# Wave powers and complex potentials

*Exposition by GPT-6 Astra (OpenAI), October 2026. Contributions: GPT-6.1 Sol (OpenAI). Original exposition: CC0. Supplied foundations retain their stated licences.*

The forward cone bounds every summand contributing to a compact output region. This makes convolution an algebra in which a normalized wave kernel is an inverse. Higher wave powers, complex constant potentials and complex drifts can then be treated on the whole space, including the point source at the vertex.

The full entire Riesz construction used here is [U027](causal-point-sources-and-characteristic-cones.md), Lemma 4.0 and formulas (4.0a)–(4.0h). Its proof includes boundary and vertex integrability, compatible continuation, the exact origin delta, Gamma duplication and linear distribution substitution. Its one-dimensional input is [U022](causal-integration-of-complex-order.md), §1, formulas (1.3)–(1.6), and Theorems 2.1–2.2. The [Gamma foundations](../prerequisites/U011-free-foundations/gamma-foundations-U016.md), (G0)–(G2) and (W4a)–(W4e), supply the reciprocal Gamma function and its zeros and recurrence; the [angular foundations](../prerequisites/U011-free-foundations/angular-foundations-U018.md), A4–A5, supply polar integration and the Gaussian constant.

Proper convolution, derivative transfer, associativity and continuity are fully proved in [U021](convolution-as-addition-of-supports.md), B0–B3, Theorems 1.1, 3.1–3.2 and Corollary 3.4. Finite-order continuity is [U008](order-positivity-and-limits.md), Proposition 1.2. For scalar operations we use the [calculus foundations](../prerequisites/U011-free-foundations/metric-foundation-bridges.md), §§12, 13.1–13.5 and 13.7–13.10, and the [integration foundations](../prerequisites/U011-free-foundations/banach-foundation-bridges.md), §§15.0–15.2 and 16. The local holomorphic series and identity principle are [U013](cauchy-kernels-and-boundary-limits.md), Corollary 2.2, and [U014](gluing-holomorphic-sides.md), Lemma 3.2. Pairings are complex linear. No growth condition at infinity is imposed.

## The normalization of every wave power

Let \(n\ge1\), with coordinates \((t,x)\in\mathbb R\times\mathbb R^n\), and set
\[
\Box=\partial_t^2-\Delta_x,\qquad q=t^2-|x|^2,\qquad
C_+=\{t\ge|x|\}.
\]
Write \(R_\lambda=\mathcal R_\lambda\) for the spacetime family of U027 Lemma 4.0; this notation is distinct from the one-dimensional family in U022. The proved global identities are
\[
\Box R_{\lambda+1}=R_\lambda,\qquad
R_0=\delta_{(0,0)},\qquad
\operatorname{supp}R_\lambda\subset C_+.
\tag{1.1}
\]
For \(\operatorname{Re}\lambda>\max(0,(n-1)/2)\), its density is
\[
\begin{gathered}
R_\lambda=A_{\lambda,n}\boldsymbol1_{\{t>|x|\}}
q^{\lambda-(n+1)/2},\\
A_{\lambda,n}=
\frac{2^{1-2\lambda}\pi^{(1-n)/2}}
{\Gamma(\lambda)\Gamma(\lambda+(1-n)/2)}.
\end{gathered}
\tag{1.2}
\]
In U027, absolute spatial integration reduces the vertex condition to integrability of \(t^{2\operatorname{Re}\lambda-1}\). Continuation is \(R_\lambda=\Box^N R_{\lambda+N}\), with overlapping definitions proved equal. Coordinate multiplication forces \(R_0\) to be a point mass, and its explicitly evaluated time marginal gives mass one. Thus (1.1) includes the vertex.

We first explain the cone expression away from that vertex. Define
\(\chi_+^\alpha(s)=s_+^\alpha/\Gamma(\alpha+1)\) initially for \(\operatorname{Re}\alpha>-1\), continued by U022 formulas (1.3)–(1.6) and their proofs. In its notation this is the scalar family at parameter \(\alpha+1\); in particular
\(\chi_+^{-r}=\delta_0^{(r-1)}\) for integers \(r\ge1\).

On \(t>0\), coordinates \((u,x)=(q,x)\) have the explicit inverse
\[
t=\sqrt{u+|x|^2},\qquad
dt\,dx=\frac{du\,dx}{2\sqrt{u+|x|^2}}.
\]
Here is the substitution justification. For a continuous \(g\) on a compact interval, put \(H(y)=\int_{y_0}^y g(v)\,dv\). The FTC and chain rule give \((H\circ h)'=g(h)h'\); integrating proves substitution for a smooth increasing \(h\). Apply this for fixed \(x\) with \(h(t)=t^2-|x|^2\), on intervals in \(t>0\), then use Fubini. For a singular but integrable power at \(u=0\), apply the same calculation off a deleted neighborhood and pass to the limit in the absolute integrals. The exponents \(\operatorname{Re}\alpha>-1\) make that passage integrable.

The positive square root is continuous by its monotonicity and the identity
\[
\sqrt{y+h}-\sqrt y=\frac{h}{\sqrt{y+h}+\sqrt y}.
\]
The denominator stays bounded below near \(y>0\). Dividing by \(h\) gives \(v'=1/(2v)\); induction through this formula proves smoothness on the positive half-line.

For \(\psi\in C_c^\infty(\{t>0\})\), extend
\[
G_\psi(u,x)=
\frac{\psi(\sqrt{u+|x|^2},x)}{2\sqrt{u+|x|^2}}
\]
by zero wherever the square root is unavailable. This extension is smooth and compactly supported: the original support has \(t\ge d>0\), so the expression is identically zero near \(u+|x|^2=0\); its coordinate image is compact. Differentiation under its compact \(x\)-integral is the parameter-integral proof in U021 B1. Define the distribution \(\chi_+^\alpha(q)\) on \(t>0\) by
\[
\chi_+^\alpha(q)(\psi)
=\chi_+^\alpha\!\left(u\longmapsto\int G_\psi(u,x)\,dx\right).
\]
The transformed test has common compact support and bounded seminorms when \(\psi\) does, so this is a distribution and is entire in \(\alpha\). In the integrable range it agrees with the ordinary composed density. Comparison with (1.2) and the scalar identity principle therefore give, on \(t>0\),
\[
R_\lambda=
\frac{2^{1-2\lambda}\pi^{(1-n)/2}}{\Gamma(\lambda)}
\chi_+^{\,\lambda-(n+1)/2}(q)
\]
for every parameter, with \(1/\Gamma\) understood as its entire reciprocal. No critical substitution at the vertex is required.

