The three-dimensional wave Cauchy problem: Kirchhoff representation, uniqueness, domain of dependence and compact support

The setting. Let $h$ also denote its $C^{\mathrm{\infty }}$ extension to ${\mathbb{R}}^3\times J$, put $\mathrm{\Omega }:={\mathbb{R}}^3\times \mathbb{R}\times J$, and let $S^2=\{n\in {\mathbb{R}}^3:⟨n,n⟩=1\}$ carry the surface measure $d\mathrm{\Omega }$ of total mass $4\pi$ with normalised probability $\nu :=d\mathrm{\Omega }\mathrm{/}4\pi$ of [F9]. For a fixed datum $\phi \in C^{\mathrm{\infty }}({\mathbb{R}}^3)$ put $m_{\phi }(x,s,v):=\frac{1}{4\pi }{\int }_{S^2}\phi (x+sn)d\mathrm{\Omega }(n)$ (independent of $v$), so that $H_{\phi }(x,t):=t{m}_{\phi }(x,ct,v_0)$ for any fixed $v_0\in J$ is the $H_{\phi }$ of the statement; for the datum $h$ put $m(x,s,v):=\frac{1}{4\pi }{\int }_{S^2}h(x+sn,v)d\mathrm{\Omega }(n)$ on $\mathrm{\Omega }$, so that $m_{\phi }=m_{h_{\phi }}$ for the spacetime datum $h_{\phi }(x,v):=\phi (x)$. Define, for $(x,t)\in {\mathbb{R}}^3\times \mathbb{R}$ and $s:=t-t_0$, $u_h(x,t):={\mathrm{\partial }}_s{H}_a(x,s)+H_b(x,s),H(x,\sigma ;v):=\sigma m(x,c\sigma ,v),$ and, for $t\in [t_0,T]$, $u_s(x,t):=c^2{\int }_{t_0}^{t}H(x,t-v;v)dv$ and $u:=u_h+u_s$; here ${\mathrm{\partial }}_s{H}_a(x,s)$ is the derivative of $H_a$ in its second argument at $(x,s)$. All surface integrals are read under the ${\mathrm{A}\mathrm{C}}_{\omega }$ convention carried by [F3] and [F5].

Properties of the family. By [F3] applied to $h$ and to each $h_{\phi }$, the families $m$ and $m_{\phi }$ are of class $C^{\mathrm{\infty }}$ on their domains in the extension sense, and every word iterated partial derivative equals the sphere integral of the corresponding derivative of the composed integrand. In particular $m(x,0,v)=h(x,v)$ and $m_{\phi }(x,0,v)=\phi (x)$, because the sphere integral of the constant $\phi (x)$ is $\phi (x)\cdot 4\pi \mathrm{/}4\pi$; consequently $H_{\phi }(\cdot ,0)=0$ and ${\mathrm{\partial }}_t{H}_{\phi }(\cdot ,0)=\phi$ by the product rule [F2]. The reflection substitution $n\mapsto -n$ with [F9] gives $m(x,-s,v)=m(x,s,v)$ and $m_{\phi }(x,-s,v)=m_{\phi }(x,s,v)$: both families are even in $s$. Hence the Euler-Poisson-Darboux identity ${\mathrm{\partial }}_s^{2}m+\frac{2}{s}{\mathrm{\partial }}_{s}m=\mathrm{\Delta }m(s\mathrm{\ne }0)$ holds on $\mathrm{\Omega }\setminus \{s=0\}$ and likewise with $m_{\phi }$ in place of $m$: for $s>0$ it is [F4] applied to the fixed smooth datum $x\mapsto h(x,v)$ (respectively $x\mapsto \phi (x)$), and for $s<0$ it follows from the evenness just proved, since for an even $C^{\mathrm{\infty }}$ function of $s$ the derivative ${\mathrm{\partial }}_s$ is odd and ${\mathrm{\partial }}_s^2$ is even while $\mathrm{\Delta }$ differentiates only the $x$-variable.

