# The simplest delay differential equation

I am trying to understand a bit about solutions of delay differential equations, so I tried analyzing one of the most simple ones: $$u'(t)=-\beta u(t-1), \text{and for } t\in [-1,0), u(t)=\phi(t), \text{for some known function \phi}.$$

One way to obtain solutions is to form the characteristic equation by substituting $u(t)=e^{\lambda t}$. Using this method, one can prove that if $\beta=e^{-1}$, then $e^{-t}$ and $te^{-t}$ satisfy $u'(t)=-\beta u(t-1)$. These solutions don't take into account the initial function $\phi$.

Another way of obtaining a solution is to start with the function $\phi$ and integrate the equation in every interval of length $1$. This method is called method of steps.

My question is, what is the relationship between the solutions one obtains from the characteristic equation method and the method of steps?

• If a solution does not take into account a part of the problem (initial condition), then it is not a solution. – user53153 Jan 23 '13 at 14:32

## 1 Answer

The general solution is $u(t) = \sum_n c_n \exp(\lambda_n t)$ where $\lambda_n$ are the solutions of the characteristic equation. The $c_n$ are obtained from $\phi$ using a Laplace transform, which is explained for example in here, but I have not read it in detail so I do not know if the relation is simple, https://www.rand.org/content/dam/rand/pubs/reports/2006/R374.pdf

This solution must be equal to the one obtained with the method of steps with the same initial condition.