Traveling Wave Solutions of Degenerate Coupled Multi-KdV Equations

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1 Traveling Wave Solutions of Degenerate Coupled Multi-KdV Equations Metin Gürses Aslı Pekcan Department of Mathematics, Science Faculty Bilkent University, 68 Ankara - Turkey Department of Mathematics, Faculty of Science Hacettepe University, 68 Ankara - Turkey arxiv:156.36v [nlin.si] Nov 16 Astract Traveling wave solutions of degenerate coupled l-kdv equations are studied. Due to symmetry reduction these equations reduce to one ODE, (f = P n (f) where P n (f) is a polynomial function of f of degree n = l+, where l 3 in this work. Here l is the numer of coupled fields. There is no known method to solve such ordinary differential equations when l 3. For this purpose, we introduce two different type of methods to solve the reduced equation and apply these methods to degenerate three-coupled KdV equation. One of the methods uses the Cheyshev s Theorem. In this case we find several solutions some of which may correspond to solitary waves. The second method is a kind of factorizing the polynomial P n (f) as a product of lower degree polynomials. Each part of this product is assumed to satisfy different ODEs. Keywords: Traveling wave solution, Degenerate KdV system, Cheyshev s theorem, Alternative method. 1 Introduction The system of degenerate coupled multi-field KdV equations is given as [1]-[6] u t = 3 uu x + q x gurses@fen.ilkent.edu.tr asli@fen.ilkent.edu.tr q t = q u x + 1 uq x + q 3 x.. qt l 1 = q l 1 u x + 1 uql 1 x + v x.. v t = 1 4 u xxx + vu x + 1 uv x, (1.1) 1

2 where q 1 = u and q l = v. In [1]-[7], it was shown that this system is also a degenerate KdV system of rank one. In a previous work [8] we focused on the equation (1.1) for l =. We reduced this equation into an ODE (f = P 4 (f) where P 4 (f) is a polynomial function of degree four. We analyzed all possile cases aout the zeros of P 4 (f). Due to this analysis we determined the cases when the solution is periodic or solitary. When the polynomial has one doule f and two simple zeros f 1, f 3 with f 1 < f < f 3, or one triple and one simple zeros, the solution is solitary. Other cases give periodic or non-real solutions. By using the Jacoi elliptic functions [9], we otained periodic solutions and all solitary wave solutions which rapidly decay to some constants, explicitly. We have also shown that there are no real asymptotically vanishing traveling wave solutions for l =. Indeed we have the following theorem for the degenerate coupled l-kdv equation, l. The degenerate coupled l-kdv equation can e reduced to the equation (f = P l+ (f) (1.) y taking l functions as u(x, t) = q 1 (x, t) = f(ξ), q (x, t) = f (ξ),..., v(x, t) = q l = f l (ξ), where ξ = x ct in (1.1). Here P l+ (f) is a polynomial of f of degree l +. If we apply the asymptotically vanishing oundary conditions to (1.), we have (f = Bf (f + c) l. (1.3) Theorem 1.1 When l = odd, we have B >, and then the degenerate coupled l-kdv equation has real traveling wave solution with asymptotically vanishing oundary conditions, ut when l = even, the constant B <. Hence the equation does not have a real traveling wave solution with asymptotically vanishing oundary conditions. Proof. Consider the degenerate coupled l-kdv equation (1.1). Let u(x, t) = q 1 (x, t) = f(ξ), q (x, t) = f (ξ),..., v(x, t) = q l = f l (ξ) where ξ = x ct. By using the first l 1 equations we otain all functions f i (ξ), i =, 3,..., l as a polynomial of f(ξ). We get f (ξ) = Q (f) = cf 3 4 f + d 1 = α f + A 1 (f) f 3 (ξ) = Q 3 (f) = 3 cf + 1 f 3 + (c d 1 )f + d = α 3 f 3 + A (f) f 4 (ξ) = Q 4 (f) = 5 16 f 4 3 ( cf c + 3 ) 4 d 1 f + ( c 3 + cd 1 d )f + d 3 = α 4 f 4 + A 3 (f). f l (ξ) = Q l (f) = ( 1) l+1 α l f l + A l 1 (f),

3 where α j > are constants, A i (f) and Q j (f) are polynomials of f of degree i and j, i = 1,,..., l 1, j =, 3,..., l, respectively. Now use the lth equation. We otain 1 4 f = ( 1) l 1 α l f f (1+ l l ) Integrating aove equation once we get +c( 1) l 1 lα l f f l 1 +f A l f A l 1(f) x 1 4 f = ( 1) l 1 α l (1 l + 1 f l+1 + l ) + R l (f). By using f as an integrating factor, we integrate once more. Finally, we otain A l 1(f). (1.4) t (f = ( 1) l 1 8α l (1 (l + 1)(l + ) f l+ + l ) + R l+1 (f) = Bf l+ + R l+1 (f), (1.5) where R i (f) is a polynomial of f of degree i, i = l, l + 1. When l is odd, the coefficient of f l+, that is B, is positive, and when l is even, B is negative. Applying asymptotically vanishing oundary conditions, we get (f = Bf (f +c) l, where sign(b) = ( 1) l 1. Hence for l = odd, the degenerate coupled l-kdv equation has real traveling wave solution with asymptotically vanishing oundary conditions, ut when l = even, it does not. In this work, we study the equation (1.1) for l = 3 which is u t = 3 uu x + v x (1.6) v t = vu x + 1 uv x + ω x (1.7) ω t = 1 4 u xxx + ωu x + 1 uω x, (1.8) in detail. It is clear from Theorem 1.1 that unlike the case l =, we have real traveling wave solution with asymptotically vanishing oundary conditions in l = 3 case. Here the system (1.6)-(1.8) reduces to a polynomial of degree five, (f = f 5 + 3cf 4 + (6c d 1 )f 3 + 4(c 3 cd 1 + d )f + 8d 3 f + 8d 4 = P 5 (f), (1.9) where f(ξ) = u(x, t), and c, d 1, d, d 3, d 4 are constants. When the degree of the polynomial in the reduced equation is equal to five or greater it is almost impossile to solve them. As far as we know there is no known method to solve these equations. We shall introduce two methods to solve such equations. The first one is ased on the Cheyshev s Theorem [1] which is used recently to solve the Einstein field equations for a cosmological model [11], [1]. By using the Cheyshev s theorem we give several solutions of the reduced equations for l = 3 and also for aritrary l. The second method is ased on the factorizing the polynomial P l+ (f) as product of lower degree polynomials. In this way we make use of the reduced equations of lower degrees. We have given all possile such solutions for l = 3. 3

