Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 1

The weight cancels the denominator in the Sturm-Liouville operator. Two integrations by parts, with real $p,q$, give the Lagrange identity for a Sturm-Liouville operatorThus the boundary form must vanish for every pair in the operator domain:Separated real Robin boundary conditions $A_{s}y(s)+B_{s}p(s)y^{\prime }(s)=0$ at $s=a,b$, with $(A_s,B_s)\ne (0,0)$, make each endpoint contribution vanish. This includes Dirichlet boundary conditions and Neumann boundary conditions. Coupled conditions can also work: periodic $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$ cancel the two endpoint contributions. For the regular problem, imposing a complete pair of such self-adjoint boundary conditions gives a self-adjoint differential operator; the displayed identity is the boundary-form criterion.

First, setting $f=g=y_n$ in the self-adjoint operator identity shows that ${\lambda }_n⟨y_n,y_n{⟩}_w$ is real, so every eigenvalue is real. For two eigenfunctions,Distinct eigenvalues therefore implyThis is orthogonality of Sturm-Liouville eigenfunctions.

Factor out $x$ from the differential expression:Comparing with $\mathcal{L}y=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ givesMultiplying all three by the same positive constant would give the same Sturm-Liouville operator.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and

Expand the solution and the forcing in the orthonormal Sturm-Liouville eigenfunction expansion. The forcing coefficients areEvery ${\lambda }_n$ is nonzero, so the solution coefficient is $b_n/{\lambda }_n$. ThusAs an independent check, solving $Y^{\prime \prime }+Y=-1$ directly with the two endpoint conditions givesIt satisfies both Dirichlet boundary conditions and the forced ordinary differential equation.