Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

Let $\mathbf{n}=\mathbf{r}/a$ be the unit normal from the electrical conductor into the vacuum. The electrostatic boundary conditions at a conductor, together with the vanishing interior electric field, giveWrite $u=\cos \theta$ in spherical coordinates, $D=|\mathbf{r}-{\mathbf{r}}_0|$ on $r=a$, and $z^{\star }=a^2/r_0$. Differentiating the electric potential at fixed angles givesHere the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density isFor positive $q$ this surface charge density is negative everywhere, with largest magnitude nearest the point charge.

The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

The net upward force is the sum of the contact point forces, since the distributed normal contact force vanishes in every regime here. Before contact it is zero. During isolated contact it is $R_0=12B(\alpha L-2H)/L^3$. For an interval of contact it is $R_{-s}+R_s=12BH/d^3$, with $d=3H/\alpha$. ThereforeThe formulae agree at both transitions. At $\alpha =3H/L$ they give $12BH/L^3$, and their first derivatives also agree there. As a check on the sign and normalization, integrating the Euler-Bernoulli beam equation over the whole elastic beam gives $F_{\mathrm{up}}=B(w^{\prime \prime \prime }(L^{-})-w^{\prime \prime \prime }(-L^{+}))$, which yields the same values.

Let the contact interval be $[-s,s]$, where $w=-H$. Its slope and curvature vanish. Continuity of the displacement field, slope and bending moment at $x=s$ givesOn the free interval $s<x<L$, the cubic polynomial satisfying these three conditions is $w=-H+C(x-s{)}^3$. Put $d=L-s$. The end boundary conditions becomeConsequentlyThis requires $\alpha \ge 3H/L$, with $s=0$ at the transition. The full displacement field isIt satisfies nonpenetration. In the open contact interval the distributed normal contact force is $R=B{w}^{\prime \prime \prime \prime }=0$, since the obstacle is flat and $p=T=0$. The normal contact force is concentrated at the two endpoints. At $x=s$, $w^{\prime \prime \prime }$ jumps from zero to $6H/d^3$; at $x=-s$, it jumps from $-6H/d^3$ to zero. Both upward point forces are thereforeAt $s=0$ they merge into the single midpoint point force of the preceding regime. These formulae belong to the small-slope model; interpreting arbitrarily large $\alpha$ as a finite physical rotation would exceed its assumptions.

Seek an even function with isolated contact at $x=0$. On $0\le x\le L$, the contact and reflection conditions impose $w(0)=-H$ and $w^{\prime }(0)=0$, while the free interval still has $w^{\prime \prime \prime \prime }=0$. WriteThe two end boundary conditions giveand therefore the full displacement field isThe jump condition at zero gives the upward point forceIt is nonnegative precisely when $\alpha \ge 2H/L$. In that range $A_3\ge 0$. Nonpenetration near zero requires $A_2\ge 0$, equivalently $\alpha \le 3H/L$. Conversely, when both inequalities hold,with strict inequality away from zero. Thus this is an admissible isolated-contact solution throughoutAt the upper endpoint the contact curvature $w^{\prime \prime }(0)=2A_2$ vanishes, and the profile remains above the obstacle for $x\ne 0$ because $A_3>0$. Increasing $\alpha$ further would make $A_2<0$, forcing penetration near zero. An interval of contact must then develop.

The admissible solution is unique within this linear model: the bending elastic energy $B\int (w^{\prime \prime }{)}^2/2$ is strictly convex on the convex set of admissible profiles with the prescribed end data. Integration by parts shows that a solution with nonnegative contact point forces satisfies the corresponding minimum condition, since every admissible variation lifts a point already on the obstacle. Reflection therefore justifies the assumed even function also after contact.

With zero tension and zero distributed load, a free interval satisfies $w^{\prime \prime \prime \prime }=0$. Before contact the entire elastic beam is free, so $w$ is a cubic polynomial. The four end boundary conditions determine it uniquely. Reflection $x\mapsto -x$ preserves those boundary conditions, so the unique solution is an even function and its odd terms vanish. Direct substitution givesFor $\alpha \ge 0$, its minimum occurs at $x=0$ and is $-\alpha L/2$. The constraint $w\ge -H$ is therefore satisfied exactly when $\alpha \le 2H/L$. The first contact is at the midpoint:At equality the normal contact force is still zero. The uniqueness used here can also be checked by integrating $v^{\prime \prime \prime \prime }v$ for the difference of two solutions with homogeneous clamped boundary conditions, which gives ${\int }_{-L}^L(v^{\prime \prime }{)}^{2}dx=0$ and hence $v=0$.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore

Because $f$ is a compactly supported distribution, the convolution of distributions with a compactly supported factor defines $E\ast f$. The convolution derivative identity and part (i) giveThe difference between any two solutions has zero ${\partial }_{\bar{z}}$ distributional derivative. Part (ii) therefore proves that the general solution of the partial differential equation isThis formula is an identity of distributions; a pointwise integral is not required for arbitrary $f$.

