= Solution
Define on $H$ the <bilinear form>
$$
a(u,v)=\int_0^1u'v'+\int_0^1\frac{uv}{x^2}-\int_0^1u'v
$$
and the linear functional
$$
\ell(v)=\int_0^1\frac fx\frac vx.
$$
The <Hardy inequality on an interval>, <Cauchy-Schwarz inequality>, and the one-sided <Poincare inequality> show that $a$ is a <bounded bilinear form> and that
$$
|\ell(v)|\leq2\|f/x\|_2\|v'\|_2\leq C\|f/x\|_2\|v\|_{H^1}.
$$
For $v\in H$, $v(x)=\int_0^xv'(t)\,dt$, so <Cauchy-Schwarz inequality>[Cauchy--Schwarz] and <Fubini's theorem> give
$$
\|v\|_2^2\leq\frac12\|v'\|_2^2.
$$
Hence
$$
\begin{aligned}
a(v,v)
&=\|v'\|_2^2+\|v/x\|_2^2-\int_0^1v'v\\
&\geq\left(1-\frac1{\sqrt2}\right)\|v'\|_2^2\\
&\geq c\|v\|_{H^1}^2.
\end{aligned}
$$
Thus $a$ is a <coercive bilinear form>. The <Lax-Milgram theorem> supplies a unique $u\in H$ satisfying $a(u,v)=\ell(v)$ for every $v\in H$. This is precisely the unique weak solution described by the <weak boundary value problem with an inverse-square potential>.
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