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LetThe integral is absolutely convergent. For a Schwartz function $\phi$, Fubini's theorem and integration by parts in $\lambda$ giveEquivalently, $v^{\prime }=-\hat{u}$ as a distributional derivative, and hence

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus

The centrifugal criterion concerns axisymmetric disturbances, so set $m=0$. The azimuthal equation givesEliminating $w^{\prime }$ with incompressibility and then $p^{\prime }$ with the axial equation yieldswhere the Rayleigh discriminant isImpermeability at the two solid walls gives $u^{\prime }(r_i)=u^{\prime }(r_o)=0$.

Multiply the equation by $r\overline{u^{\prime }}$, integrate between the walls, and use integration by parts. The boundary terms vanish and one obtainsThe denominator is positive. Therefore $\Phi \ge 0$ throughout the annulus excludes positive real ${\sigma }^2$ and gives centrifugal stability. Since $r^4{\Omega }^2=(r^2\Omega {)}^2$ is the square of the specific angular momentum, Rayleigh's circulation criterion isAn outward decrease of squared specific angular momentum permits an axisymmetric centrifugal instability.

Multiplyby $\hat{w}$ and integrate from the rigid bottom to an upper endpoint $z$ at which $\hat{w}{\hat{w}}_z=0$. The bottom term also vanishes because $\hat{w}(0)=0$. Integration by parts givesTherefore the horizontal wavenumber has the Rayleigh quotientFor a trapped wave the upper endpoint may be taken to infinity because the eigenfunction decays.

The clamped energy space isTwice applying integration by parts, with the boundary terms killed by the clamped conditions, gives the symmetric bilinear formConsequentlyfor every nonzero $u\in \mathcal{H}$: equality forces $u^{\prime \prime }=0$ almost everywhere, and the clamped boundary values then force $u=0$. Hence $L$ is symmetric and positive definite.

Strictly under the stated assumption $p\in L^2$ rather than $L^{\infty }$, the first integral need not be finite for every $u\in H_0^2$. The literal energy domain is therefore $\{u\in H_0^2:\sqrt{p}{u}^{\prime \prime }\in L^2\}$; under the usual coefficient assumption $0<p_0\le p\le p_1<\infty$, it is exactly $H_0^2$ and the form is coercive there.

Steady mass conservation gives $\dot{m}=-2\pi R\Sigma u_R$. With specific angular momentum $l=R^2\Omega$, multiply the angular-momentum equation by $2\pi$ and define the signed viscous torque in an accretion diskThenIntegration from $R_1$ to $R_2$ gives

The viscous power generated in an annulus is $2\pi R{F}_{\mathrm{diss}}dR=\mathcal{G}d\Omega$. Taking the inner torque to vanish and writing $l_{\mathrm{in}}=l(R_{\mathrm{in}})$ gives $\mathcal{G}=-\dot{m}(l-l_{\mathrm{in}})$, henceIntegration by parts yieldsFor steady circular force balance, $de=\Omega dl$; equivalently, the stated equality of gravitational- and rotational-potential differences makes the integral $e_{\mathrm{out}}-e_{\mathrm{in}}$. Therefore

When $R_{\mathrm{out}}\gg R_{\mathrm{in}}$, the outer energy and boundary term vanish. For an approximately Keplerian inner orbit, $e_{\mathrm{in}}=-R_{\mathrm{in}}^2{\Omega }_{\mathrm{in}}^2/2$, soComparison with part (c) givesA Keplerian inner flow generates the standard thin-disk power. A pressure-supported sub-Keplerian slim disk generates less through shear, and its emergent luminosity can be smaller still because photon trapping in an accretion flow carries part of that generated energy through the inner edge.

On the boundary the integral is ${\int }_{-\pi }^{\pi }\theta dy$. Integration by parts has zero endpoint term and leaves

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

For $u_0\in L^{\infty }(\mathbb{R})$, call $u\in L^{\infty }({\mathbb{R}}_{+}\times \mathbb{R})$ a weak solution when, for every $\phi \in C_c^1([0,\infty )\times \mathbb{R})$,This follows by multiplying the equation by a test function and applying integration by parts in time and space; $(F\phi {)}_x$ includes the term $F_x\phi$ because the original transport operator is not in divergence form.

If $u$ is $C^1$, test functions supported away from $t=0$ show that $u_t+F{u}_x=g$ as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give $u(0,x)=u_0(x)$. Thus a $C^1$ weak solution is the unique classical solution from part a.

Write the Inviscid Burgers equation in conservation form asA bounded function $u$ is a weak solution with initial datum $u_0$ whenfor every compactly supported $C^1$ test function $\phi$.

