- #1
member 428835
Hi PF!
I'm trying to find the eigenvalues of this ODE $$y''(x) + \lambda y = 0 : u(0)=u(1)=0$$ by using the basis functions ##\phi_i = (1-x)x^i : i=1,2,3...n## and taking inner products to formulate the matrix equation $$A_{ij} = \int_0^1 \phi_i'' \phi_j \, dx\\ B_{ij} = \int_0^1 \phi_i\phi_j\,dx :\\A+\lambda B = 0 .$$
Solving ##A+\lambda B = 0## is direct; it's a linear algebraic equation. We can compare our approximate solution to the exact solution ##(i \pi)^2 : i=1,2,3...## In this way, I know if a solution is correct or not.
Understanding the above, why is it if I choose the basis functions to begin at ##i=2## I do not get the correct solution? I assume in this case the basis does not span the solution's function space, but can someone elaborate? How do I know if a given basis function spans the solution space?
I'm trying to find the eigenvalues of this ODE $$y''(x) + \lambda y = 0 : u(0)=u(1)=0$$ by using the basis functions ##\phi_i = (1-x)x^i : i=1,2,3...n## and taking inner products to formulate the matrix equation $$A_{ij} = \int_0^1 \phi_i'' \phi_j \, dx\\ B_{ij} = \int_0^1 \phi_i\phi_j\,dx :\\A+\lambda B = 0 .$$
Solving ##A+\lambda B = 0## is direct; it's a linear algebraic equation. We can compare our approximate solution to the exact solution ##(i \pi)^2 : i=1,2,3...## In this way, I know if a solution is correct or not.
Understanding the above, why is it if I choose the basis functions to begin at ##i=2## I do not get the correct solution? I assume in this case the basis does not span the solution's function space, but can someone elaborate? How do I know if a given basis function spans the solution space?
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