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

CHAPTER 5 SOLUTIONS TO THE DIFFUSION E Q U AT I 0 N In Chapter 4 we described many of the general features of the diffusion equation and several methods of solving it when D varies in different ways. We now address in more detail methods to solve the diffusion equation for a variety of initial and boundary conditions when D is constant and therefore has the relatively simple form of Eq. 4.18; that is, dC - = DV2c at 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: nd e r’?/(4Dt) 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, v2c = 0 (5.2) Solutions to the Laplace equation are called harmonic functions. Some harmonic functions are given below for particular boundary conditions. 5.1.1 One Dimension Consider diffusion through an infinite flat plate of thickness L, with 0 < x < L, subject to boundary conditions c(0, t) = co c(L, t) = CL (5.3) 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. 5 1 STEADY-STATE SOLUTIONS 101 Integrating the one-dimensional Laplacian, d2 c /dx2 = 0, twice yields ~(z) = alz + a2 (5.4) where al and a2 are integration constants. Solving for the integration constants using the boundary conditions, Eq. 5.3, produces the one-dimensional steady-state solution, (5.5) 0 0 tz L c(x) = c - (c -c )- i.e., the concentration varies linearly across the plate as illustrated in Fig. 5.1. The flux is constant and proportional to the slope: L 0 X- Figure 5.1: Concentration, C(Z). vs. J for steady-state diffusion through a plate 5.1.2 Cylindrical Shell Consider steady-state diffusion through a cylindrical shell with inner radius r'" and outer radius rollt as in Fig. 5.2. The boundary conditions are c(T'", 8, z, t) = c'" c(rUUt. 8. z. t) = c""~ (5.7) The Laplacian operator operating on c(r, 8. z) in cylindrical coordinates is Figure 5.2: shell. Concentration, c(T), vs. T for steady-state diffusion through a cylindrical 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 c(r) = a1 lnr + a2 and applying the boundary conditions gives - Gout c(r) = cln - In( rout/rin) The flux J = -D(dc/dr) depends inversely on r: ,in - ,out 1 (5.10) (5.11) (5.12) Note that the total current of particles entering the inner surface per unit length of cylinder [I = J(ri")2min] is the same as the total current leaving the outer surface, ,in - ,out (5.13) which is a requirement for the steady state. 5.1.3 Spherical Shell 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. 5.1.4 Variable Diffusivity When steady-state conditions prevail and D varies with position (e.g., D = D(f)), the diffusion equation can readily be integrated. Equation 4.2 then takes the form In one dimension] the solution can then be obtained by integration, I: & c(z) = c(z1) + a1 (5.15) 5 2. NON-STEADY-STATE DIFFUSION 103 5.2 N 0 N - ST 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 ndz, nd,, and nd, are constants. This may be written simply c(r,t) = nd e-r2/(4Dt) (4~Dt)~~~ (5.17) where nd E nd, x nd, x nd, This result has spherical symmetry and describes the spreading of a point source into an infinite domain. Integration verifies that nd is equal to the total amount of diffusant in the system. As t -+ 0, the solution approaches a delta function form, corresponding to the initial localized source [i.e., Table 5.1: 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 d2 e-z2/(4Dt) Line source in 2D 22 c(z,t) = (4n;:)1,2 Plane source in 3D Two-Dimensional Diffusion Point source in 2D Line source in 3D Id d r drrz __ Three-Dimensional Diffusion 1 d 2d e-r2/(4Dt) Point source in 3D Tzr dr 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 22/(4Dt) (r Dt) lI2 c(x,t) = (5.18) 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 nd = 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 - - 4 *Dt (5.19) 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 s6(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 Solutions Involving the Error Function cg x>o { 0 x<o c(x,O) = (5.20) dc dx -(x = kx,t) = 0 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 nd 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 (5.21) So the solution corresponding to the conditions given by Eq. 5.20 must be the integral over all sources, I I -mt I X x=o 6 (5.22) Figure 5.5: 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 2 2 2m 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 x>a (5.24) x < -a c(x,t = 0) = 0 - dC (x = fcqt) = 0 dX 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 nd particles per unit length, the contribution of an effective point source of strength nd d< at (0, O,<) to the point (x, y, z) is nd d< e-[x2+y2+(E-z)2]/(40t) (4n Dt) 3/2 dc = (5.25) so that In cylindrical coordinates, r2 = x2 + y2 and, after integration, in agreement with the entrv in Table 5.1. (5.27) where the source strength nd has dimensions length-'. 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 System Equation Solution One dimension, z - dc dt - - Ds d2c c(a, t) = X(z)T(t) c(r, 8, z, t) = R(r)@(e)qz)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: c(z,O) = co c(0,t) = 0 c(L,t) = 0 (5.28) 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 ~eparable.~ Letting c(x, t) = X(x)T(t) and substituting into the diffusion equation gives 1 dT 1 d2X DT dt X dx2 - - (5.29) Because the left side depends only on t and the right side depends only on x, each side must be equal to the same constant. This may be understood by considering Fig. 5.6, in which f is a function of y only and g is a function of x only. Each surface 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 = -XDT (5.30) dT dt - 71f a solution is found for the initial and boundary conditions, there is a uniqueness theorem that justifies the assumption. Whether a solution can be found using separation of variables depends on whether the boundary conditions follow the symmetry of the separation variables. 