Stiff equation

Explicit numerical methods exhibiting instability when integrating a stiff ordinary differential equation

Consider the initial value problem

${\displaystyle \,y'(t)=-15y(t),\quad t\geq 0,y(0)=1.}$

(1)

The exact solution (shown in cyan) is

${\displaystyle y(t)=e^{-15t}\,}$ with ${\displaystyle y(t)\to 0}$ as ${\displaystyle t\to \infty .}$

(2)

We seek a numerical solution that exhibits the same behavior.

The figure (right) illustrates the numerical issues for various numerical integrators applied on the equation.

One of the most prominent examples of the stiff Ordinary differential equations (ODEs) is a system that describes the chemical reaction of Robertson:[1]

${\displaystyle {\dot {x}}=-0.04x+10^{4}y\cdot z}$ ${\displaystyle {\dot {y}}=0.04x-10^{4}y\cdot z-3\cdot 10^{7}y^{2}}$ ${\displaystyle {\dot {z}}=3\cdot 10^{7}y^{2}}$

(4)

If one treats this system on a short interval, for example, ${\displaystyle t\in [0,40]}$ there is no problem in numerical integration. However, if the interval is very large (10¹¹ say), then many standard codes fail to integrate it correctly.

Additional examples are the sets of ODEs resulting from the temporal integration of large chemical reaction mechanisms. Here, the stiffness arises from the coexistence of very slow and very fast reactions. To solve them, the software packages KPP and Autochem can be used.

Consider the linear constant coefficient inhomogeneous system

${\displaystyle \mathbf {y} '=\mathbf {A} \mathbf {y} +\mathbf {f} (x),}$

(5)

where ${\displaystyle \mathbf {y} ,\mathbf {f} \in \mathbb {R} ^{n}}$ and ${\displaystyle \mathbf {A} }$ is a constant, diagonalizable, ${\displaystyle n\times n}$ matrix with eigenvalues ${\displaystyle \lambda _{t}\in \mathbb {C} ,t=1,2,\ldots ,n}$ (assumed distinct) and corresponding eigenvectors ${\displaystyle \mathbf {c} _{t}\in \mathbb {C} ^{n},t=1,2,\ldots ,n}$. The general solution of (5) takes the form

${\displaystyle \mathbf {y} (x)=\sum _{t=1}^{n}\kappa _{t}\exp(\lambda _{t}x)\mathbf {c} _{t}+\mathbf {g} (x),}$

(6)

where the κ_t are arbitrary constants and ${\displaystyle \mathbf {g} (x)}$ is a particular integral. Now let us suppose that

${\displaystyle Re(\lambda _{t})<0,\qquad t=1,2,\ldots ,n,}$

(7)

which implies that each of the terms ${\displaystyle \exp(\lambda _{t}x)\mathbf {c} _{t}\rightarrow 0}$ as ${\displaystyle x\rightarrow \infty }$, so that the solution ${\displaystyle \mathbf {y} (x)}$ approaches ${\displaystyle \mathbf {g} (x)}$ asymptotically as ${\displaystyle x\rightarrow \infty }$; the term ${\displaystyle \exp(\lambda _{t}x)\mathbf {c} _{t}}$ will decay monotonically if λ_t is real and sinusoidally if λ_t is complex. Interpreting x to be time (as it often is in physical problems), ${\displaystyle \Sigma _{t=1}^{n}\kappa _{t}\exp(\lambda _{t}x)\mathbf {c} _{t}}$ is called the transient solution and ${\displaystyle \mathbf {g} (x)}$ the steady-state solution. If ${\displaystyle |Re(\lambda _{t})|}$ is large, then the corresponding term ${\displaystyle \kappa _{t}\exp(\lambda _{t}x)\mathbf {c} _{t}}$ will decay quickly as x increases and is thus called a fast transient; if ${\displaystyle |Re(\lambda _{t})|}$ is small, the corresponding term ${\displaystyle \kappa _{t}\exp(\lambda _{t}x)\mathbf {c} _{t}}$ decays slowly and is called a slow transient. Let ${\displaystyle {\overline {\lambda }},{\underline {\lambda }}\in \{\lambda _{t},t=1,2,\ldots ,n\}}$ be defined by

${\displaystyle |Re({\overline {\lambda }})|\geq |Re(\lambda _{t})|\geq |Re({\underline {\lambda }})|,\qquad t=1,2,\ldots ,n}$

(8)

so that ${\displaystyle \kappa _{t}\exp({\overline {\lambda }}x)\mathbf {c} _{t}}$ is the fastest transient and ${\displaystyle \kappa _{t}\exp({\underline {\lambda }}x)\mathbf {c} _{t}}$ the slowest. We now define the stiffness ratio as

${\displaystyle {\frac {|Re({\overline {\lambda }})|}{|Re({\underline {\lambda }})|}}.}$[2]

(9)

In this section we consider various aspects of the phenomenon of stiffness. "Phenomenon" is probably a more appropriate word than "property", since the latter rather implies that stiffness can be defined in precise mathematical terms; it turns out not to be possible to do this in a satisfactory manner, even for the restricted class of linear constant coefficient systems. We shall also see several qualitative statements that can be (and mostly have been) made in an attempt to encapsulate the notion of stiffness, and state what is probably the most satisfactory of these as a "definition" of stiffness.

