A Note on Solitary Wave Solutions of the Nonlinear Generalized Camassa-Holm Equation

Abstract

We give a simple method for applying ordinary differential equation to solve the nonlinear generalized Camassa-Holm equation 𝑢_𝑡+2𝑘𝑢_𝑥−𝑢_𝑥𝑥𝑡+𝑎𝑢^𝑚𝑢_𝑥−2𝑢_𝑥𝑢_𝑥𝑥+𝑢𝑢_𝑥𝑥𝑥=0. Furthermore we give a new ansätz. In the cases where 𝑚=1,2,3, the numerical simulations demonstrate the results.

1. Introduction

A new dispersive shallow water equation 𝑢_𝑡+𝜅𝑢_𝑥−𝑢_𝑥𝑥𝑡+3𝑢𝑢_𝑥=2𝑢_𝑥𝑢_𝑥𝑥+𝑢𝑢_𝑥𝑥𝑥 (1) known as the Camassa-Holm equation has been derived by Camassa and Holm [1]. They showed that the Camassa-Holm equation has peaked wave solitary solutions which have the first derivative discontinuity at the wave peak called “peakons.” In [2], the integral bifurcation method was used to study a nonlinearly dispersive wave equation of Camassa-Holm equation type. In [3], attractor for a coupled nonhomogeneous Camassa-Holm equation with periodic boundary condition was investigated by means of several inequalities. In [4], sufficient conditions for the modified two-component Camassa-Holm system were established. In [5], a class of nonlinear fourth order analogue of a generalized Camassa-Holm equation was studied by using sine-cosine method. In [6], He et al. studied a generalized Camassa-Holm equation. In [7], Wazwaz established solitary wave solutions to the modified forms of Degasperis-Procesi and Camassa-Holm equations. In [8], a family of Camassa-Holm equations with distinct parameters was investigated. Also many aspects of the problems were studied by researchers [9–16]. Authors in [4, 17] presented periodic wave solutions and traveling wave solutions for some equations. In [18], Khuri investigated the periodic wave and peaked solitary wave solutions of the nonlinear generalized Camassa-Holm equation and gave a ansätz for demonstrating the existence of a new class of solutions. In [19], Tian and Song derive some new exact peaked solitary wave solutions of the generalized Camassa-Holm equation and two types of new exact traveling wave solutions of the generalized weakly dissipative Camassa-Holm equations. In this paper we give a simple method for applying ordinary differential equation to solve the generalized Camassa-Holm equation and give the improved ansätz. The numerical simulation examples demonstrate that our methods are applicable.

2. Simplification of the Nonlinear Generalized Camassa-Holm Equation

Consider the following nonlinear generalized Camassa-Holm equation: 𝑢_𝑡+2𝑘𝑢_𝑥−𝑢_𝑥𝑥𝑡+𝑎𝑢^𝑚𝑢_𝑥−2𝑢_𝑥𝑢_𝑥𝑥+𝑢𝑢_𝑥𝑥𝑥=0. (2) With the velocity constant 𝑐, we seek the traveling wave solution of the form 𝑢(𝑥,𝑡)=𝑣(𝜉) of (2) where 𝜉=𝑥−𝑐𝑡. Substituting 𝑢(𝑥,𝑡)=𝑣(𝜉) into (2), we have −𝑐d𝑣d𝜉+2𝑘d𝑣dd𝜉+𝑐³𝑣d𝜉³+𝑎𝑣^𝑚d𝑣d𝜉−2d𝑣dd𝜉²𝑣d𝜉²d−𝑣³𝑣d𝜉³=0. (3) Integrating both sides of (3), we obtain ()d𝑣−𝑐²𝑣d𝜉²+12d𝑣d𝜉²=𝑎𝑣𝑚+1^𝑚+1+(2𝑘−𝑐)𝑣+𝑏, (4) where 𝑏 is an arbitrary constant. Let d𝑣/d𝜉=𝑤. Then d²𝑣/d𝜉²=𝑤(d𝑤/d𝑣). Substituting this into (4) yields d𝑤+1d𝑣2()((𝑣𝑣−𝑐𝑤=𝑎/𝑚+1))^𝑚+1+(2𝑘−𝑐)𝑣+𝑏𝑤𝑣−𝑐⁻¹. (5) Solving (4) leads to 𝑤²=(𝑣−𝑐)⁻¹2𝑎𝑑+𝑣𝑚+1^𝑚+1+(2𝑘−𝑐)𝑣+𝑏d𝑣, (6) where 𝑑 is an arbitrary constant. Therefore d𝑣d𝜉²=(𝑣−𝑐)⁻¹×𝑑+2𝑎()𝑣𝑚+1)(𝑚+2^𝑚+2+(2𝑘−𝑐)𝑣²+𝑏𝑣. (7) From (d𝜉)²=(d|𝜉|)², we have d𝑣d𝜉²=(𝑣−𝑐)⁻¹×𝑑+2𝑎()𝑣𝑚+1)(𝑚+2^𝑚+2+(2𝑘−𝑐)𝑣²+𝑏𝑣. (8) Therefore 𝑣−𝑐𝑑+(2𝑎/((𝑚+1)(𝑚+2)))𝑣^𝑚+2+(2𝑘−𝑐)𝑣²=𝜉,𝜉,+𝑏𝑣d𝑣ℎ+𝜉,ℎ−𝜉,ℎ+ℎ− (9) where ℎ is an arbitrary constant.

