cs: O ani pe ch n e A c M c Hassan, Fluid Mech Open Acc 2017, 4:2 d e i s u s l F Fluid Mechanics: Open Access DOI: 10.4172/2476-2296.1000152 ISSN: 2476-2296 Perspective Open Access Green’s Function for the Heat Equation Abdelgabar Adam Hassan* Department of Mathematics, College of Science and Arts at Tabrjal, AlJouf University, Kingdom of Saudi Arabia Abstract The solution of problem of non-homogeneous partial differential equations was discussed using the joined Fourier- Laplace transform methods in finding the Green’s function of heat equation in different situations. Keywords: Heat equation; Green’s function; Sturm-Liouville So problem; Electrical engineering; Quantum mechanics dy d22 y dp() x dy d y dy d () =+=+() () ()() px px 22px pxbx dx dx dx dx dx dx dx Introduction Thus eqn (3) can be written as: The Green's function is a powerful tool of mathematics method dy is used in solving some linear non-homogenous PDEs, ODEs. So d px() ++ qxy()()λ rxy =0 (4) Green’s functions are derived by the specially development method of dx dx separation of variables, which uses the properties of Dirac’s function. Where q(x)=p(x)c(x) and r(x)=p(x)d(x). This method was considerable more efficient than the others well known classical methods. Equation in form (4) is known as Sturm-Liouville equation. Satisfy the boundary conditions [1-5]. The series solution of differential equation yields an infinite series which often converges slowly. Thus it is difficult to obtain an insight Regular Sturm-Liouville problem: In case p(a)≠0 and p(b)≠0, into over-all behavior of the solution. The Green’s function approach p(x),q(x),r(x) are continuous, the Sturm-Liouville eqn (4) can be would allow us to have an integral representation of the solution expressed as: instead of an infinite series. L[y]=λr(x)y (5) To obtain the filed u, caused by distributed source we calculate the d dy Where L ≡ px() + qx() (6) effect of each elementary portion of source and add (integral) them all. dx dx If G(r, r ) is the field at the observers point r caused by a unit source 0 If the above equation is associated with the following boundary at the source point r , then the field at r caused by distribution f(r ) is 0 0 condition: the integral of f(r0) G(r, r0) over the whole range of r0 occupied by the course. The function G is called Green’s function. α1y(a)+α2y(a)=0 Preliminaries β1y(b)+β2y(b)=0 (7) Where +α ≠0 and β +β ≠0 Sturm-Liouville problem 1 2 1 2 Consider a linear second order differential equation: The eqn (4) and the boundary condition (7) are called regular Sturm-Liouville problem (RSLP). 2 Ax() d y+ Bx() dy ++ Cxy()()λ Dxy =0 (1) dx2 ∂x Singular Sturm-Liouville problem: Consider the equation: Where λ is a parameter to be determined by the boundary L[y]+λr(x)y=0 a<x<b (8) conditions. A(x) is positive continuous function, then by dividing every Where L is defined by eqn (6), p(x) is smooth and r(x)is positive, term by A(x), eqn (1) can be written as: then the Sturm-Liouville problem is called singular if one of the 2 following situations is occurred. d y+() dy ++()()λ = (2) 2 bx cxy dxy 0 dx ∂x (i) If p(a)=0 or p(b)=0 or both Bx() Cx() Dx() Where bx() = , cx() = and dx() = (ii)The interval (a,b) is infinite. Ax() Ax() Ax() Let us define integrating factor p(x) by: x () = ζζ px exp∫ bd() *Corresponding author: Hassan AA, Department of Mathematics, College of a Science and Arts at Tabrjal, AlJouf University, Kingdom of Saudi Arabia, Tel: Multiplying eqn (2) by p(x), we have: +966538043646; E-mail: [email protected] 2 Received March 01, 2016; Accepted March 22, 2017; Published April 02, 2017 () d y+()() dy ++()()()()λ = (3) px2 pxbx pxcxy pxdxy 0 dx ∂x Citation: Hassan AA (2017) Green’s Function for the Heat Equation. Fluid Mech Open Acc 4: 152. doi: 10.4172/2476-2296.1000152 Since: Copyright: © 2017 Hassan AA. This is an open-access article distributed under xx bdζζ bd ζζ the terms of the Creative Commons Attribution License, which permits unrestricted ∫∫ x dp() x ddaa use, distribution, and reproduction in any medium, provided the original author and = e= e∫ b()ζζ d= pxbx()() dx dx dx a source are credited. Fluid Mech Open Acc, an open access journal Volume 4 • Issue 2 • 1000152 ISSN: 2476-2296 Citation: Hassan AA (2017) Green’s Function for the Heat Equation. Fluid Mech Open