Some challenges of four-dimensional data assimilation problems
Juan Carlos De los Reyes
Centro de Modelización Matemática (MODEMAT) Escuela Politécnica Nacional de Ecuador New trends in PDE constrained optimization Linz, 2019
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Some challenges of four-dimensional data assimilation problems Juan - - PowerPoint PPT Presentation
Some challenges of four-dimensional data assimilation problems Juan Carlos De los Reyes Centro de Modelizacin Matemtica (MODEMAT) Escuela Politcnica Nacional de Ecuador New trends in PDE constrained optimization Linz, 2019 1 / 30
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p(z)
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u
l
i
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u
l
i
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u
l
i
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u
B−1 ∂y ∂t + Ay + g(y) =
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u
B−1 ∂y ∂t + Ay + g(y) =
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u
T
B−1 + ϑ
L2(Ω) ∂y ∂t + Ay + g(y) =
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0(Ω) and the nonlinear term verifies the assumption, then the semilinear
0(Ω)), for some c > 0. 12 / 30
0(Ω) → H2,1(Q), u → S(u) = y, is Gâteaux
0(Ω), is given by η ∈ H2,1(Q)
∂η ∂t + Aη + g′(y)η =
0 (Ω)), with
m m−1[. Casas-Clason-Kunisch 2013, Meyer-Susu 2017
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∂¯ y ∂t + A¯
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B−1, which is not appropriate to reconstruct sharp fronts
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B−1, which is not appropriate to reconstruct sharp fronts
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B−1, which is not appropriate to reconstruct sharp fronts
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B−1, which is not appropriate to reconstruct sharp fronts
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B−1, which is not appropriate to reconstruct sharp fronts
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w,σ∈{0,1}
N
j ) dxdt + β
N
j ) dx + βw
uj
T
2 ∇(uj − ubj)2 L2(Ω)
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w,σ∈{0,1}
N
j ) dxdt + β
N
j ) dx + βw
j , y† j
j , y† j
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0≤w,σ≤1
N
j ) dxdt + β
N
j ) dx + βw
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0≤w,σ≤1
N
j ) dxdt + β
N
j ) dx + βw
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0≤w,σ≤1
N
j ) dxdt + β
N
j ) dx + βw
x ǫ
2
2 < x ≤ 2ǫ
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0≤w,σ≤1 J(y, p, u, w) =
N
j ) dxdt + β
N
j ) dx + βw
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0≤w,σ≤1 J(y, p, u, w) =
N
j ) dxdt + β
N
j ) dx + βw
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0≤w,σ≤1 J(y, p, u, w) =
N
j ) dxdt + β
N
j ) dx + βw
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0 (Ω)) × H1 0(Ω) solution of the bilevel problem, find
0 (Ω)) × Vad × L2(Q) × L2(Q) that solves:
w,s,v Jγ(y, p, w, s, v) = J(y, p, w) + γ
L2(Q) + 1
0 :=
0(Ω).
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0 (Ω)) × Vad × L2(Q) × L2(Q) and
mr m−r;
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0 (Ω)) × Vad × L2(Q) × L2(Q) and
mr m−r;
0(Ω), for all
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N
T
k − λb k,
N
T
ns+i − λb ns+i,
r ≥ 0, λb r ≥ 0,
k ¯
k(¯
ns+i¯
ns+i(¯
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k Hk
k Hksk
k (y# k )T
k y# k
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k Hk
k Hksk
k (y# k )T
k y# k
k (y# k )T
k )Ts# k
k (s# k )T
k )Ts# k
k (s# k )T
k )Ts# k
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1 2k∇f(w0), k = 0, 1, . . .
y
y2+ε2
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