= Solution
For radial data, set
$$
f_0(r)=r\,u_0(|r|),\qquad f_1(r)=r\,u_1(|r|)
$$
on $\mathbb R$. The conditions at the origin say exactly that these are smooth odd compactly supported functions. The radial reduction $w(t,r)=ru(t,r)$ satisfies the one-dimensional wave equation, and the <D'Alembert formula> gives
$$
w(t,r)=\frac12\big(f_0(r-t)+f_0(r+t)\big)
+\frac12\int_{r-t}^{r+t}f_1(s)\,ds.
$$
In null coordinates this is
$$
w=\frac12\big(f_0(-\xi)+f_0(\eta)\big)
+\frac12\int_{-\xi}^{\eta}f_1(s)\,ds.
$$
The limits are therefore
$$
\psi_+(\xi)=
\frac12f_0(-\xi)+\frac12\int_{-\xi}^{\infty}f_1(s)\,ds,
$$
$$
\psi_-(\eta)=
\frac12f_0(\eta)-\frac12\int_{\eta}^{\infty}f_1(s)\,ds.
$$
They are smooth and compactly supported because $f_0,f_1$ are odd.
Write
$$
A(s)=\frac12f_0(s),\qquad
B(s)=\frac12\int_s^\infty f_1(q)\,dq.
$$
Then $A$ is an arbitrary odd test function and $B$ is an arbitrary even test function. Conversely, every odd $A\in C_c^\infty(\mathbb R)$ gives $f_0=2A$, and every even $B\in C_c^\infty(\mathbb R)$ gives $f_1=-2B'$. Thus both maps are injective and
$$
X_-=X_+=C_c^\infty(\mathbb R).
$$
Since
$$
\psi_-=A-B,\qquad
\psi_+=-A+B,
$$
the radial <scattering map> is
$$
S\psi=-\psi.
$$
Back to article page
