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By Etienne Emmrich, Petra Wittbold

This article features a sequence of self-contained reports at the cutting-edge in several components of partial differential equations, provided through French mathematicians. subject matters comprise qualitative homes of reaction-diffusion equations, multiscale tools coupling atomistic and continuum mechanics, adaptive semi-Lagrangian schemes for the Vlasov-Poisson equation, and coupling of scalar conservation laws.

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Extra info for Analytical and numerical aspects of partial differential equations : notes of a lecture series

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Chechkin and Andrey Yu. 10. 29). ) Conclusion. , up to the time T there is no dissipation of the kinetic energy; the kinetic energy stays constant on [0, T ). 29), we have dE < 0, dt so that the kinetic energy dissipates (on a shock wave, a part of it is transformed into heat). 1) irreversible. The readers who sometimes spend vacations at the sea are probably acquainted with this phenomenon. Near the shore, if the sea is calm and the waves are temperate, the sea temperature near the surface is almost the same as the air temperature above.

19′ ) means that the chord Ch with the endpoints (u− , f (u−)), (u+, f (u+ )) has a smaller slope (the slope is measured as the inclination of the chord with respect to the positive direction of the u-axis) than the slope of the segment joining the point (u− , f (u− )) with the point (u, f (u)), where u runs over the interval (u− , u+ )). Consequently, the point (u, f (u)) and thus the whole graph of f = f (u) on the interval (u− , u+ ) lies above the chord Ch. 20′ ) signifies that the graph of f = f (u) for u ∈ (u+ , u− ) is situated below the chord Ch.

As previously, we denote by u± = limx→x(t)±0 u(t, x) the one-sided limits (limits along the x-axis) of the solution u on the discontinuity curve. Then dS = u(t, x(t) − 0) · x˙ (t) + dt x(t) ut (t, x) dx −∞ +∞ − u(t, x(t) + 0) · x˙ (t) + ut (t, x) dx x(t) x(t) = (u− − u+ ) · x˙ (t) − +∞ f (u(t, x)) −∞ x f (u(t, x)) dx − x(t) x dx = (u− − u+ ) · x˙ (t) − f (u(t, x(t) − 0)) + f (u(t, −∞)) − f (u(t, +∞)) + f (u(t, x(t) + 0)) = (f (u+ ) − f (u−)) − (u+ − u− ) · x˙ (t). 2) itself, we took advantage of the fact that u has compact support in x, so that f (u(t, −∞)) = f (u(t, +∞)) = f (0).

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