Application of
Euler and Homotopy Operators
to Integrability Testing
Willy Hereman
Department of Applied Mathematics and Statistics
Colorado School of Mines
Golden, Colorado, U.S.A.
http://www.mines.edu/fs home/whereman/
The Ninth IMACS International Conference on
Nonlinear Evolution Equations and Wave Phenomena:
Computation and Theory
Athens, Georgia
Thursday, April 2, 2015, 4:40p.m.
Acknowledgements
Douglas Poole (Chadron State College, Nebraska)
Ünal Göktaş (Turgut Özal University, Turkey)
Mark Hickman (Univ. of Canterbury, New Zealand)
Bernard Deconinck (Univ. of Washington, Seattle)
Several undergraduate and graduate students
Research supported in part by NSF
under Grant No. CCF-0830783
This presentation was made in TeXpower
Outline
•
Motivation of the Research
•
Part I: Continuous Case
I
Calculus-based formulas for the continuous
homotopy operator (in multi-dimensions)
See: Anco & Bluman (2002); Hereman et al.
(2007); Olver (1986)
I
Symbolic integration by parts and inversion
of the total divergence operator
I
Application: symbolic computation of
conservation laws of nonlinear PDEs in
multiple space dimensions
•
Part II: Discrete Case
I
Simple formula for the homotopy operator
See: Hereman et al. (2004); Hydon &
Mansfield (2004); Mansfield & Hydon (2002)
•
I
Symbolic summation by parts and inversion
of the forward difference operator
I
Analogy: continuous and discrete formulas
I
Application: symbolic computation of
conservation laws of nonlinear DDEs
Conclusions and Future Work
Motivation of the Research
Conservation Laws for Nonlinear PDEs
•
System of evolution equations of order M
ut = F(u(M ) (x))
with u = (u, v, w, . . .) and x = (x, y, z).
•
Conservation law in (1+1)-dimensions
Dt ρ + D x J =
˙ 0
where =
˙ means evaluated on the PDE.
Conserved density ρ and flux J.
•
Conservation law in (2+1)-dimensions
D t ρ + ∇ · J = D t ρ + D x J1 + D y J2 =
˙ 0
Conserved density ρ and flux J = (J1 , J2 ).
•
Conservation law in (3+1)-dimensions
Dt ρ + ∇ · J = Dt ρ + Dx J1 + Dy J2 + Dz J3 =
˙ 0
Conserved density ρ and flux J = (J1 , J2 , J3 ).
Notation – Computations on the Jet Space
•
Independent variables: t (time), x = (x, y, z)
•
Dependent variables u = (u(1) , u(2) , . . . , u(j) , . . . , u(N ) )
In examples: u = (u, v, w, θ, h, . . .)
•
∂k u
Partial derivatives ukx = ∂xk ,
∂5u
Examples: uxxxxx = u5x = ∂x5
uxx yyyy = u2x 4y =
•
ukx ly =
∂6u
∂x2 y 4
Differential functions
Example: f = uvvx + x2 u3x vx + ux vxx
∂ k+l u
,
∂xk y l
etc.
•
Total derivatives: Dt , Dx , Dy , . . .
Example: Let f = uvvx + x2 u3x vx + ux vxx Then
∂f
∂f
∂f
Dx f =
+ ux
+ uxx
∂x
∂u
∂ux
∂f
∂f
∂f
+ vxx
+ vxxx
+vx
∂v
∂vx
∂vxx
= 2xu3x vx + ux (vvx ) + uxx (3x2 u2x vx + vxx )
+vx (uvx ) + vxx (uv + x2 u3x ) + vxxx (ux )
= 2xu3x vx + vux vx + 3x2 u2x vx uxx + uxx vxx
+uvx2 + uvvxx + x2 u3x vxx + ux vxxx
Example 1:
The Zakharov-Kuznetsov Equation
ut + αuux + β(uxx + uyy )x = 0
models ion-sound solitons in a low pressure uniform
magnetized plasma.
•
Conservation laws:
Dt u + Dx α2 u2 + βuxx + Dy βuxy =
˙ 0
3
2
2
Dt u2 + Dx 2α
u
−
β(u
−
u
x
y ) + 2βu(uxx + uyy )
3
+ Dy − 2βux uy =
˙ 0
•
More conservation laws (ZK equation):
3
3β
2
(u
x
α
+ u2y ) + Dx 3u2 ( α4 u2 + βuxx ) − 6βu(u2x + u2y )
2
2
+ 3βα (u2xx − u2yy ) − 6βα (ux (uxxx + uxyy ) + uy (uxxy + uyyy ))
2
+ Dy 3βu2 uxy + 6βα uxy (uxx + uyy ) =
˙ 0
Dt u −
3
2
2
Dt tu2 − α2 xu + Dx t( 2α
u
−
β(u
−
u
x
y ) + 2βu(uxx + uyy ))
3
2β
1
u
)
+
u
+
D
−
2β(tu
u
+
xuxy ) =
˙ 0
− x(u2 + 2β
y
x y
α xx
α x
α
Example 2:
Shallow Water Wave Equations
P. Dellar, Phys. Fluids 15 (2003) 292-297
ut + (u·∇)u + 2 Ω × u + ∇(θh) − 21 h∇θ = 0
θt + u·(∇θ) = 0
ht + ∇·(uh) = 0
In components:
ut + uux + vuy − 2 Ωv + 12 hθx + θhx = 0
vt + uvx + vvy + 2 Ωu + 21 hθy + θhy = 0
θt + uθx + vθy = 0
ht + hux + uhx + hvy + vhy = 0
where u(x, y, t), θ(x, y, t) and h(x, y, t)
•
First few densities-flux pairs of SWW system:


