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The Schwartz space consists of the smooth functions $\phi$ for whichfor all multi-indices $\alpha ,\beta$. Its topology is generated by these seminorms; ${\phi }_j\to \phi$ means convergence in every seminorm. A tempered distribution is a continuous linear functional on this space. Convergence in ${\mathcal{S}}^{\prime }$ here means weak convergence: $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every Schwartz function.

Use the unnormalized angular-frequency Fourier transformThe transform is a continuous automorphism of the Schwartz space, so this duality defines a tempered distribution. Put $D_j=-i{\partial }_j$, as required by the stated identities. Integration by parts and differentiation under the integral giveThus, keeping the transpose factor of $D=-i\partial$ explicit,andIterating gives the Fourier derivative identities for distributions

The logarithm defines a tempered distribution because it is locally integrable and grows more slowly than a polynomial. Its derivative is $u^{\prime }(x)=x/(1+x^2)$. The standard transform pair $\widehat{(1+x^2{)}^{-1}}=\pi e^{-|\lambda |}$, obtainable by the residue theorem, givesAs $\widehat{u^{\prime }}=i\lambda \hat{u}$, we have $\lambda \hat{u}=-\pi \mathrm{sgn}(\lambda )e^{-|\lambda |}$. DefineThis integral is absolutely convergent: the numerator is $O(|\lambda |)$ near zero, and the exponential controls the tails. It defines a tempered distribution satisfying the same multiplication equation. The kernel of multiplication by a coordinate therefore gives $\hat{u}=T+c{\delta }_0$.

To determine $c$, use the supplied Gaussian Fourier transform pair with ${\phi }_n(0)=1/(2\sqrt{\pi })$. ThenAlso $⟨T,{\phi }_n⟩\to 0$ by the dominated convergence theorem: the relevant integrand is bounded by a constant times $|\lambda |e^{-|\lambda |}$ for $n\ge 1$. ConsequentlyThe cutoff limit agrees with the absolutely convergent integral; this particular subtraction convention fixes the otherwise ambiguous delta term. The reusable identity is Fourier transform of the logarithm of one plus x squared.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

The Wirtinger derivative $u_z=(u_x-i{u}_y)/2$ is holomorphic, since ${\partial }_{\bar{z}}{u}_z=\Delta u/4=0$ by the Laplace equation. The conformal map $w=z^2$ takes the quadrant to the complex upper half-plane. Integrating the prescribed tangential boundary derivatives, choose a common corner value $c$ and define the real boundary functionThe decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its Poisson integral converges. A holomorphic function whose real part is that Poisson integral is supplied by the regularized Schwarz integral formula:For $u_{\ast }(z)=\mathrm{Re}W(z^2)$, the chain rule gives $(u_{\ast }{)}_z=z{W}^{\prime }(z^2)$. Apply integration by parts to obtainSubstituting $s=r^2$ and $s=-r^2$ gives the requested particular integral representation:Both integrals converge absolutely for an interior point of the quadrant. The Sokhotski–Plemelj formula gives $\mathrm{Re}(u_{\ast }{)}_z(x)=g_2(x)/2$ on the horizontal edge and $\mathrm{Im}(u_{\ast }{)}_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a holomorphic function $H$ satisfyingEquivalently, write $H(z)=ih(z^2)/z$, where $h$ is holomorphic in the complex upper half-plane with real boundary values on the nonzero real axis. An antiderivative of $2H$ has a harmonic real part with zero tangential boundary derivatives. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular harmonic function contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the Poisson integral representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

Use the weighted Neumann heat kernel with constant drift. The spatial operator has the self-adjoint formso its homogeneous Neumann boundary conditions give a regular Sturm-Liouville problem in the weighted inner productThe constant eigenfunction is ${\varphi }_0=1$, with eigenvalue ${\lambda }_0=0$ and squared norm $N_0=(e^{\beta L}-1)/\beta$. For $k\ge 1$, put $q_k=k\pi /L$. Substitution of $\varphi =e^{-\beta x/2}v$ in $\mathcal{L}\varphi =-\lambda \varphi$ gives $v^{\prime \prime }+(\lambda -{\beta }^2/4)v=0$ and the Robin boundary conditions $v^{\prime }=\beta v/2$ at both endpoints. HenceIndeed,The Sturm-Liouville eigenfunction expansion is complete and orthogonal. Its heat kernel, relative to the measure $e^{\beta y}dy$, isFor a coefficient $a_k(t)=⟨{\varphi }_k,u(\cdot ,t){⟩}_{\beta }/N_k$, two applications of integration by parts yieldSolving these scalar linear differential equations and summing gives the Neumann boundary-forcing formula:This depends only on the given data. In particular, the zero eigenvalue retains the changing weighted mean:The endpoint derivatives of the boundary integral are understood as interior limits. Differentiating each homogeneous Neumann eigenfunction at the endpoint before summing would incorrectly discard the prescribed boundary forcing: the short-time heat kernel makes that interchange invalid. The stated compatibility conditions give the classical boundary traces.

