# Calculation of Some Integrals Arising in Heat Transfer in Grinding

Authors: J. L. G. Santander; P. Castañeda Porras; Yu. L. Ratis; J. M. Isidro; P.  Fernández de Córdoba
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We calculate some integrals involved in the temperature field evaluation of
a workpiece during the grinding process. For the case of dry continuous grinding, this calculation allows a faster computation of the temperature field on surface and inside the workpiece.


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## Body

## 1. Introduction

In the resolution of the heat equation that models the heat transfer during the grinding process [1], there appear integrals [2] of the following type:(1.1)I1(a,b,c):=∫0∞[erf(aσ+bσ)-erf(a-cσ+bσ)]dσ,(1.2)I2(x,y):=|y|π∫-∞∞K0(|x-ξ|)K1(y2+ξ2)y2+ξ2dξ,
which have not been tabulated yet [3]. The first integral arises in the evaluation of the temperature on the surface of the workpiece, while the second integral is used for the evaluation of the field temperature inside the workpiece. On the one hand, despite the fact that these integrals can be computed by using the uniqueness of the solution to the Laplace equation, as done in [2], this is cumbersome, since the integrals to be calculated are subproducts of a wider framework. This framework is the resolution of the heat equation in the stationary regime in two different ways. This paper presents a straightforward proof, based on elementary integral calculus and complex variables. On the other hand, the computation of these improper integrals, which show a parametric dependence on a,b, and c for I1, and x and y for I2, is expensive in order to get the field temperature of the wokpiece being ground. The goal of this paper is the resolution of (1.1) and (1.2), so that the evaluation of the temperature field may be faster.This paper is organized as follows. Sections2 and 3 are devoted to the calculation of I1 and I2, respectively. Section 4 applies I1 and I2 to the solution of the stationary regime in dry continuous grinding within the Samara-Valencia model [1].


## 2. First Integral

In order to calculate the integral (1.1), let us solve the integral (2.1) in two different ways. Afterwards, comparing the results obtained, we will find the sought-for solution. Let us define:(2.1)I:=∫0∞{∫0cexp(-(x-a-4bs)24s)dx}dss,
where a, b, and c are parameters within the integral.


### 2.1. First Calculation

Applying Fubini's theorem to (2.1), we can exchange the integration order, obtaining(2.2)I=∫0c{∫0∞exp(-(x-a-4bs)24s)dss}dx.
Let us develop the exponential within (2.2), as follows:(2.3)-(x-a-4bs)24s=-(x-a)24s-4b2s+2b(x-a).
Calling the inner integral in (2.2) Î and taking into account (2.3), we have(2.4)Î:=e2b(x-a)∫0∞exp(-(x-a)24s-4b2s)dss.
Let us perform the change of variables σ=4b2s in (2.4), and let us define the variable z as(2.5)z:=|2b(x-a)|,
so that (2.4) becomes(2.6)Î=e2b(x-a)∫0∞exp(-z24σ-σ)dσσ.
Knowing the following integral representation of the zero order modified Bessel function [4]:(2.7)K0(z)=12∫0∞exp(-σ-z24σ)σ-1dσ,
and substituting (2.7) in (2.6), we get(2.8)Î=2e2b(x-a)K0(z).
Remembering the definition of z given in (2.5), and substituting (2.8) in (2.2), we find the following expression for I:(2.9)I=2∫0ce2b(x-a)K0(|2b(x-a)|)dx.
Now, performing in (2.9) the change of variables u=2b(x-a), we obtain(2.10)I=1b∫-2ba2b(c-a)euK0(|u|)du.
In the appendix, it is shown that,(2.11)Jg(x):=∫0xeuK0(|u|)du={xex[K0(|x|)+sign(x)K1(|x|)]-1,∀x≠0,0,∀x=0.
Straightforwardly from (2.11), (2.10) can be rewritten as(2.12)I=1b{Jg(2b(c-a))-Jg(-2ba)}.


### 2.2. Second Calculation

Let us consider now the inner integral in (2.1):(2.13)I¯:=∫0cexp(-(x-a-4bs)24s)dx.
Performing the  following change of variables in (2.13) u=(x-a-4bs)/2s, we obtain that(2.14)I¯=2s∫-(a+4bs)/2s(c-a-4bs)/2se-u2du.
We can rewrite (2.14) in terms of the error function [4], so that(2.15)I¯=πs[erf(c-a-4bs2s)+erf(a+4bs2s)].
Therefore, substituting (2.15) in (2.1), we obtain that(2.16)I=π∫0∞1s[erf(a2s+2bs)-erf(a-c2s+2bs)]ds.
Let us perform now the change of variables σ=2s in (2.16), so that (2.16) becomes(2.17)I=π∫0∞[erf(aσ+bσ)-erf(a-cσ+bσ)]dσ.


