symbolic software for soliton theory: integrability
TRANSCRIPT
.
SYMBOLIC SOFTWARE FOR SOLITON THEORY:
INTEGRABILITY, SYMMETRIES
CONSERVATION LAWS
AND EXACT SOLUTIONS
Willy Hereman
Dept. of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado
Kruskal Fest
Symposium in Applied Mathematics:
Nonlinear Waves, Dynamics, Asymtotic Analysis
and Physical Applications
Boulder, Colorado
August 3-6, 1995
I. INTRODUCTION
Symbolic Software
• Painleve test for systems of ODEs and PDEs(Macsyma & Mathematica)
• Conservation laws of systems of evolution equations(Mathematica)
• Solitons via Hirota’s method (Macsyma & Mathematica)
• Lie symmetries for systems of ODEs and PDEs(Macsyma)
Purpose of the programs
• Study of integrability of nonlinear PDEs
• Exact solutions as bench mark for numerical algorithms
• Classification of nonlinear PDEs
• Lie symmetries −→ solutions via reductions
Collaborators
• Unal Goktas, Chris Elmer, Wuning Zhuang(MS students)
• Ameina Nuseir (Ph.D student)
• Mark Coffey (CU-Boulder)
• Tony Miller & Tracy Otto (BS students)
II. SYMBOLIC SOFTWARE
Program 1 – Macsyma
Painleve Integrability Test
for Systems of ODEs and PDEs
Integrability of (a systems of) ODEs or PDEs requires that theonly movable singularities in its solution are poles
Definition: A single equation or system has the PainleveProperty if its solution in the complex plane has no worsesingularities than movable poles
Aim: Verify whether or not the system of equationssatisfies the necessary criteria to have the Painleve Prop-erty
For simplicity, consider the case of a single PDE
The solution f expressed as a Laurent series
f = gα∞∑k=0
ukgk
should only have movable poles
Steps of the Painleve Test
• Step 1:
1. Substitute the leading order term
f ∝ u0 gα
into the given equation
2. Determine the integer α < 0 by balancingthe most singular terms in g
3. Calculate u0
• Step 2:
1. Substitute the generic terms
f ∝ u0 gα + ur g
α+r
into the equation, retaining its most singular terms
2. Require that ur is arbitrary
3. Determine the corresponding values of r > 0called resonances
• Step 3:
1. Substitute the truncated expansion
f = gαR∑k=0
uk gk
into the complete equation(R represents the largest resonance)
2. Determine uk unambiguously at the non-resonance lev-els
3. Check whether or not the compatibilitycondition is satisfied at resonance levels
• An equation or system has the Painleve Property andis conjectured to be integrable if:
1. Step 1 thru 3 can be carried out consistently with α < 0and with positive resonances
2. The compatibility conditions are identicallysatisfied for all resonances
• The above algorithm does not detect essentialsingularities
Painleve Integrability Test
• Painleve test for 3rd order equations by Hajee(Reduce, 1982)
• Painleve program (parts) by Hlavaty (Reduce, 1986)
• ODE Painleve by Winternitz & Rand(Macsyma, 1986)
• PDE Painleve by Hereman & Van den Bulck(Macsyma, 1987)
• Painleve test by Conte & Musette (AMP, 1988)
• Painleve analysis by Renner (Reduce, 1992)
• Painleve test for systems of ODEs and PDEsby Hereman, Elmer and Goktas(Macsyma, 1994-96, under development)
• Painleve test for single ODEs and PDEsby Hereman, Miller and Otto(Mathematica, 1995, under development)
• Painleve test for systems of ODEs and PDEsby Hereman and Miller(Mathematica, 1995-96, planned)
Painleve Test of Systems
• System of ODEs due to Akhiezer
uxx + 2u3 − 2au + 2bv = 0
vxx + 2v3 − 2av + 2bu = 0