4 The layout of our paper is as follows: In Sec. II, we study the ehavior of the solutions in the neighorhood of the zeros of P 5 (f) and discuss the cases giving solitary wave solutions. In Sec. III, we find exact solutions of the system (1.6)-(1.8) y a new method proposed for any l 3 which uses the Cheyshev s Theorem and analyze the cases in which we may have solitary wave and kink-type solutions. In Sec. IV, we present an alternative method. Particularly, we find the solutions of the system (1.6)-(1.8). Here we otain many solutions for l = 3 including solitary wave, kink-type, periodic, and unounded solutions. We present some of them in the text ut the rest of the solutions are given in the Appendices A and B. General Waves of Permanent Form for (l = 3).1 Zeros of P 5 (f) and Types of Solutions Here we will analyze the zeros of P 5 (f) in (1.9). (i) If f 1 = f(ξ 1 ) is a simple zero of P 5 (f) we have P 5 (f 1 ) =. Taylor expansion of P 5 (f) aout f 1 gives (f = P 5 (f 1 ) + P 5(f 1 )(f f 1 ) + O((f f 1 ) = P 5(f 1 )(f f 1 ) + O((f f 1 ). From here we get f (ξ 1 ) = and f (ξ 1 ) = P 5(f 1 )/. Hence we can write the function f(ξ) as f(ξ) = f(ξ 1 ) + (ξ ξ 1 )f (ξ 1 ) + 1 (ξ ξ 1 f (ξ 1 ) + O((ξ ξ 1 ) 3 ) = f (ξ ξ 1 P 5(f 1 ) + O((ξ ξ 1 ) 3 ). (.1) Thus, in the neighorhood of ξ = ξ 1, the function f(ξ) has local minimum or maximum as P 5(f 1 ) is positive or negative respectively since f (ξ 1 ) = P 5(f 1 )/. (ii) If f 1 = f(ξ 1 ) is a doule zero of P 5 (f) we have P 5 (f 1 ) = P 5(f 1 ) =. Taylor expansion of P 5 (f) aout f 1 gives (f = P 5 (f 1 ) + P 5(f 1 )(f f 1 ) + 1 (f f 1 P 5 (f 1 ) + O((f f 1 ) 3 ) = 1 (f f 1 P 5 (f 1 ) + O((f f 1 ) 3 ). (.) To have real solution f, we should have P 5 (f 1 ) >. From the equality (.) we get f ± 1 f P 5 (f 1 ) ± 1 f 1 P 5 (f 1 ), 4

5 which gives f(ξ) f 1 + αe ± 1 P 5 (f 1)ξ, (.3) where α is a constant. Hence f f 1 as ξ. The solution f can have only one peak and the wave extends from to. (iii) If f 1 = f(ξ 1 ) is a triple zero of P 5 (f) we have P 5 (f 1 ) = P 5(f 1 ) = P 5 (f 1 ) =. Taylor expansion of P 5 (f) aout f 1 gives (f = P 5 (f 1 ) + P 5(f 1 )(f f 1 ) + 1 (f f 1 P 5 (f 1 ) (f f 1) 3 + O((f f 1 ) 4 ) = 1 6 (f f 1) 3 P 5 (f 1 ) + O((f f 1 ) 4 ). (.4) This is valid only if oth signs of (f f 1 ) 3 and P 5 (f 1 ) are same. Hence, to otain real solution f we have the following two possiilities: 1) (f f 1 ) > and P 5 (f 1 ) >, ) (f f 1 ) < and P 5 (f 1 ) <. If (f f 1 ) > and P 5 (f 1 ) > then we have which gives f ± 1 6 (f f 1 ) 3/ P 5 (f 1 ), f(ξ) f 1 + ( where α 1 is a constant. Thus f f 1 as ξ ±. Let (f f 1 ) < and P 5 (f 1 ) <. Then which yields 4 ± 1 6 P 5 (f 1 )ξ + α 1, (.5) f ± 1 6 (f 1 f) 3/ P 5 (f 1 ), f(ξ) f 1 ( where α is a constant. Thus f f 1 as ξ ±. 4 ± 1 6 P 5 (f 1 )ξ + α, (.6) (iv) If f 1 = f(ξ 1 ) is a quadruple zero of P 5 (f), then we have P 5 (f 1 ) = P 5(f 1 ) = P 5 (f 1 ) = P 5 (f 1 ) =. In this case Taylor expansion of P 5 (f) aout f 1 gives (f = 1 4 (f f 1) 4 P (4) 5 (f 1 ) + O((f f 1 ) 5 ). (.7) 5

