Solving chaotic differential equations using an - Infoscience
Transcript of Solving chaotic differential equations using an - Infoscience
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Solving chaotic differential equations using anadaptative algorithm
Master Thesis
Lionel Walter
Prof. Alfio QuarteroniDr. Erik BurmanDr. Paolo Zunino
Private Defense / January 27th 2005
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Outline
1 Main NotionsThe Cauchy ProblemThe ChaosThe Ideas to Solve ItCentral Theorem
2 Algorithm validation and improvementOur algorithmComparison AlgorithmTest CasesImprovementsResults
3 Application to PDEBurgers EquationKuramoto-Sivashinsky Equation
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Outline
1 Main NotionsThe Cauchy ProblemThe ChaosThe Ideas to Solve ItCentral Theorem
2 Algorithm validation and improvementOur algorithmComparison AlgorithmTest CasesImprovementsResults
3 Application to PDEBurgers EquationKuramoto-Sivashinsky Equation
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Outline
1 Main NotionsThe Cauchy ProblemThe ChaosThe Ideas to Solve ItCentral Theorem
2 Algorithm validation and improvementOur algorithmComparison AlgorithmTest CasesImprovementsResults
3 Application to PDEBurgers EquationKuramoto-Sivashinsky Equation
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The Cauchy Problem
Find a function X : R → Rd such that{X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
a(·, ·) and X0 are given.
Finding the exact X is usually impossible. We approximate X by X andour goal is to minimize
g(X (T ))− g(X (T ))
where g : Rd → R is a given function, called the goal function.
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The Cauchy Problem
Find a function X : R → Rd such that{X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
a(·, ·) and X0 are given.
Finding the exact X is usually impossible. We approximate X by X andour goal is to minimize
g(X (T ))− g(X (T ))
where g : Rd → R is a given function, called the goal function.
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The Cauchy Problem
Find a function X : R → Rd such that{X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
a(·, ·) and X0 are given.
Finding the exact X is usually impossible. We approximate X by X andour goal is to minimize
g(X (T ))− g(X (T ))
where g : Rd → R is a given function, called the goal function.
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Goal over the entire time interval
Assume we want to have a goal function of the following form :
h(X (T )) +
∫ T
0k(t ,X (t))dt
for given functions h : Rd → R et k : R× Rd → R.If we modify the Cauchy problem as follows
X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
y ′(t) = k(t , y(t))y(0) = 0
then we can define the goal g(X (T )) = h(X (T )) + y(T ) and we get thewanted goal.
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Goal over the entire time interval
Assume we want to have a goal function of the following form :
h(X (T )) +
∫ T
0k(t ,X (t))dt
for given functions h : Rd → R et k : R× Rd → R.If we modify the Cauchy problem as follows
X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
y ′(t) = k(t , y(t))y(0) = 0
then we can define the goal g(X (T )) = h(X (T )) + y(T ) and we get thewanted goal.
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What is Chaos ?
Chaos is used to indicate behavior of solutions to dynamical systemsthat is highly irregular and usually unexpected.
Lorenz applet.
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What is Chaos ?
Chaos is used to indicate behavior of solutions to dynamical systemsthat is highly irregular and usually unexpected.
Lorenz applet.
0 20 40 60 80 100−20
0
20
t
This intuitive notion is present in religion, philosophy, politics, physics. . .
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And in Mathematics ?
It’s rather an idea than a precise mathematical concept. But EdwardOtt gives a mathematical definition of Chaos. He says that if thedifference between the solution of an ODE and the solution of thesame ODE slightly modified grows exponentially with time then thesystem is said to be chaotic.{
X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
{Y ′(t) = a(t ,Y (t)) ∀t ∈ [0 T ]Y (0) = X0 + δ0
If
limδ0→0
‖Y (t)− X (t)‖δ0
≈ eDt
for a constant D > 0, then the system is said to be chaotic.
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Example
0 10 20 30−20
−10
0
10
20
X(t)
t
X(0)=1X(0)=1.005
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Meteorological Example
Monday Morning, 10 degrees. What will be the temperature onFriday Morning ?
We have all the laws of meteorology at hand.
We have 4 methods at our disposal to compute the temperature ofthe following day.
1 2 3 4
speed
precision
We want to have Friday’s temperature with a precision of 1 degreein the least amount of time
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Strategy A
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Strategy A
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Strategy A
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Strategy A
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Strategy A
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Chaotic problems : Thunderstorm
wed
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degrees thunderstorm
bigsmall
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Chaotic problems : Anticyclone
wed
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degrees
Anticyclone
big small
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A better strategy : Strategy B
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A better strategy : Strategy B
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A better strategy : Strategy B
wed
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A better strategy : Strategy B
wed
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A better strategy : Strategy B
wed
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A better strategy : Strategy B
wed
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A better strategy : Strategy B
wed
nes.
