Symbolic Computation of
Conservation Laws of Nonlinear
Partial Differential Equations
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, U.S.A.
whereman@mines.edu
http://inside.mines.edu/∼whereman/
Universidad Nacional de Ingenierı́a
Lima, Peru
Friday, March 13, 2009, 2:00p.m.
.
Acknowledgements
Loren ‘Douglas’ Poole
Ph.D. student, CSM
Research supported in part by NSF
under Grant CCF-0830783
This presentation was made in TeXpower
.
In Memory of
Martin D. Kruskal (1925-2006)
.
Outline
•
Conservation laws of nonlinear PDEs
•
Famous examples in historical perspective
•
Example: Shallow water wave equations (Dellar)
•
Algorithmic method for conservation laws
•
Computer demonstration
•
Tools:
• The variational derivative (testing exactness)
• The homotopy operator (inverting Dx and Div)
•
Application to shallow water wave equations
•
Additional examples
•
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 Examples in Historical Perspective
•
Example: Korteweg-de Vries (KdV) equation
∂u
∂u
∂3u
+u
+
=0
3
∂t
∂x
∂x
Diederik Korteweg
Gustav de Vries
•
First six (of ∞ many) densities-flux pairs:
!
u2
Dt (u) + Dx
+ u2x = 0
2
2 3
2
Dt u + D x
u − ux 2 + 2uu2x = 0
3
Dt u3 − 3ux 2 +
3 4
u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x = 0
Dx
4
108 2
5
2
2
2
u3x +
Dt u − 30 u ux + 36 uu2x −
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
648
216
720
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 + D x
u − ux 2 + 2uu2x = 0
3
Dt u3 − 3ux 2 +
3 4
Dx
u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x = 0
4
.
Transcendental Equations in (1+1)-Dimensions
•
Example: sine-Gordon equation
UXT = sin U
or
utt − uxx = sin u
Written as a first order system:
ut = v
vt = uxx + α sin u
•
Scaling invariance (trick!)
x t 0
(x, t, u, v, α) → ( , , λ u, λv, λ2 α)
λ λ
λ is arbitrary parameter
•
.
First few densities-flux pairs
ρ(1) = 2α cos u + v 2 + ux 2
ρ(2) = 2ux v
J(1) = −2ux v
J(2) = 2α cos u − v 2 − ux 2
ρ(3) = 12vux cos u + 2v 3 ux + 2vux 3 − 16vx u2x
ρ(4) = 2 cos2 u − 2 sin2 u + v 4 + 6v 2 ux 2 + ux 4
+ 4v 2 cos u + 20ux 2 cos u − 16vx 2 − 16u2x 2
J(3) and J(4) are not shown (too long)
.
An 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)


ρ(1) = h
J =
vh


uhθ
(2)

ρ(2) = 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) = (2Ω + vx − uy )θ


(4Ωu − 2uuy + 2uvx − hθy )θ
1
(5)