6 This is valid only if P (4) 5 (f 1 ) >. Then we have f ± 1 6 (f f 1 P (4) 5 (f 1 ), which gives 1 f(ξ) f 1, (.8) ± 1 P (4) 6 5 (f 1 )ξ + γ 1 where γ 1 is a constant. Thus f f 1 as ξ ±. (v) If f 1 = f(ξ 1 ) is a zero of multiplicity 5 of P 5 (f), then we have P 5 (f) = (f f 1 ) 5 /. This is valid only if f f 1 >. So we otain the solution f as ( 4 ) 1/3 1 f = f 1 + ( ) 9 /3, (.9) ± ξ + m 1 where m 1 is a constant. Hence f f 1 as ξ ±.. All Possile Cases Giving Solitary Wave Solutions We analyze all possile cases aout the zeros of P 5 (f) that may give solitary wave solutions. Here in each cases we will present the sketches of the graphs of P 5 (f). Real solutions of (f = P 5 (f) occur in the shaded regions. (1) One doule and three simple zeros. Figure 1: Graphs of P 5 (f) having one doule and three simple zeros In Figure 1.(), f 1, f 3, and f 4 are simple zeros and f is a doule zero. The real solution occurs when f stays etween f 1 and f or f and f 3. At f 1, P 5(f 1 ) = f (ξ 1 ) > hence graph of the function f is concave up at ξ 1. At doule zero f, f f as ξ ±. Hence we have a solitary wave solution with amplitude f 1 f <. Similarly at f 3, P 5(f 3 ) = f (ξ 3 ) <, hence graph of the function f is concave down at ξ 3. Therefore, we also have a solitary wave solution with amplitude f 3 f >. Now consider the graph (d) in Figure 1. For f 3 we have P 5(f 3 ) = f (ξ 3 ) > thus graph of the function is concave up at ξ 3. At doule zero f 4, f f 4 as ξ ±. Hence we have a solitary wave solution with amplitude f 3 f 4 <. In other cases we have periodic solutions. 6

7 () Two doule and one simple zeros. Figure : Graphs of P 5 (f) having two doule and one simple zeros In Figure.(f), f 1 and f 3 are doule zeros and f is a simple zero. The real solution occurs when f stays etween f and f 3. For f we have P 5(f ) = f (ξ ) > thus graph of the function is concave up at ξ. At doule zero f 3, f f 3 as ξ ±. Hence we have a solitary wave solution with amplitude f f 3 <. Now consider the graph (g) in Figure. Here f and f 3 are doule zeros and f 1 is a simple zero. The real solution occurs when f stays etween f 1 and f or f and f 3. For f 1 we have P 5(f 1 ) = f (ξ 1 ) > thus graph of the function is concave up at ξ 1. At doule zero f, f f as ξ ±. Hence we have a solitary wave solution with amplitude f 1 f <. The other cases give kink, anti-kink type or unounded solutions. (3) One triple and two simple zeros. Figure 3: Graphs of P 5 (f) having one triple and two simple zeros Consider the graph (h) in Figure 3. Here f and f 3 are simple zeros and f 1 is a triple zero. The real solution occurs when f stays etween f 1 and f. For f we have P 5(f ) = f (ξ ) < so graph of the function f is concave down at ξ. At triple zero f 1, f f 1 as ξ ±. Hence we may have a solitary wave solution with amplitude f f 1 >. In the graph in Figure 3.(i), f 1 and f 3 are simple zeros and f is a triple zero. The real solution occurs when f stays etween f 1 and f. For f 1 we have P 5(f 1 ) = f (ξ 1 ) > so graph of the function f is concave up at ξ 1. At triple zero f, f f as ξ ±. Hence we may have a solitary wave solution with amplitude f 1 f <. The other case gives periodic solution. 7

8 (4) One quadruple and one simple zeros. Figure 4: Graphs of P 5 (f) having one quadruple and one simple zeros In the graph Figure 4.(l), f 1 is a simple zero and f is a quadruple zero. The real solution occurs etween f 1 and f. At f 1 we have P 5(f 1 ) = f (ξ 1 ) >, so graph of the function f is concave up at ξ 1. At quadruple zero f, f f as ξ ±. Hence we may have a solitary wave solution with amplitude f 1 f <. The other case gives unounded solution. (5) One doule and one simple zeros. Figure 5: Graphs of P 5 (f) having one doule and one simple zeros Consider the graph Figure 5.(n). Here f 1 is a simple zero and f is a doule zero. The real solution occurs etween f 1 and f. At f 1 we have P 5(f 1 ) = f (ξ 1 ) >, so graph of the function f is concave up at ξ 1. At doule zero f, f f as ξ ±. Hence we have a solitary wave solution with amplitude f 1 f <. The other case gives unounded solution. To sum up, we can give the following proposition for l = 3 case. Proposition.1 Equation (1.9) may admit solitary wave solutions when the polynomial function P 5 (f) admits (i) one doule and three simple zeros (ii) two doule and one simple zeros (iii) one triple and two simple zeros (iv) one quadruple and one simple zeros (v) one doule and one simple zeros. 3 Exact Solutions y Using the Cheyshev s Theorem The Cheyshev s Theorem is given as following [1]. 8

