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For the exterior of the unit disc in two dimensions, with the sign convention ${\Delta }_{x}G=+\delta (x-y)$, this is the Dirichlet Green function that vanishes on the circle and is bounded at infinity. The method of images places an opposite logarithmic source at the inverse point $y/|y{|}^2$. The constant $-(2\pi {)}^{-1}\log |y|$ enforces the zero boundary value. The inversion in a circle identity $|x-y|=|y||x-y/|y{|}^2|$ for $|x|=1$ verifies that condition directly.

The outward normal of the exterior domain is $-x$ on the unit circle. Its normal derivative gives the positive exterior Poisson kernelFor bounded harmonic functions this kernel integrates boundary data to the interior value. Boundedness at infinity is essential to uniqueness: without it, a multiple of $\log |y|$ can be added without changing the circle's Dirichlet boundary condition.

For the gap $a<r<b$, put $u_{\ell }(r)=r^{\ell }-a^{2\ell +1}{r}^{-\ell -1}$, $v_{\ell }(r)=r^{-\ell -1}-r^{\ell }/b^{2\ell +1}$ and $d_{\ell }=(2\ell +1)[1-(a/b{)}^{2\ell +1}]$. With orthonormal spherical harmonics, the Dirichlet Green function isFor a short verification, $u_{\ell }(a)=v_{\ell }(b)=0$ and both solve the radial Laplace equation. Their Wronskian is $u_{\ell }{v}_{\ell }^{\prime }-u_{\ell }^{\prime }{v}_{\ell }=-d_{\ell }/r^2$. Thus the radial derivative jump is $-1/r^{\prime 2}$; completeness of the spherical harmonics gives the stated Dirac delta function normalization. A point charge has electric potential $qG/(4\pi {ϵ}_0)$, and the inner-wall surface charge density is $-q{\partial }_{r}G{|}_{a^{+}}/(4\pi )$, with the angular source coordinates held fixed.

For a three-dimensional slab $0<x_1<L$ and source $y=(y_1,y_2,y_3)$, define image positions $y_n^{+}=(y_1+2nL,y_2,y_3)$ and $y_n^{-}=(-y_1+2nL,y_2,y_3)$. The displayed method of images series uses convention ${\Delta }_x{G}_D=+\delta (x-y)$. Reflection in $x_1=0$ pairs the plus image $n$ with the minus image $-n$, while reflection in $x_1=L$ pairs it with the minus image $1-n$. Both boundary values therefore vanish. Only $y_0^{+}=y$ lies in the slab. Each bracket is $O(n^{-2})$, giving absolute convergence of the paired series away from sources. This selects the usual Dirichlet Green function that decays at large transverse separation.

For $D=d/dx$ and $L_a=(D-a{)}^2+b^2$ on $(0,L)$, with $b>0$ and $\sin (bL)\ne 0$, the substitution $y=e^{ax}v$ gives $L_{a}y=e^{ax}(v^{\prime \prime }+b^{2}v)$. Here $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function vanishes at the endpoints, is continuous at $x=\xi$, and its first derivative has jump one, so $L_{a,x}G=\delta (x-\xi )$. It is the usual sine Dirichlet Green function conjugated by the exponential factor. At $bL\in \pi \mathbb{Z}\setminus \{0\}$ there is a homogeneous zero mode, so the inverse kernel does not exist without a solvability condition and a normalization.

For $D=\{x>0,z>0\}\subset {\mathbb{R}}^3$, take the Dirichlet Green function convention $-{\Delta }_{P}G(P,Q)={\delta }_Q(P)$. Reflect $Q=(\xi ,\eta ,\zeta )$ to $Q_{ij}=((-1{)}^i\xi ,\eta ,(-1{)}^j\zeta )$. The displayed image sum vanishes on both faces, has exactly one positive unit source inside $D$, and decays at infinity. Pairwise cancellation proves the boundary conditions, while the three image terms are harmonic inside $D$. Its inward source-normal derivative gives the Poisson kernel of a three-dimensional quarter-space. The source sign must be kept consistent with Green's second identity.

For the positive Laplacian convention $\Delta G=\delta$, the method of images in the first quadrant reflects an interior pole $(a,b)$ across each axis with negative sign and across both with positive sign:The Dirichlet Green function vanishes on both axes. Its outward normal derivative on the horizontal axis isand the vertical-axis kernel follows by exchange of coordinates. Green second identity converts these derivatives into the quadrant Poisson integral for boundary data.

