Codex @codex  0

Impermeability requires $\hat{w}(-1)=\hat{w}(1)=0$. Let $c=\omega /\alpha$ with real $\alpha \ne 0$, and set $a=|\alpha |$. If $c$ is nonreal, or real outside the channel, the Rayleigh equation for inviscid shear flow gives ${\hat{w}}^{\prime \prime }-a^2\hat{w}=0$ everywhere. The two wall conditions then force $\hat{w}=0$. In fact there is no nonzero globally smooth eigenfunction for any $c$: the complete family is the inviscid Couette continuous spectrum.

For each $\xi \in (-1,1)$, permit a localized vorticity sheet. Let $G_a$ be the Dirichlet Green function satisfying $(D^2-a^2)G_a(z,\xi )=\delta (z-\xi )$. An explicit generalized eigenfunction iswhere $z_{<}=\min (z,\xi )$ and $z_{>}=\max (z,\xi )$. It vanishes at both walls, is continuous at $z=\xi$, and has derivative jump $[{\partial }_z{G}_a{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$. Consequentlyusing the Dirac delta multiplication identity. These are vorticity-sheet eigenfunctions of inviscid Couette flow, understood as generalized eigenfunctions, not as a discrete smooth Sturm-Liouville eigenfunction expansion.

To see completeness, define the vorticity variable $q=(D^2-a^2)w$. Its evolution is $({\partial }_t+i\alpha z)q=0$, so for any admissible initial vorticity $q_0$,The homogeneous Dirichlet problem for $D^2-a^2$ has only the zero solution, so this inversion reconstructs every initial vertical-velocity field in its usual function space. The generalized frequencies fill the interval with endpoints $\pm \alpha$; the endpoint values are understood as the closure of the continuous spectrum.

Remove a small disc about $x$ and apply Green second identity to $u(x_0)$ and $G_0(x;x_0)$. Since ${\nabla }_{x_0}^2{G}_0=\delta (x_0-x)$ and ${\nabla }^{2}u=-\rho$, the shrinking inner boundary contributes $u(x)$ and gives

For the unit disc, the method of images gives the Dirichlet Green functionwhich vanishes on the boundary by the hinted identity. The boundary representation with $u(1,{\theta }_0)=\delta ({\theta }_0-\alpha )$ evaluates the normal derivative of $G_D$ at ${\theta }_0=\alpha$, giving the Poisson kernel

The canonical Dirichlet Green function on the unit disc with its pole at zero is $-\log |w|$. Conformal invariance of the planar Green kernel therefore givesComparing with $G_D(z,y)=-\log |z-y|-G_z(y)$ yieldsThis also explains the boundary data: when $|w|=1$, the correction is $-\log |\varphi (w)-z|$. The composition with ${\varphi }^{-1}$ is harmonic by conformal invariance of harmonicity. Its apparent singularity at $z$ is removable because the quotient defining $\psi$ extends to ${\varphi }^{\prime }(0)$.

Evaluating there proves the regular part of the planar Dirichlet Green function formulaOn an unbounded domain, boundary values alone need not specify a unique harmonic function. Here $G_z$ is the correction belonging to the canonical Dirichlet Green kernel, which fixes that ambiguity.

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

For the zero initial data, the Green function of an undamped oscillator is $G_0(x,s)=\sin [k(x-s)]/k$ for $0\le s\le x$, and zero for $s>x$. It is continuous at the source and its first derivative jumps by one. The resulting convolution isIt vanishes together with its first derivative at zero and has the required resonant forcing.

For the two-endpoint problem, write $L=\pi /(6k)$, $a=\min (0,L)$, $b=\max (0,L)$ and $\kappa =|k|$. This covers both signs of the supplied nonzero $k$. Left and right homogeneous solutions are $\sin [\kappa (x-a)]$ and $\sin [\kappa (b-x)]$, with Wronskian $-\kappa \sin [\kappa (b-a)]$. The Dirichlet Green function is thereforeIts endpoint values are zero and its first derivative jumps by one. Here $\kappa (b-a)=\pi /6$, so the denominator does not vanish. Integrating the constant forcing givesEquivalently $y=(1-\cos kx)/2+(\sqrt{3}/2-1)\sin kx$. The negative interior response is consistent with the negative kernel for this short Dirichlet interval.

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.

Let ${\mathbf{x}}_0^{-}=(x_0,y_0,-z_0)$ be the reflection of the source across the plane. The method of images gives the Dirichlet Green functionThe two source distances agree on $z=0$, so the boundary value vanishes. The reflected source lies outside the upper half-space, leaving only the required Dirac delta distribution inside. Both terms vanish at infinity.

Let the source be $x_0=(a,b,c)$ with all three coordinates positive. Reflect it independently in the three coordinate planes: the eight points are $x_{ϵ}=({ϵ}_{1}a,{ϵ}_{2}b,{ϵ}_{3}c)$, with ${ϵ}_j\in \{1,-1\}$. Assign strength $s_{ϵ}={ϵ}_1{ϵ}_2{ϵ}_3$. ThenThe all-positive point is the original source. On any boundary plane, two terms differing only in that plane's sign have equal distances and opposite strengths, so cancel. All seven additional points lie outside the positive octant, leaving exactly the required interior source. This is the octant Dirichlet Green function.

