Explicit mathematical hypothesis/convention examples

Smoothness and the wave equation. The map $\pi (x,t):=x_1-ct$ is affine, hence $C^{\mathrm{\infty }}$ from [F3]; $f$ is $C^{\mathrm{\infty }}$ and $u=f\circ \pi$ is therefore $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times \mathbb{R}$ by the composition clause of [F4]. Its partial derivatives are ${\mathrm{\partial }}_{t}u=-c(f^{\mathrm{\prime }}\circ \pi )$, ${\mathrm{\partial }}_t^{2}u=c^2(f^{\mathrm{\prime }\mathrm{\prime }}\circ \pi )$ and ${\mathrm{\partial }}_{1}u=f^{\mathrm{\prime }}\circ \pi$ with ${\mathrm{\partial }}_{0}u={\mathrm{\partial }}_{2}u=0$, by the chain rule [F5] together with the scalar-multiple and linearity rules [F6] and the definitions of the partial derivatives and the gradient [F7, F8]; hence $\mathrm{\nabla }u=(0,f^{\mathrm{\prime }}\circ \pi ,0)$ and $\mathrm{\Delta }u={\mathrm{\partial }}_1{\mathrm{\partial }}_{1}u=f^{\mathrm{\prime }\mathrm{\prime }}\circ \pi$ by [F2]. Consequently $Wu=c^{-2}{\mathrm{\partial }}_t^{2}u-\mathrm{\Delta }u=f^{\mathrm{\prime }\mathrm{\prime }}\circ \pi -f^{\mathrm{\prime }\mathrm{\prime }}\circ \pi =0$ identically, with $W$ as in [F1].

The curve of part (ii). Let $c>0$, $a>0$ and $z(s):=(g(s),0,0)$ with $g(s):=\sqrt{c^2{s}^2+a^2}$. For every integer $m\ge 0$, induction using [F13] gives $(d\mathrm{/}dy{)}^m{y}^{1\mathrm{/}2}=({\prod }_{j=0}^{m-1}(1\mathrm{/}2-j))y^{1\mathrm{/}2-m}$ on $(0,\mathrm{\infty })$, with the empty product equal to $1$; each derivative is continuous by [F13], so $y\mapsto \sqrt{y}$ is $C^{\mathrm{\infty }}$ by [F3]. The polynomial $s\mapsto c^2{s}^2+a^2$ takes values in $(0,\mathrm{\infty })$ and is $C^{\mathrm{\infty }}$ by direct repeated differentiation, so $g$ is $C^{\mathrm{\infty }}$ on $\mathbb{R}$ by [F4], and $z$ is $C^{\mathrm{\infty }}$ by [F3]. The chain rule [F5] with $(\sqrt{x}{)}^{\mathrm{\prime }}=1\mathrm{/}(2\sqrt{x})$ [F13] gives $g^{\mathrm{\prime }}(s)=c^{2}s\mathrm{/}g(s)$, so $z^{\mathrm{\prime }}(s)=(c^{2}s\mathrm{/}g(s),0,0)$ [F8, F9] and $\parallel z^{\mathrm{\prime }}(s){\parallel }_2=c^2\mid s\mid \mathrm{/}g(s)$ [F10]. One has $\parallel z^{\mathrm{\prime }}(s){\parallel }_2<c$ for every $s$ because $c^2\mid s\mid <cg(s)$ is equivalent to $c^2{s}^2<c^2{s}^2+a^2$, true for $a>0$; and $c^2\mid s\mid \mathrm{/}g(s)=c\cdot \frac{c\mid s\mid }{\sqrt{c^2{s}^2+a^2}}\to c$ as $\mid s\mid \to \mathrm{\infty }$, so ${\sup }_s\parallel z^{\mathrm{\prime }}(s){\parallel }_2=c$ and there is no $v_{\ast }<c$ with $\parallel z^{\mathrm{\prime }}(s){\parallel }_2\le v_{\ast }$ for all $s$.

