I will try to describe the problem, it will necessarily be incomplete, so please if you have questions or remarks to make it more clear do not hesitate to leave them in comments.
The problem
Let $a$ be a constant, we know that a PDE (let us call it ‘the equation’) has solutions of the form
$$u(x)= \int_{S^2}\alpha(x\cdot\omega+ a,\ \omega)\,d\sigma(\omega)\quad , \ x\in \mathbb{R}^3$$
where
\begin{align*}
\alpha(s,\omega)&= \frac{\partial}{\partial s}\mathscr{R}g(s,\omega)+ \frac{\partial^2}{\partial s^2}\mathscr{R}f(s,\omega)
\end{align*}
for given functions $f,g \in C_0(\mathbb{R}^3)$.
And $\mathscr{R}:C_0(\mathbb{R}^3)\rightarrow C_0(\mathbb{R}\times S^2)$ is the Radon transform in the $3$ dimensional space.
The question is, how would we generalize $u$ to a distribution so that we include the cases when $\mathscr{R}f$ and $\mathscr{R}g$ are not differentiable (with respect to $s$)?
It is clear that $\alpha$ can be regarded as a distribution in $s$.
The trial
Here is what we can try for example, we want to generalize $u(x)$ to a distribution $U\in S'(\mathbb{R}^3)$, which is defined by
\begin{align*}
U(\phi)&= \int_{\mathbb{R}^3} u(x)\phi(x) \,dx \\
&= \int_{\mathbb{R}^3} \int_{S^2}\alpha(x\cdot\omega+a,\omega)\phi(x) \,d\sigma(\omega)\,dx \quad \text{ for all } \phi\in S(\mathbb{R}^3).
\end{align*}
To each $\omega\in S^2$ correspond two vectors $\nu, \eta\in S^2$ (infinitely many choices) such that $(\omega, \nu, \eta)$ is an orthonormal basis of $\mathbb{R}^3$, we have the following bijection (the change of coordinates) as an isometry;
\begin{align*}
\theta_\omega &: \ \mathbb{R}^3\rightarrow \mathbb{R}^3 \\
& \ \ \ x\mapsto (x\cdot\omega,\ x\cdot\nu,\ x\cdot\eta)^t
\end{align*}
Letting $s= x\cdot\omega+a$, $y_1= x\cdot\nu$ and $y_2=x\cdot\eta$, we can write
$$x= (x\cdot\omega)\omega+ y_1\nu+ y_2\eta= s\omega+ y_1\nu+ y_2\eta- a\omega $$
then for $y= (y_1, y_2)^t\in \mathbb{R}^2$ we can define the test function $\tau_y\in S(\mathbb{R})$ by $\tau_y(s)= \phi(x)$.
And we can then consider a distribution induced by $\alpha(s,\omega)$ (as a function of $s$) that we can denote $\alpha_\omega$:
\begin{align*}
U(\phi)&= \int_{\mathbb{R}^3} \int_{S^2}\alpha(s,\omega)\tau_y(s) \,d\sigma(\omega)\,dx \quad \text{ by the change of variable we want to apply } \\
&= \int_{S^2} \int_{\mathbb{R}^2}\left(\int_{\mathbb{R}}\alpha(s,\omega)\tau_y(s) \,ds\right)\,dy \,d\sigma(\omega) \quad \text{ isolating $s$ before integrating over $\mathbb{R}^2$ }\\
&= \int_{\mathbb{R}^2}\int_{S^2} \alpha_\omega(\tau_y) \,d\sigma(\omega)\,dy \quad \quad\quad \text{ here $\alpha_\omega$ is a distribution acting on $\tau_y$.}
\end{align*}
Let us assume we will find no trouble in considering that integrating the distribution over $S^2$ then over $\mathbb{R}^2$ is well defined.
Now for distributions $f,g\in S'(\mathbb{R}^3)$ of compact support, and $\psi\in S(\mathbb{R}^2)$ such that $\psi(y)= 1$, $\forall y\in P_{\mathbb{R}^2}(B_r(0)) $ (the projection of the ball containing the compact support of $f$ and $g$ on the subspace $\mathbb{R}^2$).
We want to define a ‘distributional Radon transform’ $\mathscr{R}_\omega h$ (of any function $h$):
$$\mathscr{R}_\omega h \in S'(\mathbb{R}) : \phi\mapsto h\big( (\phi\otimes\psi) \circ\theta_\omega\big) $$ where the tensor product of the two test functions is defined by $\phi\otimes\psi\begin{pmatrix} s \\ y_1 \\ y_2 \end{pmatrix}= \phi(s)\psi(y)$ for all $(s, y_1, y_2)^t\in \mathbb{R}^3$.
And then it will follow that $$\alpha_\omega= \mathscr{R}_\omega g '+\mathscr{R}_\omega f''.$$ Or $$\alpha_\omega(\tau)= -\mathscr{R}_\omega g (\tau')+ \mathscr{R}_\omega f(\tau'') \quad \forall \tau \in S(\mathbb{R}).$$
This last formula is then used to retrieve the distribution $U$.
The question
Is any of that making any sense?
The classical Radon transform $\mathscr{R}h$ captures the sum of the density $h$ over some Euclidean plane $P_{(s,\omega)}$, whereas this distributional Radon transform ‘mimics’ this but for distributions $h$!
Would you think of that problem in another way?
**double stars**, not $\textbf{TeX fakery}$$\textbf{TeX fakery}$, for bold; and actual blank lines, not$$$$, for blank lines in the rendered post. I have edited accordingly. $\endgroup$$$$$in place of a blank line if you want to, but, aside from making the page slightly slower to load, why would you want to? Anyway, I meant only to suggest, and to apprise if you didn't already know; if, having heard the suggestion, you decide to do things otherwise, then that is your right (but your posts might continue to be edited). $\endgroup$