æ
Symbolic Computation of Travelling Wave Solutions
of Nonlinear Partial Differential Equations
and Differential-Difference Equations
Prof. Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, CO-80401, U.S.A.
http://www.mines.edu/fs home/whereman/
whereman@mines.edu
Laboratory of Plasma Physics
Royal Military Academy, Brussels, Belgium
Thursday, December 23, 2004, 14:30
Collaborators: Douglas Baldwin & Ünal Göktaş (Wolfram Research, Inc.)
Research supported in part by NSF
under Grants DMS-9912293 and CCR-9901929
1
OUTLINE
Purpose & Motivation
Typical Examples of ODEs, PDEs, and DDEs
Part I: Tanh Method for PDEs
Quick Review of Tanh Algorithm
A Typical Example
Other Types of Functions
Demo of Mathematica Software: PDESpecialSolutions.m
Part II: Tanh Method for DDEs (Lattices)
Algorithm for Tanh Solutions
Typical Examples
Table with Results
Demo of Mathematica Software: DDESpecialSolutions.m
Conclusions & Future Research
Research Papers & Software
2
Purpose & Motivation
• Develop and implement various methods to find closed form solutions of nonlinear PDEs and DDEs: direct methods, Lie symmetry
methods, similarity methods, etc.
• Fully automate the hyperbolic and elliptic function methods to
compute travelling solutions of nonlinear PDEs.
• Fully automate the hyperbolic tanh method to compute travelling
wave solutions of nonlinear differential-difference equations (DDEs
or lattices).
• Class of nonlinear PDEs and DDEs solvable with such methods
includes famous evolution and wave equations, and lattices.
Examples PDEs: Korteweg-de Vries, Boussinesq, and KuramotoSivashinsky equations.
Fisher and FitzHugh-Nagumo equations.
Examples ODEs: Duffing and nonlinear oscillator equations.
Examples DDEs: Volterra, Toda, and Ablowitz-Ladik lattices.
• PDEs: Solutions of tanh (kink) or sech (pulse) type model solitary waves in fluids, plasmas, circuits, optical fibers, bio-genetics,
etc.
DDEs: discretizations of PDEs, lattice theory, queing and network
problems, solid state and quantum physics.
• Benchmark solutions for numerical PDE and DDE solvers.
3
• Research aspect: Design high-quality application packages to
compute solitary wave solutions of large classes of nonlinear evolution and wave equations and lattices.
• Educational aspect: Software as course ware for courses in nonlinear PDEs and DDEs, theory of nonlinear waves, integrability,
dynamical systems, and modeling with symbolic software.
REU projects of NSF. Extreme Programming!
• Users: scientists working on nonlinear wave phenomena in fluid
dynamics, nonlinear networks, elastic media, chemical kinetics, material science, bio-sciences, plasma physics, and nonlinear optics.
4
Typical Examples of ODEs and PDEs
• The Duffing equation:
u00 + u + αu3 = 0
Solutions in terms of elliptic functions:
q
c21 − 1
c21 − 1
u(x) = ± √
cn(c1x + ∆;
),
α
2c21
and
r
2(c21 − 1)
1 − c21
sn(c1x + ∆; 2 ).
u(x) = ± √
α
c1
• The Korteweg-de Vries (KdV) equation:
ut + 6αuux + u3x = 0.
Solitary wave solution:
8c31 − c2 2c21
tanh2 [c1x + c2t + ∆] ,
u(x, t) =
−
6αc1
α
or, equivalently,
