By Chipot M., Quittner P.
This guide is the 6th and final quantity within the sequence dedicated to desk bound partial differential equations. the subjects lined by means of this quantity comprise particularly area perturbations for boundary price difficulties, singular strategies of semi-linear elliptic difficulties, optimistic strategies to elliptic equations on unbounded domain names, symmetry of strategies, desk bound compressible Navier-Stokes equation, Lotka-Volterra structures with cross-diffusion, mounted element idea for elliptic boundary price problems.* selection of self contained, cutting-edge surveys* Written through recognized specialists within the box* Informs and updates on the entire most recent advancements
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Additional info for Handbook of Differential Equations: Stationary Partial Differential Equations
10) . 1. 10) tends to zero when M ↓ 0, while the right-hand side is at least separated away from zero. 3). 1. 8) in the form ai |Di u1 |pi −2 ∇u1 − |Di u2 |pi −2 ∇u2 ∇u dx I := Ω i c1 |u1 |σ −2 u1 − |u2 |σ −2 u2 u dx + Ω (c2 − c1 )|u2 |σ −2 u2 u dx. 11) we see that the first term on the left-hand side of this inequality is nonnegative. 11) and boundedness of u, we obtain the following chain of relations: |u|2 dx λ0 Ω I |u|2 dx λ1 L Ω ⇒ (λ0 − λ1 L) |u|2 dx Ω 0 38 S. Antontsev and S. Shmarev with the constants λ0 = c0 min σ (x) − 1 2M σ (x) (σ (x)−2)/σ (x) Ω , and λ1 = max M σ (x)−1 .
Because of the absolute continuity of the integral LΩ\Ωε (vk , η) − LΩ\Ωε (v, η) C vk V + v V η V(Ω\Ωε ) → 0 when ε → 0. On the other hand, LΩε (vk , η) − LΩε (v, η) → 0 when k → ∞ for every fixed ε. Thus, for every η ∈ PN (g(vk ) − g(v), η)N → 0, which means that g(vk ) → g(v) weakly in PN . Since the dimension of PN is finite, the weak convergence g(vk ) → g(v) implies the strong convergence: |g(vk ) − g(v)|N → 0 when k → ∞. 4. 8) satisfies the estimate u(N ) V K with a finite constant K independent of N .
U ∈ W0 p(x) dx |u| Ω λ1 = inf Ω The following example taken from  shows that λ1 may be zero. Let Ω = (−2, 2) ⊂ R, p(x) = 3 4 − |x| if 0 if 1 x 1, |x| 2, u(x) = 1 2 − |x| for 0 for 1 and 1,p(x) Obviously, au(x) ∈ W0 shows, however, that au (x) p(x) |x| |x| 1, 2. (Ω) for every a = const > 0. The straightforward computation dx = Ω 2a 2 (a − 1), ln a |au|p(x) dx 2a 3 , Ω whence lim a→∞ |Du|p(x) dx = 0. 1) becomes very complicated. 1). 1). 2) Ω which is true for p = const as well as for p(x) being a function of x.