Symbolic Computation of Scaling
Invariant Lax Pairs in Operator Form
for Integrable Systems
Willy Hereman
Department of Applied Mathematics and Statistics
Colorado School of Mines, Golden, Colorado
Colorado Nonlinear Day
University of Colorado–Colorado Springs
Saturday, November 1, 2014, 5:00p.m.
Publication:
M. Hickman, W. Hereman, J. Larue, and Ü. Göktaş
Scaling invariant Lax pairs of nonlinear evolution
equations, Applicable Analysis 91(2), 381-402 (2012)
REU students:
Sara Clifton, Jacob Rezac, Oscar Aguilar, and
Tony McCollom
Research was supported in part by NSF
under Grant CCF-0830783
Outline
•
What are Lax pairs of nonlinear PDEs?
•
Lax pairs in operator form
•
Lax pairs in matrix form
•
Reasons to compute Lax pairs
•
Quick method to find Lax pairs
•
More algorithmic approach
•
Examples of Lax pairs of nonlinear PDEs
•
Conclusions and future work
Peter D. Lax (1926-)
Seminal paper: Integrals of nonlinear equations of
evolution and solitary waves
Commun. Pure Appl. Math. 21 (1968) 467-490
What are Lax Pairs of Nonlinear PDEs?
•
Historical example: Korteweg-de Vries equation
ut + α uux + uxxx = 0
•
Key idea: Replace the nonlinear PDE with a
compatible linear system (Lax pair):
ψxx + 61 αu − λ ψ = 0
ψt + 4ψxxx + αuψx + 21 αux ψ + a(t)ψ = 0
ψ is eigenfunction; λ is constant eigenvalue
(λt = 0) (isospectral), and a(t) is an arbitrary
function. We will set a(t) = 0
Class of Equations and Notation
•
Consider a system of evolution equations:
ut = f (u, ux , uxx , . . . , uM x )
with u(x, t) = (u(1) , u(2) , . . . , u(N ) ) and where
(j)
ukx
∂ k u(j)
=
∂xk
•
In examples, the components of u are u, v, . . .
•
Define the total derivative operator as
N
M
∂ • X X ∂ • k (j) Dt • =
+
Dx ut
(j)
∂t
j=1 k=0 ∂u
kx
Lax Pairs in Operator Form
•
Replace a completely integrable nonlinear PDE
by a pair of linear equations (called a Lax pair):
Lψ = λψ
•
Dt ψ = Mψ
and
Require compatibility of both equations
Hence,
Lt ψ + LDt ψ
=
λDt ψ
Lt ψ + LMψ
=
λMψ
=
Mλψ
=
˙
MLψ
Lt ψ + (LM − ML)ψ =
˙ 0
•
Lax equation:
Lt + [L, M] =
˙ O
with commutator [L, M] = LM − ML.
Furthermore,
Lt ψ = [Dt , L]ψ = Dt (Lψ) − LDt ψ
and =
˙ means “evaluated on the PDE”
•
Example: Lax operators for the KdV equation
L = D2x + 61 αu I
M = − 4 D3x + αuDx + 12 αux I
•
Note: Lt ψ + [L, M]ψ = 61 α (ut + αuux + uxxx ) ψ
Alternate Operator Formulations
•
Define
•
Then, the Lax pair becomes
L̃ = L − λ I
L̃ψ = 0
and
M̃ = M − Dt
and
M̃ψ = 0
and the Lax equation becomes [L̃, M̃] =
˙ O
Challenge: Find commuting operators modulo the
(nonlinear) PDE
•
If S is an arbitrary invertible operator, then
L̂ = S L S −1
M̂ = S M S −1
satisfy L̂t + [L̂, M̂] =
˙ O
D̂t = S Dt S −1
Lax Pairs in Matrix Form
•
Express compatibility of
Dx Ψ = X Ψ
Dt Ψ = T Ψ

•

ψ
 1
 
 ψ2 
 , X and T are N × N matrices
where Ψ = 
.
 . 
 . 
 
