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For a radial stellar oscillation, put ${\xi }_r=r\eta$. The self-gravitating adiabatic equation is $({\Gamma }_{1}P{r}^4{\eta }^{\prime }{)}^{\prime }+r^3[(3{\Gamma }_1-4)P{]}^{\prime }\eta +\rho r^4{\omega }^2\eta =0$. It is a Sturm-Liouville problem whose Rayleigh quotient tests radial stability. The familiar constant-exponent threshold is ${\Gamma }_1=4/3$; a variable exponent requires the full integral test.

A fixed circular elastic membrane of radius $a$ has radial normal modes satisfying the displayed Sturm-Liouville problem, with weight $r$, $R(a)=0$ and regularity at the origin. The regular solutions are Bessel functions of the first kind $J_0(\sqrt{\lambda }r)$. Integration by parts gives the positive Rayleigh quotient $\lambda ={\int }_0^{a}r|R^{\prime }{|}^{2}dr/{\int }_0^{a}r|R{|}^{2}dr$. Distinct modes are orthogonal in the weighted inner product ${\int }_0^{a}r\overline{R}Sdr$.

For noninteger real $\gamma$, expand a smooth forcing as $F={\sum }_{n\ge 0}{F}_n{T}_n$ in Chebyshev polynomials. Because $(1-x^2)T_n^{\prime \prime }-x{T}_n^{\prime }=-n^2{T}_n$, division of each coefficient by ${\gamma }^2-n^2$ gives the regular solution. The coefficients use the weighted inner product with weight $(1-x^2{)}^{-1/2}$. Choosing the same endpoint domain as the regular eigenfunctions of the Sturm-Liouville problem is essential: finite values alone also allow homogeneous solutions $\cos (\gamma \arccos x)$ and $\sin (\gamma \arccos x)$, which generally have divergent endpoint derivatives. Setting $x=\cos \theta$ converts the regular endpoint condition into the Neumann boundary conditions $Z_{\theta }(0)=Z_{\theta }(\pi )=0$. The homogeneous equation then has no nonzero solution when $\gamma$ is noninteger, proving uniqueness in this domain.

These Robin boundary conditions give the spatial Sturm-Liouville operator $-d^2/d{x}^2$ one negative eigenvalue, $-1$, with eigenfunction $e^x$. The other eigenvalues are $n^2{\pi }^2$, with eigenfunctions $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$, $n\ge 1$; there is no zero eigenvalue. Hence the diffusion equation has a growing contribution $c_0{e}^{t+x}$ in addition to decaying modes.

There is no contradiction with ordinary diffusive smoothing. The outward normal derivative is $-u_x$ at the left endpoint and $u_x$ at the right endpoint, so imposing the same coordinate derivative at both ends does not impose dissipative outward-normal data at both ends. The energy identity is $\frac{1}{2}{\partial }_t\int u^2=[u{u}_x{]}_0^1-\int u_x^2$, whose boundary term is $u(1{)}^2-u(0{)}^2$ here and can supply energy. Orthogonal projection of initial data $f$ gives $c_0=2{\int }_0^{1}f{e}^{x}dx/(e^2-1)$; it vanishes precisely when the growing mode is absent.

The reflected boundary conditions $u_x(0,t)=-u(0,t)$ and $u_x(1,t)=-u(1,t)$ have the same spectrum, because reflection $x\mapsto 1-x$ reverses the coordinate derivative. Here the growing eigenfunction is $e^{-x}$ and the decaying functions are $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$, givingFor square-integrable initial data $f$, orthogonal projection givesThe norms are $(1-e^{-2})/2$ and $(n^2{\pi }^2+1)/2$. The spatial Sturm-Liouville problem has no zero eigenvalue and exactly one negative eigenvalue for $-D^2$, namely $-1$. The energy boundary term now reads $u(0{)}^2-u(1{)}^2$, so the left endpoint can supply energy. The growing mode disappears exactly when ${\int }_0^1{e}^{-x}f(x)dx=0$.

