Jtam.dvi


JOURNAL OF THEORETICAL

AND APPLIED MECHANICS

2, 39, 2001

ON INTEGRATION OF THE GREEN FUNCTION IN

A DISCONTINUOUS TEMPERATURE FIELD IN A 2D

DOMAIN

Grzegorz Jemielita

Institute of Structural Mechanics, Warsaw University of Technology

e-mail: g.jemielita@il.pw.edu.pl

The paper discusses the problem of Green’s functions application to
stress calculations in the case of a discontinuous temperature field ac-
ting in a 2D domain. It is proved that the double integral of the Green
function for the stressesdoesnot exist.Anewmethodoffinding iterrated
integrals of Green’s functions which enables obtaining correct functions
for the stress σαβ is presented as well.

Key words: Green’s functions, discontinuous temperature field

1. Introduction

Consider an infinite space with a discontinuous temperature field

θ(x1,x2,x3)=

{
θ0(x1,x2) for (x1,x2,x3)∈Ω× (−∞,+∞)

0 for (x1,x2,x3) 6∈Ω× (−∞,+∞)
(1.1)

where Ω is the temperature domain, Ω=<−a,a>×<−b,b> (Fig.1).
We seek for displacement and stress fields in the space. Solution to the

above problem can be reduced to solving the Poisson equation in the form
given by Nowacki (1986)

∇2Φ=

{
mθ0 for (x1,x2)∈Ω

0 for (x1,x2) 6∈Ω
(1.2)

where Φ represents the potential of the thermoelastic displacement, while the
displacements can be represented as follows

uα =Φ,α (1.3)



298 G.Jemielita

Fig. 1.

and

m=
ν

2µ+λ
(1.4)

where
µ,λ – Lame’s constants
ν – constant to be determined by thermal and mechanical pro-

perties of the material
∇2 – 2D Laplace’s operator.

In Eq (1.3) the coma stands for differentiation with respect to xα (α=1,2).

The stresses can be found from the formulae

σαβ =2µ(Φ,αβ −∇
2Φδαβ)

(1.5)

σ33 =

{
−2µmθ0 for (xα)∈Ω

0 for (xα) 6∈Ω

where δαβ is the Kronecker delta.

The approach presented below, based on the application of Green’s func-
tions, is widely used in the literature (cf Nowacki, 1970a,b, 1986).

Consider the infinite space with the unit temperature kernel acting along
the line {L : x1 = ζ1,x2 = ζ2,x3 ∈R}, i.e.

θ= δ(x1− ζ1)δ(x2− ζ2) (1.6)

A solution to the differential equation

∇2Φ0 =mδ(x1− ζ1)δ(x2− ζ2) (1.7)



On integration of the Green function... 299

has the following Green’s function form

G(xα;ζβ)=Φ0 =
m

2π
lnR (1.8)

where
R2 =(x1− ζ1)

2+(x2− ζ2)
2 (1.9)

Employing Eq (1.8), we can represent the potential of the thermoelastic
displacement (according to the theory of potential it is called the logarithmic
potential) as follows

Φ(xα)=
m

2π

∫

Ω

θ(ζβ)lnRdΩζ (1.10)

The following theorems are to be provedwhenmaking use of the potential
theory approach (see Giunter, 1957):

I. If the function θ(ζα) is integrable in the domain (region) Ω, then the
logarithmic potential Φ(xα) is of C

1 class in the entire space, and its
partial derivatives result directly from differentiation of the integrand,
i.e.

Φ,α =
m

2π

∫

Ω

θ(ζβ)(lnR),α dΩζ =
m

2π

∫

Ω

θ(ζβ)
xα−ζα
R2

dΩζ (1.11)

II. Poisson theorem

If the function θ(ζα) integrable in the domain Ω is of C
1 class in a

certain neighbourhood of the point (xα)∈Ω, then the potential Φ(xα)
is of C2 class at the point (xα) and Eq (1.2) is true.

It should be noted, however, that the second order derivatives of the po-
tential Φ, i.e. Φ,αβ cannot be calculated in the way used for the first order
derivative Φ,α, i.e. by differentiating the integrand.
By virtue of Eqs (1.5) and (1.8) Green’s functions for the stresses σ̂αβ can

be written in the form

σ̂11(xα;ζβ)=−G,22 =−
µm

π

(x1− ζ1)
2− (x2− ζ2)

2

R4

σ̂22 =−σ̂11 (1.12)

σ̂12(xα;ζβ)=−G,12 =−
2µm

π

(x1− ζ1)(x2− ζ2)

R4



300 G.Jemielita

If the temperature domain is represented by Eq (1.1) the stresses σαβ have
the following form (see Nowacki, 1970a,b, 1986)

σαβ(xα)=

∫

Ω

θ(ζβ)σ̂αβ(xα;ζγ) dΩζ (1.13)

When one uses the above approach the following questions arise:

• Does double integral (1.13) exist for every (xα)?

