Numerical Methods for Differential Equations Chapter 1...
Transcript of Numerical Methods for Differential Equations Chapter 1...
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Numerical Methods for Differential EquationsChapter 1: Initial value problems
Gustaf SoderlindNumerical Analysis, Lund University
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Contents V4.16
1. The initial value problem
2. The Explicit Euler method
3. Convergence
4. Order of consistency
5. The trapezoidal rule
6. Theta methods
7. Numerical tests
8. The linear test equation and numerical stability
9. Stiff equations
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1. Initial value problems
Standard formulation of a system of ODEs
y ′ = f (t, y) ; y(0) = y0
with f : R× Rm → Rm
Theorem If f (t, y) is continuous for t ∈ [0,T ] and satisfies theLipschitz condition
‖f (t, u)− f (t, v)‖ ≤ L[f ] · ‖u − v‖
for all u, v ∈ Rm with Lipschitz constant L[f ] <∞, then thereexists a unique solution to the initial value problem on [0,T ] forevery initial value y(0) = y0
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Existence and uniqueness
Some problems always satisfy Lipschitz conditions on Rm
Example A linear constant coefficient differential equation
y = Ay ; y(0) = y0
has Lipschitz constant
L[f ] = maxu 6=v
‖Au − Av‖‖u − v‖
= maxy 6=0
‖Ay‖‖y‖
= ‖A‖
The matrix norm ‖A‖ is a Lipschitz constant for f (y) = Ay
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Existence and uniqueness
Note Most nonlinear problems do not satisfy a Lipschitz conditionon all of Rm
Example The problem
y = y2; y(0) = y0 > 0
has solutiony(t) =
y01− y0t
The solution blows up at t = 1/y0 (“Finite escape time”)
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Standard form x ′ = F (t, x)
Given y ′′ = f (t, y , y ′) with y(0) = y0, y ′(0) = y ′0
Standard substitution Introduce new variables
x1 = y
x2 = y ′
to obtain a system of first order equations
x ′1 = x2
x ′2 = f (t, x1, x2)
with x1(0) = y0 and x2(0) = y ′0
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2. The Explicit Euler method (1768)
Replace y ′ in y ′ = f (t, y) by finite difference approximation
y ′(tn) ≈ y(tn + h)− y(tn)
h
Let yn denote the numerical approximation to y(tn) in
yn+1 − ynh
= f (tn, yn), y0 = y(t0)
Explicit Euler method Compute {yn} recursively from
yn+1 = yn + hf (tn, yn)
tn+1 = tn + h
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Explicit Euler Taylor series expansion
Taylor series expansion
y(t + h) = y(t) + hy ′(t) +h2
2!y ′′(ξ)
= y(t) + hf (t, y(t)) + O(h2) ⇒
y(t + h) ≈ y(t) + hf (t, y(t))
Explicit Euler method obtained by dropping higher order terms
yn+1 = yn + hf (tn, yn)
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Explicit Euler Geometric interpretation
“Take a step of size h in the direction of the tangent”
0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.7
0.75
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0.85
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1
1.05
1.1
1.15
1.2
t
y
Ending up on a different solution trajectory (dashed curves), each stepintroduces a local error, eventually accumulating a large global error
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Explicit Euler Example
y ′ = −y cos t; y(0) = 1; t ∈ [0, 2π]
Stepsize h = 2π/N with N = 24 gives h = π/12
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1.8
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The numerical solution is a sequence of points (tn, yn)
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3. Convergence
Analytical and numerical solutions at h = π/8 and h = π/128
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N=64
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N=64*16
The numerical solution approaches the exact solution as h→ 0
Definition A method is convergent if, for every ODE with aLipschitz vector field f , and every fixed T = N · h, it holds that
limN→∞
‖yN,h − y(T )‖ = 0
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Local and global errors
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1
1.05
1.1
1.15
t
y
yn+1•
yn•
yn+1
•
y(t)y(tn)• •
y(tn+1)
Global error en = yn − y(tn) and en+1 = yn+1 − y(tn+1)
Local error ln+1 = yn+1 − y(tn+1)
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The local error Explicit Euler
Insert exact data y(tn) and y(tn+1)
y(tn+1) = y(tn) + hf (tn, y(tn))− ln+1
The residual is the local error
Taylor series y(tn+1) = y(tn) + hy ′(tn) + h2
2 y′′(tn) + . . . ⇒
Explicit Euler local error ln+1 ≈ − h2
2 y ′′(tn)
The local error is evaluated along exact solution (solid curve)
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Error propagation From local to global error
Explicit Euler (numerical solution)
yn+1 = yn + hf (tn, yn)
Subtract Taylor series expansion of exact solution
y(tn+1) = y(tn) + hf (tn, y(tn)) +h2
2y ′′(tn) + . . .