**Theorem 1.1 (forward inverses of the wave powers).** For every integer \(k\ge0\), the unique \(C_+\)-supported fundamental solution of \(\Box^{k+1}\) is
\[
E_k=R_{k+1}.
\tag{1.3}
\]
On \(t>0\) it has the cone expression
\[
\begin{gathered}
E_k=\kappa_{k,n}\chi_+^{\,k+(1-n)/2}(q),\\
\kappa_{k,n}=\frac{2^{-2k-1}\pi^{(1-n)/2}}{k!}.
\end{gathered}
\tag{1.4}
\]
At the vertex, the notation means the global distribution (1.3). If \(n=2m+1\) is odd and \(k<m\), its exact support is the closed boundary \(\partial C_+\). In every other case its exact support is \(C_+\). For every cone-supported distribution \(g\), \(E_k*g\) is the unique cone-supported solution of \(\Box^{k+1}u=g\).

**Proof: normalization and support.** Iterating (1.1) gives \(\Box^{k+1}R_{k+1}=R_0=\delta\). The preceding coordinate construction and \(\Gamma(k+1)=k!\) give (1.4), with its exact coefficient.

For odd \(n=2m+1\) and \(k<m\), the scalar factor is
\(\chi_+^{k-m}=\delta_0^{(m-k-1)}\).
It is zero off \(q=0\). It is nonzero in every neighborhood of a nonvertex cone point: choose small coordinate product supports about \((u,x)=(0,x_0)\), and a test of the form
\[
\psi(t,x)=2t\,h(t^2-|x|^2)\rho(x)
\]
within that chart, where \(\int\rho=1\) and \(h^{(m-k-1)}(0)\ne0\). Its transformed integrated test is \(h\), so its pairing is a nonzero multiple of that derivative. Such supports can be arbitrarily small, because \(x_0\ne0\) and \(t_0=|x_0|>0\). A cutoff supported in \(t>0\) and equal to one on the positive inverse image makes the displayed expression a global test there. This proves that all nonvertex cone points belong to the support. There is no interior support, by the same scalar formula; off-cone support is excluded by (1.1). Closedness adds the vertex and gives exactly \(\partial C_+\).

For the remaining odd-dimensional cases the scalar exponent in (1.4) is a nonnegative integer, so the interior density is nonzero. For even \(n\) it is a half-integer. The reciprocal Gamma normalization is nonzero at that half-integer plus one, since its only zeros are the nonpositive integers, as proved in the supplied Gamma foundation. Thus the interior density is again a nonzero smooth power. A nonzero continuous density has nonzero pairing on a sufficiently small nonnegative bump after multiplication by a constant phase; hence every interior point belongs to the support. Closedness and (1.1) give \(C_+\).

**Proof: all sources and uniqueness.** Addition of any finite number of points in \(C_+\) is proper. If their sum lies in a compact set with time coordinate at most \(T\), each summand satisfies \(0\le t_j\le T\) and \(|x_j|\le t_j\). The contributing set is bounded and closed, hence compact. The triangle inequality gives \(C_++C_+=C_+\). Consequently all convolutions below exist and stay in this cone, by U021.

If \(V\) is another cone-supported fundamental solution, derivative transfer and the identity distribution give
\[
\begin{aligned}
V&=(\Box^{k+1}E_k)*V\\
 &=E_k*(\Box^{k+1}V)=E_k.
\end{aligned}
\tag{1.5}
\]
For arbitrary cone-supported \(g\), derivative transfer gives \(\Box^{k+1}(E_k*g)=g\). Any other solution \(w\) satisfies \(w=E_k*(\Box^{k+1}w)=E_k*g\) by the same calculation. This also proves uniqueness for zero source. \(\square\)

**Example 1.2 (a regular wave power on the line).** When \(n=1\), formula (1.2) applies to every \(\lambda=k+1\), and Gamma recurrence gives
\[
E_k=\frac{\boldsymbol1_{C_+}q^k}{2^{2k+1}(k!)^2}.
\]
In particular \(E_0=\boldsymbol1_{C_+}/2\). The boundary \(t=|x|\) has measure zero by Fubini, since every fixed-\(x\) time section is a singleton. Values of this density there can therefore be chosen arbitrarily without changing its distribution. The source equation holds globally.

## A convergent series for every complex potential

**Theorem 2.1 (the entire mass family).** For every \(a\in\mathbb C\), the series
\[
F_a=\sum_{j=0}^{\infty}a^jR_{j+1}
\tag{2.1}
\]
converges as distributions, locally uniformly in \(a\), also uniformly on bounded families of tests. Each test pairing is entire. Every parameter derivative is a distribution with a common finite test-order bound on fixed compact spacetime and parameter sets. The sum is the unique \(C_+\)-supported fundamental solution of \(\Box-a\). For every cone-supported \(g\), \(F_a*g\) is the unique cone-supported solution of \((\Box-a)u=g\).

**Proof: a majorant for the densities.** Set \(J=n\). For \(j\ge J\), the initial density formula applies and gives
\[
\begin{gathered}
R_{j+1}=C_{j,n}\boldsymbol1_{\{t>|x|\}}q^{\alpha_j},
\qquad \alpha_j=j+(1-n)/2,\\
C_{j,n}=\frac{2^{-2j-1}\pi^{(1-n)/2}}
{j!\Gamma(j+(3-n)/2)}.
\end{gathered}
\tag{2.2}
\]
The exponent is positive, since \(\alpha_J=(n+1)/2\); extension by zero is continuous, including at the vertex. All \(C_{j,n}\) in this range are positive, because the Gamma integral is positive on the positive real axis.

Let a compact test region \(K\) be contained in \(|t|\le T\), let \(L=\max(1,T^2)\), and bound \(|a|\le A\) with \(A\ge1\). On \(K\cap C_+\), \(0\le q\le L\). Put \(b_J=(n+3)/2\), and use the notation \((b)_0=1\), \((b)_m=b(b+1)\cdots(b+m-1)\). Gamma recurrence gives
\[
\begin{gathered}
D=A^JC_{J,n}L^{\alpha_J},\\
\sup_K|a^{J+m}R_{J+m+1}|
\le D\frac{(AL/4)^m}{(J+1)_m(b_J)_m}.
\end{gathered}
\tag{2.3}
\]
The ratio of consecutive majorant terms is
\[
\frac{AL}{4(J+m+1)(b_J+m)}\longrightarrow0.
\tag{2.4}
\]
In particular this ratio is at most \(1/2\) after some index. The remaining tail is bounded by a convergent geometric series, whose tails tend uniformly to zero. Multiplication by any fixed power of \(J+m\) preserves this conclusion: its extra successive ratio tends to one. This proves uniform convergence of the continuous density tail on \(K\) and on the parameter disk.