108 CHAPTER 5 SOLUTIONS TO THE DIFFUSION EQUATION and X X Figure 5.6: Represer1tat)ion in (z, y) space of g(z) and f(y) - = -AX d2X dx2 (5.31) Equation 5.31 has solutions of the form A sin( fix) + B cos( fix) (A > 0) x(x) = AIeG” + BIe-G” (A 0) (5.32) { A//x + B// (A = 0) 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, A,, = n 2 - 7r2 (5.33) L2 because sin fiL = sin nr = 0, where n is an integer. The A, are the linear differential equation’s eigenvalues for the boundary conditions. Because use of any A,, satisfies the boundary conditions in Eq. 5.28, each of the functions Xn(z) = a, sin nr- (5.34) ( 3 satisfies Eq. 5.31 for the boundary conditions in Eq. 5.28. The X,, are known as the eigenfunctions for the boundary conditions. The general solution to Eq. 5.30 is (5.35) o -XDt T(t) = T e where To is a constant. But A must take on the values given in Eq. 5.33, and t,herefore the time-dependent eigenfunction solutions can be written (5.36) The general solution, satisfying the boundary conditions. is then (by superposi- n2n2 Dtl Lz TTL(t) = T,” e- t’ion) a sum of the products of the eigenfunction solutions and is of the form [...]... a1 = 0: co - + u2 e P (5. 145 ) - m m transform the boundary condition given by Eq 5. 141 t o obtain -D (g) = x=o -a E(0,p) (5. 146 ) Therefore, putting Eq 5. 145 into Eq 5. 146 yields CY Cn (5. 147 ) and (5. 148 ) where h = a / D Find the desired solution by taking the inverse transform using a table o f transforms t o obtain c(x,t ) = co [erf (m)+ X eh=+h2gt erfc (& +h m ) ] (5. 149 ) (b) From Eq 5. 149 the concentration... following series expansions of erf(z): For small z , (5. 142 ) and for large z , 1 1 1 x 3 - - - + 1 X 3 X 5 erfc(z) = 1- erf(z) = 2 3 9 + ) (5. 143 ) (c) Give a physical interpretation of the results in (b) Solution (a) Use the Laplace transform method Transforming the diffusion equation along with the initial condition given by Eq 5. 141 yields the same result as Eq 5. 64: (5. 144 ) The solution is therefore... coefficients are given by (5 .40 ) (5 .41 ) I) .-( .- If u ( z ) is an odd function [ u ( z ) = and the sine expansion is applied, (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: (5 .44 ) (5 .45 ) Finally, any function can be written as a sum of an odd and an even function Using Eq 5 .43 , the coefficients, A,,... function of time to maintain the constant-flux condition (see Fig 5.7) 5.2: NON-STEADY-STATE DIFFUSION 113 Diffusion profiles necessary to maintain constant flux into a semi-infinite Figure 5.7: body Note the fixed value of dc/dz(,=o for t > 0 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... the first term of the series in Eq 5 .47 suffices to a good approximation, such that (5 .48 ) with a maximum error of about 1% The average composition C in this “long-time” regime can be obtained by integration of Eq 5 .48 : 8cO c ( t )= 7 - r 2 D t / L 2 e (5 .49 ) C therefore decays exponentially with the characteristic time r = L 2 / ( r 2 0 ) This is reasonably consistent with estimates of diffusion depths... method of superposition o f point-source solutions can be applied t o this problem Taking the number of particles in a volume dV = dXdqdC equal t o dN = co dXdqdC as a point source and integrating over all point sources in the cube using the point-source solution in Table 5.1, the concentration a t x , y, z is c ( x ,Y l 2 ,t ) co dX dqdC e - [ ( ~ - ~ ) 2 + ( y - q ) 2 + ( z - C ) z ] / ( 4 D t )... 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 problems and problems on the semi-infinite domain After a Laplace transform is performed on the original boundary-value problem, the transformed equation is often easily solved The transformed solution is then back-transformed... transform of a function f(z,) is defined as t (5.52) 111 5.2: NON-STEADY-STATE DIFFUSION The Laplace-transformed f is represented by both the operational form L{f}and the shorthand f The variable p is the transformation variables8 The key utility of the Laplace transform involves its operation on time derivatives: (5.53) Integrating the right-hand side of Eq 5.53 by parts, t=m e-Ptf(x, t ) ~ + p~~ e-ptf(x,... transforming, these become (5.153) (5.1 54) E'(L,p) = EJ'(L,p) (5.155) After determining the four constants o f integration and putting them into Eqs 5.59 and 5.65, the solutions in regions I and II are I E = 6'' co co - - - exp(-ql) + [exp(qz) exp(-qz)] 2P co = exp(-qz) [exp(-qL) - e x p ( q ~ ) l 2P P (5.156) 129 EXERCISES where q = @, Using standard tables of transforms t o transform back t o coordinates,... t ) dt = pelf} - f ( x ,t = 0 ) t=O Therefore, c {} : =pC{f} - f ( x , t = O ) - (5. 54) (5.55) 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, (5.56) The method can be demonstrated by considering diffusion into a semi-infinite body where the surface concentration c(x = 0, t ) is fixed: c(x = 0 , t ) = co d C - ( x = 03, t ) . 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. (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. A, sin (n7rz) n=l which, as seen in Eq. 5 .42 , is a Fourier sine series representation of CO. (5.38) Synopsis of Fourier Series If a function u(z) exists on the interval -L <