Non-existence of weak solution in one dimension

Question

Let $\Omega=(1,\infty).$ Then for any given $f\in L^2(\Omega),$ the equation $$ -u''=f\,\,\text{in}\,\,\Omega, $$ does not admit any weak solution in $W_{0}^{1,2}(\Omega).$

I tried the solution by contradictory argument. Indeed, suppose such a weak solution exists, say $u.$ Then for every $\phi\in H_0^{1}(\Omega),$ we have $$ \int_{\Omega}\,u'\phi'\,dx=\int_{\Omega}f\phi\,dx $$ I tried to construct $\phi_n\in C_c^{\infty}(\Omega)$ such that the above inequality become false. I think this idea will work, but I am still unable to construct such $\phi.$ If anyone can help me regarding this or with some other idea on solving this problem, it will be very grateful for me. Thanks.

Answer 1

I tried to sow some doubt about the question yesterday, by giving a couple of examples, where a weak solution actually exists in the comments.

Below is a failed attempt to show that a weak solution always exists, but as pointed out by daw in a comment I failed at properly checking the coercivity. You can consider it as an example on how not to prove something.

For every $f\in L^2(\Omega)$ the equation $$ -u'' = f $$ admits a weak solution $u\in H^1_0(\Omega)$.

By multiplying with test functions and partial integration you get the weak formulation: \begin{equation}\tag{1} a(u,v) := \langle u',v'\rangle_{L^2} = \langle f, v\rangle_{L^2}, \end{equation} for all $v \in C^\infty_0(\Omega)$. By density this statement also holds for all $v \in H^1_0(\Omega)$. One easily verifies that $a$ is a coercive (this is not true), bounded sesquilinear form on $H^1_0(\Omega)$. The theorem of Lax-Milgram shows that $u\in H^1_0(\Omega)$ such that (1) holds exists if $\langle f, \cdot \rangle_{L^2} \in (H^1_0(\Omega))'$. That $\langle f, \cdot \rangle_{L^2} \in (H^1_0(\Omega))'$ is clear by the Cauchy-Schwarz inequality.

Answer 2

Let me take $\Omega=(0,+\infty)$ for convenience. Define $$ u(x) = \min(x,x^{-1/2}). $$ Then $u\not\in L^2(\Omega)$ and $u\in L^1_{loc}(\Omega)$. It has first and second weak derivatives $u'$ and $u''$ given by $$ u'(x) = \chi_{(0,1)} -\frac12 \chi_{(1,\infty)} x^{-3/2} $$ and $$ u''(x) = \frac34 \chi_{(1,\infty)} x^{-5/2}. $$ It follows $u',u''\in L^2(\Omega)$. In addition, $u$ (rather $u'$) satisfies the weak formulation. Moreover, the weak formulation has at most one solution since $\int_\Omega (u')^2=0$ implies $u=const$ hence $u=0$.

Now take $f=-u''$. Then you have $f\in L^2(\Omega)$ with non-existent weak solution in $H^1(\Omega)$.