**Proof: differentiation and distribution bounds.** For an integer \(r\ge0\),
\[
\partial_a^r a^j=
\begin{cases}
j(j-1)\cdots(j-r+1)a^{j-r},&j\ge r,\\
0,&j<r.
\end{cases}
\]
For \(A\ge1\), its modulus on the disk is at most \(j^r A^j\) when \(j\ge r\); the finitely many small indices cause no issue. Hence (2.3) with a polynomial factor bounds every derivative series uniformly on compact disks and compact spacetime sets.

Here is the differentiation justification. Each polynomial partial sum \(S_N\), paired with a fixed test, satisfies
\[
S_N(a+h)-S_N(a)
=h\int_0^1 S_N'(a+\theta h)\,d\theta.
\]
The segment lies in a slightly larger fixed disk. Uniform convergence of both series permits passage to the limit in this finite integral. Its derivative-series limit is continuous, so division by \(h\) and passage \(h\to0\) gives the complex derivative. Apply the same argument to every derivative series. Thus every pairing is entire, with exactly the indicated derivatives.

The tail and all its parameter derivatives have pairing bounded by their uniform density bound times the volume of a fixed box containing \(K\), times \(\|\psi\|_\infty\). Each of the finitely many distributions \(R_{j+1}\), \(j<J\), has some finite order on that compact box, by U008 Proposition 1.2 (also directly by the continuation in U027). Choose their maximum order \(M\). Their finitely many polynomial coefficients and each fixed derivative are uniformly bounded on the parameter disk. Adding these estimates proves
\[
|\partial_a^r F_a(\psi)|\le C_{K,A,r}\|\psi\|_{C^M},
\qquad \operatorname{supp}\psi\subset K,\quad |a|\le A.
\]
The same \(M\) works for all \(r\), though the constant depends on \(r\). Density-tail errors tend to zero times \(\|\psi\|_\infty\), which proves convergence on bounded test families. Low-index kernels need not be densities.

**Proof: the equation retains the vertex.** At every finite \(N\), (1.1) gives the telescoping identity
\[
(\Box-a)\sum_{j=0}^Na^jR_{j+1}
=\delta_{(0,0)}-a^{N+1}R_{N+1}.
\tag{2.5}
\]
Indeed the differentiated \(j=0\) term is \(R_0\), and the differentiated \(j\)-th term for \(j\ge1\) cancels the potential term with index \(j-1\). For \(N\ge J\), the last term is bounded by \(A\) times the \(j=N\) term of (2.3), and tends to zero on compact tests uniformly in the parameter disk. Differentiation of distributions is continuous because it differentiates the fixed test. Taking the limit proves \((\Box-a)F_a=\delta\). Tests outside \(C_+\) pair to zero with every summand and therefore with the limit.

Proper convolution gives \((\Box-a)(F_a*g)=g\). If \(w\) is another cone-supported solution, then
\[
w=((\Box-a)F_a)*w=F_a*((\Box-a)w)=F_a*g.
\tag{2.6}
\]
This proves the asserted uniqueness and all source equations. \(\square\)

**Corollary 2.2 (resolvents and higher powers).** For all \(a,b\in\mathbb C\),
\[
\begin{gathered}
F_a-F_b=(a-b)F_a*F_b,\\
\partial_aF_a=F_a*F_a,\qquad
\left.\partial_a^kF_a\right|_{a=0}=k!R_{k+1}.
\end{gathered}
\tag{2.7}
\]
For every integer \(k\ge0\),
\[
G_{a,k}=\frac1{k!}\partial_a^kF_a
\tag{2.8}
\]
is the unique cone-supported fundamental solution of \((\Box-a)^{k+1}\).

**Proof.** The equation for \(F_b\) gives
\((\Box-a)F_b=\delta+(b-a)F_b\).
Convolve with \(F_a\). Derivative transfer turns the left side into \(F_b\); the right side is \(F_a+(b-a)F_a*F_b\). Rearrangement proves the first identity. For \(b\ne a\), divide by \(a-b\) and let \(b\to a\). The proved parameter derivative gives the left limit, while U021 Theorem 3.2, on the fixed cone supports, gives \(F_a*F_b\to F_a*F_a\). Termwise differentiation at zero proves the third identity.

Differentiating the source equation \(k\) times, by the established test bounds, gives
\[
(\Box-a)\partial_a^kF_a
=k\,\partial_a^{k-1}F_a\qquad(k\ge1).
\]
Iteration and \((\Box-a)F_a=\delta\) give \((\Box-a)^{k+1}\partial_a^kF_a=k!\delta\). Every parameter derivative still has cone support, because its pairings outside the cone differentiate the identically zero scalar function. The inverse-convolution argument of (2.6) with the higher operator proves uniqueness. \(\square\)

**Example 2.3 (an entire cosine without a square-root choice).** Define
\[
\mathcal C(z)=\sum_{j=0}^{\infty}\frac{z^j}{(2j)!}.
\]
The factorial ratio proves locally uniform convergence of every derivative series as above, so this is entire. Gamma recurrence and \(\Gamma(1/2)=\sqrt\pi\) give, by induction,
\[
\Gamma(j+1/2)=\frac{(2j)!\sqrt\pi}{4^j j!}.
\]
For \(n=2\), all \(R_{j+1}\), including \(j=0\), lie in the initial integrable range. Substituting this Gamma formula gives
\[
F_a=\frac{\boldsymbol1_{\{t>|x|\}}}{2\pi\sqrt q}\mathcal C(aq).
\tag{2.9}
\]
Indeed the \(j\)-th coefficient is \(q^{j-1/2}/[2\pi(2j)!]\). On a compact set the entire series is bounded uniformly for bounded \(a\). The leading \(q^{-1/2}\) density is locally integrable, including at the vertex, by U027's actual integrability proof. Dominated convergence therefore identifies the distributional sum with (2.9). For \(a=-\mu^2\), the scalar cosine series gives \(\mathcal C(aq)=\cos(\mu\sqrt q)\) inside the cone. Formula (2.9) itself uses no square root of \(a\).

## Complex drift is exponential conjugation

**Theorem 3.1 (all constant complex drift coefficients).** For arbitrary \(b_0,\ldots,b_n,c\in\mathbb C\), let
\[
P=\Box+2b_0\partial_t+2\sum_{j=1}^nb_j\partial_{x_j}+c.
\]
Set
\[
\ell(t,x)=-b_0t+\sum_{j=1}^nb_jx_j,\qquad
a=b_0^2-\sum_{j=1}^nb_j^2-c.
\tag{3.1}
\]
Then the unique \(C_+\)-supported fundamental solution is
\[
F=e^\ell F_a.
\tag{3.2}
\]
The squares in (3.1) are complex bilinear squares, without conjugation.

