Graduate Solving an ODE Eigenvalue Problem via the Ritz method

[member 428835]

Hi PF!

I want to solve $u''(x) = -\lambda u(x) : u(0)=u(1)=0$. I know solutions are $u(x) = \sin(\sqrt{\lambda} x):\lambda = (n\pi)^2$. I'm trying to solve via the Ritz method. Here's what I have:

define $A(u)\equiv d^2_x u$ and $B(u)\equiv u$. Then in operator form we have $A(u) = -\lambda B(u)$. Taking the inner product of this expression with $u$, where $(f,g) \equiv \int_0^1 fg$, yields

$$(A(u),u) = -\lambda(B(u),u).$$

A variational approach implies the solution is equivalent to solving

$$\min\left((A(u),u)\right) : (B(u),u) = 1.$$

Next make the series expansion

$$u = \sum_i\alpha_i\phi_i(x).$$

Optimizing via Lagrange multipliers with respect to $\alpha$'s implies (details omitted but straightforward)

$$\sum_i(A_{ij} +\lambda B_{ij})\alpha_i = 0,\\
A_{ij} \equiv (A(\phi_i),\phi_j),\,B_{ij} \equiv (B(\phi_i),\phi_j).
$$

This is an algebraic eigenvalue problem and solutions are direct. To solve for each harmonic $\sin(n\pi x)$, simply take $\phi_n\cdot\alpha_n$ where $\alpha_n$ is the $n$th eigenvector and $\phi_n$ is a vector of the first $n$ $\phi(x)$'s.

So, substituting $\phi_i(x) = x^i(1-x)$ into $\sum_i(A_{ij} +\lambda B_{ij})\alpha_i = 0$ and calculating eigenvectors should give me a good solution, but it's not. Any ideas what I'm doing wrong? I can supply my code if that helps (Mathematica).

Edit: this isn't HW, but perhaps it should go there? Could a Mentor please move this there (I tried but I don't know how to move this without copy-pasting, hence having a double thread)?

 

[member 428835]

Update: my eigenvalues are very accurate, matching the analytic ones very well! However, when I reconstruct the harmonics ($\sin(\pi x),\sin(2\pi x),\sin(3\pi x)$ etc) I require the $n$th eigenvector, but the eigenvector is randomly scaled. Thus, the solution I obtain for the harmonics is not scaled properly. How do I determine the scaling?

Never mind, I figured it out!