u
(1)
(1)

ρ =h
J =h
v
 
u
(2)
(2)
ρ = hθ
J = hθ 
v
 
u
(3)
2
(3)
2
ρ = hθ
J = hθ  
v


2 + v 2 + 2hθ)
u
(u

ρ(4) = h (u2 + v 2 + hθ)
J(4) = h 
v (v 2 + u2 + 2hθ)
ρ(5) = θ (2Ω + vx − uy )

J(5) = 21
θ
4Ωu − 2uuy + 2uvx − hθy
4Ωv + 2vvx − 2vuy + hθx


Generalizations:
Dt f (θ)h + Dx f (θ)hu + Dy f (θ)hv =
˙ 0
Dt g(θ)(2Ω + vx − ux )
+ Dx 21 g(θ)(4Ωu − 2uuy + 2uvx − hθy )
+ Dy 12 g(θ)(4Ωv − 2uy v + 2vvx + hθx ) =
˙ 0
for any functions f (θ) and g(θ)
Conservation Laws for Nonlinear
Differential-Difference Equations (DDEs)
•
System of DDEs
u̇n = F(· · · , un−1 , un , un+1 , · · · )
•
Conservation law in (1 + 1) dimensions
Dt ρn + ∆Jn = Dt ρn + Jn+1 − Jn =
˙ 0
conserved density ρn and flux Jn
•
Example: Toda lattice
u̇n = vn−1 − vn
v̇n = vn (un − un+1 )
•
First few densities-flux pairs for Toda lattice:
(0)
ρn = ln(vn )
(0)
Jn = un
(1)
ρn = un
(1)
Jn = vn−1
(2)
ρn = 21 u2n
+ vn
(2)
Jn = un vn−1
+ un (vn−1 + vn )
(3)
Jn = un−1 un vn−1
(3)
ρn = 31 u3n
2
+ vn−1
PART I: CONTINUOUS CASE
Problem Statement
Continuous case in 1D:
Example: For u(x) and v(x)
f = 8vx vxx −u3x sin u +2ux uxx cos u −6vvx cos u +3ux v 2 sin u
Question: Can the expression be integrated?
Z
If yes, find F = f dx (so, f = Dx F )
Result (by hand): F = 4 vx2 + u2x cos u − 3 v 2 cos u
Continuous case in 2D or 3D:
Example: For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
Question: Is there an F so that f = Div F ?
If yes, find F.
Result (by hand): F = (uvy − ux vy , −uvx + ux vx )
Can this be done without integration by parts?
Can the computation be reduced to a single integral
in one variable?
Tools from the Calculus of Variations
•
Definition:
A differential function f is a exact iff f = Div F.
Special case (1D): f = Dx F.
•
Question: How can one test that f = Div F ?
•
Theorem (exactness test):
f = Div F iff Lu(j) (x) f ≡ 0,
j = 1, 2, . . . , N.
N is the number of dependent variables.
The Euler operator annihilates divergences
•
Euler operator in 1D (variable u(x)):
M
X
k ∂
Lu(x) =
(−Dx )
∂ukx
k=0
∂
∂
∂
2 ∂
3
=
− Dx
+ Dx
− Dx
+ ···
∂u
∂ux
∂uxx
∂uxxx
•
Euler operator in 2D (variable u(x, y)):
Lu(x,y) =
My
Mx X
X
k=0
∂
(−Dx ) (−Dy )
∂ukx `y
`=0
k
`
∂
∂
∂
− Dx
− Dy
=
∂u
∂ux
∂uy
∂
∂
2 ∂
2 ∂
3
+ Dx