u(x) = gα1(x)∞∑k=0
uk gk(x)
v(x) = gα2(x)∞∑k=0
vk gk(x)
Leading orders: α1 = α2 = −1
Resonances: r = −1, r = 4
Coefficients:
u0 = v0 = i
u1 = v1 = 0
u2 = v2 = i(b− a)/3
u3 = v3 = 0
At level 4: compatibility condition is satisfiedCoefficients u4 and v4 are arbitrary and independent
The system passes the Painleve test
• Carleman system (simplified version)
ut + ux − c uv = 0
vt − vx + c uv = 0
u(x, t) = gα1(x, t)∞∑k=0
uk(x, t) gk(x, t)
v(x, t) = gα2(x, t)∞∑k=0
vk(x, t) gk(x, t)
Leading orders: α1 = α2 = −1
Resonances: r = −1, r = 1
Coefficients:
u0 =1
c(gt − gx), v0 = −1
c(gt + gx)
At level 1: compatibility condition is satisfiedHowever u1 and v1 dependent on each other
u1 =gxx − gtt + c v1(gt − gx)
c (gt + gx)
The system passes the Painleve test
• Coupled KdV Equations (Hirota-Satsuma system)
ut − 3uux + 6vvx −1
2uxxx = 0
vt + vxxx + 3uvx = 0
Use Kruskal simplification, g(x, t) = x− h(t), in
u(x, t) = gα1(x, t)∞∑k=0
uk(x, t) gk(x, t)
v(x, t) = gα2(x, t)∞∑k=0
vk(x, t) gk(x, t)
Leading orders: First Branch α1 = α2 = −2
Resonances: r = −2,−1, 3, 4, 6, 8
Coefficients:
u0 = −4, v0 = −2
u1 = v1 = 0
u2 = ht/3, v2 = 2ht/3
u5 = −htt + 30v3ht63
, v5 =htt − 12v3ht
63
u7 = −v4t + 12v3v4
12, v7 =
4hthtt + v3(8h2t − 84v4) + 21v4t
504
At level 3: compatibility condition is satisfiedHere u3 = v3 , one function is arbitary
At level 4: compatibility condition is satisfiedHere u4 = 2v4, one function is arbitrary
At level 6: compatibility condition is satisfied
Here u6 =−v3t + 24v6 − 3v3
2
12,
one function is arbitrary
At level 8: compatibility condition is satisfied
Here u8 =
−httt + 6v3htt + (12v3t + 198v23)ht − 3024v8 − 756v4
2
756,
one function is arbitrary
Leading orders: Second Branch α1 = −2, α2 = −1
Resonances: r = −1, 0, 1, 4, 5, 6
Coefficients:
u0 = −2, v0 free
u1 = 0, v1 free
u2 = −(2ht + 3v20)/6, v2 = (4v0ht + 3v3
0)/12
u3 = −v0v1, v3 =v0t + 3v2
0v1
12
At level 0: compatibility condition is satisfiedCoefficient v0 is arbitary
At level 1: compatibility condition is satisfiedCoefficient v1 is arbitary
At level 4: compatibility condition is satisfied
Here u4 = −16v0h2t + 24v3
0ht − 24v1t + 9v50 + 72u4v0
288,
one function is arbitrary
At level 5: compatibility condition is satisfied
Here
u5 =2htt − 12v0v1ht + 3v0v0t − 9v0
3v1
18one function is arbitrary
At level 6: compatibility condition is satisfied
Here
u6 =1
5760(−64v0h
3t− 144v3
0h2t +(96v1t− 108v5
0− 480u4v0)ht
+24v02v1t− 16v0tt+ 288v3
0v21− 27v7
0− 360u4v30+ 576u6v0)
one function is arbitrary
The system passes the Painleve test
Program 2 – Mathematica
Conserved Densities
• Purpose
Compute polynomial-type conservation laws ofsingle evolution equations and systems of evolution equa-tions
For simplicity, consider a single evolution equation
ut = F(u, ux, uxx, ..., unx)
Conservation law is of the form
ρt + Jx = 0
both ρ(u, ux, u2x, . . . , unx) and J(u, ux, u2x, . . . , unx)are polynomials in their arguments
Consequently
P =∫ +∞−∞ ρ dx = constant
provided J vanishes at infinity
– Conservation laws describe the conservation offundamental physical quantities such as mass,linear momentum, total energy (compare with constantsof motion in mechanics)
– For nonlinear PDEs, the existence of a sufficiently large(in principal infinite) number of conservation laws as-sures complete integrability
– Tool to test numerical integrators for PDEs
• Example
Consider the KdV equation, ut + uux + u3x = 0Conserved densities:
ρ1 = u
ρ2 = u2
ρ3 = u3 − 3u2x
...