Events on the worldline. Let $z$ be $C^1$ with $\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }\le v_{\ast }<c$ and let $x=z(t)$ for some $t\in \mathbb{R}$; then $0=t-t=\mathrm{\mid }x-z(t)\mathrm{\mid }\mathrm{/}c$, so $s=t$ solves $t-s=\mathrm{\mid }x-z(s)\mathrm{\mid }\mathrm{/}c$. Applying [F18] at the event $(x,t)=(z(t),t)$: its unique root $s(x,t)$ satisfies $s(x,t)\le t$ with equality exactly when the observation point equals $z(t)$, which is exactly the present case, so $s(x,t)=t$ and there is no root strictly below $t$. Hence, for the model of [F19], which is defined on the complement of the worldline $\mathrm{\Gamma }=\{(z(s),s)\}$, the domain restriction $x\mathrm{\ne }z(t)$ in clause (a) is not redundant: at an event with $x=z(t)$ the retarded root below $t$ does not exist.

The nonzero level set. If $f=0$, step 1.1 gives $u=0$ and $U=\mathrm{\varnothing }$; the nonemptiness and divergence claims concern only $f\mathrm{\ne }0$. Assume $f\mathrm{\ne }0$. The profile $f$ is differentiable on $\mathbb{R}$ because it is $C^{\mathrm{\infty }}$ [F3], so it is continuous at every point by [F14]; hence $U=\{s:f(s)\mathrm{\ne }0\}$, the preimage of the open set $\mathbb{R}\setminus \{0\}$, is open, and it is nonempty because $f$ is not the zero function. For every $t\in \mathbb{R}$ the equivalence $u(x,t)\mathrm{\ne }0\iff f(x_1-ct)\mathrm{\ne }0\iff x_1-ct\in U$ exhibits the level set as $\{x:u(x,t)\mathrm{\ne }0\}=\mathbb{R}\times (ct+U)\times \mathbb{R}$; since $U\mathrm{\ne }\mathrm{\varnothing }$ and the $x_0$ and $x_2$ coordinates are unrestricted, this set is nonempty and unbounded, so for every $t$ the function $u(\cdot ,t)$ does not have compact spatial support.

The spatial $L^2$ norm diverges. Assume $f\mathrm{\ne }0$; pick $s_0$ with $f(s_0)\mathrm{\ne }0$ and put $\eta :=\mid f(s_0){\mid }^2\mathrm{/}4>0$. Continuity of $f$ at $s_0$ [F14, F15] gives $\delta >0$ with $\mid f(s)-f(s_0)\mid <\mid f(s_0)\mid \mathrm{/}2$ for $\mid s-s_0\mid <2\delta$, whence $\mid f(s)\mid \ge \mid f(s_0)\mid \mathrm{/}2$ and $\mid f(s){\mid }^2\ge \eta$ on the interval $I:=[s_0-\delta ,s_0+\delta ]$. Fix $t\in \mathbb{R}$ and $R>0$ and put $K:=(-R,R)\times (ct+I)\times (-R,R)$, so that for $x\in K$ one has $x_1-ct\in I$ and therefore $\mid u(x,t){\mid }^2=\mid f(x_1-ct){\mid }^2\ge \eta {\mathbf{1}}_K(x)$. The map $x\mapsto \mid u(x,t){\mid }^2$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3$ by the composition and product clauses of [F4] applied to the affine map $x\mapsto x_1-ct$ and the smooth profile, hence continuous [F3] and Borel measurable with ${\mathbf{1}}_K$ Borel; by [F16] both functions are measurable for the Lebesgue measure space, so the monotonicity and scalar rules [F11] give ${\int }_{{\mathbb{R}}^3}\mid u(x,t){\mid }^{2}dx\ge \eta {\lambda }_3(K)=\eta \cdot 2\delta \cdot (2R{)}^2$ by the box-measure theorem [F12], and the right-hand side exceeds every prescribed real number for $R$ large. Hence ${\int }_{{\mathbb{R}}^3}\mid u(x,t){\mid }^{2}dx=+\mathrm{\infty }$ for every $t$. Finally, the Cauchy data of $u$ at $t=0$ are the smooth functions $x\mapsto f(x_1)$ and $x\mapsto -c{f}^{\mathrm{\prime }}(x_1)$, whose first member does not have compact support as a function of $x\in {\mathbb{R}}^3$; by clauses (a) and (d) of [F20] the smooth-data clause carries no global finite-energy or decay conclusion, while the compact-support clause is stated for data vanishing off compact sets, so compact support is a genuine hypothesis of clause (d) and not a consequence of smoothness. The wave energy is separately infinite: choose $b>s_0$ outside the compact support of $f$, so $f(b)=0$ and [F21] gives a point $a_1\in (s_0,b)$ with $f^{\mathrm{\prime }}(a_1)=-f(s_0)\mathrm{/}(b-s_0)\mathrm{\ne }0$. Continuity of $f^{\mathrm{\prime }}$ gives a closed interval $I_1$ of positive length and $m_1>0$ with $\mathrm{\mid }{f}^{\mathrm{\prime }}{\mathrm{\mid }}^2\ge m_1$ on $I_1$. For this travelling wave [F21] and step 1.1 give $e=c^2\mathrm{\mid }{f}^{\mathrm{\prime }}(x_1-ct){\mathrm{\mid }}^2$, so on $(-R,R)\times (ct+I_1)\times (-R,R)$ its integral is at least $c^2{m}_1\mathrm{\mid }{I}_1\mathrm{\mid }(2R{)}^2$, which is unbounded as $R\to \mathrm{\infty }$ by [F11], [F12], [F16]. Thus failure of finite wave energy follows from the derivative profile, independently of the already proved infinite $L^2$ norm of $u$. Countable Choice is used in this step through the box-measure and measurability interfaces [F12, F16, F17].

