Why order arrows are easy to draw. Multivalue methods and quadratic approximations. The Ehle and Butcher-Chipman conjectures. Riemann Surface Snakes ...
Riemann Surface Snakes and Ladders John Butcher The University of Auckland New Zealand
Riemann Surface Snakes and Ladders – p. 1/??
Riemann Surface Snakes and Ladders John Butcher The University of Auckland New Zealand
“ two thousand Beta-Minus mixed doubles were playing Riemann-surface tennis” Brave New World, A. Huxley Riemann Surface Snakes and Ladders – p. 2/??
Contents Riemann Surface Snakes and Ladders
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows Why order arrows are hard to draw
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows Why order arrows are hard to draw A differential equation satisfied by order arrows
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows Why order arrows are hard to draw A differential equation satisfied by order arrows Why order arrows are easy to draw
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Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows Why order arrows are hard to draw A differential equation satisfied by order arrows Why order arrows are easy to draw Multivalue methods and quadratic approximations Riemann Surface Snakes and Ladders – p. 3/??
Contents Riemann Surface Snakes and Ladders A two sheet Riemann surface Stability of a Runge–Kutta method A-stability Padé approximations to the exponential function Order Stars and Order Arrows Why order arrows are hard to draw A differential equation satisfied by order arrows Why order arrows are easy to draw Multivalue methods and quadratic approximations The Ehle and Butcher-Chipman conjectures Riemann Surface Snakes and Ladders – p. 3/??
Riemann Surface Snakes & Ladders
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Riemann Surface Snakes & Ladders
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Riemann Surface Snakes & Ladders
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A two sheet Riemann surface
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A two sheet Riemann surface
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A two sheet Riemann surface
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A two sheet Riemann surface
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A two sheet Riemann surface
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Stability of a Runge–Kutta method One method of approximating the solution of a differential equation y 0 (x) = f (y(x)), y(x0 ) = x0 , is to calculate Y and y1 satisfying 5 1 Y − 12 y0 f (Y ) 12 = +h 3 1 y1 f (y1 ) y0 4 4
Riemann Surface Snakes and Ladders – p. 6/??
Stability of a Runge–Kutta method One method of approximating the solution of a differential equation y 0 (x) = f (y(x)), y(x0 ) = x0 , is to calculate Y and y1 satisfying 5 1 Y − 12 y0 f (Y ) 12 = +h 3 1 y1 f (y1 ) y0 4 4 The first quantity computed is Y ≈ y(x0 + 13 h) and the second is the output value: y1 ≈ y(x0 + h).
Riemann Surface Snakes and Ladders – p. 6/??
Stability of a Runge–Kutta method One method of approximating the solution of a differential equation y 0 (x) = f (y(x)), y(x0 ) = x0 , is to calculate Y and y1 satisfying 5 1 Y − 12 y0 f (Y ) 12 = +h 3 1 y1 f (y1 ) y0 4 4 The first quantity computed is Y ≈ y(x0 + 13 h) and the second is the output value: y1 ≈ y(x0 + h). To gauge the likely performance for solving stiff problems, we calculate the stability function.
Riemann Surface Snakes and Ladders – p. 6/??
Stability of a Runge–Kutta method One method of approximating the solution of a differential equation y 0 (x) = f (y(x)), y(x0 ) = x0 , is to calculate Y and y1 satisfying 5 1 Y − 12 y0 f (Y ) 12 = +h 3 1 y1 f (y1 ) y0 4 4 The first quantity computed is Y ≈ y(x0 + 13 h) and the second is the output value: y1 ≈ y(x0 + h). To gauge the likely performance for solving stiff problems, we calculate the stability function. That is, we consider the linear differential equation y 0 = qy and find y1 /y0 as a function of z = hq. Riemann Surface Snakes and Ladders – p. 6/??
For this Runge–Kutta method, we find 5 1 Y − 12 y0 Y 12 = +z 3 1 y1 y1 y0 4 4
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For this Runge–Kutta method, we find 5 1 Y − 12 y0 Y 12 = +z 3 1 y1 y1 y0 4 4 and hence
I −z
5 12 3 4
1 − 12 1 4
Y y1
= y0
1 1
Riemann Surface Snakes and Ladders – p. 7/??
