= Solution
\b[Self-adjoint form.] Grouping the two derivative terms gives $[(1+\epsilon x)u']'-\epsilon u=-\lambda u$. Thus a regular <Sturm-Liouville operator> is
$$
\boxed{\mathcal L=-\frac d{dx}\left[(1+\epsilon x)\frac d{dx}\right]+\epsilon,
\qquad p=1+\epsilon x,\quad q=\epsilon,\quad w=1.}
$$
The <boundary conditions> are mixed <Neumann boundary conditions> at zero and <Dirichlet boundary conditions> at $\pi$. For $\epsilon\geq0$, $p>0$, $q\geq0$ and $w>0$, as required. The boundary term $[p(fg'-f'g)]_0^\pi$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.
\b[Constrained variation.] Introduce a <Lagrange multiplier> $\Lambda$ for the normalization and vary
$$
I[v]-\Lambda\left(\int_0^\pi wv^2\,dx-1\right).
$$
Its first variation is
$$
2[pv'\delta v]_0^\pi+2\int_0^\pi[-(pv')'+qv-\Lambda wv]\delta v\,dx.
$$
The endpoint term vanishes because $v'(0)=0$ and $\delta v(\pi)=0$. The <Euler-Lagrange equation> is therefore
$$
-(pv')'+qv=\Lambda wv,
$$
which is exactly the original <eigenvalue> problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions give
$$
I[v]=\int_0^\pi(pv'^2+qv^2)\,dx=\Lambda\int_0^\pi wv^2\,dx=\Lambda.
$$
Thus \b[the stationary functional values are the <eigenvalues>], with $\Lambda=\lambda$. More generally, before normalization the relevant <Rayleigh quotient> is
$$
\mathcal R_\epsilon[v]=\frac{\int_0^\pi[(1+\epsilon x)v'^2+\epsilon v^2]dx}{\int_0^\pi v^2dx}.
$$
\b[Unperturbed ground mode.] At $\epsilon=0$, the equation is $-u''=\lambda u$. The quadratic form is nonnegative, and zero <eigenvalue> would force a constant function incompatible with the endpoint at $\pi$. For positive $\lambda$, $u=A\cos(\sqrt\lambda x)+B\sin(\sqrt\lambda x)$. The derivative condition forces $B=0$, while the other endpoint gives $\cos(\pi\sqrt\lambda)=0$. Hence
$$
\lambda_n=(n+1/2)^2,\qquad n=0,1,2,\ldots.
$$
The lowest <eigenvalue> and normalized <eigenfunction> are
$$
\boxed{\lambda_0=\frac14,\qquad U_0(x)=\sqrt{\frac2\pi}\cos\frac x2,\qquad
\int_0^\pi U_0^2dx=1.}
$$
\b[Trial-function bound.] The <Rayleigh-Ritz method> says that the minimum <eigenvalue> is no larger than the quotient of any admissible trial function. The same $U_0$ still satisfies both <boundary conditions> and retains unit norm. Its derivative is $U_0'=-\sqrt{2/\pi}\sin(x/2)/2$, giving
$$
\mathcal R_\epsilon[U_0]=\frac14+\epsilon\left[1+\frac1{2\pi}\int_0^\pi x\sin^2\frac x2\,dx\right].
$$
Using $\sin^2(x/2)=(1-\cos x)/2$,
$$
\int_0^\pi x\sin^2\frac x2\,dx=\frac{\pi^2}{4}+1.
$$
Therefore the <affine-parameter Sturm-Liouville eigenvalue bound> is
$$
\boxed{\lambda_{\min}(\epsilon)\leq\frac14+\epsilon\left(1+\frac\pi8+\frac1{2\pi}\right)
\simeq\frac14+1.551854\epsilon.}
$$
This is an upper bound for every $\epsilon\geq0$, not merely a small-parameter approximation. As a lower-bound check, $p\geq1$ and the unperturbed variational minimum imply $\mathcal R_\epsilon[v]\geq1/4+\epsilon$ for every admissible normalized $v$. Thus
$$
\boxed{\frac14+\epsilon\leq\lambda_{\min}(\epsilon)\leq\frac14+\left(1+\frac\pi8+\frac1{2\pi}\right)\epsilon.}
$$
Back to article page
