Green's Functions in Mathematical Physics

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TJMM1 (2009), No. 1-2, 01-10

Green’s Functions in Mathematical Physics

WILHELMKECS

ABSTRACT. The determination of Green functions for some operators allows the effective writingof solutions to some boundary problems of mathematical physics.

2010Mathematics Subject Classication. 34B27, 42A38.

Key words and phrases.Green’s functions, Fourier transform.

1. INTRODUCTION

The solution of many problems of mathematical physics is related to the con-struction of Green’s function with the help of which the solutions of the boundary-value problems may be determined explicitly and presented in an integral form.Thus, the Green functions are closely connected to the fundamental solution of the linear differential operator corresponding to a given problem.

In the case in which the fundamental solution is a distribution, we have to dealwith Green’s distributions.

To illustrate the construction of a Green function as well as its connection to

the fundamental solution of the operator corresponding to the given boundary-value problem, we shall consider the heat conduction equation for an innite bar,the generalized Poisson equation and the vibrating string equation.

2. HEAT CONDUCTION EQUATION

We shall consider a cylindrical homogeneous and isotropic bar whose lateralsurface is isolated from the rest of the medium. We take the symmetry axis of the bar asOx -axis. We shall denote byu(x, t ), x∈

R , t ≥ 0 the temperature of the bar in the pointx at the momentt,andbyϱ , c, kthe density, the specic heat andthe thermal conductivity. We use the Fourier-Newton law regarding the amountof heat owing in the unit time across the unit cross section of the bar, then, inthe absence of internal sources of heat, the heat conduction equation is

∂u (x, t )∂t

= a2 ∂ 2u(x, t )∂x 2 , x∈

R , t > 0, (1)

where the constanta2 has the expressiona2 = k/ ϱc.We consider the parabolic equation (1) with the initial condition

u(x, 0) = φ(x), φ∈C 0(R ). (2)The equation (1) with the condition (2) represents the boundary value problem

of the heat conduction in an innite homogeneous bar.1



2 Wilhelm Kecs

The boundary value problem has unique solution for the given functionφ. Theproblem solution can be represented in the form

u(x, t ) = L(φ(x)) = ∞

−∞G(x,ξ, t )φ(ξ )dξ = ( φ(ξ ), G(x,ξ, t )) , (3)

where the kernel of the integral operatorLis G(x,ξ, t ) and is called Green func-tion.

To obtain the Green function, we will considerφ(x) = δ (x −x0), wherex0∈R

is a parameter. From (3), it resultsu(x, t ) = ( δ (ξ −x0), G(x,ξ, t )) = ( δ (ξ ), G(x, ξ + x0 , t )) = G(x, x 0 , t ), (4)

henceu(x, t ) = G(x, x

0, t ). (5)

Consequently, the Green functionG(x, x 0 , t ) represents thesolution of theequa-tion (1) with the condition

u(x, 0) = δ (x −x0). (6)Hence, the Green functionG(x,ξ, t ) satises the equation (1)

∂G (x,ξ, t )∂t

=∂ 2G(x,ξ, t )

∂x 2 (7)

and the initial conditionG(x,ξ, t )|t =0 = δ (x −ξ ). (8)

LetE (x, t )∈ D′(R ×R ) be the fundamental solution of the equation (1), hence∂E ∂t −a2 ∂ 2E

∂x 2 = δ (x, t ) = δ (x) ×δ (t). (9)

Applying the Fourier transform with respect tox to the equation (9) we obtainddt

E (α, t ) −a2(−iα)2 E (α, t ) = δ (t), (10)

where E (α, t ) = F x [E (x, t )].The equation (11) shows that the distributionE (α, t ) is the fundamental solu-

tion of the operatorddt

+ a2α 2 . (11)

