We are asked to solve the following PDE:
$$ \Delta u(x) = \phi(x) \hspace{.3 cm} x\in \mathbb{R}^3 $$ $$ \phi(x) = \begin{cases} \exp\left(-\dfrac{1}{1−x^2}\right) & |x|<1 \\ 0 & |x| \geq 1 \end{cases} $$
I know in order to get the final solution, I know I need to do the convolution of the fundamental solution with the inhomogeneous part but I genuinely do not know if the convolution it self is integrable at this point. Any help is extremely appreciated.
I'm going to use slightly different notation, because I am an applied, not pure, mathematician. We want solutions of $$\nabla^2u(\mathbf{r})=f(\mathbf{r})$$ Here $\mathbf{r}=(x,y,z)$, the position vector. We know $$ f(\mathbf{r}) = \begin{cases} \exp\left(-\dfrac{1}{1−\Vert\mathbf{r}\Vert^2}\right) & \Vert\mathbf{r}\Vert<1 \\ 0 & \Vert\mathbf{r}\Vert \geq 1 \end{cases} $$ Let's do the $\Vert\mathbf{r}\Vert\geq1$ case first. Call this solution $u_2$. Since $f$ is spherically symmetric, it makes sense to express our problem in spherical polar coordinates. We want solutions to $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u_2}{\partial r}\right)=0$$ It's fairly elementary to see that our solution is $$u_2(r)=a_1+\frac{a_2}{r}$$
Now onto the $\Vert\mathbf{r}\Vert<1$ case. Call this solution $u_1$. Now we want solutions to $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u_1}{\partial r}\right)=\exp\left(\frac{-1}{1-r^2}\right)$$ Since we've already found the solution of the homogeneous case, all we need is a particular solution of the above and we can combine it with the homogeneous solution for a full answer. Let's expand the L.H.S, $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u_1}{\partial r}\right)=\frac{1}{r^2}\left(2r\frac{\partial u_1}{\partial r}+r^2\frac{\partial^2u_1}{\partial r^2}\right)=\frac{\partial^2u_1}{\partial r^2}+\frac{2}{r}\frac{\partial u_1}{\partial r}$$ So we need to study the ODE $${u_1}''(r)+\frac{2}{r}{u_1}'(r)=\exp\left(\frac{-1}{1-r^2}\right)$$
Let $v(r)=u_1'(r).$ We now have the first order ODE $$v'(r)+\frac{2}{r}v(r)=\exp\left(\frac{-1}{1-r^2}\right)$$ This is a first order linear ODE, so $$v(r)=\frac{1}{\mu(r)}\int \mu(r)\exp\left(\frac{-1}{1-r^2}\right)\mathrm{d}r$$ Where $\mu(r)=\exp\left(\int\frac{2}{r}\mathrm{d}r\right)=\exp\left(2\ln(r)\right)=r^2$. So we need to take a look at the integral $$\int r^2\exp\left(\frac{-1}{1-r^2}\right)\mathrm{d}r$$ Unfortunately, this integral can't be expressed in terms of elementary functions, so we'll use a series expansion instead. A Maclaurin series shows us that $$\exp\left(\frac{-1}{1-x^2}\right)\approx \frac{1}{e}\left(1-x^2-\frac{x^4}{2}\right)$$
Therefore $$v(r)\approx \frac{1}{r^2}\frac{1}{e}\int r^2\left(1-r^2-\frac{r^4}{2}\right)\mathrm{d}r$$ The algebra is rather tedious, but at the end we find $$v(r)\approx \frac{1}{e\cdot r^2}\left(\frac{r^3}{3}-\frac{r^5}{5}+\frac{r^7}{14}+b_1\right)=\frac{b_1}{r^2}+\frac{r}{e}-\frac{r^3}{5e}+\frac{r^5}{14e}$$ Since $v(r)={u_1}'(r)$, $$u_1(r)\approx\int \left(\frac{b_1}{r^2}+\frac{r}{e}-\frac{r^3}{5e}+\frac{r^5}{14e}\right)\mathrm{d}r=-\frac{b_1}{r}+\frac{r^2}{2e}+\frac{e}{420}r^4(5r^2-21)+b_2$$
Thus, finally, our general answer is $$ u(r) \approx \begin{cases} u_1(r)=-\frac{b_1}{r}+\frac{r^2}{2e}+\frac{e}{420}r^4(5r^2-21)+b_2 & r<1 \\ u_2(r)=a_1+\frac{a_2}{r} & r \geq 1 \end{cases} $$ We want $u$ to be smooth at the boundary, that is $u_1(1)=u_2(1)~;~{u_1}'(1)={u_2}'(1)$ We can compute $${u_1}'(r)=\frac{b_1}{r^2}+\frac{e}{70}r^3(5r^2-14)+\frac{x}{e} ~;~ {u_2}'(r)=\frac{-a_2}{r^2}$$ Therefore $${u_1}'(1)=b_1-\frac{9e}{70}+\frac{1}{e}=-a_2={u_2}'(1)$$ And $$a_1+a_2=-b_1+\frac{1}{2e}-\frac{4e}{105}+b_2$$ Since we have two equations in four variables, we don't have sufficient information to work out $a_1,a_2,b_1,b_2$. Other IC's and BC's are needed to compute these.
Hope this helped!
EDIT: Linked below is an interactive plot of the solution curves. $b_1$ and $b_2$ are free parameters and $a_1,a_2$ will vary with these parameters to keep the curve smooth.
