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The locally integrable function $E(z)=1/(\pi z)$ on the plane is a fundamental solution for the Wirtinger derivative ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. Let $L=(4\pi {)}^{-1}\log (x^2+y^2)$ be the fundamental solution of the Laplace equation. Its first distributional derivatives agree with its ordinary ones: deleting a radius-$\epsilon$ disk in integration by parts gives an error $O(\epsilon |\log \epsilon |)$, tending to zero. Hence $E=4{\partial }_{z}L$ andThe analogous fundamental solution for ${\partial }_z$ is $1/(\pi \bar{z})$. The singular support of each kernel is exactly $\{0\}$: they are smooth elsewhere, and smoothness near zero would make their Wirtinger derivatives smooth there, contradicting the Dirac delta distribution identity.

Target: Distributional Cauchy-Riemann lemma. Parent: Cauchy-Riemann equations. Intended file: analysis.bigb.

For the quadrant $x,y>0$, the displayed Green function satisfies $\Delta G={\delta }_{(\xi ,\eta )}$, zero value on $y=0$ and zero outward normal derivative on $x=0$. The method of images uses the logarithmic fundamental solution of the Laplace equation with coefficient $1/(2\pi )$: keep the source sign when reflecting across the Neumann side and reverse it when reflecting across the Dirichlet side. The four image signs are therefore $+,+,-,-$ for sources $(\xi ,\eta ),(-\xi ,\eta ),(\xi ,-\eta ),(-\xi ,-\eta )$.

On $y=0$, equal numerator and denominator give zero. Evenness in $x$ gives the Neumann condition at $x=0$. The outward normal derivative on the Dirichlet side is the positive kernelGreen's second identity consequently gives $u(P)={\int }_{D}Gf+{\int }_{y=0}g{\partial }_{n}G$ for $\Delta u=f$ with these mixed boundary data, assuming any remote-boundary contribution vanishes. The sign convention is $\Delta G=+\delta$; changing to $-\Delta G=\delta$ changes both the Green function and the representation signs.

Let $X=(x,y)$ be the integration variable and $P=(\xi ,\eta )$ the interior source point. Green's second identity, obtained from Green's theorem by applying the divergence theorem to $u\nabla G-G\nabla u$, isThe outward normal is used throughout. To obtain the prescribed representation with its positive volume term, chooseIndeed, substituting $\Delta u=f$ makes the left side $u(P)-{\int }_{D}Gf$. On the Dirichlet portion, $u=g$ and $G=0$ leave only $g{\partial }_{n}G$. On the Neumann portion, both normal derivatives vanish. Rearranging gives exactly the desired boundary-plus-volume representation. The distributional identity can equivalently be justified by excluding a small circle around $P$ and using the logarithmic singularity there.

For the quarter-plane, the fundamental solution of the Laplace equation with this sign convention is $\Phi (X,P)=(2\pi {)}^{-1}\log |X-P|$, since $\Delta \Phi ={\delta }_P$. The method of images uses an equal-sign reflection across the Neumann boundary $x=0$ and an opposite-sign reflection across the Dirichlet boundary $y=0$. HenceEquivalently,Only $(\xi ,\eta )$ lies inside the domain, so the interior Laplacian has exactly the required unit point source. At $y=0$ numerator and denominator agree, giving $G=0$. The formula is even in $x$, giving ${\partial }_{x}G=0$ at $x=0$, and therefore zero outward normal derivative there.

On the Dirichlet boundary $y=0$, the outward normal is $-{\partial }_y$, and differentiation gives the useful checkThis is the mixed Dirichlet-Neumann Green function in a quadrant. The image solution tends to zero at spatial infinity; on the unbounded domain the representation uses data and solutions for which the boundary term at infinity in Green's second identity vanishes.

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.

