.
IMACS PDE7 — FIRST TALK – Incomplete!
SYMBOLIC SOFTWARE FOR
THE STUDY OF NONLINEAR
PARTIAL DIFFERENTIAL EQUATIONS
Willy Hereman
Dept. of Mathematical and Computer Sciences
Colorado School of Mines
Golden, CO 80401-1887
Rutgers University, New Brunswick, NJ
Monday, June 22, 1992
1:30pm
Example: The cylindrical KDV Equation
The cylindrical Korteweg-de Vries equation,
fx
+ fxxxx + 6f fxx + 6(fx)2 + ftx = 0,
(∗)
2t
has the Painlevé property. One easily determines the coefficient 2t1 in (*), by analyzing a cylindrical KdV equation with
arbitrary coefficient a(t) of fx. Integration of the compatibility
condition
a(t)t + 2a(t)2 = 0,
gives
a(t) =
1
.
2t
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − −−
PAINLEVE ANALYSIS OF EQUATION,
a(t)fx + fxxxx + 6f fxx + 6(fx)2 + ftx = 0
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − −−
SUBSTITUTE u0 g alf a FOR f IN ORIGINAL EQUATION.
MINIMUM POWERS OF g ARE [2 alf a − 2, alf a − 4]
∗ COEFFICIENT OF g 2 alf a−2 IS 6u02 alf a (2alf a − 1) (gx)2
∗ COEFFICIENT OF g alf a−4 IS u0 (alf a−3)(alf a−2)(alf a−
1) alf a (gx)4
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − −−
FOR EXPONENTS (2 alf a − 2) AND (alf a − 4) OF g, DO
WITH alf a = −2, POWER OF g is −6 −→ SOLVE FOR
u0
TERM 60 u0 (gx)2 (2(gx)2 + u0) g16 IS DOMINANT
IN EQUATION.
WITH u0 = −2(gx)2 −→ FIND RESONANCES
SUBSTITUTE u0 g alf a + ur g r+alf a FOR f IN EQUATION
TERM (gx)4(r − 6)(r − 5)(r − 4)(r + 1)ur g r−6 IS DOMINANT
IN EQUATION.
THE 3 NON-NEGATIVE INTEGRAL ROOTS ARE
[r = 4, r = 5, r = 6]
WITH MAXIMUM RESONANCE = 6 −→ CHECK RESONANCES.
SUBSTITUTE POWER SERIES P6k=0 g k−2uk FOR f IN
EQUATION.
WITH u0 = −2(gx)2
∗ COEFFICIENT OF
1
g5
IS 120(gx)4(2gxx − u1)
u1 = 2gxx
∗ COEFFICIENT OF g14 IS
−12(gx)2(4gxgxxx − 3(gxx)2 + 6u2(gx)2 + gtgx)
4gxgxxx − 3(gxx)2 + gtgx
u2 = −
6(gx)2
∗ COEFFICIENT OF
1
g3
IS
4((gx)3a(t) + (gx)2gxxxx − 4gxgxxgxxx
+3(gxx)2 − gtgxgxx − 6u3(gx)4 + gtx(gx)2)
u3 = ((gx)3a(t) + (gx)2gxxxx − 4gxgxxgxxx + 3(gxx)2 − gtgxgxx
+gtx(gx)2)/(6(gx)4)
∗ COEFFICIENT OF
1
g2
IS 0
u4 IS ARBITRARY !
COMPATIBILITY CONDITION IS SATISFIED !
∗ COEFFICIENT OF g1 IS 0
u5 IS ARBITRARY !
COMPATIBILITY CONDITION IS SATISFIED !
a(t)t + 2a(t)2
∗ COEFFICIENT OF 1 IS
6
u6 IS ARBITRARY !
a(t)t + 2a(t)2
COMPATIBILITY CONDITION:
= 0,
6
∗ ∗ ∗ CONDITION IS NOT SATISFIED .∗ ∗ ∗
∗∗∗ CHECK FOR FREE PARAMETERS OR PRESENCE
OF u0 ∗ ∗ ∗
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − −−
Example: The cylindrical KP Equation
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − − − − − −−
PAINLEVE ANALYSIS OF EQUATION, fyy b(t) + fx a(t) +
fxxxx + 6f fxx + 6(fx)2 + ftx = 0
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − − − − − −−
SUBSTITUTE u0 g alfa FOR f IN ORIGINAL EQUATION.
MINIMUM POWERS OF g ARE [2 alfa − 2, alfa − 4]
∗ COEFFICIENT OF g 2 alfa−2 IS 6u02 alfa (2alfa − 1) (gx)2
∗ COEFFICIENT OF g alfa−4 IS u0 (alfa − 3)(alfa − 2)(alfa −
1) alfa (gx)4
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − − − − − −−
FOR EXPONENTS (2 alfa − 2) AND (alfa − 4) OF g, DO
WITH alfa = −2, POWER OF g is −6 −→ SOLVE FOR
u0
TERM 60 u0 (gx)2 (2(gx)2 + u0) g16 IS DOMINANT IN
EQUATION.
