
Citation: Kasia A. Pawelek, Sarah Tobin, Christopher Griffin, Dominik Ochocinski, Elissa J. Schwartz, Sara Y. Del Valle. Impact of A Waning Vaccine and Altered Behavior on the Spread of Influenza[J]. AIMS Medical Science, 2017, 4(2): 217-232. doi: 10.3934/medsci.2017.2.217
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Nonlinear evolution equations (NLEEs) model many complex phenomena in physics including plasma, solid state, chemical and optical fibers, nonlinear optics, fluid mechanics, etc. Exploring exact traveling wave solutions plays a significant role in nonlinear physics. For this purpose, a number of techniques were developed including method of modified Khater [1,2], first integral [3,4], functional variable [5], expansions [6,7] of new generalized (G′/G) [8,9,10], new Φ6-model [11], Jacobi elliptic function [12,13], sine-Gordon [14], bifurcation [15,16], exp-function [17,18], new auxiliary equation [19], exp(-ϕ(ξ))-expansion [20,21], fan sub-equation [22,23], inverse scattering [24], generalized Kudryshov [25,26,27], Hirota's bilinear [28,29], extended direct algebraic [30], Lie group [31].
Consider the (2+1)-dimensional KK equations [32]
9ut+u5x+15uuxxx+752uxuxx+45u2ux+5σuxxy−5σ∂−1xuyy+15σuuy+15σux∂−1xuy=0. | (1.1) |
where σ2=1,∂−1x=∫dx. This equation has been widely applied in many branches of physics like plasma physics, fluid dynamics, nonlinear optics, and so forth. If we take u(x,y,t)=u(x,t), Eq (1.1) becomes the (1+1)-dimensional KK equation [32]
9ut+u5x+15uuxxx+752uxuxx+45u2ux=0, | (1.2) |
In [33], method of exp-function was applied to Eq (1.2). In [32], symmetric method was applied to the nonlinear (2+1)-KK equation.
The method of the present paper, a candid, succinct and efficient technique, considered as a generalization of (G′/G)-expansion technique [34,35,36,37] was developed in [38,39,40,41,42,43,44,45]. Main purpose of this paper is to investigate the applicability of the method to (1+1)-dimensional KK equation which was not considered in the history of research so far.
We shortly overview the method in such a fashion that maintains four remarks and five basic postulates:
Remark I. If we set up
ϕ=G′/G,ψ=1/G, | (2.1) |
in
G″(ξ)+λG(ξ)=β, | (2.2) |
then we must have the relations
ϕ′=−ϕ2+βψ−λ,φ′=−φψ, | (2.3) |
wherein λ and β are parameters.
Remark II. If λ is negative, general solution of (2.2) is:
G(ξ)=D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ. | (2.4) |
and we receive the following relation
ψ2=−λλ2α1+β2(φ2−2βψ+λ), | (2.5) |
wherein D1 and D2 are arbitrary constants and α1=D21−D22.
Remark III. If λ is positive, general solution of (2.2) is:
G(ξ)=D1sin(ξ√λ)+D2cos(ξ√λ)+βλ, | (2.6) |
consequently, we obtain
ψ2=λλ2α2−β2(φ2−2βψ+λ), | (2.7) |
wherein α2=D21+D22.
Remark IV. If λ=0, the general solution of (2.2),
G(ξ)=β2ξ2+D1ξ+D2, | (2.8) |
and therefore we get,
ψ2=φ2−2βψD21−2βD2. | (2.9) |
Now let us consider:
R(u,ut,ux,uy,utt,uxx,uyy,uxt,⋯)=0, | (2.10) |
wherein R is a polynomial function in u and ut=∂u∂t, ux=∂u∂x, uy=∂u∂y, uxx=∂2u∂x2, uyy=∂2u∂y2, uxy=∂2u∂x∂y and so on.
Postulate 1. Consider:
u(x,y,t)=u(ξ),andξ=ηx+ωy+ct, | (2.11) |
wherein η, ω and c are parameters. By traveling wave transformations (2.11), the Eq.(2.10) can be reduced to:
T(u,cu′,ηu′,ωu′,c2u″,η2u″,ω2u″,ηωu″,cηu″,⋯)=0, | (2.12) |
wherein T is a polynomial.
Postulate 2. Let us assume that the following relation is the general solution expressed by a polynomial:
u(ξ)=a0+∑Ni=1(aiφi(ξ)+biφi−1(ξ)ψ(ξ)), | (2.13) |
wherein a0, ai and bi(i=1,2,3,...,N) are the constant coefficients such that a2N+b2N≠0.
Postulate 3. By homogeneous balance, we determine N in Eq (2.13).