+Dx Dy
+Dy
−Dx
···
∂uxx
∂uxy
∂uyy
∂uxxx
Application: Testing Exactness – Continous Case
Example:
f = 8vx vxx −u3x sin u +2ux uxx cos u −6vvx cos u +3ux v 2 sin u
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 = 8vx vxx −u3x sin u +2ux uxx cos u −6vvx cos u +3ux v 2 sin u
∂f
∂f
2 ∂f
− Dx
+ Dx
≡0
Lu(x) f =
∂u
∂ux
∂uxx
∂f
∂f
2 ∂f
Lv(x) f =
− Dx
+ Dx
≡0
∂v
∂vx
∂vxx
•
Question: How can one compute F = Div−1 f ?
•
Theorem (integration by parts):
• 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
(x)
(y)
F = Div−1 f = (Hu(x,y) f, Hu(x,y) f )
The homotopy operator inverts total
derivatives and divergences!
Peter Olver’s Book
Homotopy Concept in Olver’s Book
Homotopy Formula in Olver’s Book
Zoom into Homotopy Formula in Olver’s Book
•
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
(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.
•
Homotopy Operator in 2D (variables x and y):
(x)
Hu(x,y)
(y)
Hu(x,y)
Z
f =
N
1X
0 j=1
Z
f =
(x)
(Iu(j)
N
1X
0 j=1
(y)
(Iu(j)
dλ
f )[λu]
λ
dλ
f )[λu]
λ
where for dependent variable u(x, y)

k+`−i−j−1
M
i+j
y
M
k−1
`
x X
X
XX
k−i−1

Iu(x) f =
uix jy i
k+`
k=1 `=0
(−Dx )
i=0 j=0
k−i−1
−Dy
k
`−j ∂f
∂ukx `y
Application of Homotopy Operator in 1D
Example:
f = 8vx vxx −u3x sin u +2ux uxx cos u −6vvx cos u +3ux v 2 sin u
Goal: Find
F = 4 vx2 + u2x cos u − 3 v 2 cos u
Easy to verify: f = Dx F
•
Compute
Iu f
∂f
∂f
= u
+ (ux I − uDx )
∂ux
∂uxx
= −uu2x sin u + 3uv 2 sin u + 2u2x cos u
•
Similarly,
Iv f
•
Finally,
∂f
∂f
= v
+ (vx I − vDx )
∂vx
∂vxx
= −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
Application of Homotopy Operator in 2D
•
Example: f = ux vy − uxx vy − uy vx + uxy vx
By hand: F̃ = (uvy − ux vy , −uvx + ux vx )
Easy to verify: f = Div F̃
•
Compute
Iu(x) f
∂f
∂f
= u
+ (ux I − uDx )
∂ux
∂uxx
∂f
+ 21 uy I − 21 uDy
∂uxy
= uvy + 12 uy vx − ux vy + 12 uvxy
•
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 = λ uvy + 12 uy vx − ux vy + 21 uvxy − uy v + uxy v dλ
=
0
1
uvy
2
+ 41 uy vx − 21 ux vy + 14 uvxy − 12 uy v + 12 uxy v
•
Analogously,
Z 1
dλ
(y)
(y)
(y)
F2 = Hu(x,y) f =
Iu f + Iv f [λu]
λ
0
Z 1 =
λ −uvx − 12 uvxx + 12 ux vx + λ (ux v − uxx v) dλ
0
= − 21 uvx − 41 uvxx + 14 ux vx + 21 ux v − 12 uxx v
•
So,


1  2uvy + uy vx − 2ux vy + uvxy − 2uy v + 2uxy v 
F=
4
−2uvx − uvxx + ux vx + 2ux v − 2uxx v
Let K = F̃−F then


1  2uvy − uy vx − 2ux vy − uvxy + 2uy v − 2uxy v 
K=
4
−2uvx + uvxx + 3ux vx − 2ux v + 2uxx v
then Div K = 0
•
Also, K = (Dy φ, −Dx φ) with φ = 41 (2uv − uvx − 2ux v)
(curl in 2D)
After removing the curl term K:
F̃ = F + K = (uvy − ux vy , −uvx + ux vx )
Avoid curl terms algorithmically!
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
= u +ux
+u2x
+· · ·+uM x
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
R1
Theorem (inverse operator)
= 0 g[λu] dλ
λ
Proof:
M
M
X
X
∂g[λu] dλuix
1
∂g[λu]
1
d
g[λu] =
=
uix
= Mg[λu]
dλ
∂λuix dλ
λ i=0
∂uix
λ
i=0
M−1 g(u)
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 up uqx ur3x
Hence,
M−1 g =
Z
1
(p + q + r) λp+q+r−1 up uqx ur3x dλ
0
= up uqx ur3x
λ=1
λp+q+r = 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
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
M
X
k ∂f
uLu(x) f = u
(−Dx )
∂ukx
k=0


M
M
X
X
∂f
k−1 ∂f
k−1 ∂f 

+ ux
(−Dx )
− Dx u
(−Dx )
= u
∂u
∂u
∂u
kx
kx
k=1
k=1

M
X
∂f
∂f
k−1 ∂f

= u
+ ux
− Dx u
(−Dx )
∂u
∂ux
∂ukx
k=1

M
M
X
X
k−2 ∂f 
k−2 ∂f
+ux
(−Dx )
+ u2x
(−Dx )
∂u
∂u
kx
kx
k=2
k=2
= ...
∂f
∂f
∂f
= u
+ ux
+ . . . + uM x
∂u
∂ux
∂uM x

M
M
X
X
k−2 ∂f
k−1 ∂f

+ ux
(−Dx )
−Dx u
(−Dx )
∂ukx
∂ukx
k=2
k=1

M
X
k−M ∂f 
+ . . . + u(M −1)x
(−Dx )
∂ukx
k=M


M
M
M−1
X
X
X
∂f
k−(i+1) ∂f 

(−Dx )
uix
uix
=
− Dx
∂u
∂u
ix
kx
i=0
i=0
k=i+1


M−1
M
X
X
k−(i+1) ∂f 

uix
(−Dx )
= Mf − Dx
∂ukx
i=0
k=i+1
= 0

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
∂f
k−(i+1)
−1

uix
(−Dx )
f = D x M
∂ukx
i=0
k=i+1
−1 .
Apply Dx−1 and use
the
formula
for
M


Z 1 M−1
M
X
X
dλ
k−(i+1) ∂f 
−1

uix
(−Dx )
F = Dx f =
[λu]
∂u
λ
0
kx
i=0
k=i+1




Z 1 X
M
k−1
X
dλ

k−(i+1)  ∂f 

=
uix (−Dx )