No root at the origin for the curve of part (ii). With $z$ as in step 1.2, suppose a root $s<0$ existed for the observation $(x,t)=(0,0)$: then $-s=\mid x-z(s)\mid \mathrm{/}c=\parallel z(s){\parallel }_2\mathrm{/}c=\sqrt{c^2{s}^2+a^2}\mathrm{/}c$. Since $s<0$ the left side is positive, so squaring is legitimate and gives $s^2=(c^2{s}^2+a^2)\mathrm{/}{c}^2=s^2+a^2\mathrm{/}{c}^2$, hence $a^2=0$ and $a=0$, contradicting $a>0$. So no root $s<0$ exists at $(0,0)$ even though $z$ is $C^{\mathrm{\infty }}$ with $\parallel z^{\mathrm{\prime }}(s){\parallel }_2<c$ at every $s$; therefore the existence clause of the theorem [F19], whose hypothesis $\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }\le v_{\ast }<c$ fails for this curve, cannot be applied here, and the uniform speed bound is a genuine hypothesis. The substitution denominator degenerates in the same limit: with $x=0$ one has $n(s)=(0-z(s))\mathrm{/}\parallel z(s){\parallel }_2=-e_0$ and $\beta (s)=z^{\mathrm{\prime }}(s)\mathrm{/}c\to (-1,0,0)=-e_0$ as $s\to -\mathrm{\infty }$, so $n(s)\cdot \beta (s)\to 1$ and $\kappa (s)=1-n(s)\cdot \beta (s)\to 0$ [F6, F10].

Audit. Step 1.1 proves the smoothness and $Wu=0$ claims of (i), step 2.1 the structure, nonemptiness and unboundedness of the nonzero level set, and step 2.2 the divergence of the spatial $L^2$ norm together with the stated consequence for clauses (a) and (d) of the wave theorem; steps 1.2 and 2.3 prove that the pointwise subluminal curve without a uniform bound fails the root-existence conclusion and degenerates the denominator $1-n\cdot \beta$; and step 1.3 proves that events on the worldline have no retarded root below the observation time, so the exclusion of $\mathrm{\Gamma }$ is needed. The counterexamples (ii) and (iii) are choice-free: they use only the explicit algebraic curve, the strict inequalities and the root lemma [F18]; Countable Choice is assumed only in part (i), where it enters through the box-measure and measurability interfaces [F12, F16, F17]. No measurement, physical field identification or empirical claim is made, and no quantitative physical precision is asserted.