Periodic or Finite Domains: Separation of Variables
Consider the diffusion equation on the finite one-dimensional domain, , with Dirichlet BCs:
The possible solutions are:
The BCs cannot be solved simultaneously with a non-trivial solution unless .
Applying the BCs to Eq. , , but there are an infinite number of other values that will satisfy the boundary conditions:
The general solutions (for the BCs and ICs given above) are:
However, to satisfy the initial conditions the general solution must be taken as a linear superposition of each possible solution--and with the weighting factor at chosen to satisfy the initial conditions. Therefore, seek such that:
Remarks on the Fourier Method of Solution
From the solution above, a periodic solution in can be constructed for and for all integer (use the in , etc).
Note how each term in the Fourier series decays much faster than the previous term--the rate at which amplitudes get smaller is exponential with smaller wavelengths.
This dependence on wavelengths is a useful intuitive device for understanding the diffusion equation.
For example, consider two different initial conditions on the infinite domain :
The solutions and are obviously,
Time Dependent Boundary Conditions, Semi-Infinite Domains
The final analytic method for obtaining an analytic solution to to the diffusion equation is the method of Laplace transforms. This method is applicable to semi-infinite domains and is especially effective for problems with time-dependent boundary conditions.
The method of Laplace transforms is an example of operator calculus. In this case, the operator is a transformation of time derivatives to algebraic expressions in a transformation variable. The transformed diffusion equation becomes an inhomogeneous ordinary differential equation in the spatial variable. The ordinary differential equation is solved for the transformed boundary conditions and then the transformation is reversed--usually through a table of Laplace transform pairs.
The Laplace Transform is defined as the linear operator:
The essential property of the Laplace transform is its operation on time derivatives:
The transform has no effect on spatial derivatives, therefore the transformed diffusion equation becomes:
Two Examples of the Laplace Transform Method
For this initial condition, the transformed diffusion equation, Eq. 11-13, becomes
For this problem, the homogeneous solution is:
The particular solution is:
The transformed boundary conditions become
This equation requires transformation back to the variable through the inverse of a Laplace transform. The inverses are usually much more difficult to find than the Laplace transforms. Fortunately, Laplace transforms and their inverses are usually tabulated in math handbooks. For example,
|Selected Laplace Transform Pairs|
So, the solution can be obtained through the use of the above table:
For a second example with time-dependent solutions,
For this case, the transformed diffusion equation is the same as that calculated in the previous example:
Therefore, the surface concentration changes as :
Anisotropic Diffusion Coefficients
The methods of solution that have been treated do not account for the general case of an anisotropic diffusion coefficient. However, a simple strategem can be used to reconstruct the diffusion equation from its tensor form into the scalar form that is treated above.
Recall form of the the flux relationship in a particular frame of reference:
The diffusion equation becomes:
First, find the rotation which diagonalizes :
The are the eigenvalues of and the coordinate frame of reference of have axes parallel to the eigenvalues of .
The diffusion equation is much simpler in the eigenframe:
Define an effective diffusivity :