Wave Equation Representation Formulas — Examples

Substitution. By [F1] and [F2], $u$ is $C^2$, ${\mathrm{\partial }}_{t}u=-c{F}^{\mathrm{\prime }}(x-ct)+c{G}^{\mathrm{\prime }}(x+ct)$, ${\mathrm{\partial }}_t^{2}u=c^2{F}^{\mathrm{\prime }\mathrm{\prime }}(x-ct)+c^2{G}^{\mathrm{\prime }\mathrm{\prime }}(x+ct)$ and ${\mathrm{\partial }}_x^{2}u=F^{\mathrm{\prime }\mathrm{\prime }}(x-ct)+G^{\mathrm{\prime }\mathrm{\prime }}(x+ct)$, so ${\mathrm{\partial }}_t^{2}u=c^2{\mathrm{\partial }}_x^{2}u$ on ${\mathbb{R}}^2$; at $t=0$ the displayed data are read off directly.

Check (i). If $G=0$ then $u(x,t)=F(x-ct)$; for each fixed $t$ the graph of $u(\cdot ,t)$ is the graph of $F$ translated by $ct$, so the profile moves to the right at speed $c$ with its shape unchanged.

Check (ii). The d'Alembert expression with data $(u_0,u_1)=(F+G,-c{F}^{\mathrm{\prime }}+c{G}^{\mathrm{\prime }})$ is $\frac{1}{2}(F(x-ct)+G(x-ct)+F(x+ct)+G(x+ct))+\frac{1}{2c}[-c(F(x+ct)-F(x-ct))+c(G(x+ct)-G(x-ct)\left)\right]$ by [F4], and the $F$-terms and $G$-terms collapse to $F(x-ct)+G(x+ct)=u(x,t)$.

By [F3] the converse holds on every nonempty open rectangle, so the sum of a right- and a left-moving profile is exactly the general one-dimensional solution, and the d'Alembert formula returns it from its data.

Substituting $u_0=0$ into the velocity integral gives $u(x,t)=\frac{1}{2c}{\int }_{x-ct}^{x+ct}{\mathbf{1}}_{[-a,a]}(y)dy=\frac{1}{2c}\mid [x-ct,x+ct]\cap [-a,a]\mid$, since the integral of the indicator is the overlap length. For smooth admissible approximations, the same d'Alembert expression is a classical solution by [F1].

Nonzero set and plateau. For $t>0$, the overlap has positive length exactly when $x-ct<a$ and $x+ct>-a$, that is $\mathrm{\mid }x\mathrm{\mid }<a+ct$; its closure, the topological support, is $[-a-ct,a+ct]$. At $t=0$ the moving interval is a singleton, so the overlap has length zero for every $x$ and the support is empty. The overlap is the full data interval, of length $2a$, when $\mathrm{\mid }x\mathrm{\mid }\le ct-a$ (which requires $ct\ge a$), giving value $a\mathrm{/}c$ there.

The formula therefore has expanding nonzero set $(-a-ct,a+ct)$ for $t>0$, closed support $[-a-ct,a+ct]$, zero support at $t=0$, and the stated full-data plateau when $ct\ge a$.

Forward direction. For a radial function $u(x,t)=v(r,t)$, $r=\mathrm{\mid }x\mathrm{\mid }$, the chain rule gives ${\mathrm{\partial }}_{j}u=v_r{x}_j\mathrm{/}r$ and ${\mathrm{\partial }}_j^{2}u=v_{rr}{x}_j^2\mathrm{/}{r}^2+v_r(1\mathrm{/}r-x_j^2\mathrm{/}{r}^3)$ for each $j$, so $\mathrm{\Delta }u={\sum }_{j=1}^3{\mathrm{\partial }}_j^{2}u=v_{rr}+\frac{2}{r}{v}_r=\frac{1}{r}{\mathrm{\partial }}_r^2(rv)=\frac{1}{r}{w}_{rr}$ by the product rule. Therefore $0=u_{tt}-c^2\mathrm{\Delta }u=\frac{1}{r}(w_{tt}-c^2{w}_{rr})$ and, since $r>0$, the one-dimensional equation $w_{tt}=c^2{w}_{rr}$ follows.

