Exact Solutions Of Time Fractional Generalized Burgers-Fisher Equation Using Exp function and Exponential Rational Function Methods
EExact Solutions Of Time Fractional GeneralizedBurgers-Fisher Equation Using Exp function andExponential Rational Function Methods
Ramya Selvaraj ∗ and Swaminathan Venkatraman † Department of Mathematics, Srinivasa Ramanujan Centre,SASTRA Deemed to be University, Kumbakonam - 612 001, TamilNadu, India. Discrete Mathematics Laboratory, Department of Mathematics,Srinivasa Ramanujan Centre, SASTRA Deemed to be University,Kumbakonam - 612 001, Tamil Nadu, India.
Abstract
Using modified Riemann-Liouville derivative, the Exp function andExponential rational function methods are implemented to solve the timefractional generalized Burgers-Fisher equation (TF-GBF). The TF-GBFis transformed into a nonlinear ordinary differential equation (NLODE)by applying the transformation of traveling wave. The suggested methodsare then introduced to formulate exact solutions for the resulting equation.The solutions are depicted using 2D and 3D plots.
Keywords:
Fractional derivative; Time fractional generalizedBurgers-Fisher equation; Exp function method; Exponential rationalfunction method
In the year 1695, Leibnitz introduced a fractional calculus. Severalstudies have been carried out on science and engineering related tofractional phenomena such as plasma physics, viscoelastic materials, fluidflow, biology, economics, probability and statistics, polymers, opticalfibers, etc. In recent years, fractional differential equations (FDEs) havegained a considerable popularity among scientists and engineers andfinding solutions for nonlinear FDE (NLFDE) is an indefatigable researchfield [1, 2, 3, 4, 5, 6]. ∗ [email protected] † Corresponding Author: [email protected] a r X i v : . [ n li n . S I] A ug umerous mathematical procedures have been developed andinvestigated to find the exact solutions of the NLFDE. For instance, thesine-cosine method [7, 8], the tanh method [9], Adomian decompositionmethod [10], the sub-equation method [11, 12], variational iterationmethod [13], the first integral method [14, 15] and so on [16, 17, 18].Recently, an Exp-function method, which was proposed by He andWu [19] and consistently studied in [20, 21, 22]. This method waspreviously suggested for evaluating a solution for PDEs. In addition, itwas expanded successfully to FDEs [23, 24, 25, 26, 27].In addition, the exponential rational function method (ERFM)is one of the methods to reveal the exact solutions. At first, it wasoriginated by [28]. Furthermore, it has been executed in many fieldsof engineering and science [29, 30, 31]. From the contribution ofthe prior mentioned schemes, we carry out this analysis to the timefractional generalized Burgers-Fisher equation (TF-GBFE). A nonlinearequation is called the Burgers-Fisher equation and is the combinationof reaction, convection and diffusion process. The properties of theconvective phenomenon of Burgers and Fisher’s diffusion transport aswell as reaction form characteristics are used in the nonlinear equation.The GBFE is used in the field of fluid dynamics. It was also found insome applications including gas dynamics, heat conduction, elasticity, etc.The GBFE with the order of time fractional α and an arbitraryconstants β, γ and δ is given by u αt + βu δ u x − u xx = γu (1 − u δ ) , (1)where α ∈ (0 , Recent studies have shown that the dynamics of many physical processesare precisely described using FDEs with various types of fractionalderivatives. Jumarie has given a different interpretation of the fractionalderivative with a minor modification of the Riemann-Liouville (RL) erivative [32, 33] and it is defined as D αt g ( t ) = − α ) ddt (cid:90) t ( t − ζ ) − α − [ g ( ζ ) − g (0)] dζ, < α < , ( g ( ϑ ) ( t )) ( α − ϑ ) , ϑ ≤ α < ϑ + 1 , ϑ ≥ , (2)where α denotes the order of fractional derivative and g ( t ) denotes acontinuous non differentiable function g : R → R , t → g ( t ). Some notableproperties of modified RL derivative are listed below [34, 35]:Property: 1 D αt t κ = Γ(1 + κ )Γ(1 + κ − α ) t κ − α , κ > . Property: 2 D αt ( pg ( t ) + qh ( t )) = pD αt g ( t ) + qD αt h ( t ) . where g and h are constants.Property: 3 D αt k = 0 , k = constant. For the following problem, these properties can be used.
