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Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.

Take fixed-endpoint variations $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation isThe boundary term vanishes by the endpoint conditions. The fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $f_y-d(f_{y^{\prime }})/dx=0$. Along its solutions,When $f_x=0$, this proves the Beltrami identity $f-y^{\prime }{f}_{y^{\prime }}=\text{constant}$.

For the surface $z=F(y)$, the line element is $d{s}^2=d{x}^2+[1+F^{\prime }(y{)}^2]d{y}^2$. The arc-length integrand for a graph $y(x)$ is $\mathcal{L}=\sqrt{1+[1+F^{\prime }(y{)}^2]y^{\prime 2}}$. It has no explicit $x$ dependence and satisfies $\mathcal{L}-y^{\prime }{\mathcal{L}}_{y^{\prime }}=1/\mathcal{L}$. Thus $[1+F^{\prime }(y{)}^2]y^{\prime 2}$ is constant. Choosing the fixed direction along a smooth nonconstant segment and integrating yieldsThe same result follows by defining the transverse arc coordinate $u(y)={\int }^y\sqrt{1+F^{\prime }(s{)}^2}ds$, which changes the metric to $d{s}^2=d{x}^2+d{u}^2$. Its geodesics are straight lines in $(x,u)$ coordinates. This proves the formula for geodesics on a generalized cylinder and shows it gives locally shortest paths. The additional straight lines $x=\text{constant}$ in the unrolled plane are also geodesics, but cannot be represented as graphs $y(x)$. Classical differential-geometric statements assume the surface has sufficient smoothness; the arc-length coordinate itself only requires the displayed integral to be well defined.

At a regular constrained extremum, the method of Lagrange multipliers makes $I+\lambda J$ stationary for a constant $\lambda$. For an endpoint-vanishing variation $\eta$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yieldsThis Euler-Lagrange equation must be supplemented by the two endpoint conditions and the prescribed integral constraint, which determine the constants and multiplier. Existence of the normal multiplier assumes that the first variation of $J$ is not identically zero. If that regularity fails, the general multiplier equation uses $\alpha I+\lambda J$, with $(\alpha ,\lambda )\ne (0,0)$; one cannot automatically set $\alpha =1$.

Introduce a Lagrange multiplier for the constraint $G=1$ and vary $F-\lambda G$ with $y\mapsto y+ϵ\eta$, where $\eta (a)=\eta (b)=0$. The first variation isThe fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Multiplying by $y$, integrating by parts and using the endpoint conditions gives $F=\lambda G$. Therefore the normalization implies $\lambda =F$. Assume the coefficient and function regularity needed for these variations and integrations; the condition $w>0$ makes the constraint derivative nonzero at a normalized function.

For the radial problem, set $\sigma =r\psi$. A direct differentiation gives ${\sigma }^{\prime \prime }=r{\psi }^{\prime \prime }+2{\psi }^{\prime }$, soFinite $\psi (0)$ supplies the first boundary value. The Rayleigh quotient is $R[\sigma ]={\int }_0^1{\sigma }^{\prime 2}dr/{\int }_0^1{\sigma }^{2}dr$. For the two trials, ${\sigma }_1=r(1-r)$ and ${\sigma }_2=r(1-r^2)$, the numerator-denominator pairs are $(1/3,1/30)$ and $(4/5,8/105)$, respectively. HenceThe exact eigenvalues are $n^2{\pi }^2$, so $E_{\min }={\pi }^2\simeq 9.86960$. Both trials give upper bounds by the Rayleigh-Ritz variational principle, and the first estimate is more accurate, since it is smaller. Trial admissibility uses the Dirichlet energy form for $\sigma$; the first trial's nonzero radial derivative at the origin does not invalidate this variational calculation.

To improve, add independent trial functions and minimize the quotient over their span. Here merely combining the two supplied trials gives no improvement: their two-dimensional stiffness and mass matrices areThe lowest Ritz value is still ten. Instead add the symmetric quartic $\tau =r^2(1-r{)}^2$. On the span of ${\sigma }_1,\tau$, the generalized eigenvalue problem hasIts determinant is $(E^2-112E+1008)/529200$, so the least root givesThis is strictly better than both original estimates. The symmetry obstruction in a polynomial Rayleigh-Ritz trial space explains why the cubic addition failed while the quartic helps. It is also the shifted-interval version of the polynomial Rayleigh-Ritz trial space for a one-dimensional box.

Use the real weighted inner product $⟨u,v{⟩}_w={\int }_a^{b}wuvdx$ and the energy bilinear form $B(u,v)={\int }_a^b(p{u}^{\prime }{v}^{\prime }+quv)dx$. Integration by parts, the endpoint conditions, and the Sturm-Liouville problem giveThus orthonormality in the stated expansion givesand hence the Rayleigh quotientThe expansion is understood for nonzero admissible functions with finite norm and energy.