Converse direction, regularity at the origin. Suppose $w\in C^3([0,\mathrm{\infty })\times \mathbb{R})$ solves $w_{tt}=c^2{w}_{rr}$ on $r>0$ and $w(0,t)=0$ for all $t$. By Every continuous function on an interval has a primitive; two primitives differ by a constant; and ${\int }_a^{b}f=G(b)-G(a)$ for any primitive $G$ and $w(0,t)=0$, $w(r,t)=r{\int }_0^1{w}_r(sr,t)ds$, hence $v(r,t):=w(r,t)\mathrm{/}r={\int }_0^1{w}_r(sr,t)ds$; as $w\in C^3$, Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral applied locally to each derivative permits differentiation under the integral and shows that $v\in C^2([0,\mathrm{\infty })\times \mathbb{R})$ with $v(0,t)=w_r(0,t)$ and ${\mathrm{\partial }}_{r}v(0,t)=\frac{1}{2}{w}_{rr}(0,t)=0$, the last equality because $w_{rr}(0,t)=c^{-2}{w}_{tt}(0,t)$ by continuity of the equation up to $r=0$ and $w(0,\cdot )\equiv 0$.

The equation extends to the origin. For $r>0$ the product rule gives $v_r=\frac{1}{r}{w}_r-\frac{1}{r^2}w$, $v_{rr}=\frac{1}{r}{w}_{rr}-\frac{2}{r^2}{w}_r+\frac{2}{r^3}w$ and $v_{tt}=\frac{1}{r}{w}_{tt}=\frac{c^2}{r}{w}_{rr}$, so the radial three-dimensional Laplacian of $v$ is $v_{rr}+\frac{2}{r}{v}_r=\frac{1}{r}{w}_{rr}=c^{-2}{v}_{tt}$. For the limits at $r\downarrow 0$, the integral formulas and $w_{rr}(sr,t)={\int }_0^{sr}{w}_{rrr}(a,t)da$, since $w_{rr}(0,t)=0$, give $v_{rr}(r,t)={\int }_0^1{s}^2{w}_{rrr}(sr,t)ds\to \frac{1}{3}{w}_{rrr}(0,t)$ and $v_r(r,t)={\int }_0^{1}s{w}_{rr}(sr,t)ds=\frac{r}{3}{w}_{rrr}(0,t)+o(r)$, so that $\frac{2}{r}{v}_r(r,t)\to \frac{2}{3}{w}_{rrr}(0,t)$; hence $(v_{rr}+\frac{2}{r}{v}_r)(r,t)\to w_{rrr}(0,t)$. On the other hand, differentiating $w_{tt}=c^2{w}_{rr}$ in $r$ and letting $r\downarrow 0$ gives $v_{tt}(0,t)=w_{rtt}(0,t)=c^2{w}_{rrr}(0,t)$. These limits are uniform for $t$ in compact intervals by continuity of the derivatives of $w$. For $r>0$, $u_{ij}=(v_{rr}-v_r\mathrm{/}r)x_i{x}_j\mathrm{/}{r}^2+(v_r\mathrm{/}r){\delta }_{ij}$, and both $v_{rr}$ and $v_r\mathrm{/}r$ tend to $w_{rrr}(0,t)\mathrm{/}3$, so $u_{ij}$ extends continuously with this value times ${\delta }_{ij}$. The gradient is $v_{r}x\mathrm{/}r$ and is differentiable at $x=0$ by the same limits. Also $v_r(0,t)=0$ implies $v_{rt}(0,t)=0$; hence $u_{it}=v_{rt}{x}_i\mathrm{/}r\to 0$, agreeing with the derivative in $t$ of $u_i(0,t)=0$. Together with the integral formula for $v_{tt}$, this proves $u(x,t):=v(\mathrm{\mid }x\mathrm{\mid },t)$ is $C^2$ on ${\mathbb{R}}^3\times \mathbb{R}$ and satisfies $u_{tt}=c^2\mathrm{\Delta }u$ at every point, including the origin, where both sides equal $c^2{w}_{rrr}(0,t)$.

