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A phase function is a real smooth functionthat is positively homogeneous of degree one in $\theta$ and has no critical point in all variables:The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of all $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and multi-indices $\alpha ,\beta$,

For a cutoff $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero, define the oscillatory integral byRepeated integration by parts makes the limit meaningful and independent of the cutoff.

The singular support is the complement of the largest open set on which the distribution is represented by a smooth function. Suppose $x_0$ does not belong toOn a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|{\nabla }_{\theta }\Phi |$. The differential operatorsatisfies $L{e}^{i\Phi }=e^{i\Phi }$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_{\Phi }(a)$ is smooth near $x_0$, proving

For the stated distribution on ${\mathbb{R}}^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho =|x|$. The angular integral isTherefore, for $\rho >0$,The original amplitude is not Lebesgue integrable in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional Yukawa potential, satisfying $(-\Delta +1)u=4{\pi }^2{\delta }_0$ with the normalization used in the question.

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

Define the perturbation vorticity componentTaking the curl of the linear momentum equation and using the buoyancy equation givesMultiply the first equation by $b^{\prime }$, the second by ${\zeta }^{\prime }$, and horizontally average. Periodicity makes the averaged $x$ derivatives vanish, whileby incompressibility and integration by parts. It follows thatTherefore the wave activity conservation law is

The equation can be writtenFor the squared L2 norm, periodic integration by parts and the reality of $V$ giveThus the Schrodinger equation generates a unitary flow and $\Vert u(\cdot ,t){\Vert }_2$ is constant.

The volumetric flow rate is $Q=2\pi {\int }_0^{R}u(r)rdr$. Integration by parts, using finite $u(0)$ and $u(R)=0$, givesBecause the stress magnitude is $\tau ={\tau }_{R}r/R$, change variables from $r$ to $\tau$:Therefore the required power is $n=3$. The fundamental theorem of calculus gives the pipe form of the Weissenberg–Rabinowitsch equation:Finally $\eta ({\dot{\gamma }}_R)={\tau }_R/{\dot{\gamma }}_R$, soA measured pressure-drop--flow-rate curve therefore recovers the wall viscosity without assuming a constitutive form.

The Riemann-Lebesgue lemma states that the Fourier transform of an $L^1$ function is continuous and tends to zero at infinity.

For a Schwartz function $f$, differentiation under the integral and integration by parts giveThe Riemann-Lebesgue lemma bounds the right side by an $L^1$ seminorm of $D^{\alpha }(x^{\beta }f)$. Every Schwartz seminorm of $\hat{f}$ is therefore bounded by finitely many Schwartz seminorms of $f$. Hence the Fourier transform maps $\mathcal{S}({\mathbb{R}}^n)$ continuously into itself.

Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

Choose a test function $\chi$ that equals one on the unit ball and vanishes outside the ball of radius two. Sinceintegration by parts givesThe first term on the right is supported where $1\le |x|\le 2$, where $u^2\le |x{|}^{\alpha }{u}^2$. The Cauchy-Schwarz inequality and Young inequality bound the second term byAfter absorbing the first integral,

This estimate also proves completeness. Indeed, a Cauchy sequence in $\Sigma$ is Cauchy in $H^1$ locally, while its gradients and the functions $|x{|}^{\alpha /2}{u}_n$ converge in $L^2$. The limits agree locally with a function $u$, so $u\in \Sigma$ and $u_n\to u$ in the energy norm. Thus $\Sigma$ is a Hilbert space; it is the confining-potential energy space for $V(x)=|x{|}^{\alpha }$.

WriteDifferentiating $\log J(u+th)$ at $t=0$ for a real test function $h$ givesAfter integration by parts,Rescaling the dependent and independent variables reduces this Euler-Lagrange equation to the ground-state equation for $Q$. The uniqueness of its positive radial solution and the equality cases in rearrangement show that all minimizers areThis is the classification of Weinstein-functional minimizers.

For real $h\in \mathcal{D}({\mathbb{R}}^d)$, differentiation under the integral givesThe ground-state equation $\Delta Q-Q+Q^{1+4/d}=0$ and integration by parts reduce this to

In particular, along the amplitude direction $h=Q$ the derivative is $-\Vert Q{\Vert }_2^2<0$. Hence $u_0=(1+\delta )Q$ has negative energy for every sufficiently small $\delta >0$, whilewhen $\delta$ is chosen small enough. The ground state has finite variance, so negative-energy blowup for the mass-critical focusing nonlinear Schrödinger equation shows that the corresponding solution blows up in finite time.

Write $e^x=1+xg(x)$, where $g=(e^x-1)/x$ is smooth at zero. Thenand a uniformly integrable expansion givesBy integration by parts,Therefore

Choose the Sobolev space $V$ encoding the homogeneous essential boundary conditions, setafter the appropriate integration by parts, and define the energy functionalIts first variation is $J^{\prime }(u)v=a(u,v)-\ell (v)$, so its stationary points are exactly the solutions of the weak formulationIf $a$ is bounded, symmetric, and coercive and $\ell \in V^{\prime }$, the Lax-Milgram theorem supplies a unique weak solution $u$. Moreover, for every $w\in V$,unless $w=0$. Thus $J$ is strictly convex, and $u$ is its unique global minimizer. This proves existence and uniqueness of the minimizer and of the weak solution simultaneously.

Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

For an ultrarelativistic particle, $\epsilon =pc$. Including the two spin states of an electron, counting wavevectors in a sphere gives the density of statesThe grand canonical partition function of the ideal Fermi gas is thereforeSince the grand potential satisfies $pV=k_{B}T\log \mathcal{Z}$, integration by parts givesThusThis is the equation of state of an ideal ultrarelativistic gas.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

Let $u\mapsto u+\epsilon \eta$ and $v\mapsto v+\epsilon \xi$, where the variations vanish on the boundary. The first variation isApplying integration by parts to the four derivative terms and discarding the boundary contributions givesThe variations $\eta$ and $\xi$ are independent and arbitrary in the interior. The fundamental lemma of the calculus of variations therefore gives the two Euler-Lagrange equations for two fields

The zero-boundary Sobolev space isOn it defineIf this quadratic form vanishes, then $\nabla u=0$. The Poincare inequality givesso $u=0$. It is therefore an inner product, and its norm is equivalent to the usual $H^1$ norm.

A function $u\in H_0^1(\Omega )$ is a weak solution whenThis follows from integration by parts and incorporates the homogeneous Dirichlet boundary condition through membership in $H_0^1$.

For a compactly supported variation $v+t\varphi$, differentiation of the area functional at $t=0$ gives the first variationAfter integration by parts, this is the minimal surface equation for a graphFor $v_R(x)=R^{-1}v(Rx)$ one has $D{v}_R(x)=Dv(Rx)$. The equation is invariant under this scaling, so $v_R$ solves it on $B_1$ for every $R>0$.