- #1

joshmccraney

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One way to solve a simple eigenvalue problem like

$$y''(x)+\lambda y(x) = 0,\\

y(0)=y(1)=0$$

(I realize the solution's amplitude can be however large, but my point here is not to focus on that) is to solve the inverse problem. If we say ##A[u(x)] \equiv d^2_x u(x)## and ##B[u(x)] \equiv u(x)## then we also know that $$A^{-1}[u(x)] = \int_0^1Gu(x)\, dx,\\ B^{-1}[u(x)] = u(x)$$

where ##G## is the Green's function to the original ODE and BC.

One way to solve this problem is to let ##u(x) = \sum_{i=1}^N a_i\phi_i(x)## where ##\phi_i## is predetermined function of ##x## and ##a_i## is to be determined. This problem is know to be solved via

$$(\beta - \lambda \alpha)\textbf{ a} = \textbf{ 0},\\

\beta_{ij} =\left(B^{-1}[\phi_i],\phi_j\right) ,\\

\alpha_{ij}= \left(A^{-1}[\phi_i],\phi_j\right)$$

where ##(,)## denotes inner products and ##\textbf a## denotes the series coefficients ##a_i##. After solving this algebraic eigenvalue problem we use vector components of ##\textbf a## to approximate ##u## as a series (shown above).

My question is, since the Green's function ##G## uses the boundary conditions, is there a restriction on the selection of ##\phi_i## that requires it satisfy the boundary? In other words, can ##\phi_i = x^i## or must it be something like ##\phi_i=x^i(x-1)##?

Thanks!