J =2
(4Ωv + 2vvx − 2vuy + hθx )θ
Generalizations:
Dt (f (θ)h) + Dx (f (θ)uh) + Dy (f (θ)vh) = 0
Dt g(θ)(2Ω + vx − ux )
1
g(θ)(4Ωu − 2uuy + 2uvx − hθy )
+ Dx
2
1
+ Dy
g(θ)(4Ωv − 2uy v + 2vvx + hθx ) = 0
2
for any functions f (θ) and g(θ)
.
Reasons to Compute Conservation Laws
I
Conservation of physical quantities (linear
momentum, mass, energy)
I
Verify the closure of a model
I
Testing of complete integrability and
application of Inverse Scattering Transform
I
Testing of numerical integrators
I
Study of quantitative and qualitative
properties of PDEs (Hamiltonian structure,
recursion operators)
Notation – Computations on the Jet Space
I
Independent variables x = (x, y, z)
I
Dependent variables
u = (u(1) , u(2) , . . . , u(j) , . . . , u(N ) )
In examples: u = (u, v, θ, h, . . .)
∂k u
,
∂xk
I
Partial derivatives ukx =
I
Differential functions
Example: f = uvvx + x2 u3x vx + ux vxx
Total derivative (with respect to x)
I
ukx ly =
(1)
∂
Dx =
+
∂x
(1)
Mx
M
x
X
k=0
∂ k+l u
,
∂xk y l
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)
.
I
Example: f = uvvx + x2 u3x vx + ux vxx
(1)
(2)
Here, Mx = 1 and Mx = 2
I
Total derivative with respect to x:
1
2
X
X
∂f
∂f
∂f
+
v(k+1)x
+
u(k+1)x
Dx f =
∂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 )
.
A Method to Compute Conservation Laws
I Density is linear combination of scaling
invariant terms with undetermined coefficients
I
Compute Dt ρ with total derivative operator
I
Use variational derivative (Euler operator) to
compute the undetermined coefficients
I
Use homotopy operator to compute flux J
(invert Dx or Div)
I
Use linear algebra and variational calculus
(algorithmic)
I
Work with linearly independent pieces in finite
dimensional spaces
.
Algorithm for PDEs in (1+1)-dimensions
I
Example: Density of rank 6 for the KdV
equation
ut + uux + u3x = 0
I
Step 1: Compute the dilation symmetry
Set W (Dx ) = 1 and solve
W (u) + W (Dt ) = 2W (u) + 1 = W (u) + 3
Hence,
W (u) = 2,
W (Dt ) = 3
Thus,
x t
(x, t, u) → ( , 3 , λ2 u)
λ λ
.
I
Step 2: Determine the form of the density
List powers of u, up to rank 6 :
[u, u2 , u3 ]
Introduce x derivatives to ‘adjust’ the rank
u has weight 2, introduce D4x
u2 has weight 4, introduce D2x
u3 has weight 6, no derivative needed
.
Apply the Dx derivatives
Remove total and highest derivative terms:
[u4x ] → [ ]
[ux 2 , uu2x ] → [ux 2 ]
empty list
since uu2x = (uux )x − ux 2
[u3 ] → [u3 ]
Linearly combine the “building blocks”
Candidate density:
ρ = c1 u3 + c2 ux 2
.
I
Step 3: Compute the coefficients ci
Compute
∂ρ
Dt ρ =
+ ρ0 (u)[ut ]
∂t
M
X
∂ρ
∂ρ k
Dx ut
=
+
∂t
∂u
kx
k=0
= (3c1 u2 I + 2c2 ux Dx )ut
Substitute ut by −(uux + u3x )
E = −Dt ρ = (3c1 u2 I + 2c2 ux Dx )(uux + u3x )
= 3c1 u3 ux + 2c2 u3x + 2c2 uux u2x
+3 c1 u2 u3x + 2c2 ux u4x
.
Apply the Euler operator (variational derivative)
m
X
(0)
k ∂
Lu =
(−Dx )
∂ukx
k=0
Here, E has order m = 4, thus
L(0)
u E
∂E
∂E
2 ∂E
3 ∂E
4 ∂E
=
− Dx
+ Dx
− Dx
+ Dx
∂u
∂ux
∂u2x
∂u3x
∂u4x
= −6(3c1 + c2 )ux u2x
This term must vanish!
So, c1 = − 31 c2 . Set c2 = −3, then c1 = 1
Hence, the final form density is
ρ = u3 − 3ux 2
.
I
Step 4: Compute the flux J
Method 1: Integrate by parts (simple cases)
Now,
E = 3u3 ux + 3u2 u3x − 6u3x − 6uux u2x − 6ux u4x
Integration of Dx J = E yields final form of the flux
3 4
J = u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x
4
.
Method 2: Build the form of J (cumbersome)
Note: Rank J = Rank ρ + Rank Dt − 1
Build up form of J. Compute
m
X ∂J
∂J
u(k+1)x
Dx J =
+
∂x k=0 ∂ukx
m is the order of J. Match Dx J = E
.
Method 3: Use the homotopy operator
(most powerful)
Z
Z 1
dλ
−1
(Iu E)[λu]
J = Dx E = E dx = Hu(x) E =
λ
0
with integrand


