Akitoshi Takayasu

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

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自己紹介

高安亮紀早稲田大学応用数理学科助教（大石研究室）

分野：(偏微分方程式の)数値解析，精度保証付き数値計算

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計算機を用いた非線形PDEへの解析アプローチ

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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）

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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 ∈ H10 (Ω).

Lp(Ω): the set of Lp-functions,

H1(Ω): the first order Sobolev space of L2(Ω),

H10 (Ω) := v ∈ H1(Ω) : v = 0 on ∂Ω.

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記号

A : D(A) ⊂ H10 (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.

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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∥H10:= ∥∇u∥L2 ,

and∥u∥H−1 := sup

0=v∈H10 (Ω)

∥v∥H10=1

|⟨u, v⟩| ,

where ⟨·, ·⟩ is a dual product between H10 (Ω) and H

−1(Ω)1.

1The topological dual space of H10 (Ω).

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精度保証付き数値計算

(PJ)の弱解: For t ∈ J , u(t) := u(t, ·) ∈ H10 (Ω) with the

initial function u0 such that

(∂tu(t), v)L2 + a(u(t), v) = (f(u(t)), v)L2 , ∀v ∈ H10 (Ω),

where a : H10 (Ω)×H1

0 (Ω) → R is a bilinear form:

a(u, v) :=∑

1≤i,j≤2

(aij(x)

∂u

∂xi,∂v

∂xj

)L2

satisfyinga(u, u) ≥ µ∥u∥2H1

0, ∀u ∈ H1

0 (Ω),

|a(u, v)| ≤M∥u∥H10∥v∥H1

0, ∀u, v ∈ H1

0 (Ω).

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a(u, v) :=∑

1≤i,j≤2

(aij(x)

∂u

∂xi,∂v

∂xj

)L2


0, ∀u ∈ H1

0 (Ω),

|a(u, v)| ≤M∥u∥H10∥v∥H1

0, ∀u, v ∈ H1

0 (Ω).

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a(u, v) :=∑

1≤i,j≤2

(aij(x)

∂u

∂xi,∂v

∂xj

)L2


0, ∀u ∈ H1

0 (Ω),

|a(u, v)| ≤M∥u∥H10∥v∥H1

0, ∀u, v ∈ H1

0 (Ω).

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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．

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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 linearparabolic initial-boundary value problems”, Computing94(2-4), 151–162, 2012.

M.T. Nakao, T. Kimura and T. Kinoshita,

“Constructive A Priori Error Estimates for a Full DiscreteApproximation 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 linearparabolic equations with applications to the numericalenclosure of solutions for nonlinear problems”, Numer. Math,Online First, 2013.

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力学系の観点からみた研究

P. Zgliczynski and K. Mischaikow,

“Rigorous Numerics for Partial Differential Equations: TheKuramoto―Sivashinsky Equation”, Foundations ofComputational Mathematics, 1(3), 1615–3375, 2001.

P. Zgliczynski,

“Rigorous numerics for dissipative PDEs III. An effectivealgorithm for rigorous integration of dissipative PDEs”, Topol.Methods Nonlinear Anal., 36, 197–262, 2010.

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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 theevolution equation ut + A(t)u = 0 of parabolic type”, Topicsin 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.

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Concatenation scheme

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Considered problem

J := (t0, t1] : arbitrary time interval. τ := t1 − t0.

(PJ)


u(t, x) = 0 on J × ∂Ω,

u(t0, x) = u0(x) in Ω,

where u0 is an initial function in H10 (Ω).

Vh ⊂ H10 (Ω) : a finite dimensional subspace.

Starts from: u0, u1 ∈ Vh

ω(t) = u0ϕ0(t) + u1ϕ1(t), t ∈ J,

where ϕk(t) is a piecewise linear Lagrange basis: ϕk(tj) = δkj(δkj is a Kronecker’s delta).

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Considered problem

J := (t0, t1] : arbitrary time interval. τ := t1 − t0.

(PJ)


u(t, x) = 0 on J × ∂Ω,

u(t0, x) = u0(x) in Ω,

where u0 is an initial function in H10 (Ω).

