Solve eigenvalue PDE problem

`[`

produces
the solution to the FEM formulation of the scalar PDE eigenvalue problem`v`

,`l`

] =
pdeeig(`model`

,`c`

,`a`

,`d`

,`r`

)

$$-\nabla \cdot (c\nabla u)+au=\lambda du\text{on}\Omega $$

or the system PDE eigenvalue problem

$$-\nabla \cdot (c\otimes \nabla u)+au=\lambda du\text{on}\Omega ,$$

with geometry, boundary conditions, and mesh specified in `model`

,
a `PDEModel`

object. See Solve Problems Using PDEModel Objects.

The eigenvalue PDE problem is a *homogeneous* problem,
i.e., only boundary conditions where *g* = 0 and *r* =
0 can be used. The nonhomogeneous part is removed automatically.

In the standard case *c* and *d* are
positive in the entire region. All eigenvalues are positive, and 0
is a good choice for a lower bound of the interval. The cases where
either *c* or *d* is zero are discussed
next.

If

*d*= 0 in a subregion, the mass matrix*M*becomes singular. This does not cause any trouble, provided that*c*> 0 everywhere. The pencil (*K,M*) has a set of infinite eigenvalues.If

*c*= 0 in a subregion, the stiffness matrix`K`

becomes singular, and the pencil (*K,M*) has many zero eigenvalues. With an interval containing zero,`pdeeig`

goes on for a very long time to find all the zero eigenvalues. Choose a positive lower bound away from zero but below the smallest nonzero eigenvalue.If there is a region where both

*c*= 0 and*d*= 0, we get a singular pencil. The whole eigenvalue problem is undetermined, and any value is equally plausible as an eigenvalue.

Some of the awkward cases are detected by `pdeeig`

.
If the shifted matrix is singular, another shift is attempted. If
the matrix with the new shift is still singular a good guess is that
the entire pencil (*K,M*) is singular.

If you try any problem not belonging to the standard case, you must use your knowledge of the original physical problem to interpret the results from the computation.

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