### 2.3. Comparison

Comparing (2.12) and (2.17), and the remembering the definition given in (1.1), we finally get,(2.18)I1(a,b,c)=1bπ{Jg(2b(c-a))-Jg(-2ba)},
where the function Jg(x) has been defined in (2.11).


## 2.1. First Calculation

Applying Fubini's theorem to (2.1), we can exchange the integration order, obtaining(2.2)I=∫0c{∫0∞exp(-(x-a-4bs)24s)dss}dx.
Let us develop the exponential within (2.2), as follows:(2.3)-(x-a-4bs)24s=-(x-a)24s-4b2s+2b(x-a).
Calling the inner integral in (2.2) Î and taking into account (2.3), we have(2.4)Î:=e2b(x-a)∫0∞exp(-(x-a)24s-4b2s)dss.
Let us perform the change of variables σ=4b2s in (2.4), and let us define the variable z as(2.5)z:=|2b(x-a)|,
so that (2.4) becomes(2.6)Î=e2b(x-a)∫0∞exp(-z24σ-σ)dσσ.
Knowing the following integral representation of the zero order modified Bessel function [4]:(2.7)K0(z)=12∫0∞exp(-σ-z24σ)σ-1dσ,
and substituting (2.7) in (2.6), we get(2.8)Î=2e2b(x-a)K0(z).
Remembering the definition of z given in (2.5), and substituting (2.8) in (2.2), we find the following expression for I:(2.9)I=2∫0ce2b(x-a)K0(|2b(x-a)|)dx.
Now, performing in (2.9) the change of variables u=2b(x-a), we obtain(2.10)I=1b∫-2ba2b(c-a)euK0(|u|)du.
In the appendix, it is shown that,(2.11)Jg(x):=∫0xeuK0(|u|)du={xex[K0(|x|)+sign(x)K1(|x|)]-1,∀x≠0,0,∀x=0.
Straightforwardly from (2.11), (2.10) can be rewritten as(2.12)I=1b{Jg(2b(c-a))-Jg(-2ba)}.


## 2.2. Second Calculation

Let us consider now the inner integral in (2.1):(2.13)I¯:=∫0cexp(-(x-a-4bs)24s)dx.
Performing the  following change of variables in (2.13) u=(x-a-4bs)/2s, we obtain that(2.14)I¯=2s∫-(a+4bs)/2s(c-a-4bs)/2se-u2du.
We can rewrite (2.14) in terms of the error function [4], so that(2.15)I¯=πs[erf(c-a-4bs2s)+erf(a+4bs2s)].
Therefore, substituting (2.15) in (2.1), we obtain that(2.16)I=π∫0∞1s[erf(a2s+2bs)-erf(a-c2s+2bs)]ds.
Let us perform now the change of variables σ=2s in (2.16), so that (2.16) becomes(2.17)I=π∫0∞[erf(aσ+bσ)-erf(a-cσ+bσ)]dσ.


## 2.3. Comparison

Comparing (2.12) and (2.17), and the remembering the definition given in (1.1), we finally get,(2.18)I1(a,b,c)=1bπ{Jg(2b(c-a))-Jg(-2ba)},
where the function Jg(x) has been defined in (2.11).