4c31 + c2 2c21
u(x, t) = −
+
sech2 [c1x + c2t + ∆] .
6αc1
α
Cnoidal wave solution:
4c31(1 − 2m) − c2 12m c21 2
u(x, t) =
+
cn (c1x + c2t + ∆; m),
αc1
α
modulus m.
5
• The modified Korteweg-de Vries (mKdV) equation:
ut + αu2ux + u3x = 0.
Solitary wave solution:
v
u
u
u
t
6
3
u(x, t) = ±
c1 sech c1x − c1t + ∆ .
α
• Three-dimensional modified Korteweg-de Vries equation:
ut + 6u2ux + uxyz = 0.
Solitary wave solution:
√
u(x, y, z, t) = ± c2c3 sech [c1x + c2y + c3z − c1c2c3t + ∆] .
• The combined KdV-mKdV equation:
ut + 6αuux + 6βu2ux + γu3x = 0.
Real solitary wave solution:
v
u
uγ
c1
α
t c1 sech(c1 x +
(3α2 − 2βγc21)t + ∆).
u(x, t) = − ± u
2β
β
2β
Complex solutions:
v
u
uγ
α
c1
t c1 tanh(c1 x +
u(x, t) = − ± iu
(3α2 + 4βγc21)t + ∆),
2β
β
2β
v
uγ
α
1u
t c1 ( sechξ ± i tanhξ),
u(x, t) = − + u
2β 2 β
and
v
uγ
α
1u
t c1 ( sechξ ∓ i tanhξ)
u(x, t) = − − u
2β 2 β
c1
with ξ = c1x + 2β
(3α2 + βγc21)t + ∆.
6
• The Fisher equation:
ut − uxx − u (1 − u) = 0.
Solitary wave solution:
u(x, t) =
with
1 1
1
± tanhξ + tanh2ξ,
4 2
4
5
1
ξ = ± √ x ± t + ∆.
12
2 6
• The generalized Kuramoto-Sivashinski equation:
ut + uux + uxx + σu3x + u4x = 0.
Solitary wave solutions
(ignoring symmetry u → −u, x → −x, σ → −σ) :
For σ = 4 :
u(x, t) = 9 − 2c2 − 15 tanhξ (1 + tanhξ − tanh2ξ)
with ξ =
x
2
For σ =
√12
47
u(x, t) =
+ c2t + ∆.
:
45 ∓ 4418c2
45
45
15
√
± √ tanhξ− √ tanh2ξ± √ tanh3ξ
47 47
47 47
47 47
47 47
with ξ = ± 2√147 x + c2t + ∆.
7
√
For σ = 16/ 73 :
u(x, t) =
2 (30 ∓ 5329c2)
75
60
15
√
± √ tanhξ− √ tanh2ξ± √ tanh3ξ
73 73
73 73
73 73
73 73
with ξ = ± 2√173 x + c2t + ∆.
For σ = 0 :
v
u
u
u
t
u(x, t) = −2
with ξ =
1
2
r
11
19
v
u
u
u
t
v
u
u
u
t
19
135 11
165 11
c2 −
tanhξ +
tanh3ξ
11
19 19
19 19
x + c2t + ∆.
• The Boussinesq (wave) equation:
utt − u2x + 3uu2x + 3ux2 + αu4x = 0,
or written as a first-order system (v auxiliary variable):
ut + vx = 0,
vt + ux − 3uux − αu3x = 0.
Solitary wave solution:
c21 − c22 + 8αc41
u(x, t) =
− 4αc21 tanh2 [c1x + c2t + ∆] ,
2
3c1
v(x, t) = b0 + 4αc1c2 tanh2 [c1x + c2t + ∆] .
• The Broer-Kaup system:
uty + 2(uux)y + 2vxx − uxxy = 0,
vt + 2(uv)x + vxx = 0.
Solitary wave solution:
c3
u(x, t) = −
+ c1 tanh [c1x + c2y + c3t + ∆] ,
2c1
v(x, t) = c1c2 − c1c2 tanh2 [c1x + c2y + c3t + ∆] .
8
• System of three nonlinear coupled equations (Gao & Tian, 2001):
ut − ux − 2v = 0,
vt + 2uw = 0,
wt + 2uv = 0.
Solutions:
u(x, t) = ±c2 tanhξ,
1
v(x, t) = ∓ c2(c1 − c2) sech2ξ,
2
1
w(x, t) = − c2(c1 − c2) sech2ξ,
2
and
u(x, t) = ±ic2 sechξ,
1
v(x, t) = ± ic2(c1 − c2) tanhξ sechξ,
2
1
2
w(x, t) = c2(c1 − c2) 1 − 2 sech ξ ,
4
and also
1
u(x, t) = ± ic2 ( sechξ + i tanhξ) ,
2
1
v(x, t) = ± c2(c1 − c2) sechξ ( sechξ + i tanhξ) ,
4
1
w(x, t) = − c2(c1 − c2) sechξ ( sechξ + i tanhξ)
4
with ξ = c1x + c2t + ∆.
9
• Nonlinear sine-Gordon equation (light cone coordinates):
Φxt = sin Φ.
Set u = Φx, v = cos(Φ) − 1,
uxt − u − u v = 0,
u2t + 2v + v 2 = 0.
Solitary wave solution (kink):
1
1
u = ± √ sech[ √ (x − ct) + ∆],
−c
−c
1
v = 1 − 2 sech2[ √ (x − ct) + ∆].
−c
Solution:
1
u(x, t)dx = ±4 arctan exp  √ (x − ct) + ∆ .
−c