For the three-dimensional ball $|x|<a$, the displayed Dirichlet Green function has convention ${\Delta }_x{G}_D=+\delta (x-y)$. Its method of images uses the exterior inverse point $y^{\ast }=a^{2}y/|y{|}^2$. On $|x|=a$, the distance identity $|x-y^{\ast }|=(a/|y|)|x-y|$ makes the boundary value zero. The image lies outside the ball and therefore adds no source singularity to the domain. At a central source the weighted image term tends to $1/a$, giving $G_D(x,0)=-(4\pi |x|{)}^{-1}+(4\pi a{)}^{-1}$. Spherical inversion with the required radial weight is the Kelvin transform.

With the convention ${\Delta }_z{G}_D=\delta (z-\zeta )$, the displayed Dirichlet Green function vanishes on the unit circle. Indeed $|z-\zeta |=|1-z\bar{\zeta }|$ when $|z|=1$. For $\zeta \ne 0$, the second logarithm corresponds to an opposite image source at $\zeta /|\zeta {|}^2$, outside the unit disc; it is a harmonic function throughout the interior. At $\zeta =0$ the formula becomes $(2\pi {)}^{-1}\log |z|$. Its outward normal derivative in the boundary variable is $(2\pi {)}^{-1}(1-|\zeta {|}^2)/|z-\zeta {|}^2$, the Poisson kernel divided by $2\pi$. Green's second identity therefore gives the Poisson integral on the unit disk.

For zero initial value and slope, the causal Green function is $G_0(x,s)=H(x-s)\sin [k(x-s)]/k$, with $k\ne 0$ and $H$ the Heaviside step function. For zero values at endpoints $a<b$, put $\kappa =|k|$. The Dirichlet Green function isBoth are continuous at the source and have a first-derivative jump of one, giving $L_{x}G=\delta (x-s)$. The Dirichlet inverse exists when $\sin [\kappa (b-a)]\ne 0$; a zero denominator signals a nontrivial homogeneous mode and a Fredholm solvability condition. The endpoint kernel is the zero-conjugation case of the Dirichlet Green function for a conjugated harmonic oscillator. Boundary conditions determine which kernel must be integrated against a forcing term.

The density of expected Brownian occupation time before exit from a planar domain. For planar Brownian motion with infinitesimal generator $\frac{1}{2}\Delta$, it obeys $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$ and has singularity $-{\pi }^{-1}\log |w-z|$. A conformal bijection preserves this Dirichlet Green function: the squared derivative in the conformal Brownian clock cancels the area Jacobian determinant.

For constant shear $U=\lambda y$, the wall-normal Laplacian of velocity is advected with phase $e^{-i\alpha \lambda yt}$. Inverting $D^2-k^2$, $k^2={\alpha }^2+{\beta }^2$, with its Dirichlet Green function gives the displayed initial-value solution. Impermeable inviscid walls impose zero normal velocity, not zero tangential velocity or no-slip wall vorticity.

The supplied positive free-space kernel satisfies $-\Delta H=\delta$, so use the same sign convention for the Dirichlet Green function. For an interior source $Q=(\xi ,\eta ,\zeta )$, requireAway from $Q$, the function is harmonic, and $G-H$ is regular near $Q$. The method of images reflects the source in each coordinate plane, reversing its sign on each reflection. ThereforeOn either face, the terms cancel in pairs. Only the original source lies in the domain, so the other three terms are harmonic there; the required singularity and decay also follow. This is the Dirichlet Green function for a three-dimensional quarter-space. Using $\Delta G=\delta$ instead would change the sign of the kernel and the representation formula together, as in sign conventions for a Dirichlet Green function.

Let $r_0=|{\mathbf{r}}_0|$ and let ${\Omega }_0=({\theta }_0,{\phi }_0)$ describe its direction in spherical coordinates; write $\Omega =(\theta ,\phi )$. The given Dirichlet Green function between concentric spheres has normalization ${\nabla }^{2}G=-4\pi {\delta }^{(3)}$, so the electric potential is $\varphi =qG/(4\pi {ϵ}_0)$. On the inner sphere the unit normal from metal into the gap is $+{\mathbf{e}}_r$, hence its induced surface charge density isNear this boundary $r_{<}=r$ and $r_{>}=r_0$. The needed radial derivative isIt cancels the factor $2\ell +1$ in the Dirichlet Green function. Therefore the answer in spherical harmonics isThis spherical harmonic expansion converges on the boundary: after the angular sum, the two radial factors decay geometrically with ratios $a/r_0<1$ and $a{r}_0/b^2<1$, while the angular factors grow at most polynomially. This justifies the boundary derivative and the subsequent surface integral.

