Please help with this boundary value problem I came across the following problem and I am stuck on it :  

Let $u$ be a solution of the boundary value problem $u''+\dfrac{1}{t}u'=f(t),t \in (0,1)$ and $u'(0)=a,u(1)=b.$ Define for $x^2+y^2 \leq 1,v(x,y)=u(\sqrt {x^2+y^2})$ and $g(x,y)=f(\sqrt {x^2+y^2})$.Then $v$ is a solution of the PDE $v_{xx}+v_{yy}=g$ in $\{(x,y):x^2+y^2 < 1\}$ and $v(x,y)=0$ on $\{(x,y):x^2+y^2 = 1\}$ if
  1.$\space a>0 ,b>0$
  2.$\space a>0,b=0$
  3.$\space a=0,b=0$
  4.$\space a<0,b=0$.  

I have to determine which of the above four options is/are true?   Can someone point me in the right direction? Thanks in advance for your time.

Attacking the problem backwards from Poisson equation to its radial symmetric equation is more intuitive in my opinion. 
$\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}$
$\newcommand{\pb}[2]{\frac{\partial }{\partial #2}\left(#1\right)}$
$\newcommand{\ppt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$
Consider the Poisson's equation $\Delta v = v_{xx} + v_{yy} = g$ in polar coordinates $(r,\theta)$: 
$$
r = \sqrt{x^2 + y^2} \text{ and } \theta = \arctan(y/x) + C
$$ 
We have: 
$$\pp{r}{x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r}, \text{ and } \pp{r}{y} = \frac{y}{r}
$$
$$\pp{\theta}{x} = \frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}} = -\frac{y}{r^2}, \text{ and } \pp{\theta}{y} = \frac{x}{r^2}
$$
Now 
$$
\pp{v}{x} = \pp{v}{r} \pp{r}{x} + \pp{v}{\theta} \pp{\theta}{x} = \pp{v}{r} \frac{x}{r} -  \pp{v}{\theta} \frac{y}{r^2}
$$
and 
$$
\begin{aligned}
v_{xx}= \pb{\pp{v}{x}}{x} &= \ppt{v}{r}\pp{r}{x}\frac{x}{r}  + \pp{v}{r}\pb{\frac{x}{r} }{x} - \ppt{v}{\theta}\pp{\theta}{x}\frac{y}{r^2}  - \pp{v}{\theta}\pb{\frac{y}{r^2}}{x}
\\
&= \ppt{v}{r}\frac{x^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{x^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{y^2}{r^4} + \pp{v}{\theta}\frac{2xy}{r^4}
\end{aligned}\tag{1}
$$
Similarly we have:
$$
v_{yy} = \ppt{v}{r}\frac{y^2}{r} + \pp{v}{r}\left(\frac{1}{r} - \frac{y^2}{r^3}\right)
+ \ppt{v}{\theta}\frac{x^2}{r^4} - \pp{v}{\theta}\frac{2xy}{r^4}\tag{2}
$$
Exploiting $x^2+y^2 = r^2$, (1) plus (2) gives:
$$
\Delta v =  \ppt{v}{r} + \frac{1}{r}\pp{v}{r}+\frac{1}{r^2}\ppt{v}{\theta}.
$$
If a function is radial symmetric, meaning it can be represented using only $r$ for it is a constant for fixed $r$, i.e., it is a constant on any fixed circle, and rotating does not change its value. Hence $$\displaystyle \ppt{v}{\theta} = 0.$$ And the Laplacian of a radial symmetric function is again radial symmetric for no involvement of $\theta$. Here I use $r$ for the $t$ in your question.
Let $v(x,y) =u(\sqrt{x^2 + y^2}) =u(r)$, $g(x,y) = f(r)$. Then
$$
u'' + \frac{1}{r}u' = f(r)
$$
Using your favorite method of solving ODE, here I cook up using product rule reverse direction:
$$
(ru')' = rf(r) 
$$
Integrating: 
$$ u'(r) = \frac{1}{r}(\int^r_0 sf(s) ds+ r u'(r)|_{r=0}) \tag{3}
$$
Since the limit exists when $r\to 0$:
$$
a = u'(0) = \lim_{r\to 0}\frac{\int^r_0 sf(s) ds}{r} = \lim_{r\to 0} rf(r) =0
$$
Now for $b$, since $v = 0$ when $r= 1$, hence $b = u(1) = 0$.
Hence $a = b =0$.

If you would like to get a form of the solution, integrating (3) again yields:
$$ u(r) - u(0) = \int^{r}_0 u'(\tau)d\tau =  \int^{r}_0 \frac{1}{\tau}\int^{\tau}_0 sf(s) ds d\tau.
$$