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day
mon
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frid
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thur
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A better strategy : Strategy B
wed
nes.
tues
day
mon
day
frid
ay
thur
sday
degrees
1
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
SolutionThe central theorem says that the strategy B can becomputed with the same amount of time as Strategy A
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
True Solution
Last step :
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Analysis of Strategy B
Advantage
Strategy B deals withThunderstorms andAnticyclones.
Drawbacks
Strategy B is muchslower than strategy A
Strategy B could givewrong results
SolutionCombine Strategy A and Strategy B !
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Conclusion
Combining strategy A and B and using the central theorem, we cannow estimate temperature precisely, coping with thunderstorms andanticyclones. The time we need will be at most twice the time ofstrategy A only.
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u and ψ
Definition (u)u(s,Y ) ∈ R is the value of g(X (T )) if we replace the initial conditionX (0) = X0 by X (s) = Y .
ts
Y
T
X(t) u(s,Y)
u if g(x) = x
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u and ψ
Definition (u)u(s,Y ) ∈ R is the value of g(X (T )) if we replace the initial conditionX (0) = X0 by X (s) = Y .
Definition (ψ)
Let X be the true solution of the Cauchy problem. Let W ∈ Rd .ψ : R → Rd measures the sensitivity of the goal function g around thetrue solution :(
ψ(s),W)
= limδ→0
u(s,X (s) + δW )− u(s,X (s))
δ
ψ is the gradient of u with respect to the second variable.
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Local Error
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X(t), true solution
t1 t2 t3 t4=T
e(t2)
e(t4)X(t1)
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Local Error of an approximation
Definition (Local Solution)
Assume we approximate the true solution X (t) by X (t) on a grid0 = t0 < t1 < . . . < tN = T . X (t) is defined as follows, for tn < t ≤ tn+1 :{
X ′(t) = a(t , X (t)) ∀t ∈ (tn, tn+1]
X (tn) = X (tn)
Definition (Local Error)
e(tn) = X (tn)− X (tn).
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Local Error of an approximation
Definition (Local Solution)
Assume we approximate the true solution X (t) by X (t) on a grid0 = t0 < t1 < . . . < tN = T . X (t) is defined as follows, for tn < t ≤ tn+1 :{
X ′(t) = a(t , X (t)) ∀t ∈ (tn, tn+1]
X (tn) = X (tn)
Definition (Local Error)
e(tn) = X (tn)− X (tn).
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Central Theorem (intuitive)
TheoremAssume that the Cauchy problem has a unique solution for all possibleX0. Assume that a(t , x) is differentiable in x for all t in [0 T ]. For alldifferential functions g, the global error is a weighted sum of the localerrors :
g(X (T ))− g(X (T )) =N∑
n=1
(e(tn), ψ(tn)
)and ψ satisfies −dψ(s)
ds= (a′)∗(s,X (s)) ψ(s)
ψ(T ) = ∇g(X (T ))
The above problem is called the dual problem of the Cauchy problem.
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Central Theorem (intuitive)
TheoremAssume that the Cauchy problem has a unique solution for all possibleX0. Assume that a(t , x) is differentiable in x for all t in [0 T ]. For alldifferential functions g, the global error is a weighted sum of the localerrors :
g(X (T ))− g(X (T )) =N∑
n=1
(e(tn), ψ(tn)
)and ψ satisfies −dψ(s)
ds= (a′)∗(s,X (s)) ψ(s)
ψ(T ) = ∇g(X (T ))
The above problem is called the dual problem of the Cauchy problem.
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Central Theorem (intuitive)
TheoremAssume that the Cauchy problem has a unique solution for all possibleX0. Assume that a(t , x) is differentiable in x for all t in [0 T ]. For alldifferential functions g, the global error is a weighted sum of the localerrors :
g(X (T ))− g(X (T )) =N∑
n=1
(e(tn), ψ(tn)
)and ψ satisfies −dψ(s)
ds= (a′)∗(s,X (s)) ψ(s)
ψ(T ) = ∇g(X (T ))
The above problem is called the dual problem of the Cauchy problem.
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Central Theorem (exact form)
TheoremAssume that the Cauchy problem has a unique solution for all possibleX0. Assume that a(t , x) is differentiable in x for all t in [0 T ]. For alldifferential functions g, the global error is a weighted sum of the localerrors :
g(X (T ))− g(X (T )) =N∑
n=1
(e(tn),
∫ 1
0ψ(tn, X (tn) + se(tn))ds
)and ψ satisfies −dψ(s,X (s))
ds= (a′)∗(s,X (s)) ψ(s,X (s))
ψ(T ,X (T )) = ∇g(X (T ))
The above problem is called the dual problem of the Cauchy problem.