Φ(x, t) =
Z

• ODEs from quantum field theory:
uxx = −u + u3 + auv 2,
vxx = bv + cv 3 + av(u2 − 1).
Solitary wave solutions:
v
u
u
u
u
t
a2 − c
u = ± tanh[
x + ∆],
2(a − c)
v
u
u
u
u
t
v
u
u
u
t
1−a
a2 − c
v = ±
sech[
x + ∆],
a−c
2(a − c)
s
provided b =
a2 −c
2(a−c) .
10

Typical Examples of DDEs (lattices)
• The Volterra lattice:
u̇n = un(vn − vn−1),
v̇n = vn(un+1 − un).
Travelling wave solution:
un(t) = −c1 coth(d1) + c1 tanh [d1n + c1t + δ] ,
vn(t) = −c1 coth(d1) − c1 tanh [d1n + c1t + δ] .
• The Toda lattice:
ün = (1 + u̇n) (un−1 − 2un + un+1) .
Travelling wave solution:
un(t) = a10 ± sinh(d1) tanh [d1n ± sinh(d1) t + δ] .
• The Relativistic Toda lattice:
u̇n = (1 + α un)(vn − vn−1),
v̇n = vn(un+1 − un + α vn+1 − α vn−1).
Travelling wave solution:
1
un(t) = − − c1 coth(d1) + c1 tanh [d1n + c1t + δ] ,
α
c1 coth(d1) c1
vn(t) =
− tanh [d1n + c1t + δ] .
α
α
11
• The Ablowitz-Ladik lattice:
u̇n(t) = (α + unvn)(un+1 + un−1) − 2αun,
v̇n(t) = −(α + unvn(vn+1 + vn−1) + 2αvn.
Travelling wave solution:
α sinh2(d1) 2
un(t) =
±1 − tanh d1n + 2αt sinh (d1) + δ ,
a21
2
vn(t) = a21(±1 + tanh d1n + 2α sinh (d1)t + δ ).
• 2D Toda lattice:
∂ 2 un
∂un
(x, t) = 
+ 1 (un−1 − 2un + un+1) .
∂x∂t
∂t
Travelling wave solution:


sinh2(d1)
1
2

x + c2t + δ  .
un(x, t) = a10 + sinh (d1) tanh d1n +
c2
c2


• Hybrid lattice:
u̇n(t) = (1 + αun + βu2n)(un−1 − un+1),
Travelling wave solution:


√ 2
2
−α ± α − 4β tanh(d1)
α − 4β
un(t) =
tanh d1n +
tanh(d1)t + δ  .
2β
2β
12
Algorithm for Tanh Solutions for system of PDEs
System of nonlinear PDEs of order m
∆(u(x), u0(x), u00(x), · · · u(m)(x)) = 0.
Dependent variable u has M components ui (or u, v, w, ...).
Independent variable x has N components xj (or x, y, z, ..., t).
Step T1:
• Seek solution u(x) = U(T ), with
T = tanhξ = tanh

