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The Rayleigh-Ritz variational principle bounds the lowest eigenvalue by the Rayleigh quotient of any nonzero admissible trial function. For $f=1-(r/a{)}^p$, take $p>0$, which ensures boundedness, the outer Dirichlet boundary condition, and finite radial Dirichlet energy. Direct integration givesThereforeThe derivative is $1-2/p^2$, so the unique minimizing positive exponent is $p=\sqrt{2}$. This yields the power-law variational bound for a circular membraneThe optimizing function has zero derivative at the origin and is admissible for the energy principle despite its unbounded second derivative there. For comparison, the exact radial fundamental has $\sqrt{{\lambda }_1}a\simeq 2.404826$, giving ${\nu }_1\simeq 0.382740c/a$. The bound is about $0.39\%$ above this value.

A positive matching root $x_{\ast }$ gives $s=x_{\ast }^2/R^2$. Substituting the root value requested by the question gives the intended expressionAfter the index correction in part (iii), this is indeed the lowest mode in the $\ell =1$ sector, because its condition is $j_0(x)=\sin x/x=0$ and its first positive root is $\pi$. It is also the minimum over $\ell \ge 1$.

However, the literal scalar problem includes $\ell =0$, for which matching gives $\cos x=0$. Its first positive root is $\pi /2$, so the actual minimum for the written scalar equations and stated boundary conditions isAn explicit counterexample to the larger proposed minimum is $B_{\mathrm{in}}=j_0(\pi r/(2R))$, $B_{\mathrm{out}}=2R/(\pi r)$, with no angular dependence. At $R$, both values are $2/\pi$ and both radial derivatives are $-2/(\pi R)$.

To verify minimality, put $u=r{F}_{\ell }$. The radial problem is $-u^{\prime \prime }+\ell (\ell +1)u/r^2=su$, $u(0)=0$, $u^{\prime }(R)+\ell u(R)/R=0$. Its Rayleigh quotient isFor $\ell =0$ this is the Dirichlet-Neumann interval problem, whose lowest eigenvalue is ${\pi }^2/(4R^2)$. The additional nonnegative terms increase with $\ell$, proving the scalar minimum and the minimum-at-$\ell =1$ statement for the restricted family. A monopole-free magnetic multipole interpretation would exclude $\ell =0$ and yield the intended value, but that extra restriction is not supplied by the stated scalar model. The claim about the first zero is not valid for every $f_{\ell }$ as printed.

The radial Rayleigh quotient isFor the given trial profile, the numerator is $4a^{-4}{\int }_0^a{r}^{4}dr=4a/5$. The denominator isThereforeThe profile is regular at the origin and vanishes at the wall, so it belongs to the admissible trial space. This is the variational estimate for the radial Dirichlet modes of a spherical cavity.

Assume $H$ is self-adjoint on its specified operator domain, and the admissible variations preserve that domain and any boundary conditions, so no boundary terms spoil self-adjointness. Let $N=⟨\varphi ,\varphi ⟩>0$. The first variation of the Rayleigh quotient isAt $\varphi ={\psi }_n$, the residual vanishes, so $\boxed{\delta F[{\psi }_n]=0}$. More explicitly, for a normalized ${\psi }_n$ and $\varphi ={\psi }_n+\eta$,For a fixed admissible direction scaled by a small parameter, the change is second order. This is stationarity, and is not necessarily a minimum for an excited state.

Conversely, vanishing first variation for every admissible complex direction yields the stationary equationAt the original stationary point, this is $(H-F[{\psi }_n]){\psi }_n=0$. The printed equation with an arbitrary perturbed $\varphi$ in the final factor is not generally true. Taking $\varphi ={\psi }_n+ϵ{\psi }_m$ with $E_m\ne E_n$ gives $(H-E_n)\varphi =ϵ(E_m-E_n){\psi }_m\ne 0$, even though the first-order change in $F$ is zero. Thus the valid claim is the stationary equation at the extremal state, not an exact eigenvalue equation for every variation.

For an eigenfunction in the operator domain, $H{\psi }_n=E_n{\psi }_n$. Substitution into the Rayleigh quotient givesThe cancellation only needs a finite nonzero norm; orthonormalization makes both denominator and norm equal to one.

Multiply the eigenvalue equation of the Sturm-Liouville problem by $y$ and integrate. Integration by parts givesSince $y$ is a nonzero eigenfunction and $\omega >0$, the denominator is strictly positive. Thus the Rayleigh quotient isThe endpoint term must be retained for a general Robin boundary condition; it vanishes for homogeneous Dirichlet boundary conditions but need not vanish in the present general problem.

The left boundary condition is $y(0)=0$, while the right one is $y^{\prime }(\ell )=-ky(\ell )$. Hence $[y{y}^{\prime }{]}_0^{\ell }=-ky(\ell {)}^2$. The Rayleigh quotient becomesEvery term on the right is nonnegative. If the whole expression were zero, in particular ${\int }_0^{\ell }(y^{\prime }{)}^{2}dx=0$, so $y$ would be constant. Its zero value at the left endpoint would force $y\equiv 0$, contrary to its being an eigenfunction. ThereforeThis is the positivity mechanism for Dirichlet-Robin eigenvalues on a finite interval. The sign of the boundary contribution is positive for the convention $y^{\prime }+ky=0$ with $k>0$.