For this Runge–Kutta method, we find 5 1 Y − 12 y0 Y 12 = +z 3 1 y1 y1 y0 4 4 and hence
I −z
5 12 3 4
1 − 12 1 4
Y y1
= y0
1 1
leading to 1 + 13 z y1 = 2 1 2 y0 1 − 3z + 6z
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The stability function is R(z) =
1+ 13 z . 1− 23 z+ 61 z 2
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The stability function is R(z) =
1+ 13 z . 1− 23 z+ 61 z 2
The portion of the complex plane where |R(z)| ≤ 1 is the unshaded region in the diagram:
3i
0
6
−3i Riemann Surface Snakes and Ladders – p. 8/??
To verify rigorously that |R(z)| ≤ 1 for z in the left half-plane, use the maximum modulus principle:
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To verify rigorously that |R(z)| ≤ 1 for z in the left half-plane, use the maximum modulus principle: √ The poles of R(z) are equal to 2 ± i 2.
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To verify rigorously that |R(z)| ≤ 1 for z in the left half-plane, use the maximum modulus principle: √ The poles of R(z) are equal to 2 ± i 2. Hence R is analytic in the left half-plane.
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To verify rigorously that |R(z)| ≤ 1 for z in the left half-plane, use the maximum modulus principle: √ The poles of R(z) are equal to 2 ± i 2. Hence R is analytic in the left half-plane. Hence it is sufficient to prove |R(z)| ≤ 1 for z = iy.
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To verify rigorously that |R(z)| ≤ 1 for z in the left half-plane, use the maximum modulus principle: √ The poles of R(z) are equal to 2 ± i 2. Hence R is analytic in the left half-plane. Hence it is sufficient to prove |R(z)| ≤ 1 for z = iy. This is true because |1− 23 z + 16 z 2 |2 − |1+ 13 z|2 = (1− 16 y 2 )2 + 49 y 2 − 1 − 19 y 2 1 4 = 36 y ≥0
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A-stability A Runge–Kutta method is A-stable if its stability region contains the left half-plane.
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A-stability A Runge–Kutta method is A-stable if its stability region contains the left half-plane. This means that the stability function has no poles in the left half-plane and is bounded by 1 on the imaginary axis.
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A-stability A Runge–Kutta method is A-stable if its stability region contains the left half-plane. This means that the stability function has no poles in the left half-plane and is bounded by 1 on the imaginary axis. These statements concerning R(z) are equally true of R(z) exp(−z)
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A-stability A Runge–Kutta method is A-stable if its stability region contains the left half-plane. This means that the stability function has no poles in the left half-plane and is bounded by 1 on the imaginary axis. These statements concerning R(z) are equally true of R(z) exp(−z) Drawing the “relative stability region”, based on the “relative stability function" R(z) exp(−z) leads to the “order stars” of Hairer, Nørsett and Wanner. Riemann Surface Snakes and Ladders – p. 10/??
Order Stars and Order Arrows For the special method we have been considering, we introduce its order star
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Order Stars and Order Arrows For the special method we have been considering, we introduce its order star
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Order Stars and Order Arrows For the special method we have been considering, we introduce its order star and the closely related order arrow
Riemann Surface Snakes and Ladders – p. 11/??
Order Stars and Order Arrows For the special method we have been considering, we introduce its order star and the closely related order arrow
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Why order arrows are hard to draw We try evaluating the [5, 5] Padé approximation, scaled by exp(−z), on circles z = reiθ , for r = 1.0, 0.5, 0.25.
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Why order arrows are hard to draw We try evaluating the [5, 5] Padé approximation, scaled by exp(−z), on circles z = reiθ , for r = 1.0, 0.5, 0.25. 1.0
0.5 0.25
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N (z) In the diagrams below, we plot exp(−z) D(z) − 1:
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N (z) In the diagrams below, we plot exp(−z) D(z) − 1: 10−10
r = 1.0
−10−10 0
2π
Riemann Surface Snakes and Ladders – p. 13/??
N (z) In the diagrams below, we plot exp(−z) D(z) − 1: 10−10
r = 1.0
−10−10 0
2π
0
2π
5×10−14
r = 0.5 −5×10−14
Riemann Surface Snakes and Ladders – p. 13/??
N (z) In the diagrams below, we plot exp(−z) D(z) − 1: 10−10
r = 1.0
−10−10 0
2π
0
2π
5×10−14
r = 0.5 −5×10−14 4×10−16
r = 0.25 −4×10−16 0
2π Riemann Surface Snakes and Ladders – p. 13/??
A differential equation satisfied by arrows For a Padé approximation N (z)/D(z), consider wD(z) exp(z) − N (z) = 0 with w = exp(t) exp(z + t)D(z) − N (z) = 0
(1)
I want to form an autonomous differential equation expressing the dependence of z on t.
