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cs: O ani pe ch n e A c M c Hassan, Fluid Mech Open Acc 2017, 4:2 d e i s

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F Fluid Mechanics: Open Access DOI: 10.4172/2476-2296.1000152

ISSN: 2476-2296

Perspective Open Access

Green’s for the Equation Abdelgabar Adam Hassan* Department of , 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- methods in finding the Green’s function of in different situations.

Keywords: Heat equation; Green’s function; Sturm-Liouville So problem; Electrical ; 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 , 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 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 filedu , 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 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

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

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Eigenvalue and : The Eigenvalue from eqn (4) Some important properties of defining by a Sturm-Liouville operator can be expressed as: Property (1): −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 . 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

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 Green’s Function for the Heat −−ζτs − expxsa (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. Let us find G(x, t, ζ, τ). Solution: Let us find the Green’s function for the following We begin by taking the Laplace transform of eqn (24) with respect problem: 2 to t, we have: ∂∂GG2 −a=δ( xt − ζδ) ( − τ),0 < x , ζ <∞ ,0 < t , τ (32) 2 ∂t ∂x2 sgxs,,,ζτ− gx ,0,, ζτ −=− a2 dg δ x ζ e−sτ ( ) ( ) 2 ( ) ζτ <∞ dx Subject to the boundary conditions G(0, t, ζ, τ), limx→∞ Gxt( , , , ) 2 and initial condition G(x, 0, ζ, τ) dg−s =−−δζ−sτ So 2 2 g( xe) (26) dx a From the boundary condition G(0, t, ζ, τ)=0, we deduce that G(x, t, Where g(x, s, ζ, τ) the Laplace transform of G(x, t, ζ, τ) ζ, τ) Green’s function by introducing an image source of −δ(x−ζ) and resolving eqn (25) with the source (x−ζ)δ (t−τ)−δ (x+ζ)δ (t−τ). Eqn Now by taking the of eqn (26) with respect to (30) gives the solution for each delta function and the green’s function x, so that: for eqn (32) can be written: 2 −−ikζτ s 22 −ζτ −=se ζτ − −−xxζζ −+ ( ik) G (k,s, ,) 22G ( k,s, , ) H (t −τ ) ( ) ( ) Gxs,,,ζτ = exp− exp (33) aa ( ) 2 22 −−ikζτ s 4πτat( − ) 44at( −−ττ) at( ) kG2 k,s,ζτ ,+=se G k,s, ζτ ,  ( ) aa22( ) 22 22 H (t −τ ) −−xx++22ζζ −−xx ζζ Gxs( ,,,ζτ) = exp− exp Where G (k,s,ζτ , ) is Fourier transform of g(x, s, ζ, τ), now let 2 22 4πτat( − ) 44at( −−ττ) at( ) s = b2 a2 22 −−ikζτ s −−xxζζ 22+ ζτ= e (27) exp+− (k bG) (k,s, , ) 2 22−−ττ a H (t −τ )  42at( ) at( )  Gxs( ,,ζτ , ) =  πτ2 − 22 To find g(x, s, ζ, τ), we use the inversion integral: 4 at( ) −−xxζζ exp− 22  42at( −−ττ) at( )   ix−ζ −sτ ∞  ζτ = ee (28) 22 g( x,, s , ) 2∫  22dk −+x ζ π −∞ + exp× 2 a kb 2 −τ H (t −τ ) 4at( )  Gxs,,ζτ , =  ( ) 2  πτ− − 4 at( ) xxζζ Transforming eqn (27) into a closed contour, we evaluate it by the exp− exp 22−−ττ residue theorem and find that: 22at( ) at( )

   −ζ 22 ix H (t −τ ) −+xxζζ ∞  = (34) −sτ Gxs( ,,,ζτ) expsinh = ee 2 22−−ττ g( x,, s ζτ , ) ∫ dk 4πτat( − ) 42at( ) at( ) 2πa2 −∞ (k+− ib)( k ib) In a similar manner, if the boundary condition at x=0 changes to −sτ G (0, t, ζ, τ)=0, then eqns (31) and (32) become: ζτ = e x gxs( ,, , ) 2 ∑ bi 2τa 22       −−xxζζ −+  H (t −τ )      =  +  −−−xbζζ xb − Gxs,,,ζτ exp exp =± =11 −= πτ2 − 44at22( −−ττ)  at( )  at k ib then∑ b e e e 4 at( )    22ib ib   

22 −−ζζ −+ − τ −ζ −−xζ b − sr H (t −τ ) ( xx) ( ) e s − xb1 Gxs,,,ζτ = exp− exp (35) ∴gxs,,ζτ , = e= e ( ) 2 22 ( ) 2 4πτat( − ) 44at( −−ττ) at( ) 22a b ib 

   (t −τ ) xx22+ζζ Now substituting for = s , we have: = H −   b Gxs( ,,,ζτ) exp cosh  (36) a 2 22   4πτat( − ) 42at( −−ττ)  at( ) 