J. D. Lambert defines stiffness as follows:

If a numerical method with a finite region of absolute stability, applied to a system with any initial conditions, is forced to use in a certain interval of integration a step length which is excessively small in relation to the smoothness of the exact solution in that interval, then the system is said to be stiff in that interval.

There are other characteristics which are exhibited by many examples of stiff problems, but for each there are counterexamples, so these characteristics do not make good definitions of stiffness. Nonetheless, definitions based upon these characteristics are in common use by some authors and are good clues as to the presence of stiffness. Lambert refers to these as "statements" rather than definitions, for the aforementioned reasons. A few of these are:

1. A linear constant coefficient system is stiff if all of its eigenvalues have negative real part and the stiffness ratio is large.
2. Stiffness occurs when stability requirements, rather than those of accuracy, constrain the step length.
3. Stiffness occurs when some components of the solution decay much more rapidly than others.[3]

The origin of the term "stiffness" has not been clearly established. According to Joseph Oakland Hirschfelder, the term "stiff" is used because such systems correspond to tight coupling between the driver and driven in servomechanisms.[4] According to Richard. L. Burden and J. Douglas Faires,

Significant difficulties can occur when standard numerical techniques are applied to approximate the solution of a differential equation when the exact solution contains terms of the form e^λt, where λ is a complex number with negative real part.

...

Problems involving rapidly decaying transient solutions occur naturally in a wide variety of applications, including the study of spring and damping systems, the analysis of control systems, and problems in chemical kinetics. These are all examples of a class of problems called stiff (mathematical stiffness) systems of differential equations, due to their application in analyzing the motion of spring and mass systems having large spring constants (physical stiffness).[5]

For example, the initial value problem

${\displaystyle m{\ddot {x}}+c{\dot {x}}+kx=0,\qquad x(0)=x_{0},\qquad {\dot {x}}(0)=0,}$

(10)

with m = 1, c = 1001, k = 1000, can be written in the form (5) with n = 2 and

${\displaystyle \mathbf {A} =\left({\begin{array}{rr}0&1\\-1000&-1001\end{array}}\right),}$

(11)

${\displaystyle \mathbf {f} (t)=\left({\begin{array}{c}0\\0\end{array}}\right),}$

(12)

${\displaystyle \mathbf {x} (0)=\left({\begin{array}{c}x_{0}\\0\end{array}}\right),}$

(13)

and has eigenvalues ${\displaystyle {\overline {\lambda }}=-1000,{\underline {\lambda }}=-1}$. Both eigenvalues have negative real part and the stiffness ratio is

${\displaystyle {\frac {|-1000|}{|-1|}}=1000,}$

(14)

which is fairly large. System (10) then certainly satisfies statements 1 and 3. Here the spring constant k is large and the damping constant c is even larger.[6] (Note that "large" is a vague, subjective term, but the larger the above quantities are, the more pronounced will be the effect of stiffness.) The exact solution to (10) is

${\displaystyle x(t)=x_{0}\left(-{\frac {1}{999}}e^{-1000t}+{\frac {1000}{999}}e^{-t}\right)\approx x_{0}e^{-t}.}$

(15)

Note that (15) behaves quite nearly as a simple exponential x₀e^−t, but the presence of the e^−1000t term, even with a small coefficient is enough to make the numerical computation very sensitive to step size. Stable integration of (10) requires a very small step size until well into the smooth part of the solution curve, resulting in an error much smaller than required for accuracy. Thus the system also satisfies statement 2 and Lambert's definition.

The behaviour of numerical methods on stiff problems can be analyzed by applying these methods to the test equation y' = ky subject to the initial condition y(0) = 1 with ${\displaystyle k\in \mathbb {C} }$ . The solution of this equation is y (t) = e^kt. This solution approaches zero as ${\displaystyle t\to \infty }$ when ${\displaystyle \mathrm {Re} \,(k)<0.}$ If the numerical method also exhibits this behaviour (for a fixed step size), then the method is said to be A-stable.[7] (Note that a numerical method that is L-stable (see below) has the stronger property that the solution approaches zero in a single step as the step size goes to infinity.) A-stable methods do not exhibit the instability problems as described in the motivating example.

