# Conditional Lie-Bäcklund Symmetry Reductions and Exact Solutions of a Class of Reaction-Diffusion Equations

Authors: Xinyang Wang; Junquan Song
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

The method of conditional Lie-Bäcklund symmetry is applied to solve a class of reaction-diffusion equationsut+uxx+Qxux2+Pxu+Rx=0, which have wide range of applications in physics, engineering, chemistry, biology, and financial mathematics theory. The resulting equations are either solved exactly or reduced to some finite-dimensional dynamical systems. The exact solutions obtained in concrete examples possess the extended forms of the separation of variables.


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## Body

## 1. Introduction

In this paper, we analyze a class of reaction-diffusion equations (RDEs)(1)ut+uxx+Qxux2+Pxu+Rx=0,which admit certain conditional Lie-Bäcklund symmetries (CLBSs). Equation (1) can be simplified from the following RDEs:(2)ut+Fxuxx+Qxux2+Gxux+Pxu+Rx=0,which have wide range of applications in physics, engineering, chemistry, biology, and financial mathematics theory [1–5]. Here the coefficient functions depend upon the variable x which typically represents the value of the underlying asset, such as the price of a stock upon which an option is placed. There exist several special cases of (2), say, the Black-Scholes-Merton equation, the Longstaff equation [6], the Vasicek equation [7], and the Cox-Ingersoll-Ross equation [8], in which Q(x) is zero. In order to keep the computations as simple as and consistent with this requirement, authors in [4] transform (2) into (1) by using an equivalence point transformation, namely, by means of (3)ut,x=Ut,X+gx,X=hx, with (4)h′x2=1Fx,g′x=F′x-2Gx4Qx, Equation (2) can be transformed into (5)Ut+Uxx+QxFxUx2+PxU+Rx+gxPx-Gx24Qx+F′x216Qx,-FxG′x2Qx+FxGxQ′x2Qx2-FxF′xQ′x4Qx2+FxF′′x4Qx=0, where the primes denote differentiation with respect to x. Then, with the inverse transformation x=h-1(X), this equation can be performed as (1) on reversion to lower case variables and redefinition of the coefficient functions as appropriate. When Q(x)=constant and P(x)=0, (1) becomes (6)ut+uxx+kux2+Rx=0, which can simplify the linear form of (7)vt+vxx+fxvx=0, under the transformation u=log⁡(v)/k+∫fxdx/2k+γt with 4kR(x)=-(2f′+f2+4kγ).It is known that symmetry reductions and exact solutions play important roles in the study of RDEs. The conditional Lie-Bäcklund symmetry (CLBS) method introduced by Zhdanov [9] and Fokas and Liu [10, 11] firstly has been proved to be very powerful to classify equations or specify the functions appeared in the equations and construct the corresponding group invariant solutions. Furthermore, authors have shown that CLBS is closely related to the invariant subspace; namely, exact solutions defined on invariant subspaces for equations or their variant forms can be obtained by using the CLBS method [12–24].Motivated by the form of (1), we set the following second-order nonlinear CLBSs:(8)η=uxx+a1xux2+a2xux+a3xu+a4x,which are very powerful to specify the functions appeared in (1) and construct the corresponding exact solutions. The remainder of this paper is organized as follows. In Section 2, some equations of the form (1) admitting CLBSs generated by (8) are obtained. CLBS reductions and exact solutions of two concrete examples are used to illustrate the results. Section 3 is devoted to conclusions and discussions.


