# 8.6: Derivation of the Initial-Value Theorem

Consider a physical function $$f(t)$$, with derivative $$d f / d t$$, and with Laplace transform $$L[f(t)]=F(s)$$. The initial-value theorem is:

$\lim _{t \rightarrow 0^{+} \text {from } t>0} f(t) \equiv f\left(0^{+}\right)=\lim _{s \rightarrow \infty}[s F(s)]\label{eqn:8.20}$

In general, Equation $$\ref{eqn:8.20}$$ gives the initial value $$f\left(0^{+}\right)$$ of a time function $$f(t)$$ based only on the Laplace transform $$L[f(t)]=F(s)$$, without requiring that the equation for $$f(t)$$ be available. If $$f(t)$$ is dynamic response to excitation that involves the ideal unit-impulse function $$\delta(t-0)$$, then $$f\left(0^{+}\right)$$ is the post-impulse initial value, as defined in Section 8.5; otherwise, $$f\left(0^{+}\right) \equiv f(0) \equiv f\left(0^{-}\right)$$, which is the standard initial value known to exist before excitation occurs.

Our derivation of the initial-value theorem (from a more detailed proof in Cannon, 1967, p. 569) is based upon the form of Laplace transform that can accommodate the ideal impulse function $$\delta(t-0)$$:

$L[f(t)]=\int_{t=0^{-}}^{t=\infty} e^{-s t} f(t) d t\label{eqn:8.12}$

First, we need the following Laplace transform of a derivative, the transform that is associated with definition Equation $$\ref{eqn:8.12}$$:

$L\left[\frac{d}{d t} f(t)\right]=\int_{t=0^{-}}^{t=\infty} e^{-s t} \frac{d f}{d t} d t=s F(s)-f\left(0^{-}\right)\label{eqn:8.21}$

The derivation of Equation $$\ref{eqn:8.21}$$ using integration by parts is almost identical to the derivation shown in Equation 2.2.8, the only difference being the lower limit of the integral at $$t = 0^-$$ instead of $$t = 0$$. We do not need the corresponding formula for higher-order derivatives right now, but it is appropriate here to state that the initial conditions at $$t = 0$$ in general formula Equation 2.2.10 may similarly be replaced by values at $$t=0^-1$$:

$L\left[\frac{d^{n}}{d t^{n}} f(t)\right]=s^{n} F(s)-s^{n-1} f\left(0^{-}\right)-s^{n-2} \dot{f}\left(0^{-}\right)-\cdots-\overset{(n-1)}{f}\left(0^{-}\right)\label{eqn:8.22}$

Next, taking the limit of all terms in Equation $$\ref{eqn:8.21}$$ as $$s \rightarrow \infty$$ gives

$\lim _{s \rightarrow \infty} L\left[\frac{d f}{d t}\right]=\lim _{s \rightarrow \infty}\left(\int_{t=0^{-}}^{t=0^{+}} 1 \times \frac{d f}{d t} d t+\int_{t=0^{+}}^{t=\infty} e^{-s t} \frac{d f}{d t} d t\right)=\lim _{s \rightarrow \infty}[s F(s)]-f\left(0^{-}\right)\label{eqn:8.23}$

In Equation $$\ref{eqn:8.23}$$, we separate the definite integral into two parts:

1. a part over the interval from $$t=0^{-}$$ to $$t=0^{+}$$, during which we set $$e^{-s t}=1$$ (and during which an ideal impulse including $$\delta(t-0)$$ could be acting); and
2. a part over the interval from $$t=0^{+}$$ to $$t=\infty$$. The integrand of the second part includes $$e^{-s t}$$, and since $$s \rightarrow \infty$$, we set this integral to zero: $$\lim _{s \rightarrow \infty} \int_{t=0^{+}}^{t=\infty} e^{-s t}(d f / d t) d t=0$$. Furthermore, the first integral, which is now independent of $$s$$, is evaluated indentically as $$\int_{t=0^{-}}^{t=0^{+}}(d f / d t) d t=f\left(0^{+}\right)-f\left(0^{-}\right)$$. Therefore, Equation $$\ref{eqn:8.23}$$ becomes

$\lim _{s \rightarrow \infty} L\left[\frac{d f}{d t}\right]=f\left(0^{+}\right)-f\left(0^{-}\right)=\lim _{s \rightarrow \infty}[s F(s)]-f\left(0^{-}\right)\label{eqn:8.24}$

$\Rightarrow \quad f\left(0^{+}\right)=\lim _{s \rightarrow \infty}[s F(s)]$

This is the version of initial-value theorem that was applied in Section 8.5 to re-derive result Equation 8.5.8.

If $$f(t)$$ is dynamic response to excitation that does not involves an ideal unit-impulse function $$\delta(t-0)$$, then there is no discontinuous jump at $$t=0$$, i.e. $$f\left(0^{+}\right)-f\left(0^{-}\right)=0$$. For this case, therefore, Equation $$\ref{eqn:8.24}$$ gives the more common (but less general) version of initial-value theorem:

$f\left(0^{-}\right)=\lim _{s \rightarrow \infty}[s F(s)]$