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X and ψ, its dual
Primal {X ′(t) = a(t ,X (t)) ∀t ∈ [0 T ]X (0) = X0
Dual {−ψ′(s) = (a′)∗(s,X (s)) ψ(s) ∀s ∈ [0 T ]ψ(T ) = ∇g(X (T ))
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Approximation of e(tn)
To approximate e(tn) = X (tn)− X (tn), we use another approximationX (tn) and we do Richardson extrapolation based on the differentorders of the estimation X (tn) and X (tn).
e(tn) = γ(X (tn)− X (tn))
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Approximation of ψ(tn)
We replace the system −dψ(s)
ds= (a′)∗(s,X (s)) ψ(s)
ψ(T ) = ∇g(X (T ))
by the system −dψ(s)
ds= (a′)∗(s, X (s)) ψ(s)
ψ(T ) = ∇g(X (T ))
So, we can find ψ.
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The algorithm
From the estimation of the global error
g(X (T ))− g(X (T )) ≈N∑
n=1
(e(tn), ψ(tn)
)︸ ︷︷ ︸
rn
we will construct an adaptative algorithm to solve chaotic ordinarydifferential equations. The idea is to refine the mesh at the placeswhere rn is big.
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The algorithm
Inputs
TOL, the tolerance we want on g(X (T ))− g(X (T )).
A one-step numerical method M to approximate ODE (Eulerprogressive, Runge-Kutta, . . . ).
N0 the initial number of time steps.
a(·, ·) the given function of the Cauchy Problem.
g(·) the goal function.
T the final time.
X0 the initial condition.
OutputsT , the mesh the algorithm use to find the approximation.
X the approximation of X on the mesh T that satisfiesg(X (T ))− g(X (T )) < TOL
Algorithm validation and improvement Our algorithm 23 / 65
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The algorithm
Inputs
TOL, the tolerance we want on g(X (T ))− g(X (T )).
A one-step numerical method M to approximate ODE (Eulerprogressive, Runge-Kutta, . . . ).
N0 the initial number of time steps.
a(·, ·) the given function of the Cauchy Problem.
g(·) the goal function.
T the final time.
X0 the initial condition.
OutputsT , the mesh the algorithm use to find the approximation.
X the approximation of X on the mesh T that satisfiesg(X (T ))− g(X (T )) < TOL
Algorithm validation and improvement Our algorithm 23 / 65
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The algorithm
Inputs
TOL, the tolerance we want on g(X (T ))− g(X (T )).
A one-step numerical method M to approximate ODE (Eulerprogressive, Runge-Kutta, . . . ).
N0 the initial number of time steps.
a(·, ·) the given function of the Cauchy Problem.
g(·) the goal function.
T the final time.
X0 the initial condition.
OutputsT , the mesh the algorithm use to find the approximation.
X the approximation of X on the mesh T that satisfiesg(X (T ))− g(X (T )) < TOL
Algorithm validation and improvement Our algorithm 23 / 65
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The algorithm
Inputs
TOL, the tolerance we want on g(X (T ))− g(X (T )).
A one-step numerical method M to approximate ODE (Eulerprogressive, Runge-Kutta, . . . ).
N0 the initial number of time steps.
a(·, ·) the given function of the Cauchy Problem.
g(·) the goal function.
T the final time.
X0 the initial condition.
OutputsT , the mesh the algorithm use to find the approximation.
X the approximation of X on the mesh T that satisfiesg(X (T ))− g(X (T )) < TOL
Algorithm validation and improvement Our algorithm 23 / 65
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The algorithm
Inputs
TOL, the tolerance we want on g(X (T ))− g(X (T )).
A one-step numerical method M to approximate ODE (Eulerprogressive, Runge-Kutta, . . . ).
N0 the initial number of time steps.
a(·, ·) the given function of the Cauchy Problem.
g(·) the goal function.
T the final time.
X0 the initial condition.
OutputsT , the mesh the algorithm use to find the approximation.
X the approximation of X on the mesh T that satisfiesg(X (T ))− g(X (T )) < TOL
Algorithm validation and improvement Our algorithm 23 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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The algorithm
1 T a uniform grid of N0 timesteps.
2 Compute X on the mesh T using the numerical method M.
3 Compute X on a different mesh using M.4 Use extrapolation to compute the approximation of the local error
e = γ(X − X ).5 Compute the weights ψ on the mesh T using M.6 Compute the residuals ri = (ei , ψi)
7 Compute the estimation of the error E =∑
i ri . If E < TOL stop.8 Refine the mesh where ri is big.9 Go back to 2.
We call this algorithm mstz .
Algorithm validation and improvement Our algorithm 24 / 65
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Adaptive Runge-Kutta
One of the most used method for solving ODE. Based on theDormand-Prince pair. Use a pair of Runge-Kutta methods of order4 and 5 to estimate the local error.