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

The function $U(x)=1/|x|$ is locally integrable in ${\mathbb{R}}^3$, so it defines a distribution. The claimed distributional identity means that for every $\phi \in C_c^{\infty }({\mathbb{R}}^3)$,Choose a ball containing the support and remove $B_{\epsilon }(0)$. On the punctured region $\Delta U=0$. Green's second identity givesThe outer boundary contributes zero. On the inner boundary, the outward normal for the punctured region is $-\hat{r}$, so ${\partial }_{n}U=+1/{\epsilon }^2$, fixing the sign.

The first term is $O(\epsilon )$; the second equals $-{\int }_{S^2}\phi (\epsilon \omega )d\omega \to -4\pi \phi (0)$. Local integrability allows the left side to converge to its full-space integral. Therefore $\boxed{-\Delta (1/|x|)=4\pi {\delta }_0}$. In this sign convention, $1/(4\pi |x|)$ is the fundamental solution of the Laplace equation for the operator $-\Delta$.

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

The interior elliptic regularity assertion for the Laplace equation is that a harmonic function is smooth, in fact real analytic, throughout $U$. No boundary regularity of its unspecified boundary values is implied.

First let $u\in C^2(U)$ and $\Delta u=0$. On a ball compactly contained in $U$, differentiating its spherical average and applying the divergence theorem expresses that derivative as a constant factor times ${\int }_{B_r}\Delta u$, which is zero. The spherical average tends to $u$ at the center as $r\to 0$. This proves the mean value property for harmonic functions.

Choose a radially symmetric smooth mollifier ${\eta }_r$, supported in $B_r$, with integral one. By integrating the spherical mean value property,whenever $B_r(x)⋐U$. For fixed $r$ the right side is a smooth convolution, since all derivatives can be placed on ${\eta }_r$. Thus $u$ is smooth. The same argument proves the Weyl lemma for a distributionally harmonic $u\in L_{\mathrm{loc}}^1$: first mollify $u$, apply the fixed-radius identity, and let the mollification radius tend to zero in distributions to obtain the same smooth representative.

To prove real analytic regularity, differentiating the fixed-radius convolution gives the interior derivative estimate for a harmonic functionfor any harmonic $v$. All derivatives of $u$ are harmonic. On nested balls between $B_R(a)$ and $B_{R/2}(a)$, apply this estimate $k$ times, decreasing the radius by $R/(2k)$ each time. For $|\alpha |=k$,The inequality $k^k\le e^{k}k!$ follows by integrating $\log s$ below the sum defining $\log (k!)$. Apply the one-dimensional Taylor theorem along each segment, expanding directional derivatives by the multinomial formula. The remainder is bounded by $M(A{\sum }_j|h_j|{)}^k$, so it tends to zero for sufficiently small $h$. This gives a locally convergent multivariate Taylor series. A harmonic function is real analytic in the interior.

The usual inhomogeneous elliptic regularity statement also follows: if $\Delta u=f$ and $f$ is smooth, take a cutoff $\chi$ equal to one near a given point and set $w=\Phi \ast (\chi f)$, where $\Phi$ is a fundamental solution of the Laplace equation with $\Delta \Phi =\delta$. Moving every derivative to the compactly supported smooth function $\chi f$ shows $w$ is smooth. Locally $u-w$ is harmonic, so $u$ is smooth there too.

Write $({\varphi }_{\epsilon }\ast f)(x)=\int \varphi (y)f(x-\epsilon y)dy$. Since $\int \varphi =1$,A Schwartz function is bounded and uniformly continuous. Given a tolerance, first choose $R$ so the tail integral of $|\varphi |$ is small. On $|y|\le R$, uniform continuity makes the difference uniformly small for every $x$ when $\epsilon R$ is small. On the tail it is bounded by $2\Vert f{\Vert }_{\infty }$. This proves uniform convergence, without assuming $\varphi \ge 0$, and is the approximate identity argument.

For the radial regularization of the three-dimensional fundamental solution of the Laplace equation, direct differentiation givesThe radial integral is $\int \varphi =3{\int }_0^{\infty }{r}^2(1+r^2{)}^{-5/2}dr=1$. This even approximate identity therefore gives $\int (-\Delta N_{\epsilon })f\to f(0)$ for $f\in \mathcal{S}({\mathbb{R}}^3)$.

