haemhweg
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 Jan4 awarded Yearling Dec17 awarded Caucus Dec11 revised Why there is this kind of relation between power and factorial? deleted 38 characters in body Dec2 asked Is it possible to transform inhomogeneous Robin boundary problem to homogeneous Robin boundary problem? Dec2 accepted Prove that partial differential equation has no weak solution Dec2 comment Prove that partial differential equation has no weak solution Thank you for your profound elaboration, this is exactly what I was looking for. My problem was that I could really figure out how $v$ should look like to fulfill given conditions and still stay in $H^1_0(0,1)$. Dec2 comment Variational form of boundary value problem Yes it will do, thank you. Dec1 comment Prove that partial differential equation has no weak solution I am not sure because I suppose weak solutions can differ from classical ones, so $u$ could have another form rather than one above. Dec1 comment Prove that partial differential equation has no weak solution So would it be right to tell that since it is impossible for classical solution $u(x)=\frac{c_1}{x}+c_2+\text{log}(x)$ to have $u(0)=0$, what is a necessary condition for $u$ to be in $H^1_0(0,1)$, then we can't solve our problem in $H^1_0(0,1)$? Dec1 revised Prove that partial differential equation has no weak solution deleted 105 characters in body Dec1 asked Prove that partial differential equation has no weak solution Dec1 asked Variational form of boundary value problem Nov30 accepted For which $s$ is the function $(||x||^{s-2}x_i)^2$ integrable on the unit ball of $\mathbb R^n$? Nov25 asked For which $s$ is the function $(||x||^{s-2}x_i)^2$ integrable on the unit ball of $\mathbb R^n$? Oct13 awarded Notable Question Jul30 awarded Popular Question Jul2 accepted Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$? Jul2 comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$? Thank you, nice work. Since the most work was done, I'll accept your answer. Only detail I wanted to tweak is that in my second comment here equality $f(e_1\cdot[w])=f(e_2)$ doesn't hold, but this doesn't affect surjectivity I was about to prove anyway. Jul2 comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$? In order to prove surjectivity, let $e\in E$. Pick any other $e_1 \in E$ and let $f(e_1)=e_2$. Now let $w$ be the path between $e_2$ and $e$ (exists, since $E$ path-connected). Then $e = e_2\cdot[w] = f(e_1)\cdot[w] = f(e_1 \cdot [w]) = f(e_2)$. So $f$ is surjective, and bijective as it maps into itself. Does this look sound? Jul2 comment Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$? Concerning bijectivity: in my another question where you also engaged in discussion I have defined an action of the fundamental group on $E$ (path (homotopy class) in $X$ maps each point in $E$ to the endpoint of the unique lifted path) and proved that $f(e\cdot [w]) = f(e) \cdot[w]$, where $f$ is a continuous map on $E$ having $pf=p$ and $e\cdot[w]$ is aforementioned action.