Hirota Direct Method Graham W. Griffiths 1 City University, UK. oo0oo

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1 Hirota Direct Method Graham W. Griffiths 1 City University, UK. oo0oo Contents 1 Introduction 1 Applications of the Hirota method 3.1 Single-soliton solution to the KdV equation Two-soliton solution to the KdV equation Three-soliton solution to the KdV equation Four-soliton solution to the KdV equation Two-Soliton solution to the Boussinesq equation Two-Soliton solution to the Sawada-Kotera equation Two-Soliton solution to the D Kadomtsev-Petviashvili equation Introduction The Hirota direct method was first published in a paper by Hirota in The introduction of this approach provided a direct method for finding N-soliton solutions to non-linear evolutionary equations and, by way of an example, Hirota applied this method to the Korteweg-deVries (KdV) equation [Hir-71], i.e. where u = u(x, t). u t 6uu x u xxx = 0, (1) As part of Hirota s analysis he introduced a bilinear form defined as ( D x (f g) = x ) f(x) g(x ), () x x =x or, more generally D m x D n t (f g) = ( x ) m ( x t ) n f(x) g(x ) t x =x, t =t, (3) where D is a binary operator (because it operates on a pair of functions) and is called the Hirota derivative. On expanding an mth power of the bilinear operator of eqn. (), we see that it results in a binomial expression of the form ( x ) m m ( ) m m ( ) m = ( 1) r r x r x m r x = ( 1) r x m r x m r, (4) r=0 where we have made use of the binomial coefficient, ( ) m m! r =, 0 r m. This, r!(m r) in turn, gives rise to a more compact form for the Hirota derivative, i.e. 1 graham@griffiths1.com Note: The Hirota derivative acts on a product of two functions in a similar way to the standard Calculus rule for differentiation. However, there is a sign difference, e.g. D x(f g) f xg fg x compared with x(f g) f xg fg x. i=0 Graham W Griffiths 1 10 March 01

2 D m x f g = m ( m ( 1) i r r=0 ) f (m r)x g rx. (5) This form is particularly useful for a computer algebra system (CAS) implementation, as we see below. Some examples of bilinear operations follow. D x (f g) = f x g fg x = D x (g h) Dx(f g) = f xx g f x g x fg xx = Dx(g f) D x D t (f g) = D x (f t g fg t ) = f xt g f t g x f x g t fg xt = D x D t (g f) D x D t (f f) = (ff xt f x f t ) Dx(f 4 g) = f xxxx g 4f xxx g x 6f xx g xx 4f x g xxx fg xxxx = Dx(g 4 f) Dx(f n g) = 0, (n odd) Also, fundamental to Hirota s approach was the introduction of the ancillary function f = f(x, t), which is a logarithmic transformation defined as u(x, t) = ln f(x, t) x (6) = ff xx f x f, (7) where the leading constant of was chosen so that, on application of this equation, certain terms cancel to yield a quadratic in f. The form of function f in eqn. (7) is chosen to be a perturbation expansion, defined as f(x, t) = 1 for functions f 1, f,, etc. yet to be found. ɛ n f n (x, t), (8) Now, for a given bilinear operation, say B, where we have B(f f) = 0, the coefficient of like powers of epsilon can be equated to zero to obtain the following set of equations, n=1 O(ɛ 0 ) : B(1 1) = 0, (9) O(ɛ 1 ) : B(1 f 1 f 1 1) = 0, (10) O(ɛ ) : B(1 f f 1 f 1 f 1) = 0, (11) O(ɛ 3 ) : B(1 f 3 f 1 f f f 1 f 3 1) = 0, (1) O(ɛ 4 ) : B(1 f 4 f 1 f 3 f f f 3 f 1 f 4 1) = 0, (13) ( n ) O(ɛ n ) : B f j f n j = 0, (f 0 = 1). (14) j=0 The above scheme is general and is applicable to any explicit bilinear operator expression. It can be shown that if the original PDE admits a N-soliton solution, then eqn. (8) will truncate at the n = N term provided f 1 is the sum of precisely N simple exponential terms. This leads to f(x, t) = 1 ɛ 1 f 1 :one soliton, (15) f(x, t) = 1 ɛ 1 f 1 ɛ f :two solitons, (16) f(x, t) = 1 ɛ 1 f 1 ɛ f ɛ 3 f 3 :three solitons, (17) etc. (18) Graham W Griffiths 10 March 01

3 We will now illustrate use of eqn. (10) by application of the bilinear examples in eqns. (6) above, when we see that for m = 1 and n = 1, D x D t (1 f 1 ) = D x ( 1 f 1,t ) = f 1,xt, (19) D x D t (f 1 1) = D x (f 1,t 1) = f 1,xt, (0) D x D t (1 f 1 f 1 1) = f 1,xt. (1) Similarly, for m = 4 and n = 0 we see that D 4 x(1 f 1 ) = f 1,xxxx, () D 4 x(f 1 1) = f 1,xxxx, (3) D 4 x(1 f 1 f 1 1) = f 1,xxxx. (4) Thus, we have ( Dx D t D 4 x) (1 f1 f 1 1) = (f 1,xt f 1,xxxx ). (5) A similar approach to eqn (11) yields, ( ) Dx D t Dx 4 (1 f f 1 f 1 f 1) = (f 1 f 1,xt f 1,x f 1,t ) [ (f,xt f,xxxx ) (f 1 f 1,xxxx 4f 1,x f 1,xxx 3f1,xx) ]. (6) We make use of these relationships to obtain appropriate dispersion relations and coupling coefficients in the KdV examples below. Whilst the above equations look quite complex, they do provide a route to solving evolutionary equations. This will be illustrated by obtaining detailed solutions to the above KdV equation (1) by way of example. In addition, solutions to other well known evolutionary equations will be presented. Additional expanded discussion on this method can be found in [Dra-9], [Hir-04], [Her-94] and [Waz-09]. Applications of the Hirota method.1 Single-soliton solution to the KdV equation The KdV equation (1) which we repeat here for convenience, u t 6uu x u xxx = 0, (7) can be transformed by eqn (7), the so-called Cole-Hopf transformation, to yield f 3 [ ft,3x f f 3x f t f 4f x f t,x f 4f x f t 4f 3x f x,x f 1f 3x f x 16f 3x f x f 5x f 10f 4x f x f ] = 0. (8) This equation can be written more succinctly in the following form using Hirota s bilinear operator, [ ] 1 ( ) Dx D x f t Dx 4 f f = 0. (9) Graham W Griffiths 3 10 March 01