ρ6 = u6 − 60u3u2x − 30u4
x + 108u2u22x
+720
7u3
2x −648
7uu2
3x +216
7u2
4x
...
Integrable equations have ∞ many conservation laws
• Algorithm and Implementation
Consider the scaling (weights) of the KdV
u ∼ ∂2
∂x2,
∂
∂t∼ ∂3
∂x3
Compute building blocks of ρ3
(i) Start with building block u3
Divide by u and differentiate twice (u2)2x
Produces the list of terms
[u2x, uu2x] −→ [u2
x]
Second list: remove terms that are total derivativewith respect to x or total derivativeup to terms earlier in the list
Divide by u2 and differentiate twice (u)4x
Produces the list: [u4x] −→ [ ]
[ ] is the empty list
Gather the terms:
ρ3 = u3 + c[1]u2x
where the constant c1 must be determined
(ii) Compute∂ρ3
∂t= 3u2ut + 2c1uxuxt
Replace ut by −(uux + uxxx) and uxt by −(uux + uxxx)x
(iii) Integrate the result with respect to x
Carry out all integrations by parts
∂ρ3
∂t=−[
3
4u4 + (c1−3)uu2
x + 3u2uxx− c1u2xx+ 2c1uxuxxx]x
−(c1 + 3)u3x
The last non-integrable term must vanish
Thus, c1 = −3
Result:ρ3 = u3 − 3u2
x
(iv) Expression [. . .] yields
J3 =3
4u4 − 6uu2
x + 3u2uxx + 3u2xx − 6uxuxxx
Computer building blocks of ρ6
(i) Start with u6
Divide by u and differentiate twice
(u5)2x produces the list of terms
[u3u2x, u
4u2x] −→ [u3u2x]
Next, divide u6 by u2, and compute (u4)4x
Corresponding list:
[u4x, uu
2xu2x, u
2u22x, u
2uxu3x, u3u4x] −→ [u4
x, u2u2
2x]
Proceed with (u6
u3)6x = (u3)6x, (u6
u4)8x = (u2)8x
and (u6
u5)10x = (u)10x
Obtain the lists:
[u32x, uxu2xu3x, uu
23x, u
2xu4x, uu2xu4x, uuxu5x, u
2u6x] −→[u3
2x, uu23x]
[u24x, u3xu5x, u2xu6x, uxu7x, uu8x] −→ [u2
4x]
and [u10x] −→ [ ]
Gather the terms:
ρ6 = u6 + c1u3u2
x+ c2u4x+ c3u
2u22x+ c4u
32x+ c5uu
23x+ c6u
24x
where the constants ci must be determined
(ii) Compute ∂∂tρ6
Replace ut, uxt, . . . , unx,t by −(uux + uxxx), . . .