Let us consider the NLFDE with an unknown function u , the polynomialG of u and its partial derivatives, involving the highest order derivativesand the nonlinear terms, G ( u, D αt u, D βx u, D αt D αt u, D αt D βx u, D βx D βx u, . . . ) = 0 , < α, β < . (3)With the following fractional complex transform [35] with nonzeroarbitrary constants λ and ξ , we transform fractional differential equationinto ordinary differential equation: u ( x, t ) = u ( ζ ) , ζ = λt α Γ[1 + α ] + ξx β Γ[1 + β ] , (4)Reduce Eq.(3) to the following ordinary differential equation (ODE) ofinteger order: H ( u, u (cid:48) , u (cid:48)(cid:48) , u (cid:48)(cid:48)(cid:48) , . . . ) = 0 , (5)where prime denotes a derivative of ζ .With the positive integers g, h, k, and j , also with the unknown constants a q and b p , the Exp function can be expressed in the form: u ( ζ ) = (cid:80) hq = − g a q e qζ (cid:80) jp = − k b p e pζ . (6)We may evaluate the values of h, j and g, k respectively by balancing thehighest order and the lowest order within the Exp function. .1 Exp function for the time fractionalgeneralized Burgers-Fisher equation We apply the following transformation in order to solve the TF-GBFEusing the proposed method: u ( x, t ) = u ( ζ ) , ζ = kx − λt α Γ[1 + α ] (7)Now Eq.(7) in Eq.(1), we get k u (cid:48)(cid:48) + ( λ − kβu δ ) u (cid:48) + γu (1 − u δ ) = 0 . (8)Applying the following folding transformation in Eq.(8) u ( ζ ) = v δ ( ζ ) , (9)we obtain the following NLODE: k δvv (cid:48)(cid:48) + k (1 − δ ) v (cid:48) + ( λ − kβv ) δvv (cid:48) + γδ (1 − v ) v = 0 . (10)According to Exp function method, we assume that the solution of Eq.(10)can be expressed in the form u ( ζ ) = (cid:80) hq = − g a q e qζ (cid:80) jp = − k b p e pζ = a − g e − gζ + · · · + a k e kζ b − h e − hζ + · · · + b j e jζ . (11)Balancing the highest order in Eq.(10) for vv (cid:48)(cid:48) and v v (cid:48) , we have vv (cid:48)(cid:48) = h e (2 k +3 j ) ζ + ... h e (5 j ) ζ + ... (12)and v v (cid:48) = h e (3 k + j ) ζ + . . .h e (4 j ) ζ + . . . = h e (3 k +2 j ) ζ + . . .h e (5 j ) ζ + . . . . (13)Balancing highest order of Exp function in Eq.(12) and Eq.(13), we obtain2 k + 3 j = 3 k + 2 j, k = j. (14)Likewise, balancing the lowest order in Eq.(10) for vv (cid:48)(cid:48) and v v (cid:48) , we have vv (cid:48)(cid:48) = a e − (2 g +3 h ) ζ + . . .a e ( − h ) ζ + . . . (15)and v v (cid:48) = a e − (3 g + h ) ζ + . . .a e ( − h ) ζ + . . . = a e − (3 g +2 h ) ζ + . . .a e ( − h ) ζ + . . . . (16)Balancing lowest order of Exp function in Eq.(15) and Eq.(16), we obtain2 g + 3 h = 3 g + 2 h, g = h. (17)For the choices of k = j = 1 and g = h = 1, Eq.(11) becomes, v ( ζ ) = a e ( ζ ) + a + a − e ( − ζ ) b e ( ζ ) + b + b − e ( − ζ ) . (18) ubstituting Eq.(18) into Eq.(10), we obtain the following cases withdifferent parametric choices of a − , a , a , b − , b and b .Case:1 a − = 0 , a = 0 , b − = 0 , b = 0 , λ = − k − γδ δ , k = − γδβ , Eq.(10) becomes − γδvv (cid:48)(cid:48) − β v (1 + v (cid:48) ) + β v + ( β + γ ) vv (cid:48) + γ ( δ −
1) = 0 , (19)we obtain v ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b , (20)then the exact solution is u ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b δ . (21)Case:2 a − = 0 , a = 0 , b − = 0 , a = b , γ = 2 k δ , λ = − k − γδ δ , k = − βδ δ , Eq.(10) becomes δvv (cid:48)(cid:48) + (2 + (1 + δ ) v (cid:48) ) v − ( δ − v (cid:48) − vv (cid:48) − v = 0 , (22)we obtain v ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + b , (23)then the exact solution is u ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + b δ . (24)Case:3 a − = 0 , a = 0 , b − = 0 , b = 0 , λ = β γδ + γ δβ , k = γδβ , Eq.(10) becomes γ δ vv (cid:48)(cid:48) − γ δ ( δ − v (cid:48) + β ( λ − γδv ) vv (cid:48) − β γδ ( v − v = 0 , (25)we have v ( x, t ) = a e − (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b , (26)then the exact solution is u ( x, t ) = a e − (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b δ . (27) ase:4 a − = 0 , b − = 0 , b = 0 , b = a , δ = 1 , λ = γ δβ , k = γδβ , Eq.