Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 6

When $r=p^{\prime }$, write the differential expression in divergence form,Two integrations by parts give the Lagrange identity for a Sturm-Liouville operator:The $q$ terms cancel because $q$ is real. Hence the exact boundary criterion, in the sense of self-adjointness defined in the question, isAt regular endpoints, all separated homogeneous conditions of the following type satisfy it:with real coefficients, not both zero at either endpoint. The allowed value-and-flux pairs at an endpoint are proportional, so their determinant $p(f{g}^{\prime }-f^{\prime }g)$ vanishes. This includes Dirichlet boundary conditions, Neumann boundary conditions, Robin boundary conditions and arbitrary mixtures of these at the two ends.

Coupled conditions can also work: periodic conditions $y(b)=y(a)$ and $p(b)y^{\prime }(b)=p(a)y^{\prime }(a)$ make the two endpoint expressions equal; simultaneous minus signs give the antiperiodic case. Thus requiring each endpoint term separately to vanish is sufficient, but not necessary.

More generally, for regular endpoints set $Y=(y(a),y(b){)}^{\mathsf{T}}$ and $N=(-p(a)y^{\prime }(a),p(b)y^{\prime }(b){)}^{\mathsf{T}}$. All maximal homogeneous self-adjoint boundary conditions can be written $AY+BN=0$ with real two-by-two matrices satisfying $\mathrm{rank}(AB)=2$ and $A{B}^{\mathsf{T}}=B{A}^{\mathsf{T}}$. If an endpoint is singular, the relevant limiting boundary form and admissible domain must instead be specified; the assumption $p>0$ only in the interior does not itself fix those endpoint limits.

First move the order parameter term to the right-hand side:Here $p=x^2$, $r=x$, so the integrating factor obeysChoose the irrelevant positive constant $C=1$. Multiplying the equation by $1/x$ givesThus a Sturm-Liouville theory eigenvalue formulation with the requested eigenvalue isThis is the Sturm-Liouville formulation with Bessel order as eigenvalue. The spectral parameter is the squared order of the Bessel function, with $\beta$ treated as fixed. This differs from the also familiar formulation in which ${\beta }^2$ is the eigenvalue.

The factor $1/x$ is singular at the origin. Consequently the integral defining $w$ cannot literally use lower limit zero; take a positive reference point $x_{\ast }>0$, obtain $w=x_{\ast }/x$, and absorb $x_{\ast }$ into the normalization. The differential identity holds for $x>0$. Boundary conditions and admissibility at zero and at the outer endpoint are additional ingredients needed to make a particular self-adjoint spectral problem.

Denote the unshifted differential expression by $L_{0}y=(p{y}^{\prime }{)}^{\prime }+qy$. The supplied Green's function satisfies $L_{0}G(x,\xi )=\delta (x-\xi )$, whereas the inhomogeneous equation is $L_{0}y=\lambda wy+wf$.

Multiply the latter equation by $G$ and the Green equation by $y$, subtract and integrate in $x$:The left-hand side is the boundary termIt vanishes because $G$ and $y$ satisfy the same homogeneous Robin boundary conditions. For example, where $\alpha \ne 0$, substituting $y^{\prime }=-\beta y/\alpha$ and $G_x=-\beta G/\alpha$ cancels the endpoint expression; where $\alpha =0$, both values vanish. The other endpoint works identically. HenceThe Dirac delta function contribution is valid for an interior evaluation point $\xi$; equivalently, $G$ is continuous there and has flux-derivative jump $[p{G}_x{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$.

This uses the existence of the unshifted Green's function stated in the question. Such a function is not automatic for every choice of $p,q$ and endpoint conditions: a nontrivial zero mode of $L_0$ would obstruct an ordinary inverse and would require a compatibility condition and a modified Green's function. Under the stated inverse assumption, the differential equation and the displayed integral equation are equivalent for functions in the boundary-condition domain.