## 2. Equations Admitting CLBSs and Two Examples

To consider further, we need the following proposition derived in [9, 10].Proposition 1.
RDEs (1) admit CLBSs (8) if and only if η′E=0 whenever u satisfies (1) and η=0, where the prime denotes the Gateaux derivative, that is, η′E=d/dϵ|ϵ=0η[u+ϵE] and E=-uxx-Q(x)ux2-P(x)u-R(x).A direct computation from the above proposition yields(9)η′E=2a12a1-Qux4+2a1′Q-2a1+2a1Q′+2a1a2-2Qa2ux3+2a2′Q-a1+a1-P-4Qa4+2a22+4a1a4+a23Q′-2Qa2-4a1′+a3Q-Q′′+a1′′ux2+4a1a3a1-Quux2+4a1a2a3-2a1P′+2Qa3′-4Qa2a3-4a1′a3+4Q′a3uux+2a3′-2P′+a2′′-2a1R′+2Qa4′-2a2a2′+4a4-Qa2+a1a2-a1′+Q′ux+2a32a1-Qu2+-a2P′-2a2′a3+a3′′-P′′-4Qa3a4+4a1a3a4u+a4P-2Qa4-2a2′+2a1a4+a4′′-a3R-a2R′-R′′=0.To vanish all the coefficients of (9), we have the following overdetermined system:(10)2a12a1-Q=0,2a1′Q-2a1+2a1Q′+2a1a2-2Qa2=0,2a2′Q-a1+a1-P-4Qa4+2a22+4a1a4+a23Q′-2Qa2-4a1′+a3Q-Q′′+a1′′=0,4a1a3a1-Q=0,4a1a2a3-2a1P′+2Qa3′-4Qa2a3-4a1′a3+4Q′a3=0,2a3′-2P′+a2′′-2a1R′+2Qa4′-2a2a2′+4a4-Qa2+a1a2-a1′+Q′=0,2a32a1-Q=0,-a2P′-2a2′a3+a3′′-P′′-4Qa3a4+4a1a3a4=0,a4P-2Qa4-2a2′+2a1a4+a4′′-a3R-a2R′-R′′=0.Solving this system, we can obtain the unknown functions in (1) and the corresponding CLBSs (8). From the first and seventh equations of system (10), it is apparent that the solutions can be divided into two cases including a1(x)=Q(x) and a1(x)=a3(x)=0,a1(x)≠Q(x).Case 1 (a1(x)=Q(x)=1).
Substitutinga1(x), Q(x) into the third equation of the above system (10), we can obtain a3(x)=P(x). Furthermore we can derive a2(x)=μ1/P1/2, P(x)≠0, or P(x)=0 from the eighth equation of (10) by substituting all these.
WhenP(x)=0, we can obtain R(x)=1/2a2′+a4-1/2a22+μ1 from the sixth equation of (10). Then substituting R(x) into (10) again, we obtain a2=0 or a4=1/4a22+1/2a2′+a2′2/4a22-a2′′/2a2+μ2/a22, resulting in these two cases:(i)
ut+uxx+ux2+a2′-1/4a22+a2′2/4a22-a2′′/2a2+μ2/a22+μ1=0,(11)η=uxx+ux2+a2ux+14a22+12a2′+a2′24a22-a2′′2a2+μ2a22,a2≠0;(ii)
ut+uxx+ux2+R(x)=0, (12)η=uxx+ux2+Rx-μ1.When P(x)≠0, substituting a2(x)=μ1/P1/2 into the sixth equation of system (10), it arrives at R(x)=-μ12/2P-μ1P′/4P3/2+a4+μ2. Then this system is reduced to (13)4μ1P′′′P2-P′′18μ1P′P+4μ12P3/2+μ1P′4P3-8μ12P+16P2a4+10μ12P1/2P′2+15μ1P′3+8μ12P7/2-16μ2P9/2-16μ1P3a4′=0. However, there are two unknown functions P(x), a4(x) to be determined with only one determining equation, resulting in(14)a4=116P7/2∫-8P′P3/2μ12+10P′2P-4P′′P2+8P4μ1+4P′P7/2+15P′3P1/2+4P′′′P5/2-18P′P′′P3/2-16P5μ2μ1dx+μ3P,μ1≠0. Noting that when μ1=0, it leads to η=uxx+ux2+P(x)u+R(x)=0, which reveals ut=0 and implies that u(x,t) is independent of t which we are not interested. Therefore, it is impossible to obtain the general solutions. However we can solve it explicitly for several special cases.
IfP(x)=k, we derive a4=k/2μ1-2μ2k/μ1x+kμ3. Then let μ1=sk, μ2=-sa/k-s/2, μ3=as/k2+β/k; the solution becomes(i)