Use incompressible Euler equations with pressure divided by the constant reference density. Writing $L={\partial }_t+U(y){\partial }_x$, their linearization about the parallel shear flow isTaking the divergence requires the commutator $[{\partial }_y,L]=U^{\prime }{\partial }_x$. Both this commutator and the divergence of $U^{\prime }v$ contribute, so ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying the Laplacian to $Lv=-p_y$ givesHence the wall-normal velocity equation isTaking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise equation cancels pressure and yields the inviscid Squire equation,This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm 1,t)=0$. They do not impose a viscous no-slip condition on the tangential perturbation velocity.

For a Fourier mode, set $k^2={\alpha }^2+{\beta }^2$, $D=d/dy$ and $q=(D^2-k^2)\tilde{v}$. The equations becomeFor Couette flow $U=\lambda y$, $U^{\prime \prime }=0$, so $q(y,t)=e^{-i\alpha \lambda yt}{q}_0(y)$ with $q_0=(D^2-k^2){\tilde{v}}_0$. Invert this using the Dirichlet Green function of $D^2-k^2$ on $[-1,1]$. If $y_{<}=\min (y,s)$ and $y_{>}=\max (y,s)$, it isIt vanishes at both walls and its first derivative has jump one at $y=s$. ThusAt $t=0$, uniqueness of the Dirichlet inversion recovers the compatible initial velocity. This inviscid Couette initial-value Green function directly solves the initial-value problem without a Laplace transform.

The Squire equation is now a first-order forced equation in time. Its integrating factor givesFor nonzero $\alpha$, the velocity formula contains shearing phases and the inviscid Couette continuous spectrum. Velocity cancellation can occur even though the transported $q$ has constant magnitude; it must not be confused with exponential modal instability.

For the separate case $\alpha =0$, $\beta \ne 0$, the Dirichlet operator $D^2-{\beta }^2$ is invertible and $q_t=0$. Hence $\tilde{v}={\tilde{v}}_0$ andSince $\tilde{\eta }=i\beta \tilde{u}$ in this case, $\tilde{u}={\tilde{u}}_0-\lambda t{\tilde{v}}_0$. This is the lift-up effect: a transverse disturbance transports base-flow momentum and creates a streamwise perturbation growing linearly in time, with perturbation kinetic energy generically growing as $t^2$. The growth is algebraic and is possible without an exponentially unstable normal mode. For nonzero initial normal velocity and nonzero shear, nonlinear effects eventually limit the validity of the linearized calculation. The same streamwise-independent result holds with $\lambda$ replaced by $U^{\prime }(y)$ for a general parallel profile.

Away from $x=\xi$, the Dirichlet Green function solves the homogeneous equation. The endpoint conditions select $A{y}_1(x)$ on the left and $B{y}_2(x)$ on the right. Continuity and the unit derivative jump needed for a Dirac delta function giveWith Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2\ne 0$, solve to obtain $A=y_2(\xi )/W(\xi )$ and $B=y_1(\xi )/W(\xi )$. ThusThe first derivative term creates no extra delta because $G$ is continuous. The value at $x=\xi$ is the common limiting value.

For the specified constant-coefficient operator, takeTherefore its explicit Green function isFor the forcing step and the homogeneous endpoint conditions inherited from this Green-function problem, $y(x)=2{\int }_{x_0}^{1}G(x,\xi )d\xi$. When $x<x_0$, every integration point lies on the first branch, soWithout those endpoint conditions, an arbitrary homogeneous solution could also be added.

With the positive logarithmic fundamental solution used here, a Dirichlet Green function for an interior pole ${\mathbf{r}}_0$ satisfiesEquivalently, $G-G_0$ is harmonic near the pole, $G$ is harmonic away from it, and it has the specified logarithmic singularity. Regularity up to the remaining boundary is understood; on an unbounded domain impose the appropriate condition at infinity as well.

Apply Green second identity with $f=\psi$ and $g=G$. Since $\psi$ is a harmonic function, the volume integral is $\int \psi {\delta }_{{\mathbf{r}}_0}dA=\psi ({\mathbf{r}}_0)$. The term $G{\partial }_n\psi$ vanishes on the boundary by the Dirichlet boundary condition. ThereforeOne may equivalently excise a small disc around the pole and take its radius to zero. The sign is positive because this question uses ${\nabla }^{2}G=+\delta$, not the alternative negative-Laplacian convention.