Acc 4: 152. doi: 10.4172/2476-2296.1000152 Page 2 of 6 Eigenvalue and Eigenfunction: The Eigenvalue from eqn (4) Some important properties of Dirac delta function defining by a Sturm-Liouville operator can be expressed as: Property (1): Symmetry −1 d dy λ = px( ) + qx( ) (9) δ (−x)=δ (x) (18) rx( ) dx dx Proof: Let ζ=−x, then dx=−dζ The non-trival solutions that satisfy the equation and boundary conditions are called eigenfunctions. Therefore the eigenfunction of We can write: the Sturm-Liouville problem from complete sets of orthogonal bases +∞ −∞ +∞ ∫fxdxfdfd( xf)δ(−=−) ∫∫(ζδζ) ( ) ζ= ( ζδζ) ( ) ζ= (0) (19) for the function space is which the weight function is r(x). −∞ +∞ −∞ The Dirac Delta function +∞ But, ∫ f( xf)δ ( x) dx = (0) (20) The delta function is defined as: −∞ Therefore, from eqns (19) and (20), we conclude that δ (−x)=δ (x). 0 x≠ζ δζ( x−=) (10) Property (2): Scaling ∞=x ζ δδ(ax) = 1 ( x) (21) But such that the integral of δ (x−ξ) is normalized to unit; a +∞ Proof: Let ζ=ax, then dx= 1 dζ ∫ δζ( x−=) dx 1 (11) a −∞ If a>0, then, In fact the first operator where Dirac used the delta function is the +∞ +∞ ζζ+∞ ( x)δ( ) = δζ1 ζ= 1 δζ ζ=101ff = (0) integration: ∫∫f ax dx f( ) d ∫ f( ) d −∞ −∞ a a a−∞ a aaa +∞ ∫ f( x)δζ( x− ) dx (12) +∞ 11+∞ 1 −∞ Since, ∫∫f(ζδ) ( xxf) dx= f( ) δ( x) dx = (0) −∞ aa−∞ a Where f(x) is a continuous function, we have to find the value of Therefore: δδ= 1 the integration eqn (12). (ax) a ( x) Since δ (x−ξ) is zero for x≠ξ, the limit of integration may be change Similarly for a<0, δδ(ax) = 1 ( x) , then, to ζ− and ζ+ε, where ε is a small positive number, f(x) is continuous at −a x−ξ, it’s values within the interval (ζ−ε,ζ−) will not different much from We can write: δδ(ax) = 1 ( x) f(ζ), approximately that: a +∞ ζζ++εε Property (3) ∫∫f( xx)δζ( x−=) dx f( ) δζ( x −≈) dx f( ζδζ) ∫( x −) dx (13) −∞ ζζ−εε− δ22−=1 δδ ++ ( −) ( x a) ( xa) xa (22) With the approximation improving as ε approaches zero. 2a From eqn (11), we have: Proof: The argument of this function goes to zero when x=a and x=−a, wherefore: +∞ ζ +ε I = ∫∫δζ( x−=) dx δζ( x −=) dx 1 (14) +∞ +∞ −∞ ζ −ε 2 2 ∫∫f(ζδ) ( xa−=) dx f( x)δ( x+− a)( x a) dx From all values ofε, then by letting ε→0, we can exactly have: −∞ −∞ +∞ ∫ f( xf)δζ( x−=) dx ( ζ) (15) Only at the zero of the argument of the delta function that is: −∞ +∞ −aa +εε+ ∫∫∫f( x)δ( xa2−=2) dx f( x) δδ(xa 22 −+ 22) dx f( x) (xa −) dx (23) Despite the delta function considered as fundamental role in −∞ −aa −εε− electrical engineering and quantum mechanics, but no conventional 2 2 could be found that satisfies eqns (10) and (11), then the delta function Near the two zeros x −a can be approximated as: sought to be view as the limit of the sequence of strongly peaked −2axa( +) ,x →− a 22 = x−=−+ a) ( x ax)( a) function δ (x) such that: ( +2axa( −) ,x →+ a n δδ( xx) = n ( ) limn→∞ (16) In the limit as ε0 the integral eqn (23) becomes: As: +∞ −aa +εε+ 2 2 ∫∫f( x)δδ(x− a) dx = f( x) ( −+22axa)( ) dx + ∫ f( x) δ ( axa)( −) dx δ ( ) = n −∞ −aa −εε− n x 22 −+aaεε+ π + 11 (1 nx) =∫∫f( x)δδ(xa++) dx f( x) (xa−) dx 22aa−−aaεε− 22 +∞ n −nx 1 δ ( ) = e (17) = −+ − n x ∫ f( x) δδ(xa) ( xa) dx π −∞ 2a 2 sin (nx) δ ( x) = ,etc 221 n 2 Therefore: δxa− = δδ( xa −+) ( xa −) nxπ ( ) 2a Fluid Mech Open Acc, an open access journal Volume 4 • Issue 2 • 1000152 ISSN: 2476-2296 Citation: Hassan AA (2017) Green’s Function for the Heat Equation. Fluid Mech Open Acc 4: 152. doi: 10.4172/2476-2296.1000152 Page 3 of 6 Green’s Function for the Heat −−ζτs − expxsa (29) gxs,,ζτ , = Heat equation over infinite or semi-infinite domains ( ) 2as Consider one dimensional heat equation: Taking Laplace transform of eqn (29) we obtain: 2 ∂∂uu2 2 −=−a f xt, (24) 2 ( ) H (t −τ ) −−( x ζ ) ∂∂tx Gxs,,,ζτ = exp ( ) 2 2 (30) aatπτ( − ) aa( t −τ ) Subject to boundary conditions |u(x,t)|<∞ as |x|<∞ and initial condition. Let G(x, t, ζ, τ) be the Green’s function for the one- Example (5.1.1) Consider one dimensional heat equation: dimensional heat equation, then: ∂∂2 uu−−a2 = f( xt, ) ∂∂2 ∂t ∂x2 (31) GG−=2 δ ( xt−ζτ) ∂( −), −∞< x , ζ <∞ ,0 < t , τ (25) a 2 ∂∂tx Subject to boundary condition u(0,t)=0, limuxt( , ) <∞ and initial x→0 Subject to the boundary condition |G(x, t, ζ, τ)|<∞ as |x|<∞, and the condition U(x,0)=0 initial condition G(x, t, ζ, τ)=0.