To determine the distributionE (α, t ), we consider the homogeneous equationdV (t)

dt+ a2α 2V (t) = 0 , (12)

the general solution of which is

V (t) = C e−a 2 α 2 t . (13)Imposing the conditionV (0) = 0 , we obtainC = 1 and consequently the

fundamental solution of the operator (11) is

V (t) = H (t)e−a 2 α 2 t = E (α, t ), (14)



Green’s Functions in Mathematical Physics 3

whereH represents the Heaviside functionH (x) = {0, x < 0,

1, x ≥0. (15)

Applying the inverse Fourier transform, we yield

E (x, t ) = F −1 E (α, t ) =H (t)

2a√ πtexp −

x2

4a2 t, x∈

R , t∈R , (16)

becauseF−1[e−b2 x 2] = π

|b|exp −x 2

4b2 .Consideringt as parameter, from (14), we obtain, fort > 0

E t (x) =1

2a√ πtexp −

x2

4a2t. (17)

We remark thatE t (x), t > 0 is an element of the test spaceS (R ) of indenitelydifferentiable functions which, for|x| → ∞, approach zero together with all theirderivative of any order, faster than any power of |x|−

1 . It satises the relation

∫ R E t (x)dx = 1 , because∫ R e−u 2du = √ π .

By direct calculation it is veried thatE t is the solution of the equation (1) andsatises the initial condition

E t (x)|t =+0 = δ (x). (18)Indeed, taking into account (14), we have

limt →+0

Fx [E (x, t )] = F x [limt →+0

E t (x)]= limt→+0

E (α, t ) = 1 = F x [δ (x)], (19)

hence limt →+0 E t (x) = δ (x). (20)

Taking account of (7), (8) and (17), we obtain

G(x,ξ, t ) = E t (x −ξ ) =exp (−

( x − ξ ) 2

4 a 2 t )2a √ πt , t > 0,

E (x −ξ ) = H (t)G(x,ξ, t ).

(21)

These relations show the dependence between the Green function and the fun-damental solution of the heat conduction equation in a homogeneous innitely bar.

Consequently, the solution of the boundary value problem (1), (2) is given by

the convolution with respect tox∈R

u(x, t ) = E t (x)∗φ(x) = ∞

−∞E t (x −ξ )φ(ξ )dξ = ∞

−∞G(x,ξ, t )φ(ξ )dξ, (22)

where φ is considered to be a bounded function andlimt →+0

u(x, t ) = φ(x) in allcontinuity points of the functionφ.

All the results can be generalized toR n . Thus, the heat conduction equation inR n is

∂u (x, t )∂t

= a2∆u (x, t ), t > 0, a > 0, x∈R n , (23)



4 Wilhelm Kecs

where∆ = ∂ 2

∂x 21 + ... + ∂

2

∂x 2n represents the Laplace operator.

The initial condition isu(x, 0) = φ(x), φ∈C 0(R n ). (24)

TheGreenfunctionG(x,ξ, t ), x ,ξ ∈R n , t > 0 for theboundary value problem

(23), (24) represents the solution of the equation (23) with the initial conditionG(x,ξ, t )|t =+0 = δ (x −ξ ). (25)

LetE t (x), x∈R n , t > 0, be the solution of the equation (23) with the initial

conditionE t (x)|t =+0 = δ (x). (26)

Then, the Green functionG(x,ξ, t ) is given by the relationG(x,ξ, t ) = E

t(x

−ξ ). (27)

The boundary value problem (23), (24) is

u(x, t ) = E t (x)∗φ(x) = R n E t (x −ξ )φ(ξ )dξ = R nG(x,ξ, t )φ(ξ )dξ (28)

whereφ is considered a bounded function andlimt →+0

u(x, t ) = φ(x) in all continu-ity points of the functionφ.

We remark that the solutionE t (x) of the equation (23) with the condition (26)can be obtained applying the Fourier transform with respect to the variablex∈R n . Thus, applying the Fourier transform to the equation

∂ E t (x)∂t

= a2∆ E t (x) (29)

and the condition (26), we obtainddt

Ê t (α) + a2∥α∥

2 Ê t (α) = 0 ,ddt

Ê t (α )t =+0

= 1 , (30)

where Ê t (α) = F [ E t (x)] (α ), α∈R n .

