# Variational form of boundary value problem

Given Dirichlet boundary value problem

$$\begin{array}[t]{rcl} -\Delta u&=f &\text{ in }\Omega\\ u&=0 &\text{ on }\partial\Omega, \end{array}$$

we can use Green's theorem and transform it into variational form:

$$l(v):=\int_\Omega fv\text{ dx}=\int_\Omega -\Delta u f\text{ dx}=\int_\Omega \nabla u\cdot\nabla v \text{ dx}=: a(u,v),$$

so that the problem boils down to find $u\in H^1(\Omega)$ (Sobolev space) such that:

$$a(u,v)=l(v) \,\,\,\,\forall v \in H^1(\Omega).$$

But now I am given a task that goes in opposite direction:

Which boundary value problem is solved by the problem in variational form:

$$\begin{array}[t]{rcl} a(u,v)&=&\int_0^1 x^2u'(x)v'(x)\text{ dx}\\ l(v)&=&\int_0^1 v(x)\text{ dx}. \end{array}$$

So we got different $a$ with $x^2$ on the right hand side and this must be some another common PDE, and I can't see which.

• What about $(x^2 u')' = 1$ with $u(0) = u(1) = 0$? Commented Dec 2, 2014 at 12:04
Clearly, if you plug in test function you fine out that your variational formulation is equivalent to $$x^2u''+2xu'= (x^2u')' = 1, ~~~in ~~~~(0,1)~~~~and ~~~~ u(0)=u(1)=0.$$