- 1. 解析半群を用いた半線形放物型 方程式の解に対する精度保証付 き数値計算法とその応用 高安 亮紀 早稲田大学 基幹理工学部 応用数理学科 第 3 回 数理人セミナー＠早稲田大学西早稲田キャンパス 2015 年 1 月 15 日 1/55
- 4. 共同研究者 ▶ 水口 信 （早稲田大学 基幹理工学研究科 数学応用数理専攻） ▶ 久保 隆徹 （筑波大学 数理物質系） ▶ 大石 進一 （早稲田大学 基幹理工学部 応用数理学科） 4/55
- 5. 半線形放物型方程式 Let Ω be a bounded polygonal domain in R2 . (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(t0, x) = u0(x) in Ω, ▶ J := (t0, t1], 0 ≤ t0 < t1 < ∞ or J := (0, ∞), ▶ f : twice diﬀerentiable nonlinear mapping, ▶ u = 0 on ∂Ω is the trace sense, ▶ u0 ∈ H1 0 (Ω). Lp (Ω): the set of Lp -functions, H1 (Ω): the ﬁrst order Sobolev space of L2 (Ω), H1 0 (Ω) := {v ∈ H1 (Ω) : v = 0 on ∂Ω}. 5/55
- 6. 記号 A : D(A) ⊂ H1 0 (Ω) → L2 (Ω) is deﬁned by A := − ∑ 1≤i,j≤2 ∂ ∂xj ( aij(x) ∂ ∂xi ) , where aij(x) = aji(x) is in W1,∞ (Ω) and satisﬁes ∑ 1≤i,j≤2 aij(x)ξiξj ≥ µ|ξ|2 , ∀x ∈ Ω, ∀ξ ∈ R2 with µ > 0. 6/55
- 7. 記号 We endow L2 (Ω) with the inner product: (u, v)L2 := ∫ Ω u(x)v(x)dx. Use the usual norms: ∥u∥L2 := √ (u, u)L2 , ∥u∥H1 0 := ∥∇u∥L2 , and ∥u∥H−1 := sup 0̸=v∈H1 0 (Ω) ∥v∥H1 0 =1 |⟨u, v⟩| , where ⟨·, ·⟩ is a dual product between H1 0 (Ω) and H−1 (Ω)1 . 1 The topological dual space of H1 0 (Ω). 7/55
- 8. 精度保証付き数値計算 (PJ ) の弱解: For t ∈ J, u(t) := u(t, ·) ∈ H1 0 (Ω) with the initial function u0 such that (∂tu(t), v)L2 + a(u(t), v) = (f(u(t)), v)L2 , ∀v ∈ H1 0 (Ω), where a : H1 0 (Ω) × H1 0 (Ω) → R is a bilinear form: a(u, v) := ∑ 1≤i,j≤2 ( aij(x) ∂u ∂xi , ∂v ∂xj ) L2 satisfying a(u, u) ≥ µ∥u∥2 H1 0 , ∀u ∈ H1 0 (Ω), |a(u, v)| ≤ M∥u∥H1 0 ∥v∥H1 0 , ∀u, v ∈ H1 0 (Ω). 8/55
- 9. 精度保証付き数値計算 (PJ ) の弱解: For t ∈ J, u(t) := u(t, ·) ∈ H1 0 (Ω) with the initial function u0 such that (∂tu(t), v)L2 + a(u(t), v) = (f(u(t)), v)L2 , ∀v ∈ H1 0 (Ω), where a : H1 0 (Ω) × H1 0 (Ω) → R is a bilinear form: a(u, v) := ∑ 1≤i,j≤2 ( aij(x) ∂u ∂xi , ∂v ∂xj ) L2 satisfying a(u, u) ≥ µ∥u∥2 H1 0 , ∀u ∈ H1 0 (Ω), |a(u, v)| ≤ M∥u∥H1 0 ∥v∥H1 0 , ∀u, v ∈ H1 0 (Ω). 8/55
- 10. 精度保証付き数値計算 (PJ ) の弱解: For t ∈ J, u(t) := u(t, ·) ∈ H1 0 (Ω) with the initial function u0 such that (∂tu(t), v)L2 + a(u(t), v) = (f(u(t)), v)L2 , ∀v ∈ H1 0 (Ω), where a : H1 0 (Ω) × H1 0 (Ω) → R is a bilinear form: a(u, v) := ∑ 1≤i,j≤2 ( aij(x) ∂u ∂xi , ∂v ∂xj ) L2 satisfying a(u, u) ≥ µ∥u∥2 H1 0 , ∀u ∈ H1 0 (Ω), |a(u, v)| ≤ M∥u∥H1 0 ∥v∥H1 0 , ∀u, v ∈ H1 0 (Ω). 8/55
- 11. 精度保証付き数値計算 問題 (PJ ) の弱解の存在と一意性を計算機を援用し証明する． すなわち X を J × Ω 上のある Banach 空間とし，弱解を数 値解 ω を中心とする閉球: BJ (ω, ρ) := {v ∈ X : ∥v − ω∥X ≤ ρ}. 内に数学的に厳密に包み込む． 9/55
- 12. 精度保証付き数値計算に関する先行研究 M.T. Nakao, T. Kinoshita and T. Kimura, “On a posteriori estimates of inverse operators for linear parabolic initial-boundary value problems”, Computing 94(2-4), 151–162, 2012. M.T. Nakao, T. Kimura and T. Kinoshita, “Constructive A Priori Error Estimates for a Full Discrete Approximation of the Heat Equation”, Siam J. Numer. Anal., 51(3), 1525–1541, 2013. T. Kinoshita, T. Kimura and M.T. Nakao, “On the a posteriori estimates for inverse operators of linear parabolic equations with applications to the numerical enclosure of solutions for nonlinear problems”, Numer. Math, Online First, 2013. 10/55
- 13. 力学系の観点からみた研究 P. Zgliczy´nski and K. Mischaikow, “Rigorous Numerics for Partial Diﬀerential Equations: The Kuramoto―Sivashinsky Equation”, Foundations of Computational Mathematics, 1(3), 1615–3375, 2001. P. Zgliczy´nski, “Rigorous numerics for dissipative PDEs III. An eﬀective algorithm for rigorous integration of dissipative PDEs”, Topol. Methods Nonlinear Anal., 36, 197–262, 2010. 11/55