The homogeneous blocks solve the wave equation with the prescribed data. Fix $\phi \in C^{\mathrm{\infty }}({\mathbb{R}}^3)$. By [F2] the function $H_{\phi }(x,t)=t{m}_{\phi }(x,ct,v_0)$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times \mathbb{R}$. Differentiating in $t$ and using the identity of step 1.2 at $s=ct$, one has ${\mathrm{\partial }}_t^2{H}_{\phi }=2c{\mathrm{\partial }}_s{m}_{\phi }+c^{2}t{\mathrm{\partial }}_s^2{m}_{\phi }$ and $c^2\mathrm{\Delta }{H}_{\phi }=c^{2}t\mathrm{\Delta }{m}_{\phi }=c^{2}t{\mathrm{\partial }}_s^2{m}_{\phi }+2c{\mathrm{\partial }}_s{m}_{\phi }$ for $t\mathrm{\ne }0$; at $t=0$ both sides vanish, since $H_{\phi }(\cdot ,0)=0$ gives $\mathrm{\Delta }{H}_{\phi }(\cdot ,0)=0$ and ${\mathrm{\partial }}_s{m}_{\phi }(\cdot ,0)=0$ by evenness. Hence ${\mathrm{\partial }}_t^2{H}_{\phi }=c^2\mathrm{\Delta }{H}_{\phi }$ on ${\mathbb{R}}^3\times \mathbb{R}$. Differentiating this equation in $t$ and commuting ${\mathrm{\partial }}_t$ with $\mathrm{\Delta }$ through [F2] gives the homogeneous equation also for $K_{\phi }:={\mathrm{\partial }}_t{H}_{\phi }$; its data are $K_{\phi }(\cdot ,0)=\phi$ by step 1.2 and ${\mathrm{\partial }}_t{K}_{\phi }(\cdot ,0)=c^2\mathrm{\Delta }{H}_{\phi }(\cdot ,0)=0$.

Smoothness of the source term and its extension to a neighbourhood. Put $\theta :=t-t_0$ and, for $\theta$ in an open interval $({\theta }_1,{\theta }_2)$ containing $[0,T-t_0]$ and small enough that $t_0+\theta (1-z)\in J$ for all $z\in [0,1]$, define ${\tilde{u}}_s(x,t):=c^2{\theta }^2{\int }_0^{1}zm(x,c\theta z,t_0+\theta (1-z))dz\mathrm{.}$ By step 1.2 and [F2] the integrand $A(x,\theta ,z):=c^{2}zm(x,c\theta z,t_0+\theta (1-z))$ extends to a $C^{\mathrm{\infty }}$ function on ${\mathbb{R}}^3\times ({\theta }_1,{\theta }_2)\times [0,1]$ whose word derivatives are continuous there. Iterating [F5] in the parameter directions $x_j$ and $\theta$ over the fixed compact set $[0,1]$ shows that every word derivative in $(x,\theta )$ of ${\int }_0^{1}A(x,\theta ,z)dz$ exists and equals ${\int }_0^1{D}^{w}A(x,\theta ,z)dz$; each of these integrals is jointly continuous in $p=(x,\theta )$: on a closed spatial-time box $Q\times K$ inside the parameter domain, $D^{w}A$ is continuous on the compact product $Q\times K\times [0,1]$, so [F7] gives uniform continuity there and $\mid {\int }_0^1{D}^{w}A(p,z)dz-{\int }_0^1{D}^{w}A(p_0,z)dz\mid \le {\sup }_{z\in [0,1]}\mathrm{\mid }{D}^{w}A(p,z)-D^{w}A(p_0,z)\mathrm{\mid }⟶0(p\to p_0)\mathrm{.}$ Every parameter point has such a box with that point in its interior by openness, and compactness is [F8]. Hence ${\int }_0^{1}Adz$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times ({\theta }_1,{\theta }_2)$, and so is ${\tilde{u}}_s={\theta }^2{\int }_0^{1}Adz$ by [F2]. For $\theta >0$ the substitution $v=t_0+\theta (1-z)$, $dv=-\theta dz$, in the defining integral of $u_s$ gives ${\tilde{u}}_s=u_s$, while ${\tilde{u}}_s(x,t_0)=0=u_s(x,t_0)$; hence $\tilde{u}:=u_h+{\tilde{u}}_s$ is a $C^{\mathrm{\infty }}$ extension of $u$ to the open neighbourhood ${\mathbb{R}}^3\times (t_0-\delta ,T+\delta )$ of the closed slab, and $u$ is of class $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense.