9 Theorem 3.1 Let a,, c, α, β e given real numers and αβ. The antiderivative I = x a (α + βx ) c dx (3.1) is expressile y means of the elementary functions only in the three cases: (1) + c Z, () Z, (3) c Z. (3.) The term x a (α + βx ) c is called a differential inomial. Note that the differential inomial may e expressed in terms of the incomplete eta function and the hypergeometric function. Let us define u = βx /α. Then we have I = 1 α a+1 = +c β a a α a+1 +c β a+1 B y ( 1 + a u 1+a F ), c 1. (, c; 1 + a + ; u ), (3.3) where B y is the incomplete eta function and F (τ, κ; η; u) is the hypergeometric function. Our aim is to transform the system (1.1) to (y = P l+ (y) y taking f(ξ) = γ + ᾱy( βξ). We can apply Cheyshev s Theorem to this equation if we assume that P l+ (y) reduces to the form P l+ (y) = Ay a (α + βy ) c, where a c + = l +. For l = 3, let u(x, t) = f(ξ) = γ + ᾱy( βξ) in (1.9), then the equation ecomes where α 1 = ᾱ3 β, α = 1 β α 3 = 1 β ( α 4 = 1 β (y = P 5 (y) = α 1 y 5 + α y 4 + α 3 y 3 + α 4 y + α 5 y + α 6, (3.4) ( 5ᾱ γ ) + 3ᾱ c, ( 5ᾱγ + 6ᾱc ᾱd 1 + 1ᾱcγ ) 4cd 1 + 4d + 18cγ 6d 1 γ + 5γ 3 + 4c c γ, α 5 = 1 ( ) ᾱ β 8d 3 + 1cγ 3 + 5γ4 + 18c γ 8cd 1 γ + 8c 3 γ + 8d γ 6d 1 γ, α 6 = 1 ( ) ᾱ β 8d 3 γ + 6c γ 3 + γ5 + 8d 4 4cd 1 γ d 1 γ 3 + 4c 3 γ + 4d γ. To apply the Cheyshev s Theorem 3.1 we assume that P 5 (y) reduces to the following form, ), P 5 (y) = Cy a (α + βy ) c, (3.5) where a,, c are the constants in Theorem 3.1, a c = 5, and C is a constant. Here we present the cases mentioned in the Proposition.1. Other cases are given in Appendix A. 9

10 1) Let P 5 (y) = y(α+βy. This form corresponds to the case of one simple or one simple and two doule zeros. We have y = ±y 1/ (α + βy ) so y 1/ (α + βy ) 1 dy = ±ξ + A. (3.6) Here a = 1/, =, and c = 1. Hence (1) + c = 3/4 / Z, () = 1/4 / Z, (3) c = 1 Z. For (3), from (3.6), y letting u = βy /α we otain ( 1 α 3/4 β 1/4 u 1/4 F 4, 3; 5 ) 4 ; u = ±ξ + A. (3.7) Here choose α = 1, β = 1, the integration constant A =, plus sign in (3.6), we have arctanh( y) + arctan( y) = (ξ + ), (3.8) and the graph of this solution for ξ is given in Figure 6 Figure 6: Graph of a solution of (y = y( 1 + y This is a kink-type solution. ) Let P 5 (y) = y(α + βy) 4. This form corresponds to the case of one simple and one quadruple zeros. We have y = ±y 1/ (α + βy so y 1/ (α + βy) dy = ±ξ + A. (3.9) Here a = 1/, = 1, and c =. Hence (1) + c = 3/ / Z, () = 1/ / Z, (3) c = Z. For (3), from (3.9), y letting u = βy /α we otain ( ) β y βα ( 1 α 3/ β 1/ u 1/ F, 4; 3 ) ; u = y arctan α(βy + α) + α βα 1 = ±ξ + A. (3.1)

11 Here choose α = 1, β = 1, the integration constant A =, plus sign in (3.9) we have y 1 + y + arctan( y) = ξ +, (3.11) and the graph of this solution for ξ < π/ is given in Figure 7 Figure 7: Graph of a solution of (y = y(1 + y) 4 3) Let P 5 (y) = y (α + βy 3 ). This form corresponds to the case of one simple and one doule zeros. We have y = ±y(α + βy 3 ) 1/ so y 1 (α + βy 3 ) 1/ dy = ±ξ + A. (3.1) Here a = 1, = 3, and c = 1/. Hence (1) + c = 1/ / Z, () For (), from (3.1), we otain y = = Z, (3) c = 1/ / Z. ( α ( ( 3 ) 1/3. tanh α(a ± ξ) 1)) (3.13) β Here choose α = 4, β = 4, the integration constant A =, and plus sign in (3.1) we otain y(ξ) = (sech(3ξ + 6)/3 (3.14) and the graph of this solution is given in Figure 8 Figure 8: Graph of a solution of (y = y (4 + 4y 3 ) This is clearly a solitary wave solution. 11

12 4) Let P 5 (y) = y (α + βy) 3. This form corresponds to the case of one doule and one triple zeros. We have y = ±y(α + βy) 3/ so y 1 (α + βy) 3/ dy = ±ξ + A. (3.15) Here a = 1, = 1, and c = 3/. Hence (1) + c = 3/ / Z, () = Z, (3) c = 3/ / Z. For (), from (3.15), we otain ( βy+α ) arctanh α βy + α α = ±ξ + A. (3.16) α 3/ Here choose α = 1, β = 1, A =, and plus sign in (3.15) we have y + 1 arctanh( y + 1) = ξ +, (3.17) and the graph of this solution is given in Figure 9 Figure 9: Graph of a solution of (y = y (1 + y) 3 5) Let P 5 (y) = y 4 (α + βy). This form corresponds to the case of one simple and one quadruple zeros. We have y = ±y (α + βy) 1/ so y (α + βy) 1/ dy = ±ξ + A. (3.18) Here a =, = 1, and c = 1/. Hence (1) + c = 3/ / Z, () For (), from (3.18), y letting u = βy /α we otain = 1 Z, (3) c = 1/ / Z. ( βy+α ) α ( α 3/ βu 1 F 1, 5 ) βy + α βarctanh ; ; u = + = ±ξ + A. (3.19) αy α 3/ 1