WITH u0 = −2(gx)2 −→ FIND RESONANCES
SUBSTITUTE u0 g alfa + ur g r+alfa FOR f IN EQUATION
TERM (gx)4(r − 6)(r − 5)(r − 4)(r + 1)ur g r−6 IS DOMINANT IN EQUATION.
THE 3 NON-NEGATIVE INTEGRAL ROOTS ARE [r =
4, r = 5, r = 6]
WITH MAXIMUM RESONANCE = 6 −→ CHECK RESONANCES.
SUBSTITUTE POWER SERIES P6k=0 g k−2uk FOR f IN
EQUATION.
WITH u0 = −2(gx)2
∗ COEFFICIENT OF g15 IS 120(gx)4(2gxx − u1)
u1 = 2gxx
∗ COEFFICIENT OF g14 IS −12 (gx)2 ((gy )2 b(t)+4gxgxxx−
3(gxx)2 + 6u2(gx)2 + gtgx)
(gy )2 b(t) + 4gxgxxx − 3(gxx)2 + gtgx
u2 = −
6(gx)2
∗ COEFFICIENT OF
1
g3
IS
4 ((gx)2gyy b(t) − gxx(gy )2 b(t) + (gx)3 a(t) + (gx)2gxxxx − 4gxgxxgxx
− gtgxgxx − 6u3(gx)4 + gtx(gx)2)
u3 = ((gx)2gyy b(t) − gxx(gy )2 b(t) + (gx)3 a(t) + (gx)2gxxxx − 4gxgx
− gtgxgxx + gtx(gx)2)/(6(gx)4)
∗ COEFFICIENT OF
1
g2
IS 0
u4 IS ARBITRARY !
COMPATIBILITY CONDITION IS SATISFIED !
∗ COEFFICIENT OF
1
g
IS 0
u5 IS ARBITRARY !
COMPATIBILITY CONDITION IS SATISFIED !
∗ COEFFICIENT OF 1 IS
(gx)2gyy b(t)t + gxx(gy )2 b(t)t − 2gxgxy gy b(t)t + (gx)3 a(t)t + 4(gx)2
+4gxx(gy )2 a(t)b(t) − 8gxgxy gy a(t)b(t) + 2(gx)3 a2(t))/(6(gx)3)
u6 IS ARBITRARY !
COMPATIBILITY CONDITION:
(gx)2gyy b(t)t + gxx(gy )2 b(t)t − 2gxgxy gy b(t)t + (gx)3 a(t)t + 4(gx)2
+4gxx(gy )2 a(t)b(t) − 8gxgxy gy a(t)b(t) + 2(gx)3 a2(t))/(6(gx)3) = 0,
∗ ∗ ∗ CONDITION IS NOT SATISFIED .∗ ∗ ∗
∗ ∗ ∗ CHECK FOR FREE PARAMETERS OR PRESENCE
OF u0 ∗ ∗ ∗
−−−−−−−−−−−−−−−−−−−−−−−−−−
− − − − − − − − − − − − − − −−
Further Analysis
The variable coefficient equation
(ut + 6uux + uxxx)x + a(t)ux + b(t)uyy = 0
will posses the Painlevé property if at resonance level r = 6 the compatibility condition
[at + 2a2]gx3 + [bt + 4ab](gx2gyy + gxxgy 2 − 2gxgxy gy ) = 0
is satisfied
Thus
at + 2a2 = 0
bt + 4ab = 0
First case:
a(t) = 0
b(t) = k
For k = 0 one gets the Korteweg-de Vries equation
ut + uux + uxxx = 0
For k = ±1 one obtains the Kadomtsev-Petviashvili equation
(ut + uux + uxxx)x ± uyy = 0
Second case:
1
2(t − t0)
k
b(t) =
(t − t0)2
a(t) =
k and t0 integration constants (set t0 = 0)
For k = 0 one obtains to the cylindrical Korteweg-de Vries equation
u
ut + 6uux + uxxx + = 0
2t
For k 6= 0 one gets the cylindrical Kadomtsev-Petviashvili equation
1
3σ 2
(ut + 6uux + uxxx)x + ux + 2 uyy = 0
2t
t
(σ 2 = ±1)
Example 1: The Harry Dym Equation
ut − u3uxxx = 0
• one equation
• two independent variables t and x
• one dependent variable u
∂
∂
∂
+ η t ∂t
+ ϕu ∂u
• vector field α = η x ∂x
Format for SYMMGRP.MAX
• variables x[1] = x,
x[2] = t,
u[1] = u
• equation e1 : u[1, [0, 1]] − u[1]3 ∗ u[1, [3, 0]]
• variable to be eliminated v1 : u[1, [0, 1]]
• coefficients of vectorfield in SYMMGRP.MAX:
eta[1] = η x, eta[2] = η t and phi[1] = ϕu
Format for PDELIE and SYM DE
• variables: [x, t] and [u]