Vh ⊂ H10 (Ω) : a finite dimensional subspace.

Starts from: u0, u1 ∈ Vh

ω(t) = u0ϕ0(t) + u1ϕ1(t), t ∈ J,


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Considered problem

Let the initial function satisfy

∥u0 − u0∥H10≤ ε0.

We rigorously enclose the solution in a Banach space2,

L∞ (J ;H1

0 (Ω)):=

u(t) ∈ H1

0 (Ω) : ess supt∈J

∥u(t)∥H10<∞

.

Namely, we compute a radius ρ > 0 of the ball:

BJ(ω, ρ) :=y ∈ L∞ (

J ;H10 (Ω)

): ∥y − ω∥L∞(J ;H1

0 (Ω)) ≤ ρ.

2∥u∥L∞(J;H10 (Ω)) := ess supt∈J ∥u(t)∥H1

0

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Considered problem


∥u0 − u0∥H10≤ ε0.


L∞ (J ;H1

0 (Ω)):=

u(t) ∈ H1


∥u(t)∥H10<∞

.


BJ(ω, ρ) :=y ∈ L∞ (

J ;H10 (Ω)

): ∥y − ω∥L∞(J ;H1

0 (Ω)) ≤ ρ.


0

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Considered problem


∥u0 − u0∥H10≤ ε0.


L∞ (J ;H1

0 (Ω)):=

u(t) ∈ H1


∥u(t)∥H10<∞

.


BJ(ω, ρ) :=y ∈ L∞ (

J ;H10 (Ω)

): ∥y − ω∥L∞(J ;H1

0 (Ω)) ≤ ρ.


0

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Analytic semigroup

The weak form of A, which is denoted by −A3, generates theanalytic semigroup e−tAt≥0 over H−1(Ω). The followingabstract problem has an unique solution:

∂tu+Au = 0, u(0, x) = u0 =⇒ ∃u = e−tAu0.

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−tAt≥0 generated by −A. Proofs are found in several textbooks.

3A : H10 (Ω) → H−1(Ω) s.t. ⟨Au, v⟩ := a(u, v), ∀v ∈ H1

0 (Ω).16/55

Theorem

Assume that the initial function u0 satisfies ∥u0 − u0∥H10≤ 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 ;H10 (Ω)),

where ∀φ, ψ ∈ BJ(ω, ρ0) ⊂ L∞ (J ;H10 (Ω)). If

M

µε0 +

2

µ

√Mτ

eLρρ+ δ < ρ,

then the weak solution u(t), t ∈ J of (PJ) uniquely exists inthe ball BJ(ω, ρ).

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Theorem



t0


∥∥∥∥L∞(J ;H1

0 (Ω))≤ δ.




M

µε0 +

2

µ

√Mτ

eLρρ+ δ < ρ,


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Theorem



t0


∥∥∥∥L∞(J ;H1

0 (Ω))≤ δ.




M

µε0 +

2

µ

√Mτ

eLρρ+ δ < ρ,


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Theorem



t0


∥∥∥∥L∞(J ;H1

0 (Ω))≤ δ.




M

µε0 +

2

µ

√Mτ

eLρρ+ δ < ρ,


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Theorem



t0


∥∥∥∥L∞(J ;H1

0 (Ω))≤ δ.




M

µε0 +

2

µ

√Mτ

eLρρ+ δ < ρ,


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Sketch of proofLet 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 bythe definition of the natural embedding L2(Ω) → H−1(Ω),(ψ, v)L2 = ⟨ψ, v⟩ holds for ψ ∈ L2(Ω).

Define S : L∞ (J ;H10 (Ω)) → L∞ (J ;H1

0 (Ω)) using theanalytic semigroup e−tA as

S(z) := e−(t−t0)A(u0 − u0) +

∫ t

t0

e−(t−s)Ag(z(s))ds.

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Sketch of proofLet 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 bythe definition of the natural embedding L2(Ω) → H−1(Ω),(ψ, v)L2 = ⟨ψ, v⟩ holds for ψ ∈ L2(Ω).

