Example of Ritz method with Bessel functions for trial function

[member 428835]

Hi PF!

Do you know of any examples of the Ritz method which use Bessel functions as trial functions? I’ve seen examples with polynomials, Legendre polynomials, Fourier modes. However, all of these are orthogonal with weight 1. Bessel functions are different in this way.

Any advice on an example (published journal, notes, book, etc) is SO appreciated!

thanks so much!

 

[pasmith]

I don't think there's anything more conecptually difficult about evaluating $$\frac{\langle L(f), f \rangle}{\langle f, f \rangle}$$ where $$\langle f, g \rangle = \int_a^b f(x)g(x)w(x)\,dx$$ for non-constant $w$ than there is for $w = 1$.

Given your eigenvalue problem $L(f) = \lambda f$ subject to self-adjoint boundary conditions on $f$, you approximate $f$ as a linear combination of basis functions which satisfy the boundary conditions: $$f = \sum_{n = 1}^N a_n \phi_n.$$ Then since $L$ is linear, $$L(f) = \sum_{n=1}^N a_n L(\phi_n) = \sum_{n=1}^N a_n\sum_{m=1}^N M_{nm}\phi_m$$ where $$L(\phi_n) = \sum_{m=1}^N M_{nm} \phi_m.$$ If you let $L(f) = \sum b_m\phi_m(x)$ then you have now $$\mathbf{b}^T = \mathbf{a}^T M$$ and taking the inner product with $f$ you get $$\sum_{m} \sum_{n} b_m a_n \langle \phi_m, \phi_n \rangle = \sum_n a_nb_n \|\phi_n\|^2$$ since the basis functions are orthogonal with respect to $\langle \cdot, \cdot \rangle$. This gives you the approximation $$\lambda = \frac{\langle L(f), f \rangle}{\langle f, f \rangle} \approx<br /> \frac{\mathbf{b}^TD\mathbf{a}}{\mathbf{a}^TD\mathbf{a}}<br /> = \frac{\mathbf{a}^TMD\mathbf{a}}{\mathbf{a}^TD\mathbf{a}}$$ where $$D = \operatorname{diag}(\|\phi_1\|^2, \dots, \|\phi_N\|^2).$$ The difficulty with using Bessel functions rather than trigonometric or polynomials is not that the weight function is not constant (there are systems of orthogonal polynomials which also have non-constant weight functions) but in determining $$M_{nm} = \langle L(\phi_n), \phi_m \rangle = \int_a^b x L(\phi_n)(x) \phi_m(x)\,dx$$ and $$\|\phi_n\|^2 = \int_a^b x \phi_n(x)^2\,dx.$$ This is one of the reasons why Bessel functions are less likely to be used in numerical methods than Chebyshev polynomials or finite elements.

If you do want to attempt it, then the relevant chapters of Abramowitz & Stegun are probably a good
place to start.

 

[member 428835]

If you let $L(f) = \sum b_m\phi_m(x)$ then you have now $$\mathbf{b}^T = \mathbf{a}^T M$$ and taking the inner product with $f$ you get $$\sum_{m} \sum_{n} b_m a_n \langle \phi_m, \phi_n \rangle = \sum_n a_nb_n \|\phi_n\|^2$$ since the basis functions are orthogonal with respect to $\langle \cdot, \cdot \rangle$. This gives you the approximation $$\lambda = \frac{\langle L(f), f \rangle}{\langle f, f \rangle} .$$

Okay, I think this makes sense, but if we instead had the generalized eigenvalue problem $L(f) = \lambda K(f)$, then the results would change, right? Now instead of orthogonality with respect to $\langle \cdot, \cdot \rangle$ we would require orthogonality with respect to $\langle K[\cdot], \cdot \rangle$, right? Then the approximation would be $$\lambda = \frac{\langle L(f), f \rangle}{\langle M(f), f \rangle}$$

 

[member 428835]

Also @pasmith, so let me highlight the process and you can tell me if I'm understanding it correctly:

$$L(f) = \lambda K(f) \implies L(\sum_j a_j \phi_j) = \lambda K(\sum_j a_j \phi_j) \implies L(\sum_j a_j \phi_j) = \lambda K(\sum_j a_j \phi_j) $$

and then taking inner products we have

$$\left(L(\sum_j a_j \phi_j),\sum_i a_i \phi_i\right)= \lambda \left(K(\sum_j a_j \phi_j),\sum_i a_i \phi_i\right) \implies \sum_j \sum_i a_ja_i(L(\phi_j),\phi_i) = \lambda \sum_j \sum_i a_ja_i(K(\phi_j),\phi_i)$$

but from here I'm unsure how you proceeded. Could you explain the next step on how one ultimately arrives at the matrix equation $L = \lambda K$?

 

[pasmith]

You can set $L_{ij} = (L(\phi_i),\phi_j)$, and then $$\sum_j \sum_i a_i a_j L_{ij} = \sum_j \sum_i a_i L_{ij} a_j = \mathbf{a}^T\mathbf{L}\mathbf{a}$$

 

[member 428835]

You can set $L_{ij} = (L(\phi_i),\phi_j)$, and then $$\sum_j \sum_i a_i a_j L_{ij} = \sum_j \sum_i a_i L_{ij} a_j = \mathbf{a}^T\mathbf{L}\mathbf{a}$$

Right, but ultimately shouldn't we arrive at $$\mathbf{L} \mathbf{a} = \lambda \mathbf{K} \mathbf{a}$$ where $\mathbf L = (L(\phi_i),\phi_j)$ and $\mathbf K = (K(\phi_i),\phi_j)$ and $\mathbf a$ the eigenvectors? Isn't Rayleigh-Ritz a variational method? But we've not used any calculus to this point (aside from inner products).