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Jan
26
comment Show that elements $u \in W^{1,\infty}(U)$ have continuous representatives
possible duplicate of relation between $W^{1,\infty}$ and $C^{0,1}$
Jan
24
reviewed Approve Help me understand procedure of integrals.
Jan
21
comment why $ \nabla v_n \to \nabla v \ \ (a.e.)$ and $ v_n \to v $
You are welcome @shi'looo
Jan
21
revised $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
added 15 characters in body
Jan
21
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
@Voliar, I think we should explain that, the result proposed by OP is not true. What do you think?
Jan
21
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
Sure @Voliar, you can do as you wish.
Jan
21
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
@Voliar, I see. Well, so I proved another thing. Have your time to edit my answer and add this observation? Thank you
Jan
21
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
@Voliar, can you see any error in my calculations above?
Jan
20
comment why $ \nabla v_n \to \nabla v \ \ (a.e.)$ and $ v_n \to v $
You can use Calderón-Zygmund estimates to prove that $\|u_n\|_{3,q}$ is bounded, which implies the a.e. convergence of $\nabla v_n$, however, it is worth to note that the a priori estimate $\|v_n\|_{1,q}\le C\|f_n\|_1$ is true, so you can avoid all this problem and conclude directly that $v_n\to v$ strongly. Take a look here and references therein.
Jan
20
comment Video lectures on Partial Differential Equations
@Akitirija, take a look in Impa's lectures. Most of them are in portuguese.
Jan
20
revised Video lectures on Partial Differential Equations
edited body
Jan
20
comment Video lectures on Partial Differential Equations
You are right @Jose27, I will fixt it, thank you.
Jan
19
answered Video lectures on Partial Differential Equations
Jan
18
comment if $\Delta u = |\nabla u|^{3/2}$ then $u \in C^{\infty}$
Here is an idea: your problem is in the points where $\nabla u=0$, however, in these points, you must have $\Delta u=0$, hence, your function is harmonic in these points. We know that harmonicity implies regularity, however, this is done when $u$ is harmonic in some open domain. Maybe you can adapt the proof for the case of a point?
Jan
18
comment The obivious “why”-questions
Why so many questions?
Jan
18
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
@Winther, only for the case where $\beta=1$. In this case, Fredholm theory does apply and we can find a sequence of eigenvalues.
Jan
18
revised $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
deleted 258 characters in body
Jan
18
comment $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
Have your read my answer @Winther? I have proved it for all $\alpha>0$. You are right about your first comment, I will delete my first paragraph.
Jan
17
comment This Sobolev function is continuous?
What is $r$? Is it the radius of the ball $B$?
Jan
17
revised $-\Delta u - \alpha u^{1/3} = 0$ implies $u \equiv 0$ if $\alpha$ is small
edited tags