**Proof.** For a smooth \(h\), define \(hT(\psi)=T(h\psi)\). The product rule and compact bounds on derivatives of \(h\) preserve the finite test estimate, so this is a distribution. Directly on tests,
\[
\partial_j(hT)(\psi)=-T(h\partial_j\psi)
=\bigl(h\partial_jT+(\partial_jh)T\bigr)(\psi).
\]
The scalar exponential identities give \(e^\ell e^{-\ell}=1\) and its usual derivatives, even for complex coefficients. Applying the proved product rule twice yields
\[
\begin{aligned}
e^{-\ell}P(e^\ell T)
&=(\partial_t-b_0)^2T-\sum_j(\partial_{x_j}+b_j)^2T\\
&\quad+2b_0(\partial_t-b_0)T
+2\sum_jb_j(\partial_{x_j}+b_j)T+cT\\
&=(\Box-a)T.
\end{aligned}
\tag{3.3}
\]
The time first derivatives cancel, as do the spatial first derivatives. The remaining scalar is \(c-b_0^2+\sum_jb_j^2=-a\). Thus \(P(e^\ell F_a)=e^\ell\delta=\delta\), since \(\ell(0)=0\).

Multiplication by a nowhere-zero smooth function preserves support: multiplication cannot enlarge support, and the inverse multiplier proves the reverse inclusion. Every other forward inverse becomes an inverse of \(\Box-a\) after multiplication by \(e^{-\ell}\), so Theorem 2.1 proves uniqueness. This includes \(a=0\) and imposes no boundedness condition on the exponential. \(\square\)

## Time slices and the full smooth Cauchy problem

The flat Fourier–Laplace and Volterra arguments below use the programme proof of AN-03, *Causal kernels, initial data, and short-time geometry*, Sections 2, 2.1 and 4. The half-space convolution argument uses *The wave Cauchy problem and Kirchhoff's formula*, Theorem 1. Here the inputs are the already constructed \(R_\lambda\) and the proved convolution rules above. The normalization remains \(E_k=R_{k+1}\); the AN-03 convention for its index \(k\) is \(k!R_{k+1}\).

For Fourier inversion we use the supplied [Schwartz and Fourier foundations](../prerequisites/U011-free-foundations/schwartz-fourier-foundations-U017.md), F1–F5, with transform \(e^{-ix\cdot\xi}\). All sphere integrals use the surface measure of the angular foundation.

**Lemma 4.1 (compact spatial slices).** Fix \(c>0\), put
\[
 P_c=c^{-2}\partial_t^2-\Delta_x,\qquad
 C_c=\{(t,x):t\ge0,\ |x|\le ct\},\qquad
 K_c(t,x)=cE_0(ct,x).
 \tag{T1}
\]
Then \(P_cK_c=\delta_{(0,0)}\). On \(t\ge0\), \(K_c\) is a smooth one-sided family of compactly supported spatial distributions, with
\[
 \begin{split}
 \operatorname{supp}K_c(t)&\subset\{|x|\le ct\},\\
 K_c^{(2j)}(0)&=0,\\
 K_c^{(2j+1)}(0)&=c^{2j+2}\Delta_x^j\delta_0\quad(j\ge0).
 \end{split}
 \tag{T2}
\]
Every fixed time derivative has a common finite spatial test-order bound on a bounded time interval. The family is smooth in the strong topology of \(\mathcal E'\), not just pointwise on individual tests. Its extension by zero to negative time is exactly the spacetime distribution in (T1).

**Proof of the Fourier identification.** We first retain the full transform computation for the existing family. For real \(s>0\), its initial integrable range in (1.2) gives
\[
 \mathcal F_x\mathcal L_t R_\lambda(s,\xi)
          =(s^2+|\xi|^2)^{-\lambda}.
 \tag{T3}
\]
At \(\xi=0\), polar integration, (4.0a)–(4.0b) of U027 and the gamma integral give \(s^{-2\lambda}\): the spatial integral leaves
\(2^{1-2\lambda}\sqrt\pi/[\Gamma(\lambda)\Gamma(\lambda+1/2)]\),
and duplication identifies it with \(1/\Gamma(2\lambda)\).
To include frequency, replace \(-i\xi\) temporarily by a real \(\eta\) with \(|\eta|<s\). Rotate \(\eta\) to the first axis. In the \((t,x_1)\) plane put
\[
 t=a u+b v,\qquad x_1=b u+a v,\qquad
 a=\frac{s}{\sqrt{s^2-|\eta|^2}},\quad
 b=\frac{|\eta|}{\sqrt{s^2-|\eta|^2}} .
 \tag{T4}
\]
This change has determinant one, preserves \(t^2-|x|^2\) and the forward cone, and changes \(st-\eta\cdot x\) into \(\sqrt{s^2-|\eta|^2}\,u\). The cone is preserved because \(a>|b|\), and its inverse has the same property. The integral is therefore \((s^2-\eta\cdot\eta)^{-\lambda}\).

For complex \(\eta=A+iB\) with \(|A|<s\),
\[
 \operatorname{Re}(s^2-\eta\cdot\eta)
                 =s^2-|A|^2+|B|^2>0.
 \tag{T5}
\]
This fixes the right-half-plane logarithm. On each compact subset of this tube, the integrand and every fixed parameter derivative have an integrable bound by the original cone power times a polynomial in \(t\) and \(e^{-\epsilon t}\). Near zero the initial positive power margins and their logarithmic estimates are the ones in U027. Dominated difference quotients prove holomorphy. Successive one-variable identity theorems on product disks, followed along a finite chain covering any compact path in this convex tube, extend the real identity to \(\eta=-i\xi\). This proves (T3).

For a complex time parameter \(p=s+i\sigma\), \(s>0\), \(p^2+|\xi|^2\) avoids the nonpositive real axis: if its imaginary part \(2s\sigma\) is zero, then \(\sigma=0\) and it is positive. The same exponential bounds and one-variable identity theorem extend (T3) to \(p\). Consequently
\[
 \mathcal F_{t,x}(e^{-st}R_\lambda)(\sigma,\xi)
       =\big((s+i\sigma)^2+|\xi|^2\big)^{-\lambda}.
 \tag{T6}
\]
Here damping is defined on tempered distributions by replacing \(e^{-st}\) outside \(t\ge-1\) by \(\theta(t)e^{-st}\), where \(\theta=0\) for \(t\le-2\). This function and all its derivatives are bounded, so the full product rule makes it a Schwartz multiplier. It agrees with the exponential near the cone; different choices have identical action on \(R_\lambda\).

The continuation \(R_\lambda=\Box^N R_{\lambda+N}\) retains (T6), since on cone-supported distributions
\[
 e^{-st}\Box T=((\partial_t+s)^2-\Delta_x)(e^{-st}T).
 \tag{T7}
\]
Thus Fourier differentiation contributes precisely the polynomial in (T6). All parameter continuations are justified on a compact parameter set by one choice of \(N\) and the existing uniform Schwartz bounds. We use (T6) at \(\lambda=1\).