13 Figure 1: Graph of a solution of (y = y 4 (1 + y) Take α = 1, β = 1, A =, and plus sign in (3.18) we get arctanh( y + 1 y + 1) = ξ +, (3.) y and the graph of this solution for ξ is given in Figure 1 4 An Alternative Method to Solve (f = P n (f), When n 5: Factorization of Polynomials When we have (f = P n (f) for the reduced equation it ecomes quite difficult to solve such equations for n 5. For this purpose, we shall introduce a new method which is ased on the factorization of the polynomial P n (f) = P l+ (f), n 5. Let the polynomial P l+ (f) l+ have l + real roots i.e. P l+ (f) = B (f f i ), B is a constant. Define a new function ρ(ξ) i=1 so that f = f(ρ(ξ)). Hence we have (f = = κ dρ ( dρ = µ dξ ( dρ. By taking dρ dξ N 1 i=1 N i=1 (f f i ), (f f i ), where N 1 + N = l +, B = κµ, we get a system of ordinary differential equations. Solving this system gives the solution of the degenerate coupled l-kdv equation. For illustration, we start with a differential equation where we know the solution. Consider (y = α(y y 1 )(y y )(y y 3 )(y y 4 ). (4.1) Let y = y(ρ(ξ)) so ( dy ( dρ = α(y y1 )(y y )(y y 3 )(y y 4 ). dρ dξ 13

14 Take ( dy = κ(y y 1 )(y y ) (4.) dρ ( dρ = µ(y y 3 )(y y 4 ), (4.3) dξ where κµ = α. We start with solving the equation (4.). We have dy (y y1 )(y y ) = dy ( ) = ± κdρ, y a 1 where a = (y 1 +y )/ and = (y 1 y )/. Let (y a)/ = cosh z. This gives us dz = ± κdρ. After taking the integral we otain ( y a ) arccosh = ± κρ + C 1, C 1 constant, so that y = h(ρ) = a + cosh( κρ + C ), C constant. (4.4) Now we insert this solution into Eq.(4.3). Let cosh( κρ + C ) = v. Hence we get dv κ v 1 (v A B = ± µξ + C 3, (4.5) where A = (y 3 +y 4 y 1 y )/(y 1 y ), B = (y 3 y 4 )/(y 1 y ), and C 3 is a constant. Solving (4.5) and using (4.4) gives y = y 1(y 3 y ) + y (y 1 y 3 )sn ((1/) κ(y 3 y )(y 4 y 1 )( µξ + C 3 ), k) (y 3 y ) + (y 1 y 3 )sn ((1/) κ(y 3 y )(y 4 y 1 )(, (4.6) µξ + C 3 ), k) which was otained in [8]. Here sn is the Jacoi elliptic function and k is the elliptic modulus satisfying k = (f f 4 )(f 1 f 3 )/(f f 3 )(f 1 f 4 ). Now we apply our method to the case when the polynomial P n (f) is of degree n = 5. We have several possile cases, ut here we shall give the case when the polynomial (1.9) has five real roots. Other cases are presented in Appendix B. Case 1. If (1.9) has five real roots we can write it in the form (f = α(f f 1 )(f f )(f f 3 )(f f 4 )(f f 5 ), (4.7) where f 1, f, f 3, f 4, and f 5 are the zeros of the polynomial function P 5 (f). Now define a new function ρ(ξ) so that f = f(ρ(ξ)). Hence (4.7) ecomes (f = ( dρ = α(f f1 )(f f )(f f 3 )(f f 4 )(f f 5 ). (4.8) dρ dξ 14

15 i. Take = (f f 1 )(f f )(f f 3 )(f f 4 ) (4.9) dρ ( dρ = α(f f 5 ). (4.1) dξ In [8] we have found the solutions of Eq.(4.9). One of the solutions is f = h(ρ) = f 4(f 3 f 1 ) + f 3 (f 1 f 4 )sn ((1/) (f 1 f 3 )(f f 4 )ρ, k) (f 3 f 1 ) + (f 1 f 4 )sn ((1/), (4.11) (f 1 f 3 )(f f 4 )ρ, k) where k is the elliptic modulus satisfying k = (f 3 f )(f 4 f 1 )/(f 3 f 1 )(f 4 f ). We use this solution in the equation (4.1) and get where ρ f5 h(ˆρ) = A + Bsn (ωˆρ, k) C + Dsn (ωˆρ, k) = ± αξ + C 1, C 1 constant, (4.1) A = f 3 f 1, B = f 1 f 4, C = (f 5 f 4 )(f 3 f 1 ), D = (f 5 f 3 )(f 1 f 4 ), and ω = (1/) (f 1 f 3 )(f f 4 ). For particular values; f 1 = f = 1, f 3 = 3, f 4 =, f 5 =, α = 1, C 1 =, and choosing plus sign in (4.1) we get the graph of the solution f(ξ) given in Figure 11 Figure 11: Graph of the solution (4.11) with (4.1) for particular parameters This is a solitary wave solution. ii.take = κ(f f 1 )(f f )(f f 3 ) (4.13) dρ ( dρ = µ(f f 4 )(f f 5 ), (4.14) dξ 15