A typical example of the strategy A.
Implemented in MATLAB under the name ode45 .
Need to give to the solver the tolerance ε we want on the localerror.
As there is no estimation of the global error with ode45 , we decided togive to the solver the exact solution X (T ) and we decrease ε until wehave g(X (T ))− g(X (T )).
Algorithm validation and improvement Comparison Algorithm 25 / 65
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Adaptive Runge-Kutta and global error
1 Set the tolerance on the local error ε =TOLN0
.
2 Compute X using ode45 .3 Compute
E = g(X (T ))− g(X (T ))
If E < TOL, stop. Else set ε = ε/10 and go back to 2
We call this algorithm RK45.
Algorithm validation and improvement Comparison Algorithm 26 / 65
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Adaptive Runge-Kutta and global error
1 Set the tolerance on the local error ε =TOLN0
.
2 Compute X using ode45 .3 Compute
E = g(X (T ))− g(X (T ))
If E < TOL, stop. Else set ε = ε/10 and go back to 2
We call this algorithm RK45.
Algorithm validation and improvement Comparison Algorithm 26 / 65
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Adaptive Runge-Kutta and global error
1 Set the tolerance on the local error ε =TOLN0
.
2 Compute X using ode45 .3 Compute
E = g(X (T ))− g(X (T ))
If E < TOL, stop. Else set ε = ε/10 and go back to 2
We call this algorithm RK45.
Algorithm validation and improvement Comparison Algorithm 26 / 65
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Strategy A
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Algorithm validation and improvement Comparison Algorithm 27 / 65
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Strategy A
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Algorithm validation and improvement Comparison Algorithm 27 / 65
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Strategy A
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Algorithm validation and improvement Comparison Algorithm 27 / 65
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Strategy A
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Algorithm validation and improvement Comparison Algorithm 27 / 65
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Strategy A
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Algorithm validation and improvement Comparison Algorithm 27 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
wed
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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A better strategy : Strategy B
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Algorithm validation and improvement Comparison Algorithm 28 / 65
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Test Cases
We will have 6 test cases :
The Exponential
A non-linear problem with blow-up
A stiff problem
A problem with a discontinuity in the derivative
A model of the transition to the turbulence
The Lorenz problem
The last 2 problems are chaotic. For the numerical method we use aRunge-Kutta method of order 5. The coefficients are taken from theDormand-Prince pair.
Algorithm validation and improvement Test Cases 29 / 65
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The Exponential
(a)
X ′(t) = X (t) ∀t ∈ [0 3]X (0) = 1g(x) = xTOL = 10−8
N0 = 5
solution : X (t) = et
0 1 2 30
10
20
30X(t)
t0 1 2 3
0
10
20
30ψ(t)
t0 1 2 3
−1.5
−1
−0.5
0x 10−11 e(t)
t0 1 2 3
0.0375
0.0375
0.0375
0.0375
t
taille des pas de temps
Algorithm validation and improvement Test Cases 30 / 65
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Non-linear with blow-up
(b)
X ′(t) = 2(t + 1)X 2(t) ∀t ∈ [0 0.4]X (0) = 1g(x) = x2
TOL = 0.1N0 = 5
solution : X (t) =−1
t2 + 2t − 1
0 0.2 0.40
10
20
30X(t)
t0 0.2 0.4
0
1
2
3
4x 104 ψ(t)
t0 0.2 0.4
−4
−2
0
2x 10−4 e(t)
t0 0.2 0.4
0.01
0.02
0.03
0.04
0.05
t
taille des pas de temps
Algorithm validation and improvement Test Cases 31 / 65
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Stiff problem
(c)
X ′(t) = t(1− X (t)) + (1− t)e−t ∀t ∈ [0 10]X (0) = 1g(x) = xTOL = 10−8
N0 = 5
solution : X (t) = e−t2/2 − e−t + 1.
