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For homogeneous corrections at the square boundary, integration by parts gives $⟨{\psi }_0\Delta {\psi }_1⟩=⟨{\psi }_1\Delta {\psi }_0⟩$ and $⟨{\theta }_0\Delta {\theta }_1⟩=⟨{\theta }_1\Delta {\theta }_0⟩$. The first-derivative cross terms give $⟨{\psi }_0{\theta }_{1x}⟩=-⟨{\psi }_{0x}{\theta }_1⟩$ and $⟨{\theta }_0{\psi }_{1x}⟩=-⟨{\theta }_{0x}{\psi }_1⟩$. Using $\Delta {\psi }_0=q_c^2{\theta }_{0x}$ and $\Delta {\theta }_0=-{\psi }_{0x}$, these four contributions cancel in pairs. This proves exactly the supplied solvability condition, including the weight $q_c^2$ on its second row. Equivalently $({\psi }_0,q_c^2{\theta }_0)$ is the relevant adjoint eigenfunction for this coupled system.

At first order in the detuning, the inhomogeneous equations areInsert these right-hand sides into the verified solvability condition. One obtainsA further integration by parts, using the leading heat equation, givesHence the growth-rate solvability for conducting-square Darcy convection yieldsThe PDF prints the reciprocal of the required integral ratio. With its declared ${\theta }_t=ϵ\lambda \theta$, the displayed reciprocal does not follow and is false for $q_1\ne 0$. Both integrals are positive for these nonzero modes. The square's Poincare inequality gives their ratio at least $2{\pi }^2$, so it cannot equal its reciprocal. For $q_1=1$, the two explicit parities give approximately $\lambda =11.34770$ for odd $\theta$ and $\lambda =7.30164$ for even $\theta$; the printed expression instead gives about $0.00446439$ and $0.00693825$. These values provide direct mode-based counterexamples.

The corrected growth rate is positive for $q_1>0$ and negative for $q_1<0$, as expected at a convection threshold. Because the critical eigenspace is two-dimensional, an arbitrary superposition generally splits into two different first-order growth rates. The specified parity modes remain independent under the detuning: the temperature linear operator preserves reflection parity, so the cross-parity projection vanishes. This justifies applying the scalar solvability condition to either listed eigenfunction; it should not be applied as a single common eigenvalue to an arbitrary mixture.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincare inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincare inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.

Use the downward-positive displacement and the filament bending modulus $B$ and load density $w$ defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is $-\delta E/\delta h=-B{h}_{xxxx}+w$. In resistive-force theory, transverse motion has drag per unit length $-{\zeta }_{\perp }{h}_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.

At steady state, $B{h}_s^{\prime \prime \prime \prime }=w$. Integrating first from the free end gives $h_s^{\prime \prime \prime }=(w/B)(x-L)$ and $h_s^{\prime \prime }=(w/2B)(x-L{)}^2$. Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is $\delta =h_s(L)=w{L}^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses $A$ as Young's modulus; if $A$ itself is flexural rigidity, the same result reads $k_{\mathrm{eff}}=8A/L^3$. This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.

For dimensional analysis, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[{\zeta }_{\perp }]=$ force times time/length squared makes ${\zeta }_{\perp }{h}_t$, $B{h}_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $w{L}^3/B\ll 1$, since $h_s^{\prime }(L)=w{L}^3/(6B)$.

Reset $t=0$ when gravity is switched off. The subsequent overdamped relaxation of a clamped--free filament obeys ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}$ with the same homogeneous endpoint conditions. Separate a normal mode as $h=\varphi (x/L)e^{-\Gamma t}$. Writing $s=x/L$ givesThe two clamp conditions reduce the general hyperbolic/trigonometric solution toThe two free-end conditions have matrixIts determinant is $2(1+\cosh \alpha \cos \alpha )$. Hence the clamped--free bending mode values satisfyThere is no zero mode: at zero eigenvalue $\varphi$ is cubic, and the four clamp/free conditions force all coefficients to vanish. With the following normalization a mode isThe fourth-order operator is self-adjoint and positive under these endpoint conditions, since $\int f{g}^{\prime \prime \prime \prime }=\int f^{\prime \prime }{g}^{\prime \prime }$. Its orthogonal eigenfunction expansion givesFor the previously attained steady profile, the first coefficient is nonzero: $h_s$ is positive and the first mode can be chosen positive in the interior. Two applications of integration by parts, using both functions' endpoint conditions, also giveThus the dominant long-time answer, including its shape and amplitude, isThe next mode decays much faster. For other initial shapes with an accidentally zero first-mode projection, the lowest mode with $c_n\ne 0$ determines the asymptotic behaviour. The decay time has the correct units and exhibits the strong $L^4$ dependence characteristic of bending opposed by local viscous drag.

Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

With periodic or sufficiently decaying boundary conditions, integration by parts givesAssume $\kappa \ge 0$ for a stable quadratic free energy; the diffusive smoothing discussed below requires $\kappa >0$. In a periodic box of volume $V$, use the unitary convention ${\varphi }_{\mathbf{q}}=V^{-1/2}{\int }_V\varphi (\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}d\mathbf{r}$. A real field has ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. Since the Fourier transform of a derivative converts ${\nabla }^2$ to $-q^2$, each Fourier mode obeysIt is an Ornstein-Uhlenbeck process. Multiplication by the integrating factor $e^{r(q)t}$ givesThe noise integral is understood as an Itô integral with a deterministic kernel; writing $\eta ds$ is shorthand for the corresponding Wiener increment. Direct differentiation in the mild stochastic sense checks both the equation and the initial value. The stated Kronecker-delta noise covariance corresponds to this finite-volume normalization. Unlike a conserved order parameter, the zero wavenumber mode decays at the nonzero rate $\Gamma a$.

First ${\int }_{\mathbb{R}}{e}^{-x^2/2}dx=\sqrt{2\pi }$: square the positive integral and use polar coordinates to obtain $2\pi {\int }_0^{\infty }r{e}^{-r^2/2}dr=2\pi$. Scaling therefore normalizes the one-dimensional Gaussian function $g_t$.

Let $I(\xi )={\int }_{\mathbb{R}}(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}{e}^{-i\xi x}dx$. Differentiation under the integral is justified by the integrable majorant $|x|g_t(x)$. Since $g_t^{\prime }(x)=-x{g}_t(x)/t$, integration by parts, with zero endpoint terms, gives $I^{\prime }(\xi )=-t\xi I(\xi )$. With $I(0)=1$, this yields $I(\xi )=e^{-t{\xi }^2/2}$. Multiplication of the one-dimensional integrals givesApplying the same computation with reciprocal scale $1/t$ evaluates its inverse integral:For $f\in L^1$, Fubini's theorem gives $\widehat{f\ast g_t}=\hat{f}{e}^{-t|\xi {|}^2/2}$, which is integrable because $\hat{f}$ is bounded. Its inverse integral isAbsolute integrability of $|f(y)|e^{-t|\xi {|}^2/2}$ justifies this second application of Fubini's theorem. Thus Fourier inversion holds for every Gaussian convolution, in fact pointwise for this continuous representative.

The horizontal contribution is $B(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{e}^{-ix\sin z}{e}^{i\nu z}dz$. The two nondegenerate stationary points of the phase $-\sin z$ are $\pi /2$ and $-\pi /2$, with second derivatives $1$ and $-1$. The stationary phase method gives their contributionswith remainder $O(x^{-3/2})$ for smooth fixed amplitudes localized near the stationary points. On the remaining portions, use integration by parts. The endpoint term isCutoff endpoints cancel between pieces; a second integration gives the other nonstationary contributions $O(x^{-2})$. Thus this endpoint term cancels the leading vertical term. AltogetherThis stationary phase endpoint cancellation in a half-strip contour is necessary: omitting it would leave an erroneous $x^{-1}$ remainder for noninteger $\nu$.

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.

Differentiating the grand potential givesA reversible adiabatic change in an isolated gas keeps $S$ and $N$ fixed. To infer constant $y$, one must check that this ratio is one-to-one, rather than merely assuming it.