 [λu]
∂ukx
λ
0
i=0
k=1
= Hu(x) f
Application 1: Zakharov-Kuznetsov Equation
Computation of Conservation Laws
ut + αuux + β(uxx + uyy )x = 0
•
Step 1: Compute the dilation invariance
ZK equation is invariant under scaling symmetry
t x y 2
(t, x, y, u) → ( 3 , , , λ u)= (t̃, x̃, ỹ, ũ)
λ λ λ
λ is an arbitrary parameter.
•
Hence, the weights of the variables are
W (u) = 2, W (Dt ) = 3, W (Dx ) = 1, W (Dy ) = 1.
•
A conservation law is invariant under the scaling
symmetry of the PDE.
W (u) = 2, W (Dt ) = 3, W (Dx ) = 1, W (Dy ) = 1.
For example,
2
2 α 2
2
2
2
+u
)
+
D
3u
(
+
u
Dt u3 − 3β
(u
u
+βu
)
−6βu(u
x
xx
x
y
x
y)
α
4
2
2
+ 3βα (u2xx −u2xy ) − 6βα (ux (uxxx +uxyy ) + uy (uxxy +uyyy ))
2
6β
2
+ Dy 3βu uxy + α uxy (uxx + uyy ) =
˙ 0
Rank (ρ) = 6,
Rank (J) = 8.
Rank (conservation law) = 9.
Compute the density of selected rank, say, 6.
•
Step 2: Construct the candidate density
For example, construct a density of rank 6.
Make a list of all terms with rank 6:
{u3 , u2x , uuxx , u2y , uuyy , ux uy , uuxy , u4x , u3xy , u2x2y , ux3y , u4y }
Remove divergences and divergence-equivalent
terms.
Candidate density of rank 6:
ρ = c1 u3 + c2 u2x + c3 u2y + c4 ux uy
•
Step 3: Compute the undetermined coefficients
Compute
∂ρ
Dt ρ =
+ ρ0 (u)[ut ]
∂t
My
Mx X
X
∂ρ
∂ρ
=
+
Dkx D`y ut
∂t
∂ukx `y
k=0 `=0
= 3c1 u2 I + 2c2 ux Dx + 2c3 uy Dy + c4 (uy Dx + ux Dy ) ut
Substitute ut = − αuux + β(uxx + uyy )x .
E = −Dt ρ = 3c1 u2 (αuux + β(uxx + uxy )x )
+ 2c2 ux (αuux + β(uxx + uyy )x )x + 2c3 uy (αuux
+ β(uxx + uyy )x )y + c4 (uy (αuux + β(uxx + uyy )x )x
+ ux (αuux + β(uxx + uyy )x )y )
Apply the Euler operator (variational derivative)
My
Mx X
X
k
` ∂E
Lu(x,y) E =
(−Dx ) (−Dy )
∂ukx `y
k=0 `=0
= −2 (3c1 β + c3 α)ux uyy + 2(3c1 β + c3 α)uy uxy
+2c4 αux uxy +c4 αuy uxx +3(3c1 β +c2 α)ux uxx
≡0
Solve a parameterized linear system for the ci :
3c1 β + c3 α = 0, c4 α = 0, 3c1 β + c2 α = 0
Solution:
3β
,
c
=
−
, c4 = 0
c1 = 1, c2 = − 3β
3
α
α
Substitute the solution into the candidate density
ρ = c1 u3 + c2 u2x + c3 u2y + c4 ux uy
Final density of rank 6:
3β 2
ρ=u −
(ux + u2y )
α
3
•
Step 4: Compute the flux
Use the homotopy operator to invert Div:
(x)
(y)
J = Div−1 E = Hu(x,y) E, Hu(x,y) E
Z 1
where
dλ
(x)
(x)
(Iu E)[λu]
Hu(x,y) E =
λ
0
with
Iu(x) E =
My k−1 `
Mx X
X
XX
k=1 `=0
(−Dx )
i+j k+`−i−j−1
k−i−1
uix jy i
k+`
i=0 j=0
k
k−i−1
`−j ∂E
−Dy
∂ukx `y
(y)
(y)
Similar formulas for Hu(x,y) E and Iu E.
Let A = αuux + β(uxxx + uxyy ) so that
E = 3u2 A −
6β
u A
α x x
−
6β
u A
α y y
Then,
(x)
(y)
J = Hu(x,y) E, Hu(x,y) E
2
4
2
2
+
u
u
+
βu
(3u
+
2u
)
−
2βu(3u
= 3α
xx
yy
x
y)
4
+
−
+
−
3β 2
β2
7
u(u
+
u
)
−
u
(
u
+ 6uxxx )
2x2y
4y
4α
α x 2 xyy
β2
β2
2
3
5 2
3 2
u
(4u
+
u
)
+
(3u
+
u
+
u )
y
xxy
yyy
xx
α
2
α
2 xy
4 yy
5β 2
2
u
u
,
βu
uxy − 4βuux uy
xx
yy
4α
3β 2
β2
u(ux3y + u3xy ) − 4α ux (13uxxy + 3uyyy )
4α
− 4α uy (uxxx + 3uxyy ) + 4α uxy (uxx + uyy )
5β 2
9β 2
Application 2: Shallow Water Wave Equations
Computation of Conservation Laws
Quick Recapitulation
•
Conservation law in (2+1) dimensions
D t ρ + ∇ · J = D t ρ + D x J1 + D y J2 =
˙ 0
conserved density ρ and flux J = (J1 , J2 )
•
Example: Shallow water wave (SWW) equations
ut + uux + vuy − 2 Ωv + 12 hθx + θhx = 0
vt + uvx + vvy + 2 Ωu + 21 hθy + θhy = 0
θt + uθx + vθy = 0
ht + hux + uhx + hvy + vhy = 0
•
Typical density-flux pair:
ρ(5) = θ (vx − uy + 2Ω)