Across a straight discontinuity $x=\sigma t$, integration by parts on its two sides shows that the boundary terms cancel exactly when the Rankine-Hugoniot condition holds:when $u_{-}\ne u_{+}$. For every $c>0$, defineThe three jumps have left and right states $(0,-c)$, $(-c,c)$, and $(c,0)$, so their Rankine-Hugoniot speeds are respectively $-c/2$, $0$, and $c/2$, exactly the speeds of the displayed lines. Hence each $u_c$ satisfies the weak equation away from the origin and across every jump. Moreover, its nonzero support at time $t$ has length $ct$, so $u_c(t,\cdot )\to 0$ in $L_{\mathrm{loc}}^1$ as $t\downarrow 0$; its initial datum is therefore zero in the weak identity.

The zero function and all the distinct functions $u_c$ are bounded weak solutions with the same zero initial datum. Thus weak solutions are not unique. The central jump from $-c$ to $c$ is an expansion shock, which an entropy condition would exclude.

For the p-energythe first variation in the direction $\psi \in C_c^{\infty }(\Omega )$ isAn integration by parts therefore gives the Euler-Lagrange equationwhich is the p-Laplacian equation. In the notation of the question one takes $a^{ij}(x,z,\eta )=|\eta {|}^{p-2}{\delta }^{ij}$.

For $\eta \ne 0$, the principal coefficient matrix isIts eigenvalue in directions orthogonal to $\eta$ is $|\eta {|}^{p-2}$, while its eigenvalue parallel to $\eta$ is $(p-1)|\eta {|}^{p-2}$. The coefficients are $C^{1,\alpha }$ away from $\eta =0$, and the condition number there is at most $p-1$. On every region where $0<m\le |Du|\le M$, this gives uniform ellipticity with constants depending on $m$, $M$, and $p$. At $Du=0$ all principal eigenvalues vanish, so the operator is degenerate there and is not strictly elliptic on a domain containing a critical point.

First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

Because $v$ is a divergence-free vector field, the product rule givesThe integral of a divergence over the periodic domain vanishes. Hence integration by parts yieldsThis is the skew-symmetry of incompressible transport.

Let $q=2p+2$ and multiply the vorticity equation by ${\omega }^{q-1}$, first for an odd power as in the hint. Periodic integration by parts and $\nabla \cdot u=0$ giveThe diffusion term is nonnegative, while the Holder inequality bounds the right-hand side by $\Vert g{\Vert }_q\Vert \omega {\Vert }_q^{q-1}$. ConsequentlyOn the finite-volume torus, $L^q$ norms increase to the essential supremum as $q\to \infty$. Hence the damped-vorticity maximum estimate gives

Let $⟨F⟩=(4\pi {)}^{-1}\int Fd\Omega$ and $f=\bar{f}(ϵ)-ϵ{\bar{f}}^{\prime }(ϵ)\Theta (\mathbf{e})$ with comoving energy $ϵ=aE$. If ${ϵ}^4\bar{f}$ vanishes at both integration endpoints, integration by parts gives $-\int {ϵ}^4{\bar{f}}^{\prime }dϵ=4\int {ϵ}^3\bar{f}dϵ$. The kinetic stress-energy tensor then gives ${\delta }_{\gamma }=4⟨\Theta ⟩$, $v_{\gamma }^i=3⟨\Theta e^i⟩$, $\delta P=\bar{\rho }{\delta }_{\gamma }/3$, and ${\pi }^{ij}=4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3)⟩$. If $\Pi =-\pi$ is used, the last expression changes sign. Frequency independence of $\Theta$ is essential to this common temperature description.

The temporal-temporal component of the photon stress-energy tensor in a local orthonormal tetrad. For an isotropic one-particle distribution function expressed using comoving energy $ϵ=aE$, it is ${\bar{\rho }}_{\gamma }=4\pi a^{-4}\int dϵ{ϵ}^3\bar{f}(ϵ)$. A thermal photon temperature perturbation gives a first-order directional energy perturbation four times its fractional temperature change. Boundary conditions on the distribution justify this factor through integration by parts.

A Pohozaev identity is an integral identity for a nonlinear partial differential equation, obtained by multiplying the equation by a scaling vector field such as $x\cdot \nabla u$ and applying integration by parts. It relates kinetic, potential, and nonlinear energies.

For a mode with nonreal phase velocity $c$, let $V=U-c$, choose a continuous branch of $V^a$, and put $\hat{w}=V^{a}q$. Under impermeable boundary conditions, multiplying the transformed Taylor–Goldstein equation by $V^{2a}\overline{q}$ and applying integration by parts givesChoosing $a=1/2$ gives the Miles–Howard theorem; choosing $a=1$ makes the real phase velocity of an unstable mode a weighted mean of $U$.