(iii) Integrate the result with respect to x
Carry out all integrations by parts
Require that non-integrabe part vanishes
Set to zero all the coefficients of the independentcombinations involving powers of u andits derivatives with respect to x
Solve the linear system for unknowns c1, c2, . . . , c6
Result:
ρ6 = u6 − 60u3u2x − 30u4
x + 108u2u22x
+720
7u3
2x −648
7uu2
3x +216
7u2
4x
(iv) Flux J6 can be computed by substitutingthe constants into the integrable part of ρ6
Conserved Densities
• Conserved densities by Ito & Kako(Reduce, 1985, 1994)
• Conserved densities in DELiA by Bocharov(Pascal, 1990)
• Conserved densities by Gerdt (Reduce, 1993)
• Conserved densities by Roelofs, Sanders and Wang(Reduce 1994, Maple 1995)
• Conserved densities by Hereman and Goktas(Mathematica, 1993-1995)
• Conservation laws by Wolf (Reduce, 1995)
A Class of Fifth-order Evolution Equations
ut + αu2ux + βuxu2x + γuu3x + u5x = 0
Special cases:
α = 30 β = 20 γ = 10 Lax
α = 5 β = 5 γ = 5 Sawada Kotera
or Caudry−Dodd−Gibbon
α = 20 β = 25 γ = 10 Kaup−Kuperschmidt
α = 2 β = 6 γ = 3 Ito
Under what conditions for the parameters α, β, and γ doesthis equation admit ∞ many conservation laws?
• If α = 310γ
2 and β = 2γ then there is a sequence(without gaps!) of conserved densities (Lax case)
• If α = 15γ
2 and β = γ then there is a sequence(with gaps!) of conserved densities (SK case)
• If α = 15γ
2 and β = 52γ then there is a sequence
(with gaps!) of conserved densities (KK case)
• If α = −2β2+7βγ−4γ2
45 or β = 2γ then there is a conserveddensity of degree 4
Combining both conditions gives α = 2γ2
9 and β = 2γ (Itocase)
Conserved Densities of Systems of Evolution Eqs.
• Coupled KdV Equations (Hirota-Satsuma system)
ut − a(uxxx + 6uux)− 2bvvx = 0
vt + vxxx + 3uvx = 0
u ∼ ∂2
∂x2, v ∼ ∂2
∂x2
ρ1 = u
ρ2 = u2 +2
3bv2
ρ3 = (1 + a)(u3 − 1
2u2x) + b(uv2 − v2
x)
and e.g.
ρ4 = u4 − 2uu2x +
1
5u2xx
+4
5b(u2v2 +
2
3uvvxx +
8
3uv2
x −2
3v2xx) +
4
15b2v4
provided a = 12
There are infinitely many more conservation laws
• The Ito system
ut − uxxx − 6uux − 2vvx = 0
vt − 2uxv − 2uvx = 0
u ∼ ∂2
∂x2, v ∼ ∂2
∂x2
ρ1 = c1u + c2v
ρ2 = u2 + v2
ρ3 = 2u3 + 2uv2 − u2x
ρ4 = 5u4 + 6u2v2 + v4 − 10uu2x + 2v2u2x + u2
2x
and there are infinitely many more conservation laws
• The Drinfel’d-Sokolov system
ut + 3vvx = 0
vt + vux + 2uvx + 2v3x = 0
u ∼ ∂2
∂x2, v free, choose v ∼ ∂2
∂x2
ρ1 = u
ρ2 = v2
ρ3 =−2
9u3 + uv2 +
1
6u2x −
3
2v2x
ρ4 =1
2u4 − 1
6u2v2 − 1
8v4 − 1
6uu2
x + uv2x
− 5
12v2u2x +
1
36u2
2x −3
4v2
2x
and there are presumably infinitely many moreconservation laws
• The dispersiveless long-wave system