1  4Ωu − 2uuy + 2uvx − hθy 
(5)
J
=
θ
2
4Ωv + 2vvx − 2vuy + hθx
•
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
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 = ρ(5) = θ (2Ω − uy + vx )
•
Step 3: Compute the flux J
E = θ(ux vx + uvxx + vx vy + vvxy + 2Ωux
+ 21 θx hy − ux uy − uuxy − uy vy − uyy 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
∂E
∂E
+ 21 uy I − 21 uDy
Iu(x) E = u
∂ux
∂uxy
= uvx θ + 2Ωuθ + 21 u2 θy − uuy θ
Similarly, compute
Iv(x) E = vvy θ + 12 v 2 θy + uvx θ
(x)
Iθ E
(x)
Ih E
=
1 2
θ hy
2
+ 2Ωuθ − uuy θ + uvx θ
= − 12 θθy h
Next,
(x)
J1 = Hu(x,y) E
Z 1
Iu(x) E
=
0
Z 1
=
0
dλ
(x)
(x)
(x)
+ Iv E + Iθ E + Ih E [λu]
λ
4λΩuθ + λ2 3uvx θ + 12 u2 θy − 2uuy θ + vvy θ
+ 12 v 2 θy + 12 θ2 hy − 12 θθy h dλ
= 2Ωuθ− 32 uuy θ+ uvx θ+ 13 vvy θ+ 16 u2 θy
+61 v 2 θy − 16 hθθy + 61 hy θ2
Analogously,
J2 =
(y)
Hu(x,y) E
= 2Ωvθ + 23 vvx θ − vuy θ − 31 uux θ − 16 u2 θx − 16 v 2 θx
+ 16 hθθx − 61 hx θ2
Hence,
J
= J(5) =

112Ωuθ−4uuy θ+6uvx θ+2vvy θ+(u2 +v 2 )θy −hθθy +hy θ2
6 12Ωvθ+4vvx θ−6vuy θ−2uux θ−(u2 +v 2 )θx +hθθx −hx θ2
There are curl terms in J
Indeed, subtract K where Div K = 0
Here


1  −(2uuy θ+2vvy θ+u2 θy +v 2 θy +2hθθy +hy θ2 ) 
K=
6
2vvx θ+2uux θ+u2 θx +v 2 θx +2hθθx +hx θ2
Note that K = (Dy φ, −Dx φ) with
φ = −(hθ2 + u2 θ + v 2 θ) (i.e., curl in 2D).
After removing the curl term K,


1  4Ωu − 2uuy + 2uvx − hθy 
(5)
J̃ = J̃ = θ
2
4Ωv + 2vvx − 2vuy + hθx
PART II: DISCRETE CASE
Problem Statement
Discrete case in 1D:
Example:
2
+un+3 vn+2 −un+1 vn
fn = −un un+1 vn −vn2 +un+1 un+2 vn+1 +vn+1
Question: Can the expression be summed by parts?
If yes, find Fn = ∆−1 fn (so, fn = ∆Fn = Fn+1 − Fn )
Result (by hand):
Fn = vn2 + un un+1 vn + un+1 vn + un+2 vn+1
How can this be done algorithmically?
Can this be done as in the continuous case?
Tools from the Discrete Calculus of Variations
•
Definitions:
D is the up-shift (forward or right-shift) operator
DFn = Fn+1 = Fn |n→n+1
D−1 the down-shift (backward or left-shift) operator
D−1 Fn = Fn−1 = Fn |n→n−1
∆ = D − I is the forward difference operator
∆Fn = (D − I)Fn = Fn+1 − Fn
•
Problem to be solved: Given fn .
Find Fn = ∆−1 fn (so fn = ∆Fn = Fn+1 − Fn )
Analogy Continuous & Discrete Cases
Euler Operators
Continuous Case
Lu(x)
M
X
∂
=
(−Dx )
∂uk x
k=0
k
Discrete Case
Lun =
M
X
k=0
−k
D
∂
∂un+k
M
X
∂
=
D−k
∂un k=0
Analogy Continuous & Discrete Cases
Homotopy Operators & Integrands
Continuous Case
Z
Hu(x) f=
0
Iu(j) f =
Discrete Case
N
1X
Z 1X
N
dλ
dλ
(Iu(j) f )[λu] Hun fn=
(Iu(j) fn )[λun ]
n
λ
λ
0
j=1
j=1
(j)
M
X
Iu(j) fn =
(j)
M
X
n
k=1
k=1