Define S : L∞ (J ;H10 (Ω)) → L∞ (J ;H1

0 (Ω)) using theanalytic semigroup e−tA as

S(z) := e−(t−t0)A(u0 − u0) +

∫ t

t0

e−(t−s)Ag(z(s))ds.

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Sketch of proof

For ρ > 0, Z := z : ∥z∥L∞(J ;H10 (Ω)) ≤ ρ ⊂ L∞ (J ;H1

0 (Ω)).

On the basis of Banach’s fixed-point theorem, we show asufficient 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 − u0)∥∥H1

0≤ µ−1

∥∥A e−(t−t0)A(ζ − u0)∥∥H−1

≤ M

µe−(t−t0)λminε0.

Then ∥∥e−(t−t0)A(ζ − u0)∥∥L∞(J ;H1

0 (Ω))≤ M

µε0.

4µ∥u∥H10≤ ∥Au∥H−1 ≤ M∥u∥H1

0is used.

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Sketch of proof


0 (Ω)).




0≤ µ−1

∥∥A e−(t−t0)A(ζ − u0)∥∥H−1

≤ M


Then ∥∥e−(t−t0)A(ζ − u0)∥∥L∞(J ;H1

0 (Ω))≤ M

µε0.

4µ∥u∥H10≤ ∥Au∥H−1 ≤ M∥u∥H1

0is used.

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Sketch of proof


0 (Ω)).




0≤ µ−1

∥∥A e−(t−t0)A(ζ − u0)∥∥H−1

≤ M


Then ∥∥e−(t−t0)A(ζ − u0)∥∥L∞(J ;H1

0 (Ω))≤ M

µε0.

4µ∥u∥H10≤ ∥Au∥H−1 ≤ M∥u∥H1

0is used.

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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)−12 e−

12(t−s)λminds,

supt∈J

ν(t) ≤ supt∈J

∫ t

t0

(t− s)−12ds = 2

√τ .

Furthermore, ready an inequality

µ12∥u∥L2 ≤ ∥A

12u∥H−1 ≤M

12∥u∥L2 .

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Sketch of proof



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)−12 e−

12(t−s)λminds,

supt∈J

ν(t) ≤ supt∈J

∫ t

t0

(t− s)−12ds = 2

√τ .


µ12∥u∥L2 ≤ ∥A

12u∥H−1 ≤M

12∥u∥L2 .

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Sketch of proof



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)−12 e−

12(t−s)λminds,

supt∈J

ν(t) ≤ supt∈J

∫ t

t0

(t− s)−12ds = 2

√τ .


µ12∥u∥L2 ≤ ∥A

12u∥H−1 ≤M

12∥u∥L2 .

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Sketch of proof

The term of g1(s):∥∥∥∥∫ t

t0

e−(t−s)Ag1(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 12 e−(t−s)AA 1

2 (f(ω(s) + z(s))− f(ω(s)))∥∥∥H−1

ds

≤ µ−1e−12

∫ t

t0

(t− s)−12 e−

12 (t−s)λmin

∥∥∥A 12 (f(ω(s) + z(s))− f(ω(s)))

∥∥∥H−1

≤ µ−1M12 e−

12

∫ t

t0

(t− s)−12 e−

12 (t−s)λmin ∥f(ω(s) + z(s))− f(ω(s))∥L2 ds

≤ µ−1M12 e−

12 ν(t) ∥f(ω + z)− f(ω)∥L∞(J;L2(Ω)).

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Sketch of proof

Then ∥∥∥∥∫ t

t0

e−(t−s)Ag1(s)ds

∥∥∥∥L∞(J ;H1

0 (Ω))≤ 2

µ

√Mτ

eLρρ.

The term of g2(s) is nothing but the residual of theapproximate solution, which is estimated by δ.

Then it follows

∥S(z)∥L∞(J ;H10 (Ω)) ≤

M

µε0 +

2

µ

√Mτ

eL(ρ)ρ+ δ.

∥S(z)∥L∞(J ;H10 (Ω)) < ρ holds from the condition of the

theorem. It implies that S(z) ∈ Z.

