Poisson's formula in two dimensions by descent

Descent. Extend $u_0,u_1$ to $U_0,U_1$ on ${\mathbb{R}}^3$ by $U_j(\xi ,z):=u_j(\xi )$; these are $C^3$ respectively $C^2$. By [F1] the function $V(\xi ,z,t):={\mathrm{\partial }}_t[t{M}_{U_0}^{(3)}((\xi ,z),ct)]+t{M}_{U_1}^{(3)}((\xi ,z),ct)$ is $C^2$ on ${\mathbb{R}}^3\times (0,\mathrm{\infty })$ with $V_{tt}=c^2{\mathrm{\Delta }}_{3}V$. Since $U_j(\xi +ct{\omega }^{\mathrm{\prime }},z+ct{\omega }_3)=u_j(\xi +ct{\omega }^{\mathrm{\prime }})$ does not depend on $z$, the sphere means of $U_0,U_1$ at centre $(\xi ,z)$ are independent of $z$; hence $V$ is independent of $z$, so ${\mathrm{\partial }}_z^{2}V=0$, ${\mathrm{\Delta }}_{3}V={\mathrm{\Delta }}_{2}V$, and the restriction $u(x,t):=V(x,0,t)$ is a $C^2$ function on ${\mathbb{R}}^2\times (0,\mathrm{\infty })$ with $u_{tt}=c^2{\mathrm{\Delta }}_{2}u$.

The weighted-ball form. By [F2] with $n=2$ and $G=U_j$, $t{M}_{U_j}^{(3)}((x,0),ct)=\frac{1!!}{c}{W}_{u_j}(x,ct)=\frac{1}{c}{W}_{u_j}(x,ct)$; therefore $u(x,t)=\frac{1}{c}[{\mathrm{\partial }}_t{W}_{u_0}(x,ct)+W_{u_1}(x,ct)]$. With $n=2$, $n!!V_n=2\pi$, [F3] reads $W_f(x,ct)=\frac{1}{2\pi }{\int }_{B_{ct}(x)}f(y)(c^2{t}^2-\mathrm{\mid }y-x{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy$, so $u$ is the displayed Poisson expression; the derivative ${\mathrm{\partial }}_t$ acts on the $C^1$ function $t\mapsto W_{u_0}(x,ct)$, since $W_{u_0}(x,ct)=ct{M}_{U_0}^{(3)}((x,0),ct)$ by [F2], and the spherical mean is $C^3$, and not by differentiating a singular integrand.

The integrated equivalent form. For fixed $x$ put $w(z):=(1-\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$ on $B_1\subset {\mathbb{R}}^2$ and $A(x,t):={\int }_{B_1}{u}_0(x+ctz)w(z)dz$. By [F3], $w\in L^1(B_1)$. To differentiate in $t$, fix compact sets $K\subset {\mathbb{R}}^2$ and $J\subset (0,\mathrm{\infty })$ for $x$ and $t$. Choose a closed ball $Q$ containing $x+c(t+s)z$ for $x\in K$, $t\in J$, $\mathrm{\mid }s\mathrm{\mid }$ sufficiently small and $\mathrm{\mid }z\mathrm{\mid }\le 1$. By [F6], $C:={\sup }_Q\mathrm{\mid }\mathrm{\nabla }{u}_0\mathrm{\mid }<\mathrm{\infty }$. The difference quotients of $u_0(x+ctz)$ in $t$, for these parameters and sufficiently small increments $h$, are bounded in absolute value by $cC\mathrm{\mid }z\mathrm{\mid }$ by [F7]. After multiplication by $w(z)$ they are dominated by the locally uniform integrable function $cC\mathrm{\mid }z\mathrm{\mid }w(z)$. They converge pointwise to $c\mathrm{\nabla }{u}_0(x+ctz)\cdot zw(z)$, so [F4] gives ${\mathrm{\partial }}_{t}A(x,t)=c{\int }_{B_1}\mathrm{\nabla }{u}_0(x+ctz)\cdot zw(z)dz$. The same argument for spatial difference quotients, and dominated convergence applied to convergent parameter sequences with the same compact-set majorant, shows that these first derivatives are continuous locally; in particular $A$ is $C^1$ in $(x,t)$. Now [F5] gives $I(x,t):={\int }_{B_{ct}(x)}{u}_0(y)(c^2{t}^2-\mathrm{\mid }y-x{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy=ctA(x,t)$, hence ${\mathrm{\partial }}_{t}I=cA+ct{\mathrm{\partial }}_{t}A$. The same change of variables gives ${\int }_{B_{ct}(x)}\mathrm{\nabla }{u}_0(y)\cdot (y-x)(c^2{t}^2-\mathrm{\mid }y-x{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy=(ct{)}^2{\int }_{B_1}\mathrm{\nabla }{u}_0(x+ctz)\cdot zw(z)dz=(ct{)}^2{\mathrm{\partial }}_{t}A\mathrm{/}c$. Therefore $\frac{1}{2\pi c}[{\mathrm{\partial }}_{t}I+{\int }_{B_{ct}(x)}{u}_1(y)(c^2{t}^2-\mathrm{\mid }y-x{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}dy]=\frac{1}{2\pi ct}{\int }_{B_{ct}(x)}\frac{u_0(y)+\mathrm{\nabla }{u}_0(y)\cdot (y-x)+t{u}_1(y)}{\sqrt{c^2{t}^2-\mathrm{\mid }y-x{\mathrm{\mid }}^2}}dy$, which is the equivalent form.

Both displays therefore define the same $C^2$ solution of the two-dimensional homogeneous wave equation on ${\mathbb{R}}^2\times (0,\mathrm{\infty })$; the limit at $t\downarrow 0$ and the attainment of the data are the subject of the data-attainment lemma below.