The function $N=1/(4\pi |x|)$ is locally integrable and determines a tempered distribution. We have $0<N_{\epsilon }\le N$, and $N|\Delta f|$ is integrable near zero and infinity for a Schwartz test function. Integration by parts and the dominated convergence theorem giveThus $-\Delta N={\delta }_0$, proving that $N$ is a fundamental solution of a linear differential operator. In this last part the test-function space is $\mathcal{S}({\mathbb{R}}^3)$; the printed ${\mathbb{R}}^n$ in the three-dimensional integral is understood with $n=3$.

For $n\ge 3$, the radial functionis harmonic on ${\mathbb{R}}^n\setminus \overline{B_1(0)}$, vanishes on the unit sphere, and is positive in the domain. It therefore violates the weak maximum principle for elliptic operators. In two dimensions the corresponding counterexample is $u(x)=\log |x|$, since the fundamental solution of the Laplace equation changes from a power to a logarithm.

Use Green's second identity in the source coordinate:Take $u({\mathbf{x}}^{\prime })=\psi ({\mathbf{x}}^{\prime })$ and $v({\mathbf{x}}^{\prime })=-1/[4\pi |\mathbf{x}-{\mathbf{x}}^{\prime }|]$, the fundamental solution of the Laplace equation satisfying ${\Delta }^{\prime }v=\delta ({\mathbf{x}}^{\prime }-\mathbf{x})$. With localized charge and the decaying potential, the boundary integral tends to zero on expanding spheres. The volume identity then gives $\psi (\mathbf{x})+(1/{ϵ}_0)\int vqd{V}^{\prime }=0$, hence the Coulomb potentialConvergence of this integral and a condition fixing the potential at infinity are needed: mere pointwise decay of $q$ is not enough, and a harmonic function could otherwise be added. A bounded, compactly supported charge distribution is a sufficient setting for this derivation.

For the disk put $\sigma =Q/(\pi a^2)$. On its axis, every ring of radius $\rho$ is at distance $\sqrt{{\rho }^2+z^2}$, soFor $z>0$ subtract $z$; for $z<0$ add $z$. This is the electrostatic potential of a uniformly charged disk with zero potential at infinity. The absolute value is essential for reflection symmetry and the correct behavior on the lower axis.

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.

Write $\rho =|\mathbf{x}-\mathbf{y}|$. A rotationally symmetric fundamental solution of the Laplace equation is harmonic for $\rho >0$. The Laplace equation in polar coordinates reduces to $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. Integrate $-\Delta G={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ over a disk of radius $ϵ$ around the source. The divergence theorem givesTaking $B=0$ therefore givesChanging the reference length inside the logarithm merely adds a constant. Notice the sign convention: the potential equation in this question is $\Delta V=q{\delta }^{(2)}$, whereas this Green function satisfies $-\Delta G={\delta }^{(2)}$. Thus the point-source potential used below is $-qG=+(q/2\pi )\log \rho$.

For the two insulating walls, successive Euclidean reflections in $x=b$ and $x=-b$ generate sources of strength $q$ at $(2nb,0)$ for every integer $n$. Only $n=0$ lies inside the physical strip. This is the Neumann image lattice for a point source in a strip. SetThe raw sum of logarithmic potentials diverges by additive constants. A well-defined potential is obtained by pairing opposite images and subtracting their constant reference terms:Each paired summand is $O(n^{-2})$ on compact subsets away from the source. The series therefore has local uniform convergence, as do the required derivative series. It givesIn the corresponding $x$-derivative, the images pair as $n$ and $1-n$ at $x=b$, and as $n$ and $-1-n$ at $x=-b$. Their opposite horizontal separations make the derivatives cancel. Thus both homogeneous Neumann boundary conditions hold. The constructed solution is even in $y$; without a condition at infinity the side conditions alone would allow further homogeneous harmonic functions.