ψN
Lax equation (zero-curvature equation):
Dt X − Dx T + [X, T] =
˙ 0
with commutator [X, T] = XT − TX
•
Example: Lax pair for the KdV equation


0
1

X=
λ − 16 αu 0

T=
−4λ2 +
1
αux
6
1
1 2 2
αλu
+
α u
3
18
+ 61 αu2x

−4λ − 13 αu

− 16 αux
Substitution into the Lax equation yields

Dt X − Dx T + [X, T] =
− 16 α 
0
ut + αuux + u3x
0


0
Equivalence under Gauge Transformations
•
Lax pairs are equivalent under a gauge
transformation:
If (X, T) is a Lax pair then so is (X̃, T̃) with
X̃ = G X G−1 + Dx (G) G−1
T̃ = G T G−1 + Dt (G) G−1
G is arbitrary invertible matrix and Ψ̃ = GΨ.
Thus,
X̃t − T̃x + [X̃, T̃] =
˙ 0
•
Example:

For the KdV equation


0
1
−ik
 and X̃ = 
X=
λ − 16 αu 0
−1
Here,
X̃ = G X G−1
and
T̃ = G T G−1
with

−i k

G=
−1
where λ = −k2

1

0

1
αu
6

ik
Reasons to Compute a Lax Pair
•
Compatible linear system is the starting point for
application of the IST and the Riemann-Hilbert
method for boundary value problems
•
Confirm the complete integrability of the PDE
•
Zero-curvature representation of the PDE
•
Compute conservation laws of the PDE
•
Discover families of completely integrable PDEs
Question: How to find a Lax pair of a completely
integrable PDE?
Answer: There is no completely systematic method
Dilation Invariance and Weights
•
KdV equation is invariant under dilation symmetry
(x, t, u) → (κ−1 x, κ−3 t, κ2 u) = (x̃, t̃, ũ)
where κ is an arbitrary parameter. Indeed,
ut + αuux + uxxx
•
The weight W of a variable is the exponent of κ
in the symmetry. Thus, W (x) = −1, W (t) = −3, or
W (Dx ) = 1,
•
1
= 0 → 5 (ũt̃ + αũũx̃ + ũx̃x̃x̃ ) = 0
κ
W (Dt ) = 3,
W (u) = 2
The total weight of the KdV equation is 5
because each monomial scales with κ5
Key Observation
•
The Lax operators for the KdV equation are
scaling invariant.
Indeed,
L = D2x + 61 αu I
is uniform of weight 2.
M = − 4D3x + αuDx + 12 αux I
is uniform of weight 3
•
Furthermore, Lψ = λψ and Dt ψ = Mψ are
uniform in weight if W (λ) = W (L) = 2 and
W (M) = W (Dt ) = 3
Elementary Method to Compute Lax Pairs
Using the KdV equation as an example
•
Select W (L) = 2. Here W (M) = 3. In general,
W (L) ≥ W (u) and W (M) = W (Dt ).
•
Build L and M as linear combinations of scaling
invariant terms with undetermined coefficients:
L = D2x + c1 u I
M = c2 D3x + c3 uDx + c4 ux I
•
Substitute into Lt + [L, M] =
˙ O, and replace
ut by −(αuux + u3x )
•
Set the coefficients of D2x , Dx , and I equal to zero
•
Set the coefficients of like monomial terms in
u, ux , uxx , etc. equal to zero
•
Reduce the nonlinear algebraic system
2c3 − 3c1 c2 = 0,
c1 (c3 + α) = 0,
2c4 + c3 − 3c1 c2 = 0,
c1 − c4 + c1 c2 = 0
with the Gröbner basis method into
•
c1 (6c1 − α) = 0,
c1 (c2 + 4) = 0,
c1 (2c4 + α) = 0,
6c1 + c3 = 0,
Solve: c1 = 61 α,
c2 = −4,
c1 (c3 + α) = 0,
3c1 + c4 = 0
c3 = −α,
c4 = − 12 α
•
Substitute the coefficients into L and M :
L = D2x + 61 αu I
M = − 4D3x + αuDx + 21 αux I
•
In complicated cases the nonlinear algebraic
systems are long and hard to solve (too many
solution branches)
•
A divide and conquer strategy is needed
Algorithm to Compute Lax Pairs
Using the KdV equation as an example
