Kinetics of Materials - R. Balluff S. Allen W. Carter (Wiley 2005) WW Part 4 ppt

100 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION boundary-value problems can therefore be adopted as solutions to corresponding mass diffusion problems.2 For problems with relatively

dC

- at = DV2c

This equation is a second-order linear partial-differential equation with a rich math- ematical literature [l] For a large class of initial and boundary conditions, the solu- tion has theorems of uniqueness and existence as well as theorems for its maximum and minimum values

Many texts, such as Crank’s treatise on diffusion [2], contain solutions in terms

of simple functions for a variety of conditions-indeed, the number of worked prob- lems is enormous As demonstrated in Section 4.1, the differential equation for

the “diffusion” of heat by thermal conduction has the same form as the mass diffusion equation, with the concentration replaced by the temperature and the mass diffusivity replaced by the thermal diffusivity, K Solutions to many heat-flow

‘If the diffusivity is imaginary, the diffusion equation has the same form as the time-dependent Schrodinger’s equation at zero potential Also, Eq 4.18 implies that the velocity of the diffusant can be infinite Schrodinger’s equation violates this relativistic principle

99

Kinetics of Materials By Robert W Balluffi, Samuel M Allen, and W Craig Carter

Copyright @ 2005 John Wiley & Sons, Inc

100 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION

boundary-value problems can therefore be adopted as solutions to corresponding

mass diffusion problems.2

For problems with relatively simple boundary and initial conditions, solutions can probably be found in a library However, it can be difficult to find a closed-form solution for problems with highly specific and complicated boundary conditions In such cases, numerical methods could be employed For simple boundary conditions, solutions to the diffusion equation in the form of Eq 4.18 have a few standard forms, which may be summarized briefly

For various instantaneous localized sources diffusing out into an infinite medium, the solution is a spreading Gaussian distribution:

2d (.rrDt)d/2

C ( F , t ) = where d is the dimensionality of the space in which matter is diffusing and nd is the source strength introduced in Section 4.2.3 When the initial condition can be represented by a distribution of sources, one simply superposes the solutions for the individual sources by integration, as in Section 4.2.3 When the boundaries are planar orthonormal surfaces, solutions to the diffusion equation have the form of trigonometric series For diffusion in a cylinder, the trigonometric series is replaced

by a sum over Bessel functions For diffusion with spherical symmetry, trigonomet- ric functions apply All such solutions can be obtained by the separation-of-variables method, which is described below

A third method-solution by Laplace transforms-can be used to derive many of the results already mentioned It is a powerful method, particularly for complicated problems or those with time-dependent boundary conditions The difficult part

of using the Laplace transform is back-transforming to the desired solution, which usually involves integration on the complex domain Fortunately, Laplace transform tables and tables of integrals can be used for many problems (Table 5.3)

5.1 STEADY-STATE SOLUTIONS

A particularly simple case occurs when the diffusion is in a steady state and the composition profile is therefore not a function of time Steady-state conditions are often achieved for constant boundary conditions in finite samples at very long times.3 Then dc/dt = 0, all local accumulation (divergence) vanishes, and the diffusion equation reduces to the Laplace equation,

2Carslaw and Jaeger’s treatise on heat flow is a primary source [3]

3Estimates of times required for “nearly steady-state’’ conditions are addressed in Section 5.2.6

102 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION

Because the boundary conditions are independent of 0 and z , the solution will also

be independent of these variables The solution must therefore satisfy

Integrating twice produces

The Laplacian operator operating on c(r, 8,#) in spherical coordinates is

The steady-state solution for diffusion through spherical shells with boundary con- ditions dependent only on r may be obtained by integrating twice and determining the two constants of integration by fitting the solution to the boundary conditions

5 2 NON-STEADY-STATE DIFFUSION 103 5.2 N 0 N - S T E A DY - STAT E (TI M E- D E P E N D E N T) D I F F U S I 0 N

When the diffusion profile is time-dependent, the solutions to Eq 4.18 require considerably more effort and familiarity with applied mathematical methods for solving partial-differential equations We first discuss some fundamental-source solutions that can be used to build up solutions to more complicated situations by means of superposition