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A differential equation satisfied by arrows For a Padé approximation N (z)/D(z), consider wD(z) exp(z) − N (z) = 0 with w = exp(t) exp(z + t)D(z) − N (z) = 0
(1)
I want to form an autonomous differential equation expressing the dependence of z on t. Differentiate (1)
exp(z+t)(z 0 (t)(D 0 (z)+D(z))+D(z))−z 0 (t)N 0 (z) = 0
(2)
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A differential equation satisfied by arrows For a Padé approximation N (z)/D(z), consider wD(z) exp(z) − N (z) = 0 with w = exp(t) exp(z + t)D(z) − N (z) = 0
(1)
I want to form an autonomous differential equation expressing the dependence of z on t. Differentiate (1)
exp(z+t)(z 0 (t)(D 0 (z)+D(z))+D(z))−z 0 (t)N 0 (z) = 0
(2)
Now eliminate exp(z + t) from (1) and (2) to yield the differential equation z 0 (t)( D(z)N 0 (z) − D 0 (z)N (z) − D(z)N (z) ) = N (z)D(z) Riemann Surface Snakes and Ladders – p. 14/??
Why order arrows are easy to draw It is interesting that the factor in this equation with a frame around it is always exactly a real number multiplied by z p .
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Why order arrows are easy to draw It is interesting that the factor in this equation with a frame around it is always exactly a real number multiplied by z p . This can be used to draw order arrows accurately.
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Why order arrows are easy to draw It is interesting that the factor in this equation with a frame around it is always exactly a real number multiplied by z p . This can be used to draw order arrows accurately. In a special case we have considered the rational function is 1 3 1 1 z + 1008 z 4 + 30240 z5 1 + 12 z + 19 z 2 + 72 R(z) = 1 3 1 1 1 − 12 z + 19 z 2 − 72 z + 1008 z 4 − 30240 z5
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Why order arrows are easy to draw It is interesting that the factor in this equation with a frame around it is always exactly a real number multiplied by z p . This can be used to draw order arrows accurately. In a special case we have considered the rational function is 1 3 1 1 z + 1008 z 4 + 30240 z5 1 + 12 z + 19 z 2 + 72 R(z) = 1 3 1 1 1 − 12 z + 19 z 2 − 72 z + 1008 z 4 − 30240 z5
Order arrows constructed using this approach are shown in the next diagram Riemann Surface Snakes and Ladders – p. 15/??
Riemann Surface Snakes and Ladders – p. 16/??
Multivalue methods and quadratic approximations The Runge–Kutta method we discussed earlier can be generalized by basing the result on a linear combination of the previous two step values. 8 2 1 3 Y f (Y ) yn−1 − 7 yn−2 −7 7 7 = 8 +h 2 4 1 yn f (yn ) y − y 7 n−1 7 n−2 7 7
Riemann Surface Snakes and Ladders – p. 17/??
Multivalue methods and quadratic approximations The Runge–Kutta method we discussed earlier can be generalized by basing the result on a linear combination of the previous two step values. 8 2 1 3 Y f (Y ) yn−1 − 7 yn−2 −7 7 7 = 8 +h 2 4 1 yn f (yn ) y − y 7 n−1 7 n−2 7 7 Consider stability by writing z = hq where f (y) = qy. (1 − 67 z + 27 z 2 )yn − 87 yn−1 + 17 yn−2 = 0
Riemann Surface Snakes and Ladders – p. 17/??
Multivalue methods and quadratic approximations The Runge–Kutta method we discussed earlier can be generalized by basing the result on a linear combination of the previous two step values. 8 2 1 3 Y f (Y ) yn−1 − 7 yn−2 −7 7 7 = 8 +h 2 4 1 yn f (yn ) y − y 7 n−1 7 n−2 7 7 Consider stability by writing z = hq where f (y) = qy. (1 − 67 z + 27 z 2 )yn − 87 yn−1 + 17 yn−2 = 0 leading to the stability function Φ(w, z) = (1 − 67 z + 27 z 2 )w2 − 87 w +
1 7
Riemann Surface Snakes and Ladders – p. 17/??
This is an example of a quadratic Padé approximation.
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This is an example of a quadratic Padé approximation. Given a triple of non-negative integers, [k, l, m], a function Φ(w, z) = P (z)w 2 + Q(z)w + R(z)
(1)
is a quadratic approximation to the exponential function if deg(P ) ≤ k, deg(Q) ≤ l, deg(R) ≤ m and Φ(exp(z), z) = O(z k+l+m+2 )
(2)
Riemann Surface Snakes and Ladders – p. 18/??