For a general endpoint-preserving variation $y+\epsilon \eta$, explicitlysoThe fundamental lemma of the calculus of variations shows that stationarity is exactly the eigenvalue equation with $\lambda =\Lambda$. Conversely an eigenfunction has vanishing first variation and quotient equal to its eigenvalue. Completeness gives the same weighted-average formula for any nonzero admissible expansion, even one with zero ground-state coefficient. Since every ${\lambda }_i\ge {\lambda }_0$,This is the Rayleigh-Ritz variational principle.

For the interval problem, the normalized Dirichlet eigenfunctions and their eigenvalues, with the paper's zero-based indexing, areFor $u=x(1-x)$, direct integration gives $F={\int }_0^1(1-2x{)}^{2}dx=1/3$ and $G={\int }_0^1{x}^2(1-x{)}^{2}dx=1/30$. ThereforeThe sine expansion and Rayleigh quotient of a parabolic Dirichlet trial follow from the coefficient integralConsequently the requested expansion hasEquivalently $u=(8/{\pi }^3){\sum }_{n\ge 1,n\text{ odd}}\sin (n\pi x)/n^3$. In particular $c_1=0$: the first excited eigenfunction $\sqrt{2}\sin (2\pi x)$ is antisymmetric about $1/2$, whereas the trial function is symmetric. Since $10-{\pi }^2>0$, the final ratio is exactlyThe first nonzero correction is $c_2=1/27$, not $c_1$. Reindexing only the nonzero terms would change the meaning of the requested ratio and is not used here.

One suitable choice is homogeneous Dirichlet boundary conditions, $\psi (\alpha )=\psi (\beta )=0$, with variations vanishing there. Assume real $q$ and the stated positive $p,w$. Introduce a Lagrange multiplier $\mu$ for $N[\psi ]=1$. For real variations, integration by parts givesThe boundary term vanishes, so the fundamental lemma of the calculus of variations yieldsMultiplying this equation by $\psi$, integrating, and using the boundary conditions gives $Q[\psi ]=\mu N[\psi ]=\mu$. Thus the multiplier, the eigenvalue, and the constrained stationary value coincide. Conversely every normalized eigenfunction makes the first variation vanish for all tangent variations satisfying $\delta N=0$.

Homogeneous Neumann boundary conditions, $p{\psi }^{\prime }=0$ at both endpoints, or mixed Dirichlet and Neumann conditions also work with this energy form. General Robin boundary conditions require the corresponding boundary terms in the form; they cannot simply be inserted into the displayed unmodified integral.

The boundary term $[f_{y^{\prime }}\eta {]}_a^b$ vanishes, so stationarity and the fundamental lemma of the calculus of variations give the Euler-Lagrange equation

For a variation $y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$, the first variation of the functional isThe boundary term vanishes by the fixed endpoints. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

For a variation $q\mapsto q+ϵ\eta$ with $\eta (t_0)=\eta (t_1)=0$, the first variation isThe endpoint term vanishes. Since the interior variation is arbitrary, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equationAlong such a trajectory,If $L$ has no explicit time dependence, the first integral is thereforeIts negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian $L=T-V$ with quadratic velocity dependence, this conserved quantity is the total mechanical energy $T+V$.

For fixed endpoint values and a smooth variation vanishing at the endpoints, the first variation isThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationIf $f$ is independent of $y$, then $y$ is a cyclic coordinate, and integration givesThis is conservation of the corresponding generalized momentum.

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.

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.

For fixed initial and final coordinates, the two Euler-Lagrange equations areIndeed the first variation of the functional isIntegration by parts moves the time derivatives from the variations; their endpoint values are zero. The fundamental lemma of the calculus of variations then gives the two equations above.

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 fundamental lemma of the calculus of variations says that if $h\in L_{\mathrm{loc}}^1(\Omega )$ and $\int h\phi =0$ for every $\phi \in C_0^{\infty }(\Omega )$, then $h=0$ almost everywhere.

If $v_1,v_2$ both represent $D^{\alpha }u$, subtracting their defining identities gives $\int (v_1-v_2)\phi =0$ for every test function. The lemma gives $v_1=v_2$ almost everywhere, proving uniqueness. If $u\in C^1$, part (i) gives exactly the defining identity with $v={\partial }_{j}u$. Since this classical derivative is locally integrable, it is the unique weak derivative $D_{j}u$, up to equality almost everywhere.