5.2.1 Instantaneous Localized Sources in Infinite Media

Equation 4.40 gives the solution for one-dimensional diffusion from a point source

on an infinite line, an infinite thin line source on an infinite plane, and a thin planar source in an infinite three-dimensional body (summarized in Table 5.1) Corresponding solutions for two- and three-dimensional diffusion can easily be ob- tained by using products of the one-dimensional solution For example, a solution for three-dimensional diffusion from a point source is obtained in the form

where n d z , n d , , and n d , are constants This may be written simply

in One-, Two-, and Three-Dimensional Infinite Media Fundamental Solutions for Instantaneous, Localized Sources

Solution Type Symmetric Part of V2 Fundamental Solution

One-Dimensional Diffusion Point source in 1D

Line source in 2D 2 2 c ( z , t ) = (4n;:)1,2

Plane source in 3D

Two-Dimensional Diffusion Point source in 2D

Line source in 3D r d r r z _ _ I d d

Three-Dimensional Diffusion

Point source in 3D T z r d r C ( T , t ) = (4n;:)3,2

104 CHAPTER 5 : SOLUTIONS TO THE DIFFUSION EQUATION

C(T, t = 0 ) = b ( ~ ) ] ~ Corresponding results for two-dimensional diffusion are given

in Table 5.1

The form of the solution for one-dimensional diffusion is illustrated in Fig 5.3

The solution c(x, t) is symmetric about x = 0 (i.e., c(x, t ) = c(-x, t)) Because the flux at this location always vanishes, no material passes from one side of the plane

to the other and therefore the two sides of the solution are independent Thus the general form of the solution for the infinite domain is also valid for the semi-infinite domain (0 < x < m) with an initial thin source of diffusant at x = 0 However, in the semi-infinite case, the initial thin source diffuses into one side rather than two and the concentration is therefore larger by a factor of two, so that

nd e - - 2 2 / ( 4 D t )

(r Dt) lI2

Figure 5.3: Spreading of point, line, and planar diffusion sources with increasing time

according to the one-dimensional solution in Table 5.1 Curves were calculated from Eq 5.18

for times shown and n d = 1, D = and -1 < 3: < 1 (all units arbitrary)

Equation 5.18 offers a convenient technique for measuring self-diffusion coeffi-

cients A thin layer of radioactive isotope deposited on the surface of a flat specimen serves as an instantaneous planar source After the specimen is diffusion annealed, the isotope concentration profile is determined With these data, Eq 5.18 can be

written

X2

In *c = constant - -

and *D can be determined from the slope of a In *c vs x2 plot, as shown in Fig 5.4.5

4A delta function, 6(F'),is a distribution that equals zero everywhere except where its argument is

zero, where it has an infinite singularity It has the property s j(F')6(F- Fo)dr'= f ( r ' 0 ) ; so it also follows that s 6 ( F - Fo)dr'= 1 The singularity of 6 ( F - TO) is located at Fo

5This technique can be used to measure the diffusivity in anisotropic materials, as described in Section 4.5 Measurements of the concentration profile in the principal directions can be used to

determine the entire diffusion tensor

5 2 NON-STEADY-STATE DIFFUSION 105

,,-+Slope = -1/(4 *Dt)

\

Figure 5.4:

planar source is used and Eq 5.18 applies

Plot of In *c vs xz used to determine self-diffusivity when an iiistaiitaneous

5.2.2

The instantaneous local-source solutions in Table 5.1 can be used to build up solu- tions for general initial distributions of diffusant by using the method of superpo- sition (see Section 4.2.3)

Section 4.2.2 shows how to use the scaling method to obtain the error function solution for the one-dimensional diffusion of a step function in an infinite medium given by Eq 4.31 The same solution can be obtained by superposing the one- dimensional diffusion from a distribution of instantaneous local sources arrayed to

simulate the initial step function The boundary and initial conditions are

The initial distribution is simulated by a uniform distribution of point, line, or planar sources placed along x > 0 as in Fig 5.5 The strength, or the amount

of diffusant contributed by each source, must be co dx The superposition can be achieved by replacing n d in Table 5.1 with c(Z)dV [c(x)dx in one dimension] and integrating the sources from each point

Consider the contribution at a general position x from a source at some other position 5 The distance between the general point x and the source is 5 - x, thus

souice of strength cg d< located at [

Diagram used to determine the contribution at the general point z of a local

106 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION

or by transforming the integration variable by using u = (< - x ) / m and the properties of an even integrand,

c ( x , t ) = - e-u2 du

(5.23)

