Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 7

For the given weighted inner product, a self-adjoint operator $L$ obeys $L=L^{†}$, including equality of its domain with the adjoint domain. In the differential-equation calculation this requiresfor all functions in a specified common domain, with endpoint conditions that remove the boundary form. Equality of these inner products by itself describes symmetry; the domain conditions complete the self-adjoint realization.

Divide the equation by $\sqrt{1-x^2}$ to obtain its Sturm-Liouville form:Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. For $L=-w^{-1}(pd/dx{)}^{\prime }$, the Lagrange identity for a Sturm-Liouville operator gives the endpoint form $[p({\bar{f}}^{\prime }g-\bar{f}{g}^{\prime }){]}_{-1}^1$. A natural self-adjoint choice isUnder $x=\cos \theta$, these become the ordinary zero-derivative conditions at $\theta =0,\pi$ for $-d^2/d{\theta }^2$. They therefore define the realization whose Neumann eigenfunctions of the Chebyshev differential operator are $T_n(x)=\cos (n\arccos x)$, with $n=0,1,\dots$.

Boundedness alone does not choose these endpoint conditions: $\sin (n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $p{y}^{\prime }$. Requiring bounded $y^{\prime }$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $z(x)=e^{-2x}u(x)$. It gives $u^{\prime \prime }+\lambda u=0$ with $u(0)=u(1)=0$. Positive $\lambda =k^2$ requires $\sin k=0$; zero or negative $\lambda$ yields only the trivial solution under both endpoint conditions. HenceMultiplication of the original equation by $e^{4x}$ gives $-(e^{4x}{z}^{\prime }{)}^{\prime }-4e^{4x}z=\lambda e^{4x}z$. Thus the appropriate weighted inner product has weight $e^{4x}$, andMultiplying each eigenfunction by $\sqrt{2}$ gives weighted norm one.