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seen Nov 20 at 21:44

Jun
12
comment Convergence weak* in $L^{\infty}([0,T];L^2(\mathbb{T}^2))$ implies convergence weak* a.e in $L^2(\mathbb{T}^2)$?
Why does $x_n$ tend weakly to zero? I see that this should be true, but I don't see the proof now. Please, can you elaborate?
Jun
4
accepted A bound in Sobolev spaces of negative order
Jun
3
asked A bound in Sobolev spaces of negative order
May
15
awarded  Yearling
May
6
comment Bounded data means bounded solution to parabolic PDE
Concerning you comment on the sign of the laplacian: the approximate solution will be as smooth as you want, isn't it? However, the problem with the boundary is pertinent. Even if you mollify the initial $L^\infty$ data (which gives you $C^1$ function with the appropriate $L^\infty$ bound) there is a problem close to the boundary.
May
6
comment Bounded data means bounded solution to parabolic PDE
Well, I don't know it it's seems unlikely or not. You can NOT establish differentiability. What you do is to prove that M(t) is Lipschitz, ergo, it is differentiable a.e.. Once you have this, you prove that $M'(t)=\partial_t u(x_t,t)$. Of course that $M(t)$ may be attained in more than one point, however, this is not relevant since at every point will decay. This "overcomplicated" approach is widely spread (check this, for instance, uam.es/personal_pdi/ciencias/acordoba/documentos/articulos/…)
Apr
30
comment Bounded data means bounded solution to parabolic PDE
Actually you need $C^1([0,T]\times\Omega)$ for the approximated solutions.
Apr
30
answered Bounded data means bounded solution to parabolic PDE
Apr
29
answered PDE Evans Chapter 7 problem 16
Apr
17
answered Intregation by parts
Apr
15
answered Solving the reaction-diffusion equation for a single species
Mar
6
asked Bound on the inverse laplacian
Feb
22
answered Behaviour of nonlinear heat equation
Feb
13
comment PDE problem for 1D
By Sobolev embedding I mean the inequality $\|u\|_{L^\infty}\leq c\|u\|_{H^1}$
Feb
13
answered PDE problem for 1D
Feb
10
answered Getting a bound on solution of PDE in $L^\infty(0,T;L^2(\Omega))$?
Feb
10
revised Understanding what people mean in PDEs
added 459 characters in body
Feb
10
revised Understanding what people mean in PDEs
added 459 characters in body
Feb
10
answered Understanding what people mean in PDEs
Jan
26
comment Extending weak solution to global weak solution of parabolic PDE
Well, as long as the galerkin coefficients exist, you will have a weak solution. However, notice that a solution such that $\limsup_{t\rightarrow\infty}|u(t)|_{V}=\infty$ is also a global solution that can't be in $L^2((0,\infty),V)$