4 Hence, integrating once with respect to x and taking the integration functions to be zero, the bilinear equation equivalent to eqn. (7) is ( ) Dx D t Dx 4 f f = 0. (30) Thus, if f is a solution of eqn. (30), then u from eqn. (7) is a solution to eqn. (7). See [Rog-8, p69-71] for a detailed discussion and derivation. The problem has now changed to finding the auxiliary function f that satisfies eqn. (30) which, when inserted into eqn. (7), will yield u(x, t). If we try the ansatz f = 1 and substitute this into eqn. (7), it leads to the trivial solution u = 0. However, if we try the ansatz f 1 = e ξ, where ξ = k(x ct), k = wavenumber and c = wave velocity, then on substituting f = 1 ɛf 1 into eqn. (7), we obtain the solution u = k e (k(x ct) (1 e (k(x ct) ) = 1 k sech where we have set ɛ = 1, without loss of generality. [ ] 1 k(x ct), (31) We can simplify this solution further by substituting f 1 into eqn. (10), i.e. (D x D t D xxxx ) (1 f 1 f 1 1) = 0, (3) ( k c k 4) e k(x ct) = 0, (33) where we have used B = (D x D t D xxxx ) along with eqns. (1) and (4). This leads directly to the dispersion relation 3 for the KdV equation, c = k. (34) Thus, we observe that c = c(k) and k cannot be determined independently for the KdV equation. With this result, we have finally arrived at the situation where the complete solution is known, i.e. u = 1 [ ] 1 k sech k(x k t). (35) This is the well known single-soliton solution for the KdV equation. A plot of this solution is given in Fig (1). Manual implementation of the above calculations can be rather tedious and is prone to error, so employing a symbolic computer program is highly recommended. A Maple program that solves this single-soliton example and generates an animation is given in Listing (1). Listing 1: Maple program that uses the Hirota method to derive a single-soliton solution to KdV eqn (7) # Hirota bilinear method # Single - soliton solution to KdV equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), f=f(x,t)): # KdV equation pde1 := diff (u,t) 6* u* diff (u,x) diff (u,x,x,x); 3 The dispersion relation characterizes the mathematical [ ] relationship between temporal and spacial variations. For π example, for a harmonic wave of the form A sin (x ct) we have wavenumber k = π πc and frequency ω = λ λ λ, where λ represents wavelength, giving the dispersal relation ω(k) = ck. Graham W Griffiths 4 10 March 01

5 # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; v :=* diff (ln(f),x,x); declare (f): pde := eval ( subs (u=v, pde1 )) =0; # Check bilinear transform correct pde3 := simplify (pde, size ); pde3_chk := simplify ( diff (( Hirota_Dxt (f,f) Hirota_Dxm (f,f,4) )/(f^),x),size ) =0; testeq ( lhs ( pde3 )= lhs ( pde3_chk )); # Ansatz for f_1 f1 := exp (k*(x-c*t)); B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (1, ff1,4) Hirota_Dxm (ff1,1,4) =0; eqn1 := B1(f1); # B (1. f1 f1.1) ; sol1 := solve ( simplify (eqn1, symbolic ),c); c:= sol1 ; # Single soliton ancilliary solution F :=1 f1; # Single soliton solution soln :=u= normal (* diff (ln(f),x,x)); # Check solution solution pdetest (soln, pde1 ); # Animate result zz1 := subs ({k= sqrt () }, rhs ( soln )); animate (zz1,x = , t = -0..0, numpoints =300, frames =50, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" KdV Equation : Single - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; Figure 1: Initial animation profile for KdV equation solution, u(x, t) = 1 k sech [ 1 k(x k t) ], at t = 0, for k = 1. The soliton moves left to right.. Two-soliton solution to the KdV equation This problem is similar to the single-soliton problem except that the auxilliary function f has an additional term, so we now need to find f 1 and f. Because we are seeking a two-soliton solution, we use a two-term form of f 1 that was successful for the single soliton case, i.e. f 1 = e ξ 1 e ξ, along with the ansatz f = a 1 e (ξ 1ξ ), where a 1 is Graham W Griffiths 5 10 March 01

6 a coupling constant yet to be determined. This enables us to construct the auxilliary function, f = 1 ɛf 1 ɛ f. The form for f is not an arbitrary choice but, rather, is suggested by eqn (11) which evaluates to B(1 f f.1) B(f 1 f 1 ) = B(1 f f 1) 6k 1 k (k 1 k ) e ξ 1ξ = 0. (36) For eqn (36) to be equal to zero, clearly, the first term must be of the form, B(1 f f 1) = a 1 e ξ 1ξ. (37) We now substitute the auxilliary function f into eqn. (7) to obtain the solution u = ( k1e ξ 1 ke ξ a 1 (k 1 k ) e ) (ξ 1ξ ) (1 e ξ 1 e ξ a1 e (ξ 1ξ ), ) ( k 1 e ξ 1 k e ξ a 1 (k 1 k )e (ξ 1ξ ) ) (1 e ξ 1 e ξ a1 e (ξ 1ξ ) ). (38) Where, again without loss of generality, we have set ɛ = 1. As for the single-soliton solution, we can obtain the dispersion relations by substituting, in turn, the 1 st and nd terms of f 1 into eqn. (10), to obtain c 1 = k 1, c = k. (39) Also, on substituting f 1 and f into eqn. (11) we obtain an expression which can now be solved for the coupling constant in terms of wave numbers, k 1 and k. This yields, a 1 = (k 1 k ) (k 1 k ). (40) We now have a fully defined two-soliton solution for KdV eqn (7). A plot of this solution is given in Fig (). A Maple program that solves this two-soliton example and generates an animation is given in Listing (). Listing : Maple program that uses the Hirota method to derive a two-soliton solution to KdV eqn (7) # Hirota bilinear method # Two - soliton Solution to KdV equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), f=f(x,t)): # KdV equation pde1 := diff (u,t) 6* u* diff (u,x) diff (u,x,x,x); # Convert to potential form pde := subs (u= diff (v,x), pde1 ); # Integrate once wrt x pde3 := int (pde,x); # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Graham W Griffiths 6 10 March 01

7 Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; v1 :=* diff (ln(f),x); # equivalent to ==>> u =* diff (ln(f),x,x) declare (f): pde4 := eval ( subs (v=v1, pde3 )) =0; # Check bilinear transform correct pde5 := simplify ( pde4 *(f ^/),size ); pde5_chk :=( Hirota_Dxt (f,f) Hirota_Dxm (f,f,4) ) /=0; testeq ( lhs ( pde5 )= lhs ( pde5_chk )); # Ansatz for f_1 f11 := exp (k [1]*(x-c [1]* t)); f1 := exp (k []*(x-c []* t)); f1 := f11 f1 ; B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (ff1,1,4) Hirota_Dxm (1, ff1,4) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b }, symbolic ),{c[1],c []}) ; assign ( sol1 ); # == > > Dispersion relations # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 f :=a [1]* combine ( f11 * f1 ); # Two - soliton ancilliary function F :=1 f1f; B :=( ff1, ff ) -> Hirota_Dxt (1, ff ) Hirota_Dxt (ff,1) Hirota_Dxm (1, ff,4) Hirota_Dxm (ff,1,4) Hirota_Dxt (ff1, ff1 ) Hirota_Dxm (ff1,ff1,4) ; eqn := simplify (B(f1,f),size ) =0; a [1]:= factor ( simplify ( solve (eqn,a [1]), symbolic )); # Two - soliton solution soln :=u= simplify (* diff (ln(f),x,x),size );; # Check solution pdetest (soln, pde1 ); # Animate result zz1 := subs ({k[1]=1, k []= sqrt () }, rhs ( soln )); animate (zz1,x = , t = -0..0, numpoints =300, frames =100, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" KdV Equation : Two - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; Figure : Initial animation profile for KdV equation two-soliton solution, eqn. (38), at t = 0, for k = 1 and k =. Solitons move left to right with the taller faster soliton overtaking the shorter slower soliton. Graham W Griffiths 7 10 March 01