## 3. Second Integral

In order to calculate the second integral defined in (1.2), let us perform the translation of coordinates, ξ′=x-ξ, thus,(3.1)I2(x,y)=|y|π∫-∞∞K0(|ξ′|)K1(y2+(ξ′+x)2)y2+(ξ′+x)2dξ′.
Let us define the complex integral along the contour C, as presented in Figure 1:(3.2)IC(x,y):=|y|π∫CK0(|ξ′|)K1(y2+(ξ′+x)2)y2+(ξ′+x)2dξ′.
Observe that we can divide the integration path as follows:(3.3)IC(x,y)=IR-(x,y)+IR+(x,y)+ICR(x,y)+ICϵ(x,y),
where we have defined:(3.4)IR±(x,y):=∓|y|π∫±R±ϵK0(|ξ′|)K1(y2+(ξ′+x)2)y2+(ξ′+x)2dξ′,(3.5)ICR,ϵ(x,y):=|y|π∫CR,ϵK0(|ξ′|)K1(y2+(ξ′+x)2)y2+(ξ′+x)2dξ′,
and where CR and Cϵ are the semicircumferences appearing in Figure 1. Taking absolute values and performing in (3.5) the changes of variables ξ′=Reiθ and ξ′=ϵeiθ along ICR and ICϵ, respectively, we obtain that(3.6)|ICR(x,y)|=|yRπK0(R)∫0πK1(y2+(Reiθ+x)2)y2+(Reiθ+x)2dθ|,(3.7)|ICϵ(x,y)|=|yϵπK0(ϵ)∫0πK1(y2+(ϵeiθ+x)2)y2+(ϵeiθ+x)2dθ|.
Taking limits in (3.6),(3.8)limR→∞|ICR(x,y)|=limR→∞|yRπK0(R)∫0πK1(R)Rdθ|=limR→∞|yK0(R)K1(R)|=0.
Taking into account the asymptotic formula [4], (3.9)K0(z)~z→0+log(2z),
we can take limits in (3.7), so that(3.10)limϵ→0+|ICϵ(x,y)|=limϵ→0+|yϵπK0(ϵ)∫0πK1(y2+x2)y2+x2dθ|=|yK1(y2+x2)y2+x2||limϵ→0+ϵlog(2ϵ)|=0.
Observe that from (3.4) and (3.10), the following limit is satisfied:(3.11)limR→∞,ϵ→0+IR-(x,y)+IR+(x,y)=I2(x,y).
Therefore, taking limits in (3.3) and applying the results (3.11), (3.8), we can conclude that(3.12)limR→∞,ϵ→0+IC(x,y)=I2(x,y).
Since the integrand in (3.2) contains an unique singularity inside the contour C at the point,(3.13)ξ0=-x+i|y|,
applying the residue theorem [5], we can contract the contour C to the contour C0, the latter surrounding the neighbourhood of the point ξ0, as shown in Figure 1. Therefore, according to (3.12),(3.14)IC0(x,y)=limR→∞,ϵ→0+IC(x,y)=I2(x,y),
where(3.15)IC0(x,y):=|y|π∫C0K0(|ξ′|)K1(y2+(ξ′+x)2)y2+(ξ′+x)2dξ′.
Taking a circular contour C0 around ξ0, we can perform in (3.15) the change of variables ξ′=ξ0+ηeiθ(3.16)IC0(x,y)=iη|y|π∫02πK0(|ξ0+ηeiθ|)K1(y2+(ξ0+ηeiθ+x)2)y2+(ξ0+ηeiθ+x)2eiθdθ.
Taking now into account the asymptotic formula [4],(3.17)K1(z)~z→0+1z,
we can take limits in (3.16),(3.18)IC0(x,y)=limη→0+IC0(x,y)=i|y|πK0(|ξ0|)limη→0+η∫02πeiθdθy2+(ξ0+ηeiθ+x)2.
Substituting the value of ξ0 given in (3.13), and simplifying,(3.19)IC0(x,y)=i|y|πK0(|ξ0|)limη→0+∫02πdθηeiθ+2i|y|=K0(|ξ0|)=K0(x2+y2).
Finally, from (3.14) and (3.19) we conclude that(3.20)I2(x,y)=K0(x2+y2).Figure 1


## 4. Application to the Grinding Process

### 4.1. Samara-Valencia Model

The Samara-Valencia model setup is depicted in Figure2. The workpiece moves at a constant speed vd and is assumed to be infinite along Ox and Oz, and semiinfinte along Oy. The plane y=0 is the surface being ground. The contact area between the wheel and the workpiece is an infinitely long strip of width δ located parallel to the Oz axis and on the plane y=0. Both the wheel and the workpiece are assumed rigid.Figure 2The Samara-Valencia model [1] solves the convection heat equation,(4.1)∂tT(t,x,y)=k[∂xxT(t,x,y)+∂yyT(t,x,y)]-vd∂xT(t,x,y),
subject to the initial condition(4.2)T(0,x,y)=0,
and the boundary condition,(4.3)k0∂yT(t,x,0)=b(t,x)T(t,x,0)+d(t,x),
where -∞<x<∞and t,y≥0. The first term of (4.3) models the application of coolant over the workpiece surface considering b(t,x) as the heat transfer coefficient. The second term, d(t,x), represents the heat flux entering into the workpiece. This heat flux is generated on the surface by friction between the wheel and the workpiece. The solution of the Samara-Valencia model (4.1)–(4.3) may be presented as the sum of two terms,(4.4)T(t,x,y):=T(0)(t,x,y)+T(1)(t,x,y),
where(4.5)T(0)(t,x,y):=-14πk0∫0tdssexp(-y24ks)×∫-∞∞dx′d(t-s,x′)exp(-(x′-x-vds)24ks),T(1)(t,x,y):=14π∫0tdssexp(-y24ks)∫-∞∞dx′(y2ks-b(t-s,x′)k0)×T(t-s,x′,0)exp(-(x′-x-vds)24ks).
Notice that T(0) contains the friction function d(t,x), and T(1) contains the temperature field on the surface and the heat transfer coefficient b(t,x).


