Have a look at this multiple select question:
Q. Let $u$ be a solution of the boundary value problem $$\begin{array}{c}u^{\prime \prime}+\frac{1}{t} u^{\prime}=f(t), t \in(0,1), \\ u^{\prime}(0)=a, u(1)=b,\end{array}~~~~~~~~~~~(*) $$ Define for $x^2+y^2 \leq 1, v(x, y)=u\left(\sqrt{x^2+y^2}\right)$ and $g(x, y)=f\left(\sqrt{x^2+y^2}\right)$, then $v$ is the solution of the PDE $$v_{x x}+v_{y y}=g,~~~~~~~~~~~~~~~~~(**)$$ in $\left\{(x, y): x^2+y^2<1\right\}, v(x, y)=0$ on $\left\{(x, y): x^2+y^2=1\right\}$ if
(1) $a>0$ and $b>0$
(2) $a>0$ and $b=0$
(3) $a=0$ and $b=0$
(4) $a<0$ and $b=0$
Observe that $(*)$ is equivalent to $(**)$ for $t=\sqrt{x^2+y^2} \in (0,1)$, and for instance, I got that $a=0$ is necessary for the trivial choice $f=0$ as the general solution $$u(t)=c_1+c_2 \ln t,$$ for $(*)$ must be realistic towards $u'(0)$. That is, option (3) seems true as long as $f(t)=0$ however I couldn't verify for arbitrary $f$.
I hope you can help me, thanks in advance.