8 .3 Three-soliton solution to the KdV equation This problem is similar to the two-soliton problem except that the auxilliary function f has an additional term, so we now need to find f 1, f and f 3. Because we are seeking a three-soliton solution, we use three-term forms of f 1 and f that were successful for the two-soliton case, i.e. f 1 = e ξ 1 e ξ e ξ 3 and f = a 1 e (ξ 1ξ ) a 13 e (ξ 1ξ 3 ) a 3 e (ξ ξ 3 ), along with the ansatz f 3 = b 13 e ξ 1ξ ξ 3, where coupling constants a 1, a 13, a 3 and b 13 are yet to be determined. This enables us to construct the auxilliary function, f = 1 ɛf 1 ɛ f ɛ 3 f 3. The a i,j auxillary coupling constants are determined as for the two-soliton case. On substituting in turn all combinations of two terms from f, i.e. 1 st and nd ; 1 st and 3 rd ; nd and 3 rd ; into eqn. (11), we obtain expressions which can be solved for these coupling constants in terms of wave numbers, k 1, k and k 3. Similarly to f in the two-soliton case, the form for f 3 is not an arbitrary choice but, rather, is suggested by eqn (1) which evaluates to B(1 f 3 f 3 1)B(f 1 f 1 f f ) = B(1 f 3 f 3 1) 6 (k k 3 ) (k 1 k 3 ) (k 1 k ) (k 1 k k 3 ) e ξ 1ξ ξ 3 = 0. (41) (k 1 k ) (k 1 k 3 ) (k k 3 ) For eqn (41) to be equal to zero, clearly, the first term must be of the form, B(1 f 3 f 3 1) = b 13 e ξ 1ξ ξ 3. (4) We now substitute the auxilliary function f into eqn. (7) to obtain the solution where u = N 1 D 1 ( N D ), (43) N 1 =a 1 (k 1 k ) e ξ 1ξ a 13 (k 1 k 3 ) e ξ 1ξ 3 a 3 (k k 3 ) e ξ ξ 3 k 1 e ξ 1 k e ξ k 3 e ξ 3 b 13 (k 1 k k 3 ) e ξ 1ξ ξ 3, (44) N =a 1 (k 1 k ) e ξ 1ξ a 13 (k 1 k 3 ) e ξ 1ξ 3 a 3 (k k 3 ) e ξ ξ 3. b 13 (k 1 k k 3 ) e ξ 1ξ ξ 3 k 1 e ξ 1 k e ξ k 3 e ξ 3 ), (45) D 1 =1 e ξ 1 e ξ e ξ 3 a 1 e ξ 1ξ a 13 e ξ 1ξ 3 a 3 e ξ ξ 3 b 13 e ξ 1ξ ξ 3, (46) D =D 1. (47) Again, without loss of generality, we have set ɛ = 1 in eqn. (8). Similarly to the two-soliton solution, we can obtain the dispersion relations by substituting, in turn, the 1 st, nd and 3 rd terms of f 1 into eqn. (10), to obtain c 1 = k 1, c = k, c 3 = k 3. (48) We now substitute pairs of f 1 terms along with the coresponding term of f into eqn. (11), i.e. terms 1 and of f 1 plus term 1 of f inserted into eqn. (11) yields an expression Graham W Griffiths 8 10 March 01

9 that can be solved for a 1. Coupling constants a 13 and a 3 are found in the same way to yield, a 1 = (k 1 k ) (k 1 k ), a 13 = (k 1 k 3 ) (k 1 k 3 ), a 3 = (k k 3 ) (k k 3 ). (49) Similarly, on substituting f 1, f and f 3 into eqn. (1) we obtain an expression which can be solved for the coupling constant b 13 in terms of wave numbers, k 1, k and k 3. This yields, b 13 = (k k 3 ) (k 1 k 3 ) (k 1 k ) (k k 3 ) (k 1 k 3 ) (k 1 k ). (50) We now have a fully defined three-soliton solution for KdV eqn (7). solution is given in Fig (3). A plot of this A Maple program that solves this three-soliton example and generates an animation is given in Listing (3). Listing 3: Maple program that uses the Hirota method to derive a three-soliton solution to KdV eqn (7) # Hirota bilinear method # Three - soliton Solution to KdV equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), f=f(x,t)): # KdV equation pde1 := diff (u,t) 6* u* diff (u,x) diff (u,x,x,x); # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; v :=* diff (ln(f),x,x); declare (f): pde := eval ( subs (u=v, pde1 )) =0; # Check bilinear transform correct pde3 := simplify (pde, size ); pde3_chk := simplify ( diff (( Hirota_Dxt (f,f) Hirota_Dxm (f,f,4) )/(f^),x),size ) =0; testeq ( lhs ( pde3 )= lhs ( pde3_chk )); # Ansatz for f_1 f11 := exp (k [1]*(x-c [1]* t)); f1 := exp (k []*(x-c []* t)); f13 := exp (k [3]*(x-c [3]* t)); f1 := f11 f1 f13 ; B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (ff1,1,4) Hirota_Dxm (1, ff1,4) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; eqn1c := B1(f13 ); # B (1. f3 f3.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b, eqn1c }, symbolic ),{c[1],c[],c [3]}) ; assign ( sol1 ); # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 f1 :=a [1]* combine ( f11 * f1 ); f :=a [13]* combine ( f11 * f13 ); f3 :=a [3]* combine ( f1 * f13 ); f := f1 f f3 ; B :=( ff1, ff ) -> Hirota_Dxt (1, ff ) Hirota_Dxt (ff,1) Hirota_Dxm (1, ff,4) Hirota_Dxm (ff,1,4) Hirota_Dxt (ff1, ff1 ) Hirota_Dxm (ff1,ff1,4) ; eqna := simplify (B(f11 f1, f1 ),size ); eqnb := simplify (B(f11 f13, f ),size ); eqnc := simplify (B(f1 f13, f3 ),size ); a [1]:= factor ( simplify ( solve ( eqna,a [1]), symbolic )); a [13]:= factor ( simplify ( solve ( eqnb,a [13]), symbolic )); a [3]:= factor ( simplify ( solve ( eqnc,a [3]), symbolic )); Graham W Griffiths 9 10 March 01

