# What is the general solution of $\frac{df(x)}{dx} = f(x-a)$? [duplicate]

Let $$a \in \mathbb{R}$$ be a constant. What is the general solution of the following delay differential equation (DDE)? $$\frac{df(x)}{dx} = f(x-a)$$

For example, for $$a = - \frac{\pi}{2}$$,

$$\begin{split} \frac{d(\sin x)}{dx} &= \cos x \\ &= \sin\left(x + \frac{\pi}{2}\right) \end{split}$$

• This is a en.wikipedia.org/wiki/Delay_differential_equation , and the field is enormously complicated. I don't believe the general solution is known even for this simple case. Feb 10, 2022 at 17:15
• If $a$ is a constant, it should not be too bad. Feb 10, 2022 at 17:47
• Each branch $W_n$ of the Lambert $W$ function makes multiples of $\exp(xW_n(a)/a)$ a solution, but whether that can be parlayed into a general solution is another matter.
– J.G.
Feb 10, 2022 at 17:57
• If you have an explicit initial condition $f(t)=\phi(t),-a<t<0$ you can use method of steps to build an explicit solution, but it will ugly in most cases. Feb 10, 2022 at 18:00
• Some time ago I helped to solve one delayed ODE via Laplace transforms. It may be helpful, but some considerations have to be made math.stackexchange.com/questions/4363539/… Feb 10, 2022 at 20:04

$$f'(x)$$ 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 $$S_a[f],$$ where $$S_a$$ is the shift operator, another linear operator. Now, $$S_a$$ can be written in terms of $$D.$$ Think about this: you can expand $$f(x+a)$$ as a Taylor series centered at $$x,$$ hence $$f(x+a)=\sum_{m=0}^{\infty}\frac{(D^m[f])(x)[(x+a)-x]^m}{m!}=\sum_{m=0}^{\infty}\frac{a^mD^m[f]}{m!}(x).$$ Since this is true for any analytic function $$f,$$ it means that $$S_a=\sum_{m=0}^{\infty}\frac{a^mD^m}{m!}=\exp(aD)=e^{aD}.$$ This is important for the equation solving. Now, the equation can be rewritten simply as $$D[f]=S_a[f],$$ which is equivalent to $$D[f]-S_a[f]=0,$$ which is equivalent to $$(D-S_a)[f]=(D-e^{aD})[f]=0.$$ Thus, solving your equation is equivalent to finding the null space, also called the kernel, of the operator $$D-e^{aD}.$$ Now, consider the eigenvalue equations $$D[f]=\lambda{f}$$ and $$e^{aD}[f]=e^{a\lambda}f,$$ where $$\lambda\in\mathbb{C}.$$ Subtract the latter from the former, and you have that $$(D-e^{aD})[f]=(\lambda-e^{a\lambda})f,$$ giving an eigenvalue equation for $$D-e^{aD}.$$ Solving $$(D-e^{aD})[f]=0$$ is then equivalent to finding the eigenvectors (eigenfunctions) of $$D-e^{aD}$$ such that their eigenvalue is $$0.$$ This amounts to finding the eigenfunctions of $$D$$ that correspond to the eigenvalues $$\lambda$$ such that $$\lambda-e^{a\lambda}=0.$$ We know the eigenfunctions of $$D$$ are given by $$Ce^{\lambda{x}},$$ so we merely want those precise exponentials with $$\lambda$$ solving the equation $$\lambda-e^{a\lambda}=0.$$
Now onto solving the latter equation. Notice that it is equivalent to $$\lambda=e^{a\lambda},$$ which is equivalent to $$\lambda{e}^{-a\lambda}=1,$$ which is equivalent to $$-a\lambda{e}^{-a\lambda}=-a.$$ Taking into account the complex branches of the Lambert W, this means $$W_n(-a)=-a\lambda,$$ meaning that $$\lambda=-\frac{W_n(-a)}{a}.$$ Therefore, for some sequence of $$C_n,$$ $$f(x)=\sum_{n\in\mathbb{Z}}C_ne^{-\frac{W_n(-a)}{a}x}$$ is the complete solution family to the equation $$f'(x)=f(x+a).$$
In the special case that $$a=\frac{\pi}2,$$ we get that $$W_0\left(-\frac{\pi}2\right)=\frac{i\pi}2,$$ hence $$-\frac{W_0\left(-\frac{\pi}2\right)}{\frac{\pi}2}=-i,$$ 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.