For a Dirichlet problem $$\begin{array}[t]{rcl}Lu&=&-\frac{\text{d}}{\text{d}x}(x^2\cdot u'(x))=1=:f\,\,\,\text{ in }(0, 1)\\ u&=&0\,\,\,\text{ in }\{0,1\} \end{array}$$
there is a corresponding variational form of this problem for obtention of weak solution $u\in H^1_0(0,1)$:
$$\int_0^1x^2u'(x)v'(x)\text{d}x=\int_0^1v(x)\text{d}x \,\,\,\forall v\in H^1_0(0,1).$$
The task is to prove that no such $u\in H^1_0(0,1)$ exists, i.e. the aforementioned weak equation has no solution in $H^1_0(0,1)$.
Notice that $$u(0)=u(1)=v(0)=v(1)=\int_0^1v'(x)\text{d}x=\int_0^1u'(x)\text{d}x=0.$$
I thought it might work by assuming the statement is true and finding appropriate $v$ for counterexample, but all my effort until now was in vain.
Thank you in advance.
First of all, let us define: $$a(f,g)=\int_0^1{x^2f'g'\mathrm{d}x},\\F(v)=\int_0^1{v\mathrm{d}x}$$
Now, since $a$ is a continuous bilinear form, $F$ is continuous, and $H_0^1(0,1)$ happens to be a Hilbert space, the Lemma of Lax-Milgram would guarantee us a solution $u$ if only $a$ was coercive.
Meaning, any counterexample we want to think of must make use of that fact.
If we look at $a$, the intuitive reason it is not coercive is that we can can take any series of functions with constant norm and small support, and shift them closer and closer to the left bound of the interfal - and since $x^2$ goes to $0$, so will our bilinear form. (Obviously, this is far from any kind of proof, just to illustrate - all we need is the idea.)
So, we want to find a series of functions $v_n$, such that, for any $u$, $F(v_n)=c$ for some constant $c$, whereas $a(u,v_n)\to 0$.
To get a lower bound of $a(u,v_n)$ and get rid of what we cannot choose freely - namely, $u$ - we use the Cauchy inequality:
$|a(u,v_n)|=|\int_0^1{x^2u'v_n'\mathrm{d}x}|\leq \|u'\|_{L^2}\|x^2v_n'\|_{L^2}$
and since, given a solution $u$, the left factor is constant, all we need to do is find $v_n$ such that $F(v_n)=c, \|x^2v'_n\|_{L^2}\to 0$.
Now, I'm not sure how much you have tried until now. In case you didn't really know what to look for, I suggest stopping at this point and trying to think of something for a while. If this is no more than you were already aware of or you're running out of ideas, feel free to continue. (And if anyone could tell me how to put spoilers onto formulas, I'd be very grateful. Could be exceptionally useful here.)
Continued solution
Since all we want is a $H_0^1$-function, which does not have to be differentiable, we can work with fairly simple functions. Constant doesn't quite work yet, thus we work with linear ones. Take a look at $$v_n(x) = \begin{cases} n^2x, & 0\leq x<\frac{1}{n} \\ n-n^2(x-\frac{1}{n}), &\frac{1}{n}\leq x<\frac{2}{n}\\ 0 & \text{else} \end{cases}$$
which is a series of increasingly narrower hat-functions with $F(v_n)=1$.
Now, just hoping that this series fulfills what we want it to:$$\|x^2v'_n\|_{L^2}^2=\int_0^1{x^2v'_n\mathrm{d}x}=\int_0^\frac{1}{n}{x^4n^4\mathrm{d}x}+\int_\frac{1}{n}^\frac{2}{n}{x^4n^4\mathrm{d}x}=n^4\int_0^\frac{2}{n}{x^4\mathrm{d}x}=n^4\frac{32}{5n^5}=\frac{32}{5n}\to 0$$
and, behold, it does.
To summarize:
We have found a series of functions $v_n\in H_0^1(0,1)$, such that $F(v_n)=1$, and $a(u,v_n)\leq \|u'\|_{L^2}\|x^2v'\|_{L^2}\to 0$.
However, given any solution $u$, we would need $F(v_n)=1=a(u,v_n) \quad\forall v_n\in H_0^1(0,1)$, and we have our contradiction.