10 # Ansatz for f3 suggested by equation : B (1. f3f1.ff.f1f3.1) =0 f3 :=b [13]* combine ( f11 * f1 * f13 ); # Three - soliton ancilliary function F :=1 f1ff3; B3 :=( ff1,ff, ff3 ) -> Hirota_Dxt (1, ff3 ) Hirota_Dxt (ff3,1) Hirota_Dxm (1, ff3,4) Hirota_Dxm (ff3,1,4) Hirota_Dxt (ff1, ff ) Hirota_Dxm (ff1,ff,4) Hirota_Dxt (ff, ff1 ) Hirota_Dxm (ff,ff1,4) ; F := B3(f1,f,f3): F3 := solve (F,b [13]) : b [13]:= factor ( simplify (F3, symbolic )); # b [13]:= a [1]* a [13]* a [3]; # Three - soliton solution soln :=u= simplify (* diff (ln(f),x,x),size ); # Check solution - Can take a long time!! pdetest (soln, pde1 ); # Animate result zz1 := simplify ( subs ({k[1]=1, k []= sqrt (),k [3]=}, rhs ( soln )),size ); animate (zz1,x = , t = , numpoints =300, frames =100, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" KdV Equation : Three - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; Figure 3: Initial animation profile for KdV equation three-soliton solution, eqn. (43), at t = 10, for k = 1, k = and k 3 =. Solitons move left to right with the taller faster solitons overtaking the shorter slower solitons..4 Four-soliton solution to the KdV equation This problem is similar to the three-soliton problem except that the auxilliary function f has an additional term, f 4. So now we need to find f 1, f, f 3 and f 4. Because we are seeking a four-soliton solution, we use four-term forms of f 1, f and f 3 that were successful for the three-soliton case, i.e. f 1 = e ξ 1 e ξ e ξ 3 e ξ 4, f = a 1 e (ξ 1ξ ) a 13 e (ξ 1ξ 3 ) a 14 e (ξ 1ξ 4 ) a 3 e (ξ ξ 3 ) a 4 e (ξ ξ 4 ) a 34 e (ξ 3ξ 4 ), f 3 = b 13 e ξ 1ξ ξ 3 b 14 e ξ 1ξ ξ 4 b 134 e ξ 1ξ 3 ξ 4 b 34 e ξ ξ 3 ξ 4 along with the ansatz f 4 = c 13 e ξ 1ξ ξ 3 ξ 4, where coupling constants a 1, a 13, a 14, a 3, a 4, b 13, b 14, b 134, b 34 and c 134 are yet to be determined. This enables us to construct the auxilliary function, f = 1 ɛf 1 ɛ f ɛ 3 f 3 ɛ 4 f 4. The a i,j auxillary coupling constants and the b ijk auxillary coupling constants are determined as for the three-soliton case. On substituting in turn all combinations of three terms from f 3, into eqn. (1), i.e. 1 st, nd and 3 rd ; 1 st, nd and 4 th ; 1 st, 3 rd and 4 th ; nd, 3 rd and 4 th ; we obtain expressions which can be solved for these coupling constants in Graham W Griffiths March 01

11 terms of wave numbers, k 1, k, k 3 and k 4. Similar to f 3 in the three-soliton case, the form for f 4 is not an arbitrary choice but, rather, is suggested by eqn (13) which evaluates to B(1 f 4 f 4 1) B(f 1 f 3 f f f 3 f 1 ) = B(1 f 4 f 4 1) 6 Q e ξ 1ξ ξ 3 ξ 4 (k 1 k ) (k 1 k 3 ) (k 1 k 4 ) (k k 3 ) (k k 4 ) = 0, (51) (k 3 k 4 ) where Q is a complicated expression involving wave numbers, k i, i = 1,, 4. For eqn (51) to be equal to zero, clearly, the first term must be of the form, B(1 f 4 f 4 1) = c 134 e ξ 1ξ ξ 3 ξ 4. (5) We now substitute the auxilliary function f into eqn. (7) to obtain the solution u = N ( ) 1 N, (53) D 1 where N 1, N, D 1 and D are defined below. A plot of ths solution is given in Fig (4). N 1 =k 1 e ξ 1 k e ξ k 3 e ξ 3 k 4 e ξ 4 (k 1 k ) e ξ 1ξ (k 1 k 3 ) e ξ 1ξ 3 (k 1 k 4 ) e ξ 1ξ 4 (k k 3 ) e ξ ξ 3 (k k 4 ) e ξ ξ 4 (k 3 k 4 ) e ξ 3ξ 4 b 13 (k 1 k k 3 ) e ξ 1ξ ξ 3 b 14 (k 1 k k 4 ) e ξ 1ξ ξ 4 b 134 (k 1 k 3 k 4 ) e ξ 1ξ 3 ξ 4 b 34 (k k 3 k 4 ) e ξ ξ 3 ξ 4 c 134 (k 1 k k 3 k 4 ) e ξ 1ξ ξ 3 ξ 4. (54) D 1 =1 e ξ 1 e ξ e ξ 3 e ξ 4 a 1 e ξ 1ξ a 13 e ξ 1ξ 3 a 14 e ξ 1ξ 4 a 3 e ξ ξ 3 a 4 e ξ ξ 4 a 34 e ξ 3ξ 4 b 13 e ξ 1ξ ξ 3 b 14 e ξ 1ξ ξ 4 D b 134 e ξ 1ξ 3 ξ 4 b 34 e ξ ξ 3 ξ 4 c 134 e ξ 1ξ ξ 3 ξ 4. (55) N =k 1 e ξ 1 k e ξ k 3 e ξ 3 k 4 e ξ 4 (k 1 k ) e ξ 1ξ (k 1 k 3 ) e ξ1ξ3 k 1 k k 1 k 3 (k 1 k 4 ) e ξ 1ξ 4 (k k 3 ) e ξξ3 (k k 4 ) e ξξ4 (k 3 k 4 ) e ξ3ξ4 k 1 k 4 k k 3 k k 4 k 3 k 4 b 13 (k 1 k k 3 ) e ξ 1ξ ξ 3 b 14 (k 1 k k 4 ) e ξ 1ξ ξ 4 b 134 (k 1 k 3 k 4 ) e ξ 1ξ 3 ξ 4 b 34 (k k 3 k 4 ) e ξ ξ 3 ξ 4 c 134 (k 1 k k 3 k 4 ) e ξ 1ξ ξ 3 ξ 4. (56) D =1 e ξ 1 e ξ e ξ 3 e ξ 4 a 1 e ξ 1ξ a 13 e ξ 1ξ 3 a 14 e ξ 1ξ 4 a 3 e ξ ξ 3 a 4 e ξ ξ 4 a 34 e ξ 3ξ 4 b 13 e ξ 1ξ ξ 3 b 14 e ξ 1ξ ξ 4 b 134 e ξ 1ξ 3 ξ 4 b 34 e ξ ξ 3 ξ 4 c 134 e ξ 1ξ ξ 3 ξ 4. (57) Where, again without loss of generality, we have set ɛ = 1 in eqn. (8). As for the single-soliton solution, we can obtain the dispersion relations by substituting, in turn, the four terms of f 1 into eqn. (10), to obtain c 1 = k 1, c = k, c 3 = k 3, c 4 = k 4. (58) Graham W Griffiths March 01