• For which (xα) is Eq (1.13) true?

• Whatway should σαβ be calculated fromEq (1.13) for the results to be
correct?

Thepresent paper aims at answering the above questions.Wewill consider
the case of a constant temperature field acting upon the thermally insulated
domain Ω.

2. Displacements

Consider a 2D problem in a space with the temperature field

θ=

{
θ0 for (x1,x2)∈Ω

0 for (x1,x2) 6∈Ω
(2.1)

where θ0 =const.

From Eqs (1.3), (1.11) the following formulae for the displacements uα
yield

u1(xα)=
mθ0

2π

{
(x1−a)

(
arctan

x2− b

x1−a
−arctan

x2+ b

x1−a

)
−

−(x1+a)
(
arctan

x2− b

x1+a
−arctan

x2+ b

x1+a

)
+(x2− b)ln

r1

r2
− (x2+ b)ln

r3

r4

}

(2.2)

u2(xα)=
mθ0

2π

{
(x2− b)

(
arctan

x1−a

x2−b
−arctan

x1+a

x2− b

)
−

−(x2+ b)
(
arctan

x1−a

x2+ b
−arctan

x1+a

x2+ b

)
+(x1−a)ln

r1

r2
− (x1+a)ln

r2

r4

}



On integration of the Green function... 301

where

r21 =(x1−a)
2+(x2− b)

2 r22 =(x1+a)
2+(x2− b)

2

r23 =(x1−a)
2+(x2+ b)

2 r24 =(x1+a)
2+(x2+ b)

2
(2.3)

Eqs (2.2) are true at each point of the space and it can be easily shown
that the functions uα are continuous in the entire space.

3. Stresses

We can rewrite the formula for the thermoelastic potential Φ in the follo-
wing, more suitable form

Φ(xα)=
m

4π

∫

Ω

θ(ζβ)lnR
2 dΩζ (3.1)

In can be noted that the integrand in Eq (1.11) for (xα) 6∈Ω does not reveal
any singular points and the second order derivatives Φ,αβ can be calculated
directly by differentiation of the integrand

Φ,αβ =
m

4π

∫

Ω

θ(ζγ)(lnR
2),αβ dΩζ (3.2)

Slight obstacles can be encountered, however, when one wants to prove
that for (xα)∈Ω the second order derivatives Φ,αβ can be determined from
the formula

Φ,αβ =−
m

4π

[∫

Γ

θ(ζγ)(lnR
2),β cos(n,ζα) dΓζ −

∫

Ω

θ(ζγ),α(lnR
2),β dΩζ

]
(3.3)

where Γ = ∂Ω denotes the boundary of Ω.
Substituting Eq (3.3) into Eq (1.5), the following formulae for the stresses

σαβ in Ω yield

σ11(xα)=

=−
mµθ0

π

(
arctan

x1−a

x2− b
+arctan

x1+a

x2+ b
−arctan

x1−a

x2+ b
−arctan

x1+a

x2− b

)

σ22(xα)= (3.4)

=−
mµθ0

π

(
arctan

x2− b

x1−a
+arctan

x2+ b

x1+a
−arctan

x2+ b

x1−a
−arctan

x2− b

x1+a

)

σ12(xα)=
mµθ0

2π
ln
r21r
2
4

r22r
2
3



302 G.Jemielita

It can be seen that the stresses σαβ calculated fromEq (3.2) for (xα) 6∈Ω are
the same as those represented by Eqs (3.4), therefore the following conclusion
can be drawn: the formulae for stresses (3.4) are true at each point of the
space.
Upon analysing the stresses in the vicinity of Γ it can be easily seen that

the stress σ11 reveals discontinuity for x2 =±b, σ22 reveals discontinuity for
x1 = ±a, while the stress σ12 = σ21 tends to infinity when approaching the
corners of Ω (Fig.1). The same conclusion for the first time was formulated
byGoodier (1937). He also determined the jump,which in our case equals (see
Fig.2)

∆σ=σ(i)ss −σ
(e)
ss =−2µm(θ

(i)−θ(e))=−2µmθ0 (3.5)

while
σ(i)nn =σ

(e)
nn (3.6)

where
σ
(i)
ss ,σ

(i)
nn – stresses on the domain boundary, when approaching the

discontinuity from the inside of Ω

σ
(e)
ss ,σ

(e)
nn – stresses on the domain boundary, when approaching the

discontinuity from the outside.

Fig. 2.

