Define Lagrangian using struct

New to Julia
1

I’m new in Julia and I want to define a dynamical Lagrangian which is a difference between the kinetic energy and the potential energy L(x,v)=K(v)-U(x). I use this code in order to do trick for example if we but a=Lagrangian and a.K give me the kinetic energy.

mutable struct Lagrangian{T<:Float64}
          x::T
           v::T
            function K{T}(v::T) where T<:Float64
             return  0.5*v^2
            end
     function U{T}(x::T) where T<:Float64
             return  0.5*x^2
            end
        end

UndefVarError: K not defined

Stacktrace:
[1] top-level scope
@ In[3]:1

2

In Julia, structs are only really for data, and all functions are defined separately (maybe an exception for a constructor). So for your example:

mutable struct Lagrangian{T}
    x::T
    v::T
end

kintetic(L::Lagrangian) = L.v^2/2
potential(L::Lagrangian) = L.x^2/2

There is an option to override some functions like getproperty to do what you want but I would recommend the above method instead.

2 Likes
3

Thank you @jmair for your reply. That was clear, but how can I evaluate kintetic(L::Lagrangian) = L.v^2/2 let say for v=1.5

4

I see in this example that it could use a function with structs like

struct Point{T<:Real}
           x::T
           y::T
           Point{T}(x,y) where {T<:Real} = x^2+y^2
       end

and then use Point{Int}(1,1) for the evaluation which gives 2.

5

Just call like a regular function:

julia> L = Lagrangian(0.0, 2.0);
julia> kinetic(L)==2.0
true

This is what I would call a “functor” but again, I would avoid this unless there is a good reason.

mutable struct Point{T}
    x::T
    y::T
end
function (point::Point)()
    return point.x^2+point.y^2
end

Although I am not sure what reason you have for the struct at all, these are all probably better done with normal plain functions.

1 Like
6

Great @jmair thank you!