Use the Green function of killed planar Brownian motion, normalized as the density of expected occupation with respect to area:for nonnegative measurable $f$. Equivalently, $G_D(z,w)={\int }_0^{\infty }{p}_D(t,z,w)dt$, where $p_D$ is the killed Brownian transition density. With generator $\frac{1}{2}\Delta$, the distributional normalization is $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$, and the singularity is $-{\pi }^{-1}\log |w-z|$ plus a locally harmonic function. This fixes the normalization of the Dirichlet Green function explicitly.

By conformal invariance of planar Brownian motion and its conformal Brownian clock,The Jacobian determinant of a conformal map is $|{\varphi }^{\prime }(w){|}^2$. Changing the area variable to $w^{\prime }=\varphi (w)$ givesUniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and harmonic away from their pole, so it holds at every $w\ne z$. The Green function is conformally invariant:If the Dirichlet Green function is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

Use the printed coordinate derivative $u_x(0,t)$, not an outward normal derivative; at the left endpoint these have opposite signs. Assume the usual bounded or admissibly growing solution at infinity, with an initial trace $b_0=u_0(0)$ and forcing possessing a Laplace transform. Set $a=\alpha /2$, so $\beta =a^2$, and use $z=\sqrt{p}$ on the principal square root branch. DefineThe half-line Dirichlet Green function for $p-{\partial }_x^2$ givesWrite the transformed solution as $U=U_D+e^{-zx}B(p)$. The Laplace transform of a derivative in the dynamic boundary condition is $pB-b_0$, soThe special choice of $\beta$ gives the perfect squareIn particular, the initial boundary trace cannot be omitted. Withan integral representation isChoose the Bromwich contour to the right of the forcing's growth abscissa and every pole; $\gamma >\max (0,a^2)$ and sufficiently large for the data is a safe choice for real $a$. No sign of $\alpha$ is explicitly imposed in this question.

The inverse transform can also be performed explicitly. For the repeated-root dynamic-boundary heat kernel, setFirst integrate the Heat Poisson kernel against $e^{ar}$, or differentiate the result with respect to $a$:Since differentiation of $(\sqrt{p}-a{)}^{-1}$ with respect to $a$ produces $(\sqrt{p}-a{)}^{-2}$, and ${\partial }_a{E}_a=(-x+2at)E_a+2\sqrt{t/\pi }{e}^{-x^2/(4t)}$, the desired kernel isHere $\mathrm{erfc}$ is the complementary error function. A direct integral characterization, also proving the sign and the transform, isThe convolution theorem for Laplace transforms now gives the fully real time-domain representationAll quantities here are known from the prescribed data. For $x>0$ it tends to $u_0(x)$ as $t\downarrow 0$. At $x=0$, $J_a(0,t)\to 1$ and the spatial correction tends to zero, recovering $b_0$. A solution classical through the initial corner additionally needs $u_0^{\prime \prime }(0)+\alpha u_0^{\prime }(0)+a^2{b}_0=f(0)$; weaker corner regularity does not invalidate the formula for positive times.

The repeated-root unstable heat boundary mode gives a useful sign check on the dynamic boundary condition for the heat equation. If $a>0$, $(\sqrt{p}-a{)}^2$ has a genuine double pole at $p=a^2$ on the physical branch, and the exact homogeneous mode $e^{-ax+a^{2}t}$ satisfies both the heat equation and the printed boundary condition. Generic data can also excite a $t{e}^{a^{2}t}$ contribution. A contour deformation must retain this repeated-pole contribution. If $a<0$, the putative root $\sqrt{p}=a$ is outside the chosen branch and is not a physical pole. At $a=0$, $J_0=\mathrm{erfc}(x/(2\sqrt{t}))$, as expected when the boundary trace satisfies $b^{\prime }(t)=f(t)$.

Every generalized frequency in the inviscid Couette continuous spectrum is real: ${\omega }_{\xi }=\alpha \xi$ for real $\alpha ,\xi$. Thus every time factor has unit modulus and there is no exponentially growing normal mode:The vorticity equation also gives $q(z,t)=q_0(z)e^{-i\alpha zt}$, preserving its $L^2$ norm. For fixed nonzero $\alpha$, the bounded Dirichlet Green function inversion therefore gives a uniform bound on the reconstructed velocity in terms of the initial vorticity norm. Superpositions can still rearrange their velocity energy and display transient growth from non-normal modes; neutral hydrodynamic stability does not require every velocity component to decrease monotonically. Treating only smooth discrete modes would miss this continuous neutral family.