769 reputation
17
bio website
location
age 27
visits member for 1 year, 2 months
seen Jun 25 at 16:45

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
15
comment Short time existence and Maximum Principle for an explicit parabolic PDE
You define $M(t)=max_x f(x,t)=f(x_t,t)$. If everything is smooth, $M(t)$ is lipschitz. Then, using Rademacher, $M'(t)$ exists a.e.. Moreover, you can verify that $$ M'(t)=\partial_t f(x_t,t). $$
Feb
15
comment Short time existence and Maximum Principle for an explicit parabolic PDE
You can substract $h(x)=kx/D$. Now $\Phi(x)=\psi(x)+h(x)$ verifies the boundari conditions $\Phi(0)=0, \Phi(D)=0$. The equation for $\Phi$ remains almost the same: The only change comes from $$ \psi\psi'=(\psi+h-h)(\psi'+k/D-k/D)=(\Phi-h)(\Phi'-k/D), $$ so the nonlinear term reads $$ (\Phi-h)(\Phi'-k/D)=\Phi\Phi'-k\Phi/D-h\Phi'+hk/D $$ The evolution for the $\max_x\Phi$ satisfies $$ \frac{d}{dt}\|\Phi(t)\|_{L^\infty}\leq k(h(x_t)-\|\Phi(t)\|_{L^\infty})/D $$
Feb
15
comment Short time existence and Maximum Principle for an explicit parabolic PDE
You can use Galerkin method to show the existence. Also notice that if $V=k=0$ you have a maximum principle for $\|u(t)\|_{L^\infty}$ (use Rademacher theorem for $\max_x u(x,t)=u(x_t,t)$). In fact, if I understand well, your equation is called Burgers' equation and should have a very long literature, right?
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