Exponential response formula

The exponential response formula lets us find a particular solution to a linear inhomogeneous differential equation.

Let

p(r)

be the characteristic polynomial, such that the ODE is

p(D)x = Ae^{kt}

where

k

is real or complex. Here

D=\frac{d}{dt}

is the differential operator.

Note that

p(D) e^{\theta x} = p(\theta) e^{\theta x}

since for each derivative another

\theta

term appears (expand the polynomial and distribute if this isn’t clear). We can then write:

Generalized ERF

p

is a polynomial and

k

is a real such that

p(k_0)=p'(k_0)=\cdots= p^{(m-1)}(k_0)=0

, then

p(D)(t^m e^{k_0t}) = p^{(m)}(k_0) e^{k_0t}

Proof

We want to solve $p(D) y = e^{k_0t}$ for $y$ . We know that $p(D) e^{k_0t} = p(r) e^{k_0t}$ , but since $p(k_0)=0$ we cannot divide by it. Instead we will consider what happens near $k_0$ . We substitute $k$ for $k_0$ and differentiate with respect to $k$ .

\begin{align*} p(D) e^{kt} &= p(k) e^{kt} \\ \frac{\partial}{\partial k} p(D) e^{kt} &= \frac{\partial}{\partial k} P(k) e^{kt}. \end{align*}

Since $D=\frac{d}{dt}$ and $p$ is a polynomial (i.e. it is well behaved), by linearity we can say $\frac{\partial}{\partial k} p(D) = p(D) \frac{\partial}{\partial k}$ .

\begin{align*} p(D) \frac{\partial}{\partial k} e^{kt} = p'(k) e^{kt} + p(k) t e^{kt}. \end{align*}

$k_0$ is a root of $p$ with multiplicity $m$ . In other words, $p(k_0)=p'(k_0)=\cdots= p^{(m-1)}(k_0)=0$ . We differentiate the equation above $m$ times and evaluate at $k=k_0$ to find:

\begin{align*} p(D) (t^m e^{kt}) &= (p^{(m)}(k) + p^{(m-1)}(k) t + \cdots + t^m p(k)) e^{kt} \\ p(D) (t^m e^{kt}) &= p^{(m)}(k_0) e^{k_0t}. \end{align*}