When 𝑚=1, (9) is the case of the Camassa-Holm equation. We take 2𝑘≠𝑐, 𝑎𝑐−3𝑐+2𝑘=0, 𝑑=0, and 𝑏=0, then 3𝑎𝑐ln𝑣=𝜉,𝜉.ℎ+𝜉,ℎ−𝜉,ℎ+ℎ− (10) So 𝑣=𝑐e^{(𝑎/3)(ℎ+𝜉)},𝑐e^{(𝑎/3)(ℎ−𝜉)},𝑐e^{(𝑎/3)(ℎ+|𝜉|)},𝑐e^{(𝑎/3)(ℎ−|𝜉|)}. (11) It can be checked that 𝑢(𝑥,𝑡)=𝑐e^{ℎ+(𝑎/3)(𝑥−𝑐𝑡)}, 𝑢(𝑥,𝑡)=𝑐e^{ℎ−(𝑎/3)(𝑥−𝑐𝑡)}, and 𝑢(𝑥,𝑡)=𝑐e^{ℎ+(𝑎/3)|𝑥−𝑐𝑡|} are infinitely great solutions of the Camassa-Holm equation without asymptotic behavior [20]. Only 𝑢(𝑥,𝑡)=𝑐e^{ℎ−(𝑎/3)|𝑥−𝑐𝑡|} (12) are the solitary wave solutions of the Camassa-Holm equation.

When 𝑚=2, we take 2𝑘<𝑐, 𝑎𝑐²−6𝑐+12𝑘=0, 𝑑=0, and 𝑏=0. In the similar way in the case 𝑚=1, we only choose 6𝑎1𝑣√𝜉,𝑣+𝑐d𝜑=ℎ− (13) then −24𝑎𝑐atanh𝑣+𝑐𝑐=ℎ−𝜉. (14) From (14), we have 𝑣=𝑐tanh²𝑎𝑐𝜉24−ℎ−𝑐. (15) Therefore we have the solitary wave solutions of the generalized Camassa-Holm equation as 𝑢(𝑥,𝑡)=𝑐tanh²𝑎𝑐|24(|𝑥−𝑐𝑡−ℎ)−𝑐. (16)