The Duhamel identity for a smooth datum. Let $g$ be smooth on ${\mathbb{R}}^3\times J_g$ with $J_g$ an open interval containing $t_0$, and let $m_g$ be its signed-radius spherical-mean family. Set $H_g(x,\sigma ;v):=\sigma m_g(x,c\sigma ,v)$ and, for $\tau >0$ with $t_0+\tau \in J_g$, $J_g^{\ast }(x,\tau ):={\int }_0^{\tau }{H}_g(x,\tau -r;t_0+r)dr={\int }_0^{\tau }\sigma m_g(x,c\sigma ,t_0+\tau -\sigma )d\sigma \mathrm{.}$ The equality follows by $\sigma =\tau -r$. For each fixed $v$, step 2.1 applied to $g(\cdot ,v)$ gives ${\mathrm{\partial }}_{\sigma }^2{H}_g=c^2{\mathrm{\Delta }}_x{H}_g$, $H_g(x,0;v)=0$ and ${\mathrm{\partial }}_{\sigma }{H}_g(x,0;v)=g(x,v)$. At fixed $x$, the integrand $F(\tau ,r)=H_g(x,\tau -r;t_0+r)$ is smooth on $\mathbb{R}\times (J_g-t_0)$ by [F3] and [F2], including negative $\tau -r$. Locally at each admissible ${\tau }_{\ast }>0$, choose an open interval $I$ about ${\tau }_{\ast }$ with $\inf I>0$ and $[0,\sup I]\subset J_g-t_0$; thus [F6] applies on a genuine rectangular neighbourhood with endpoints $0$ and $\tau$. Two applications give ${\mathrm{\partial }}_{\tau }{J}_g^{\ast }={\int }_0^{\tau }{\mathrm{\partial }}_{\sigma }{H}_g(x,\tau -r;t_0+r)dr$ and ${\mathrm{\partial }}_{\tau }^2{J}_g^{\ast }=g(x,t_0+\tau )+{\int }_0^{\tau }{\mathrm{\partial }}_{\sigma }^2{H}_g(x,\tau -r;t_0+r)dr$. At each fixed $\tau$, [F5] permits two spatial derivatives through the fixed $r$-integral, hence ${\mathrm{\partial }}_{\tau }^2{J}_g^{\ast }-c^2{\mathrm{\Delta }}_x{J}_g^{\ast }=g(x,t_0+\tau )=m_g(x,0,t_0+\tau )$. The positive-radius interval was arbitrary; no time derivative of the source is needed for these two endpoint differentiations.

The initial data. At $t=t_0$ one has $\tilde{u}(x,t_0)={\mathrm{\partial }}_s{H}_a(x,0)+H_b(x,0)+0=a(x)+0+0$, since ${\mathrm{\partial }}_s{H}_a(x,0)=m_a(x,0,v_0)=a(x)$ and $H_b(x,0)=0$ by step 2.1, while ${\tilde{u}}_s(x,t_0)=0$ by step 2.2. Likewise ${\mathrm{\partial }}_t\tilde{u}(x,t_0)={\mathrm{\partial }}_s^2{H}_a(x,0)+{\mathrm{\partial }}_s{H}_b(x,0)+{\mathrm{\partial }}_t{\tilde{u}}_s(x,t_0)=0+b(x)+0$: here ${\mathrm{\partial }}_s^2{H}_a(x,0)=c^2\mathrm{\Delta }{H}_a(x,0)=0$ and ${\mathrm{\partial }}_s{H}_b(x,0)=b(x)$ by step 2.1, and ${\mathrm{\partial }}_t{\tilde{u}}_s(x,t_0)=0$ by the product rule [F2] applied to ${\theta }^2{\int }_0^{1}A(x,\theta ,z)dz$ at $\theta =0$ in step 2.2. This proves the initial-data clause of (a).