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich 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 giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is

Unperturbed ground mode. At $ϵ=0$, the equation is $-u^{\prime \prime }=\lambda u$. 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 $\cos (\pi \sqrt{\lambda })=0$. HenceThe lowest eigenvalue and normalized eigenfunction are

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 $U_0^{\prime }=-\sqrt{2/\pi }\sin (x/2)/2$, givingUsing ${\sin }^2(x/2)=(1-\cos x)/2$,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every $ϵ\ge 0$, not merely a small-parameter approximation. As a lower-bound check, $p\ge 1$ and the unperturbed variational minimum imply ${\mathcal{R}}_{ϵ}[v]\ge 1/4+ϵ$ for every admissible normalized $v$. Thus

For constant diffusivity, the trial Rayleigh quotient simplifies toIts derivative is $1-2/q^2$, so its only positive critical point is $q=\sqrt{2}$. The second derivative is $4/q^3>0$; also the quotient diverges as $q\downarrow 0$ or $q\to \infty$. Therefore this is the global optimum within the trial family:The two requested profiles areThe initial radial graph decreases from one at the centre to zero at the rim, has horizontal tangent at the centre, and is concave downward for $0<r<1$. The central temperature decreases exponentially from one to zero, with initial time derivative $-(3+2\sqrt{2})$. These are trial profiles, not exact solutions of the full heat equation. Although the radial trial function is not twice differentiable at the centre, it has finite weighted energy and belongs to the variational trial class used by the Rayleigh-Ritz variational principle.

Enlarge the trial space while retaining the optimized Gaussian. For example, usenormalize it and minimize the Rayleigh quotient over its coefficients, optionally optimizing $\alpha$ too. This even trial space respects the even ground-state parity of the confining symmetric potential and adds independent shape freedom beyond a Gaussian width.

The original Gaussian remains included by setting the later coefficients to zero, so the optimized upper bound cannot increase. Finite-basis optimization is the Rayleigh-Ritz method, with matrix entries given by the Hamiltonian and overlap integrals. Increasing the basis size systematically improves the available approximation.

Move the differential terms to the left in the sign convention of part (a). Then $\tilde{a}=1+x$, $\tilde{c}=0$, and $\tilde{b}=(3+2x)-{\tilde{a}}^{\prime }=2(1+x)$. The integrating factor is therefore $w=e^{2x}$, up to a positive constant. The Sturm-Liouville problem isThe trial $y=x{e}^{-x}$ is admissible because $y(0)=0$ and $y^{\prime }=e^{-x}(1-x)$ vanishes at one. Its Rayleigh quotient has numeratorand denominatorConsequently the Rayleigh-Ritz variational principle givesThe numerator is positive for every nonzero admissible function, so the lowest eigenvalue is positive. The trial is not an exact eigenfunction: substituting it would require $x^2+2x-1=\lambda x$ for every $x$. Thus the upper inequality is actually strict. This example underlies the exponential-polynomial trial space for a mixed Sturm-Liouville problem.

The added sine vanishes at zero. Its derivative at one is $B\cos B$, so nontrivial admissible choices areNegative values give the same trial directions after changing the sign of $C$; $B=0$ adds nothing. For each choice, minimize the Rayleigh quotient of $x{e}^{-x}+C\sin (Bx)$ over $C$, or optimize over several coefficients simultaneously. The original trial is included, so the best bound is no larger than $5/4$. Enlarging the trial subspace successively is the Rayleigh-Ritz method; the boundary-compatible half-integer sines provide additional independent shapes.

For the added trial term $q_n=x^n{e}^{-Ax}$, the left Dirichlet boundary condition is automatic for a positive integer $n$. At the right endpoint,To preserve $y^{\prime }(1)=0$ for an arbitrary variational coefficient $C$, choose $\boxed{A=n}$. Then minimize the Rayleigh quotient of $x{e}^{-x}+C{x}^n{e}^{-nx}$ over $C$.

Since $C=0$ remains available, this optimization cannot worsen the original upper bound. For $n=1$, the added function is a multiple of the first trial, so the quotient does not change: an independent choice $n\ge 2$ is needed for extra freedom. More generally, several such functions form the exponential-polynomial trial space for a mixed Sturm-Liouville problem, and the Rayleigh-Ritz method reduces minimization to a small generalized eigenvalue problem.

Choose homogeneous Dirichlet boundary conditions $\psi (a)=\psi (b)=0$, and take a nonzero real admissible trial $\psi$. Write the numerator as $F[\psi ]$ and denominator as $G[\psi ]>0$. For an endpoint-zero variation $\eta$, integration by parts givesThus the first variation of the Rayleigh quotient isBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiplying an eigenfunction equation by $\psi$ and integrating gives $F=\lambda G$, so its first variation vanishes. This establishes the equivalence for the chosen Sturm-Liouville problem. Homogeneous Neumann boundary conditions, or a mixed Dirichlet-Neumann pair, also cancel the boundary term; general Robin boundary conditions require adding the corresponding boundary contribution to the numerator.

