By Nassif Ghoussoub
ISBN-10: 0387848967
ISBN-13: 9780387848969
This textual content is meant for a starting graduate direction on convexity equipment for PDEs. The generality selected by way of the writer places this less than the class of “functional analysis”. The functions, although, require a good wisdom of classical research and PDEs that's had to make really apt offerings of functionality areas the place the self-dual variational ideas must be utilized, and those offerings unavoidably require past wisdom of the predicted regularity of the (weak) recommendations. whereas this article includes many new effects, it's the author’s desire that this fabric will quickly develop into average for all graduate students.
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Additional info for Self-dual Partial Differential Systems and Their Variational Principles
Sample text
4 (3), to get that both Lλ1 = L Mλ and (Lλ1 )∗ = (L Mλ )∗ = L∗ ⊕Mλ = (L∗ )2λ are self-dual and uniformly convex in both variables. 12) above – that Jλ (x, p) = x − λ ∂1 Lλ (x, p) is Lipschitz as well. E. More on regularizations of self-dual Lagrangians 1. Assume p > 1 and (x, r) ∈ X × X ∗ 1 p + 1 q = 1. 13) and Lλ2 ,p (x, r) = inf L(x, s) + r−s λq q + λ p−1 x p p ;s ∈ X∗ . 3. 2. Consider a self-dual Lagrangian of the form L(x, p) = ϕ (x) + ϕ ∗ (Ax + p), where ϕ is a convex lower semicontinuous function on a reflexive Banach space X and A : X → X ∗ is a bounded skew-adjoint operator.
7. Assume that L : H × H → R ∪ {+∞} is a Lagrangian on a Hilbert space H such that both L and L∗ are uniformly convex in both variables. Then, for all x, u ∈ H, there exists a unique v ∈ H, denoted v = R(u, x), such that x = ∂2 L(u, v). Moreover, the map (u, x) → R(u, x) is jointly Lipschitz on H × H. 2 2 Proof. Since L is uniformly convex, then L(x, p) = M(x, p)+ ε x2 + p2 , where M is convex lower semicontinuous in such a way that x = ∂2 L(u, v) if and only if 0 ∈ ∂2 M(u, v) + ε v − x if and only if v is the solution to the minimization problem 58 3 Self-dual Lagrangians on Phase Space min M(u, p) + p ε p 2 2 − x, p .
Proof. Assume first that the pair (B, Γ ) is antisymmetric and that L is a Lagrangian that is continuous on X in the first variable. We first consider the case where y ∈ D(Γ ). Indeed, write LΓ∗ (B∗ q, By) = sup { By, p + x, B∗ q − L(Bx, Γ x + p); x ∈ D(Γ ), p ∈ X ∗ } . Substituting r = Γ x + p and since for y ∈ D(Γ ) we have By, Γ x = − Bx, Γ y , and using that L is continuous in the first variable, we obtain LΓ∗ (B∗ q, By) = sup { By, r − Γ x + Bx, q − L(Bx, r)} x∈D(Γ ) r∈X ∗ = sup x∈D(Γ ) r∈X ∗ Bx, Γ y + By, r + Bx, q − L(Bx, r) = sup Bx, Γ y + q + By, r − L(Bx, r); x ∈ D(Γ ), r ∈ X ∗ = sup z, Γ y + q + By, r − L(z, r); z ∈ X, r ∈ X ∗ = L∗ (Γ y + q, By).
Self-dual Partial Differential Systems and Their Variational Principles by Nassif Ghoussoub
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