ut + vux + uvx = 0
vt + ux + vvx = 0
u free, v free, but u ∼ 2v
choose u ∼ ∂
∂xand 2v ∼ ∂
∂x
ρ1 = v
ρ2 = u
ρ3 = uv
ρ4 = u2 + uv2
ρ5 = 3u2v + uv3
ρ6 =1
3u3 + u2v2 +
1
6uv4
ρ7 = u3v + u2v3 +1
10uv5
ρ8 =1
3u4 + 2u3v2 + u2v4 +
1
15uv6
always the same set irrespective the choice of weights
Further Examples
• A generalized Schamel equation
n2ut + (n + 1)(n + 2)u2nux + uxxx = 0
n positive integer
ρ1 = u, ρ2 = u2
ρ3 =1
2u2x −
n2
2u2+2
n
no further conservation laws
• The Boussinesq equation
utt − buxx + 3uuxx + 3u2x + au4x = 0
a and b are real constants
Write as system of evolution equations
ut + vx = 0
vt + bux − 3uux − au3x = 0
u ∼ ∂2
∂x2, b ∼ ∂2
∂x2, v free, take v ∼ ∂3
∂x3
ρ1 = u
ρ2 = u
ρ3 = bu
ρ4 = bc1u + c2uv
ρ5 = b2c1u + c2(bu2 − u3 + v2 + au2
x)
and there are infinitely many more conservation laws
• A modified vector derivative NLS equation
∂B⊥
∂t+
∂
∂x(B2
⊥B⊥) + αB⊥0B⊥0 ·∂B⊥
∂x+ ex ×
∂2B⊥
∂x2= 0
Replace vector equation by
ut +(u(u2 + v2) + βu− vx
)x
= 0
vt +(v(u2 + v2) + ux
)x
= 0
u and v denote the components of B⊥ paralleland perpendicular to B⊥0 and β = αB2
⊥0
2u ∼ ∂
∂x, 2v ∼ ∂
∂x, β ∼ ∂
∂x
The first 6 conserved densities are:
ρ1 = c1u + c2v
ρ2 = u2 + v2
ρ3 =1
2(u2 + v2)2 − uvx + uxv + βu2
ρ4 =1
4(u2 + v2)3 +
1
2(u2
x + v2x)− u3vx + v3ux +
β
4(u4 − v4)
ρ5 =1
4(u2 + v2)4 − 2
5(uxvxx − uxxvx) +
4
5(uux + vvx)
2
+6
5(u2 + v2)(u2
x + v2x)− (u2 + v2)2(uvx − uxv)
+β
5(2u2
x − 4u3vx + 2u6 + 3u4v2 − v6) +β2
5u4
ρ6 =7
16(u2 + v2)5+
1
2(u2
xx + v2xx)
− 5
2(u2 + v2)(uxvxx−uxxvx) + 5(u2 + v2)(uux + vvx)
2
+15
4(u2 + v2)2(u2
x + v2x)−
35
16(u2 + v2)3(uvx − uxv)
+β
8(5u8 + 10u6v2 − 10u2v6 − 5v8 + 20u2u2
x
− 12u5vx + 60uv4vx − 20v2v2x)
+β2
4(u6 + v6)
Table 1 Conserved Densities for Sawada-Kotera and Lax equations
Density Sawada-Kotera equation Lax equation
ρ1 u u
ρ2 ---- 12u2
ρ313u3 − u2
x13u3 − 1
6u2x
ρ414u4 − 9
4uu2x + 3
4u22x
14u4 − 1
2uu2x + 1
20u22x
ρ6 ---- 15u5 − u2u2
x + 15uu2
2x − 170u2
3x
ρ616u6 − 25
4 u3u2x − 17
8 u4x + 6u2u2
2x16u6 − 5
3u3u2x − 5
36u4x + 1
2u2u22x
+2u32x − 21
8 uu23x + 3
8u24x + 5
63u32x − 1
14uu23x + 1
252u24x
ρ717u7 − 9u4u2
x − 545 uu4
x + 575 u3u2
2x17u7 − 5
2u4u2x − 5
6uu4x + u3u2
2x
+64835 u2
xu22x + 489
35 uu32x − 261
35 u2u23x +1
2u2xu2
2x + 1021uu3
2x − 314u2u2
3x
−28835 u2xu2
3x + 8135uu2
4x − 935u2
5x − 542u2xu2
3x + 142uu2
4x − 1924u2
5x
ρ8 ---- 18u8 − 7
2u5u2x − 35
12u2u4x + 7
4u4u22x
+72uu2
xu22x + 5
3u2u32x + 7
24u42x + 1
2u3u23x
−14u2
xu23x − 5
6uu2xu23x + 1
12u2u24x
+ 7132u2xu2
4x − 1132uu2
5x + 13432u2
6x
Table 2 Conserved Densities for Kaup-Kuperschmidt and Ito equations
