Stiff equation

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In mathematics, a stiff equation is a differential equation for which certain numerical methods for solving the equation are numerically unstable, unless the step size is taken to be extremely small. It has proven difficult to formulate a precise definition of stiffness, but the main idea is that the equation includes some terms that can lead to rapid variation in the solution.

1 Motivating example
2 Characterization of stiffness
3 A-stability
4 Runge–Kutta methods
- 4.1 Example: The Euler and trapezoidal methods
- 4.2 General theory
5 Multistep methods
- 5.1 Example: The second-order Adams–Bashforth method
- 5.2 General theory
6 See also
7 Notes
8 References

[edit] Motivating example

Consider the initial value problem

$\,y'(t)=-15y(t),\quad t \ge 0, y(0)=1.$

The exact solution is $y (t) = e - 15 t$ , and clearly $y (t)$ → $0$ as $t$ → ∞. We want the numerical solutions to exhibit the same behaviour.

Figure 1. Explicit numerical methods experiencing instability on a stiff equation

Figure 1 illustrates the numerical issues for various numerical integrators applied on the equation. First, Euler's method with a step size of $h = 1 / 4$ oscillates wildly and quickly exits the figure. Halving the step size and rerunning the integration with $h = 1 / 8$ the result is within the figure boundaries, but oscillates and by no means represents the qualitative solution.

The trapezoidal method, that is, the two-stage Adams–Moulton method, is given by

$y_{n+1}=y_n+{1\over 2}h\left(f(t_n,y_n)+f(t_{n+1},y_{n+1})\right).$

Applying this method instead of Euler's method gives a much better result. As the figure shows, the numerical results decrease monotonically to zero, just like the exact solution.

[edit] Characterization of stiffness

Certain types of problems can be characterized as stiff:

problems of the form

$y' = ky + f(t),\,$

where

k

∈C and

| k |

is large.

systems of the form

$\mathbf{y}' = K\mathbf{y}+\mathbf{f}(t),$

where

K

is a square matrix having at least one eigenvalue

l

∈C and

| l |

is large.

systems of the form

$\mathbf{y}' = \mathbf{f}(\mathbf{y}, t),$

with the Jacobian of f having at least one eigenvalue

m

∈C and

| m |

is large.

[edit] A-stability

The behaviour of numerical methods on stiff problems can be analyzed by applying these methods to the test equation $y' = k y$ with k ∈ C. The solution of this equation is $y (t) = e k t$ . This solution approaches zero as $t\to\infty$ when Re $k$ < 0. If the numerical method also exhibits this behaviour, then the method is said to be A-stable.^[1] A-stable methods do not exhibit the instability problems as described in the motivating example.

[edit] Runge–Kutta methods

Runge–Kutta methods applied to the test equation $y' = k y$ take the form $y n + 1 = φ(h k) y n$ , and by induction, $y n = φ(h k) n y 0$ (the function φ is called the stability function). Thus, the condition that $y n$ → 0 as $n$ → ∞ is equivalent to $| φ(h k) | < 1$ . This motivates the definition of the region of absolute stability (sometimes referred to simply as stability region), which is the set { $z$ ∈ C | $| φ(z) | < 1$ }. The method is A-stable if the region of absolute stability contains the set { $z$ ∈ C | Re( $z$ ) < 0 }, that is, the left half plane.

[edit] Example: The Euler and trapezoidal methods

The pink disk shows the stability region for the Euler method.

Consider both the Euler and trapezoidal methods above. The Euler method applied to the test equation $y' = k y$ is

$y_{n+1} = y_n + hf(t_n, y_n) = y_n + h(ky_n) = y_n + hky_n = (1+hk)y_n. \,$

Hence, $y n = (1 + h k) n y 0$ with $φ(z) = 1 + z$ . The region of absolute stability for this method is thus { $z$ ∈ C | |1+z| < 1 } which is the disk depicted on the right. The Euler method is not A-stable.

The motivating example had $k = - 15$ . The value of z when taking step size $h = 1 / 4$ is $z = - 3.75$ , which is outside the stability region. Indeed, the numerical results do not converge to zero. However, with step size $h = 1 / 8$ , we have $z = - 1.875$ which is just inside the stability region and the numerical results converge to zero, albeit rather slowly.

The pink region is the stability region for the trapezoidal method.

The trapezoidal method

$y_{n+1} = y_n + \tfrac12h \left(f(t_n,y_n)+f(t_{n+1},y_{n+1})\right),$

when applied to the test equation $y' = k y$ , is

$y_{n+1} = y_n + \tfrac12h \left(ky_n+ky_{n+1})\right).$

Solving for $y n + 1$ yields

$y_{n+1}={1+{1\over 2}hk \over 1-{1\over 2}hk}y_n.$

Thus, the stability function is

$\phi(z)={1+{1\over 2}z \over 1-{1\over 2}z}$

and the region of absolute stability is

$\left\{ z \in \Bbb{C} \left|\ \left| {1+{1\over 2}z \over 1-{1\over 2}z} \right| < 1 \right.\right\}.$

This region contains the left-half plane, so the trapezoidal method is A-stable. In fact, the stability region is identical to the left-half plane, and thus the numerical solution of $y' = k y$ converges to zero if and only if the exact solution does. Nevertheless, the trapezoidal method does not have perfect stability: it does damp all decaying components, but rapidly decaying components are damped only very mildly, because $\phi(z) \to 1$ as $z \to -\infty$ . This led to the concept of L-stability: a method is L-stable if it is A-stable and $\phi(z) \to 0$ as $|z| \to \infty$ . The trapezoidal method is not L-stable. The implicit Euler method is an example of an L-stable method.^[2]

[edit] General theory

The stability function of a Runge–Kutta method with coefficients A and b is given by

$\phi(z) = \frac{\det(I-zA+zeb^T)}{\det(I-zA)},$

where e denotes the vector with ones. This is a rational function (one polynomial divided by another).