Runge–Kutta methods applied to the test equation ${\displaystyle y'=k\cdot y}$ take the form ${\displaystyle y_{n+1}=\phi (hk)\cdot y_{n}}$, and, by induction, ${\displaystyle y_{n}=\left(\phi (hk)\right)^{n}\cdot y_{0}}$. The function ${\displaystyle \phi }$ is called the stability function. Thus, the condition that ${\displaystyle y_{n}\to 0}$ as ${\displaystyle n\to \infty }$ is equivalent to ${\displaystyle |\phi (hk)|<1}$. This motivates the definition of the region of absolute stability (sometimes referred to simply as stability region), which is the set ${\displaystyle \{z\in \mathbb {C} |\,|\phi (z)|<1\}}$. The method is A-stable if the region of absolute stability contains the set ${\displaystyle \{z\in \mathbb {C} |\mathrm {Re} (z)<0\}}$, that is, the left half plane.

Example: The Euler methods

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

Consider the Euler methods above. The explicit Euler method applied to the test equation ${\displaystyle y'=k\cdot y}$ is

${\displaystyle y_{n+1}=y_{n}+h\cdot f(t_{n},y_{n})=y_{n}+h\cdot (ky_{n})=y_{n}+h\cdot k\cdot y_{n}=(1+h\cdot k)y_{n}.}$

Hence, ${\displaystyle y_{n}=(1+hk)^{n}\cdot y_{0}}$ with ${\displaystyle \phi (z)=1+z}$. The region of absolute stability for this method is thus ${\displaystyle \{z\in \mathbb {C} ||1+z|<1\}}$ which is the disk depicted on the right. The Euler method is not A-stable.

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

Example: Trapezoidal method

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

Consider the trapezoidal method

${\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{2}}h\cdot \left(f(t_{n},y_{n})+f(t_{n+1},y_{n+1})\right),}$

when applied to the test equation ${\displaystyle y'=k\cdot y}$, is

${\displaystyle y_{n+1}=y_{n}+{\tfrac {1}{2}}h\cdot \left(ky_{n}+ky_{n+1}\right).}$

Solving for ${\displaystyle y_{n+1}}$ yields

${\displaystyle y_{n+1}={\frac {1+{\frac {1}{2}}hk}{1-{\frac {1}{2}}hk}}\cdot y_{n}.}$

Thus, the stability function is

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

and the region of absolute stability is

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

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 ${\displaystyle y'=k\cdot y}$ converges to zero if and only if the exact solution does. Nevertheless, the trapezoidal method does not have perfect behavior: it does damp all decaying components, but rapidly decaying components are damped only very mildly, because ${\displaystyle \phi (z)\to 1}$ as ${\displaystyle z\to -\infty }$. This led to the concept of L-stability: a method is L-stable if it is A-stable and ${\displaystyle |\phi (z)|\to 0}$ as ${\displaystyle z\to \infty }$. The trapezoidal method is A-stable but not L-stable. The implicit Euler method is an example of an L-stable method.[8]

Linear multistep methods have the form

${\displaystyle y_{n+1}=\sum _{i=0}^{s}a_{i}y_{n-i}+h\sum _{j=-1}^{s}b_{j}f(t_{n-j},y_{n-j}).}$

Applied to the test equation, they become

${\displaystyle y_{n+1}=\sum _{i=0}^{s}a_{i}y_{n-i}+hk\sum _{j=-1}^{s}b_{j}y_{n-j},}$

which can be simplified to

${\displaystyle (1-b_{-1}z)y_{n+1}-\sum _{j=0}^{s}(a_{j}+b_{j}z)y_{n-j}=0}$

where z = hk. 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

${\displaystyle \Phi (z,w)=w^{s+1}-\sum _{i=0}^{s}a_{i}w^{s-i}-z\sum _{j=-1}^{s}b_{j}w^{s-j}.}$

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 ${\displaystyle z\in \mathbb {C} }$ 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.

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

${\displaystyle y_{n+1}=y_{n}+h\left({\tfrac {3}{2}}f(t_{n},y_{n})-{\tfrac {1}{2}}f(t_{n-1},y_{n-1})\right).}$

The characteristic polynomial is

${\displaystyle \Phi (w,z)=w^{2}-(1+{\tfrac {3}{2}}z)w+{\tfrac {1}{2}}z=0}$

which has roots

${\displaystyle w={\tfrac {1}{2}}{\Big (}1+{\tfrac {3}{2}}z\pm {\sqrt {1+z+{\tfrac {9}{4}}z^{2}}}{\Big )},}$

thus the region of absolute stability is

${\displaystyle \left\{z\in \mathbb {C} \left|\ \left|{\tfrac {1}{2}}{\Big (}1+{\tfrac {3}{2}}z\pm {\sqrt {1+z+{\tfrac {9}{4}}z^{2}}}{\Big )}\right|<1\right.\right\}.}$

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.

• Condition number
• Differential inclusion, an extension of the notion of differential equation that allows discontinuities, in part as way to sidestep some stiffness issues
• Explicit and implicit methods

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