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For a radial function, the equation isTherefore, assuming the population density and radial flux decay at infinity and are regular at the origin,Thus $N$ is conserved; it is the total population obtained by integrating density over the plane. This is the global form of the equation's conservation law.

SetThenwhile the radial diffusion term has the same prefactor and equalsEquating them and multiplying by $x$ givesso the proposed ansatz is a similarity solution.

Regularity at $x=0$ makes the constant of integration zero. Wherever $f>0$,for some $x_0>0$. The nonnegative weak solution continues as $f=0$ once this parabola reaches zero:This is the two-dimensional Barenblatt profile for quadratic porous-medium diffusion; its degenerate diffusivity permits compact support.

Indeed, conservation requiresso $x_0=(16/\pi {)}^{1/4}$ if desired. At time $t$ the populated disk has radius $x_0(NDt{)}^{1/4}$ and hence area

SetThe ground-state transform for a weighted Dirichlet energy, obtained by expanding the square and using integration by parts, isThe boundary term vanishes because $h$, and hence $g$, has compact support.

Since $V(x)\to \ell >0$, choose $R_0$ so that $V\ge \ell /2$ when $|x|\ge R_0$, and setFor every $R\ge R_0$, discard the nonnegative $|g^{\prime }{|}^2$ term and the nonnegative contribution from $R_0\le |x|\le R$ to obtainBecause $|g{|}^2=|h{|}^2{e}^{-\Phi }$, this is exactly the required estimate.

Extend $u$ by zero to ${\mathbb{R}}^n$; its compact support inside the ball makes the extension smooth. The Fourier transform of a derivative givesThe assumption says that $L_0$ is a uniformly elliptic operator, soMoreover,The Plancherel theorem therefore yieldsAll integrands vanish outside the original support where appropriate, so

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

Applying the distributional Leibniz rule twice givesEliminating the middle term provesChoose $\rho \in \mathcal{D}(\mathbb{R})$ with $\rho (0)=1$. Parts i and ii and the identity just proved, with $\phi =\rho$ and $u=xH$, yieldThus explicit continuous functions of compact support are

Solve the adjoint transport equationbackward with terminal value zero. Along the characteristic flow map $Z_{t,s}(x)=e^{s-t}x$, the required solution isDifferentiation under the integral verifies the equation. If $\psi$ has compact support in $[0,T]\times [-R,R]$, then $\phi$ vanishes for $t>T$ and for $|x|>R$, so $\phi \in C_c^1$ as required.

When the initial datum is zero, inserting this $\phi$ into the weak formulation givesfor every $\psi \in C_c^1$. Thus $u=0$ almost everywhere. The difference of two bounded weak solutions has zero initial datum, so this proves uniqueness.

SetThe scalar conservation law is $u_t+a(t)f^{\prime }(u)u_x=0$. Its characteristic curve issuing from $\xi$ satisfiesThe Jacobian of the one-dimensional characteristic map isBefore characteristic crossing, differentiation with respect to $\xi$ givesBecause $u_0$ has compact support, $m$ is continuous and vanishes outside a compact set. It therefore attains its minimumby the hypothesis. Since $a(t)>\epsilon$, the function $A$ is strictly increasing and tends to infinity. There is consequently a unique first time $T_{\ast }>0$ satisfyingAt a minimizer of $m$, the numerator $u_0^{\prime }$ is nonzero because $f^{\prime \prime }$ is nonzero, while the denominator tends to zero as $t\uparrow T_{\ast }$. Hence the classical solution has gradient blow-up:

For a nonnegative $f$, first set $g_n(x)=\min (f(x),n)1_{(1/n,n)}(x)$ and $F_n(x)={\int }_0^x{g}_n(t)dt$. These bounded functions vanish near zero and have compact support. Their averages have finite Lp norms: they vanish near zero and decay as $1/x$ beyond $n$. Integration by parts givesThe boundary term $x^{1-p}{F}_n(x{)}^p$ vanishes at zero because $F_n$ is zero there, and at infinity because $F_n$ is eventually constant and $p>1$. The inequality is Hölder's inequality with conjugate exponents $p$ and $p/(p-1)$. Divide when $I_n>0$; the zero case is immediate. Since $g_n\uparrow f$ and $H{g}_n\uparrow Hf$, the monotone convergence theorem proves the full bound without presupposing that $Hf$ has finite Lp norm. Apply it to $|f|$ for the signed or complex case.^[ref]

Nesting in an orthonormal multiresolution analysis expresses a scaling function in the finer-scale orthonormal basis. The associated low-pass symbol is $m(\xi )=2^{-1/2}{\sum }_k{h}_k{e}^{-ik\xi }$, and $\hat{\phi }(2\xi )=m(\xi )\hat{\phi }(\xi )$. If $\phi$ has compact support, only finitely many coefficients $h_k$ are nonzero. With $\hat{\phi }(0)=1$, iteration gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$ when the limit exists.