Density Kaup-Kuperschmidt equation Ito equation
ρ1 u u
ρ2 ---- u2
2
ρ3u3
3 − 18u2
x ----
ρ4u4
4 − 916uu2
x + 364u2
2xu4
4 − 94uu2
x + 34u2
2x
ρ5 ---- ----
ρ6u6
6 − 3516u3u2
x − 31256u4
x + 5164u2u2
2x ----
+ 37256u3
2x − 15128uu2
3x + 3512u2
4x
ρ7u7
7 − 278 u4u2
x − 369320uu4
x + 6940u3u2
2x ----
+26194480u2
xu22x + 2211
2240uu32x − 477
1120u2u23x
−171640u2xu2
3x + 27560uu2
4x − 94480u2
5x
ρ8 ---- ----
Example 3 – Macsyma/Mathematica
Solitons – Hirota’s Method
• Hirota’s Direct Methodallows to construct soliton solutions of
– nonlinear evolution equations
– wave equations
– coupled systems
• Test conditions for existence of soliton solutions
• Examples:
– Korteweg-de Vries equation (KdV)
ut + 6uux + u3x = 0
– Kadomtsev-Petviashvili equation (KP)
(ut + 6uux + u3x)x + 3u2y = 0
– Sawada-Kotera equation (SK)
ut + 45u2ux + 15uxu2x + 15uu3x + u5x = 0
Hirota’s Method
Korteweg-de Vries equation
ut + 6uux + u3x = 0
Substitute
u(x, t) = 2∂2 ln f (x, t)
∂x2
Integrate with respect to x
ffxt − fxft + ff4x − 4fxf3x + 3f 22x = 0
Bilinear form
B(f ·f ) def=(DxDt +D4
x
)(f ·f ) = 0
Introduce the bilinear operator
Dmx D
nt (f ·g) = (∂x− ∂x′)
m(∂t− ∂t′)
nf (x, t) g(x′, t′)|x′=x,t′=t
Use the expansion
f = 1 +∞∑n=1
εn fn
Substitute f into the bilinear equation
Collect powers in ε (book keeping parameter)
O(ε0) : B(1·1) = 0
O(ε1) : B(1·f1 + f1·1) = 0
O(ε2) : B(1·f2 + f1·f1 + f2·1) = 0
O(ε3) : B(1·f3 + f1·f2 + f2·f1 + f3·1) = 0
O(ε4) : B(1·f4 + f1·f3 + f2·f2 + f3·f1 + f4·1) = 0
O(εn) : B(n∑j=0
fj·fn−j) = 0 with f0 = 1
Start with
f1 =N∑i=1
exp(θi) =N∑i=1
exp (ki x− ωi t + δi)
ki, ωi and δi are constantsDispersion law
ωi = k3i (i = 1, 2, ..., N)
If the original PDE admits a N-soliton solutionthen the expansion will truncate at level n = N
Consider the case N=3Terms generated by B(f1, f1) determine
f2 = a12 exp(θ1 + θ2) + a13 exp(θ1 + θ3) + a23 exp(θ2 + θ3)
= a12 exp [(k1 + k2)x− (ω1 + ω2) t + (δ1 + δ2)]
+ a13 exp [(k1 + k3)x− (ω1 + ω3) t + (δ1 + δ3)]
+ a23 exp [(k2 + k3)x− (ω2 + ω3) t + (δ2 + δ3)]
Calculate the constants a12, a13 and a23
aij =(ki − kj)
2
(ki + kj)2 i, j = 1, 2, 3
Terms from B(f1·f2 + f2·f1) determine
f3 = b123 exp(θ1 + θ2 + θ3)
= b123 exp [(k1+k2+k3)x−(ω1+ω2+ω3)t+(δ1+δ2+δ3)]
with
b123 = a12 a13 a23 =(k1 − k2)
2 (k1 − k3)2 (k2 − k3)
2
(k1 + k2)2 (k1 + k3)
2 (k2 + k3)2
Subsequently, fi = 0 for i > 3
Set ε = 1
f = 1 + exp θ1 + exp θ2 + exp θ3
+ a12 exp(θ1 + θ2) + a13 exp(θ1 + θ3) + a23 exp(θ2 + θ3)
+ b123 exp(θ1 + θ2 + θ3)
Return to the original u(x, t)
u(x, t) = 2∂2 ln f (x, t)
∂x2
Single soliton solution
f = 1 + eθ , θ = kx− ωt + δ
k, ω and δ are constants and ω = k3
