- 5.10.1 The physical problem
- 5.10.2 The mathematical problem
- 5.10.3 Step 0: Fix the boundary conditions
- 5.10.4 Step 1: Find the eigenfunctions
- 5.10.5 Step 2: Solve the problem
- 5.10.6 Summary of the solution
- 5.10.7 An alternative procedure

5.10 More general eigenfunctions

In the simplest problems, the eigenfunctions are sines and cosines. That includes the examples so far. But it is quite easy to get different eigenfunctions. In this example, they will turn out to be products of sines and exponentials.

5.10.1 The physical problem

Find the unsteady temperature distribution in the moving bar below for arbitrary position and time if the initial distribution at time zero and the temperatures of the ends are known.

5.10.2 The mathematical problem

- Finite domain :
- Unknown temperature
- Parabolic
- One initial condition
- Two Dirichlet boundary conditions
- Constant

Try separation of variables:

5.10.3 Step 0: Fix the boundary conditions

The -boundary conditions are inhomogeneous:

So we try finding a satisfying these boundary conditions:

A linear expression works:

This can be solved to find

To get rid of the inhomogeneous boundary conditions, we subtract
from . That will produce homogeneous boundary conditions for
the remainder . Indeed, if you plug
into the boundary conditions, you get

And since satisfies the inhomogeneous boundary conditions, that becomes

Substitute into the partial differential equation
to
get

where

Substitute into the initial condition :

The problem for is therefor:

5.10.4 Step 1: Find the eigenfunctions

Substitute into the *homogeneous* partial differential equation
:

Separate:

The Sturm-Liouville problem for is now:

This is a constant coefficient ordinary differential equation, with a characteristic polynomial:

The fundamentally different cases are now two real roots (discriminant positive), a double root (discriminant zero), and two complex conjugate roots (discriminant negative.) We do each in turn.

*Case :*Roots and real and distinct:

Boundary conditions:

No nontrivial solutions since the roots are different.*Case :*Since :

Boundary conditions:

No nontrivial solutions.*Case :*For convenience, we will write the roots of the characteristic polynomial more concisely as:

where according to the solution of the quadratic

Since it can be confusing to have too many variables representing the same thing, let's agree that is our ``representative'' for , and our ``representative'' for . In terms of these representatives, the solution is, after clean-up,

Boundary conditions:

Nontrivial solutions can only occur if

which gives us our eigenvalues, by substituting in for :

Also, choosing each :

5.10.5 Step 2: Solve the problem

Expand all variables in the problem for in a Fourier series:

We want to first find the Fourier coefficients of the known functions
and . Unfortunately, the ordinary differential equation found in the previous section,

is

We want that the second coefficient is the derivative of the first:

This is a simple ordinary differential equation for the we are trying to find, and a valid solution is:

Having found , we can write the orthogonality
relationships for the generalized Fourier coefficients of and
(remember that
):

The integrals in the bottoms equal .

Expand the partial differential equation
in a generalized
Fourier series:

Because of the choice of the ,
:

So, the ordinary differential equation for the generalized Fourier coefficients of becomes:

Expand the initial condition
in a generalized Fourier series:

so

Solve this ordinary differential equation and initial condition for :

Homogeneous equation:

Inhomogeneous equation:

Initial condition:
.

5.10.6 Summary of the solution

Total solution:

5.10.7 An alternative procedure

Define a new unknown by
. Put this
in the partial differential equation for and choose and so that the
and terms drop out. This requires:

Then:

No fun! Note that the generalized Fourier series coefficients for become normal Fourier coefficients for .