The Power of the Homotopy Operator
Willy Hereman
Department of Applied Mathematics and Statistics
Colorado School of Mines, Golden, Colorado
Colorado Nonlinear Day
University of Colorado–Colorado Springs
Saturday, April 30, 2016, 5:00p.m.
Acknowledgements
Douglas Poole (Chadron State College, Nebraska)
Bernard Deconinck (Univ. of Washington, Seattle)
Ünal Göktaş (Turgut Özal University, Turkey)
Mark Hickman (Univ. of Canterbury, New Zealand)
Several undergraduate and graduate students
Research was supported in part by NSF
under Grant No. CCF-0830783
This presentation was made in TeXpower
Outline
•
Motivation, problem statement, and examples
•
Calculus-based formulas for the homotopy
operator
See: Anco & Bluman (2002); Hereman et al.
(2007); Olver (1986)
•
Symbolic integration and inversion of the
total divergence operator
•
Application: symbolic computation of
conservation laws of nonlinear PDEs
•
Conclusions and Future Work
Motivation, Problem Statement, Examples
Example 1: Computation of conservation laws
of nonlinear PDEs
Conservation law in (1+1)-dimensions
Dt ρ + Dx J =
˙ 0
where =
˙ means evaluated on the PDE.
Conserved density ρ and flux J.
Example: Short pulse equation (SPE)
(nonlinear optics)
uxt = u + (u3 )xx = u + 6uu2x + 3u2 uxx
for u(x, t) is completely integrable.
First of infinitely many non-polynomial conservation
laws:
p
p
1 + 6u2x − Dx 3u2 1 + 6u2x =
˙ 0
Dt
p
6ux uxt
2
Dt
1 + 6ux = p
1 + 6u2x
−6ux (u + 6uu2x + 3u2 uxx )
p
=
1 + 6u2x
Let
6ux (u + 6uu2x + 3u2 uxx )
p
f =
1 + 6u2x
Question: Can the expression be integrated?
Z
If yes, find F = f dx (so, f = Dx F )
Result (by hand): F =
3u2
p
1 + 6u2x
Mathematica cannot compute this integral!
Example 2: Integrating an expression in 2 variables
While investigating the sine-Gordon equation
ut = v,
vt = uxx + sin u
we obtained
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
Mathematica cannot compute this integral!
Example 3: Inverting a total divergence
For u(x, y) and v(x, y)
f = ux vy − u2x vy − uy vx + uxy vx
Question: Is there an F = (F1 , F2 ) so that
f = Div F = Dx F1 + Dy F2 ?
If yes, find a F̃.
not unique: F̃ = F + (Dy φ, −Dx φ) for any φ
Result (by hand): F̃ = (uvy − ux vy , −uvx + ux vx )
Mathematica has no function to compute Div−1
Can this be done without integration by parts?
Can the computation be reduced to a single
integral in one variable?
Example 4: Conservation law of nonlinear PDEs in
multiple space dimensions
•
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 ).
Example: 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
Dt u3 −
3β
2
(u
x
α
−
6βu(u2x
−
6β 2
+
+ u2y ) + Dx 3u2 ( α4 u2 + βuxx )
u2y )
+
3β 2
2
(u
xx
α
− u2yy )
(ux (uxxx + uxyy ) + uy (uxxy + uyyy ))
2
6β
2
+ Dy 3βu uxy + α uxy (uxx + uyy ) =
˙ 0
α
Mathematica has no function to compute Div−1
The Homotopy Operator
•
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
F = Div
−1
f =
(x)
(Hu(x,y) f,
(y)
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
i=0

(j)
uix
k−(i+1) 
(−Dx )
∂f
(j)
∂ukx
(Iu(j) f )[λu] means that in Iu(j) f one replaces
u → λu, ux → λux , etc.
More general: u → u0 + λ(u − u0 ) = (1 − λ)u0 + λu
ux → ux 0 + λ(ux − ux 0 ) = (1 − λ)ux 0 + λux ,
etc.
Homotopy Concept
u
λ=1
u0
λ=0
T (u0 , u) = u0 + λ(u − u0 ) = (1 − λ)u0 + λu
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
+ (ux I − uDx )
= u
∂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
0
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)
(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)

My
i+j k+`−i−j−1
M
k−1
`
x
XX XX
k−i−1
(x)