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Sketch of proof


t0

e−(t−s)Ag1(s)ds

∥∥∥∥L∞(J ;H1

0 (Ω))≤ 2

µ

√Mτ

eLρρ.


Then it follows

∥S(z)∥L∞(J ;H10 (Ω)) ≤

M

µε0 +

2

µ

√Mτ

eL(ρ)ρ+ δ.



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Sketch of proof


t0

e−(t−s)Ag1(s)ds

∥∥∥∥L∞(J ;H1

0 (Ω))≤ 2

µ

√Mτ

eLρρ.


Then it follows

∥S(z)∥L∞(J ;H10 (Ω)) ≤

M

µε0 +

2

µ

√Mτ

eL(ρ)ρ+ δ.



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Sketch of proof


t0

e−(t−s)Ag1(s)ds

∥∥∥∥L∞(J ;H1

0 (Ω))≤ 2

µ

√Mτ

eLρρ.


Then it follows

∥S(z)∥L∞(J ;H10 (Ω)) ≤

M

µε0 +

2

µ

√Mτ

eL(ρ)ρ+ δ.



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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−1M12 e−

12ν(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 ;H10 (Ω)) ≤

2

µ

√Mτ

eLρ∥z1 − z2∥L∞(J ;H1

0 (Ω)).

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Sketch of proof


S(z1)− S(z2) =

∫ t

t0

e−(t−s)A(f(z1 + ω)− f(z2 + ω))ds


t0

e−(t−s)A(f(z1 + ω)− f(z2 + ω))ds

∥∥∥∥H1

0

≤ µ−1M12 e−

12ν(t)∥f(z1 + ω)− f(z2 + ω)∥L∞(J ;L2(Ω)).


∥S(z1)− S(z2)∥L∞(J ;H10 (Ω)) ≤

2

µ

√Mτ


0 (Ω)).

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Sketch of proof


S(z1)− S(z2) =

∫ t

t0

e−(t−s)A(f(z1 + ω)− f(z2 + ω))ds


t0

e−(t−s)A(f(z1 + ω)− f(z2 + ω))ds

∥∥∥∥H1

0

≤ µ−1M12 e−

12ν(t)∥f(z1 + ω)− f(z2 + ω)∥L∞(J ;L2(Ω)).


∥S(z1)− S(z2)∥L∞(J ;H10 (Ω)) ≤

2

µ

√Mτ


0 (Ω)).

23/55

Sketch of proof

The condition of theorem also implies

2

µ

√Mτ

eL (ρ) < 1.

Therefore, S is a contraction mapping. Banach’s fixed pointtheorem yields that there uniquely exists a fixed-point in Z.

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Theorem (A posteriori error estimate)

Assume that existence and local uniqueness of the weaksolution u(t), t ∈ J , is proved in BJ(ω, ρ). Assume also thatω 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)− u1∥H10≤ M

µe−τλminε0 +

2

µ

√Mτ

eLρρ+ δ =: ε1.

25/55

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)


u(t, x) = 0 on J × ∂Ω,

u(0, x) = u0(x) in Ω,

where u0 ∈ H10 (Ω) is a given initial function satisfies

∥u0 − u0∥H10≤ ε0.

26/55

Approximate solution (Backward Euler)

Find uhkk≥0 ⊂ Vh such that(uhk − uhk−1

τ, vh

)L2

+ a(uhk, vh)L2 = (f(uhk), vh)L2

and(uh0 , vh

)L2 = (u0, vh)L2 for ∀vh ∈ Vh. Numerically

compute each approximation uk (≈ uhk) ∈ Vh.

From the data uk(≈ uhk) ∈ Vh, we construct ω(t):

ω(t) :=n∑

k=0

ukϕk(t), t ∈ T,


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Approximate solution (Backward Euler)

Find uhkk≥0 ⊂ Vh such that(uhk − uhk−1

τ, vh

)L2

+ a(uhk, vh)L2 = (f(uhk), vh)L2

and(uh0 , vh

)L2 = (u0, vh)L2 for ∀vh ∈ Vh. Numerically

compute each approximation uk (≈ uhk) ∈ Vh.

