Can we use LinearExponential with Turing.jl model?

ChrisRackauckas · October 9, 2022, 7:16pm

Cool, I was expecting something like that could happen.

SciML/OrdinaryDiffEq.jl/blob/master/src/perform_step/linear_perform_step.jl#L735


      
              integrator.k[2] = integrator.fsallast
          end
          
          
function perform_step!(integrator, cache::LinearExponentialCache, repeat_step = false)
              @unpack t, dt, uprev, u, f, p = integrator
              @unpack tmp, KsCache = cache
              alg = unwrap_alg(integrator, nothing)
              A = f.f # assume f to be an ODEFunction wrapped around a linear operator
          
          
    if alg.krylov == :off
                  E = exp(dt * Matrix(A))
                  mul!(tmp, E, u)
              elseif alg.krylov == :simple
                  Ks, expv_cache = KsCache
                  arnoldi!(Ks, A, u; m = min(alg.m, size(A, 1)),
                           opnorm = integrator.opts.internalopnorm, iop = alg.iop)
                  expv!(tmp, dt, Ks; cache = expv_cache)
              else
                  expv_timestep!(tmp, dt, A, u; adaptive = true, caches = KsCache,
                                 m = min(alg.m, size(A, 1)), iop = alg.iop,
                                 opnorm = integrator.opts.internalopnorm,

exp isn’t differentiable. That’s the issue. But as mentioned/shown above, the new ones in ExponentialUtilities.jl are. So:

https://exponentialutilities.sciml.ai/dev/matrix_exponentials/#Matrix-Exponentials

swap that to exponential!(dt * Matrix(A)) and it should be good. Give it a quick try?

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Can we use LinearExponential with Turing.jl model?

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