12 The coupling constants a, b and c are determined as described above and yield, a 1 = (k 1 k ) (k 1 k ), a 13 = (k 1 k 3 ) (k 1 k 3 ), a 14 = (k 1 k 4 ) a 3 = (k k 3 ) (k k 3 ), a 4 = (k k 4 ) (k k 4 ), a 34 = (k 3 k 4 ) (k 1 k 4 ). (59) (k 3 k 4 ). (60) b 13 = (k k 3 ) (k 1 k 3 ) (k 1 k ) (k k 3 ) (k 1 k 3 ) (k 1 k ), b 14 = (k k 4 ) (k 1 k 4 ) (k 1 k ) (k k 4 ) (k 1 k 4 ) (k 1 k ). (61) b 134 = (k 3 k 4 ) (k 1 k 4 ) (k 1 k 3 ) (k 3 k 4 ) (k 1 k 4 ) (k 1 k 3 ), b 34 = (k 3 k 4 ) (k k 4 ) (k k 3 ) (k 3 k 4 ) (k k 4 ) (k k 3 ). (6) c 134 = (k 3 k 4 ) (k k 4 ) (k k 3 ) (k 1 k 4 ) (k 1 k 3 ) (k 1 k ) (k 3 k 4 ) (k k 4 ) (k k 3 ) (k 1 k 4 ) (k 1 k 3 ) (k 1 k ). (63) We now have a fully defined four-soliton solution for KdV eqn (7). A plot of this solution is given in Fig (4). A Maple program that solves this four-soliton example and generates D and 3D animations, is given in Listing (4). Listing 4: Maple program that uses the Hirota method to derive a four-soliton solution to KdV eqn (7) # Hirota bilinear method # Four - soliton Solution to KdV equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), f=f(x,t)): # KdV equation pde1 := diff (u,t) 6* u* diff (u,x) diff (u,x,x,x); # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; v :=* diff (ln(f),x,x); declare (f): pde := eval ( subs (u=v, pde1 )) =0; # Check bilinear transform correct pde3 := simplify (pde, size ); pde3_chk := simplify ( diff (( Hirota_Dxt (f,f) Hirota_Dxm (f,f,4) )/(f^),x),size ) =0; testeq ( lhs ( pde3 )= lhs ( pde3_chk )); # Ansatz for f_1 f11 := exp (k [1]*(x-c [1]* t)); f1 := exp (k []*(x-c []* t)); f13 := exp (k [3]*(x-c [3]* t)); f14 := exp (k [4]*(x-c [4]* t)); f1 := f11 f1 f13 f14 ; B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (ff1,1,4) Hirota_Dxm (1, ff1,4) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; eqn1c := B1(f13 ); # B (1. f3 f3.1) ; eqn1d := B1(f14 ); # B (1. f3 f4.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b, eqn1c, eqn1d }, symbolic ), {c[1],c[],c[3],c [4]}) ; assign ( sol1 ); # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 Graham W Griffiths 1 10 March 01

13 f1 :=a [1]* combine ( f11 * f1 ); f :=a [13]* combine ( f11 * f13 ); f3 :=a [14]* combine ( f11 * f14 ); f4 :=a [3]* combine ( f1 * f13 ); f5 :=a [4]* combine ( f1 * f14 ); f6 :=a [34]* combine ( f13 * f14 ); f := f1 f f3 f4 f5 f6 ; B :=( ff1, ff ) -> Hirota_Dxt (1, ff ) Hirota_Dxt (ff,1) Hirota_Dxm (1, ff,4) Hirota_Dxm (ff,1,4) Hirota_Dxt (ff1, ff1 ) Hirota_Dxm (ff1,ff1,4) ; eqna := simplify (B(f11 f1, f1 ),size ); eqnb := simplify (B(f11 f13, f ),size ); eqnc := simplify (B(f11 f14, f3 ),size ); eqnd := simplify (B(f1 f13, f4 ),size ); eqne := simplify (B(f1 f14, f5 ),size ); eqnf := simplify (B(f13 f14, f6 ),size ); a [1]:= factor ( simplify ( solve ( eqna,a [1]), symbolic )); a [13]:= factor ( simplify ( solve ( eqnb,a [13]), symbolic )); a [14]:= factor ( simplify ( solve ( eqnc,a [14]), symbolic )); a [3]:= factor ( simplify ( solve ( eqnd,a [3]), symbolic )); a [4]:= factor ( simplify ( solve ( eqne,a [4]), symbolic )); a [34]:= factor ( simplify ( solve ( eqnf,a [34]), symbolic )); # Ansatz for f3 suggested by equation : B (1. f3f1.ff.f1f3.1) =0 f31 :=b [13]* combine ( f11 * f1 * f13 ); f3 :=b [14]* combine ( f11 * f1 * f14 ); f33 :=b [134]* combine ( f11 * f13 * f14 ); f34 :=b [34]* combine ( f1 * f13 * f14 ); f3 := f31 f3 f33 f34 ; B3 :=( ff1,ff, ff3 ) -> Hirota_Dxt (1, ff3 ) Hirota_Dxt (ff3,1) Hirota_Dxm (1, ff3,4) Hirota_Dxm (ff3,1,4) Hirota_Dxt (ff1, ff ) Hirota_Dxm (ff1,ff,4) Hirota_Dxt (ff, ff1 ) Hirota_Dxm (ff,ff1,4) ; eqn3a := simplify (B3(f11 f1 f13, f1 f f4, f31 ),symbolic ): eqn3b := simplify (B3(f11 f1 f14, f1 f3 f5, f3 ),symbolic ): eqn3c := simplify (B3(f11 f13 f14, f f3 f6, f33 ),symbolic ): eqn3d := simplify (B3(f1 f13 f14, f4 f5 f6, f34 ),symbolic ): #t :=0; x :=0; sol_13 := factor ( simplify ( solve ({ eqn3a },{b [13]}),size )); sol_14 := factor ( simplify ( solve ({ eqn3b },{b [14]}),size )); sol_134 := factor ( simplify ( solve ({ eqn3c },{b [134]}),size )); sol_34 := factor ( simplify ( solve ({ eqn3d },{b [34]}),size )); assign ( sol_13, sol_14, sol_134, sol_34 ); # Ansatz for f4 suggested by equation : B (1. f4f1.f3f.ff3.f1f4.1) =0 f4 :=c [134]* combine ( f11 * f1 * f13 * f14 ); F :=1 f1ff3f4; B4 :=( ff1,ff,ff3, ff4 ) -> Hirota_Dxt (1, ff4 ) Hirota_Dxm (1, ff4,4) Hirota_Dxt (ff4,1) Hirota_Dxm (ff4,1,4) Hirota_Dxt (ff1, ff3 ) Hirota_Dxm (ff1,ff3,4) Hirota_Dxt (ff3, ff1 ) Hirota_Dxm (ff3,ff1,4) Hirota_Dxt (ff, ff ) Hirota_Dxm (ff,ff,4) ; F := B4(f1,f,f3,f4): F3 := solve (F,c [134]) : # c [134]:= a [1]* a [13]* a [14]* a [3]* a [4]* a [34]; c [134]:= factor ( simplify (F3, symbolic )); # Four - soliton solution soln :=u=* diff (ln(f),x,x); # Animate result zz1 := simplify ( subs ({k[1]=1, k []= sqrt (),k [3]=, k [4]=3}, rhs ( soln )),size ): animate (zz1,x = , t = -5..5, numpoints =300, frames =30, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" KdV Equation : Four - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; animate3d (zz1,x = , y = , t = -5..5, frames =10, axes = framed, grid =[100,100], gridstyle = triangular, style = patchnogrid, shading = ZHUE, title ="1D-KdV Equation : Four - soliton solution ", Graham W Griffiths March 01

