A Lie Symmetry Classification of a Nonlinear Fin Equation in Cylindrical Coordinates

and Applied Analysis 3 f (x) N 2 x 2 φ + N 2 ux 2 ξ(f (x)) x + x 2 φ t − f (x) N 2 ux 2 φ u − k (u) xφ x + f (x) N 2 ux 2 τ t − k (u) x 2 φ xx − k (u) φ yy = 0, (17) x 2 η x + ξ y = 0, (18) φk(u) u − 2k (u) ξ x + k (u) τ t = 0, (19) −2k (u) ξ + xφk(u) u − 2xk (u) η y + k (u) xτ t = 0. (20) To determine the unknown functions, η, τ, and φ, we solve the above coupled system of differential equations by first considering (19). Differentiating this equation twice with respect to u leads to the following expression:


Introduction
A heat exchange in industrial applications such as compressors, air conditioners, and air craft engines is achieved through the surfaces called fins.Fins which are of different shapes are described by a variety of mathematical models [1].Moitsheki [2] has recently discussed the problem of temperature profiles and heat transfer per fin length by considering 2-dimensional Laplace equation given in the following form: In this problem, the author uses the method of separation of variables and the Newton Raphson method for computing the temperature profile and heat transfer.More recently, Pakdemirli and Sahin [3,4] have studied the problem where () is conductivity,  is the fin constant, and () is heat transfer coefficient using the Lie symmetries of the above governing partial differential equation (2).This method was introduced by Lie [5] to find exact solutions of a number of linear and nonlinear PDEs arising in engineering, mathematical, and biological sciences.A good account of the Lie approach and its applications to the differential equations can be found in [6][7][8][9].Bokhari et al. [10] have recently studied the above nonlinear fin equation (2) using Lie symmetry approach and give some new interesting exact solutions.Moitsheki et al. [11] obtained some exact solutions of (2) by considering a power law form of the thermal conductivity.Moitsheki [2] has also considered a radial onedimensional steady state heat transfer problem by assuming the fin equation in the following form: where  and ℎ are the nonuniform thermal conductivity and heat transfer coefficients, respectively, depending on the temperature.Some exact solutions are obtained for thermal diffusion fin with a rectangular profile and a hyperbolic profile.Continuing their investigation, Moitsheki [12] has also studied a steady heat transfer problem of a longitudinal fin with triangular and parabolic shapes by considering the following problem: where   represents the profile area, () represents fin profile, and  and ℎ, respectively, represent nonuniform thermal conductivity and heat transfer coefficient depending on the temperature.
Vaneeva et al. [13] have performed a Lie group classification of the nonlinear (1 + 1) dimensional fin equation and presented exact solutions considering the steady state heat transfer problem for a rectangular fin.Moitsheki and Rowjee [14] have discussed the (1 + 1) dimensional problem in which  is the dimensionless temperature,  is the internal heat generation function, and  is the thermal conductivity.
Employing the Lie symmetry analysis to classify the internal heat generating function, they obtained some reductions of the fin equation as well as exact solutions.Some authors have also considered the two-dimensional problem with  = 0 in (5) by assuming constant thermal conductivity [15,16].
In the same series of studies Moitsheki and Harley [17] have considered a two-dimensional pin fin equation with length  and radius , having the form which can be rewritten in the form where  and , respectively, represent radial and angular coordinates.Also , , , and  in (8), respectively, represent the dimensionless temperature, the thermal conductivity, the heat transfer, and the fin parameter.Using Lie symmetry analysis, we present a complete classification of () and () and obtain exact solutions in cases of interest.The results presented in this paper are more general than previously studied cases in the literature [2-4, 10-12, 14-17].The plan of this paper is as follows.In the next section, we perform a complete symmetry analysis of the fin equation.In Section 3, we present classification of  and  according to Lie point symmetries.In Section 4, we present exact solutions whenever possible and in other cases reduce the fin equation to ODE.A brief summary of this work is given in the last section.

