Laplace Transform

Initial Value Problem (IVP)

First seen in MATH213.

In Lecture 7, we’ve defined:

Definition 6: Piecewise Continuous

A function $f(x)$ is piecewise-continuous on a given finite interval if

1. $f$ has a finite number of discontinuities in that interval and
2. for each discontinuity $x_0$ both the left and right hand limits exits (note they can be different values)

Theorem 2: Initial Value Theorem

If $f(t)$ is a piecewise-continuous and ${\int }_0^{\infty }|f(t)|e^{-\alpha t}$ converges for some $\alpha \in \mathbb{R}$ then

$f(0^{+)=}{\lim }_{s\to +\infty }sF(s)$

Understand this theorem:

• The above gives us a way of computing the IC of $f(t)$ from $F(s)$ without computing the inverse Laplace transform and instead evaluating a limit in the complex plane
• $s$ is generally complex so the limit is in the complex plane NOT the real number line thus:
  • $Re(s)$ tends to infinity
  • $Im(s)$ can do anything (stay 0, oscillate, go to infinity, do random things, etc.)
• For many problems we can treat the limit as the standard limit you are used to working with.