Exponential Integrators
Time-stepping methods that solve the linear stiff part of a problem exactly using the matrix exponential, freeing the step size from stiffness.
Solving the linear part exactly
For a problem du/dt = A u + N(u), where A is a stiff linear operator and N is a milder nonlinearity, the linear part alone has the exact solution given by the matrix exponential of A times t. Exponential integrators use this exact solution as the backbone of the step, treating only the nonlinear remainder by an explicit-style correction. Because the stiff linear dynamics are handled exactly, the time step is limited by accuracy of the nonlinear part, not by stability of the stiff part.
The phi functions
The schemes are built from the exponential and a family of related functions called phi functions, which arise from integrating the exponential against polynomial terms. An exponential Rosenbrock or exponential Runge-Kutta method combines several phi-function evaluations at intermediate stages to reach high order while retaining exact treatment of the linear operator.
Computing the matrix exponential action
The central cost is applying the matrix exponential (or phi functions) to a vector, not forming the dense exponential explicitly. For large sparse A this is done with Krylov subspace approximations: Arnoldi builds a small subspace, the exponential is computed on the small projected matrix, and the result is mapped back. This makes exponential integrators viable for the large systems produced by discretized PDEs.
- Exact for purely linear problems regardless of step size
- Excellent for highly oscillatory or strongly diffusive stiff systems
- Cost concentrated in Krylov matrix-exponential evaluations
- Effective when the stiffness is dominated by a linear operator
Comparison to alternatives
Exponential integrators and IMEX schemes both target stiff-plus-nonstiff problems. IMEX solves the stiff part implicitly (one linear solve per step); exponential integrators solve it exactly (one matrix-exponential action per step). The better choice depends on whether the linear operator is easier to exponentiate or to invert.