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

Use integration by parts with a complex conjugate, so that the argument also covers complex-valued eigenfunctions. The coefficient functions are understood to be real. For a nonzero eigenfunction $u$ with Dirichlet boundary conditions,The denominator is strictly positive and the numerator is real. Consequently the eigenvalue satisfiesThis is the Rayleigh quotient for the sign convention used here. Positivity of $a$ is unnecessary for this particular reality argument; real coefficients and disappearance of the endpoint term are what matter.

The Lagrange identity for a Sturm-Liouville operator givesIn the intended class of functions whose values and first derivatives stay finite at the endpoints, the endpoint factor $a$ makes this entire boundary term vanish. Thus no separately prescribed endpoint values are needed to eliminate the boundary form. The same integration by parts argument then proves that the eigenvalues are real.

The admissible endpoint behaviour is part of this conclusion: it is the vanishing endpoint coefficient and the Sturm-Liouville boundary form, together with endpoint regularity, that replaces explicit boundary conditions. If singular functions are allowed, a vanishing coefficient alone need not suffice: $a(x)=x(1-x)$ and $v(x)=\log x$ have $a(x)v^{\prime }(x)\to 1$ as $x\downarrow 0$. A self-adjoint differential operator therefore still needs a specified domain of admissible functions.

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.