- 14. 離散半群を用いた数値スキームの研究 H. Fujita, “On the semi-discrete ﬁnite element approximation for the evolution equation ut + A(t)u = 0 of parabolic type”, Topics in numerical analysis III, Academic Press, 143–157, 1977. H. Fujita and A. Mizutani, “On the ﬁnite element method for parabolic equations, I; approximation of holomorphic semi-groups”, J. Math. Soc. Japan, 28, 749–771, 1976. H. Fujita, N. Saito and T. Suzuki, “Operator theory and numerical methods”, Elsevier(Holland), 308pages, 2001. 12/55
- 16. Considered problem J := (t0, t1] : arbitrary time interval. τ := t1 − t0. (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(t0, x) = u0(x) in Ω, where u0 is an initial function in H1 0 (Ω). Vh ⊂ H1 0 (Ω) : a ﬁnite dimensional subspace. Starts from: ˆu0, ˆu1 ∈ Vh ω(t) = ˆu0ϕ0(t) + ˆu1ϕ1(t), t ∈ J, where ϕk(t) is a piecewise linear Lagrange basis: ϕk(tj) = δkj (δkj is a Kronecker’s delta). 14/55
- 17. Considered problem J := (t0, t1] : arbitrary time interval. τ := t1 − t0. (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(t0, x) = u0(x) in Ω, where u0 is an initial function in H1 0 (Ω). Vh ⊂ H1 0 (Ω) : a ﬁnite dimensional subspace. Starts from: ˆu0, ˆu1 ∈ Vh ω(t) = ˆu0ϕ0(t) + ˆu1ϕ1(t), t ∈ J, where ϕk(t) is a piecewise linear Lagrange basis: ϕk(tj) = δkj (δkj is a Kronecker’s delta). 14/55
- 18. Considered problem Let the initial function satisfy ∥u0 − ˆu0∥H1 0 ≤ ε0. We rigorously enclose the solution in a Banach space2 , L∞ ( J; H1 0 (Ω) ) := { u(t) ∈ H1 0 (Ω) : ess sup t∈J ∥u(t)∥H1 0 < ∞ } . Namely, we compute a radius ρ > 0 of the ball: BJ (ω, ρ) := { y ∈ L∞ ( J; H1 0 (Ω) ) : ∥y − ω∥L∞ (J;H1 0 (Ω)) ≤ ρ } . 2 ∥u∥L∞(J;H1 0 (Ω)) := ess supt∈J ∥u(t)∥H1 0 15/55
- 19. Considered problem Let the initial function satisfy ∥u0 − ˆu0∥H1 0 ≤ ε0. We rigorously enclose the solution in a Banach space2 , L∞ ( J; H1 0 (Ω) ) := { u(t) ∈ H1 0 (Ω) : ess sup t∈J ∥u(t)∥H1 0 < ∞ } . Namely, we compute a radius ρ > 0 of the ball: BJ (ω, ρ) := { y ∈ L∞ ( J; H1 0 (Ω) ) : ∥y − ω∥L∞ (J;H1 0 (Ω)) ≤ ρ } . 2 ∥u∥L∞(J;H1 0 (Ω)) := ess supt∈J ∥u(t)∥H1 0 15/55
- 20. Considered problem Let the initial function satisfy ∥u0 − ˆu0∥H1 0 ≤ ε0. We rigorously enclose the solution in a Banach space2 , L∞ ( J; H1 0 (Ω) ) := { u(t) ∈ H1 0 (Ω) : ess sup t∈J ∥u(t)∥H1 0 < ∞ } . Namely, we compute a radius ρ > 0 of the ball: BJ (ω, ρ) := { y ∈ L∞ ( J; H1 0 (Ω) ) : ∥y − ω∥L∞ (J;H1 0 (Ω)) ≤ ρ } . 2 ∥u∥L∞(J;H1 0 (Ω)) := ess supt∈J ∥u(t)∥H1 0 15/55
- 21. Analytic semigroup The weak form of A, which is denoted by −A3 , generates the analytic semigroup {e−tA }t≥0 over H−1 (Ω). The following abstract problem has an unique solution: ∂tu + Au = 0, u(0, x) = u0 =⇒ ∃u = e−tA u0. Fact Let x ∈ D(A) and λ0 be a positive number. A satisﬁes ⟨−Ax, x⟩ ≤ 0, R(λ0I + A) = H−1 (Ω). Then, there exists an analytic semigroup {e−tA }t≥0 generated by −A. Proofs are found in several textbooks. 3 A : H1 0 (Ω) → H−1 (Ω) s.t. ⟨Au, v⟩ := a(u, v), ∀v ∈ H1 0 (Ω). 16/55
- 22. Theorem Assume that the initial function u0 satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0; Assume that ω satisﬁes the following estimate: ∫ t t0 e−(t−s)A (∂tω(s) + Aω(s) − f(ω(s)))ds L∞ (J;H1 0 (Ω)) ≤ δ. Assume that, for ∀ρ0 ∈ (0, ρ] with a certain ρ 0, f satisﬁes ∥f(φ) − f(ψ)∥L∞(J;L2(Ω)) ≤ Lρ0 ∥φ − ψ∥L∞ (J;H1 0 (Ω)), where ∀φ, ψ ∈ BJ (ω, ρ0) ⊂ L∞ (J; H1 0 (Ω)). If M µ ε0 + 2 µ √ Mτ e Lρρ + δ ρ, then the weak solution u(t), t ∈ J of (PJ ) uniquely exists in the ball BJ (ω, ρ). 17/55
- 23. Theorem Assume that the initial function u0 satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0; Assume that ω satisﬁes the following estimate: ∫ t t0 e−(t−s)A (∂tω(s) + Aω(s) − f(ω(s)))ds L∞ (J;H1 0 (Ω)) ≤ δ. Assume that, for ∀ρ0 ∈ (0, ρ] with a certain ρ 0, f satisﬁes ∥f(φ) − f(ψ)∥L∞(J;L2(Ω)) ≤ Lρ0 ∥φ − ψ∥L∞ (J;H1 0 (Ω)), where ∀φ, ψ ∈ BJ (ω, ρ0) ⊂ L∞ (J; H1 0 (Ω)). If M µ ε0 + 2 µ √ Mτ e Lρρ + δ ρ, then the weak solution u(t), t ∈ J of (PJ ) uniquely exists in the ball BJ (ω, ρ). 17/55