The wave equation on the whole neighbourhood. Let $\tilde{H}$ denote the $C^{\mathrm{\infty }}$ extension of $h$ to ${\mathbb{R}}^3\times J$ and let $\tilde{u}=u_h+{\tilde{u}}_s$ be the extension of step 2.2, and put $R(x,t):={\mathrm{\partial }}_t^2\tilde{u}-c^2\mathrm{\Delta }\tilde{u}-c^2\tilde{H}(x,t)$ on ${\mathbb{R}}^3\times (t_0-\delta ,T+\delta )$; $R$ is continuous there by step 2.2 and [F2]. For $\theta >0$, [F5] gives $\mathrm{\Delta }{u}_s=c^2{\int }_{t_0}^t{\mathrm{\Delta }}_{x}H(x,t-v;v)dv$ and step 3.1 with $g=h$ and $J_g^{\ast }=u_s\mathrm{/}{c}^2$ gives ${\mathrm{\partial }}_t^2{u}_s=c^{2}h+c^2\mathrm{\Delta }{u}_s$ on ${\mathbb{R}}^3\times (t_0,T+\delta )$; the homogeneous part satisfies ${\mathrm{\partial }}_t^2{u}_h=c^2\mathrm{\Delta }{u}_h$ on all of ${\mathbb{R}}^3\times \mathbb{R}$ by step 2.1 and [F2], so $R=0$ on ${\mathbb{R}}^3\times (t_0,T+\delta )$. For $\theta <0$ put $\mu :=-\theta >0$ and $\widehat{h}(x,v):=h(x,2t_0-v)$, an extension-sense smooth datum on an interval around $t_0$, with family $\widehat{m}(x,s,v):=m(x,s,2t_0-v)$; the substitution $\sigma =\mu z$, $z\in [0,1]$, together with the evenness of step 1.2 gives ${\tilde{u}}_s(x,t_0-\mu )=c^2{\int }_0^{\mu }\sigma \widehat{m}(x,c\sigma ,t_0+\mu -\sigma )d\sigma$, so step 3.1 applied to $\widehat{h}$ gives ${\mathrm{\partial }}_t^2{\tilde{u}}_s-c^2\mathrm{\Delta }{\tilde{u}}_s=c^2\widehat{m}(x,0,t_0+\mu )=c^{2}h(x,t_0-\mu )$ on ${\mathbb{R}}^3\times (t_0-\delta ,t_0)$, while the homogeneous part satisfies the homogeneous equation there by step 2.1; hence $R=0$ on ${\mathbb{R}}^3\times (t_0-\delta ,t_0)$. The two open pieces cover every time except $t_0$; continuity gives $R(x,t_0)=0$ as well, so ${\mathrm{\partial }}_t^2\tilde{u}=c^2\mathrm{\Delta }\tilde{u}+c^2\tilde{H}$, that is $W\tilde{u}=\tilde{H}$, on the open neighbourhood ${\mathbb{R}}^3\times (t_0-\delta ,T+\delta )$; restricting gives $Wu=h$ on ${\mathbb{R}}^3\times (t_0,T)$ and indeed on the open neighbourhood of the closed time interval on which the extension of $h$ is defined.

Uniqueness. Let $v$ be of class $C^2$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense with $Wv=h$ on ${\mathbb{R}}^3\times (t_0,T)$, $v(\cdot ,t_0)=a$ and ${\mathrm{\partial }}_{t}v(\cdot ,t_0)=b$, and form the function $\tilde{v}$ equal to $v$ for $t\ge t_0$ and to $u$ for $t\le t_0$: the pieces agree in value and in all derivatives through order two at $t=t_0$: their spatial derivatives are those of $a$, their time derivatives equal $b$, and their mixed derivatives equal ${\mathrm{\partial }}_{i}b$; the second time derivative there being ${\mathrm{\partial }}_t^{2}v(\cdot ,t_0)={\mathrm{\partial }}_t^{2}u(\cdot ,t_0)=c^2\mathrm{\Delta }a+c^{2}h(\cdot ,t_0)$ by the equation of step 4.1 and the data, and each second partial derivative is continuous on its side and has the stated common value, so $\tilde{v}$ is $C^2$ on a neighbourhood of the closed slab; then $w:=u-\tilde{v}$ is $C^2$ on that neighbourhood, satisfies ${\mathrm{\square }}_{c}w=0$ on each of the open half-slabs $t>t_0$ (where both $u$ and $\tilde{v}$ solve $Wu=h$) and $t<t_0$ (where the two pieces coincide), and since ${\mathrm{\square }}_{c}w$ is continuous it vanishes at $t=t_0$ as well, so ${\mathrm{\square }}_{c}w=0$ on ${\mathbb{R}}^3\times (t_0-\delta ,T)$ for some $\delta >0$; no equation for an arbitrary extension of $v$ beyond $T$ is used; also $w(\cdot ,t_0)=0$ and ${\mathrm{\partial }}_{t}w(\cdot ,t_0)=0$ on all of ${\mathbb{R}}^3$. Fix $(x_{\ast },t_{\ast })$ with $t_0<t_{\ast }<T$ and apply [F12] in the translated time coordinates $\tau :=t-t_0$ to the pair $(w,0)$ on the closed backward cone with vertex $(x_{\ast },t_{\ast })$: the vertex time is $t_{\ast }-t_0>0$, both functions are $C^2$ on a neighbourhood of the closed cone (contained in the neighbourhood of the closed slab on which $w$ is defined), the sources agree on the cone, and the Cauchy data agree on the base ball ${\bar{B}}_{c(t_{\ast }-t_0)}(x_{\ast })$; hence $w=0$ on the cone, so $u(x_{\ast },t_{\ast })=\tilde{v}(x_{\ast },t_{\ast })=v(x_{\ast },t_{\ast })$. Since $(x_{\ast },t_{\ast })$ is arbitrary in ${\mathbb{R}}^3\times (t_0,T)$, this gives $u=v$ on the open slab. The initial data already give equality at $t_0$, and continuity at $T$ gives equality there by taking $t\uparrow T$ at each fixed $x$. Thus $u=v$ on the whole closed slab, which is clause (b).