For fixed $y>0$, uniform convergence on $[-b,b]$ permits integration term by term. The $n$th term has integralHere $\arctan$ is the ordinary real inverse trigonometric function, taking values in $(-\pi /2,\pi /2)$. Summing from $-N$ to $N$ makes a telescoping series:The source flux splits equally between the upper and lower parts of the strip. The next two parts give the single-wall constructions that underlie the image lattice.

For $\Delta G=+{\delta }^{(3)}$, the fundamental solution of the Laplace equation in three dimensions isIts sign follows from the flux: on a small sphere of radius $\epsilon$, its outward radial derivative is $1/(4\pi {\epsilon }^2)$, with total flux one.

Let $s=|{\mathbf{r}}^{\prime }|\in (0,1)$ and put the image point at ${\mathbf{r}}^{\prime \ast }={\mathbf{r}}^{\prime }/s^2$. The method of images gives the Dirichlet Green function inside a sphere:The image lies outside the ball, so its potential is harmonic inside. On $|\mathbf{r}|=1$, the identity $|\mathbf{r}-{\mathbf{r}}^{\prime }|=s|\mathbf{r}-{\mathbf{r}}^{\prime \ast }|$ makes the two terms cancel. The original source retains unit strength. For a source at the centre, take the limit:

Use the sign convention in which a fundamental solution of the Laplace equation satisfiesin the sense of distributions. It is a harmonic function away from the source. Writing $\rho =|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial equation is $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. The divergence theorem on a small circle about the source fixes its total flux:ConsequentlyThe arbitrary constant is a harmonic function. For the alternative convention $-\Delta G={\delta }^{(2)}$, the logarithm has the opposite sign; subsequent Green's function formulas must use that convention consistently.

The three-dimensional fundamental solution of the Laplace equation, with $\Delta F=+\delta$, isIt is harmonic away from the source, and its gradient has total outward flux one through a small source-centred sphere.

For $s=|{\mathbf{r}}^{\prime }|\in (0,1)$, place the image at ${\mathbf{r}}^{\ast }={\mathbf{r}}^{\prime }/s^2$, outside the unit ball. On $|\mathbf{r}|=1$, expansion of squared distances gives $|\mathbf{r}-{\mathbf{r}}^{\ast }|=|\mathbf{r}-{\mathbf{r}}^{\prime }|/s$. The method of images therefore gives the Dirichlet Green function inside a sphereThe boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ there. At the central source, take the finite limit of the weighted image term:

A fundamental solution of the Laplace equation satisfies ${\Delta }_{\mathbf{r}}{G}_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ in the sense of distributions, without a boundary condition. Choose it rotationally symmetric about the source. Away from the source, with $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial Laplace equation is $(s{G}_0^{\prime }{)}^{\prime }=0$, giving $G_0=A\log s+B$.

Integrating over a small disc around the source and using the divergence theorem fixes the normalization:ThereforeThe additive constant is arbitrary; adding other entire harmonic functions is also possible if rotational symmetry is not imposed. The positive sign here follows from the convention $\Delta G_0=+\delta$.

For the rotationally invariant free-space fundamental solution of the Laplace equation, put $s=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}$. Away from the source, the radial Laplacian gives $(s{G}^{\prime }{)}^{\prime }=0$, hence $G=A\log s^2+C$. Integrate its Poisson equation over a disk of radius $r$ about the source. The divergence theorem gives $2\pi r(2A/r)=-1/{ϵ}_0$, soA fixed reference length can be inserted inside the logarithm and absorbed into $C$. The resulting electric field decays like $1/s$.

Strictly, the stated pointwise decay as $y\to \pm \infty$ alone does not justify “most general”: one can also add the harmonic function $\mathrm{Re}{e}^{(x+iy{)}^2}$, whose gradient has that decay for every fixed $x$. This is the same growth issue as a zero-boundary harmonic function with pointwise vertical decay. The displayed answer is the usual isotropic free-space choice; requiring the field to vanish uniformly in every direction at spatial infinity removes these extra harmonic additions. An additive constant remains undetermined.