In practice, condition (7.43) is not very manageable as a way of determining if a given scheme is stable since it involves computing the norm of the power of a matrix. A simpler condition based on the eigenvalues of as opposed to the norm of is von Neumann’s one:

This condition is necessary for numerical stability: if is an eigenvalue of , is an eigenvalue of and That is, which, in order to be valid for all , implies Eq. (7.44).As already mentioned, in order to analyze numerical stability, one can drop lower-order terms. Doing so typically leads to depending on and only through a quotient (the CFL factor) of the form (with for hyperbolic equations)

Then for Eq. (7.44) to hold for all while keeping the CFL factor fixed (in particular, for small ), the following condition has to be satisfied: and one has a stronger version of the von Neumann condition, which is the one encountered in Example 33; see Eq. (7.18).

We return to the periodic scalar case, such as the schemes discussed in Examples 33, 34, 35, and 36 with some more generality. Suppose then, in addition to the linearity and time-independent assumptions of the continuum problem, that the initial data and discretization (7.40, 7.41) are periodic on the interval . Through a Fourier expansion we can write the grid function corresponding to the initial data as

where . The approximation becomes Assuming that is diagonal in the basis , such that as is many times the case, we obtain, using Parseval’s identity, If for some constant then and stability follows. Conversely, if the scheme is stable and (7.52) holds, (7.54) has to be satisfied. Take then or for arbitrary , which implies (7.54). Therefore, provided the condition (7.52) holds, stability is equivalent to the requirement (7.54) on the eigenvalues of .

However, as mentioned, the von Neumann condition is not sufficient for stability, neither in its original form (7.44) nor in its strong one (7.49), unless, for example, can be uniformly diagonalized. This means that there exists a matrix such that

is diagonal and the condition number of with respect to the same norm, is bounded for some constant independent of resolution (an example is that of being normal, ). In that case andNext, we discuss two examples where the von Neumann condition is satisfied but the resulting scheme is unconditionally unstable. The first one is for a well-posed underlying continuum problem and the second one for an ill-posed one.

Example 38. An unstable discretization, which satisfies the von Neumann condition for a trivially–well-posed problem [228].

Consider the following system on a periodic domain with periodic initial data

discretized as with given by Eq. (7.32). The Fourier transform of the amplification matrix and its -th power are The von Neumann condition is satisfied, since the eigenvalues are . However, the discretization is unstable for any value of . For the unit vector , for instance, we have which grows without bound as is increased for .The von-Neumann condition is clearly not sufficient for stability in this example because the amplification matrix not only cannot be uniformly diagonalized, but it cannot be diagonalized at all because of the Jordan block structure in (7.65).

Example 39. Ill-posed problems are unconditionally unstable, even if they satisfy the von Neumann condition. The following example is drawn from [107].

Consider the periodic Cauchy problem

where , is a constant matrix, and the following discretization. The right-hand side of the equation is approximated by a second-order centered derivative plus higher (third) order numerical dissipation (see Section 8.5) where is the identity matrix, an arbitrary parameter regulating the strength of the numerical dissipation and are first-order forward and backward approximations of , The resulting system of ordinary differential equations is marched in time (method of lines, discussed in Section 7.3) through an explicit method: the iterated Crank–Nicholson (ICN) one with an arbitrary but fixed number of iterations (see Example 37).If the matrix is diagonalizable, as in the scalar case of Example 37, the resulting discretization is numerically stable for and , even without dissipation. On the other hand, if the system (7.68) is weakly hyperbolic, as when the principal part has a Jordan block,

one can expect on general grounds that any discretization will be unconditionally unstable. As an illustration, this was explicitly shown in [107] for the above scheme and variations of it. In Fourier space the amplification matrix and its -th power take the form with coefficients depending on such that for an arbitrary small initial perturbation at just one gridpoint, the solution satisfies and is therefore unstable regardless of the value of and . On the other hand, the von Neumann condition is satisfied if and only if Notice that, as expected, the addition of numerical dissipation cannot stabilize the scheme independent of its amount. Furthermore, adding dissipation with a strength parameter violates the von Neumann condition (7.75) and the growth rate of the numerical instability worsens.
Living Rev. Relativity 15, (2012), 9
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