### 4.2. Stationary Regime in Dry Continuous Grinding

Let us consider a wrokpiece being ground with thermal diffusivityk(m2s-1) and thermal conductivity k0(Wm-1K-1). In the stationary regime, t→∞, let us define, (4.6)T(0)(x,y):=limt→∞T(0)(t,x,y),T(1)(x,y):=limt→∞T(1)(t,x,y).
For the case of dry grinding, b(t,x)=0, and a constant flux q(Wm-2) entering into the workpiece, we can rewrite (4.6) as, [2](4.7)T(0)(X,Y)=𝒯∫X-ΔXe-uK0(Y2+u2)du,(4.8)T(1)(X,Y)=Y2π∫-∞∞T(X-u,0)e-uK1(Y2+u2)Y2+u2du,
where we have used dimensionless variables,(4.9)Y:=vd2ky,X:=vd2kx,Δ:=vd2kδ,
and a normalized  temperature,(4.10)𝒯:=qkπk0vd.
Notice that the numerical solution of the temperature field(4.11)T(X,Y)=T(0)(X,Y)+T(1)(X,Y),
is cumbersome since (4.8) involves the temperature field at the workpiece surface inside the integrand. We will see in the following section that this integral equation can be solved by using I2(x,y).


### 4.3. Application ofI2

For simplicity, let us define the functions, forY≥0,(4.12)n(u,Y):=Ye-uK1(Y2+u2)Y2+u2,m(u,Y):=e-uK0(Y2+u2).
so that we can rewrite (1.2), (4.7), and (4.8) into a simpler form,(4.13)m(X,Y)=1π∫-∞∞m(X-u,0)n(u,Y)du,(4.14)T(0)(X,Y)=𝒯∫X-ΔXm(u,Y)du,(4.15)T(1)(X,Y)=12π∫-∞∞T(X-u,0)n(u,Y)du.
Defining the Fourier transform as (4.16)ℱ[f(x)](ω):=f̂(ω):=12π∫-∞∞f(x)eiωxdx,
we may apply the convolution theorem of the Fourier transform to (4.13) and (4.15) [3], obtaining(4.17)m̂(ω,Y)=2m̂(ω,0)n̂(ω,Y),(4.18)T̂(1)(ω,Y)=T̂(ω,0)n̂(ω,Y).
Performing the derivative with respect to X in (4.14),(4.19)∂T(0)∂X=𝒯[m(X-Δ,Y)-m(X,Y)],
and applying the Fourier transform to (4.19), we eventually obtain(4.20)T̂(0)(ω,Y)=h(ω)m̂(ω,Y),
where we have defined (4.21)h(ω):=𝒯1-eiΔωiω.
Multiplying by h(ω) (4.17) and taking into account (4.20), we get,(4.22)T̂(0)(ω,Y)=2T̂(0)(ω,0)n̂(ω,Y).
Let us divide now (4.18) and (4.22) for Y=0,(4.23)T̂(1)(ω,0)T̂(0)(ω,0)=T̂(ω,0)2T̂(0)(ω,0).
But straightforwardly from (4.11), (4.24)T̂(ω,0)=T̂(0)(ω,0)+T̂(1)(ω,0),
so that (4.23) can be rearranged, leading to, (4.25)T̂(1)(ω,0)=T̂(0)(ω,0),
thus,(4.26)T̂(ω,0)=2T̂(0)(ω,0).
Substituting (4.26) into (4.18) and comparing the result to (4.22), (4.27)T̂(1)(ω,Y)=2T̂(0)(ω,0)n̂(ω,Y)=T̂(0)(ω,Y),
so that,(4.28)T̂(ω,Y)=2T̂(0)(ω,Y).
Performing now the Fourier antitransformation in (4.28),(4.29)T(X,Y)=2T(0)(X,Y),
and according to (4.7),(4.30)T(X,Y)=2𝒯∫X-ΔXe-uK0(Y2+u2)du.
Equation (4.29) agrees with [6]. Notice that the numerical computation of T(X,Y) in (4.30) is now quite more simple.