Substituting f into
u(x, t) = 2∂2 ln f (x, t)
∂x2
= 2(fxxf − f 2
x
f 2)
Take k = 2K
u = 2K2sech2K(x− 4K2t + δ)
Two-soliton solution
f = 1 + eθ1 + eθ2 + a12eθ1+θ2
θi = kix− ωit + δi
with ωi = k3i , (i = 1, 2) and a12 = (k1−k2)
2
(k1+k2)2
Select
eδi =c2ikiekix−ωit+∆i
f =1
4fe−
12(θ1+θ2)
θi = kix− ωit + ∆i
c2i =k2 + k1
k2 − k1
kiReturn to u(x, t)
u(x, t) = u(x, t) = 2∂2 ln f (x, t)
∂x2
=
k22 − k2
1
2
k2
2cosech2 θ22 + k2
1sech2 θ1
2
(k2 coth θ22 − k1 tanh θ1
2 )2
Program 4 – MacsymaLie-point Symmetries
• System of m differential equations of order k
∆i(x, u(k)) = 0, i = 1, 2, ...,m
with p independent and q dependent variables
x = (x1, x2, ..., xp) ∈ IRp
u = (u1, u2, ..., uq) ∈ IRq
• The group transformations have the form
x = Λgroup(x, u), u = Ωgroup(x, u)
where the functions Λgroup and Ωgroup are to be determined
• Look for the Lie algebra L realized by the vector field
α =p∑i=1ηi(x, u)
∂
∂xi+
q∑l=1ϕl(x, u)
∂
∂ul
Procedure for finding the coefficients
• Construct the kth prolongation pr(k)α of the vector field α
• Apply it to the system of equations
• Request that the resulting expression vanisheson the solution set of the given system
pr(k)α∆i |∆j=0 i, j = 1, ...,m
• This results in a system of linear homogeneous PDEsfor ηi and ϕl, with independent variables x and u( determining equations)
• Procedure thus consists of two major steps:
deriving the determining equationssolving the determining equations
Procedure for Computing the Determining Equations
• Use multi-index notation J = (j1, j2, ..., jp) ∈ INp,to denote partial derivatives of ul
ulJ ≡∂|J |ul
∂x1j1∂x2
j2...∂xpjp,
where |J | = j1 + j2 + ... + jp
• u(k) denotes a vector whose components are all the partialderivatives of order 0 up to k of all the ul
• Steps:
(1) Construct the kth prolongation of the vector field
pr(k)α = α +q∑l=1
∑JψJl (x, u(k))
∂
∂ulJ, 1 ≤ |J | ≤ k
The coefficients ψJl of the first prolongation are:
ψJil = Diϕl(x, u)−p∑j=1
ulJjDiηj(x, u),
where Ji is a p−tuple with 1 on the ith position and zeroselsewhere
Di is the total derivative operator
Di =∂
∂xi+
q∑l=1
∑JulJ+Ji
∂
∂ulJ, 0 ≤ |J | ≤ k
Higher order prolongations are defined recursively:
ψJ+Jil = Diψ
Jl −
p∑j=1
ulJ+JjDiη
j(x, u), |J | ≥ 1
(2) Apply the prolonged operator pr(k)α to eachequation ∆i(x, u(k)) = 0
Require that pr(k)α vanishes on the solution set of the sys-tem
pr(k)α ∆i |∆j=0 = 0 i, j = 1, ...,m
(3) Choose m components of the vector u(k),say v1, ..., vm, such that:
(a) Each vi is equal to a derivative of a ul (l = 1, ..., q)with respect to at least one variable xi (i = 1, ..., p).