Postulate 4. To convert the left-hand-side of Eq (2.12) into a polynomial function in ψ and ϕ, we write Eq (2.13) into Eq (2.12) with Eq (2.3) and Eq (2.5). By solving polynomial, we obtain the system: in a0, ai, bi(i=1,2,3,...,N), λ(<0), β, η, ω, c, D1 and D2. We solve this system with Mathematica. Setting values of above algebraic constants in Eq (2.13), solutions by hyperbolic functions in Eq (2.12) are obtained.
Postulate 5. Similar to Postulate 4, substituting Eq (2.13) into Eq (2.12), using Eq (2.3) and Eq (2.5) (or Eq (2.3) and Eq (2.7)), we obtain the exact traveling wave solutions of Eq (2.12) demonstrated by trigonometric functions.
Let us consider transformation:
u(x,t)=u(ξ),ξ=x+ct, | (3.1) |
wherein c is a parameter, which reduces Eq (1.2) to:
9cu′+u(5)+15uu‴+752u′u″+45u2u′=0. | (3.2) |
According to postulate 2, the positive number N=2 is obtained by balancing between u(5) and u2u′, thus general solutions of Eq (3.2) is:
u(ξ)=a0+a1φ(ξ)+a2φ2(ξ)+b1ψ(ξ)+b2φ(ξ)ψ(ξ), | (3.3) |
whereina0, ai and bi(i=1,2) are constant coefficients such thata2N+b2N≠0(N=1,2), ϕ(ξ) and ψ(ξ) are satisfied by the Eq (2.3). Now, there are three categories of solutions of Eq (3.2):
Category 1: When λ<0 (solutions by hyperbolic functions):
Writing Eq (3.3) with Eq (2.3) and Eq (2.5) into Eq (3.2), Eq (3.2) forms a polynomial in ψ(ξ) and ϕ(ξ). Solving this polynomial, we obtain a system: a0, a1, a2, b1, b2, λ(<0), β, c and α1. Solving this system with Mathematica, we obtain the values of a0 a1, a2, b1, b2, β and c as:
Result 1:
a0=−10λ3,a1=0,a2=−4,b1=4β,b2=±4√β2+λ2α1√−λ,c=−11λ29,β=β. | (3.4) |
Writing these constants from Eq (3.4) into (3.3) and by Eq (2.1) and Eq (2.4), we obtain explicit solutions of Eq (1.2):
u(ξ)=−10λ3+4λ{D1cosh(ξ√−λ)D2sinh(ξ√−λ)}2{D1sinh(ξ√−λ+D2cosh(ξ√−λ)+βλ}2+4β{D1sinh(ξ√−λ+D2cosh(ξ√−λ)+βλ}±4√β2+λ2α1λ{D1cosh(ξ√−λ)D2sinh(ξ√−λ)}2{D1sinh(ξ√−λ+D2cosh(ξ√−λ)+βλ}2 | (3.5) |
wherein ξ=x−11λ2t9 and α1=D21−D22.
In particular, if we choose D1≠0, D2=0 and β=0 in Eq (3.5), we get:
u(x,t)=−10λ3+4λcoth(√−λ(x−11λ2t9)){coth(√−λ(x−11λ2t9))±csch(√−λ(x−11λ2t9))}. | (3.6) |
Similarly, if we choose D2≠0, D1=0 and β=0 in Eq (3.5), we get:
u(x,t)=−10λ3+4λtanh(√−λ(x−11λ2t9)){tanh(√−λ(x−11λ2t9))±isech(√−λ(x−11λ2t9))}, | (3.7) |
wherein i=√−1.
Result 2:
a0=−5λ12,a1=0,a2=−12, b1=β2,b2=±√β2+λ2α12√−λ,c=−λ2144,β=β. | (3.8) |
Explicit solutions of Eq (1.2) are given by:
u(ξ)=−5λ12+λ{D1cosh(ξ√−λ)+D2sinh(ξ√−λ)}22{D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ}2+β2{D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ}±√β2+λ2α1{D1cosh(ξ√−λ)+D2sinh(ξ√−λ)}2{D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ}2, | (3.9) |
wherein ξ=x−λ2t144 and α1=D21−D22.
In particular, if we choose D1≠0, D2=0 and β=0 in Eq (3.9), we get:
u(x,t)=−5λ12+λ2coth(√−λ(x−λ2t144)){coth(√−λ(x−λ2t144))±csch(√−λ(x−λ2t144))}. | (3.10) |
Similarly, if we choose D2≠0, D1=0 and β=0 in Eq (3.9), we get:
u(x,t)=−5λ12+λ2tanh(√−λ(x−λ2t144)){tanh(√−λ(x−λ2t144))±isech(√−λ(x−λ2t144))}, | (3.11) |
wherein i=√−1.
Result3:
a0=−11λβ2+8λ3α112(β2+λ2α1),a1=0,a2=−1,b1=β,b2=0,c=−λ2(β4−28λ2β2α1+16λ4α21)144(β2+λ2α1)2,β=β. | (3.12) |
wherein β2+λ2α1≠0.