k−1
k
X
X
∂f
∂f
n
(j)
(j)
k−(i+1)
−i




uix (−Dx )
D
un+k (j)
(j)
∂ukx i=1
∂un+k
i=0
Euler Operators Side by Side
Continuous Case
(for component u)
∂
∂
2 ∂
3 ∂
Lu =
− Dx
+ Dx
− Dx
+ ···
∂u
∂ux
∂u2x
∂u3x
Discrete Case
Lun
(for component un )
∂
∂
∂
∂
−1
−2
−3
+D
+D
+D
+ ···
=
∂un
∂un+1
∂un+2
∂un+3
∂
=
(I + D−1 + D−2 + D−3 + · · · )
∂un
Homotopy Operators Side by Side
Continuous Case
(for components u and v)
Z 1
dλ
Hu(x) f =
(Iu f + Iv f ) [λu]
λ
0
with
Iu f =
(1)
M
X

k−1
X


uix (−Dx )
k−(i+1) 
i=0
k=1
∂f
∂ukx
and
M (2)
Iv f =

X
k−1
X

k=1
i=0

k−(i+1) 
vix (−Dx )
∂f
∂vkx
Discrete Case
(for components un and vn )
Z 1
dλ
Hun fn =
(Iun fn + Ivn fn ) [λun ]
λ
0
with