For a symmetric stress-energy tensor with compact support and ${\partial }_b{T}^{ab}=0$ in Minkowski spacetime, let $I^{ij}=\int T^{00}{x}^i{x}^j{d}^{3}x$. Two integrations by parts give ${\dot{I}}^{ij}=\int (T^{0i}{x}^j+T^{0j}{x}^i)d^{3}x$ and the displayed identity. Compact support removes the boundary fluxes. The support should stay bounded over the time interval used when differentiating the integrals.

In the trap frame, a particle under linear drag has relative speed $F(y)/\zeta -v_T$. A uniform strictly negative relative speed guarantees finite passage through a compact support. For a bounded continuous force, $v_T>\sup F/\zeta$ supplies that condition. For ordinary smooth forces, an encountered zero of the relative speed is an equilibrium or a passage bottleneck, not a finite crossing. Negative forces do not impede forward trap passage.

For distributions $u$ on ${\mathbb{R}}^m$ and $v$ on ${\mathbb{R}}^n$, their tensor product acts on a joint test function by $⟨u\otimes v,\Phi ⟩=⟨u_x,⟨v_y,\Phi (x,y)⟩⟩$. The inner function is smooth with compact support in the projection of $\mathrm{supp}\Phi$, and finite order of a distribution estimates prove continuity. Reversing the pairings gives the same value: finite sums of product kernels are dense in the required smooth seminorms, and the two orders agree on product kernels. Its support is contained in the Cartesian product of the two supports. This provides a rigorous replacement for formal integration in separate distribution variables.

A test function is a smooth function with compact support. Weak and distributional equations are identities obtained by integrating against every test function.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]

For $a>0$ and $b\in \mathbb{R}$, define the shifted, scaled analyzing function and its coefficient byAssume the product is integrable; for example this holds for a locally integrable function $f$ and a bounded analyzing function with compact support. The bar is complex conjugation. This normalization preserves the L2 norm of the analyzing function. If $p$ is a polynomial of degree of a polynomial less than $q$, then $p(b+au)$ is a sum of monomials $u^r$ with $r<q$. Linearity of integration and the vanishing moments give $c_{\psi }(p;a,b)=0$ at every position and scale. Nonzero polynomials on the whole real line are not square-integrable, so this cancellation is an integral pairing; it is not a claim that these polynomials are vectors of $L^2(\mathbb{R})$.

Here is a quantitative reason that smooth regions yield small details. Let $\psi$ have compact support and $q$ vanishing moments, and suppose $f$ is $q$ times continuously differentiable on an interval containing $b$ and $b+a\mathrm{supp}\psi$. Subtract its degree-$(q-1)$ Taylor polynomial at $b$. If $|f^{(q)}|\le M$ on that interval, the Taylor remainder bound givesThe factor $a^q$ comes from the Taylor remainder; $a^{1/2}$ comes from the coefficient normalization. Thus the analyzing function measures the departure from the local polynomial trend, rather than the height of $f$ itself. The wavelet coefficient decay for Hölder functions uses the same argument with a Hölder-Taylor remainder bound. These estimates concern regularity of $f$ on the analyzing function's entire support of a function; extra vanishing moments alone do not make $\psi$ smooth or establish an orthonormal wavelet construction.^[ref]

A wave packet is a localized superposition of Fourier modes. In a narrow packet, nearby wavenumbers interfere to produce a slowly varying wave envelope around a rapid carrier wave. Its leading translation is governed by group velocity; the carrier's wave phase follows phase velocity. Localized does not necessarily mean compact support: a Gaussian function decays rapidly but extends over the whole line. The Gaussian envelope in a Klein-Gordon medium gives an original example with both translation and spreading.