Global error recursion
en+1 = en + hf (tn, y(tn) + en)− hf (tn, y(tn)) + ln+1
shows how local errors accumulate into global error
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Error propagation . . .
en+1 = en + hf (tn, y(tn) + en)− hf (tn, y(tn)) + ln+1
Take norms and use Lipschitz condition
‖en+1‖ ≤ ‖en‖+ hL[f ] · ‖en‖+ ‖ln+1‖
Lemma Assume that a non-negative sequence {an} satisfies
an+1 ≤ (1 + h µ)an + ch2
with a0 = 0 and µ > 0. Then, for all n ≥ 0,
an ≤ch
µ[(1 + h µ)n − 1] ≤ ch
eµnh − 1
µ
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Convergence of the Euler method
Theorem The explicit Euler method is convergent
Proof Suppose f is sufficiently differentiable. Given h > 0 and afixed T = Nh, let en,h = yn,h − y(tn)
Apply the lemma to global error recursion, to get
‖en,h‖ ≤c
L[f ]h [(1 + h L[f ])n − 1] ≤ ch
eTL[f ] − 1
L[f ]
withc = max
n‖ln‖/h2 ≈ max
t‖y ′′‖/2
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Convergence of the Euler method . . .
‖en,h‖ ≤c
L[f ]h (eTL[f ] − 1) = C (T ) · h
implies the method is convergent, as limh→0 ‖en,h‖ = 0
Note
1) The global error can be made arbitrarily small
2) The error bound is way too large for practical purposes
3) Better error bounds can be obtained
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Theoretical error bound . . . is hopeless!
Exampley ′ = −100y , y(0) = 1
Then L[f ] = 100 and the exact solution is y(t) = e−100t withy ′′(t) = 1002e−100t , so c = 1002/2, with bound
‖en,h‖ ≤1002
2 · 100h (e100T − 1)
Error estimate at T = 1 is ‖en,h‖ ≤ 50 h e100 ≈ 1.4 · 1045h !
Actual error As yn = (1− 100h)n, the error at T = 1 for h < 1/50is ‖en,h‖ = |(1− 100/N)N − e−100| ≤ 3.7 · 10−44h !
The error is overestimated by at least 89 orders of magnitude
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Computational test y = λ(y − sin t) + cos t
λ = 0.2, with initial condition y(π/4) = 1/√
2
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.5
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y
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t
y
h = π/10 h = π/20
Note Local error O(h2), global error O(h)!