0 5 100.9
1
1.1
1.2
1.3X(t)
t0 5 10
0
0.5
1ψ(t)
t0 5 10
−1
−0.5
0
0.5
1x 10−6 e(t)
t0 5 10
0.1
0.15
0.2
0.25
t
taille des pas de temps
Algorithm validation and improvement Test Cases 32 / 65
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Singularity
(d)
X ′(t) =X (t)√
t − 5/3 + π10−8∀t ∈ [0 10]
X (0) = e−2√
53−π10−8
g(x) = xTOL = 0.1N0 = 5
sol. : X (t) = exp(2 · sign(t − 5/3 + π10−8) ·√|t − 5/3 + π10−8|)
0 2 40
10
20
30X(t)
t0 2 4
0
100
200
300ψ(t)
t0 2 4
−2
0
2
4
6x 10−3 e(t)
t0 2 4
0
0.5
1
t
taille des pas de temps
Algorithm validation and improvement Test Cases 33 / 65
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Transition to turbulence
(t)
X ′(t) =
(−R−1 1
0 −R−1
)X (t) + ||X (t)||
(0 −11 0
)X (t)
X (0) =δ√2(1,1) δ > 0
g(X ) = x1 (the first component)TOL = 10−6
N0 = 500
0 50010−10
10−5
100
105||X(t)||
t0 500
10−5
100
105||ψ(t)||
t0 500
10−20
10−15
10−10
10−5||e(t)||
t0 500
0
0.5
1
t
taille des pas de temps
Algorithm validation and improvement Test Cases 34 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6 X t=20
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6 X t=40
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6 X t=60
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6 X t=80
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=100
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=120
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=140
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=160
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=180
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=200
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=220
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=240
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=260
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=280
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=300
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=320
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=340
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=360
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=380
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=400
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=420
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=440
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=460
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=480
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.4
2 4 6
x 10−4
5
10
15
x 10−6X t=500
0 0.5 10
0.2
0.4
0.6
0.8
ψ
Algorithm validation and improvement Test Cases 35 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=10
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=20
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=30
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=40
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=50
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=60
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=70
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=80
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=90
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=100
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=110
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=120
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=130
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=140
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=150
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=160
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=170
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=180
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=190
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=200
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=210
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=220
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=230
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=240
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=250
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=260
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=270
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=280
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=290
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=300
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=310
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=320
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=330
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=340
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=350
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=360
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=370
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=380
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=390
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=400
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=410
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=420
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=430
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=440
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=450
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=460
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=470
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=480
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=490
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Transition to turbulence ‖X0‖ = 10−5.2
−0.5 0 0.5
0.2
0.4
0.6
0.8
1
1.2
1.4
X t=500
−10000−5000 0 5000−6
−4
−2
0
x 104 ψ
Algorithm validation and improvement Test Cases 36 / 65
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Lorenz Problem
(l)
x ′1(t) = −σx1(t) + σx2(t)x ′2(t) = rx1(t)− x2(t)− x1(t)x3(t)x ′3(t) = x1(t)x2(t)− bx3(t)
∀t ∈ [0 10] avec σ = 10, b = 8/3, r = 28X (0) = (1,0,0)g(X ) = x1
TOL = 0.1 et 0.01N0 = 300
0 10 20 30−20
−10
0
10
20
x1(t)
t0 10 20 30
0
2
4
6x 106 ||ψ(t)||
t0 10 20 30
0
2
4
6x 10−5 ||e(t)||
t0 10 20 30
0
0.02
0.04
0.06
t
taille des pas de temps
Algorithm validation and improvement Test Cases 37 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=1
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=2
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=3
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=4
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=5
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=6
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=7
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=8
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=9
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=10
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=11
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=12
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=13
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=14
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=15
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=16
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=17
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=18
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=19
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=20
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=21
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=22
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=23
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=24
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=25
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=26
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=27
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=28
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=29
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Lorenz with Dual
020
40−100
10
−20
0
20
X t=30
−20
2x 106
−20
2x 106
−4
−2
0
2
4
x 106ψ
Algorithm validation and improvement Test Cases 38 / 65
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Methods implemented
The brut method. Divide the timesteps by 2 when ri >TOLN
.
The basique method. Divide the timesteps by
M =
⌊(|ri |TOLN
) 1p+1
⌋
when ri >TOLN
.
The demi-pas method. Same as basique, except that we estimatethe error using a coarser approximation of X rather than a finer asbefore. The dual is solved on a mesh 2 times coarser than theprimal.