If $u$ were a smooth function, its Wirtinger derivative ${\partial }_{\bar{z}}u=f$ would also be smooth, contradicting $\mathrm{singsupp}f\ne \varnothing$. In fact the singular support is exactly preserved here:One inclusion follows because a differential operator cannot create a singularity where $u$ is smooth. For the reverse inclusion, $E$ is smooth off zero, and its singular support is exactly $\{0\}$: if it were smooth near zero, ${\partial }_{\bar{z}}E={\delta }_0$ would be smooth there. The support and singular support rules for convolution give $\mathrm{singsupp}(E\ast f)\subseteq \{0\}+\mathrm{singsupp}f$, and adding the smooth entire function $h$ leaves this singular support unchanged. Thus there is no globally smooth solution when the given singular support is nonempty.

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

Let $u,v\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$, with at least one a compactly supported distribution, so their convolution is defined. The support and singular support rules for convolution areHere $A+B=\{a+b:a\in A,b\in B\}$ is the Minkowski sum. The compact-factor hypothesis also ensures that these sums of closed sets are closed. More generally, the same rules hold when addition is a proper map on $\mathrm{supp}u\times \mathrm{supp}v$, which is another sufficient condition for defining the convolution. The second inclusion explains why a compactly supported smooth function smooths every distribution under convolution.

Define two open sets byThe support of a distribution and its singular support areBoth are relatively closed because the removed sets are open. The locality of distributions gives $u{|}_{Z(u)}=0$. On an overlap of two open smoothness regions, the two smooth functions represent the same distribution, so their difference is zero almost everywhere and hence everywhere by continuity. They therefore join to one smooth function on $G(u)$. The same partition of unity argument shows that it represents $u$ there.

For $u=1_{\omega }$, openness of $\omega$ impliesIndeed every neighborhood of a point in this relative closure meets $\omega$ in an open set containing a nonnegative test function with positive integral; outside the closure the indicator function is identically zero near the point.

The singular support of an indicator function needs a measure-sensitive boundary. For a Lebesgue measurable set $A\subseteq \Omega$, define its essential closure byThen the general answer isOutside this intersection, some ball has either no points of $\omega$ or a complement of Lebesgue measure zero. The regular distribution is respectively the constant zero or one on that ball and is smooth. Conversely, at a point of the intersection, every sufficiently small ball has positive Lebesgue measure in both sets. If a smooth function represented the indicator function there, its values would be in $\{0,1\}$ everywhere: any other value, by continuity, would give a ball of positive Lebesgue measure on which equality almost everywhere fails. On a connected ball such a continuous two-valued function is constant, contradicting the two positive measures.

In particular, the answer need not be the full topological boundary. For $\Omega ={\mathbb{R}}^n$ and $\omega ={\mathbb{R}}^n\setminus \{0\}$, the indicator function is the constant one almost everywhere, so its singular support is empty although the boundary is $\{0\}$. The full relative boundary is the answer when every neighborhood of each boundary point contains a positive Lebesgue measure portion of the complement, as for an open set with smooth boundary.

If $\varphi \in \mathcal{D}(\omega )$, let $\tilde{\varphi }$ be its extension by zero to $\Omega$. Its compact support is contained in $\omega$, so this extension remains a test function. The restriction of a distribution isExtension by zero preserves convergence in the space of test functions, so this defines a distribution on $\omega$.

Choose a partition of unity $({\chi }_j)$ subordinate to the neighborhoods ${\omega }_x$ from the hypothesis. For an arbitrary $\varphi \in \mathcal{D}(\Omega )$, local finiteness and compactness of $\mathrm{supp}\varphi$ give a finite decomposition $\varphi ={\sum }_j{\chi }_j\varphi$. Each term is a test function supported in one ${\omega }_x$. ThusSince this holds for every test function, $v=0$. This is the locality of distributions.

