Does there exist unique $u \in V$ satisfying integral equation? Set$$V = \{v \in H^1(0, 1) : v(0) = 0\}.$$Given $f \in L^2(0, 1)$ such that ${1\over x}f(x) \in L^2(0, 1)$, does there exist a unique $u \in V$ satisfying$$\int_0^1 u'(x)v'(x)\,dx + \int_0^1 {{u(x)v(x)}\over{x^2}}\,dx = \int_0^1 {{f(x)v(x)}\over{x^2}}\,dx \text{ for all }v \in V?$$
 A: PARTIAL ANSWER: 
The main difficulty here is (IMHO) to prove that all integrals make sense. Once this is accomplished, the problem SHOULD be standard and solvable, e.g., via Lax-Milgram's theorem. 
Now in the case of 
$$\int_0^1 \frac{f(x)v(x)}{x^2}\, dx$$
I would proceed as follows. The task is to show that $\frac{v(x)}{x}\in L^2$. Use the fact that
$$v(x)=\int_0^x v'(y)\, dy.$$
So 
$$
\begin{split}
\int_0^1 v(x)\frac1x\, dx &= \int_0^1 \left(\int_0^x v'(y)\, dy\right)\, d\log x \\ 
& = \int_0^1 -v'(x)\log x\, dx +\log 1 \int_0^1 v'(y)\, dy - \lim_{x\to 0} \log x \int_0^x v'(y)\, dy.
\end{split}
$$
By l'Hôpital's rule the last limit is $0$. So the integral is finite and this shows that $\frac{v(x)}{x}\in L^2$. 
The same reasoning shows that the term $\frac{u(x)}x\in L^2$ too. We are therefore dealing with the weak formulation of the Sturm-Liouville problem 
$$
\begin{cases} u'' +\frac{1}{x^2}u' = F, & F\overset{\text{def}}{=}\frac{f}{x^2} \\
u(0)=0 \\
u'(1)=0
\end{cases}
$$
Now I should check if Lax-Milgram's theorem is directly applicable or if we need some more ideas. The fact that $F$ is not necessarily in $L^2$ makes me tend towards the second scenario.
But I have ran out of time, so I am just writing down this community wiki hoping that it can be of some help.
A: If your weak equation holds, then it holds for all $v\in C^{\infty}_c(0,1)$ (compactly supported,) which implies that $u'$ must be absolutely continuous and satisfy
$$
                -u''+\frac{1}{x^2}u = \frac{1}{x^2}f.
$$
The solutions of $-u''+\frac{1}{x^2}u=0$ have the form $x^{p}$ where $p(p-1)-1=0$. So,
$$
               (p-1/2)^2-5/4=0,\\
                 p = \frac{1}{2}\pm\frac{\sqrt{5}}{2},\\
           p_1 = \frac{1+\sqrt{5}}{2},\;\;p_2=\frac{1-\sqrt{5}}{2}.
$$
The Wronskian of the solutions $x^{p_1}$, $x^{p_2}$ is
$$
            W(x^{p_1},x^{p_2})=x^{p_1+p_2-1}(p_1-p_2)=\sqrt{5}.
$$
Consider the function $g$ given by
$$
        -\sqrt{5}g(x) = x^{p_2}\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt+x^{p_1}\int_{x}^{1}t^{p_2}\frac{f(t)}{t^2}dt.
$$
This function is defined on $(0,1)$ because
$$
       t^{p_1}\frac{f(t)}{t^2}=t^{p_1-1}\frac{f(t)}{t}\in L^1.
$$
$g$ is a Green function solution with
\begin{align}
    -\sqrt{5}g'(x) & = (x^{p_2})'\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt
     +(x^{p_1})'\int_{x}^{1}(t^{p_2}-t^{p_1})\frac{f(t)}{t^2}dt \\
   -\sqrt{5}g''(x) & = (x^{p_2})''\int_{0}^{x}t^{p_1}\frac{f(t)}{t^2}dt
   +(t^{p_1})''\int_{x}^{1}t^{p_1}\frac{f(t)}{t^2}dt \\
    & + W(x^{p_1},x^{p_2})\frac{f(x)}{x^2}
\end{align}
Therefore,
$$
       \sqrt{5}\left(-g''+\frac{1}{x^2}g\right) = \sqrt{5}\frac{f(x)}{x^2}.