•
Step 1:
Compute the weights
W (Dx ) = 1,
•
Step 2:
W (Dt ) = 3,
W (u) = 2
Build a candidate Lax pair
Select W (L) = 2. Here W (M) = 3.
The candidate Lax pair is
L = D2x + f1 Dx + f0 I
M = c3 D3x + g2 D2x + g1 Dx + g0 I
with undetermined functions f0 , f1 , g0 , g1 , g2 and
undetermined constant coefficient c3
•
Step 3:
Substitute into the Lax equation
Lt + [L, M] =
2Dx g2 − 3c3 Dx f1 D3x
+ D2x g2 − 3c3 D2x f1 + f1 Dx g2 + 2Dx g1 − 2g2 Dx f1
−3c3 Dx f0 D2x
+ Dt f1 − c3 D3x f1 + D2x g1 − g2 D2x f1 − 3c3 D2x f0
+f1 Dx g1 + 2Dx g0 − g1 Dx f1 − 2g2 Dx f0 Dx
+ Dt f0 − c3 D3x f0 + D2x g0 − g2 D2x f0 + f1 Dx g0 − g1 Dx f0 I
•
Step 4: Solve the kinematic constraints
(i.e., equations not involving Dt )
Equate the coefficients of D3x and D2x to zero and
solve, yielding
g2 =
g1 =
3
c f ,
2 3 1
3
c D f
4 3 x 1
+ 38 c3 f12 + 23 c3 f0
•
The candidate M operator reduces to
M = c3 D3x + 32 c3 f1 D2x + 38 c3 2Dx f1 + f12 + 4f0 Dx + g0 I
•
The candidate L remains unchanged
•
Step 5: Solve the dynamical equations
(i.e., equations that do involve Dt )
The coefficients of I and Dx yield
Dt f1 + 2Dx g0 − 81 c3 Dx 2D2x f1 + 12Dx f0
−f13 + 12f1 f0 = 0
Dt f0 + D2x g0 + f1 Dx g0 − c3 D3x f0 + 32 f1 D2x f0
+ 34 Dx f1 Dx f0 + 83 f12 Dx f0 + 32 f0 Dx f0 = 0
•
Because W (L) = 2 one has f1 = 0. Thus,
2Dx g0 − 23 c3 D2x f0 = 0
Dt f0 + D2x g0 − c3 D3x f0 + 23 f0 Dx f0 = 0
•
Step 5: continued
Solving these equations gives
g0 = 34 c3 Dx f0
and
f0 = b0 u
•
Replace ut by −(αuux + u3x ),
α + 23 c3 b0 uux + 1 + 14 c3 u3x = 0
•
Hence,
c3 = −4, b0 = 61 α, f0 = 61 αu, f1 = 0, g0 = − 21 αux
•
Step 6: Substitute the coefficients into the
undetermined functions and these into the
candidate pair.
Thus,
L = D2x + 61 αu I
and
M = − 4 D3x + αu Dx + 12 αux I
form a Lax pair for the KdV equation
Algorithm for Computing Lax Pairs
•
Compute the scaling symmetry of the PDE
•
Select W (L) = l ≥ 1.
From the Lax equation: W (M) = W (∂t ) = m
•
Build a candidate Lax pair of the form
L = Dlx + fl−1 Dxl−1 + . . . + f0 I
m−1
M = cm Dm
+
g
D
+ . . . + g0 I
m−1
x
x
for a constant cm
•
Substitute into the Lax equation
•
Separate into kinematic constraints and dynamical
equations
•
Solve the kinematic equations
•
Solve the dynamical equations
•
Substitute the coefficients into undetermined
functions and these into the candidate Lax pair
•
Test the Lax pair
•
Example 1: The modified KdV (mKdV) equation
ut + αu2 ux + u3x = 0
has weights of W (u) = W (Dx ) = 1 and W (Dt ) = 3
•
Selecting W (L) = 1 gives a trivial Lax pair
•
Select W (L) = 2, as in the KdV case, yields
L = D2x + f1 Dx + f0 I
M = c3 D3x + g2 D2x + g1 Dx + g0 I
•
Requiring uniform weights gives
f1 = b0 u, f0 = b1 u2 + b2 ux , g0 = a1 u3 + a2 uux + a3 uxx
•
Example 1: The mKdV equation – continued
•
Solving the kinematic constraints and dynamical
equations gives the Lax pair
√
2
2
2
1
L = Dx + 2uDx + 6
6 + α u + 6 ± −6α ux I
M = −4D3x − 12uD2x
√
2
2
12 + α u + 12 ± −6α ux Dx
−
√
3
3
2
2
− 4 + 3 α u + 12 ± −6α + α uux
√
+ 3 ± 12 −6α uxx I
[M. Wadati, J. Phys. Soc. Jpn., 1972-1973]
•
Example 2: The Boussinesq system