N
X

j j
j

c x + δ  .
• Observe tanh0ξ = 1 − tanh2ξ or T 0 = 1 − T 2. Hence, all derivative
of T are polynomial in T. For example, T 00 = −2T (1 − T 2), etc.
• Repeatedly apply the operator rule
∂•
∂ξ dT d•
d•
=
= cj (1 − T 2)
∂xj ∂xj dξ dT
dT
Produces a nonlinear system of ODEs
∆(T, U(T ), U0(T ), U00(T ), . . . , U(m)(T )) = 0.
NOTE: Compare with the ultra-spherical (linear) ODE:
(1 − x2)y 00(x) − (2α + 1)xy 0(x) + n(n + 2α)y(x) = 0
with integer n ≥ 0 and α real. Includes:
* Legendre equation (α = 21 ),
* ODE for Chebeyshev polynomials of type I (α = 0),
* ODE for Chebeyshev polynomials of type II (α = 1).
13
• Example: For the Boussinesq system
ut + vx = 0,
vt + ux − 3uux − αu3x = 0,
after cancelling common factors 1 − T 2,
c2U 0 + c1V 0 = 0,
c2V 0 + c1U 0 − 3c1U U 0
3
2
0
2
00
2 2 000
+αc1 2(1 − 3T )U + 6T (1 − T )U − (1 − T ) U = 0.
Step T2:
• Seek polynomial solutions
Ui(T ) =
M
Xi
j=0
aij T j .
Determine the highest exponents Mi ≥ 1.
Substitute Ui(T ) = T Mi into the LHS of ODE.
Gives polynomial P(T ).
For every Pi consider all possible balances of the highest exponents
in T.
Solve the resulting linear system(s) for the unknowns Mi.
• Example: Balance highest exponents for the Boussinesq system
M1 − 1 = M2 − 1,
2M1 − 1 = M1 + 1.
So, M1 = M2 = 2.
Hence,
U (T ) = a10 + a11T + a12T 2,
V (T ) = a20 + a21T + a22T 2.
14
Step T3:
• Derive algebraic system for the unknown coefficients aij by setting
to zero the coefficients of the power terms in T.
• Example: Algebraic system for Boussinesq case
a11 c1 (3a12 + 2α c21) = 0,
a12 c1 (a12 + 4α c21) = 0,
a21 c1 + a11 c2 = 0,
a22 c1 + a12 c2 = 0,
a11 c1 − 3a10 a11 c1 + 2αa11 c31 + a21 c2 = 0,
−3a211 c1 + 2 a12 c1 − 6a10 a12 c1 + 16α a12 c31 + 2a22 c2 = 0.
Step T4:
• Solve the nonlinear algebraic system with parameters.
• Example: Solution for Boussinesq system
c21 − c22 + 8αc41
, a11 = 0,
a10 =
3c21
a12 = −4αc21, a20 = free,
a21 = 0, a22 = 4αc1c2.
Step T5:
• Return to the original variables. Test the final solution(s) of PDE.
Reject trivial solutions.