- 24. Theorem Assume that the initial function u0 satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0; Assume that ω satisﬁes the following estimate: ∫ t t0 e−(t−s)A (∂tω(s) + Aω(s) − f(ω(s)))ds L∞ (J;H1 0 (Ω)) ≤ δ. Assume that, for ∀ρ0 ∈ (0, ρ] with a certain ρ 0, f satisﬁes ∥f(φ) − f(ψ)∥L∞(J;L2(Ω)) ≤ Lρ0 ∥φ − ψ∥L∞ (J;H1 0 (Ω)), where ∀φ, ψ ∈ BJ (ω, ρ0) ⊂ L∞ (J; H1 0 (Ω)). If M µ ε0 + 2 µ √ Mτ e Lρρ + δ ρ, then the weak solution u(t), t ∈ J of (PJ ) uniquely exists in the ball BJ (ω, ρ). 17/55
- 25. Theorem Assume that the initial function u0 satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0; Assume that ω satisﬁes the following estimate: ∫ t t0 e−(t−s)A (∂tω(s) + Aω(s) − f(ω(s)))ds L∞ (J;H1 0 (Ω)) ≤ δ. Assume that, for ∀ρ0 ∈ (0, ρ] with a certain ρ 0, f satisﬁes ∥f(φ) − f(ψ)∥L∞(J;L2(Ω)) ≤ Lρ0 ∥φ − ψ∥L∞ (J;H1 0 (Ω)), where ∀φ, ψ ∈ BJ (ω, ρ0) ⊂ L∞ (J; H1 0 (Ω)). If M µ ε0 + 2 µ √ Mτ e Lρρ + δ ρ, then the weak solution u(t), t ∈ J of (PJ ) uniquely exists in the ball BJ (ω, ρ). 17/55
- 26. Theorem Assume that the initial function u0 satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0; Assume that ω satisﬁes the following estimate: ∫ t t0 e−(t−s)A (∂tω(s) + Aω(s) − f(ω(s)))ds L∞ (J;H1 0 (Ω)) ≤ δ. Assume that, for ∀ρ0 ∈ (0, ρ] with a certain ρ 0, f satisﬁes ∥f(φ) − f(ψ)∥L∞(J;L2(Ω)) ≤ Lρ0 ∥φ − ψ∥L∞ (J;H1 0 (Ω)), where ∀φ, ψ ∈ BJ (ω, ρ0) ⊂ L∞ (J; H1 0 (Ω)). If M µ ε0 + 2 µ √ Mτ e Lρρ + δ ρ, then the weak solution u(t), t ∈ J of (PJ ) uniquely exists in the ball BJ (ω, ρ). 17/55
- 27. Sketch of proof Let z(t) ∈ H1 0 (Ω) for t ∈ J. We put u(t) = ω(t) + z(t). For any v ∈ H1 0 (Ω), (∂tz(t), v)L2 + a(z(t), v) = (f(u(t)), v)L2 − ((∂tω(t), v)L2 + ⟨Aω(t), v⟩) =: ⟨g(z(t)), v⟩ , where g(z(t)) = f(u(t)) − (∂tω(t) + Aω(t)). Note that by the deﬁnition of the natural embedding L2 (Ω) → H−1 (Ω), (ψ, v)L2 = ⟨ψ, v⟩ holds for ψ ∈ L2 (Ω). Deﬁne S : L∞ (J; H1 0 (Ω)) → L∞ (J; H1 0 (Ω)) using the analytic semigroup e−tA as S(z) := e−(t−t0)A (u0 − ˆu0) + ∫ t t0 e−(t−s)A g(z(s))ds. 18/55
- 28. Sketch of proof Let z(t) ∈ H1 0 (Ω) for t ∈ J. We put u(t) = ω(t) + z(t). For any v ∈ H1 0 (Ω), (∂tz(t), v)L2 + a(z(t), v) = (f(u(t)), v)L2 − ((∂tω(t), v)L2 + ⟨Aω(t), v⟩) =: ⟨g(z(t)), v⟩ , where g(z(t)) = f(u(t)) − (∂tω(t) + Aω(t)). Note that by the deﬁnition of the natural embedding L2 (Ω) → H−1 (Ω), (ψ, v)L2 = ⟨ψ, v⟩ holds for ψ ∈ L2 (Ω). Deﬁne S : L∞ (J; H1 0 (Ω)) → L∞ (J; H1 0 (Ω)) using the analytic semigroup e−tA as S(z) := e−(t−t0)A (u0 − ˆu0) + ∫ t t0 e−(t−s)A g(z(s))ds. 18/55
- 29. Sketch of proof For ρ 0, Z := {z : ∥z∥L∞ (J;H1 0 (Ω)) ≤ ρ} ⊂ L∞ (J; H1 0 (Ω)). On the basis of Banach’s ﬁxed-point theorem, we show a suﬃcient condition of S having a ﬁxed-point in Z. S(Z) ⊂ Z Since the analytic semigroup e−tA is bounded, the ﬁrst term of S(z) is estimated4 by e−(t−t0)A (ζ − ˆu0) H1 0 ≤ µ−1 A e−(t−t0)A (ζ − ˆu0) H−1 ≤ M µ e−(t−t0)λmin ε0. Then e−(t−t0)A (ζ − ˆu0) L∞(J;H1 0 (Ω)) ≤ M µ ε0. 4 µ∥u∥H1 0 ≤ ∥Au∥H−1 ≤ M∥u∥H1 0 is used. 19/55
- 30. Sketch of proof For ρ 0, Z := {z : ∥z∥L∞ (J;H1 0 (Ω)) ≤ ρ} ⊂ L∞ (J; H1 0 (Ω)). On the basis of Banach’s ﬁxed-point theorem, we show a suﬃcient condition of S having a ﬁxed-point in Z. S(Z) ⊂ Z Since the analytic semigroup e−tA is bounded, the ﬁrst term of S(z) is estimated4 by e−(t−t0)A (ζ − ˆu0) H1 0 ≤ µ−1 A e−(t−t0)A (ζ − ˆu0) H−1 ≤ M µ e−(t−t0)λmin ε0. Then e−(t−t0)A (ζ − ˆu0) L∞(J;H1 0 (Ω)) ≤ M µ ε0. 4 µ∥u∥H1 0 ≤ ∥Au∥H−1 ≤ M∥u∥H1 0 is used. 19/55