14 labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16], labels =[" x","y","u(x,t)"], orientation =[55,55]) ; # Check Solution - takes a long time on Maple 16!! pdetest (soln, pde1 ); Figure 4: Initial animation profile for KdV equation four-soliton solution, eqn. (53), at t = 5, for k = 1, k =, k 3 = and k 4 = 3. Solitons move left to right with the taller faster solitons overtaking the shorter slower solitons..5 Two-Soliton solution to the Boussinesq equation If we substitute eqn. (7) into eqn. (64) we obtain, u tt u xx 3(u ) xx u 4x = 0. (64) 1 f 4 [ f6x f 3 1f 5x f x f ( 36f x f f 3 6f xx f ) f 4x (65) f ttxx f 3 4f 3x f ( 48f x 3 8f x f ) f 3 x 4f x f ttx f 4f txx f t f 1f xx 3 f ( 6f 36f x ) f xx f ( 1f x f t f tt f ) f xx 16f x f tx f t f 1f x f t 1f x 4 4f tx f 4f x f tt f ] = 0. Then, using Hirota s bilinear operator, the result can be expressed in the following succinct bilinear form [ ] 1 ( ) D x f t Dx Dx 4 (f f) = 0. (66) Hence, integrating twice with respect to x and taking the integration functions to be zero, the bilinear equation equivalent to eqn (64) is B(f f) = ( D t D x D 4 x) (f f) = 0. (67) Thus, if f is a solution of eqn. (67), then u from eqn. (7) is a solution to eqn. (64). See [Rog-8, p69-71] for a detailed discussion and derivation. If we now follow the same procedure as for the two-soliton solution to the KdV equation we obtain the following solution, u = [( k e ξ 1 k 1 e ξ (k 1 k ) ) a 1 e ξ 1ξ ( (k 1 k ) e ξ k ) ] 1 e ξ 1 k e ξ (1 e ξ 1 e ξ a1 e ξ 1ξ ). (68) Graham W Griffiths March 01

15 As for the KdV two-soliton solution, we obtain the dispersion relations by substituting, in turn, the two terms of f 1 into eqn. (10), to obtain c 1 = ± 1 k 1, c = ± 1 k (69) We choose the positive solutions. Then on substituting f 1 and f into eqn. (11) we obtain the following expression for the coupling constant, a 1 = 1 k1 1 k 1 k 1 k 3k 1 k 1 k1 1 k 1 k 1 3k 1 k k. (70) We now have a fully defined two-soliton solution to Boussinesq eqn (64). A plot of this solution is given in Fig (5). A Maple program that solves this two-soliton example and generates an animation is given in Listing (5). Listing 5: Maple program that uses the Hirota method to derive a two-soliton solution to Boussinesq eqn (64) # Hirota bilinear method # Two - soliton Solution to Boussinesq equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), f=f(x,t)): # Bousinesq equation pde1 := diff (u,t,t)-diff (u,x,x) -3* diff (u^,x,x)-diff (u,x,x,x,x); # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; v :=* diff (ln(f),x,x); declare (f): pde := eval ( subs (u=v, pde1 )) =0; # Check bilinear transform correct pde3 := simplify (pde, size ); pde3_chk := simplify ( diff (( Hirota_Dtm (f,f,) - Hirota_Dxm (f,f,) - Hirota_Dxm (f,f,4) )/f^,x,x),size ) =0; testeq ( lhs ( pde3 )= lhs ( pde3_chk )); # Ansatz for f_1 f11 := exp (k [1]*(x-c [1]* t)); f1 := exp (k []*(x-c []* t)); f1 := f11 f1 ; B1 :=( ff1 ) -> Hirota_Dtm (1, ff1,) Hirota_Dtm (ff1,1,) - ( Hirota_Dxm (1, ff1,) Hirota_Dxm (ff1,1,) )- ( Hirota_Dxm (1, ff1,4) Hirota_Dxm (ff1,1,4) ) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b }, symbolic ),{c[1],c []}) ; # Note : Boussinesq eqn has / - velocity values. # Therefore waves can travel in opposite directions! sol11 := allvalues ( sol1 [1]) ; sol1 := allvalues ( sol1 []) ; assign ( sol11 [1], sol1 [1]) ; c [1]; c []; # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 f :=a [1]* combine ( f11 * f1 ); # Two - soliton ancilliary function F :=1 f1f; B :=( ff1, ff ) ->( Hirota_Dtm (1, ff,) Hirota_Dtm (ff,1,) )- Graham W Griffiths March 01

16 ( Hirota_Dxm (1, ff,) Hirota_Dxm (ff,1,) )- ( Hirota_Dxm (1, ff,4) Hirota_Dxm (ff,1,4) ) ( Hirota_Dtm (ff1,ff1,) - Hirota_Dxm (ff1,ff1,) - Hirota_Dxm (ff1,ff1,4) ); f1;f; eqn := simplify (B(f1,f),size ) =0; a [1]:= factor ( simplify ( solve (eqn,a [1]), symbolic )); # Two - soliton solution soln :=u= simplify (* diff (ln(f),x,x),size );; # Animate result zz1 := simplify ( subs ({k[1]=1, k []=1/}, rhs ( soln )),symbolic ); animate (zz1,x = , t = , numpoints =1000, frames =50, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" Boussinesq Equation : Two - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; # Check Solution - takes a long time with Maple 16 simplify ( subs (soln, pde1 ),symbolic ); Figure 5: Initial animation profile for Boussinesq equation two-soliton solution, eqn (68), at t = 130, for k = 1 and k = 1/. Solitons move left to right with the taller faster soliton overtaking the shorter slower soliton..6 Two-Soliton solution to the Sawada-Kotera equation The Sawada-Kotera (SK) equation is given by, u t 45u u x 15u x u xx 15uu 3x u 5x = 0. (71) This equation is also known as the Caudrey-Dodd-Gibbon equation (CDGE). To demonstrate that the bilinear form is correct, we derive the potential form of the Sawada-Kotera equation by substituting u = w/ x into eqn (71) to obtain Also, we note that eqn (7) becomes w x,t 45w x w x,x 15w x,x w 3x 15w x w 4x w 6x = 0. (7) ln f(x, t) w(x, t) = x which, on substituting into eqn (7), we obtain = f x f f x f t 1f 5x f x 30f 4x f x,x f t,x f f 6x f 0f 3x (73) f = 0. (74) Graham W Griffiths March 01