The solution of the problem (30) is the functionÊ t (α ) = exp −a2

∥α∥2 t , t > 0. (31)

According to the formula

F exp −b2∥x∥

2 =√ π

b

n

exp −∥α∥

2

4b2 , b > 0, (32)

we obtain

E t (x) = F −1 Ê t (α ) (x) = F −1 exp(−a2∥α∥

2 t)

=1

(2a√ πt )nexp −∥

x∥2

4a2 t. (33)

Hence, the Green function corresponding to the boundary value problem of the heat conduction inR n is

G(x,ξ, t ) = E t (x −ξ ) =1

(2a√ πt )nexp −∥

x −ξ ∥2

4a2 t. (34)




The formulae (28), (34) generalize those obtained for the casen = 1 . Finally,we observe thatE t (x)∈ S (R n ), t > 0, and we have∫ R n E t (x)dx = 1 .LetE (x, t )∈ D′(R n +1 ) be the fundamental solution of the equation (23), hence

of the operator∂ / ∂t −a2∆ of heat conduction inR n . Proceeding as in the casen = 1 , we can establish the relation

E (x, t ) = H (t)E t (x). (35)Thus, we can write

∂E (x, t )∂t −a2∆E (x, t ) = δ (x, t ) = δ (x) ×δ (t). (36)

3. GENERALIZEDPOISSON EQUATION

The generalized Poisson equation is∆u (x,y,z ) −k2u(x,y,z ) = f (x,y,z ), (37)

where∆ is the Laplace operator,k = const , and f is a distribution fromD′(R 3).Fork = 0 we obtain the Poisson equation

∆u (x,y,z ) = f (x,y,z ). (38)By denition, theGreen function corresponding to theequation (37) represents

the solution of the equation∆u (x,y,z ) −k2u(x,y,z ) = δ (x −ξ, y −η, z −τ ), (39)

whereξ, η, τ are parameters.Let E (x,y,z ) ∈ D′(R 3) be the fundamental solution of the operator∆ −k2;

hence, we have

∆E (x,y,z ) −k2

E (x,y,z ) = δ (x,y,z ). (40)The Green functionG(x,y,z ; ξ ,η ,τ ) satises the equation∆G (x,y,z ; ξ ,η ,τ ) −k2G(x,y,z ; ξ ,η ,τ ) = δ (x −ξ, y −η, z −τ ) (41)

and, consequently, the dependence betweenG and E isG(x,y,z ; ξ ,η ,τ ) = E (x −ξ, y −η, z −τ ). (42)

This relation shows that the Green function is obtained by a translation of thedistributionE (x,y,z ) to the point(ξ ,η ,τ ).

Applying the Fourier transform to the equation (40) inR 3 , we obtain

−(α 2 + β 2 + γ 2)E (α,β,γ ) −k2 E (α,β,γ ) = 1 (43)whereF [E (x,y,z )] = E (α,β,γ ) and α, β, γ are real variables.

From (43) it results

E (α,β,γ ) = −1

k2 + α 2 + β 2 + γ 2. (44)

We shall denote byFx [ ], Fy [ ], F z [ ] the Fourier transform inR , with respectto the variablesx, y, z , respectively; we have

F x [1r

exp(−kr )]= 2 K 0 √ α 2 + k2√ y2 + z2 , (45)

whereK 0 represents the modied Bessel function of second kind and zero orderand r = √ x2 + y2 + z2 is the radius vector.