16 where κµ = α. Eq.(4.13) has a solution f = h(ρ) = (f f 1 )sn ((1/) f 3 f 1 ( κρ + C 1 ), k) + f 1, (4.15) where k satisfies k Eq.(4.14) and get = (f 1 f )/(f 1 f 3 ) and C 1 is a constant. We use this solution in ρ (h(ˆρ) f4 )(h(ˆρ) f 5 ) = Asn4 (ωˆρ + δ, k) + Bsn (ωˆρ + δ, k) + C = ± µξ +C, where C is a constant and (4.16) A = (f f 1, B = (f f 1 )(f 1 f 4 f 5 ), C = (f 1 f 4 )(f 1 f 5 ), ω = (1/) κ(f 3 f 1 ), and δ = (1/) f 3 f 1 C 1. For particular values; f 1 = 1, f =, f 3 = 3, f 4 = 1, f 5 =, κ = µ = 1, C 1 =, and choosing plus sign in (4.16) we get the graph of the solution f(ξ) given in Figure 1 Figure 1: Graph of the solution (4.15) with (4.16) for particular parameters This solution is periodic. iii. Take = κ(f f 1 )(f f ) (4.17) dρ ( dρ = µ(f f 3 )(f f 4 )(f f 5 ), (4.18) dξ where κµ = α. Consider first the equation (4.17). We have df (f f1 )(f f ) = ± κdρ. (4.19) 16

17 This equality can e reduced to df ( ) = ± κdρ, (4.) f a 1 where a = (f 1 + f )/, = (f 1 f )/. By making the change of variales (f a)/ = u, we get du u 1 = ± κρ + C 1, (4.1) where C 1 is a constant. Thus we otain which yields ( f a ) arccoshu = arccosh = ± κρ + C 1, (4.) f = h(ρ) = a + cosh( κρ + C 1 ), a = (f 1 + f )/, = (f 1 f )/. (4.3) Now we use this result in (4.18), = (h(ˆρ) f3 )(h(ˆρ) f 4 )(h(ˆρ) f 5 ) = A cosh 3 ( κˆρ + C 1 ) + B cosh ( κˆρ + C 1 ) + C cosh( κˆρ + C 1 ) + D = ± µξ + C, where C is a constant and (4.4) A = 3, B = (3a f 3 f 4 f 5 ), C = {(a f 3 )(a f 4 )+(a f 3 )(a f 5 )+(a f 4 )(a f 5 )}, and D = (a f 3 )(a f 4 )(a f 5 ). For particular values; f 1 = 5, f = 1, f 3 = f 4 = 3, f 5 = 4, κ = 4, µ = 1, C 1 = C =, and choosing plus sign in (4.4), we get the graph of the solution f(ξ) given in Figure 13 Figure 13: Graph of the solution (4.3) for particular parameters 17

18 iv. Take = κ(f f 1 ) (4.5) dρ ( dρ = µ(f f )(f f 3 )(f f 4 )(f f 5 ), (4.6) dξ where κµ = α. Consider the Eq.(4.5). We have df f f1 = ± κdρ. (4.7) Integrating oth sides gives ( f = h(ρ) = ± We use this result in the equation (4.6) = (h(ˆρ) f )(h(ˆρ) f 3 )(h(ˆρ) f 4 )(h(ˆρ) f 5 ) = κ ρ + C 1 + f1, C 1 constant. (4.8) ( κ 4 ˆρ ± κc 1 ˆρ + A)( κ 4 ˆρ ± κc 1 ˆρ + B)( κ 4 ˆρ ± κc 1 ˆρ + C)( κ 4 ˆρ ± κc 1 ˆρ + D) = ± µξ + C, where C is a constant and A = C 1 + f 1 f, B = C 1 + f 1 f 3, C = C 1 + f 1 f 4, D = C 1 + f 1 f 5. (4.9) For particular values, f 1 = 1, f = f 3 =, f 4 = 3, f 5 = 4, κ = 4, µ = 1, and C 1 = C =, and choosing plus sign in (4.9) we get the graph of the solution f(ξ) given in Figure 14 This is a solitary wave solution. Figure 14: Graph of the solution (4.8) for particular parameters 18

19 5 Conclusion We study the degenerate l-coupled KdV equations. We reduce these equations into an ODE of the form (f = P l+ (f), where P l+ is a polynomial function of f of degree l +, l 3. We give a general approach to solve the degenerate l-coupled equations y introducing two new methods that one of them uses Cheyshev s Theorem and the other one is an alternative method, ased on factorization of P l+ (f), l 3. Particularly, for the degenerate three-coupled KdV equations we otain solitary-wave, kink-type, periodic, or unounded solutions y using these methods. 6 Acknowledgment This work is partially supported y the Scientific and Technological Research Council of Turkey (TÜBİTAK). APPENDIX A. Other Cases Using Cheyshev s Theorem for Degenerate Three- Coupled KdV Equation Here we present other cases that we use Cheyshev s theorem to solve degenerate threecoupled KdV equation. 1) Let P 5 (y) = y(α + βy 4 ). This form corresponds to the case of one simple zero or three simple zeros. We have y = ±y 1/ (α + βy 4 ) 1/ so y 1/ (α + βy 4 ) 1/ dy = ±ξ + A. Here a = 1/, = 4, and c = 1/. So we have (1) + c = 3/8 / Z, () = 1/8 / Z, (3) c = 1/ / Z. Hence we cannot otain a solution through the Cheyshev s theorem. ) Let P 5 (y) = y 3 (α + βy ). This form corresponds to the case of one triple or two simple and one triple zeros. We have y = ±y 3/ (α + βy ) 1/ so y 3/ (α + βy ) 1/ dy = ±ξ + A. Here a = 3/, =, and c = 1/. So we have (1) + c = 3/4 / Z, () = 1/4 / Z, (3) c = 1/ / Z. Hence we cannot otain a solution through the Cheyshev s theorem. 19

