Perturbation Methods: Guarantee a boundary layer?

[Master1022]

Homework Statement

A small parameter multiplying the highest derivative does not guarantee that the solution will have a boundary layer for small values of $\epsilon$. This may be due to the form of the differential equation, or the particular boundary conditions used in the problem. Can you:
(a) think of an example where this might be true, and
(b) why this might be the case

Relevant Equations

differential equations

Hi,

I was working on the following problem:

Question:
A small parameter multiplying the highest derivative does not guarantee that the solution will have a boundary layer for small values of $\epsilon$. This may be due to the form of the differential equation, or the particular boundary conditions used in the problem. Can you:
(a) think of an example where this might be true, and
(b) why this might be the case

Attempt:
I am really not sure where to start with this one... Any advice would be appreciated because I don't know how to think of either the more specific case or a more general principle.

 

[lippyka]

Answer:A good example of this is a boundary-value problem with a linear differential equation and a Dirichlet boundary condition at one end. In this case, if you introduce a small parameter multiplying the highest derivative, the solution will not have a boundary layer for small values of $\epsilon$ because the solution is already linear. This is because the Dirichlet boundary condition forces the solution to be constant at one end, which means that the solution will not have any kind of singularity or boundary layer.