### 4.4. Application ofI1

According to the integral representation of theK0 Bessel function (2.7), we can rewrite (4.30) as(4.31)T(X,Y)=𝒯∫X-ΔXe-udu∫0∞exp(-t-Y2+u24t)dtt.
By Fubini's theorem, we may rearrange (4.31) as(4.32)T(X,Y)=𝒯∫0∞exp(-Y24t)dtt∫X-ΔXexp[-(u2t+t)2]du.
The inner integral in (4.32) can be expressed in terms of the error function [4], (4.33)T(X,Y)=π𝒯∫0∞exp(-Y24t)×[erf(X2t+t)-erf(X-Δ2t+t)]dtt.
Performing the change of variables σ=2t we get (4.34)T(X,Y)=π𝒯∫0∞exp(-Y2σ2)×[erf(Xσ+σ2)-erf(X-Δσ+σ2)]dσ,
so that the temperature field on the surface, Y=0, is(4.35)T(X,0)=π𝒯∫0∞[erf(Xσ+σ2)-erf(X-Δσ+σ2)]dσ.
Applying now (2.18), taking a=X,b=1/2, and c=Δ, we may calculate (4.35), (4.36)T(X,0)=2𝒯[Jg(Δ-X)-Jg(-X)].


## 4.1. Samara-Valencia Model

The Samara-Valencia model setup is depicted in Figure2. The workpiece moves at a constant speed vd and is assumed to be infinite along Ox and Oz, and semiinfinte along Oy. The plane y=0 is the surface being ground. The contact area between the wheel and the workpiece is an infinitely long strip of width δ located parallel to the Oz axis and on the plane y=0. Both the wheel and the workpiece are assumed rigid.Figure 2The Samara-Valencia model [1] solves the convection heat equation,(4.1)∂tT(t,x,y)=k[∂xxT(t,x,y)+∂yyT(t,x,y)]-vd∂xT(t,x,y),
subject to the initial condition(4.2)T(0,x,y)=0,
and the boundary condition,(4.3)k0∂yT(t,x,0)=b(t,x)T(t,x,0)+d(t,x),
where -∞<x<∞and t,y≥0. The first term of (4.3) models the application of coolant over the workpiece surface considering b(t,x) as the heat transfer coefficient. The second term, d(t,x), represents the heat flux entering into the workpiece. This heat flux is generated on the surface by friction between the wheel and the workpiece. The solution of the Samara-Valencia model (4.1)–(4.3) may be presented as the sum of two terms,(4.4)T(t,x,y):=T(0)(t,x,y)+T(1)(t,x,y),
where(4.5)T(0)(t,x,y):=-14πk0∫0tdssexp(-y24ks)×∫-∞∞dx′d(t-s,x′)exp(-(x′-x-vds)24ks),T(1)(t,x,y):=14π∫0tdssexp(-y24ks)∫-∞∞dx′(y2ks-b(t-s,x′)k0)×T(t-s,x′,0)exp(-(x′-x-vds)24ks).
Notice that T(0) contains the friction function d(t,x), and T(1) contains the temperature field on the surface and the heat transfer coefficient b(t,x).


## 4.2. Stationary Regime in Dry Continuous Grinding

Let us consider a wrokpiece being ground with thermal diffusivityk(m2s-1) and thermal conductivity k0(Wm-1K-1). In the stationary regime, t→∞, let us define, (4.6)T(0)(x,y):=limt→∞T(0)(t,x,y),T(1)(x,y):=limt→∞T(1)(t,x,y).
For the case of dry grinding, b(t,x)=0, and a constant flux q(Wm-2) entering into the workpiece, we can rewrite (4.6) as, [2](4.7)T(0)(X,Y)=𝒯∫X-ΔXe-uK0(Y2+u2)du,(4.8)T(1)(X,Y)=Y2π∫-∞∞T(X-u,0)e-uK1(Y2+u2)Y2+u2du,
where we have used dimensionless variables,(4.9)Y:=vd2ky,X:=vd2kx,Δ:=vd2kδ,
and a normalized  temperature,(4.10)𝒯:=qkπk0vd.
Notice that the numerical solution of the temperature field(4.11)T(X,Y)=T(0)(X,Y)+T(1)(X,Y),
is cumbersome since (4.8) involves the temperature field at the workpiece surface inside the integrand. We will see in the following section that this integral equation can be solved by using I2(x,y).