When 𝑚=3, we take 2𝑘≠𝑐, 𝑎𝑐³−10𝑐+20𝑘=0, 𝑑=0, and 𝑏=0 then 10𝑎1√𝑣²+𝑐𝑣+𝑐²d𝑣=ℎ−𝜉. (17) Then −10𝑎𝑐²𝑐()atanh𝑣+2𝑐|𝑐|√𝑣²+𝑐𝑣+𝑐²=ℎ−𝜉. (18) Solving (18), we obtain 𝑣=−2𝑐1−tanh²𝑎𝑐²𝜉|𝑐|310−ℎ±21−tanh²𝑎𝑐²𝜉10−ℎ×tanh²𝑎𝑐²𝜉10−ℎ^1/2×1−4tanh²𝑎𝑐²𝜉10−ℎ⁻¹. (19) Therefore 𝑢(𝑥,𝑡)=−2𝑐1−tanh²𝑓(|𝑐|𝑥−𝑐𝑡)+231−tanh²𝑓(𝑥−𝑐𝑡)tanh²𝑓()×𝑥−𝑐𝑡1−4tanh²𝑓(𝑥−𝑐𝑡)⁻¹ (20) or 𝑢(𝑥,𝑡)=−2𝑐1−tanh²𝑓(|𝑐|𝑥−𝑐𝑡)−231−tanh²𝑓(𝑥−𝑐𝑡)tanh²𝑓()×𝑥−𝑐𝑡1−4tanh²𝑓(𝑥−𝑐𝑡)⁻¹, (21) where 𝑓(𝑥−𝑐𝑡)=(𝑎𝑐²/10)(|𝑥−𝑐𝑡|−ℎ).

3. Ansätz for the Generalized Camassa-Holm Equation

From [18], we have the following ordinary differential equation in 𝐹: 2𝐹³d²𝐹d𝜉²+4𝐹²d𝐹d𝜉²=𝑎𝑚+1^𝑚+1_𝑘=0𝑘𝐹𝑚+1_2𝑘𝑐^{𝑚+1−𝑘}+(2𝑘−𝑐)𝐹²+2𝑘𝑐−𝑐²+𝑏, (22) where 𝐹=√𝑣−𝑐.

Let d𝐹/d𝜉=𝐺. Then d²𝐹/d𝜉²=𝐺(d𝐺/d𝐹). Substituting this into (22) gives the following first order Bernoulli’s ordinary differential equation: d𝐺+2d𝐹𝐹𝑎𝐺=𝑚+1^𝑚+1_𝑘=0𝑘𝐹𝑚+1_2𝑘𝑐^{𝑚+1−𝑘}+(2𝑘−𝑐)𝐹²+2𝑘𝑐−𝑐²×+𝑏2𝐹³⁻¹𝐺⁻¹. (23) Solving (23) leads to 𝐺²=𝐹⁻⁴𝑎𝑔+𝑚+1^𝑚+1_𝑘=0𝑘𝐹𝑚+1_2𝑘𝑐^{𝑚+1−𝑘}+(2𝑘−𝑐)𝐹²+2𝑘𝑐−𝑐²+𝑏𝐹d𝐹. (24) Therefore 𝐹d𝐹d𝜉²=𝑔𝐹⁻²+𝑎𝑚+1^𝑚+1_𝑘=01𝑘𝐹2𝑘+2𝑚+1_2𝑘𝑐^{𝑚+1−𝑘}+14(2𝑘−𝑐)𝐹²+122𝑘𝑐−𝑐²+𝑏, (25) where 𝑔 is an arbitrary constant. We observe that 𝐹(d𝐹/d𝜉) hardly becomes a polynomial in 𝐹 unless in the particular cases. So this is a new ansätz compared with the ansätz in [18]. Similarly, we have 𝐹d𝐹d𝜉²=𝑔𝐹⁻²+𝑎𝑚+1^𝑚+1_𝑘=01𝑘𝐹2𝑘+2𝑚+1_2𝑘𝑐^{𝑚+1−𝑘}+14(2𝑘−𝑐)𝐹²+122𝑘𝑐−𝑐²+𝑏. (26) Seemingly, (26) is more difficult than (8). On the contrary, (26) is much easier to be solved than (8). This can be seen in the following.

