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For a smooth function $u$, the distributional derivative of the product of $u$ with the Heaviside function is $D(Hu)=H{u}^{\prime }+u(0){\delta }_0$. Induction gives the displayed formula. It records all boundary sources, including derivatives of the Dirac delta distribution, and determines which initial derivatives must vanish in a retarded fundamental solution.

Use $G=c=1$, select the retarded source field with no added homogeneous radiation, and work to leading order in the weak-field approximation and long-wavelength source approximation. The retarded fundamental solution of the Linearized Einstein equations in Lorenz gauge in linearized gravity isFor $r\gg d$, the denominator is $r$ to leading order. The delay is $t-r+\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(d^2/r)$, where $\mathbf{n}=\mathbf{x}/r$. In the usual slowly evolving source regime, its characteristic time $\tau$ obeys $d/\tau \ll 1$, so the source-size part of the delay is negligible:The stress-energy conservation equation ${\partial }_{\mu }{T}^{\mu \nu }=0$ is required at the approximation order used; it follows from the divergence of the gauge-fixed field equation. Compact support removes the integration by parts surface terms. Since $T^{00}=T_{00}$ and $T^{ij}=T_{ij}$ in this signature,Combining these identities gives the retarded quadrupole fieldFor an ordinary nonrelativistic bound source, $\tau \sim d/v$ makes $d/\tau \sim v\ll 1$. More generally the size relative to the variation timescale must also be small; speed alone does not exclude a rapidly varying small-amplitude motion.

The radiation boundary condition for linearized gravity is essential for a statement about the full field. Without it, the displayed implication is false: take $T_{\mu \nu }=0$ and add ${\bar{h}}_{11}=A\cos (k(t-z))$, ${\bar{h}}_{22}=-{\bar{h}}_{11}$, all other components zero. This weak plane gravitational wave satisfies both the homogeneous wave equation and Lorenz gauge in linearized gravity, while every $I_{ij}$ is zero. The proved formula is for the retarded source contribution, with the standard slow-source qualification. Its transverse–traceless projection gives the physical radiative strain.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

Take $K>0$. For $t>0$, the diffusion fundamental solution of a linear differential operator $F$ satisfies $({\partial }_t-K{\partial }_x^2)F=0$ and has initial value $\delta (x)$ in the sense of distributions:for every smooth compactly supported test function $\psi$. With the usual spatial decay it has unit mass. Equivalently the causal extension $\Theta (t)F(x,t)$ is a retarded fundamental solution satisfying $({\partial }_t-K{\partial }_x^2)(\Theta F)=\delta (x)\delta (t)$.

The causal Green function is zero for $t<\tau$ and satisfiesFor $t>\tau$ it solves the homogeneous diffusion equation, and its initial jump is $G(x,{\tau }^{+};y,\tau )=\delta (x-y)$, again distributionally. The natural spatial decay and a standard bounded or integrable solution class specify the usual kernel, avoiding unrestricted growing solutions. The PDE acts on $x,t$; $y,\tau$ label the location and time of the unit impulse. Values of the Heaviside step function exactly at the jump do not affect these distribution identities.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

For each one-dimensional factor $P_j(D_j)$, choose its retarded fundamental solution $E_j=H(x_j)g_j(x_j)$. Partial-fraction decomposition of the reciprocal polynomial giveswhere each ${\alpha }_{rj}$ is a polynomial whose degree is one less than the multiplicity of the associated root. Constants, including powers of $i$ from $D=-i\partial$, can be absorbed into the polynomials and exponents.

Take the tensor productIt vanishes unless every $x_j>0$, has the required polynomial-exponential form there, and satisfies

For $P(z)=a_N{z}^N+\cdots +a_0$ with $N\ge 1$ and $a_N\ne 0$, the unique retarded fundamental solution is $E=Hu$, where $P(D)u=0$, $u^{(j)}(0)=0$ for $j<N-1$, and $u^{(N-1)}(0)=1/a_N$. The distributional jump formula for a Heaviside product gives $P(D)E={\delta }_0$. Its explicit expression isThe residue theorem gives the initial derivatives from the coefficients at infinity. Equivalently, the inverse Fourier transform is integrated below all its poles, producing support of a distribution in $[0,\infty )$. Uniqueness follows because the difference of two retarded solutions is an analytic homogeneous solution vanishing on the negative half-line.