Gen. Math. Notes, Vol. 23, No. 2, August 2014, pp. 108-115 ISSN 2219-7184; Copyright © ICSRS Publication, 2014 www.i-csrs.org Available free online at http://www.geman.in Greens' Function and Dual Integral Equations Method to Solve Heat Equation for Unbounded Plate N.A. Hoshan Tafila Technical University, Tafila, Jordan E-mail: [email protected] (Received: 16-3-13 / Accepted: 14-5-14) Abstract In this paper we present Greens function for solving non-homogeneous two dimensional non-stationary heat equation in axially symmetrical cylindrical coordinates for unbounded plate high with discontinuous boundary conditions. The solution of the given mixed boundary value problem is obtained with the aid of the dual integral equations (DIE), Greens function, Laplace transform (L-transform) , separation of variables and given in form of functional series. Keywords: Greens function, dual integral equations, mixed boundary conditions. 1 Introduction Several problems involving homogeneous mixed boundary value problems of a non- stationary heat equation and Helmholtz equation were discussed with details in monographs [1, 4-7]. In this paper we present the use of Greens function and DIE for solving non- homogeneous and non-stationary heat equation in axially symmetrical cylindrical coordinates for unbounded plate of a high h. The goal of this paper is to extend the applications DIE which is widely used for solving mathematical physics equations of the 109 N.A. Hoshan elliptic and parabolic types in different coordinate systems subject to mixed boundary conditions, to solve non-homogeneous heat conduction equation, related to both first and second mixed boundary conditions acted on the surface of unbounded cylindrical plate. The solution of the non-homogeneous mixed problem is based on earlier results received in [5, 7] and emphasizes that the weight function, unknown function, and free term depend on a parameter of a L- transform parameter. The DIE solution discussed below is reduced to some type of singular integral with unknown function kernel and free term depend on a parameter of L-transform. The obtained integral equation is solved by expanding its unknown function as a functional series of a Laplace transform parameter. The use of Green's function in the solution with unmixed boundary conditions with different coordinates and applications can be found for example in [3, 9]. 2 Problem Formulations Find a temperature distribution function u= urz(, ,τ ) , of unbounded solid plate 0 ≤z < h ≤ < ∞ τ > 0 r , 0 with heat generation function, the initial temperature distribution function is constant and satisfies mixed boundary value problem −2 ∇ 2 = ut a u grt( , ) (1) ur( ,0,τ )= fr ( , τ ), r ∈Ω (2) 1 τ= τ ∈Ω urz ( ,0, ) fr2 ( , ), r (3) The initial condition urz(, ,0)=θ (, rz ,0) − T , T constant 0 0 τ = The functions fi ( r , ), i 1,2 known and integrable functions, with respect to < < ∞ τ τ > Ω={ << = } Ω={ <<∞ = } θ = ∂θ ∂ r 0, r ,and , 0 , r,0 r Rz , 0 , rR, r , z 0 , z / z , a is a heat diffusivity coefficient, constant, g( r , t ) is known function,(heat generating function). On a surface z= h , a linear combination of unmixed boundary condition is given αu+ α u = 0 (4) 1 2 z α α Where 1, 2 constants. There are three particular cases of (4) α= α = = = τ Case (i): If 11, 2 0, a long the boundary surface z hu, urh (,,) is kept at zero temperature satisfies a homogeneous boundary conditions first kind urh(,,)τ = 0, r ≥ 0 (5) 110 N.A. Hoshan Applying (L-transform) to the boundary-value problem (1)-(5) with respect to τ , then by separating variables for (1) under the assumption that r, z bounded at zero and infinity, we obtain the solution of the homogeneous part of the problem in form of improper integral ∞ sh[( h− z ) p2 + k ] urzs(,,)= CpsJpr (,)( ) dp (6) 1∫ 0 2 0 ch[ h p+ k ] Where sh x)( and ch x)( are sine and cosine hyperbolic functions, J 0 ( pr ) is a Bessel function first kind of order zero, s: parameter of L-transform [2] , a is a heat diffusivity coefficient (constant) k= s/ a and C sp ),( is an unknown function. Substitution a mixed boundary conditions (1) and (2) into (6), yields the following a pair of DIE to determine the unknown function C sp ),( : ∞ 