Algorithm validation and improvement Improvements 39 / 65
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Results
a b c d t l l 0.010
20
40
60
80
100 RK45 .
Algorithm validation and improvement Results 40 / 65
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Results
a b c d t l l 0.010
20
40
60
80
100 RK45 .brut
Algorithm validation and improvement Results 40 / 65
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Results
a b c d t l l 0.010
20
40
60
80
100 RK45 .brutbasique
Algorithm validation and improvement Results 40 / 65
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Results
a b c d t l l 0.010
20
40
60
80
100 RK45 .brutbasiquedemi−pas
Algorithm validation and improvement Results 40 / 65
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Exponential, Non-linear with blow-up and Stiff problem
TOL True error Est. errorr # eval. a # eval. (a′)∗ CPU N itbrut 1e-08 -1.16e-09 -1.16e-09 2790 930 0.89 80 5basique 1e-08 -3.86e-10 -3.86e-10 3240 1080 0.59 100 4demi-pas 1e-08 -4.8e-09 -4.38e-09 900 300 0.17 30 3RK45 1e-08 -1.33e-09 - 776 0 0.93 79 2
The Exponential
TOL True error Est. errorr # eval. a # eval. (a′)∗ CPU N itbrut 0.1 0.0125 0.00629 810 270 0.18 16 4basique 0.1 0.0125 0.00629 594 198 0.12 16 3demi-pas 0.1 0.0194 -0.0122 360 120 0.07 8 3RK45 0.1 0.0986 - 237 0 0.06 13 3
Non-linear with blow-up
TOL True error Est. errorr # eval. a # eval. (a′)∗ CPU N itbrut 1e-08 1.39e-09 1.41e-09 2160 720 0.41 45 5basique 1e-08 4.77e-10 4.87e-10 1944 648 0.35 53 3demi-pas 1e-08 3.68e-10 8.41e-10 990 330 0.18 50 2RK45 1e-08 5.92e-10 - 895 0 0.13 146 1
Stiff problem
Algorithm validation and improvement Results 41 / 65
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Singularity and Transition to Turbulence
TOL True error Est. errorr # eval. a # eval. (a′)∗ CPU N itbrut 0.1 0.0484 0.0976 4320 1440 0.81 24 16basique 0.1 0.0484 0.0976 4320 1440 0.79 24 16demi-pas 0.1 -1.1 0.0858 810 270 0.16 10 6RK45 0.1 0.0055 - 1061 0 0.19 71 5
Singularity
TOL True error Est. error # eval. a # eval. (a′)∗ CPU N itbrut 1e-06 5.49e-08 1.55e-07 184716 61572 44.82 2531 7basique 1e-06 6.31e-08 1.7e-07 113652 37884 27.4 2514 4demi-pas 1e-06 9.16e-08 3.19e-07 50832 16944 12.23 1335 3RK45 1e-06 -9.69e-07 - 140045 0 24.04 9569 5
Transition to turbulence
Algorithm validation and improvement Results 42 / 65
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Lorenz, 2 tolerances
TOL True error Est. error # eval. a # eval. (a′)∗ CPU N itbrut 0.1 0.0623 0.0674 256734 85578 57.91 6461 6basique 0.1 -0.0251 -0.0261 157680 52560 34.43 5789 3demi-pas 0.1 -0.0146 -0.0182 94716 31572 20.7 3176 3RK45 0.1 -0.0147 - 168110 0 25.29 10513 8
Lorenz problem
TOL Vraie erreur Est. erreur # eval. a # eval. (a′)∗ CPU N itbrut 0.01 0.000979 0.000998 472788 157596 108.75 10463 7basique 0.01 -0.0037 -0.00365 198108 66036 44.29 7949 3demi-pas 0.01 -0.00428 -0.00521 115434 38478 25.39 3985 3RK45 0.01 -0.0015 - 264932 0 40.37 16651 8
Lorenz problem
Algorithm validation and improvement Results 43 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : 7.56, RK45 : 1.18
iteration 1, mstz : 300 timesteps, RK45 : 678 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : 0.87, RK45 : 8.75
iteration 2, mstz : 2128 timesteps, RK45 : 1087 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 2.73
iteration 3, mstz : 3985 timesteps, RK45 : 1656 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 17.48
iteration 4, mstz : 3985 timesteps, RK45 : 2696 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 0.12
iteration 5, mstz : 3985 timesteps, RK45 : 4173 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 0.13
iteration 6, mstz : 3985 timesteps, RK45 : 6637 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 0.015
iteration 7, mstz : 3985 timesteps, RK45 : 10513 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Size of time steps (Lorenz problem, TOL= 0.01)
0 5 10 15 20 25 300
0.05
0.1
0.15
0.2
erreur, mstz : −0.0042, RK45 : 0.0015
iteration 8, mstz : 3985 timesteps, RK45 : 16651 timesteps
mstzRK45
Algorithm validation and improvement Results 44 / 65
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Comparison with Szepessy and al.
TOL |True error| N Ntot
mstz demi-pas 0.1 0.02 6352 10524mstz Szepessy 0.1 0.01 6000 20000mstz demi-pas 0.01 0.004 7970 12826mstz Szepessy 0.01 0.003 9000 34000
Algorithm validation and improvement Results 45 / 65
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Application to PDE
We now use the algorithm mstz to solve semi-discretization of partialdifferential equations. We use the heat equation to validate our code.
1 We study the properties of the Burgers Equation (a non-chaoticequation) with the help of mstz . We use a centered finitedifference scheme for the spatial discretization.
2 We compare RK45 and mstz on the Kuramoto-SivashinskyEquation, a chaotic PDE. We use the pseudo-spectral method forthe spatial discretization.
Application to PDE 46 / 65
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Application to PDE
We now use the algorithm mstz to solve semi-discretization of partialdifferential equations. We use the heat equation to validate our code.
1 We study the properties of the Burgers Equation (a non-chaoticequation) with the help of mstz . We use a centered finitedifference scheme for the spatial discretization.