$$
Take a look at the behavior of $g$ near $x=0$. The first integral term for $\sqrt{5}g$ is bounded using Cauchy-Schwarz by
$$
         x^{p_2}\left(\int_{0}^{x}t^{2p_1-2}\right)^{1/2}\left(\int_{0}^{x}\left(\frac{f(t)}{t}\right)^{2}dt\right)^{1/2} \\
   \le x^{p_2}\frac{x^{p_1-1/2}}{\sqrt{2p_1-1}}k(x) \\
    = \frac{\sqrt{x}}{\sqrt{2p_1-1}}k(x),
$$
where $k(x)=\|\chi_{[0,x]}(t)f(t)/t\|\rightarrow 0$ as $x\rightarrow 0$. The second integral term for $\sqrt{5}g$ is bounded by
$$
          x^{p_1}\left(\int_{x}^{1}(t^{p_2-1})^2dt\right)^{1/2}\|f(t)/t\| \\
        \le C x^{p_1}x^{p_2-1/2}=C\sqrt{x}.
$$
Therefore $g$ vanishes at $x=0$.
Any other twice absolutely continuous solution of $-u''+\frac{1}{x^{2}}u=\frac{1}{x^2}f$ on $(0,1)$ must have the form
$$
            u = g + Ax^{p_1}+Bx^{p_2},
$$
However $B=0$ must hold in order for $u$ to vanish at $x=0$. By choosing $B$ appropriately, you can arrange for $u'(1)=0$. Then, when you integrate your weak equation by parts, you get
$$
    \int_{0}^{1}(-u'')vdx = \left.\int_{0}^{1}u'v'dx+u'v\right|_{0}^{1}
   = \int_{0}^{1}u'v'dx,\;\;\; v\in V.
$$
(You will need to estimate $u'$ near $0$ and use the properties of $v\in V$, but the expression is given above, and the techniques are the same.) So this $u$ is a weak solution. And you can see that it is unique by the classical construction given above. This may not be what you hoped for, but I think it's a full solution. :) The answer is: Yes, there is a unique solution as posed.
A: The above methods were so called "indirect" methods. But if you know something about "direct" methods, which are also called variational methods, the problem can be solved very quickly, since a particular construction for a solution like above is not needed: we define the energy functional
\begin{equation}
J(x,u,u'):=\int_0^1 (u')^2+\frac{u^2}{x^2}-\frac{fu}{x^2}dx
\end{equation}
and 
\begin{equation}
f(x,u,A):=(A)^2+\frac{u^2}{x^2}-\frac{fu}{x^2}.
\end{equation}
First of all, for fixed $x$ and $u$, $f$ is convex in $A$, so $J$ is weakly lower semi continous; secondly, using Poincare's and Hölder's inequality you will conclude that $J$ is coercive; thirdly, using $u=0$ you know that the supremum of the functional is finite. Now applying the abstract existence theorem you know that this functional has a minimizer. The Euler-Lagrange equation in the solution gives you the equation that you want to have. Notice also that for fixed $x$, f is strictly convex in $(u,A)$, therefore the minimizer is also unique. This implies the uniqueness of the solution.
A: Consider the weighted Sobolev space $V_1=\{v\in H^1(0,1):v(1)=0,\int_0^1\frac{v}{x^2}<\infty\}$, endowed with the norm $\|v\|_{V_1}^2=\int_0^1\frac{v}{x^2}+|v|_{H^1(0,1)}^2$. This space is a Hilbert space when equipped with the analogous inner product.
The bilinear form $a(u,v)=\int_0^1u'v'+\int_0^1\frac{uv}{x^2}$ is coercive and bounded on $V_1$, and the right hand side defines a continuous linear functional on $V_1$ if $f$ satisfies the assumption you gave.
The Lax-Milgram Theorem gives us existence and uniqueness in $V_1$. Note that $V_1\subset V$, to prove uniqueness, let $u_1$ and $u_2$ be two solutions and let $w=u_1-u_2$, then we see that $\|w\|_{V_1}=0\Rightarrow w\equiv0\Rightarrow u_1\equiv u_2$.
