A Rayleigh Ritz and inner products

[member 428835]

Hi PF!

Say we have the variational problem $$M[\phi(x)] = \lambda K[\phi(x)]$$ where $\lambda$ is the eigenvalue and $M,K$ are linear integro-differential operators. Now given a set of basis functions $\phi = \sum_j c_j \phi_j$, is this simple enough to conclude that eigenvalues $\lambda$ can be approximated via the Rayleigh-Ritz variational technique as the eigenvalues of the augmented matrix equation (in component form) as $$\left( M[\phi_i],\phi_j\right) = \lambda \left( K[\phi_i],\phi_j\right)$$

where $(f,g) \equiv \int f g$? Or does the inner product get more complicated given basis functions that are orthogonal w.r.t. different weights?

 

[pasmith]

You sohuld use the inner product with respect to which the basis functions are orthogonal. That leaves you with the generalized eigenvalue problem <br /> T(\lambda)\mathbf{c} = 0 for some N \times N matrix T. How you solve that eigenvalue problem is up to you. However, if your basis functions do not satisfy the boundary conditions of your problem then the eigenvalue \lambda won't appear in every row of T (one or more rows must enforce the boundary condition rather than the operator equation) so methods which assume that will not work.

Whether the eigenvalues you get are a good approximation to those of the original problem is uncertain. Some of them will be; some will not. Boyd, Chebyshev and Fourier Spectral Methods (2nd ed, Dover 2001) discusses this issue in Chapter 7.

 

[member 428835]

Can you show me exactly why the inner product needs to be orthogonal with the basis functions? I believe you, but can’t convince myself this needs to be the case.

 

[pasmith]

Ignoring complications from the boundary conditions, it doesn't matter which inner product you use, and the natural choice is the one which makes the basis functions orthogonal.

Suppose the \phi_i are orthogonal with respect to (\cdot,\cdot)_1. Then if \tilde{M} is the matrix representation of M where \tilde{M}_{ji} =(M(\phi_i),\phi_j)_1 we have for a different inner product (\cdot,\cdot)_2 that <br /> (M(\phi_i),\phi_j)_2 = (\phi_k,\phi_j)_2\tilde{M}_{ki} = P_{jk}\tilde{M}_{ki} <br /> so changing the inner product amounts to left-multiplication by P. But P is invertible (see below), so the eigenvalues of P\tilde{M} are those of \tilde{M}. (The eigenvectors, of course, will differ.)

To show P is invertible, let V be the finite dimensional space spanned by the \phi_i. Then we can find a basis \{\psi_i\} for V which is orthonormal with respect to (\cdot,\cdot)_2, and there exists an invertible matrix B such that \phi_i = B_{ji}\psi_j. Then
P_{ji} = (\phi_i,\phi_j)_2 = B_{ki}B_{lj}(\psi_k,\psi_l)_2 = B_{ki}B_{kj}. But P = B^{T}B is invertible, since B and thus B^T are invertible.

Having to use rows of T to enforce boundary conditions might result in getting different eigenvalues, but they are only approximations to the eigenvalues of the original problem anyway. However it is best to ensure that every function in V satisfies the boundary conditions.

 

[member 428835]

Ignoring complications from the boundary conditions, it doesn't matter which inner product you use, and the natural choice is the one which makes the basis functions orthogonal.

Suppose the \phi_i are orthogonal with respect to (\cdot,\cdot)_1. Then if \tilde{M} is the matrix representation of M where \tilde{M}_{ji} =(M(\phi_i),\phi_j)_1 we have for a different inner product (\cdot,\cdot)_2 that <br /> (M(\phi_i),\phi_j)_2 = (\phi_k,\phi_j)_2\tilde{M}_{ki} = P_{jk}\tilde{M}_{ki}<br /> so changing the inner product amounts to left-multiplication by P. But P is invertible (see below), so the eigenvalues of P\tilde{M} are those of \tilde{M}. (The eigenvectors, of course, will differ.)

I'm trying to understand this, but I think I'm missing something. If We define $M = d/dx$ and let $\phi_{1,2} = (1-x)x, (1-x)x^2$ with (\cdot,\cdot)_1 = \int_0^1 \cdot \cdot then \tilde{M} = 0,-1/60; 1/60,0. If we let (\cdot,\cdot)_2 = \int_0^1 x \cdot \cdot so that it's a different inner product then we get $(M(\phi_i),\phi_j)_2 = -1/60,-1/60; 0, -1/210$. But $(\phi_k,\phi_j)_2\tilde{M}_{ki} = -1/6300, -1/10080; 1/3600, 1/6300$. So it seems the two different inner products give different results and don't uphold the equality. Am I missing something?

 

[member 428835]

This might be easier to see

 

[pasmith]

Suppose the \phi_i are orthogonal with respect to (\cdot,\cdot)_1. Then if \tilde{M} is the matrix representation of M where \tilde{M}_{ji} =(M(\phi_i),\phi_j)_1 we have for a different inner product (\cdot,\cdot)_2 that <br /> (M(\phi_i),\phi_j)_2 = (\phi_k,\phi_j)_2\tilde{M}_{ki} = P_{jk}\tilde{M}_{ki} <br /> so changing the inner product amounts to left-multiplication by P. But P is invertible (see below), so the eigenvalues of P\tilde{M} are those of \tilde{M}. (The eigenvectors, of course, will differ.)

I of course meant that the values of \lambda for which \det(\tilde{M} - \lambda \tilde{K}) = 0 are the same as those for which \det(P\tilde{M} - \lambda P\tilde{K})) = 0, since \det(P) \neq 0. Naturally if P is not the identity then P\tilde M \neq \tilde M.

 

[member 428835]

I of course meant that the values of \lambda for which \det(\tilde{M} - \lambda \tilde{K}) = 0 are the same as those for which \det(P\tilde{M} - \lambda P\tilde{K})) = 0, since \det(P) \neq 0. Naturally if P is not the identity then P\tilde M \neq \tilde M.

Wow, I feel stupid. Like REALLY stupid. But seriously, thank you so much!

Do you know Mathematica? Would you be interested in helping me out with a very tough problem that is closely related to this? I've done most of the heavy lifting, but something is wrong and I can't figure out what it is.

I can write up a neat description for you in LaTeX and I have very clear comments on the Mathematica code (you almost don't even need to know the language to read my steps, since it's Mathematica is pretty user friendly)

Edit: it's a fluid dynamics problem, which seems like something you're interested in from your bio.

 

[member 428835]

Solved it, FINALLY! Thanks for the interest everyone!