Prove Eigenfunctions Corresponding to Different Eigenvalues Are Orthogonal
$\Begingroup$ I'm Considering the Eigenvalue Problem $L(\Phi) = -\Lambda \Sigma(X) \Phi$, Subject to a Given Set of Homogeneous Boundary Conditions. Assume...
I'm considering the eigenvalue problem $L(\phi) = -\lambda \sigma(x) \phi$, subject to a given set of homogeneous boundary conditions. Assume $\sigma(x) \gt 0$ on an interval $[a,b]$.
Suppose that $\int_{a}^{b} [uL(v) - vL(u)]dx = 0$ for all functions $u$ and $v$ satisfying the same homogeneous boundary conditions.
How can I prove that two eigenfunctions corresponding to different eigenvalues $\lambda_1 \ne \lambda_2$ are orthogonal with respect to some choice of weight?
1 Answer
It may be helpful to understand the similar fact about left and right eigenvectors of a matrix, in order to appreciate the role being played here by self-adjointness of the operator $L$.
In any case the computation is straightforward, applying what we know in the case that $u,v$ are eigenfunctions (satisfying the unspecified but requisite homogeneous boundary conditions) corresponding to distinct eigenvalues $\lambda_1 \neq \lambda_2$ respectively:
$$ L(u) = -\lambda_1 \sigma u $$
$$ L(v) = -\lambda_2 \sigma v $$
where multiplier function $\sigma(x)$ on $[a,b]$ must be assumed positive (a.e.) there if we want the natural bilinear function to be an inner product, with the corresponding sense of "orthogonality":
$$ 0 = \int_a^b u L(v) - v L(u) \mathrm{d} x = (\lambda_1 - \lambda_2) \int_a^b u v \; \sigma(x) \mathrm{d} x$$
Since the eigenvalues are unequal, the last integral is zero, and thus $u,v$ are orthogonal with respect to this weighted inner product on $[a,b]$.
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