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
\[\laplace \left(\delta(t-c)\right) (s) = \ee^{-sc}\]
Give your answer as a function of the variable #s# for #s\gt 0#.
\[\begin{array}{rcl}\laplace{(\delta(t-2))}(s) &=&\displaystyle \int_0^{\infty}\left( \delta (t-2) \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-2) \right)\\
&&\phantom{xx}\color{blue}{\delta(t-c)\,\dd t = \dd u(t-c)}\\
&=&\displaystyle \euler^ {- 2 s }\\
&&\phantom{xx}\color{blue}{ \int_0^{\infty}f(t)\,\dd u(t-c)= f(c)}
\end{array}\]
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