(10) becomes γ ( v (cid:48)(cid:48) + v (cid:48) ) − β ( v (cid:48) − v − β v = 0 , (28)we have v ( x, t ) = 1 + a e − (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) a , (29)then the exact solution is u ( x, t ) = a e − (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) a δ . (30)Case:5 a − = 0 , a = b , a = 0 , b − = 0 , λ = − k + kβ + γδ, k = βδ δ , Eq.(10) becomes β δvv (cid:48)(cid:48) − β ( δ − v (cid:48) +(1+ δ )( γ + γδ − β v (cid:48) ) v +( β + γ (1+ δ )) vv (cid:48) − γ (1+ δ ) v = 0 , (31)we obtain v ( x, t ) = b b + b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (32)then the exact solution is u ( x, t ) = (cid:34) b b + b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) (cid:35) δ . (33)Case:6 a − = 0 , a = b , b − = 0 , b = 0 , δ = 1 , λ = − k + kβ + γδ, k = − γδβ , Eq.(10) becomes γ ( v (cid:48)(cid:48) − v (cid:48) ) + β (1 + v (cid:48) ) v − β v = 0 , (34)we acquire v ( x, t ) = 1 + a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b , (35)then the exact solution is u ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b δ . (36) ase:7 a − = 0 , a = 0 , b = 0 , b = 0 , λ = − k − γδ δ , k = − γδ β Eq.(10) becomes γδvv (cid:48)(cid:48) − γ ( δ − v (cid:48) − β + γ ) vv (cid:48) + 2 β (2 + v (cid:48) ) v − β v = 0 , (37)we attain v ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b − , (38)then the exact solution is u ( x, t ) = a e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b − δ . (39)Case:8 a − = 0 , a = 0 , a = b , b = 0 , λ = − k − γδ δ , γ = 2 k (2 + δ ) δ , k = − βδ δ )Eq.(10) becomes δvv (cid:48)(cid:48) +2(2+ δ +(1+ δ ) v (cid:48) ) v − ( δ − v (cid:48) − (4+ δ ) vv (cid:48) − δ ) v = 0 , (40)we get v ( x, t ) = b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + b − , (41)then the exact solution is u ( x, t ) = b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + b − δ . (42)Case:9 a = 0 , a = b , b − = 0 , b = 0 , δ = 1 , λ = γ δ β , k = γδ β Eq.(10) becomes γ ( v (cid:48)(cid:48) + 2 v (cid:48) ) − β ( v (cid:48) − v − β v = 0 , (43)we have v ( x, t ) = 1 + a − e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b , (44)then the exact solution is u ( x, t ) = a − e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) b δ . (45) The Exponential Rational FunctionMethod (ERFM)
We start by considering an NLFDE with the polynomial G of u and itspartial derivatives, that includes the derivatives of the highest order andnonlinear terms B ( u, D αt u, D βx u, D αt D αt u, D αt D βx u, D βx D βx u, . . . ) = 0 , < α, β < . (46)where the role of u is unknown.He and Wu [19] suggested a fractional complex transformation to turnthe fractional differential equation into ordinary differential equation.Using the following fractional complex transform with nonzero arbitraryconstants λ and ξu ( x, t ) = u ( ζ ) , ζ = λt α Γ[1 + α ] + ξx β Γ[1 + β ] , (47)we reduce Eq.(3) to the following ordinary differential equation (ODE) ofinteger order: F ( u, u (cid:48) , u (cid:48)(cid:48) , u (cid:48)(cid:48)(cid:48) , . . . ) = 0 , (48)where prime is the derivative with respect to ζ .The Exp rational function can be expressed as u ( ζ ) = N (cid:88) q =0 a q (1 + e ζ ) q . (49)where constants a q ( a q (cid:54) = 0) to be determined later. Using balancingprincipal, N can be determined.Replace the Eq.(49) in Eq.(48) and compile all the coefficients withthe same power of e mζ ( m = 1 , , . . . , e mζ coefficient on the left side of the ODE, together. The unknown parametersfor a q can be achieved by setting e mζ ( m = 1 , , . . . ,
6) to zero in the listof algebraic equations. To solve the system of equation, an exact solutionis to be developed for non-linear ODE.
