What is the general solution ofdf(x)dx=f(x−a)?

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Keith Steele · Accepted Answer

Step 1f′(x) can be written in terms of operators as D[f], where D is the derivative operator, a linear operator. f(x+a) can be written in terms of operators as Sa[f], where Sa is the shift operator, another linear operator. Now, Sa can be written in terms of D. Think about this: you can expand f(x+a) as a Taylor series centered at x, hencef(x+a)=∑m=0∞(Dm[f])(x)[(x+a)−x]mm!=∑m=0∞amDm[f]m!(x)Since this is true for any analytic function f, it means thatSa=∑m=0∞amDmm!=exp⁡(aD)=eaD.This is important for the equation solving. Now, the equation can be rewritten simply as D[f]=Sa[f],, which is equivalent to D[f]−Sa[f]=0,, which is equivalent to (D−Sa)[f]=(D−eaD)[f]=0..Thus, solving your equation is equivalent to finding the null space, also called the kernel, of the operator D−eaD.. Now, consider the eigenvalue equationsD[f]=λ{f}andeaD[f]=eaλf,where λ∈C.. Subtract the latter from the former, and you have that(D−eaD)[f]=(λ−eaλ)f,giving an eigenvalue equation for D−eaD.. Solving (D−eaD)[f]=0is then equivalent to finding the eigenvectors (eigenfunctions) ofD=eaD such that their eigenvalue is 0. This amounts to finding the eigenfunctions of D that correspond to the eigenvalues λ such that.λ−eaλ=0.. We know the eigenfunctions of D are given by Ceλ{x},, so we merely want those precise exponentials with λ solving the equationλ−eaλ=0.Now onto solving the latter equation. Notice that it is equivalent toλ=eaλ, which is equivalent to λ{e}−aλ=1, which is equivalent to −aλ{e}−aλ=−a. Taking into account the complex branches of the Lambert W, this meansWn(−a)=−aλ,meaning thatλ=−Wn(−a)a.Therefore, for some sequence of Cn,f(x)=∑n∈ZC≠−Wn(−a)axis the complete solution family to the equationf′(x)=f(x+a).In the special case that a=π2, we get thatW0(−π2)=iπ2,hence−W0(−π2)π2=−i,so we have that f(x)=Ce−ix is one of the solutions given. This means sin⁡(x) and cos⁡(x) are solutions, as expected. There are other solutions, but there is no nice way of expressing them, due to the nature of the Lambert W map.

What is the general solution of \(\displaystyle{\frac{{{d}{f{{\left({x}\right)}}}}}{{{\left.{d}{x}\right.}}}}={f{{\left({x}-{a}\right)}}}\)?

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