how does the basis of an inner product space change when the domain is deformed

Assume we have a complete orthogonal system on a domain $D$, given by the eigenfunctions of the Laplacian on $D$. For example, the set $\{e^{int}\}$ on $[-\pi, \pi]$, or the spherical harmonics on the unit ball. Now consider a domain $D'$, which is "close" to D in some sense (the boundary of $D$ is close to the boundary of $D'$ in some suitable norm).

Are the eigenfunctions of the Laplacian on $D$ close, in some sense, to the eigenfunctions of the Laplacian on $D'$? Does knowing the basis of $D$ help approximate the basis of $D'$ ? Any known results along these or similar lines appreciated.

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This is touching on something interesting, but I think you need to be more precise, because there is no such thing as the basis for $L^2(D')$. It seems like you have some preferred basis in mind, but if so you need to pose your question more precisely – Yemon Choi Jan 24 '10 at 22:53
What do you mean by L^2 distances between boundaries of domains? – Jonas Meyer Jan 24 '10 at 22:55
I think fredjalves is thinking about the preferred basis for L^2(D) and L^2(D') from eigenfunctions of the Laplacian. At least that's what the two examples suggest. – j.c. Jan 24 '10 at 22:55
good point Yemon and jc. Yes, I was thinking of the basis from eigenfunctions of the Laplacian, not just any basis. Jonas: assuming $D$ and $D'$ are subsets of $R^n$, then their boundaries are functions on $R^{n-1}$. I'd like to say that those boundaries are close, with respect to some suitable norm (different from the the norm induced by the inner product on $R^n$) – fredjalves Jan 24 '10 at 23:30