6 Wilhelm Kecs

Also, we can write

F y K 0 √ α 2 + k2√ y2 + z2 =π exp −|z|√ α 2 + β 2 + k2

√ α 2 + β 2 + k2, (46)

F z

exp −|z|√ α 2 + β 2 + k2

√ α 2 + β 2 + k2=

2α 2 + β 2 + γ 2 + k2 . (47)

BecauseF[ ] represents the Fourier transform inR 3 , with respect to all vari-ables, we have

F [1r

exp(−kr )]=4π

α 2 + β 2 + γ 2 + k2 , (48)

wherefromF−1 [ 1

α 2 + β 2 + γ 2 + k2]=1

4πrexp(−kr ), (49)

r being the radius vector.Taking into account (44), we obtain

E (x,y,z ) = F −1 E (α,β,γ ) = −1

4πrexp(−kr ); (50)

in accordance with the relation (42), the Green function corresponding to the gen-eralized Poisson equation has the expression

G(x,y,z ; ξ ,η ,τ ) = −1

4πρexp(−kρ), (51)

whereρ = √ (x −ξ )2 + ( y −η)2 + ( z −τ )2 . (52)

Particularly, fork = 0 , we obtain the Green function corresponding to theequation (38) in the form

G(x,y,z ; ξ ,η ,τ ) = E 1(x −ξ, y −η, z −τ ) = −1

4πρ, (53)

where E 1(x,y,z ) = −1/4 πr represents the fundamental solution of the Laplaceoperator∆ in R 3 , hence

∆E 1(x,y,z ) = δ (x,y,z ). (54)With the help of the Green function (51), the solution of the generalized Pois-

son equation is

u(x,y,z ) = E (x,y,z )∗f (x,y,z ), (55)wheref ∈ D′(R 3) is a distribution with compact support.If f is a distributionof function type forwhich theconvolution productE (x,y,z )∗f (x,y,z ) exists, then, developing the convolution product (55), we obtain

u(x,y,z ) = R 3E (x −ξ, y −η, z −τ )f (ξ ,η ,τ )dξ dηdτ

= R 3G(x,y,z ; ξ ,η ,τ )f (ξ ,η ,τ )dξ dηdτ = L(f (x,y,z )) , (56)

where the kernel of the integral operatorLis the Green function.




Remark 1. . The fundamental solution E (x,y,z ) ∈ D′(R 3) of the operator∆ + k2can be obtained using the derivative rule of the homogeneous functions having at theorigin, x = 0 , a singular point. Then, according with [21], p.126, the functionh(x) =exp(i kr )r 2−n , x∈

R n − {0}, r = ∥x∥, k∈C , n ≥3, is locally integrable, and the

functions x j r 1−n , xj r −n are homogeneous functions. It is shown the relation

∆h (x) + k2h(x) = −ik(n −3)h(x)

r −(n −2)S 1δ (x), (57)

where S 1 = 2 π n /2

/Γ (n/2) represents the area of the unit sphere fromR n . Particularly, forn = 3 , S 1 = 4 π and considering k = i k1 , we obtain the relation

∆h 1(x,y,z ) −k21 h1(x,y,z ) = −4πδ (x,y,z ), (58)

whereh1(x,y,z ) = 1

rexp(−k1r ) , r = √ x2 + y2 + z2 . (59)

From (58), it results

∆ −h1

4π −k21 −

h1

4π= δ (x,y,z ); (60)

hence, the fundamental solution of the operator∆ −k21 is

E (x,y,z ) = −h1(x,y,z )

4π= −

14πr

exp(−k1r ) . (61)

4. GREEN’S FUNCTION FOR THE VIBRATING STRING

For small vibrations of an homogeneous string, in the absence of the externalforces, the motion equation of hyperbolic type is

∂ 2u(x, t )∂t 2 −a2 ∂ 2u(x, t )

∂x 2 = 0 , (62)

where the constanta2 has the expressiona2 = T / ϱ, whereT is the tension in thestring, andϱ is the density of the string per unit of length.