- 31. Sketch of proof For ρ 0, Z := {z : ∥z∥L∞ (J;H1 0 (Ω)) ≤ ρ} ⊂ L∞ (J; H1 0 (Ω)). On the basis of Banach’s ﬁxed-point theorem, we show a suﬃcient condition of S having a ﬁxed-point in Z. S(Z) ⊂ Z Since the analytic semigroup e−tA is bounded, the ﬁrst term of S(z) is estimated4 by e−(t−t0)A (ζ − ˆu0) H1 0 ≤ µ−1 A e−(t−t0)A (ζ − ˆu0) H−1 ≤ M µ e−(t−t0)λmin ε0. Then e−(t−t0)A (ζ − ˆu0) L∞(J;H1 0 (Ω)) ≤ M µ ε0. 4 µ∥u∥H1 0 ≤ ∥Au∥H−1 ≤ M∥u∥H1 0 is used. 19/55
- 32. Sketch of proof Decompose g(z(s)) ∈ H−1 (Ω) into two parts: g(z(s)) = f(ω(s) + z(s)) − (∂tω(s) + Aω(s)) = g1(s) + g2(s), g1(s) := f(ω(s) + z(s)) − f(ω(s)), g2(s) := f(ω(s)) − (∂tω(s) + Aω(s)) . Put ν(t) := ∫ t t0 (t − s)−1 2 e−1 2 (t−s)λmin ds, sup t∈J ν(t) ≤ sup t∈J ∫ t t0 (t − s)− 1 2 ds = 2 √ τ. Furthermore, ready an inequality µ 1 2 ∥u∥L2 ≤ ∥A 1 2 u∥H−1 ≤ M 1 2 ∥u∥L2 . 20/55
- 33. Sketch of proof Decompose g(z(s)) ∈ H−1 (Ω) into two parts: g(z(s)) = f(ω(s) + z(s)) − (∂tω(s) + Aω(s)) = g1(s) + g2(s), g1(s) := f(ω(s) + z(s)) − f(ω(s)), g2(s) := f(ω(s)) − (∂tω(s) + Aω(s)) . Put ν(t) := ∫ t t0 (t − s)−1 2 e−1 2 (t−s)λmin ds, sup t∈J ν(t) ≤ sup t∈J ∫ t t0 (t − s)− 1 2 ds = 2 √ τ. Furthermore, ready an inequality µ 1 2 ∥u∥L2 ≤ ∥A 1 2 u∥H−1 ≤ M 1 2 ∥u∥L2 . 20/55
- 34. Sketch of proof Decompose g(z(s)) ∈ H−1 (Ω) into two parts: g(z(s)) = f(ω(s) + z(s)) − (∂tω(s) + Aω(s)) = g1(s) + g2(s), g1(s) := f(ω(s) + z(s)) − f(ω(s)), g2(s) := f(ω(s)) − (∂tω(s) + Aω(s)) . Put ν(t) := ∫ t t0 (t − s)−1 2 e−1 2 (t−s)λmin ds, sup t∈J ν(t) ≤ sup t∈J ∫ t t0 (t − s)− 1 2 ds = 2 √ τ. Furthermore, ready an inequality µ 1 2 ∥u∥L2 ≤ ∥A 1 2 u∥H−1 ≤ M 1 2 ∥u∥L2 . 20/55
- 35. Sketch of proof The term of g1(s): ∫ t t0 e−(t−s)A g1(s)ds H1 0 = ∫ t t0 e−(t−s)A (f(ω(s) + z(s)) − f(ω(s)))ds H1 0 ≤ µ−1 ∫ t t0 A e−(t−s)A (f(ω(s) + z(s)) − f(ω(s))) H−1 ds = µ−1 ∫ t t0 A 1 2 e−(t−s)A A 1 2 (f(ω(s) + z(s)) − f(ω(s))) H−1 ds ≤ µ−1 e− 1 2 ∫ t t0 (t − s)− 1 2 e− 1 2 (t−s)λmin A 1 2 (f(ω(s) + z(s)) − f(ω(s))) H−1 ≤ µ−1 M 1 2 e− 1 2 ∫ t t0 (t − s)− 1 2 e− 1 2 (t−s)λmin ∥f(ω(s) + z(s)) − f(ω(s))∥L2 ds ≤ µ−1 M 1 2 e− 1 2 ν(t) ∥f(ω + z) − f(ω)∥L∞(J;L2(Ω)). 21/55
- 36. Sketch of proof Then ∫ t t0 e−(t−s)A g1(s)ds L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρρ. The term of g2(s) is nothing but the residual of the approximate solution, which is estimated by δ. Then it follows ∥S(z)∥L∞ (J;H1 0 (Ω)) ≤ M µ ε0 + 2 µ √ Mτ e L(ρ)ρ + δ. ∥S(z)∥L∞ (J;H1 0 (Ω)) ρ holds from the condition of the theorem. It implies that S(z) ∈ Z. 22/55
- 37. Sketch of proof Then ∫ t t0 e−(t−s)A g1(s)ds L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρρ. The term of g2(s) is nothing but the residual of the approximate solution, which is estimated by δ. Then it follows ∥S(z)∥L∞ (J;H1 0 (Ω)) ≤ M µ ε0 + 2 µ √ Mτ e L(ρ)ρ + δ. ∥S(z)∥L∞ (J;H1 0 (Ω)) ρ holds from the condition of the theorem. It implies that S(z) ∈ Z. 22/55
- 38. Sketch of proof Then ∫ t t0 e−(t−s)A g1(s)ds L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρρ. The term of g2(s) is nothing but the residual of the approximate solution, which is estimated by δ. Then it follows ∥S(z)∥L∞ (J;H1 0 (Ω)) ≤ M µ ε0 + 2 µ √ Mτ e L(ρ)ρ + δ. ∥S(z)∥L∞ (J;H1 0 (Ω)) ρ holds from the condition of the theorem. It implies that S(z) ∈ Z. 22/55
- 39. Sketch of proof Then ∫ t t0 e−(t−s)A g1(s)ds L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρρ. The term of g2(s) is nothing but the residual of the approximate solution, which is estimated by δ. Then it follows ∥S(z)∥L∞ (J;H1 0 (Ω)) ≤ M µ ε0 + 2 µ √ Mτ e L(ρ)ρ + δ. ∥S(z)∥L∞ (J;H1 0 (Ω)) ρ holds from the condition of the theorem. It implies that S(z) ∈ Z. 22/55