2 We compare RK45 and mstz on the Kuramoto-SivashinskyEquation, a chaotic PDE. We use the pseudo-spectral method forthe spatial discretization.
Application to PDE 46 / 65
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Application to PDE
We now use the algorithm mstz to solve semi-discretization of partialdifferential equations. We use the heat equation to validate our code.
1 We study the properties of the Burgers Equation (a non-chaoticequation) with the help of mstz . We use a centered finitedifference scheme for the spatial discretization.
2 We compare RK45 and mstz on the Kuramoto-SivashinskyEquation, a chaotic PDE. We use the pseudo-spectral method forthe spatial discretization.
Application to PDE 46 / 65
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Study of Burgers Equation
For x ∈ [−1 1] and t ∈ [0 1], we have∂
∂tu(t , x) +
∂
∂xu2(t , x)
2= 0
u(0, x) = u0(x)
Application to PDE Burgers Equation 47 / 65
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Study of Burgers Equation
For x ∈ [−1 1] and t ∈ [0 1], we have∂
∂tu(t , x) +
∂
∂xu2(t , x)
2= ε
∂2
∂x2 u(t , x)
u(0, x) = u0(x)
Viscous term
Application to PDE Burgers Equation 47 / 65
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Centered Finite Difference Scheme
uhj (t) = −
(uh
j+1(t))2 −
(uh
j−1(t))2
4h+ ε
uhj+1(t)− 2uh
j (t) + uhj−1(t)
h2
Application to PDE Burgers Equation 48 / 65
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Initial conditions
0x
Shock Wave
Application to PDE Burgers Equation 49 / 65
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Initial conditions
0x
Shock Wave
Application to PDE Burgers Equation 49 / 65
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Initial conditions
0x
Shock Wave
0x
Rarefaction Wave
Application to PDE Burgers Equation 49 / 65
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Initial conditions
0x
Shock Wave
0x
Rarefaction Wave
Application to PDE Burgers Equation 49 / 65
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Initial conditions
0x
Shock Wave
0x
Rarefaction Wave
Which one is the hardest to solve ?
Application to PDE Burgers Equation 49 / 65
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Goal Function
For g, we select a point x ∈ [−1 1]. Our goal is to have a very precisevalue of u(T , x). To have a smoother function in the x-domain, we takea weighted sum around the point x . The weights are give by thenormal density.
g(u(T )) =M−1∑j=0
wju(T , xj) =M−1∑j=0
1√2πσ
e− 1
2
(xj−x
σ
)2
u(T , xj)
(∇g(u(T )))j = wj =1√2πσ
e− 1
2
(xj−x
σ
)2
Application to PDE Burgers Equation 50 / 65
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Parameters
ε=0.01
h=0.025
∆t=0.005 (initial)
x = 0.5, σ = 0.1.
TOL=0.005
Numerical method M : Euler progressive
Application to PDE Burgers Equation 51 / 65
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Shock Wave
Application to PDE Burgers Equation 52 / 65
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Rarefaction Wave
Application to PDE Burgers Equation 53 / 65
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Comparison
Shock Wave (down)
0 0.5 10
0.005
0.01
t
size of time steps
Number of timestepsneeded : 523
Number of evaluationsneeded : 3168
Rarefaction Wave (up)
0 0.5 10
1
2x 10−4
t
time steps
Number of timestepsneeded : 8565
Number of evaluationsneeded : 48750
Application to PDE Burgers Equation 54 / 65
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Explanation
Shock Wave (down)
t
0
x0
The perturbationstravel quicker than thewave. They are eatenby the shock
The important parts arethe parts just prior T
Rarefaction Wave (up)
t
0
0 x
The perturbations areamplified along thetime interval
It’s important to makegood calculations at thebegining
Application to PDE Burgers Equation 55 / 65
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A chaotic PDE
The Kuramoto-Sivashinsky Equation. x ∈ R, t ∈ [0 120].∂u(t , x)
∂t= − ∂2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂xu(0, x) = u0(x)
u(t , x) = u(t , x + l)
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A chaotic PDE
The Kuramoto-Sivashinsky Equation. x ∈ R, t ∈ [0 120].∂u(t , x)
∂t= − ∂2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂xu(0, x) = u0(x)
u(t , x) = u(t , x + l)
Inverse Heat term. Amplifies the perturbations
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A chaotic PDE
The Kuramoto-Sivashinsky Equation. x ∈ R, t ∈ [0 120].∂u(t , x)
∂t= − ∂2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂xu(0, x) = u0(x)
u(t , x) = u(t , x + l)
Beam term. Reduces the perturbations
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A chaotic PDE
The Kuramoto-Sivashinsky Equation. x ∈ R, t ∈ [0 120].∂u(t , x)
∂t= − ∂2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂xu(0, x) = u0(x)
u(t , x) = u(t , x + l)
Burgers term. Non-linear transport term
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A chaotic PDE
The Kuramoto-Sivashinsky Equation. x ∈ R, t ∈ [0 120].∂u(t , x)
∂t= − ∂2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂xu(0, x) = u0(x)
u(t , x) = u(t , x + l)
All together : the solution remain bounded !
