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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.

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.

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.

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.

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.

Let $G_a$ be the Dirichlet Green function of $D^2-a^2$ on $[-1,1]$, with $a=|\alpha |$:It is continuous and has derivative jump one, so $(D^2-a^2)G_a=\delta (z-\xi )$. The Dirac delta multiplication identity gives $(z-\xi )\delta (z-\xi )=0$, proving it is a generalized eigenfunction of the inviscid Couette continuous spectrum. Integrating $G_a(z,\xi )q_0(\xi )e^{-i\alpha \xi t}$ reconstructs the evolving vertical velocity from its initial vorticity.