= - co + -erf co (-) X

which is consistent with the solution given by Eq 4.31

the general method of Green's functions

conditions for a triangular source are

Summations over point-, line-, or planar-source solutions are useful examples of

For instance, the boundary and initial

A solution to this boundary-value problem can be obtained by using Eq 4.40 with

a position-dependent point-source density (this method is useful for solving Exer- cise 5.7)

As a last example, the solution for two-dimensional diffusion from a line source lying along z in three dimensions can be obtained by integrating over a distribution

of point sources lying along the z-axis If the point sources are distributed so that the source strength along the line is n d particles per unit length, the contribution

of an effective point source of strength n d d< at (0, O,<) to the point ( x , y, z ) is

n d d< e-[x2+y2+(E-z)2]/(40t) (4n D t ) 3/2

6Green's functions arise in the general solution to many partial-differential equations They are generally obtained from the fundamental solution for a point, line, or planar source Subsequently,

an integral equation for a general solution is obtained by integrating over all the source terms; the fundamental solution becomes the kernel to the integral equation, which is the term that multiplies the source density in the integrand

5.2: NON-STEADY-STATE DIFFUSION 107

5.2.3 Method of Superposition

In Section 4.2.3 we described application of the method of superposition to infinite and semi-infinite systems The method can also be applied, in principle, to finite systems, but it often becomes unwieldy (see Crank’s discussion of the reflection method [2])

5.2.4 Method of Separation of Variables: Diffusion on a Finite Domain

A standard method to solve many partial-differential equations is to assume that the solution can be written as a product of functions, each a function of one of the independent variables Table 5.2 provides several functional forms of such solutions

Table 5.2:

Cartesian and Cylindrical Coordinates

Product Solutions for the Separation-of-Variables Method in

One dimension, z - dc d t - Ds d 2 c c(a, t ) = X ( z ) T ( t )

c(r, 8, z , t ) = R ( r ) @ ( e ) q z ) T ( t )

Three dimensions, (z, y , z ) = DV2c c(a, y , z , t ) = X ( z ) Y ( y ) Z ( z ) T ( t )

dc - Cylindrical, ( T , 8, z ) - dt - ~o~~

The following example illustrates the method Consider a one-dimensional dif- fusion problem with the initial and boundary conditions for the domain 0 < x < L:

This situation may represent the diffusion of a high-vapor-pressure dopant out of a thin film (thickness L , initial dopant concentration cg) of silicon when placed in a vacuum Assume that the variables are ~ e p a r a b l e ~ Letting c(x, t) = X ( x ) T ( t ) and substituting into the diffusion equation gives

is a “ruled” surface; that is, the surface contains lines of constant value running in one direction If the two functions are equal as in the separation equation (Eq 5.29),

the surface must be flat in both variables Thus, if the two functions are equal, they are constant

Let that constant be -A Then

108 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION

Equation 5.31 has solutions of the form

A sin( fix) + B cos( fix) (A > 0)

x ( x ) = { A I e G ” A//x + B// + B I e - G ” (A 0) (A = 0) (5.32)

where A and B are constants that must satisfy t,he boundary conditions For the

particular boundary conditions specified in Eq 5.28, nontrivial solutions to Eq 5.31

exist only if A > 0 and B = 0 However, there is no nonzero A that can satisfy the boundary conditions for a general X > 0, so X must take on values appropriate to the boundary conditions Therefore,

t,herefore the time-dependent eigenfunction solutions can be written

(5.36)

The general solution, satisfying the boundary conditions is then (by superposi-

n 2 n 2 D t l L z TTL(t) = T,” e-

t’ion) a sum of the products of the eigenfunction solutions and is of the form

Synopsis of Fourier Series

If a function u ( z ) exists on the interval -L < z < L, u ( z ) can be represented as

u(z) = < + [ansin (n.2) + bncos

n=l where the coefficients are given by

If u ( z ) is an odd function [ u ( z ) = I -.) -.( and the sine expansion is applied,

(5.39)

(5.40) (5.41)

(5.42)

(5.43)

Similarly, if u ( z ) is an even function [u(z) = u ( - z ) ] , all an will vanish and u ( z ) can

be written as a cosine expansion only:

Finally, any function can be written as a sum of an odd and an even function

(5.44) (5.45)

Using Eq 5.43, the coefficients, A,, are then given by

(5.46)