For the two sine modes, orthogonality givesThe diagonal potential integrals vanish, while the mixed one isThus ${\int }_0^1\cos (\pi x){\tilde{\psi }}^{2}dx=\alpha /2$, giving the two-sine variational bound for a Dirichlet Mathieu equation:Its derivative vanishes when $1+6{\pi }^2\alpha -{\alpha }^2=0$. Hence the two stationary parameters are $3{\pi }^2\pm \sqrt{9{\pi }^4+1}$. To identify the minimum, use the normalized sine basis: the restricted self-adjoint operator has real symmetric matrixIts smaller eigenvalue is the minimum of the Rayleigh quotient, attained at the negative stationary parameter. ThereforeThe estimate improves on the single-sine value ${\pi }^2$. The Rayleigh-Ritz variational principle remains valid here even though $q(x)=\cos (\pi x)$ changes sign: on this finite interval the Sturm-Liouville operator is self-adjoint and bounded below, which is sufficient for the upper bound.

The original PDF has a typographical omission: its displayed equation has $\lambda w(x)$ rather than $\lambda w(x)y(x)$ on the right. Taken literally, it is an inhomogeneous boundary value problem, not the stated Sturm-Liouville problem. The quotient and the later eigenfunction expansion identify the intended equation, which the following derivation produces explicitly.

Choose homogeneous Dirichlet boundary conditions $y(a)=y(b)=0$ and a nonzero real admissible $y$. Then $G[y]>0$ because $w>0$. For an endpoint-zero variation $\eta$, integration by parts givesThe first variation of the Rayleigh quotient is thusBy the fundamental lemma of the calculus of variations, stationarity is equivalent toConversely, multiply this corrected eigenvalue equation by $y$ and integrate. The same boundary cancellation gives $F[y]=\lambda G[y]$, so the first variation vanishes. This proves the requested equivalence for the intended equation. Homogeneous Neumann boundary conditions, or a Dirichlet condition at one end and a Neumann condition at the other, also eliminate the boundary term. General Robin boundary conditions would require the corresponding boundary contribution in the numerator.

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

For an admissible nonzero real trial function, integration by parts and the zero-flux boundary conditions reduce the Rayleigh quotient toA nonzero constant trial function attains zero. Positivity of $p$ also shows that equality requires $y^{\prime }=0$. HenceIf the conserved total amount is $M={\int }_{-L}^{L}cdx$, the normalized stationary solution isIts diffusive flux vanishes everywhere. The Boltzmann distribution fixes the spatial profile, while the diffusivity affects the relaxation toward it. Boundedness of $V$ makes the normalization finite on this finite interval.

Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

The power iteration starts from $v_0\ne 0$, normally scaled to unit Euclidean norm, and repeatsNormalization prevents the rapidly growing or decaying magnitudes of $A^k{v}_0$ from obscuring its direction. The iteration requires $w_{k+1}\ne 0$.

Because $A$ is a diagonalizable matrix, choose an eigenbasis $u_1,\dots ,u_n$, with corresponding eigenvalues ${\lambda }_1=\lambda ,{\lambda }_2,\dots ,{\lambda }_n$, and expand $v_0={\sum }_j{c}_j{u}_j$. First suppose the dominant eigenvalue is simple and $c_1\ne 0$. ThenForthe sum in parentheses after $u_1$ is $O(q^k)$ in Big O notation. Consequently the power iteration direction error is $O(q^k)$: the iterates approach the line spanned by $u_1$. For a real matrix, a uniquely dominant eigenvalue is real, since otherwise its complex conjugate would have the same modulus. If $\lambda >0$, $v_k$ approaches one unit eigenvector; if $\lambda <0$, successive $v_k$ alternate signs. In either case,This is the sign and phase in power iteration qualification. Also, the necessary assumption is a nonzero coefficient $c_1$ in the eigenbasis, rather than merely a nonzero ordinary inner product with $u_1$: those tests can differ for a nonsymmetric matrix. If $c_1=0$, the method cannot generate the missing component.

If the dominant value $\lambda$ has a higher-dimensional eigenspace, the same argument replaces $c_1{u}_1$ by the nonzero sum of all starting components in that eigenspace. The limiting direction is then that of this sum, which is still an eigenvector for $\lambda$. The usual algorithm assumes $\lambda \ne 0$; a strict dominance gap guarantees this whenever another eigenvalue is present.

An estimate retaining the sign of $\lambda$ is the Rayleigh quotientContinuity of this quotient at a nonzero eigenvector proves the limit even when $A$ is a nonsymmetric matrix. Alternatively, $(A{v}_k{)}_i/(v_k{)}_i\to \lambda$ for a coordinate whose limiting eigenvector component is nonzero. The scaling factor $\Vert A{v}_k{\Vert }_2$ alone tends only to $|\lambda |$.