When 𝑚=1, let 𝑔=0. For 𝐹²=𝜃, we have 𝑎3𝜃²+14()𝜃+12𝑘−𝑐+4𝑎𝑐22𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏^−1/2d𝜃=𝑟−𝜉, (27) where 𝑟 is an arbitrary constant. So 3𝑎3()ln𝜃+2𝑘−𝑐+4𝑎𝑐+8𝑎𝜃²+3(2𝑘−𝑐+4𝑎𝑐)34𝑎𝜃+2𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏𝜉.2𝑎=𝑟− (28) Solving (28), we have that when 3(2𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏)/(2𝑎)−9(2𝑘−𝑐+4𝑎𝑐)²/(16𝑎²)>0, 𝜃=32𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏−9()2𝑎2𝑘−𝑐+4𝑎𝑐²16𝑎²×sinh3𝑎𝜉−3()𝑟−2𝑘−𝑐+4𝑎𝑐,8𝑎 (29) when 3(2𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏)/(2𝑎)−9(2𝑘−𝑐+4𝑎𝑐)²/(16𝑎²)<0, 𝜃=32𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏−9()2𝑎2𝑘−𝑐+4𝑎𝑐²16𝑎²×cosh3𝑎𝜉−3()𝑟−2𝑘−𝑐+4𝑎𝑐.8𝑎 (30) Equations (29) and (30) have no asymptotic behavior. When 3(2𝑘𝑐−𝑐²+2𝑎𝑐²+𝑏)/(2𝑎) = 9(2𝑘−𝑐+4𝑎𝑐)²/(16𝑎²), we obtain 1𝜃=2e^{(𝑎/3)(𝑟−|𝜉|)}−3(2𝑘−𝑐+4𝑎𝑐).16𝑎 (31) So 𝑢()=1𝑥,𝑡2e^{(𝑎/3)(𝑟−|𝑥−𝑐𝑡|)}−6𝑘−3𝑐−4𝑎𝑐.16𝑎 (32) These solitary solutions have the first derivative discontinuity at the wave peak.

When 𝑚=2, let 𝑔=0. Consider the following:𝑎6𝜃³+2𝑎𝑐3𝜃²+142𝑘−𝑐+4𝑎𝑐²𝜃+122𝑘−𝑐²+4𝑎𝑐³3+𝑏^−1/2d𝜃=𝑟−𝜉. (33) Take 2𝑘−𝑐+4𝑎𝑐²=0 and 2𝑘−𝑐²+4𝑎𝑐³/3+𝑏=0, then −6𝑎𝑐atanh𝜃+4𝑐𝜉.4𝑐=𝑟− (34) From (34), we have 𝜃=4𝑐tanh²𝑎𝑐6𝜉−𝑟−4𝑐. (35) Therefore, we have the solitary wave solutions of the generalized Camassa-Holm equation as 𝑢(𝑥,𝑡)=4𝑐tanh²𝑎𝑐6(|𝑥−𝑐𝑡|−𝑟)−3𝑐. (36) Take 2𝑘−𝑐+4𝑎𝑐²=8𝑎𝑐²/3 and 2𝑘−𝑐²+4𝑎𝑐³/3+𝑏=0, then 12𝑎𝑐atan𝜃𝜉.2𝑐=𝑟− (37) Similarly, we have 𝑢(𝑥,𝑡)=2𝑐tan²𝑎𝑐(|12𝑟−𝑥−𝑐𝑡|)+𝑐. (38) This result is the same as (28) in [18]. But it is easily to obtain.

When 𝑚=3, let 𝑔=0. Consider the following:𝑎𝜃10⁴+𝑎𝑐2𝜃³+𝑎𝑐²𝜃²+142𝑘−𝑐+4𝑎𝑐³𝜃+122𝑘−𝑐²+𝑎𝑐⁴+𝑏^−1/2d𝜃=𝑟−𝜉. (39) From (39) and 𝑎,𝑐>0, we have seen that 𝐹(d𝐹/d𝜉) is not a polynomial in 𝐹 (compared with Equation (35) in [18]). Take 2𝑘−𝑐+4𝑎𝑐³=0 and 2𝑘−𝑐²+𝑎𝑐⁴+𝑏=0, then −1𝑎𝑐²√atanh10𝑐𝜃+4𝑐²4|𝑐|𝜃²+5𝑐𝜃+10𝑐²=𝑟−𝜉. (40) Solving (40), we obtain 𝜃=−20𝑐1−tanh²𝑝(𝜉|𝑐|)±4151−tanh²𝑝(𝜉)tanh²𝑝(𝜉)×5−8tanh²𝑝(𝜉)⁻¹, (41) where 𝑝(𝜉)=√𝑎|𝑐|(|𝜉|−𝑟). Therefore 𝑢(𝑥,𝑡)=−20𝑐1−tanh²𝑝(|𝑐|×𝑥−𝑐𝑡)+4151−tanh²𝑝(𝑥−𝑐𝑡)tanh²𝑝()×𝑥−𝑐𝑡5−8tanh²𝑝(𝑥−𝑐𝑡)⁻¹+𝑐, (42) or 𝑢(𝑥,𝑡)=−20𝑐1−tanh²𝑝(|𝑐|×𝑥−𝑐𝑡)−4151−tanh²𝑝(𝑥−𝑐𝑡)tanh²𝑝()×𝑥−𝑐𝑡5−8tanh²𝑝(𝑥−𝑐𝑡)⁻¹+𝑐, (43) where 𝑝(𝑥−𝑐𝑡)=√𝑎|𝑐|(|𝑥−𝑐𝑡|−𝑟).

