# 8.6: Derivation of the Initial-Value Theorem

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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:

- 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
- 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:

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