20 3) Let P 5 (y) = y 3 (α + βy. This form corresponds to the case of one doule and one triple zeros. We have y = ±y 3/ (α + βy) so y 3/ (α + βy) 1 dy = ±ξ + A. (6.1) Here a = 3/, = 1, and c = 1. Hence (1) + c = 3/ / Z, () For (3), from (6.1), y letting u = βy /α we otain = 1/ / Z, (3) c = 1 Z. ( βy ) ( α 3/ β 1/ u 1/ F 1, 3; 1 ) ; u = β arctan α y + α = ±ξ + A. (6.) α 3/ 4) Let P 5 (y) = (α + βy) 5. This form corresponds to the case of a zero with multiplicity five. We have y = ±(α + βy) 5/ so (α + βy) 5/ dy = ±ξ + A. (6.3) Here a =, = 1, and c = 5/. Hence (1) + c = 3/ / Z, () For (), from (6.3), y letting u = βy /α we otain α 3/ β 1 uf (1, 9 ) ; ; u = 1 Z, (3) c = 5/ / Z. = = ±ξ + A. (6.4) 3β(βy + α) 3/ Here we get y = 1 β [( 4 ) 9β (±ξ + A ] α. (6.5) 5) Let P 5 (y) = (α + βy 5 ). This form corresponds to the case of one simple zero. We have y = ±(α + βy 5 ) 1/ so (α + βy 5 ) 1/ dy = ±ξ + A. Here a =, = 5, and c = 1/. We have (1) + c = 3/1 / Z, () = 1/5 / Z, (3) c = 1/ / Z. Hence we cannot otain a solution through the Cheyshev s theorem. APPENDIX B. Other Cases Using Alternative Method for Degenerate Three- Coupled KdV Equation

21 Here we present the other cases for the method of factorization of the polynomial P 5 (f). Case. If (1.9) has three real roots we can write it in the form (f = α(f f 1 )(f f )(f f 3 )(af + f + c), (6.6) where f 1, f, and f 3 are the zeros of the polynomial function P 5 (f) of degree five and 4ac <. Now define a new function ρ(ξ) so that f = f(ρ(ξ)). Hence (6.6) ecomes (f = ( dρ = α(f f1 )(f f )(f f 3 )(af + f + c), (6.7) dρ dξ i. Take = κ(f f 1 )(f f )(f f 3 ) (6.8) dρ ( dρ = µ(af + f + c), (6.9) dξ where κµ = α. Consider the first equation aove. We have df (f f1 )(f f )(f f 3 ) = ± κdρ. (6.1) We know from Case 1.ii that Eq.(6.8) has a solution f = h(ρ) = (f f 1 )sn ((1/) f 3 f 1 ( κρ + C 1 )) + f 1, (6.11) where C 1 is a constant. We use this solution in (6.9) and we get ρ ah (ˆρ) + h(ˆρ) + c = Asn4 (ωˆρ + δ) + Bsn (ωˆρ + δ) + C = ± µξ+c, (6.1) where C is a constant and A = a(f f 1, B = (af 1 + )(f f 1 ), C = (af 1 + f 1 + c), ω = (1/) κ(f 3 f 1 ), and δ = (1/) f 3 f 1 C 1. ii. Take = κ(f f 1 )(f f ) (6.13) dρ ( dρ = µ(f f 3 )(af + f + c), (6.14) dξ where κµ = α. Consider first the equation (6.13). It has a solution f = h(ρ) = a cosh( κρ + C 1 ), a 1 = (f 1 + f )/, 1 = (f 1 f )/, (6.15) 1

22 as it is found in Case 1.iii. We use this solution in (6.14) and get = (h(ˆρ) f3 )(ah (ˆρ) + h(ˆρ) + c) = A cosh 3 ( κˆρ + C 1 ) + B cosh ( κˆρ + C 1 ) + C cosh( κˆρ + C 1 ) + D = ± µξ + C, where C is a constant and A = a 3 1, B = 1(3aa 1 + f 3 ), C = 1 (3aa 1 aa 1 f 3 + a 1 f 3 + c), and D = (a 1 f 3 )(a 1a + a 1 + c). (6.16) iii. Take = κ(f f 1 ) (6.17) dρ ( dρ = µ(f f )(f f 3 )(af + f + c), (6.18) dξ where κµ = α. Consider the equation (6.17). It has a solution ( κ f = h(ρ) = ± ρ + C 1 + f1, C 1 constant (6.19) as it is otained in Case 1.iv. We use this result in Eq.(6.18) = (h(ˆρ) f )(h(ˆρ) f 3 )(ah (ˆρ) + h(ˆρ) + c) = ( κ 4 ˆρ ± κc 1 ˆρ + A)( κ 4 ˆρ ± κc 1 ˆρ + B)( aκ 16 ˆρ4 ± aκ3/ C 1 ˆρ 3 + C ˆρ + Dˆρ + E) = ± µξ + C, where C is a constant and A = C 1 + f 1 f, B = C 1 + f 1 f 3, C = κ( + 6aC 1 + af 1 )/4, (6.) and D = ± κc 1 (af 1 + ac 1 + ), E = (f 1 + C 1)(af 1 + ac 1 + ) + c. iv. Take = κ(af + f + c) (6.1) dρ ( dρ = µ(f f 1 )(f f )(f f 3 ), (6.) dξ