Symmetry Analysis of the Fin Equation
In order to classify solutions of the Fin equation, we use the well-known Lie symmetry method [8].This method is based upon finding Lie point symmetries of the PDEs that leave them invariant.In order to derive symmetry generators of the Fin equation, we consider one parameter Lie point transformation that leaves it invariant.The transformation [16] x =   +   (, , , ) +  ( 2 ) ,  = 1, . . ., 4, ( where   = x  /| =0 defines the symmetry generator associated with (9) given by The prolonged symmetry generator associated with (8) has the following form: where 0 represents  and 1 and 2, respectively, represent  and , and the coefficients,   and   , of the derivatives with respect to dependent variables in (11) are evaluated using the expressions At this stage we use the Lie symmetry criterion by requiring that ( 8) is invariant under the prolonged symmetry generator given in (11) modulu Equation ( 8) itself.Mathematically, this requirement is given by Using ( 12) and comparing terms involving derivatives of the dependent function , we obtain the following determining equations: To determine the unknown functions, , , and , we solve the above coupled system of differential equations by first considering (19).Differentiating this equation twice with respect to  leads to the following expression: Using ( 14) into (21), it reduces to We proceed from above equation to obtain complete classification of both  and  as shown in the next section.

Classification
In order to find a complete classification of solutions of (8), we note that the following three cases arise from ( 22): (1) For obtaining a complete classification, we consider all the three cases one by one.Since procedure of classification in all the three cases is similar, we give detailed procedure in the first case and only give results in the remaining cases.To begin the classification, we proceed as follows.
Case 1. Solving equation (/  )  = 0, for () instantly yields where , , and  are integration constants.Using (23) into (19) immediately gives Using ( 23) and ( 24) in (20), we obtain a differential relation in  and  given by Differentiating (24) twice and (25) once with respect to  and using (18) in the resulting expressions give At this stage we use ( 27) and ( 26) into (20), to get Differentiating (28) with respect to ", " while keeping (23) in mind, we obtain Keeping in mind that , , and  in the above equation are nonzero, we conclude that the above equation is satisfied only when (the case  = 1 is not to be considered as it becomes a special case of (1.3) that is dealt with later).Note that (30) is a separable DE and can be easily solved to find  given by To require consistency of  found above, we use (31) into (28).This suggests that (28) is satisfied when the following differential constraint is met: The above equation implies that   (, ) = () and (, ) = ().Using these results into (32) gives To determine  we use (33) into (27).This leads to the following second-order differential constraint: To solve the above equation, we differentiate it with respect to  to get We then put (35) into (34), to get a second-order linear differential equation, The above equation is solved to get At this stage we use ( 35) and (37) into (33), to infer that Having determined  completely, we now find .For this purpose, we use (38) there.This yields an expression for  as given below: To determine  1 , we use (39) and ( 38) into (25) to obtain Using above value of  1 into (39) and (24), respectively, becomes  (, , ) = − 3 sin  +  4 cos  +  1  ln  +  1 () , (41) We now use all the above results into (4), which suggests that it is satisfied if the following condition is met: From the above equation, we immediately infer that  1 () =  5 and  1 = 0. Therefore, (38), (41), and (42) take the form To determine consistency, we now use the last of the determining equations, that is, (17).This requirement gives  ()  2 (2 5 −   ) +  2   (− 3 sin  +  4 cos  +  5 ) In order for (45) to be satisfied, we proceed as follows: comparing coefficients of (), this equation gives Differentiation of the above equation with respect to  gives The above equation is a separable DE in  and  and can be written as solving (48) implies that () =  ( a constant) whereas the  becomes To require consistency, we use (49) with () =  ̸ = 0 in (46), to obtain From (50) four cases arise, namely, The commutation relations for each of the above symmetry generators are listed in Table 1.