17 This equation can be written more succinctly in the following form using Hirota s bilinear operator 1 ( ) Dx D f t Dx 6 (f f) = 0. (75) Hence the bilinear equation equivalent to eqns (71) and (7) is [Dra-9] B(f f) = ( D x D t D 6 x) (f f) = 0. (76) Thus, if f is a solution of eqn. (76), then w from eqn. (73) is a solution to eqn. (7) and u from eqn. (7) is a solution to eqn. (71). If we now follow the same procedure as for the two-soliton solution to the KdV equation we obtain the following solution, u = (( k e ξ 1 k 1 e ξ (k 1 k ) ) a 1 e ξ 1ξ ( (k 1 k ) e ξ k ) ) 1 e ξ 1 k e ξ (1 e ξ 1 e ξ a1 e ξ 1ξ ). which has the same form as the two-soliton solution to the Boussinesq equation. However, the Sawada-Kotera equation has different dispersion relations which we obtain by substituting, in turn, the two terms of f 1 into eqn. (10), to obtain (77) c 1 = k 4 1, c = k 4. (78) Also, on substituting f 1 and f into eqn. (11) we obtain the following expression for the coupling constant, ( k1 k 1 k k ) (k 1 k ) a 1 = ( k1 k 1 k k ) (k 1 k ). (79) Again, this coupling constant is different to that of the Boussinesq eqn. We now have a fully defined two-soliton solution to Sawada-Kotera eqn (71). A plot of this solution is given in Fig (6). A Maple program that solves this two-soliton example and generates an animation is given in Listing (6). Listing 6: Maple program that uses the Hirota method to derive a two-soliton solution to Sawada-Kotera eqn (71) # Hirota bilinear method # Two - soliton Solution to Sawada - Kotera equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,t), v=v(x,t), w=w(x,t), f=f(x,t)): # SW equation pde1 := diff (u,t) 45* u ^* diff (u,x) 15* diff (u,x)* diff (u,x,x) 15* u* diff (u,x,x,x) diff (u,x,x,x,x,x) =0; eqn1 :=u= diff (w,x); declare (w); # Derive potential version of pde1 pde1a := subs (eqn1, pde1 ); # manually integrate pde1a pde1b := diff (w, t) 15* diff (w, x) ^315* diff (w, x)* diff (w, x, x, x) diff (w, x, x, x, x, x) = 0; # Differentiate pde1b pde1c := diff ( pde1b,x); # Check that integration is correct testeq ( lhs ( pde1a )= lhs ( pde1c )); # Define Hirota Derivatives Graham W Griffiths March 01

18 Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; # Hirota logrithmic transformation ( Cole - Hopf ) lnfunc :=w=* diff (ln(f),x); # equivalent to: u :=* diff (ln(f),x,x) declare (f): pde := eval ( subs ( lnfunc, pde1b )); # Check bilinear transformation is correct pde3 := simplify (pde, size ); pde3_chk := normal ( Hirota_Dxt (f,f) Hirota_Dxm (f,f,6) )/f ^=0; testeq ( lhs ( pde3 )-lhs ( pde3_chk )); # Ansatz for f_1 f11 := exp (k [1]*(x-c [1]* t)); f1 := exp (k []*(x-c []* t)); f1 := f11 f1 ; # B1=B ((1. f f.1) ) B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (ff1,1,6) Hirota_Dxm (1, ff1,6) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b }, symbolic ),{c[1],c []}) ; assign ( sol1 ); # == > > Dispersion relations # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 f :=a [1]* combine ( f11 * f1 ); # Two - soliton ancilliary function F :=1 f1f; B :=( ff1, ff ) -> Hirota_Dxt (1, ff ) Hirota_Dxt (ff,1) Hirota_Dxm (1, ff,6) Hirota_Dxm (ff,1,6) Hirota_Dxt (ff1, ff1 ) Hirota_Dxm (ff1,ff1,6) ; eqn := simplify (B(f1,f),size ); a [1]:= factor ( simplify ( solve (eqn,a [1]), symbolic )); # Two - soliton solution soln :=u= simplify (* diff (ln(f),x,x),size );; # Animate result zz1 := simplify ( subs ({k[1]=1, k []=1/}, rhs ( soln )),symbolic ); animate (zz1,x = , t = , numpoints =1000, frames =50, axes = framed, labels =[" x","u(x,t)"], thickness =3, title =" Sawada - Kotera Equation : Two - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16]) ; # Check Solution pdetest (soln, pde1 );.7 Two-Soliton solution to the D Kadomtsev-Petviashvili equation The D Kadomtsev-Petviashvili (KP) equation is given by, If we substitute eqn. (7) into eqn. (80) we obtain (u t 6uu x u 3x ) x ± 3u yy = 0. (80) f 4 [ 6fx f t,x f 6f x 3 f t 18f 4x f x f 4f 3x f x 3 18f xx f x f xx f y f 6f x f y 6f 5x f x f 3f 4x f xx f f xxy f y f f xx f y,y f f x f xyy f f x f yy f 3f txx f x f 3f xx f tx f f 3x f t f 6f xx f t f x f 8f x f xy f y f f 3x f 6f xx 3 f f xy f f t3x f 3 f 6x f 3 f xxyy f 3] = 0 (81) Graham W Griffiths March 01

19 Figure 6: Initial animation profile for Sawada-Kotera equation two-soliton solution, eqn. (77), at t = 30, for k = 1 and k = 1/. Solitons move left to right with the taller faster soliton overtaking the shorter slower soliton. Then, using Hirota s bilinear operator, the result can be expressed in the following succinct bilinear form [ ] 1 ( ) Dx D x f t Dx 4 ± 3Dy (f f) = 0. (8) Hence, integrating twice with respect to x and taking the integration functions to be zero, the bilinear equation equivalent to eqn. (80) is [Her-91] B(f f) = ( D x D t D 4 x ± 3D y) (f f) = 0. (83) Thus, if f is a solution of eqn. (83), then u from eqn. (7) is a solution to eqn. (80). In our analysis we elect to use 3u yy in eqn. (80) and 3D y in eqn. (83). This problem is the same as the KdV two-soliton problem except that it is two-dimensional and has a different bilinear equation. Because we are seeking a two-soliton solution, we use the two-term form of f 1 that was successful for the KdV two-soliton example, i.e. f 1 = e ξ 1 e ξ, along with the ansatz f = a 1 e (ξ 1ξ ), where a 1 is a coupling constant yet to be determined. This enables us to construct the auxilliary function, f = 1 ɛf 1 ɛ f. However, because we are dealing with a D system, we use the transformations ξ 1 = k 1 x l 1 x ω 1 t and ξ = k x l x ω t. Again, the form for f is not an arbitrary choice but, rather, is suggested by eqn (11) which evaluates to B(1 f f.1) B(f 1 f 1 ) = B(1 f f 1) 6 (k l 1 k k 1 k 1 k k 1 l ) ( k l 1 k k 1 k 1 k k 1 l ) k 1 k e ξ 1ξ = 0. (84) For eqn (84) to be equal to zero, clearly, the first term must be of the form, B(1 f f 1) = a 1 e ξ 1ξ. (85) We now substitute the auxilliary function f into eqn. (7) to obtain the solution u = [ ( a 1 k 1 e ξ ke ξ 1 (k 1 k ) ) e ξ 1ξ ( )] (k 1 k ) e ξ k1)e ξ 1 ke ξ (a 1 e ξ 1ξ e ξ 1 e ξ 1). (86) Where, again without loss of generality, we have set ɛ = 1. Graham W Griffiths March 01