23 where κµ = α. Consider the first equation aove. We have f where W = 4ac /a. Let d ˆf a ( ˆf+ a W + 1 = ± κρ + C 1, ( ) f + /W = tan θ. Hence the aove equality ecomes a which gives θ sec ˆθdˆθ = a(± κρ + C 1 ), ln sec θ + tan θ = a(± κρ + C 1 ). Hence after some simplifications and taking the integration constant zero we get We use this result in (6.) = where C is a constant and f = h(ρ) = ± 4ac sinh( aκρ)/a. (6.3) (h(ˆρ) f1 )(h(ˆρ) f )(h(ˆρ) f 3 ) = A sinh 3 ( aκˆρ) + B sinh ( aκˆρ) + C sinh( aκˆρ) + D = ± µξ + C, (6.4) A = ±(4ac ) 3/ /8a 3, B = ( 4ac)(f 1 + f + f 3 )/4a, and C = ± 4ac (f 1 f 3 + f f 3 + f 1 f )/a, D = f 1 f f 3. Case 3. If (1.9) has just one real root we can write it in the form (f = α(f f 1 )(f 4 + a 3 f 3 + a f + a 1 f + a ), (6.5) where f 1 is the zero of the polynomial function P 5 (f) of degree five and the constants a i, i =, 1,, 3 are so that f 4 + a 3 f 3 + a f + a 1 f + a for real f. Now define a new function ρ(ξ) so that f = f(ρ(ξ)). Hence (6.5) ecomes (f = ( dρ = α(f f1 )(f 4 + a 3 f 3 + a f + a 1 f + a ). (6.6) dρ dξ 3

24 i. Take = κ(f f 1 ) (6.7) dρ ( dρ = µ(f 4 + a 3 f 3 + a f + a 1 f + a ), (6.8) dξ where κµ = α. Consider the first equation aove. It has a solution ( κ f = h(ρ) = ± ρ + C 1 + f1, C 1 constant (6.9) as it is otained in Case 1.iv. We use this result in (6.8) and get h4 (ˆρ) + a 3 h 3 (ˆρ) + a h (ˆρ) + a 1 h(ˆρ) + a = ± µξ + C, (6.3) where C is a constant. Here we do not get a simpler expression than the original equation. Hence it is meaningless to use the method of factorization of the polynomial for this case. ii. Take = κ(f 4 + a 3 f 3 + a f + a 1 f + a ) (6.31) dρ ( dρ = µ(f f 1 ), (6.3) dξ where κµ = α. Here for simplicity, we will use the form f 4 +f +c instead of the polynomial function of degree four aove. Since f 4 +f +c = (f +/ /4+c, we will take c > /4 ( to get an irreducile polynomial. If c = df /4, then take >. From dρ) = κ(f 4 + f + c) f d ˆf ˆf 4 + ˆf + c = ± κρ + C 1, C 1 constant. We have where k = f(ρ) = c sn(( /)(C 1 ± κρ) + 4c +, k), + 4c + c + + 4c +. Then we use this function f in (6.3) and get c A c sn((a /)(C1 ± κρ), k) Af 1 = ± µξ + C, C constant, (6.33) where A = + 4c +. 4

25 Now take c = /4 with >. In this case we have f d ˆf ˆf + / = ± κρ + C 1, C 1 constant, which gives the solution f = h(ρ) = ± Now use the equation (6.3). We have ± tan ( ( tan ( κρ + C ) 1 ). (6.34) ( κˆρ + C ) = ± µξ + C, 1 ) f 1 where C is a constant. Solving the aove equation gives ±4 { [ (A1 A + A ] ( κ(a + A 3)A 3 4 A 3 A 1 A )} 3 3 ln arctan = ± µξ +C (A 1 + A + A 3 A 3 A 1 + A, where A 1 = ± tan ( ( κρ + C ) 1 ) 4f 1, A = 4f1 + f 1, and A 3 = 4f f 1. References [1] Gürses, M., Karasu, A., Degenerate Svinolupov systems, Phys. Lett. A 14, 1-6 (1996). [] Gürses, M., Karasu, A., Integrale coupled KdV systems, J. Math. Phys. 39, (1998). [3] Alonso, L. M., Schrödinger spectral prolems with energy-dependent potentials as sources of nonlinear Hamiltonian evolution equations, J. Math. Phys. 1, (198). [4] Antonowicz, M., Fordy, A. P., A family of completely integrale multi-hamiltonian systems, Phys. Lett. A 1, (1987). [5] Antonowicz, M., Fordy, A. P., Coupled KdV equations with multi-hamiltonian structures, Physica D 8, (1987). [6] Antonowicz, M., Fordy, A. P., Factorization of energy-dependent Schrödinger operators: Miura maps and modified systems, Comm. Math. Phys. 14, (1989). 5

26 [7] Gürses, M., Integrale hierarchy of degenerate coupled KdV equations, (unpulished), e-print arxiv: [8] Gürses, M., Pekcan A., Traveling Wave Solutions of the Degenerate Coupled Korteweg-de Vries Equation, J. of Math. Phys. 55, 9151 (14). [9] Bradurry, T. C., Theoretical Mechanics, (R. E. Krieger Pulishing Co., Malaar, FL, 1981). [1] Tcheichef, M. P., L intégration des Différentielles Irrationnelles, J. Math Pures Appl. 18, (1853). [11] Chen, S., Gions, G. W., Yang, Y., Friedmann s Equations in All Dimensions and Cheyshev s Theorem, J. Cosm. Astropart. Phys. 1, 35 (14). [1] Chen, S., Gions, G. W., Yang, Y., Explicit Integration of Friedmann s Equation with Nonlinear Equations of State, J. Cosm. Astropart. Phys. 15, (15). 6