(1)
M
k
X X
∂fn
−i 

un+k
Iun fn =
D
∂un+k
i=1
k=1
and


(2)
M
k
X X
∂f
n
−i


Ivn fn =
D
vn+k
∂vn+k
i=1
k=1
Analogy of Definitions & Theorems
Continuous Case (PDE) Semi-discrete Case (DDE)
ut = F(u, ux , u2x , · · · )
u̇n = F(· · · , un−1 , un , un+1 , · · · )
Dt ρ + D x J = 0
Dt ρn + ∆ Jn = 0
•
Definition: fn is exact iff fn = ∆Fn = Fn+1 − Fn
•
Theorem (exactness test): fn = ∆Fn iff Lun fn ≡ 0
•
Theorem (summation with homotopy operator):
If fn is exact then Fn = ∆−1 fn = Hun (fn )
Testing Exactness – Discrete Case
Example:
2
+un+3 vn+2 −un+1 vn
fn = −un un+1 vn −vn2 +un+1 un+2 vn+1 +vn+1
•
fn is exact
•
After summation by parts (done by hand):
Fn = vn2 + un un+1 vn + un+1 vn + un+2 vn+1
Easy to verify: fn = ∆Fn = Fn+1 − Fn
•
Exactness test with Euler operator:
For component un (highest shift 3):
∂ Lun fn =
I + D−1 + D−2 + D−3 fn
∂un
= −un+1 vn − un−1 vn−1 + un+1 vn −vn−1
+ un−1 vn−1 + vn−1
≡ 0
Similarly,
Lvn fn
∂ =
I + D−1 fn
∂vn
= un un+1 + 2vn − un un+1 − 2vn
≡ 0
Application of Discrete Homotopy Operator
Example:
2
+un+3 vn+2 −un+1 vn
fn=−un un+1 vn −vn2 +un+1 un+2 vn+1 +vn+1
Here, M (1) = 3 and M (2) = 2.
Compute
Iun fn = (D
−1
) un+1
+ (D
−1
∂fn
∂un+1
+D
−2
) un+2
∂fn
∂un+2
∂fn
+ (D + D + D ) un+3
∂un+3
= 2un un+1 vn + un+1 vn + un+2 vn+1
−1
−2
−3
Ivn fn = (D
−1
) vn+1
+ (D
−1
∂fn
∂vn+1
+D
−2
) vn+2
∂fn
∂vn+2
= un un+1 vn + 2vn2 + un+1 vn + un+2 vn+1
Finally,
Z 1
dλ
Fn =
(Iun fn + Ivn fn ) [λun ]
λ
0
Z 1
2λvn2 + 3λ2 un un+1 vn + 2λun+1 vn + 2λun+2 vn+1 dλ
=
=
0
vn2 +
un un+1 vn + un+1 vn + un+2 vn+1
Application: Computation of Conservation Laws
•
System of DDEs
u̇n = F(· · · , un−1 , un , un+1 , · · · )
•
Conservation law
Dt ρn + ∆Jn =
˙ 0
conserved density ρn and flux Jn
•
Example: Toda lattice
u̇n = vn−1 − vn
v̇n = vn (un − un+1 )
•
Typical density-flux pair:
ρ(3)
=
n
1 3
u
3 n
+ un (vn−1 + vn )
2
Jn(3) = un−1 un vn−1 + vn−1
Computation Conservation Laws for Toda Lattice
Step 1: Construct the form of the density
The Toda lattice is invariant under scaling symmetry
(t, un , vn ) → (λ−1 t, λun , λ2 vn )
Construct a candidate density, for example,
ρn = c1 u3n + c2 un vn−1 + c3 un vn
which is scaling invariant under the symmetry
Step 2: Determine the constants ci
Compute En = Dt ρn and remove time derivatives
En = (3c1 − c2 )u2n vn−1 + (c3 − 3c1 )u2n vn + (c3 − c2 )vn−1 vn
2
− c3 un un+1 vn − c3 vn2
+c2 un−1 un vn−1 + c2 vn−1
Compute Ẽn = DEn to remove negative shift n − 1
Require that Lun Ẽn = Lvn Ẽn ≡ 0
Solution: c1 = 31 , c2 = c3 = 1 gives
ρn = ρ(3)
n =
1
3
u3n + un (vn−1 + vn )
Step 3: Compute the flux Jn
2
Ẽn = DEn = un un+1 vn + vn2 − un+1 un+2 vn+1 − vn+1
Apply the homotopy operator
J˜n = DJn = −∆−1 (Ẽn ) = −Hun (Ẽn )
Compute
Iun Ẽn
∂ Ẽn
∂ Ẽn
−1
−2
= (D ) un+1
+ (D + D ) un+2
∂un+1
∂un+2
= −2un un+1 vn
−1
Likewise,
Ivn Ẽn = (D
−1
) vn+1
∂ Ẽn
= −(un un+1 vn + 2vn2 )
∂vn+1
Next, compute
J˜n = −
Z
1
Iun Ẽn + Ivn Ẽn
0
Z
=
dλ
[λun ]
λ
1
(3λ2 un un+1 vn + 2λvn2 ) dλ
0
= un un+1 vn + vn2
Finally, backward shift Jn = D−1 (J˜n ) given
2
Jn = Jn(3) = un−1 un vn−1 + vn−1
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 uxx
R
R 2
f dx = uv + u uxx dx
•
Use other homotopy formulas (moving terms
amongst the components of the flux; prevent curl
terms)
•
Homotopy operator approach pays off for
computing irrational fluxes
Example: short pulse equation (nonlinear optics)
uxt = u + (u3 )xx = u + 6uu2x + 3u2 uxx
with non-polynomial conservation law
p
p
Dt
1 + 6u2x − Dx 3u2 1 + 6u2x =
˙ 0
•
Continue the implementation in Mathematica
•
Software: http://inside.mines.edu/∼whereman
Thank You
Publications
1. D. Poole and W. Hereman, Symbolic computation
of conservation laws for nonlinear partial
differential equations in multiple space dimensions,
J. Symb. Comp. 46(12), 1355-1377 (2011).
2. 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.
3. 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.
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).
5. W. Hereman, Symbolic computation of
conservation laws of nonlinear partial differential
equations in multi-dimensions, Int. J. Quan.
Chem. 106(1), 278-299 (2006).