4. Numerical Simulation Examples

Example 1. In (12), take 𝑐=1, ℎ=0, and 𝑎=3. Then the solitary wave solution is 𝑢(𝑥,𝑡)=𝑒^{−|𝑥−𝑡|}.

We want to show figures with peakon feature. But to save space, omitting figures we only give MATLAB program: “𝑥=−10 : 0.001 : 10; 𝑢=exp(−abs(𝑥));plot(𝑥,𝑢)" for 𝑡=0 and “for 𝑛=1 : 101 𝑥(𝑛)=(𝑛−1)/5−10;𝑡(𝑛)=𝑥(𝑛); end [𝑥(𝑛),𝑡(𝑛)]=meshgrid(𝑥(𝑛),𝑡(𝑛)); for 𝑛=1 : 101 for 𝑚=1 : 101 𝑢(𝑛,𝑚)=exp(−abs(𝑥(𝑛)−𝑡(𝑚))); end end mesh(𝑢)" for the 3-dimensional case.

Example 2. In (16), take 𝑐=1, ℎ=0, and 𝑎=24. Then solitary wave solution is 𝑢(𝑥,𝑡)=tanh²(|𝑥−𝑡|−1)−1.

The figures with peakon feature will be constructed by “𝑥=−10 :0.001 : 10;𝑢=(tanh(abs(𝑥)−1)). ^∧2−1;plot(𝑥,𝑢)” for 𝑡=0 and “for 𝑛=1 : 101 𝑥(𝑛)=(𝑛−1)/5−10;𝑡(𝑛)=𝑥(𝑛); end [𝑥(𝑛),𝑡(𝑛)]=meshgrid(𝑥(𝑛),𝑡(𝑛)); for 𝑛=1 : 101 for 𝑚=1 : 101 𝑢(𝑛,𝑚)=(tanh(abs(𝑥(𝑛)−𝑡(𝑚))−1)). ^∧2−1; end end mesh(𝑢)” for the 3-dimensional case.

Example 3. In (20), take 𝑐=1, ℎ=1, and 𝑎=10. Then symmetrically solitary wave solution is𝑢()=𝑥,𝑡−21−tanh²(|𝑥−𝑡|−1)+231−tanh²(|𝑥−𝑡|−1)tanh²|×(|𝑥−𝑡−1)1−4tanh²(|𝑥−𝑡|−1)⁻¹ (44)

For 𝑡=0, when (𝑎=−2,𝑏=2), (𝑎=2.31,𝑏=2.32), and (𝑎=2.32,𝑏=5), respectively, the figures are with peakon feature. 𝑥=  a : 0.0001 : b; 𝑦=(−2.∗(1−tanh(abs(𝑥)−1).^∧2) + 2.∗sqrt(3.∗(1−tanh(abs(𝑥)−1).^∧2).∗tanh(abs(𝑥))))./(1−4.∗(1−tanh(abs(𝑥)−1).^∧2));plot(𝑥,𝑦). Replace 𝑥 by 𝑥(𝑛)−𝑡(𝑚) and the others are similar to Example 2 program for the 3-dimensional case.

5. Conclusions

In this paper, we make simplification of the nonlinear generalized Camassa-Holm equation and give the improved ansatz for the generalized Camassa-Holm equation. The three numerical simulation examples with mATLAB programs demonstrate solitary wave solutions with peakon feature which show our method applicable. This method may be applied to many other nonlinear equations.

Acknowledgments

This work was supported partially by National Natural Science Foundation of China (Grant nos. 10871056 and 10971150) and by Science Research Foundation in Harbin Institute of Technology (Grant no. HITC200708).