We apply the following transformation in order to solve the TF-GBFEusing the proposed method u ( x, t ) = u ( ζ ) , ζ = kx − λt α Γ[1 + α ] , (50)in Eq.(1), we get k u (cid:48)(cid:48) + ( λ − kβu δ ) u (cid:48) + γu (1 − u δ ) = 0 . (51) he following folding transformation is implemented in Eq.(8) u ( ζ ) = v δ ( ζ ) , (52)We obtain NLODE as follows: k δvv (cid:48)(cid:48) + k (1 − δ ) v (cid:48) + ( λ − kβv ) δvv (cid:48) + γδ (1 − v ) v = 0 (53)Balancing vv (cid:48)(cid:48) and v , we obtain N = 2. In accordance with ERFM, wepresume that the solution of Eq.(10) can be formulated as v ( ζ ) = a + a e ζ + a (1 + e ζ ) . (54)We have the following cases when we replace Eq.(54) with Eq.(10) andcollect all terms with the same power with e mζ ( m = 1 , , . . . , a = 1 , a = 0 , a = − , β = 2 k ( δ − δ , γ = − k δ ,λ = 4 k + 2 kβ − γδ , δ = 1 , Eq.(10) turns into v (cid:48)(cid:48) + 5 v (cid:48) + 6 v − v = 0 , (55)we obtain v ( x, t ) = 2 e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (56)then the exact solution is u ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) δ . (57)Case: 2 a = 1 , a = − , a = − − a , γ = − β δ (1 + δ ) , λ = k + kβ + γδ, k = − βδ δ , Eq.(10) transforms into δvv (cid:48)(cid:48) + (1 + δ ) v v (cid:48) + (2 δ − vv (cid:48) − ( δ − v (cid:48) + 2 δv ( v −
1) = 0 , (58)we get v ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα α ] (cid:1) cosh (cid:16) kx Γ[1+ α ] − λt α α ] (cid:17) , (59)then the exact solution is u ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα α ] (cid:1) cosh (cid:16) kx Γ[1+ α ] − λt α α ] (cid:17) δ . (60) ase: 3 a = 1 , a = − , a = − − a , β = − k − k δ + γδ kδ , γ = − k (2 + δ ) δ ,λ = k + kβ + γδ, k = − βδ δ , Eq.(10) becomes δvv (cid:48)(cid:48) − (4 + δ ) vv (cid:48) − ( δ − v (cid:48) − δ )(1 − v ) v = 0 , (61)we obtain v ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (62)then the exact solution is u ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) δ . (63)Case: 4 a = 0 , a = − a , a = − k (2 + δ ) k − βδ , β = 0 , γ = − k δ , λ = k + γδ δ , Eq.(10) becomes δvv (cid:48)(cid:48) − ( δ − v (cid:48) + v − v = 0 , (64)we obtain v ( x, t ) = 4 e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + 4 δe (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (65)then the exact solution is u ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + 4 δe (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) δ . (66)Case: 5 a = 0 , a = 1 − a , a = 0 , γ = β δ , λ = k + γδ δ , k = βδ δ , Eq.(10) becomes − δvv (cid:48)(cid:48) − (2 + δ ) vv (cid:48) + ( δ − v (cid:48) + (1 + δ ) v (cid:48) v − (1 + δ )(1 + v ) v = 0 , (67)we obtain v ( x, t ) = e − (cid:0) kx Γ[1+ α ] − λtα α ] (cid:1) cosh e (cid:16) kx Γ[1+ α ] − λtα α ] (cid:17) , (68) hen the exact solution is u ( x, t ) = e − (cid:0) kx Γ[1+ α ] − λtα α ] (cid:1) cosh e (cid:16) kx Γ[1+ α ] − λtα α ] (cid:17) δ . (69)Case: 6 a = 0 , a = 1 − a , a = − , β = − k ( δ − δ , γ = − k ( k + 2 kδ − βδ ) δ ,λ = k + γδ δ , δ = 1 , Eq.