Therefore, the final solution is given by

(5.47) The coefficients of the higher-order (shorter-wavelength) terms in Eq 5.47 de-

crease as l/n Not only do the shorter-wavelength terms start out smaller but they

110 CHAPTER 5 : SOLUTIONS TO THE DIFFUSION EQUATION

also decay exponentially at rates that scale inversely as the square of the wave- length Thus, even at times as short as L2/(250), the first term of the series in

Eq 5.47 suffices to a good approximation, such that

of the system is higher [e.g., c(z, y, z , t ) ] , a separate Fourier series must be obtained for each of the three separate functions in the product X ( x ) Y ( y ) Z ( z )

boundary conditions and initial conditions have cylindrical symmetry (see Eqs 5.7 and 5.8) If c ( r , t ) = R(r)T(t), the resulting ordinary differential equation for R(r)

is

d2R 1dR

- dr2 + r dr + a2R = 0 This equation has general solutions

(5.50)

(5.51)

where Jo and YO are Bessel functions of order zero of the first and second kind The

an are solved by matching boundary conditions, and the coefficients a, and bn are determined by matching the initial conditions in a Bessel function series expansion See Carslaw and Jaeger for examples [3]

5.2.5 Method of Laplace Transforms

The Laplace transform method is a powerful technique for solving a variety of partial-differential equations, particularly time-dependent boundary condition prob- lems and problems on the semi-infinite domain After a Laplace transform is per- formed on the original boundary-value problem, the transformed equation is often easily solved The transformed solution is then back-transformed to obtain the desired solution

The Laplace transform of a function f(z, t ) is defined as

(5.52)

The Laplace transform of a spatial derivative of f is seen from Eq 5.52 to be

equal to the spatial derivative of f; that is,

112 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION

Solving for the coefficients in Eq 5.59 leads to the solution

(5.61)

This solution is then inversely transformed by use of a table of Laplace transforms

(see Table 5.3) to obtain the desired solution:

c(x, t ) = co [ 1 - erf (&)I - = coerfc ( 7 3 (5.62)

where erfc(z) = 1 - erf(z) is known as the complementary error function Note that this solution could have been deduced directly from Eq 5.23 (the solution for the

step-function initial conditions for an infinite system) because in that solution, the plane 2 = 0 always maintains a constant composition

Table 5.3: Selected Laplace Transform Pairs

1

pyil v > - 1 r ( v t" + 1)

w w2 + w 2 sin w t

cos w t

Example with Time-Dependent Boundary Conditions

constant flux, Jo, is imposed on the surface of a semi-infinite sample: Consider the case where a

-(x dX = 0 , t ) = D =constant

c(x = 0 , t ) = co (5.63) c(x, t = 0 ) = co for 0 5 x < 00

This boundary condition might apply for solute absorption with its rate moderated

by some thin passive surface layer Note that the surface concentration at x = 0

must be a function of time to maintain the constant-flux condition (see Fig 5.7)

5.2: NON-STEADY-STATE DIFFUSION 113

Figure 5.7:

body Note the fixed value of dc/dz(,=o for t > 0

Diffusion profiles necessary to maintain constant flux into a semi-infinite

Using the Laplace transform,

(5.64) Equation 5.64 is an inhomogeneous ordinary differential equation and its solution

is therefore the sum of the solution of its homogeneous form (i.e., Eq 5.59) and a particular solution (i.e., E = c o / p ) Therefore,

The transformed boundary conditions are

%(x = 0 , p ) = 5,

CO

P E(x = C q p ) = - Solving for the coefficients a1 and a2,

114 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION

An estimate of the penetration distance for the error-function solution (Eq 5.23)

is the distance where c(x, t ) = ~ 0 / 8 , or equivalently, e r f [ x / ( 2 m ) ] = -3/4: which corresponds to

A reasonable estimate for the penetration depth is therefore again 2m

To estimate the time at which steady-state conditions are expected, the required penetration distance is set equal to the largest characteristic length over which diffusion can take place in the system If L is the characteristic linear dimension

of a body, steady state may be expected to apply at times r >> L2/Dmin, where

Dmin is the smallest value of the diffusivity in the body Of course, there are many physical situations where steady-state conditions will never arise, such as when the boundary conditions are time dependent or the system is infinite or semi-infinite