Set \(S(t,\omega)=\sin(t\omega)/\omega\), with \(S(t,0)=t\). Two integrations by parts in its differential equation give
\[
 \int_0^\infty e^{-pt}S(t,\omega)\,dt
                =(p^2+\omega^2)^{-1}\qquad(\operatorname{Re}p>0).
 \tag{T8}
\]
Indeed \(S(0,\omega)=0\), \(S_t(0,\omega)=1\), \(S_{tt}=-\omega^2S\); the exponential removes the infinite endpoint. For real \(t,\omega\),
\[
 S(t,\omega)=t\int_0^1\cos(rt\omega)\,dr .
 \tag{T9}
\]
All time derivatives on \(|t|\le T\) have bounds \(C_{j,T}(1+|\omega|)^j\). Inverse spatial Fourier pairing with a Schwartz test is consequently absolutely convergent, and differentiation of every fixed order is justified by a common integrable envelope. This defines a smooth distribution-valued family \(k(t)\) on the whole real line. Its causal extension, obtained by integrating those pairings over \(t\ge0\), is tempered; the bound \(|S|\le t\) suffices.

Damping this causal extension gives the transform in (T6) at \(\lambda=1\), by (T8) and absolute integration against Schwartz tests. Full spacetime Fourier injectivity identifies it with \(R_1=E_0\), including time zero. No term supported on that plane has been omitted. The transform of \(k(t)\) is therefore \(S(t,|\xi|)\). Its initial even derivatives vanish, and its derivative of order \(2j+1\) has transform \((-1)^j|\xi|^{2j}\), the transform of \(\Delta^j\delta_0\).

**Proof of the support and topology assertions.** The spacetime support of \(E_0\) is \(C_+\). If a spatial test is supported outside the ball of radius \(t_0>0\), the joint distribution pairs to zero against its product with every time test in a small interval around \(t_0\). Continuity of the scalar pairing with \(k(t)\) forces that pairing to vanish there: a continuous nonzero scalar value would have a small test of matching phase with nonzero integral. Localizing the spatial test proves \(\operatorname{supp}k(t_0)\subset\{|x|\le t_0\}\). The assertion at zero follows from the computed trace. The same argument on an open interval disjoint from the cone proves this support bound for each time derivative as well.

On \(0\le t\le T\), all supports lie in a fixed ball. Choose a compact smooth cutoff equal to one on a larger ball. Pairing with any smooth function equals pairing with this cutoff times that function. Repeated integration by parts in the latter's Fourier transform bounds it by \(C_M(1+|\xi|)^{-M}\) times finitely many derivatives on the cutoff support. Choose \(M>n+j\). Together with (T9) this gives a common finite-order bound for \(k^{(j)}(t)\), uniform in \(t\), and dominated differentiation. Taylor's formula with the next two time derivatives makes its difference-quotient remainder tend to zero uniformly on every bounded set of smooth functions. This is precisely strong \(\mathcal E'\) smoothness, also one-sided at zero.

Finally \(K_c(t)=c k(ct)\), so its time derivative of order \(j\) is \(c^{j+1}k^{(j)}(ct)\). The linear distributional chain rule and \(\delta(ct)=c^{-1}\delta(t)\) give (T1)–(T2) with all speed factors. \(\square\)

**Lemma 4.2 (uniqueness on a time half-space).** For every distribution \(g\) supported in \(H=\{t\ge0\}\), there is exactly one distribution \(u\) supported in \(H\) with \(P_cu=g\), namely \(u=K_c*g\). In particular \(K_c\) is the only half-space-supported fundamental solution.

**Proof.** Addition is proper on \(C_c\times H\). If \((a,b)+(s,y)\) lies in a compact set, then \(a,s\ge0\) and their bounded sum bounds both times. The cone bound \(|b|\le ca\) bounds \(b\), and the bounded spatial sum bounds \(y\). The contributing set is closed and bounded, hence compact. U021's properly supported convolution therefore exists, and its support stays in \(H\). Derivative transfer gives \(P_c(K_c*g)=g\). For any half-space-supported \(u\) the same properness gives
\[
 u=(P_cK_c)*u=K_c*(P_cu).
 \tag{T10}
\]
This proves uniqueness without any condition at spatial infinity. Reflection \(t\mapsto-t\), whose absolute determinant is one, proves the backward version. \(\square\)

**Theorem 4.3 (smooth data and forcing).** Let \(u_0,u_1\in C^\infty(\mathbb R^n)\) and \(f\in C^\infty(H)\), where all derivatives have continuous one-sided boundary values. Then the unique smooth solution on \(H\) of
\[
 P_cu=f,\qquad u(0)=u_0,\qquad \partial_tu(0)=u_1
 \tag{T11}
\]
is
\[
 \begin{split}
 u(t)&=c^{-2}K_c'(t)*u_0+c^{-2}K_c(t)*u_1\\
     &\quad+\int_0^t K_c(t-s)*f(s)\,ds .
 \end{split}
 \tag{T12}
\]
These convolutions are spatial. Compact support or growth assumptions on the data are unnecessary.

**Proof of smoothness.** For \(0\le t\le T\) and \(x\) in a compact set \(B\), every spatial pairing in (T12) uses data in the fixed compact set \(B+\overline B(0,cT)\). Insert a cutoff on a larger compact neighborhood. The finite-order estimates and strong smoothness of Lemma 4.1, with the smooth dependence of translated data on \(s,x\), permit differentiation of each pairing of every fixed order, uniformly on this compact parameter set. This can also be checked directly by the product difference quotient: the distribution difference is uniformly controlled on bounded smooth tests, and the test difference converges in every required finite norm. For the integral, write \(s=tr\), \(0\le r\le1\). Its smooth tested integrand on the fixed compact parameter domain proves smoothness through \(t=0\) by dominated differentiation. The argument only uses one-sided derivatives of \(f\).

For its forcing term \(v\), differentiation with the original integration endpoints gives
\[
 \begin{split}
 v'(t)&=\int_0^t K_c'(t-s)*f(s)\,ds,\\
 v''(t)&=c^2f(t)+\int_0^t K_c''(t-s)*f(s)\,ds .
 \end{split}
 \tag{T13}
\]
The first boundary term is zero because \(K_c(0)=0\); the second is \(c^2f(t)\) because \(K_c'(0)=c^2\delta\). For \(t\ge0\) the smooth family satisfies \(K_c''=c^2\Delta K_c\): it holds for \(t>0\) by the source equation and extends to zero by continuity. Thus \(P_cv=f\), and the two data terms in (T12) solve the homogeneous equation. Equations (T2) give \(u(0)=u_0\), \(u'(0)=u_1\), since \(K_c''(0)=0\), \(v(0)=v'(0)=0\).

