ExponentialAction.jl implements the action of the Matrix exponential.

For details, see the docstring of expv.

Exports

• expv: compute the action of the matrix exponential

expv(t, A, B; shift=true, tol)

Compute $\exp(tA)B$ without computing $tA$ or the matrix exponential $\exp(tA)$.

Computing the action of the matrix exponential is significantly faster than computing the matrix exponential and then multiplying it when the second dimension of $B$ is much smaller than the first one. The "time" $t$ may be real or complex.

In short, the approach computes

\[F = T_m(tA / s)^s B,\]

where $T_m(X)$ is the Taylor series of $\exp(X)$ truncated to degree $m = m^*$. The term $s$ determines how many times the Taylor series acts on $B$. $m^*$ and $s$ are chosen to minimize the number of matrix products needed while maintaining the required tolerance tol.

The algorithm is described in detail in Algorithm 3.2 in ^{[AlMohyHigham2011]}.

Keywords

• shift=true: Expand the Taylor series about the $n \times n$ matrix $A-μI=0$ instead of $A=0$, where $μ = \operatorname{tr}(A) / n$ to speed up convergence. See §3.1 of ^{[AlMohyHigham2011]}.
• tol: The relative tolerance at which to compute the result. Defaults to the tolerance of the eltype of the result.

expv_sequence(t::AbstractVector, A, B; kwargs...)

Compute $\exp(t_i A)B$ for the (sorted) sequence of (real) time points $t=(t_1, t_2, \ldots)$.

At each time point, the result $F_i$ is computed as

\[F_i = \exp\left((t_i - t_{i-1}) A\right) F_{i - 1}\]

using expv, where $t_0 = 0$ and $F_0 = B$. For details, see Equation 5.2 of ^{[AlMohyHigham2011]}.

Because the cost of computing expv is related to the operator 1-norm of $t_i A$, this incremental computation is more efficient than computing expv separately for each time point.

See expv for a description of acceptable kwargs.

expv_sequence(t::AbstractRange, A, B; kwargs...)

Compute expv over the uniformly spaced sequence.

This algorithm takes special care to avoid overscaling and to save and reuse matrix products and is described in Algorithm 5.2 of ^{[AlMohyHigham2011]}.

expv_taylor(t, A, B, degree_max; tol)

Compute $\exp(tA)B$ using the truncated Taylor series with degree $m=$ degree_max.

Instead of computing the Taylor series $T_m(tA)$ of the matrix exponential directly, its action on $B$ is computed instead.

The series is truncated early if

\[\frac{\lVert \exp(t A) B - T_m(tA) B \rVert_1}{\lVert T_m(tA) B \rVert_1} \le \mathrm{tol},\]

where $\lVert X \rVert_1$ is the operator 1-norm of the matrix $X$. This condition is only approximately checked.

expv_taylor_cache(t, A, B, degree_max, k, Z; tol)

Compute $\exp(tkA)B$ using the truncated Taylor series with degree $m=$ degree_max.

This method stores all matrix products in a cache Z, where $Z_p = \frac{1}{(p-1)!} (t A)^{p-1} B$. This cache can be reused if $k$ changes but $t$, $A$, and $B$ are unchanged.

Z is a vector of arrays of the same shape as B and is not mutated; instead the (possibly updated) cache is returned.

Returns

• F::AbstractMatrix: The action of the truncated Taylor series
• Z::AbstractVector: The cache of matrix products of the same shape as F. If the cache is updated, then this is a different object than the input Z.

See expv_taylor.

parameters(t, A, ncols_B; kwargs...) -> (degree_opt, scale)

Compute Taylor series parameters needed for $\exp(tA)B$.

This is Code Fragment 3.1 from ^{[AlMohyHigham2011]}.

Keywords

• tol: the desired relative tolerance
• degree_max=55: the maximum degree of the truncated Taylor series that will be used. This is $m_{\mathrm{max}}$ in ^{[AlMohyHigham2011]}, where they recommend a value of 55 in §3.
• ℓ=2: the number of columns in the matrix that is multiplied for norm estimation (note: currently only used for control flow.). Recommended values are 1 or 2.

Returns

• degree_opt: the degree of the truncated Taylor series that will be used. This is $m^*$ in ^{[AlMohyHigham2011]},
• scale: the amount of scaling $s$ that will be applied to $A$. The truncated Taylor series of $\exp(t A / s)$ will be applied $s$ times to $B$.