(b) None of the vi is the derivative of another one in theset.
(c) The system can be solved algebraically for the vi interms of the remaining components of u(k), which we de-noted by w:
vi = Si(x,w), i = 1, ...,m.
(d) The derivatives of vi,
viJ = DJSi(x,w),
where DJ ≡ Dj11 D
j22 ...D
jpp , can all be expressed in terms
of the components of w and their derivatives, without everreintroducing the vi or their derivatives.
For instance, for a system of evolution equations
uit(x1, ..., xp−1, t) = F i(x1, ..., xp−1, t, u(k)), i = 1, ...,m,
where u(k) involves derivatives with respect to the variablesxi but not t, choose vi = uit.
(4) Eliminate all vi and their derivatives from the ex-pression prolonged vector field, so that all the remainingvariables are independent
(5) Obtain the determining equations for ηi(x, u) andϕl(x, u) by equating to zero the coefficients of the remain-ing independent derivatives ulJ .
Example: The Korteweg-de Vries Equation
ut + auux + uxxx = 0
• one equation (parameter a)
• two independent variables t and x
• one dependent variable u
• vector field α = ηx ∂∂x + ηt ∂∂t + ϕu ∂∂u
Format for SYMMGRP.MAX
• variables x[1] = x, x[2] = t, u[1] = u
• equation e1 : u[1, [0, 1]] + a ∗ u[1] ∗ u[1, [1, 0]] + u[1, [3, 0]]
• variable to be eliminated v1 : u[1, [0, 1]]
• coefficients of vectorfield in SYMMGRP.MAX:eta[1] = ηx, eta[2] = ηt and phi[1] = ϕu
There are only eight determining equations
∂eta[2]
∂u[1]= 0
∂eta[2]
∂x[1]= 0
∂eta[1]
∂u[1]= 0
∂2phi[1]
∂u[1]2= 0
∂2phi[1]
∂u[1]∂x[1]− ∂2eta[1]
∂x[1]2= 0
∂phi[1]
∂x[2]+∂3phi[1]
∂x[1]3+ u[1]
∂phi[1]
∂x[1]= 0
3∂3phi[1]
∂u[1]∂x[1]2− ∂eta[1]
∂x[2]− ∂3eta[1]
∂x[1]3
+ 2 u[1]eta[1]
∂x[1]+ phi[1] = 0
u[1]∂eta[2]
∂x[2]+ 3
∂3phi[1]
∂u[1]∂x[1]2− ∂eta[1]
∂x[2]− ∂3eta[1]
∂x[1]3
− u[1]eta[1]
∂x[1]+ phi[1] = 0
The solution in the original variables
ηx = k1 + k3 t− k4 x
ηt = k2 − 3k4 t
ϕu = k3 + 2 k4 u
The four infinitesimal generators are
G1 = ∂xG2 = ∂tG3 = t ∂x + ∂uG4 = x ∂x + 3 t ∂t − 2 u ∂u
Equation is invariant under:
• translations G1 and G2
• Galilean boost G3
• scaling G4
Computation of the flows corresponding to G1 thru G4 shows thatfor any solution u = f (x, t) of the KdV equation the transformedsolutions
u = f (x− ε, t)
u = f (x, t− ε)
u = f (x− ε, t) + ε
u = e−2εf (e−εx, e−3εt)
will solve the KdV equation
Note that ε is the parameter of the transformation group
III. PLANS FOR THE FUTURE
Extension of Symbolic Software Packages(Macsyma/Mathematica)
• Lie symmetries of differential-difference equations
• Solver for systems of linear, homogeneous PDEs(Hereman)
• Painleve test for systems of PDEs(Elmer, Goktas & Coffey)
• Solitons via Hirota’s method for bilinear equations (Zhuang)
• Simplification of Hirota’s method (Hereman & Nuseir)
• Conservation laws of PDEs with variable coefficients (Goktas)
• Lax pairs, special solutions, ...
New Software
• Wavelets (prototype/educational tool)
• Other methods for Differential Equations