ut − vx = 0
vt − βux + 3uux + αu3x = 0
has W (Dx ) = 1, W (Dt ) = W (u) = W (β) = 2, W (v) = 3
•
Select W (L) = 3. Then,
L = D3x + f1 Dx + f0 I
M = c2 D2x + g0 I
•
The kinematic constraint yields g0 = 32 c2 f1 + c0 β
The dynamical equations
then become
Dt f1 = c2 2Dx f0 − D2x f1
Dt f0 = c2 D2x f0 − 23 D3x f1 − 23 f1 Dx f1
•
Example 2: The Boussinesq system – continued
•
The uniform weight ansatz gives
f1 = a1 u + a2 β
2
2
f0 = a3 ux + D−1
a
u
+
a
βu
+
a
v
+
a
β
4
5
6 x
7
x
•
Solving the dynamical equations gives
√
1
L = D3x + 4α
(3u−β) Dx + 8α3 2 αux ± 13 3αv I
√
√
3α
M = ± 3α D2x ± 2α
uI
[V. E. Zakharov, Sov. Phys. JETP, 1974]
•
Example 3: The coupled KdV system (Hirota &
Satsuma)
ut − 6βuux + 6vvx − βu3x = 0
vt + 3uvx + v3x = 0
has W (Dx ) = 1, W (Dt ) = 3, W (u) = W (v) = 2.
•
Select W (L) = 4. If β = 21 , then
L = D4x + 2uD2x + 2(ux − vx )Dx
+ (u2 − v 2 + u2x − v2x ) I
M = 2D3x + 3uDx + 3 21 ux − vx I
[R. K. Dodd & A. Fordy, Phys. Lett. A, 1982]
•
Example 4: The Drinfel’d-Sokolov-Wilson system
ut + 3vvx = 0,
vt + 2uvx + αux v + 2v3x = 0
has W (Dx ) = 1, W (Dt ) = 3, W (u) = W (v) = 2.
•
Select W (L) = 6. If α = 1, then
L = D6x + 2uD4x + (4ux −3vx ) D3x
+ 29 (u2x −v2x )−u2 − v 2 D2x
+ 52 (u3x −v3x ) + 2 (uux −vvx ) + ux v−uvx Dx
+ 12 (u4x −v4x ) + 12 (u+v)(u2x −v2x ) + 14 (u2x −vx2 ) I
M = D3x + uDx − 21 (3vx −ux ) I
[G. Wilson, Phys. Lett. A, 1974]
•
Example 5: Class of fifth-order KdV equations
ut + αu2 ux + βux uxx + γuu3x + u5x = 0
includes several completely integrable equations:
Parameter ratios
α , β
2 γ
Commonly used values
3 , 2)
( 10
(30, 20, 10), (120, 40, 20),
γ
Equation name
(α, β, γ)
Lax
(270, 60, 30)
1 , 1)
(5
(5, 5, 5), (180, 30, 30),
Sawada-Kotera
(45, 15, 15)
1, 5)
(5
2
(20, 25, 10)
Kaup-Kupershmidt
•
Example 5: Fifth-order equations – continued
•
For W (L) = 2, only Lax’s equation has a Lax pair
L = D2x +
1
γu I
10
M = −16 D5x − 4γu D3x − 6γux D2x − γ 5uxx +
3
γuux I
− γ 32 u3x + 10
2
3
γu
10
[P. Lax, Commun. Pure Appl. Math., 1968]
Dx
•
Example 5: Fifth-order equations – continued
•
For W (L) = 3, the Sawada-Kotera and
Kaup-Kupershmidt equations have Lax pairs
•
For the Kaup-Kupershmidt equation:
L = D3x + 15 γu Dx +
1
γux
10
I
M = 9 D5x + 3γu D3x + 92 γux D2x +
+ 15 γ 2 uux + γu3x I
1 2 2
γ u
5
+ 72 γuxx
[A. Fordy & J. Gibbons, J. Math. Phys., 1980]
•
Example 5: Fifth-order equations – continued
•
For the Sawada-Kotera equation with W (L) = 3:
L = D3x + 15 γu Dx
M = 9 D5x + 3γu D3x + 3γux D2x +
1 2 2
γ u
5
+ 2γu2x Dx
[R. K. Dodd & J. D. Gibbon, Proc. R. Soc.
Lond. A, 1978]
Computations also resulted in:
L̃ = D3x + 15 γu Dx + 51 γux I = Dx L D−1
x
M̃ = 9 D5x + 3γu D3x + 6γux D2x + 51 γ 2 u2 + 5γu2x Dx
+ 52 γ 2 uux + 2γu3x I = Dx M D−1
x
Conclusions and Future Work
•
Scaling invariant Lax pairs in operator form are
fairly easy to construct
•
Scaling invariant Lax pairs in matrix form are hard
to construct
•
Gauge equivalence: Which Lax pairs are useful,
which ones are not?
•
Compare with Wahlquist & Estabrook method,
pseudo-differential operator method, etc.
•
Implementation in Mathematica
Thank You