M
k−1
X
X
k−(i+1)  ∂E

Iu E =
uix (−Dx )
∂ukx
i=0
k=1
Here M = 4, thus
∂E
∂E
Iu E = (uI)(
) + (ux I − uDx )(
)
∂ux
∂u2x
∂E
2
+(u2x I − ux Dx + uDx )(
)
∂u3x
+(u3x I − u2x Dx +
ux D2x
−
uD3x )(
∂E
)
∂u4x
= (uI)(3u3 + 18u2x − 6uu2x − 6u4x )
+(ux I − uDx )(−6uux )
+(u2x I − ux Dx + uD2x )(3u2 )
+(u3x I − u2x Dx + ux D2x − uD3x )(−6ux )
= 3u4 − 18uu2x + 9u2 u2x + 6u22x − 12ux u3x
Note: correct terms but incorrect coefficients!
Finally,
Z
J = Hu(x) E =
0
=
Z 1
1
dλ
(Iu E)[λu]
λ
3λ3 u4 − 18λ2 uu2x + 9λ2 u2 u2x + 6λu22x
0
−12λux u3x ) dλ
3 4
= u − 6uu2x + 3u2 u2x + 3u22x − 6ux u3x
4
Final form of the flux:
3 4
J = u − 6uux 2 + 3u2 u2x + 3u2x 2 − 6ux u3x
4
.
Computer Demonstration
.
I
Review of Vector Calculus
Definition: F is conservative if F = ∇f
I
Definition: F is irrotational or curl free if
∇×F=0
I
Theorem: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
I
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 ?
I
.
I
Review of Vector Calculus
Definition: F is conservative if F = ∇f
I
Definition: F is irrotational or curl free if
∇×F=0
I
Theorem: F = ∇f iff ∇ × F = 0
The curl annihilates gradients!
I
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!
I
.
I
Tools from the Calculus of Variations
Definition:
A differential function f is exact iff f = Dx F
I
Theorem (exactness test):
(0)
f = Dx F iff Lu(j) (x) f ≡ 0, j = 1, 2, . . . , N
I
Definition:
A differential function f is a divergence if
f = Div F
I
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
k=0
∂
(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
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!
•
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!
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
k−(i+1) 
(−Dx )
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,
Iv f
•
Finally,
∂f
∂f
= v
+ (vx I − vDx )
∂vx
∂v2x
= −6v 2 cos u + 8vx2
Z
F = Hu(x) f =
0
=
Z 1
1
dλ
(Iu f + Iv f ) [λu]
λ
3λ2 uv 2 sin(λu) − λ2 uu2x sin(λu)
0
+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
(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 + 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 for PDEs in (2+1)-dimensions
•
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:
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 + uxxx )x + σ 2 uyy = 0
parameter α ∈ IR and σ 2 = ±1.
Equation be written as a conservation law
Dt (ux ) + Dx (αuux + uxxx ) + Dy (σ 2 uy ) = 0.
Exchange y and t and set ut = v
ut = v
vt
1
= − 2 (uxy + αu2x + αuuxx + uxxxx )
σ
•
Examples of conservation laws for KP equation
(explicitly dependent on t, x, and y)
Dt xux +Dx 3u2 −uxx −6xuux +xuxxx +Dy αxuy = 0
Dt yux +Dx y(αuux + uxxx ) +Dy σ 2 (yuy − u) = 0
√ √
2y2
2y2
√
σ
σ
Dt
tu + Dx α2 tu2 + tuxx + √ ut + √ uxxx
4 t
4 t
√
√
√
ασ 2 y 2
+ √ uux − x tut − αx tuux − x tuxxx
4 t
√
y 2 uy
yu +Dy x tuy + √ − √ = 0
4 t
2 t
•
More general conservation laws for KP equation:
Dt f u + Dx f ( α2 u2 + uxx )
2
+( σ2 f 0 y 2 − f x)(ut + αuux + u3x )
+Dy ( 21 f 0 y 2 − σ 2 f x)uy − f 0 yu = 0