20 As for the KdV two-soliton solution, we can obtain the dispersion relations by substituting, in turn, terms 1 and of f 1 into eqn. (10), to obtain ω 1 = k4 1 3l 1 k 1, ω = k4 3l k. (87) Also, on substituting f 1 and f into eqn. (11) we obtain an expression which can be solved for the coupling constant in terms of wave numbers, k 1 and k. This yields, a 1 = (k l 1 k k 1 k 1 k k 1 l ) ( k l 1 k k 1 k 1 k k 1 l ) ( k l 1 k k 1 k 1 k k 1 l ) (k l 1 k k 1 k 1 k k 1 l ). (88) We now have a fully defined two-soliton solution to the D KP eqn (80). A plot of this solution is given in Fig (7). A Maple program that solves this two-soliton example and generates an animation is given in Listing (7). Listing 7: Maple program that uses the Hirota method to derive a two-soliton solution to D Kadomtsev- Petviashvili eqn (80) # Hirota bilinear method # Two - soliton Solution to D Kadomtsev - Petviashvili ( KP) equation restart ; with ( PDEtools ): with ( PolynomialTools ): with ( plots ): alias (u=u(x,y,t), f=f(x,y,t)): # KP equation pde1 := diff ( diff (u,t) 6* u* diff (u,x) diff (u,x,x,x),x) diff (u,y,y); # Define Hirota Derivatives Hirota_Dxm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,x$(m-r)) * diff (g,x$r ),r =0.. m); Hirota_Dym :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,y$(m-r)) * diff (g,y$r ),r =0.. m); Hirota_Dtm :=(f,g,m) ->sum (( -1) ^r* binomial (m,r)* diff (f,t$(m-r)) * diff (g,t$r ),r =0.. m); Hirota_Dxt :=(f,g) ->f* diff (g,x,t)-diff (f,x)* diff (g,t) -diff (f,t)* diff (g,x) diff (f,x,t)*g; u1 :=* diff (ln(f),x,x); # logrithmic transformation declare (f): pde := eval ( subs (u=u1, pde1 )) =0: # Check bilinear transform correct pde3 := simplify (pde, symbolic ); pde3_chk := simplify ( diff (( Hirota_Dxt (f,f) Hirota_Dxm (f,f,4) Hirota_Dym (f,f,) )/f^,x,x),symbolic ) =0; testeq ( lhs ( pde3 )= lhs ( pde3_chk )); # Ansatz for f_1 f11 := exp (k [1]* xl [1]*y- omega [1]* t); f1 := exp (k []* xl []*y- omega []* t); f1 := f11 f1 ; B1 :=( ff1 ) -> Hirota_Dxt (1, ff1 ) Hirota_Dxt (ff1,1) Hirota_Dxm (ff1,1,4) Hirota_Dxm (1, ff1,4) 3*( Hirota_Dym (1, ff1,) Hirota_Dym (ff1,1,) ) =0; eqn1a := B1(f11 ); # B (1. f1 f1.1) ; eqn1b := B1(f1 ); # B (1. f f.1) ; sol1 := solve ( simplify ({ eqn1a, eqn1b }, symbolic ),{ omega [1], omega []}) ; assign ( sol1 ); # == > > Dispersion relations # Ansatz for f suggested by equation : B (1. ff1.f1f.1) =0 f :=a [1]* combine ( f11 * f1 ); # Two - soliton ancilliary function F :=1 f1f; B :=( ff1, ff ) -> Hirota_Dxt (1, ff ) Hirota_Dxt (ff,1) Hirota_Dxm (1, ff,4) Hirota_Dxm (ff,1,4) 3*( Hirota_Dym (1, ff,) Hirota_Dym (ff,1,) ) Graham W Griffiths 0 10 March 01

21 (a) Top view, t = 0 (b) Side view, t = 0 Figure 7: Profiles of D Kadomtsev-Petviashvili equation two-soliton solution, eqn. (86), for k = 1, k =, l 1 = 1/5 and l = 5/6. Solitons move right to left with the taller right hand soliton overtaking the smaller left hand soliton. Hirota_Dxt (ff1, ff1 ) Hirota_Dxm (ff1,ff1,4) 3* Hirota_Dym (ff1,ff1,) ; eqn := simplify (B(f1,f),size ) =0; a [1]:= factor ( simplify ( solve (eqn,a [1]), symbolic )); # Two - soliton solution soln :=u= simplify (* diff (ln(f),x,x),size );; # Animate result zz1 := subs ({k[1]=1, k []= sqrt (),l [1]=1/5, l []=5/6}, rhs ( soln )); animate3d (zz1,x = , y = , t = -0..0, frames =10, axes = framed, grid =[100,100], gridstyle = triangular, style = patchnogrid, shading = ZHUE, title ="D-KP Equation : Two - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16], labels =[" x","y","u(x,t)"], orientation =[90,0]) ; animate3d (zz1,x = , y = , t = -0..0, frames =10, axes = framed, grid =[100,100], gridstyle = triangular, style = patchnogrid, shading = ZHUE, title ="D-KP Equation : Two - soliton solution ", labelfont =[ TIMES, ROMAN, 16], axesfont =[ TIMES, ROMAN, 16], titlefont =[ TIMES, ROMAN, 16], labels =[" x","y","u(x,t)"], orientation =[11,81]) ; # Check Solution - takes a long time with Maple 16!! # simplify ( subs (soln, pde1 ),symbolic ); References [Dra-9] Drazin, P. G. and R. S. Johnson (199). Solitons: an introduction, Cambridge University Press, Cambridge. [Hir-71] Hirota, R. (1971). Exact solution of the Korteweg-de Vries equation for multiple collisions of solitons, Phys. Review Letters, 7, [Hir-04] Hirota, R. (004). Direct Method in Soliton Theory, Cambridge University Press, Cambridge. Graham W Griffiths 1 10 March 01

22 [Her-91] Hereman, W., Zhuang, W. (1991). A MACSYMA Program for the Hirota Method. In 13th World Congress on Computation and Applied Mathematics, IMACS91 (Vichnevetsky, R., Miller, J.J.H., eds.) Vol., Criterion Press, Dublin, Ireland, [Her-94] Hereman, W. and W. Zhuang (1994). Symbolic Computation of Solitons via Hirota s Bilinear Method, Report-Department of Mathematical and Computer Sciences, Colorado School of Mines, August. [Rog-8] Rogers, C. and W. F. Shadwick (198), Bäcklund transformations and their applications, Academic Press, New York. [Waz-09] Wazwaz, Abdul-Majid (009). Partial Differential Equations and Solitary Waves Theory, Springer-Verlag, Berlin. Graham W Griffiths 10 March 01