• Example: Solitary wave solution for Boussinesq system:
c21 − c22 + 8αc41
u(x, t) =
− 4αc21 tanh2 [c1x + c2t + δ] ,
2
3c1
v(x, t) = a20 + 4αc1c2 tanh2 [c1x + c2t + δ] .
15
Other Types of Solutions for PDEs
Function ODE (y 0 = dy
dξ )
tanh(ξ) y 0 = 1 − y 2
√
0
sech(ξ) y = −y 1 − y 2
tan(ξ)
y0 = 1 + y2
exp(ξ)
y0 = y
r
0
cn(ξ; m) y = −
0
Chain Rule
∂•
2 d•
∂xj = cj (1 − T ) dT
√
∂•
2 d•
∂xj = −cj S 1 − S dS
r
(1−y 2)(1−m+m y 2)
sn(ξ; m) y = (1 −
y 2)(1
−
m y2)
16
∂•
∂xj
∂•
∂xj
∂•
∂xj
∂•
∂xj
= cj (1 + TAN2) dd•
TAN
= cj E d•
dE
r
=−cj (1− CN2)(1−m+m CN2) dd•
CN
r
= cj (1 − SN2)(1 − mSN2) dd•SN
Algorithm for Tanh Solutions for System of DDEs
Nonlinear differential-difference equations (DDEs) of order m
∆(un+p1 (x), un+p2 (x), · · · , un+pk (x), u0n+p1 (x), u0n+p2 (x), · · · , u0n+pk (x), · · · ,
(r)
(r)
(r)
un+p1 (x), un+p2 (x), · · · , un+pk (x)) = 0.
Dependent variable un has M components ui,n (or un, vn, wn, · · ·)
Independent variable x has N components xi (or t, x, y, · · ·).
Shift vectors pi ∈ ZQ.
u(r)(x) is collection of mixed derivatives of order r.
Simplest case for independent variable (t), and one lattice point (n):
(r)
(r)
∆(..., un−1, un, un+1, ..., u̇n−1, u̇n, u̇n+1, ..., un−1, u(r)
n , un+1 , ...) = 0.
Step D1:
• Seek solution un(x) = Un(Tn), with Tn = tanh(ξn),
ξn =
Q
X
i=1
dini +
N
X
j=1
cj xj + δ = d · n + c · x + δ.
• Repeatedly apply the operator rule
d•
∂ξn dTn d•
d•
=
= cj (1 − Tn2)
,
dxj ∂xj dξn dTn
dTn
transforms DDE into
∆(Un+p1 (Tn), · · · , Un+pk (Tn), U0n+p1 (Tn), · · · , U0n+pk (Tn), · · ·
(r)
(r)
Un+p1 (Tn), · · · , Un+pk (Tn)) = 0.
Note: Un+ps is function of Tn not of Tn+ps .
17
• Example: Toda lattice
ün = (1 + u̇n) (un−1 − 2un + un+1)
transforms into
c21(1−Tn2)
2TnUn0 −
(1−Tn2)Un00 +
1+c1(1−Tn2)Un0
[Un−1 −2Un + Un+1] = 0.
Step D2:
• Seek polynomial solutions
Ui,n(Tn) =
Use tanh(x + y) =
M
Xi
j=0
aij Tnj .
tanh x + tanh y
to deal with the shift:
1 + tanh x tanh y
Tn+ps =
Tn + tanhφs
,
1 + Tn tanhφs
where
φs = ps · d = ps1d1 + ps2d2 + · · · + psQdQ,
Substitute Ui,n(Tn) = TnMi , and
Mi
T
+
tanhφ
n
s