(10) becomes v (cid:48)(cid:48) − v (cid:48) + 6 v − v = 0 , (70)we obtain v ( x, t ) = 1 + 2 e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (71)then the exact solution is u ( x, t ) = e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) δ . (72)Case: 7 a = 0 , a = 0 , a = 1 , β = 4 k + k δ − δλkδ , λ = k (4 + δ ) δ , Eq.(10) becomes δvv (cid:48)(cid:48) + (4 + δ ) vv (cid:48) − ( δ − v (cid:48) − δ )( v − v = 0 , (73)we obtain v ( x, t ) = 11 + 2 e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) , (74)then the exact solution is u ( x, t ) = (cid:34)
11 + 2 e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) + e (cid:0) kx Γ[1+ α ] − λtα Γ[1+ α ] (cid:1) (cid:35) δ . (75) The exact solutions of the TF-GBF have been generated by the Expfunction method. The 2D and 3D plots show the exact solutions throughFigures (1)–(9) with α = 0 . , . , . , and 0 .
78 for different values ofspace variable x and time t . These figures show when x and t increase, thesolution u decreases for the corresponding exact solutions. Fig (9) shows, -
10 10 20 x × × × × u Figure 1: 2D and 3D shape of Eq.(21) - -
10 10 20 x u Figure 2: 2D and 3D shape of Eq.(24) when x and t increase, the solution u decreases for the exact solution ofEq. (45).The ERFM was used to obtain the exact solution of the TF-GBF.The 2D and 3D plots show the exact solutions via Figures (10)–(16) with α = 0 . , . , . , . , . , . , and 0 .
7, for various space variable x and time t . Fig (10) and (12) display the 2D and 3D plots of Eq.(57) andEq.(63) respectively which shows the solution u increases, when x and t increase. Fig (11), (14), (15) and (16) display the 2D and 3D plots ofEq.(60), Eq.(69), Eq.(72) and Eq.(75) respectively as x and t increases, thesolution u decreases. Fig.(13) shows the 2D and 3D plot of Eq. (66) when x and t increase or decrease, the solution u also increases or decreases. In this study, the Exp function and exponential rational function methodsare used to find the exact solution for the time fractional GBF equation.This generates quite few coefficients and some have been taken intoaccount in providing various analytical solutions. We verified all of theTF-GBF equation’s analytic solutions ousing the coefficients and the 2D -
10 10 20 x u Figure 3: 2D and 3D shape of Eq.(27) - -
10 10 20 x u Figure 4: 2D and 3D shape of Eq.(30) - -
20 20 40 x u Figure 5: 2D and 3D shape of Eq.(33)13 - -
10 10 20 30 x × × u Figure 6: 2D and 3D shape of Eq.(36) - - -
10 10 20 30 x × × × × u Figure 7: 2D and 3D shape of Eq.(39) - - -
10 0 10 20 30 x u Figure 8: 2D and 3D shape of Eq.(42)14 - -
10 10 20 30 x u Figure 9: 2D and 3D shape of Eq.(45) - -
10 10 20 x u Figure 10: 2D and 3D shape of Eq.(57) - -
10 10 20 x u Figure 11: 2D and 3D shape of Eq.(60)15 -
10 10 20 x u Figure 12: 2D and 3D shape of Eq.(63) - -
10 10 20 x u Figure 13: 2D and 3D shape of Eq.(66) - - -
10 10 20 30 x u Figure 14: 2D and 3D shape of Eq.(69)16 - -
10 10 20 30 x × - × - × - × - × - × - u Figure 15: 2D and 3D shape of Eq.(72) - - -
10 10 20 30 x u Figure 16: 2D and 3D shape of Eq.(75)17 nd 3D plots are also included.
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