We get explicit solutions of Eq (1.2) as:
u(ξ)=−11λβ2+8λ3α112(β2+λ2α1)+λ{D1cosh(ξ√−λ)+D2sinh(ξ√−λ)}2{D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ}2+β{D1sinh(ξ√−λ)+D2cosh(ξ√−λ)+βλ}, | (3.13) |
wherein ξ=x−λ2t(β4−28λ2β2α1+16λ4α21)144(β2+λ2α1)2 and α1=D21−D22.
In particular, if we choose D1≠0, D2=0 and β=0 in Eq (3.13), we get:
u(x,t)=−2λ3+λcoth2(√−λ(x−λ2t9)). | (3.14) |
Similarly, if we choose D2≠0, D1=0 and β=0 in Eq (3.14), we get:
u(x,t)=−2λ3+λtanh2(√−λ(x−λ2t9)). | (3.15) |
Category 2: For λ>0, (i.e. trigonometric functions),
According to Postulate 5, if we execute as the category 1, we attain the values of a0, a1, a2, b1, b2, β and c as the following results:
Result 1:
a0=−10λ3,a1=0,a2=−4,b1=4β,b2=±4√−β2+λ2α1√λ,c=−11λ29,β=β. | (3.16) |
Writing constants in Eq (3.16) into Eq (3.3) and by Eq (2.1) and Eq (2.6), we get explicit solutions of Eq (1.2):
u(ξ)=−10λ3−4λ{D1cos(ξ√λ)−D2sin(ξ√λ)}2{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}2+4βD1sin(ξ√λ)+D2cos(ξ√λ)+βλ±4√−β2+λ2α2{D1cos(ξ√λ)−D2sin(ξ√λ)}{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}2, | (3.17) |
wherein ξ=x−11λ2t9 and α2=D21+D22.
Result 2:
a0=−5λ12,a1=0,a2=−12,b1=β2,b2=±√−β2+λ2α12√λ,c=−λ2144,β=β. | (3.18) |
We get explicit solutions of Eq (1.2) as:
u(ξ)=−5λ12−λ{D1cos(ξ√λ)−D2sin(ξ√λ)}22{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}2+β2{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}±√−β2+λ2α2{D1cos(ξ√λ)−D2sin(ξ√λ)}2{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}2,. | (3.19) |
wherein ξ=x−λ2t144 and α2=D21+D22.
Result 3:
a0=11λβ2−8λ3α212(−β2+λ2α2),a1=0,a2=−1,b1=β,b2=0,c=−λ2(β4+28λ2β2α2+16λ4α22)144(−β2+λ2α2)2,β=β. | (3.20) |
wherein −β2+λ2α2≠0.
We get explicit solutions of Eq (1.2) as:
u(ξ)=11λβ2−8λ3α212(−β2+λ2α2)−λ{D1cos(ξ√λ)−D2sin(ξ√λ)}2{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}2+β{D1sin(ξ√λ)+D2cos(ξ√λ)+βλ}, | (3.21) |
wherein ξ=x−λ2t(β4+28λ2β2α2+16λ4α22)144(−β2+λ2α2)2 and α2=D21+D22.
Category 3: For λ=0, (i.e.rational functions),
According to Postulate 5, if we execute as the category 1, we attain the values of a0, a1, a2, b1, b2, β and c as the following results:
a0=−β24(−D21+2βD2),a1=0,a2=−1,b1=β,b2=0,c=−5β416(−D21+2βD)2,β=β. | (3.22) |
We get explicit solutions of Eq (1.2) as:
u(ξ)=−β24(−D21+2βD2)−(βξ+D1)2(β2ξ2+D1ξ+D2)2+β(β2ξ2+D1ξ+D2), | (3.23) |
wherein ξ=x−5β4t16(−D21+2βD)2 and −D21+2βD≠0.
If we set up the particular values of the arbitrary constants if we choose D1, D2 and βin the above Eq (3.17), Eq (3.19), Eq (3.21) and Eq (3.23), we attain abundant new explicit wave solutions of KK equation which are unexposed for minimalism of length of the paper.
We obtained new explicit solutions for the (1+1)-dimensional KK equation. We achieved solitary wave solutions for analogous traveling wave solutions of Eq (1.2). These affluent solutions including bell and anti-bell solitons, kink and anti-kink solitons, periodic and rational functions of KK equation indicate that double (G′/G,1/G)-expansion technique is more powerful than the method of (G′/G,1/G)-expansion. Comparing the solutions with the ones in [33], we presume that all the solutions are renewed which are un-indicted elsewhere. Our mentioned method is more powerful and also an offering method to demonstrate many higher order nonlinear PDEs. We will investigate the applicability of the method to (2+1)-dimensional KK equation in a future extension of the present work.
The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485).
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
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