For uniqueness let \(w\) be the difference of two solutions, with zero forcing and initial traces, and extend it by zero to \(t<0\). Two integrations by parts in time, against a compact spacetime test, give in general
\[
 P_c(\boldsymbol1_Hw)
   =\boldsymbol1_H P_cw
       +c^{-2}\delta(t)w_t(0,x)
       +c^{-2}\delta'(t)w(0,x).
 \tag{T14}
\]
Each boundary coefficient occurs once. Spatial derivatives have no boundary term on a time plane. In the present case the right side is zero, and Lemma 4.2 forces the zero extension to vanish. The same argument proves uniqueness already among \(C^2(H)\) solutions. \(\square\)

## Reading a spatial slice of the cone

The earlier cone formula and Lemma 4.1 give, for \(t>0\),
\[
 K_c(t,x)=\frac c2\pi^{(1-n)/2}
                   \chi_+^{(1-n)/2}(c^2t^2-|x|^2).
 \tag{T15}
\]
This spatial composition is legitimate. Near its zero sphere the argument is a submersion; near its only critical point \(x=0\) the argument is positive and the scalar distribution is smooth. These two definitions agree on their overlap by the ordinary submersion construction. Integrating against a time test recovers the existing spacetime formula in positive time, so continuity of the slices identifies it with \(K_c(t)\).

For \(n=1,2\) this directly yields the integrable formulas
\[
 \begin{split}
 \langle K_c(t),\psi\rangle
   &=\frac c2\int_{|x|<ct}\psi(x)\,dx &&(n=1),\\
 \langle K_c(t),\psi\rangle
   &=\frac c{2\pi}\int_{|x|<ct}
         \frac{\psi(x)}{\sqrt{c^2t^2-|x|^2}}\,dx &&(n=2).
 \end{split}
 \tag{T16}
\]
For \(n=2m+1\), \(m\ge1\), define near \(r=ct>0\)
\[
 \widetilde\psi(r)=r^{n-2}\int_{S^{n-1}}\psi(r\omega)\,d\omega .
\]
Polar integration and \(z=c^2t^2-r^2\) give a factor \(1/2\); the scalar factor is \(\delta^{(m-1)}(z)\). Its distributional sign cancels the sign from differentiating \(c^2t^2-z\). Hence
\[
 \langle K_c(t),\psi\rangle
   =\frac{c}{4\pi^m}
       \left(\frac{d}{da}\right)^{m-1}
          \widetilde\psi(\sqrt a)\Big|_{a=c^2t^2}.
 \tag{T17}
\]
Only a neighborhood of the positive radius is differentiated; no singular substitution at \(r=0\) is involved. In particular for \(n=3\),
\[
 (K_c(t)*g)(x)
       =\frac{c^2t}{4\pi}\int_{S^2}g(x+ct\omega)\,d\omega.
 \tag{T18}
\]
Reflection of the sphere changes the convolution's minus displacement to plus. For odd \(n>1\) only the sphere is sampled; in two dimensions the disk interior also contributes. The initial traces in (T2) govern the common limit as that sphere or disk collapses.

![The compact portion of a backward cone that determines a wave at a fixed event.](../figures/wave-cauchy-domain.png)

*A one-dimensional spatial section with \(c=2\), observation time \(t=1\) and observation point \(x=0\). At intermediate time \(s\), forcing is paired only in \(|y|\le2(1-s)\); initial data are paired in \([-2,2]\). Theorem 4.3 and Solution 10 prove the support statement in every dimension. The shaded triangle describes a possible dependence region; it does not assert that an odd-dimensional wave kernel has an interior density. [Reproducible figure source](../figures/wave-cauchy-domain.py).*

## Exercises

**Exercise 1 (basic: an oscillating interior).** In two spatial dimensions find the forward fundamental solution of \(\Box+4\) and determine its exact support.

**Exercise 2 (intermediate: a third wave power).** In one spatial dimension compute the forward inverse of \(\Box^3\), and compare it with \(\left.\partial_a^2F_a\right|_{a=0}\).

**Exercise 3 (intermediate: a convolution resolvent).** Express \(F_3*F_{-2}\) without a convolution integral. Find the coincident-parameter limit.

**Exercise 4 (advanced: two singular cone layers).** In five spatial dimensions compute the two singular layers of \(F_a\) off the vertex and its complete regular tail. Determine the exact support for zero and nonzero \(a\).

**Exercise 5 (intermediate: complex squares).** In two spatial dimensions set \(b_0=i\), \(b_1=1+i\), \(b_2=2\), \(c=3i\). Find the forward inverse and the exact source error obtained by replacing the coefficient squares with absolute squares.

**Exercise 6 (intermediate: an anisotropic cone).** Find the forward fundamental solution of
\[
\partial_t^2-4\partial_{x_1}^2-9\partial_{x_2}^2-a
\]
with support in \(t\ge\sqrt{x_1^2/4+x_2^2/9}\). Include its point-source Jacobian and an explicit density.

**Exercise 7 (advanced: convolution of wave powers).** For integers \(p,q\ge0\), prove \(E_p*E_q=E_{p+q+1}\). In one spatial dimension compute \(E_0*E_0\) directly in characteristic coordinates.

**Exercise 8 (advanced: a third massive power).** In two spatial dimensions compute the unique forward inverse of \((\Box-a)^3\) in terms of \(\mathcal C\), and verify its normalization at \(a=0\).

**Exercise 9 (basic: spatially constant forcing).** Take \(u_0=u_1=0\) and \(f(t,x)=h(t)\), with \(h\) smooth on the closed half-line. Evaluate (T12) and check both initial traces and the equation.

**Exercise 10 (intermediate: a local domain of dependence).** Suppose two data triples agree on a neighborhood of \(\{|y-x_*|\le cT\}\) at time zero and on a relative neighborhood in \(H\) of \(\{0\le s\le T,\ |y-x_*|\le c(T-s)\}\) for their forcing. Prove that their solutions agree at \((T,x_*)\). Explain why a neighborhood, rather than equality on a singular supporting surface alone, is appropriate for a distributional kernel.

**Exercise 11 (advanced: every initial jet).** Express all time derivatives of the solution at zero recursively in \(u_0,u_1\) and the boundary derivatives of \(f\). Compute the first four derivatives after \(u_0\), keeping every power of \(c\). Relate the homogeneous recursion to (T2).

## Complete solutions

**Solution 1.** Taking \(a=-4\) in (2.9) gives
\[
F_{-4}=\frac{\boldsymbol1_{\{t>|x|\}}}{2\pi\sqrt q}\cos(2\sqrt q).
\]
Local integrability and the global source equation follow from Theorem 2.1 and Example 2.3.