- 40. Sketch of proof For any z1, z2 in Z, S(z1) − S(z2) = ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds holds. We have ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds H1 0 ≤ µ−1 M 1 2 e−1 2 ν(t)∥f(z1 + ω) − f(z2 + ω)∥L∞(J;L2(Ω)). Here, zi + ω ∈ B(ω, ρ) (i = 1, 2) holds. Then ∥S(z1) − S(z2)∥L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρ∥z1 − z2∥L∞ (J;H1 0 (Ω)). 23/55
- 41. Sketch of proof For any z1, z2 in Z, S(z1) − S(z2) = ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds holds. We have ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds H1 0 ≤ µ−1 M 1 2 e−1 2 ν(t)∥f(z1 + ω) − f(z2 + ω)∥L∞(J;L2(Ω)). Here, zi + ω ∈ B(ω, ρ) (i = 1, 2) holds. Then ∥S(z1) − S(z2)∥L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρ∥z1 − z2∥L∞ (J;H1 0 (Ω)). 23/55
- 42. Sketch of proof For any z1, z2 in Z, S(z1) − S(z2) = ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds holds. We have ∫ t t0 e−(t−s)A (f(z1 + ω) − f(z2 + ω))ds H1 0 ≤ µ−1 M 1 2 e−1 2 ν(t)∥f(z1 + ω) − f(z2 + ω)∥L∞(J;L2(Ω)). Here, zi + ω ∈ B(ω, ρ) (i = 1, 2) holds. Then ∥S(z1) − S(z2)∥L∞ (J;H1 0 (Ω)) ≤ 2 µ √ Mτ e Lρ∥z1 − z2∥L∞ (J;H1 0 (Ω)). 23/55
- 43. Sketch of proof The condition of theorem also implies 2 µ √ Mτ e L (ρ) 1. Therefore, S is a contraction mapping. Banach’s ﬁxed point theorem yields that there uniquely exists a ﬁxed-point in Z. 24/55
- 44. Theorem (A posteriori error estimate) Assume that existence and local uniqueness of the weak solution u(t), t ∈ J, is proved in BJ (ω, ρ). Assume also that ω satisﬁes ∫ t1 t0 e−(t1−s)A (∂tω(s) + Aω(s) − f(ω(s))) ds H1 0 ≤ ˜δ. Then, the following a posteriori error estimate holds: ∥u(t1) − ˆu1∥H1 0 ≤ M µ e−τλmin ε0 + 2 µ √ Mτ e Lρρ + ˜δ =: ε1. 25/55
- 45. On several intervals For n ∈ N, 0 = t0 t1 · · · tn ∞. Jk := (tk−1, tk], τk := tk − tk−1, and J = ∪ Jk. (k=1,2,...,n) (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(0, x) = u0(x) in Ω, where u0 ∈ H1 0 (Ω) is a given initial function satisﬁes ∥u0 − ˆu0∥H1 0 ≤ ε0. 26/55
- 46. Approximate solution (Backward Euler) Find {uh k}k≥0 ⊂ Vh such that ( uh k − uh k−1 τ , vh ) L2 + a(uh k, vh)L2 = (f(uh k), vh)L2 and ( uh 0, vh ) L2 = (u0, vh)L2 for ∀vh ∈ Vh. Numerically compute each approximation ˆuk (≈ uh k) ∈ Vh. From the data ˆuk(≈ uh k) ∈ Vh, we construct ω(t): ω(t) := n∑ k=0 ˆukϕk(t), t ∈ T, where ϕk(t) is a piecewise linear Lagrange basis: ϕk(tj) = δkj (δkj is a Kronecker’s delta). 27/55
- 47. Approximate solution (Backward Euler) Find {uh k}k≥0 ⊂ Vh such that ( uh k − uh k−1 τ , vh ) L2 + a(uh k, vh)L2 = (f(uh k), vh)L2 and ( uh 0, vh ) L2 = (u0, vh)L2 for ∀vh ∈ Vh. Numerically compute each approximation ˆuk (≈ uh k) ∈ Vh. From the data ˆuk(≈ uh k) ∈ Vh, we construct ω(t): ω(t) := n∑ k=0 ˆukϕk(t), t ∈ T, where ϕk(t) is a piecewise linear Lagrange basis: ϕk(tj) = δkj (δkj is a Kronecker’s delta). 27/55
- 48. Veriﬁcation scheme 0 t u0 t1 t2 tk ... ... 28/55
- 49. Veriﬁcation scheme 0 t u0 t1 t2 tk ... ... 28/55
- 50. Veriﬁcation scheme 0 t u0 t1 t2 tk ... ... 28/55
- 51. Veriﬁcation scheme 0 t u0 t1 t2 tk ... ... 28/55
- 52. Veriﬁcation scheme 0 t u0 t1 t2 tk ... ... 28/55
- 53. ここまでのまとめ ▶ 各 t ∈ Jk において対象問題 (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(0, x) = u0(x) in Ω, に対する弱解の存在と一意性を逐次的に数値解の近傍 BJk (ω, ρk) に包み込む． ▶ (PJ ) の弱解は各 k = 1, 2, ..., n について B(ω) := { y ∈ L∞ ( J; H1 0 (Ω) ) : y(t) ∈ BJk (ω, ρk), t ∈ Jk } の中に一意存在する事が計算機援用証明できる． 29/55
- 54. ここまでのまとめ ▶ 各 t ∈ Jk において対象問題 (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(0, x) = u0(x) in Ω, に対する弱解の存在と一意性を逐次的に数値解の近傍 BJk (ω, ρk) に包み込む． ▶ (PJ ) の弱解は各 k = 1, 2, ..., n について B(ω) := { y ∈ L∞ ( J; H1 0 (Ω) ) : y(t) ∈ BJk (ω, ρk), t ∈ Jk } の中に一意存在する事が計算機援用証明できる． 29/55