In the case of an innite string the Cauchy problem consists in the determina-tion of the functionu(x, t )∈C 2(R ×R + ) which satises the equation (62), as wellas the initial conditions

u(x, t )|t =0 = φ(x),∂u (x, t )

∂t t =0= ψ(x), (63)

whereφ, ψ

∈

C 0(R ).We remark that the vibrating string (62) as well as the initial condition (63) can

be considered in the distributions space. The solution of the Cauchy problem will be thus the distributionu(x, t )∈ D′(R ) depending on the parametert ≥ 0; φ, ψare distributions fromD′(R ).

The Green function corresponding to the Cauchy problem (62) and (63) forthe vibrating string isG(x,ξ, t ) ∈C 2(R ×R + ), x ∈

R , t ≥ 0, whereξ ∈R is a

parameter, satisfying the conditions∂ 2G∂t 2 −

∂ 2G∂x 2 = 0 , x∈

R , t > 0, (64)



8 Wilhelm Kecs

G(x,ξ, t )|t =+0 = 0 , ∂ ∂t G(x,ξ, t )t =+0

= δ (x −ξ ). (65)

LetE t (x) = E (x, t )∈C 2(R ×R + ) be the solution of the equation (62), hence∂ 2

∂t 2 E t (x) −a2 ∂ 2

∂x 2 E t (x) = 0 , (66)

which satises the conditions

E t (x)|t =+0 = 0 ,∂ ∂t E t (x)

t =+0= δ (x). (67)

In this way, between the functionsG(x,ξ, t ) and E t we haveg(x,ξ, t ) = E t (x −ξ ). (68)

With the help of Green’s function, the solution of the Cauchy problem (62) and(63) for the vibrating string can be write in the form

u(x, t ) = E t (x)∗ψ(x) +∂ ∂t

(E t (x)∗φ(x))

= R E t (x −ξ )ψ(ξ )dξ +∂ ∂t R E t (x −ξ )φ(ξ )dξ. (69)

Taking into account (68), we have

u(x, t ) = RG(x,ξ, t )ψ(ξ )dξ +

∂ ∂t R

G(x,ξ, t )φ(ξ )dξ. (70)

We shall denote byP(D) = P ( ∂ t , ∂ x ) the operator corresponding to the vibrat-ing string, namely

P(D) = ∂ 2

∂t 2 −a2 ∂ 2

∂x 2 . (71)Because

P(D) E t (x) =∂ 2E t (x)

∂t 2 −a2 ∂ 2E t (x)∂x 2 = 0 ,

on the basis of differentiation rule of the convolution product we obtain

P(D) u(x, t ) = P(D) E t (x)∗ψ(x) +∂ ∂t

[P(D) E t (x)∗φ(x)] = 0. (72)

From here it results that the two terms of the solution (69),E t (x)∗ψ(x) and∂ (E t (x)∗ϕ(x))/ ∂t are solutions of the vibrating string equation (62).

Taking into account the conditions (67) we obtain

u(x, t )

|t =+0=

E t (x)

|t =+0∗

ψ(x) +∂ E t (x)

∂t t =+0∗

φ(x)

= 0∗ψ(x) + δ (x)∗φ(x) = φ(x), (73)

∂u (x, t )∂t t =+0

=∂E t (x)

∂t t =+0 ∗ψ(x) +

∂ 2E t (x)∂t 2

t =+0 ∗φ(x)

= δ (x)∗ψ(x) + 0∗φ(x) = ψ(x), (74) because

∂ 2E t (x)∂t 2

t =+0= 0 .




Indeed, from the equation∂ 2E t (x)

∂t 2 −a2 ∂ 2E t (x)∂x 2 = 0 , (75)

by passing to the limit, we obtain

limt →+0

∂ 2E t (x)∂t 2 = a2 ∂ 2

∂x 2 [limt →+0 E t (x)]= a2 ∂ 2

∂x 2 (0) = 0 . (76)

We shall determine now the expression of the functionE t (x), x∈R , t > 0.

