### Differential equations and Laplace transforms: Laplace-transformations

### Laplace transforms of delta functions

In theory *Differentials* we have seen that a differential is an expression of the form #f(t)\,\dd g(t)#, where both #f(t)# as #g(t)# are functions #t#. It is neither a number nor a function, but an expression that indicates an "infinitely small" (infinitesimal) change depending on the change in #g(t)#. One of the useful properties of a differential is that it defines an integral: #\int f(t)\,\dd g(t)#, where #f# is a real function.

If #g# is differentiable, then #f(t)\,\dd g(t) =f(t)\cdot g'(t) \dd t# and we can interpret the definite integral #\int_a^b f(t)\,\dd g(t)# as #\int_a^b f(t)\cdot g'(t) \dd t# by means of Riemann sums on the interval #\ivoo{a}{b}#. Here #a# and #b# are real numbers (or plus or minus infinity) with #a\lt b#. Using *Riemann-Stieltjes integration*, we can interpret these expressions even if #g# is not differentiable. An important example is the Heaviside step function #u_c#, where #c# is a real number, which satisfies

\[\int_0^\infty f(t)\,\dd u_c(t) =f(c)\]

If #g# is not differentiable, then we could use the same notation, in which #g'(t)\,\dd t# is seen as synonymous to #\dd g(t)#. This suggests that #g'(t)# is a function of #t#, but this is not always the case. Still, the suggestion is useful, especially for #g(t) = u_c(t)#, the Heaviside step function. In that case, we usually write #\delta_c(t) # rather than #u_c'(t)#. We will often write #\delta(t-c)# instead of #\delta_c(t)#, just as we write #u(t-c)# rather than #u_c(t)#.

By the viewing the derivative as a limit, we can intuitively understand #\delta# as the limit for #\varepsilon\to0# of the following functions #\Delta_{\varepsilon}#.

Although we write #\delta(t-c)#, this expression makes sense only in the presence of #\dd t#. Nevertheless, we can treat #\delta# almost like a regular function by interpreting it as the limit for #\varepsilon\to0# of the following real functions #\Delta_{\varepsilon}#:

\[\Delta_\varepsilon (x) = \dfrac{u_{-\varepsilon}(x)-u_{\varepsilon}(x)}{2\varepsilon}\]

We call #\delta(t)# the **Dirac impulse function**.

In the course of defining #\delta#, we have found the following two properties that enable us to calculate with Dirac impulse functions.

Properties of the impulse function The Dirac impulse function #\delta# behaves as a function to the extent that, for every real number #c# and each piece-wise continuous function #f#,

- #\delta(x-c) = 0# if #x\ne c#, and,
- \(\int_{-\infty}^\infty f(x)\delta(x-c)\,\dd x =f(c)\)

In particular, we have

\[\laplace \left(\delta(t-c)\right) (s) = \ee^{-sc}\]

*generalized function*\[\delta(t-6)\] on #\ivco{0}{\infty}#. Here, for #c# a non-negative real number, #\delta(t-c)# is the delta function #\delta_c(t)# on #\ivco{0}{\infty}#.

Give your answer as a function of the variable #s# for #s\gt 0#.

\[\begin{array}{rcl}\laplace{(\delta(t-6))}(s) &=&\displaystyle \int_0^{\infty}\left( \delta (t-6) \right)\cdot \ee^{-st} \,\dd t\\

&&\phantom{xx}\color{blue}{\text{\(\laplace\bigl(f(t)\bigr)(s) := \int_0^\infty f(t)\, \e^{-st} \,\dd t\)}}\\

&=&\displaystyle\int_0^{\infty} \ee^{-st} \,\dd\left( u (t-6) \right)\\

&&\phantom{xx}\color{blue}{\delta(t-c)\,\dd t = \dd u(t-c)}\\

&=&\displaystyle \euler^ {- 6 s }\\

&&\phantom{xx}\color{blue}{ \int_0^{\infty}f(t)\,\dd u(t-c)= f(c)}

\end{array}\]

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