Let ${\lambda }_n$ be the transverse eigenvalues of a Sturm-Liouville problem for the two Robin boundary conditions. For a harmonic strip solution decaying in the longitudinal direction, projection of the prescribed end Neumann boundary condition onto every nonpositive eigenvalue must vanish. Positive modes give $q_n(x)=-g_n{e}^{-\sqrt{{\lambda }_n}x}/\sqrt{{\lambda }_n}$. Zero modes are affine and negative modes are oscillatory, so neither can contribute to a decaying solution. The condition also establishes uniqueness within the decay class.

An eigenfunction solving a Sturm-Liouville problem, including its prescribed boundary conditions.

A second-order differential equation in the displayed divergence form. Under the usual regular spectral assumptions, $p>0$, $w>0$, the coefficients have suitable regularity, and separated self-adjoint boundary conditions specify a Sturm-Liouville problem. Multiplying by $y$ and using zero endpoint values gives the Rayleigh quotient $\lambda =\int (p{y}^{\prime 2}+q{y}^2)dx/\int w{y}^{2}dx$ for a nonzero real eigenfunction. Sign conventions for the potential and eigenvalue vary; they must be read from the stated equation.

The weighted operator $L=W^{-1}[-(P\xi \prime )\prime +Q\xi ]$, with real coefficients and $P,W>0$ in the interior, defines a Sturm-Liouville problem. Its realization as a self-adjoint differential operator requires a domain whose endpoint conditions annihilate the boundary form $[P(\eta {\prime }^{\ast }\xi -{\eta }^{\ast }\xi \prime )]$. Its Rayleigh quotient uses the weight $W$ in the denominator.

For a real one-dimensional Sturm-Liouville problem with separated self-adjoint boundary conditions and discrete simple spectrum, the eigenfunction for the $n$th eigenvalue has exactly $n$ interior zeros, numbering from zero. For a confining real Schrödinger operator on the line, the same ordering makes the ground state nodeless and the first excited state have one node. If the potential is even, uniqueness and reflection symmetry imply that these states are respectively even and odd.

For $-{\psi }^{\prime \prime }+\cos (\pi x)\psi =\lambda \psi$ on $(0,1)$ with zero Dirichlet boundary conditions, use $\psi =\sin (\pi x)+\alpha \sin (2\pi x)$. Its Rayleigh quotient is $({\pi }^2+\alpha +4{\pi }^2{\alpha }^2)/(1+{\alpha }^2)$. In the normalized sine basis the restricted operator has matrix $\left(\begin{array}{cc}{\pi }^2& 1/2\\ 1/2& 4{\pi }^2\end{array}\right)$. Its smaller eigenvalue gives the displayed upper bound, attained within this trial space at $\alpha =3{\pi }^2-\sqrt{9{\pi }^4+1}$. The other stationary parameter gives the larger restricted eigenvalue. This is a finite-dimensional Rayleigh-Ritz method for a regular Sturm-Liouville problem with a sign-changing potential.

Vector normalization converts a nonzero vector in a normed vector space into a unit vector. Absolute homogeneity of a norm givesThe direction is unchanged by multiplying by a positive real scalar. For $c>0$, $\Vert cv\Vert =c\Vert v\Vert$ and $\widehat{cv}=\hat{v}$; for a negative real scalar, $\widehat{cv}=-\hat{v}$. Thus $(3,4)$ and $(6,8)$ normalize to the same unit vector, while $(-3,-4)$ normalizes to its negative. This decomposition is useful for specifying a direction independently of a force, velocity or displacement magnitude.

In a complex normed vector space, a nonzero complex scalar $c$ instead gives $\widehat{cv}=(c/|c|)\hat{v}$. The remaining complex argument is not removed by normalization. For a pure quantum state, vectors differing only by this overall global phase represent the same physical state; vector normalization alone does not identify them.

The zero vector cannot be normalized this way. The construction also applies in a function space: an eigenfunction of a Sturm-Liouville problem can be scaled to unit weighted norm byThe positive weight $w$ specifies the weighted inner product. Multiplication by this nonzero constant preserves the eigenvalue equation and its homogeneous boundary conditions.