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The solution and its first derivative are continuous at $x=1$: a jump would create a Dirac delta distribution or its derivative, absent from the forcing. Put $r=x-1$, $A=a+1$ and $B=b$. The outer expansions, fixed by the respective endpoint boundary conditions, areBalancing diffusion and advection near $r=0$ gives the interior layer at a simple zero of advection with inner variable $z=r/\epsilon$. The leading inner expansion solves $Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$, so matching to $A$ and $B$ givesFor the next term define the Gaussian drift primitivesWriting $H$ for the Heaviside step function, let $F(z)=zH(-z)+(\sqrt{\pi }/4)\mathrm{erf}|z|$. Its value and first derivative match at zero and $(D^2+2zD)F=2zH(-z)$. Thuswhere the logarithmic overlaps fixHere $C_1,C_2$ are the constants in the supplied large-positive-$z$ limits of $E_0,E_1$. The terms $\epsilon \log \epsilon$ are essential switchback terms; discarding them would not give accuracy through $O(\epsilon )$.

Subtracting the common overlap from the inner expansion and outer expansions gives the additive composite expansionThe standard overlap subtraction leaves $\epsilon rH(-r)$. Replacing $H(-r)$ by $(1-\mathrm{erf}z)/2$ changes the value only by $O({\epsilon }^2)$ uniformly and gives a composite with continuous first derivative. The last term restores the endpoint values through the retained order. When $b=a+1$, $d=0$ and the order-one error-function jump disappears. An $O(\epsilon )$ interior adjustment remains to accommodate the leading outer derivative mismatch, together with logarithmic matching when $A\ne 0$.

Reversing the diffusion sign changes the leading inner equation to $-Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$. Its nonconstant solution grows like the integral of $e^{z^2}$, so bounded matching forces the same leading value $C$ on both sides. The bulk is $C+r$ on the left and $C$ on the right; both endpoint conditions are instead supplied by decaying endpoint layers for reversed diffusion, of width ${\epsilon }^2$. A weaker width-$\epsilon$ interior adjustment matches derivatives and selects $C$. Indeed, its first-order equation has a non-growing solution only ifThe leading structure is consequentlywith the smaller interior correction understood. Unlike the positive-diffusion case, the endpoint values are not transported into two distinct order-one inner limits.

For $t>2$, the Heaviside step function contributes a constant unit forcing. Solving the resulting first-order linear differential equation from the value at $t=2$ gives

Since $(x^2-1)(-{\delta }_0)={\delta }_0$, write $u^{\prime }=-{\delta }_0+w$, where $(x^2-1)w=0$. The polynomial has two simple zeros. The distribution annihilated by a function with simple zeros lemma givesIntegrating with the Heaviside step function yieldsThis includes every solution: its derivative has exactly the required delta terms, and any two primitives differ only by a constant distribution.

For $f(t)=\sin \omega t$,The Bromwich inversion formula givesWhen $t>x/c$, close the contour to the left. The residues at $p=\pm i\omega$ sum to $\sin (\omega (t-x/c))$. When $t<x/c$, close it to the right, where there are no poles, and the integral vanishes. HenceEquivalently, $y(x,t)=\Theta (t-x/c)\sin (\omega (t-x/c))$ in terms of the Heaviside step function. The boundary oscillation travels to the right at wave speed $c$ without reflection or dispersion; a point at $x$ begins moving after the propagation delay $x/c$.

The causal Green function for the heat operator iswhere $H$ is the Heaviside step function. It vanishes for $t<\tau$, satisfies the homogeneous heat equation away from the source, and approaches $\delta (x-\xi )$ as $t\downarrow \tau$. Superposing the responses to all infinitesimal sources gives Duhamel principle:The lower limit and causality give the required homogeneous initial data.

After separating the conserved horizontal factor as $\psi =e^{ipx}f(z)$, the Helmholtz equation becomeswhere $H$ is the Heaviside step function. The outgoing Green function for $d^2/d{z}^2+q_0^2$ isThe Born approximation at first order replaces $f$ on the right-hand side by the incident profile $f_i(z^{\prime })=e^{-i{q}_0{z}^{\prime }}$. For $z>0$ this givesThe integral is understood with the usual outgoing-wave convergence factor. Sincewe obtainTherefore the Born reflected field is

Letwhere $\mathcal{H}$ is the Heaviside step function. Thenso the Eliassen equation for residual circulation isTake no normal residual flow at $y=0,L$ and $z=0$, decay as $z\to \infty$, and choose the streamfunction constant on the connected rigid boundary to be zero. Thus

The required Fourier series in sine modes isDefineFor each mode, the vertical equation isThe Dirac delta function requiresThe solution satisfying the boundary conditions is thereforewhere

The momentum equation givesThe jump of ${\overline{\chi }}_a^{\ast }$ supplies an equal positive delta function in $f_0{\overline{\chi }}_{a,z}^{\ast }$, so the singular terms cancel. The regular acceleration iswhereFinally, the transformed density equation givesThese exponentially decaying modes are the balanced mean response to wave-activity deposition at $z=H$.

For $y_{tt}=c^2{y}_{xx}$ on $0<x<L$, initially at rest with zero displacement, apply $y(0,t)=a$ for $t>0$ and keep $y(L,t)=0$. A geometric series in the Laplace transform givesHere $H$ is the Heaviside step function, $L,c>0$, and only finitely many terms contribute at any finite time. The second step in each pair is a sign-reversed reflection from the fixed endpoint. The formula solves the wave equation classically away from the fronts and distributionally across them; the midpoint is a square wave of period $2L/c$.

Integrating the smooth positive-time impulse kernel gives the zero-data response to the Heaviside step function. For nonzero $\omega$,At zero frequency and nonzero $k$ the limit is $[1-e^{-kt}(1+kt)]/k^2$; when both parameters vanish it is $t^2/2$. The integral representation avoids ambiguous divisions in these limiting cases.