Let (xα)∈Ω, we can calculate then the following iterated integrals



On integration of the Green function... 303

A11(xα)= θ0

b∫

−b

( a∫

−a

σ̂11 dζ1

)
dζ2 =−2µθ0

b∫

−b

( a∫

−a

G,22 dζ1

)
dζ2 =

=−
mµθ0

π

(
arctan

x2− b

x1−a
+arctan

x2+ b

x1+a
−arctan

x2+ b

x1−a
−arctan

x2− b

x1+a

)

(3.7)

B11(xα)= θ0

a∫

−a

( b∫

−b

σ̂11 dζ2

)
dζ1 =−2µθ0

a∫

−a

( b∫

−b

G,22 dζ2

)
dζ1 =

=−
mµθ0

π

(
arctan

x1−a

x2− b
+arctan

x1+a

x2+ b
−arctan

x1−a

x2+ b
+arctan

x1+a

x2− b

)

It can be seen from the above formulae that for (xα)∈Ω,A(xα) 6=B(xα),
on the grounds of the Fubini theorem of the double integral existence, the
double integral does not exist ∀xα ∈Ω. Therefore, Eqs (1.13) are not true.
To find the reason why for (xα)∈Ω Eq (1.13) does not hold let us notice

that by virtue of Eqs (1.5) and (1.13) we have

σ11(xα)=

∫

Ω

θ(ζβ)σ̂(xα;ζγ) dΩζ =−2µ

∫

Ω

θ(ζβ)G(xα;ζβ),22 dΩζ (3.8)

however, from Eqs (1.5) and (1.10) it results

σ11(xα)=−2µΦ,22 =−2µ
(∫

Ω

θ(ζβ)G(xα;ζγ) dΩζ
)

,22
(3.9)

Having two equations at our disposal (Eqs(3.8) and (3.9)) we can conclu-
de that the latter one is true since in Eq (3.8) the second derivative of the
thermoelastic potential is calculated by differentiating directly the integrand,
which can be done only for (xα) 6∈Ω.
Comparing the stresses σ11 calculated from Eq (3.4)1 with the stresses

B11 yields

σ11(xα)=B11(xα)= θ0

a∫

−a

( b∫

−b

σ̂11 dζ2

)
dζ1 =−2µθ0

a∫

−a

( b∫

−b

G,22 dζ2

)
dζ1

(3.10)
In an analogous way

σ22(xα)=A11(xα)= θ0

b∫

−b

( a∫

−a

σ̂22 dζ1

)
dζ2 =−2µθ0

b∫

−b

( a∫

−a

G,11 dζ1

)
dζ2

(3.11)



304 G.Jemielita

It can be proved that the following equations are true

σ12(xα)=σ21(xα)= 2µθ0

b∫

−b

( a∫

−a

G,12 dζ1

)
dζ2 =2µθ0

a∫

−a

( b∫

−b

G,21 dζ2

)
dζ1

(3.12)

4. Conclusions

The considerations presented in Section 3 supply answers to the questions
posed in the introduction. In the considered case of constant temperature we
have proved that:

• Double integral (1.13) does not exist for (xα)∈Ω

• Eq (1.13) yields correct results for (xα) 6∈Ω

• Eq (1.13) can be employed only in the case when the double integral is
iterated,with the integration performedover the variablewith respect to
which the potential G(xα;ζβ) is differentiated, and then over the second
variable, i.e. if the potential G(xα;ζβ) is differentiated with respect to
x1 we integrate it over ζ1, and then over ζ2.

The question whether all the above conclusions are true for an arbitrary
function θ(xα) remains unanswered.

References

1. Giunter N.M., 1957,Teoria potencjału, PWNWarszawa

2. Goodier J.N., 1937, On the integration of the thermoelastic equations,Phil.
Mag., 7, 23

3. Krzyżański M., 1957, Równania różniczkowe cząstkowe rzędu drugiego, Cz.
1, PWNWarszawa

4. Nowacki W., 1970a, Teoria sprężystości, PWNWarszawa

5. Nowacki W., 1970b, Thermal Stresses in a Micropolar Body Induced by the
Action of aDiscontinuousTemperatureField,Bull. Acad. Polon. Sci., Ser. Sci.
Techn.,XVIII, 3, 125-131

6. Nowacki W., 1986, Thermoelasticity, PWN Warszawa, Pergamon Press,
Oxford



On integration of the Green function... 305

O całkowaniu funkcji Greena przy działaniu nieciągłego pola temperatury

Streszczenie

W pracy rozpatrzono problem zastosowania funkcji Greena dla naprężeń w przy-
padku działania nieciągłego pola temperatury w 2-wymiarowymobszarze Ω.Wyka-
zano, że całka podwójna z funkcji Greena dla naprężeń nie istnieje. Podano sposób
obliczania całek iterowanych z funkcji Greena taki, za pomocą którego uzyskujemy
poprawne wzory na naprężenia σαβ.

Manuscript received December 20, 2000; accepted for print February 2, 2001