Dt f yu + Dx f y( α2 u2 + uxx )
+y(
σ2
+Dy
6
f 0 y 2 − f x)(ut + αuux + u3x )
y( 61 f 0 y 2 − σ 2 f x)uy +(σ 2 f x − 12 f 0 y 2 )u = 0
where f (t) is arbitrary function.
.
•
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
Various generalizations exist
.
•
Example: Khoklov-Zabolotskaya Equation
(describes e.g. sound waves in nonlinear media)
(ut − uux )x − u2y = 0
•
Examples of conservation laws (with f (t)):
Dt (ux ) + Dx (−uux ) + Dy (−uy ) = 0
1 2
1 0 2
Dt (f u) + Dx −(f x + 2 f y )(ut − uux ) − f u
2
+Dy (f x + 21 f 0 y 2 )uy − f 0 yu = 0
1 0 3
1
Dt (f yu) + Dx −(f xy + f y )(ut − uux ) − f yu2
6
2
1 0 2
1 0 3
+Dy (f xy + f y )uy − (f x + f y )u = 0
6
2
.
•
Example: Zakharov-Kuznetsov Equation
(describes e.g. ion acoustic solitons in magnetic
plasma)
ut + αuux + β∇2 ux = 0
in 2-D,
•
∇2
=
∂2
∂x2
+
∂2
∂y 2
Examples of conservation laws:
α 2
Dt (u) + Dx
u + βu2x + Dy βuxy = 0
2
2α 3
2
u − β(u2x − u2y ) + 2βu(u2x + u2y )
Dt u + D x
3
+Dy −2βux uy = 0
.
3β 2
3α 4
3
2
Dt u −
(ux + uy ) + Dx
u + 3βu2 u2x
α
4
2
3β
−6βu(u2x +u2y )+
(u22x −u22y )
α
!
6β 2
(ux (u3x +ux2y )+uy (u2xy +u3y ))
−
α
!
2
6β
+Dy 3βu2 uxy +
uxy (u2x + u2y ) = 0
α
2α 3
2
2
Dt tu − xu + Dx t( u − β(u2x − u2y )
α
3
2 α 2
2β
+2βu(u2x + u2y )) − x( u + βu2x ) +
ux
α 2
α
1
−Dy 2β(tux uy + xuxy ) = 0
α
.
•
Example: Navier’s Equation
(describes e.g. wave motion in elastic solids)
∂2u
ρ 2 = (λ + µ)∇(∇ · u) + µ∆u
∂t
where u = (u, v, w), λ and µ are Lamé’s constants
In components,
ρu2t = (λ + µ) u2x + vxy + wxz + µ u2x + u2y + u2z
ρv2t = (λ + µ) uxy + v2y + wyz + µ v2x + v2y + v2z
ρw2t = (λ + µ) uxz + vyz + w2z + µ w2x + w2y + w2z
.
•
Examples of densities (fluxes are long):
ρ(1) = ρut
ρ(2) = ρvt
ρ(3) = ρwt
ρ(4) = ux (vtz − wty ) − vx (utz − wtx ) + wx (uty − vtx )
ρ(5) = uy (vtz − wty ) + vx wty − vy utz − wx vty + wy uty
ρ(6) = (uy − vx )wtz − (uz − wx )vtz + (vz − wy )utz
ρ(7) = ρ(vt utz −wt (uty −vtx ))+µ uy wzz +vx (uxz −wyy −wzz )
+ vy uyz + vz uzz + wx (vxx − uxy ) − wy uyy
and many more....
.
Conclusions and Future Work
•
The power of Euler and homotopy operators:
I
Testing exactness
I
−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
•
Treat broader class of PDEs (other than those of
evolution type)
•
Full implementation in Mathematica
.
Thank You
Software packages in Mathematica
Codes are available via the Internet:
URL: http://inside.mines.edu/∼whereman/
Publications
1. W. Hereman, P. J. Adams, H. L. Eklund, M. S.
Hickman, and B. M. Herbst, Direct Methods and
Symbolic Software for Conservation Laws of
Nonlinear Equations, In: Advances of Nonlinear
Waves and Symbolic Computation, Ed.: Z. Yan,
Nova Science Publishers, New York (2009),
Chapter 2, pp. 19-79.
2. 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.
3. W. Hereman, Symbolic computation of
conservation laws of nonlinear partial differential
equations in multi-dimensions, Int. J. Quan.
Chem. 106(1), 278-299 (2006).
4. 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).