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 Mathematical and Statistical Sciences
University of Kwa-Zulu Natal, Durban, South Africa
Monday, April 7, 2008, 12:30
.
Acknowledgements – Collaborators
Mark Hickman (Univ. of Canterbury, New Zealand)
Bernard Deconinck (University of Washington, Seattle)
Loren ‘Douglas’ Poole (Ph.D. student, CSM)
Research supported in part by NSF
under Grant CCR-9901929
This presentation was made in TeXpower
.
In Memory of
Martin D. Kruskal (1925-2006)
.
Outline
•
Conservation laws of nonlinear PDEs
•
Famous example in historical perspective
•
Example: Shallow water wave equations (Dellar)
•
Algorithmic methods for conservation laws
•
Computer demonstration
•
Tools:
• The Euler operator (testing exactness)
• The homotopy operator (invert Dx and Div)
•
Application to shallow water wave equations
•
Conclusions and future work
.
Conservation Laws
•
Conservation law in (1+1) dimensions
Dt ρ + Dx J = 0 (on PDE)
conserved density ρ and flux J
•
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 )
.
Famous Example in Historical Perspective
•
Example: Korteweg-de Vries (KdV) equation
ut + uux + u3x = 0
Diederik Korteweg
Gustav de Vries
•
Examples of conservation laws
!
u2
Dt (u) + Dx
+ u2x = 0
2
2 3
2
Dt u + Dx
u − ux 2 + 2uu2x = 0
3
Dt u3 − 3ux 2 +
3 4
Dx
u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x = 0
4
108 2
5
2
2
2
Dt u − 30 u ux + 36 uu2x −
u3x +
7
5 6
216
3 2
Dx
u − 40u ux − . . . −
u3x u5x = 0
6
7
Dt u6 − 60 u3 ux 2 − 30 ux 4 + 108 u2 u2x 2
720
648
216
3
2
+
u2x −
uu3x +
u4x 2 +
7
7
7
6 7
432
4 2
Dx
u − 75u ux − . . . +
u4x u6x = 0
7
7
•
Third conservation law: Gerald Whitham, 1965
•
Fourth and fifth: Norman Zabusky, 1965-66
•
Sixth: algebraic mistake, 1966
•
Seventh (sixth thru tenth): Robert Miura, 1966
.
Robert Miura
.
•
First five: IBM 7094 computer with FORMAC
(1966) — storage space problem!
•
First eleven densities: AEC CDC-6600 computer
(2.2 seconds) — large integers problem!
•
2006 Leroy P. Steele Prize (AMS): Gardner,
Greene, Kruskal, and Miura
•
National Medal of Science, von Neuman Prize
(SIAM), . . .: Martin Kruskal
.
•
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.
•
Examples of conservation laws
2 3
2
Dt u + Dx
u − ux 2 + 2uu2x = 0
3
Dt u3 − 3ux 2 +
3 4
Dx
u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x = 0
4
.
Example in (2+1) Dimensions
•
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
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 =
+
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 )
.
Review of Vector Calculus
•
Definition: F is conservative if F = ∇f
•
Definition: F is irrotational or curl free if
∇×F=0
•
Theorem: 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: 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: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem: 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: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem: 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: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem: 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: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
•
Definition: F is incompressible or divergence free
if ∇ · F = 0
•
Theorem: 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):
(0)
f = Dx F iff Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
•
Definition:
A differential function f is a divergence if
f = Div F
•
Theorem (exactness test):
(0)
f = Div F iff 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):
(0)
f = Dx F iff Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
•
Definition:
A differential function f is a divergence if
f = Div F
•
Theorem (exactness test):
(0)
f = Div F iff 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 =
0
N
1X
dλ
(Iu(j) f )[λu]
λ
j=1
with integrand
(j)
Mx
Iu(j) f =

X
k−1
X

k=1

(j)
uix
(−Dx )
k−(i+1) 
i=0
∂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
= u
+ (ux I − uDx )
∂ux
∂u2x
= 3uv 2 sin u − uu2x sin u + 2u2x cos u
.
•
Similarly,
∂f
∂f
= v
+ (vx I − vDx )
∂vx
∂v2x
= −6v 2 cos u + 8vx2
Iv f
•
Finally,
Z
F = Hu(x) f =
0
=
1
dλ
(Iu f + Iv f ) [λu]
λ
Z 1
0
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
=
N
1X
0 j=1
(x)
(Iu(j) f )[λu]
dλ
λ
(y)
(Iu(j) f )[λu]
dλ
λ
where
(j)
Mx
(x)
Iu(j) f
=
(j)
My
X X

ky
kX
x −1 X

kx =1 ky =0
(−Dx )
ix +iy uix x iy y ix
ix =0 iy =0
kx −ix −1
−Dy
ky −iy kx +ky −ix −iy −1
kx −ix −1
kx +ky kx
∂f
(j)
∂ukx x ky y
.
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
+ (ux I − uDx )
= u
∂ux
∂u2x
1
1
∂f
+
uy I − uDy
2
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)
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 + 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 + 43 ux vx − 12 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 + 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:
J = (J1 , J2 ) = Div
−1
E=
(x)
(y)
(Hu(x,y) E, Hu(x,y) E)
.
Compute
Iu(x) E
∂E
=u
+
∂ux
1
1
∂E
uy I − uDy
2
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


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
√ Dt
tu + Dx
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
.
•
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
Dt (ux ) + Dx
αux − u22x + 2ux u3x − σ 2 u2y
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).