Bibliography

1 P.M Morse and H Feshbach Methods of Theoretical Physics, Vols 1 and 2 McGraw-

2 J Crank The Mathematics of Diffusion Oxford University Press, Oxford, 2nd edition,

3 H.S Carslaw and J.C Jaeger Conduction of Heat i n Solids Oxford University Press,

Hill, New York, 1953

1975

Oxford, 2nd edition, 1959

EXERCISES

5.1 A flat bilayer slab is composed of layers of material A and B , each of thickness

L A component is diffusing through the bilayer in the steady state under conditions where its concentration is maintained at c = co = constant at one surface and at c = 0 at the other Its diffusivity is equal to the constants D A

and D B in the two layers, respectively No other components in the system diffuse significantly

Does the flux through the bilayer depend on whether the concentration is maintained at c = co at the surface of the A layer or the surface of the B

layer? Assume that the concentration of the diffusing component is continuous

at the A / B interface

Solution Solve for the difFusion in each layer and match the solutions across the A / B

interface Assume that c = co at the surface of the A layer and let c = c ~ / ~ be the concentration at the A / B interface Using Eq 5.5, the concentration in the A layer in the interval 0 < x < L is

(5.72)

(5.73) For the B slab in the interval L 5 x 5 2 L ,

(5.75)

Solution The gradient operator in cylindrical coordinates is

Therefore, the steady-state radially-symmetric difFusion equation becomes

which can be integrated twice t o give

C(T = R'" + AR) = coUt Will the total diffusion current through the cylinder

be the same if the materials that make up the inner and outer shells are exchanged? Assume that the concentration of the diffusant is the same in the inner and outer layers at the bilayer interface

Solution The concentration profile at the bilayer interface will not have continuous derivatives Break the problem into separate difFusion problems in each layer and then impose the continuity of flux at the interface Let the concentration at the bilayer interface be

Inner region: R'" 5 r 5 Ri" +

116 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION

Using Eq 5.82,

The flux at the bilayer interface is

Outer region: R'" + 5 r 5 Rin + AR

C ~ u t - ci/o

R'"+AR

In ( R 1 " + A R / 2 ) In ( Rin + AR/2 COUt(T) =

The flux at the bilayer interface is

Setting the fluxes at the interfaces equal and solving for ci/O yields

C y ~ u t out + ainCin aout + Cyin

ci/o = c where

5.4 Suppose that a very thin planar layer of radioactive Au tracer atoms is placed between two bars of Au to produce a thin source of diffusant as illustrated

in Fig 5.8 A diffusion anneal will cause the tracer atoms to spread by self-diffusion as illustrated in Fig 5.3 (A mathematical treatment of this spreading out is presented in Section 4.2.3.) Suppose that the diffusion ex-

EXERCISES 117

Thin source

periment is now carried out with a constant electric current passing through the bars along x

(a) Using the statement of Exercise 3.10, describe the difference between the

way in which the tracer atoms spread out when the current is present and when it is absent

(b) Assuming that DVCV is known, how could you use this experiment to determine the electromigration parameter p for Au?

Inert

em bedded marker

Figure 5.9: (a) The initially thin distribution of tracer atoms that, subsequently, will

spread due to diffusion and drift due to electromigration (b) The electromigration has caused the distribution to spread out and to be translated bodily by Ax = ( V A ) t relative to the fixed marker

118 CHAPTER 5: SOLUTIONS TO THE DIFFUSION EQUATION

This may be shown by choosing an origin at the initial position of the source in

a coordinate system fixed with respect t o the marker The diffusion equation is then

5.5 Obtain the instantaneous plane-source solution in Table 5.1 by representing the plane source as an array of instantaneous point sources in a plane and integrating the contributions of all the point sources

Solution Assume an infinite plane containing m point sources per unit area each of strength nd The plane is located in the ( y ~ ) plane a t x = 0 All the point sources

in the plane lying within a thin annular ring of radius r and thickness dr centered on the z-axis will contribute a concentration a t the point P located along the x-axis at a distance, x , given by

n d e-(z2+r2)/(4Dt)

(47rDt)3I2

where the point-source solution in Table 5.1 has been used The total concentration is

then obtained by integrating over all the point sources in the plane, so that

where M = mnd is the total strength of the planar source per unit area

5.6 Consider an infinite bar extending from -cc to +cc along x Starting at

t = 0, heat is generated at a constant rate in the x = 0 plane Show that the temperature distribution along the bar is

(5.100)