Put $q=y/k$, $n(x)=1/(1+e^{x-q})$, $F_j={\int }_0^{\infty }{x}^{j}n(x)dx$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(x)(1-n(x))dx$ for $j=0,1,2$. Integration by parts gives $M_1=b{F}_a$ and $M_2=(b+1)F_{a+1}$, andThe strict inequality is the Cauchy-Schwarz inequality for the positive weight $x^{a}n(x)(1-n(x))$, since $x$ is not constant on its support. This proves strict monotonicity of Fermi gas entropy per particle and hence $\mu /T$ is constant. Then $N=kV{T}^b{f}^{\prime }(y)$ gives $V{T}^b$ constant, and $P=k{T}^{b+1}f(y)$ givesThis adiabatic scaling of a Fermi gas with power-law density of states concerns positive temperature; the zero-temperature limit is a separate limiting case.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

Dominated convergence makes $\hat{f}$ continuous, and $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. Smooth compactly supported functions are dense in $L^1$. Their Fourier transforms decay at infinity by integration by parts, so uniform approximation proves the Riemann-Lebesgue lemma: $\hat{f}\in C_0$.

For density, smooth compactly supported functions are also uniformly dense in $C_0$ by cutoff and mollification. Each such target $g$ is a Schwartz function; its inverse Fourier transform is another Schwartz function, hence belongs to $L^1$, and Fourier inversion recovers $g$. Therefore the image of a function of $L^1$ is dense in $C_0$.

The $L^1$ Fourier inversion theorem says that if $f,\hat{f}\in L^1$, then $f$ agrees almost everywhere with the inverse integral of $\hat{f}$. If $\hat{f}=0$, this applies and gives $f=0$ almost everywhere. Hence the Fourier transform is injective as a map of $L^1$ equivalence classes, with dense image in $C_0$.

The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

For $-\beta /{\ell }^2<V\le 0$, there is one positive stationary wavenumber $k_0\in (0,\ell ]$ satisfying $\beta (k_0^2-{\ell }^2)/(k_0^2+{\ell }^2{)}^2=V$, and its negative. Since ${\omega }^{\prime \prime }(k_0)>0$, their paired stationary-phase contribution isAt $V=0$ the finite stationary points are $\pm \ell$ and are nondegenerate, so the same formula applies, assuming sufficient decay to control the large-wavenumber tail.

For $V>\beta /(8{\ell }^2)$ or $V<-\beta /{\ell }^2$, there are no real stationary points: no group-velocity ray reaches those slopes. Integration by parts gives much smaller contributions; for a Schwartz amplitude they decay faster than every inverse power of $t$ at a fixed such $V$. For less regular data the rate depends on the amplitude, and the printed even function assumption alone does not specify it. At the excluded endpoint values of the group velocity stationary points coalesce and an ordinary $t^{-1/2}$ formula fails, typically requiring an Airy-type transition rather than the expressions above.

Substitute the Laplace contour-integral method for a linear differential equation into the differential equation and use $z{e}^{zt}={\partial }_t{e}^{zt}$. Integration by parts around a closed contour avoiding singularities givesIt suffices to solve $(1+t^2)f^{\prime }+4tf=0$, givingNonzero contour integrals arise by enclosing one or both double poles $t=\pm i$. At $t=i$, the residue isThe conjugate pole gives the conjugate expression for real $z$. Taking real and imaginary linear combinations yields the real solutionsTheir derivatives are $y_1^{\prime }=z\sin z$, $y_2^{\prime }=z\cos z$, and their Wronskian is $-z^2$, so they are linearly independent on every interval avoiding zero. Both extend analytically across zero and remain independent as functions; the vanishing Wronskian at zero reflects the singular leading coefficient of the printed differential equation there.

For a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is

Repeated integration by parts, or the definition of the $m$th distributional derivative, gives

Repeated integration by parts inproduces the first $k+1$ terms of the Taylor polynomial and leaves $f(x)$, proving the integral remainder in Taylor theoremApply the linear functional ${\lambda }_k$ to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

For the midpoint quadrature rule on $[0,1]$, $e[1]=e[x]=0$, so $k=1$. Its Peano kernel isFor $f(x)=x^2$, the formula givesagreeing with ${\int }_0^1{x}^{2}dx-(1/2{)}^2=1/3-1/4$.