=
,
1 + Tn tanhφs

Mi
Ui,n+ps (Tn) = Tn+p
s

and balance the highest exponents in Tn to determine Mi.
Note: Ui,n+0(Tn) = TnMi is of degree Mi in Tn.
Ui,n+ps (Tn) =
Tn + tanhφs Mi
1+Tn tanhφs
is of degree zero in Tn.
18
• Example: Balance of exponents for Toda lattice
2M1 + 1 = M1 + 2.
So, M1 = 1. Hence,
Un(Tn) = a10 + a11Tn,
Un±1(T (n ± 1)) = a10 + a11T (n ± 1) = a10 + a11
Tn ± tanh(d1)
.
1 ± Tn tanh(d1)
Step D3:
• Determine the algebraic system for the unknown coefficients aij by
setting to zero the coefficients of the powers in Tn.
• Example: Algebraic system for Toda lattice
c21 − tanh2(d1) − a11c1 tanh2(d1) = 0,
c1 − a11 = 0.
Step D4:
• Solve the nonlinear algebraic system with parameters.
• Example: Solution of algebraic system for Toda lattice
a10 = free,
a11 = ± sinh(d1),
c1 = ± sinh(d1).
Step D5:
• Return to the original variables. Test solution(s) of DDE.
Reject trivial ones.
• Example: Solitary wave solution for Toda lattice:
un(t) = a10 ± sinh(d1) tanh [d1n ± sinh(d1) t + δ] .
19
Example of System of DDEs: Relativistic Toda Lattice
u̇n = (1 + αun)(vn − vn−1),
v̇n = vn(un+1 − un + αvn+1 − αvn−1).
Change of variables
un(t) = Un(Tn),
vn(t) = Vn(Tn),
with
Tn(t) = tanh [d1n + c1t + δ] .
gives
c1(1 − T 2)Un0 − (1 + αUn)(Vn − Vn−1) = 0,
c1(1 − T 2)Vn0 − Vn(Un+1 − Un + αVn+1 − αVn−1) = 0.
Seek polynomial solutions
Un(Tn) =
M
X1
j=0
a1j Tnj ,
Vn(Tn) =
M
X2
j=0
a2j Tnj .
Balance the highest exponents in Tn to determine M1, and M2 :
M 1 + 1 = M 1 + M 2 , M2 + 1 = M 1 + M 2 .
So, M1 = M2 = 1. Hence,
Un = a10 + a11Tn,
Vn = a20 + a21Tn.
20
Algebraic system for aij :
−a11 c1 + a21 tanh(d1) + α a10 a21 tanh(d1) = 0,
a11 tanh(d1) (α a21 + c1) = 0,
−a21 c1 + a11 a20 tanh(d1) + 2α a20 a21 tanh(d1) = 0,
tanh(d1) (a11 a21 + 2α a221 − a11 a20 tanh(d1)) = 0,
a21 tanh2(d1) (c1 − a11) = 0.
Solution of the algebraic system
1
a10 = − − c1 coth(d1),
α
a11 = c1,
c1 coth(d1)
,
a20 =
α
c1
a21 = − .
α
Solitary wave solution in original variables:
1
un(t) = − − c1 coth(d1) + c1 tanh [d1n + c1t + ∆] ,
α
c1 coth(d1) c1
vn(t) =
− tanh [d1n + c1t + ∆] .
α
α
21
Multi-dimensional Example: 2D Toda Lattice
2D Toda lattice:
∂ 2 yn
= exp (yn−1 − yn) − exp (yn − yn+1),
∂x∂t
yn(x, t) is displacement from equilibrium of the n-th unit mass under
an exponential decaying interaction force between nearest neighbors.
Set
∂un
= exp(yn−1 − yn) − 1.
∂t
(∗)
Then,
exp(yn−1 − yn) =
∂un
+ 1,
∂t
and the 2D-Toda lattice becomes
∂ 2 yn
∂un
∂un+1
∂un ∂un+1
=
+1−
+ 1 =
−
.
∂x∂t
∂t
∂t
∂t
∂t
Integrate with respect to t to get


∂yn
= un − un+1.
∂x
Differentiate (*) with respect to x and
∂ 2 un
∂
=
(exp(yn−1 − yn) − 1)
∂x∂t
∂x


∂y
∂y
n−1
n
,
= exp(yn−1 − yn) 
−
∂x
∂x


∂un
+ 1 [(un−1 − un) − (un − un+1)],
= 
∂t


∂u
n
= 
+ 1 (un−1 − 2un + un+1) .
∂t
22
So, the 2D Toda lattice is written in polynomial form:
∂ 2 un
∂un
(x, t) = 
+ 1 (un−1 − 2un + un+1) .
∂x∂t
∂t
Travelling wave solution:


