bio | website | |
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visits | member for | 1 year, 7 months |
seen | 8 hours ago | |
stats | profile views | 281 |
Dec 17 |
awarded | Popular Question |
Nov 6 |
accepted | Reference request: Wasserstein metric spaces for non linear weights/mobility? |
Nov 6 |
comment |
Reference request: Wasserstein metric spaces for non linear weights/mobility?
A quick look indicates that this is exactly what I needed, thank you Nicola. |
Nov 6 |
revised |
Reference request: Wasserstein metric spaces for non linear weights/mobility?
added 15 characters in body |
Nov 6 |
asked | Reference request: Wasserstein metric spaces for non linear weights/mobility? |
Aug 12 |
comment |
Reference request: density of $C_c^{\infty}(\mathbb R^d)$ in $L^2(\mathbb R^d,d\rho)$
OK, thanks @Christian Remling. Any specific reference I could cite? I was suspecting that, but for example I'm having a hard time figuring out what $L^2(\mathbb R^d,\delta_0)$ looks like (Dirac mass)... |
Aug 12 |
asked | Reference request: density of $C_c^{\infty}(\mathbb R^d)$ in $L^2(\mathbb R^d,d\rho)$ |
Aug 4 |
comment |
Spectrum of this ODE
what about the pairwise eigenfunctions, do they have any remarkable "pairwise" behaviour (odd/even, or whatever...)? |
Aug 1 |
comment |
Weak convergence of a sequence
Indeed, my argument really shows that $\nabla u_k\rightharpoonup \nabla u$ in $L^2L^2$, I just kept using $\nabla \Psi$ as test functions in order to stay consistent with the OP's setting. But I disagree with $\Delta u \in L^2(0,T; H^1_0)$, the above inequality gives $\Delta u \in L^2(0,T; H^{-1})$! |
Aug 1 |
answered | Weak convergence of a sequence |
Jul 11 |
comment |
Want to show rigorously $\frac{d}{dt}\int_{\Omega}|u(t)|^r = r\langle u_t(t), |u(t)|^{r-2}u(t)\rangle_{H^{-1}(\Omega), H^1(\Omega)}$
what about $C_c^{\infty}((0,T)\times\Omega)$? |
Jul 11 |
comment |
Want to show rigorously $\frac{d}{dt}\int_{\Omega}|u(t)|^r = r\langle u_t(t), |u(t)|^{r-2}u(t)\rangle_{H^{-1}(\Omega), H^1(\Omega)}$
You should specify the range of $r$ (which I guess is $r\in[2,\infty)$). This should work by approximation. I would recommend writing $|u|^{r-1}\text{sgn}(u)=|u|^{r-2}u$, since the sign function is typically not very well behaved in Sobolev spaces (e.g. $\text{sgn}(u)\notin H^1$). |
Jul 4 |
comment |
$L^p$ norm of solution to porous medium equation decreases in time: how to make formal calculation rigorous?
For non-negative solutions you can use approximation with standard $L^1$-contractivity. For signed solutions I have no idea, I don't think the argument is so straightforward... |
Jul 2 |
awarded | Curious |
Jun 30 |
comment |
$b_n \rightharpoonup b$ in $L^q(Q) \forall q < \infty$, $b_n \to b$ in $C^0([0,T];H^{-1})$ implies $b_n(t) \rightharpoonup b(t)$ in $L^q(\Omega)$
perhaps you should tell us what is the connection between $u_n$ and $b_n$? |
Jun 27 |
comment |
Getting existence for $L^1$ data given existence for $L^\infty$ data and $L^1$ continuous dependence result
By the strong $L^1(Q)$ convergence $F(u_n)\to v$ you get $F(u_n)(t,x)\to v(t,x)$ a.e. $t,x$. By continuity of $F^{-1}(z)=z^m$ you see that $u_n(t,x)\to u(t,x)=F^{-1}(v)(t,x)$ a.e., so all you need now is prove that $u_n\to u$ in $L^1$. By dominated convergence it should be enough to prove some uniform $L^1(Q)$ bounds. Have you tried taking $\varphi=u$ as a test function in your strong formulation? I guess it should give you an $L^{\infty}(0,T;L^{1+1/m})$ estimate since formally $u\partial_t F(u)= C\partial_t(u^{1+1/m})$. But that's just a suggestions... |
Jun 23 |
comment |
Comparison principle using truncation for porous medium equation
Well, typically you may want to assume something about $\nabla \Phi(u)$ rather than for $\nabla u$, and the information one usually either assumes or gets somehow is $\nabla\Phi(u)\in L^2(0,T;L^2)$. This means that you're actually considering weak energy solutions, not just very weak solutions. That's why I was saying that dual methods seem to be in order here (at least to the best of my knowledge). |
Jun 23 |
comment |
Want to show $\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle = \int_\Omega |u(T)| - \int_\Omega |u(0)|$
It's really like saying that $\frac{d}{dt}|u|^2_{L^2}=2<u_t,u>_{H^{-1},H^1_0}$. Indeed approximation should work but this is classical. The key argument is that $T_{\epsilon}$ is lipschitz so with your assumptions $S_{\epsilon}'(u)=T_{\epsilon}(u)/\epsilon$ is in $L^2(0,T;H^1_0)$. But note carefully that, in my chain rule, I'm not integrating by parts in time as you wrote in your OP, this is the point (apply the chain rule to $S_{\epsilon}$, not $T_{\epsilon}$). |
Jun 23 |
revised |
Want to show $\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle = \int_\Omega |u(T)| - \int_\Omega |u(0)|$
deleted 28 characters in body |
Jun 23 |
answered | Want to show $\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle = \int_\Omega |u(T)| - \int_\Omega |u(0)|$ |