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visits member for 2 years, 11 months
seen Dec 17 at 10:15

Dec
17
awarded  Caucus
Dec
11
revised Why there is this kind of relation between power and factorial?
deleted 38 characters in body
Dec
2
asked Is it possible to transform inhomogeneous Robin boundary problem to homogeneous Robin boundary problem?
Dec
2
accepted Prove that partial differential equation has no weak solution
Dec
2
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 $H^1_0(0,1)$.
Dec
2
comment Variational form of boundary value problem
Yes it will do, thank you.
Dec
1
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.
Dec
1
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 $H^1_0(0,1)$?
Dec
1
revised Prove that partial differential equation has no weak solution
deleted 105 characters in body
Dec
1
asked Prove that partial differential equation has no weak solution
Dec
1
asked Variational form of boundary value problem
Nov
30
accepted For which $s$ is the function $(||x||^{s-2}x_i)^2$ integrable on the unit ball of $\mathbb R^n$?
Nov
25
asked For which $s$ is the function $(||x||^{s-2}x_i)^2$ integrable on the unit ball of $\mathbb R^n$?
Oct
13
awarded  Notable Question
Jul
30
awarded  Popular Question
Jul
2
accepted Is continuous map from covering space to itself homeomorphism assumed both cover and base path-connected and $pf=p$?
Jul
2
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.
Jul
2
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?
Jul
2
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.
Jul
2
awarded  Curious