How to add other arguments to function turned to LinearOperator?

General Usage
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1

Trying to write a matrix-free solver using LinearOperators.jl and Krylov.jl
However, from this: Introduction to Linear Operators
I’m seeing that you can only use functions that follow the 5-argument (and 3) mul! operation.

But I would like to create a linear operator that is able to take in more variables such as:

function customfunc!(out, in, a, b, dt, nu)
for i in 1:length(in)
out[i] = in[i]*dt*(nu[i]+nu[i+1])/2.0
end

In LinearMaps.jl it would be something like:

customfunc = (N, dt, nu) -> LinearMap(N; ismutating=true) do Z,U

for i in 1:N
Z[i] = U[i]*dt*(nu[i]+nu[i+1])/2.0
end
return Z
end

CFunc = customfunc(100, 0.01, [1.0,2.0,3.0])

Then I could use CFunc in IterativeSolvers.jl.

Any suggestions?

2

This seems exactly like you would solve this problem. What is your issue with it?

If you don’t like using an anonymous function to create a closure with a lot of captured variables or you want to change some paramteres without making a new LinearMap, you can of course convert it to a proper type from which you construct the LinearMap.

3

What I’m asking is to use LinearOperators.jl rather than LinearMaps.jl. The tutorials show that matrix-free functions should follow the 5-argument mul! which wouldn’t allow for extra variables to define a linear operator.

Are you saying that LinearOperators.jl is useless when it comes to changing operators and one should stick to LinearMaps.jl?

4

I figured it out.

You can use an anonymous function inside another anonymous function such as:

customfunc = (N, dt, nu) -> (Z,U,alpha,beta) -> begin

for i in 1:N
Z[i] = U[i]*dt*(nu[i]+nu[i+1])/2.0
end
return Z
end

Then you initialize with your parameters:

N =12
dt = 0.5
nu=0.01
custmul! = customfun(N,dt,nu)

Then you can put this into a linear operator because this is now a 4 argument function that works as:

custmul!(Out, In, alpha, beta)