Case 1.2 (𝑘(𝑢) = 𝛾(𝛼𝑢)
Accordingly, the  symmetry generators associated with above infinitesimals are given by The commutation relations for these generators are given in Table 2.
Case 1.3 (() = ( + ) 1/ and () = ).Using the conditions with the values of  and  of this case into (49), (45), and (44), the expressions for , , , and  take the forms and the corresponding generators are As before the commutation relations form a closed algebra and are represented in Table 3. ( With the above infinitesimals there are five generators associated: The commutation relation satisfied by the above five generators is given in Table 4. Case 2 (2  −   = 0).In this case the system of determining equations given by ( 14)-(20) becomes Following the procedure adopted in Case 1, we easily find that The above infinitesimals satisfy all the equations in the system (59) except (iv).Using (60) in ( 59)-(iv), we obtain, From (61) two cases arise: (2.1) Case 2.1.In this case () and () in system (59) are arbitrary functions and the general expressions of , , , and  take the forms The two commuting symmetry generators in this case are Case 2.2.Here () is an arbitrary function and () = / 2 .The general expressions of , , , and  are The three symmetry generators associated with (64) are The commutation relation satisfied by three generators is presented in Table 5.

Reduction under Two-Dimensional Subalgebra and Exact Invariant Solutions
Case 1.In this section, we present solutions of (8) via reductions.These reductions are obtained by the similarity variables obtained through symmetry generators.To perform reductions of (8), we first consider two symmetry generators from Table 1, X 1 and X 2 , that span an abelian subalgebra.
To start reduction, we first consider X 1 .The characteristic equation corresponding to this generator is Solving the above equation, it is straight forward [6] to find that it yields the similarity variables  =  cos  and  =  with (, ) = .Replacing  in (8) To proceed further, we first transform X 2 in terms of new variables , , and .Thus, X2 = /.The similarities corresponding to this generator are  =  and V() = .This reduces (69) to a first-order differential equation given by Solving this equation, we immediately find that V() = exp(− 2 ), which in original coordinates becomes Case 2. Here, we first consider the generators X 1 and X 3 given in Table 3, satisfying [X 1 , X 3 ] = 0. Following procedure followed in the previous case, the generator X 1 reduces ( 8) to (69).In the light of X 1 , the X 3 transforms to X3 = (1/ 2  4 ) exp(− 2 )(/)+(−/ 2 ) exp(− 2 )(/), which gives  =  with  = −exp(− 2 )V().In the light of these similarity variables, (69) reduces to the following ODE: Choosing  =  2 , the above solution takes the form Writing above in original coordinates, it becomes The graphical profile of the above solution is given in Figure 1.
For constant  the same solution is plotted and the solution depicts a saddle point behavior as shown in Figure 2.
Case 3. In this case, we consider the two generators X 3 and X 6 that satisfy [X 3 , X 6 ] = 0 as shown in Table 2. Since the two generators commute, we can start reduction by either X 3 or X 6 .First considering X 3 , the characteristic equation becomes The similarity variables corresponding to above equation become  = ,  = , and  =  2 .These variables reduce (8) to a PDE of the form Using similarity variables transformation obtained from X 3 , X 6 transforms to X6 = /.This leads to the new coordinates  =  and V() = .In the light of these similarities, (76) transforms to Choosing  2  = 1,  = 1, and  = −1, the above equation takes the form giving exact solution (, , ) = exp 2.
Case 5.In this case, we consider the two symmetry generators X 1 , X 3 which satisfy a commutative relationship [X 1 , X 3 ] = 0 as shown in Table 3.First considering X 1 , we obtain the similarity variables  = ,  =  −1/2 , and  = .Equation (8)  (84) First writing  3 into X3 = / and solving the resulting characteristic equation the similarity variables are given by  =  and V() = .These variables can be used to recast (84) to an ODE, Choosing (V) = 1, the solution of (85) becomes Reduction in all the remaining cases is given in the form of Tables 6, 7, and 8.

Summary and Discussion
A complete classification of the Lie point symmetries of the nonlinear fin equation in cylindrical coordinates according to thermal diffusivity and heat transfer coefficient is obtained.Using an exhaustive procedure, the determining equations