Domain of dependence. Let $\tilde{a},\tilde{b}\in C^{\mathrm{\infty }}({\mathbb{R}}^3)$, let $\tilde{h}$ be of class $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense, and let $\tilde{u}$ be the solution produced by the construction of steps 2.2 and 4.1 applied to $(\tilde{a},\tilde{b},\tilde{h})$; by those steps $\tilde{u}$ is $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense, satisfies $W\tilde{u}=\tilde{h}$ on an open neighbourhood of the closed slab and has data $\tilde{a},\tilde{b}$ at $t_0$. Suppose $\tilde{a}=a$ and $\tilde{b}=b$ on the closed ball ${\bar{B}}_{c(t_{\ast }-t_0)}(x_{\ast })$ and $\tilde{h}=h$ on the closed cone $K(x_{\ast },t_{\ast })$ of [F11]. If $t_{\ast }=t_0$, the base ball is $\{x_{\ast }\}$ and the initial values give $u(x_{\ast },t_0)=a(x_{\ast })=\tilde{a}(x_{\ast })=\tilde{u}(x_{\ast },t_0)$. Now assume $t_{\ast }>t_0$. Then $w:=u-\tilde{u}$ is $C^2$ on a neighbourhood of the closed slab with ${\mathrm{\square }}_{c}w=c^2(h-\tilde{h})$ on each open half-slab, hence on the whole neighbourhood because both sides are continuous, it vanishes together with its time derivative at $t_0$ on the base ball, and $h-\tilde{h}=0$ on the cone; applying [F12] in translated time coordinates to the pair $(w,0)$ therefore gives $w=0$ on the closed backward cone, and in particular $u(x_{\ast },t_{\ast })=\tilde{u}(x_{\ast },t_{\ast })$. This is clause (c).

Compact support. Assume that $a$ vanishes off a compact $K_a\subseteq {\mathbb{R}}^3$, that $b$ vanishes off a compact $K_b$, that $h(\cdot ,v)$ vanishes off a fixed compact $K_h$ for every $v\in [t_0,T]$, and that $K_a\cup K_b\cup K_h\subseteq {\bar{B}}_{\rho }(0)$ for some $\rho >0$, which exists because every compact subset of ${\mathbb{R}}^3$ is bounded by [F8]. Fix $t\in [t_0,T]$ and $x$ with $\mathrm{\mid }x\mathrm{\mid }>\rho +c(t-t_0)$ and compare $u$ with the solution $\tilde{u}$ of the construction for the data $\tilde{a}=\tilde{b}=0$, $\tilde{h}=0$, namely $\tilde{u}=0$. Every $y\in {\bar{B}}_{c(t-t_0)}(x)$ satisfies $\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }y-x\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-c(t-t_0)>\rho$ by the reverse triangle inequality [F10], so $y\mathrm{\notin }{K}_a\cup K_b$ and $a(y)=b(y)=0=\tilde{a}(y)=\tilde{b}(y)$; similarly every $(y,v)\in K(x,t)$ satisfies $\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }y-x\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-c(t-v)\ge \mathrm{\mid }x\mathrm{\mid }-c(t-t_0)>\rho$, so $y\mathrm{\notin }{K}_h$ and $h(y,v)=0=\tilde{h}(y,v)$. The hypotheses of clause (c), proved in step 5.2, are therefore met at the vertex $(x,t)$ and give $u(x,t)=\tilde{u}(x,t)=0$. This is clause (d).

Conclusion. Steps 2.2 and 4.1 show that $u$ is of class $C^{\mathrm{\infty }}$ on ${\mathbb{R}}^3\times [t_0,T]$ in the extension sense and that $Wu=h$ on ${\mathbb{R}}^3\times (t_0,T)$, indeed on the open neighbourhood of the closed time interval on which the extension of $h$ is defined; step 3.2 gives $u(\cdot ,t_0)=a$ and ${\mathrm{\partial }}_{t}u(\cdot ,t_0)=b$, completing clause (a); clause (b) is step 5.1, clause (c) is step 5.2 and clause (d) is step 6.1. The argument assumed no boundary condition on a finite spatial domain, no finite-total-energy hypothesis and no spatial decay or compact support of the data; only local bounds on compact parameter sets were used.