From the data uk(≈ uhk) ∈ Vh, we construct ω(t):

ω(t) :=n∑

k=0

ukϕk(t), t ∈ T,


27/55

Verification scheme

0 t

u0

t1 t2 tk...

...

28/55

Verification scheme

0 t

u0

t1 t2 tk...

...

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Verification scheme

0 t

u0

t1 t2 tk...

...

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Verification scheme

0 t

u0

t1 t2 tk...

...

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Verification scheme

0 t

u0

t1 t2 tk...

...

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ここまでのまとめ

各 t ∈ Jkにおいて対象問題

(PJ)


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 ;H10 (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．

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ここまでのまとめ

各 t ∈ Jkにおいて対象問題

(PJ)


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 ;H10 (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．

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Computational results 1

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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.

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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

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0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t

ρ k

gamma = 1

gamma = 10

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0 0.1 0.2 0.3 0.4 0.510

−3

10−2

10−1

100

101

102

103

t

ρ k

gamma = 30

gamma = 50

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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.

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Computational results (h = 2−4)

0 0.1 0.2 0.3 0.4 0.50

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

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Computational results (τ ≪ h)

0 0.1 0.2 0.3 0.4 0.50

0.1

0.2

0.3

0.4

0.5

t

ρ k

h = 1/4

h = 1/8

h = 1/16

h = 1/32

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Global existence proof

using verified computations

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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) ∈ H10 (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．

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L∞ ((0,∞);H1

0 (Ω))


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L∞ ((0,∞);H1

0 (Ω))


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大域解の存在に関する先行研究

S. Cai,

“A computer-assisted proof for the pattern formation onreaction-diffusion systems”, 学位論文, Graduate School ofMathematics, 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．

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Considered problem

Let Ω be a bounded polygonal domain in R2 and J := (0,∞).

(PJ)


u(t, x) = 0 on J × ∂Ω,

u(0, x) = u0(x) in Ω,

where A = −∆, u0 ∈ H10 (Ω) is an initial function, and

f : R → R is a twice Frechet differentiable nonlinear mapping.

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Considered problem

Let Ω be a bounded polygonal domain in R2 and J := (0,∞).

(PJ)


u(t, x) = 0 on J × ∂Ω,

u(0, x) = u0(x) in Ω,

where A = −∆, u0 ∈ H10 (Ω) is an initial function, and

f : R → R is a twice Frechet differentiable nonlinear mapping.

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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 ∈ H10 (Ω) satisfies ∥η − un∥H1

0≤ εn for a certain

εn > 0.

We enclose a solution for t ∈ (t′,∞) in a neighborhood of astationary solution ϕ ∈ D(A) of (PJ) such that

Aϕ = f(ϕ) in Ω,

ϕ = 0 on ∂Ω.

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記号

For ρ > 0, v ∈ L∞((t′,∞);H10 (Ω)), define a ball

B(v, ρ) :=y ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥y − v∥L∞((t′,∞);H1

0 (Ω)) ≤ ρ.

The Frechet derivative of f at w is denoted by

f ′[w] : L∞((t′,∞);H10 (Ω)) → L∞((t′,∞);L2(Ω)).

For y ∈ B(v, ρ), we assume that there exists a non-decreasingfunction L : R → R such that

∥f ′[y]u∥L∞(J ;L2(Ω)) ≤ L(ρ)∥u∥H10, u ∈ H1

0 (Ω).

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記号

Define a function space Xλ: for a fixed λ > 0,

Xλ :=

u ∈ L∞((t′,∞);H1

0 (Ω)) : ess supt∈(t′,∞)

e(t−t′)λ∥u(t)∥H10<∞

,

which becomes a Banach space with the norm

∥ · ∥Xλ:= ess sup

t∈(t′,∞)

e(t−t′)λ∥u(t)∥H10.

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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 ϕ,

For a un ∈ Vh, εn > 0, the initial function satisfies∥η − un∥H1

0< εn.

For a fixed λ satisfying 0 ≤ λ < λmin/2, if ρ > 0 satisfies

∥η − ϕ∥H10+ L(ρ)ρ

√2π

e(λmin − 2λ)< ρ.