ut+uxx+ux2+ku+ax+β=0,(15)η=uxx+ux2+sux+ku+ax+β+sak.If P(x)=eλx, then a4=-2μ2/λμ1e3/2λx+1/4λx+μ3eλx-μ1/λe1/2λx-μ1λ/4e-1/2λx+μ12/4e-λx-λ2/16. For λ≠0, consequently η=0 can not be easily solved. If P(x)=xλ (λ≠0,±2), then a4=λ/4xλ[ln⁡(x)+μ3]+μ12/4x-λx-2μ2/(λ+2)μ1x3/2λ+1-μ1/λ-2x1/2λ+1-μ1λ/4x-1/2λ-1-λ2+4λ/16x2 leading to the fact that η=0 also can not be easily solved.
IfP(x)=x2, then a4=-μ2/2μ1x4+2μ1+1x2ln⁡(x)+(μ1+1)(μ1-3)/4x2+μ3x2. To make η solvable, we let μ1=-1, μ2=2a, μ3=β; then the solution becomes(i)
ut+uxx+ux2+x2u+x4+2a+βx2=0,(16)η=uxx+ux2-1xux+x2u+ax2+βx2.If P(x)=x-2, then a4=μ12/4x2+3μ1/4-1/2+μ2/μ1ln⁡(x)/x2+μ3/x2. To make η solvable, we let μ1=a, μ2=-1/2a, μ3=β; then the solution becomes(i)
ut+uxx+ux2+x-2u-aax2-3/4+β/x2=0,(17)η=uxx+ux2+axux+x-2u+aax2+34+βx2.Case 2 (a1(x)=Q(x)≠costant).
By similar calculation, we can geta2(x)=μ2a1/a1′, a3(x)=μ1a1′2/a12, and P(x)=-μ2+μ1a1′2/a12, then system (10) is transformed into the following equations: (18)μ2a1a1′′′a1′-2μ2a1a1′′2+μ2a1′′a1′2+2a1R′a1′3-2a1a4′a1′3+2μ22a1a1′2-2μ22a12a1′′=0,2μ2a13a1′′a4-μ1a1′4R+μ1a1′4a4-R′′a12a1′2+a4′′a12a1′2-3μ2a12a1′2a4-μ2a13a1′R′=0. This system is so complicated that we can not even obtain general formations. Consequently, special solutions of the above system are given as follows:(i)
ut+uxx+x-2ux2+ku+3/4kx2+a=0,(19)η=uxx+x-2ux2+k2xux+116k2x4+34kx2;(ii)
ut+uxx+x2ux2+ku+k4x2-k2ln⁡x/4+a=0,(20)η=uxx+x2ux2-k2xux+k4x2+k216;(iii)
ut+uxx+csc2λxux2+2λ2u+a=0,(21)η=uxx+csc2λxux2+λtan⁡λxux,λ≠0;(iv)
ut+uxx+sec2λxux2+2λ2u+a=0,(22)η=uxx+sec2λxux2-λcot⁡λxux,λ≠0;(v)
ut+uxx+eλxux2-λ2u+a=0,(23)η=uxx+eλxux2+λux,λ≠0.Case 1 (a1(x)=a3(x)=0,P(x)=costant(=k)).
In this case, after a lengthy calculation, we have(24)2Qa2′-Q′′+3a2Q′-2Qa22=0,-4Qa2a4+4Q′a4+2Qa4′-2a2′a2+a2′′=0,-2Qa42-R′′+a4′′+ka4-a2R′-2a2′a4=0. Special solutions of the above system are given as follows:(i)
ut+uxx+x-2ux2+ku+a+βx2=0,(25)η=uxx-x-1ux;(ii)
ut+uxx+x2ux2-3/4x4ln2x-1/x4ln⁡x+5/4x4-γ2/4ln2x+γ/x2ln2x+aln2x+β=0,(26)η=uxx+ln⁡x-1xlnxux+2ln2x-ln⁡x-2+2γx22x4ln2x;(iii)
ut+uxx+xλux2+1/16λ3λ+4x-2-λ+ax-1/2λ+1+β=0,(27)η=uxx+λ2xux+182λ+λ2x-2-λ,λ≠0,±2;(iv)
ut+uxx+xλux2+λ+3λ-1/4x-2-λ+ax2-λ+β=0,(28)η=uxx+λ-1xux+λ-1x-2-λ,λ≠0,±2;(v)
ut+uxx+ux2+ku-γ2-2γ/4x2-1/2γklnx+ax2+β=0,(29)η=uxx-1xux+γx2.Case 2 (a1(x)=a3(x)=0,P(x)≠costant).
By similar calculation, we can obtaina2(x)=-P′′/P′ and Q(x)=μ1P+μ2/P′2, then system (10) is reduced to (30)2P′μ1P+μ2a4′+4μ1P′2-μ1PP′′-μ2P′′a4+P′′′P′′P′-P′′′′P′2-2P′4=0,P′2a4′′-2μ1P+μ2a42+2P′′′P′-2P′′2+PP′2a4+P′′P′R′-P′2R′′=0. We solve out the result with special cases listed in the following:(i)
ut+uxx+ux2+ax+βu-1/12a2x4+1/6aβ-3γx3+1/2γβ-2γx2+θ1x+θ2=0,(31)η=uxx+ax+γ;(ii)
ut+uxx+ux2+kx2u-γ2-2γ/4x2-1/2γkx2ln⁡x+ax2+β=0,(32)η=uxx-1xux+γx2+k2x2;(iii)