Applying the Fourier transform with respect to the variablex∈R to the equa-

tion (75) and the conditions (67), we obtaind 2

dt 2Ê t (α) + a2α 2 Ê t (α ) = 0 ,

Ê t (α )t =+0

= 0 , ddt

Ê t (α)t =+0

= 1 . (77)

The solution of this Cauchy problem is

Ê t (α ) =sin(aαt )

aα. (78)

Applying the inverse Fourier transform, we obtain

E t (x) = F −1x

Ê t (α) = {1

2a , |x| ≤at,0, |x| > at. (79)

Indeed, we can write

Fx [

E t (x)] =

R

eiαx

E t (x)dx =

1

2a at

−at

eiαx dx =1

2iαxeiαx at

−at

=1

2iαa (eiαta −e−iαta

)=1

aαsin(atα ). (80)

Taking into account (68), we shall obtain for Green’s function correspondingto the Cauchy problem of the vibrating spring the expression

G(x,ξ, t ) = E t (x −ξ ) = {1

2a , |x −ξ | ≤at0, |x −ξ | > at = {

12a , ξ ∈[x −at,x + at ],0, ξ /∈[x −at,x + at ].

(81)Thus, (70) becomes

u(x, t ) =1

2a x + at

x −atψ(ξ )dξ +

12a

∂ ∂t

x + at

x −atφ(ξ )dξ. (82)

Applying the Leibniz formula of differentiation of integrals depending on aparameter we obtain

u(x, t ) =φ(x + at ) + φ(x −at )

2+

12a

x + at

x −atψ(ξ )dξ, (83)

which represents the d’Alembert’s formula.We remark that the formula (69) take place in the case when the equation (62)

is considered in the distributions spaceD′(R ), whereu(x, t ) is a distribution fromD′(R ), t > 0 being the parameter, andφ, ψ from the initial conditions (63) beingdistributions with compact support.



10 Wilhelm Kecs

LetE (x, t )∈ D′(R ×R ) be the fundamental solution of the operator (71), hence∂ 2E (x, t )

∂t 2 −a2 ∂ 2E (x, t )∂x 2 = δ (x, t ) = δ (x) ×δ (t). (84)

The expression of E (x, t ) is

E (x, t ) =1

2aH (at − |x|) =

0, t < 0,1

2a , 0 ≤ |x| ≤at,0, |x| > at ≥0,

orE (x, t ) = H (t)E t (x). (85)

Indeed, applying the Fourier transform with respect to the variablex∈R , we

obtaind2

dt2 E (α, t ) + a2α

2

E (α, t ) = δ (t), (86)where E (α, t ) = F x [E (x, t )].

This relation shows that the distributionE (α, t ) is the fundamental solution of the operatord2

/dt2 + a2α 2.But, the fundamental solution of this operator is the distribution

E (α, t ) = H (t)sin(aαt )

aα. (87)

Applying the inverse Fourier transform we obtain

E (x, t ) = F −1x E (α, t ) =

12a

H (at − |x|) , (88)this because

F [H (R − |x|)] = ∫ R

−R eiαx

dx = 2sin( Rα )

α ,

H (R − |x|) = {0, R < |x| ,1, |x| ≤R.

(89)

From (85) and (68), it resultsE (x −ξ, t) = H (t)E t (x −ξ ) = H (t)G(x,ξ, t ), (90)

fact which shows the dependence between the fundamental solutionE (x, t ) of the vibrating spring equation (62) and the Green’s function corresponding toCauchy problem (62), (63).

REFERENCES

[1] Schilov, G.E.,Generalized Functions and Partial Differential Equations, Gordon and Breach Science

Publishers, 1968[2] Kecs, W.W.,Teoria distribut, iilor s, i aplicat, ii, Editura Academiei Române, Bucharest, 2003[3] Nowacki, W.,Dinamica sistemelor elastice, Editura Tehnic˘a, Bucures, ti, 1969

UNIVERSITY OFPETROS, ANIDEPARTMENT OFMATHEMATICSUNIVERSITAT, II NO 20, 332006, PETROS, ANI, ROMÂNIA

Green's Functions in Mathematical Physics

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