Then a solution u(t) for t ∈ (t′,∞) uniquely exists in

Uϕ :=u ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥u− ϕ∥Xλ

≤ ρ.

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Assume that





0< εn.


∥η − ϕ∥H10+ L(ρ)ρ

√2π



Uϕ :=u ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥u− ϕ∥Xλ

≤ ρ.

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Assume that





0< εn.


∥η − ϕ∥H10+ L(ρ)ρ

√2π



Uϕ :=u ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥u− ϕ∥Xλ

≤ ρ.

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Assume that





0< εn.


∥η − ϕ∥H10+ L(ρ)ρ

√2π



Uϕ :=u ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥u− ϕ∥Xλ

≤ ρ.

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Assume that





0< εn.


∥η − ϕ∥H10+ L(ρ)ρ

√2π



Uϕ :=u ∈ L∞ (

(t′,∞);H10 (Ω)

): ∥u− ϕ∥Xλ

≤ ρ.

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Sketch of proof

Let z ∈ Uϕ. A nonlinear operatorS : L∞ ((t′,∞);H1

0 (Ω)) → L∞ ((t′,∞);H10 (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 acondition of S having a fixed-point in Uϕ.For s ∈ (t′,∞) and ψ1, ψ2 ∈ Uϕ, the mean-value theoremstates 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)∥H10.

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How to get ∥η − ϕ∥H10?

Since ∥η − un∥H10< εn and a stationary solution ϕ encloses in

a neighborhood of a numerical solution, it follows

∥η − ϕ∥H10

≤ ∥η − un∥H10+ ∥un − ϕ∥H1

0+ ∥ϕ− ϕ∥H1

0

≤ εn + ∥un − ϕ∥H10+ ρ′.

We need to estimate

∥u(t′)− un∥H10≤ εn.

This can be obtained by the concatenation scheme!

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∥η − ϕ∥H10

≤ ∥η − un∥H10+ ∥un − ϕ∥H1

0+ ∥ϕ− ϕ∥H1

0

≤ εn + ∥un − ϕ∥H10+ ρ′.

We need to estimate

∥u(t′)− un∥H10≤ εn.


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∥η − ϕ∥H10

≤ ∥η − un∥H10+ ∥un − ϕ∥H1

0+ ∥ϕ− ϕ∥H1

0

≤ εn + ∥un − ϕ∥H10+ ρ′.

We need to estimate

∥u(t′)− un∥H10≤ εn.


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∥η − ϕ∥H10

≤ ∥η − un∥H10+ ∥un − ϕ∥H1

0+ ∥ϕ− ϕ∥H1

0

≤ εn + ∥un − ϕ∥H10+ ρ′.

We need to estimate

∥u(t′)− un∥H10≤ εn.


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Computational results 2

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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.

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Table: 時間大域解の検証例（N = 8, λ = 1/40, τk = 2−7）

γ n t′ ρ0.01 5 0.046875 0.010850.011 5 0.046875 0.0119360.0121 6 0.054688 0.0116050.01331 7 0.0625 0.0112740.014641 7 0.0625 0.0124030.016105 8 0.070312 0.0120380.017716 8 0.070312 0.0132440.019487 9 0.078125 0.0128450.021436 10 0.085938 0.0124480.023579 10 0.085938 0.0136950.025937 11 0.09375 0.0132630.028531 11 0.09375 0.0145930.031384 12 0.10156 0.014123

...

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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 0.0299452.5164 41 0.32812 0.02962.768 42 0.33594 0.0294463.0448 42 0.33594 0.0340853.3493 43 0.34375 0.0346573.6842 44 0.35156 0.0359174.0527 45 0.35938 0.0384194.4579 45 0.35938 0.0505114.9037 46 0.36719 0.0664555.3941 47 0.375 0.17656

このとき∥u(t)∥H1

0≤ ρe−

(t−t′)40 , t ∈ (tn,∞).

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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.

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Fig. The numerical solution ϕ.

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.

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Fig. The numerical solution ϕ.

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.

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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

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Thank you for kind attention!

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Science

Akitoshi Takayasu