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1
vote
Finding Free surface elevation in semi-infinite channel
What I seem to understand is that the bottom is at $y=-h+v(x-ct)$ where $v$ has compact support. That corresponds to an obstacle moving at a constant speed towards the infinite end. But it might mean …
2
votes
Stationary Navier-Stokes solutions
Tai-Peng Tsai's book Lectures on Navier-Stokes Equations (2018) cites as Theorem 8.3 (p.149) a theorem of V. Sverak (2011) that excludes the existence of minus one homogeneous solutions on $\mathbb R^ …
6
votes
1
answer
423
views
Stationary Navier-Stokes solutions
Are there known nontrivial ($u\neq0$) stationary solutions to Navier-Stokes equations in $\mathbb R^3$ ? Not square integrable of course (that's impossible), but with self-similar amplitudes of Fourie …
2
votes
Reformulation of the classical Navier-Stokes equation as a semilinear evolution equation and...
You are right, $(4)$ has to be understood in the special sense that the semigroup $S(t)$ (and the (Leray?) $L^2$-projection operator to $H$, as well) extend to a wider space, that of "distributional d …