Better error report for dividing by FunctionOperator with no inverse

[KevinDCarlson]

Is your feature request related to a problem? Please describe.

If you try to write F \ v for F::FunctionOperator with F.op_inverse not defined, you get the opaque error message ERROR: MethodError: objects of type Nothing are not callable.

Describe the solution you’d like

Write a clearer error message.

[divital-coder]

[Potential Solution]

We should add a check before attempting to use op_inverse to provide a more helpful error message.

function Base.:\(L::FunctionOperator{iip, true}, v::AbstractArray) where {iip}
    if L.op_inverse === nothing
        throw(ArgumentError("Cannot divide by FunctionOperator with no inverse function defined. Please provide 'op_inverse' when constructing the operator."))
    end
    
    _sizecheck(L, nothing, v)
    _, V, vec_output = _unvec(L, nothing, v)

    W = L.op_inverse(V, L.u, L.p, L.t; L.traits.kwargs...)

    vec_output ? vec(W) : W
end

[divital-coder]

[ Issue reproduction ]
A minimal working example to test this would be :

using SciMLOperators, LinearAlgebra
using Test

# Create a simple function operator without an inverse
N = 4
f(v, u, p, t) = 2.0 * v  # A simple scaling operation
f(w, v, u, p, t) = w .= 2.0 * v  # In-place version

# Create vectors for testing
v = rand(N)
result = zeros(N)

# Create a FunctionOperator without specifying op_inverse
F = FunctionOperator(f, zeros(N), zeros(N); islinear=true)


try
    # This should fail with "objects of type Nothing are not callable"
    result = F \ v
    println("Division unexpectedly succeeded")
catch e
    println("Error type: ", typeof(e))
    println("Error message: ", e)
end

# Now try a version with op_inverse defined for comparison
finv(v, u, p, t) = 0.5 * v  # Inverse operation
finv(w, v, u, p, t) = w .= 0.5 * v

F_with_inverse = FunctionOperator(f, zeros(N), zeros(N); 
    op_inverse=finv, islinear=true)

# This should work
result = F_with_inverse \ v
println("\nDivision with proper inverse: ", result[1:min(3,length(result))], "...")