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4. Order of consistency
Always insert exact data and find the residual
If yn+1 = Φh(f , tn, yn, yn−1, . . . ), then the local error is
y(tn+1) = Φh(f , tn, y(tn), y(tn−1), . . . )− ln+1
Definition The order of consistency is p if
y(tn+1)− Φh(f , h, y(tn), y(tn−1), . . . ) = O(hp+1)
as h→ 0, for every analytic f . The local error is then O(hp+1)
Alternative The order of consistency is p if the formula is exactfor all polynomials y = P(t) of degree p or less
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Order of consistency Explicit Euler
Example Expanding in Taylor series,
y(tn+1)− [y(tn) + h f (tn, y(tn))] = O(h2)
so the method’s consistency order is one
Alternatively, suppose y(t) = 1 with f = y ′ = 0. Theny(tn+1) = y(tn) + hf (tn, y(tn)) = 1 + h · 0 = 1, so exact
Next take 1st degree polynomial y(t) = t with f = y ′ = 1. Thenh = y(tn+1) = y(tn) + hf (tn, y(tn)) = 0 + h · 1 = h
For 2nd degree polynomial y(t) = t2 with f = y ′ = 2t we geth2 = y(tn+1) = y(tn) + hf (tn, y(tn)) = 0 + h · 0 = 0 6= h2
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5. The trapezoidal rule
Explicit Euler linearizes at tn with slope y ′(tn). Instead, takeaverage of y ′(tn) and y ′(tn+1) and approximate
y ′(tn) ≈ 1
2[f (tn, y(tn)) + f (tn+1, y(tn+1))]
This gives the trapezoidal rule
yn+1 = yn +h
2[f (tn, yn) + f (tn+1, yn+1)]
The method is implicit
Nonlinear equation solving is required on each step
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Trapezoidal rule Order and convergence
Insert exact solution and expand in Taylor series
y(tn+1)− {y(tn) +h
2[f (tn, y(tn)) + f (tn+1, y(tn+1))]}
= y(tn) + h y ′(tn) +h2
2y ′′(tn) +
h3
6y ′′′(tn) + O(h4)
− {y(tn) +h
2
(y ′(tn) + [y ′(tn) + h y ′′(tn) +
h2
2y ′′′(tn)]
)}
= −h3
12y ′′′(tn) + O(h4)
Theorem The trapezoidal rule is convergent of order two(No proof given here)
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The dramatic impact of 2nd order convergence
y ′ = −y cos t; y(0) = 1; t ∈ [0, 8π]
Stepsize h = 8π/N with N = 96 gives h = π/12
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N=96 Trapezoidal rule
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N=96 Explicit Euler
Numerical solutions with Trapezoidal rule and Explicit Euler
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Implicit methods Implicit Euler
Implicit Euler yn+1 = yn + h f (tn+1, yn+1)
We need to solve a nonlinear equation to compute yn+1
The extra cost is motivated if we can take larger steps
There are some problems where implicit methods can takeenormously large time steps without losing accuracy!
We will return to how to solve nonlinear equations
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Implicit methods Implicit Midpoint Rule
We can also approximate the derivative y ′(t) by
y ′(t) ≈ f
(tn + tn+1
2,yn + yn+1
2
), t ∈ [tn, tn+1]
resulting in the 2nd order Implicit Midpoint Method
yn+1 = yn + h f
(tn + tn+1
2,yn + yn+1
2
)
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6. Theta methods
Approximate y ′ by a convex combination of y ′(tn) and y ′(tn+1)
yn+1 = yn + h [θf (tn+1, yn+1) + (1− θ)f (tn, yn)], θ ∈ [0, 1]
• Explicit Euler θ = 0
• Trapezoidal rule (implicit) θ = 1/2
• Implicit Euler θ = 1 ⇒ yn+1 = yn + h f (tn+1, yn+1)
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Theta methods Order and convergence
Use Taylor series expansion to get
y(tn+1)− y(tn)− h [θf (tn+1, y(tn+1)) + (1− θ)f (tn, y(tn))]
= −(θ − 1
2)h2y ′′(tn)− 1
2(θ − 2
3)h3y ′′′(tn) + O(h4)
If θ = 1/2 the method is of order 2; otherwise it is of order 1
Theorem (without proof) The Theta-methods are convergent
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7. Numerical tests Explicit Euler
y = λ(y − sin t) + cos t with λ = −0.2
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y
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y
h = π/10 h = π/20
Local error O(h2), global error O(h)
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Numerical tests Implicit Euler
y = λ(y − sin t) + cos t with λ = −0.2
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1.2
t
y
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0.6
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0.8
0.9
1
1.1
1.2
t
y
h = π/10 h = π/20
Local error O(h2), global error O(h)
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Numerical tests Implicit Euler
y = λ(y − sin t) + cos t with λ = −10
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.5
0.6
0.7
0.8
0.9
1
1.1
1.2
t
y
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.5
0.6
0.7
0.8
0.9
1
1.1
1.2
t
y
h = π/10 h = π/20
Local error O(h2), global error O(h)
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Numerical tests Explicit Euler
y = λ(y − sin t) + cos t with λ = −10
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.5
0.6
0.7
0.8
0.9
1
1.1
1.2
t
y
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.5
0.6
0.7
0.8
0.9
1
1.1
1.2
t
y
h = π/10 h = π/20
Numerical instability!
