EXACT SOLUTIONS OF A GENERALIZED BRATU EQUATION

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1 EXACT SOLUTIONS OF A GENERALIZED BRATU EQUATION LAZHAR BOUGOFFA Department of Mathematics, Faculty of Science, Al Imam Mohammad Ibn Saud Islamic University (IMSIU), P.O. Box 90950, Riyadh 1163, Saudi Arabia bougoffa@hotmail.com, lbbougoffa@imamu.edu.sa Received October 4, 016 Abstract. In this short article, we derive the closed-form solutions of a generalized Bratu equation, which arises in many fields of science and engineering such as radiative heat transfer, combustion theory, and nanotechnology. Key words: Bratu equation; generalized Bratu equation; exact solution. 1. INTRODUCTION Bratu solved in 1914 the following boundary value problem in one-dimensional planar coordinates u + λe u = 0, 0 < x < 1 (1) and u(0) = u(1) = 0, () which is a basic one-dimensional (1-D) model in combustion theory [1] that was revisited by Gelfand in 1963 []. Equations (1)-() arises in many fields of science and engineering such as radiative heat transfer, combustion theory, and nanotechnology. For λ > 0, the exact solution of Eqs. (1)-() is given by [3] ) u(x) = ln cosh( (x 1 ) θ cosh( θ 4 ), (3) where θ satisfies θ = ( ) θ λcosh. (4) 4 The Bratu problem has zero, one or two solutions when λ > λ c, λ = λ c, and λ < λ c, respectively, where the critical value λ c satisfies the equation 1 = 4 1 ( ) θ λc sinh, (5) 4 and the numerical value of λ c is computed and given by λ c = In this short article, we are interested in the general nonlinear differential equations, which contains the classical Bratu equation as a special case, Romanian Journal of Physics 6, 110 (017) u + e au = f(x), (6) v..0* #6597f710

2 Article no. 110 Lazhar Bougoffa where > 0, f(x) are two functions and a is a constant. The purpose of this work is to present new exact solutions of this generalized Bratu equation under suitable conditions on the functions and f(x). Thus We make the change of variables. THE EXACT SOLUTIONS.1. THE FIRST SOLUTION au(x) = ln ( ) v(x). (7) au (x) = v (x) v(x) λ (x). (8) Eq. (6) becomes (x) ( λ ) (x) + av(x) = f(x). (9) v(x) ( ) If we assume that f(x) + λ (x) = 0, that is, f(x) + (ln) = 0, then by the change of variables w(x) = v (x) v(x), (10) and in view of v(x) = v (x), Eq.(9) reduces to w(x) w (x) + av (x) w(x) = 0, (11) which is a separable ordinary differential equation (ODE). It follows that w(x)w (x) + av (x) = 0. (1) Integrating Eq. (1) with respect to x, we obtain where c 1 is a constant of integration. So that Hence w(x) = ± av(x) + c 1, (13) v (x) v(x) = ± av(x) + c 1. (14) dv v av + c 1 = ±dx. (15)

3 3 Exact solutions of a generalized Bratu equation Article no. 110 Consequently, we have where c is another constant of integration. Thus we have Consequently, ( ) tanh 1 c1 av c1 = ±x + c, (16) c1 v(x) = c 1 a c 1 a tanh v(x) = c 1 a Now using the given boundary conditions () we get c 1a 1 ) = λ 0, ( c1 c1 c x ). (17) ( 1 cosh ). (18) c1 c1 c x cosh ( c1 c c 1a 1 cosh ( c1 c1 c ) = λ 1, where λ 0 = λ(0) and λ 1 = λ(1). c1 We choose θ 1 = and θ = ±c, then cosh(θ 1 θ ) = aλ 0 θ 1, cosh(θ 1 + θ 1 θ ) = aλ 1 θ 1. Thus we have cosh(θ 1 + θ 1 θ ) cosh(θ 1 θ ) = (19) (0) λ0 λ 1, θ 1 0. (1) We have thus proved Theorem 1 Let λ be a positive function. If we assume that f(x) + (ln) = 0, then the exact solution to the general boundary value problem (6) with () is given as u(x) = 1 a lnv(x) 1 ln, () a where v(x) = θ 1 1 a cosh (3) (θ 1 (x + θ )) and the constants θ 1 and θ satisfy Eq. (1). Now we consider the simplest case of Theorem 1, namely the case when = λ > 0 and a = 1. Thus f(x) = 0, and the condition of this theorem is satisfied. Hence ( ) θ u(x) = ln 1 ln(cosh(θ 1 (x + θ ))). (4) λ

4 Article no. 110 Lazhar Bougoffa 4 A simple calculation leads to u(x) = ln cosh(θ 1(x + θ )) λ θ 1. (5) From (1) we get θ = 1 that could make Eq. (1) true and solvable. Thus u(x) = ln cosh( θ 1 (x 1 )) λ θ 1, (6) where θ 1 satisfies (0), that is cosh( θ 1 ) = θ 1 λ. Replacing the constant θ instead of θ 1 into this solution, we obtain the same Bratu s solution... THE SECOND SOLUTION WHEN f(x) > 0 We make the change of variables ( ) f(x)v(x) au(x) = ln, (7) where f(x) is a positive differentiable function. Equation (6) reduces to (x) ( + ln( f(x) ) v(x) ) + af(x)v(x) = af(x). (8) ( ) If we assume that ln( f(x) ) = 0, then Eq. (8) becomes (x) + af(x)v(x) = af(x). (9) v(x) Clearly, v(x) = 1 is a solution of Eq. (9). We then get from Eq. (7) u(x) = 1 ( ) f(x) a ln, f(x) > 0 (30) and u(x) = 0 as long as f(x) = at x = 0 and x = 1. (31) Thus we have proved ( ) Theorem Let f and λ be two positive functions. If we assume that ln( f(x) ) = 0, then the exact solution to the general boundary value problem (6) with () is given by (30)-(31).

5 5 Exact solutions of a generalized Bratu equation Article no CONCLUSION In this paper, we have considered a generalization of the Bratu equation. When the functions and f(x) in Eq. (6) are replaced by the constant λ and by 0, respectively, then Eq. (6) becomes the standard Bratu equation. We have demonstrated that the closed-form solutions of this problem can be obtained in a straightforward manner by a direct method under suitable conditions on the functions and f(x). REFERENCES 1. G. Bratu, Bull. M. Soc. France 4, (1914).. I. M. Gelfand, AMS Trans. Ser. (9) (1963). 3. A. M. Wazwaz, Rom. J. Phys. 61, (016).