Both directions are proved: a radial three-dimensional solution corresponds to a one-dimensional solution $w=rv$ on the half-line, and a one-dimensional solution vanishing at $r=0$ gives back a $C^2$ radial three-dimensional solution, so radial three-dimensional data may be propagated by the one-dimensional formula.

Means and expression. By [F1] the means are constant, $M_{g_0}\equiv g_0$ and $M_{v_0}\equiv v_0$, so the Kirchhoff expression becomes ${\mathrm{\partial }}_t[t{g}_0]+t{v}_0=g_0+t{v}_0$; this is $C^{\mathrm{\infty }}$, satisfies ${\mathrm{\partial }}_t^{2}u=0=c^2\mathrm{\Delta }u$ and has the prescribed values $u(\cdot ,0)=g_0$, ${\mathrm{\partial }}_{t}u(\cdot ,0)=v_0$.

Normalisation check. A constant has mean itself on the sphere, so any unnormalised sphere integral ${\int }_{\mathrm{\partial }{B}_{ct}(x)}{u}_{0}dS$ would equal $4\pi c^2{t}^2$ times the mean for $u_0$ constant on the sphere of radius $ct$; the constant-data check therefore detects exactly that factor, confirming the normalisation $1\mathrm{/}(4\pi )$ in the Kirchhoff expression.

At $x=0$, $t=1$, [F1] gives $u(0,1)=\frac{1}{2\pi }{\int }_{B_1(0)}{u}_1(y)(1-\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy$. The integrand is nonnegative, the weight $(1-\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$ is strictly positive and bounded below by $1$ (and above by $2\mathrm{/}\sqrt{3}$) on the support $B_{1\mathrm{/}2}(0)$ of $u_1$, and $u_1$ is positive on a set of positive measure; hence the integral is strictly positive.

The support of $u_1$ lies strictly inside $B_1(0)$ and is disjoint from $\mathrm{\partial }{B}_1(0)$, so the value $u(0,1)$ is produced by data at distance at most $1\mathrm{/}2$ from the origin, strictly behind the wavefront of radius $1$; by [F2] this is exactly the two-dimensional interior tail, in contrast with the $(n\ge 3)$ odd-dimensional evaluation, which reads the data near the sphere.

Constant displacement. Take $u_0\equiv 1$, $u_1\equiv 0$. The correct solution is $u\equiv 1$ by [F1], whereas replacing the surface integral by a ball integral in the actual Kirchhoff expression gives ${\mathrm{\partial }}_t[t(4\pi c^2{t}^2{)}^{-1}(4\pi c^3{t}^3\mathrm{/}3)]={\mathrm{\partial }}_t[c{t}^2\mathrm{/}3]=2ct\mathrm{/}3$. Its displacement limit is zero for every $c>0$, so it fails to attain $u_0=1$. It also has velocity limit $2c\mathrm{/}3$, instead of zero.

Constant velocity. Take $u_0\equiv 0$, $u_1\equiv 1$. The correct solution is $u=t$ by [F1], whereas the refuted expression gives $t(4\pi c^2{t}^2{)}^{-1}(4\pi c^3{t}^3\mathrm{/}3)=c{t}^2\mathrm{/}3$. Its velocity limit is zero, not one, and its second time derivative is $2c\mathrm{/}3$ while its spatial Laplacian is zero. Thus it fails both the Cauchy data and the homogeneous wave equation for every $c>0$.

The sphere-area normalisation converts a surface integral into the spherical mean. A ball integral divided by that same area instead returns $ct\mathrm{/}3$ times the datum when it is constant. Keeping the factors $t$ of Kirchhoff's formula then gives the two incorrect functions above. Hence sphere area and ball volume cannot be interchanged in that formula.