• equation: 0diff(u, t) − u3 ∗ 0diff(u, x, 3) = 0
• variable to be eliminated: 0diff(u, t)
• coefficients of vectorfield in PDELIE:
GX = η x, GT = η t and FU = ϕu
• coefficients of vectorfield in SYM DE:
X1 = η x, X2 = η t and Y1 = ϕu
Format for SPDE and LIE
• variables X1 = x,
X2 = t,
U (1) = u
• equation e1 : U (1, 2) − U (1)3 ∗ U (1, 1, 1, 1)
• variable to be eliminated U (1, 2)
• coefficients of vectorfield in SPDE:
XI(1) = η x, XI(2) = η t and ETA(1) = ϕu
• coefficients of vectorfield in LIE:
XXX#1 = η x, XXX#2 = η t and UUU#1 = ϕu
There are only eight determining equations,
in SYMMGRP.MAX notation
∂eta[2]
= 0
∂u[1]
∂eta[2]
= 0
∂x[1]
∂eta[1]
= 0
∂u[1]
∂ 2phi[1]
= 0
∂u[1]2
∂ 2phi[1]
∂ 2eta[1]
−
= 0
∂u[1]∂x[1]
∂x[1]2
3
∂phi[1]
3 ∂ phi[1]
− u[1]
= 0
3
∂x[2]
∂x[1]
3
∂ 3phi[1]
∂eta[1]
3 ∂ eta[1]
3u[1]
+
− u[1]
= 0
∂u[1]∂x[1]2
∂x[2]
∂x[1]3
3
u[1]
∂eta[2]
∂eta[1]
− 3u[1]
+ 3 phi[1] = 0
∂x[2]
∂x[1]
The solution in the original variables
η x = k1 + k3 x + k5 x 2
η t = k2 − 3k4 t
ϕu = (k3 + k4 + 2k5 x) u
The five infinitesimal generators are
G1
G2
G3
G4
G5
=
=
=
=
=
∂x
∂t
x∂x + u∂u
−3t∂t + u∂u
x2∂x + 2xu∂u
Equation is invariant under:
• translations G1 and G2
• scaling G3 and G4
• what else ?
Computation of the flow corresponding to G5
dx̃
= x̃2
d
x̃(0) = x
dt̃
= 0
d
t̃(0) = t
(1)
dũ
ũ(0) = u
= 2x̃ũ
d
is the parameter of the transformation group
One obtains
x
x̃() =
1 − x
t̃() = t
u
ũ() =
(1 − x)2
Conclusion: for any solution u = f (x, t) of the Harry Dym
equation the transformed solution
x̃
ũ(x̃, t̃) = (1 + x̃)2f (
, t̃)
1 + x̃
will solve
ũt̃ − ũ3ũx̃x̃x̃ = 0
Example 2: The Nonlinear Schrödinger Equation
iut + uxx + 2|u|2u = 0
Split in real and imaginary parts via u = v + iw
vt + vxx + 2(v 2 + w2)w = 0
wt − wxx − 2(v 2 + w2)v = 0
• two coupled equations
• two independent variables t and x
• two dependent variable v and w
∂
∂
∂
∂
• vector field α = η x ∂x
+ η t ∂t
+ ϕv ∂v
+ ϕw ∂w
Format for SYMMGRP.MAX
• variables x[1] = x,
x[2] = t,
u[1] = u,
u[2] = w
• equations
e1 : u[1, [0, 1]] + u[2, [2, 0]] + 2 ∗ u[2] ∗ (u[1]2 + u[2]2)
e2 : u[2, [0, 1]] − u[1, [2, 0]] − 2 ∗ u[1] ∗ (u[1]2 + u[2]2)
• variables to be eliminated v1 : u[1, [0, 1]] and v2 : u[2, [0, 1]]
• coefficients of vectorfield in SYMMGRP.MAX:
eta[1] = η x, eta[2] = η t, phi[1] = ϕv and phi[2] = ϕw
There are 20 determining equations
The solution in the original variables
ηx
ηt
ϕv
ϕw
=
=
=
=
k1 + k4 x + 2 k5 t
k2 + 2k4 t
−k3 w − k4 v − k5 x w
k3 v − k4 w + k5 x v
The five infinitesimal generators are
G1
G2
G3
G4
G5
=
=
=
=
=
∂x
∂t
v ∂w − w ∂v
2 t ∂t + x ∂x − v ∂v − w ∂w
2 t ∂x − x w ∂v + x v ∂w
Equation is invariant under:
• translations G1 and G2
• rotation G3
• dilation or scaling G4
• Galilean boost or inversion G5