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

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Heat Equation within a finite Cartesian Domain Liouville problem, we will encounter Bessel functions. In this section we find Green’s function for the heat equation Example (5.3.1) within finite Cartesian domains. These solutions can be written as Find the solution of the following problem by construction the series involving orthonormal eigenfunction form regular Sturm- Green’s function: Liouville problem [6-10]. ∂2 ∂∂f( rt, ) Example (5.2.1) Ua− U =, <ρτ << (50) r0 r , bt ,0 , ∂tr ∂∂ rr2π r Here we find the Green’s function for the one-dimensional heat limU( rt , ) <∞ equation over the interval 0

With α1g(0,t,ζ,τ)+β1g (0,t,ζ,t)=0, (43) k ρ kr And α g(L,t,ζ,τ)+β g (L,t,ζ,t)=0 (44) JJn n 2 2 ′  −sτ ∞ 00 So that 1 ddg se bb (55) rg− =− ∑ Applying the technique of the eigenfunction expansions, we have r 2 π 2 2 2 ′ dr dra a b n=1 Jk1 ( n ) that: ∞ φnn( ζφ) ( x) k ρ kr gxs,,ζτ , = e−st (45) JJn n ( ) ∑ 22 2 −sτ ∞ 00 + d g dg bb n=1 s akn +−1 seg =− 22222 ∑ drr draJπ a b n=1 Where φn(x) is the nth orthonormal eigenfunction to the regular 1 Sturm-Liouville problem: By applying Fourier transform: φ(x)+k2φ (x)=0 (46)    knnρ  kr  JJ   −sτ ∞ 00   Subject to the boundary conditions: 22sG e bb   ′ −−kG+=− ikG ∑ 2 2 2 2  a πabn=1 Jkn α1φ(0)+β1φ (x)=0 (47) 1 

And α φ(L)+β φ (L)=0 (48)  2 ′ 2 ρ JJkn krn −sτ ∞ 00 Taking the inverse′ of eqn (45), we have that: 2 se bb Gk+ =− 2 2 2 ∑ 2 ∞ φ ζφ( ) a πabn=1 nn( ) x −kn22at( −τ ) Jk1 ( n ) ζτ = −τ (49) Gxt( ,, , ) ∑ 22e Ht( ) n=1 s+ ak n ρ  JJkn krn −sτ ∞ 00bb Heat equation within cylinder Grs( ,,ρτ , ) = e ∑  (56) π 2 2 + 22 2 In this section, we turn our attention to cylindrical domains. The a n=1(sb knn a) J1 ( k ) techniques used here have much in common with those used in the Taking the inverse of eqn (56) and applying second shifting previous section. However, in place of sines cosines form in Sturm- theorem,

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

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 β β knρ krn 2 = 1 22+ 2 JJ And ∫ y( krrdrn) r y( kr nn) y( kr) (69) ∞ 00 α 0 2 01α ( ρτ) = Ht( −τ ) bb (57) Grs,, , 2 ∑ 2 with the relationship: πb n=1 Jk 1 ( n ) − J( zy) ( z) −= J( zy) ( z) 2 (70) If we modify the boundary condition at r=b, 01 10π z We find that: Gx(b,t,ρ,τ)+hG(b,t,ρ,τ)=0 (58) β 2 α Where h0, our analysis now leads to: 2 1 Jk0 ( n ) ∫φn (r) rdr = −1 (71)  α 2π 22k Jk2 β knρ krn 0 ( n ) ∞ JJ0022 Ht( −τ ) bb−−akn ( t τ ) (59) ρτ = exp ∞ 22 β φ ρφ( τ) Grs( ,, , ) ∑  4 − τ kJn0 ( k nn) ( ) n π 22= 22  ( ρτ) = s bbn 1 JkJk( nn) ( )  Therefore grs,, , e ∑  (72) 01 π n=1 2αβ− 2 + 22 Jk00( nn) Jk( ) ( sak n) Where k are the positive roots of kJ (k)+hbJ (k)=0. n 1 0 Inverting the Laplace transform, we obtain: Problem (3.3.2) ∞ 22 β φ ρφ( τ) 4 kJn0 ( k nn) ( ) n Grs( ,,ρτ , ) =−H t τ exp −−ak22 t τ (73) Find the solution for the heat equation in cylindrical by construction π ( ) ∑ 22 n ( ) n=1 Jk( nnαβ) − Jk( ) the Green’s function: 00 In similar manner, we can solve the general problem eqn (61) with ∂2 ∂∂f( rt, ) Ua−=< U , (60) Robin boundary conditions: rt0 ∂tr ∂∂ rr2π r a1Gr(α,t, ρ,τ)−a2G(α,t,ρ,τ)=0 (74) Subject to the boundary conditions U(α,t)=U(β,t)=0, and the initial condition U(r, 0)=0 And b1Gr(β,t, ρ,τ)−b2G(β,t,ρ,τ)=0 (75)