1
sinh2(d1)
2

un(x, t) = a10 + sinh (d1) tanh d1n +
x + c2t + δ  .
c2
c2


Complicated Example: Ablowitz-Ladik Lattice
The Ablowitz-Ladik lattice:
u̇n(t) = (α + unvn)(un+1 + un−1) − 2αun,
v̇n(t) = −(α + unvn(vn+1 + vn−1) + 2αvn.
Travelling wave solution:
α sinh2(d1) 2
±1 − tanh d1n + 2αt sinh (d1) + δ ,
un(t) =
a21
2
vn(t) = a21(±1 + tanh d1n + 2α sinh (d1)t + δ ).
23
Analyzing and Solving Nonlinear Parameterized Systems
Assumptions:
• All ci 6= 0 and di 6= 0 (and modulus m 6= 0).
• Parameters (α, β, γ, ...). Otherwise the maximal exponents Mi may
change.
• All Mi ≥ 1.
• All ai Mi 6= 0. Highest power terms in Ui must be present, except in
mixed sech-tanh-method.
• Solve for aij , then ci, tanh(di), and m. Then find conditions on
parameters.
Strategy followed by hand:
• Solve all linear equations in aij first (cost: branching). Start with
the ones without parameters. Capture constraints in the process.
• Solve linear equations in ci, tanh(di), m if they are free of aij .
• Solve linear equations in parameters if they free of aij , ci, tanh(di), m.
• Solve quasi-linear equations for aij , ci, tanh(di), m.
• Solve quadratic equations for aij , ci, tanh(di), m.
• Eliminate cubic terms for aij , ci, tanh(di), m, without solving.
• Show remaining equations, if any.
Alternatives:
• Use (adapted) Gröbner bases techniques.
• Use Ritt-Wu characteristic sets method.
• Use combinatorics on coefficients aij = 0 or aij 6= 0.
24
Implementation Issues – Software Demo – Future Work
• Demonstration of Mathematica package for hyperbolic and elliptic
function methods for PDEs and tanh function for DDEs.
• Long term goal: Develop PDESolve and DDESolve for analytical
solutions of nonlinear PDEs and DDEs.
• Implement various methods: Lie symmetry methods, etc.
• Look at other types of explicit solutions involving
– other hyperbolic and elliptic functions sinh, cosh, dn, ....
– complex exponentials combined with sech or tanh.
• Other applications (of the nonlinear algebraic solver):
Computation of conservation laws, symmetries, first integrals, etc.
leading to linear parameterized systems for unknowns coefficients
(see InvariantsSymmetries by Göktaş and Hereman).
25
• Papers:
D. Baldwin, Ü. Göktaş, W. Hereman, L. Hong, R. Martino, and
J.C. Miller, Symbolic computation of exact solutions expressible
in hyperbolic and elliptic functions for nonlinear PDEs,
Journal of Symbolic Computation 37 (2004) 669–705.
D. Baldwin, Ü. Göktaş, and W. Hereman, Symbolic computation
of exact tanh solutions of nonlinear differential-difference equations, Computer Physics Communications 162 (2004) 203–217.
• Software:
D. Baldwin, Ü. Göktaş, W. Hereman, L. Hong, R. Martino, and
J.C. Miller, PDESpecialSolutions.m: A Mathematica program
for the symbolic computation of exact solutions expressible in hyperbolic and elliptic functions for systems of nonlinear partial differential equations (2001-2003).
Available via anonymous FTP from mines.edu in directory
pub/papers/math cs dept/software/PDESpecialSolutions;
or via Internet URL: http://www.mines.edu/fs home/whereman/
