AND APPROXIMATION

methods for solving partial differential equations which are discussed in Chapter 4, the

similarity method may be used with nonlinear as well as linear problems. The technique is illustrated by an example in this section and by several other applications throughout the book.

A similarity solution is developed here for the short-time, transient diffusion prob-

lem described in Example 3.4-1, involving a step change in concentration at and

= Using the dimensionless concentration the governing equations are

One requirement of the similarity method is seen to be satisfied, in that this problem contains no fixed characteristic length. That is, there is not a boundary condition im- posed at some finite value of such as = L.

Assume now that can be expressed as a function of a single independent vari- able where

and is a scale factor which is to be determined. The only difference between and the penetration depth in section 3.4 is that, whereas was used only to describe orders of magnitude, will have specific numerical values. Although our knowledge of indicates that we will show how to derive without such prior information.

The derivatives in (3.5-1) are expressed now in terms of In this introductory example, subscripts are used as reminders of which variable is being held constant in each partial derivative. The derivatives are given by

terms of the similarity variable, (3.5-1) becomes

assumption that = is correct, then neither t nor can appear separately in equation and other conditions for 8. Because g this will be true

for (3.5-9) only if the product is a constant. The exact value of the constant is

immaterial, except that a positive constant is needed to have and The sim- plest equation for is obtained by setting

=

With this choice, Eq. (3.5-9) reduces to

,

. Integrating 1) once yields

,

where a is a constant. Notice that the particular constant chosen in Eq. (3.5-10) has

simplified the argument of the exponential in Eq. (3.5-12).

Turning now to the initial and boundary conditions, it is evident that Eqs. (3.5-2)

and (3.5-4) will be equivalent to one another if

That is, t = and = will both correspond to = if Eq. (3.5-1 3) is satisfied. Requir- ing this, the boundary conditions for Eq. (3.5-1 1) become

Thus, only two boundary conditions must be satisfied, which is the number that can be accommodated by a second-order, ordinary differential equation such as

The similarity transformation has evidently been successful.

Integrating Eq. (3.5-12) and applying the boundary conditions yields

where and are the and complementary respectively.

functions are plotted in Fig. As varies from to increases from to 1, whereas decreases from 1 to The properties of the error function and related functions are discussed in many mathematical handbooks, including

(1970). Values of the functions are tabulated in such books, and they are available also from a variety of software packages spreadsheet programs for per-

sonal computers).

To complete the solution, we need to determine Equation (3.5-10) is nonlinear but noticing that = it is a linear differential equation for namely,

AND APPROXIMATION

methods for solving partial differential equations which are discussed in Chapter 4, the

similarity method may be used with nonlinear as well as linear problems. The technique is illustrated by an example in this section and by several other applications throughout the book.

A similarity solution is developed here for the short-time, transient diffusion prob-

lem described in Example 3.4-1, involving a step change in concentration at and

= Using the dimensionless concentration the governing equations are

One requirement of the similarity method is seen to be satisfied, in that this problem contains no fixed characteristic length. That is, there is not a boundary condition im- posed at some finite value of such as = L.

Assume now that can be expressed as a function of a single independent vari- able where

and is a scale factor which is to be determined. The only difference between and the penetration depth in section 3.4 is that, whereas was used only to describe orders of magnitude, will have specific numerical values. Although our knowledge of indicates that we will show how to derive without such prior information.

The derivatives in (3.5-1) are expressed now in terms of In this introductory example, subscripts are used as reminders of which variable is being held constant in each partial derivative. The derivatives are given by

terms of the similarity variable, (3.5-1) becomes

assumption that = is correct, then neither t nor can appear separately in equation and other conditions for 8. Because g this will be true

for (3.5-9) only if the product is a constant. The exact value of the constant is

immaterial, except that a positive constant is needed to have and The sim- plest equation for is obtained by setting

=

With this choice, Eq. (3.5-9) reduces to

,

. Integrating 1) once yields

,

where a is a constant. Notice that the particular constant chosen in Eq. (3.5-10) has

simplified the argument of the exponential in Eq. (3.5-12).

Turning now to the initial and boundary conditions, it is evident that Eqs. (3.5-2)

and (3.5-4) will be equivalent to one another if

That is, t = and = will both correspond to = if Eq. (3.5-1 3) is satisfied. Requir- ing this, the boundary conditions for Eq. (3.5-1 1) become

Thus, only two boundary conditions must be satisfied, which is the number that can be accommodated by a second-order, ordinary differential equation such as

The similarity transformation has evidently been successful.

Integrating Eq. (3.5-12) and applying the boundary conditions yields

where and are the and complementary respectively.

functions are plotted in Fig. As varies from to increases from to 1, whereas decreases from 1 to The properties of the error function and related functions are discussed in many mathematical handbooks, including

(1970). Values of the functions are tabulated in such books, and they are available also from a variety of software packages spreadsheet programs for per-

sonal computers).

To complete the solution, we need to determine Equation (3.5-10) is nonlinear but noticing that = it is a linear differential equation for namely,

TECHNIQUES