Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 1

For the weighted inner product, the factor $w$ cancels the $w^{-1}$ in the Sturm-Liouville operator. Assuming the coefficients are real, two applications of integration by parts giveThis is the boundary form in the Lagrange identity for a Sturm-Liouville operator. The potential terms cancel; if $q$ were complex, an additional integral of $(q-q^{\ast })f^{\ast }g$ would remain, so suitable real coefficients are part of formal self-adjointness. The endpoint restrictions in part (a) eliminate the displayed form for every pair of functions in the domain.

The necessary boundary identity for a self-adjoint operator isAt regular endpoints, a standard full family of separated self-adjoint conditions iswith real coefficients and neither pair $({\alpha }_a,{\beta }_a)$, $({\alpha }_b,{\beta }_b)$ zero. For example, if ${\beta }_a\ne 0$, both functions obey $p{f}^{\prime }=-{\alpha }_{a}f/{\beta }_a$ at $a$, so $p{f}^{\prime \ast }g-p{f}^{\ast }{g}^{\prime }=0$ there. The cases ${\beta }_a=0$ and the endpoint $b$ follow directly. These include Dirichlet boundary conditions, Neumann boundary conditions and Robin boundary conditions.

Separate vanishing at each endpoint is sufficient but not necessary. Coupled conditions such as $y(b)=e^{i\chi }y(a)$ and $p(b)y^{\prime }(b)=e^{i\chi }p(a)y^{\prime }(a)$, with real $\chi$, make the two endpoint forms equal and hence their difference zero. More generally the allowed endpoint data must be a maximal subspace on which the boundary form vanishes; this ensures that the adjoint has the same domain. Merely imposing extra endpoint restrictions until the boundary term vanishes proves symmetry, but can overrestrict the domain and fail to give self-adjointness.

Let $L{y}_n={\lambda }_n{y}_n$. By self-adjointness,Taking $m=n$ first shows that each eigenvalue is real: $⟨y_n,y_n{⟩}_w>0$ for a nonzero eigenfunction, since $w>0$. ThereforeFor distinct eigenvalues this proves orthogonality of Sturm-Liouville eigenfunctions:The weight is essential: this need not be orthogonality under the unweighted integral.

The useful coordinate for this scale-dependent differential operator is $t=\ln x$. With $Y(t)=y(e^t)$, the chain rule givesThe interval becomes $0\le t\le 1$, and the endpoint values become $Y(0)=Y(1)=0$. This logarithmic change of variable in a Sturm-Liouville problem reduces the calculation to a constant-coefficient Dirichlet problem. Also $dx/x=dt$, so the transformation preserves the appropriate weighted inner product.

Divide the eigenvalue equation by $x$ and collect the first two derivative terms into a single derivative:Thus one choice of Sturm-Liouville form isIndeed $w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]=-x^2{y}^{\prime \prime }-x{y}^{\prime }-y$. Both $p$ and $w$ are positive and regular throughout this interval. The zero endpoint values make the boundary form vanish.

In logarithmic coordinates the eigenvalue problem becomes $-Y^{\prime \prime }=(\lambda +1)Y$. A nonzero function with zero values at both ends requires $\lambda +1>0$: multiplying by $Y^{\ast }$ and integrating gives $(\lambda +1){\int }_0^1|Y{|}^{2}dt={\int }_0^1|Y^{\prime }{|}^{2}dt>0$. Write $\kappa =\sqrt{\lambda +1}$. The first endpoint gives $Y=A\sin \kappa t$, and the second requires $\kappa =n\pi$, $n=1,2,\dots$.

For vector normalization, ${\int }_1^e|y{|}^{2}dx/x=|A{|}^2{\int }_0^1{\sin }^2(n\pi t)dt=|A{|}^2/2$. A positive real normalization constant therefore givesThe sine orthogonality identity also gives ${\int }_1^e{y}_m{y}_{n}dx/x={\delta }_{mn}$. Completeness is the usual completeness of the Fourier sine series on $(0,1)$ under the logarithmic coordinate transformation.

Expand the forcing and the solution in the normalized eigenfunctions: $1={\sum }_n{b}_n{y}_n$ and $y={\sum }_n{c}_n{y}_n$. Their coefficients areSince all ${\lambda }_n=n^2{\pi }^2-1$ are nonzero, the equation gives $c_n=b_n/{\lambda }_n$. Hence the required Sturm-Liouville eigenfunction expansion isThis is constant forcing in a logarithmic Dirichlet Sturm-Liouville problem. The coefficients are $O(n^{-3})$, so the solution series converges uniformly and has zero endpoint values. Applying $L$ is justified in the weighted square-integrable sense; the forcing sine series equals one at interior points, while its endpoint sums are zero. The equation is imposed in the interior, so this does not contradict the boundary conditions.

As an independent check, solving $-Y^{\prime \prime }-Y=1$ directly givesIt has both zero endpoint values and satisfies $Ly=1$. The invertibility of the Dirichlet operator makes it the same solution as the expansion.