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/
Department of Mathematics
Colorado State University, Fort Collins, Colorado
Thursday, September 17, 2009, 1:00p.m.
.
Acknowledgements
Mark Hickman (Univ. of Canterbury, New Zealand)
Bernard Deconinck(University of Washington, Seattle)
Loren ‘Douglas’ Poole (Ph.D. Student, CSM)
Several undergraduate and graduate students
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!
IBM 7094 Computer
•
First eleven densities: Control Data Computer
CDC-6600 computer (2.2 seconds) — large
integers problem!
Control Data CDC-6600
Control Data CDC-6600 in Museum
•
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, . . .)
I
Partial derivatives ukx =
I
Differential functions
∂k u
,
∂xk
ukx ly =
Example: f = uvvx + x2 u3x vx + ux vxx
∂ k+l u
,
∂xk y l
etc.
.
I
Total derivative (with respect to x)
(2)
(1)
Dx
∂
+
=
∂x
M
x
X
k=0
∂
+
u(k+1)x
∂ukx
M
x
X
k=0
∂
v(k+1)x
∂vkx
∂
∂
∂
∂
=
+ ux
+ uxx
+ u3x
+ ...
∂x
∂u
∂ux
∂uxx
∂
∂
∂
+vx
+ vxx
+ v3x
+ ...
∂v
∂vx
∂vxx
(1)
Mx
(2)
Mx
is the order of f in u (with respect to x)
is the order of f in v (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!
•
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
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).