Reading the general solution on the line $\xi =0$: if $u=F(\xi )+G(\eta )$, then $u(0,\eta )=F(0)+G(\eta )$, ${\mathrm{\partial }}_{x}u(0,\eta )=F^{\mathrm{\prime }}(0)+G^{\mathrm{\prime }}(\eta )$ and ${\mathrm{\partial }}_{t}u(0,\eta )=-c{F}^{\mathrm{\prime }}(0)+c{G}^{\mathrm{\prime }}(\eta )$. Thus the line data determine the function $G$ up to an additive constant and the number $F^{\mathrm{\prime }}(0)$; $F(0)$ retains the common additive-shift freedom, but leave the function $F$ away from $\xi =0$ completely free; by [F1] every $C^2$ solution near the line has this form.

The witness pair. Both $u_1=0+\sin (\eta )$ and $u_2={\xi }^3+\sin (\eta )$ are $C^2$ sums of a function of $\xi$ and a function of $\eta$, hence $C^2$ solutions by [F1]. On $\xi =0$ their traces agree: $u_1(0,\eta )=\sin \eta =u_2(0,\eta )$, ${\mathrm{\partial }}_x{u}_1=\cos \eta$ and ${\mathrm{\partial }}_x{u}_2=3{\xi }^2+\cos \eta$ coincide at $\xi =0$, and ${\mathrm{\partial }}_t{u}_1=c\cos \eta$ and ${\mathrm{\partial }}_t{u}_2=-3c{\xi }^2+c\cos \eta$ coincide there as well.

However $u_2-u_1={\xi }^3$ is nonzero for every $\xi \mathrm{\ne }0$, so the two solutions differ at points of every neighbourhood of the line $\xi =0$; hence data on the characteristic line do not determine the solution. This exhibits failure of uniqueness for these characteristic line data.

Therefore the displayed statement is refuted: the same values of $u,{\mathrm{\partial }}_{x}u,{\mathrm{\partial }}_{t}u$ on the single characteristic line are shared by two $C^2$ solutions that disagree on every neighbourhood of it.

Fubini on a fixed product. By [F1], $I_{\epsilon }:=\int \phi (x)u_{\epsilon }(x,t)dx=\frac{t}{4\pi }{\int }_{{\mathbb{R}}^3}{\int }_{S^2}\phi (x){\rho }_{\epsilon }(x+ct\omega )d\sigma (\omega )dx$. The integrand vanishes unless $\mathrm{\mid }x\mathrm{\mid }\le ct+\epsilon$, where $\phi$ is bounded; the absolute integrand is bounded by $C\mathrm{\parallel }{\rho }_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathbf{1}}_{{\bar{B}}_{ct+\epsilon }(0)\times S^2}$, with $C={\sup }_{\mathrm{\mid }x\mathrm{\mid }\le ct+\epsilon }\mathrm{\mid }\phi (x)\mathrm{\mid }<\mathrm{\infty }$. This majorant is integrable because the product rectangle has finite measure. Thus [F3] applies to the fixed product ${\mathbb{R}}^3\times S^2$. Set $y=x+ct\omega$ in the inner Euclidean integral, then reflect $\omega \mapsto -\omega$ using Reflection invariance and vanishing first moment of the sphere measure. This gives $I_{\epsilon }=\frac{t}{4\pi }\int {\rho }_{\epsilon }(y){\int }_{S^2}\phi (y+ct\omega )d\sigma (\omega )dy=\frac{t}{4\pi c^2{t}^2}\int {\rho }_{\epsilon }(y)\mathrm{\Psi }(y)dy$, with the last equality supplied by sphere-measure scaling Agreement with the existing polar sphere measure.

By [F4], $\int {\rho }_{\epsilon }\mathrm{\Psi }\to \mathrm{\Psi }(0)$. Therefore $I_{\epsilon }\to \frac{t}{4\pi c^2{t}^2}{\int }_{\mathrm{\partial }{B}_{ct}(0)}\phi dS=\frac{t}{4\pi }{\int }_{S^2}\phi (ct\omega )d\sigma (\omega )$. The limiting measure has total mass $t$ and constant surface density $t\mathrm{/}(4\pi c^2{t}^2)$; dividing the measure by $t$ gives the uniform probability measure on that sphere.