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Pseudo-spectral method
The 4th order term is to heavy to discretize using finite differences. Weuse the pseudo-spectral method based on Fourier theory. This allowsas well an easy way of dealing with periodic boundary conditions. Wewrite :
u(t , x) =+∞∑
n=−∞cn(t)e2iπn x
l
with cn(t) =1l
∫ l
0u(t , x)e−2iπn x
l dx
Taking a finite number of Fourier coefficients and computing them withthe Fast Fourier Transform gives the pseudo-spectral scheme.
Application to PDE Kuramoto-Sivashinsky Equation 57 / 65
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The semi-discretization
We obtain the following ODE (q = 2πl ):
cn(t) = (n2q2 − n4q4)cn(t)−12
inqdn(t) ∀n = −m, . . . ,m
where dn(t) are the Fourier coefficients of u2(t , x). We compute themusing the inverse Fourier transform of all the cn, then squaring all thecomponents and then taking the Fourier transform again.
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The semi-discretization
We obtain the following ODE (q = 2πl ):
cn(t) = (n2q2 − n4q4)cn(t)−12
inqdn(t) ∀n = −m, . . . ,m
where dn(t) are the Fourier coefficients of u2(t , x). We compute themusing the inverse Fourier transform of all the cn, then squaring all thecomponents and then taking the Fourier transform again.
Inverse Heat term. Amplifies the perturbations
Application to PDE Kuramoto-Sivashinsky Equation 58 / 65
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The semi-discretization
We obtain the following ODE (q = 2πl ):
cn(t) = (n2q2 − n4q4)cn(t)−12
inqdn(t) ∀n = −m, . . . ,m
where dn(t) are the Fourier coefficients of u2(t , x). We compute themusing the inverse Fourier transform of all the cn, then squaring all thecomponents and then taking the Fourier transform again.
Beam term. Reduces the perturbations
Application to PDE Kuramoto-Sivashinsky Equation 58 / 65
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The semi-discretization
We obtain the following ODE (q = 2πl ):
cn(t) = (n2q2 − n4q4)cn(t)−12
inqdn(t) ∀n = −m, . . . ,m
where dn(t) are the Fourier coefficients of u2(t , x). We compute themusing the inverse Fourier transform of all the cn, then squaring all thecomponents and then taking the Fourier transform again.
Burgers term. Non-linear transport term
Application to PDE Kuramoto-Sivashinsky Equation 58 / 65
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Initial condition
Following Wanner and Hairer, the initial condition has the followingshape :
0 50 100 150 200 2500
0.5
1
x
u(0,
x)
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Number of coefficients
To get a solution in a reasonable amount of time, we decided to use150 Fourier coefficients. The problem is that we can not solve theorginal Kuramoto-Sivashinsky Equation with 150 coefficients. Thenumerical scheme is unstable. We modify the Kuramoto-Sivashinskyequation as follows :
∂u(t , x)
∂t= −λ∂
2u(t , x)
∂x2 − ∂4u(t , x)
∂x4 − 12∂u2(t , x)
∂x
For λ = 1, we have the original equation. For λ < 1, the perturbationsare less amplified by the inverse heat term, so the equation is easier tosolve. We will be able to solve it for λ = 0.93.
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Solution
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Solution of the dual
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Results
TOL E t E t# eval. a # eval. (a′)∗ CPU N it
mstz 0.0001 2.09e-05 3.41e-05 18666 6222 115.94 637 2RK45 0.0001 3.55e-05 - 29499 0 55.49 2382 3mstz 1e-05 -8.01e-06 -8.61e-06 21078 7026 133.53 771 2RK45 1e-05 4.2e-06 - 45915 0 86.2 3726 3mstz 1e-06 4.61e-08 -6.26e-07 25776 8592 158.04 1032 2RK45 1e-06 8.22e-07 - 71853 0 138.16 5855 3mstz 1e-07 4.15e-07 3.24e-09 79254 26418 484.8 2524 3RK45 1e-07 No solution !
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Summary
Using an algorithm which monitors the global error instead of onewhich monitors the local error is really important for chaotic differentialequations. It provides :
An indicator of the accuracy of the solution. With local errorsalgorithms, we have no garanty that we are close to the truesolution even if the local tolerance is really small.
A better efficiency. We get the solution faster or for a smallertolerance.
Computing the solution of the dual less accurately than thesolution of the primal gives a really efficient algorithm
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Questions
Summary 65 / 65