This is an example of a quadratic Padé approximation. Given a triple of non-negative integers, [k, l, m], a function Φ(w, z) = P (z)w 2 + Q(z)w + R(z)
(1)
is a quadratic approximation to the exponential function if deg(P ) ≤ k, deg(Q) ≤ l, deg(R) ≤ m and Φ(exp(z), z) = O(z k+l+m+2 )
(2)
If Φ(w, 0) has only a single zero at 1, (2) is equivalent to (1) having a “principal solution” equal to w(z) = exp(z) + O(z k+l+m+2 ) Riemann Surface Snakes and Ladders – p. 18/??
The Ehle and Butcher-Chipman conjectures Because the ability of a numerical method to solve stiff problems hinge on its possible A-stability, it is important to know which stability functions have the required property.
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The Ehle and Butcher-Chipman conjectures Because the ability of a numerical method to solve stiff problems hinge on its possible A-stability, it is important to know which stability functions have the required property. In particular it is important to know which Padé approximations to exp(z) are acceptable.
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The Ehle and Butcher-Chipman conjectures Because the ability of a numerical method to solve stiff problems hinge on its possible A-stability, it is important to know which stability functions have the required property. In particular it is important to know which Padé approximations to exp(z) are acceptable. Byron Ehle conjectured that a [d, n] approximation is acceptable only if d−n≤2
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The Ehle and Butcher-Chipman conjectures Because the ability of a numerical method to solve stiff problems hinge on its possible A-stability, it is important to know which stability functions have the required property. In particular it is important to know which Padé approximations to exp(z) are acceptable. Byron Ehle conjectured that a [d, n] approximation is acceptable only if d−n≤2 The proof of this conjecture was the first major success of the order star approach Riemann Surface Snakes and Ladders – p. 19/??
The situation is much more complicated for the quadratic case than for the linear case.
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The situation is much more complicated for the quadratic case than for the linear case. However, extensive numerical searching led Fred Chipman and myself to suggest that A-stability might be impossible if k >l+m+3
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The situation is much more complicated for the quadratic case than for the linear case. However, extensive numerical searching led Fred Chipman and myself to suggest that A-stability might be impossible if k >l+m+3 This suggestion has become known as the “Butcher–Chipman conjecture”
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Both the Ehle and Butcher-Chipman conjectures can be proved once it can be shown that every pole is reached by an up-arrow from zero.
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Both the Ehle and Butcher-Chipman conjectures can be proved once it can be shown that every pole is reached by an up-arrow from zero. This is easy for the Ehle result because of the simple one-sheet structure of the corresponding Riemann surface.
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Both the Ehle and Butcher-Chipman conjectures can be proved once it can be shown that every pole is reached by an up-arrow from zero. This is easy for the Ehle result because of the simple one-sheet structure of the corresponding Riemann surface. But it seems to be much more difficult for the B-C conjecture because of the more complicated topology of two-sheet Riemann surfaces.
Riemann Surface Snakes and Ladders – p. 21/??
Both the Ehle and Butcher-Chipman conjectures can be proved once it can be shown that every pole is reached by an up-arrow from zero. This is easy for the Ehle result because of the simple one-sheet structure of the corresponding Riemann surface. But it seems to be much more difficult for the B-C conjecture because of the more complicated topology of two-sheet Riemann surfaces. Work is proceeding.
Riemann Surface Snakes and Ladders – p. 21/??
Both the Ehle and Butcher-Chipman conjectures can be proved once it can be shown that every pole is reached by an up-arrow from zero. This is easy for the Ehle result because of the simple one-sheet structure of the corresponding Riemann surface. But it seems to be much more difficult for the B-C conjecture because of the more complicated topology of two-sheet Riemann surfaces. Work is proceeding. But before I close, I would like to look at a typical case of the Ehle result Riemann Surface Snakes and Ladders – p. 21/??
1 The order arrow for R(z) = 1 − z + 12 z 2 − 16 z 3
Riemann Surface Snakes and Ladders – p. 22/??
1 The order arrow for R(z) = 1 − z + 12 z 2 − 16 z 3
Riemann Surface Snakes and Ladders – p. 22/??
1 The order arrow for R(z) = 1 − z + 12 z 2 − 16 z 3
Riemann Surface Snakes and Ladders – p. 22/??
