vx
∂v2x
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
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 )
•
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
i=0
(j)
(j)
uix
M
x
X
(−Dx )
k−(i+1)
k=i+1
∂f
(j)
∂ukx
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.
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
•
Homotopy Operator in 2D (variables x and y):
(x)
Hu(x,y) f
(y)
Hu(x,y) f
Z
=
N
1X
0 j=1
Z
=
(x)
(Iu(j) f )[λu]
N
1X
0 j=1
(y)
(Iu(j) f )[λu]
dλ
λ
dλ
λ
where
(x)
Iu(j) f
=
(j)
(j)
(j)
Mx
My
X X
X
X
(j)
Mx −1
My
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
∂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)
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)
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)
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
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).