Explicit Runge–Kutta methods have a strictly lower triangular coefficient matrix A and thus, their stability function is a polynomial. It follows that explicit Runge–Kutta methods cannot be A-stable.

The stability function of implicit Runge–Kutta methods is often analyzed using order stars. The order star for a method with stability function $φ$ is defined to be the set { $z$ ∈ C | $| φ(z) | > e z$ }. A method is A-stable if and only if its stability function has no poles in the left-hand plane and its order star contains no purely imaginary numbers.^[3]

[edit] Multistep methods

Linear multistep methods have the form

$y_{n+1}=\sum_{i=0}^s a_i y_{n-i}+h\sum_{j=-1}^s b_j f(t_{n-j},y_{n-j}).$

Applied to the test equation, they become

$y_{n+1}=\sum_{i=0}^s a_i y_{n-i}+hk\sum_{j=-1}^s b_jy_{n-j},$

which can be simplified to

$(1-b_{-1}z)y_{n+1}-\sum_{j=0}^s (a_j+b_jz)y_{n-j}=0$

where $z = h k$ . This is a linear recurrence relation. The method is A-stable if all solutions ${y n}$ of the recurrence relation converge to zero when Re z < 0. The characteristic polynomial is

$\Phi(z, w) = w^{s+1}-\sum_{i=0}^s a_iw^{s-j} - z\sum_{j=-1}^s b_jw^{s-j}.$

All solutions converge to zero for a given value of $z$ if all solutions $w$ of $Φ(z, w) = 0$ lie in the unit circle..

The region of absolute stability for a multistep method of the above form is then the set of all $z$ ∈ C for which all $w$ such that $Φ(z, w) = 0$ satisfy $| w | < 1$ . Again, if this set contains the left-half plane, the multi-step method is said to be A-stable.

[edit] Example: The second-order Adams–Bashforth method

The pink region is the stability region for the second-order Adams–Bashforth method.

Let us determine the region of absolute stability for the two-step Adams–Bashforth method

$y_{n+1} = y_n + h \left( \tfrac32 f(t_n, y_n) - \tfrac12 f(t_{n-1}, y_{n-1}) \right) .$

The characteristic polynomial is

$\Phi(w,z) = w^2 - (1+\tfrac32z)w + \tfrac12 z = 0$

which has roots

$w = \tfrac12 \Big(1 + \tfrac32 z \pm\sqrt{1 + z + \tfrac94 z^2}\Big),$

thus the region of absolute stability is

$\left\{ z \in \Bbb{C} \left| \ \left| \tfrac12 \Big(1 + \tfrac32 z \pm\sqrt{1 + z + \tfrac94 z^2}\Big) \right| < 1 \right.\right\}.$

This region is shown on the right. It does not include all the left half-plane (in fact it only includes the real axis between $z = - 1$ and $z = 0$ ) so the Adams–Bashforth method is not A-stable.

[edit] General theory

Explicit multistep methods can never be A-stable, just like explicit Runge–Kutta methods. Implicit multistep methods can only be A-stable if their order is at most 2. The latter result is known as the second Dahlquist barrier; it restricts the usefulness of linear multistep methods for stiff equations. An example of a second-order A-stable method is the trapezoidal rule mentioned above, which can also be considered as a linear multistep method.^[4]

[edit] See also

Explicit and implicit methods

[edit] Notes

^ This definition is due to Dahlquist (1963).
^ The definition of L-stability is due to Ehle (1969).
^ The definition is due to Wanner, Hairer & Nørsett (1978); see also Iserles & Nørsett (1991).
^ See Dahlquist (1963).

[edit] References

Dahlquist, Germund (1963), “A special stability problem for linear multistep methods”, BIT 3: 27–43 .
Ehle, B. L. (1969), On Padé approximations to the exponential function and A-stable methods for the numerical solution of initial value problems, Report 2010, University of Waterloo .
Hairer, Ernst & Wanner, Gerhard (1996), Solving ordinary differential equations II: Stiff and differential-algebraic problems (second ed.), Berlin: Springer Verlag, ISBN 978-3-540-60452-5 .
Iserles, Arieh & Nørsett, Syvert (1991), Order Stars, Chapman and Hall, ISBN 978-0-412-35260-7 .
Wanner, Gerhard; Hairer, Ernst & Nørsett, Syvert (1978), “Order stars and stability theory”, BIT 18: 475–489 .

Categories: Numerical differential equations

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Stiff equation

From Wikipedia, the free encyclopedia

Contents

[edit] Motivating example

[edit] Characterization of stiffness

[edit] A-stability

[edit] Runge–Kutta methods

[edit] Example: The Euler and trapezoidal methods

[edit] General theory

[edit] Multistep methods

[edit] Example: The second-order Adams–Bashforth method

[edit] General theory

[edit] See also

[edit] Notes

[edit] References

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