We justify the exact support, including the nodal issue. A nonzero entire function cannot vanish on a real interval. If it did, at an interior point of that interval its local power series could have no first nonzero coefficient: such a coefficient would factor the function into a power times a function nonzero nearby, contradicting accumulating zeros. Thus it would vanish in a disk, and U014's identity principle would make it identically zero. The local series is proved in U013 Corollary 2.2.

Here the entire function \(r\mapsto\mathcal C(-4r)\) has value one at zero, so it vanishes on no interval. Every spacetime open set meeting the cone interior contains an interval of \(q\)-values: fix its spatial coordinate and vary \(t>0\) slightly, so \(q=t^2-|x|^2\) strictly increases. Some point in that open set has nonzero density, hence some smaller bump has nonzero pairing. Every interior point is therefore in the distributional support, and its closure is \(C_+\). Nodal hypersurfaces do not remove any open subset from this support.

**Solution 2.** Example 1.2 at \(k=2\) gives
\[
E_2=\frac{\boldsymbol1_{C_+}q^2}{128},
\qquad \Box^3E_2=\delta.
\]
Equation (2.7) gives
\[
\left.\partial_a^2F_a\right|_{a=0}
=2!R_3=2E_2=\frac{\boldsymbol1_{C_+}q^2}{64}.
\]
The factor two is precisely the parameter factorial. Both densities are nonzero everywhere in the cone interior, so their exact support is \(C_+\).

**Solution 3.** Cone addition is proper, and the resolvent identity gives
\[
F_3*F_{-2}=\frac{F_3-F_{-2}}5.
\]
More generally the divided difference \((F_a-F_b)/(a-b)\) tends to \(\partial_aF_a\) as \(b\to a\). The fixed-support continuity used in Corollary 2.2 identifies this with \(F_a*F_a\). The parameter limit holds as distributions and on bounded test families by the derivative estimates and the segment-FTC argument in Theorem 2.1.

**Solution 4.** For \(n=5\), the exponent in (1.4) is \(k-2\). Consequently on \(t>0\),
\[
R_1=\frac{\delta'(q)}{2\pi^2},\qquad
R_2=\frac{\delta(q)}{8\pi^2},\qquad
R_3=\frac{\boldsymbol1_{C_+}}{64\pi^2}.
\]
The first two mean the explicitly defined noncritical coordinate distributions there. Globally they mean the already constructed \(R_1,R_2\), including their vertex behavior.

For the remaining indices write \(j=m+2\). Formula (1.2) and \(\Gamma(m+1)=m!\) give the density
\[
R_{m+3}
=\frac{2^{-2m-5}}{\pi^2(m+2)!m!}
\boldsymbol1_{C_+}q^m.
\]
Hence the full global decomposition is
\[
\begin{gathered}
F_a=R_1+aR_2+
\frac{a^2\boldsymbol1_{C_+}}{32\pi^2}\mathcal B_2(aq/4),\\
\mathcal B_2(z)=\sum_{m=0}^{\infty}\frac{z^m}{m!(m+2)!}.
\end{gathered}
\]
The ratio for successive absolute terms is \(|z|/[(m+1)(m+3)]\). The same geometric-tail and derivative argument as in Theorem 2.1 proves entire dependence and local uniform convergence. The regular tail is a locally bounded density, beginning with \(a^2\boldsymbol1_{C_+}/(64\pi^2)\).

If \(a\ne0\), its interior profile \(r\mapsto a^2\mathcal B_2(ar/4)/(32\pi^2)\) is entire and nonzero at \(r=0\). Solution 1's interval argument shows that every cone-interior open set meets a nonzero density point. The singular layers vanish in the interior, so cannot cancel this density. The exact support is \(C_+\). For \(a=0\), only \(R_1\) remains, whose exact boundary support is Theorem 1.1.

**Solution 5.** Compute the actual complex squares:
\[
\begin{gathered}
a=i^2-(1+i)^2-2^2-3i=-5-5i,\\
\ell=-it+(1+i)x_1+2x_2.
\end{gathered}
\]
The inverse is \(e^\ell F_{-5-5i}\), explicitly
\[
F=\frac{e^{-it+(1+i)x_1+2x_2}\boldsymbol1_{\{t>|x|\}}}
{2\pi\sqrt q}\mathcal C((-5-5i)q).
\]
Absolute squares instead give \(\widetilde a=1-2-4-3i=-5-3i\). For \(\widetilde F=e^\ell F_{\widetilde a}\), the true conjugation formula gives
\[
\begin{aligned}
P\widetilde F
&=e^\ell(\Box-a)F_{\widetilde a}\\
&=\delta+(\widetilde a-a)\widetilde F
=\delta+2i\widetilde F.
\end{aligned}
\]
This error is nonzero: \(F_{\widetilde a}\) has full cone support by the interval argument of Solution 1, since \(\mathcal C(0)=1\); the exponential is nowhere zero. The erroneous candidate therefore fails the point-source equation.

**Solution 6.** Let \(A(t,x_1,x_2)=(t,x_1/2,x_2/3)\), whose absolute determinant is \(1/6\). U027's proved linear distribution formula gives
\[
F_G=\tfrac16 A^*F_a,\qquad
F_G(\psi)=F_a\bigl((t,y)\longmapsto\psi(t,2y_1,3y_2)\bigr).
\]
This last pairing also verifies normalization directly. The chain rule transforms the adjoint test operator into \(\partial_t^2-\Delta_y-a\), so applying the anisotropic operator to \(F_G\) gives \(\psi(0)\). Equivalently \(A^*\delta=6\delta\), canceled by \(1/6\).

With
\[
q_G=t^2-x_1^2/4-x_2^2/9,\qquad
C_G=\{t\ge\sqrt{x_1^2/4+x_2^2/9}\},
\]
the density is
\[
F_G=\frac{\boldsymbol1_{\{t>\sqrt{x_1^2/4+x_2^2/9}\}}}
{12\pi\sqrt{q_G}}\mathcal C(aq_G).
\]
It is locally integrable by the invertible linear substitution in (2.9). The same map sends \(C_G\) onto \(C_+\), so properness of addition transfers to \(C_G\). The derivative-transfer inverse argument proves uniqueness on that cone. Its support is exactly \(C_G\), by the entire-profile argument and the homeomorphism \(A\).

**Solution 7.** Proper cone addition defines the convolution and keeps its support in \(C_+\). Transferring each group of derivatives to the designated factor gives
\[
\begin{aligned}
\Box^{p+q+2}(E_p*E_q)
&=(\Box^{p+1}E_p)*(\Box^{q+1}E_q)\\
&=\delta*\delta=\delta.
\end{aligned}
\]
Theorem 1.1 proves \(E_p*E_q=E_{p+q+1}\).