2 + = ∈Ω ∫CpsJ(,)()(0 prthh p kdp ) f1 (,), rs r (7) 0 ∞ 2 + = ∈Ω ∫Cps(,) p kJ0 () prdp f 2 (,), rs r (8) 0 where thx= shx/ chx . Hyperbolic tangent function. Now to solve DIE (7) and (8) ,the unknown function C sp ),( should be replaced by another unknown function ϕ(t , s ) with the help of the relation [1] R p 2 Cps(,)=ϕ (,)cos( ts tpkdt + ) (9) 2 ∫ p+ k 0 Where ϕ(t , s ) = L[ϕ (, t τ )] differentiable with respect to a variable t and analytical with respect to a parameter s, also ϕ t τ ),( should be continuous or piecewise continuous in any τ < τ < τ interval 1 2 , then use the expansion [5] ∞ ϕ= − ϕ n /2− 1 (,)ts exp( Rk )∑ n () ts (10) n =0 The inverse Laplace transform of (10) always exists [2] ∞ 2 ∞ − − 2 R H− ( R /2 a τ ) LRkts1 exp(− )ϕ ()n /2 1 = exp( − ) ϕ () t n 1 ∑n τ ∑ n n τ n /2 n =0 π 4a n = 0 2 111 N.A. Hoshan Substitute (9) into (6), use (10) we obtain a general solution of (1) in L-transform image R ∞ chh((− z ) p2 + k ) = ϕ p 2 + urzs1 (,,)∫ (,) ts ∫ Jpr0 ()cos ( tpk) dpdt 2 + 2 + 0 0 ak shhpk( ) ∞ ∞ ∞ R − +2 + 2 1 ν+ µ − exp(cR ( aµ r ) =( − 1) 1ϕ ()tsn /2 dt (11) ∑∑∑ ∫ n 2 2 ν= µ = 2 n 0 0 0 aµ + r a=(21)ν +−−− hhzita ( ) , = (21) ν ++−− hhzit ( ) , 1 2 =ν +−−+ = ν ++−+ a3 (21) hhzita ( ) ,4 (21) hhzit ( ) . ϕ = i (t ), i 1,...., n are the solution of an integral equation of the second kind [5]. Apply the inverse L-transform to (11), a general solution of the homogeneous part of the problem (1)-(6) for the first particular case when g(, r τ )= 0 R τ= ϕ τ urz1(,,)∫ n () tGrzt 1 (,,,,0) dt (12) 0 2 2 2 ∞ ∞ ∞ (R+ aµ + r ) a ν+ µ − 1 G(,,,,0) r z t τ =() − 1 exp − x 1 π ∑∑∑ τ n =0υ = 0 µ = 0 4a (13) +2 + 2 τ HRarn −1((µ )/24 a n n /2 2 2 2 τ aµ + r Based on (12),(13), we construct the solution of the non-homogeneous part of the problem when g(,) r τ ≠ 0 R τ R τϕ= τ + 1 ξ τξξ urz(,,)∫n ()(,,,,0) tGrzt1 dt ∫∫ gt (,)(,,,,) Grzt1 dtd (14) 0 a 0 0 τ ξ Where G1 zr ),,,( is the Green's function given by the expression 2 2 2 ∞ ∞ ∞ (R+ aµ + r ) a ν+ µ − 1 G(,,,,) r z t τ ξ =() − 1 exp − x 1 π ∑∑∑ τ− ξ n =0υ = 0 µ = 0 4a ( ) (15) +2 + 2 τ − ξ HRarn −1((µ )/24( a )) n n /2 2 2 2 (τ− ξ ) aµ + r 112 N.A. Hoshan α= α = Case (ii): The second particular case if 10, 2 1, on a surface of cylindrical coordinates z = ,rh ≥ 0 in (4) a normal derivative is prescribed unmixed homogeneous boundary condition of the second kind (an insulated boundary condition) in L-transform field θ z shr ),,( = ,0 r ≥ 0 (16) θ = θ τ τ = z shr ),,( L[ z hr )],,( .The general solution of the heat equation (2.1) when g(, r ) 0 in L-transform with regard to (16), under the assumption that θ zr τ ),,( is limited 2 2 as z+ r → ∞ and r = 0 , the medium is initially at constant temperature, by applying (L-transform) and separation of variables for (1), we have ∞ ch(( h− z ) p2 + k ) = urzs2 (,,) CpsJpr (,)(0 ) dp (17) ∫ 2 0 sh( h p+ k ) Use a mixed boundary conditions (2) and (3) (17) ,a pair DIE were obtained to determine the unknown function C sp ),( ∞ 2 + = ∈Ω ∫CpsJpr(,)(0 )coth( hp kdp ) frs1 (,), r 0 ∞ (18) 2 + = ∈Ω ∫Cps(,) p kJ0 () prdp f 2 (,), rs r 0 where coth x = chx / shx hyperbolic cotangent function. The solution of the dual equations (18) is obtained with the help of equalities (9) and (10) in L-transform image. Repeat the same procedure of case (i), the general solution of the homogeneous part of the problem for the second particular case when g(, r τ )= 0 is ∞ R p chh[(− z ) p2 + sa / ] =ϕ 2 + 2 urzs2 (,,)∫ (,) ts ∫ Jpr0 ()cos ( tpc) dpdt 2+ 2 2 + 0 0 ac shhpsa[ / ] ∞ ∞ ∞ R − +2 + 2 1 µ− exp(kR ( aµ r )) =( − 1) 1ϕ ()tsn /2 dt (19) ∑∑∑ ∫ n 2 2 ν= µ = 2i n 0 0 0 aµ + r Applying the inverse L-transform for the homogeneous part of the problem (19) when g(,) r τ ≠ 0 , the solution of the non-homogeneous mixed boundary value problem(1)-(5) and (16) is 113 N.A. Hoshan R τ R τϕ= τ + ξ τξξ urz(,,)∫n () tGrzt2 (,,,,0) dt ∫∫ gt (,) Grzt2 (,,,,) dtd (20) 0 0 0 τ ξ Where G2 (, r z ,, ) is the Green's function given by the expression 2 2 2 ∞ ∞ ∞ (R+ aµ + r ) a µ− G(,,,,) r z t τ ξ = (1) −1 exp − x 2 π ∑∑∑ τ− ξ i n =0υ = 0 µ = 0 4a ( ) (21) +2 + 2 τ − ξ HRarn −1((µ )/24( a )) n n /2 2 2 2 (τ− ξ ) aµ + r µ = = − where aµ , 4,3,2,1 i 1 in (21) have the same values given in (13).