doc

src/derivatives.jl

@@ -16,7 +16,6 @@ function FirstDerivative(x::AbstractVector, y::AbstractVector; bc = (0, 0), dire
 	FirstDerivative{eltype(y)}(x, y, bc, direction)
 end
 
-
 Base.size(d::FirstDerivative) = (length(d.x), 1)
 
 Base.IndexStyle(d::FirstDerivative) = IndexLinear()
@@ -56,7 +55,6 @@ function SecondDerivative(x::AbstractVector, y::AbstractVector; bc = (0, 0))
 	SecondDerivative{eltype(y)}(x, y, bc)
 end
 
-
 Base.size(d::SecondDerivative) = (length(d.x), 1)
 
 Base.IndexStyle(d::SecondDerivative) = IndexLinear()

src/feynman_kac.jl

@@ -5,12 +5,18 @@ With direction = :backward
 Solve the following PDE:
 u(x, t[end]) = ψ(x)
 0 = u_t + 𝕋u - v(x, t)u + f(x, t)
-
+Equivalently, in integral form, 
+u(x, t) = E[∫_t^T e^{-∫_t^s v(x_u) du} f(x_s)ds + \int_t^t e^{-\int_t^T v(x_u)du} ψ(x_T)|x_t = x]
+(notations are from the wikipedia article for Feynman–Kac formula)
 
 With direction = :forward
 Solve the following PDE:
 u(x, t[1]) = ψ(x)
 u_t = 𝕋u - v(x, t)u + f(x, t)
+Equivalently, in integral form, 
+u(x, t) = E[∫_0^t e^{-∫_0^s v(x_u) du} f(x_s)ds + \int_0^t e^{-\int_0^t v(x_u)du} ψ(x_t)|x_0 = x]
+
+The function returns a matrix of size(length(f), length(t))
 """
 function feynman_kac(𝕋; 
     t::AbstractVector = range(0, 100, step = 1/12),