- 56. 藤田型方程式 Ω = (0, 1)2 : Square domain ∂tu − ∆u = u2 in (0, ∞) × Ω, u(t, x) = 0 on (0, ∞) × ∂Ω, u(0, x) = u0 in Ω, Let γ 0 be an parameter of the initial function: u0(x) = γx1(1 − x1)x2(1 − x2). h: spatial mesh size (P2 element), τ: time step of B.E. method. 31/55
- 57. Computational results Table: h = 2−4 , τ = 2−8 , γ = 1. Tk = (tk−1, tk] εk ρk (0,0.0039062] 0.020155 0.037646 (0.0039062,0.0078125] 0.030051 0.041554 (0.0078125,0.011719] 0.038089 0.049313 (0.011719,0.015625] 0.044657 0.055699 (0.015625,0.019531] 0.050001 0.060873 ... ... ... (0.48047,0.48438] 0.00013041 0.00014184 (0.48438,0.48828] 0.00012186 0.00013255 (0.48828,0.49219] 0.00011388 0.00012386 (0.49219,0.49609] 0.00010641 0.00011573 (0.49609,0.5] 9.9431E-5 0.00010813 32/55
- 58. Computational results 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t ρk gamma = 1 gamma = 10 33/55
- 59. Computational results 0 0.1 0.2 0.3 0.4 0.5 10 −3 10 −2 10 −1 10 0 10 1 10 2 10 3 t ρ k gamma = 30 gamma = 50 34/55
- 60. 半線形放物型方程式 Ω = (0, 1)2 : Square domain ∂tu − ∆u = u − u3 in (0, ∞) × Ω, u(t, x) = 0 on (0, ∞) × ∂Ω, u(0, x) = u0 in Ω. We set the initial function: u0(x) = x1(1 − x1)x2(1 − x2). h: spatial mesh size (P2 element), τ: time step of B.E. method. 35/55
- 61. Computational results (h = 2−4 ) 0 0.1 0.2 0.3 0.4 0.5 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 t ρ k tau = 1/16 tau = 1/32 tau = 1/64 tau = 1/128 tau = 1/256 tau = 1/512 36/55
- 62. Computational results (τ ≪ h) 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 t ρk h = 1/4 h = 1/8 h = 1/16 h = 1/32 37/55
- 63. Global existence proof using veriﬁed computations 38/55
- 64. 時間大域解の証明 本講演では t ∈ (0, ∞) で存在する (PJ ) の解 u(t) ∈ H1 0 (Ω) を 時間大域解といい，以下の 2 つのステップで証明を試みる． t′ 0 をある時刻として ▶ (t′ , ∞) で定常解まわりに大域的に存在する範囲を計算 機で導く．（Global existence proof） ▶ ある時刻 t′ までの解を数値解の近傍に包み込む． （Concatenation scheme） 上記の方法によって，(PJ ) の時間大域解を関数空間 L∞ ( (0, ∞); H1 0 (Ω) ) で一意存在することが計算機を用いて証明できる． 39/55
- 65. 時間大域解の証明 本講演では t ∈ (0, ∞) で存在する (PJ ) の解 u(t) ∈ H1 0 (Ω) を 時間大域解といい，以下の 2 つのステップで証明を試みる． t′ 0 をある時刻として ▶ (t′ , ∞) で定常解まわりに大域的に存在する範囲を計算 機で導く．（Global existence proof） ▶ ある時刻 t′ までの解を数値解の近傍に包み込む． （Concatenation scheme） 上記の方法によって，(PJ ) の時間大域解を関数空間 L∞ ( (0, ∞); H1 0 (Ω) ) で一意存在することが計算機を用いて証明できる． 39/55
- 66. 時間大域解の証明 本講演では t ∈ (0, ∞) で存在する (PJ ) の解 u(t) ∈ H1 0 (Ω) を 時間大域解といい，以下の 2 つのステップで証明を試みる． t′ 0 をある時刻として ▶ (t′ , ∞) で定常解まわりに大域的に存在する範囲を計算 機で導く．（Global existence proof） ▶ ある時刻 t′ までの解を数値解の近傍に包み込む． （Concatenation scheme） 上記の方法によって，(PJ ) の時間大域解を関数空間 L∞ ( (0, ∞); H1 0 (Ω) ) で一意存在することが計算機を用いて証明できる． 39/55
- 67. 大域解の存在に関する先行研究 S. Cai, “A computer-assisted proof for the pattern formation on reaction-diﬀusion systems”, 学位論文, Graduate School of Mathematics, Kyushu University (2012) 71 pages. ▶ 反応拡散方程式のあるクラスの定常解に対する精度保 証付き数値計算法を示している． ▶ (t′ , ∞) で定常解まわりに大域的に存在する範囲を L∞ (Ω) × L∞ (Ω) 上で生成された解析半群を用いて，計 算している． 40/55
- 68. Considered problem Let Ω be a bounded polygonal domain in R2 and J := (0, ∞). (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(0, x) = u0(x) in Ω, where A = −∆, u0 ∈ H1 0 (Ω) is an initial function, and f : R → R is a twice Fr´echet diﬀerentiable nonlinear mapping. 41/55
- 69. Considered problem Let Ω be a bounded polygonal domain in R2 and J := (0, ∞). (PJ ) ∂tu + Au = f(u) in J × Ω, u(t, x) = 0 on J × ∂Ω, u(0, x) = u0(x) in Ω, where A = −∆, u0 ∈ H1 0 (Ω) is an initial function, and f : R → R is a twice Fr´echet diﬀerentiable nonlinear mapping. 41/55