Let G(b,t,ρ,τ) Green’s function, then: Where a1, a2, b1, b2≥0. The solution for this problem is: 2 kn  ∂2 ∂∂δ(rt−− ρδ) ( τ) ∞ bknn J kββ−× b J k n Ga− G =, <<<ρβ τ (61) 4 1 1( ) 20( ) r 0rt , ,0 , H Fkn (76) ∂tr ∂∂ rr 2π r Grs( ,,ρτ , ) =( t − τ) ∑ ( ) π n=1 φ( ρφ) ( )exp −−22 τ Subject to the boundary condition that G(α,t,ρ,τ)= G(β,t,ρ,τ)=0, nnr ak n( t ) and the initial condition G(r,0,ρ,τ)=0.  r ββ+ J0( kn)  ak 1 nn y 1( k) a 20 y( k n)  We begin by taking Laplace transform of eqn (61), we obtain: Where φ (r) =  (77) n  − r αα+  −sτ y0( kn)  ak 1 nn J 1( k) a 20 J( k n)  1 dg erδρ( − )  ds− =− (62) rg22 r dr dr a2π ar 22 2 (b kn+ b)  ak nn J( kαα) + a J( k n)  δρ( − ) 1 2 1 1 20 r And Fkn =  (78) Now, we express r as a by considering the ( )  −ykr akykββ+ ayk regular Sturm-Liouville problem: 0( n)  1 nn 1( ) 20( n)   Heat equation within a sphere 1 dφ d + 2φ= φα( ) = φα( ) = 0 rk0, (63) r dr dr Let us find Green’s function for the radically symmetric heat equation within a sphere of radius b. mathematically, we must solve: The eigenfunction that satisfy eqn (63) are: ∂∂Ga22 δ(rt−− ρδ) ( τ) φ ( x) = ykαα Jkrr− Jk yk (64) −(rG ∂=) ,0

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

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2  ∞  2 ∂∂rG δ−− ρδ τ δρ()r − nrπ ∂ ()rt( ) (87) = ∑ sin Ga− =,αρ

22b  πρ With the boundary conditions G(α,t,ρ,τ)=G(β,t,ρ,τ)=0, and the = sinnrπ dr = sin n Where Crn ∫δρ()−   (83) b0  bbρ  b initial condition G(r,0,ρ,τ)=0. Therefore the Green’s function in this particular case is: Therefore we write eqn (81) as: Ht( −τ ) ∞ nπρ()−− α nr π() α ()rt,,,ρτ = ∑ sin sin × G  ()βα−− ()βα  2 −sτ ∞ nπρ π  2πρ()βα− r n=1    dU− s= e sin  sin n r  (84) (88) 2 22U ∑    22 2 dr a4π arn=1 b  b  −−anπτ( t ) exp  βα− 2 () ππ−sτ ∞ nπρ  and in− s= e sin sin n r References 22U ∑  baa4πρn=1 bb 1. Al-Gwaiz MA (2008) Sturm-Liouville theory& its Application, Verlag London 22π 2 limited, UK. an + s 2 −sτ ∞ nπρ π 2. Blakledge G, Evans J, Yardley P (2000) Analytic methods for Partial differential b =− e sin sin nr 22U ∑  Equations, Sprenger London Limited. aa4πρn=1 bb   3. Kwong-Tin T (2007) Mathematical methods for Engineers and Scientists 3, − τ ∞ Verlag Berlin Heidelbarg. s 1 nπρ π ()rt,,,ρτ =− e sin sin nr U ∑ 22 2  4πρ n=1s+ anπ bb (85) 4. Gerald HB (1992) and Application, Wodsworth. Inc. Belmon b2 California. Because this particular solution also satisfies the boundary 5. Kwong-Tin T (2007) Mathematical methods for Engineers and Scientists 2, condition, we do not require any homogeneous solution so that the Verlag Berlin Heidelbarg. sum of the particular and homogeneous solution satisfies the boundary 6. Neta B (2002) Partial differential equations, lecture notes, Naval postgraduate conditions. School, California.

Taking the inverse of eqn (85), using the second shifting theorem 7. Pinchover Y, Rubinstein J (2005) An introduction to Partial differential and substituting for U(r,t,ρ,τ), we finally obtain: equations, Cambridge University. 22 2 Ht()−τ∞ πρ   −−anπτ() t 8. Roach GF (1970) Green’s function introductory theory with Application, New ρτ n nrπ  G ()rt,,, = ∑ sin  sin  exp  (86) York Toronto Melbourne.    2 2πbrn=1 b b b 9. Addison W (1984) , Monterey California. For the case of a sphere α

Fluid Mech Open Acc, an open access journal Volume 4 • Issue 2 • 1000152 ISSN: 2476-2296