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8. The linear test equation
Definition The linear test equation is
y ′ = λy , y(0) = 1, t ≥ 0, λ ∈ C
As y(t) = eλt , we have
|y(t)| ≤ K ⇔ Re(λ) ≤ 0
Mathematical stability Bounded solutions if Re(λ) ≤ 0
When does a numerical method have this property?
Does Re(λ) ≤ 0 imply numerical stability?
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Numerical stability Stability region
Definition The stability region D of a method is the set of allhλ ∈ C such that |yn| ≤ K when the method is applied to the testequation
Example For Euler’s method, yn+1 = (1 + hλ)yn, implying thatyn remains bounded if and only if |1 + hλ| ≤ 1
−4 −3 −2 −1 0 1 2 3 4−4
−3
−2
−1
0
1
2
3
4
DEuler = {z ∈ C : |1 + z | ≤ 1}
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A-stability
Definition A method is called A-stable if C− ⊂ D
For the trapezoidal rule
DTR =
{z ∈ C :
∣∣∣∣∣1 + 12z
1− 12z
∣∣∣∣∣ ≤ 1
}≡ C−
so Re(z) ≤ 0 implies that yn remains bounded for any h > 0
The explicit Euler method is not A-stable but the implicit Eulermethod and the trapezoidal rule are A-stable
Common idea – “If the original problem is stable, then an A-stablemethod will replicate that behavior numerically”
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Relevance of linear test equation
Assume that y ′ = Ay is diagonalizable, with AT = TΛ
Then u = T−1y satsifies u′ = Λu (scalar equations)
If the explicit Euler method is applied to y ′ = Ay , we get
yn+1 = (I + hA)yn
Putting un = T−1yn leads to
un+1 = (I + hΛ)un
T diagonalizes the differential equation and its discretization
Interpret λ in linear test equation as eigenvalue of A
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9. Stiff ODEs
Example Solve y = λ(y − sin t) + cos t with λ = −500
Solution
Particular yP(t) = sin tHomogeneous yH(t) = eλt
General y(t) = eλ(t−t0)(y(t0)− sin t0) + sin t
Study the flow of this equation and numerical solutions
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Flow (solution trajectories) λ = −50
0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.40.4
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1
1.1
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t
y
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Stiffness
Stiff differential equations are characterized by homogeneoussolutions being strongly damped
Example y = λ(y − sin t) + cos t with λ� −1
Explicit methods have bounded stability regions putting strongstability restrictions on the step size h
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Stiffness Explicit methods
With Explicit Euler the solution approaches sin t as t →∞ if andonly if h < 1/250
The step size h must be kept small, not to keep errors small, butto maintain numerical stability
The Trapezoidal Rule solution approaches sin t as t →∞for every h > 0
The step size must only be kept small to keep errors small
For A-stable methods, there is no stability restriction, only anaccuracy restriction
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Stiffness Implicit methods
Implicit methods with unbounded stability regions put no stabilityrestrictions on h
The stepsize is only restricted by accuracy requirements
Example The implicit Euler applied to the problem above onlyrequires
ln ≈h2y ′′
2≈ h2
2sin tn
to be sufficiently small, independently of λ
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