- 70. Aim of this part Let Ω be a bounded polygonal domain in R2 . (PG) ∂tu + Au = f(u) in (t′ , ∞) × Ω, u(t, x) = 0 on (t′ , ∞) × ∂Ω, u(t′ , x) = η in Ω, where η ∈ H1 0 (Ω) satisﬁes ∥η − ˆun∥H1 0 ≤ εn for a certain εn 0. We enclose a solution for t ∈ (t′ , ∞) in a neighborhood of a stationary solution ϕ ∈ D(A) of (PJ ) such that { Aϕ = f(ϕ) in Ω, ϕ = 0 on ∂Ω. 42/55
- 71. 記号 For ρ 0, v ∈ L∞ ((t′ , ∞); H1 0 (Ω)), deﬁne a ball B(v, ρ) := { y ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥y − v∥L∞ ((t′,∞);H1 0 (Ω)) ≤ ρ } . The Fr´echet derivative of f at w is denoted by f′ [w] : L∞ ((t′ , ∞); H1 0 (Ω)) → L∞ ((t′ , ∞); L2 (Ω)). For y ∈ B(v, ρ), we assume that there exists a non-decreasing function L : R → R such that ∥f′ [y]u∥L∞(J;L2(Ω)) ≤ L(ρ)∥u∥H1 0 , u ∈ H1 0 (Ω). 43/55
- 72. 記号 Deﬁne a function space Xλ: for a ﬁxed λ 0, Xλ := { u ∈ L∞ ((t′ , ∞); H1 0 (Ω)) : ess sup t∈(t′,∞) e(t−t′)λ ∥u(t)∥H1 0 ∞ } which becomes a Banach space with the norm ∥ · ∥Xλ := ess sup t∈(t′,∞) e(t−t′)λ ∥u(t)∥H1 0 . 44/55
- 73. Theorem (Global existence) Assume that ▶ a solution of (P) is enclosed until t′ 0, ▶ a stationary solution ϕ ∈ D(A) uniquely exists around a numerical solution ˆϕ, ▶ For a ˆun ∈ Vh, εn 0, the initial function satisﬁes ∥η − ˆun∥H1 0 εn. For a ﬁxed λ satisfying 0 ≤ λ λmin/2, if ρ 0 satisﬁes ∥η − ϕ∥H1 0 + L(ρ)ρ √ 2π e(λmin − 2λ) ρ. Then a solution u(t) for t ∈ (t′ , ∞) uniquely exists in Uϕ := { u ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥u − ϕ∥Xλ ≤ ρ } . 45/55
- 74. Theorem (Global existence) Assume that ▶ a solution of (P) is enclosed until t′ 0, ▶ a stationary solution ϕ ∈ D(A) uniquely exists around a numerical solution ˆϕ, ▶ For a ˆun ∈ Vh, εn 0, the initial function satisﬁes ∥η − ˆun∥H1 0 εn. For a ﬁxed λ satisfying 0 ≤ λ λmin/2, if ρ 0 satisﬁes ∥η − ϕ∥H1 0 + L(ρ)ρ √ 2π e(λmin − 2λ) ρ. Then a solution u(t) for t ∈ (t′ , ∞) uniquely exists in Uϕ := { u ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥u − ϕ∥Xλ ≤ ρ } . 45/55
- 75. Theorem (Global existence) Assume that ▶ a solution of (P) is enclosed until t′ 0, ▶ a stationary solution ϕ ∈ D(A) uniquely exists around a numerical solution ˆϕ, ▶ For a ˆun ∈ Vh, εn 0, the initial function satisﬁes ∥η − ˆun∥H1 0 εn. For a ﬁxed λ satisfying 0 ≤ λ λmin/2, if ρ 0 satisﬁes ∥η − ϕ∥H1 0 + L(ρ)ρ √ 2π e(λmin − 2λ) ρ. Then a solution u(t) for t ∈ (t′ , ∞) uniquely exists in Uϕ := { u ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥u − ϕ∥Xλ ≤ ρ } . 45/55
- 76. Theorem (Global existence) Assume that ▶ a solution of (P) is enclosed until t′ 0, ▶ a stationary solution ϕ ∈ D(A) uniquely exists around a numerical solution ˆϕ, ▶ For a ˆun ∈ Vh, εn 0, the initial function satisﬁes ∥η − ˆun∥H1 0 εn. For a ﬁxed λ satisfying 0 ≤ λ λmin/2, if ρ 0 satisﬁes ∥η − ϕ∥H1 0 + L(ρ)ρ √ 2π e(λmin − 2λ) ρ. Then a solution u(t) for t ∈ (t′ , ∞) uniquely exists in Uϕ := { u ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥u − ϕ∥Xλ ≤ ρ } . 45/55
- 77. Theorem (Global existence) Assume that ▶ a solution of (P) is enclosed until t′ 0, ▶ a stationary solution ϕ ∈ D(A) uniquely exists around a numerical solution ˆϕ, ▶ For a ˆun ∈ Vh, εn 0, the initial function satisﬁes ∥η − ˆun∥H1 0 εn. For a ﬁxed λ satisfying 0 ≤ λ λmin/2, if ρ 0 satisﬁes ∥η − ϕ∥H1 0 + L(ρ)ρ √ 2π e(λmin − 2λ) ρ. Then a solution u(t) for t ∈ (t′ , ∞) uniquely exists in Uϕ := { u ∈ L∞ ( (t′ , ∞); H1 0 (Ω) ) : ∥u − ϕ∥Xλ ≤ ρ } . 45/55
- 78. Sketch of proof Let z ∈ Uϕ. A nonlinear operator S : L∞ ((t′ , ∞); H1 0 (Ω)) → L∞ ((t′ , ∞); H1 0 (Ω)) is deﬁned by S(z) := e−(t−t′)A (η − ϕ) + ∫ t t′ e−(t−s)A (f(z(s)) − f(ϕ)) ds. On the basis of Banach’s ﬁxed-point theorem, we show a condition of S having a ﬁxed-point in Uϕ. For s ∈ (t′ , ∞) and ψ1, ψ2 ∈ Uϕ, the mean-value theorem states that there exists y ∈ Uϕ such that ∥f(ψ1(s)) − f(ψ2(s))∥L2 = ∥f′ [y(s)](ψ1(s) − ψ2(s))∥L2 . Since y ∈ Uϕ ⊂ B(ϕ, ρ) holds, we obtain ∥f(ψ1(s)) − f(ψ2(s))∥L2 ≤ L(ρ)∥ψ1(s) − ψ2(s)∥H1 0 . 46/55