Let $\Omega \subset {\mathbb{R}}^n$ be an open set and $(U_{\alpha }{)}_{\alpha \in A}$ an open cover of $\Omega$. There exist nonnegative smooth functions $({\chi }_j{)}_{j\in J}$ and choices $\alpha (j)$ such thatand the family of supports is locally finite: every point has a neighborhood meeting only finitely many of them. This is a smooth partition of unity subordinate to the cover, allowing a refinement of its indexing. In Euclidean open sets, one can choose $J$ countable and the individual supports compact in their assigned cover members. Local finiteness ensures that the sum is smooth and that only finitely many terms meet any given compact set.

Let $M={\sup }_j{j}^{-m}\Vert g_j{\Vert }_{L^1}<\infty$. For each $j$, apply the periodic primitives with zero mean construction of part (iii) to $g_j$, writing $h_{j,0}=g_j$ andThe estimates are independent of $j$: $\Vert h_{j,r}{\Vert }_{L^1}\le \Vert g_j{\Vert }_{L^1}$. With $N=m+1$, the same repeated integration by parts gives, for every test function $\varphi \in \mathcal{D}(0,1)$,This is precisely weak convergence of distributions, so $\boxed{u_j\to 0\text{ in }{\mathcal{D}}^{\prime }(0,1)}$. The extra integration by parts beyond the growth exponent $m$ is what ensures decay.

To repeat integration by parts across whole periods, choose periodic primitives with zero mean. Set $h_0=f$, and recursively defineEvery $h_r$ has zero mean. Part (i) gives $h_{r+1}^{\prime }=-h_r$ almost everywhere, and $h_{r+1}(0)=h_{r+1}(1)$ because $R{h}_r$ vanishes at both endpoints. Its periodic extension is consequently locally an absolutely continuous function, with no jump at the integers. Part (ii) gives the uniform estimateThe weak derivative identity on the whole line is $D{\overline{h}}_{r+1}=-{\overline{h}}_r$. After scaling by the integer $j$ and repeating integration by parts $N$ times,The boundary terms vanish because $\varphi$ has compact support in $(0,1)$. The interval contains exactly $j$ periods, soTaking $N=m+1$ therefore yieldsThis rapid distributional decay of periodic oscillations uses only integrability of $f$ and smoothness of the test function.

Using $(Rf{)}^{\prime }={\int }_0^{1}f-f$ and integration by parts, with $\varphi$ a test function, givesFor each $k\ge 1$, apply the same identity to $R^{k}f$ and the test function ${\varphi }^{(k)}$. Since ${\int }_0^1{\varphi }^{(k)}=0$, it givesStarting from $k=1$ and iterating provesThe intermediate means disappear because derivatives of a test function have zero integral; no assumption that $R^{k}f$ has zero mean is needed.

For the bound, writeThe triangle inequality and Fubini's theorem implyThus $\boxed{c=\frac{1}{2}\text{ is valid}}$, and every iterate $R^{k}f$ is in the L1 space.

Put $A={\int }_0^{1}f$. ThenThe theorem on differentiation of an indefinite Lebesgue integral makes this an absolutely continuous function on $[0,1]$, withBoth $A-f$ and $Rf$ belong to the L1 space: for example, $|Rf(x)|\le |A|+\Vert f{\Vert }_{L^1}\le 2\Vert f{\Vert }_{L^1}$. The ordinary derivative of an absolutely continuous function is its weak derivative, as integration by parts against a test function shows. Therefore $\boxed{Rf\in W^{1,1}(0,1)}$, the indicated Sobolev space.

The L1-normalized dilation of a distribution isFor a regular distribution this is the function $u_{\epsilon }(x)={\epsilon }^{-1}u(x/\epsilon )$. In terms of the wiki's dilation of a distribution convention, $u_{\epsilon }={\epsilon }^{-1}{\delta }_{1/\epsilon }u$; the normalization preserves the L1 space norm.

If $u\in L^1(\mathbb{R})$, the dominated convergence theorem, with dominating integrable function $|u(y)|\Vert \varphi {\Vert }_{\infty }$, givesConsequentlyAn arbitrary distribution need not have this behavior. Take the constant regular distribution $u=1$. Then $u_{\epsilon }={\epsilon }^{-1}$, so for a test function with integral one,Hence $\boxed{u=1\text{ gives no distributional limit}}$.