Past exam of the differential equations 2 course of the University of Oxford / 2017 / 2 / b

The useful change of independent variable in a second-order ODE is $t=\log x$, followed by $y(x)=xv(t)$. The chain rule givesThus the eigenvalue equation becomes $v_{tt}=\lambda v$ on $0<t<1$, and the endpoint values become $v(0)=v(1)=0$. At the specified eigenvalues,The first endpoint forces $A=0$; the second holds for every positive integer $k$. Choosing a convenient normalization yieldsEach nonzero constant multiple is the same eigenfunction up to normalization. This logarithmic conjugacy of a repeated-root Euler operator turns the Cauchy-Euler equation into the familiar Fourier sine series problem on a unit interval.

Match the coefficient of $y^{\prime }$ with the derivative of the coefficient of $y^{\prime \prime }$. For Sturm-Liouville form this requiresA positive integrating factor is therefore $r(x)=x^{-3}$, with any positive constant multiple equally valid. Direct calculation givesThis uses the plus-divergence sign convention of the question. The usual minus-divergence Sturm-Liouville operator is its negative; no extra sign is inserted into the requested eigenvalue equation.

Because $r$ never vanishes on the interval, multiplying the original eigenvalue equation by $r$ leaves the requested weighted eigenvalue problem unchanged:The weight in the inner product is essential. The substitution $t=\log x$ givesThus orthogonality of Sturm-Liouville eigenfunctions holds for the weighted inner product with weight $x^{-3}$, and the squared norm is $N_k=1/2$.

Use the normalization from part (ii), and writeThe Sturm-Liouville eigenfunction expansion coefficient is $c_k=N_k^{-1}{\int }_1^{e}ry{\hat{y}}_{k}dx$. Although ${\hat{y}}_k$ has zero endpoint values, $y$ generally does not. Retain the endpoint term in the Lagrange identity for a Sturm-Liouville operator:The positive sign comes from subtracting the contribution at $x=1$. The differential equation also gives $\hat{L}y=r{e}^x-(\beta -1)ry$. Substitution into the preceding identity yieldsHence, away from resonance in a differential equation,This is the effect of boundary data in a Sturm-Liouville eigenfunction expansion: omitting the endpoint contribution would discard the dependence on $\alpha$.

If $\beta =1+n^2{\pi }^2$ for a positive integer $n$, the Fredholm solvability condition isWhen it fails, there is no solution. When it holds, $c_n$ is arbitrary and the other coefficients retain the displayed formula, giving infinitely many solutions differing by a multiple of ${\hat{y}}_n$. Sufficiency follows by subtracting the smooth lift $h(x)=\alpha x(1-\log x)$, which has the required endpoint values and satisfies $Lh=0$, and applying the Fredholm alternative to the resulting homogeneous Dirichlet boundary conditions.

Finally, for $\alpha \ne 0$, the displayed Sturm-Liouville eigenfunction expansion represents the solution on $1<x<e$ and in the weighted inner product sense. Every term is zero at $x=1$, so that series cannot give the nonzero prescribed value by substitution at the endpoint. The boundary trace must be imposed separately. Equivalently, expand $y-h$ with homogeneous Dirichlet boundary conditions and add $h$ back afterwards.