- 79. How to get ∥η − ϕ∥H1 0 ? Since ∥η − ˆun∥H1 0 εn and a stationary solution ϕ encloses in a neighborhood of a numerical solution, it follows ∥η − ϕ∥H1 0 ≤ ∥η − ˆun∥H1 0 + ∥ˆun − ˆϕ∥H1 0 + ∥ˆϕ − ϕ∥H1 0 ≤ εn + ∥ˆun − ˆϕ∥H1 0 + ρ′ . We need to estimate ∥u(t′ ) − ˆun∥H1 0 ≤ εn. This can be obtained by the concatenation scheme! 47/55
- 80. How to get ∥η − ϕ∥H1 0 ? Since ∥η − ˆun∥H1 0 εn and a stationary solution ϕ encloses in a neighborhood of a numerical solution, it follows ∥η − ϕ∥H1 0 ≤ ∥η − ˆun∥H1 0 + ∥ˆun − ˆϕ∥H1 0 + ∥ˆϕ − ϕ∥H1 0 ≤ εn + ∥ˆun − ˆϕ∥H1 0 + ρ′ . We need to estimate ∥u(t′ ) − ˆun∥H1 0 ≤ εn. This can be obtained by the concatenation scheme! 47/55
- 81. How to get ∥η − ϕ∥H1 0 ? Since ∥η − ˆun∥H1 0 εn and a stationary solution ϕ encloses in a neighborhood of a numerical solution, it follows ∥η − ϕ∥H1 0 ≤ ∥η − ˆun∥H1 0 + ∥ˆun − ˆϕ∥H1 0 + ∥ˆϕ − ϕ∥H1 0 ≤ εn + ∥ˆun − ˆϕ∥H1 0 + ρ′ . We need to estimate ∥u(t′ ) − ˆun∥H1 0 ≤ εn. This can be obtained by the concatenation scheme! 47/55
- 82. How to get ∥η − ϕ∥H1 0 ? Since ∥η − ˆun∥H1 0 εn and a stationary solution ϕ encloses in a neighborhood of a numerical solution, it follows ∥η − ϕ∥H1 0 ≤ ∥η − ˆun∥H1 0 + ∥ˆun − ˆϕ∥H1 0 + ∥ˆϕ − ϕ∥H1 0 ≤ εn + ∥ˆun − ˆϕ∥H1 0 + ρ′ . We need to estimate ∥u(t′ ) − ˆun∥H1 0 ≤ εn. This can be obtained by the concatenation scheme! 47/55
- 84. 藤田型方程式 Let Ω := (0, 1)2 be an unit square domain in R2 . (F) ∂tu − ∆u = u2 in (0, ∞) × Ω, u(t, x) = 0 on (0, ∞) × ∂Ω, u(0, x) = u0(x) in Ω, where u0(x) = γ sin(πx) sin(πy). ▶ Vh := {∑N k,l=1 ak,l sin(kπx) sin(lπy) : ak,l ∈ R } ; ▶ Crank-Nicolson scheme is employed; ▶ we ﬁxed λ = 1/40 in the global existence theorem. 49/55
- 85. Table: 時間大域解の検証例（N = 8, λ = 1/40, τk = 2−7 ） γ n t′ ρ 0.01 5 0.046875 0.01085 0.011 5 0.046875 0.011936 0.0121 6 0.054688 0.011605 0.01331 7 0.0625 0.011274 0.014641 7 0.0625 0.012403 0.016105 8 0.070312 0.012038 0.017716 8 0.070312 0.013244 0.019487 9 0.078125 0.012845 0.021436 10 0.085938 0.012448 0.023579 10 0.085938 0.013695 0.025937 11 0.09375 0.013263 0.028531 11 0.09375 0.014593 0.031384 12 0.10156 0.014123 ... 50/55
- 86. Table: 時間大域解の検証例（N = 8, λ = 1/40, τk = 2−7 ） γ n t′ ρ ... 2.2876 40 0.32031 0.029945 2.5164 41 0.32812 0.0296 2.768 42 0.33594 0.029446 3.0448 42 0.33594 0.034085 3.3493 43 0.34375 0.034657 3.6842 44 0.35156 0.035917 4.0527 45 0.35938 0.038419 4.4579 45 0.35938 0.050511 4.9037 46 0.36719 0.066455 5.3941 47 0.375 0.17656 このとき ∥u(t)∥H1 0 ≤ ρe− (t−t′) 40 , t ∈ (tn, ∞). 51/55
- 87. 半線形放物型方程式 Let Ω := (0, 1)2 be an unit square domain in R2 . ∂tu − ∆u = f(u) in (0, ∞) × Ω, u(t, x) = 0 on (0, ∞) × ∂Ω, u(0, x) = u0(x) in Ω, where u0(x) = sin(πx) sin(πy). ▶ f(u) = u2 + 4 ∑ 1≤k,l≤3 sin(kπx) sin(lπy); ▶ Vh := {∑N k,l=1 ak,l sin(kπx) sin(lπy) : ak,l ∈ R } ; ▶ Crank-Nicolson scheme is employed; ▶ we ﬁxed λ = 1/40 in the global existence theorem. 52/55
- 88. Fig. The numerical solution ˆϕ. 時間大域解の検証は N = 10, λ = 1/40, τk = 2−8 で成功 して， ρ = 0.04035, t′ = 0.2578125. 53/55
- 89. Fig. The numerical solution ˆϕ. 時間大域解の検証は N = 10, λ = 1/40, τk = 2−8 で成功 して， ρ = 0.04035, t′ = 0.2578125. 53/55
- 90. まとめ ▶ 解析半群 { e−tA } t≥0 を用いる精度保証付き数値計算手法 ▶ Concatenation scheme（数値解のまわりに包み込む） ▶ 精度保証付き数値計算を用いた時間大域解の存在証明 （定常解のまわりに包み込む） 今後の課題 ▶ 方程式の拡張（多種の反応拡散方程式，波動方程式等） ▶ 無限次元力学系との関連 ▶ 藤田型方程式の爆発時刻の精度保証付き数値計算 54/55
- 91. Thank you for kind attention! 55/55