For fixed endpoint values, the Euler-Lagrange equation isIt follows by differentiating the functional along $y+\epsilon \eta$ with $\eta (0)=\eta (\pi )=0$, integrating the $F_{y^{\prime }}{\eta }^{\prime }$ term by parts and applying the fundamental lemma of the calculus of variations.

For this quadratic integrand it becomes $y^{\prime \prime }+ky=0$. If $k>0$, write $\nu =\sqrt{k}$; the condition at zero gives $y=B\sin (\nu x)$, and that at $\pi$ requires $B\sin (\nu \pi )=0$. Consequently $y=0$ is the only extremal when $\sqrt{k}$ is not an integer. If $\sqrt{k}=n$ is a positive integer, all extremals areThe integer case $n=0$ must be handled separately: $y^{\prime \prime }=0$ with both endpoints zero has only $y=0$. If negative real $k$ is allowed, the hyperbolic-function solutions also yield only zero under these endpoints. These statements concern stationary functions; no minimum is assumed.

Let $\eta$ be a smooth variation with $\eta (a)=\eta (b)=0$. Assuming the usual continuous second derivatives of $F$, stationarity of the functional under $y+t\eta$ gives its first variationIntegration by parts removes the derivative of $\eta$; its boundary term vanishes because the endpoint values are fixed. Thusfor every such variation. The fundamental lemma of the calculus of variations implies the Euler-Lagrange equation $F_y=(d/dx)F_{y^{\prime }}$.

For arc length, $F=\sqrt{1+(y^{\prime }{)}^2}$, soThe function $s\mapsto s/\sqrt{1+s^2}$ is strictly increasing; hence $y^{\prime }$ is constant. With endpoints $(a,A)$ and $(b,B)$, $a<b$, the extremal isIt is also the global shortest curve: integrating the unit displacement vector dotted with the curve velocity gives the endpoint distance, which is at most the integrated speed by Cauchy-Schwarz inequality; the straight segment attains equality. A vertical segment is described parametrically or after rotating axes, since it cannot be represented by $y(x)$ on a nontrivial horizontal interval.

Take a smooth variation $u+\epsilon \eta$ with $\eta$ vanishing on the boundary; compactly supported test functions suffice for the interior equation. At an extremum, the first variation is zero:Integration by parts removes the derivatives from $\eta$. The boundary terms vanish, givingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange equationThis is the classical equation for sufficiently smooth admissible extrema; for weaker regularity the first-variation identity supplies its weak form.

For the squared-gradient integrand, the equation is $2(u_{xx}+u_{yy}+u_{zz})=0$, namely the Laplace equation. Put $u_0=ax+by+cz$ and write any admissible function as $u=u_0+v$, with $v=0$ on the boundary. The energy identity isBy the divergence theorem, the middle integral is ${\int }_{\partial V}v(a,b,c)\cdot ndS=0$; equivalently it vanishes by weak integration by parts for zero-trace functions. The final term is nonnegative and vanishes at $v=0$. Thus this is a global minimum, not merely a stationary value:

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

Take independent smooth variations $y+\epsilon \eta$ and $z+\epsilon \zeta$. Differentiating the functional and using integration by parts gives its first variationFirst choose variations vanishing at the endpoints. To use the stated form of the fundamental lemma of the calculus of variations, take $\eta =(x-a)(b-x)\chi$ with arbitrary smooth $\chi$, and similarly for $\zeta$. The lemma makes $(x-a)(b-x)(F_y-d{F}_{y^{\prime }}/dx)$ vanish, and hence makes the unweighted coefficient vanish in the interior; continuity extends this to the endpoints. Applying the same argument to $z$ gives the Euler-Lagrange equationsIf either endpoint value of $y$ or $z$ is free, the corresponding variation there is arbitrary. The remaining boundary term gives the natural boundary conditions for a free endpoint: $F_{y^{\prime }}=0$ for a free $y$ value and $F_{z^{\prime }}=0$ for a free $z$ value at that endpoint. With both coordinates free at both endpoints, all four conditions apply.

For the specified integrand these equations reduce toand the natural conditions are $y^{\prime }=z^{\prime }=0$ at each free endpoint. Eliminating $z$ givesIts four independent modes give the full general solutionAt zero, $y^{\prime }(0)=C+2D$ and $z^{\prime }(0)=2C+D$. Vanishing forces $C=D=0$, givingAt $x=\pi /6$ the two derivative conditions become $A/2+\sqrt{3}B=0$ and $A+\sqrt{3}B/2=0$. Their coefficient determinant is nonzero, so $A=B=0$. Thus the unique stationary extremal with all four natural boundary conditions is $y=z=0$. The word extremal here means a stationary point of the functional; the indefinite integrand does not require this point to be a minimum.