OK, I found the answer to this from another source. It means that if I linearize about the solution, all of the resulting eigenvalues are negative.

Dear Bradley

I think you meant that the real parts of the eigenvalues are negative .

In fact there are more cases to be discussed concerning your problem

where all the eigenvalues are pure imaginary and this will bring you to read

some elementary facts about qualitative D.E.

best

@Issam: You're right --- I should have said that the real parts of the eigenvalues are negative.

This problem has several issues. First, your solution should be considered in the setting containing both PDE and their boundary conditions. If all this stuff allows you to define a semi-group of operators along the solutions, the stability condition can be expressed in the language of the infinitesimal generator spectrum. This spectrum must lie in some half plane of C - the complex plane - defined by e.g. Re(z) less than some strictly negative real number. But the spectrum is not limited to eigenvalues but usually it contains also the continuum spectrum. Therefore - be careful!