ut+uxx+eλxux2+a+βe-λxu+β+aβ/4λ2e-2λx-γ2/λ2eλx+γx/λa+βe-λx+θ1e-λx+θ2=0,(33)η=uxx+λux+12βe-2λx+γ,λ≠0;(iv)
ut+uxx+eλxux2+a+βe-1/2λxu+3/2+-6γ+2a/3λ2βe-3/2λx-β2/3λ2e-2λx+2aγ-4γ2/λ2+2γe-λx+θ1e-1/2λx+θ2=0,(34)η=uxx+λ2ux+βe-3/2λx+γe-λx,λ≠0;(v)
ut+uxx+sec2λxux2+a+βsin⁡λxu+2γ+β2+6γ2-3aγ/6λ2cos2λx-β2/12λ2cos4λx+3/2+3γ-a/6λ2βsin⁡λxcos2λx+θ1sin⁡λx+θ2=0,(35)η=uxx+λtan⁡λxux+βsin⁡λxcos2λx+γcos2λx,λ≠0;(vi)
ut+uxx+csc2λxux2+a+βcos⁡λxu+2γ+β2+6γ2-3aγ/6λ2sin2λx-β2/12λ2sin4λx+3/2+3γ-a/6λ2βsin2λxcos⁡λx+θ1cos⁡λx+θ2=0,(36)η=uxx-λcot⁡λxux+βsin2λxcos⁡λx+γsin2λx,λ≠0.Thus we have obtained 21 classes of equations (11)-(36) with form (1) which admit certain second-order CLBSs. To reduce and solve equations by means of corresponding CLBSs, one solves η=0 to obtain u as a function of x with x-independent integration constants and then substitutes this solution into (1) to determine the time evolution of these constants. Next, we only present two examples to illustrate this approach.Example 1.
Equation(37)ut+uxx+x-2ux2+ku+34kx2+a=0 admits the CLBS (38)η=uxx+x-2ux2+k2xux+k216x4+34kx2. The corresponding solutions are given by (39)ux,t=-116kx4+C1tx+C12tln⁡x-C1t+C2t, where C1(t) and C2(t) satisfy the finite-dimensional dynamical system (40)C1′=-k2C1,C2′=-kC2+k2C12-a. Exact solutions can be obtained as (41)C1=c1e-k/2t,C2=12c12kt-akekt+c2e-kt with two arbitrary constants c1 and c2.Example 2.
Equation(42)ut+uxx+x-2ux2+ku+a+βx2=0 admits the CLBS (43)η=uxx-1xux. The corresponding solutions are given by (44)ux,t=C1t+C2tx2, where C1(t) and C2(t) satisfy the system (45)C1′=-kC1-2C2-4C22-a,C2′=-kC2-β. This dynamical system can be solved and exact solutions are (46)C1=1k34c22k2e-2kt+8c2tβ-2c2tk+c1kk2e-kt-ak2+2βk-4β2,C2=-βk+c2e-kt with two arbitrary constants c1 and c2. Here the solutions u(x,t) preserve the forms of the separation of variables u=ϕ1(t)+ϕ2(t)ψ(x), which are associated with the invariant subspace L{1,x2} mentioned in [3] and the references therein.


## 3. Conclusions and Discussions

In this paper, we have discussed RDEs (1) by means of CLBS with characteristic (8). The key for this method is to determine presumably the form of the CLBS. For (1), we found that nonlinear CLBS (8) is very effective, which can yield some interesting symmetry reductions and exact solutions. Two examples are considered to illustrate this method in terms of the compatibility of CLBSs and the governing equations. Generally speaking, the obtained solutions cannot be derived within the framework of Lie’s classical method and nonclassical method.In addition, it must be pointed out that, for the corresponding equations with certain fractional derivative [25], (47)∂α∂tαu+uxx+Qxux2+Pxu+Rx=0,0<α<1, we can do similar work including CLBS classification, reductions, and exact solutions.



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