For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

The mass matrix is a positive-definite matrix, and integration by parts with periodic boundary conditions gives $C^T=-C$. HenceThis is energy conservation for semidiscrete Galerkin advection: $U^{T}MU=\Vert u_h{\Vert }_{L^2(0,1)}^2$. Thus the semidiscretization is stable and conserves its finite element L2 norm.

For a mesh-independent comparison with the grid norm, the Fourier symbol of $M$ is $h(2+\cos \theta )/3$, between $h/3$ and $h$. ConsequentlyThe conserved energy therefore gives a uniform bound in $\sqrt{h{\sum }_m|U_m{|}^2}$ as well. Equivalently, each Fourier mode evolves with the purely imaginary exponent $\sigma (\theta )=3i\sin \theta /[h(2+\cos \theta )]$. This concerns the semidiscrete method; a chosen time integrator must separately control those imaginary eigenvalues.

Set the noise strengths to the model A fluctuation-dissipation relation and the fluctuation-dissipation relation for a conserved flux:The preceding logarithmic ratio becomesThe last two terms combine as ${\partial }_i({\mu }_j{W}_{ij})$. They integrate to zero under the periodic boundary conditions, by integration by parts. The remaining term is $F_2-F_1$ by the functional chain rule, henceAfter including the equilibrium initial weights, the forward and reversed histories have equal probability. Thus microscopic reversibility is satisfied by the two channels together, with each channel's own thermal noise strength.

LetSince $d(e^{-t^2})=-2t{e}^{-t^2}dt$, repeated integration by parts givesThe final remainder estimate follows by one further integration by parts, or by bounding the remaining positive integral by its leading boundary term. Therefore the requested first three terms are

Suppose a normalizable solution satisfies $L\psi =-{\kappa }^2\psi$. Multiplication by $\overline{\psi }$, integration over $\mathbb{R}$, and integration by parts giveThe boundary term vanishes for a normalizable eigenfunction. The left side is negative unless $\psi =0$, so no nontrivial normalizable solution exists.

Differentiating under the contour integration sign givesSince ${\partial }_t{e}^{zt}=z{e}^{zt}$, integration by parts turns this intoThe integral vanishes whenHence, up to an irrelevant multiplicative constant,The remaining endpoint condition isThus each open end of $\gamma$ must lie where $\mathrm{Re}(zt-1/t)\to -\infty$; a closed contour also makes the boundary term vanish when the integrand is single-valued along it.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

Assume for contradiction that $\pi =a/b$ with positive integers $a,b$. DefineandThe coefficient of $x^{n+j}$ in $F_n$ isConsequently every derivative $F_n^{(k)}(0)$ is an integer: it is zero unless $n\le k\le 2n$, and in that range the factor $k!/n!$ clears the denominator. Since $F_n(\pi -x)=F_n(x)$, every $F_n^{(k)}(\pi )$ is also an integer.

Repeated integration by parts, ending when derivatives above degree $2n$ vanish, givesThus $I_n$ is an integer. It is strictly positive because $F_n(x)\sin x>0$ for $0<x<\pi$.

On the other hand, $x(\pi -x)\le {\pi }^2/4$, soThe factorial-over-power series implies that the right-hand side tends to zero. For all sufficiently large $n$ this says that the positive integer $I_n$ is less than one, a contradiction. Therefore the Irrationality of pi is established:

Put $a=1+ϵ$ and, for $m\ge 2$, defineOne integration by parts givesStarting with $I=I_2$ and iterating $N+1$ times yieldswhereSince $0<I_{N+3}(ϵ)\le (1+ϵ{)}^{-1}$, the remainder is $O({ϵ}^{N+1})$, which is little-$o$ of the last retained term. ThereforeThe displayed terms form an asymptotic sequence because the ratio of successive magnitudes tends to zero.

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

Multiplication by $(1-x^2{)}^k$ puts the equation in self-adjoint form:Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leavingIf $Q_k$ is not identically zero, the integral on the right without ${\lambda }_k$ is positive, whereas the left side is nonnegative. Therefore

The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives