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/
MAA Rocky Mountain Section Meeting
Colorado School of Mines, Golden, Colorado
Friday, April 17, 2009, 4:20p.m.
.
Acknowledgements
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
.
Outline
•
What are Conservation laws of nonlinear PDEs?
•
Famous example in historical perspective
•
Application to shallow water wave equation
•
Computer demonstration
•
Conclusions
.
What are Conservation Laws of Nonlinear PDEs
•
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
•
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
.
Reasons to Compute Conservation Laws
•
Conservation of physical quantities (linear
momentum, mass, energy)
•
Verify the closure of a model
•
Testing of complete integrability and application
of Inverse Scattering Transform
•
Testing of numerical integrators
•
Study of quantitative and qualitative properties of
PDEs (Hamiltonian structure, recursion operators)
.
•
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
.
A Method 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, calculus, and variational
calculus (algorithmic!)
•
Work with linearly independent pieces in finite
dimensional spaces
.
Algorithm for PDEs in (1+1)-dimensions
•
Example: Density of rank 6 for the KdV equation
ut + uux + u3x = 0
•
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
.
•
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
.
•
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
k ∂
Lu =
(−Dx )
∂u
kx
k=0
Here, E has order m = 4, thus
∂E
∂E
2 ∂E
3 ∂E
4 ∂E
Lu 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
.
•
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: Use the homotopy operator
(most powerful)
J=
D−1
x E
Z
=
Z
E dx = Hu(x) E =
0
1
dλ
(Iu E)[λu]
λ
with integrand
Iu E =
M
X

k−1
X

k=1
i=0

k−(i+1) 
uix (−Dx )
∂E
∂ukx
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
.
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)
.
Algorithm for PDEs in (2+1)-dimensions
Computation of Conservation Laws for SWW
We will show how to compute
ρ(5) = vx θ − uy θ + 2Ωθ


1  4Ωuθ − 2uuy θ + 2uvx θ − hθθy 
(5)
J
=
2
4Ωvθ + 2vvx θ − 2vuy θ + hθθx
•
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)
.
Computer Demonstration
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
•
Treat broader class of PDEs (other than those of
evolution type)
•
Full implementation in Mathematica
.
Thank You