For the direct line calculation write \(\xi=t-x\), \(\eta=t+x\), so \(dt\,dx=\tfrac12d\xi\,d\eta\), \(q=\xi\eta\), and the cone is \(\xi,\eta\ge0\). The integrand for \(E_0*E_0\) is \(1/4\) where both characteristic pairs are nonnegative. For output \(\xi,\eta\ge0\), the intermediate pair ranges over \([0,\xi]\times[0,\eta]\). Thus
\[
(E_0*E_0)(t,x)
=\tfrac14\cdot\tfrac12\,\xi\eta
=\frac{\boldsymbol1_{C_+}q}{8}=E_1
\]
inside the cone, and it is zero outside. The bounded rectangle makes the integral finite; its density agrees with the distributional convolution by U021's tensor integration formula. Boundary values have measure zero and do not alter this identity.

**Solution 8.** Differentiate (2.9) twice. The derivatives are justified on compact tests by the summable series bounds already proved; equivalently the differentiated scalar series is dominated by a constant times \(q^{-1/2}\) on every compact cone region. This gives
\[
G_{a,2}=\tfrac12\partial_a^2F_a
=\frac{\boldsymbol1_{\{t>|x|\}}q^{3/2}}{4\pi}\mathcal C''(aq).
\]
Corollary 2.2 proves the global equation \((\Box-a)^3G_{a,2}=\delta\) and uniqueness with cone support.

The coefficient of \(z^2\) in \(\mathcal C\) is \(1/4!\), so \(\mathcal C''(0)=2/4!=1/12\). Hence
\[
G_{0,2}=\frac{\boldsymbol1_{C_+}q^{3/2}}{48\pi}=R_3.
\]
Indeed \(\Gamma(3)=2\), \(\Gamma(5/2)=3\sqrt\pi/4\), and (1.2) gives the same coefficient. This is the inverse with point-source coefficient one.

**Solution 9.** Pairing \(K_c(t)\) with the constant one is legitimate by compact support. Its mass \(M(t)\) satisfies \(M''(t)=c^2\langle\Delta K_c(t),1\rangle=0\); use one cutoff equal to one near all supports on each bounded time interval to justify that pairing. Equations (T2) give \(M(0)=0\), \(M'(0)=c^2\). Thus \(M(t)=c^2t\), and
\[
 u(t,x)=c^2\int_0^t(t-s)h(s)\,ds .
 \tag{T19}
\]
The first derivative is \(c^2\int_0^t h(s)\,ds\); the second is \(c^2h(t)\). The spatial Laplacian is zero and both initial traces vanish. For \(h=1\), \(u=c^2t^2/2\); removing the factor \(c^2\) would fail the normalization of \(P_c\).

**Solution 10.** Apply (T12) to the differences of the data. The supports of \(K_c(T)\) and \(K_c'(T)\) lie in \(\overline B(0,cT)\), so their pairings with the translated initial differences vanish. For each \(s\in[0,T]\), the support of \(K_c(T-s)\) lies in \(\overline B(0,c(T-s))\). The translated forcing difference vanishes on a neighborhood of this support, so the pairing is zero. Integrating gives zero forcing contribution. Compactness of the closed backward cone permits the common neighborhoods and seminorm bounds used in the proof of (T12). A distribution can differentiate a test in the normal direction: a delta derivative pairs nontrivially with a function that vanishes on its supporting surface but has nonzero normal derivative there. Vanishing in a neighborhood kills every such derivative, which is the support argument actually used.

**Solution 11.** Put \(g_j=\partial_t^ju(0)\) and \(f_j=\partial_t^jf(0)\). Smoothness through the boundary and the equation give
\[
 g_0=u_0,\quad g_1=u_1,\qquad
 g_{j+2}=c^2(\Delta g_j+f_j)\quad(j\ge0).
 \tag{T20}
\]
In particular
\[
 \begin{split}
 g_1&=u_1,\\
 g_2&=c^2\Delta u_0+c^2f_0,\\
 g_3&=c^2\Delta u_1+c^2f_1,\\
 g_4&=c^4\Delta^2u_0+c^4\Delta f_0+c^2f_2 .
 \end{split}
 \tag{T21}
\]
Induction gives the full formulas
\[
 \begin{split}
 g_{2m}&=c^{2m}\Delta^mu_0+
     \sum_{r=0}^{m-1}c^{2(m-r)}\Delta^{m-1-r}f_{2r},\\
 g_{2m+1}&=c^{2m}\Delta^mu_1+
     \sum_{r=0}^{m-1}c^{2(m-r)}\Delta^{m-1-r}f_{2r+1},
 \end{split}
 \tag{T22}
\]
where an empty sum is zero. Substituting either expression into (T20) shifts the previous terms by one Laplacian and adds the last forcing term, proving the induction. When \(f=0\), differentiating the two initial-data terms in (T12) and using (T2) gives exactly the first term in each line. These are smooth boundary jets; no convergence of a Taylor series of arbitrary smooth data is asserted.

## Sources and exact proof dependencies

- Christian Bär, Nicolas Ginoux and Frank Pfäffle, *Wave Equations on Lorentzian Manifolds and Quantization*, freely accessible author version, [arXiv:0806.1036](https://arxiv.org/abs/0806.1036), §1.2, Lemma 1.2.2 and Proposition 1.2.4, printed pp.10–17. Their parameter \(\alpha\) is \(2\lambda\) here, and their dimension is spacetime dimension \(n+1\). U027 Lemma 4.0 supplies the complete programme proof, including the scalar identities for which the source refers elsewhere.
- [U027](causal-point-sources-and-characteristic-cones.md), Lemma 4.0 and formulas (4.0a)–(4.0h), proves the global family, all constants, origin delta and linear substitution. [U022](causal-integration-of-complex-order.md), §1, formulas (1.3)–(1.6), and Theorems 2.1–2.2, proves the scalar causal family and Beta input. [U021](convolution-as-addition-of-supports.md), Theorems 1.1, 3.1–3.2 and Corollary 3.4, proves the used convolution and continuity statements.
- The nonvertex cone coordinates, exact supports, uniform potential series, all parameter derivatives, resolvent and higher-power identities, exponential conjugation and all eleven solutions are proved here. No external citation replaces these proofs.
- Lars Hörmander, *The Analysis of Linear Partial Differential Operators I*, second edition, §6.2, Theorems 6.2.3–6.2.4 and formulas (6.2.7)–(6.2.9), for the causal wave kernel and smooth Cauchy problem.
- Programme proof contributions in Lemmas 4.1–4.2: AN-03 course-writing task and OpenAI Codex, *Causal kernels, initial data, and short-time geometry*, Sections 2, 2.1 and 4 (CC0); GPT-6.1 Sol (OpenAI), *The wave Cauchy problem and Kirchhoff’s formula*, Theorem 1 (CC0). The needed arguments and exact earlier proof inputs are supplied above.