D. Baldwin, Ü. Göktaş, W. Hereman, L. Hong, R. Martino, and J.C.
Miller, DDESpecialSolutions.m: A Mathematica program for
the symbolic computation of tanh solutions for systems of nonlinear
differential-difference equations (2001).
Available via anonymous FTP from mines.edu in directory
pub/papers/math cs dept/software/DDESpecialSolutions;
or via Internet URL: http://www.mines.edu/fs home/whereman/
26
Appendix: A Complicated Case
Class of fifth-order evolution equations with parameters:
ut + αγ 2u2ux + βγuxu2x + γuu3x + u5x = 0.
Well-Known Special cases
Lax case: α =
3
10 , β
= 2, γ = 10. Two solutions:
u(x, t) =
4c21
−
6c21 tanh2
c1 x −
56c51t
+∆ ,
and
u(x, t) = a0 −
2c21 tanh2
c1 x −
2(15a20c1
−
40a0c31
+
28c51)t
+∆ ,
where a0 is arbitrary.
Sawada-Kotera case: α = 51 , β = 1, γ = 5. Two solutions:
u(x, t) =
8c21
−
12c21 tanh2
c1 x −
16c51t
+∆ ,
and
u(x, t) = a0 −
6c21 tanh2
c1 x −
(5a20c1
−
40a0c31
+
76c51)t
+∆ ,
where a0 is arbitrary.
Kaup-Kupershmidt case: α = 15 , β = 52 , γ = 10. Two solutions:
3 2
2
2
5
u(x, t) = c1 − c1 tanh c1x − c1t + ∆
2
and
2
2
2
5
u(x, t) = 8c1 − 12c tanh c1x − 176c1t + ∆ .
No free constants!
Ito case: α = 92 , β = 2, γ = 3. One solution:
u(x, t) =
20c21
−
30c21 tanh2
27
c1 x −
96c51t
+∆ .
What about the General case?
Q1: Can we retrieve the special solutions?
Q2: What are the condition(s) on the parameters α, β, γ for solutions
of tanh-type to exist?
Tanh solutions:
u(x, t) = a0 + a1 tanh [c1x + c2t + ∆] + a2 tanh2 [c1x + c2t + ∆] .
Nonlinear algebraic system must be analyzed, solved (or reduced!):
a1(αγ 2a22 + 6γa2c21 + 2βγa2c21 + 24c41) = 0,
a1(αγ 2a21 + 6αγ 2a0a2 + 6γa0c21 − 18γa2c21 − 12βγa2c21 − 120c41) = 0,
αγ 2a22 + 12γa2c21 + 6βγa2c21 + 360c41 = 0,
2αγ 2a21a2 + 2αγ 2a0a22 + 3γa21c21 + βγa21c21 + 12γa0a2c21
−8γa22c21 − 8βγa22c21 − 480a2c41 = 0,
a1(αγ 2a20c1 − 2γa0c31 + 2βγa2c31 + 16c51 + c2) = 0,
αγ 2a0a21c1 + αγ 2a20a2c1 − γa21c31 − βγa21c31 − 8γa0a2c31 + 2βγa22c31
+136a2c51 + a2c2 = 0.
Unknowns: a0, a1, a2.
Parameters: c1, c2, α, β, γ.
Solve and Reduce cannot be used on the whole system!
28
Actual Solution: Two major cases:
CASE 1: a1 = 0, two subcases
Subcase 1-a:
3
a2 = − a0,
2
c2 = c31(24c21 − βγa0),
where a0 is one of the two roots of the quadratic equation:
αγ 2a20 − 8γa0c21 − 4βγa0c21 + 160c41 = 0.
Subcase 1-b: If β = 10α − 1, then
a2 = −
6 2
c,
αγ 1
1
c2 = − (α2γ 2a20c1 − 8αγa0c31 + 12c51 + 16αc51),
α
where a0 is arbitrary.
CASE 2: a1 6= 0, then
α=
1
(39 + 38β + 8β 2)
392
and
a2 = −
168
c21,
γ(3 + 2β)
provided β is root of
(104β 2 + 886β + 1487)(520β 3 + 2158β 2 − 1103β − 8871) = 0.
29
1
Subcase 2-a: If β 2 = − 104
(886β + 1487), then
α = −
2β + 5
,
26
49c21(9983 + 4378β)
a0 = −
,
26γ(8 + 3β)(3 + 2β)2
336c21
a1 = ±
,
γ(3 + 2β)
168c21
,
a2 = −
γ(3 + 2β)
364 c51 (3851 + 1634β)
c2 = −
.
6715 + 2946β
Subcase 2-b: If β 3 =
1
520 (8871
+ 1103β − 2158β 2), then
39 + 38β + 8β 2
,
α =
392
28 c21 (6483 + 5529β + 1066β 2)
a0 =
,
(3 + 2β)(23 + 6β)(81 + 26β)γ
a21
28224 c41 (4β − 1)(26β − 17)
,
=
(3 + 2β)2(23 + 6β)(81 + 26β)γ 2
168c21
a2 = −
,
γ(3 + 2β)
8 c51 (1792261977 + 1161063881β + 188900114β 2)
c2 = −
.
959833473 + 632954969β + 105176786β 2
30