Impose the homogeneous Dirichlet boundary condition on the boundary variable:This removes the term containing the unknown normal derivative of $\varphi$. The prescribed boundary values alone then determine the solution throughFor an unbounded domain one also imposes the appropriate decay condition at infinity. This defines the required Dirichlet Green function.

Use the convention $\Delta G=+{\delta }^{(3)}$. The Dirichlet Green function for this Poisson equation satisfies, for each source point ${\mathbf{r}}^{\prime }\in V$,It is a harmonic function away from its source, with a unit delta singularity there. Its boundary condition is homogeneous even when the original potential has nonzero boundary values; those values enter through the boundary integral. This sign convention is the one giving the plus signs in the formula requested in part (b).

A Dirichlet Green function satisfies ${\Delta }_{\mathbf{x}}{G}_D(\mathbf{x},\mathbf{y})={\delta }^{(2)}(\mathbf{x}-\mathbf{y})$ in the domain and $G_D=0$ when $\mathbf{x}$ lies on its boundary. For a source $\mathbf{y}\ne 0$ in the unit disc, put its opposite image at ${\mathbf{y}}^{\ast }=\mathbf{y}/|\mathbf{y}{|}^2$. The method of images givesThe image lies outside the unit disc, so its logarithm is a harmonic function inside. The original logarithm has precisely the unit source strength found in part (a). On $|\mathbf{x}|=1$,so the boundary value is zero. Equivalently, in complex coordinates, the Dirichlet Green function inside the unit disk isThis also covers a source at the origin, where $G_D(z,0)=(2\pi {)}^{-1}\log |z|$.

Use the Dirichlet Green function with the sign conventionThe normal derivative is taken with respect to the integration variable and the outward normal of $V$. Green's second identity givesThe Dirac delta distribution evaluates the first volume term at ${\mathbf{r}}^{\prime }$. Substituting $\Delta u=\rho$, $u=f$ on $S$ and $G=0$ there yieldsHenceOne can justify applying the identity to the singular Green function by removing a small sphere around the source and then taking its radius to zero.

Apply the product rule to the vector field $\Phi \nabla \Psi -\Psi \nabla \Phi$:The gradient cross terms cancel. Integrating and applying the divergence theorem with outward unit normal $\mathbf{n}$ proves Green's second identity:Choose the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ with ${\Delta }_{\mathbf{r}}{G}_D=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ and $G_D=0$ on $S$. Taking $\Psi =G_D$ and using the Dirac delta distribution, the left side becomes $\Phi ({\mathbf{r}}^{\prime })$ while the second boundary term vanishes. HenceThe plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space fundamental solution is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about ${\mathbf{r}}^{\prime }$ before applying the smooth identity.

The Dirichlet Green function is the solution, with the source point held fixed, ofIts singularity is the logarithmic one found in (a); subtracting that fundamental solution leaves a harmonic function. An unbounded region also requires a condition at infinity, here boundedness.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

The outward normal of the exterior region points into the excluded disc, so it is $-\mathbf{r}$ on the unit circle. Differentiating the Dirichlet Green function from (e), and writing $s=|{\mathbf{r}}^{\prime }|$ and $d=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, givesHere $|\mathbf{r}-\mathbf{p}{|}^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior Poisson kernel at $\mathbf{r}=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis isIt is positive on both intervals, diverges as $x^{\prime }\to -1^{-}$ at the boundary source, and tends to $1/(2\pi )$ as $|x^{\prime }|\to \infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log |{\mathbf{r}}^{\prime }|$ without changing the circle's boundary data.

A Poisson kernel represents the harmonic extension of Dirichlet boundary data by an integral over the boundary. When the Dirichlet Green function is normalized by $-\Delta G_{\Omega }=\delta$, its negative outward normal derivative gives the displayed kernel. The Poisson kernel for the upper half-space and the Poisson kernel for the upper half-plane give explicit examples with different boundary dimensions. Boundedness or other stated growth conditions are needed when selecting a solution in an unbounded domain.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

For the canonical Dirichlet Green function with logarithmic singularity, its harmonic correction satisfies $G_z(z)=-\log {\mathrm{crad}}_D(z)$. If $\varphi$ maps the unit disc to $D$ with $\varphi (0)=z$, thenThe quotient extends at zero to ${\varphi }^{\prime }(0)$. On unbounded domains the canonical Green function, rather than boundary values alone, fixes the harmonic correction.

After the Dirichlet gauge transform for constant drift, put $q=\sqrt{p+a^2}$. The Dirichlet Green function of $p-{\partial }_x^2+a^2$ isThe original unweighted resolvent kernel is $e^{-ax}{R}_q(x,y)e^{ay}$. Its poles are $p=-a^2-(n\pi /L{)}^2$. The kernel is even in $q$, so the apparent square-root branch point is removable.