## 4.3. Application ofI2

For simplicity, let us define the functions, forY≥0,(4.12)n(u,Y):=Ye-uK1(Y2+u2)Y2+u2,m(u,Y):=e-uK0(Y2+u2).
so that we can rewrite (1.2), (4.7), and (4.8) into a simpler form,(4.13)m(X,Y)=1π∫-∞∞m(X-u,0)n(u,Y)du,(4.14)T(0)(X,Y)=𝒯∫X-ΔXm(u,Y)du,(4.15)T(1)(X,Y)=12π∫-∞∞T(X-u,0)n(u,Y)du.
Defining the Fourier transform as (4.16)ℱ[f(x)](ω):=f̂(ω):=12π∫-∞∞f(x)eiωxdx,
we may apply the convolution theorem of the Fourier transform to (4.13) and (4.15) [3], obtaining(4.17)m̂(ω,Y)=2m̂(ω,0)n̂(ω,Y),(4.18)T̂(1)(ω,Y)=T̂(ω,0)n̂(ω,Y).
Performing the derivative with respect to X in (4.14),(4.19)∂T(0)∂X=𝒯[m(X-Δ,Y)-m(X,Y)],
and applying the Fourier transform to (4.19), we eventually obtain(4.20)T̂(0)(ω,Y)=h(ω)m̂(ω,Y),
where we have defined (4.21)h(ω):=𝒯1-eiΔωiω.
Multiplying by h(ω) (4.17) and taking into account (4.20), we get,(4.22)T̂(0)(ω,Y)=2T̂(0)(ω,0)n̂(ω,Y).
Let us divide now (4.18) and (4.22) for Y=0,(4.23)T̂(1)(ω,0)T̂(0)(ω,0)=T̂(ω,0)2T̂(0)(ω,0).
But straightforwardly from (4.11), (4.24)T̂(ω,0)=T̂(0)(ω,0)+T̂(1)(ω,0),
so that (4.23) can be rearranged, leading to, (4.25)T̂(1)(ω,0)=T̂(0)(ω,0),
thus,(4.26)T̂(ω,0)=2T̂(0)(ω,0).
Substituting (4.26) into (4.18) and comparing the result to (4.22), (4.27)T̂(1)(ω,Y)=2T̂(0)(ω,0)n̂(ω,Y)=T̂(0)(ω,Y),
so that,(4.28)T̂(ω,Y)=2T̂(0)(ω,Y).
Performing now the Fourier antitransformation in (4.28),(4.29)T(X,Y)=2T(0)(X,Y),
and according to (4.7),(4.30)T(X,Y)=2𝒯∫X-ΔXe-uK0(Y2+u2)du.
Equation (4.29) agrees with [6]. Notice that the numerical computation of T(X,Y) in (4.30) is now quite more simple.


## 4.4. Application ofI1

According to the integral representation of theK0 Bessel function (2.7), we can rewrite (4.30) as(4.31)T(X,Y)=𝒯∫X-ΔXe-udu∫0∞exp(-t-Y2+u24t)dtt.
By Fubini's theorem, we may rearrange (4.31) as(4.32)T(X,Y)=𝒯∫0∞exp(-Y24t)dtt∫X-ΔXexp[-(u2t+t)2]du.
The inner integral in (4.32) can be expressed in terms of the error function [4], (4.33)T(X,Y)=π𝒯∫0∞exp(-Y24t)×[erf(X2t+t)-erf(X-Δ2t+t)]dtt.
Performing the change of variables σ=2t we get (4.34)T(X,Y)=π𝒯∫0∞exp(-Y2σ2)×[erf(Xσ+σ2)-erf(X-Δσ+σ2)]dσ,
so that the temperature field on the surface, Y=0, is(4.35)T(X,0)=π𝒯∫0∞[erf(Xσ+σ2)-erf(X-Δσ+σ2)]dσ.
Applying now (2.18), taking a=X,b=1/2, and c=Δ, we may calculate (4.35), (4.36)T(X,0)=2𝒯[Jg(Δ-X)-Jg(-X)].



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*Source: 535801-2010-09-16.xml*