Iu 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 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!
Computer Demo
Why does this work?
Sketch of Derivation and Proof
(in 1D with variable x, and for one component u)
Recall: Euler’s theorem for homogeneous functions
If f (x1 , x2 , . . . , xn ) is homogeneous of degree p, i.e.,
f (λx1 , λx2 , . . . , λxn ) = λp f (x1 , x2 , . . . , xn )
then, defining g and P, as follows
n
X
∂f
g = Pf =
xi
= pf
∂xi
i=1
Proof: Differentiate both sides with respect to λ:
n
n
X
X
∂f ∂(λxi )
∂f
=
= pλp−1 f
xi
∂(λxi ) ∂λ
∂(λxi )
i=1
i=1
What is the inverse of P?
R1
−1
f = P g(x1 , x2 , . . . , xn ) = 0 g(λx1 , λx2 , . . . , λxn ) dλ
λ
Proof:
Z
1
dλ
g(λx1 , λx2 , . . . , λxn )
λ
0
Z 1
dλ
=
pf (λx1 , λx2 , . . . , λxn ))
λ
0
Z 1
dλ
p
=
pλ f (x1 , x2 , . . . , xn ))
λ
0
Z 1
= f
pλp−1 dλ = f λp |10 = f
0
Definition: Degree operator M
M
X
∂f
∂f
∂f
∂f
∂f
Mf =
uix
= u +ux
+u2x
+· · ·+uM x
∂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) M−1 g(u) = 0 g[λu] dλ
λ
Proof:
M
M
X
X
d
∂g[λu] d(λuix )
1
∂g[λu]
1
g[λu] =
=
uix
= Mg[λu]
dλ
∂(λuix ) dλ
λ i=0
∂uix
λ
i=0
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
=M
g[λu]
=
Mg[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 +
g[λu] = (p + q +
q r
p
r) u ux u3x ,
r)λp+q+r
then
q r
p
u ux u3x
Hence,
M−1 g =
Z
0
1
(p + q + r) λp+q+r−1 up uqx ur3x dλ
= up uqx ur3x
λ=1
p+q+r λ
= 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 )
∂u
kx
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
∂f
∂f
k−1
k−1
 + ux
= u
(−Dx )
− Dx u
(−Dx )
∂u
∂ukx
∂ukx
k=1
k=1

M
X
∂f
∂f
k−1 ∂f

= u
+ ux
− Dx u
(−Dx )
∂u
∂ux
∂u
kx
k=1

M
M
X
X
k−2 ∂f 
k−2 ∂f
+ u2x
+ux
(−Dx )
(−Dx )
∂ukx
∂ukx
k=2
k=2
= ...
∂f
∂f
∂f
+ ux
+ . . . + uM x
= u
∂u
∂ux
∂uM x

M
M
X
X
∂f
∂f
k−2
k−1
+ 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−1
M
X
X
X
∂f
∂f
k−(i+1)

=
uix
− Dx 
uix
(−Dx )
∂uix
∂ukx
i=0
i=0
k=i+1


M−1
M
X
X
k−(i+1) ∂f 

= Mf − Dx
uix
(−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
−1
k−(i+1)

f = D x M
uix
(−Dx )
∂ukx
i=0
k=i+1
−1 .
Apply Dx−1 and use
the
formula
for
M


Z 1 M−1
M
X
X
dλ
−1
k−(i+1) ∂f 

F = Dx f =
uix
(−Dx )
[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
Early work by Kruskal and collaborators
Zoom into the formula for computation of fluxes
Application: Zakharov-Kuznetsov Equation
Computation of Conservation Laws
ut + αuux + β(uxx + uyy )x = 0
•
Assume that the density if already computed
3β 2
3
ρ=u −
(ux + u2y )
α
•
Compute
∂ρ
Dt ρ =
+ ρ0 (u)[ut ]
∂t
My
Mx X
X
∂ρ
∂ρ
=
+
Dkx D`y ut
∂t
∂u
kx `y
k=0 `=0
= 3u2 I − 6β
(ux Dx + uy Dy ) ut
α
•
Substitute ut = − αuux + β(uxx + uyy )x .
Then, E = −Dt ρ = 3u2 A −
6β
u A
α x x
−
6β
u A
α y y
where A = αuux + β(uxxx + uxyy )
•
Use the homotopy operator to invert Div:
(x)
(y)
−1
J = Div E = Hu(x,y) E, Hu(x,y) E
Z 1
where
dλ
(x)
(x)
Hu(x,y) E =
(Iu E)[λu]
λ
0
with
My k−1 `
i+j k+`−i−j−1
M
x
XX XX
k−i−1
(x)
uix jy i
Iu E =
k+`
k=1 `=0
i=0 j=0
(−Dx )k−i−1
k
`−j ∂E
−Dy
∂ukx `y
(y)
(y)
•
Similar formulas for Hu(x,y) E and Iu E.
•
Flux:
(x)
(y)
J = Hu(x,y) E, Hu(x,y) E
2
4
2
2
= 3α
+
u
u
+
βu
(3u
+
2u
)
−
2βu(3u
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 )
xxy
xx
α y
2 yyy
α
2 xy
4 yy
5β 2
2
u
u
,
βu
uxy − 4βuux uy
4α xx yy
3β 2
β2
u(ux3y + u3xy ) − 4α ux (13uxxy + 3uyyy )
4α
− 4α uy (uxxx + 3uxyy ) + 4α uxy (uxx + uyy )
5β 2
9β 2
•
After removal of curl in 2D:
J=
3u2 ( α4 u2
−
+ βuxx ) −
6β 2
(ux (uxxx
α
3βu2 uxy
6βu(u2x
+
u2y )
+
3β 2
2
(u
xx
α
+ uxyy ) + uy (uxxy + uyyy )),
2
+ 6βα uxy (uxx + uyy )
− u2yy )
Conclusions and Future Work
•
•
The power of the homotopy operator:
I
Integration (by parts): D−1
x
I
Computation of Div−1
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
•
Further investigate the conservation laws of short
pulse equation
•
Continue the implementation in Mathematica
•
Software: http://inside.mines.edu/∼whereman
Thank You